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10800	https://statisticseasily.com/glossario/what-is-bernoullis-inequality-explained/	What is: Bernoulli's Inequality Explained Skip to content LEARN STATISTICS EASILY Learn Data Analysis Now! Home Articles Statistical Analysis ANCOVA ANOVA Chi-Square Generalized Linear Models Kendall Correlation Kruskal-Wallis Test Linear Regression Logistic Regression Mann-Whitney U Test MANOVA Normality Test Pearson Correlation Spearman Correlation Student’s t-test Descriptive Statistics Absolute Mean Deviation Kurtosis Mean Median Mode Skewness Standard Deviation Data Science Artificial Intelligence Deep Learning Machine Learning Bayesian Statistics Books General Generators Graphs Important Concepts Multivariate statistics Results Reporting Sample Size Softwares eBooks Books About Us Contact Us . Click Here to Master Data Analysis! LEARN STATISTICS EASILY Learn Data Analysis Now! What is: Bernoulli’S Inequality What is Bernoulli’s Inequality? Bernoulli’s Inequality is a fundamental theorem in mathematics, particularly in the fields of statistics and probability theory. It states that for any real number x greater than -1 and any integer n greater than or equal to 0, the inequality (1 + x)^n ≥ 1 + nx holds true. This inequality is particularly useful in various applications, including data analysis and statistical modeling, as it provides a way to estimate the growth of functions and the behavior of random variables. Advertisement Simplify Data Analysis! Don’t let statistics intimidate you. Learn to analyze your data with ease and confidence. Read more! Data Analysis Consulting Statistical Analysis Tools Learn More Understanding the Mathematical Expression The expression (1 + x)^n represents the expansion of a binomial expression, where x can be any real number. The term 1 + nx is a linear approximation of this expression when x is small. Bernoulli’s Inequality highlights that the exponential growth of (1 + x)^n will always be greater than or equal to the linear growth represented by 1 + nx, provided that the conditions of the inequality are met. This property is essential for understanding the behavior of sequences and series in statistical contexts. Applications in Data Science In data science, Bernoulli’s Inequality is often applied in the analysis of algorithms and performance metrics. For instance, when estimating the expected value of a random variable, this inequality can help in bounding the probabilities involved. By providing a lower bound for the expected outcomes, data scientists can make more informed decisions based on statistical evidence, enhancing the reliability of their models. Bayesian Statistics Course Data Science Books Relation to Other Mathematical Concepts Statistical Analysis Tools Bernoulli’s Inequality is closely related to other mathematical concepts such as the binomial theorem and Taylor series. The binomial theorem provides a way to expand expressions of the form (a + b)^n, while Bernoulli’s Inequality can be seen as a specific case that emphasizes the growth properties of these expansions. Understanding these relationships is crucial for statisticians and data analysts who rely on mathematical foundations to derive insights from data. Proof of Bernoulli’s Inequality The proof of Bernoulli’s Inequality can be approached using mathematical induction. For the base case, when n = 0, the inequality holds true as both sides equal 1. Assuming the inequality holds for n = k, we can show that it also holds for n = k + 1. By manipulating the expressions and applying the inductive hypothesis, we can derive the required result, thereby confirming the validity of Bernoulli’s Inequality across all integers n ≥ 0. Advertisement Analyze Data like a Professional! Unravel the secrets of statistical analysis and enhance your scientific projects. Don’t miss this chance! Data Science Books Statistical Analysis Tools Learn More Implications for Risk Assessment In the context of risk assessment, Bernoulli’s Inequality provides a framework for evaluating potential losses and gains. By applying this inequality, analysts can estimate the worst-case scenarios in probabilistic models, allowing for better risk management strategies. This is particularly important in fields such as finance and insurance, where understanding the bounds of potential outcomes is critical for decision-making. Limitations of Bernoulli’s Inequality While Bernoulli’s Inequality is a powerful tool, it is essential to recognize its limitations. The inequality only holds for real numbers x greater than -1, which means that it cannot be applied universally across all mathematical scenarios. Additionally, the inequality provides a lower bound, which may not always be sufficient for certain analyses that require precise estimations of growth or decay in data trends. Data Science Books Bernoulli’s Inequality in Machine Learning Machine Learning Platform Bayesian Statistics Course In machine learning, Bernoulli’s Inequality can be utilized to analyze the performance of algorithms, particularly in scenarios involving probabilistic models. By leveraging this inequality, practitioners can derive bounds on the expected error rates of classifiers and regression models. This understanding aids in model selection and tuning, ensuring that the chosen algorithms perform optimally under various conditions. Conclusion on the Importance of Bernoulli’s Inequality Bernoulli’s Inequality serves as a cornerstone in the study of statistics and data analysis. Its applications span various domains, from theoretical mathematics to practical data science. By providing insights into the behavior of functions and random variables, this inequality enables statisticians and data scientists to make informed decisions based on rigorous mathematical foundations. Advertisement Simplify the Analysis of Your Data! Gain mastery of statistics and analyze your data with confidence. Start your journey right now! 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10801	https://www.youtube.com/watch?v=cljdQ2tgIBY	Converting a circle into standard form and then graphing Brian McLogan 1590000 subscribers 1803 likes Description 141570 views Posted: 3 May 2013 Learn how to graph the equation of a circle by completing the square. Completing the square will allow us to transform the equation of a circle from general form to standard form. When the equation is in standard form we can identify the center and radius of the circle to graph to then graph the circle. conicsections #circleconicsections 115 comments Transcript: so uh noah what you're going to look at on this one is i'm going to ask you to graph this all right now this one's not going to be so bad for you noah for what we're going to do so to graph a problem like this all right we know that a circle is there we talked about the standard form of circle right the standard form of a circle says x minus h squared plus y minus k squared equals r squared yes okay so what i'm going to want to do ava in this problem is take this and rewrite it in this format now remember in parabolas when we didn't have a quadratic that was in our vertex form we had to complete the square once right we had to complete the square to produce our perfect square binomial well now our purpose to a trinomial but now you notice for a circle how many binomial squares do we have two so it might be possible and it is possible in this case that we're gonna have to alex alex alex so in this case what we're going to do is we're going to complete the square twice so what we're going to do to complete the square we first have to arrange our variables together so we have x squared minus 8x plus y squared plus 12y equals 0. now i can complete the square for each of these separately fortunately for us we don't have any coefficients right so i can just do this one is negative 8 divided by 2 squared which equals 16. this one's 12 divided by 2 squared which equals 36. therefore i can write this as x squared minus 8x plus 16 plus y squared plus 12y plus 36. now remember since i'm adding a 16 i have to i could subtract 16 on the same side but i don't really want anything else except for my binomial squares on the left side so i can put it on the other side so if i add 16 on the right side i have to add to the other side all right now i can rewrite this as a binomial squared so by factoring this by factoring this i write this as x minus 4 squared plus y plus 6 squared equals 42. 52. okay so now we need to determine what is my radius right um oh crap or wrote the problem wrong oops it's okay this is going to be important because is it going to be kind of hard to graph a radius when you have it as 52 yeah so let me uh go and rewrite the problem again that i had i think it i did write this so it's plus 12 at the end that's an alex can you guys find two different seats just not even that row just totally two different seats so therefore this ends up equaling 64. so find two different seats not on that row anywhere i don't care just anywhere in the anywhere so now what we can do is now that we're looking at this okay what we can now simply look at is say all right now i've determined what now i can determine what the center is center is going to be 4 comma negative 6 and then my radius is what 8. so now is it going to be possible for me to be able to graph this of course so what i'll be able to do is let's find a center so my center is going to have 4 negative 6 1 2 3 4 negative 6. 1 2 3 4 5 6. okay then let's find the radius so the radius would be eight so that means i could go to the left eight one two three four five six seven eight that'd be an end point and then up eight one two three four five six seven eight and obviously you could be able to determine you obviously you could be able to determine different points on there all right but i'll just kind of keep this simple and i'll show you those two points by using that all right because it's the opposite of h and opposite of k
10802	https://math.stackexchange.com/questions/1603906/placing-the-integers-1-2-ldots-n-on-a-circle-for-n1-in-some-special	combinatorics - Placing the integers ${1,2,\ldots,n}$ on a circle ( for $n>1$) in some special order - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Placing the integers {1,2,…,n}{1,2,…,n} on a circle ( for n>1 n>1) in some special order Ask Question Asked 9 years, 8 months ago Modified9 years, 4 months ago Viewed 1k times This question shows research effort; it is useful and clear 33 Save this question. Show activity on this post. For which integer n>1 n>1 can we place the integers {1,2,…,n}{1,2,…,n} on a circle (say boundary of S 1 S 1 ) in some order such that for each s∈{1,2,…,n(n+1)2}s∈{1,2,…,n(n+1)2} , there exist a connected subset of the circle on which the sum of the integers placed is exactly s s? combinatorics number-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 14, 2016 at 3:03 Dan Brumleve 17.7k 6 6 gold badges 51 51 silver badges 102 102 bronze badges asked Jan 8, 2016 at 1:51 user228168 user228168 12 2 @YogUrt I don't understand your example. We're supposed to put the integers 1 through n in some order--how does your example do that?Joey Zou –Joey Zou 2016-01-09 00:13:56 +00:00 Commented Jan 9, 2016 at 0:13 7 I don't know why this is tagged "general-topology" and "connectedness". The role of the "circle" here is completely superficial; this is just a question about labellings of finite cyclically ordered sets.Eric Wofsey –Eric Wofsey 2016-01-14 01:40:35 +00:00 Commented Jan 14, 2016 at 1:40 3 I looked for solutions with the additional property that no sum appears more than twice, again by generating random permutations of each size and testing them, and I found examples up to n=11 n=11. Maybe some progress could be made by looking at this stronger conjecture.Dan Brumleve –Dan Brumleve 2016-01-14 02:57:59 +00:00 Commented Jan 14, 2016 at 2:57 5 For what it's worth, recursively inserting n n into a solution to the n−1 n−1 problem doesn't generate enough solutions to answer the question. This strategy only reaches two solutions for n=16 n=16: (1,6,7,16,12,13,4,11,5,15,3,14,2,10,9,8)(1,6,7,16,12,13,4,11,5,15,3,14,2,10,9,8) and (1,7,6,11,4,5,14,3,16,9,10,8,12,2,13,15)(1,7,6,11,4,5,14,3,16,9,10,8,12,2,13,15). It doesn't find a solution for n=17 n=17, although other solutions do exist.Chris Culter –Chris Culter 2016-01-14 21:06:41 +00:00 Commented Jan 14, 2016 at 21:06 4 @SaunDev: Because if some consecutive set covers a sum s s then its complement (also consecutive) covers a sum s′=n(n+1)/2−s s′=n(n+1)/2−s. So you need to check only the (say) lower half. Further, the range 1⋯n 1⋯n is trivially fulfilled.leonbloy –leonbloy 2016-01-17 12:32:53 +00:00 Commented Jan 17, 2016 at 12:32 \|Show 7 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. I think I have got an answer : The arrangement is possible for any n≥1 n≥1 . Proof : c a s e 1––––––c a s e 1 _ , n n is even : n n is even , so pair the numbers 1,2,...,n 1,2,...,n into n/2 n/2 pairs with the two numbers in each pair adding to n+1 n+1 i.e. the pairing is like {1,n};{2,n−1};...;{n 2,n 2+1}{1,n};{2,n−1};...;{n 2,n 2+1} . We claim that any circular arrangement of the numbers 1,2,...,n 1,2,...,n that keeps the numbers in each pair adjacent to one another gives all possible sum values from 1 1 to n(n+1)2 n(n+1)2 . To see this , consider any s s as 1≤s≤n(n+1)2 1≤s≤n(n+1)2 , by division algorithm , there are integers q,r q,r such that s=q(n+1)+r s=q(n+1)+r , where 0≤r≤n 0≤r≤n and 0≤q≤n/2 0≤q≤n/2 . If r=0 r=0 then we choose any q q consecutive pairs , as numbers in each pair are adjacent , this gives a connected subset with sum ( since each pair has sum n+1 n+1) q(n+1)=q(n+1)+r(=0)=s q(n+1)=q(n+1)+r(=0)=s ; if r>0 r>0 ( and so that q<n/2)q<n/2) , we choose the connected subset to be beginning with r r and then q q consecutive pairs clockwise or anticlock wise as appropriate , obtaining a sum equal to r+q(n+1)=s r+q(n+1)=s [Note :In s=(n+1)q+r s=(n+1)q+r , the inequality 0≤r≤n 0≤r≤n comes from division algorithm but the bounds 0≤q≤n/2 0≤q≤n/2 does not come directly from division algorithm ; it is due to as follows : q(n+1)=s−r≤s≤n(n+1)2 q(n+1)=s−r≤s≤n(n+1)2 so q≤n/2 q≤n/2 , and q(n+1)=s−r≥1−r≥1−n=−(n−1)>−(n+1)q(n+1)=s−r≥1−r≥1−n=−(n−1)>−(n+1) , so that q>−1 q>−1 , then since q q is an integer , we get q≥0 q≥0 ] c a s e 2––––––c a s e 2 _ , n n is odd : n n is odd ,then we form n+1 2 n+1 2 pairs each with sum n n , thinking of the singleton n n as a degenerate pair ; i.e. {1,n−1};{2,n−2};...;{n−1 2,n+1 2};{n}{1,n−1};{2,n−2};...;{n−1 2,n+1 2};{n} are the required pairs . Here also , any arrangement that keeps the numbers in each pair adjacent to one another gives all possible sum , the justification being same as that of the previous one . Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 17, 2016 at 8:55 answered May 1, 2016 at 7:03 user228168 user228168 3 1 Wow, very nice one, congratulations for a beatiful proof!jpvee –jpvee 2016-05-01 08:29:44 +00:00 Commented May 1, 2016 at 8:29 @jpvee : Thank you very much for taking the trouble to look at the proof :)user228168 –user228168 2016-05-01 08:43:44 +00:00 Commented May 1, 2016 at 8:43 2 Such a cute and simple solution! It also gives a nice lower bound 2⌊n/2⌋−1⌊(n−1)/2⌋!2⌊n/2⌋−1⌊(n−1)/2⌋! for the number of solutions (not counting rotations and reflections).zhoraster –zhoraster 2016-05-01 08:46:48 +00:00 Commented May 1, 2016 at 8:46 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Note that this is not meant as an answer to the OP's question, but the space in the comment fields is too limited, so I put my comment here. Using brute (computer) force, I managed to find solutions for all n≤26 n≤26, and it is quite likely that solutions for larger n n can be found. I still can't make up my mind, though, whether or not I believe that there is an integer N N for which the problem has no solutions: I modified my program to count the number of possible solutions out of all permutations of the integers 1,…,n 1,…,n on the circle (not counting rotations and reflections) and looked at the ratios of these numbers: 2: 1 solution(s) found (out of 1 possible). Ratio is 1.0 3: 1 solution(s) found (out of 1 possible). Ratio is 1.0 4: 2 solution(s) found (out of 3 possible). Ratio is 0.6666666666666666 5: 10 solution(s) found (out of 12 possible). Ratio is 0.8333333333333334 6: 41 solution(s) found (out of 60 possible). Ratio is 0.6833333333333333 7: 126 solution(s) found (out of 360 possible). Ratio is 0.35 8: 537 solution(s) found (out of 2520 possible). Ratio is 0.2130952380952381 9: 3956 solution(s) found (out of 20160 possible). Ratio is 0.19623015873015873 10: 19776 solution(s) found (out of 181440 possible). Ratio is 0.10899470899470899 11: 76340 solution(s) found (out of 1814400 possible). Ratio is 0.04207451499118166 12: 388047 solution(s) found (out of 19958400 possible). Ratio is 0.019442791005291005 13: 2775155 solution(s) found (out of 239500800 possible). Ratio is 0.011587247307733419 14: 15013424 solution(s) found (out of 3113510400 possible). Ratio is 0.004822024683135794 The program is currently busy to get the figures for n=15 n=15. I'm sure it can be somewhat optimized, but I don't expect that many additional values can be computed in acceptable time, so there probably won't be much more insight than what's available already: It is clear to see that while the absolute numbers of admissible permutations grow with n n, the ratios with respect to the total numbers of permutations decrease, but they do so quite irregularly, note that the ratios for n=8 n=8 and n=9 n=9 are quite close whereas in other places there is a drop by more than 50%, so it is not only size that matters, but possibly some number theoretic constellations. So, this limited evidence supports both possible propositions: The growing number of solutions suggests that there is a solution for every integer, and the decreasing ratios as well as the irregularity in the numbers of solutions could be a sign that eventually there might be a number N N for which no solution exists. Can anyone provide some additional thoughts on the problem? I think that it is quite a gem, but I don't really have an idea how to tackle it. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 24, 2016 at 9:31 jpveejpvee 3,683 2 2 gold badges 24 24 silver badges 33 33 bronze badges 3 Addendum: Meanwhile the sequence of distinct solutions has been submitted to the OEIS as A272135. It also gives the numbers for n=15 n=15 and n=16 n=16 which yield the ratios 0.001243163963278050 0.001243163963278050 and 0.000416230553507339 0.000416230553507339.jpvee –jpvee 2016-04-22 09:16:20 +00:00 Commented Apr 22, 2016 at 9:16 I have posted an answer , could you please take a look ...?user228168 –user228168 2016-05-01 08:06:16 +00:00 Commented May 1, 2016 at 8:06 1 Looks correct to me, very nice!jpvee –jpvee 2016-05-01 08:30:29 +00:00 Commented May 1, 2016 at 8:30 Add a comment\| You must log in to answer this question. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 2Existence of a special type of injection f:N→S 1 f:N→S 1 such that for every n∈N n∈N relating to connectedness Related 14Arc sums for a circle of k k positive integers whose total sum is n n 7Does {1,2,…,3000}{1,2,…,3000} contain a subset of 2000 2000 integers with no member twice another? 2Existence of a special type of injection f:N→S 1 f:N→S 1 such that for every n∈N n∈N relating to connectedness 3Rearranging rows of a matrix 1Give a proof by contradiction to show if the odd integers 1,3,5,7,9,11,13,15,17,19,1,3,5,7,9,11,13,15,17,19, are placed around a circle... 4Axioms for The Integers 19A finite set of distinct positive numbers is special if each integer in the set divides the sum of all integers within the set. 3For which integers m m does an infinite string of characters S=c 1⋅c 2⋅c 3⋅c 4⋅c 5…S=c 1⋅c 2⋅c 3⋅c 4⋅c 5… exist Hot Network Questions Can induction and coinduction be generalized into a single principle? 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10803	https://pubs.rsc.org/en/content/articlelanding/1968/rr/rr9680100062	Maintenance work is planned on Thursday 2nd October 2025 from 09:00 to 10:00 BST. During this time the performance of our website may be affected - searches may run slowly, some pages may be temporarily unavailable, and you may be unable to log in or to access content. If this happens, please try refreshing your web browser or try waiting two to three minutes before trying again. We apologise for any inconvenience caused and thank you for your patience. Royal Institute of Chemistry, Reviews Structure and properties of water Abstract The first page of this article is displayed as the abstract. Please wait while we load your content... Article information Download Citation Permissions Structure and properties of water D. J. G. Ives and T. H. Lemon, R. Inst. Chem., Rev., 1968, 1, 62 DOI: 10.1039/RR9680100062 To request permission to reproduce material from this article, please go to the Copyright Clearance Center request page. If you are an author contributing to an RSC publication, you do not need to request permission provided correct acknowledgement is given. If you are the author of this article, you do not need to request permission to reproduce figures and diagrams provided correct acknowledgement is given. If you want to reproduce the whole article in a third-party publication (excluding your thesis/dissertation for which permission is not required) please go to the Copyright Clearance Center request page. Read more about how to correctly acknowledge RSC content. Search articles by author Fetching data from CrossRef. This may take some time to load. Fetching data from CrossRef. This may take some time to load. Loading related content Spotlight Advertisements Journals, books & databases Journals Books Databases More
10804	https://fusioncrossmedia.com/products/steam-tables-book?srsltid=AfmBOopdWUE08iAhJJMCSr-jI6eFUuKuMZbtQGTRlvkHbeavsOjPgq6y	Steam Tables Book - FUSION - \| / Save up to %Save %Save up to Save Sale Sold out In stock Phone: 860-647-8367 Hours: M-F 8:30 AM - 4:30 PM SearchLogin 0 Cart Shopping Cart 0 Your Cart is Empty Continue Shopping Subtotal: $0.00 USD Taxes and shipping calculated at checkout Checkout Go to cart Creative Digital Print Commercial Print SolutionsPosters Signs & Visual Graphics Our Story ON-THE-SHELF-PRINTS STEAM TABLE BOOKS Contact & Send File SearchLogin 0 Cart Shopping Cart 0 Your Cart is Empty Continue Shopping Subtotal: $0.00 USD Taxes and shipping calculated at checkout Checkout Go to cart Creative Digital Print Commercial Print SolutionsPosters Signs & Visual Graphics Our Story ON-THE-SHELF-PRINTS STEAM TABLE BOOKS Contact & Send File [x] Creative [x] Digital [x] Print Commercial Print Solutions Posters Signs & Visual Graphics [x] Our Story [x] ON-THE-SHELF-PRINTS STEAM TABLE BOOKS [x] Contact & Send File Login Translation missing: en.general.currency.dropdown_label CAD CAD USD INR GBP AUD EUR JPY Search 0 Cart Shopping Cart 0 Your Cart is Empty Continue Shopping Subtotal: $0.00 USD Taxes and shipping calculated at checkout Checkout Go to cart Menu title This section doesn’t currently include any content. Add content to this section using the sidebar. Add description and links to your promotion Shop now Your headline Image caption appears here Your product's name $68.29 CAD Add your deal, information or promotional text Home All Steam Tables Book Steam Tables Book $20.91 CAD Steam Tables - Properties of Saturated and Superheated Steam is a 46-page softcover reference book with tables adopted by The Sixth International Conference on the Properties of Steam for the ASME Research Committee. Reference includes .08865 to 15,500 LB Per Sq In. absolute pressure System International (SI Metric Units) and Conversion Factors Includes a fold-out 11.25" x 19" Mollier Diagram Email address Notify me when this product is available: Add to cart More payment options This item is a recurring or deferred purchase. By continuing, I agree to the cancellation policy and authorize you to charge my payment method at the prices, frequency and dates listed on this page until my order is fulfilled or I cancel, if permitted. Pickup available at 99 Loomis Street Usually ready in 5+ days View store information 15+ Years of Customer Success High-Quality Quick Turnaround Recycled Products 10,000 sq. ft. of Capability Local Delivery & Nationwide Shipping Main menu Creative Digital Print Our Story ON-THE-SHELF-PRINTS Contact & Send File 99 Loomis Street, Manchester, CT 06042 860-647-8367 Translation missing: en.general.currency.dropdown_label CAD CAD USD INR GBP AUD EUR JPY © 2025 FUSION. 2025 FUSION CROSS-MEDIA LLC Search What are you looking for? Subscribe to our mailing list Be the first to read about our new arrivals, latest interviews, and newsletter exclusives by signing up below. First Name Email Sign Up What are you looking for?
10805	https://www.researchgate.net/publication/288124128_Collusion_in_a_Bertrand_duopoly_model_with_decreasing_returns_and_product_differentiation	(PDF) Collusion in a Bertrand duopoly model with decreasing returns and product differentiation Home Methodology Laboratory Techniques Laboratory Techniques and Procedures Weights and Measures Article PDF Available Collusion in a Bertrand duopoly model with decreasing returns and product differentiation June 2008 Estudios de economía 35(1):19-31 License CC BY-NC-SA Authors: Roberto Contreras Metropolitan University of Technology Nikolaos Georgantzis Burgundy School of Business Miguel Gines Miguel Gines This person is not on ResearchGate, or hasn't claimed this research yet. Download full-text PDFRead full-text Download full-text PDF Read full-text Download citation Copy link Link copied Read full-textDownload citation Copy link Link copied Citations (1)References (25) Abstract We study cartel stability in a differentiated price-setting duopoly with returns to scale. We show that a cartel may be equally stable in the presence of lower differentiation, provided that the decreasing returns parameter is high. In addition we demonstrate that for a given factor of discount, there are technologies that can have decreasing returns to scale where the cartel always is stable independent of the differentiation degree. Discover the world's research 25+ million members 160+ million publication pages 2.3+ billion citations Join for free Public Full-text 1 Content uploaded by Nikolaos Georgantzis Author content All content in this area was uploaded by Nikolaos Georgantzis on Jan 18, 2016 Content may be subject to copyright. Colusión en un duopolio de Bertrand… / R. Contreras, N. Geor gantzis, M. Ginés 19 Estudios de Economía. V ol. 35 - Nº 1, Junio 2008. Págs. 19-31 Colusión en un duopolio de Bertrand Con rendimientos deCreCientes y produCtos diferenCiados COLLUSION IN A BER TRAND DUOPOL Y MODEL WITH DECREASING RETURNS AND PR ODUCT DIFFERENTIATION R C N G M G Resumen Estudiamos las implicaciones que tienen sobre la estabilidad del cartel los rendimientos de escala en presencia de productos diferenciados cuando las empresas compiten en precios. Mostramos que el cartel puede ser igualmente estable en presencia de un menor grado de diferenciación siempre que las deseconomías de escala sean más intensas. Además, encontramos que para un determinado valor del factor de descuento crítico, siempr e será posible hallar unas determinadas deseconomías de escala donde los acuerdos colusivos son estables independiente del grado de diferenciación de los pr oductos. Palabras clave: Colusión, producto diferenciado, retornos decrecientes a escala. Abstract W e study cartel stability in a differentiated price-setting duopoly with returns to scale. W e show that a cartel may be equally stable in the presence of lower differentiation, provided that the decreasing r eturns parameter is high. In addi- tion we demonstrate that for a given factor of discount, there are technologies that can have decreasing returns to scale where the cartel always is stable independent of the differentiation de gree. Key words: Collusion, pr oduct differentiation, decreasing returns to scale. JEL Classification: C70; D21; D43. Deseamos agradecer a los participantes del XXXI Simposio de Análisis Económicos (Universidad de Oviedo, diciembre de 2006) por sus comentarios. Los errores u omisiones son exclusiva responsabilidad nuestra. Contreras desea agradecer la financiación recibi- da por la Uni versidad Tecnológica Metropolitana (Proyecto N° 287/2007, Resolución N° 04844 Exenta). Georgantzis agradece la hospitalidad del Depto. de Economía de la Universidad de Chipre y financiación recibida por el Ministerio de Educación y Ciencia (SEJ 2005-07544/ECON y PR 2007-0153). Departamento de Economía, Uni versidad Tecnológica Metropolitana, FAE, Santiago, Chile. E-mail: roberto.contreras@utem.cl Departamento de Economía y LEE, Universitat Jaume I, Campus Riu Sec 12071 Castellón, España. E-mails: georgant@eco.uji.es; mgines@eco.uji.es Estudios de Economía, V ol. 35 - Nº 1 20 I n t r o d u c c I ó n La literatura que analiza la estabilidad de un cartel, cuando las empresas compiten en precios, es bastante amplia. Así en un contexto espacial podemos mencionar los trabajos de Chang (1991) y Häckner (1996) en los que el grado de diferenciación se representa por la distancia que existe entre dos empresas dada su localización sobre la ciudad lineal de Hotelling (1929). Tales empresas deciden el precio al que venden sus productos en un juego que se repite infini- tamente, concluyendo que una mayor diferenciación entre el producto ofrecido por las dos empresas relaja la competencia y facilita la colusión. La diferencia entre ambos trabajos es que el primero utiliza como castigos la estrategia “tipo gatillo” propuesta por Friedman (1971), mientras que el segundo utiliza castigos simétricos óptimos propuestos por Abreu (1986). Chang (1992) analiza el comportamiento colusi vo de las firmas en mercados caracterizados por la diferenciación horizontal de productos, donde la elección del producto es endogenizada y las firmas pueden rediseñar sus productos in- tertemporalmente. Este autor encuentra que cuanto más flexible sea el diseño del producto (cuanto más bajo es el costo de relocalización) más difícil se hace sostener la colusión.En cambio, Häckner (1995) supone que el costo de redi- señar los productos es despreciable. Así obtiene que, si el factor de descuento respecto a los beneficios futuros es alto, las empresas escogerán un grado de diferenciación intermedio. Por otra parte, cuanto más bajo es el factor de des- cuento, más alta debe ser la diferenciación para que la colusión sea estable, por tanto, la diferenciación relaja la competencia y facilita la colusión. Georgantzis y Sabater (2002) siguiendo el criterio de Hendel y Figueiredo 1 y utilizando como marco de estudio el modelo de Salop (1979) con costos de transporte lineales, en el que la localización simétrica de las empresas sobre la ciudad circular está dada, analizan los efectos de publicitar la especificación del producto sobre la estabilidad del cartel. Se demuestra que publicitar la especificación del producto antes de su puesta en venta se convierte en una estrategia dominante para las empresas cuyo objetivo es coludir, debido a que el diseño del producto acordado se convierte en una señal para la empresa rival de la voluntad de cumplir el acuerdo colusivo. De esta forma, si el diseño del producto se publicita antes de su puesta en venta, es imposible engañar de forma total a la empresa ri val. Así, encuentran que las empresas diferencian su producto lo suficiente y ofrecen publicidad de sus modelos futuros para poder coludir de forma sostenible. Ross (1992) analiza la estabilidad del cartel y la diferenciación de productos utilizando para este análisis dos modelos diferen- tes. En el primero se utiliza un modelo no espacial, en el que encuentra que existe una relación no monótona entre el tipo de descuento crítico necesario para sostener el cartel y la diferenciación de productos compitiendo en precios. Así el cartel sería menos estable mientras más homogéneos sean los productos hasta cierto nivel de homogeneidad, a partir del cual se invierte la relación. En 1 Consideran que la elección de la especificación del producto se realiza con antelación a la decisión del precio del producto; el juego que desarrollan consta de tres etapas: en la primera etapa las empresas deciden si entran a producir, en la segunda etapa deciden el diseño de su producto y en la última etapa deciden el precio de su producto. Colusión en un duopolio de Bertrand… / R. Contreras, N. Geor gantzis, M. Ginés 21 el segundo se utiliza un modelo espacial donde las empresas compiten en pre- cios. En este caso, el autor determina que niveles más altos de diferenciación implican niveles más bajos de tipo de descuento crítico. Por lo tanto, el cartel que produce bienes heterogéneos es más estable que el cartel que produce bienes homogéneos. Rothschild (1997) también encuentra que existe una relación no monótona entre el tipo de descuento crítico necesario para sostener la colusión y la diferenciación de productos para un modelo de competencia en precios, pero, a diferencia del caso anterior, encuentra que la estabilidad del cartel es consistente con incrementos de homogeneidad sobre casi todo el rango cubierto por el parámetro diferenciador. La conclusión de estos dos trabajos es que la visión más tradicional, que establece que mientras más diferenciados sean los productos más fácil es sostener la colusión, resulta algo más confusa. Häckner (2000) en una nota sobre competencia en precios y cantidades en oligopolios diferenciados muestra que los resultados desarrollados en Singh y Vi ves (1984)2 son sensibles al tipo de duopolio que se asume. Si son más de dos empresas, los precios pueden ser mayores bajo competencia en precio que bajo competencia en cantidades. Este podría ser el caso en que las diferencias de calidad fueran grandes. Así no resultaría tan e vidente cuál de las dos maneras de competir (en precios o cantidades) es más eficiente. En resumen, cuando las empresas compiten en precios, generalmente se concluye que mientras más diferenciados son los productos, más fácil es sostener la colusión. Este hallazgo contrasta con los resultados de los estudios empíricos que llegan a una conclusión diferente, esto es, que mientras más homogéneos son los productos, más estable es el cartel. Así, se puede apreciar en los trabajos de Hay y Kelley (1974) y Symeonidis (2003). Debemos indicar que todos estos trabajos suponen rendimientos de escala constantes. Sin embargo, tanto la teoría económica como la evidencia empírica nos informan sobre el hecho de que la adopción de una determinada escala de producción por las empresas es el resultado de posibles deseconomías que aparecen a partir de la mínima escala eficiente. En términos teóricos, esto co- rresponde a la existencia de rendimientos decrecientes a escala que limitan la capacidad productiva de cada empresa puesto que una expansión excesiv a de su output hace que su costo marginal y medio superen los posibles beneficios de vender más unidades de producto. Intuitivamente, esto afecta los incenti vos individuales que las empresas integrantes de un cartel tienen para desviarse de un pacto de precios colusivos, puesto que la expansión del output resultante de una desviación conlleva pérdidas de eficiencia productiva como consecuencia de los rendimientos decrecientes. En concreto, pensamos que una desviación del precio colusivo en presencia de rendimientos decrecientes a escala no será tan rentable, ya que el aumento de demanda conseguido por la empresa que se desvía poniendo un precio inferior al pactado conlleva un aumento del costo marginal que merma los beneficios de la desviación, situación que no ocurre 2 Estos autores encuentran que, cuando los bienes son sustitutos, los beneficios de las empresas son mayores cuando compiten à la Cournot, mientras que cuando los bienes son complementarios los beneficios son mayores cuando compiten à la Bertrand. Así las empresas tendrán una estrategia dominante: elegir cantidad como variable estratégica, cuando los bienes son sustitutos y precio cuando los bienes son complementos. Estudios de Economía, V ol. 35 - Nº 1 22 cuando los costos marginales son constantes. Así, el objeto de este artículo es formalizar esta intuición. d e s c r I p c I ó n d e l M o d e l o Utilizaremos un modelo simétrico no espacial o modelo de Chamberlin, donde la producción por parte de las empresas tendrá unos costos (c i) que serán función no lineal del output de cada una de las empresas. Consideramos una industria compuesta por dos firmas i y j. Donde cada firma produce una variedad de un producto diferenciado que se enfrenta a la siguiente función de demanda 3: (1) qp p ii j =+−−+− 1 1 1 11 22 22 γ γ γ γ Así q i, p i representan respectivamente las cantidades y los precios de cada una de las firmas. Por otra parte, γ sirve como una medida del grado de dife- renciación entre los productos; si el valor de γ es uno, entonces los bienes son sustitutos perfectos. Por el contrario, si el valor de γ es cero, son bienes con funciones de demanda independientes (máxima diferenciación). Desde esa perspectiva los bienes son sustitutos imperfectos 4 si 0 ≤ γ < 1. La función de costos de la firma i viene representada por: cc qd q ii i =+ 1 2 2 (2) c c p p d i i j =+−−+−     + 1 1 1 1 1 1 2 1 1 2 2 () ()() ( γ γ γ γ ++ −−+−      γ γ γ γ )()() 1 1 1 2 2 2 p p i j Para que los costos, los precios, las cantidades y los beneficios sean positivos, para todos los subjuegos del juego analizado, suponemos 5 que: (3) dc≥∈    20 1 y, 3 Esta función de demanda para un modelo en donde la variable estratégica es el precio, es su equivalente a la función inversa de demanda p i= 1 – q i – γ q 2, donde la variable estratégica es la cantidad. 4 T ambién se puede indicar que los bienes son complementarios, si −1 < γ ≤ 0. Pero nosotros descartamos este caso para su análisis. 5 Aunque nuestro análisis se puede extender al caso de rendimientos a escala constantes y crecientes, nos concentramos en el caso de rendimientos decrecientes (d ≥ 2) para e vitar complicaciones innecesarias, relacionadas con la existencia y estabilidad de los equilibrios del juego. Colusión en un duopolio de Bertrand… / R. Contreras, N. Geor gantzis, M. Ginés 23 Como d es mayor que cero, nos encontramos en la zona donde los rendi- mientos de escala son decrecientes. Por otra parte, siguiendo la nomenclatura habitual, 1– c puede considerarse como el tamaño del mercado. c o l u s I ó n e n u n J u e g o r e p e t I d o La política de la competencia de las economías modernas impide celebrar contratos legales sobre acuerdos colusivos. Por tanto, la colusión debe ser equilibrio de Nash perfecto en subjuegos (NPS) de un juego repetido infinitas veces para que sea sostenible. En general, la colusión puede sostenerse si los beneficios presentes de respetar el acuerdo colusivo son mayores o iguales que los beneficios presentes de jugar el equilibrio no cooperativo de Bertrand. La estrategia colusiva que utilizaremos en nuestro modelo es la de Friedman (1971) quien define una estrategia “tipo gatillo” para un juego repetido. Definiremos π ic como el beneficio colusiv o de la empresa i si las empresas adhieren a la estrategia colusiv a. En el caso de una empresa que decide eng añar a sus socios de cartel, π id es el beneficio de la empresa i que se desvía de la estra- tegia colusi va (beneficio de hacer trampas). Y π ie es el beneficio de la empresa j que sufre el engaño mientras ella respeta el pacto. Finalmente π in es el beneficio que la empresa i gana en el equilibrio no cooperativo de Bertrand. 3.1. Cooperación en precios (equilibrio colusivo) en un período En nuestro modelo la solución cooperativa para las empresas involucra la maximización de los beneficios conjuntos respecto a los precios. De ellas se obtienen las condiciones de primer orden y de la solución del sistema se obtie- nen los precios de equilibrio a partir de los cuales obtendremos los beneficios colusivos de cada una de las empresas, esto es: (4) p cd d q c d ic ic i =+++ ++ ⇒= − ++ ⇒ ()() () () 11 21 1 21 γ γγ π c c c d =− ++ () (( )) 1 22 1 2 γ 3.2. Competencia en precios (equilibrio Bertrand-Nash) El proceso es similar al anterior, pero viene de la solución simultánea de la maximización de los beneficios individuales respecto a los precios, esto es: (5) p c d d q c d in in =++−+ ++ −⇒= − ++ ()() () ( 1 1 2 1 1 2 1 γ γ γ γ γ −− ⇒= −+− ++ − γ π γ γ γ ) ()() (()) in c d d 1 2 2 22 1 2 2 2 Estudios de Economía, V ol. 35 - Nº 1 24 3.3. Desviación unilateral del precio pactado (una de las empresas hace trampas) Una de las empresas puede calcular el precio para fijar la desviación óptima del acuerdo colusivo, mientras la otra empresa respeta el acuerdo (empresa engañada), suponiendo inicialmente que es la empresa 1 la que se desvía del acuerdo colusivo y la empresa 2 respeta el acuerdo (el resultado es simétrico cuando la situación se plantea a la inversa). p d d c id =++−++−++++ 2 2 2 2 3 2 1 2 3 1 2()()()()( γ γ γ γ γ γ d d d d +− − ++ +−⇒ γ γ γ γ γ 2 2 2 2 2 2 3 2 ) ()() (6) q c d d d id id =−++− +++− ⇒= ()() ()() (1 2 2 2 2 2 1 2 2 γ γ γ γ π −−++− +++− ⇒= −c d d d q ie )( ) ()() ( 2 2 2 2 2 2 2 2 2 2 2 1 γ γ γ γ c c d d d )( ) ()() 2 3 2 2 2 2 2 3 2 +− − ++ +− γ γ γ γ π γ γ γ γ γ γ ie c d d =−++− ++ −++() ()()(1 6 2 4 41 2 4 2 3 2 2 3 2 3 3 2 2 4 2 1 1 2 1 6 2 2 2 2 2 )()( ) ()( ++ −+−    ++ + d d γ γ γ γ γ d d −2 2 2 γ ) 3.4. Estructura de pagos del juego Tras simplificar los beneficios en cada una de las situaciones posibles, ob- tenemos los siguientes resultados: π γ π γ γγ ic in d d d =++ =+− ++− ( ) 1 2 2 2 2 2 2 2 2 () () π γγ γ γ id d d d =++− ++ +− ( ) () )( 2 2 2 2 2 22 2 2 (7) π γ γ γ γ γ γ ie d d d =++−++−+−+ 3 2 2 2 2 4 2 3 2 1 12 16 2()()()4 41 2 4 2 3 2 3 2 2 ()()+−++    − ( ) γ γ γ γ γ k h k Así, las empresas implicadas en el acuerdo sobre fijación de precios en común se enfrentan a la decisión de “Respetar” o “No Respetar” el acuerdo. De la decisión que cada una de las empresas tome, se generan determinados beneficios que denominaremos “Pagos”, los cuales cumplen las siguientes propiedades: (8) ππ ππ id ic in ie La primera desigualdad significa que la empresa tiene un incenti vo a deser- tar de la estrategia colusiva. La segunda desigualdad indica que la reversión a la estrategia de Bertrand es costosa, comparada con la adhesión a la estrategia colusiva. Finalmente, la tercera desigualdad nos indica que la empresa tiene un Colusión en un duopolio de Bertrand… / R. Contreras, N. Geor gantzis, M. Ginés 25 incentivo a aplicar la estrategia de Bertrand cuando es engañada y por lo tanto el castigo es creíble. F a c t o r d e d e s c u e n t o c r í t I c o Definimos un factor de descuento crítico similar al utilizado por Rothschild (1997), donde la colusión será una opción factible si el factor de descuento (α) es mayor o igual que el factor de descuento crítico (α). Esto es, si: (9) αα ππ ππ γγ γ ≥= − −=++− ++ () ()( id ic id in d d 2 21 8 22 22 −−+ ++−84 2 22 γγ γγ )( )d En adelante utilizaremos el factor de descuento crítico como medida de la estabilidad del cartel. Entonces, para valores cercanos a la unidad del factor de descuento crítico (α), menos estable es el cartel y para valores cercanos a un medio de (α ) más fácil será sostener la colusión. Eso es así, porque cuanto más alto es α, menor es el rango de valores de α para los que el cartel es factible. Después de operar y simplificar se puede reescribir la ecuación (9) de la siguiente forma: (10) αα γγ γγ γγ ≥=++ ++ −+ ++ () ()()( 1 2 1 22 18 84 2 22 22 2 dd−−    γ 2) Dado que d≥ 2 y que 0 ≤ γ <1, entonces el f actor de descuento crítico será mayor que 0.5, lo que es coherente con la literatura de la colusión. r e s u l t a d o s A partir de la expresión anterior encontramos una relación entre el paráme- tro de los rendimientos de la producción y la homogeneidad de los productos para un determinado valor del factor de descuento crítico, representado por la siguiente expresión: (11) 24 2 12 88 2 22 22 2 dd++−+ +−+−−   () () ()() γγ γα γγ γ  −21 α Esto implica que para un determinado valor del factor de descuento crítico (α) encontraremos una relación cuadrática entre los rendimientos de la pro- ducción y el parámetro de homogeneidad. Sin embargo, solo una de las raíces es factible como solución 6: 6 La ecuación tiene dos raíces, estas son: d 1 1 2 2 4 2 1=++−−− −    ()( γ α γ α γ γ α α /( ), ()(/(2 1 1 2 2 4 2 1 2 2 α γ αγ α γ γ α α −=++−−+−    d α α −1) . Sólo d 2 cumple la restricción impuesta en . Estudios de Economía, V ol. 35 - Nº 1 26 (12) d=++−−+− − ()()12 24 21 21 γα γα γγ αα α Así podemos decir que dado un valor crítico de α existe una relación cuadrática y directa entre los rendimientos de la producción y el parámetro diferenciador. Lo que nos conduce a la siguiente proposición: Proposición 1: En el duopolio de Bertrand con pr oductos diferenciados y ren- dimientos decrecientes (con d > 2) la facilidad de sostener acuerdos colusivos permanece constante, mientras un mayor grado de sustituibilidad entre los productos de los duopolistas (mayor γ ) viene compensado por un mayor nivel de deseconomías de escala (mayor d), como indica la expresión (12). En la Figura 1 que se muestra a continuación se puede observ ar la relación que existe entre los rendimientos de escala y la diferenciación de productos para distintos valores del factor de descuento crítico que garantiza la sostenibilidad del pacto colusivo. T odas las tecnologías de producción que combinan distintos grados de homo- geneidad de productos con determinados rendimientos de producción localizados sobre cada una de las curvas respectivas presentan un factor de descuento mayor que el factor de descuento crítico y donde el cartel o la colusión es estable para todas esas situaciones posibles. Se puede apreciar que existe alto grado de sustituibilidad entre los rendimientos de la producción (d) y la diferenciación de productos ( γ ) necesarios para sostener FIGURA 1 RELACIÓN DIFERENCIACIÓN Y RENDIMIENTOS DE PRODUCCIÓN 0,6 0,8 1 g -2 2 4 6 8 10 12 d 0,2 0,4 (α= 0,505, 0,51, 0,52, 0,53, 0,54, 0,55, 0,56, 0,57 y d = 2). Colusión en un duopolio de Bertrand… / R. Contreras, N. Geor gantzis, M. Ginés 27 un cartel. En otras palabras, para un factor de descuento bastante cercano a 0,5 se podrá coludir con un producto muy homogéneo siempre que las deseconomías de escala sean bastante intensas. Además, a medida que el f actor de descuento crítico se incrementa, disminuyen las deseconomías de escala necesarias para sostener la colusión con un producto muy homogéneo. En resumen, se puede constatar que, dado un determinado factor de descuento crítico a partir del cual es posible mantener la colusión, podremos encontrar carteles sostenibles en industrias con un menor o mayor grado de diferencia- ción dependiendo de los rendimientos de escala que éstas tengan. Finalmente, podemos indicar que la pendiente de la recta que relaciona la diferenciación de los productos con los rendimientos de escala decrece a medida que el factor de descuento crítico (α) se incrementa, lo que indica que, cuando el factor de descuento crítico aumenta ante un aumento del grado de homogeneidad del producto, se necesitará una variación menor de los rendimientos de escala para poder sostener la colusión. Por otra parte, evaluamos qué ocurre con el factor de descuento crítico necesario para sostener la colusión, con un grado de diferenciación constante, a medida que incrementamos las deseconomías de escala. (13) ∂ ∂=− +++− ++ − αγ γγ γ γγ ()() (()( d d d 41 2 22 18 8 22 2 22 +++ +− γγ γ 22 2 42)( ))d Se puede apreciar que existe una variación negativa entre los rendimientos de escala y el factor de descuento crítico. Corolario 1: A medida que las deseconomías de escala son más intensas (mientras mayor es d),el factor de descuento crítico necesario para sostener la colusión disminuye y, por tanto, se facilita la colusión. La Figura 2 muestra esta relación, donde cada una de las curv as representa la variación del factor de descuento crítico necesaria para que, aumentando las deseconomías de escala, se pueda mantener un determinado nivel de homogeneidad. Mientras más alejada del origen se encuentre la curva, más homogéneos son los bienes. En cada curva podemos observar cómo el factor de descuento disminuye a medida que se incrementa el parámetro (d), indicando esta relación negativa entre el f actor de descuento crítico y los rendimientos de producción. Otro de los aspectos que es importante analizar es saber qué ocurre con el factor de descuento crítico necesario para sostener la colusión con unos deter- minados rendimientos de la producción constantes a medida que incrementamos la homogeneidad de los productos, lo que podría ser válido para rendimientos constantes como sugieren los trabajos analizados en la literatura o con rendi- mientos decrecientes de producción. (14) ∂ ∂=+++− +++ + α γ γγ γγ γγ ()()(()) ( 21 22 21 2 22 2 dd d d(()()()())18 84 21 22 2 +−++ −+ γγ γγ γ d Estudios de Economía, V ol. 35 - Nº 1 28 Se puede apreciar que existe una variación positiva entre el grado de ho- mogeneidad de los productos y el factor de descuento crítico necesario para sostener la colusión. Corolario 2: P ara empresas con unos determinados rendimientos de escala encontramos que, mientras más homogéneos son los productos, más difícil es sostener el cartel. Lo anterior concuerda con los trabajos de Chang (1991), Ross (1992) y Häckner (1996) que analizan la facilidad para coludir en un modelo de com- petencia en precios, sin embargo, estos trabajos consideran rendimientos de producción constantes (d = 0). Finalmente, analizamos lo que ocurre con el factor de descuento crítico necesario para sostener la colusión cuando variamos los rendimientos de escala y el grado de homogeneidad de los productos. Por razones de espacio reempla- zaremos (1 + γ) por h. (15) ∂ ∂∂=− ++ ++−+ 2 3 2 2 3 2 4 6 4 9 3 2 2 α γ γ γ γ γ γ ()() d h d h d dh 2 2 2 3 3 2 3 4 24 12 11 2 3 2 8 4 5 2()()+−++−+−+   γ γ γ γ γ γ γ h  +−+++−    2 8 8 4 2 2 2 2 2 3 d h d()() γ γ γ γ Una consecuencia directa del signo negativo de la e xpresión (15) es: (16) γγ αα γγ 12 12 <⇒ ∂ ∂>∂ ∂ ,dd , que nos lleva a la siguiente proposición: FIGURA 2 RENDIMIENTOS DE PRODUCCIÓN Y FACTOR DE DESCUENT O CRÍTICO 4 6 8 10 12 d 0,50 5 0,51 0,51 5 0,52 0,52 5 0,53 a ( γ = 0,1, 0,3, 0,5, 0,7, 0,9, 1). Colusión en un duopolio de Bertrand… / R. Contreras, N. Geor gantzis, M. Ginés 29 Proposición 2: El efecto positivo de los rendimientos decrecientes sobre la sostenibilidad del cartel es monotónicamente cr eciente en el grado de sustitui- bilidad entre los pr oductos de los duopolistas. Intuitivamente, cuanto mayor es la sustituibilidad entre los productos de los duopolistas, menor es el efecto positivo (descrito en el Corolario 1) que los ren- dimientos decrecientes tienen sobre la sostenibilidad del cartel. Alternativamente, según la expresión (15), a medida que se incrementan las deseconomías de escala, la variación del factor de descuento crítico respecto a la homogeneidad de los productos disminuye y, por tanto, se incrementa la estabilidad del cartel. Por el contrario, cuanto menor es d, la variación del factor de descuento crítico respecto a la homogeneidad de los productos se incrementa y, por consiguiente, más difícil es sostener la colusión. Para entender el mecanismo que genera nuestro resultado central se debe tener en cuenta la fuerza principal que conlleva al colapso de un cartel, que es la rentabilidad de una desviación unilateral. De la teoría microeconómica y la definición del equilibrio Bertrand-Nash sabemos que desviarse unilateralmente de un pacto colusiv o es rentable. Sin embargo, la rentabilidad de dicha desvia- ción, que en gran medida determina lo inestable que es un cartel, depende de la sustituibilidad de los bienes que fabrican las empresas y de los rendimientos de producción. En concreto, una desviación del precio colusivo en presencia de rendimientos decrecientes a escala no será tan rentable, ya que el aumento de demanda conseguido por la empresa que desvía poniendo un precio inferior al pactado conlleva un aumento del costo marginal que merma los beneficios de la desviación. Eso explica intuitivamente el efecto positivo de las deseco- nomías a escala sobre la sostenibilidad del cartel enunciado en el Corolario 1 y la expresión (13). Consideremos ahora la rentabilidad de una desviación en presencia de menor diferenciación entre los productos fabricados por los integrantes de un cartel. Como es lógico, cuanto más sustitutivos sean los productos mayor será la rentabilidad de una desviación ya que con un recorte determinado del precio respecto al precio pactado se consigue atraer un mayor número de clientes de la otra empresa. Resumiendo estas dos observaciones sobre la intuición detrás de nuestros resultados, nótese que una desviación respecto al precio colusivo es más ren- table cuanto más sustitutivos sean los productos de los integrantes de un cartel y mientras los rendimientos decrecientes de la producción no alteren excesi- vamente el costo marginal como consecuencia del aumento de la demanda de la empresa desviante. En otras palabras, la diferenciación de los productos y las deseconomías de escala favorece la sostenibilidad de un cartel porque su presencia conjunta merma los beneficios generados de posibles desviaciones respecto al precio pactado. Es más, la ausencia o baja incidencia de una de las dos condiciones puede ser compensada por la presencia suficientemente fuerte de la otra, restaurando el clima necesario para la creación y supervivencia de pactos colusivos. Las implicaciones de este hallazgo para la política de la competencia es que en sectores en los que la escala mínima de producción es alta debido a una menor presencia de deseconomías a escala, podríamos observar menos carteles, a no ser que las empresas implicadas vendan productos muy diferenciados. Bajo este resultado, no sería suficiente aportar la no existencia de deseconomías o Estudios de Economía, V ol. 35 - Nº 1 30 incluso la existencia de economías de escala como prueba de la dificultad de pactar y mantener precios colusiv os en una industria, si no se demuestra que los productos de las empresas sospechadas por comportamientos anticompetiti vos son lo suficientemente sustituti vos entre sí. Sólo la ausencia de ambos factores, deseconomías de escala y diferenciación de los productos, puede generar el entorno de mercado más propicio para la adopción de estrategias pro competitivas. c o n c l u s I o n e s Como anticipamos, cuando las empresas compiten en precios existe una interacción entre los rendimientos de escala y el grado de diferenciación de los productos en la facilidad para coludir. En concreto, para un determinado grado de dificultad para sostener un cartel obtenemos una relación cuadrática y positiva entre el parámetro de los rendimientos decrecientes a escala y el grado de sustituibilidad de los productos.Desde esta perspectiva, se podrá encontrar industrias que teniendo la misma facilidad para coludir lo podrán hacer con un producto más homogéneo que otras, siempre que las deseconomías de escala sean más intensas. Una forma alternativ a de enunciar nuestros resultados sería observar que para un determinado nivel de diferenciación, cuanto más intensas son las deseconomías de escala, menor es el factor de descuento crítico y, por tanto, más fácil será sostener un cartel. Nuestros resultados confirman la intuición de que la existencia de rendimien- tos decrecientes reduce los incentiv os de desviarse de un pacto colusivo, debido al aumento del costo marginal que merma los beneficios de la desviación. En cuanto a la aplicabilidad de los resultados, nuestro análisis propone que sólo la ausencia de ambos factores, deseconomías de escala y diferenciación de pro- ductos, puede generar un entorno de mercado más propicio para la adopción de estrategias pro competitivas. Finalmente, podemos indicar que cuando las empresas compiten en pre- cios nuestro trabajo genera explicaciones razonables sobre las diferencias que existen entre los estudios teóricos, que en general concluyen que, mientras más diferenciados sean los productos, más fácil es sostener la colusión, por una parte, y por otra los estudios empíricos,que concuerdan en que mientras más homogéneos sean los productos, más fácil es sostener la colusión. Nosotros demostramos que ambas situaciones son posibles y que esto dependerá de los rendimientos de escala. r e F e r e n c I a s Abreu, D. (1986). “External Equilibria of Oligopolistic Supergames”. Journal of Economic Theory, 39: 191-225. Brock, W. y Scheinkman, J. (1985). “Price Setting Supergames with Capacity Constraints”. Review of Economic Studies, 52: 371-382. Chamberlin, E. (1951). “Monopolistic Competition Revisited”. Economica, New Series, XVIII, November, 343-362. Chang, M.-H. (1991). “The Effects of Product Differentiation on Collusive Pricing”. International Journal of Industrial Organization, 3: 453-470. Colusión en un duopolio de Bertrand… / R. Contreras, N. Geor gantzis, M. Ginés 31 Chang, M.-H. (1992). “Intertemporal Product Choice and Its Effects on Collusive Firm Behavior”. International Economic Review, 4: 773-793. Deneckere, R. (1983). “Duopoly Supergames with Product Differentiation”. Economics Letters, 11: 37-42. Friedman, J.W. (1971). “A Noncooperati ve Equilibrium for Supergames”. Review of Economic Studies, 38: 1-12. Georgantzis, L. y Sabater, G. (2002). “Market Transparenc y and Collusion: On the UK Agricultural Tractor Registration Exchange”. Eur opean Journal of Law and Economics, 14 (2), 129-150. Häckner, J. (1995). “Endogenous Product Design in an Infinitely Repeated Game”. International Journal of Industrial Organization, 13: 277-299. Häckner, J. (1996). “Optional Symmetric Punishments in a Bertrand Differentiated Products Duopoly”. International Journal of Industrial Organization, 14: 611-630. Häckner, J. (2000). “A Note on Price and Quantity Competition in Differentiated Oligopolies”. Journal of Economic Theory, 93: 233-239. Hay, G. y K elley, D. (1974). “An Empirical Survey of Price Fixing Conspiracies”. Journal of Law and Economics, 17: 13-38. Hendel, I. y Figueiredo J.N. (1997). “Product Differentiation and Endogenous Discounting”. International Journal of Industrial Organization, 16: 63-79. Hotelling, H. (1929). “Stability in Competition”. Economic Journal, 39: 41-57. Lambertini, L., Poddar, S. y Sasaki, D. (1998). “Standardization and the Stability of Collusion”.Economics Letters, 58: 303-310. Lambertini L. y Sasaki, D. (2001). “Marginal Costs and Collusive Sustainability”. Economics Letters, 72: 117-120. Lambertini, L. y Schultz, C. (2003). “Price or Quantity in Tacit Collusion?”. Economics Letters, 78: 131-137. Lancaster, K. (1966). “A New Approach to Consumer Theory”. Journal of P olitical Economy 74: 132-57. Martin, S. (1993). Advanced Industrial Economics. Oxford, UK Cambridge, Mass, USA Blackwell. Ross, T. (1992). “Cartel Stability and Product Differentiation”. International Journal of Industrial Organization, 10: 1-13. Rothschild, R. (1992). “On the Sustainability of Collusion in Differentiated Duopolies”. Economics Letters,40: 33-37. Rothschild, R. (1997). “Product Differentiation and Cartel Stability: Chamberlin versus Hotelling”. Annals of Regional Science, 31: 259-271. Salop, S. (1979). “Monopolistic Competition with Outside Goods”. Bell Journal of Economics, 10: 141-156. Symeonidis, G. (2003). “In which Industries Is Collusion More Likely? Evidence from the UK”. The Journal of Industrial Economics, 1: 45-74. Citations (1) References (25) ... Vasconcelos (2005) in a quantity setting oligopoly model assumes asymmetry by modelling that firms have different shares of a specific asset and shows that the sustainability of perfect collusion crucially depends on the most inefficient firm in the agreement, which represents the main obstacle to the enforcement of collusion. More recently, Miklós-Thal (2011) shows that in a Bertrand supergame some collusion is also sustainable under cost asymmetry whenever collusion is sustainable under cost symmetry and Contreras et al. (2008) have shown that with differentiated products a cartel may also be stable provided that returns to scale are high enough. Summarising, both the literature on static cartel stability and the dynamic models of tacit collusion suggest that collusion is unlikely to be observed in the presence of substantial competitive advantage, and therefore, a prior step before studying collusion sustainability when costs are heterogeneous and firms agree on output quotas is to consider whether collusion is viable. ... Imperfect collusion in an asymmetric duopoly Article Full-text available Jun 2018 Marc Escrihuela-Villar Carlos Gutierrez-Hita Using the coefficient of cooperation, we analyse the effect of cost asymmetries on collusive agreements when firms are able to coordinate on distinct output levels than the unrestricted joint profit maximization outcome. In this context, we first investigate the extent to which collusive agreements are feasible. Secondly, we focus on collusion sustainability in an infinitely repeated game. We show that, regardless of the degree of cost asymmetry, at least some collusion is always sustainable. Finally, the degree of collusion is also endogeneised to show that cooperation has an upper bound determined by the most inefficient firm. View Show abstract Monopolistic Competition Revisited Article Nov 1951 Edward H. Chamberlin View Advanced Industrial Economics Article Jan 1993 Stephen Martin View Product Differentiation and Endogenous Disutility Article Nov 1997 INT J IND ORGAN Igal Hendel João Neiva de Figueiredo This paper models the choice of degree of focus (or general purposeness) available to firms by endogenizing transportation costs in an address model of horizontal differentiation. The formulation is in three stages: entry, focus or design competition and price competition. The strategic effect of product design is analyzed. The equilibrium level of general purposeness is shown to depend critically on ‘neighbor exclusivity’. The latter and, more generally, market structure and product diversity are shown to depend on the cost of producing general purpose products. If general purposeness is ‘free’—to design and produce—only two firms enter the market and set large transportation costs, underproviding product diversity. View Show abstract Marginal costs and collusive sustainability Article Jul 2001 ECON LETT Luca Lambertini Dan Sasaki In an oligopoly supergame, firms’ actions in prices and quantities are subject to non-negativity constraints. These constraints can obstruct the practicability of some of the tacitly collusive subgame perfect equilibria, such as single-period optimal punishment which relies indispensably upon firms’ ability to charge prices strictly below marginal costs, i.e. temporarily loss making pricing. Thereby under the presence of positive price constraints, marginal costs can serve as a ‘fudge’ to materialise such penal pricing. In this paper we briefly shed light on the effects of profit–cost ratios (or mark-ups) on the sustainability of tacit collusion. View Show abstract Price or Quantity in Tacit Collusion? Article Jan 2003 ECON LETT Luca Lambertini Christian Schultz We investigate the choice of market variable, price or quantity, of an optimal implicit cartel. If the discount factor is high, the cartel can realize the monopoly profit in both cases. Otherwise, it is optimal for the cartel to rely on quantities in the collusive phase if goods are substitutes and prices if goods are complements. The reason is that this minimizes the gains from deviations from collusive play. View Show abstract Standardization and the Stability of Collusion Article Jul 1997 ECON LETT Luca Lambertini S. Poddar Dan Sasaki We characterize the interplay between firms' decision in terms of product standardization and the nature of their ensuing market behaviour. We prove the existence of a nonmonotone relationship between firms' decision at the product stage and their intertemporal preferences. View Show abstract The Effects of Product Differentiation on Collusive Pricing Article Sep 1991 INT J IND ORGAN Myong-Hun Chang Using the spatial competition framework of Hotelling (1929), this paper explores the relationship between the degree of product differentiation and the ability of firms to collude with respect to price. We show that the minimum discount factor required to support the joint profit maximum as an equilibrium outcome monotonically increases as products become better substitutes. When the joint profit maximum cannot be supported, the optimal collusive price is shown to decline as products become more substitutable. These findings suggest that firms producing stronger substitutes tend to find it tougher to collude in terms of their product price. View Show abstract Cartel Stability and Product Differentiation Article Feb 1990 INT J IND ORGAN Thomas W. Ross This paper employs a supergame-theoretic model of collusion to analyze the effects of different levels of product differentiation on cartel stability. In contrast to earlier work, the ability of the parties to reach an agreement is assumed. Here the focus is on how greater product homogeneity increases both the gain to cheating on a collusive agreement and the magnitude of the punishments that follow a defection. The total effect on the likelihood a cartel is stable is therefore unclear. Two particular differentiated products models are then used to demonstrate that, contrary to the conventional view, greater homogeneity can reduce cartel stability. View Show abstract Duopoly supergames with product differentiation Article Dec 1983 ECON LETT Raymond Deneckere When goods are complements or very close substitutes, more tacit collusion is supported by trigger strategies in price setting supergames than in quantity setting supergames. For moderate or poor substitutes, the situation is reversed. This contrasts with the static situation, where price setting strategies always lead to higher welfare. View Show abstract A New Approach to Consumer Theory Article Apr 1966 Kelvin J. Lancaster The research agendas of psychologists and economists now have several overlaps, with behavioural economics providing theoretical and experimental study of the relationship between behaviour and choice, and hedonic psychology discussing appropriate measures of outcomes of choice in terms of overall utility or life satisfaction. Here we model the relationship between values (understood as principles guiding behaviour), choices and their final outcomes in terms of life satisfaction, and use data from the BHPS to assess whether our ideas on what is important in life (individual values) are broadly connected to what we experience as important in our lives (life satisfaction). View Show abstract Show more Recommended publications Discover more about:Weights and Measures Article On the Macroscopic Asymptotic Stability of an Interconnected System with Time-Delay January 1974 · Transactions of the Society of Instrument and Control Engineers Hidekatsu Tokumaru Norihiko Adachi Takashi AMEMIYA In a preceeding paper the concept of macroscopic stability was introduced, and a sufficient condition for the macroscopic asymptotic stability of a large scale interconnected system is given. The concept of macroscopic stability is introduced from the points of view that there must be certain variables representing the global characteristics of a large scale system and that in the analysis of a ... [Show full abstract] large scale system the behavior of these variables is important. The sufficient condition was given on the basis of the comparison function for each subsystem with regard to its input-output relation. In this paper the macroscopic stability of interconnected systems with time delay is considered. For this purpose the Matrosov's comparison theorem is generalized to the functional differential inequalities. With the aid of this theorem and comparison functions, the macroscopic asymptotic stability of interconnected systems with time delay between subsystems is studied, and it is shown that the theorem in the previous paper for a system with no time delay is also applicable to this system without any modification. Furthermore a more general comparison function is introduced for a subsystem with time delay and by means of this function and comparison theorem, conditions for the macroscopic stability of the more general interconnected systems with time delay are studied. The relation between the Lyapunov functional and the comparison function is presented for a subsystem with time delay. Read more Article Exponential stability and stabilizability in mean square of large-scale stochastic systems September 1999 · Stochastic Analysis and Applications Christophe Boulanger This paper deals with decomposition techniques for nonlinear large ndash scale stochastic systems which have the feature that the interactions between the various subsystems are nonadditive. We consider the system differential equations into a hierarchical form (see Vidyasagar §). We prove that the overall system is exponentially stable (respectively stabilizable) in mean square if and only if ... [Show full abstract] each of the subsystems is exponentially stable (respectively stabilizable) in mean square. Several examples are presented to illustrate the various theorems Read more Article The effect of differential shift costs on capital utilization March 1976 · Journal of Development Economics Mary Ann Baily This paper presents the results of a micro study of the influence of the shift differential factor on utilization rates in Kenyan manufacturing. Qualitative information about the nature of technology and of shift differentials is incorporated into models of the choice of utilization rate under the alternative assumptions of constant returns to scale and increasing returns to scale. Econometric ... [Show full abstract] tests are then used to show that 1971 data is consistent with the theoretical predictions, although the results suggest that the quantitative significance of the shift differential factor in causing excess capacity should not be overstated. Read more Article Full-text available Charged-Higgs-boson production at the LHC: Next-to-leading-order supersymmetric QCD corrections June 2009 · Physical Review D Stefan Dittmaier M. Kraemer Michael Spira Manuel Walser The dominant production process for heavy charged-Higgs bosons at the LHC is the associated production with heavy quarks. We have calculated the next-to-leading-order supersymmetric QCD corrections to charged-Higgs production through the parton processes qq̅ ,gg→tbH± and present results for total cross sections and differential distributions. The QCD corrections reduce the renormalization and ... [Show full abstract] factorization scale dependence and thus stabilize the theoretical predictions. We present a comparison of the next-to-leading-order results for the inclusive cross section with a calculation based on bottom-gluon fusion gb→tH± and discuss the impact of the next-to-leading-order corrections on charged-Higgs searches at the LHC. 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10806	https://marcjoye.github.io/papers/JY00sd2r.pdf	Optimal Left-to-right Binary Signed-Digit Recoding [Published in IEEE Transactions on Computers 49(7):740–748, 2000.] Marc Joye1 and Sung-Ming Yen2 1 Gemplus Card International Parc d’Activit´ es de G´ emenos, B.P. 100, 13881 G´ emenos, France marc.joye@gemplus.com 2 Laboratory of Cryptography and Information Security (LCIS) National Central University, Chung-Li, Taiwan 320, R.O.C. yensm@csie.ncu.edu.tw Abstract. This paper describes new methods for producing optimal binary signed-digit representations. This can be useful in the fast com-putation of exponentiations. Contrary to existing algorithms, the digits are scanned from left to right (i.e., from the most signiﬁcant position to the least signiﬁcant position). This may lead to better performances in both hardware and software. Keywords. Computer arithmetic, converter, signed-digit representation, re-dundant number representation, SD2 left-to-right recoding, canonical/non-adjacent/minimum-weight form, exponentiation, elliptic curves, smart-cards, cryptography. 1 Introduction Methodology using signed-digit (sd) representations, also called redundant num-ber representations, for fast parallel arithmetic were considered in the late 1950’s by Avizienis . More recently, algorithms using signed-digit representations with the digit set {−1, 0, 1} (in this paper, we call it the sd2 representation) accompanied with their applications to eﬃcient methods for addition, multipli-cation, division, and their vlsi chip designs are presented in [2–6]. In general, after the computations in the sd2 domain, a converter is required to transform a number from its sd2 representation into the conventional binary representation. Such converters may be found in [7, Section 1.5], . For many cryptosystems, (modular) exponentiation is one of the most time-consuming operations. Therefore, eﬃcient algorithms to perform this operation are crucial in the performance of the resulting cryptographic protocols. Basically, when computing αr, two types of exponentiations may be distinguished. – The ﬁrst type involves exponentiations with a ﬁxed exponent r, such as in the rsa cryptosystem [9, 10]. The goal is then to quickly compute αr for 2 Marc Joye and Sung-Ming Yen randomly chosen α. This is usually achieved thanks to addition chains [11, Section 4.6.3]. The problem of ﬁnding the shortest addition chain was shown to be NP-hard by Downey, Leong and Sethi ; but good heuristics are known . – In the second type of exponentiation, the base α is ﬁxed and the exponent r varies. Examples include the ElGamal cryptosystem and its numerous variations [15, Section 11.5]. In that case, good performances are obtained by the basic square-and-multiply technique (see Section 2). If larger amount of storage is available, this can be further improved via precomputations . In this paper, we are mainly concerned with the second type of exponenti-ation. However, we note that our methods may lead to some advantages in the ﬁrst type, too. Another application is when inverses can be virtually computed for free, as for elliptic curves . The basic idea is to recode the exponent in a representation which has fewer nonzero digits, namely the sd2 representation. Already in 1951, this was successfully exploited by Booth to eﬃciently multiply two numbers . An optimal version (in terms of the number of zero digits of the recoding) was later given by Reitwiesner . In , Jedwab and Mitchell re-discovered Reitwiesner’s algorithm and slightly generalized it by taking as input any sd2 representation of the exponent (instead of the binary representation). Our algorithms also produce optimal outputs but scan the digits of the expo-nent from left to right, i.e., from the most signiﬁcant digit (msd) to the least signiﬁcant digit (lsd). This brings some advantages, especially in the hardware realization or for memory-constrained environments like smart-cards. The rest of this paper is organized as follows. In Section 2, we review the square-and-multiply methods for fast exponentiation and extend them to expo-nents given in the sd2 representation. Section 3 presents Reitwiesner’s algorithm. New exponent recoding algorithms are proposed in Section 4. Their hardware implementation is given in Section 5. Section 6 discusses further advantages of the proposed methods. Finally, we conclude in Section 7. Notations If r = Pm−1 i=0 ri 2i denotes the binary expansion of r, then we represent r as the vector (rm−1, . . . , r0)2. The bit-length of r is denoted by \|r\|. By abuse of notations, we do not make the distinction between the value of r and its repre-sentation and write r = (rm−1, . . . , r0)2. For signed-digit systems, we sometimes write ¯ 1 for −1. Moreover, if r = Pm i=0 r′ i 2i denotes the binary signed expan-sion of r (that is, r′ i ∈{¯ 1, 0, 1}), then we also abuse the notations and write r = (r′ m, . . . , r′ 0)sd2. Let S be a string. Then ⟨S⟩k means S, S, . . . , S (k times); for example, (⟨0, 1⟩2, ¯ 1)sd2 represents (0, 1, 0, 1, ¯ 1)sd2. If t is a real number, then ⌊t⌋is the largest integer ≤t and ⌈t⌉is the least integer ≥t. Throughout this paper, the multiplicative notation is used. However, the described techniques also apply to additively written sets (e.g., additive group Optimal Left-to-right Binary Signed-Digit Recoding 3 of integers, points on an elliptic curve over a ﬁeld). Exponentiation has then to be understood as multiplication. 2 Binary Algorithms for Fast Exponentiation The most commonly used algorithms for computing αr are the binary methods [11, Section 4.6.3]. The binary methods (also called square-and-multiply methods) scan the bits of exponent r either from right to left or from left to right (Fig. 1). At each step, a squaring is performed and depending on whether the scanned bit-value is equal to 1, a multiplication is also performed. Let r = Pm−1 i=0 ri 2i (with rm−1 = 1) be the binary expansion of r. The right-to-left (rl) algorithm is based on the observation that αr = α20r0 α21r1 · · · α2m−1rm−1, while the left-to-right (lr) algorithm follows from αr = · · · (αrm−1)2 αrm−22 αrm−32 · · · αr12 αr0 . INPUT: α, r = (rm−1, . . . , r0)2 OUTPUT: M = αr M ←1; S ←α for i from 0 to m −1 do if (ri = 1) then M ←M · S S ←S2 od (a) Right-to-left (rl). INPUT: α, r = (rm−1, . . . , r0)2 OUTPUT: M = αr M ←1 for i from m −1 down to 0 do M ←M 2 if (ri = 1) then M ←M · α od (b) Left-to-right (lr). Fig. 1. Square-and-multiply algorithms. We remark that the lr algorithm (Fig. 1 (b)) requires 2 registers (for α and for M) and that the rl algorithm (Fig. 1 (a)) requires one more register (for S). However, we note that S can be used in place of α if the value of α is not needed thereafter. The rl algorithm presents the advantage to be parallelizable: one multiplier performs the multiplications M ←M ·S and another one performs the squarings S ←S2. However, if only one multiplier is available, the lr algorithm may be preferred because the multiplications are always done by the ﬁxed value α, M ←M·α. So, if α has a special structure, these multiplications may be easier than multiplying two arbitrary numbers (see [15, Note 14.81] or [21, pp. 9–10] for examples of application). Let ω(r) denote the Hamming weight of r (that is, the number of 1’s in the binary representation of r). Both algorithms require ω(r)−1 multiplications and m−1 squarings (we do not count multiplications by 1, nor 1·1, nor the last squar-ing in Fig. 1 (a)) to compute αr. It is well-known that m−1 is a lower bound for the number of squarings. However, the number of subsequent multiplications can 4 Marc Joye and Sung-Ming Yen be further reduced by using a recoding algorithm [20, 22]. For example, to com-pute α15, the lr algorithm will successively evaluate α, α2, α3, α6, α7, α14, α15, that is, it performs 3 squarings and 3 multiplications. If the value of α−1 is supplied along with α (or if α−1 can cheaply be computed, as for elliptic curves ), then α15 is more quickly evaluated as α16 · α−1 = (((α2)2)2)2 · α−1, which requires 4 squarings and 1 multiplication. If we allow the digits of the exponent to be in {¯ 1, 0, 1}, then the binary methods to compute αr are easily modiﬁed as depicted in Fig. 2. Note that the sd2 representation of r may require an extra digit, r′ m, we refer to Section 3 for an explanation. INPUT: α, r = (r′ m, . . . , r′ 0)sd2 OUTPUT: M = αr M ←1; S ←α for i from 0 to m do if (r′ i = 1) then M ←M · S if (r′ i = ¯ 1) then M ←M · S−1 S ←S2 od (a) Modiﬁed right-to-left (mrl). INPUT: α, r = (r′ m, . . . , r′ 0)sd2 OUTPUT: M = αr M ←1 for i from m down to 0 do M ←M 2 if (r′ i = 1) then M ←M · α if (r′ i = ¯ 1) then M ←M · α−1 od (b) Modiﬁed left-to-right (mlr). Fig. 2. Square-and-multiply-or-divide algorithms. We see that the modiﬁed right-to-left (mrl) algorithm (Fig. 2 (a)) now works more diﬀerently and less eﬃciently. Indeed, when r′ i = ¯ 1, the inverse of S has to be computed (note that the value of S varies at each step). Most commonly used cryptosystems work in the multiplicative group of a ﬁnite ﬁeld or ring. Inversion is then usually achieved via the extended Euclidean algorithm [15, Algorithm 2.142] or Fermat’s Theorem [15, Fact 2.127] (see also for a spe-cialized implementation). This is a rather costly operation; therefore the beneﬁts resulting from the reduced number of multiplications may be annihilated. The modiﬁed left-to-right (mlr) algorithm (Fig. 2 (b)) only needs the ﬁxed value of α−1, which can be precomputed. Consequently, in the sequel, we will only con-sider the mlr algorithm. We note, however, that the mrl algorithm may be useful when the inverse is available at no cost, as for elliptic curves or for the additive group of integers. With the sd2 representation, the (minimal) Hamming weight ω(r) of r (i.e., the number of nonzero digits in the sd2 expansion of r) is equal to (m+1)/3, on average . The computation of αr can thus be performed with 4 3 m+O(1) mul-tiplications plus squarings with the mlr algorithm, while the (standard) binary algorithms need 3 2 m + O(1) multiplications plus squarings (see Fig. 1), on aver-age. Assuming that a squaring is approximatively as costly as a multiplication, then we can roughly expect a gain of ( 3 2 −4 3)/ 3 2 ≈11.11 % over the (standard) binary methods. Optimal Left-to-right Binary Signed-Digit Recoding 5 The next section presents an algorithm to convert the exponent r from its binary representation into its sd2 representation. This algorithm, due to Re-itwiesner, is optimal in the sense that it gives an sd2 output with minimal Hamming weight. Unfortunately, Reitwiesner’s algorithm scans the bits of the exponent from right to left, while we have seen that only the modiﬁed left-to-right algorithm may bring some advantages. These heterogeneous modes of operation require to ﬁrst recode the exponent r into its sd2 representation (r′ m, . . . , r′ 0)sd2 and to temporarily store it for its latter usage in the mlr exponentiation algo-rithm. Noting that each digit in sd2 representation is encoded with 2 bits, twice the memory space taken by r is needed to store its sd2 representation. These shortcomings are alleviated in Section 4, where optimal left-to-right recoding algorithms are presented. 3 Reitwiesner’s Method In binary signed-digit notation1 (i.e., using the digits {¯ 1, 0, 1}), a number is not uniquely represented. Two representations (aℓ, aℓ−1, . . . , a0)sd2 and (bℓ, bℓ−1, . . . , b0)sd2 of a same number are equivalent if they have both the same length and the same Hamming weight; this equivalence will be denoted (aℓ, aℓ−1, . . . , a0)sd2 ≡ (bℓ, bℓ−1, . . . , b0)sd2. A binary signed-digit representation is said to be canonical (or sparse) if no two adjacent digits are nonzero. For that reason, some authors sometimes call it the nonadjacent form (naf) of a number . The canonical recoding was studied by Reitwiesner . He proved that this representation is unique (if the binary representation is viewed as padded with an initial 0). Following Hwang [7, pp. 150–151], Reitwiesner’s method to con-vert a number r = (rm−1, . . . , r0)2 with ri ∈{0, 1} into its canonical form r = (r′ m, r′ m−1, . . . , r′ 0)sd2 with r′ i ∈{¯ 1, 0, 1} is given by the algorithm depicted in Fig. 3. This is also known as Booth canonical recoding algorithm; however, Booth’s method does not present the naf property (see Footnote 2). INPUT: (rm−1, . . . , r0)2 OUTPUT: (r′ m, r′ m−1, . . . , r′ 0)sd2 c0 ←0; rm+1 ←0; rm ←0 for i from 0 to m do ci+1 ←⌊(ci + ri + ri+1)/2⌋ r′ i ←ci + ri −2ci+1 od Fig. 3. Reitwiesner’s canonical recoding algorithm. Reitwiesner’s algorithm is very eﬃcient. It can be done by using the following look-up table (‘X’ stands for 0 or 1, that is, the output is independent of this value). 1 Also called ternary balanced notation [11, p. 190]. 6 Marc Joye and Sung-Ming Yen Table 1. Right-to-left sd exponent recoding. ci ri ri+1 ci+1 r′ i 0 0 X 0 0 0 1 0 0 1 0 1 1 1 ¯ 1 1 0 0 0 1 1 0 1 1 ¯ 1 1 1 X 1 0 At ﬁrst glance, it is not so obvious that this algorithm eﬀectively yields the canonical representation of a number. However, if we closely observe how it works, we see that this algorithm comes down to subtract r from 3r (with the additional rule 0 −1 = ¯ 1) and then to discard the last (i.e., least signiﬁcant) 0 . 2r = (rm−1, rm−2 , rm−3, . . . , r1, r0, 0 )2 + r = (rm−1, rm−2, . . . , r2, r1, r0)2 3r = (sm, sm−1 , sm−2 , sm−3, . . . , s1, s0, r0)2 −r = (rm−1, rm−2, . . . , r2, r1, r0)2 2r = (r′ m, r′ m−1 , r′ m−2 , r′ m−3, . . . , r′ 1, r′ 0, 0 )sd2 Fig. 4. A simple explanation of Reitwiesner’s method. Indeed, if r0 + Pm i=0 si 2i+1 denotes the binary expansion of 3r, then the conventional pencil-and-paper method to add nonnegative integers [11, p. 251] gives si = (ci +ri +ri+1) mod 2 = ci +ri +ri+1 −2⌊(ci +ri +ri+1)/2⌋where ci is the carry-in. Moreover, since the carry-out ci+1 is equal to 1 if and only if there are two or three 1’s among ci, ri and ri+1, we can write ci+1 = ⌊(ci+ri+ri+1)/2⌋. Hence, si = ci + ri + ri+1 −2ci+1 and thus r′ i = si −ri+1 = ci + ri −2ci+1. To see that the output is sparse, it suﬃces to remark that the algorithm scans the bits from right to left and replaces a consecutive block of several 1’s by a block of 0’s and ¯ 1 according to (⟨1⟩a)2 7→(1, ⟨0⟩a−1, ¯ 1)sd2.2 Furthermore, if two blocks of 1’s are separated by an isolated 0, the algorithm implicitly uses the fact that (¯ 1, 1)sd2 ≡(0, ¯ 1)sd2. For example, (⟨1⟩a, 0, ⟨1⟩b)2 is replaced by (1, ⟨0⟩a, ¯ 1, ⟨0⟩b−1, ¯ 1)sd2—and not by (1, ⟨0⟩a−1, ¯ 1, 1, ⟨0⟩b−1, ¯ 1)sd2. A more formal (but less intuitive) proof of the sparseness property is given in the next lemma. Lemma 1. r′ i · r′ i+1 = 0 for all 0 ≤i ≤m −1. 2 This is the transformation initially proposed by Booth . Optimal Left-to-right Binary Signed-Digit Recoding 7 Proof. Suppose r′ i ̸= 0. Since r′ i = ci + ri −2ci+1, we must have ci + ri = 1, whence ci+1 = ⌊(1 + ri+1)/2⌋= ri+1 and thus r′ i+1 = 2(ri+1 −ci+2) = 0. ⊓ ⊔ The main advantage of Reitwiesner’s algorithm is that, in some sense, it is optimal. Indeed, Reitwiesner proved that: Proposition 1. Among the sd2 representations, the canonical representation has minimal Hamming weight. ⊓ ⊔ Although, this does not rule out the existence of other minimal representa-tions. For example, (1, 0, 1, 1)sd2 and (1, 0, ¯ 1, 0, ¯ 1)sd2 are both minimal represen-tations for 11. The general case was later addressed by Clark and Liang . They present a minimal representation for any signed-radix b. In that case, Arno and Wheeler proved that the average proportion of nonzero digits is equal to (b−1)/(b+1). This has to be compared with the average proportion (b −1)/b of nonzero digits in the standard radix b representation. So, we see that exponent recoding is mostly interesting for binary signed-digit representation (b = 2) because the savings rapidly go down. 4 Proposed Methods In Section 2, we pointed out that a left-to-right recoding algorithm might be desirable for fast exponentiation. Designing such an algorithm is not as straight-forward as it appears and is even considered as a hard problem by some au-thors . Our ﬁrst algorithm is an adaptation of Reitwiesner’s algorithm. As in the right-to-left algorithm, it also presents the naf property. Unfortunately, look-up tables cannot be used. Our second algorithm does not have the naf property but is equally eﬃcient as Reitwiesner’s algorithm. Moreover, it enables the use of a look-up table. 4.1 A Simple Left-to-right Recoding Algorithm The interpretation of the naf given in the previous section suggests a simple way to construct a left-to-right recoding algorithm. Let r = (rm−1, . . . , r0)2 and 3r = (sm, . . . , s0, r0)2, then the naf for r is (r′ m, r′ m−1, . . . , r′ 0)sd2 where r′ i = si−ri+1. So, if we have at our disposal an algorithm to add r = (rm−1, . . . , r0)2 and 2r = (rm−1, . . . , r0, 0)2 from left to right, then we can compute 3r, subtract r (in sd2 representation), discard the last 0 and obtain the canonical representation of r. Fortunately, such algorithms exist. A left-to-right addition algorithm is presented in Fig. 5 (see [11, Exercise 4.3.1.6]). Taking as input u = (0, rm−1, . . . , r1)2 and v = (rm−1, rm−2, . . . , r0)2, we get w = (sm, . . . , s0)2; we have now to subtract (rm−1, . . . , r1)2 (with the rule 0 −1 = ¯ 1). However, some technical diﬃculties occur: when we enter in the while loop (Lines 5–7 and 12–14 in Fig. 5), the algorithm may output several 8 Marc Joye and Sung-Ming Yen INPUT: u = (um−1, . . . , u0)2 and v = (vm−1, . . . , v0)2 OUTPUT: u + v = (wm, wm−1, . . . , w0)2 j ←m for i from m −1 down to 0 do if (ui = vi) then wj ←vi while (j > i + 1) do j ←j −1; wj ←1 −vi od j ←j −1 fi od wj ←0 while (j > 0) do j ←j −1; wj ←1 od Fig. 5. Left-to-right addition algorithm. sj’s (with j > i) wherefrom, for each sj, we have to subtract rj+1 in order to obtain the corresponding r′ j. This means that the rj+1’s with j > i (i.e., ri+2, ri+3, . . . ) must be available. On the other hand, we remark that at step i = I, variable j is only decremented when rI+1 = rI. So, if J denotes the value of j before entering in the while loop, we see that this loop is only executed after a block (rJ+1, rJ, . . . , rI+1, rI) either of the form (B1) (0, ⟨0, 1⟩(J−i)/2, 1) or (0, ⟨0, 1⟩(J−i−1)/2, 0, 0) , or of the form (B2) (1, ⟨1, 0⟩(J−i)/2, 0) or (1, ⟨1, 0⟩(J−i−1)/2, 1, 1) . We therefore introduce an additional variable b to distinguish between the two cases. Because of the alternation of 0 and 1 inside a block, we do not have to know the value of the rj+1’s inside the while loop; only the value of the two ﬁrst consecutive equal bits (i.e., rJ+1 and rJ) is necessary: we use the variable b to keep track of this value, that is, before entering the loop, b contains the value of rJ (b = 0 in Case (B1) and b = 1 in Case (B2)). Next, inside the while loop, we alternately subtract 0 or 1, starting with 0 or 1 depending on the value of b. Putting all together, we ﬁnally our ﬁrst algorithm (Fig. 6). While this algorithm yields the canonical representation, it is not fully satisfy-ing. It looks quite cumbersome compared to the original Reitwiesner’s algorithm (see Fig. 3). The next paragraph considers another minimal representation which leads to a very elegant left-to-right recoding algorithm. Optimal Left-to-right Binary Signed-Digit Recoding 9 INPUT: (rm−1, . . . , r0)2 OUTPUT: (r′ m, r′ m−1, . . . , r′ 0)sd2 j ←m; b ←0; rm ←0 for i from m −1 down to 0 do if (ri+1 = ri) then r′ j ←ri −b while (j > i + 1) do j ←j −1; r′ j ←1 −ri −b b ←1 −b od b ←ri; j ←j −1 fi od r′ j ←−b while (j > 0) do j ←j −1; r′ j ←1 −b b ←1 −b od Fig. 6. Canonical left-to-right recoding algorithm. 4.2 Minimum-weight Left-to-right Recoding Algorithm The main diﬃculty in the previous algorithm comes from the fact that it can only “decide” what the output will be after two consecutive equal bits. In other words, when entering in an alternate block of ⟨0, 1⟩k (or ⟨1, 0⟩k), the algorithm must know a priori if this block will end with two consecutive bits equal to 0 or 1. The next lemma enables to give a minimal (albeit not sparse) output whatever the ending bits of an alternate block. The while loop’s in the canonical algorithm (Fig. 6) can therefore be removed. Lemma 2. In sd2 representation, we have the following equivalences: (a) (⟨0, 1⟩k, 1)sd2 ≡(1, ⟨0, ¯ 1⟩k)sd2; (b) (⟨0, ¯ 1⟩k, ¯ 1)sd2 ≡(¯ 1, ⟨0, 1⟩k)sd2. Proof. Let Sk = Pk i=1 22(i−1) = (22k −1)/3. In signed-digit representation, (⟨0, 1⟩k, 1)sd2 represents the number N = 1 + Pk i=1 22(i−1)+1. However since N = 1 + 2Sk = (3Sk + 1) −Sk = 22k −Pk i=1 22(i−1), it may also be represented as (1, ⟨0, ¯ 1⟩k)sd2. Noting that (⟨0, ¯ 1⟩k, ¯ 1)sd2 and (¯ 1, ⟨0, 1⟩k)sd2 are both represen-tations of −N, the second equivalence follows from the ﬁrst one. ⊓ ⊔ Elementary Blocks [This paragraph explains how the recoding algorithm was found. The reader uniquely interested by the algorithm itself may skip it.] The exponent r to be recoded may be viewed as a succession of elementary blocks (rJ+1, rJ, . . . , rI+1, rI) with J > I, rJ+1 = rJ and rI+1 = rI, and whose form is given by one of the four following possibilities. 10 Marc Joye and Sung-Ming Yen (E1) (0, ⟨0, 1⟩(J−I)/2, 1), (E2) (0, ⟨0, 1⟩(J−I−1)/2, 0, 0), (E3) (1, ⟨1, 0⟩(J−I)/2, 0), (E4) (1, ⟨1, 0⟩(J−I−1)/2, 1, 1). For each of these forms, our left-to-right canonical recoding algorithm (Fig. 6) will respectively output the corresponding block (r′ J, . . . , r′ I+1) given by (E1’) (1, ⟨0, ¯ 1⟩(J−I−2)/2, 0), (E2’) (0, ⟨1, 0⟩(J−I−1)/2), (E3’) (¯ 1, ⟨0, 1⟩(J−I−2)/2, 0), (E4’) (0, ⟨¯ 1, 0⟩(J−I−1)/2). Now, using Lemma 2, we may respectively replace these outputs by (E1) (⟨0, 1⟩(J−I−2)/2, 1, 0), (E2) (⟨0, 1⟩(J−I−1)/2, 0), (= E2’) (E3) (⟨0, ¯ 1⟩(J−I−2)/2, ¯ 1, 0), (E4) (⟨0, ¯ 1⟩(J−I−1)/2, 0). (= E4’) Example 1. Suppose that r = (1, 1, 1, 0, 1, 0, 0, 1)2. By adding artiﬁcial beginning and ending 0’s, we decompose r into 4 elementary blocks. We then apply our transformation to each elementary block to ﬁnally obtain r = (1, 0, 0, 0, ¯ 1, ¯ 1, 0, 0, 1)sd2. r = ( 0 0 1 1 1 0 1 0 0 1 . 0 0 )2 (E1) 0 0 1 1 (E4) 1 1 1 (E3) 1 1 0 1 0 0 (E2) 0 0 1 . 0 0 (E1) 1 0 (E4) 0 (E3) 0 ¯ 1 ¯ 1 0 (E2) 0 1 . 0 →r = ( 1 0 0 0 ¯ 1 ¯ 1 0 0 1 . 0 )sd2 Although not sparse, the resulting representation has the same Hamming weight as the canonical representation (1, 0, 0, ¯ 1, 0, 1, 0, 0, 1)sd2. Both representations be-ing equivalent, (1, 0, 0, 0, ¯ 1, ¯ 1, 0, 0, 1)sd2 is thus a minimal representation for r.♦ Consider an elementary block (rJ+1, rJ, . . . , rI+1, rI) and the corresponding signed-digit block (r′ J, . . . , r′ I+1). Let bi denote the value of variable b when bits ri and ri−1 are scanned, namely bi = 0 or 1 depending on whether ri and ri−1 belong to an elementary block beginning with two zeros or two 1’s. Suppose that rJ+1 = rJ = 0. Then we have bi = 0 for all J + 1 ≥i ≥I + 2, and bI+1 = 0 if rI+1 = rI = 0 or bI+1 = 1 if rI+1 = rI = 1. Similarly, if rJ+1 = rJ = 1, then Optimal Left-to-right Binary Signed-Digit Recoding 11 bi = 1 for all J + 1 ≥i ≥I + 2, and bI+1 = 0 if rI+1 = rI = 0 or bI+1 = 1 if rI+1 = rI = 1. Because of the alternation of 0 and 1 inside an elementary block, we have ri + ri−1 = 1 for all J ≥i ≥I + 2. Hence, if we set bi = ⌊(bi+1 + ri + ri−1)/2⌋, (1) we have bi = ⌊(bi+1 + 1)/2⌋= bi+1 for all J ≥i ≥I + 2 and thus by induction bi = bJ+1 for all J ≥i ≥I + 2 as required. Moreover, since rI+1 = rI, we also have bI+1 = ⌊(bI+2 + 2rI+1 + rI)/2⌋= rI+1 as required. Finally, we remark that if bi = bi−1 = 0 then r′ i = ri, and that if bi = bi−1 = 1 then r′ i = ri −1. Consequently, when bi = bi−1, we can write r′ i = ri −bi. If bi ̸= bi−1, then (1) if bi = 1 and bi−1 = 0 then r′ i = ¯ 1; (2) if bi = 0 and bi−1 = 1 then r′ i = 1. So, when bi ̸= bi−1, we can write r′ i = bi−1 −bi. Noting that if bi ̸= bi−1 then bi = ri, we can see that r′ i = (ri −bi) + (bi−1 −bi) = ri −2bi + bi−1 (2) is a valid expression for r′ i. Our Second Algorithm From the simple formulations for bi and r′ i (see Eqs. (1) and (2)), we obtain an elegant and eﬃcient left-to-right exponent re-coding algorithm. INPUT: (rm−1, . . . , r0)2 OUTPUT: (r′ m, r′ m−1, . . . , r′ 0)sd2 bm ←0; rm ←0; r−1 ←0; r−2 ←0 for i from m down to 0 do bi−1 ←⌊(bi + ri−1 + ri−2)/2⌋ r′ i ←−2bi + ri + bi−1 od Fig. 7. Minimum-weight left-to-right recoding algorithm. Similarly as Reitwiesner’s algorithm, our second algorithm can be performed still more eﬃciently thanks to table look-up. The corresponding look-up table is given hereafter (as aforementioned ‘X’ stands for 0 or 1). Note that entries (bi, ri, ri−1, ri−2) = (0, 1, 1, X) and (1, 0, 0, X) are missing; see the discussion before Lemma 3 (next paragraph) for an explanation. Main Theorem From its construction, the proposed algorithm (Fig. 7) pro-duces an sd2 representation with minimal Hamming weight. However, for com-pleteness, we now give a formal proof of this assertion. We ﬁrst consider three auxiliary lemmas. Lemma 3 implies that the cases (bi, ri, ri−1, ri−2) = (0, 1, 1, X) and (bi, ri, ri−1, ri−2) = (1, 0, 0, X) never oc-cur (see Table 2). Lemma 4 shows that if bi = ri then the output is sparse. 12 Marc Joye and Sung-Ming Yen Table 2. Left-to-right sd exponent recoding. bi ri ri−1 ri−2 bi−1 r′ i 0 0 0 X 0 0 0 0 1 0 0 0 0 0 1 1 1 1 0 1 0 X 0 1 1 0 1 X 1 ¯ 1 1 1 0 0 0 ¯ 1 1 1 0 1 1 0 1 1 1 X 1 0 Finally, Lemma 5 proves that our algorithm eﬀectively yields an sd2 represen-tation. Lemma 3. If ri = ri−1 then bi = ri. Proof. Straightforward because ri = ri−1 yields bi = ⌊(bi+1 + ri + ri−1)/2⌋= ⌊bi+1/2 + ri⌋= ri. ⊓ ⊔ Lemma 4. If bi = ri then r′ i · r′ i−1 = 0. Proof. Suppose ﬁrst that ri−1 + ri−2 = 2(1 −bi). Then ri−1 = ri−2 = 1 −bi. So, by Lemma 3, bi−1 = 1 −bi and thus bi−2 = ⌊(bi−1 + ri−2 + ri−3)/2⌋= ⌊(1 −bi) + ri−3/2⌋= 1 −bi. Hence, r′ i−1 = −2bi−1 + ri−1 + bi−2 = 0 and r′ i · r′ i−1 = 0. If ri−1 + ri−2 ̸= 2(1 −bi), then ri−1 + ri−2 ≤1 when bi = 0 and ri−1 + ri−2 ≥1 when bi = 1. So, bi−1 = ⌊(bi + ri−1 + ri−2)/2⌋= 0 when bi = 0 and bi−1 = 1 when bi = 1; or equivalently, bi−1 = bi. Therefore, since bi = ri, we have r′ i = −2bi + ri + bi−1 = 0 and r′ i · r′ i−1 = 0. ⊓ ⊔ Lemma 5. Let r = (rm−1, . . . , r0)2 be the binary representation of r. Then the output (r′ m, . . . , r′ 0)sd2 given by the minimum-weight left-to-right recoding algorithm is an sd2 representation of r. Proof. From the look-up table (which enumerates all the possible cases), we have r′ i ∈{¯ 1, 0, 1}. Hence, it remains to prove that Pm i=0 r′ i 2i = r. The algorithm gives m P i=0 r′ i 2i = m P i=0 (−2bi + ri + bi−1) 2i = m P i=0 ri 2i + m P i=0 (bi−1 2i −bi 2i+1) = rm 2m + r + (b−1 −bm 2m+1) = r since rm = bm = 0 (by deﬁnition) and b−1 = 0 (because r−1 = r−2 = 0). ⊓ ⊔ Optimal Left-to-right Binary Signed-Digit Recoding 13 Theorem 1. As Reitwiesner’s algorithm, our left-to-right recoding algorithm produces an sd2 representation with minimal Hamming weight. Proof. Proposition 1 says that a sparse method gives a minimal output. So, from Lemmas 3 and 4, it remains to prove that the following inputs lead to optimal outputs. 1. (bi, ri, ri−1, ri−2) = (0, 1, 0, 0) Hence, (bi−1, ri−1, ri−2, ri−3) = (0, 0, 0, ri−3). So, again from Table 2, we have r′ i−1 = 0 and thus r′ i · r′ i−1 = 0. 2. (bi, ri, ri−1, ri−2) = (0, 1, 0, 1) Then we have (bi−1, ri−1, ri−2, ri−3) = (0, 0, 1, ri−3) and r′ i = 1. (a) If ri−3 = 0 then, from Table 2, r′ i−1 = 0 and thus r′ i · r′ i−1 = 0. (b) Otherwise if ri−3 = 1 then bi−2 = 1 and r′ i−1 = 1. Furthermore, since bi−1 = ri−1, it follows from Lemma 4 that r′ i−2 = 0. We now examine the outputs on the left. Let k ≥1 be the least integer such that ri+2k = 0. Hence ri+2j = 1 for all 0 ≤j ≤k −1. Deﬁne ⃗ Yj = (bi+2j, ri+2j, ri+2j−1). We can prove by induction that ⃗ Yj = (bi+2j, ri+2j, ri+2j−1) = (0, 1, 0) () for all 0 ≤j ≤k −1. If j = 0 then ⃗ Y0 = (0, 1, 0). Suppose now that ⃗ Yn = (bi+2n, ri+2n, ri+2n−1) = (0, 1, 0) for some 0 ≤n ≤k −2. We must then have bi+2n+1 = 0 because bi+2n+1 = 1 implies bi+2n = 1, a contradiction. This, in turn, implies ri+2n+1 = 0 (otherwise, by Lemma 3, we would get bi+2n+1 = ri+2n+1 = 1). We also have ri+2n+2 = 1 because ri+2j = 1 for all 0 ≤j ≤k −1. Finally, since bi+2n+1 = 0, we must have bi+2n+2 = 0. Consequently, ⃗ Yn+1 = (bi+2n+2, ri+2n+2, ri+2n+1) = (0, 1, 0) and Eq. () is proven. Moreover, from Eq. (), we have bi+2j−1 = 0 for all 0 ≤j ≤k −1. Simi-larly, from ⃗ Yk−1, we can prove that bi+2k = bi+2k−1 = 0 and ri+2k−1 = 0. In short, when ri−3 = 1, we have: ri+3 ri+2 ri+1 ri ri−1 ri−2 r′ i+3 r′ i+2 r′ i+1 r′ i r′ i−1 r′ i−2 · · 0 1 0 1 · · 0 1 1 0 Since (ri+2k, . . . , ri−1, ri−2) = (0, ⟨0, 1⟩k+1) and bi+j = 0 for all 0 ≤ j ≤2k, we obtain (r′ i+2k, . . . , r′ i) = (0, ⟨0, 1⟩k). Moreover, r′ i−1 = 1 and r′ i−2 = 0. Therefore, the output is (r′ i+2k, . . . , r′ i−1, r′ i−2) = (0, ⟨0, 1⟩k, 1, 0) ≡sd2 (0, 1, ⟨0, ¯ 1⟩k, 0) by Lemma 2. Since the latter is sparse (note the beginning 0 and the ending 0), the output (r′ i+2k, . . . , r′ i−1, r′ i−2) is optimal by Proposition 1. 3. (bi, ri, ri−1, ri−2) = (1, 0, 1, 0) In this case, we have (bi−1, ri−1, ri−2, ri−3) = (1, 1, 0, ri−3) and r′ i = ¯ 1. 14 Marc Joye and Sung-Ming Yen (a) If ri−3 = 1 then r′ i−1 = 0 and thus r′ i · r′ i−1 = 0. (b) If ri−3 = 0, analogously to Case (2.b), we obtain: ri+3 ri+2 ri+1 ri ri−1 ri−2 r′ i+3 r′ i+2 r′ i+1 r′ i r′ i−1 r′ i−2 · · 1 0 1 0 · · 0 ¯ 1 ¯ 1 0 Using similar arguments to those used in Case (2.b), we can prove that the output must be of the form (r′ i+2k, . . . , r′ i−1, r′ i−2) = (0, ⟨0, ¯ 1⟩k, ¯ 1, 0) for some k ≥1. From Lemma 2, we see that this output is optimal since equivalent to the sparse representation (0, ¯ 1, ⟨0, 1⟩k, 0). 4. (bi, ri, ri−1, ri−2) = (1, 0, 1, 1) Similarly to Case (1.), (bi−1, ri−1, ri−2, ri−3) = (1, 1, 1, ri−3); so r′ i−1 = 0 and thus r′ i · r′ i−1 = 0. This concludes the proof. ⊓ ⊔ 5 Hardware Implementation Evidently, the proposed minimum-weight signed-digit converter (Fig. 7) is much more regular and simpler for hardware implementation. To implement the trans-formation algorithm, we encode r′ i (r′ i ∈{0, 1, ¯ 1}) as {0 △ = (X, 0)2; 1 △ = (0, 1)2; −1 △ = (1, 1)2}, and we denote r′ i by a two-bit representation (r′ i,H, r′ i,L)2. After some logic manipulations and minimizations, the following Boolean equations for bi−1, r′ i,H and r′ i,L can be obtained. The outputs (bi−1, r′ i) corre-sponding to the missing entries (bi, ri, ri−1, ri−2) = (0, 1, 1, X) and (1, 0, 0, X) of Table 2 were optimally assigned to (0, 1) and (1, ¯ 1), respectively.    bi−1 = bi · ri−1 · ri−2 + bi · ri−1 + bi · ri−2 r′ i,H = bi r′ i,L = bi · ri−1 · ri−2 + bi · ri + bi · ri + bi · ri−1 · ri−2 . A hardware realization of these equations is given in Figure 8. Initially, the shift-left registers {ri, ri−1, ri−2} are loaded with {0, rm−1, rm−2} and the latch is reset to logic “0”. At the end of each iteration, the outputs r′ i,H and r′ i,L are used to encode the transformed sd2 digit r′ i. At this moment, the output bi−1 is fed into the latch, the lower bit ri−3 is prepared to be fed into the shift-left registers, and the registers shift one bit to the left. Based on this minimum-weight signed-digit transformer, an hardware archi-tecture for fast exponentiation, αr, is sketched in Figure 9. In that exponentiation hardware, the outputs of the signed-digit transformer, r′ i,H and r′ i,L, are used to control the computational data ﬂow and the register fetching. Signal r′ i,H is used Optimal Left-to-right Binary Signed-Digit Recoding 15 ? ? ri−1 ri−2 ri ri−3 bi−1 Latch bi r′ i,H r′ i,L \| {z } r′ i Fig. 8. Hardware implementation of the proposed transformer. to determine which of α or α−1 has to be selected by the multiplexer MUX-1, and signal r′ i,L determines whether or not the multiplication A ←A × B (where A means the accumulator and B means the output of MUX-1) needs to be performed. In the case of r′ i,L = 1 (and thus M ′ = “0∥1′′, where ∥means con-catenation), both the squaring and the multiplication are required. On the other hand, when r′ i.L = 0 (and thus M ′ = 0), only the squaring operation is required. Using this architecture, the computation of αr needs, on average, 4 3\|r\| multiplica-tions (if we assume that multiplication and squaring are equally time-consuming operations). For example, αr is evaluated after only 684 multiplications for a 512-bit integer r. 6 Other Considerations For some special cases, the proposed algorithms may bring further advantages. 1. One of the main concerns of the proposed algorithms was to reduce the Ham-ming weight of exponent r for the computation of αr. While the number of squarings still remains the same (i.e., \|r\|), the average number of multipli-cations was reduced from \|r\|/2 to \|r\|/3. Over GF(2k), squaring operations can be achieved via a simple circular shift if the elements are represented in the so-called normal basis . Their computational costs can therefore be ignored comparing to multiplications. In this case, the real expected gain over the binary methods is then ( 1 2 −1 3)/ 1 2 ≈66.66 % instead of 11.11% as aforementioned in Section 2. 2. Because the expression of variable b, bi−1 ←⌊(bi + ri−1 + ri−2)/2⌋(see Fig. 7), is similar to that of carry c in Reitwiesner’s algorithm (Fig. 3), ci+1 ← 16 Marc Joye and Sung-Ming Yen Control signalM ′ of MUX-2 of MUX-1 Control signalr′ i,H L L ? ? ? -? ? -? ? MUX-2 MUX-1 α−1 α ”0” ”1” B ”1” Control Binary to sd2 transformer r′ i,H r′ i,L r Multiplier A: Accumulator ”0” r′ i,L = 0 ⇒M′ = 0 r′ i,L = 1 ⇒M′ = 0∥1 Fig. 9. Hardware architecture for the computation of αr. ⌊(ci+ri+ri+1)/2⌋, the parallel recoding method proposed by Ko¸ c readily extends to our minimum-weight left-to-right recoding algorithm. 7 Conclusions To improve the performance of the square-and-multiply exponentiation for the evaluation of αr, the Hamming weight of exponent r should be reduced. This can be achieved by adopting a signed-digit representation for exponent r. Assuming that α−1 is provided along with α (or can be cheaply evaluated, as for elliptic curves), a minimum-weight signed-digit representation for r can improve quite a large the computation of αr. However, in order to produce the minimum-weight signed-digit representation, existing solutions must initiate the recoding process from right to left while only the modiﬁed left-to-right (mlr) exponentiation al-gorithm is suitable. The heterogeneous processings bring some time and memory ineﬃciencies for hardware realization. Indeed, the recoding has ﬁrst to be per-formed, the resulting representation for r must be saved (which requires twice more memory space than its binary representation). Then and only then, the exponentiation process may start in order to obtain αr. In this paper, new and homogeneous approaches for minimum-weight signed-digit representation are presented. Our methods are homogeneous in the sense that both the proposed recoding algorithms and the mlr exponentation algo-rithm initiate from the most signiﬁcant position of the exponent. The signed-digit representation of exponent r is consequently obtained in real-time and does not need to be stored. This better memory usage is especially useful for small devices like smart-cards. Our ﬁrst algorithm gives the canonical representation for the Optimal Left-to-right Binary Signed-Digit Recoding 17 exponent. This algorithm is then modiﬁed into another minimum-weight recod-ing algorithm so that table look-up is possible. A hardware implementation of this second converter and the corresponding architecture for exponentiation are also presented. Acknowledgments We are grateful to Dan Gordon for sending a preprint of . Many thanks also go to the anonymous reviewers for their excellent work. This work was partly supported by the National Science Council of the Re-public of China under contracts NSC89-2213-E-008-049, NSC87-2213-E-032-012, and NSC87-2811-E-032-0001. References 1. A. Avizienis, “Signed-digit number representations for fast parallel arithmetic,” IRE Trans. Electronic Computers, vol. EC-10, pp. 389–400, 1961. Reprinted in [30, vol. II, pp. 54–65]. 2. I. Koren, Computer arithmetic algorithms, Prentice-Hall, 1993. 3. N. Takagi, H. Yasuura, and S. Yajima, “High-speed VLSI multiplication algorithm with a redundant binary addition tree,” IEEE Trans. Computers, vol. C-34, no. 9, pp. 789–796, 1985. 4. S. Kuninobu, T. Nishiyama, H. Edamatsu, T. Taniguchi, and N. Takagi, “Design of high speed MOS multiplier and divider using redundant binary representation,” in Proc. 8th Symp. Computer Arithmetic, pp. 80–86, 1987. 5. Y. Harata, Y. Nakamura, H. Nagase, M. Takigawa, and N. Takagi, “A high-speed multiplier using a redundant binary adder tree,” IEEE J. Solid-State Circuits, vol. 22, no. 1, pp. 28–34, 1987. 6. A. Vandemeulebroecke, E. Vanzieleghem, T. Denayer, and P.G.A. Jespers, “A new carry-free division algorithm and its application to a single-chip 1024-b RSA pro-cessor,” IEEE J. Solid-State Circuits, vol. 25, no. 3, pp. 748–755, 1990. 7. K. Hwang, Computer arithmetic, principles, architecture and design, John Wiley & Sons, 1979. 8. S.M. Yen, C.S. Laih, C.H. Chen, and J.Y. Lee, “An eﬃcient redundant-binary number to binary number converter,” IEEE J. Solid-State Circuits, vol. 27, no. 1, pp. 109–112, 1992. 9. R.L. Rivest, A. Shamir, and L.M. Adleman, “A method for obtaining digital signa-tures and public-key cryptosystems,” Commun. ACM, vol. 21, no. 2, pp. 120–126, 1978. 10. C ¸.K. Ko¸ c, “High-speed RSA implementations,” Tech. Rep. TR 201, RSA Labora-tories, Nov. 1994. 11. D.E. Knuth, The art of computer programming/Seminumerical algorithms, vol. 2, Addison-Wesley, 2nd edition, 1981. 12. P. Downey, B. Leong, and R. 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Booth, “A signed binary multiplication technique,” The Quaterly J. Mechan-ics and Applied Mathematics, vol. 4, pp. 236–240, 1951. Reprinted in [30, vol. I, pp. 100–104]. 19. G.W. Reitwiesner, “Binary arithmetic,” Advances in Computers, vol. 1, pp. 231– 308, 1960. 20. J. Jedwab and C. Mitchell, “Minimum weight modiﬁed signed-digit representations and fast exponentiation,” Electronics Letters, vol. 25, pp. 1171–1172, 1989. 21. H. Cohen, A course in computational algebraic number theory, Springer-Verlag, 1993. 22. ¨ O. Eˇ gecioˇ glu and C ¸.K. Ko¸ c, “Exponentiation using canonical recoding,” Theoret-ical Computer Science, vol. 129, no. 2, pp. 407–417, 1994. 23. B.S. Kaliski, “The Montgomery inverse and its applications,” IEEE Trans. Com-puters, vol. C-44, no. 8, pp. 1064–1065, 1995. 24. S. Arno and F.S. Wheeler, “Signed digit representations of minimal Hamming weight,” IEEE Trans. Computers, vol. C-42, no. 8, pp. 1007–1010, 1993. 25. D.M. 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10807	https://math.stackexchange.com/questions/2178974/division-by-zero-and-proofs-that-the-slope-is-defined	Division by Zero and proofs that the slope is defined. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Division by Zero and proofs that the slope is defined. Ask Question Asked 8 years, 6 months ago Modified8 years, 6 months ago Viewed 3k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Most people think that when they encounter a fraction number as such: n 0 as being an error due to division of 0 and it is considered undefined. I have expressed this or a similar question with the use of points, vectors, and trig functions that vertical slope should be defined found here: and I want to re ask this question but in a different context using the same principles that I've applied before that involves the use of: slope m=y 2−y 1 x 2−x 1=δ y δ x=sin θ cos θ=tan θ But this time I want to consider not just division by 0, but also the infamous fraction of 0 0 and what it should equal. First let me state that 0 does not have a value and that 0 is actually a N A N that means it is a place holder and represents either the null or the empty set. Now when we have a set, the set it self is something tangible and when we divide anything tangible by itself the result is 1 do to the identity property. For example consider this with the set without taking any of its elements into consideration: Set a when divided by itself as in a a=1 should be a valid statement? Let's consider the linear equation y=x that has an understood slope of 1 which makes the slope also 1 1=sin θ cos θ=tan θ where θ=45° and for every coordinate pair that belongs to this line has a slope of 1 at that point on the line. This set of coordinate pairs is the set of values of the domain and the range to this function, equation or expression. I will use a few of points belonging to y=x where slope m here is understood as 1: x y−3−3−2−2−1−1−0+0(-) is comging from and (+) is going to+1+1+2+2+3+3 Since slope is defined as r i s e r u n or δ y δ x or sin θ cos θ or tan θ, and knowing that we can take the coordinate pairs and make them as slope since it is understood as 1 have the form of y x when y=x? So does this not generate this sequence of values: −3−3=−2−2=−1−1=0 0=1 1=2 2=3 3⟹1 Does the point at (0,0) on this line not have a slope of 1? Yes! However the input and output are both 0. The reason that 0 0=1 works is because of 2 main reasons: First is the Identity Property, anything divided by itself is of course itself! Second is the fact that the point at (0,0) with a (+) slope of 1 is a point of reflection or point of symmetry, an origin of rotation, and a point in which the function has as a starting location that it expands out from making this the ROOT of the function. In the natural form of y=x, its graph in the first quadrant is approaching +∞ while it is approaching −∞ in the third quadrant. So here the coordinate point for slope m that has a value of 1 is represented by the coordinate point at the origin of (0,0). Thus this does define 0 0 to have a value of 1. So if 0 0 is defined by this then why is that n 0 should be or must be undefined? A line never looses it's slope! A lines slope has a range of ±∞ and for the domain we will use domains of the sin and cos independently of each other that has a domain of R instead of referring to the tan that is typically claimed to be undefined at an angle of 90° or π 2 due to the current assumption that division by 0 is undefined. So how is it that "vertical slope" is Undefined? When the change in height = 0 this is horizontal slope and it is defined. So when the opposite or the perpendicularity occurs where we do have change in y but the change in x stops we have a problem. This labeling of division by 0 as being undefined is made from a long list of wrong assumptions. Does this not also support and back up the proofs I showed from before why this is valid although everyone else is hell-bent on claiming that it isn't? proof-verification slope Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 13, 2017 at 12:21 CommunityBot 1 asked Mar 9, 2017 at 11:01 Francis CuglerFrancis Cugler 177 2 2 silver badges 10 10 bronze badges 12 1 With your reasoning, what would the slope be in the point (0,0) if we take y=2 x?user304329 –user304329 2017-03-09 11:05:19 +00:00 Commented Mar 9, 2017 at 11:05 n 0 is considered "impossible" because usually, division has the following property: a b=c⟹a=b c. Applying this to 0 n, we see that n 0=x⟹n=0⋅x=0. This shows that there exists no real number, not even a complex number for that matter, such that x=n 0 IF n≠0. The reason we consider 0 0 "impossible" too, is because it doesn't represent one thing. Since all numbers satisfy n=0⋅x, the expression n 0 is equal to all numbers simultaneously. This isn't usable and things should only mean one thing.user304329 –user304329 2017-03-09 11:12:10 +00:00 Commented Mar 9, 2017 at 11:12 "everyone else is hell-bent on claiming that it isn't" because they have proved it isn't (as @vrugtehagel did), and you just argued, with many flaws in your argument.Eclipse Sun –Eclipse Sun 2017-03-09 11:14:38 +00:00 Commented Mar 9, 2017 at 11:14 "[0/0] represents either the null or the empty set." - it doesn't. It's just some meaningless symbols.Patrick Stevens –Patrick Stevens 2017-03-09 11:37:58 +00:00 Commented Mar 9, 2017 at 11:37 @vrugtehagel The slope you shown is 2 not 1 however if tan(45°)=1 where then this would be 2 tan(45°) The slope would be 2 since this is a different line then y=x. It is still a root that contains both intercepts. The point (0,0) is unique because without any direct linear equation then the point itself arbitrarily which is 0 Dimensional contains all slopes of all lines that can pass through the origin. And since you can rotate a vector 360° or 2 π radians around this point in a full circle. Then it still holds. For your first question.Francis Cugler –Francis Cugler 2017-03-09 11:42:40 +00:00 Commented Mar 9, 2017 at 11:42 \|Show 7 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Okay so let's repeat a definition here Slope is calculated by finding the ratio of the "vertical change" to the "horizontal change" between (any) two distinct points on a line. So first of all, you cannot get the slope of a line by taking two equal points and finding the ratio between their distance (which would be 0 0). With your reasoning, you find the gradient of y=x at the point (0,0) by taking (0,0) as reference point, and then calculating 0−0 0−0, after which you proceed to argue that that should be equal to 1. However, the line y=2 x for example, has slope 2 (yes, also in (0,0)), yet with your reasoning, it should be 0−0 0−0, which you said should be 1. Second of all, vertical lines have no "horizontal change", and as such, the ratio between the "vertical change" and the "horizontal change" cannot exist, since we'd be dividing by 0. Why can we not divide by 0? In the real numbers (this works for complex numbers too, but we'll just go with R), we have the following property (for all b≠0): a b=c⟺a=b c When you say that division by 0 should be possible, you implicitly say that the above rule shouldn't only work for all b≠0, but also for b=0. This makes no sense however, since with that, we have: a 0=c⟺a=0⋅b But it is known, even without division by 0, that 0⋅c=0 for all c, and so that is still true when division by zero is possible; but now we encounter a problem. The property with b=0 now states that for all a,c, we must have a 0=c⟺a=0 This is problematic, because, if division by zero was possible, that would mean that if you divide any number n by 0, that number must've been 0; moreover, we have 0 0=c for all c. See how that produces problems? To sum it up, if you allow division by 0, then all numbers suddenly become equal, and we're dealing with {0} instead of the usual, much more useful R. This is why we don't do division by 0. As a sidenote, scepticism makes a good mathematician; however, don't underestimate the mathematicians that've existed and thought about all sorts of things over the ages. With such basic results as this, you can assume very talented and respected mathematicians have thought about this, and especially if they all agree, they're probably right. Edit It seems like you view real numbers more like quantifiers than a set of mathematical objects; this view is fine, but often doesn't work. You say 0 is "just a placeholder", but it really doesn't work that way. The set of real numbers, as well as the set of complex numbers, the set of fractions, the integers, and many more, are sets with elements that follow specific rules; those rules work for all numbers in the set, and so 0 is being treated, as should be treated, as just "one of the elements" of said set. 0 does not mean "nothing" nor "NaN", even though real-life applications suggest the nothingness of 0. You have to keep in mind mathematics is purely theoretical, and there's applications in the real world that mathematics wouldn't fully agree with (take for example physics, which does a lot of things like "rounding off", "discarding terms because they're insignificant" which are fine for the real world, and it works, but mathematically speaking, some things would be considered non-rigorous). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Mar 9, 2017 at 12:01 answered Mar 9, 2017 at 11:35 user304329 user304329 6 If a coordinate point belongs to a line with a given slope, then it is not suffice to say that that point also has that slope with respect to the line it is on? Also doesn't that point also belong to an Infinite about of other lines which also has an infinite amount of other slopes? And if the linear equations do not have any horizontal or vertical translations to where they do pass the origin at (0,0) and if 0 0 and be understood as 1 through the Identity property, then y=0 0 x is still y=x then same for y=0 0(n)(x)⟹y=(1)(m)(x)=m x Francis Cugler –Francis Cugler 2017-03-09 11:54:02 +00:00 Commented Mar 9, 2017 at 11:54 You should be aware of the fact that points don't have slopes, but lines have slopes in points.user304329 –user304329 2017-03-09 12:03:18 +00:00 Commented Mar 9, 2017 at 12:03 That is not true points do when they are associated to a specific line, because that is the progression of the points in a specified direction. A single arbitrary point without any relative association to anything else is just locale and that is 0D space. When you have a series of points that follow a pattern that pattern exist within those points. It is due to this that we can even plot and make graphs of functions in the first place. A single point is arbitrary, but once you have two points you have line segment, and from that a vector.Francis Cugler –Francis Cugler 2017-03-09 12:06:07 +00:00 Commented Mar 9, 2017 at 12:06 Also, what about this: y=x=0 0 x=0⋅x 0=0 0=1. I employ you to see reason; the evidence that division by 0 is impossible is incontrovertible.user304329 –user304329 2017-03-09 12:06:55 +00:00 Commented Mar 9, 2017 at 12:06 I repeat; points don't have slopes, lines have slopes in points. Points aren't associated by lines to give a slope to a point, lines are associated with points to give a slope to a line user304329 –user304329 2017-03-09 12:08:32 +00:00 Commented Mar 9, 2017 at 12:08 \|Show 1 more comment This answer is useful 0 Save this answer. Show activity on this post. I just rewrite your argument word by word, using y=2 x instead of y=x. All typos are intensionally leaved unchanged. I will use a few of points belonging to y=2 x where slope m here is understood as 2: x y−3−6−2−4−1−2−0+0(-) is comging from and (+) is going to+1+2+2+4+3+6 ... So does this not generate this sequence of values: −6−3=−4−2=−2−1=0 0=2 1=4 2=6 3⟹2Does the point at (0,0)on this line not have a slope of 2? Yes! However the input and output are both 0... You can indeed assume that 0/0=1, but you cannot avoid others to assume that 0/0=m for any m. Mathematics avoids ambiguity. So normally we say that 0/0 is undefined. You said that 0 does not have a value, it is a placeholder or the empty set. That's correct. More correctly, NOTHING in mathematics has a value. Everything in mathematics is nothing but sets. For example, we use 0 (or more commonly, ∅) to denotes the empty set, and 1={0},2={0,1},3={0,1,2},… By the way, a/b is not defined by slopes, but slopes is defined by a/b. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 10, 2017 at 2:57 Eclipse SunEclipse Sun 9,458 22 22 silver badges 45 45 bronze badges 1 Comments are not for extended discussion; this conversation has been moved to chat.Daniel Fischer –Daniel Fischer 2017-03-12 10:49:32 +00:00 Commented Mar 12, 2017 at 10:49 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions proof-verification slope See similar questions with these tags. 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10808	https://opened.cuny.edu/courseware/lesson/602/overview	Preview Please log in to save materials. Log in Report Details Subject: : Applied Science Material Type: : Module Provider: : Rice University Provider Set: : OpenStax College Tags: : - Consumption Function - Expenditure-output Model - Inflationary Gap - Keynesian Cross Diagram - Marginal Propensity to Consume (MPC) - Marginal Propensity to Import (MPI) - Marginal Propensity to Save (MPS) - Multiplier Effect - National Income - Potential GDP - Real GDP - Recessionary Gap Log in to add tags to this item. License: : Creative Commons Attribution Language: : English Show More Show Less Principles of Macroeconomics 2e The Expenditure-Output Model Preface Welcome to Economics! Choice in a World of Scarcity Demand and Supply Labor and Financial Markets Elasticity The Macroeconomic Perspective Economic Growth Unemployment Inflation The International Trade and Capital Flows The Aggregate Demand/Aggregate Supply Model The Keynesian Perspective The Neoclassical Perspective Money and Banking Monetary Policy and Bank Regulation Exchange Rates and International Capital Flows Government Budgets and Fiscal Policy The Impacts of Government Borrowing Macroeconomic Policy Around the World International Trade Globalization and Protectionism The Use of Mathematics in Principles of Economics The Expenditure-Output Model The Expenditure-Output Model The Expenditure-Output Model The Expenditure-Output Model (This appendix should be consulted after first reading The Aggregate Demand/Aggregate Supply Model and The Keynesian Perspective.) The fundamental ideas of Keynesian economics were developed before the AD/AS model was popularized. From the 1930s until the 1970s, Keynesian economics was usually explained with a different model, known as the expenditure-output approach. This approach is strongly rooted in the fundamental assumptions of Keynesian economics: it focuses on the total amount of spending in the economy, with no explicit mention of aggregate supply or of the price level (although as you will see, it is possible to draw some inferences about aggregate supply and price levels based on the diagram). The Axes of the Expenditure-Output Diagram The expenditure-output model, sometimes also called the Keynesian cross diagram, determines the equilibrium level of real GDP by the point where the total or aggregate expenditures in the economy are equal to the amount of output produced. The axes of the Keynesian cross diagram presented in Figure show real GDP on the horizontal axis as a measure of output and aggregate expenditures on the vertical axis as a measure of spending. Remember that GDP can be thought of in several equivalent ways: it measures both the value of spending on final goods and also the value of the production of final goods. All sales of the final goods and services that make up GDP will eventually end up as income for workers, for managers, and for investors and owners of firms. The sum of all the income received for contributing resources to GDP is called national income (Y). At some points in the discussion that follows, it will be useful to refer to real GDP as “national income.” Both axes are measured in real (inflation-adjusted) terms. The Potential GDP Line and the 45-degree Line The Keynesian cross diagram contains two lines that serve as conceptual guideposts to orient the discussion. The first is a vertical line showing the level of potential GDP. Potential GDP means the same thing here that it means in the AD/AS diagrams: it refers to the quantity of output that the economy can produce with full employment of its labor and physical capital. The second conceptual line on the Keynesian cross diagram is the 45-degree line, which starts at the origin and reaches up and to the right. A line that stretches up at a 45-degree angle represents the set of points (1, 1), (2, 2), (3, 3) and so on, where the measurement on the vertical axis is equal to the measurement on the horizontal axis. In this diagram, the 45-degree line shows the set of points where the level of aggregate expenditure in the economy, measured on the vertical axis, is equal to the level of output or national income in the economy, measured by GDP on the horizontal axis. When the macroeconomy is in equilibrium, it must be true that the aggregate expenditures in the economy are equal to the real GDP—because by definition, GDP is the measure of what is spent on final sales of goods and services in the economy. Thus, the equilibrium calculated with a Keynesian cross diagram will always end up where aggregate expenditure and output are equal—which will only occur along the 45-degree line. The Aggregate Expenditure Schedule The final ingredient of the Keynesian cross or expenditure-output diagram is the aggregate expenditure schedule, which will show the total expenditures in the economy for each level of real GDP. The intersection of the aggregate expenditure line with the 45-degree line—at point E0 in Figure—will show the equilibrium for the economy, because it is the point where aggregate expenditure is equal to output or real GDP. After developing an understanding of what the aggregate expenditures schedule means, we will return to this equilibrium and how to interpret it. Building the Aggregate Expenditure Schedule Aggregate expenditure is the key to the expenditure-income model. The aggregate expenditure schedule shows, either in the form of a table or a graph, how aggregate expenditures in the economy rise as real GDP or national income rises. Thus, in thinking about the components of the aggregate expenditure line—consumption, investment, government spending, exports and imports—the key question is how expenditures in each category will adjust as national income rises. Consumption as a Function of National Income How do consumption expenditures increase as national income rises? People can do two things with their income: consume it or save it (for the moment, let’s ignore the need to pay taxes with some of it). Each person who receives an additional dollar faces this choice. The marginal propensity to consume (MPC), is the share of the additional dollar of income a person decides to devote to consumption expenditures. The marginal propensity to save (MPS) is the share of the additional dollar a person decides to save. It must always hold true that: For example, if the marginal propensity to consume out of the marginal amount of income earned is 0.9, then the marginal propensity to save is 0.1. With this relationship in mind, consider the relationship among income, consumption, and savings shown in Figure. (Note that we use “Aggregate Expenditure” on the vertical axis in this and the following figures, because all consumption expenditures are parts of aggregate expenditures.) An assumption commonly made in this model is that even if income were zero, people would have to consume something. In this example, consumption would be $600 even if income were zero. Then, the MPC is 0.8 and the MPS is 0.2. Thus, when income increases by $1,000, consumption rises by $800 and savings rises by $200. At an income of $4,000, total consumption will be the $600 that would be consumed even without any income, plus $4,000 multiplied by the marginal propensity to consume of 0.8, or $ 3,200, for a total of $ 3,800. The total amount of consumption and saving must always add up to the total amount of income. (Exactly how a situation of zero income and negative savings would work in practice is not important, because even low-income societies are not literally at zero income, so the point is hypothetical.) This relationship between income and consumption, illustrated in Figure and Table, is called the consumption function. The pattern of consumption shown in Table is plotted in Figure. To calculate consumption, multiply the income level by 0.8, for the marginal propensity to consume, and add $600, for the amount that would be consumed even if income was zero. Consumption plus savings must be equal to income. \| Income \| Consumption \| Savings \| --- \| $0 \| $600 \| –$600 \| \| $1,000 \| $1,400 \| –$400 \| \| $2,000 \| $2,200 \| –$200 \| \| $3,000 \| $3,000 \| $0 \| \| $4,000 \| $3,800 \| $200 \| \| $5,000 \| $4,600 \| $400 \| \| $6,000 \| $5,400 \| $600 \| \| $7,000 \| $6,200 \| $800 \| \| $8,000 \| $7,000 \| $1,000 \| \| $9,000 \| $7,800 \| $1,200 \| The Consumption Function However, a number of factors other than income can also cause the entire consumption function to shift. These factors were summarized in the earlier discussion of consumption, and listed in Table. When the consumption function moves, it can shift in two ways: either the entire consumption function can move up or down in a parallel manner, or the slope of the consumption function can shift so that it becomes steeper or flatter. For example, if a tax cut leads consumers to spend more, but does not affect their marginal propensity to consume, it would cause an upward shift to a new consumption function that is parallel to the original one. However, a change in household preferences for saving that reduced the marginal propensity to save would cause the slope of the consumption function to become steeper: that is, if the savings rate is lower, then every increase in income leads to a larger rise in consumption. Investment as a Function of National Income Investment decisions are forward-looking, based on expected rates of return. Precisely because investment decisions depend primarily on perceptions about future economic conditions, they do not depend primarily on the level of GDP in the current year. Thus, on a Keynesian cross diagram, the investment function can be drawn as a horizontal line, at a fixed level of expenditure. Figure shows an investment function where the level of investment is, for the sake of concreteness, set at the specific level of 500. Just as a consumption function shows the relationship between consumption levels and real GDP (or national income), the investment function shows the relationship between investment levels and real GDP. The appearance of the investment function as a horizontal line does not mean that the level of investment never moves. It means only that in the context of this two-dimensional diagram, the level of investment on the vertical aggregate expenditure axis does not vary according to the current level of real GDP on the horizontal axis. However, all the other factors that vary investment—new technological opportunities, expectations about near-term economic growth, interest rates, the price of key inputs, and tax incentives for investment—can cause the horizontal investment function to shift up or down. Government Spending and Taxes as a Function of National Income In the Keynesian cross diagram, government spending appears as a horizontal line, as in Figure, where government spending is set at a level of 1,300. As in the case of investment spending, this horizontal line does not mean that government spending is unchanging. It means only that government spending changes when Congress decides on a change in the budget, rather than shifting in a predictable way with the current size of the real GDP shown on the horizontal axis. The situation of taxes is different because taxes often rise or fall with the volume of economic activity. For example, income taxes are based on the level of income earned and sales taxes are based on the amount of sales made, and both income and sales tend to be higher when the economy is growing and lower when the economy is in a recession. For the purposes of constructing the basic Keynesian cross diagram, it is helpful to view taxes as a proportionate share of GDP. In the United States, for example, taking federal, state, and local taxes together, government typically collects about 30–35 % of income as taxes. Table revises the earlier table on the consumption function so that it takes taxes into account. The first column shows national income. The second column calculates taxes, which in this example are set at a rate of 30%, or 0.3. The third column shows after-tax income; that is, total income minus taxes. The fourth column then calculates consumption in the same manner as before: multiply after-tax income by 0.8, representing the marginal propensity to consume, and then add $600, for the amount that would be consumed even if income was zero. When taxes are included, the marginal propensity to consume is reduced by the amount of the tax rate, so each additional dollar of income results in a smaller increase in consumption than before taxes. For this reason, the consumption function, with taxes included, is flatter than the consumption function without taxes, as Figure shows. \| Income \| Taxes \| After-Tax Income \| Consumption \| Savings \| --- --- \| $0 \| $0 \| $0 \| $600 \| –$600 \| \| $1,000 \| $300 \| $700 \| $1,160 \| –$460 \| \| $2,000 \| $600 \| $1,400 \| $1,720 \| –$320 \| \| $3,000 \| $900 \| $2,100 \| $2,280 \| –$180 \| \| $4,000 \| $1,200 \| $2,800 \| $2,840 \| –$40 \| \| $5,000 \| $1,500 \| $3,500 \| $3,400 \| $100 \| \| $6,000 \| $1,800 \| $4,200 \| $3,960 \| $240 \| \| $7,000 \| $2,100 \| $4,900 \| $4,520 \| $380 \| \| $8,000 \| $2,400 \| $5,600 \| $5,080 \| $520 \| \| $9,000 \| $2,700 \| $6,300 \| $5,640 \| $660 \| The Consumption Function Before and After Taxes Exports and Imports as a Function of National Income The export function, which shows how exports change with the level of a country’s own real GDP, is drawn as a horizontal line, as in the example in Figure (a) where exports are drawn at a level of $840. Again, as in the case of investment spending and government spending, drawing the export function as horizontal does not imply that exports never change. It just means that they do not change because of what is on the horizontal axis—that is, a country’s own level of domestic production—and instead are shaped by the level of aggregate demand in other countries. More demand for exports from other countries would cause the export function to shift up; less demand for exports from other countries would cause it to shift down. Imports are drawn in the Keynesian cross diagram as a downward-sloping line, with the downward slope determined by the marginal propensity to import (MPI), out of national income. In Figure (b), the marginal propensity to import is 0.1. Thus, if real GDP is $5,000, imports are $500; if national income is $6,000, imports are $600, and so on. The import function is drawn as downward sloping and negative, because it represents a subtraction from the aggregate expenditures in the domestic economy. A change in the marginal propensity to import, perhaps as a result of changes in preferences, would alter the slope of the import function. Using an Algebraic Approach to the Expenditure-Output Model In the expenditure-output or Keynesian cross model, the equilibrium occurs where the aggregate expenditure line (AE line) crosses the 45-degree line. Given algebraic equations for two lines, the point where they cross can be readily calculated. Imagine an economy with the following characteristics. Y = Real GDP or national income T = Taxes = 0.3Y C = Consumption = 140 + 0.9(Y – T) I = Investment = 400 G = Government spending = 800 X = Exports = 600 M = Imports = 0.15Y Step 1. Determine the aggregate expenditure function. In this case, it is: Step 2. The equation for the 45-degree line is the set of points where GDP or national income on the horizontal axis is equal to aggregate expenditure on the vertical axis. Thus, the equation for the 45-degree line is: AE = Y. Step 3. The next step is to solve these two equations for Y (or AE, since they will be equal to each other). Substitute Y for AE: Step 4. Insert the term 0.3Y for the tax rate T. This produces an equation with only one variable, Y. Step 5. Work through the algebra and solve for Y. This algebraic framework is flexible and useful in predicting how economic events and policy actions will affect real GDP. Step 6. Say, for example, that because of changes in the relative prices of domestic and foreign goods, the marginal propensity to import falls to 0.1. Calculate the equilibrium output when the marginal propensity to import is changed to 0.1. Step 7. Because of a surge of business confidence, investment rises to 500. Calculate the equilibrium output. For issues of policy, the key questions would be how to adjust government spending levels or tax rates so that the equilibrium level of output is the full employment level. In this case, let the economic parameters be: Y = National income T = Taxes = 0.3Y C = Consumption = 200 + 0.9(Y – T) I = Investment = 600 G = Government spending = 1,000 X = Exports = 600 Y = Imports = 0.1(Y – T) Step 8. Calculate the equilibrium for this economy (remember Y = AE). Step 9. Assume that the full employment level of output is 6,000. What level of government spending would be necessary to reach that level? To answer this question, plug in 6,000 as equal to Y, but leave G as a variable, and solve for G. Thus: Step 10. Solve this problem arithmetically. The answer is: G = 1,240. In other words, increasing government spending by 240, from its original level of 1,000, to 1,240, would raise output to the full employment level of GDP. Indeed, the question of how much to increase government spending so that equilibrium output will rise from 5,454 to 6,000 can be answered without working through the algebra, just by using the multiplier formula. The multiplier equation in this case is: Thus, to raise output by 546 would require an increase in government spending of 546/2.27=240, which is the same as the answer derived from the algebraic calculation. This algebraic framework is highly flexible. For example, taxes can be treated as a total set by political considerations (like government spending) and not dependent on national income. Imports might be based on before-tax income, not after-tax income. For certain purposes, it may be helpful to analyze the economy without exports and imports. A more complicated approach could divide up consumption, investment, government, exports and imports into smaller categories, or to build in some variability in the rates of taxes, savings, and imports. A wise economist will shape the model to fit the specific question under investigation. Building the Combined Aggregate Expenditure Function All the components of aggregate demand—consumption, investment, government spending, and the trade balance—are now in place to build the Keynesian cross diagram. Figure builds up an aggregate expenditure function, based on the numerical illustrations of C, I, G, X, and M that have been used throughout this text. The first three columns in Table are lifted from the earlier Table, which showed how to bring taxes into the consumption function. The first column is real GDP or national income, which is what appears on the horizontal axis of the income-expenditure diagram. The second column calculates after-tax income, based on the assumption, in this case, that 30% of real GDP is collected in taxes. The third column is based on an MPC of 0.8, so that as after-tax income rises by $700 from one row to the next, consumption rises by $560 (700 × 0.8) from one row to the next. Investment, government spending, and exports do not change with the level of current national income. In the previous discussion, investment was $500, government spending was $1,300, and exports were $840, for a total of $2,640. This total is shown in the fourth column. Imports are 0.1 of real GDP in this example, and the level of imports is calculated in the fifth column. The final column, aggregate expenditures, sums up C + I + G + X – M. This aggregate expenditure line is illustrated in Figure. \| National Income \| After-Tax Income \| Consumption \| Government Spending + Investment + Exports \| Imports \| Aggregate Expenditure \| --- --- --- \| \| $3,000 \| $2,100 \| $2,280 \| $2,640 \| $300 \| $4,620 \| \| $4,000 \| $2,800 \| $2,840 \| $2,640 \| $400 \| $5,080 \| \| $5,000 \| $3,500 \| $3,400 \| $2,640 \| $500 \| $5,540 \| \| $6,000 \| $4,200 \| $3,960 \| $2,640 \| $600 \| $6,000 \| \| $7,000 \| $4,900 \| $4,520 \| $2,640 \| $700 \| $6,460 \| \| $8,000 \| $5,600 \| $5,080 \| $2,640 \| $800 \| $6,920 \| \| $9,000 \| $6,300 \| $5,640 \| $2,640 \| $900 \| $7,380 \| National Income-Aggregate Expenditure Equilibrium The aggregate expenditure function is formed by stacking on top of each other the consumption function (after taxes), the investment function, the government spending function, the export function, and the import function. The point at which the aggregate expenditure function intersects the vertical axis will be determined by the levels of investment, government, and export expenditures—which do not vary with national income. The upward slope of the aggregate expenditure function will be determined by the marginal propensity to save, the tax rate, and the marginal propensity to import. A higher marginal propensity to save, a higher tax rate, and a higher marginal propensity to import will all make the slope of the aggregate expenditure function flatter—because out of any extra income, more is going to savings or taxes or imports and less to spending on domestic goods and services. The equilibrium occurs where national income is equal to aggregate expenditure, which is shown on the graph as the point where the aggregate expenditure schedule crosses the 45-degree line. In this example, the equilibrium occurs at 6,000. This equilibrium can also be read off the table under the figure; it is the level of national income where aggregate expenditure is equal to national income. Equilibrium in the Keynesian Cross Model With the aggregate expenditure line in place, the next step is to relate it to the two other elements of the Keynesian cross diagram. Thus, the first subsection interprets the intersection of the aggregate expenditure function and the 45-degree line, while the next subsection relates this point of intersection to the potential GDP line. Where Equilibrium Occurs The point where the aggregate expenditure line that is constructed from C + I + G + X – M crosses the 45-degree line will be the equilibrium for the economy. It is the only point on the aggregate expenditure line where the total amount being spent on aggregate demand equals the total level of production. In Figure, this point of equilibrium (E0) happens at 6,000, which can also be read off Table. The meaning of “equilibrium” remains the same; that is, equilibrium is a point of balance where no incentive exists to shift away from that outcome. To understand why the point of intersection between the aggregate expenditure function and the 45-degree line is a macroeconomic equilibrium, consider what would happen if an economy found itself to the right of the equilibrium point E, say point H in Figure, where output is higher than the equilibrium. At point H, the level of aggregate expenditure is below the 45-degree line, so that the level of aggregate expenditure in the economy is less than the level of output. As a result, at point H, output is piling up unsold—not a sustainable state of affairs. Conversely, consider the situation where the level of output is at point L—where real output is lower than the equilibrium. In that case, the level of aggregate demand in the economy is above the 45-degree line, indicating that the level of aggregate expenditure in the economy is greater than the level of output. When the level of aggregate demand has emptied the store shelves, it cannot be sustained, either. Firms will respond by increasing their level of production. Thus, the equilibrium must be the point where the amount produced and the amount spent are in balance, at the intersection of the aggregate expenditure function and the 45-degree line. Finding Equilibrium Table gives some information on an economy. The Keynesian model assumes that there is some level of consumption even without income. That amount is $236 – $216 = $20. $20 will be consumed when national income equals zero. Assume that taxes are 0.2 of real GDP. Let the marginal propensity to save of after-tax income be 0.1. The level of investment is $70, the level of government spending is $80, and the level of exports is $50. Imports are 0.2 of after-tax income. Given these values, you need to complete Table and then answer these questions: What is the consumption function? What is the equilibrium? Why is a national income of $300 not at equilibrium? How do expenditures and output compare at this point? \| National Income \| Taxes \| After-tax income \| Consumption \| I + G + X \| Imports \| Aggregate Expenditures \| --- --- --- \| $300 \| $236 \| \| $400 \| \| $500 \| \| $600 \| \| $700 \| Step 1. Calculate the amount of taxes for each level of national income(reminder: GDP = national income) for each level of national income using the following as an example: Step 2. Calculate after-tax income by subtracting the tax amount from national income for each level of national income using the following as an example: Step 3. Calculate consumption. The marginal propensity to save is given as 0.1. This means that the marginal propensity to consume is 0.9, since MPS + MPC = 1. Therefore, multiply 0.9 by the after-tax income amount using the following as an example: Step 4. Consider why the table shows consumption of $236 in the first row. As mentioned earlier, the Keynesian model assumes that there is some level of consumption even without income. That amount is $236 – $216 = $20. Step 5. There is now enough information to write the consumption function. The consumption function is found by figuring out the level of consumption that will happen when income is zero. Remember that: Let C represent the consumption function, Y represent national income, and T represent taxes. Step 6. Use the consumption function to find consumption at each level of national income. Step 7. Add investment (I), government spending (G), and exports (X). Remember that these do not change as national income changes: Step 8. Find imports, which are 0.2 of after-tax income at each level of national income. For example: Step 9. Find aggregate expenditure by adding C + I + G + X – I for each level of national income. Your completed table should look like Table. \| National Income (Y) \| Tax = 0.2 × Y (T) \| After-tax income (Y – T) \| Consumption C = $20 + 0.9(Y – T) \| I + G + X \| Minus Imports (M) \| Aggregate Expenditures AE = C + I + G + X – M \| --- --- --- \| $300 \| $60 \| $240 \| $236 \| $200 \| $48 \| $388 \| \| $400 \| $80 \| $320 \| $308 \| $200 \| $64 \| $444 \| \| $500 \| $100 \| $400 \| $380 \| $200 \| $80 \| $500 \| \| $600 \| $120 \| $480 \| $452 \| $200 \| $96 \| $556 \| \| $700 \| $140 \| $560 \| $524 \| $200 \| $112 \| $612 \| Step 10. Answer the question: What is equilibrium? Equilibrium occurs where AE = Y. Table shows that equilibrium occurs where national income equals aggregate expenditure at $500. Step 11. Find equilibrium mathematically, knowing that national income is equal to aggregate expenditure. Since T is 0.2 of national income, substitute T with 0.2 Y so that: Solve for Y. Step 12. Answer this question: Why is a national income of $300 not an equilibrium? At national income of $300, aggregate expenditures are $388. Step 13. Answer this question: How do expenditures and output compare at this point? Aggregate expenditures cannot exceed output (GDP) in the long run, since there would not be enough goods to be bought. Recessionary and Inflationary Gaps In the Keynesian cross diagram, if the aggregate expenditure line intersects the 45-degree line at the level of potential GDP, then the economy is in sound shape. There is no recession, and unemployment is low. But there is no guarantee that the equilibrium will occur at the potential GDP level of output. The equilibrium might be higher or lower. For example, Figure (a) illustrates a situation where the aggregate expenditure line intersects the 45-degree line at point E0, which is a real GDP of $6,000, and which is below the potential GDP of $7,000. In this situation, the level of aggregate expenditure is too low for GDP to reach its full employment level, and unemployment will occur. The distance between an output level like E0 that is below potential GDP and the level of potential GDP is called a recessionary gap. Because the equilibrium level of real GDP is so low, firms will not wish to hire the full employment number of workers, and unemployment will be high. What might cause a recessionary gap? Anything that shifts the aggregate expenditure line down is a potential cause of recession, including a decline in consumption, a rise in savings, a fall in investment, a drop in government spending or a rise in taxes, or a fall in exports or a rise in imports. Moreover, an economy that is at equilibrium with a recessionary gap may just stay there and suffer high unemployment for a long time; remember, the meaning of equilibrium is that there is no particular adjustment of prices or quantities in the economy to chase the recession away. The appropriate response to a recessionary gap is for the government to reduce taxes or increase spending so that the aggregate expenditure function shifts up from AE0 to AE1. When this shift occurs, the new equilibrium E1 now occurs at potential GDP as shown in Figure (a). Conversely, Figure (b) shows a situation where the aggregate expenditure schedule (AE0) intersects the 45-degree line above potential GDP. The gap between the level of real GDP at the equilibrium E0 and potential GDP is called an inflationary gap. The inflationary gap also requires a bit of interpreting. After all, a naïve reading of the Keynesian cross diagram might suggest that if the aggregate expenditure function is just pushed up high enough, real GDP can be as large as desired—even doubling or tripling the potential GDP level of the economy. This implication is clearly wrong. An economy faces some supply-side limits on how much it can produce at a given time with its existing quantities of workers, physical and human capital, technology, and market institutions. The inflationary gap should be interpreted, not as a literal prediction of how large real GDP will be, but as a statement of how much extra aggregate expenditure is in the economy beyond what is needed to reach potential GDP. An inflationary gap suggests that because the economy cannot produce enough goods and services to absorb this level of aggregate expenditures, the spending will instead cause an inflationary increase in the price level. In this way, even though changes in the price level do not appear explicitly in the Keynesian cross equation, the notion of inflation is implicit in the concept of the inflationary gap. The appropriate Keynesian response to an inflationary gap is shown in Figure (b). The original intersection of aggregate expenditure line AE0 and the 45-degree line occurs at $8,000, which is above the level of potential GDP at $7,000. If AE0 shifts down to AE1, so that the new equilibrium is at E1, then the economy will be at potential GDP without pressures for inflationary price increases. The government can achieve a downward shift in aggregate expenditure by increasing taxes on consumers or firms, or by reducing government expenditures. The Multiplier Effect The Keynesian policy prescription has one final twist. Assume that for a certain economy, the intersection of the aggregate expenditure function and the 45-degree line is at a GDP of 700, while the level of potential GDP for this economy is $800. By how much does government spending need to be increased so that the economy reaches the full employment GDP? The obvious answer might seem to be $800 – $700 = $100; so raise government spending by $100. But that answer is incorrect. A change of, for example, $100 in government expenditures will have an effect of more than $100 on the equilibrium level of real GDP. The reason is that a change in aggregate expenditures circles through the economy: households buy from firms, firms pay workers and suppliers, workers and suppliers buy goods from other firms, those firms pay their workers and suppliers, and so on. In this way, the original change in aggregate expenditures is actually spent more than once. This is called the multiplier effect: An initial increase in spending, cycles repeatedly through the economy and has a larger impact than the initial dollar amount spent. How Does the Multiplier Work? To understand how the multiplier effect works, return to the example in which the current equilibrium in the Keynesian cross diagram is a real GDP of $700, or $100 short of the $800 needed to be at full employment, potential GDP. If the government spends $100 to close this gap, someone in the economy receives that spending and can treat it as income. Assume that those who receive this income pay 30% in taxes, save 10% of after-tax income, spend 10% of total income on imports, and then spend the rest on domestically produced goods and services. As shown in the calculations in Figure and Table, out of the original $100 in government spending, $53 is left to spend on domestically produced goods and services. That $53 which was spent, becomes income to someone, somewhere in the economy. Those who receive that income also pay 30% in taxes, save 10% of after-tax income, and spend 10% of total income on imports, as shown in Figure, so that an additional $28.09 (that is, 0.53 × $53) is spent in the third round. The people who receive that income then pay taxes, save, and buy imports, and the amount spent in the fourth round is $14.89 (that is, 0.53 × $28.09). \| \| \| --- \| \| Original increase in aggregate expenditure from government spending \| 100 \| \| Which is income to people throughout the economy: Pay 30% in taxes. Save 10% of after-tax income. Spend 10% of income on imports. Second-round increase of… \| 70 – 7 – 10 = 53 \| \| Which is $53 of income to people through the economy: Pay 30% in taxes. Save 10% of after-tax income. Spend 10% of income on imports. Third-round increase of… \| 37.1 – 3.71 – 5.3 = 28.09 \| \| Which is $28.09 of income to people through the economy: Pay 30% in taxes. Save 10% of after-tax income. Spend 10% of income on imports. Fourth-round increase of… \| 19.663 – 1.96633 – 2.809 = 14.89 \| Calculating the Multiplier Effect Thus, over the first four rounds of aggregate expenditures, the impact of the original increase in government spending of $100 creates a rise in aggregate expenditures of $100 + $53 + $28.09 + $14.89 = $195.98. Figure shows these total aggregate expenditures after these first four rounds, and then the figure shows the total aggregate expenditures after 30 rounds. The additional boost to aggregate expenditures is shrinking in each round of consumption. After about 10 rounds, the additional increments are very small indeed—nearly invisible to the naked eye. After 30 rounds, the additional increments in each round are so small that they have no practical consequence. After 30 rounds, the cumulative value of the initial boost in aggregate expenditure is approximately $213. Thus, the government spending increase of $100 eventually, after many cycles, produced an increase of $213 in aggregate expenditure and real GDP. In this example, the multiplier is $213/$100 = 2.13. Calculating the Multiplier Fortunately for everyone who is not carrying around a computer with a spreadsheet program to project the impact of an original increase in expenditures over 20, 50, or 100 rounds of spending, there is a formula for calculating the multiplier. The data from Figure and Table is: Marginal Propensity to Save (MPS) = 30% Tax rate = 10% Marginal Propensity to Import (MPI) = 10% The MPC is equal to 1 – MPS, or 0.7. Therefore, the spending multiplier is: A change in spending of $100 multiplied by the spending multiplier of 2.13 is equal to a change in GDP of $213. Not coincidentally, this result is exactly what was calculated in Figure after many rounds of expenditures cycling through the economy. The size of the multiplier is determined by what proportion of the marginal dollar of income goes into taxes, saving, and imports. These three factors are known as “leakages,” because they determine how much demand “leaks out” in each round of the multiplier effect. If the leakages are relatively small, then each successive round of the multiplier effect will have larger amounts of demand, and the multiplier will be high. Conversely, if the leakages are relatively large, then any initial change in demand will diminish more quickly in the second, third, and later rounds, and the multiplier will be small. Changes in the size of the leakages—a change in the marginal propensity to save, the tax rate, or the marginal propensity to import—will change the size of the multiplier. Calculating Keynesian Policy Interventions Returning to the original question: How much should government spending be increased to produce a total increase in real GDP of $100? If the goal is to increase aggregate demand by $100, and the multiplier is 2.13, then the increase in government spending to achieve that goal would be $100/2.13 = $47. Government spending of approximately $47, when combined with a multiplier of 2.13 (which is, remember, based on the specific assumptions about tax, saving, and import rates), produces an overall increase in real GDP of $100, restoring the economy to potential GDP of $800, as Figure shows. The multiplier effect is also visible on the Keynesian cross diagram. Figure shows the example we have been discussing: a recessionary gap with an equilibrium of $700, potential GDP of $800, the slope of the aggregate expenditure function (AE0) determined by the assumptions that taxes are 30% of income, savings are 0.1 of after-tax income, and imports are 0.1 of before-tax income. At AE1, the aggregate expenditure function is moved up to reach potential GDP. Now, compare the vertical shift upward in the aggregate expenditure function, which is $47, with the horizontal shift outward in real GDP, which is $100 (as these numbers were calculated earlier). The rise in real GDP is more than double the rise in the aggregate expenditure function. (Similarly, if you look back at Figure, you will see that the vertical movements in the aggregate expenditure functions are smaller than the change in equilibrium output that is produced on the horizontal axis. Again, this is the multiplier effect at work.) In this way, the power of the multiplier is apparent in the income–expenditure graph, as well as in the arithmetic calculation. The multiplier does not just affect government spending, but applies to any change in the economy. Say that business confidence declines and investment falls off, or that the economy of a leading trading partner slows down so that export sales decline. These changes will reduce aggregate expenditures, and then will have an even larger effect on real GDP because of the multiplier effect. Read the following Clear It Up feature to learn how the multiplier effect can be applied to analyze the economic impact of professional sports. How can the multiplier be used to analyze the economic impact of professional sports? Attracting professional sports teams and building sports stadiums to create jobs and stimulate business growth is an economic development strategy adopted by many communities throughout the United States. In his recent article, “Public Financing of Private Sports Stadiums,” James Joyner of Outside the Beltway looked at public financing for NFL teams. Joyner’s findings confirm the earlier work of John Siegfried of Vanderbilt University and Andrew Zimbalist of Smith College. Siegfried and Zimbalist used the multiplier to analyze this issue. They considered the amount of taxes paid and dollars spent locally to see if there was a positive multiplier effect. Since most professional athletes and owners of sports teams are rich enough to owe a lot of taxes, let’s say that 40% of any marginal income they earn is paid in taxes. Because athletes are often high earners with short careers, let’s assume that they save one-third of their after-tax income. However, many professional athletes do not live year-round in the city in which they play, so let’s say that one-half of the money that they do spend is spent outside the local area. One can think of spending outside a local economy, in this example, as the equivalent of imported goods for the national economy. Now, consider the impact of money spent at local entertainment venues other than professional sports. While the owners of these other businesses may be comfortably middle-income, few of them are in the economic stratosphere of professional athletes. Because their incomes are lower, so are their taxes; say that they pay only 35% of their marginal income in taxes. They do not have the same ability, or need, to save as much as professional athletes, so let’s assume their MPC is just 0.8. Finally, because more of them live locally, they will spend a higher proportion of their income on local goods—say, 65%. If these general assumptions hold true, then money spent on professional sports will have less local economic impact than money spent on other forms of entertainment. For professional athletes, out of a dollar earned, 40 cents goes to taxes, leaving 60 cents. Of that 60 cents, one-third is saved, leaving 40 cents, and half is spent outside the area, leaving 20 cents. Only 20 cents of each dollar is cycled into the local economy in the first round. For locally-owned entertainment, out of a dollar earned, 35 cents goes to taxes, leaving 65 cents. Of the rest, 20% is saved, leaving 52 cents, and of that amount, 65% is spent in the local area, so that 33.8 cents of each dollar of income is recycled into the local economy. Siegfried and Zimbalist make the plausible argument that, within their household budgets, people have a fixed amount to spend on entertainment. If this assumption holds true, then money spent attending professional sports events is money that was not spent on other entertainment options in a given metropolitan area. Since the multiplier is lower for professional sports than for other local entertainment options, the arrival of professional sports to a city would reallocate entertainment spending in a way that causes the local economy to shrink, rather than to grow. Thus, their findings seem to confirm what Joyner reports and what newspapers across the country are reporting. A quick Internet search for “economic impact of sports” will yield numerous reports questioning this economic development strategy. Multiplier Tradeoffs: Stability versus the Power of Macroeconomic Policy Is an economy healthier with a high multiplier or a low one? With a high multiplier, any change in aggregate demand will tend to be substantially magnified, and so the economy will be more unstable. With a low multiplier, by contrast, changes in aggregate demand will not be multiplied much, so the economy will tend to be more stable. However, with a low multiplier, government policy changes in taxes or spending will tend to have less impact on the equilibrium level of real output. With a higher multiplier, government policies to raise or reduce aggregate expenditures will have a larger effect. Thus, a low multiplier means a more stable economy, but also weaker government macroeconomic policy, while a high multiplier means a more volatile economy, but also an economy in which government macroeconomic policy is more powerful. Key Concepts and Summary The expenditure-output model or Keynesian cross diagram shows how the level of aggregate expenditure (on the vertical axis) varies with the level of economic output (shown on the horizontal axis). Since the value of all macroeconomic output also represents income to someone somewhere else in the economy, the horizontal axis can also be interpreted as national income. The equilibrium in the diagram will occur where the aggregate expenditure line crosses the 45-degree line, which represents the set of points where aggregate expenditure in the economy is equal to output (or national income). Equilibrium in a Keynesian cross diagram can happen at potential GDP, or below or above that level. The consumption function shows the upward-sloping relationship between national income and consumption. The marginal propensity to consume (MPC) is the amount consumed out of an additional dollar of income. A higher marginal propensity to consume means a steeper consumption function; a lower marginal propensity to consume means a flatter consumption function. The marginal propensity to save (MPS) is the amount saved out of an additional dollar of income. It is necessarily true that MPC + MPS = 1. The investment function is drawn as a flat line, showing that investment in the current year does not change with regard to the current level of national income. However, the investment function will move up and down based on the expected rate of return in the future. Government spending is drawn as a horizontal line in the Keynesian cross diagram, because its level is determined by political considerations, not by the current level of income in the economy. Taxes in the basic Keynesian cross diagram are taken into account by adjusting the consumption function. The export function is drawn as a horizontal line in the Keynesian cross diagram, because exports do not change as a result of changes in domestic income, but they move as a result of changes in foreign income, as well as changes in exchange rates. The import function is drawn as a downward-sloping line, because imports rise with national income, but imports are a subtraction from aggregate demand. Thus, a higher level of imports means a lower level of expenditure on domestic goods. In a Keynesian cross diagram, the equilibrium may be at a level below potential GDP, which is called a recessionary gap, or at a level above potential GDP, which is called an inflationary gap. The multiplier effect describes how an initial change in aggregate demand generated several times as much as cumulative GDP. The size of the spending multiplier is determined by three leakages: spending on savings, taxes, and imports. The formula for the multiplier is: An economy with a lower multiplier is more stable—it is less affected either by economic events or by government policy than an economy with a higher multiplier. Self-Check Questions Sketch the aggregate expenditure-output diagram with the recessionary gap. The following figure shows the aggregate expenditure-output diagram with the recessionary gap. Sketch the aggregate expenditure-output diagram with an inflationary gap. The following figure shows the aggregate expenditure-output diagram with an inflationary gap. An economy has the following characteristics: Y = National income Taxes = T = 0.25Y C = Consumption = 400 + 0.85(Y – T) I = 300 G = 200 X = 500 M = 0.1(Y – T) Find the equilibrium for this economy. If potential GDP is 3,500, then what change in government spending is needed to achieve this level? Do this problem two ways. First, plug 3,500 into the equations and solve for G. Second, calculate the multiplier and figure it out that way. First, set up the calculation. Then insert Y for AE and 0.25Y for T. If full employment is 3,500, then one approach is to plug in 3,500 for Y throughout the equation, but to leave G as a separate variable. A G value of 331.25 is an increase of 131.25 from its original level of 200. Alternatively, the multiplier is that, out of every dollar spent, 0.25 goes to taxes, leaving 0.75, and out of after-tax income, 0.15 goes to savings and 0.1 to imports. Because (0.75)(0.15) = 0.1125 and (0.75)(0.1) = 0.075, this means that out of every dollar spent: 1 –0.25 –0.1125 –0.075 = 0.5625. Thus, using the formula, the multiplier is: To increase equilibrium GDP by 300, it will take a boost of 300/2.2837, which again works out to 131.25. Table represents the data behind a Keynesian cross diagram. Assume that the tax rate is 0.4 of national income; the MPC out of the after-tax income is 0.8; investment is $2,000; government spending is $1,000; exports are $2,000 and imports are 0.05 of after-tax income. What is the equilibrium level of output for this economy? \| National Income \| After-tax Income \| Consumption \| I + G + X \| Minus Imports \| Aggregate Expenditures \| --- --- --- \| \| $8,000 \| $4,340 \| \| $9,000 \| \| $10,000 \| \| $11,000 \| \| $12,000 \| \| $13,000 \| The following table illustrates the completed table. The equilibrium is level is italicized. \| National Income \| After-tax Income \| Consumption \| I + G + X \| Minus Imports \| Aggregate Expenditures \| --- --- --- \| \| $8,000 \| $4,800 \| $4,340 \| $5,000 \| $240 \| $9,100 \| \| $9,000 \| $5,400 \| $4,820 \| $5,000 \| $270 \| $9,550 \| \| $10,000 \| $6,000 \| $5,300 \| $5,000 \| $300 \| $10,000 \| \| $11,000 \| $6,600 \| $5,780 \| $5,000 \| $330 \| $10,450 \| \| $12,000 \| $7,200 \| $6,260 \| $5,000 \| $360 \| $10,900 \| \| $13,000 \| $7,800 \| $46,740 \| $5,000 \| $4,390 \| $11,350 \| The alternative way of determining equilibrium is to solve for Y, where Y = national income, using: Y = AE = C + I + G + X – M Solving for Y, we see that the equilibrium level of output is Y = $10,000. Explain how the multiplier works. Use an MPC of 80% in an example. The multiplier refers to how many times a dollar will turnover in the economy. It is based on the Marginal Propensity to Consume (MPC) which tells how much of every dollar received will be spent. If the MPC is 80% then this means that out of every one dollar received by a consumer, $0.80 will be spent. This $0.80 is received by another person. In turn, 80% of the $0.80 received, or $0.64, will be spent, and so on. The impact of the multiplier is diluted when the effect of taxes and expenditure on imports is considered. To derive the multiplier, take the 1/1 – F; where F is equal to percent of savings, taxes, and expenditures on imports. Review Questions What is on the axes of an expenditure-output diagram? What does the 45-degree line show? What determines the slope of a consumption function? What is the marginal propensity to consume, and how is it related to the marginal propensity to import? Why are the investment function, the government spending function, and the export function all drawn as flat lines? Why does the import function slope down? What is the marginal propensity to import? What are the components on which the aggregate expenditure function is based? Is the equilibrium in a Keynesian cross diagram usually expected to be at or near potential GDP? What is an inflationary gap? A recessionary gap? What is the multiplier effect? Why are savings, taxes, and imports referred to as “leakages” in calculating the multiplier effect? Will an economy with a high multiplier be more stable or less stable than an economy with a low multiplier in response to changes in the economy or in government policy? How do economists use the multiplier? Critical Thinking Questions What does it mean when the aggregate expenditure line crosses the 45-degree line? In other words, how would you explain the intersection in words? Which model, the AD/AS or the AE model better explains the relationship between rising price levels and GDP? Why? What are some reasons that the economy might be in a recession, and what is the appropriate government action to alleviate the recession? What should the government do to relieve inflationary pressures if the aggregate expenditure is greater than potential GDP? Two countries are in a recession. Country A has an MPC of 0.8 and Country B has an MPC of 0.6. In which country will government spending have the greatest impact? Compare two policies: a tax cut on income or an increase in government spending on roads and bridges. What are both the short-term and long-term impacts of such policies on the economy? What role does government play in stabilizing the economy and what are the tradeoffs that must be considered? If there is a recessionary gap of $100 billion, should the government increase spending by $100 billion to close the gap? Why? Why not? What other changes in the economy can be evaluated by using the multiplier? References Joyner, James. Outside the Beltway. “Public Financing of Private Sports Stadiums.” Last modified May 23, 2012. Siegfried, John J., and Andrew Zimbalist. “The Economics of Sports Facilities and Their Communities.” Journal of Economic Perspectives. no. 3 (2000): 95-114.
10809	https://math.stackexchange.com/questions/2308538/why-the-inverse-function-of-2x-is-log-and-not-root	logarithms - Why the inverse function of $2^x $ is log and not root? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Why the inverse function of 2 x 2 x is log and not root? Ask Question Asked 8 years, 3 months ago Modified8 years, 3 months ago Viewed 340 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I'm sure this is a very basic question, but while reading about functions and their inverse I'm having a hard time understanding why for x 2 x 2 or x 3 x 3 the inverse are x−−√x and x−−√3 x 3, but for 2 x 2 x is log 2 x log 2⁡x. logarithms roots inverse-function Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Jun 3, 2017 at 18:57 CarloCarlo 123 3 3 bronze badges 1 Because 2 x 2 x and x 2 x 2 don't commute for most x x.Michael Hoppe –Michael Hoppe 2017-06-03 18:59:45 +00:00 Commented Jun 3, 2017 at 18:59 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. There's two separate questions here: Why islog 2 x log 2⁡x the inverse of 2 x 2 x? Why isn'tx−−√x the inverse of 2 x 2 x? The answer to the first one is just the definition of logarithm: log 2 x log 2⁡x means "The number 2 2 must be raised to to get x x," so 2 log 2 x 2 log 2⁡x means "2 2 raised to (the number 2 2 must be raised to to get x x)," which is clearly x x itself. (Exercise: convince yourself that also log 2(2 x)=x log 2⁡(2 x)=x.) Basically, the logarithm is defined to be the inverse of 2 x 2 x. For an example of how to use this definition, let's say you were asked to calculate log 2 8 log 2⁡8. You'd ask yourself, "What do I need to raise 2 2 to to get 8 8?," and quickly see that the answer is "3 3" - so log 2 8=3 log 2⁡8=3. Now obviously understanding what log 2(π)log 2⁡(π) is is going to be much more complicated, but the basic idea is the same. OK,now what about x−−√x? It's easy to find a counterexample that shows that x−−√x isn't the inverse of 2 x 2 x: namely, take x=64 x=64. Then we have 2 x√=2 8=256≠x.2 x=2 8=256≠x. So it's clear that x−−√x can't be the inverse of 2 x 2 x. But this isn't very satisfying, since it doesn't explain why. To understand why x−−√x isn't the inverse of x x, think about what it means: x−−√x is "the thing we need to square to get x x." So x−−√x is the inverse of x 2 x 2: (the thing we need to square to get x x) squared is obviously just x x itself. I'm being a bit glib about square roots here - there's actually a serious issue, namely that the function x 2 x 2 isn't bijective and so doesn't have a genuine inverse. Basically, the function x−−√x isn't really "the thing we need to square to get x x," but rather "the nonnegative thing we need to square to get x x." The function −x−−√−x also has the same nice behavior that (−x−−√)2=x(−x)2=x. Really, x−−√x and −x−−√−x are right inverses of x 2 x 2. But this is a bit of an aside from the general picture here. The moral of the story: look at how the functions you're thinking about are defined. These definitions will often tell you directly what is an (almost) inverse of what; it's not always what you'd expect, but it can be figured out by thinking about what the functions mean. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 3, 2017 at 19:15 Noah SchweberNoah Schweber 262k 22 22 gold badges 390 390 silver badges 674 674 bronze badges 1 this is a really great a complete answer, and even more. The way I (wrongly) reasoned was basically this: I have 2 x 2 x and I start thinking about the values x can be. So then I think: if x is 2, or 3, then the inverse can just be the square or the cube root Carlo –Carlo 2017-06-03 19:34:19 +00:00 Commented Jun 3, 2017 at 19:34 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Recall that by definition of inverse function, we must have that: f∘f−1(x)=i d(x)=x f∘f−1(x)=i d(x)=x Define f(x)=2 x f(x)=2 x , g(x)=x−−√g(x)=x Clearly: f∘g(x)=f(x−−√)=2 x√≠x f∘g(x)=f(x)=2 x≠x Take, for example x=64 x=64. Then we have 2 64√=2 8=256≠64 2 64=2 8=256≠64 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jun 3, 2017 at 19:33 answered Jun 3, 2017 at 19:04 user370967 user370967 0 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. To find the inverse we follow the following procedure: The inverse of the function x 2 x 2: Let y=x 2 y=x 2 (we need to write x x in terms of y y) y=x 2⟹x=y√y=x 2⟹x=y Finally we change the places of x x and y y. Therefore the inverse of the function y=x 2 y=x 2 is y=x−−√y=x. The inverse of the function 2 x 2 x: Let y=2 x y=2 x (we need to write x x in terms of y y) y=2 x⟹log y=x log 2⟹x=log y log 2=log 2 y y=2 x⟹log⁡y=x log⁡2⟹x=log⁡y log⁡2=log 2⁡y Finally we change the places of x x and y y. Therefore the inverse of the function y=2 x y=2 x is y=log 2 x y=log 2⁡x. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 3, 2017 at 19:10 AhmedAhmed 1,356 8 8 silver badges 12 12 bronze badges 3 it makes sense, but why when you have y=2 x y=2 x you can't go to y√x=2–√x x y x=2 x x ?Carlo –Carlo 2017-06-03 19:12:46 +00:00 Commented Jun 3, 2017 at 19:12 1 As I wrote, at the end of the day, we need to write x x in terms of y y. So, after what you wrote you have to take the log log of the two sides to reach what I have already got.Ahmed –Ahmed 2017-06-03 19:20:20 +00:00 Commented Jun 3, 2017 at 19:20 1 I just had one of these "a-ha!" or "oooh" moments. Thanks!Carlo –Carlo 2017-06-03 19:27:03 +00:00 Commented Jun 3, 2017 at 19:27 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. One of the reasons to explain this is if I take 2^x=y and find its inverse function I have to switch the x and y's and solve for y. ``` 2^x=y 2^y=x I have switched the x's and y's \log_2 x=y I solved for y ``` As a result, we see that the inverse of 2^x is a logarithmic function Whereas, for x^3 or x^2 you can simply do the same thing. ``` x^2=y y^2=x Again, I switch the x's and y's y=\sqrtx Again, I solve for y ``` This shows that the inverse function of 2^x is, indeed, a logarithmic function and the inverses of any power functions such as x^3=y is a radical function. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 3, 2017 at 19:17 гамма функциягамма функция 1 2 2 bronze badges Add a comment\| You must log in to answer this question. 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10810	https://oeis.org/A020911	A020911 - OEIS login The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A020911 Number of digits in the base 3 representation of n-th Fibonacci number. 1 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 9, 10, 10, 11, 11, 12, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 23, 24, 24, 25, 25, 26, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 30 (list; graph; refs; listen; history; text; internal format) OFFSET 1,4 LINKS Table of n, a(n) for n=1..70. FORMULA a(n) = A081604(A000045(n)). - Andrew Howroyd, Jan 01 2020 MATHEMATICA IntegerLength[Fibonacci[Range], 3] ( Harvey P. Dale, Aug 13 2018 ) PROG (PARI) a(n)={1 + logint(fibonacci(n), 3)} \ Andrew Howroyd, Jan 01 2020 (Magma) [#Intseq(Fibonacci(n), 3):n in [1..70]]; // Marius A. Burtea, Jan 02 2020 CROSSREFS Cf. A000045 (Fibonacci numbers), A081604 (base-3 digits of n). Sequence in context: A114540A183142A121830 A336861A029125A074990 Adjacent sequences: A020908A020909A020910 A020912A020913A020914 KEYWORD nonn,base AUTHOR Clark Kimberling EXTENSIONS Offset corrected and terms a(50) and beyond from Andrew Howroyd, Jan 01 2020 STATUS approved LookupWelcomeWikiRegisterMusicPlot 2DemosIndexWebCamContributeFormatStyle SheetTransformsSuperseekerRecents The OEIS Community Maintained by The OEIS Foundation Inc. Last modified September 28 20:58 EDT 2025. Contains 388815 sequences. License Agreements, Terms of Use, Privacy Policy
10811	https://www.physicsforums.com/threads/finding-the-wavelength-on-a-sinusoidal-wave-on-a-string.806018/	Finding the wavelength on a sinusoidal wave on a string • Physics Forums Insights Blog-- Browse All Articles --Physics ArticlesPhysics TutorialsPhysics GuidesPhysics FAQMath ArticlesMath TutorialsMath GuidesMath FAQEducation ArticlesEducation GuidesBio/Chem ArticlesTechnology GuidesComputer Science Tutorials ForumsIntro Physics Homework HelpAdvanced Physics Homework HelpPrecalculus Homework HelpCalculus Homework HelpBio/Chem Homework HelpEngineering Homework Help Trending Log inRegister What's new Intro Physics Homework Help Advanced Physics Homework Help Precalculus Homework Help Calculus Homework Help Bio/Chem Homework Help Engineering Homework Help Menu Log in Register Navigation More options Style variation SystemLightDark Contact us Close Menu Forums Homework Help Introductory Physics Homework Help Finding the wavelength on a sinusoidal wave on a string Thread starter Andrew Jacobson Start date Mar 31, 2015 TagsSinusoidalStringWaveWavelength AI Thread Summary The discussion focuses on solving a problem related to a sinusoidal wave on a string, specifically determining the amplitude, period, wavelength, and wave speed. The amplitude is identified as 4mm and the period as 0.04s based on the provided graph. There is confusion regarding the relationship between the two points at x=0m and x=0.09m, with participants debating whether the wavelength could be 0.09m or larger. The relationship between wave speed, wavelength, and frequency is highlighted, emphasizing the formula v=λf. Clarification is sought on how to accurately determine the wavelength given the conditions of the problem. Mar 31, 2015 1 Andrew Jacobson 6 0 Homework Statement Hi, this is a problem regarding mathematical descriptions of waves. I've attached an imagine of the picture but I'll also type out the problem for clarity. "A sinusoidal wave is propagating along a stretched string that lies on the x-axis. The distplacement of the string as a function of time is graphed in Fig.E11 for particles x=0m and at x=0.09m. (a) What is the amplitude of the wave? (b) What is the period of the wave? (c) You are told that the two points x=0 and x=0.09 are within one wavelength of each other. If the wave is moving in the +x-direction, determine the wavelength and wave speed." Homework Equations k=2\pi/\lambda y(x,t)=Acos(kx+\omega t) The Attempt at a Solution (a) from reading off the graph you can see at the amplitude is 4mm or 0.004m (b) from reading off the graph you can see that the period = 0.04s (c) this is where I got confused. I thought about making simultaneous equations using x=0 and x=0.09. I figure that the answer involves using the fact that k=2\pi/\lambda but I'm a little unsure. I started playing with the idea that when y=0 and t=0 0.004cos(kx)=0 and therefore cos(kx)=0 and kx=n\pi/2 however that's not particularly helpful since I don't know the value of x. If anyone could give me a hint or put me on the right track it'd be much appreciated. Discover more Scientist Job Board Science Magazines Purchase engineering software license Science Education Courses Math math Buy academic advising services Physics Tutors Online Math Problem Solvers Mathematical Attachments 20150331_130354.jpg 46.8 KB · Views: 589 Physics news on Phys.org New adaptive optics system promises sharper gravitational-wave observations Physics-informed AI learns local rules behind flocking and collective motion behaviors New perspectives on light-matter interaction: How virtual charges influence material responses Mar 31, 2015 2 lep11 380 7 Andrew Jacobson said: Homework Statement Hi, this is a problem regarding mathematical descriptions of waves. I've attached an imagine of the picture but I'll also type out the problem for clarity. "A sinusoidal wave is propagating along a stretched string that lies on the x-axis. The distplacement of the string as a function of time is graphed in Fig.E11 for particles x=0m and at x=0.09m. (a) What is the amplitude of the wave? (b) What is the period of the wave? (c) You are told that the two points x=0 and x=0.09 are within one wavelength of each other. If the wave is moving in the +x-direction, determine the wavelength and wave speed." Wouldn't this mean the wavelength is 0.09m? How are frequency, wavelength and speed related? Andrew Jacobson said: (a) from reading off the graph you can see at the amplitude is 4mm or 0.004m Correct Andrew Jacobson said: (b) from reading off the graph you can see that the period = 0.04s Correct Mar 31, 2015 3 Andrew Jacobson 6 0 lep11 said: Wouldn't this mean the wavelength is 0.09m? How are frequency, wavelength and speed related?Correct Correct Well it just says that they're within one wavelength, so the wavelength could be bigger than 0.09m? I know that v=\lambda f but this doesn't help since I only know the frequency. Discover more Laboratory equipment Sciences Physics simulation software Technical writing guides STEM career coaching STEM Career Advice Educational software Mathematics Help Scientist Networking Platform Math Articles Mar 31, 2015 4 lep11 380 7 Andrew Jacobson said: Well it just says that they're within one wavelength, so the wavelength could be bigger than 0.09m? I know that v=\lambda f but this doesn't help since I only know the frequency. Oh, my bad, I thought the points are one wavelength away from each other. Anyway, v=λf=λ/T still holds. Last edited: Mar 31, 2015 Thread 'I've come across another fluid pressure problem I don't understand' Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000101(1+4)=50000 N; F1=ρgsh=1000100.14=4000 N; F3=50000-4000=46000 N? View full post » By vdance Friday, 4:31 AM Replies: 40 Thread 'Variable mass system : water sprayed into a moving container' Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}... View full post » By Su6had1p Friday, 10:41 AM Replies: 27 Thread 'Minimum mass of a block' Here we know that if block B is going to move up or just be at the verge of moving up will act downwards and maximum static friction will act downwards Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach... 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10812	https://www.goodreads.com/author/show/18221818.K_S_Narayan_Reddy	K.S. Narayan Reddy (Author of The Essentials of Forensic Medicine and Toxicology) , Home My Books Browse ▾ Recommendations Choice Awards Genres Giveaways New Releases Lists Explore News & Interviews Genres Art Biography Business Children's Christian Classics Comics Cookbooks Ebooks Fantasy Fiction Graphic Novels Historical Fiction History Horror Memoir Music Mystery Nonfiction Poetry Psychology Romance Science Science Fiction Self Help Sports Thriller Travel Young Adult More Genres Community ▾ Groups Quotes Ask the Author Sign In Join Sign up View profile Profile Friends Groups Discussions Comments Reading Challenge Kindle Notes & Highlights Quotes Favorite genres Friends’ recommendations Account settings Help Sign out Home My Books Browse ▾ Recommendations Choice Awards Genres Giveaways New Releases Lists Explore News & Interviews Genres Art Biography Business Children's Christian Classics Comics Cookbooks Ebooks Fantasy Fiction Graphic Novels Historical Fiction History Horror Memoir Music Mystery Nonfiction Poetry Psychology Romance Science Science Fiction Self Help Sports Thriller Travel Young Adult More Genres Community ▾ Groups Quotes Ask the Author Discover new books on Goodreads See if your friends have read any of K.S. Narayan Reddy's books Sign in with Facebook Sign in options Join Goodreads K.S. Narayan Reddy’s Followers (5) K.S. Narayan Reddy edit data Combine Editions K.S. Narayan Reddy’s books K.S. Narayan Reddy Average rating: 4.14 · 79 ratings · 5 reviews · 5 distinct works The Essentials of Forensic Medicine and Toxicology 4.36 avg rating — 50 ratings — 3 editionsWant to Read saving… Want to Read Currently Reading Read Error rating book. Refresh and try again. Rate this book Clear rating 1 of 5 stars2 of 5 stars3 of 5 stars4 of 5 stars5 of 5 stars MCQs in Forensic Medicine and Toxicology 6ED 3.35 avg rating — 20 ratings Want to Read saving… Error rating book. Refresh and try again. Rate this book Clear rating 1 of 5 stars2 of 5 stars3 of 5 stars4 of 5 stars5 of 5 stars The Synopsis of Forensic Medicine and Toxicology 4.40 avg rating — 10 ratings — 4 editionsWant to Read saving… Error rating book. Refresh and try again. Rate this book Clear rating 1 of 5 stars2 of 5 stars3 of 5 stars4 of 5 stars5 of 5 stars Medico Legal Manual really liked it 4.00 avg rating — 1 rating Want to Read saving… Error rating book. Refresh and try again. Rate this book Clear rating 1 of 5 stars2 of 5 stars3 of 5 stars4 of 5 stars5 of 5 stars The Essentials of Forensic Medicine and Toxicology (English) 33rd Edition(Paperback) by K.S. Narayan Reddy, O.P. Murty 0.00 avg rating — 0 ratings Want to Read saving… Error rating book. Refresh and try again. Rate this book Clear rating 1 of 5 stars2 of 5 stars3 of 5 stars4 of 5 stars5 of 5 stars More books by K.S. Narayan Reddy… Is this you? Let us know. If not, help out and invite K.S. to Goodreads. Company About us Careers Terms Privacy Interest Based Ads Ad Preferences Help Work with us Authors Advertise Authors & ads blog Connect © 2025 Goodreads, Inc. Mobile version Welcome back. Just a moment while we sign you in to your Goodreads account.
10813	https://www.cuemath.com/euclidean-distance-formula/	Euclidean Distance Formula Before going to learn the Euclidean distance formula, let us see what is Euclidean distance. In coordinate geometry, Euclidean distance is the distance between two points. To find the two points on a plane, the length of a segment connecting the two points is measured. We derive the Euclidean distance formula using the Pythagoras theorem. Let us learn the Euclidean distance formula along with a few solved examples. What Is Euclidean Distance Formula? The Euclidean distance formula, as its name suggests, gives the distance between two points (or) the straight line distance. Let us assume that ((x_1,y_1)) and ((x_2,y_2)) are two points in a two-dimensional plane. Here is the Euclidean distance formula. Euclidean Distance Formula The Euclidean distance formula says: d = √[ (x(_2) – x(_1))2 + (y(_2) – y(_1))2] where, (x(_1), y(_1)) are the coordinates of one point. (x(_2), y(_2)) are the coordinates of the other point. d is the distance between (x(_1), y(_1)) and (x(_2), y(_2)). Euclidean Distance Formula Derivation To derive the Euclidean distance formula, let us consider two points A (x(_1), y(_1)) and B (x(_2), y(_2)) and let us assume that d is the distance between them. Join A and B by a line segment. To derive the formula, we construct a right-angled triangle whose hypotenuse is AB. For this, we draw horizontal and vertical lines from A and B which meet at C as shown below. Now we will apply the Pythagoras theorem to the triangle ABC. Then we get, AB2 = AC2 + BC2 d2 = (x(_2) – x(_1))2 + (y(_2) – y(_1))2 Taking the square root on both sides, d = √[ (x(_2) – x(_1))2 + (y(_2) – y(_1))2] Hence the Euclidean distance formula is derived. Want to find complex math solutions within seconds? Use our free online calculator to solve challenging questions. With Cuemath, find solutions in simple and easy steps. Book a Free Trial Class We will see more applications of Euclidean distance formula in the section below. Examples Using Euclidean Distance Formula Example 1:Find the distance between points P(3, 2) and Q(4, 1). Solution: Given: P(3, 2) = ((x_1,y_1)) Q(4, 1) = ((x_2,y_2)) Using Euclidean distance formula, d = √[(x(_2) – x(_1))2 + (y(_2) – y(_1))2] PQ = √[(4 – 3)2 + (1 – 2)2] PQ = √[(1)2 + (-1)2] PQ = √2 units. Answer: The Euclidean distance between points A(3, 2) and B(4, 1) is √2 units. Example 2:Prove that points A(0, 4), B(6, 2), and C(9, 1) are collinear. Solution: To prove the given three points to be collinear, it is sufficient to prove that the sum of the distances between two pairs of points is equal to the distance between the third pair. We will find the distance between every pair of points using the Euclidean distance formula. AB = √[(6 – 0)2 + (2 – 4)2] = √[36 + 4] = √40 = 2√10 BC = √[(9 – 6)2 + (1 – 2)2] = √[9 + 1] = √10 CA = √[(0 – 9)2 + (4 – 1)2] = √[81 + 9] = √90 = 3√10 Here, we can see that AB + BC = CA (This is because 2√10 + √10 = 3√10). Answer:We proved that A, B, and C are collinear. Example 3:Check that points A(√3, 1), B(0, 0), and C(2, 0) are the vertices of an equilateral triangle. Solution: Three vertices A, B, and C are vertices of an equilateral triangle if and only if AB = BC = CA. Given: A(√3, 1) = ((x_1,y_1)) B(0, 0) = ((x_2,y_2)) C(2, 0) = ((x_3,y_3)) Using Euclidean distance formula, AB = √[(x(_2) – x(_1))2 + (y(_2) – y(_1))2] = √[(0 – √3)2 + (0-1)2] = √(3 + 1) = √4 = 2 BC = √[(x(_3) – x(_2))2 + (y(_3) – y(_2))2] = √[(2-0)2 + (0-0)2] = √(4 + 0) = √4 = 2 CA = √[(x(_3) – x(_1))2 + (y(_3) – y(_1))2] = √[(2 - √3)2+ (0 – 1 )2] = √(4 + 3 - 4√3 + 1) = √ (8 - 4√3) = √ (8 - 2√12) = √ (√6 - √2)2 = √6 - √2 Here AB = BC ≠ CA. Answer: A, B, and C are NOT the vertices of an equilateral triangle. FAQs on Euclidean Distance Formula What Is Euclidean Distance Formula? The Euclidean distance formula is used to find the distance between two points on a plane. This formula says the distance between two points (x(_1), y(_1)) and (x(_2), y(_2)) is d = √[(x2 – x1)2 + (y2 – y1)2]. How To Derive Euclidean Distance Formula? To derive the Euclidean distance formula, consider two points A(x(_1), y(_1)) and B(x(_2), y(_2)) and join them by a line segment. Then draw horizontal and vertical lines from A and B to meet at C. Then ABC is a right-angled triangle and hence we can apply the Pythagoras theorem to it. Then we get AB2 = AC2 + BC2 d2 = (x(_2) – x(_1))2 + (y(_2) – y(_1))2 Taking the square root on both sides, d = √[ (x(_2) – x(_1))2 + (y(_2) – y(_1))2] For detailed derivation, click here. What Are the Applications of Euclidean Distance Formula? The Euclidean distance formula is used to find the length of a line segment given two points on a plane. Finding distance helps in proving the given vertices form a square, rectangle, etc (or) proving given vertices form an equilateral triangle, right-angled triangle, etc. What Is the Difference Between Euclidean Distance Formula and Manhattan Distance Formula? For any two points (x(_1), y(_1)) and (x(_2), y(_2)) on a plane, The Euclidean distance formula says, the distance between the above points is d = √[ (x(_2) – x(_1))2 + (y(_2) – y(_1))2]. Manhattan distance formula says, the distance between the above points is d = \|x(_2) - x(_1)\| + \|y(_2) - y(_1)\|. 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10814	https://www.brickworkssupply.com/inspiration-resources/articles/tools-of-the-trade-bricklaying-like-a-pro	Bricklaying Tools: A Beginners Guide \| Brickworks Supply Skip to main content Choose my preferred store Menu Search Locations Contact Brick Match Events Search × Search Brick View All Brick Brick By Color Thin Brick Klaycoat Handmade Glazed Structural Glass Stone Natural Stone Manufactured Stone Hardscapes Pavers Outdoor Living Concrete Walls And Accessories Lighting Natural Stone Products Artificial Turf Advanced Cladding Systems SK1N Metal Cladding Thin Tech Tru-Brix CUPACLAD Façade Systems DuPont™ ArmorWall™ System™ Tools & Supplies Masonry Accessories Masonry Tools Supplies Inspiration & Resources Project Galleries Hardscape Inspiration Articles How-To And Technical Videos Catalogs Literature And Tech Notes Brick Calculator Locations Contact Brick Match Events Brick View All Brick Brick By Color Thin Brick Klaycoat Handmade Glazed Structural Glass Stone Natural Stone Manufactured Stone Hardscapes Pavers Outdoor Living Concrete Walls And Accessories Lighting Natural Stone Products Artificial Turf Advanced Cladding Systems SK1N Metal Cladding Thin Tech Tru-Brix CUPACLAD Façade Systems DuPont™ ArmorWall™ System™ Tools & Supplies Masonry Accessories Masonry Tools Supplies Inspiration & Resources Project Galleries Hardscape Inspiration Articles How-To And Technical Videos Catalogs Literature And Tech Notes Brick Calculator Breadcrumb Home Inspiration & Resources Articles Tools of The Trade: Bricklaying Like a Pro Tools of the Trade: Bricklaying Like a Pro June 5, 2024 Welcome to the world of bricklaying! Before you dive into the intricate techniques, let's lay down the foundation for success with the essential tools of the trade. Think of it as building a house – you wouldn't start with the roof, right? Let's explore these tools that will elevate your bricklaying game. The Margin Trowel: Precision in Your Hands The margin trowel is your trusted companion for spreading and shaping mortar with precision. Featuring a rectangular-shaped blade, typically measuring around 5 to 6 inches in length and 2 inches in width, it ensures smooth and even mortar application. Wielded with finesse, it's akin to a conductor leading an orchestra. Its versatility makes it indispensable for various tasks, whether you're applying mortar to small areas, spreading adhesive, or cleaning excess mortar from bricks or stones. Another common brick trowel is the Philadelphia brick trowel, distinguished by its tapered, pointed blade, which offers several advantages in bricklaying tasks. This design allows for precise control when spreading mortar and laying bricks, enabling masons to achieve accurate placements and consistent joint thickness. The pointed tip of the blade is particularly useful for navigating tight spaces and corners, ensuring thorough mortar application even in confined areas. Additionally, the versatility of the Philadelphia trowel extends to tasks beyond mortar application, such as shaping bricks and finishing joints. Its pointed blade enables masons to manipulate mortar and bricks with finesse, facilitating intricate detailing and adjustments as needed. The London brick trowel features a rectangular-shaped blade with a flat edge, offering distinct benefits for bricklaying projects. The larger surface area of the blade allows for more mortar to be held and spread with each scoop, increasing efficiency in mortar application. This feature is particularly advantageous when working on larger bricklaying projects or laying bricks over expansive areas, as it minimizes the need for frequent reloading of mortar. Furthermore, the flat edge of the London trowel is well-suited for achieving straighter joints and smoother finishes on brickwork. By providing a consistent surface for mortar application, the flat edge helps masons maintain uniformity in joint thickness and brick alignment, resulting in aesthetically pleasing and structurally sound brick structures. In summary, while both the Philadelphia and London brick trowels serve essential roles in bricklaying, each offers unique features tailored to specific tasks and preferences. The precision and versatility of the Philadelphia trowel make it ideal for intricate detailing and precise mortar application, while the efficiency and consistency provided by the London trowel's larger blade surface and flat edge are well-suited for spreading mortar over larger areas and achieving straighter joints. Ultimately, masons may choose between these trowel types based on the demands of their projects and their individual comfort and familiarity with each tool. The Brick Hammer: Shaping Bricks with Ease Also known as a mason's hammer, the brick hammer is your go-to tool for cutting and shaping bricks or stones. With a flat face for setting bricks and a chisel end for shaping, it allows you to sculpt bricks to fit precisely. Each strike chips away excess material or splits them along a scored line. Its sturdy construction ensures durability in demanding construction environments, making it indispensable for precision work. The Level: Ensuring Straightness and Alignment No bricklayer's toolkit is complete without a trusty level by their side. Whether you're laying bricks or blocks, the level ensures each element is perfectly straight and level. It serves multiple functions: ensuring horizontal alignment crucial for maintaining structural integrity and aesthetic appeal, and vertical plumbness, ensuring walls or structures are straight and true without leaning or tilting. By placing the level across multiple bricks, you can detect any high or low spots, ensuring the surface of the brickwork is even and uniform. The Jointer: Finishing Touches for Clean Joints Don't underestimate the humble jointer. It's essential for creating clean and uniform joints between bricks or stones, adding the finishing touches to your brickwork. After laying bricks or stones, mortar is typically applied between them to fill gaps and create a solid bond. The jointer trowel is used to smooth and finish these mortar joints, ensuring they're flush with the surface of the bricks and have a consistent appearance. Jointer trowels come in various shapes and sizes, allowing masons to create different joint profiles, whether it's concave, convex, V-shaped, or flush joints, each serving specific purposes in terms of water resistance, strength, or visual appeal. The Mortar Mixer: Streamlining Your Mixing Process Mixing mortars by hand is a thing of the past. Embrace the mortar mixer for efficiency and consistency, a true game-changer in brick masonry and construction projects. The mortar mixer ensures thorough mixing of mortar ingredients, such as cement, sand, and water, resulting in a consistent and uniform mixture. This consistency is essential for achieving uniformity in the mortar's strength, color, and workability. By automating the mixing process, mortar mixers reduce the need for manual labor and physical exertion, saving time and effort while providing greater control over mixing parameters. Whether it's adjusting mixing speed or duration, mortar mixers offer flexibility to customize mortar mixtures according to specific project requirements. Clean Up: Environmental Waste Disposal A useful tool to use at the end of the day while cleaning gear is the Slurry Tub. The Slurry Tub is a heavy-duty recyclable plastic tub, lined with a biodegradable paper filter. The system captures and filters the wet trade waste leaving visibly clear water to drain within the designated washout areas or be recycled. When dry enough, simply dispose of the hardened waste, along with the biodegradable filter into the work site skip or other approved disposal method. The Line and Pins: Guiding Lights for Straight Courses Line pins, twigs, and blocks are indispensable for ensuring straight and level courses of bricks or blocks, providing a reliable reference point for accuracy and precision. Line pins, typically made of durable steel or aluminum, anchor the mason's line at each end of the wall, ensuring it remains taut and straight. Line twigs, small T-shaped tools, hold the mason's line in place between courses, maintaining alignment as bricks or blocks are laid. Line blocks, larger and typically made of wood, plastic, or metal, provide additional stability and support for the mason's line, particularly on larger or more complex projects. Together, these tools work in tandem with the mason's line to ensure straight and level construction. The Story Pole: Consistency in Height and Layout A story pole is a long, straight piece of wood or metal marked with measurements and reference points, ensuring consistent course heights and layout in bricklaying projects. Before laying bricks or blocks, masons use a story pole to plan the wall layout and determine the placement of key features such as windows, doors, and architectural details. By marking these elements on the story pole, masons can visualize the overall design and ensure proper alignment and spacing. Throughout construction, the story pole aids in checking for levelness and plumbness, providing a consistent reference point for laying courses of bricks or blocks. Its use streamlines the bricklaying process, minimizing errors and waste while ensuring accuracy and precision in construction. As you embark on your journey as a bricklayer, remember that these tools are more than mere instruments – they are extensions of yourself, helping you shape the world one brick at a time. Embrace them, master them, and let them guide you on your path to greatness. Find Your Store Product availability varies by location. Enter a 5-digit zip code or select your current location to find your nearest store for most accurate results. Use my current location De-select your store location You're currently browsing: IA \| Des Moines 4601 NW Urbandale Drive, Suite 112 Urbandale, Iowa 50322 mi Phone: 515-219-7842 Hours: - Set As My Store View Store Details You're currently browsing: IL \| Bridgeview 7542 West 73rd Street Bridgeview, Illinois 60455 mi Phone: 708-607-9430 Hours: - Set As My Store View Store Details You're currently browsing: IL \| Champaign 3200 W. 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10815	https://www.ck12.org/flexi/algebra-ii/inverse-properties-of-logarithms/can-you-tell-me-some-of-the-logarithm-rules/	Flexi answers - Can you tell me some of the logarithm rules? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Algebra II Inverse Properties of Logarithms Question Can you tell me some of the logarithm rules? Flexi Says: Sure, here are some basic rules of logarithms: 1. Product Rule: log b⁡(M N)=log b⁡(M)+log b⁡(N) 2. Quotient Rule: log b⁡(M N)=log b⁡(M)−log b⁡(N) 3. Power Rule: log b⁡(M p)=p⋅log b⁡(M) 4. Change of Base Rule: log b⁡(M)=log a⁡(M)log a⁡(b) 5. Identity Rule: log b⁡(b)=1 and log b⁡(1)=0 Analogy / Example Try Asking: Let a, b, and c be positive real numbers such that b ≠ 1. The logarithm of a with base b is denoted by log_b(a), and is defined as c = log_b(a) if and only if a = b^c. Explain.Enter the correct answer in the box. Rewrite this logarithmic equation as an exponential equation. Unexpected text node: ''Rewrite in logarithmic form 125 = 5^3. How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
10816	https://www.mdpi.com/1996-1073/16/5/2281	Next Article in Journal Multi-Array Design for Hydrokinetic Turbines in Hydropower Canals Previous Article in Journal Fault Location for Distribution Smart Grids: Literature Overview, Challenges, Solutions, and Future Trends Active Journals Find a Journal Journal Proposal Proceedings Series ## Topics For Authors For Reviewers For Editors For Librarians For Publishers For Societies For Conference Organizers Open Access Policy Institutional Open Access Program Special Issues Guidelines Editorial Process Research and Publication Ethics Article Processing Charges Awards Testimonials ## Author Services Sciforum MDPI Books Preprints.org Scilit SciProfiles Encyclopedia JAMS Proceedings Series Overview Contact Careers News Press Blog Sign In / Sign Up Notice clear Notice You are accessing a machine-readable page. In order to be human-readable, please install an RSS reader. 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Get Information clear JSmol Viewer clear first_page Download PDF settings Order Article Reprints Font Type: Arial Georgia Verdana Font Size: Aa Aa Aa Line Spacing:    Column Width:    Background: Open AccessArticle Generalised Isentropic Relations in Thermodynamics by Pim Nederstigt Pim Nederstigt SciProfiles Scilit Preprints.org Google Scholar ,† and Rene Pecnik Rene Pecnik SciProfiles Scilit Preprints.org Google Scholar Department of Process & Energy, Delft University of Technology, 2628 CD Delft, The Netherlands Author to whom correspondence should be addressed. † A former researcher in Department of Process & Energy TU Delft. Energies 2023, 16(5), 2281; Submission received: 21 November 2022 / Revised: 9 February 2023 / Accepted: 20 February 2023 / Published: 27 February 2023 (This article belongs to the Section J2: Thermodynamics) Download keyboard_arrow_down Download PDF Download PDF with Cover Download XML Download Epub Browse Figures γ and the real gas exponents , , and in the -plane evaluated for HO (a), CO (b), and pentane (c), using the Span–Wagner equation of state [23,24] with NIST RefProp . The white solid lines indicate the isentropes. The black dashed line in (c) indicates . " href=" γ P v , the white lines indicate the isobars, and the black dashed lines indicate . The dash-dotted line indicates where the fundamental derivative of gas dynamics is ; (b) Expansions 1 and 2 plotted as a function of Mach number. Expansion 1 shows the distributions of the isentropic exponents and the flow work, while expansion 2 shows the temperature and nozzle area distributions obtained with RefProp compared to the results obtained with the generalised isentropic relations. " href=" Versions Notes Abstract Isentropic processes in thermodynamics are fundamental to our understanding of numerous physical phenomena across different scientific and engineering fields. They provide a theoretical reference case for the evaluation of real thermodynamic processes and observations. Yet, as analytical relations for isentropic transformations in gas dynamics are limited to ideal gases, the inability to analytically describe isentropic processes for non-ideal gases is a fundamental shortcoming. This work presents generalised isentropic relations in thermodynamics based on the work by Kouremenos et al., where three isentropic exponents , and are introduced to replace the ideal gas isentropic exponent to incorporate the departure from the non-ideal gas behaviour. The general applicability of the generalised isentropic relations is presented by exploring its connections to existing isentropic models for ideal gases and incompressible liquids. Generalised formulations for the speed of sound, the Bernoulli equation, compressible isentropic flow transformations, and isentropic work are presented thereafter, connecting previously disjoint theories for gases and liquids. Lastly, the generalised expressions are demonstrated for practical engineering examples, and their accuracy is discussed. Keywords: isentropic relations; real gas thermodynamics; speed of sound; compressible fluid flows; compressibility; isentropic work 1. Introduction Isentropic processes describe idealized processes without irreversibilities such as friction or heat losses and are therefore used as a theoretical reference case for the evaluation of real thermodynamic processes and observations. For this reason, isentropic relations are encountered in many fields of science and engineering applications. In fluid dynamics, where flows are approximated as isentropic flows outside the viscous boundary layer, the isentropic relations are an intrinsic part of modelling fluid compressibility and are, therefore, an underlying assumption in aerodynamics. In energy engineering, the isentropic relations find their way into evaluating the isentropic work of gas compression and expansion systems and are also fundamental to turbo-machinery design. Driven by increasing energy and fuel efficiency, state-of-art energy conversion systems, such as the supercritical CO cycle [1,2,3], Organic Rankine Cycles , and high-pressure industrial heat pumps , seek to exploit the non-ideal behaviour of unconventional working fluids where the ideal gas approximation is no longer valid. As the ideal gas equation is used to relate isentropic transformations, their application is limited to ideal gases only. The conventional classification between “ideal” and “non-ideal” is distinctive for the apparent lacking means to conveniently describe isentropic processes in the general sense. Where isentropic processes for ideal gases are conveniently modelled by the isentropic exponent —the often assumed constant ratio of the specific isobaric and isochoric heat capacities and —our capabilities are limited in the general case, where engineers and scientists are forced to resort to thermodynamic libraries, look-up tables, and equations of state of (semi)-empirical nature. Although this may not be a problem with today’s computing power and access to powerful (open-source) libraries, such as CoolProp and RefProp , the lacking ability to describe isentropic processes in a general way is a gap in our understanding of thermodynamics, as isentropic transformations under the ideal gas model cannot simply be projected on non-ideal gas applications as their gas-dynamic behaviour is vastly different. In a series of papers published in the 1980s, on which this work is based, Kouremenos et al. proposed a set of generalised isentropic relations to model isentropic processes in a general way [8,9,10]. They introduced three isentropic exponents to replace the adiabatic coefficient , based on the mathematical argument that the form of a generalised model should adhere to its ideal gas counterpart. Compressibility effects and departure from the ideal gas behaviour are then included in three alternative isentropic exponents. Although their derivation of the general isentropic model appeared successful, they did not pursue to extend their analysis to existing isentropic transformations, but instead performed empirical evaluations of the isentropic exponents using an equations of state [9,11]. A similar model was later proposed by Baltadijev, who made efforts to derive additional isentropic flow relations . Nederstigt further extended this approach to develop generalised isentropic relations for several process quantities. Recently, the non-ideal gas isentropic exponents also found their way into the field of computational fluid dynamics . This work seeks to introduce the generalised isentropic relations proposed by Kouremenos et al. to a broader audience and to complete the analytical framework by addressing previously unexplored connections with existing isentropic models for ideal gases and liquids—connecting previously disjoint theories for speed of sound, isentropic flows, and isentropic work between gases and liquids. Finally, the applicability of the generalised isentropic relations is demonstrated for practical engineering examples, and their accuracy is discussed. 2. Isentropic Exponents for the Real Gas Thermodynamic Region 2.1. Generalised Isentropic Relations First, the generalised isentropic relations proposed by Kouremenos et al. [8,9,10] are presented. The model is based on the isentropic relations of ideal gases where the ratio of the specific heats is replaced by exponents , and . The subscripts refer to the pressure–volume, temperature–volume, and pressure–temperature isentrope governed by each of the exponents, respectively, summarised aswhere P is the pressure, v—the specific volume, and T—the temperature. Consequently, the pressure ratio, temperature ratio, and density ratios in any isentropic transformation can be related bywhere subscripts 1 and 2 refer to the respective thermodynamic states along an isentrope. The generalised isentropic exponents are then to be expressed in terms of other thermodynamic state variables, which will be demonstrated to be a function of the specific heat capacities and partial derivatives in pressure, volume, and temperature. The derivation of exponent is shown here as an example . Let the entropy s be defined as a function of pressure and temperature, such that . Consequently, the change in entropy can be expressed as the exact differentialwhere for an isentropic process. Rearranging the derivatives yields The left-hand side of Equation (4) can be evaluated by differentiation of the assumed isentropic pressure–temperature relation, Equation (1c), yielding The right-hand side of Equation (4) can be transformed using Maxwell’s relationswhere . Finally, equating Equations (5) and (6), an expression for is obtained in terms of pressure P, the isobaric specific heat capacity , and the partial derivative , given as The exponents and for the temperature-volume and pressure-volume isentrope can be derived in a similar fashion from Equations (1a) and (1b) . Together, the three generalised isentropic relations can be summarized aswhere is the thermal expansion coefficient and —the isothermal compressibility factor . As, according to Gibbs’ phase rule, the thermodynamic state of a (pseudo)-pure single-phase substance is determined by two state variables, only two of the generalised isentropic exponents are independent. Following from Equations (8a)–(8c), the thermal expansion coefficient and isothermal compressibility factor can be expressed in terms of the exponents and as Moreover, a reciprocity can be observed between the exponents , , and through their partial derivatives, related by the triple product ruleThe isentropic exponents are thus related by 2.2. Limits of the Generalised Isentropic Exponents for Ideal Gases and Incompressible Liquids Thus far, no assumptions have been made on the thermodynamic state in the derivation of the generalised isentropic relations, nor has any equation of state been introduced to relate pressure, temperature, density, and fluid compressibility. In general, the term real gas itself is ambiguous to any thermodynamic state and broadly covers the entire range from dilute gases to dense gases and compressible liquids. Regardless of the equation of state used, the generalised isentropic relations must agree with existing isentropic models for liquids and ideal gases. In the case of liquids, that means that the isentropic relations Equations (8a)–(8c) should adopt the form of the incompressible fluid model, accounting for negligible changes in density with changing pressure under constant entropy. Note, for a van der Waals fluid, the incompressible limit is reached as the specific volume , with being the volume at the thermodynamic liquid-vapour critical point . On the other hand, for ideal gases, the generalised isentropic relations should reduce to the familiar isentropic expressions for ideal gases. The connections to these models will now be discussed. Ideal gas region: In the case of ideal gases, it can be shown that the generalised isentropic exponents reduce to the adiabatic coefficient , defined as the ratio of the specific heats /. Evaluating the partial derivatives in Equations (8a)–(8c), using the ideal gas model gives ,where the universal gas constant R is related to the specific heat capacities by in the ideal gas case. Consequently, the real gas exponents , , and are shown to be identical to their ideal gas counterparts This is ultimately a mathematical requirement by adopting the ideal gas solution as the starting point of the derivation for the non-ideal isentropic exponents. Liquid phase region: At the limit of an incompressible fluid model, the changes in fluid density with pressure are negligible. Consequently, , , and the exponent approaches infinity, see Equation (8a). Likewise, from Equation (8b), we find that as , and the exponent becomes infinite as well. Note, the thermal expansion coefficient . Subsequently, eliminating using the relation between the specific heat capacities and in Equation (8c), gives The right-hand side of Equation (8c) can be shown to go to zero. Consequently, we find that for incompressible substances. The incompressible limits can therefore be summarized asfor which the generalised isentropic exponents become Liquid–vapour coexistence region: As the specific heat capacities and are undefined in the two-phase region, neither are the real isentropic exponents , , and . It follows from here that the non-ideal isentropic exponents are only generally defined for single-phase substances. Real gas region: The conditions for thermal and mechanical stability of a single-phase substance require that the isochoric specific heat capacity and the isothermal compressibility [15,16]. The latter condition implies that the partial derivative . By reciprocity between the partial derivatives with respect to pressure, temperature, and density in Equation (10), the conditions for mechanical stability can be expressed as [15,16] As neither the specific heat capacities, nor the pressure, temperature, and specific volume can be negative, combining the inequalities in Equation (17) with Equations (8a)–(8c), it can be shown thatfor single-phase substances. A notable consequence of Equation (18), compared to the isentropic exponent for ideal gases, is that values of are permissible under the conditions for thermal and mechanical stability for single-phase substances. For instance, pentane shows a region where , as shown in Figure 1c. This gives new characteristics to isentropic transformations derived from . The value of is, in fact, directly related to the fundamental derivative of gas dynamics , which is a non-dimensional quantity that governs the dynamic behaviour of gases. The fundamental derivative of gas dynamics is defined as the derivative of the speed of sound with respect to volume at constant entropy, or alternatively, the second derivative—or curvature—of the pressure–volume isentrope, expressed as [17,18] The isentropic relations Equations (8a)–(8c) are hyperbolic functions that describe the isentropes in the pressure–volume, temperature–volume, and pressure–temperature plane. Their shape—and hence curvature—along any point of the isentrope is governed by the local value of the isentropic exponents, which are continuously varying functions along the isentrope. In the case of the fundamental derivative, substitution of Equation (8a) yields [13,19]where the derivative is small compared to the first term in Equation (20) and may be omitted, Equation (20) is approximated by , equivalent to the value of for ideal gases, [17,18]. As non-classical behaviour is observed in dense gasses for , and cannot be negative, non-classical behaviour gas behaviour can only occur where the derivative term in Equation (20) is larger than . The theoretical limits of the real isentropic relations are summarized in Table 1. 2.3. Isentropic Exponents Plotted in the Pv-Plane for Water, Carbon Dioxide, and Pentane Figure 1 shows the specific heat ratio and the three generalised isentropic exponents for water, carbon dioxide, and pentane in the -plane obtained with RefProp version 10.0 . The corresponding equations of state are given in refs. [20,21]. The three substances were chosen due to their many practical applications in the field of engineering and due to their difference in molecular size, complexity, and polarity. First of all, the value of the ideal gas exponent is of a similar order of magnitude between the gaseous and liquid regions in Figure 1 for all three substances. Only around the critical point, a rise in the value of is seen due to the increase of the isobaric heat capacity around that point. Comparing the generalised isentropic exponents, demonstrates the highest degree of variation, steeply increasing in value beyond the critical point for increasingly higher densities in the liquid region, ultimately approaching infinity. Though also tends to infinity for high densities, it quickly drops off, showing a milder progression in value change with decreasing density into the gaseous region. Exponent shows a smaller variation throughout the -plane, ranging from unity for high densities in the liquid region to the ratio of the specific heats for ideal gases. Between the three fluids, it is observed that HO and CO—with a lower molecular complexity—show a stronger variation of all three the isentropic exponents with respect to pentane, for which variations are fewer. The most notable difference between the three substances is the presence of a region where of pentane round the liquid–vapour coexistence region. Pentane has a dry liquid–vapour dome, where the concave shape of the vapour line results in dry expansion. The ratio of the specific heats has been found to control the skewness of the liquid–vapour dome in other research . Whilst evaluating the contours of the isentropic exponents for various substances with RefProp, fluids with similar dry liquid–vapour domes were also found to possess a region where . Figure 1. Contours of specific heat ratio and the real gas exponents , , and in the -plane evaluated for HO (a), CO (b), and pentane (c), using the Span–Wagner equation of state [23,24] with NIST RefProp . The white solid lines indicate the isentropes. The black dashed line in (c) indicates . Figure 1. Contours of specific heat ratio and the real gas exponents , , and in the -plane evaluated for HO (a), CO (b), and pentane (c), using the Span–Wagner equation of state [23,24] with NIST RefProp . The white solid lines indicate the isentropes. The black dashed line in (c) indicates . 3. Generalised Speed of Sound, Isentropic Flows Transformations, and Isentropic Work The generalised isentropic exponents are now used to derive generalised isentropic relations for practical engineering applications, such as the speed of sounds and isentropic flows through nozzles, compressors, and turbine stages. 3.1. Generalised Speed of Sound in Fluids The speed of sound is defined as where is the density of the substance. Using the isentropic pressure–volume relation for real gases, Equation (8a), an expression for the speed of sound can be derived in terms of the exponent . The speed of sound is then given as [26,27] Here, the benefit of adhering to the ideal gas notation for the isentropic relations becomes apparent as the departure from ideal gas conditions in Equation (22) is conveniently included in the exponent without losing the familiarity of the ideal gas notation. In the case of liquids, the speed of sound is related to the bulk modulus —a measure for the elasticity of the fluid—known as the Newton–Laplace equation [28,29] The bulk modulus of ideal gases is related to the ratio of the specific heats and the pressure through [25,28]. In its generalised form, the bulk modulus is related to the isentropic exponent through Between gasses and liquids, the isentropic bulk modulus varies several orders of magnitude, typically ranging from kilopascals to megapascals for gases, to the order of gigapascals in the case of liquids. We can now relate this to the value of , which between gases and liquids of equal pressure is demonstrated in Figure 1 and Table 1 to change several orders of magnitude between gases and liquids as the derivative becomes large due to small fluid compressibility. 3.2. Isentropic Transformations for Non-Ideal Compressible Isentropic Flows The generalised isentropic exponents are now used to derive transformations for non-ideal isentropic fluid flows from the fundamental conservation laws of mass, momentum, and energy. Though their derivation is similar to the ideal gas case, a different approach must be followed as, unlike ideal gases, the total enthalpy can no longer be related to the temperature using constant specific heats only. We start with the energy equation of a steady one-dimensional adiabatic flow where u the velocity, g—the gravitational constant, and —the change in elevation of the flow volume. In the case of ideal gases, the change in enthalpy may be directly related to the change in temperature using . In the general case, the change in enthalpy can be expressed in terms of pressure and the specific volume as , or alternatively, as , where is the fluid density. Substituting the latter in Equation (25), we obtain Although it is possible to use the differential form of the Bernoulli equation to study isentropic transformations, it is more convenient to integrate Equation (26) for direct calculations. However, certain assumptions must be made to integrate in Equation (26). Assuming that is constant and that , the integration can be performed using the pressure–volume isentropic relation, Equation (8a), to give Note that for certain cases, may be less than 1, e.g., as shown in Figure 1c. Therefore, it is possible for isentropic transformations to obtain , such that the right-hand side of (27) becomes singular. It is possible to integrate Equation (26) but this case will not be considered separately herein. The generalised form of the Bernoulli equation for non-ideal fluid flows is thenwhich is applicable for the entire single-phase region for stable substances where . The ideal gas form of the Bernoulli equation is easily recognized, for which reduces to the adiabatic coefficient , as demonstrated in Equation (12). In the case of liquids, becomes large due to fluid incompressibility, and consequently , such that Equation (28) reduces to the classic form of the Bernoulli equation for incompressible fluid flows. With the generalised Bernoulli equation for gases and liquids, the local flow properties can be related to the upstream fluid at rest as a reference state of the flow, with and denoting the stagnation pressure and density. Ignoring body forces, the downstream flow can be related to the stagnation conditions bywhere a constant value of is assumed throughout the isentropic flow field. The velocity can be expressed in terms of local Mach number as with as the local speed of sound Equation (22). Gathering the pressure terms on the left-hand side, and eliminating the density ratio , using the pressure–density isentropic relation in Equation (8a), gives the familiar expression for the total pressure in an isentropic flow where departure from ideal behaviour is incorporated in the isentropic exponent , Equivalently, the stagnation temperature and stagnation density can be derived from the stagnation pressure using the isentropic relations Equations (8c) and (8b), yielding Subsequently, the stagnation speed of sound and the stagnation compressibility factor are derived usingobtaining The generalised form of Equations (30)–(33) should once again be highlighted. The familiar ideal gas equivalents are easily obtained as the generalised isentropic relations reduce to the adiabatic coefficient for ideal gases. In the case of liquids, where the generalised isentropic exponents take on the limits listed in Table 1, note that the speed of sound c becomes large as following Equation (22) and hence, the Mach number . Therefore, in the case of incompressible liquids, the term enclosed by the brackets in Equations (30), (31) and (33) become one and the stagnation properties for an incompressible flow are then demonstrated to reduce to the trivial statements Critical property ratios for choked flow conditions can be derived in addition to the stagnation properties using the choked flow condition, indicated by superscript ∗, as a reference instead of the upstream stagnant flow condition. These are obtained by setting Ma to one in Equations (30), (31) and (33). Finally, the critical flow area and critical mass flow rate can be obtained using the critical property ratios in Equation (35). The critical flow area ratio is expressed as Using a combination of the critical property ratios Equation (35) and isentropic transformations Equation (2), the critical flow area ratio can be expressed as The critical mass flow rate defined ascan be expressed aswith a classic textbook on gas dynamics at hand, numerous variations on isentropic flow relations can be derived using the generalised isentropic relations presented in this work [13,25]. These equations may be used to model gases and liquids alike regardless of the thermodynamic state under the assumption of constant values of , , and throughout the isentropic flow field. Providing such additional equations is not the aim of this work and is therefore not further considered herein. 3.3. Isentropic Work in Compression and Expansion We conclude this section with the theoretical isentropic work during compression and expression. The theoretical isentropic work for isentropic expansion or compression between states 1 and 2 is defined as the integralevaluated in Equation (27) as part of the total energy of a compressible isentropic flow. Using the previous result, the work for isentropic compression and expansion can be expressed as The sign of the change in enthalpy in Equation (41) is positive for compression and negative for expansion. Where the similarity between the generalised relation for the isentropic work and the ideal gas notation is obtained by setting equal to the ratio of the specific heats , in the case of liquids, where becomes large due to fluid incompressibility, the ratio and the exponent . From here, it can be demonstrated that Equation (41) reduces to the isentropic work relation for fluid compression or expansion The generalised isentropic transformations derived in this section are summarized in Table A1. 4. Considerations for Non-Ideal Isentropic Relations in Practical Applications The applicability of the generalised relations will now be discussed using three practical examples of isentropic transformations. Propane is selected as a working fluid, as it exhibits large variations of and possesses a region where (see Figure 1) in the dense vapour region where non-ideal gas effects are at play. In particular, the following processes will be considered: Compression in the liquid region; Expansion 1 in the dense vapour region close to the vapour saturation line, starting from the critical pressure to a low pressure while crossing the line; Expansion 2 in the vapour region, again crossing the line, but with a lower pressure ratio. These processes shown in Figure 2 serve to illustrate the applicability of the isentropic relations. Compression. As summarized in Table 2, pentane is compressed from an initial temperature of C and a pressure of bar to a final pressure of bar (values in bold). RefProp is used to obtain the other thermodynamic states and the values of the isentropic exponents. Note, shows values close to 1, while shows large values in the compressed liquid region. Using the inlet values for and with the generalised isentropic relations to evaluate the temperature and density in state 2 shows the accuracy of this approach, even if changes from 15.35 at state 1 to 11.25 at state 2. The relative error of temperature and density in state 2 is 0.6% and 1.56%, respectively. Likewise, the enthalpy change across the isentropic compression can be evaluated using the generalised isentropic relations and, in addition, also with the incompressible model. Comparing those estimates with the evaluation of using RefProp indicates the low relative error of 0.7% with the generalised isentropic relations, while the incompressible estimate yields a relative error of 6.1%. Expansion 1. For this case, pentane is expanded (without internal irreversibilities) through an adiabatic supersonic nozzle from a total pressure of bar and a total temperature of K to a Mach number of . The solution of the expansion is obtained using a conventional root finding algorithm of the total enthalpy, , conservation equation, given aswhere is the constant entropy along the expansion. Expanding to results in an outlet pressure of bar. Figure 2b (Expansion 1) shows the distributions of the isentropic exponents as a function of Mach number, evaluated using RefProp. The following observations can be made. First, the isentropic exponents show large differences at the beginning of the expansion (non-ideal gas behaviour), while they show almost the same values at the end of the expansion, indicating ideal gas behaviour at the end of the expansion. Second, remains nearly constant throughout the expansion, indicating that the temperature–volume isentropic relation will be the most accurate—when assuming a constant value of —along this expansion in the dense vapour region. The final and most significant observation is related to , which crosses one at . In general, during an expansion, the enthalpy is converted into an increase in kinetic energy. Per definition, the enthalpy is the sum of internal energy e and flow work , as . The value of the flow work is displayed in the lower left plot in Figure 2b. While for an expansion of an ideal gas with , both the internal energy e and the flow work decrease. For an expansion in the non-ideal gas region, where , the flow work increases since , which is an increasing function for . For an expansion in a non-ideal thermodynamic region where , the increase of kinetic energy is thus reduced as the flow work increases due to the strong expansions of the gas. As such, serves as an insightful parameter to indicate the non-ideal behaviour in isentropic transformations. Expansion 2. The second example illustrates the accuracy of the isentropic relations for an expansion in which also crosses the line where , but with a smaller variation compared to Expansion 1. Here, pentane is expanded from a pressure of bar and a total temperature of K to a Mach number of . Again, at the inlet and at the outlet of the nozzle. Even if assuming a constant value of during the expansion, it can be seen that the results with the isentropic relations are in good agreement with the exact solution obtained with RefProp. This indicates that for engineering design it is possible to choose a reference value of (in this case, at the inlet of the expansion) and still obtain relatively accurate results. Alternatively, it is also possible to choose an average value of between the inlet and outlet or even split the expansion into steps to increase the accuracy. 5. Conclusions Generalised isentropic relations were presented in this work, based on the work by Kouremenos et al., where three isentropic exponents , , and are introduced to replace the ideal gas adiabatic coefficient to incorporate departure from non-ideal gas behaviour. The generalised isentropic exponents were derived from the exact differential in entropy and expressed in terms of other thermodynamic state variables using Maxwell relations, without the need for an assumption of an equation of state. Hence, the generalised isentropic relations, and any derived functions, are generally applicable throughout the thermodynamic domain of single-phase substances. When the ideal gas or incompressible liquid model is assumed, it was found that their existing isentropic models can be recovered. Between the two idealized limits, the generalised isentropic exponents can be used to relate isentropic states for real gases under the assumption of local constant values of , , and . The most notable deviation with respect to the isentropic exponent for ideal gases is that values of are permissible by the conditions for thermal and mechanical stability of single-phase substances, providing new additional insights to isentropic transformations derived from it. The generalised isentropic relations were then used to derive general formulations for the speed of sound, the Bernoulli equation, compressible isentropic flow transformations, and isentropic work, where for each case, the connection between the generalised model and the respective applications for ideal gases and liquids was highlighted. It was shown that the speed of sound for ideal gases can be obtained from the generalised form, as well as the Newton–Laplace equation for the speed of sound in liquids. Similarly, the Bernoulli equation can be derived from the generalised form presented in this paper, as well as the Bernoulli equation for ideal gases. Likewise, other generalised transformations for compressible isentropic flows were obtained, under the assumption of constant values of the isentropic exponents , , and throughout the isentropic flow field. Applicability, and the error resulting from this assumption, were then demonstrated in three examples concerning isentropic compression and isentropic expansion. Relatively small errors occurred during highly non-ideal transformations that could not otherwise have been approached with the ideal gas assumption. Even for the compression in the liquid region, the generalised model was found to be an improvement over the incompressible substance model. Author Contributions Conceptualization, P.N. and R.P.; methodology, P.N.; software, P.N. and R.P.; validation, P.N. and R.P.; formal analysis, P.N.; writing—original draft preparation, P.N.; writing—review and editing, P.N. and R.P.; visualization, P.N. and R.P. All authors have read and agreed to the published version of the manuscript. Funding RP would like to acknowledge the support from the European Research Council grant no. ERC-2019-CoG-864660. Data Availability Statement Not applicable. Conflicts of Interest The authors declare no conflict of interest. Appendix A Table A1. An overview of the isentropic relations for non-ideal compressible flows. The generalised isentropic exponents , , and are introduced to replace their ideal gas counterparts, based on the model proposed by Kouremenos et al. . The generalised form of the equations corresponds with opposing theoretical models for ideal gases and incompressible liquids. Table A1. An overview of the isentropic relations for non-ideal compressible flows. The generalised isentropic exponents , , and are introduced to replace their ideal gas counterparts, based on the model proposed by Kouremenos et al. . The generalised form of the equations corresponds with opposing theoretical models for ideal gases and incompressible liquids. \| Liquids \| Real Gases \| Ideal Gas \| --- \| Limits of the isentropic exponents \| \| \| \| \| \| \| \| \| \| \| \| \| \| Generalised speed of sound relation \| \| \| \| \| \| Generalised Bernoulli equation \| \| \| \| \| \| Generalised isentropic relations \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Isentropic compression and expansion work \| \| \| \| \| References Nan, X.; Himeno, T.; Watanabe, T. The Real Gas Effect on the Stagnation Properties for Supercritical Carbon Dioxide Flows. Int. J. Gas Turbine Propuls. Power Syst. 2020, 11, 1–8. [Google Scholar] [CrossRef] Li, X.; Zhao, Y.; Yao, H.; Zhao, M.; Liu, Z. A New Method for Impeller Inlet Design of Supercritical CO2 Centrifugal Compressors in Brayton Cycles. Energies 2020, 13, 5049. [Google Scholar] [CrossRef] Uysal, S.C.; Liese, E. 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Expansion 1 shows the distributions of the isentropic exponents and the flow work, while expansion 2 shows the temperature and nozzle area distributions obtained with RefProp compared to the results obtained with the generalised isentropic relations. Figure 2. Three exemplary isentropic transformations in different single-phase regions of pentane; (a) T-s diagram of pentane with indicated isentropic transformations. The coloured contour shows , the white lines indicate the isobars, and the black dashed lines indicate . The dash-dotted line indicates where the fundamental derivative of gas dynamics is ; (b) Expansions 1 and 2 plotted as a function of Mach number. Expansion 1 shows the distributions of the isentropic exponents and the flow work, while expansion 2 shows the temperature and nozzle area distributions obtained with RefProp compared to the results obtained with the generalised isentropic relations. Table 1. Limits of the exponents , , and for liquids, ideal gases, and real gases in the general case. The isentropic exponents are demonstrated to reduce to the incompressible fluid model for liquids and the ratio of the specific heats for ideal gases. Table 1. Limits of the exponents , , and for liquids, ideal gases, and real gases in the general case. The isentropic exponents are demonstrated to reduce to the incompressible fluid model for liquids and the ratio of the specific heats for ideal gases. \| Liquids \| Real Gases \| Ideal Gases \| --- \| Limits of derivatives and isentropic exponents \| \| \| \| \| \| \| \| Table 2. Exemplary compression in the compressed liquid region of pentane. Bold values refer to input reference state. Table 2. Exemplary compression in the compressed liquid region of pentane. Bold values refer to input reference state. \| (a) Thermodynamic states \| \| \| RefProp \| Isentropic transformations Equation (2) with and (rel. error) \| \| State \| 1 \| 2 \| 2 \| \| P [bar] \| 34 \| 128 \| \| \| T [K] \| 443.15 \| 453.27 \| 450.60 (0.6%) \| \| [kg/m] \| 433.76 \| 480.36 \| 472.89 (1.56%) \| \| \| 1.47 \| 1.28 \| \| \| \| 1.012 \| 1.023 \| \| \| \| 1.193 \| 1.250 \| \| \| (b) Enthalpy change \| \| \| RefProp \| (rel. error) \| Isentropic Work Equation (41) with (rel. error) \| \| [kJ/kg] \| 20.43 \| 21.67 (6.1%) \| 20.57 (0.69%) \| \| \| \| Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. \| © 2023 by the authors. Licensee MDPI, Basel, Switzerland. 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10817	https://www.naturalnumbers.org/sparticle.html	\| \| \| --- \| \| NATURAL NUMBERS \| \| \| exploring the undesigned intelligence of the numberverse \| \| \| \| \| \| home \| \| \| Direction and Location in the Number Spiral The compass points are a convenient way to understand orientation in the spiral. Direction in the spiral has particular significance because it determines the distribution of every number in the spiral. If you know a number, you know its location. The hemispheres of the spiral can be called north, south, east, and west, while the quadrants of the spiral are northeast, northwest, southwest, southeast. (The use of geographic terms is purely for ease of understanding and is not meant to infer any deeper significance.) None of the axes are straight lines, but are actually slightly curved. Their position on the number line is what determines their position in the spiral. East The spiral begins and completes each rotation on the designated "0°" axis. The numbers on this axis are the perfect squares, so the formula is simply: The sequence begins: 1, 4, 9, 16, 25, 36. North Due north represents a quarter turn of the spiral. The formula that describes the numbers on this approximately 90° axis is: The sequence begins: 5, 18, 39, 68, 105, 150. West Due west represents a half-turn of the spiral. The numbers on this approximately 180° axis are the pronic numbers (see below): n(n+1) The sequence begins: 2, 6, 12, 20, 30, 42. South Due south represents a quarter turn of the spiral. The formula that describes all numbers on this approximately 270° axis is: n(4n-1) The sequence begins: 3, 14, 33, 60, 95, 138. (The reason that the north and south progressions grow at twice the rate of the east and west progressions is that each north/south number is separated by two turns of the spiral.) What are pronic, or oblong, numbers? The so-called pronic numbers are of particular significance in the spiral because the curves that are derived from the pronic formula with a positive or negative constant produce statistically and predictably more prime numbers than other curves. The pronic number are simply those formed by multiplying an integer by itself-plus-one. Thus the formula is n(n+1), and the first few examples are: 1(1+1)=22(2+1)=63(3+1)=124(4+1)=205(5+1)=306(6+1)=42 In terms of the Sacks Number Spiral, the westbound equatorial line is comprised solely of pronic numbers as illustrated. Product Curves Product curves are polynomials conforming to one of the equations given above. For example, n^2 - n + 41 is a pronic curve. After circling the center of the spiral, westbound "P" curves flatten out and run parallel to the pronic axis, eastbound "S" curves flatten out and run parallel to the square axis. Both aspects of product curves' behavior have strikingly simple explanations that were identified by Sacks. Rotation and alignment Number of rotations This is determined by the offset rule, which states: "A single rule determines the number of rotations for both even and odd curves: the number of revolutions is equal to the average of the offset's squarest integral factors. For example, the curve with offset 12 makes 3.5 rotations, and 3.5 is the average of 3 and 4, the squarest factors of 12." (Robert Sacks) Horizontal alignment In the spiral, east is a zero turn, north is a quarter turn, west is a half turn, and south is a three-quarter turn. 0 represents the point of origin for a rotations and 1/2 represents the halfway point in the rotation. The square root of n^2 is n. So the fractional part of eastbound curves approaches 0. The square root of n(n+1) is n.5 So the fractional part of westbound curves approaches 1/2. You can test this fact in RadiusTest with any square (eastbound) or pronic (westbound) curve. This example illustrates that the fractional part of the square root of a pronic curve approaches one-half once the curve aligns with the pronic axis. The constant part of the polynomial (+ or - C) becomes less and less significant as the value of N increases. As the effect of the constant becomes more and more negligible, so does the curve more and more closely approximate the straight line defined by the square root or pronic number progression. \| \| \| \| \| Symmetry and Asymmetry in the Number Spiral: See the Numbers: By Density and By Ratios \| \| \| \| --- \| \| The Sacks Number Spiral \| \| \| Introduction The Spiral Plane Product Curves and Polynomials With the center of the spiral to the right of the image, the broad sweep of prime-rich westbound curves can be easily seen. (Generated with Vortex: © 2007 Robert Sacks) Introduction In 1994, Robert Sacks, a software engineer, devised an original method for representing the classical number line of positive integers. He published his findings on the web in 2003. In this method, an Archimedean spiral centered on zero and making one counterclockwise rotation for each perfect square produces a remarkably organized distribution of prime and composite numbers. In some respects, this 2-dimensional "number sphere" can be thought of as a periodic table of numbers - because of the orderly patterns and progressions it reveals. The Sacks Number Spiral is both visually arresting and intellectually compelling. It seems likely that it can provide deeper insights into prime number patterns than the well-known Ulam spiral because, in effect, it joins together the broken lines of Stanislaw Ulam's pseudo-spiral. Sacks' work has focused on product curves, lines that originate from the spiral's center, or near to it - and traverse the spiral's arms at varying angles. His graphing demonstrates that there are multiple orderly factor and prime number progressions. Since the spiral can be extended outward infinitely, the product curves themselves may be considered infinite also. It's an open question whether or not the spiral has any predictive value. For example, can the number spiral's product curves be used to predict the decomposition of very large numbers into prime factors? The Spiral Plane Like the Cartesian plane and the Complex plane, the number spiral represents a plane in which points can occupy one location only. Unlike the traditional planes, all numbers are natural numbers: the positive integers. (However, a spiral of negative integers with negative perfect squares can be visualized as sharing the zero-point center and rotating clockwise in a parallel plane. This is conceptually useful because many polynomial sequences begin with negative integers, then enter the positive spiral, before returning to the negative region.) Quite unlike the Cartesian and Complex planes, the Spiral plane does not require a coordinate system to identify a point. It is a unipolar positioning system. Just knowing the number reveals its location in the spiral - its position relative to every other number in the spiral, its distance from the previous and the next perfect squares, and, potentially, all the curves it can occupy relative to other numbers in the spiral: its polynomial classification. The parabolic curve is a well-known attribute of polynomial equations in classical algebra. In fact, solving the roots of a polynomial means finding the points along the X axis of the Cartesian plane that the parabola intersects. So curviness appears to be rooted in mathematical axioms, but is shown in a new light by the Spiral plane. The fact that the roots of many polynomials, even simple ones, cannot be solved in the Cartesian plane – that is, they do not intersect with the X axis – is one of the intellectual justifications for the Complex plane and Complex numbers. Underlying the distribution of numbers in the spiral plane is the alignment of perfect squares. The progression of perfect squares determines the expansion of the spiral. The first complete rotation of the spiral necessarily comprises three poles - 1, 2, and 3 (a perfect square and two primes). However, the spiral, taken as a whole, aligns along four axes, which can be designated, for convenience, as north, south, west, and east. Zero degrees (aligned due east in the Sacks spiral) is the primary axis and corresponds with the perfect squares: 1 (11), 4 (22), 9 (33), 16 (44), 25 (55), and so on. Due west is the axis of pronic numbers, with the progression 2 (11+1), 6 (22+2), 12 (33+3), 20 (44+4), 30 (55+5), and so on. Polynomial sequences that comprise a pronic number are the number progressions that are by far the richest in prime numbers. The plus or minus sign of the polynomial's constant determines whether the path of the curve lies before or after the spiral crosses the east-west axis. The constant is simply an offset quantity from the basic formula: n^2 for east, n(n+1) for west. For example, a curve that is -17 occurs either before the spiral crosses the pronic axis or before the spiral crosses the square axis. Hence a negative-constant curve can occupy the northwest or southeast quadrant of the spiral. One of the most striking aspects of the Sacks Number Spiral is the predominance of major prime curves in the western hemisphere (opposing side from the perfect squares). For example, one of those curves, heading southwest, contains the numbers of the form n(n - 1) + 41, which is a famous prime-generating formula discovered by Leonhard Euler in 1774. In the number spiral, Sacks is able to make the striking assertion that a prime number is "a positive integer that lies on only one product curve." Elsewhere on this site, these curves have been termed "proximate-prime polynomials". They are simply 2nd-degree polynomials that can be described by four consecutive primes (such as 41, 43, 47, and 53, the first four integers of Euler's sequence). A significant proportion of such polynomials (46 of the first 333) have greater than 50% prime values in the first 1,000 numbers of the series. Product Curves and Polynomials Sacks describes product curves as representing "products of factors with a fixed difference between them". In other words, every curve can be represented by a quadratic equation - a second-degree polynomial - no coincidence given the primacy of the perfect square in the structure of the spiral. Curves can be almost straight but, more typically, perform partial, complete, or multiple clockwise revolutions - counter to the spiral itself - around the origin before straightening out at a particular offset from the east-west axis. The number of revolutions is the direct product of the offset. The greater the offset, the greater the number of revolutions a curve must make before pulling free from its initial "orbit". This relationship is formalized in the Sacks Offset Rule. Product curves can contain composites or primes, the relative proportion of each being predicated on the all-important offset. Simple observation shows that curves with prime number offsets are reliably richer in primes than other odd-number composite offsets. This observation is the basis for the work on this site related to proximate-prime polynomials. Why do the curves straighten out anyway? It turns out the answer is quite simple and has to do with how polynomials are related to perfect squares. A polynomial curve straightens out when its values align with perfect squares or the median values of perfect squares. The point at which a polynomial curve aligns with the horizontal axis is entirely predicated on the third coefficient of a 2nd-degree polynomial equation. A product curve for a given polynomial (of type n² +/- n + x, where x is the offset) travels clockwise, counter to the spiral itself, before intersecting with the squared value of the offset number on the spiral. This is the point of pronic alignment. An offset of 41 literally means the curve will align with the pronic axis at 4141 - the composite number 1681. From that point forward, the curve is aligned with the pronic axis. (See Polynomial Pronic Alignment.) Is the curvy portion of a curve different somehow than the (infinite) straight portion? Certainly the greatest concentration of primes for many polynomials occurs within the pre-alignment portion of their paths across the number spiral. Michael M. Ross \| \| \| The Geometry of the Spiral The spiral, which is termed Archimedean, originates from the center and rotates counterclockwise starting and completing each rotation with due east. This axis is generally designated as 0°. The number line is superimposed on the spiral, and it contains every counting number, beginning with 0, which is the center of the spiral. 1 is the unit of the number line and the number spiral. Each counterclockwise revolution of the spiral begins with a perfect square: 1, 4, 9, 16, and so on. The spiral's product curves are revealed by showing prime and composite dots with different shades or sizes. Another way to think about the spiral is as a "unit circle". In the unit circle you can see the degrees of the circle begin with "due east", just as the spiral does - and that this is 0°. In the unit circle, every position is shown as some fraction of the simple XY coordinate system. (This diagram also shows fractions of Pi, each segment being 2 radians - another way to divide a circle.) The Number Spiral in the Real and Complex Planes Every number on the spiral plane has an equivalent XY coordinate in the unit circle. Since the unit circle can exist in the Z, or complex plane, it is possible that the number spiral can be used to bridge the theory of prime number distribution for real numbers (namely, the Prime Number Theorem) and complex numbers (namely, the Riemann Hypothesis). Of particular interest is the fact that curves with the highest prime density follow the -1,0 coordinate. The Riemann hypothesis for curves over finite fields states that the roots of a projective line have an absolute value q - 1/2. \| \| \| \| Most images on this page were generated with Vortex: © 2007 Robert Sacks For more illustrations of the Sacks Number Spiral and to explore the spiral yourself, with Vortex, go to NumberSpiral.com Also see Number Spirals Related Articles Proximate-prime polynomials Perfect prime polynomial Prime Probability: Catch the Wave \| \| © 2007- 2012 Michael M. Ross \| \| \| \| \| \| \|
10818	https://www.postgresql.org/docs/7.4/sql-altertable.html	PostgreSQL: Documentation: 7.4: ALTER TABLE - [x] Home About Download Documentation Community Developers Support Donate Your account September 25, 2025: PostgreSQL 18 Released! Documentation → PostgreSQL 7.4 Supported Versions: Current (18) / 17 / 16 / 15 / 14 / 13 Development Versions: devel Unsupported versions: 12 / 11 / 10 / 9.6 / 9.5 / 9.4 / 9.3 / 9.2 / 9.1 / 9.0 / 8.4 / 8.3 / 8.2 / 8.1 / 8.0 / 7.4 / 7.3 / 7.2 / 7.1 This documentation is for an unsupported version of PostgreSQL. You may want to view the same page for the current version, or one of the other supported versions listed above instead. \| PostgreSQL 7.4.30 Documentation \| \| Prev \| Fast Backward \| \| Fast Forward \| Next \| ALTER TABLE Name ALTER TABLE--change the definition of a table Synopsis ALTER TABLE [ ONLY ] name [ ] ADD [ COLUMN ] column type [ column_constraint [ ... ] ] ALTER TABLE [ ONLY ] name [ ] DROP [ COLUMN ] column [ RESTRICT \| CASCADE ] ALTER TABLE [ ONLY ] name [ ] ALTER [ COLUMN ] column { SET DEFAULT expression \| DROP DEFAULT } ALTER TABLE [ ONLY ] name [ ] ALTER [ COLUMN ] column { SET \| DROP } NOT NULL ALTER TABLE [ ONLY ] name [ ] ALTER [ COLUMN ] column SET STATISTICS integer ALTER TABLE [ ONLY ] name [ ] ALTER [ COLUMN ] column SET STORAGE { PLAIN \| EXTERNAL \| EXTENDED \| MAIN } ALTER TABLE [ ONLY ] name [ ] SET WITHOUT OIDS ALTER TABLE [ ONLY ] name [ ] RENAME [ COLUMN ] column TO new_column ALTER TABLE name RENAME TO new_name ALTER TABLE [ ONLY ] name [ ] ADD table_constraint ALTER TABLE [ ONLY ] name [ ] DROP CONSTRAINT constraint_name [ RESTRICT \| CASCADE ] ALTER TABLE name OWNER TO new_owner ALTER TABLE name CLUSTER ON index_name Description ALTER TABLE changes the definition of an existing table. There are several subforms: ADD COLUMN This form adds a new column to the table using the same syntax as CREATE TABLE. DROP COLUMN This form drops a column from a table. Indexes and table constraints involving the column will be automatically dropped as well. You will need to say CASCADE if anything outside the table depends on the column, for example, foreign key references or views. SET/DROP DEFAULT These forms set or remove the default value for a column. The default values only apply to subsequent INSERT commands; they do not cause rows already in the table to change. Defaults may also be created for views, in which case they are inserted into INSERT statements on the view before the view's ON INSERT rule is applied. SET/DROP NOT NULL These forms change whether a column is marked to allow null values or to reject null values. You can only use SET NOT NULL when the column contains no null values. SET STATISTICS This form sets the per-column statistics-gathering target for subsequent ANALYZE operations. The target can be set in the range 0 to 1000; alternatively, set it to -1 to revert to using the system default statistics target. SET STORAGE This form sets the storage mode for a column. This controls whether this column is held inline or in a supplementary table, and whether the data should be compressed or not. PLAIN must be used for fixed-length values such as integer and is inline, uncompressed. MAIN is for inline, compressible data. EXTERNAL is for external, uncompressed data, and EXTENDED is for external, compressed data. EXTENDED is the default for all data types that support it. The use of EXTERNAL will, for example, make substring operations on a text column faster, at the penalty of increased storage space. SET WITHOUT OIDS This form removes the oid column from the table. Removing OIDs from a table does not occur immediately. The space that the OID uses will be reclaimed when the row is updated. Without updating the row, both the space and the value of the OID are kept indefinitely. This is semantically similar to the DROP COLUMN process. RENAME The RENAME forms change the name of a table (or an index, sequence, or view) or the name of an individual column in a table. There is no effect on the stored data. ADD table_constraint This form adds a new constraint to a table using the same syntax as CREATE TABLE. DROP CONSTRAINT This form drops constraints on a table. Currently, constraints on tables are not required to have unique names, so there may be more than one constraint matching the specified name. All such constraints will be dropped. OWNER This form changes the owner of the table, index, sequence, or view to the specified user. CLUSTER This form marks a table for future CLUSTER operations. You must own the table to use ALTER TABLE; except for ALTER TABLE OWNER, which may only be executed by a superuser. Parameters name The name (possibly schema-qualified) of an existing table to alter. If ONLY is specified, only that table is altered. If ONLY is not specified, the table and all its descendant tables (if any) are updated. can be appended to the table name to indicate that descendant tables are to be altered, but in the current version, this is the default behavior. (In releases before 7.1, ONLY was the default behavior. The default can be altered by changing the configuration parameter sql_inheritance.) column Name of a new or existing column. type Data type of the new column. new_column New name for an existing column. new_name New name for the table. table_constraint New table constraint for the table. constraint_name Name of an existing constraint to drop. new_owner The user name of the new owner of the table. index_name The index name on which the table should be marked for clustering. CASCADE Automatically drop objects that depend on the dropped column or constraint (for example, views referencing the column). RESTRICT Refuse to drop the column or constraint if there are any dependent objects. This is the default behavior. Notes The key word COLUMN is noise and can be omitted. In the current implementation of ADD COLUMN, default and NOT NULL clauses for the new column are not supported. The new column always comes into being with all values null. You can use the SET DEFAULT form of ALTER TABLE to set the default afterward. (You may also want to update the already existing rows to the new default value, using UPDATE.) If you want to mark the column non-null, use the SET NOT NULL form after you've entered non-null values for the column in all rows. The DROP COLUMN form does not physically remove the column, but simply makes it invisible to SQL operations. Subsequent insert and update operations in the table will store a null value for the column. Thus, dropping a column is quick but it will not immediately reduce the on-disk size of your table, as the space occupied by the dropped column is not reclaimed. The space will be reclaimed over time as existing rows are updated. To reclaim the space at once, do a dummy UPDATE of all rows and then vacuum, as in: UPDATE table SET col = col; VACUUM FULL table; If a table has any descendant tables, it is not permitted to add or rename a column in the parent table without doing the same to the descendants. That is, ALTER TABLE ONLY will be rejected. This ensures that the descendants always have columns matching the parent. A recursive DROP COLUMN operation will remove a descendant table's column only if the descendant does not inherit that column from any other parents and never had an independent definition of the column. A nonrecursive DROP COLUMN (i.e., ALTER TABLE ONLY ... DROP COLUMN) never removes any descendant columns, but instead marks them as independently defined rather than inherited. Changing any part of a system catalog table is not permitted. Refer to CREATE TABLE for a further description of valid parameters. Chapter 5 has further information on inheritance. Examples To add a column of type varchar to a table: ALTER TABLE distributors ADD COLUMN address varchar(30); To drop a column from a table: ALTER TABLE distributors DROP COLUMN address RESTRICT; To rename an existing column: ALTER TABLE distributors RENAME COLUMN address TO city; To rename an existing table: ALTER TABLE distributors RENAME TO suppliers; To add a not-null constraint to a column: ALTER TABLE distributors ALTER COLUMN street SET NOT NULL; To remove a not-null constraint from a column: ALTER TABLE distributors ALTER COLUMN street DROP NOT NULL; To add a check constraint to a table: ALTER TABLE distributors ADD CONSTRAINT zipchk CHECK (char_length(zipcode) = 5); To remove a check constraint from a table and all its children: ALTER TABLE distributors DROP CONSTRAINT zipchk; To add a foreign key constraint to a table: ALTER TABLE distributors ADD CONSTRAINT distfk FOREIGN KEY (address) REFERENCES addresses (address) MATCH FULL; To add a (multicolumn) unique constraint to a table: ALTER TABLE distributors ADD CONSTRAINT dist_id_zipcode_key UNIQUE (dist_id, zipcode); To add an automatically named primary key constraint to a table, noting that a table can only ever have one primary key: ALTER TABLE distributors ADD PRIMARY KEY (dist_id); Compatibility The ADD COLUMN form conforms with the SQL standard, with the exception that it does not support defaults and not-null constraints, as explained above. The ALTER COLUMN form is in full conformance. The clauses to rename tables, columns, indexes, views, and sequences are PostgreSQL extensions of the SQL standard. ALTER TABLE DROP COLUMN can be used to drop the only column of a table, leaving a zero-column table. This is an extension of SQL, which disallows zero-column tables. PrevHomeNext ALTER SEQUENCEUpALTER TRIGGER Policies \| Code of Conduct \| About PostgreSQL \| Contact Copyright © 1996-2025 The PostgreSQL Global Development Group
10819	https://math.stackexchange.com/questions/1974976/relating-the-conditional-probability-formula-to-its-definition	Relating the conditional probability formula to its definition - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Relating the conditional probability formula to its definition Ask Question Asked 8 years, 11 months ago Modified3 years, 6 months ago Viewed 326 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Conditional probability is usually defined via: P(A\|B):=P(A∩B)P(B) P(A\|B):=P(A∩B)P(B) That's fine. P(A\|B)P(A\|B) is then given the rather tendentious name of "the probability of A A given B B". That in itself is also fine; it's just a name. But then — in the texts I have read, at least — P(A\|B)P(A\|B) may actually be directly calculated in accordance with this name — that is, one assumes that B B has occurred, and then derives the probability that A A also occurs. This to me seems like a leap which requires justification. Those two concepts are not the same, and yet they are being equated for the purposes of a substantive calculation. Let me give the following basic example: An urn contains three red balls and two black balls. Two balls are sampled from the urn, without replacement. What is the probability that both balls are red? This would typically be formalised with the sample space Ω={(ball 1,ball 2)}Ω={(ball 1,ball 2)}. The requisite probability could then be calculated by counting permutations as 3⋅2 5⋅4 3⋅2 5⋅4. But you could also calculate it more elegantly via conditional probabilities; one rearranges the conditional probability formula as P(ball 1 is red∩ball 2 is red)=P(ball 2 is red\|ball 1 is red)⋅P(ball 1 is red) P(ball 1 is red∩ball 2 is red)=P(ball 2 is red\|ball 1 is red)⋅P(ball 1 is red) The latter term in the product is easily seen to be 3 5. Crucially, the former term may also be easily determined if it is interpreted as "the probability that the second ball is red given that the first ball was red" — it is 2 4 (again giving 3⋅2 5⋅4 as the correct answer). My problem is that this did not actually calculate conditional probability as defined by the formula — indeed, it can't have, because the entire purpose was to calculate P(A∩B), which is required to evaluate said formula. It instead used the fact that the outcome of this formula — that is, the conditional probability P(A\|B) — can instead be calculated by "pretending" that event B has occurred, and then calculating the probability that A also occurs. My questions is: surely this requires demonstration? If so, how is it demonstrated? I note that this example was about classical probability. Does it also require separate demonstrations for frequentist or subjective probability? I feel like they are different, because I couldn't think of an example of a frequentist or subjective probability question where the equivalence of these two concepts actually has any substantive application, as it did in the example above. I think I need some general clarification. Apologies if the question is too vague. I will do my best to clarify if that is the case. probability probability-theory statistics conditional-probability Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Mar 24, 2022 at 23:22 Tortar 4,020 2 2 gold badges 15 15 silver badges 30 30 bronze badges asked Oct 19, 2016 at 0:35 DenziloeDenziloe 139 5 5 bronze badges 2 1 @Tortar, what is the bounty for?zhoraster –zhoraster 2022-03-29 18:03:27 +00:00 Commented Mar 29, 2022 at 18:03 @zhoraster, this was something I always thought I didn't undestand well enough and so It stroke me when I saw that someone else had the same problem. In the end I tried to wrote myself an answer, I'd like to have a feedback by some expert like you :-)Tortar –Tortar 2022-03-30 13:57:49 +00:00 Commented Mar 30, 2022 at 13:57 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 1 Save this answer. +50 This answer has been awarded bounties worth 50 reputation by Tortar Show activity on this post. If we are following your sample space set up, we have 20 elements inside (they are ordered pair): S={(a,b)\|a,b∈{R 1,R 2,R 3,B 1,B 2},a≠b} Simply take the power set as the sigma algebra, and assign 1/20 as the probability of all the singletons. Now if you want to write every thing out, A={(a,b)\|a∈{R 1,R 2,R 3},b∈{R 1,R 2,R 3,B 1,B 2},a≠b} B={(a,b)\|a∈{R 1,R 2,R 3,B 1,B 2},b∈{R 1,R 2,R 3},a≠b} And you can proceed with the definition. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 19, 2016 at 3:52 BGMBGM 7,861 1 1 gold badge 16 16 silver badges 20 20 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I have same question Fallacy on using interpretation instead of definition in computing conditional probability? (using multiplication law circularly?) I think its best not to define conditional probability this way. Define it more basically and regard P(A\|B):=P(A∩B)/P(B) as a result. A tentative definition could be found in Wikipedia's conditional probability -- formal derivation Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 30, 2017 at 18:43 Y. SiY. Si 145 4 4 bronze badges 3 "Define it more basically" but what is more basic than a ratio? I also observe that you do not propose any other more basic definition, and none is found on the link you give.Michael –Michael 2022-03-30 08:37:55 +00:00 Commented Mar 30, 2022 at 8:37 @Michael Isn't the very first sentence in the linked article section a more basic definition?Denziloe –Denziloe 2022-03-30 22:12:21 +00:00 Commented Mar 30, 2022 at 22:12 No. If that were the case, you would be able to tell me what that definition actually is. At best that first sentence simply states the well known fact that if we have a probability space (Ω,F,P) and an event B with P[B]>0, then the function μ:F→R given by μ(A)=P[A∩B]P[B] is itself a probability measure (it satisfies the 3 axioms of probability). This holds even when Ω is uncountably infinite, even though that paragraph seems to only explain the special case when Ω is finite or countably infinite.Michael –Michael 2022-03-30 23:45:21 +00:00 Commented Mar 30, 2022 at 23:45 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The general principle I think you are using is the classical interpretation of probability, on Wiki If a random experiment can result in N mutually exclusive and equally likely outcomes and if N A of these outcomes result in the occurrence of the event A, the probability of A is defined by P(A)=N A N Indeed applying it to the definition of conditional probability gives P(A\|B)=P(A∩B)P(B)=N A∩B N N B N=N A∩B N B Starting with the simpler example of the rolling of a six-sided unbiased die one can say that P(A=rolling a 6\|B=rolling a even number)=N A∩B N B=1 3 Instead for what regards your problem: An urn contains three red balls and two black balls. Two balls are sampled from the urn, without replacement. What is the probability that both balls are red? As @BGM wrote you can represent the sample space in mutually exclusive and equally likely outcomes as S={(a,b)\|a,b∈{R 1,R 2,R 3,B 1,B 2},a≠b} and so the classical interpretation is applicable. Now we can calculate the conditional probability the same way as before P(A=second ball is red\|B=first ball is red)=N A∩B N B=3⋅2 3⋅4=2 4 So "pretending that event B has occurred, and then calculating the probability that A also occurs" as you said is possible because the three simplifies in the above calculation, that is to say that any sample space S={(a\|a∈{R i,R j,B 1,B 2},i≠j} where P(the ball is red)=2 4 can be used to calculate P(A\|B). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 30, 2022 at 14:15 answered Mar 30, 2022 at 13:55 TortarTortar 4,020 2 2 gold badges 15 15 silver badges 30 30 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. An urn contains three red balls and two black balls. Two balls are sampled from the urn, without replacement. Let the sample space be {R 1 R 2,R 1 B 2,B 1 R 2,B 1 B 2}. I note that this example was about classical probability. Does it also require separate demonstrations for frequentist or subjective probability? Because P(B 1 B 2)=1 10, this probability experiment does not have equally-likely elementary events, so it is not about classical probability. In contrast, BGM's answer frames/interprets this exercise that way. "the probability that the second ball is red given that the first ball was red" My problem is that this did not actually calculate conditional probability as defined by the formula P(A\|B):=P(A∩B)P(B) From the above definition, probability that the second ball is red given that the first ball is red=P(the first and second balls are both red)/P(the first ball is red)=P(R 1 R 2)P(R 1 R 2,R 1 B 2)=3 5×2 4 3 5×2 4+3 5×2 4=1 2 and probability that the first ball is red given that the second ball is red=P(the first and second balls are both red)/P(the second ball is red)=P(R 1 R 2)P(R 1 R 2,B 1 R 2)=3 5×2 4 3 5×2 4+2 5×3 4=1 2. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 29, 2022 at 17:34 ryangryang 1 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability probability-theory statistics conditional-probability See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 3Fallacy on using interpretation instead of definition in computing conditional probability? (using multiplication law circularly?) 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10820	https://www.youtube.com/watch?v=_0C7XXWjOpQ	Imine and Enamine Formation Reactions With Reductive Amination The Organic Chemistry Tutor 9690000 subscribers 1433 likes Description 127746 views Posted: 5 May 2018 This organic chemistry video tutorial provides a basic introduction into imines and enamines. An imine can form in the reaction of a ketone with a primary amine and enamine can be produced by mixing a ketone with a secondary amine. The imine and enamine can be reduced to amines by reductive amination with NaBH3CN. Organic Chemistry - Video Lessons: 41 comments Transcript: in this video we're going to go over the formation of an enamine and an enemy so let's start with a ketone if we react a ketone with let's say a primary amine it's going to give us the immune product and to draw it all you need to do is remove water so if you take out h2o and then connect the remaining parts you're going to get an amine which looks like this now let's say if you have a ketone and you decide to react it with a secondary amine in this case you're going to get an enamine which looks like this now if you have an unsymmetrical ketone the double bond is going to be on the side with the methyl group to give us the most substituted or the most stable alkene so the preference is to put the double bond on the right side then the left side because if we put it on the left side this double bond is less stable and so it's better to put it on the right side keep that in mind once we have the imine we could reduce it into an amine in a process known as reductive emanation so if we use sodium cyanoborohydride looks like that we can turn this into an amine and so we're going to have a nitrogen we're still going to have a methyl group there's going to be a hydrogen and a lone pair but notice that the double bond between the carbon and the nitrogen atom has been reduced to a single bond so basically we're adding a hydrogen on the nitrogen and on this carbon and so here's the two hydrogen atoms that were added now let's say if we have an enamine if we use sodium cyanoborohydride we can reduce it into a tertiary amine and so the product will look like this so that's another example of a reductive emanation process now let's go over the mechanism for the formation of an immune so let's start with the ketone the formation of an emmy it's favorable under mildly acidic conditions it's best around a ph of somewhere between four and five so therefore the first step in this reaction that we're going to do is pronation and then after protonation we're going to use ammonia as our nucleophile ammonia is going to attack the carbonyl carbon and so we're going to get an oh group and we have a nitrogen with three hydrogen atoms now i'm going to use another ammonia molecule as a weak base to take off the hydrogen atom you can also use water too so just keep that in mind you don't always have to use nh3 you can use another base in the solution as well so right now i'm going to have this o h and nh2 so now the nh3 is now an nh4 plus ion so i'm going to use the ammonium ion to protonate the oh group to make it a good leaving group so right now i have the nh2 group with a lone pair and i also have the oh2 group so i'm going to take a lone pair from nitrogen to form a pi bond and expel h2o at the same time and so what i have now is an iminium ion with a positive charge now the next thing we need to do is use a weak base to remove a hydrogen it could be ammonia or water so take your pick i'm going to use nh3 and so that's how we can form the amine and so this is the end of the mechanism so as you can see it's not too bad now let's go over the mechanism for the formation of the enamine so let's start with acetone again and just like before the first step is going to be protonation since the solution is mildly acidic now the mechanism for the enamine is going to be very similar for the amine however towards the end you're going to see a difference so now let's react this with the secondary amine so the nitrogen atom is going to attack the carbonyl carbon and so right now we're going to have a nitrogen atom with four bonds attached to it and it's going to have a positive charge and now we have an o h group so we're going to follow the same process we're going to use a weak base to basically take that hydrogen and then transfer it to the oh group so let's use another amine molecule to do that since amines are more basic than water so this nitrogen is going to take the hydrogen pushing two electrons back on that nitrogen atom and so now the nitrogen atom doesn't have a positive charge anymore it has lone pair so now in the next step we're going to transfer a hydrogen to the o h group turning it into a good leaving group so now we're going to take the lone pair from the nitrogen atom form a pi bond and we're going to expel h2o so once again we have an iminium ion and so here's where everything will differ at this point now let's compare this part with the mechanism for the formation of the image so i believe we had this particular iminium ion notice that there are no hydrogen atoms on the nitrogen atom so when forming the amine the base which let's just say nh3 is the base the base will take a hydrogen forming the immune product however for the mechanism for the formation of an enemy there are no hydrogen atoms that are attached to the nitrogen atom so we can't do what we did here and that's where the mechanism will differ so instead the base is going to take a hydrogen from the carbon since there are none attached to the nitrogen so this base is going to grab the hydrogen the carbon hydrogen bond is going to break we're going to form a pi bond and then the pi bond between the carbon and the nitrogen that's going to break and the electrons in that bond is going to be pulled towards the more electronegative nitrogen atom and so that is the mechanism for the formation of an enamine so now you see how it differs from the mechanism of the formation of an image now what about the reduction of an amine group so immediate they tend to exist around the ph of four to five so let's see what's going to happen if we react with sodium cyanoborohydride so the first thing that's going to happen is this is going to be pronated since the solution is still mildly acidic so let's say the solution is around a ph of four to five so it's not strongly acidic now in the next step we're going to react this with sodium borohydride so the boron has three hydrogen atoms attached to it and a cyanide group so here we have a nucleophilic hydride ion and it's going to attack the carbon that is bonded to the nitrogen so at this point we're going to have a nitrogen with a lone pair on it and so right now we have a primary amine and here's the hydrogen that we added and so that's a simple way in which you can show the reduction of an immune into an amine now what about the enamine so let's go back to this structure how can we show the mechanism for the reduction of the enamine into a tertiary amine so let's assume that the ph is still between four and five because that's when the immune and the enemy will exist in good yields now you don't want to protonate the nitrogen atom it's just not going to work very well instead what you want to do is use the lump here to form the imidium ion and you want to protonate the double bond basically this carbon that's where the hydrogen is going to go now once you have the iminium ion it's going to be very easy to reduce the carbon nitrogen bond in this format so right now there are two hydrogens here but now we have a ch3 group and so we could take sodium borohydride i mean sodium cyanoborohydride rather and now we can reduce the double bond into a single bond and so that's how we can reduce the enamine into a tertiary amine and so that's the mechanism for that reaction you
10821	https://engineering.stackexchange.com/questions/49885/convert-from-rpm-to-per-second	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Engineering Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Convert from RPM to per second [closed] Ask Question Asked Modified 3 years, 7 months ago Viewed 2k times 0 $\begingroup$ Is this correct please for my conversion from RPM to per second? I've included my working below: (500 / 60) 2 Ï = 500 rpm to just s-1. unit Share Improve this question edited Feb 21, 2022 at 16:06 Transistor 12.3k22 gold badges2222 silver badges3333 bronze badges asked Feb 21, 2022 at 14:18 HerbertHerbert 122 bronze badges $\endgroup$ 7 3 $\begingroup$ I assume you're a younger poster, but you can easily find this by just Googling instead of needing to specifically ask. But no, it is not correct. You asked for rotations per second, not radians per second. $\endgroup$ DKNguyen – DKNguyen 2022-02-21 14:20:58 +00:00 Commented Feb 21, 2022 at 14:20 $\begingroup$ Sorry I dont mean rotations per second, but just per second $\endgroup$ Herbert – Herbert 2022-02-21 14:52:23 +00:00 Commented Feb 21, 2022 at 14:52 $\begingroup$ You can't say "just per second". You must say what kind of event or phenomenon you are counting per second, along with appropriate units. $\endgroup$ Elliot Alderson – Elliot Alderson 2022-02-21 14:54:38 +00:00 Commented Feb 21, 2022 at 14:54 $\begingroup$ "Just per second" is meaningless. Even Hz is technically cycles per second. In this case one rotation would be one cycle. $\endgroup$ DKNguyen – DKNguyen 2022-02-21 14:54:49 +00:00 Commented Feb 21, 2022 at 14:54 1 $\begingroup$ @Transistor That's just Hz which is cycles per second, since cycles is often omitted because...reasons(?), but doesn't change the fact OP does not really understand what it is they are reading or asking for. $\endgroup$ DKNguyen – DKNguyen 2022-02-21 15:33:12 +00:00 Commented Feb 21, 2022 at 15:33 \| Show 2 more comments 2 Answers 2 Reset to default 1 $\begingroup$ 60 RPM = 1 s-1. To convert RPM to s-1 you divide by 60. You only need Ï if you are dealing in radians. Share Improve this answer answered Feb 21, 2022 at 14:44 TransistorTransistor 12.3k22 gold badges2222 silver badges3333 bronze badges $\endgroup$ 4 $\begingroup$ Hi, sorry im confused, is this still correct even if I want to get rid of the revolutions and just make it per second please? $\endgroup$ Herbert – Herbert 2022-02-21 14:51:32 +00:00 Commented Feb 21, 2022 at 14:51 $\begingroup$ Think! If something rotates once per second how many times will it rotate in a minute? In two minutes? In ten minutes? (RPM = revolutions per minute, not radians per minute.) $\endgroup$ Transistor – Transistor 2022-02-21 14:56:29 +00:00 Commented Feb 21, 2022 at 14:56 $\begingroup$ @Herbert That's impossible. Your problem is the understanding of units, not any conversion calculation. $\endgroup$ DKNguyen – DKNguyen 2022-02-21 14:57:45 +00:00 Commented Feb 21, 2022 at 14:57 $\begingroup$ The $s^{-1}$ is the same as $/s$ and implies revolutions / second. $\endgroup$ Transistor – Transistor 2022-02-21 16:04:32 +00:00 Commented Feb 21, 2022 at 16:04 Add a comment \| 0 $\begingroup$ So for 1020rpm 1020/60 revs per second ie 17 revs per second Share Improve this answer answered Feb 21, 2022 at 14:21 Solar Mike♦Solar Mike 16.2k11 gold badge2727 silver badges3333 bronze badges $\endgroup$ Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions unit See similar questions with these tags. The Overflow Blog Off with your CMSs head! 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10822	https://artofproblemsolving.com/wiki/index.php/Distance_formula?srsltid=AfmBOorM4ZFJmkWlmN1n2kmR-LNQ-vN4snKd2n1Jg6-vp4UuW2AaaJ0k	Art of Problem Solving Distance formula - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Distance formula Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Distance formula The distance formula is a direct application of the Pythagorean Theorem in the setting of a Cartesian coordinate system. In the two-dimensional case, it says that the distance between two points and is given by . In the -dimensional case, the distance between and is . Shortest distance from a point to a line the distance between the line and point is Proof The equation can be written as Thus, the perpendicular line through is: where is the parameter. will be the distance from the point along the perpendicular line to . So and This meets the given line , where: , so: Therefore the perpendicular distance from to the line is: This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
10823	https://brightchamps.com/en-us/math/math-formulas/math-formula-for-unit-rate	Table Of Contents Summarize this article: ChatGPT Perplexity Last updated on August 5, 2025 Math Formula for Unit Rate In mathematics, a unit rate is a way to compare quantities in different units. It gives the amount of one unit per another unit. In this topic, we will learn the formula for calculating the unit rate. List of Math Formulas for Unit Rate Unit rate is used to compare two different quantities by dividing them. Let’s learn the formula to calculate the unit rate. Math Formula for Unit Rate The unit rate is calculated by dividing the first quantity by the second quantity. The formula for unit rate is: Unit Rate = Quantity1 / Quantity2 Importance of Unit Rate Formula In math and real life, we use the unit rate formula to understand the relationship between quantities. Here are some important uses of the unit rate: Tips and Tricks to Memorize Unit Rate Math Formula Students may find math formulas tricky. Here are some tips and tricks to master the unit rate formula: Real-Life Applications of Unit Rate Math Formula In real life, the unit rate plays a major role in making comparisons and decisions. Here are some applications of the unit rate formula: Common Mistakes and How to Avoid Them While Using Unit Rate Math Formula Students make errors when calculating unit rates. Here are some mistakes and ways to avoid them. Students make errors when calculating unit rates. Here are some mistakes and ways to avoid them. Mistake 1 Confusing the order of division Confusing the order of division Students sometimes confuse which quantity should be divided by which. To avoid this error, remember that the unit rate is often expressed as "per" something, like "miles per hour," meaning miles divided by hours. Students sometimes confuse which quantity should be divided by which. To avoid this error, remember that the unit rate is often expressed as "per" something, like "miles per hour," meaning miles divided by hours. Mistake 2 Incorrect units Incorrect units Using incorrect units in the division can lead to wrong results. Ensure that quantities are in consistent units before calculating the unit rate. Using incorrect units in the division can lead to wrong results. Ensure that quantities are in consistent units before calculating the unit rate. Mistake 3 Assuming unit rates are always whole numbers Assuming unit rates are always whole numbers Students might think unit rates should always result in whole numbers. This is not true, as unit rates can be fractions or decimals as well. Calculate accurately regardless of the result type. Students might think unit rates should always result in whole numbers. This is not true, as unit rates can be fractions or decimals as well. Calculate accurately regardless of the result type. Mistake 4 Ignoring contextual relevance Ignoring contextual relevance Sometimes students calculate unit rates without considering the context, which can lead to irrelevant results. Always interpret unit rates within the given context to make meaningful comparisons. Sometimes students calculate unit rates without considering the context, which can lead to irrelevant results. Always interpret unit rates within the given context to make meaningful comparisons. Examples of Problems Using Unit Rate Math Formula Problem 1 A car travels 300 miles in 5 hours. What is the speed in miles per hour? The speed is 60 miles per hour. The speed is 60 miles per hour. Explanation To find the speed, divide the total miles by the total hours: 300 miles / 5 hours = 60 miles per hour. To find the speed, divide the total miles by the total hours: 300 miles / 5 hours = 60 miles per hour. Problem 2 A store sells 12 apples for $3. What is the cost per apple? The cost per apple is $0.25. The cost per apple is $0.25. Explanation To find the cost per apple, divide the total cost by the number of apples: $3 / 12 apples = $0.25 per apple. To find the cost per apple, divide the total cost by the number of apples: $3 / 12 apples = $0.25 per apple. Problem 3 A recipe requires 4 cups of flour for 8 servings. How much flour is needed per serving? 0.5 cups of flour per serving. 0.5 cups of flour per serving. Explanation To find the flour per serving, divide the total cups of flour by the number of servings: 4 cups / 8 servings = 0.5 cups per serving. To find the flour per serving, divide the total cups of flour by the number of servings: 4 cups / 8 servings = 0.5 cups per serving. Problem 4 A cyclist travels 180 kilometers in 6 hours. What is the speed in kilometers per hour? The speed is 30 kilometers per hour. The speed is 30 kilometers per hour. Explanation To find the speed, divide the total kilometers by the total hours: 180 kilometers / 6 hours = 30 kilometers per hour. To find the speed, divide the total kilometers by the total hours: 180 kilometers / 6 hours = 30 kilometers per hour. FAQs on Unit Rate Math Formula 1.What is the unit rate formula? The formula to find the unit rate is: Unit Rate = Quantity1 / Quantity2 The formula to find the unit rate is: Unit Rate = Quantity1 / Quantity2 2.How do you calculate unit rate? To calculate the unit rate, divide the first quantity by the second quantity to determine the amount of one unit per another unit. To calculate the unit rate, divide the first quantity by the second quantity to determine the amount of one unit per another unit. 3.What is an example of a unit rate? An example of a unit rate is speed, such as miles per hour (mph), which indicates how many miles are traveled in one hour. An example of a unit rate is speed, such as miles per hour (mph), which indicates how many miles are traveled in one hour. Glossary for Unit Rate Math Formulas Explore More math-formulas Important Math Links IconPrevious to Math Formula for Unit Rate Important Math Links IconNext to Math Formula for Unit Rate Jaskaran Singh Saluja About the Author Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles. Fun Fact : He loves to play the quiz with kids through algebra to make kids love it.
10824	https://mightycalculator.com/ratio-simplifier/	Ratio Simplifier \| Simplify Ratios Calculator ↑ [x] Home Search Ratio Simplifier Use our simplify ratio calculator to reduce any ratio to its simplest form. With full step-by-step working out shown for A:B style ratios. Simplify Ratio Calculator Simplify Ratio Calculator Enter whole numbers, decimals or fractions to simplify the ratio to its lowest form Enter the ratio: : Calculate Clear Solution Related Ratio & Proportion Calculators Ratios and Proportions Navigation Ratio CalculatorRatio to Fraction CalculatorRatio SimplifierGolden Ratio Calculator See all maths calculators Table of Contents Simplify Ratio Calculator Related Ratio & Proportion Calculators Ratio Simplifier: Reduce Ratios to Their Simplest Form What is a Ratio? Why Simplify Ratios? How to Use the Ratio Simplifier The Step-by-Step Simplification Process Examples of Ratio Simplification Advanced Features of Our Ratio Simplifier Common Questions About Ratio Simplification Conclusion Ratio Simplifier: Reduce Ratios to Their Simplest Form This ratio calculator helps you reduce ratios to their simplest form, providing a step-by-step breakdown of the process. Whether you're working with whole numbers, decimals, or fractions, our tool will guide you through the simplification journey. What is a Ratio? A ratio is a comparison between two or more numbers, typically expressed in the form A:B or A/B. Ratios are used in various fields, including mathematics, science, cooking, and finance, to describe relationships between quantities. Why Simplify Ratios? Simplifying ratios is essential for several reasons: Clarity: Simplified ratios are easier to understand and interpret. Comparison: Simplified ratios make it easier to compare different ratios. Calculations: Working with simplified ratios often leads to simpler calculations. Problem-solving: Many mathematical and real-world problems require simplified ratios for efficient solutions. How to Use the Ratio Simplifier Our Ratio Simplifier is designed to be user-friendly and versatile. Follow these simple steps to simplify your ratios: Enter the first value (A) in the ratio Enter the second value (B) in the ratio Click "Simplify Ratio" to see the result The tool accepts various input formats: Whole numbers: e.g., 8, 12, 100 Decimals: e.g., 2.5, 3.75, 0.125 Fractions: e.g., 1/2, 3/4, 5/8 Mixed numbers: e.g., 1 1/2, 2 3/4, 3 1/4 The Step-by-Step Simplification Process Our Ratio Simplifier doesn't just give you the final answer; it shows you the entire simplification process. Here's what happens behind the scenes: Conversion: Any decimals or fractions are converted to whole numbers. GCF Calculation: The Greatest Common Factor (GCF) of A and B is determined. Division: Both A and B are divided by the GCF. Final Expression: The simplified ratio is presented in its final form. Examples of Ratio Simplification Let's walk through some examples to illustrate how the Ratio Simplifier works: Example 1: Whole Numbers Input: 24:36 Steps: Find the GCF of 24 and 36 Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 GCF = 12 Divide both numbers by the GCF: 24 ÷ 12 = 2 36 ÷ 12 = 3 Result: 2:3 Example 2: Decimals Input: 0.75:1.25 Steps: Multiply both numbers by 100 to convert to whole numbers: 0.75 × 100 = 75 1.25 × 100 = 125 Find the GCF of 75 and 125 Factors of 75: 1, 3, 5, 15, 25, 75 Factors of 125: 1, 5, 25, 125 GCF = 25 Divide both numbers by the GCF: 75 ÷ 25 = 3 125 ÷ 25 = 5 Result: 3:5 Example 3: Fractions Input: 2/3:5/6 Steps: Find a common denominator (LCD of 3 and 6 is 6): 2/3 = 4/6 5/6 remains the same Now we have 4:5 (numerators only) The GCF of 4 and 5 is 1, so the ratio is already in its simplest form Result: 4:5 Advanced Features of Our Ratio Simplifier Our Ratio Simplifier goes beyond basic simplification: Fraction to Decimal Conversion: See your simplified ratio in both fraction and decimal form. Percentage Representation: Understand the proportion each part of the ratio represents as a percentage. Graphical Representation: Visualize your ratio with an intuitive bar graph. Ratio Applications: Learn how your simplified ratio applies to real-world scenarios. Common Questions About Ratio Simplification Q: Can a ratio have zero as one of its terms? A: While mathematically possible, a ratio with zero is generally not meaningful in practical applications. Our simplifier will alert you if you enter zero as one of the terms. Q: What if my ratio can't be simplified further? A: Some ratios are already in their simplest form. In these cases, our tool will confirm that no further simplification is possible. Q: How does the Ratio Simplifier handle very large numbers? A: Our tool is designed to handle large numbers efficiently. However, for extremely large numbers, the calculation might take a bit longer. Conclusion Understanding and simplifying ratios is a fundamental skill in mathematics and many real-world applications. Our Ratio Simplifier tool makes this process easy, educational, and even fun! Whether you're a student, teacher, or professional, this tool will help you master the art of ratio simplification. Remember, practice makes perfect. The more you use the Ratio Simplifier, the better you'll become at understanding and working with ratios. Happy simplifying! Free online calculators, helpful tools & converters Social Facebook X/Twitter © Free Online Calculators & Converters \| MightyCalculator.com 2025. All rights reserved You’ve successfully subscribed to Free Online Calculators & Converters \| MightyCalculator.com Welcome back! You’ve successfully signed in. Great! You’ve successfully signed up. Success! Your email is updated. Your link has expired Success! Check your email for magic link to sign-in. Please enter at least 3 characters 0 Results for your search
10825	https://www.tsc.fl.edu/media/divisions/learning-commons/resources-by-subject/math/college-algebra/Circles,-Midpoint-and-Distance.pdf	Circles, Midpoint and Distance A circle is defined as a two-dimensional figure, which is round in shape where all the points on the surface of the circle are equidistant from the center point is called “C”. The distance from point C to the outside of the circle is called the radius of the circle (r). We have 3 formulas that we use to find the pieces to develop a formula for a circle. Circle formula: r (radius), center(h,k) Distance formula Midpoint formula: (center at (h,k)) 𝑟2 = (𝑥−ℎ)2 + (𝑦−𝑘)2 d=√(𝑥2 −𝑥1)2 + (y2 −𝑦1)2 (𝑥1 + 𝑥2 2 , 𝑦1 + 𝑦2 2 ) Example 1. Given the center and the radius If you are given the center (5, -2) and the radius of 3, then you just need to substitute them into the circle formula. In the formula, the center is designated as (h,k) and the radius is r. 𝒓𝟐= (𝒙−𝒉)𝟐+ (𝒚−𝒌)𝟐 (3)2 = ( 𝑥−5)2 + ( 𝑦−(−2))2, then simplify 𝟗= ( 𝒙−𝟓)𝟐+ (𝒚+ 𝟐)𝟐 Example 2. Given the center and one point on the circle. If you are given the center and one point on the circle, you have a two-step process to get the equation of the circle. You always need the radius and the center. Since we are given the center we need to determine the radius. Since the radius is the distance from the center to any point on the circle, we can use the distance formula. You are given that the center is at (3,4) and a point on the circle is (5,5). Thus, we can label the first point as (x1, y1) and the second as (x2, y2) to help us as we substitute the values into the distance formula. d=√(𝒙𝟐−𝒙𝟏)𝟐+ (𝐲𝟐−𝒚𝟏)𝟐 d = √(5 −3)2 + (5 −4)2 , then simplify d=√(2)2 + (1)2 d=√5 this distance is our radius. Now we have both the center and radius so we can create our circle formula that fits the information we were given: (√5)2 = (𝑥−3)2 + (𝑦−4)2, or simplified 5 = (𝒙−𝟑)𝟐+ (𝒚−𝟒)𝟐 Example 3. Given two points that are the endpoints of the diameter If you are given two points on the circle that lie on the endpoints of the diameter, you will need to calculate both the center (h,k) and the radius (r). To calculate the center, you can use the midpoint formula, since the radius is exactly in the middle between two points on the circle. Before you start substituting the points it is a good practice to label the coordinates as (x1, y1) and (x2, y2). If we start with the points (4,5) and (-2, -3) as points on the circle, we would have (𝒙𝟏+ 𝒙𝟐 𝟐 , 𝒚𝟏+ 𝒚𝟐 𝟐 ) ( 4+(−2) 2 , 5+(−3) 2 ), then simplify (1, 1) so this is our center (h,k) Next, we need the radius (r). Since this is a distance, we can utilize the distance formula. However, the distance between two points on the circle gives us the diameter so we will need to use the fact that r = ½ d. d=√(𝒙𝟐−𝒙𝟏)𝟐+ (𝐲𝟐−𝒚𝟏)𝟐 d=√(−𝟐−𝟒)𝟐+ (−𝟑−𝟓)𝟐 ; then simplify d=√(−𝟔)𝟐+ (−𝟖)𝟐 d=√𝟑𝟔+ 𝟔𝟒 d = √𝟏𝟎𝟎 d = 10 thus, our radius = ½ (10) or 5. Now we have enough information to write the equation of the circle using the information we were given and what we calculated. r = 5 and the center (h,k) = (1,1) 52 = (𝑥−1)2 + (𝑦−1)2 . simplified we have 𝟐𝟓= (𝒙−𝟏)𝟐+ (𝒚−𝟏)𝟐 Example 4. Given the center and Tangent to an axis You have a circle with a center of (-3,5) and it is tangent to the y- axis, and you need the equation of the circle. The best strategy is to graph the information given to determine the radius. The you can see that it is 3 steps from the y-axis. Since being tangent means, it only touches at 1 point, you can use this as your radius. Now we have all the information we need to create the equation of the circle. 𝑟2 = (𝑥−ℎ)2 + (𝑦−𝑘)2 (3)2 = (𝑥−(−3))2 + (𝑦−5)2, simplified 𝟗= (𝒙+ 𝟑)𝟐+ (𝒚−𝟓)𝟐 Find the equation of the circle for each of the following: 1. Center at (0, 3) and radius = 5 2. Center at (1,2) and another point on the circle at (5,3) 3. Endpoints of a diameter at (3,5) and (7,3). 4. Center at (2, -3) and tangent to the x-axis. Answers: 1. 𝑥2 + (𝑦−3)2 = 25 2. (𝑥−1)2 + (𝑦−2)2 = 17 3. (𝑥−5)2 + (𝑦−4)2 = 20 4. (𝑥−2)2 + (𝑦+ 3)2 = 4
10826	https://www.freemathhelp.com/forum/threads/what-does-it-mean-by-an-and-n-in-the-equation.112001/	What does it mean by 'an' and 'n' in the equation? \| Free Math Help Forum Home ForumsNew postsSearch forums What's newNew postsLatest activity Log inRegister What's newSearch Search [x] Search titles only By: SearchAdvanced search… New posts Search forums Menu Log in Register Install the app Install Forums Free Math Help Intermediate/Advanced Algebra What does it mean by 'an' and 'n' in the equation? Thread starterIndranil Start dateAug 2, 2018 I Indranil Junior Member Joined Feb 22, 2018 Messages 220 Aug 2, 2018 #1 What does it mean by 'an' and 'n' in the equation below? and could you explain 'the sum up sign' in an easy way? Attachments LearnMeanAbsoluteErrormeaningconceptsformulasthroughStudyMaterialNotesGÇôEmbibe.com-1.jpg 67.2 KB · Views: 16 lev888 Elite Member Joined Jan 16, 2018 Messages 2,995 Aug 2, 2018 #2 Looks like somebody forgot to finish editing the page before publishing. Explanation: Dr.Peterson Elite Member Joined Nov 12, 2017 Messages 16,844 Aug 2, 2018 #3 Indranil said: What does it mean by 'an' and 'n' in the equation below? and could you explain 'the sum up sign' in an easy way? Click to expand... It looks like what happens when you try to copy from some pages, and it gives you both the formatted and unformatted versions of the same thing. Ignore the second copy of each of the three formatted bits (the list, and the two equations for "amean".) I Indranil Junior Member Joined Feb 22, 2018 Messages 220 Aug 2, 2018 #4 Dr.Peterson said: It looks like what happens when you try to copy from some pages, and it gives you both the formatted and unformatted versions of the same thing. Ignore the second copy of each of the three formatted bits (the list, and the two equations for "amean".) Click to expand... Still, I don't understand what 'n' and 'an'? could you simplify it, please? D Denis Senior Member Joined Feb 17, 2004 Messages 1,707 Aug 2, 2018 #5 1 2 3 4 ...........n 7 3 8 6 ...........4 a(3) = 8, a(n) = 4 Dr.Peterson Elite Member Joined Nov 12, 2017 Messages 16,844 Aug 2, 2018 #6 Indranil said: Still, I don't understand what 'n' and 'an'? could you simplify it, please? Click to expand... Here is what it says: Suppose the values obtained in several measurements are a 1,a 2,a 3,...,a n\displaystyle a_1, a_2, a_3, ..., a_n a 1,a 2,a 3,...,a n. The arithmetic mean of these values is taken as the best possible value of the quantity under the given conditions of measurement as: a m e a n=(a 1+a 2+a 3+...+a n)n\displaystyle \displaystyle a_{mean} = \frac{(a_1 + a_2 + a_3 + ... + a_n)}{n}a m e a n=n(a 1+a 2+a 3+...+a n) or a m e a n=∑i=1 n a i n\displaystyle \displaystyle a_{mean} = \sum_{i=1}^{n} \frac{a_i}{n}a m e a n=i=1∑nn a i n is the number of measurements. They are called a 1, a 2, and so on, up to the last (nth) one, a n. That is, the subscript (1, 2, ..., n) distinguishes the first, second, ..., nth number. The mean is the sum of all n measurements, divided by n, the number of measurements. Is that clear? mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Aug 2, 2018 #7 Indranil said: Still, I don't understand [the meaning of] 'n' and 'an' … Click to expand... Symbol a n represents the last number in a sequence of measurements. (The sequence is named a.) a = {a 1, a 2, a 3, … a n} a 1 is the 1st measurement in sequence a a 2 is the 2nd measurement a 3 is the 3rd measurment The dots indicate that some measurements are not listed. (We don't write complete lists because they're too long. We show only the first few elements and the last one.) a n is the last measurement in the sequence (we call it the 'nth element' or 'nth term') Therefore, symbol n represents the number of measurements. The arithmetic mean is an average measurement. We calculate this average by adding all the measurements and dividing their total by the number of measurements. a mean=a 1+a 2+a 3+…+a n n\displaystyle a_{\text{mean}} = \dfrac{a_1 + a_2 + a_3 + … + a_n}{n}a mean=n a 1+a 2+a 3+…+a n a mean=∑i=1 n(a i n)\displaystyle \displaystyle a_{\text{mean}} \; = \; \sum_{i=1}^{n} \bigg(\dfrac{a_i}{n} \bigg) \;a mean=i=1∑n(n a i)=1 n⋅∑i=1 n(a i)\displaystyle \displaystyle = \; \dfrac{1}{n} \cdot \sum_{i=1}^{n} \bigg( a_i \bigg)=n 1⋅i=1∑n(a i) Note the additional part (shown in green). The summation in black shows each measurement being divided by n (then those fractions are added). But, there is a property of summations that allows us to factor out 1/n, so we can add the elements first and then divide by n once (at the end). :cool: I Indranil Junior Member Joined Feb 22, 2018 Messages 220 Aug 3, 2018 #8 mmm4444bot said: Symbol a n represents the last number in a sequence of measurements. (The sequence is named a.) a = {a 1, a 2, a 3, … a n} a 1 is the 1st measurement in sequence a a 2 is the 2nd measurement a 3 is the 3rd measurment The dots indicate that some measurements are not listed. (We don't write complete lists because they're too long. We show only the first few elements and the last one.) a n is the last measurement in the sequence (we call it the 'nth element' or 'nth term') Therefore, symbol n represents the number of measurements. The arithmetic mean is an average measurement. We calculate this average by adding all the measurements and dividing their total by the number of measurements. a mean=a 1+a 2+a 3+…+a n n\displaystyle a_{\text{mean}} = \dfrac{a_1 + a_2 + a_3 + … + a_n}{n}a mean=n a 1+a 2+a 3+…+a n a mean=∑i=1 n(a i n)\displaystyle \displaystyle a_{\text{mean}} \; = \; \sum_{i=1}^{n} \bigg(\dfrac{a_i}{n} \bigg) \;a mean=i=1∑n(n a i)=1 n⋅∑i=1 n(a i)\displaystyle \displaystyle = \; \dfrac{1}{n} \cdot \sum_{i=1}^{n} \bigg( a_i \bigg)=n 1⋅i=1∑n(a i) Note the additional part (shown in green). The summation in black shows each measurement being divided by n (then those fractions are added). But, there is a property of summations that allows us to factor out 1/n, so we can add the elements first and then divide by n once (at the end). :cool: Click to expand... Thanks a lot for all of your kind efforts for me to learn math in an easy way. Thank you all again. I Indranil Junior Member Joined Feb 22, 2018 Messages 220 Aug 3, 2018 #9 How to learn summation in an easy way because it gives me a headache? Last edited by a moderator: Aug 3, 2018 mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Aug 3, 2018 #10 Study lots of examples, and practice. :cool: Google kewords like summation lessons examples, and look for resources that you find easy to understand. If one page is too hard, try another. Here are two search results; the first is a short video introduction, and the second is a PDF file containing several examples. I Indranil Junior Member Joined Feb 22, 2018 Messages 220 Aug 3, 2018 #11 mmm4444bot said: The arithmetic mean is an average measurement. We calculate this average by adding all the measurements and dividing their total by the number of measurements. a mean=a 1+a 2+a 3+…+a n n\displaystyle a_{\text{mean}} = \dfrac{a_1 + a_2 + a_3 + … + a_n}{n}a mean=n a 1+a 2+a 3+…+a n a mean=∑i=1 n(a i n)\displaystyle \displaystyle a_{\text{mean}} \; = \; \sum_{i=1}^{n} \bigg(\dfrac{a_i}{n} \bigg) \;a mean=i=1∑n(n a i)=1 n⋅∑i=1 n(a i)\displaystyle \displaystyle = \; \dfrac{1}{n} \cdot \sum_{i=1}^{n} \bigg( a_i \bigg)=n 1⋅i=1∑n(a i) Note the additional part (shown in green). The summation in black shows each measurement being divided by n (then those fractions are added). But, there is a property of summations that allows us to factor out 1/n, so we can add the elements first and then divide by n once (at the end). :cool: Click to expand... It would be very helpful for me if you kindly explain the summation above in an easy way. Dr.Peterson Elite Member Joined Nov 12, 2017 Messages 16,844 Aug 3, 2018 #12 Indranil said: It would be very helpful for me if you kindly explain the summation above in an easy way. Click to expand... It will be very helpful for us if you could tell us what part of what has been explained is not "easy" for you. You might tell us what you think it means, in your own words, so we can see how much you understand, and also tell us specifically which parts you are unsure of. Keep in mind that mathematics is not easy, and nothing anyone does can make it not require mental effort. But we can make that effort easier for you if we know how you think, and can adjust what we say to that. What is "easier" depends on the person. This is why specific questions are much more likely to get good answers. mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Aug 4, 2018 #13 Indranil said: … kindly explain the summation above in an easy way. Click to expand... Which part do you need help understanding, in the Khan Academy introduction video? Please post the timestamp and your specific question. :cool: I Indranil Junior Member Joined Feb 22, 2018 Messages 220 Aug 9, 2018 #14 mmm4444bot said: a mean=a 1+a 2+a 3+…+a n n\displaystyle a_{\text{m}\text{ean}} = \dfrac{a_1 + a_2 + a_3 + … + a_n}{n}a m ean=n a 1+a 2+a 3+…+a n a mean=∑i=1 n(a i n)\displaystyle \displaystyle a_{\text{m}\text{ean}} \; = \; \sum_{i=1}^{n} \bigg(\dfrac{a_i}{n} \bigg) \;a m ean=i=1∑n(n a i)=1 n⋅∑i=1 n(a i)\displaystyle \displaystyle = \; \dfrac{1}{n} \cdot \sum_{i=1}^{n} \bigg( a_i \bigg)=n 1⋅i=1∑n(a i) Click to expand... What does it mean by '1/n', 'n' on the top of the summation sign and 'i=1' on the bottom of the summation sign and 'ai' in the bracket? Last edited by a moderator: Aug 9, 2018 D Denis Senior Member Joined Feb 17, 2004 Messages 1,707 Aug 9, 2018 #15 No chalk/blackboard here; have a look here: If you'd like more sites, google "introduction to summation formula". J JeffM Elite Member Joined Sep 14, 2012 Messages 7,875 Aug 9, 2018 #16 Indranil said: What does it mean by '1/n', 'n' on the top of the summation sign and 'i=1' on the bottom of the summation sign and 'ai' in the bracket? Click to expand... Are you seriously unsure of what 1 n\displaystyle \dfrac{1}{n}n 1 means in mathematics? If so, look up fraction by clicking on the word in blue. Σ\displaystyle \Sigma Σ is the capital Greek letter sigma, equivalent to S in English, and stands for "sum." (Π\displaystyle \Pi Π is the capital Greek letter pi, equivalent to P, and stands for "product.") ∑j=k n\displaystyle \displaystyle \sum_{j=k}^n j=k∑n means that we are to sum (add) n expressions, each of which will contain the variable j, with j starting at k and increasing by 1 each time. ∑j=5 8(2 j∗x j−4)=\displaystyle \displaystyle \sum_{j=5}^8 \left ( 2^j x_{j-4} \right ) =j=5∑8(2 j∗x j−4)= 2 5∗x 5−4+2 6∗x 6−4+2 7∗x 7−4+2 8∗x 8−4=32 x 1+64 x 2+128 x 3+256 x 4.\displaystyle 2^5 x_{5-4} + 2^6 x_{6-4} + 2^7 x_{7-4} + 2^8 x_{8-4} = 32x_1 + 64x_2 + 128x_3 + 256x_4.2 5∗x 5−4+2 6∗x 6−4+2 7∗x 7−4+2 8∗x 8−4=3 2 x 1+6 4 x 2+1 2 8 x 3+2 5 6 x 4. It is a very compact notation if n is unknown or if n is large. n=3⟹1 n∗∑j=1 n(x j)=1 3∗(x 1+x 2+x 3)=x 1+x 2+x 3 3\displaystyle n = 3 \implies \displaystyle \dfrac{1}{n} \sum_{j=1}^n \left ( x_j \right ) = \dfrac{1}{3} (x_1 + x_2 + x_3) = \dfrac{x_1 + x_2 + x_3}{3}n=3⟹n 1∗j=1∑n(x j)=3 1∗(x 1+x 2+x 3)=3 x 1+x 2+x 3 If n = 500, writing it out the latter way is inefficient. Last edited: Aug 9, 2018 mmm4444bot Super Moderator Joined Oct 6, 2005 Messages 10,962 Aug 9, 2018 #17 Indranil said: What does it mean by '1/n', 'n' on the top of the summation sign and 'i=1' on the bottom of the summation sign and 'ai' in the bracket? Click to expand... n represents how many numbers are added. If we add three numbers, then n=3. If we add 4,527 numbers, then n=4527. Multiplication by 1/n is the same as division by n. Therefore, the factor 1/n shows that we are dividing the total by n. i represents the index. The index counts the numbers and shows their position in the set. We start counting at 1 (the first number in set a), and the notation i=1 shows this. a i is a symbol that represents each number being added. When i=1, symbol a i becomes a 1, and a 1 represents the first number being added. When i=2, the symbol a i becomes a 2, and it represents the second number being added. The index continues counting each number, until it reaches the number n. At that point, we have symbol a n, and a n represents the last number being added (often called the "nth number"). Let's add the three numbers in this set: a = {33, 62, 85} 33 + 62 + 85 We could also use symbols to represent these numbers. The first number is 33, so let's call it a 1 The second number is 62, so let's call it a 2 The third number is 85, so let's call it a 3 Therefore a 1 + a 2 + a 3 means 33 + 62 + 85 The subscripts (in red) show us the index i counting each number (first number, second number, third number). In other words, the index starts at i=1 and it ends at i=3. In this example, n=3 because we're adding three numbers. The nth number is a 3 What if we wanted to add 3,000 numbers, instead? Nobody wants to write 3,000 numbers, so we need a shorter way to write sums. Sigma notation gives us a shorter way. Using my set {33, 62, 85} we understand that a 1=33, a 2=62, and a 3=85. We write their sum easily using Sigma notation, like this: ∑i=1 3(a i)\displaystyle \displaystyle \sum_{i=1}^{3} \bigg( a_i \bigg)i=1∑3(a i) You can see the count starts at i=1. You can also see that n=3 (this is what tells the reader that we're adding three numbers). You can see the symbol representing numbers being added (a i). i=1 tells us to start the sum with the first number a 1 33 + Next, the index increases from i=1 to i=2. Sigma notation tells us to add the next number a 2 33 + 62 + Next, the index increases from i=2 to i=3. Sigma notation tells us to add the next number a 3 33 + 62 + 85 At this point, the index has reached n, so all numbers have been added. ∑i=1 3(a i)=33+62+85=180\displaystyle \displaystyle \sum_{i=1}^{3} \bigg( a_i \bigg) = 33 + 62 + 85 = 180 i=1∑3(a i)=3 3+6 2+8 5=1 8 0 If we want to find the average (m͏ean) of set a={33, 62, 85}, then we divide the total by the count. 180/3 = 60 The mean is 60. Dividing by 3 is the same as multiplying by 1/3. Therefore, we could also write the calculation like this: 1/3 ∙ (33 + 62 + 85) This is why the Sigma notation for the mean of a i looks like this: a mean=1 3⋅∑i=1 3(a i)\displaystyle \displaystyle a_{\text{mean}} = \dfrac{1}{3} \cdot \sum_{i=1}^{3} \bigg( a_i \bigg)a mean=3 1⋅i=1∑3(a i) We could also write ∑i=1 3(a i)3\displaystyle \displaystyle \dfrac{\sum_{i=1}^{3} (a_i)}{3}3∑i=1 3(a i) but most people write it the first way (multiplying by 1/n instead of writing ratio form). Here is a final example: ∑i=1 208(x i)\displaystyle \displaystyle \sum_{i=1}^{208} \bigg( x_i \bigg)i=1∑2 0 8(x i) I can tell by looking at this notation that we are adding 208 numbers in a set called x, and subscripted symbol x i is a generic variable used to represent the numbers. When index i starts counting (1,2,3,…), then x i represents the individual numbers in set x added one by one (in order of listing). The Sigma notation is much easier than writing the sum like this: 5+35+74+53+29+72+17+82+98+43+183+71+163+8+31+63+32+1+53+73+13+28+31+52+158+3+41+5+532+223+75+52+651+987+451+121+851+451+151+625+403+322+358+233+453+532+305+23+342+145+472+312+544+293+454+1023+453+405+984+268+234+957+2034+9670+9345+7093+4867+5394+750+923+4709+3750+9232+5703+8975+6039+5730+9572+3095+7093+857+9325+7034+9793+4754+385+345+832+570+934+759+35+7+1389+757+235+74+97+54+3+95+72+34+9+5+35+74+53+29+72+17+82+98+43+183+71+163+8+31+63+32+1+53+73+13+28+31+52+158+3+41+5+532+223+75+52+651+987+451+121+851+451+151+625+403+322+358+233+453+532+305+23+342+145+472+312+544+293+454+1023+453+405+984+268+234+957+2034+9670+9345+7093+4867+5394+750+923+4709+3750+9232+5703+8975+6039+5730+9572+3095+7093+857+9325+7034+9793+4754+385+345+832+570+934+759+35+7+1389+757+235+74+97+54+3+95+72+34+9 :cool: Last edited: Aug 15, 2018 You must log in or register to reply here. 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10827	http://tasks.illustrativemathematics.org/content-standards/tasks/397	Illustrative Mathematics Typesetting math: 100% Engage your students with effective distance learning resources. ACCESS RESOURCES>> The Lighthouse Problem No Tags Alignments to Content Standards:G-MG.A.1 Student View Task Suppose that even in perfect visibility conditions, the lamp at the top of a lighthouse is visible from a boat on the sea at a distance of up to 32 km (if it is farther than that, then it is obscured by the horizon). If the "distance" in question is the straight-line distance from the lamp itself to the boat, what is the height above sea level of the lamp on top of the lighthouse? What are two other interpretations of the distance being investigated in this problem? Describe how to solve the alternate versions. IM Commentary In addition to the purely geometric and trigonometric aspects of the task, this problem asks students to model phenomena on the surface of the earth. This first step, observing that the bound of 32km on the visibility of the lighthouse is due to the curvature of the earth, will likely be a difficult part of the modeling process. Instructors should expect to provide guidance for this part, and possibly also the diagram in the solution. The instructor can also choose whether or not to provide the radius of the earth, or to have students discover its relevance and then look it up in an available resource. Students may discover that since the earth is not a perfect sphere, the "radius" of the earth varies from one location to the next. As such, students may end up with one of many possible such values, with which the calculations in the solution could be repeated verbatim. This problem is freely translated from the math and physics problem column in La Presse de Tunisie (October, 2006). It asked only the first question. La Presse is a general circulation newspaper in Tunis and this was the easiest of the three math problems. Presumably the first interpretation of â€˜distanceâ€™ was intended in the original problem. But the variants give an opportunity for students to see the differences in the three values and use several trigonometric approaches. It is tempting in b.ii to actually solve for θ and divide by cos θ. Not only is this more work but it might introduce round off error. One should note that the small differences in the three answers are dwarfed by the approximations implicit in estimating the radius of the earth and the rounding off to even kilometers in stating the problem. In particular, the calculated differences between part a and b.i of less than 1/1000th of a meter and between b.i and b.ii of 1/100th of meter are negligible. Solution Consider the triangle with vertices C, the center of the earth, A, the lamp, and B, the boat. The radius of the earth can be found on the internet as 6,371 km. (Note that the diagram is drawn to a highly disproportionate scale to aid with visibility.) With notation as above, we have A C=6371+h, B C=6371 and A B=32. Since we are told that the lighthouse can be seen from up to 32 km, C B A is a right-angle. By the Pythagorean theorem, 6371 2+32 2=(6371+h)2=6371 2+2(6371)h+h 2, which simplifies to h 2+2(6371)h=32 2, which in turn can be solved by the quadratic formula: h=−12742±12742 2−4(1)(−(32 2)−−−−−−−−−−−−−−−−−√2=.080363643 km or 80.36 meters. 2. For the sake of measuring the discrepancy in our answer coming from different interpretations of "distance," we maintain a grossly unreasonable level of precision for this part of the problem. Let D be the base of the lighthouse, R the radius of the earth and continue the notation of part a). Let θ be the angle between A D C and B C. Then R R+h=cos θ so h=R cos θ−R. The other two interpretations are given below. 1. Take the distance to be the length of the circular arc from D to B. Then θ 360=32 2 π 6371 or θ=360⋅32 2 π 6371=.2877829139 degrees. Now the height of the tower is 6371 cos(.2877829139)−6371=.0803649948 km, or approximately 80 meters. 2. Take the distance to be the straight-line distance from the boat to the base of the lighthouse, so B D=32. We calculate the angle θ using the law of cosines applied to the triangle B C D. We have B D 2+D C 2=2(B C)(B D)cos θ. So, cos θ=B D 2−(B C 2+D C 2)−2(B C)(B D) or in our case cos θ=32 2−(2 R 2)−2 R 2=.9999873859 The height of the lighthouse is 6371.9999873859−6371=.0803654448 km or 80.37 meters. For comparison purposes, we continue to find θ. Substituting, cos−1(θ)=.9999873859 we see θ=.2877832164 degrees which agrees to 4 places with the arcsin calculation. The Lighthouse Problem Suppose that even in perfect visibility conditions, the lamp at the top of a lighthouse is visible from a boat on the sea at a distance of up to 32 km (if it is farther than that, then it is obscured by the horizon). If the "distance" in question is the straight-line distance from the lamp itself to the boat, what is the height above sea level of the lamp on top of the lighthouse? What are two other interpretations of the distance being investigated in this problem? Describe how to solve the alternate versions. Print Task Typeset May 4, 2016 at 18:58:52. Licensed by Illustrative Mathematics under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
10828	https://blog.csdn.net/wizardforcel/article/details/136719377	普林斯顿算法讲义（三）_有向图传播-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作普林斯顿算法讲义（三）绝不原创的飞龙于 2024-03-14 19:12:37 发布阅读量940收藏 7 点赞数 8 CC 4.0 BY-SA版权分类专栏：算法文章标签：算法 License CC BY-NC-SA 4.0 / 自豪地采用谷歌翻译本文链接：算法专栏收录该内容 4 篇文章订阅专栏原文：普林斯顿大学算法课程译者：飞龙协议：CC BY-NC-SA 4.0 4.2 有向图原文：algs4.cs.princeton.edu/42digraph 译者：飞龙协议：CC BY-NC-SA 4.0 有向图。一个有向图（或有向图）是一组顶点和一组有向边，每条边连接一个有序对的顶点。我们说一条有向边从该对中的第一个顶点指向该对中的第二个顶点。对于 V 个顶点的图，我们使用名称 0 到 V-1 来表示顶点。术语表。这里是我们使用的一些定义。自环是连接顶点到自身的边。如果两条边连接相同的顶点对，则它们是平行的。一个顶点的 outdegree 是指指向它的边的数量。一个顶点的 indegree 是指指向它的边的数量。子图是构成有向图的一部分边（和相关顶点）的子集。在有向图中，有向路径是一个顶点序列，其中每个顶点到其后继顶点有一条（有向）边，且没有重复的边。一个有向路径是简单的，如果它没有重复的顶点。一个有向循环是一条有向路径（至少有一条边），其第一个和最后一个顶点相同。如果一个有向循环没有重复的顶点（除了第一个和最后一个顶点的必要重复），那么它是简单的。一条路径或循环的长度是指它的边数。我们说一个顶点 w 是从顶点 v可达的，如果存在一条从 v 到 w 的有向路径。如果两个顶点 v 和 w 是强连通的，那么它们是相互可达的：从 v 到 w 有一条有向路径，从 w 到 v 也有一条有向路径。如果每个顶点到每个其他顶点都有一条有向路径，那么有向图是强连通的。一个非强连通的有向图由一组强连通分量组成，这些分量是最大的强连通子图。一个有向无环图（或 DAG）是一个没有有向循环的有向图。有向图数据类型。我们实现了以下有向图 API。关键方法 adj() 允许客户端代码遍历从给定顶点邻接的顶点。我们使用以下输入文件格式准备测试数据 tinyDG.txt。图的表示。我们使用邻接表表示法，其中我们维护一个以顶点为索引的列表数组，其中包含与每个顶点通过边连接的顶点。 Digraph.java 使用邻接表表示法实现了有向图 API。AdjMatrixDigraph.java 使用邻接矩阵表示法实现了相同的 API。有向图中的可达性。深度优先搜索和广度优先搜索是基本的有向图处理算法。单源可达性: 给定一个有向图和源 s，是否存在一条从 s 到 v 的有向路径？如果是，找到这样的路径。DirectedDFS.java 使用深度优先搜索来解决这个问题。多源可达性: 给定一个有向图和一组源顶点，是否存在一条从集合中的任意顶点到 v 的有向路径？DirectedDFS.java 使用深度优先搜索来解决这个问题。单源有向路径: 给定一个有向图和源 s，是否存在一条从 s 到 v 的有向路径？如果是，找到这样的路径。DepthFirstDirectedPaths.java 使用深度优先搜索来解决这个问题。单源最短有向路径：给定一个有向图和源点s，是否存在从 s 到 v 的有向路径？如果有，找到一条最短的这样的路径。BreadthFirstDirectedPaths.java 使用广度优先搜索来解决这个问题。循环和 DAG。在涉及处理有向图的应用中，有向循环尤为重要。输入文件 tinyDAG.txt 对应于以下 DAG：有向环检测：给定一个有向图，是否存在有向环？如果有，找到这样的环。DirectedCycle.java 使用深度优先搜索来解决这个问题。深度优先顺序：深度优先搜索每个顶点恰好一次。在典型应用中，有三种顶点排序是感兴趣的：前序：在递归调用之前将顶点放入队列。后序：在递归调用后将顶点放入队列。逆后序：在递归调用后将顶点放入栈。 DepthFirstOrder.java 计算这些顺序。拓扑排序：给定一个有向图，按顶点顺序排列，使得所有的有向边都从顺序中较早的顶点指向顺序中较晚的顶点（或报告无法这样做）。Topological.java 使用深度优先搜索来解决这个问题。值得注意的是，在 DAG 中的逆后序提供了一个拓扑顺序。命题。有向图具有拓扑顺序当且仅当它是 DAG。命题。 DAG 中的逆后序是拓扑排序。命题。使用深度优先搜索，我们可以在时间上将 DAG 进行拓扑排序，时间复杂度为 V + E。强连通性。强连通性是顶点集合上的等价关系：自反性：每个顶点 v 与自身强连通。对称性：如果 v 与 w 强连通，则 w 也与 v 强连通。传递性：如果 v 与 w 强连通，且 w 与 x 强连通，则 v 也与 x 强连通。强连通性将顶点划分为等价类，我们简称为强连通分量。我们试图实现以下 API：令人惊讶的是，KosarajuSharirSCC.java 仅通过在 CC.java 中添加几行代码就实现了该 API，如下所示：给定一个有向图 G，使用 DepthFirstOrder.java 来计算其反向图 G^R 的逆后序。在 G 上运行标准 DFS，但考虑刚刚计算的顺序中的未标记顶点，而不是标准的数字顺序。从构造函数中对递归dfs()的调用到达的所有顶点都在一个强连通分量中（！），因此像在 CC 中一样识别它们。命题。 Kosaraju-Sharir 算法使用预处理时间和空间与 V + E 成比例，以支持有向图中的常数时间强连通性查询。传递闭包。有向图 G 的传递闭包是另一个有向图，具有相同的顶点集，但如果且仅当在 G 中从 v 到 w 可达时，有一条从 v 到 w 的边。 TransitiveClosure.java 通过从每个顶点运行深度优先搜索并存储结果来计算有向图的传递闭包。这种解决方案非常适合小型或密集的有向图，但不适用于我们在实践中可能遇到的大型有向图，因为构造函数使用的空间与 V² 成比例，时间与 V (V + E) 成比例。练习为 Digraph 创建一个复制构造函数，该函数以有向图 G 作为输入，并创建和初始化有向图的新副本。客户端对 G 所做的任何更改都不应影响新创建的有向图。有向图在第 591 页上有多少个强连通分量？解决方案: 10. 输入文件是 mediumDG.txt。有向无环图（DAG）的强连通分量是什么？解决方案: 每个顶点都是自己的强连通分量。真或假：有向图的反向的逆后序与有向图的逆后序相同。解决方案: 假。真或假：如果我们修改 Kosaraju-Sharir 算法，在有向图 G 中运行第一个深度优先搜索（而不是反向有向图 G^R），并在 G^R 中运行第二个深度优先搜索（而不是 G），那么它仍然会找到强连通分量。解决方案. 是的，有向图的强连通分量与其反向的强连通分量相同。真或假：如果我们修改 Kosaraju-Sharir 算法，用广度优先搜索替换第二次深度优先搜索，那么它仍然会找到强连通分量。解决方案. 真。计算具有 V 个顶点和 E 条边的Digraph的内存使用情况，根据第 1.4 节的内存成本模型。解决方案. 56 + 40V + 64E。MemoryOfDigraph.java 根据经验计算，假设没有缓存Integer值—Java 通常会缓存-128 到 127 之间的整数。创造性问题有向欧拉回路。有向欧拉回路是一个包含每条边恰好一次的有向循环。编写一个有向图客户端 DirectedEulerianCycle.java 来查找有向欧拉回路或报告不存在这样的回路。提示: 证明一个有向图 G 有一个有向欧拉回路当且仅当 G 中的每个顶点的入度等于出度，并且所有具有非零度的顶点属于同一个强连通分量。强连通分量。描述一个计算包含给定顶点 v 的强连通分量的线性时间算法。基于该算法，描述一个简单的二次时间算法来计算有向图的强连通分量。部分解决方案: 计算包含 s 的强连通分量找到从 s 可达的顶点集找到可以到达 s 的顶点集取两个集合的交集，使用这个作为子程序，你可以在时间比例为 t（E + V）的情况下找到所有强连通分量，其中 t 是强连通分量的数量。 DAG 中的哈密顿路径。给定一个 DAG，设计一个线性时间算法来确定是否存在一个访问每个顶点恰好一次的有向路径。解决方案: 计算一个拓扑排序，并检查拓扑顺序中每对连续顶点之间是否有边。唯一拓扑排序。设计一个算法来确定一个有向图是否有唯一的拓扑排序。提示: 一个有向图有一个唯一的拓扑排序当且仅当拓扑排序中每对连续顶点之间存在一个有向边（即，有向图有一个哈密顿路径）。如果有向图有多个拓扑排序，那么可以通过交换一对连续顶点来获得第二个拓扑排序。 2-可满足性。给定一个布尔公式，其合取范式中有 M 个子句和 N 个文字，每个子句恰好有两个文字，找到一个满足的赋值（如果存在）。解决方案草图: 用 2N 个顶点（每个文字及其否定一个）形成蕴含有向图。对于每个子句 x + y，从 y’到 x 和从 x’到 y 包括边缘。声明：如果没有变量 x 与其否定 x’在同一个强连通分量中，则公式是可满足的。此外，核心 DAG 的拓扑排序（将每个强连通分量缩减为单个顶点）产生一个满足的赋值。基于队列的拓扑排序算法。开发一个非递归的拓扑排序实现 TopologicalX.java，该实现维护一个顶点索引数组，用于跟踪每个顶点的入度。在一次遍历中初始化数组和源队列，就像练习 4.2.7 中那样。然后，执行以下操作，直到源队列为空：从队列中移除一个源并标记它。减少入度数组中与已移除顶点的边的目标顶点对应的条目。如果减少任何条目使其变为 0��则将相应的顶点插入源队列。最短有向循环。给定一个有向图，设计一个算法来找到具有最少边数的有向循环（或报告图是无环的）。你的算法在最坏情况下的运行时间应该与 E V 成正比。应用：给出一组需要肾移植的患者，每个患者都有一个愿意捐赠肾脏但类型不匹配的家庭成员。愿意捐赠给另一个人，前提是他们的家庭成员得到肾脏。然后医院进行“多米诺手术”，所有移植同时进行。解决方案：从每个顶点 s 运行 BFS。通过 s 的最短循环是一条边 v->s，再加上从 s 到 v 的最短路径。ShortestDirectedCycle.java。奇数长度的有向循环。设计一个线性时间算法，以确定一个有向图是否有一个奇数长度的有向循环。解决方案。我们声称，如果一个有向图 G 有一个奇数长度的有向循环，那么它的一个（或多个）强连通分量作为无向图时是非二分的。如果有向图 G 有一个奇数长度的有向循环，则此循环将完全包含在一个强连通分量中。当强连通分量被视为无向图时，奇数长度的有向循环变为奇数长度的循环。回想一下，无向图是二分的当且仅当它没有奇数长度的循环。假设 G 的一个强连通分量是非二分图（当作无向图处理时）。这意味着在强连通分量中存在一个奇数长度的循环 C，忽略方向。如果 C 是一个有向循环，那么我们完成了。否则，如果边 v->w 指向“错误”的方向，我们可以用指向相反方向的奇数长度路径替换它（这保留了循环中边数的奇偶性）。要了解如何做到这一点，请注意存在一条从 w 到 v 的有向路径 P，因为 v 和 w 在同一个强连通分量中。如果 P 的长度为奇数，则我们用 P 替换边 v->w；如果 P 的长度为偶数，则这条路径 P 与 v->w 组合在一起就是一个奇数长度的循环。 DAG 中可达的顶点。设计一个线性时间算法，以确定一个 DAG 是否有一个顶点可以从每个其他顶点到达。解决方案。计算每个顶点的出度。如果 DAG 有一个出度为 0 的顶点 v，那么它可以从每个其他顶点到达。有向图中可达的顶点。设计一个线性时间算法，以确定有向图是否有一个顶点可以从每个其他顶点到达。解决方案。计算强连通分量和核 DAG。将练习 4.2.37 应用于核 DAG。网络爬虫。编写一个程序 WebCrawler.java，使用广度优先搜索来爬取网络有向图，从给定的网页开始。不要显式构建网络有向图。网络练习符号有向图。修改 SymbolGraph.java 以创建一个实现符号有向图的程序 SymbolDigraph.java。组合电路。给定输入，确定组合电路的真值是一个图可达性问题（在有向无环图上）。权限提升。如果 A 可以获得 B 的权限，则在用户类 A 到用户类 B 之间包含一个数组。找出所有可以在 Windows 中获得管理员访问权限的用户。 Unix 程序 tsort。跳棋。将跳棋规则扩展到一个 N×N 的跳棋棋盘。展示如何确定一个跳棋在当前移动中是否可以变成国王。（使用 BFS 或 DFS。）展示如何确定黑方是否有获胜的着法。（找到一个有向欧拉路径。）优先附着模型。网络具有无标度特性，并遵循幂律。新页面倾向于优先附着到受欢迎的页面上。从指向自身的单个页面开始。每一步中，一个新页面出现，出度为 1。以概率 p，页面指向一个随机页面；以概率(1-p)，页面指向一个现有页面，概率与页面的入度成比例。子类型检查。给定单继承关系（一棵树），检查 v 是否是 w 的祖先。提示：v 是 w 的祖先当且仅当 pre[v] <= pre[w]且 post[v] >= post[w]。子类型检查。重复上一个问题，但使用有向无环图而不是树。有根树的 LCA。给定一个有根树和两个顶点 v 和 w，找到顶点 v 和 w 的最低共同祖先（lca）。顶点 v 和 w 的 lca 是离根最远的共同祖先。根树上最基本的问题之一。可以在 O(1)的查询时间内解决，预处理时间为线性时间（Harel-Tarjan，Bender-Coloton）。找到一个有向无环图，其中最短的祖先路径通向一个不是 LCA 的共同祖先 x。九个字母的单词。找到一个九个字母的英文单词，使得在适当的顺序中依次删除每个字母后仍然是一个英文单词。使用单词和顶点构建一个有向图，如果一个单词可以通过添加一个字母形成另一个单词，则在两个单词之间添加一条边。答案：一个解决方案是 startling -> starting -> staring -> string -> sting -> sing -> sin -> in -> i。电子表格重新计算。希望没有循环依赖。使用公式单元格图的拓扑排序来确定更新单元格的顺序。嵌套箱子。一个维度为 d 的箱子，其尺寸为(a1, a2, …, ad)，如果第二个箱子的坐标可以重新排列，使得 a1 < b1, a2 < b2, …, ad < bd，则该箱子嵌套在第二个箱子内。给出一个有效的算法，用于确定一个 d 维箱子嵌套在另一个箱子内的位置。提示：排序。证明嵌套是传递的：如果箱子 i 嵌套在箱子 j 内部，箱子 j 又嵌套在箱子 k 内部，那么箱子 i 也嵌套在箱子 k 内部。给定一组 n 个 d 维箱子，给出一个有效的算法，找到可以同时嵌套最多箱子的方法。提示：创建一个有向图，如果箱子 i 嵌套在箱子 j 内部，则从箱子 i 到箱子 j 添加一条边。然后运行拓扑排序。 Warshall 的传递闭包算法。 WarshallTC.java 算法适用于稠密图。依赖于 AdjMatrixDigraph.java。暴力强连通分量算法。 BruteSCC.java 通过首先计算传递闭包来计算强连通分量。时间复杂度为 O(EV)，空间复杂度为 O(V²)。 Tarjan 的强连通分量算法。 TarjanSCC.java 实现了 Tarjan 算法来计算强连通分量。 Gabow 的强连通分量算法。 GabowSCC.java 实现了 Gabow 算法来计算强连通分量。有向图生成器。 DigraphGenerator.java 生成各种有向图。有限马尔可夫链. 回归状态：一旦在状态开始，马尔可夫链将以概率 1 返回。瞬时状态：有些概率它永远不会返回（某个节点 j，i 可以到达 j，但 j 无法到达 i）。不可约马尔可夫链=所有状态都是回归的。马尔可夫链是不可约的当且仅当它是强连通的。回归组件是核 DAG 中没有离开边的组件。马尔可夫链中的通信类是强连通分量。定理. 如果 G 是强连通的，则存在唯一的稳态分布 pi。此外，对于所有 v，pi(v) > 0。定理. 如果 G 的核 DAG 具有单个没有离开边的超节点，则存在唯一的稳态分布 pi。此外，对于所有回归的 v，pi(v) > 0 且对于所有瞬时的 v，pi(v) = 0。后代引理. [R. E. Tarjan] 将 pre[v]和 post[v]分别表示为 v 的前序和后序编号，nd[v]表示 v 的后代数（包括 v）。证明以下四个条件是等价的。顶点 v 是顶点 w 的祖先。 pre[v] <= pre[w] < pre[v] + nd(v). post[v] - nd [v] < post[w] <= post[v] pre[v] <= pre[w]且 post[v] >= post[w]（嵌套引理）边引理. [R. E. Tarjan] 证明边(v, w)是以下四种之一： w 是 v 的子节点：(v, w)是一条树边。 w 是 v 的后代但不是子节点：(v, w)是一条前向边。 w 是 v 的祖先：(v, w)是一条后向边 w 和 v 无关且 pre[v] > pre[w]：(v, w)是一条交叉边。路径引理. [R. E. Tarjan] 证明从 v 到 w 的任何路径，其中 pre[v] < pre[w]，都包含 v 和 w 的共同祖先。证明如果(v, w)是一条边且 pre[v] < pre[w]，则 v 是 DFS 树中 w 的祖先。后序引理. [R. E. Tarjan] 证明如果 P 是一条路径，最后一个顶点 x 在后序中最高，则路径上的每个顶点都是 x 的后代（因此与 x 有一条路径）。解. 证明通过对 P 的长度进行归纳（或通过反证法）。设(v, w)是一条边，其中 w 是 x 的后代且 post[v] < post[x]。由于 w 是 x 的后代，我们有 pre[w] >= pre[x]。如果 pre[v] >= pre[x]，那么 v 是 x 的后代（通过嵌套引理）。如果 pre[v] < pre[x]，那么 pre[v] < pre[w]，这意味着（通过前一个练习）v 是 w 的祖先，因此与 x 有关。但是 post[v] < post[x]意味着 v 是 x 的后代。前拓扑排序. 设计一个线性时间算法来找到一个前拓扑排序：一种顶点的排序，使得如果从 v 到 w 有一条路径且 w 在排序中出现在 v 之前，则从 w 到 v 也必须有一条路径。提示：反向后序是一种前拓扑排序。这是 Kosaraju-Sharir 算法正确性证明的关键。 Wordnet.使用 WordNet 测量形容词的语义取向. 垃圾收集. 在像 Java 这样的语言中进行自动内存管理是一个具有挑战性的问题。分配内存很容易，但发现程序何时完成对内存的使用（并回收它）更加困难。引用计数：不适用于循环链接结构。标记-清除算法。根=局部变量和静态变量。从根运行 DFS，标记所有从根引用的变量，依此类推。然后，进行第二遍：释放所有未标记的对象并取消标记所有标记的对象。或者复制垃圾收集器将所有标记的对象移动到单个内存区域。每个对象使用一个额外的位。JVM 在进行垃圾收集时必须暂停。碎片化内存。应用：C 泄漏检测器（泄漏=不可达的，未释放的内存）。有向循环检测应用。应用：检查非法继承循环，检查死锁。目录是文件和其他目录的列表。符号链接是对另一个目录的引用。在列出目录中的所有文件时，需要小心避免跟随符号链接的循环！拓扑排序应用。应用：课程先修条件、大型计算机程序组件的编译顺序、因果关系、类继承、死锁检测、时间依赖性、计算作业的管道、检查符号链接循环、电子表格中的公式求值。强连通分量应用。应用于 CAD、马尔可夫链（不可约）、蜘蛛陷阱和网络搜索、指针分析、垃圾回收。单向街定理。实现一个算法来定向无向图中的边，使其成为强连通图。罗宾斯定理断言，当且仅当无向图是双边连通的（没有桥）时，这是可能的。在这种情况下，一种解决方案是运行深度优先搜索（DFS），并将 DFS 树中的所有边定向远离根节点，将所有剩余的边定向朝向根节点。定向混合图中的边以使其无环。混合图是具有一些有向边和一些无向边的图。设计一个线性时间算法来确定是否可以定向无向边，使得结果有向图是无环的。应用：老城区的狭窄道路希望使每条道路单向通行，但仍允许城市中的每个交叉口可从其他城市到达。定向混合图中的边以形成有向循环。混合图是具有一些有向边和一些无向边的图。设计一个线性时间算法来确定是否可以定向无向边，使得结果有向图具有有向循环。应用：确定最大流是否唯一。解决方案：一个算法。后序引理变种。设 S 和 T 是有向图 G 中的两个强连通分量。证明如果存在一条从 S 中的一个顶点到 T 中的一个顶点的边 e，则 S 中顶点的最高后序编号高于 T 中顶点的最高后序编号。 DAG 中路径的数量。给定一个有向无环图（DAG）和两个特定顶点 s 和 t，设计一个线性时间算法来计算从 s 到 t 的有向路径数量。提示：拓扑排序。 DAG 中长度为 L 的路径。给定一个有向无环图（DAG）和两个特定顶点 s 和 t，设计一个算法来确定是否存在一条从 s 到 t 的路径，其中恰好包含 L 条边。核心顶点。给定一个有向图 G，如果从顶点 v 可以到达 G 中的每个顶点，则顶点 v 是一个核心顶点。设计一个线性时间算法来找到所有核心顶点。提示：创建 G 的强连通分量并查看核心 DAG。强连通分量和二分图匹配。给定一个二分图 G，一个未匹配边是指不出现在任何完美匹配中的边。设计一个算法来找到所有未匹配边。提示：证明以下算法可以胜任。在 G 中找到一个完美匹配；将匹配中的边从双分区的一侧定向到另一侧；将剩余的边定向到相反方向；在不在完美匹配中的边中，返回那些端点在不同强连通分量中的边。有向图的传递闭包。有向图的传递闭包是具有与原始有向图相同传递闭包的边数最少的有向图。设计一个 V（E + V）算法来计算有向图的传递闭包。请注意，有向图中的传递闭包不一定是唯一的，也不一定是原始有向图的子图。（有向无环图中的传递闭包是唯一的且是原始有向图的子图。）奇长度路径。给定一个有向图 G 和一个源顶点 s，设计一个线性时间算法，确定通过具有奇数边数的路径（不一定简单）从 s 可达的所有顶点。解决方案：为 G 中的每个顶点 v 创建一个新的有向图 G’，其中包含两个顶点 v 和 v’。对于 G 中的每条边 v->w，包括两条边：v->w’和 w->v’。现在，在 G’中从 s 到 v’的任何路径对应于 G 中从 s 到 v 的奇长度路径。运行 BFS 或 DFS 以确定从 s 可达的顶点。找到一个有向无环图（DAG）的拓扑排序，无论深度优先搜索（DFS）以何种顺序选择起始顶点，都无法计算为 DFS 的逆后序。展示出 DAG 的每一个拓扑排序都可以被计算为 DFS 的逆后序，只要 DFS 可以任意选择构造函数中起始顶点的顺序。非递归 DFS。编写一个��序 NonrecursiveDirectedDFS.java，使用显式栈而不是递归来实现深度优先搜索。编写一个程序 NonrecursiveDirectedCycle.java，在不使用递归的情况下找到一个有向环。非递归拓扑排序。将基于队列的拓扑排序算法 TopologicalX.java 从练习 4.2.39 扩展到在有向图存在有向环时找到该有向环。将程序命名为 DirectedCycle.java。 Cartalk 难题。在字典中找到一个具有以下特性的最长单词：您可以一次删除一个字母（从任一端或中间），结果字符串也是字典中的单词。例如，STRING 是一个具有此特性的 6 字母单词（STRING -> STING -> SING -> SIN -> IN -> I）。逆后序与前序。真或假：有向图的逆后序与有向图的前序相同。 Kosaraju–Sharir 算法中的逆后序与前序。假设您在 Kosaraju–Sharir 算法中使用有向图的前序而不是逆后序。它是否仍会产生强连通分量？答案：不会，运行 KosarajuSharirPreorderSCC.java 在tinyDG.txt上。 4.3 最小生成树原文：algs4.cs.princeton.edu/43mst 译者：飞龙协议：CC BY-NC-SA 4.0 最小生成树。带权重的图是一种我们为每条边关联权重或成本的图。带权重图的最小生成树（MST）是其边权重之和不大于任何其他生成树的生成树。假设。为了简化演示，我们采用以下约定：图是连通的。我们定义的生成树条件意味着图必须是连通的才能存在 MST。如果图不连通，我们可以调整算法以计算其每个连通分量的 MST，统称为最小生成森林。边的权重不一定是距离。几何直觉有时是有益的，但边的权重可以是任意的。边的权重可能为零或负数。如果边的权重都是正数，则定义最小生成树为连接所有顶点的总权重最小的子图即可。边的权重都不同。如果边可以具有相同的权重，则最小生成树可能不唯一。做出这种假设简化了我们一些证明，但我们的所有算法即使在存在相同权重的情况下也能正常工作。基本原理。我们回顾树的两个定义性质：添加连接树中两个顶点的边会创建一个唯一的循环。从树中移除一条边会将其分成两个独立的子树。图的切割是将其顶点划分为两个不相交集合。跨越边是连接一个集合中的顶点与另一个集合中的顶点的边。我们假设为简单起见，所有边的权重都是不同的。在此假设下，MST 是唯一的。定义切割和循环。以下性质导致多种 MST 算法。命题。（切割性质）在带权重图中的任何切割中（所有边权重不同），最小权重的跨越边在图的 MST 中。切割性质是��们考虑 MST 问题的算法的基础。具体来说，它们是贪心算法的特例。命题。（贪心 MST 算法）以下方法将所有连接的带权重图的 MST 中的所有边涂黑：从所有边都涂灰色开始，找到没有黑色边的切割，将其最小权重的边涂黑，继续直到涂黑 V-1 条边。最小生成树问题](…/Images/fce4a44e5b52cd8391fb6ea99f7fa182.png) 带权重图数据类型。我们使用以下 API 表示带权重的边： either() 和 other() 方法用于访问边的顶点；compareTo() 方法通过权重比较边。Edge.java 是一个直接的实现。我们使用以下 API 表示带权重的图：我们允许平行边和自环。EdgeWeightedGraph.java 使用邻接表表示法实现 API。 MST API. 我们使用以下 API 计算带权重图的最小生成树：我们准备了一些测试数据： tinyEWG.txt 包含 8 个顶点和 16 条边 mediumEWG.txt 包含 250 个顶点和 1,273 条边 1000EWG.txt 包含 1,000 个顶点和 8,433 条边 10000EWG.txt 包含 10,000 个顶点和 61,731 条边 largeEWG.txt 包含一百万个顶点和 7,586,063 条边 Prim 算法。 Prim 算法通过在每一步将新边附加到单个增长树上来工作：从任何顶点开始作为单个顶点树；然后向其添加 V-1 条边，始终取下一个（着色为黑色）连接树上顶点与尚未在树上的顶点的最小权重边（对于由树顶点定义的切割的跨越边）。 Prim 算法的一句描述留下了一个关键问题：我们如何（高效地）找到最小权重的跨越边？懒惰实现. 我们使用优先队列来保存跨越边并找到最小权重的边。每次我们将一条边添加到树中时，我们也将一个顶点添加到树中。为了维护跨越边的集合，我们需要将从该顶点到任何非树顶点的所有边添加到优先队列中。但我们必须做更多的事情：连接刚刚添加的顶点到已经在优先队列中的树顶点的任何边现在变得不合格（它不再是跨越边，因为它连接了两个树顶点）。懒惰实现将这样的边留在优先队列中，推迟不合格测试到我们删除它们时。 LazyPrimMST.java 是这种懒惰方法的实现。它依赖于 MinPQ.java 优先队列。急切实现. 为了改进 Prim 算法的懒惰实现，我们可以尝试从优先队列中删除不合格的边，以便优先队列只包含跨越边。但我们可以消除更多的边。关键在于注意到我们唯一感兴趣的是从每个非树顶点到树顶点的最小边。当我们将顶点 v 添加到树中时，与每个非树顶点 w 相关的唯一可能变化是，添加 v 使 w 比以前更接近树。简而言之，我们不需要在优先队列中保留所有从 w 到树顶点的边 - 我们只需要跟踪最小权重的边，并检查是否添加 v 到树中需要我们更新该最小值（因为边 v-w 的权重更低），我们可以在处理 s 邻接列表中的每条边时做到这一点。换句话说，我们只保留优先队列中的一条边用于每个非树顶点：连接它与树的最短边。 PrimMST.java 是这种急切方法的实现。它依赖于 IndexMinPQ.java 索引优先队列来执行减少键操作。命题。 Prim 算法计算任何连通的边权重图的最小生成树。Prim 算法的懒惰版本使用空间与 E 成比例，时间与 E log E 成比例（在最坏情况下）来计算具有 E 条边和 V 个顶点的连通边权重图的最小生成树；急切版本使用空间与 V 成比例，时间与 E log V 成比例（在最坏情况下）。 Kruskal 算法。 Kruskal 算法按照它们的权重值（从小到大）的顺序处理边，每次添加不与先前添加的边形成循环的边作为 MST（着色为黑色），在添加 V-1 条边后停止。黑色边形成逐渐演变为单一树 MST 的树林。要实现 Kruskal 算法，我们使用优先队列按权重顺序考虑边，使用并查集数据结构标识导致循环的边，使用队列收集最小生成树边。程序 KruskalMST.java 按照这些方式实现了 Kruskal 算法。它使用了辅助的 MinPQ.java、UF.java 和 Queue.java 数据类型。命题。 Kruskal 算法使用额外空间与 E 成正比，时间与 E log E 成正比（在最坏情况下）来计算具有 E 条边和 V 个顶点的任何连通边权图的最小生成树。练习证明，通过给所有权重加上一个正常数或将它们全部乘以一个正常数，不会影响最小生成树。解决方案. Kruskal 算法只通过 compareTo() 方法访问边权重。给每个权重添加一个正常数（或乘以一个正常数）不会改变 compareTo() 方法的结果。证明，如果一个图的边都有不同的权重，那么最小生成树是唯一的。解决方案. 为了推导矛盾，假设图 G 有两个不同的最小生成树，称为 T1 和 T2。设 e = v-w 是 G 中在 T1 或 T2 中的最小权重边，但不在两者中都存在。假设 e 在 T1 中。将 e 添加到 T2 中会创建一个循环 C。C 中至少有一条边，假设为 f，不在 T1 中（否则 T1 就是循环的）。根据我们选择的 e，w(e) ≤ w(f)。由于所有边的权重都不同，w(e) < w(f)。现在，在 T2 中用 e 替换 f 会得到一棵权重小于 T2 的新生成树（与 T2 的最小性相矛盾）。如何找到边权图的最大生成树？解决方案. 反转每条边的权重（或在 compareTo() 方法中反转比较的意义）。为 EdgeWeightedGraph.java 实现从输入流读取边权图的构造函数。确定 EdgeWeightedGraph.java 用于表示具有 V 个顶点和 E 条边的图所使用的内存量，使用第 1.4 节的内存成本模型。解决方案. 56 + 40V + 112E。MemoryOfEdgeWeightedGraph.java 通过假设没有缓存 Integer 值来进行经验计算—Java 通常会缓存 -128 到 127 的整数。给定边权图 G 的最小生成树，假设删除一个不会使 G 断开的边。描述如何在与 E 成正比的时间内找到新图的最小生成树。解决方案. 如果边不在最小生成树中，则旧的最小生成树是更新后图的最小生成树。否则，从最小生成树中删除边会留下两个连通分量。添加一个顶点在每个连通分量中的最小权重边。给定边权图 G 的最小生成树和一个新边 e，描述如何在与 V 成正比的时间内找到新图的最小生成树。解决方案. 将边 e 添加到最小生成树会创建一个唯一的循环。删除此循环上的最大权重边。为 EdgeWeightedGraph.java 实现 toString()。假设你实现了 Prim 算法的急切版本，但是不使用优先队列来找到下一个要添加到树中的顶点，而是扫描 distTo[] 数组中的所有 V 个条目，找到具有最小值的非树顶点。在具有 V 个顶点和 E 条边的图上，Prim 算法的急切版本的最坏情况运行时间的增长顺序是多少？如果有的话，什么时候这种方法是合适的？为什么？请解释你的答案。解决方案. Prim 算法的运行时间将与 V² 成正比，这对于稠密图是最佳的。为 PrimMST.java 实现 edges()。创意问题最小生成森林。开发 Prim 和 Kruskal 算法的版本，计算不一定连通的边权图的最小生成森林。解决方案。 PrimMST.java 和 KruskalMST.java 实现了这一点。认证。编写一个名为check()的方法，使用以下割优化条件来验证提议的边集是否实际上是最小生成树（MST）：如果一组边是一棵生成树，并且每条边都是通过从树中移除该边定义的割的最小权重边，则这组边就是 MST。你的方法的运行时间增长率是多少？解决方案。 KruskalMST.java。实验 Boruvka 算法。开发 Boruvka 算法的实现 BoruvkaMST.java：通过将边添加到不断增长的树森林中来构建 MST，类似于 Kruskal 算法，但是分阶段进行。在每个阶段，找到将每棵树连接到另一棵树的最小权重边，然后将所有这样的边添加到 MST 中。假设边的权重都不同，以避免循环。提示：维护一个顶点索引数组，以标识连接每个组件到其最近邻居的边，并使用并查集数据结构。备注。由于每个阶段树的数量至少减少一半，所以最多有 log V 个阶段。这种方法高效且可以并行运行。网页练习最小瓶颈生成树。图 G 的最小瓶颈生成树是 G 的一棵生成树，使得生成树中任意边的最大权重最小化。设计一个算法来找到最小瓶颈生成树。解决方案。每个 MST 都是最小瓶颈生成树（但不一定反之）。这可以通过割性质来证明。最小中位数生成树。图 G 的最小中位数生成树是 G 的一棵生成树，使得其权重的中位数最小化。设计一个高效的算法来找到最小中位数生成树。解决方案。每个 MST 都是最小中位数生成树（但不一定反之）。迷宫生成。使用随机化的 Kruskal 或 Prim 算法创建迷宫。唯一 MST。设计一个算法来确定给定图 G 的 MST 是否唯一。随机生成树。给定图 G，均匀随机生成 G 的一棵生成树。使用 Aldous 和 Broder 的以下显著定理：从任意顶点 s 开始，并进行随机游走，直到每个顶点都被访问过（在所有相邻边中均匀随机选择一条出边）。如果一个顶点以前从未被访问过，则将边添加到该顶点以形成生成树 T。那么 T 是图 G 的均匀随机生成树。预期的运行时间受限于 G 的覆盖时间，最多与 EV 成比例。最小权重反馈边集。图的反馈边集是包含图中每个循环中至少一条边的子集。如果删除反馈边集的边，则结果图将是无环的。设计一个高效的算法，在具有正边权的加��图中找到最小权重的反馈边集。两个 MST 中边权重的分布。假设加权有向图有两个 MST T1 和 T2。证明如果 T1 有权重为 w 的 k 条边，则 T2 也有权重为 w 的 k 条边。美国计算奥林匹克问题。在一个城市中有 N 栋房子，每栋房子都需要供水。在第 i 栋房子建造井的成本为 w[i]美元，在第 i 和第 j 栋房子之间建造管道的成本为 c[i][j]。如果一栋房子建有井或者有一条管道路径通向有井的房子，那么这栋房子就可以接收水。设计一个算法来找到供应每栋房子所需的最小金额。解决方案.: 创建一个带有 N+1 个顶点的边权图（每个房子一个顶点加上一个源顶点 x）。包括每对顶点 i 和 j 之间的成本 c[i][j] 的边（表示潜在的管道）。包括源 s 和每个房子 i 之间成本为 w[i] 的边（表示潜在的开放井）。在这个边权图中找到一个最小生成树。恰好有 k 条橙色边的生成树。给定一个边缘着色为橙色或黑色的图，设计一个线性对数算法来找到一个包含恰好 k 条橙色边的生成树（或报告不存在这样的生成树）。最小方差生成树。给定一个连通的边权重图，找到一个最小生成树，使其边权重的方差最小化。 4.4 最短路径原文：algs4.cs.princeton.edu/44sp 译者：飞龙协议：CC BY-NC-SA 4.0 最短路径。加权有向图是一个有向图，其中我们为每条边关联权重或成本。从顶点 s 到顶点 t 的最短路径是从 s 到 t 的有向路径，具有没有更低权重的其他路径的属性。属性。我们总结了几个重要的属性和假设。路径是有方向的。最短路径必须遵守其边的方向。权重不一定是距离。几何直觉可能有所帮助，但边的权重可能代表时间或成本。并非所有顶点都需要可达。如果 t 从 s 不可达，则根本没有路径，因此从 s 到 t 的最短路径也不存在。负权重引入了复杂性。目前，我们假设边的权重是正数（或零）。最短路径通常是简单的。我们的算法忽略形成循环的零权重边，因此它们找到的最短路径没有循环。最短路径不一定是唯一的。从一个顶点到另一个顶点可能有多条最低权重的路径；我们满足于找到其中任何一条。并行边和自环可能存在。在文本中，我们假设不存在并行边，并使用符号 v->w 来表示从 v 到 w 的边，但我们的代码可以轻松处理它们。加权有向图数据类型。我们使用以下 API 表示加权边： from()和to()方法对于访问边的顶点很有用。DirectedEdge.java 实现了这个 API。我们使用以下 API 表示加权有向图： EdgeWeightedDigraph.java 使用邻接表表示实现了该 API。最短路径 API。我们使用以下 API 计算加权有向图的最短路径：我们准备了一些测试数据： tinyEWD.txt 包含 8 个顶点和 15 条边 mediumEWD.txt 包含 250 个顶点和 2,546 条边 1000EWG.txt 包含 1,000 个顶点和 16,866 条边 10000EWG.txt 包含 10,000 个顶点和 123,462 条边 largeEWG.txt 包含一百万个顶点和 15,172,126 条边。单源最短路径的数据结构。给定一个加权有向图和一个指定的顶点 s，最短路径树（SPT）是一个子图，包含 s 和所有从 s 可达的顶点，形成以 s 为根的有向树，使得每条树路径都是图中的最短路径。我们用两个顶点索引数组表示最短路径：最短路径树上的边：edgeTo[v]是从 s 到 v 的最短路径上的最后一条边。到源的距离：distTo[v]是从 s 到 v 的最短路径的长度。松弛。我们的最短路径实现基于一种称为松弛的操作。我们将distTo[s]初始化为 0，对于所有其他顶点 v，将distTo[v]初始化为无穷大。边松弛。对边 v->w 进行松弛意味着测试从 s 到 w 的已知最佳路径是否是从 s 到 v，然后沿着从 v 到 w 的边，如果是，则更新我们的数据结构。 java private void relax(DirectedEdge e) { int v = e.from(), w = e.to(); if (distTo[w] > distTo[v] + e.weight()) { distTo[w] = distTo[v] + e.weight(); edgeTo[w] = e; } } AI写代码 java 运行 1 2 3 4 5 6 7 8 顶点松弛。我们所有的实现实际上都会松弛从给定顶点指向的所有边。 java private void relax(EdgeWeightedDigraph G, int v) { for (DirectedEdge e : G.adj(v)) { int w = e.to(); if (distTo[w] > distTo[v] + e.weight()) { distTo[w] = distTo[v] + e.weight(); edgeTo[w] = e; } } } AI写代码 java 运行 1 2 3 4 5 6 7 8 9 10 迪杰斯特拉算法。戴克斯特拉算法将dist[s]初始化为 0，将所有其他distTo[]条目初始化为正无穷。然后，它重复地放松并将具有最低distTo[]值的非树顶点添加到树中，继续直到所有顶点都在树上或没有非树顶点具有有限的distTo[]值。 DijkstraSP.java 是戴克斯特拉算法的高效实现。它使用 IndexMinPQ.java 作为优先队列。命题。戴克斯特拉算法使用额外空间与 V 成正比，时间与 E log V 成正比（在最坏情况下）解决了带非负权重的带权有向图中的单源最短路径问题。无环带权有向图。我们使用术语带权有向无环图来指代无环带权有向图。带权有向无环图中的单源最短路径问题。我们现在考虑一种用于查找最短路径的算法，对于带权有向无环图而言，它比戴克斯特拉算法更简单且更快。它在线性时间内解决了单源问题。它处理负边权重。它解决了相关问题，如查找最长路径。该算法将顶点放松与拓扑排序结合起来。我们将distTo[s]初始化为 0，将所有其他distTo[]值初始化为无穷大，然后按照拓扑顺序放松顶点。AcyclicSP.java 是这种方法的实现。它依赖于这个版本的 Topological.java，扩展以支持带权有向图。带权有向无环图中的单源最长路径问题。我��可以通过将distTo[]值初始化为负无穷大并在relax()中改变不等式的意义来解决带权有向无环图中的单源最长路径问题。AcyclicLP.java 实现了这种方法。关键路径法。我们考虑并行的有前置约束的作业调度问题：给定一组指定持续时间的作业，其中有前置约束规定某些作业必须在某些其他作业开始之前完成，我们如何在相同数量的处理器上安排这些作业，以便它们在最短的时间内完成，同时仍然遵守约束条件？通过将问题制定为带权有向无环图中的最长路径问题，可以解决此问题：创建一个带权有向无环图，其中包含一个源 s，一个汇 t，以及每个作业的两个顶点（一个起始顶点和一个结束顶点）。对于每个作业，从其起始顶点到其结束顶点添加一条权重等于其持续时间的边。对于每个前置约束 v->w，从对应于 v 的结束顶点到对应于 w 的开始顶点添加一条零权重边。还从源到每个作业的起始顶点和从每个作业的结束顶点到汇添加零权重边。现在，根据从源到达的最长路径的长度安排每个作业的时间。 CPM.java 是关键路径法的实现。命题。通过按拓扑顺序放松顶点，我们可以在时间复杂度为 E + V 的情况下解决带权有向无环图的单源最短路径和最长路径问题。一般带权有向图中的最短路径。如果(i)所有权重为非负或(ii)没有循环，则可以解决最短路径问题。负循环。负循环是一个总权重为负的有向循环。如果存在负循环，则最短路径的概念是没有意义的。因此，我们考虑没有负循环的加权有向图。贝尔曼-福特算法。将distTo[s]初始化为 0，将所有其他distTo[]值初始化为无穷大。然后，以任意顺序考虑有向图的边，并放松所有边。进行 V 次这样的遍历。 java for (int pass = 0; pass < G.V(); pass++) for (int v = 0; v < G.V(); v++) for (DirectedEdge e : G.adj(v)) relax(e); AI写代码 java 运行 1 2 3 4 5 我们不详细考虑这个版本，因为它总是放松 V E 条边。基于队列的贝尔曼-福特算法。可能导致distTo[]变化的唯一边是那些离开上一轮中distTo[]值发生变化的顶点的边。为了跟踪这样的顶点，我们使用一个 FIFO 队列。BellmanFordSP.java 通过维护两个额外的数据结构来实现这种方法：一个要放松的顶点队列一个顶点索引的布尔数组onQ[]，指示哪些顶点在队列上，以避免重复负循环检测。在许多应用中，我们的目标是检查并提取负循环。因此，我们向 API 添加以下方法：当且仅当在所有边的第 V 次遍历后队列非空时，从源可达负循环。此外，我们edgeTo[]数组中的边子图必须包含一个负循环。因此，为了实现negativeCycle()，BellmanFordSP.java 从edgeTo[]中的边构建一个加权有向图，并在该图中查找循环。为了找到循环，它使用 EdgeWeightedDirectedCycle.java，这是第 4.3 节中 DirectedCycle.java 的一个版本，适用于加权有向图。我们通过仅在每次第 V 次边放松后执行此检查来分摊此检查的成本。套汇检测。考虑一个基于商品交易的金融交易市场。rates.txt 中的表显示了货币之间的转换率。文件的第一行是货币 V 的数量；然后文件每行给出货币的名称，然后是转换为其他货币的汇率。套汇机会是一个有向循环，使得交换率的乘积大于 1。例如，我们的表格显示，1000 美元可以购买 1000.00 × .741 = 741 欧元，然后我们可以用我们的欧元购买 741 × 1.366 = 1,012.206 加拿大元，最后，我们可以用我们的加拿大元购买 1,012.206 × .995 = 1,007.14497 美元，获得 7.14497 美元的利润！ \| \| \| \| --- 为了将套汇问题制定为负循环检测问题，将每个权重替换为其对数的负值。通过这种改变，在原问题中通过乘以边权重来计算路径权重对应于在转换后的问题中将它们相加。Arbitrage.java 通过解决相应的负循环检测问题来识别货币兑换网络中的套汇机会。命题。在加权有向图中，从 s 到 v 存在最短路径当且仅当从 s 到 v 存在至少一条有向路径，并且从 s 到 v 的任何有向路径上的顶点都不在负循环上。命题。贝尔曼-福特算法解决了给定源 s 的单源最短路径问题（或找到从 s 可达的负循环）对于具有 V 个顶点和 E 条边的任意加权有向图，在最坏情况下，时间复杂度为 E V，额外空间复杂度为 V。问与答 Q. Dijkstra 算法能处理负权重吗？ A. 是和否。有两种已知的最短路径算法称为 Dijkstra 算法，取决于一个顶点是否可以多次入队到优先队列。当权重为非负时，这两个版本是相同的（因为没有顶点会多次入队）。DijkstraSP.java 中实现的版本（允许一个顶点多次入队）在存在负边权（但没有负环）时是正确的，但其最坏情况下的运行时间是指数级的。（我们注意到 DijkstraSP.java 如果边权重为负数，则会抛出异常，以便程序员不会对这种指数级行为感到惊讶。）如果我们修改 DijkstraSP.java 以使一个顶点不能多次入队（例如，使用marked[]数组标记那些已经被松弛的顶点），那么算法保证在 E log V 时间内运行，但当存在负权边时可能产生错误结果。练习真或假。将每个边权重增加一个常数不会改变单源最短路径问题的解决方案。解决方案。假。为 EdgeWeightedDigraph.java 提供toString()的实现。使用第 1.4 节的内存成本模型确定 EdgeWeightedDigraph.java 用于表示具有 V 个顶点和 E 条边的图所使用的内存量。解决方案。 56 + 40V + 72E。MemoryOfEdgeWeightedDigraph.java 根据经验计算，假设没有缓存Integer值 - Java 通常缓存-128 到 127 的整数。从第 4.2 节中的DirectedCycle和Topological类中使用本节的EdgeweightedDigraph和DirectedEdgeAPI，从而实现 EdgeWeightedDirectedCycle.java 和 Topological.java。假设我们通过为EdgeWeightedGraph中的每个Edge创建两个DirectedEdge对象（分别在每个方向上）来将EdgeWeightedGraph转换为EdgeWeightedDigraph，然后使用贝尔曼-福特算法。解释为什么这种方法会失败得惊人。解决方案：即使带权图不包含负权重环，这可能会引入负成本循环。如果在贝尔曼-福特算法的同一遍历中允许一个顶点被多次入队会发生什么？答案：算法的运行时间可能呈指数增长。例如，考虑所有边权重均为-1 的完全带权有向图会发生什么。创意问题有向无环图中的最长路径。开发一个实现 AcyclicLP.java 的程序，可以解决带权有向无环图中的最长路径问题。线上的所有对最短路径。给定一个加权线图（无向连通图，所有顶点的度为 2，除了两个端点的度为 1），设计一个算法，在线性时间内预处理图，并能在常数时间内返回任意两个顶点之间最短路径的距离。部分解决方案。找到一个度为 1 的顶点 s，并运行广度优先（或深度优先）搜索以找到其余顶点出现的顺序。然后，计算从 s 到每个顶点 v 的最短路径长度，称为dist[v]。顶点 v 和 w 之间的最短路径是\|dist[v] - dist[w]\|。单调最短路径。给定一个带权有向图，找到从 s 到每个其他顶点的单调最短路径。如果路径上每条边的权重要么严格递增要么严格递减，则路径是单调的。部分解决方案：按升序松弛边并找到最佳路径；然后按降序松弛边并找到最佳路径。 Dijkstra 算法的懒惰实现。开发一个实现 LazyDijkstraSP.java 的 Dijkstra 算法的懒惰版本，该版本在文本中有描述。 Bellman-Ford 队列永不为空。证明如果在基于队列的 Bellman-Ford 算法中从源可达到一个负循环，那么队列永远不会为空。解决方案：考虑一个负循环，并假设对于循环 W 上的所有边，distTo[w] <= distTo[v] + length(v, w)。对循环上的所有边进行这个不等式求和意味着循环的长度是非负的。 Bellman-Ford 负循环检测。证明如果在通用 Bellman-Ford 算法的第 V 次遍历中有任何边被松弛，那么edgeTo[]数组中就有一个有向循环，并且任何这样的循环都是负循环。解决方案：待定。网络练习最优子结构性质。证明从 v 到 w 的最短路径上的每个子路径也是两个端点之间的最短路径。唯一最短路径树。假设从 s 到每个其他顶点都有唯一的最短路径。证明 SPT 是唯一的。没有负循环。证明如果通用算法终止，则从 s 可达的地方没有负循环。提示：在终止时，从 s 可达的所有边都满足distTo[w] <= distTo[v] + e.weight()。将这个不等式对沿循环的所有边相加。前驱图。真或假。在没有负循环的边权重有向图中执行 Bellman-Ford 时，遵循edgeTo[]数组总是会回到 s 的路径。对 Dijkstra 算法重复这个问题。 Yen 对 Bellman-Ford 的改进。 [参考] 将边分为两个 DAGs A 和 B：A 由从较低索引顶点到较高索引顶点的边组成；B 由从较高索引顶点到较低索引顶点的边组��。在 Bellman-Ford 的一个阶段中遍历所有边时，首先按顶点编号的升序（A 的拓扑顺序）遍历 A 中的边，然后按顶点编号的降序（B 的拓扑顺序）遍历 B 中的边。在遍历 A 中的边时，SPT 中从具有正确distTo[]值的顶点开始并且仅使用 A 中的边的任何路径都会得到正确的distTo[]值；B 也是如此。所需的遍历次数是路径上 A-B 交替的最大次数，最多为(V+1)/2。因此，所需的遍历次数最多为(V+1)/2，而不是 V。替换路径。给定具有非负权重和源 s 以及汇 t 的边权重有向图，设计一个算法，找到从 s 到 t 的最短路径，该路径不使用每条边 e。你的算法的增长顺序应为 E V log V。道路网络数据集。来自DIMACS 挑战。这里是每个州的所有道路。 rome99.txt 是罗马的有向道路网络的大部分数据，来自DIMACS 挑战。该图包含 3353 个顶点和 8870 条边。顶点对应道路交叉口，边对应道路或道路段。边的权重是以米为单位的距离。 NYC.txt 是纽约市的无向道路网络。该图包含 264346 个顶点和 733846 条边。它是连通的，包含平行边，但没有自环。边的权重是旅行时间，且严格为正。互联网路由。OSPF（开放最短路径优先）是互联网路由中广泛使用的协议，使用了迪杰斯特拉算法。RIP（路由信息协议）是另一种基于贝尔曼-福特算法的路由协议。具有跳过一条边的最短路径。给定具有非负权重的边权重有向图，设计一个 E log V 算法，用于找到从 s 到 t 的最短路径，其中您可以将任意一条边的权重更改为 0。解决方案。计算从 s 到每个其他顶点的最短路径；计算从每个顶点到 t 的最短路径。对于每条边 e = (v, w)，计算从 s 到 v 的最短路径长度和从 w 到 t 的最短路径长度的和。最小的这样的和提供了最短的这样的路径。无向图中的最短路径。编写一个程序 DijkstraUndirectedSP.java，使用迪杰斯特拉算法解决非负权重的无向图中的单源最短路径问题。弗洛伊德-沃舍尔算法。 FloydWarshall.java 实现了弗洛伊德-沃舍尔算法，用于全对最短路径问题。其时间复杂度与 V³ 成正比，空间复杂度与 V² 成正比。它使用了 AdjMatrixEdgeWeightedDigraph.java。随机贝尔曼-福特算法。 [参考资料] 假设我们在 Yen 算法中均匀随机选择顶点顺序（其中 A 包含所有从排列中较低顶点到较高顶点的边）。证明预期的通过次数最多为(V+1)/3。苏尔巴勒算法。给定具有非负边权重和两个特殊顶点 s 和 t 的有向图，找到从 s 到 t 的两条边不相交的路径，使得这两条路径的权重之和最小。解决方案。这可以通过巧妙地应用迪杰斯特拉算法来实现，即苏尔巴勒算法。 5. 字符串原文：algs4.cs.princeton.edu/50strings 译者：飞龙协议：CC BY-NC-SA 4.0 概述。我们通过交换字符串来进行通信。我们考虑经典算法来解决围绕以下应用程序的基本计算挑战： 5.1 字符串排序包括 LSD 基数排序、MSD 基数排序和用于对字符串数组进行排序的三向基数快速排序。 5.2 Trie 描述了用于实现具有字符串键的符号表的 R-way trie 和三向搜索 trie。 5.3 子字符串搜索描述了在大段文本中搜索子字符串的算法，包括经典的 Knuth-Morris-Pratt、Boyer-Moore 和 Rabin-Karp 算法。 5.4 正则表达式介绍了一种称为 grep 的基本搜索工具，我们用它来搜索不完全指定的子字符串。 5.5 数据压缩介绍了数据压缩，我们试图将字符串的大小减少到最小。我们介绍了经典的 Huffman 和 LZW 算法。游戏规则。为了清晰和高效，我们的实现是基于 Java String 类表达的。我们简要回顾它们最重要的特性。字符.String 是字符序列。字符的类型是 char，可以有 2¹⁶ 种可能的值。几十年来，程序员们一直关注编码为 7 位 ASCII 或 8 位扩展 ASCII 的字符，但许多现代应用程序需要 16 位 Unicode。不可变性.String 对象是不可变的，因此我们可以在赋值语句中使用它们，并且作为方法的参数和返回值，而不必担心它们的值会改变。索引.charAt() 方法以常数时间从字符串中提取指定字符。长度.length() 方法以常数时间返回字符串的长度。子字符串.substring() 方法通常以常数时间提取指定的子字符串。警告：从 Oracle 和 OpenJDK Java 7，更新 6 开始，substring() 方法在提取的子字符串大小上需要线性时间和空间。由于我们没有预料到这种 drastical 变化，我们的一些字符串处理代码将受到影响。String API 对其任何方法，包括 substring() 和 charAt()，都不提供性能保证。教训是自行承担风险。查看这篇文章获取更多细节。连接.+ 运算符执行字符串连接。我们避免逐个字符附加形成字符串，因为在 Java 中这是一个二次时间的过程。（Java 有一个 StringBuilder 类用于这种用途。）字符数组. Java 的 String 不是原始类型。标准实现提供了上述操作，以便于客户端编程。相比之下，我们考虑的许多算法可以使用低级表示，比如一个 char 值数组，许多客户端可能更喜欢这种表示，因为它占用更少的空间并且耗时更少。字母表。一些应用程序涉及从受限字母表中获取的字符串。在这种应用程序中，使用具有以下 API 的 Alphabet.java 类通常是有意义的：构造函数以 R 个字符的字符串作为参数，该字符串指定了字母表；toChar()和toIndex()方法在常数时间内在字符串字符和介于 0 和 R-1 之间的int值之间进行转换。R()方法返回字母表或基数中的字符数。包括一些预定义的字母表： Count.java 是一个客户端程序，它在命令行上指定一个字母表，读取该字母表上的一系列字符（忽略不在字母表中的字符），计算每个字符出现的频率，本章中的 Java 程序。以下是本章中的 Java 程序列表。单击程序名称以访问 Java 代码；单击参考号以获取简要描述；阅读教科书以获取全面讨论。 \| REF \| 程序 \| 描述 / JAVADOC \| --- Alphabet.java \| 字母表 \| Count.java \| 字母表客户端 \| \| 5.1 \| LSD.java \| LSD 基数排序 \| \| 5.2 \| MSD.java \| MSD 基数排序 \| InplaceMSD.java \| 原地 MSD 基数排序¹ \| \| 5.3 \| Quick3string.java \| 三向字符串快速排序 \| AmericanFlag.java \| 美国国旗排序¹ \| AmericanFlagX.java \| 非递归美国国旗排序¹ \| \| 5.4 \| TrieST.java \| 多向字典树符号表 \| TrieSET.java \| 多向字典树集合 \| \| 5.5 \| TST.java \| 三向单词查找树 \| \| 5.6 \| KMP.java \| 子字符串查找（Knuth–Morris–Pratt） \| \| 5.7 \| BoyerMoore.java \| 子字符串查找（Boyer–Moore） \| \| 5.8 \| RabinKarp.java \| 子字符串查找（Rabin–Karp） \| \| 5.9 \| NFA.java \| 正则表达式的 NFA \| GREP.java \| grep \| BinaryDump.java \| 二进制转储 \| HexDump.java \| 十六进制转储 \| PictureDump.java \| 图片转储 \| Genome.java \| 基因组编码 \| RunLength.java \| 数据压缩（行程长度编码） \| \| 5.10 \| Huffman.java \| 数据压缩（赫夫曼） \| \| 5.11 \| LZW.java \| 数据压缩（Lempel–Ziv–Welch） \| Q + A Q. 什么是 Unicode。 A. Unicode（通用字符编码）= 复杂的 21 位代码，用于表示国际符号和其他字符。 Q. 什么是 UTF-16。 A. UTF-16（Unicode 转换格式）= 复杂的 16 位可变宽度代码，用于表示 Unicode 字符。大多数常见字符使用 16 位（一个char）表示，但代理对使用一对char值表示。如果第一个char值在D800和DFFF之间，则与下一个char（在相同范围内）组合形成代理对。没有 Unicode 字符对应于D800到DFFF。例如，007A表示小写字母 Z，6C34表示中文水的符号，D834 DD1E表示音乐的 G 大调。 Unicode 参考。 Q. 什么是子字符串陷阱？ A. 字符串方法调用s.substring(i, j)返回 s 从索引 i 开始到 j-1 结束的子字符串（而不是在 j 结束，正如你可能会怀疑的那样）。 Q. 如何更改字符串的值？ A. 在 Java 中无法修改字符串，因为字符串是不可变的。如果你想要一个新的字符串，那么你必须使用字符串连接或返回新字符串的字符串方法之一，如toLowerCase()或substring()来创建一个新的字符串。网页练习挤压空格。编写一个程序 Squeeze.java，该程序接受一个字符串作为输入，并删除相邻的空格，最多保留一个空格。删除重复项。给定一个字符串，创建一个新字符串，其中删除所有连续的重复项。例如，ABBCCCCCBBAB变为ABCBAB。 N 个 x 的字符串。描述以下函数返回的字符串，给定一个正整数N？ java public static String mystery(int N) { String s = ""; while(N > 0) { if (N % 2 == 1) s = s + s + "x"; else s = s + s; N = N / 2; } return s; } AI写代码 java 运行 1 2 3 4 5 6 7 8 9 10 回文检查。编写一个函数，该函数以字符串作为输入，并在字符串是回文时返回true，否则返回false。回文是指字符串从前往后读和从后往前读是相同的。 Watson-Crick 互补回文检查。编写一个函数，该函数以字符串作为输入，并在字符串是 Watson-Crick 互补回文时返回true，否则返回false。Watson-Crick 互补回文是指 DNA 字符串等于其反向的互补（A-T，C-G）。 Watson-Crick 互补。编写一个函数，该函数以 A、C、G 和 T 字符的 DNA 字符串作为输入，并返回以其互补替换所有字符的反向字符串。例如，如果输入是 ACGGAT，则返回 ATCCGT。完美洗牌。给定长度相同的两个字符串s和t，以下递归函数返回什么？ java public static String mystery(String s, String t) { int N = s.length(); if (N <= 1) return s + t; String a = mystery(s.substring(0, N/2), t.substring(0, N/2)); String b = mystery(s.substring(N/2, N), t.substring(N/2, N)); return a + b; } AI写代码 java 运行 1 2 3 4 5 6 7 8 二叉树表示。编写一个名为TreeString.java的数据类型，使用二叉树表示不可变字符串。它应该支持在常数时间内进行连接，并在与字符数成比例的时间内打印出字符串。反转字符串。编写一个递归函数来反转一个字符串。不要使用任何循环。提示：使用 String 方法substring()。 java public static String reverse(String s) { int N = s.length(); if (N <= 1) return s; String a = s.substring(0, N/2); String b = s.substring(N/2, N); return reverse(b) + reverse(a); } AI写代码 java 运行 1 2 3 4 5 6 7 8 你的方法效率如何？我们的方法具有线性对数运行时间。随机字符串。编写一个递归函数，创建一个由字符’A’和’Z’之间的随机字符组成的字符串。 java public static String random(int N) { if (N == 0) return ""; if (N == 1) return 'A' + StdRandom.uniform(26); return random(N/2) + random(N - N/2); } AI写代码 java 运行 1 2 3 4 5 6 子序列。给定两个字符串s和t，编写一个程序 Subsequence.java，确定s是否是t的子序列。也就是说，s的字母应该按照相同的顺序出现在t中，但不一定是连续的。例如accag是taagcccaaccgg的子序列。最长互补回文。在 DNA 序列分析中，互补回文是一个等于其反向互补的字符串。腺嘌呤（A）和胸腺嘧啶（T）是互补的，胞嘧啶（C）和鸟嘌呤（G）也是互补的。例如，ACGGT 是一个互补回文。这样的序列作为转录结合位点，并与基因扩增和遗传不稳定性相关。给定一个长度为 N 的文本输入，找到文本的最长互补回文子串。例如，如果文本是 GACACGGTTTTA，那么最长的互补回文是 ACGGT。提示：将每个字母视为奇数长度可能回文的中心，然后将每对字母视为偶数长度可能回文的中心。 DNA 转 RNA。编写一个函数，该函数接受一个 DNA 字符串（A、C、G、T）并返回相应的 RNA 字符串（A、C、G、U）。 DNA 互补。编写一个函数，该函数以 DNA 字符串（A、C、G、T）作为输入，并返回互补的碱基对（T、G、C、A）。DNA 通常以双螺旋结构存在。两条互补的 DNA 链以螺旋结构连接在一起。从十六进制转换为十进制。 Hex2Decimal.java 包含一个函数，该函数接受一个十六进制字符串（使用 A-F 表示数字 11-15）并返回相应的十进制整数。它使用了一些字符串库方法和霍纳方法。 java public static int hex2decimal(String s) { String digits = "0123456789ABCDEF"; s = s.toUpperCase(); int val = 0; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); int d = digits.indexOf(c); val = 16val + d; } return val; } AI写代码 java 运行 1 2 3 4 5 6 7 8 9 10 11 12 替代方案：Integer.parseInt(String s, int radix)。更加健壮，并且适用于负整数。 5.1 字符串排序原文：algs4.cs.princeton.edu/51radix 译者：飞龙协议：CC BY-NC-SA 4.0 本节正在大规模施工中。 LSD 基数排序。程序 LSD.java 实现了用于固定长度字符串的 LSD 基数排序。它包括一种用于对待每个整数作为 4 字节字符串处理的 32 位整数进行排序的方法。当 N 很大时，这种算法比系统排序快 2-3 倍。 MSD 基数排序。程序 MSD.java 实现了 MSD 基数排序。三向字符串快速排序。程序 Quick3string.java 实现了三向字符串快速排序。问与答练习频率计数。读入一个字符串列表并打印它们的频率计数。算法：将字符串读入数组，使用三向基数快速排序对它们进行排序，并计算它们的频率计数。加速奖励：在三向分区期间计算计数。缺点：使用空间存储所有字符串。备选方案：TST。对均匀分布数据进行排序。给定 N 个来自 [0, 1] 区间的随机实数，考虑以下算法对它们进行排序：将 [0, 1] 区间分成 N 个等间距子区间。重新排列（类似于累积计数）这 N 个元素，使每个元素都在其适当的桶中。对每个桶中的元素进行插入排序（或者等效地，只对整个文件进行插入排序）。也就是说，对一个级别进行 MSD 基数排序，然后切换到插入排序。[尝试原地进行？] 解决方案：平均总共需要 O(N) 的时间。设 n_i 是桶 i 中的元素数量。插入排序所有桶的预期时间是 O(n)，因为 E[sum_i (n_i)²] <= 2n。给定一个包含 N 个不同长度的十进制整数的数组，描述如何在 O(N + K) 的时间内对它们进行排序，其中 K 是所有 N 个整数的总位数。美国国旗排序。（原地键索引计数）给定一个包含 N 个介于 0 和 R-1 之间的不同值的数组，以线性时间和 O® 的额外空间对它们进行升序排列。导致（本质上）原地字符串排序。提示：计算 count[] 数组，告诉你键需要放置的位置。扫描输入数组。取第一个键，找到它应该属于的桶，并将其交换到相应的位置（更新相应的 count[] 条目）。重复第二个键，但要小心跳过已知属于其位置的键。网络练习 2-sum. 给定一个包含 N 个 64 位整数的数组 a[] 和一个目标值 T，确定是否存在两个不同的整数 i 和 j，使得 a[i] + a[j] 等于 T。你的算法应该在最坏情况下线性时间运行。解决方案。在线性时间内对数组进行基数排序。从左到右扫描指针 i 和从右到左扫描指针 j：考虑 a[i] + a[j]。如果它大于 T，则推进 j 指针；如果它小于 T，则推进 i 指针；如果它等于 T，则我们找到了所需的索引。注意，整数数组可以使用 Franceschini、Muthukrishnan 和 Patrascu 的高级基数排序算法在线性时间和常数额外空间内进行基数排序。在排序的字符串数组中进行二分查找。实现一个用于排序字符串数组的二分查找版本，它跟踪查询字符串与 lo 和 hi 端点之间已知相同字符的数��。利用这些信息在二分查找过程中避免字符比较。比较此算法与调用 compareTo() 的版本的性能。（compareTo() 方法的优点是它不需要调用 charAt()，因为它是作为 String 数据类型的实例方法实现的。） 5.2 查找树原文：algs4.cs.princeton.edu/52trie 译者：飞龙协议：CC BY-NC-SA 4.0 本节正在大规模建设中。具有字符串键的符号表。可以使用标准符号表实现。而是利用字符串键的附加结构。为字符串（以及其他以数字表示的键）定制搜索算法。目标：像哈希一样快速，比二叉搜索树更灵活。可以有效地支持额外的操作，包括前缀和通配符匹配，例如，IP 路由表希望转发到 128.112.136.12，而实际上转发到 128.112 是它已知的最长匹配前缀。附带好处：快速且占用��间少的字符串搜索。 R 向查找树。程序 TrieST.java 使用多向查找树实现了一个字符串符号表。三向查找树。程序 TST.java 使用三向查找树实现了一个字符串符号表。参考：快速排序和搜索的算法作者 Bentley 和 Sedgewick。属性 A.（Bentley-Sedgewick）给定一个输入集，无论字符串插入的顺序如何，其 TST 中的节点数都是相同的。证明。在集合中，TST 中每个不同字符串前缀都有一个唯一的节点。节点在 TST 中的相对位置可能会根据插入顺序而改变，但节点数是不变的。高级操作。通配符搜索，前缀匹配。R 向查找树和 TST 实现包括用于通配符匹配和前缀匹配的代码。惰性删除 = 更改单词边界位。急切删除 = 清理任何死亡父链接。应用：T9 手机文本输入。用户使用手机键盘键入；系统显示所有对应的单词（并在唯一时自动完成）。如果用户键入 0，系统会显示所有可能的自动完成。问答练习编写 R 向查找树字符串集和 TST 的非递归版本。长度为 L 的唯一子字符串。编写一个程序，从标准输入中读取文本并计算其包含的长度为 L 的唯一子字符串的数量。例如，如果输入是cgcgggcgcg，那么长度为 3 的唯一子字符串有 5 个：cgc、cgg、gcg、ggc和ggg。应用于数据压缩。提示：使用字符串方法substring(i, i + L)提取第 i 个子字符串并插入符号表。另一种解决方案：使用第 i 个子字符串的哈希值计算第 i+1 个子字符串的哈希值。在第一千万位数的π或者第一千万位数的π上测试它。唯一子字符串。编写一个程序，从标准输入中读取文本并计算任意长度的不同子字符串的数量。（可以使用后缀树非常高效地完成。）文档相似性。要确定两个文档的相似性，计算每个三字母组（3 个连续字母）的出现次数。如果两个文档的三字母组频率向量的欧几里德距离很小，则它们相似。拼写检查。编写一个程序 SpellChecker.java，它接受一个包含英语词汇的字典文件的名称，然后从标准输入读取字符串并打印出不在字典中的任何单词。使用一个字符串集。垃圾邮件黑名单。将已知的垃圾邮件地址插入到存在表中，并用于阻止垃圾邮件。按国家查找 IP。使用数据文件ip-to-country.csv来确定给定 IP 地址来自哪个国家。数据文件有五个字段（IP 地址范围的开始，IP 地址范围的结束，两个字符的国家代码，三个字符的国家代码和国家名称。请参阅IP-to-country 网站。IP 地址不重叠。这样的数据库工具可用于：信用卡欺诈检测，垃圾邮件过滤，网站上语言的自动选择以及 Web 服务器日志分析。 Web 的倒排索引。给定一个网页列表，创建包含网页中包含的单词的符号表。将每个单词与出现该单词的网页列表关联起来。编写一个程序，读取一个网页列表，创建符号表，并通过返回包含该查询单词的网页列表来支持单词查询。 Web 的倒排索引。扩展上一个练习，使其支持多词查询。在这种情况下，输出包含每个查询词至少出现一次的网页列表。带有重复项的符号表。密码检查器。编写一个程序，从命令行读取一个字符串和从标准输入读取一个单词字典，并检查它是否是一个“好”密码。在这里，假设“好”意味着（i）至少有 8 个字符长，（ii）不是字典中的单词，（iii）不是字典中的单词后跟一个数字 0-9（例如，hello5），（iv）不是由一个数字分隔的两个单词（例如，hello2world）。反向密��检查器。修改上一个问题，使得（ii）-（v）也适用于字典中单词的反向形式（例如，olleh 和 olleh2world）。巧妙的解决方案：将每个单词及其反向形式插入符号表中。随机电话号码。编写一个程序，接受一个命令行输入 N，并打印 N 个形式为（xxx）xxx-xxxx 的随机电话号码。使用符号表避免多次选择相同的号码。使用这个区号列表来避免打印虚假的区号。使用 R 向 Trie。包含前缀。向StringSET添加一个方法containsPrefix()，接受字符串 s 作为输入，并在集合中存在包含 s 作为前缀的字符串时返回 true。子字符串匹配。给定一个（短）字符串列表，您的目标是支持查询，其中用户查找字符串 s，您的任务是报告列表中包含 s 的所有字符串。提示：如果您只想要前缀匹配（字符串必须以 s 开头），请使用文本中描述的 TST。要支持子字符串匹配，请将每个单词的后缀（例如，string，tring，ring，ing，ng，g）插入 TST 中。 Zipf 定律。哈佛语言学家乔治·齐普夫观察到，包含 N 个单词的英文文本中第 i 个最常见单词的频率大致与 1/i 成比例，其中比例常数为 1 + 1/2 + 1/3 + … + 1/N。通过从标准输入读取一系列单词，制表它们的频率，并与预测的频率进行比较来测试“齐普夫定律”。打字猴和幂律。（Micahel Mitzenmacher）假设一个打字猴通过将每个 26 个可能的字母以概率 p 附加到当前单词来创建随机单词，并以概率 1 - 26p 完成单词。编写一个程序来估计生成的单词长度的频率分布。如果“abc”被生成多次，则只计算一次。打字猴和幂律。重复上一个练习，但假设字母 a-z 出现的概率与以下概率成比例，这是英文文本的典型概率。 \| CHAR \| FREQ \| \| CHAR \| FREQ \| \| CHAR \| FREQ \| \| CHAR \| FREQ \| \| CHAR \| FREQ \| --- --- --- --- --- --- --- \| \| A \| 8.04 \| \| G \| 1.96 \| \| L \| 4.14 \| \| Q \| 0.11 \| \| V \| 0.99 \| \| B \| 1.54 \| \| H \| 5.49 \| \| M \| 2.53 \| \| R \| 6.12 \| \| W \| 1.92 \| \| C \| 3.06 \| \| I \| 7.26 \| \| N \| 7.09 \| \| S \| 6.54 \| \| X \| 0.19 \| \| D \| 3.99 \| \| J \| 0.16 \| \| O \| 7.60 \| \| T \| 9.25 \| \| Y \| 1.73 \| \| E \| 12.51 \| \| K \| 0.67 \| \| P \| 2.00 \| \| U \| 2.71 \| \| Z \| 0.09 \| \| F \| 2.30 \| \| \| \| \| \| \| \| \| \| \| \| \| 19. 书的索引。编写一个程序，从标准输入中读取一个文本文件，并编制一个按字母顺序排列的索引，显示哪些单词出现在哪些行，如下所示的输入。忽略大小写和标点符号。 ```java It was the best of times, it was the worst of times, it was the age of wisdom, it was the age of foolishness, age 3-4 best 1 foolishness 4 it 1-4 of 1-4 the 1-4 times 1-2 was 1-4 wisdom 4 worst 2 ``` AI写代码 java 运行 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 熵。我们定义一个包含 N 个单词的文本语料库的相对熵为 E = 1 / (N log N) sum (p[i] log(k) - log(p[i]), i = 1…k)，其中 p_i 是单词 i 出现的次数的比例。编写一个程序，读取一个文本语料库并打印出相对熵。将所有字母转换为小写，并将标点符号视为空格。最长前缀。真或假。二进制字符串 x 在符号表中的最长前缀要么是 x 的下取整，要么是 x 的上取整（如果 x 在集合中则两者都是）。错误。在 { 1, 10, 1011, 1111 } 中，1100 的最长前缀是 1，而不是 1011 或 1111。创意练习网页练习 5.3 �� 子字符串搜索原文：algs4.cs.princeton.edu/53substring 译者：飞龙协议：CC BY-NC-SA 4.0 本节正在大规模施工中。在长字符串中搜索 - 在线。这个网站是一个关于精确字符串搜索算法的重要资源。 Java 中的高性能模式匹配用于一般字符串搜索，带通配符的搜索和带字符类的搜索。程序 Brute.java 是暴力字符串搜索。基本上等同于 SystemSearch.java。拉宾卡普。程序 RabinKarp.java 实现了拉宾卡普随机指纹算法。 Knuth-Morris-Pratt。程序 KMP.java 是 Knuth-Morris-Pratt 算法。KMPplus.java 是一个改进版本，时间和空间复杂度与 M + N 成正比（与字母表大小 R 无关）。 Boyer-Moore。程序 BoyerMoore.java 实现了 Boyer-Moore 算法的坏字符规则部分。它不实现强好后缀规则。入侵检测系统。需要非常快速的字符串搜索，因为这些部署在网络的瓶颈处。应用问答练习设计一个从右到左扫描模式的暴力子字符串搜索算法。展示 Brute-Force 算法的跟踪，样式类似于图 XYZ，用于以下模式和文本字符串。 AAAAAAAB; AAAAAAAAAAAAAAAAAAAAAAAAB ABABABAB; ABABABABAABABABABAAAAAAAA 确定以下模式字符串的 KMP DFA。 AAAAAAAB AACAAAB ABABABAB ABAABAAABAAAB ABAABCABAABCB 假设模式和文本是在大小为 R >= 2 的字母表上的随机字符串。证明字符比较的期望次数为(N - M + 1) (1 - R^-M) / (1 - R^-1) <= 2 (N - M + 1)。构造一个例子，其中 Boyer-Moore 算法（仅使用坏字符规则）性能较差。如何修改拉宾卡普算法以搜索给定模式，并附加条件中间字符是一个“通配符”（任何文本字符都可以匹配它）。如何修改拉宾卡普算法以确定文本中是否存在 k 个模式子集中的任何一个（比如，所有长度相同）？解决方案。计算 k 个模式的哈希值，并将哈希值存储在一个集合中。如何修改拉宾卡普算法以在 N×N 文本中搜索 M×M 模式？或者在 N×N 文本中搜索其他不规则形状的模式？蒙特卡洛与拉斯维加斯拉宾卡普。在线回文检测。逐个读入字符。报告每个瞬间当前字符串是否是回文。提示：使用 Karp-Rabin 哈希思想。串联重复。在字符串 s 中，基本字符串 b 的串联重复是由至少一个连续的基本字符串 b 的副本组成的子字符串。给定 b 和 s，设计一个算法，在 s 中找到 b 的最大长度的串联重复。运行时间应与 M + N 成正比，其中 M 是 b 的长度，N 是 s 的长度。解决方案。这个问题是子字符串搜索的一般化（s 中是否至少有一个连续的 b 的副本？），所以我们需要一个泛化的子字符串搜索算法。创建 k 个 b 的连接副本的 Knuth-Morris-Pratt DFA，其中 k = n/m。现在，在输入 s 上模拟 DFA 并记录它达到的最大状态。从中，我们可以识别最长的串联重复。后缀前缀匹配。设计一个线性时间算法，找到一个字符串 a 的最长后缀，恰好匹配另一个字符串 b 的前缀。循环旋转。设计一个线性时间算法来确定一个字符串是否是另一个字符串的循环旋转。如果字符串 a 是字符串 b 的循环旋转，那么 a 和 b 具有相同的长度，a 由 b 的后缀和前缀组成。循环字符串的子串。设计一个线性时间算法来确定一个字符串 a 是否是循环字符串 b 的子串。最长回文子串。给定一个字符串 s，找到最长的回文子串（或 Watson-crick 回文串）。解决方案：可以使用后缀树或Manacher’s algorithm在线性时��内解决。这里有一个通常在线性对数时间内运行的更简单的解决方案。首先，我们描述如何在线性时间内找到长度恰好为 L 的所有回文子串：使用 Karp-Rabin 迭代地形成每个长度为 L 的子串（及其反转）的哈希值，并进行比较。由于你不知道 L，重复将你对 L 的猜测加倍，直到你知道最佳长度在 L 和 2L 之间。然后使用二分查找找到确切的长度。解决方案。 Manacher.java 是 Manacher 算法的实现。重复子串。 [ Mihai Patrascu] 给定一个整数 K 和长度为 N 的字符串，找到至少出现 K 次的最长子串。一个解决方案。假设你知道重复字符串的长度 L。对长度为 L 的每个子串进行哈希处理，并检查任何哈希是否出现 K 次或更多。如果是，检查以确保你没有运气不佳。由于你不知道 L，重复将你对 L 的猜测加倍，直到你知道最佳长度在 L 和 2L 之间。然后使用二分查找找到正确的值。最长公共子串。给定两个（或三个）字符串，找到在所有三个字符串中都出现的最长子串。提示：假设你知道最长公共子串的长度 L。对长度为 L 的每个子串进行哈希处理，并检查任何哈希桶是否包含每个字符串的（至少）一个条目。所有匹配。修改 KMP 以在线性时间内找到所有匹配（而不是最左匹配）。斐波那契字符串。 KMP 的有趣案例。F(1) = B, F(2) = A, F(3) = AB, F(4) = ABA, F(5) = ABAAB, F(N) = F(N-1) F(N-2)。假设 x 和 y 是两个字符串。设计一个线性时间算法来确定是否存在整数 m 和 n 使得 x^m = y^n（其中 x^m 表示 x 的 m 个副本的连接）。解决方案。只需检查 xy = yx 的位置（这个事实并不平凡 - 它来自于 Lyndon-Schutzenberger 定理）。字符串的周期。让 s 为一个非空字符串。如果对于所有 i = 0, 1, …, N-p-1 都有 s[i] = s[i+p]，则整数 p 被称为 s 的周期。字符串 s 的周期是是 s 的周期中最小的整数 p（可以是 N）。例如，ABCABCABCABCAB 的周期是 3。设计一个线性时间算法来计算字符串的周期。字符串的边界。给定一个非空字符串 s，如果 s = yw = wz 对于一些字符串 y、z 和 w 且 \|y\| = \|z\| = p，则我们将字符串 w 定义为 s 的边界，即 w 是 s 的既是前缀又是后缀的一个合适子串。字符串的边界是 s 的最长合适边界（可以为空）。例如，ABCABCABCABCAB 的边界是 w = ABCABCAB（其中 y = ABC，z = CAB，p = 3）。设计一个线性时间算法来计算字符串的边界。变位词子串搜索。给定长度为 N 的文本字符串 txt[] 和长度为 M 的模式字符串 pat[]，确定 pat[] 或其任何变位词（其 M! 种排列之一）是否出现在文本中。提示：在文本中维护长度为 M 的给定子串的字母频率直方图。 5.4 正则表达式原文：algs4.cs.princeton.edu/54regexp 译者：飞龙协议：CC BY-NC-SA 4.0 本节正在大力整理中。正则表达式。 NFA.java, DFS.java, Digraph.java, 和 GREP.java. 运行时间。 M = 表达式长度，N = 输入长度。正则表达式匹配算法可以在 O(M)时间内创建 NFA，并在 O(MN)时间内模拟输入。库实现。 Validate.java。大多数正则表达式库实现使用回溯算法，在某些输入上可能需要指数级的时间。这样的输入可能非常简单。例如，确定长度为 N 的字符串是否与正则表达式(a\|aa)b匹配，如果选择字符串得当，可能需要指数级的时间。下表展示了 Java 1.4.2 正则表达式的失败情况。 java java Validate "(a\|aa)b" aaaaaaaaaaaaaaaaaaaaaaaaaaaaaac 1.6 seconds java Validate "(a\|aa)b" aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac 3.7 seconds java Validate "(a\|aa)b" aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac 9.7 seconds java Validate "(a\|aa)b" aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac 23.2 seconds java Validate "(a\|aa)b" aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac 62.2 seconds java Validate "(a\|aa)b" aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac 161.6 seconds AI写代码 java 运行 1 2 3 4 5 6 7 java java Validate "(a)\|b" aaaaaaaaaaaaaaaaaaaaac 1.28 java Validate "(a)\|b" aaaaaaaaaaaaaaaaaaaaaac 2.45 java Validate "(a)\|b" aaaaaaaaaaaaaaaaaaaaaaac 4.54 java Validate "(a)\|b" aaaaaaaaaaaaaaaaaaaaaaaac 8.84 java Validate "(a)\|b" aaaaaaaaaaaaaaaaaaaaaaaaac 17.74 java Validate "(a)\|b" aaaaaaaaaaaaaaaaaaaaaaaaaac 33.77 java Validate "(a)\|b" aaaaaaaaaaaaaaaaaaaaaaaaaaac 67.72 java Validate "(a)\|b" aaaaaaaaaaaaaaaaaaaaaaaaaaaac 134.73 AI写代码 java 运行 1 2 3 4 5 6 7 8 9 上述示例是人为的，但它们展示了大多数正则表达式库中的一个令人担忧��缺陷。在实践中确实会出现不良输入。根据Crosby 和 Wallach的说法，以下正则表达式出现在 SpamAssassin 的一个版本中，这是一个功能强大的垃圾邮件过滤程序。 java [a-z]+@[a-z]+([a-z\.]+\.)+[a-z]+ AI写代码 java 运行 1 2 它试图匹配某些电子邮件地址，但在许多正则表达式库中，包括 Sun 的 Java 1.4.2 中，匹配某些字符串需要指数级的时间。 java java Validate "[a-z]+@[a-z]+([a-z\.]+\.)+[a-z]+" spammer@x...................... AI写代码 java 运行 1 2 这尤其重要，因为垃圾邮件发送者可以使用一种病态的返回电子邮件地址来拒绝服务攻击一个运行 SpamAssassin 的邮件服务器。这个特定的模式现在已经修复，因为 Perl 5 正则表达式使用内部缓存来在回溯过程中在相同位置短路重复匹配。这些缺陷不仅限于 Java 的实现。例如，GNU regex-0.12 对于匹配形式为aaaaaaaaaaaaaac的字符串与正则表达式(a)\|b需要指数级的时间。Sun 的 Java 1.4.2 同样容易受到这个问题的影响。此外，Java 和 Perl 正则表达式支持反向引用 - 对于这些扩展正则表达式的正则表达式模式匹配问题是NP 难的，因此在某些输入上这种指数级的增长似乎是固有的。这是我实际写的一个，用来找到字符串NYSE之前的最后一个单词：regexp = “([\w\s]+).NYSE”; 参考：正则表达式匹配可以简单快速（但在 Java、Perl、PHP、Python、Ruby 等中很慢）。比较了 Thompson NFA 和回溯方法。包含了一些针对 Thompson NFA 的性能优化。还有一些历史注释和参考资料。 Q + A Q. Java 正则表达式库的文档？ A. 这里是 Oracle 关于使用正则表达式的指南。它包括更多操作，我们不会探索。还请参阅String方法matches()、split()和replaceAll()。这些是使用Pattern和Matcher类的简写。这里有一些常见的正则表达式模式。 Q. 用于电子邮件地址、Java 标识符、整数、小数等的工业级别正则表达式？ A. 这里有一个有用的正则表达式库，提供了工业级别的模式，用于匹配电子邮件地址、URL、数字、日期和时间。试试这个正则表达式工具。 Q. 我困惑为什么(a \| b)匹配所有的 a 和 b 的字符串，而不仅仅是所有 a 的字符串或所有 b 的字符串？ A. 操作符复制正则表达式（而不是匹配正则表达式的固定字符串）。因此，上述等同于ε \| (a\|b) \| (a\|b)(a\|b) \| (a\|b)(a\|b)(a\|b) \| … Q. 历史？ A. 在 1940 年代，沃伦·麦卡洛克和沃尔特·皮茨将神经元建模为有限自动机来描述神经系统。1956 年，史蒂夫·克利纳发明了一种数学抽象称为正则集来描述这些模型。神经网络和有限自动机中事件的表示，《自动机研究》，3-42 页，普林斯顿大学出版社，新泽西州普林斯顿，1956 年。 Q. 有哪些可视化正则表达式的工具？ A. 尝试Debuggerx。练习为以下每组二进制字符串编写正则表达式。只使用基本操作。 0 或 11 或 101 只有 0 答案：0 \| 11 \| 101, 0 为以下每组二进制字符串编写正则表达式。只使用基本操作。所有二进制字符串所有二进制字符串，除了空字符串以 1 开头，以 1 结尾以 00 结尾包含至少三个 1 答案：(0\|1), (0\|1)(0\|1), 1 \| 1(0\|1)1, (0\|1)00, (0\|1)1(0\|1)_1(0\|1)\_1(0\|1)\_或 0\_ 10\ 10_ 1(0\|1)_。编写一个正则表达式描述字母表{a, b, c}上按排序顺序的输入。答案：a b c。为以下每组二进制字符串编写正则表达式。只使用基本操作。包含至少三个连续的 1 包含子串 110 包含子串 1101100 不包含子串 110 答案：(0\|1)111(0\|1), (0\|1)110(0\|1), (0\|1)1101100(0\|1), (0\|10)1。最后一个是最棘手的。为至少有两个 0 但不连续的 0 的二进制字符串编写正则表达式。为以下每组二进制字符串编写正则表达式。只使用基本操作。至少有 3 个字符，并且第三个字符为 0 0 的数量是 3 的倍数以相同字符开头和结尾奇数长度以 0 开头且长度为奇数，或以 1 开头且长度为偶数长度至少为 1 且最多为 3 答案：(0\|1)(0\|1)0(0\|1), 1 \| (1 01 01 01), 1(0\|1)_1 \| 0(0\|1)\_0 \| 0 \| 1, (0\|1)((0\|1)(0\|1))\, 0((0\|1)(0\|1))_ \| 1(0\|1)((0\|1)(0\|1))_, (0\|1) \| (0\|1)(0\|1) \| (0\|1)(0\|1)(0\|1)。对于以下每个问题，指出有多少长度为 1000 的位字符串与正则表达式匹配：0(0 \| 1)1，0101，(1 \| 01)。编写一个正则表达式，匹配字母表{a, b, c}中包含的所有字符串：以 a 开头且以 a 结尾最多一个 a 至少有两个 a 偶数个 a a 的数量加上 b 的数量为偶数找出字母按字母顺序排列的长单词，例如，almost和beefily。答案：使用正则表达式’^a b c d e f g h i j k l m n o p q r s t u v w x y z$'。编写一个 Java 正则表达式，匹配电话号码，带有或不带有区号。区号应为(609) 555-1234 或 555-1234 的形式。找出所有以nym结尾的英语单词。找出所有包含三连字母bze的英语单词。答案：subzero。找出所有以 g 开头，包含三连字母pev且以 e 结尾的英语单词。答案：grapevine。找出所有包含三个 r 且至少有两个 r 的英语单词。找出可以用标准键盘顶行写出的最长英语单词。答案：proprietorier。找出所有包含字母 a、s、d 和 f 的单词，不一定按照顺序。解决方案：cat words.txt \| grep a \| grep s \| grep d \| grep f。给定一个由 A、C、T 和 G 以及 X 组成的字符串，找到一个字符串，其中 X 匹配任何单个字符，例如，CATGG 包含在 ACTGGGXXAXGGTTT 中。编写一个 Java 正则表达式，用于 Validate.java，验证形式为 123-45-6789 的社会安全号码。提示：使用\d表示任何数字。答案：[0-9]{3}-[0-9]{2}-[0-9]{4}。修改上一个练习，使-成为可选项，这样 123456789 就被视为合法输入。编写一个 Java 正则表达式，匹配包含恰好五个元音字母且元音字母按字母顺序排列的所有字符串。答案:[^aeiou]a[^aeiou]e[^aeiou]i[^aeiou]o[^aeiou]u[^aeiou] 编写一个 Java 正则表达式，匹配有效的 Windows XP 文件名。这样的文件名由除了冒号以外的任意字符序列组成。 java / \ : ? " < > \| AI写代码 java 运行 1 2 此外，它不能以空格或句号开头。编写一个 Java 正则表达式，描述有效的 OS X 文件名。这样的文件名由除冒号以外的任意字符序列组成。此外，它不能以句点开头。给定一个代表 IP 地址的名称为s的字符串，采用 dotted quad 表示法，将其分解为其组成部分，例如，255.125.33.222。确保四个字段都是数字。编写一个 Java 正则表达式，描述形式为 Month DD, YYYY 的所有日期，其中 Month 由任意大写或小写字母字符串组成，日期是 1 或 2 位数字，年份正好是 4 位数字。逗号和空格是必需的。编写一个 Java 正则表达式，描述形式为 a.b.c.d 的有效 IP 地址，其中每个字母可以表示 1、2 或 3 位数字，句点是必需的。是：196.26.155.241。编写一个 Java 正则表达式，匹配以 4 位数字开头并以两个大写字母结尾的车牌。编写一个正则表达式，从 DNA 字符串中提取编码序列。它以 ATG 密码子开头，以停止密码子（TAA、TAG 或 TGA）结尾。参考编写一个正则表达式来检查序列 rGATCy：即，它是否以 A 或 G 开头，然后是 GATC，最后是 T 或 C。编写一个正则表达式来检查一个序列是否包含两个或更多次重复的 GATA 四核苷酸。修改 Validate.java 使搜索不区分大小写。提示: 使用(?i)嵌入式标志。编写一个 Java 正则表达式，匹配利比亚独裁者穆阿迈尔·卡扎菲姓氏的各种拼写，使用以下模板：(i)以 K、G、Q 开头，(ii)可选地跟随 H，(iii)后跟 AD，(iv)可选地跟随 D，(v)可选地跟随 H，(vi)可选地跟随 AF，(vii)可选地跟随 F，(vii)以 I 结尾。编写一个 Java 程序，读取类似(K\|G\|Q)[H]AD[D][H]AF[F]I的表达式，并打印出所有匹配的字符串。这里的符号[x]表示字母x的 0 或 1 个副本。为什么s.replaceAll("A", "B");不会替换字符串s中所有出现的字母 A 为 B？答案：使用s = s.replaceAll("A", "B");代替。replaceAll方法返回结果字符串，但不会改变s本身。字符串是不可变的。编写一个程序 Clean.java，从标准输入中读取文本并将其打印出来，在一行上去除任何尾随空格，并用 4 个空格替换所有制表符。提示: 使用replaceAll()和正则表达式\s匹配空格。编写一个正则表达式，匹配在文本a href ="和下一个"之间的所有文本。答案:href=\"(.?)\"。?使.变得不贪婪而是懒惰。在 Java 中，使用Pattern.compile("href=\\\"(.?)\\\"", Pattern.CASE_INSENSITIVE)来转义反斜杠字符。使用正则表达式提取在<title>和<\title>标签之间的所有文本。(?i)是另一种使匹配不区分大小写的方法。$2指的是第二个捕获的子序列，即title标签之间的内容。 java String pattern = "(?i)(<title.?>)(.+?)(</title>)"; String updated = s.replaceAll(pattern, "$2"); AI写代码 java 运行 1 2 3 编写一个正则表达式来匹配在和标签之间的所有文本。答案：<TD[^>]>([^<])</TD> 创意练习 FMR-1 三联重复区域。 “人类 FMR-1 基因序列包含一个三联重复区域，在该区域中序列 CGG 或 AGG 重复多次。三联体的数量在个体之间高度变化，增加的拷贝数与脆性 X 综合征相关，这是一种导致 2000 名儿童中的一名智力残疾和其他症状的遗传疾病。”（参考：Durbin 等人的《生物序列分析》）。该模式由 GCG 和 CTG 括起来，因此我们得到正则表达式 GCG (CGG \| AGG) CTG。广告拦截。Adblock 使用正则表达式来阻止 Mozilla 和 Firebird 浏览器下的横幅广告。解析文本文件。一个更高级的例子，我们想要提取匹配输入的特定部分。这个程序代表了解析科学输入数据的过程。 PROSITE 到 Java 正则表达式。编写一个程序，读取 PROSITE 模式并打印出相应的 Java 正则表达式。PROSITE 是蛋白质家族和结构域的“第一个和最著名”的数据库。其主要用途是确定从基因组序列翻译而来的未知功能蛋白质的功能。生物学家使用PROSITE 模式语法规则在生物数据中搜索模式。这是CBD FUNGAL（访问代码 PS00562）的原始数据。每行包含各种信息。也许最有趣的一行是以 PA 开头的行 - 它包含描述蛋白质基序的模式。这些模式很有用，因为它们通常对应于功能或结构特征。 java PA C-G-G-x(4,7)-G-x(3)-C-x(5)-C-x(3,5)-[NHG]-x-[FYWM]-x(2)-Q-C. AI写代码 java 运行 1 2 每个大写字母对应一个氨基酸残基。字母表由对应于 2x 氨基酸的大写字母组成。连字符-表示连接。例如，上面的模式以 CGG（Cys-Gly-Gly）开头。符号x扮演通配符的角色 - 它匹配任何氨基酸。这对应于我们符号中的.。括号用于指定重复：x(2)表示恰好两个氨基酸，x(4,7)表示 4 到 7 个氨基酸。这对应于 Java 符号中的.{2}和.{4,7}。花括号用于指定禁止的残基：{CG}表示除 C 或 G 之外的任何残基。星号具有其通常的含义。文本转语音合成。 grep 的原始动机。“例如，如何处理发音多种不同的二连音 ui：fruit, guile, guilty, anguish, intuit, beguine？” 具有挑战性的正则表达式。为以下每组二进制字符串编写一个正则表达式。只使用基本操作。除了 11 或 111 之外的任何字符串每个奇数符号是 1 包含至少两个 0 和最多一个 1 没有连续的 1s 二进制可被整除。为以下每组二进制字符串编写一个正则表达式。只使用基本操作。以二进制数解释的比特串可被 3 整除以二进制数解释的比特串可被 123 整除波士顿口音。编写一个程序，将所有的 r 替换为 h，将句子翻译成波士顿版本，例如将“Park the car in Harvard yard”翻译为波士顿版本的“Pahk the cah in Hahvahd yahd”。文件扩展名。编写一个程序，以文件名作为命令行参数，并打印出其文件类型扩展名。扩展名是跟在最后一个.后面的字符序列。例如，文件sun.gif的扩展名是gif。提示：使用split("\\.")；请记住.是一个正则表达式元字符，因此您需要转义它。反向子域。为了进行网络日志分析，方便地根据子域（如wayne.faculty.cs.princeton.edu）组织网络流量。编写一个程序来读取域名并以反向顺序打印出来，如edu.princeton.cs.faculty.wayne。银行抢劫。你刚刚目睹了一起银行抢劫案，并且得到了逃跑车辆的部分车牌号。它以ZD开头，中间有一个3，以V结尾。帮助警官写出这个车牌的正则表达式。排列的正则表达式。找到 N 个元素的所有排列集合的最短正则表达式（仅使用基本操作），其中 N = 5 或 10。例如，如果 N = 3，则语言是 abc，acb，bac，bca，cab，cba。答案：困难。解决方案的长度与 N 呈指数关系。解析带引号的字符串。读取一个文本文件并打印出所有带引号的字符串。使用类似"[^"]"的正则表达式，但需要担心转义引号。解析 HTML。一个>，可选地跟随空格，后跟a，后跟空格，后跟href，可选地跟随空格，后跟=，可选地跟随空格，后跟" java < \s a \s+ href \s = \s \\" \\" \s > AI写代码 java 运行 1 2 子序列。给定一个字符串s，确定它是否是另一个字符串t的子序列。例如，abc 是 achfdbaabgabcaabg 的一个子序列。使用正则表达式。现在不使用正则表达式重复这个过程。答案：(a) a.b.c.，(b) 使用贪婪算法。亨廷顿病诊断。导致亨廷顿病的基因位于染色体 4 上，并且具有可变数量的 CAG 三核苷酸重复。编写一个程序来确定重复次数并打印不会患 HD，如果重复次数少于 26，则打印后代有风险，如果数字为 37-35，则打印有风险，如果数字在 36 和 39 之间，则打印将患 HD。这就是遗传测试中识别亨廷顿病的方式。基因查找器。基因是基因组的一个子字符串，以起始密码子（ATG）开始，以终止密码子（TAG，TAA，TAG 或 TGA）结束，并由除起始或终止密码子之外的密码子序列（核苷酸三联体）组成。基因是起始和终止密码子之间的子字符串。重复查找器。编写一个程序Repeat.java，它接受两个命令行参数，并查找指定由第二个命令行参数指定的文件中第一个命令行参数的最大重复次数。字符过滤器。给定一个包含坏字符的字符串t，例如t = "!@#$%^&()-_=+"，编写一个函数来读取另一个字符串s并返回删除所有坏字符后的结果。 java String pattern = "[" + t + "]"; String result = s.replaceAll(pattern, ""); AI写代码 java 运行 1 2 3 通配符模式匹配器。不使用 Java 内置的正则表达式，编写一个程序 Wildcard.java 来查找与给定模式匹配的字典中的所有单词。特殊符号匹配任意零个或多个字符。因此，例如模式"w ard"匹配单词"ward"和"wildcard"。特殊符号.匹配任何一个字符。您的程序应将模式作为命令行参数读取，并从标准输入读取单词列表（由空格分隔）。通配符模式匹配器。重复上一个练习，但这次使用 Java 内置的正则表达式。警告：在通配符的上下文中，的含义与正则表达式不同。搜索和替换。文字处理器允许您搜索给定查询字符串的所有出现并用另一个替换字符串替换每个出现。编写一个程序 SearchAndReplace.java，它接受两个字符串作为命令行输入，从标准输入读取数据，并用第一个字符串替换所有出现的第一个字符串，并将结果发送到标准输出。提示：使用方法String.replaceAll。密码验证器。假设出于安全原因，您要求所有密码至少包含以下字符之一 java ~ ! @ # $ % ^ & \| AI写代码 java 运行 1 2 为String.matches编写一个正则表达式，如果密码包含所需字符之一，则返回true。答案：“[~!@# %^&\|]+ ” 字母数字过滤器。编写一个程序 Filter.java，从标准输入中读取文本，并消除所有不是空格或字母数字的字符。答案这是关键行。 java String output = input.replaceAll("[^\\s0-9a-zA-Z]", ""); AI写代码 java 运行 1 2 将制表符转换为空格。编写一个程序，将 Java 源文件中的所有制表符转换为 4 个空格。解析分隔文本文件。存储数据库的一种流行方式是将其存储在一个文本文件中，每行一个记录，每个字段由称为分隔符的特殊字符分隔。 java 19072/Narberth/PA/Pennsylvania 08540/Princeton/NJ/New Jersey AI写代码 java 运行 1 2 3 编写一个程序 Tokenizer.java，它读取两个命令行参数，一个是分隔符字符，另一个是文件名，并创建一个标记数组。解析分隔文本文件。重复上一个练习，但使用String库方法split()。检查文件格式。拼写错误。编写一个 Java 程序，验证这个常见拼写错误列表中只包含形式为的行 java misdemenors (misdemeanors) mispelling (misspelling) tennisplayer (tennis player) AI写代码 java 运行 1 2 3 4 第一个单词是拼写错误，括号中的字符串是可能的替换。有趣的英语单词 DFA 的大小与 RE 的大小呈指数关系。给出一个 RE，用于表示所有最后一个字符为 1 的比特串集合。RE 的大小应该与 k 成线性关系。现在，给出同一组比特串的 DFA。它使用了多少个状态？提示：对于这组比特串，每个确定有限自动机（DFA）至少需要有 2^k 个状态。 5.5 数据压缩原文：algs4.cs.princeton.edu/55compression 译者：飞龙协议：CC BY-NC-SA 4.0 本节正在大规模施工中。数据压缩：将文件大小缩小以节省空间存储和在传输时节省时间。摩尔定律：芯片上的晶体管数量每 18-24 个月翻一番。帕金森定律：数据会扩张以填满可用空间。文本、图像、声音、视频等。维基百科提供公共转储以供学术研究和再发布。使用 bzip 和 SevenZip 的 LZMA。对 300GB 数据进行压缩可能需要一周的时间。古代思想。摩尔斯电码，十进制数系统，自然语言，旋转电话（较低的号码拨号速度更快，所以纽约是 212，芝加哥是 312）。二进制输入和输出流。我们使用 BinaryStdIn.java、BinaryStdOut.java、BinaryDump.java、HexDump.java 和 PictureDump.java。固定长度编码。需要 ceil(lg R) 位来指定 R 个符号中的一个。Genome.java。使用 Alphabet.java。运行长度编码。 RunLength.java。变长编码。希望有唯一可解码的编码。实现这一目标的一种方法是向每个码字附加一个特殊的停止符号。更好的方法是前缀无码：没有字符串是另一个字符串的前缀。例如，{ 01, 10, 0010, 1111 } 是前缀无码，但 { 01, 10, 0010, 1010 } 不是，因为 10 是 1010 的前缀。给出传真机的例子。 Huffman 编码。构建最佳前缀无码的特定方式。由 David Huffman 在 1950 年在 MIT 时发明。Huffman.java 实现了 Huffman 算法。属性 A. 没有前缀无码使用更少的比特。 LZW 压缩。使用 TST.java 中的前缀匹配代码，LZW.java 实现了 LZW 压缩。现实世界：Pkzip = LZW + Shannon-Fano，GIF，TIFF，V.42bis 调制解调器，Unix 压缩。实际问题：将所有内容编码为二进制。限制符号表中元素的数量（GIF = 丢弃并重新开始，Unix 压缩 = 不起作用时丢弃）。最初字典有 512 个元素（其中填充了 256 个 ASCII 字符），因此我们每个整数传输 9 位。当填满时，我们将其扩展到 1024 并开始每个整数传输 10 位。只遍历树一次（可能会破坏我们的字符串表抽象）。实际问题：限制符号表中元素的数量。总结。 Huffman：固定长度符号的变长编码。LZW：变长字符串的固定长度编码。通用压缩算法。不可能压缩所有文件（通过简单计数论证）。直观论证：压缩莎士比亚的生平作品，然后压缩结果，再次压缩结果。如果每个文件都严格缩小，最终将只剩下一个比特。参考文献。卡内基梅隆大学的 Guy Blelloch 在数据压缩方面有一章非常出色。错误校正/检测。假设用于发送信息的信道存在噪声，每个比特以概率 p 翻转。发送每个比特 3 次；解码时取 3 个比特的大多数。解码比特的正确概率为 3p² - 2p³。这小于 p（如果 p < 1/2）。可以通过多次发送每个比特来减少解码比特错误的概率，但这在传输速率方面是浪费的。 Reed-Solomon 编码。参考资料。用于大容量存储系统（CD 和 DVD）和卫星传输（旅行者号探测器，火星探路者）当错误是突发性的时候。将要发送的数据视为一个度为 d 的多项式。只需要 d+1 个点来唯一指定多项式。发送更多点以实现纠错/检测错误。如果我们要发送的编码是 a0，a1，…，am-1（每个元素在有限域 K 上），将其视为多项式 p(x) = a0 + a1x + … + am-1 x^m-1。发送 p(0)，p(b)，p(b²)，…，其中 b 是 K 上的乘法循环群的生成元。香农编码定理。大致来说，如果信道容量为 C，则我们可以以略低于 C 的速率发送比特，使用编码方案将解码错误的概率降低到任意所需水平。证明是非构造性的。问答练习以下哪些编码是前缀自由的？唯一可解码的？对于那些唯一可解码的编码，给出编码为 1000000000000 的编码。 java code 1 code 2 code 3 code 4 A 0 0 1 1 B 100 1 01 01 C 10 00 001 001 D 11 11 0001 000 AI写代码 java 运行 1 2 3 4 5 6 给出一个不是前缀自由的唯一可解码编码的例子。解决方案。任何无后缀编码都是唯一可解码的，例如，{ 0, 01 }。给出一个不是前缀自由或无后缀的唯一可解码编码的例子。解决方案。 { 0011, 011, 11, 1110 }或{ 01, 10, 011, 110 }。 { 1, 100000, 00 }，{ 01, 1001, 1011, 111, 1110 }和{ 1, 011, 01110, 1110, 10011 }是唯一可解码的吗？如果不是，找到一个具有两个编码的字符串。解决方案。第一组编码是唯一可解码的。第二组编码不是唯一可解码的，因为 111-01-1110-01 和 1110-111-1001 是 11101111001 的两种解码方式。第三组编码不是唯一可解码的，因为 01110-1110-011 和 011-1-011-10011 是 011101110011 的两种解码方式。唯一可解码性测试。实现 Sardinas-Patterson 算法，用于测试一组编码词是否是唯一可解码的：将所有编码词添加到一个集合中。检查所有编码词对，看看是否有一个是另一个的前缀；如果是，提取悬挂后缀（即，长字符串中不是短字符串前缀的部分）。如果悬挂后缀是一个编码词，则编码不是唯一可解码的；否则，将悬挂后缀添加到列表中（前提是它尚未存在）。重复此过程直到没有剩余的新悬挂后缀为止。该算法是有限的，因为添加到列表中的所有悬挂后缀都是有限一组编码词的后缀，并且悬挂后缀最多只能添加一次。 { 0, 01, 11 }。编码词 0 是 01 的前缀，因此添加悬挂后缀 1。{ 0, 01, 11, 1 }。编码词 0 是 01 的前缀，但悬挂后缀 1 已经在列表中；编码词 1 是 11 的前缀，但悬挂后缀 1 已经在列表中。没有其他悬挂后缀，因此得出该集合是唯一可解码的结论。 { 0, 01, 10 }。编码词 0 是 01 的前缀，因此将悬挂后缀 1 添加到列表中。{ 0, 01, 10, 1 }。编码词 1 是 10 的前缀，但悬挂后缀 0 是一个编码词。因此，得出该编码不是唯一可解码的结论。 Kraft-McMillan 不等式。考虑一个具有长度为 n1, n2, …, nN 的 N 个编码词的编码 C。证明如果编码是唯一可解码的，则 K© = sum_i = 1 to N 2^(-ni) ≤ 1。 Kraft-McMillan 构造。假设我们有一组满足不等式 sum_i = 1 to N 2^(-ni) ≤ 1 的整数 n1, n2, …, nN。证明总是可以找到一个编码长度为 n1, n2, …, nN 的前缀自由编码。因此，通过将注意力限制在前缀自由编码上（而不是唯一可解码编码），我们不会失去太多。 Kraft-McMillan 最优前缀自由编码等式。证明如果 C 是一个最优前缀自由编码，那么 Kraft-McMillan 不等式是一个等式：K© = sum_i = 1 to N 2^(-ni) = 1。假设所有符号概率都是 2 的负幂次方。描述哈夫曼编码。假设所有符号频率相等。描述哈夫曼编码。找到一个哈夫曼编码，其中概率为 pi 的符号的长度大于 ceil(-lg pi)。解决方案. .01 (000), .30 (001), .34 (01), .35 (1)。码字 001 的长度大于 ceil(-lg .30)。真或假。任何最优前缀自由编码都可以通过哈夫曼算法获得。解决方案. 错误。考虑以下符号和频率集合（A 26, B 24, C 14, D 13, E 12, F 11）。 java C1 C2 C3 A 26 01 10 00 B 24 10 01 01 C 14 000 111 100 D 13 001 110 101 E 12 110 001 110 F 11 111 000 111 AI写代码 java 运行 1 2 3 4 5 6 7 8 在任何哈夫曼编码中，字符 A 和 B 的编码必须以不同的位开始，但是代码 C3 没有这个属性（尽管它是一个最优前缀自由编码）。以下输入的 LZW 编码是什么？ T O B E O R N O T T O B E Y A B B A D A B B A D A B B A D O O A A A A A A A A A A A A A A A A A A A A A 描述 LZW 编码中的棘手情况。解决方案. 每当遇到 cScSc，其中 c 是一个符号，S 是一个字符串，cS 在字典中但 cSc 不在字典中。作为 N 的函数，编码 N 个符号 A 需要多少位？N 个序列 ABC 需要多少位？让 F(i) 为第 i 个斐波那契数。考虑 N 个符号，其中第 i 个符号的频率为 F(i)。注意 F(1) + F(2) + … + F(N) = F(N+2) - 1。描述哈夫曼编码。解决方案. 最长的码字长度为 N-1。显示对于给定的 N 个符号集合，至少有 2^(N-1) 种不同的哈夫曼编码。解决方案. 有 N-1 个内部节点，每个节点都可以任意选择其左右子节点。给出一个哈夫曼编码，其中输出中 0 的频率远远高于 1 的频率。解决方案. 如果字符 ‘A’ 出现一百万次，字符 ‘B’ 出现一次，那么 ‘A’ 的码字将是 0，‘B’ 的码字将是 1。证明有关哈夫曼树的以下事实。两个最长的码字长度相同。如果符号 i 的频率严格大于符号 j 的频率，则符号 i 的码字长度小于或等于符号 j 的码字长度。描述如何在一组符号 { 0, 1, …, N-1 } 上传输哈夫曼编码（或最优前缀自由编码），使用 2N - 1 + N ceil(lg N) 位。提示：使用 2N-1 位来指定相应 trie 的结构。假设在一个扩展 ASCII 文件（8 位字符）中，最大字符频率最多是最小字符频率的两倍。证明固定长度的 8 位扩展 ASCII 码是最优的。香农-范诺编码。证明哈夫曼算法的以下自顶向下版本不是最优的。将码字集合 C 分成两个子集 C1 和 C2，其频率（几乎）相等。递归地为 C1 和 C2 构建树，从 0 开始为 C1 的所有码字，从 1 开始为 C2 的所有码字。为了实现第一步，香农和范诺建议按频率对码字进行排序，并尽可能地将集合分成两个子数组。解决方案. S 32, H 25, A 20, N 18, O 5。 LZMW 编码（米勒-韦格曼 1985）。 LZ 变种：在字典中搜索最长的已经存在的字符串（当前匹配）；将前一个匹配与当前匹配的连接添加到字典中。字典条目增长更快。当字典填满时，也可以删除低频率条目。难以实现。 LZAP 编码。类似于 LZMW：不仅添加前一个匹配与当前匹配的连接，还添加前一个匹配与当前匹配的所有前缀的连接。比 LZMW 更容易实现，但字典条目更多。确定一个不是前缀自由的最优编码。提示：只需要 3 个具有相等频率的符号。确定对于相同输入的两个最优前缀自由编码，其码字长度分布不同。提示：只需要 4 个符号。最小方差 Huffman 编码。由于与打破平局相关的不确定性，Huffman 算法可能生成具有不同码字长度分布的编码。在生成压缩流时传输，希望以（近）恒定速率传输比特。找到最小化 sum_i (p_i (l_i - l_average(T)) ²) 的 Huffman 编码。解决方案。在组合 tries 时，通过选择具有最小概率的最早生成的 trie 来打破平局。用于 Huffman 编码的双队列算法。证明以下算法计算出 Huffman 编码（如果输入符号已按频率排序，则在线性时间内运行）。维护两个 FIFO 队列：第一个队列包含输入符号，按频率升序排列，第二个队列包含组合权重的内部节点。只要两个队列中有超过一个节点，就通过检查两个队列的前端出队两个权重最小的节点。创建一个新的内部节点（左右子节点 = 两个节点，权重 = 两个节点的权重之和）并将其加入第二个队列。要获得最小方差的 Huffman 编码，通过从第一个队列中选择节点来打破平局。提示：证明第二个队列按频率升序排列。兄弟属性。如果（i）每个节点（除了根节点）都有一个兄弟节点，且（ii）二叉树可以按概率的非递增顺序列出，使得在列表中所有兄弟节点都相邻，则二叉树具有兄弟属性。证明二叉树表示 Huffman 树当且仅当它具有兄弟属性。相对编码。不是压缩图像中的每个像素，而是考虑像素与前一个像素之间的差异并对差异进行编码。直觉：通常像素变化不大。与颜色表字母上的 LZW 一起使用。可变宽度 LZW 编码。在第 2^p 个码字插入表后，将表的宽度从 p 增加到 p+1。与颜色表字母一起使用。自适应 Huffman 编码。一次通过算法，不需要发送前缀自由码。根据迄今为止读入的字符的频率构建 Huffman 树。在读入每个字符后更新树。编码器和解码器需要协调处理平局的约定。香农熵。具有可能值 x1, …, xN 且以概率 p1, …, pN 出现的离散随机变量 X 的熵 H 定义为 H(X) = -p1 lg p1 - p2 lg p2 - … - pN lg pN，其中 0 lg 0 = 0 与极限一致。一个公平硬币的熵是多少？一个硬币的熵是什么，其中两面都是正面？一个六面骰子的熵是多少？解决方案。 -lg (1/6) 大约为 2.584962。两个公平骰子的和的熵是多少？给定一个取 N 个值的随机变量。什么分布使熵最大化？熵是信息论中的一个基本概念。香农的源编码定理断言，要压缩来自一系列独立同分布随机变量流的数据，至少需要每个符号 H(X) 位。例如，发送一系列公平骰子投掷结果至少需要每次骰子投掷 2.584962 位。经验熵。经验熵是通过计算每个符号出现频率并将其用作离散随机变量的概率来获得的一段文本的熵。计算你最喜欢小说的经验熵。将其与 Huffman 编码实现的数据压缩率进行比较。香农实验。进行以下实验。给一个主体一段文本（或 Leipzig 语料库）中的 k 个字母序列，并要求他们预测下一个字母。估计主体在 k = 1, 2, 5, 100 时答对的比例。真或假。固定长度编码是��一可解码的。解决方案。真，它们是前缀自由的。给出两棵不同高度的 Huffman 树字符串 ABCCDD。前缀自由编码。设计一个高效的算法来确定一组二进制码字是否是前缀自由的。提示：使用二进制 trie 或排序。唯一可解码编码。设计一个唯一可解码的编码，它不是前缀自由编码。提示：后缀自由编码 = 前缀自由编码的反向。后缀自由编码的反向是前缀自由编码 -> 可以通过以相反顺序读取压缩消息来解码。不太方便。哈夫曼树。修改 Huffman.java，使得编码器打印查找表而不是先序遍历，并修改解码器以通过读取查找表构建树。真或假。在最佳前缀自由三进制编码中，出现频率最低的三个符号具有相同的长度。解答。 False. 三进制哈夫曼编码。将哈夫曼算法推广到三进制字母表（0, 1 和 2）上的码字，而不是二进制字母表。也就是说，给定一个字节流，找到一个使用尽可能少的三进制位（0、1 和 2）的前缀自由三进制编码。证明它产生最佳前缀自由三进制编码。解答。在每一步中合并最小的 3 个概率（而不是最小的 2 个）。当有 3 + 2k 个符号时，这种方法有效。为了将其减少到这种情况，添加概率为 0 的 1 或 2 个虚拟符号。（或者，如果符号数量不是 3 + 2k，则在第一步中合并少于 3 个符号。）例如：{ 0.1, 0.2, 0.2, 0.5 }。非二进制哈夫曼编码。将哈夫曼算法扩展到 m 进制字母表（0, 1, 2, …, m-1）上的码字，而不是二进制字母表。考虑以下由 3 个 a、7 个 c、6 个 t 和 5 个 g 组成的 21 个字符消息。 java a a c c c c a c t t g g g t t t t c c g g AI写代码 java 运行 1 2 以下的 43 位是否是上述消息的可能哈夫曼编码？ java 0000001111000101010010010010101010111001001 AI写代码 java 运行 1 2 尽可能简洁而严谨地证明你的答案。解答。对于一条消息的哈夫曼编码会产生使用最少位数的编码，其中 2 位二进制码 a = 00, c = 01, g = 10, t = 11 是一个使用 21 2 = 42 位的前缀自由编码。因此，哈夫曼编码将使用少于 43 位。如果一个二叉树是满的，则除了叶子节点外的每个节点都有两个子节点。证明与最佳前缀自由编码对应的任何二叉树都是满的。提示：如果内部节点只有一个子节点，请用其唯一子节点替换该内部节点。 Move-to-front 编码（Bentley, Sleator, Tarjan 和 Wei 1986）。编写一个名为 MoveToFront 的程序，实现 move-to-front 编码和解码。维护符号字母表的列表，其中频繁出现的符号位于前面。一个符号被编码为列表中在它之前的符号数。编码一个符号后，将其移动到列表的前面。参考 Move-ahead-k 编码。与 move-to-front 编码相同，但将符号向前移动 k 个位置。等待-c-并移动。与 move-to-front 编码相同，但只有在符号在上次移动到前面后遇到 c 次后才将其移动到前面。双哈夫曼压缩。找到一个输入，对该输入应用 Huffman.java 中的 compress() 方法两次比仅应用 compress() 一次导致输出严格较小。合并 k 个排序数组。你有 k 个已排序的列表，长度分别为 n1、n2、…、nk。假设你可以执行的唯一操作是 2 路合并：给定长度为 n1 的一个已排序数组和长度为 n2 的另一个已排序数组，用长度为 n = n1 + n2 的已排序数组替换它们。此外，2 路合并操作需要 n 个单位的时间。合并 k 个已排序数组的最佳方法是什么？解决方案. 将列表长度排序，使得 n1 < n2 < … < nk。重复地取最小的两个列表并应用 2 路合并操作。最优性的证明与哈夫曼编码的最优性证明相同：重复应用 2 路合并操作会产生一棵二叉树，其中每个叶节点对应于原始排序列表中的一个，每个内部节点对应于一个 2 路合并操作。任何原始列表对总体成本的贡献是列表长度乘以其树深度（因为这是其元素参与 2 路合并的次数）。 6. 上下文原文：algs4.cs.princeton.edu/60context 译者：飞龙协议：CC BY-NC-SA 4.0 本章节正在大规模施工中。概述。本章中的 Java 程序。以下是本章节中的 Java 程序列表。点击程序名称以访问 Java 代码；点击参考编号以获取简要描述；阅读教材以获取详细讨论。 \| REF \| 程序 \| 描述 / JAVADOC \| --- \| 6.1 \| CollisionSystem.java \| 碰撞系统 \| Particle.java \| 粒子 \| \| 6.2 \| BTree.java \| B 树 \| \| 6.3 \| SuffixArray.java \| 后缀数组（后缀排序） \| SuffixArrayX.java \| 后缀数组（优化） \| LongestRepeatedSubstring.java \| 最长重复子串 \| KWIK.java \| 上下文关键词 \| LongestCommonSubstring.java \| 最长公共子串 \| \| 6.4 \| FordFulkerson.java \| 最大流-最小割 \| FlowNetwork.java \| 带容量网络 \| FlowEdge.java \| 带流量的容量边 \| GlobalMincut.java \| 全局最小割（Stoer-Wagner）⁵ \| BipartiteMatching.java \| 二分图匹配（交替路径） \| HopcroftKarp.java \| 二分图匹配（Hopcroft-Karp） \| AssignmentProblem.java \| 加权二分图匹配 \| LinearProgramming.java \| 线性规划（单纯形法） \| TwoPersonZeroSumGame.java \| 双人零和博弈 \| 确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用绝不原创的飞龙关注关注 8点赞踩 7 收藏觉得还不错? 一键收藏 0评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报专栏目录 C#实现强连通分量查找算法——Kosaraju- sharir 算法 2301_79366332的博客 08-13 176 在图论中，有向图的强连通分量是一组顶点，这些顶点互相可达。Kosaraju-sharir 算法是一种基于深度优先搜索（DFS）的强连通分量查找算法，它具有高效、简单的特点，被广泛应用于实际工程领域。第47-57行：实现DFS函数，其中参数分别为翻转图的邻接表、是否被访问过、当前强连通分量、当前顶点。递归遍历所有未访问的相邻顶点，并将当前顶点加入当前强连通分量。第35-45行：实现DFS函数，其中参数分别为邻接表、是否被访问过、栈、当前顶点。第4-9行：定义邻接表数组G和GT，用于存储原图和翻转图中的边信息。 Algorithms 普林斯顿算法课程笔记（一） zpznba的博客 01-12 1753 本节将从动态连接性算法（并查集问题的模型）入手，引入算法分析和设计的整体思路和优化方法，为整个课程的引子部分。主要内容包括 Quick Find和Quick union 算法，以及这些算法的改进。动态连接性对于连接做如下定义：自反：p 连接于自身对称：若p连接于q,则q连接于p 传递：若p连接q，q连接r那么p连接r 我们的算法需要满足上述关于连接的定义。另外，引出了另一个概念... 参与评论您还未登录，请先登录后发表或查看评论寻找有向图网络中每个节点的最大传播路径长度_networkx 计算节点的... 9-24 博客主要围绕有向图展开,旨在寻找每个节点的最大传播路径长度。指出无邻居节点的传播路径长度为0,且有向图可能存在闭环导致死循环。还给出用Python的networkx构建有向图的示例及输出结果。针对有向图 G,寻找每个节点的最大传播路径长度。如果一个节点没有邻居节点,则传播路径长度为0; 有向图,可能存在闭环路径,会造成在《Graph Learning》\| 图传播算法 ( 上 ) 9-17 另外,停止条件一般选取一个最大的迭代轮数或者Rvi不再变化。下面,我们结合4个经典的图传播算法来看看该范式是怎么发挥实际作用的。第一个:PageRank 如果将每一个网页都抽象成一个节点,网页A有链接指向B,则存在一条有向边从节点A指向节点B,那么整个Web网页就抽象成为一张有向图。PageRank 算法要做的就是对图中... 有向图的传递闭包实现三种实现 ( Warshall+DFS+BFS ) qq_40691051的博客 11-30 2372 一、传递闭包：定义：有n个顶点的有向图的传递闭包可以定义为一个n阶布尔矩阵T[n][n],如果i能到j则T[i][j] =1,否则T[i][j] =0 意义: 对于无向图，我们可以通过并查集来实现两个顶点是否有通路的快速查询。但有向图不能有并查集，所以现在可以用传递闭包了 ( 当然无向图也可以用 ) 通过Floyd-Warshall 算法算法导论第二十五章：有向图的传递闭包 dllTimes 07-21 7848 已知一有向图 G=,顶点集合V={1,2,...,n},我们可能希望确定对所有顶点对i,j ∈ V，图G中事发后都存在一条从i到 j 的路径。G的传递闭包定义为图，其中：在Θ ( n^3 ) 时间内计算出图的传递闭包的一种方法是对E中每条边赋以权值1，然后运行Floyd-Wars 前向传播算法 ( Forward propagation ) 与反向传播算法 ( Back propagation ) 以... 9-13 1.前向传播如图所示,这里讲得已经很清楚了,前向传播的思想比较简单。举个例子,假设上一层结点i,j,k,…等一些结点与本层的结点w有连接,那么结点w的值怎么算呢?就是通过上一层的i,j,k等结点以及对应的连接权值进行加权和运算,最终结果再加上一个偏置项 ( 图中为了简单省略了 ),最后在通过一个非线性函数 ( 即... 深度学习反向传播 9-26 比如上面的电路图中,其实每一个『门』在拿到输入之后,都能计算2个东西: 输出值对应输入和输出的局部梯度而且很明显,每个门在进行这个计算的时候是完全独立的,不需要对电路图中其他的结构有了解。然而,在整个前向传输过程结束之后,在反向传播过程中,每个门却能逐步累积计算出它在整个电路输出上的梯度。『链式法则... 传递闭包（Transitive Closure） POJ 3275 weixin_30699443的博客 04-20 518 有时候我们需要知道有向图 G = （V, E）的顶点之间是否存在路径。那么怎样确定有向图中每对顶点之间是否存在路径呢？这涉及到一个非常重要的概念—— 有向图的传递闭包。有向图 G的传递闭包定义为 G' = ( V, E'), 其中 E' = {( u, v ) \| 图 G 中存在一条从顶点 u 到顶点 v 的路径}。我们可在O ( n^3 )（n为图G中的顶点数）的时间复杂度内计算出有... 普林斯顿算法课纲要及笔记 Princeton University Algorithms zjy997的博客 05-26 409 5/25/2023 新建了这篇笔记（呜呜呜不知道什么时候才能开始写下一个字） BP网络公式推导及理解——qjzcy的博客_参考小节8.2.2,自行计算单隐层bp... 9-19 神经网络的核心在BP网络,其主要分为正向传播和方向传播两个过程。一、正向传播过程公式见图,14~17,从输入xi得到输出yo 二、反向传播过程, 反向传播主要是为了调整各传播层的W值公式推导见图18~26 三、迭代过程 1、我们希望知道在每一个w向哪个方向调整才能减少误差,于是我们根据误差对w求偏导,偏导数的方向... 最小信号传播时间:有向图中的节点可达性, 9-25 有N个网络节点,标记为1到N。给定一个二维数组times[M],表示信号经过有向边的传递时间。times[i] = {u, v, w}, 其中u是源节点,v是目标节点,w是一个信号从源节点传递到目标节点的时间,即二维数组中的一行表示一条带权有向边。现在,我们向当前的节点K 发送一个信号。最少需要多长时间才能使... 普林斯顿算法第一部分 07-03 普林斯顿算法课第一部分，有课程视频讲义以及字幕，可以搭配booksite 使用普林斯顿 COS521 高级算法设计讲义 02-01 ### 普林斯顿 COS521 高级算法设计讲义 #### 课程概述与目标本课程——普林斯顿大学COS521高级算法设计，旨在为学生提供一个全面深入的学习平台，帮助他们掌握多种算法设计技巧。与传统本科算法课程不同的是，本... 机器学习 ( 第9章概率图模型 ) _目标概率图 9-22 消去x1也就表明可以在有向图种消去x1,然后按照上面消去的方法迭代到x5: 最终结果为: 针对无向图,则是使用势函数代替上面式子中的条件概率: 优缺点: 为了避免重复计算,我们显然可以将计算结果保存下来;这就是信念传播对其的改进。 ( 2 ) 信念传播其中,信念或者说消息,与方向性是无关的,并不是说,m35 ( x5 )=m53 ( x3... 图神经网络变体与应用 9-6 有向图 ( Directed Graphs ) 图形的第一个变体是有向图。无向边可以看作是两个有向边,表明两个节点之间存在着关系。然而,有向边比无向边能带来更多的信息。例如,在一个知识图中,边从 head 实体开始到 tail 实体结束,head 实体是 tail 实体的父类,这表明我们应该区别对待父类和子类的信息传播过程。有向图的... coursera 普林斯顿算法 4th公开课 03-27 【标题】"coursera 普林斯顿算法 4th公开课"所指的是一项在线教育课程，由世界知名学府普林斯顿大学提供，并在Coursera这个全球知名的在线学习平台上进行授课。该课程聚焦于算法的第四版内容，旨在教授学生如何设计、... 普林斯顿 COS521高级算法设计讲义：探索多样化的算法世界 "普林斯顿大学的COS521高级算法设计课程讲义，涵盖了算法设计的多样性和分析，强调研究生级别的算法与本科生算法的区别，以及计算机科学领域随着时间发展而引入的新问题和挑战。" 普林斯顿大学的COS521高级算法设计... 【数据挖掘】神经网络简介 ( 有向图本质 \| 拓扑结构 \| 连接方式 \| 学 ... 9-9 1 . 神经网络组成 :由一组连接的输入和输出单元组成, 每个连接都有一个权值 ( 系数 ); 2 . 神经网络本质 :神经网络本质是一种特殊的有向图, 有向图由节点和有向弧组成, 节点就是神经元, 有向弧就是神经元单元之间的连接; 3 . 神经元分层 :神经网络中的神经元由多层组成 , 层间的神经元单元没有... 有向图模型与马尔科夫网络:概率推理与结构学习 9-28 图模型主要分为有向图和无向图,有向图用于表示变量之间的条件独立关系,也可以用来表示因果关系。当图中的边带有方向时,该图就是有向图,反之则为无向图。二、有向无环图 ( DAGs ) 记有向图 G=( V,E ) G=( V, E ) G=( V,E ),其中VVV表示顶点集,EEE表示边集,由于边是有向的,采用有序的顶点对来表示。这里介绍... 普林斯顿算法 alg4 part 2 百度云 07-03 普林斯顿算法课第二部分，有课程视频讲义以及字幕，可以搭配booksite 使用算法第四版 ( 普林斯顿大学 ) PPT 12-12 算法第四版 ( 普林斯顿大学 ) PPT“ I will, in fact, claim that the difference between a bad programmer and a good one is whether he considers his code or his data structures more important. Bad programmers worry about the code. Good programmers worry about data structures and their relationships. ” 13.1 Sharir-Kosaraju 算法你的指尖有改变世界的力量 04-06 942 从概念定义、到算法详解、Python实现讲解Kosaraju 算法普林斯顿算法讲义（四）最新发布龙哥盟 03-14 1592 原文：普林斯顿大学算法课程译者：飞龙协议：CC BY-NC-SA 4.0 6.1 事件驱动模拟原文：algs4.cs.princeton.edu/61event 译者：飞龙协议：CC BY-NC-SA 4.0 本章节正在建设中。根据弹性碰撞的法则使用事件驱动模拟模拟 N 个碰撞粒子的运动。这种模拟在分子动力学（MD）中被广泛应用，以理解和预测粒子级别的物理系统的性质。这包括气体中分子的运动，化学反应的动力学，原子扩散，球体堆积，围绕土星的环的稳定性，铈和铯的相变，一维自引力系统以及前沿 Kosaraju 算法 Steve Wang's blog 11-20 1339 SCC SCC = strong connected component. 即强连通分量。 In the mathematical theory of directed graphs, a graph is said to be strongly connected or diconnected if every vertex is reachable from every other vert... 普林斯顿大学算法第一周个人总结1 热门推荐 revilwang的专栏 09-01 1万+ 来自普林斯顿大学的 Coursera 课程《算法，第一部分》，课程地址：普林斯顿算法-Percolation ( 渗透问题 ) weixin_38206454的博客 07-02 2903 本文参考了：xiewen99 , 周烨恒, tengyuan93 的博文底层算法都打包好了，可下载直接用，问题一下子就能应用在判断block时，用一个数组表示，头尾各增加一个所谓的“隐藏节点” n=3n=3n=3 -> Percolation.java import edu.princeton.cs.algs4.StdIn; import edu.princeton... 动态规划之有向图传递闭包的计算warshall 算法图解详 weixin_43342105的博客 05-03 3151 了解动态规划有向图传递闭包的计算-warshall 算法传递闭包的具体计算过程图解 warshall 算法的核心内容算法的求解过程伪代码具体代码实现 #include #include #include #include using na... 算法-强连通分量和Kosaraju 算法 weixin_34392435的博客 08-09 179 有向图中，连通性比较好理解，如果两个顶点V和顶点W是可达的，可以称之为强连通的，即存在路径A→B，同时也存在一条有向路径B→A.从之前的有向环的判定过程中其实我们可以得到一个结论就是两个是强连通的当且仅当它们都在一个普通的有向环中。强连通将所有的顶点分为了不同的集合，每个集合都是由相互均为强连通性的顶点的最大子集组成的，我们将这些集合称之为强连通分量。基础概念一般来说技术服务于生活，如果将... 关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司绝不原创的飞龙博客等级码龄14年人工智能领域优质创作者博客专家认证 1万+原创30万+点赞 32万+收藏 4万+粉丝关注私信热门文章 Sklearn 与 TensorFlow 机器学习实用指南第二版 254322 利用 Python 进行数据分析（Python 数据分析）· 第 2 版 243883 渗透攻击红队百科全书 242752 布客·ApacheCN 编程/后端/大数据/人工智能学习资源 2020.9 239779 ApacheCN 数据科学译文集 2020.8 239498 分类专栏默认分类8458篇 MLM3715篇人工智能579篇 pandas164篇财富2篇技术分享3篇 SRE5篇 yolov845篇 VKDoc2045篇计算机视觉19篇 FreeLearning2681篇网赚34篇深度学习113篇 transformers310篇 123 sympy61篇 scipy81篇 numba6篇 cython5篇 mnumpy1篇 mysql138篇 pdf14篇资源分享29篇 quant68篇安全46篇区块链53篇 kali1篇算法4篇数据分析7篇系统设计3篇 TypeScript3篇自媒体1篇操作系统1篇 devops8篇技术评论16篇文档自动化13篇创业课堂4篇 Go3篇 chatgpt1篇 opencv41篇灵性科学8篇神秘学2篇 pko8篇 cs1篇编程1篇 web1篇 util1篇 dotnet1篇数据库3篇 basicgames2篇翻译6篇 succinctly3篇 geeksforgeeks1篇架构2篇两性1篇笔记1篇网络安全1篇 C#1篇 jQuery1篇 React1篇 Vue1篇 Angular1篇 ASP.NET1篇 NodeJS1篇 Golang1篇其他2篇 tensorflow16篇 pytorch1篇 Matplotlib3篇 zetcode15篇 beginnersbook13篇 django82篇 csapp labs5篇 c++25篇 java82篇 c#/.net4篇资料整理30篇 javascript40篇 php9篇溢出33篇渗透176篇 android28篇 python290篇 numpy150篇逆向21篇 theano19篇机器学习162篇数据科学108篇 Linux6篇数据结构18篇面试2篇其它11篇商业4篇 TF_HOWTO10篇 ApacheCN228篇大数据36篇环材化生劝退44篇生产力观察12篇展开全部收起上一篇：普林斯顿算法讲义（四）下一篇：沃顿商学院发布的教学提示词库最新评论 SELinux 系统管理指南（三） 2504_93548464:$ runcon user_u:user_r:user_t:s0 /usr/local/pgsql/bin/psql -d db_test -U testuser psql (16.7) Type "help" for help. db_test=> SELECT FROM tb_users; uid \| name \| mail \| address \| salt \| phash -----+----------------+------------------+-----------------------------+--------+------------ 1 \| Sven Vermeulen \| some@example.com \| Some Place 10001, Somewhere \| abc123 \| f5ba94...3 2 \| Lisa McCarthy \| lisa@example.com \| Lisa Place 15, Someplace \| def456 \| ba53f2...0 (2 rows) db_test=> select sepgsql_getcon(); sepgsql_getcon ------------------------- user_u:user_r:user_t:s0 (1 row) 为什么从常规用户会话（user_t）登录也还是可以访问数据啊 SciPy 1.12 中文文档（八十）想想_:作者你好，请问scipy.maxentropy已经弃用，那应该如何修改曾经的代码呢？马士兵 MCA 架构师课程笔记（七十六） 2401_89201857:哥图片都裂了 Rust 学习指南（四） CSDN-Ada助手:恭喜你这篇博客进入【CSDN每天值得看】榜单，全部的排名请看数据科学的统计学（一）山峰哥:大神写的不错，认真学习一下！也希望得到你的帮助大家在看计算机毕业设计springboot宠物医院预约系统基于SpringBoot的宠物诊疗预约服务平台设计与实现 SpringBoot框架下的宠物健康管理中心在线预约系统 Kafka 深度剖析：架构演进、核心概念与设计精髓 1627 微信机器人-通过ntchat获取聊天记录自动发送聊天信息 1 Mysql杂志（二十七）——数据库状态 RabbitMQ （二）参数最新文章安卓 UI 开发实用指南（二）安卓 UI 开发实用指南（一） HTML5 和 CSS3 响应式 Web 设计秘籍（二） 2025 09月 2727篇 08月 475篇 07月 1194篇 06月 405篇 05月 5篇 03月 17篇 01月 500篇 2024年 13597篇 2023年 290篇 2022年 106篇 2021年 50篇 2020年 61篇 2019年 343篇 2018年 249篇 2017年 292篇 2016年 131篇 2015年 107篇 2014年 42篇 2013年 11篇目录 4.2 有向图有向图。术语表。有向图数据类型。图的表示。有向图中的可达性。循环和 DAG。命题。命题。命题。强连通性。命题。传递闭包。 4.3 最小生成树最小生成树。假设。基本原理。命题。（切割性质）命题。（贪心 MST 算法）带权重图数据类型。 MST API. Prim 算法。命题。 Kruskal 算法。命题。 4.4 最短路径最短路径。属性。加权有向图数据类型。最短路径 API。单源最短路径的数据结构。松弛。迪杰斯特拉算法。命题。无环带权有向图。命题。一般带权有向图中的最短路径。命题。命题。 5. 字符串概述。游戏规则。字母表。本章中的 Java 程序。 5.1 字符串排序 LSD 基数排序。 MSD 基数排序。三向字符串快速排序。 5.2 查找树具有字符串键的符号表。高级操作。 5.3 �� 子字符串搜索拉宾卡普。 Knuth-Morris-Pratt。 Boyer-Moore。入侵检测系统。 5.4 正则表达式正则表达式。运行时间。库实现。 5.5 数据压缩古代思想。二进制输入和输出流。固定长度编码。运行长度编码。变长编码。 Huffman 编码。 LZW 压缩。总结。通用压缩算法。参考文献。错误校正/检测。 Reed-Solomon 编码。香农编码定理。 6. 上下文概述。本章中的 Java 程序。展开全部收起目录 4.2 有向图有向图。术语表。有向图数据类型。图的表示。有向图中的可达性。循环和 DAG。命题。命题。命题。强连通性。命题。传递闭包。 4.3 最小生成树最小生成树。假设。基本原理。命题。（切割性质）命题。（贪心 MST 算法）带权重图数据类型。 MST API. Prim 算法。命题。 Kruskal 算法。命题。 4.4 最短路径最短路径。属性。加权有向图数据类型。最短路径 API。单源最短路径的数据结构。松弛。迪杰斯特拉算法。命题。无环带权有向图。命题。一般带权有向图中的最短路径。命题。命题。 5. 字符串概述。游戏规则。字母表。本章中的 Java 程序。 5.1 字符串排序 LSD 基数排序。 MSD 基数排序。三向字符串快速排序。 5.2 查找树具有字符串键的符号表。高级操作。 5.3 �� 子字符串搜索拉宾卡普。 Knuth-Morris-Pratt。 Boyer-Moore。入侵检测系统。 5.4 正则表达式正则表达式。运行时间。库实现。 5.5 数据压缩古代思想。二进制输入和输出流。固定长度编码。运行长度编码。变长编码。 Huffman 编码。 LZW 压缩。总结。通用压缩算法。参考文献。错误校正/检测。 Reed-Solomon 编码。香农编码定理。 6. 上下文概述。本章中的 Java 程序。展开全部收起上一篇：普林斯顿算法讲义（四）下一篇：沃顿商学院发布的教学提示词库分类专栏默认分类8458篇 MLM3715篇人工智能579篇 pandas164篇财富2篇技术分享3篇 SRE5篇 yolov845篇 VKDoc2045篇计算机视觉19篇 FreeLearning2681篇网赚34篇深度学习113篇 transformers310篇 123 sympy61篇 scipy81篇 numba6篇 cython5篇 mnumpy1篇 mysql138篇 pdf14篇资源分享29篇 quant68篇安全46篇区块链53篇 kali1篇算法4篇数据分析7篇系统设计3篇 TypeScript3篇自媒体1篇操作系统1篇 devops8篇技术评论16篇文档自动化13篇创业课堂4篇 Go3篇 chatgpt1篇 opencv41篇灵性科学8篇神秘学2篇 pko8篇 cs1篇编程1篇 web1篇 util1篇 dotnet1篇数据库3篇 basicgames2篇翻译6篇 succinctly3篇 geeksforgeeks1篇架构2篇两性1篇笔记1篇网络安全1篇 C#1篇 jQuery1篇 React1篇 Vue1篇 Angular1篇 ASP.NET1篇 NodeJS1篇 Golang1篇其他2篇 tensorflow16篇 pytorch1篇 Matplotlib3篇 zetcode15篇 beginnersbook13篇 django82篇 csapp labs5篇 c++25篇 java82篇 c#/.net4篇资料整理30篇 javascript40篇 php9篇溢出33篇渗透176篇 android28篇 python290篇 numpy150篇逆向21篇 theano19篇机器学习162篇数据科学108篇 Linux6篇数据结构18篇面试2篇其它11篇商业4篇 TF_HOWTO10篇 ApacheCN228篇大数据36篇环材化生劝退44篇生产力观察12篇展开全部收起登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
10829	http://www.chem.ualberta.ca/~chem33x/332/PDFs/04_organometallic_chem.pdf	18- Electron Rule. Recall that for MAIN GROUP elements the octet rule is used to predict the formulae of covalent compounds. This rule assumes that the central atom in a compound will make bonds such that the total number of electrons around the central atom is 8. THIS IS THE MAXIMUM CAPACITY OF THE s and p orbitals. This rule is only valid for Period 2 nonmetallic elements. The 18-electron Rule is based on a similar concept. The central TM can accommodate electrons in the s, p, and d orbitals. s (2) , p (6) , and d (10) = maximum of 18 This means that a TM can add electrons from Lewis Bases (or ligands) in addition to its valence electrons to a total of 18. This is also known Effective Atomic Number (EAN) Rule Note that it only applies to metals with low oxidation states. 18 Electron Rule cont’d Example 1. [Co(NH3)6]+3 Oxidation state of Co? Electron configuration of Co? Electrons from Ligands? Electrons from Co? Total electrons? Example 2. [Fe(CO)5] Oxidation state of Fe? Electron configuration of Fe? Electrons from Ligands? Electrons from Fe? Total electrons? What can the EAN rule tell us about [Fe(CO)5]? It can’t occur…… 20-electron complex. EAN Summary 1. Works well only for d-block metals. It does not apply to f-block metals. 2. Works best for compounds with TMs of low ox. state. 3. Ligands which are good σ-donors and π-acceptors utilize all the valence orbitals and thus such compounds obey this rule. 4. Complexes which contain a combination of σ-donors and π-acceptors conform to this rule. (e.g. Cr(NH3)3(CO)3 , Cr(η6-C6H6)(CO)3). 5. Compounds which obey this rule are kinetically inert to substitution reactions. 6. Exceptions to the rule occur at the two ends of the transition series where nd, (n+1)s, and (n+1)p valence orbitals are less well matched in energy. Let’s talk about electron counting briefly. Sandwich Compounds Obeying EAN Let’s draw some structures and see some new ligands. Each of these ligands is π-bonded above and below the metal center. Ferrocene is an interesting example. Half-Sandwich Compounds Obeying EAN Let’s draw some more structures. CO, NO, H, and PR3 can be brought together in combination to give 18 electrons. Some other cool ligands. These cyclic ligands need not be planar. Here are some examples of compounds of cyclooctatetraene. Can a reaction involve only compounds which obey the 18 electron rule? YES. Compounds and the EAN Rule We can divide compounds into three groups. 1. Electronic configurations are completely unrelated to the EAN rule. The central metal may have >, <, = 18 electrons. 2. Electron configurations follow the EAN rule and never have >18 electrons, but may have less. 3. A group that follows EAN rule rigorously. How can we understand this? Chemistry and “Magic Numbers” The Octet Rule: Period 2 nonmetallic elements tend to form compounds resulting in eight electrons around the central atom. You have been told this is because elements desire a pseudo-noble gas configuration. This is a VAST simplification. Stable Fullerenes: The allotrope of Carbon known as fullerenes (C60 or “Bucky-ball” is the most famous) take on a cage structure and it has been observed that particular numbers of C atoms yield more stable compounds. C60, C70, C76, C84, C90, C94 Nanoparticles: Metal Nanoparticle are really COOL! It has been observed that “magic numbers” of atoms preferentially come together to form stable structures. Bonding in TM Complexes: Many TM complexes will form with 18 electrons around the central metal atom. It was first observed by Sedgwick in 1927. 18- Electron Rule. Recall that for MAIN GROUP elements the octet rule is used to predict the formulae of covalent compounds. Think about Na+ and Cl-This rule assumes that the central atom in a compound will make bonds such that the total number of electrons around the central atom is 8. THIS IS THE MAXIMUM CAPACITY OF THE s and p orbitals. This rule is only valid for Period 2 nonmetallic elements. The 18-electron Rule is based on a similar concept. The central TM can accommodate electrons in the s, p, and d orbitals. s (2) , p (6) , and d (10) = maximum of 18 This means that a TM can add electrons from Lewis Bases (or ligands) in addition to its valence electrons to a total of 18. This is also known Effective Atomic Number (EAN) Rule Simple Examples of the 18 Electron Rule Example 1. [Co(NH3)6]+3 Oxidation state of Co? Electron configuration of Co? Electrons from Ligands? Electrons from Co? Total electrons? Example 2. [Fe(CO)5] Oxidation state of Fe? Electron configuration of Fe? Electrons from Ligands? Electrons from Fe? Total electrons? What can the EAN rule tell us about [Fe(CO)5]? It can’t occur…… 20-electron complex. Approach 1 to counting Oxidation State Electron Count. Ligands are viewed as “close-shelled” entities. (No radicals). This is what we did in the earlier examples. We dissect the structure When neutral Lewis base ligands (like NH3) are considered they are viewed as neutral molecules with 2 electrons for donation to the metal. Ligands like methyl (CH3 and Cl) are viewed as anions….NOT AS NEUTRAL RADICALS. (By definition H is viewed as H-) After removal of the ligands the metal is assigned a formal charge. [Ni(CO)4] Ni0 10 e-, CO 2 e- each (8) = 18 [PtCl2(PMe3)2] Pt2+ 8 e-, Cl- 2 e- each (4), PMe3 2 e- each (4) = 16 [Ta(Me)5] Ta5+ 0 e-, Me- 2 e- each (10) = 10 Fe(η5-C5H5)2 Fe2 6 e-, η5-C5H5 6e- each (12) = 18 Ferrocene Approach 2 to counting Neutral Atom Counting. The general premise to this approach is: REMOVE ALL THE LIGANDS FROM THE METAL AS NEUTRAL SPECIES. This approach results in no difference for neutral ligands like NH3 or CO. BUT For ligands such as methyl we remove the ligand as a radical. It is therefore a single electron donor in this model. Furthermore, in this model both the ligand and the metal must donate an electron to the bond. This method provides NO information about the metal oxidation state. Electron Counting Examples 7 Mn 9 Co Look at CO complexes of Mn You may expect to have the following structure for a CO complex of Mn. Mn CO OC OC CO CO Mn 7 3 CO Terminal 10 Total 17 electrons Prediction of Structure. (metal carbonyls) You may expect to have the following structure for a CO complex of Mn. Co CO OC OC CO What about ? Co 9 3 CO Terminal 6 2 CO Bridging 2 1 Co-Co 1 Co OC OC OC O C Co CO C O CO CO Is this the only possible structure for bis[tetracarbonylcobalt]? The EAN Rule cannot differentiate structures of compounds but it CAN provide possibilities for investigation. Compounds and the EAN Rule We can divide compounds into three groups. 1. Electronic configurations are completely unrelated to the EAN rule. The central metal may have >, <, = 18 electrons. 2. Electron configurations follow the EAN rule and never have >18 electrons, but may have less. 3. A group that follows EAN rule rigorously. (This is what I have shown you so far) How can we understand this? Group I p s d σ filled M ML6 6L ∆o Weak sigma interaction and NO pi interaction by 6L (d-electrons, valence) TiCl4(THF)2 (O,12) Ti(H2O)6 3-(1 ,13) V(urea)6 3-(2 ,14) CrCl6 3-(3 ,15) CrI2(DMSO)4 Mn(H2O)6 2+ CoF6 3-CuCl5 3-Ni(H20)6 2+ Cu(H20)6 2+ ZnCl2(biuret)2 You figure out. Valence electrons from 12 to 22. Ligands are weak field, ∆o is small. Little or no pi interaction between metals and ligands. Energy of the t2g orbitals is the same as the free metal. There are 6 low energy bonding MO’s, 5 medium energy MO’s and and 4 strongly antibonding MO’s (too high energy to be occupied). 12 electrons from the ligands fill the lowest energy orbitals (blue). Up to 6 metal electrons reside in the t2g set (nonbonding) without any destabilization of bonding. ∆o is so small that up to 4 electrons can be put into the eg set with only a small penalty. Group II p s d σ filled M ML6 6L ∆o Strong sigma interaction and NO pi interaction by 6L (d-electrons, valence) Zr(CH3)2 2-(O,12) Ti(en)3 3--(1 ,13) Re(NCS)6 -(2 ,14) Mo(NCS)6 3-(3 ,15) Os(SO3)6 8-Ir(NH3)4Cl2 2+ ReH9 2-You figure out. Valence electrons equal to 12 to 18. Strong sigma donation increases eg energy and increases ∆o . Little or no pi interaction between metals and ligands. Energy of the t2g orbitals is the same as the free metal. Their occupation has no impact on the stability of the complex. There are 6 low energy bonding MO’s, 3 medium energy MO’s and and 6 strongly antibonding MO’s (too high energy to be occupied). 12 electrons from the ligands fill the lowest energy orbitals (blue). Up to 6 metal electrons reside in the t2g set (nonbonding) without any destabilization of bonding. ∆o is so large that electrons cannot be put into the eg set without large penalty. Group III p s d σ filled M ML6 6L ∆o Strong sigma interaction and strong pi acceptor interaction by 6L. t2g vacant π (d-electrons, valence) Ti(cp)2(CO)2 (4, 18) V(CO)5NO (5 ,18) Cr(C6H6)2 (6 ,18) MnH(CO)5 (7 ,18) Fe(NO)2(CO)2 Co(NO)(CO)3 Ni(CO)4 You figure out. Valence electrons always equal to 18. Strong sigma donation increases eg energy Pi accepting ligands lower t2g energy. BOTH increase ∆o . There are 9 low energy bonding MO’s, 9 strongly antibonding MO’s (too high energy to be occupied). 12 electrons from the ligands and 6 metal electrons in the t2g orbitals fill the lowest energy orbitals (blue). Removal of the d electrons from the t2g set would destabilize the bonding. ∆o is so large that electrons cannot be put into the eg set without large penalty. Examples include CO and NO ligands which are at the top of the SCS. EAN Summary 1. Works well only for d-block metals. It does not apply to f-block metals. 2. Works best for compounds with TMs of low ox. state. 3. Ligands which are good σ-donors and π-acceptors utilize all the valence orbitals and thus such compounds obey this rule. 4. Complexes which contain a combination of σ-donors and π-acceptors conform to this rule. (e.g. Cr(NH3)3(CO)3 , Cr(η6-C6H6)(CO)3). 5. Compounds which obey this rule are kinetically inert to substitution reactions. 6. Exceptions to the rule occur at the two ends of the transition series where nd, (n+1)s, and (n+1)p valence orbitals are less well matched in energy. This Rule allows for prediction of structures, reactivity, and reaction mechanisms. Bridging or Terminal CO Fe Fe OC CO O C C O Fe OC I CO Terminal CO bonding at 2021.5 cm-1 and 1975.7 cm-1 also, because of very small symmetry differences between carbon monoxides. Terminal CO bond1887 cm-1 Bridging CO bond at 1770 cm-1 Bonding in TM Carbonyls Filled Filled Filled C O HOMO p s d σ filled M ML6 6L ∆o Strong sigma interaction and strong pi acceptor interaction by 6L. t2g vacant π CO bonding-the orbital picture 10 valence electrons C (4), O(6) A cartoon of M-CO bonding. The HOMO in carbon monoxide is the high energy σNB which is primarily derived from a carbon 2p orbital. This means a lone pair of electrons is residing on the C atom. The LUMO on CO is the π2p which are antibonding orbitals with significant 2p character. CO acts as a Lewis Base and a Lewis Acid. The back bond appearing in this systems is known as a synergistic effect. Reactions of Metal Carbonyls. i) Substitution of CO by other L (L is often a π-acid or Soft Lewis base; L= PR3, polyolefins, SR2, CH3CN) Recall that TM carbonyls obey the 18 electron rule. This means two things…. They are inert toward substitution. Reactions must proceed via a Dissociative mechanism (via M-CO bond cleavage) This provides a basis for photochemistry: If light of a suitable energy is supplied such that σ⇒σ can occur some interesting things happen. ∆E hv=∆E σ σ σ σ G.S. E.S. M-CO Photochemistry ∆E hv=∆E σ σ σ σ G.S. E.S. B.O.= 0 This negates the M-CO σ bond. Bond Order = 1/2 (electrons in bonding orbitals - electrons in anti-bonding orbitals) CO is photoejected! LnM L' 18 electrons LnM-CO 18 electrons hv slow - CO [LnM] 16 electrons High energy reactive intermediate. L' fast ∆E~390nm In theory, by filtering the excitation light it should be possible to remove only 1 CO. This is not simple given the broad nature of the UV-vis bands. M-CO photochemistry Examples (CO)4 Ru (OC)4Ru Ru(CO)4 hv, λ>370nm L (L= olefin) 3 LRu(CO)4 Orange, colour arises from σ⇒σ(Ru-Ru) λ~390nm Another example involving Fe and an 18 electron transition state Fe OC OC CO CO 18 electrons alkyne 2e- donor hv - CO Fe OC OC CO 18 electrons alkyne 4e- donor This intermediate is not 16e- and is stabilized by a 4e- donor alkyne. It substitutes 1013x faster than Fe(CO)5. L Reduction of TM Carbonyls What will happen if electrons are added to 18e- TM carbonyls? High energy 19 or 20 electron systems will result and CO will be ejected. (This can be viewed as the two electrons taking the place of the CO or breaking M-M bonds) Fe(CO)4 2-2Mn(CO)5 -2Na/Hg 2Na/Hg These anions are of significant importance. They are nucleophiles and react further to form M-C and M-H bonds. Formation of M-H and M-C bonds Mn(CO)5 -RX Mn CO CO OC CO CO R Mn CO CO OC CO CO C R O B RCOX The difference between A and B is the presence of CO between M and R. CO and heat Mn CO CO OC CO CO CO Mn CO CO OC CO CO R Mn CO CO OC CO CO R R CO A Empty bonding site. This is referred to as “CO insertion” although the mechanism involves migration of R. Mn(CO)5 - + H+ H-Mn(CO)5 -Collman’s Reagent Application of “carbonylmetallates” in organic synthesis. Fe OC OC CO R CO -Fe OC OC CO CO CO -R R R' O R OH O R X O D+ RX RCOCl O2 O2 X2 X2 HNR2 Na2Fe(CO)4 RD CO R'X R'X H2O R'OH Disodium tetracarbonylferrate is useful in the functionalization of organic halides. Oxidation of TM Carbonyls Oxidation weakens the M-CO or M-M bonds and results in CO elimination or M-M cleavage with the formation of TM carbonyl halides. + X2 Fe(CO)4X2 + CO 18 electrons 18 electrons + X2 2Mn(CO)5X Mn Mn OC CO X X OC OC CO CO CO CO 18 electrons heat Mn(CO)5X RLi RMgX Mn(CO)5R Special Case. Oxidative Addition (4-coordinate Vaska’s Compound 1961, 16 electron species) Ir Ph3P H PPh3 CO H Cl Ir Ph3P Cl PPh3 CO R X H2 RX Ir Ph3P Cl PPh3 CO Ir(I), d8, 16e-(activation of H2) Ir(III), d6, 18e-Reactions of Coordinated M-CO The attachment of CO to a TM makes the C electrophillic and may be attacked by a nucleophile) (CO)5Co CO R-(CO)5Co R C O-(CO)5Co R C O R' R'X This is a carbene complex; E.O. Fischer discovered this type of molecule and shared the Nobel Prize with Wilkinson. The “(CO)5Co” structural unit acts as an electron withdrawing; It is a pseudo ester. (CO)5Co R C OR' HNR2'' (CO)5Co R C NR2'' Trans-esterification The Mond Process Nickel carbonyl, a gas formed from carbon monoxide and metallic nickel. Scientific Serendipity In 1890 Ludwig Mond, was investigating the rapid corrosion of nickel valves used in apparatus for the Solvay process, and discovered Ni(CO)4. In contrast to many nickel compounds which are usually green solids, Ni(CO)4 is a colourless, volatile, toxic liquid with a very "organic character". He used it as the basis of a method to purify nickel, called the "Mond process". Ni reacts with CO (leaving the impurities behind), to form Ni(CO)4. The Ni(CO)4 is passed through a tower filled with nickel pellets at a high velocity and 400 K. Pure Ni plates out on the pellets. A commercial process for the manufacture of Na2CO3. NH3 and CO2 are passed into a sat’d NaCl(aq) solution to form soluble (NH4)(HCO3), which reacts with the NaCl to form soluble NH4Cl and solid NaHCO3 if the reactor temperature is maintained below 15°C. The NaHCO3 is filtered off and heated to produce Na2CO3. Hemoglobin and Heme
10830	https://readingfeynman.org/2015/07/21/maxwell-boltzmann-bose-einstein-and-fermi-dirac-statistics/	Skip to content Reading Feynman Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics Jean Louis Van Belle Mathematics, Physics 13 Minutes Pre-scriptum added much later: We have advanced much in our understanding since we wrote this post. If you are reading it because you want to understand more about the boson-fermion distinction, then you shouldn’t be here. The general distinction between bosons and fermions is a useless theoretical generalization which actually preventsyou from understanding what is really going on. I am keeping this post online for documentation purposes only. It is interesting from a math point of view but you are not here to learn math, are you? Jean Louis Van Belle, 20 May 2020 Original post: I’ve discussed statistics, in the context of quantum mechanics, a couple of times already (see, for example, my post on amplitudes and statistics). However, I never took the time to properly explain those distribution functions which are referred to as the Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac distribution functions respectively. Let me try to do that now—without, hopefully, getting lost in too much math! It should be a nice piece, as it connects quantum mechanics with statistical mechanics, i.e. two topics I had nicely separated so far. 🙂 You know the Boltzmann Law now, which says that the probabilities of different conditions of energy are given by e−energy/kT = 1/eenergy/kT. Different ‘conditions of energy’ can be anything: density, molecular speeds, momenta, whatever. The point is: we have some probability density function f, and it’s a function of the energy E, so we write: f(E) = C·e−energy/kT= C/eenergy/kT C is just a normalization constant (all probabilities have to add up to one, so the integral of this function over its domain must be one), and k and T are also usual suspects: T is the (absolute) temperature, and k is the Boltzmann constant, which relates the temperate to the kinetic energy of the particles involved. We also know the shape of this function. For example, when we applied it to the density of the atmosphere at various heights (which are related to the potential energy, as P.E. = m·g·h), assuming constant temperature, we got the following graph. The shape of this graph is that of an exponential decay function (we’ll encounter it again, so just take a mental note of it). A more interesting application is the quantum-mechanical approach to the theory of gases, which I introduced in my previous post. To explain the behavior of gases under various conditions, we assumed that gas molecules are like oscillators but that they can only take on discrete levels of energy. [That’s what quantum theory is about!] We denoted the various energy levels, i.e. the energies of the various molecular states, by E0, E1, E2,…, Ei,…, and if Boltzmann’s Law applies, then the probability of finding a molecule in the particular state Ei is proportional to e−Ei /kT. We can then calculate the relativeprobabilities, i.e. the probability of being in state Ei, relative to the probability of being in state E0, is: Pi/P0 = e−Ei /kT/e−E0 /kT = e−(Ei–E0)/kT= 1/e(Ei–E0)/kT Now, Pi obviously equals ni/N, so it is the ratio of the number of molecules in state Ei (ni) and the total number of molecules (N). Likewise, P0 = n0/N and, therefore, we can write: ni/n0= e−(Ei−E0)/kT= 1/e(Ei–E0)/kT This formulation is just another Boltzmann Law, but it’s nice in that it introduces the idea of a ground state, i.e. the state with the lowest energy level. We may or may not want to equate E0 with zero. It doesn’t matter really: we can always shift all energies by some arbitrary constant because we get to choose the reference point for the potential energy. So that’s the so-called Maxwell-Boltzmann distribution. Now, in my post on amplitudes and statistics, I had jotted down the formulas for the other distributions, i.e. the distributions when we’re not talking classical particles but fermions and/or bosons. As you know, fermions are particles governed by the Fermi exclusion principle: indistinguishable particles cannot be together in the same state. For bosons, it’s the other way around: having one in some quantum state actually increases the chance of finding another one there, and we can actually have an infinite number of them in the same state. We also know that fermions and bosons are the realworld: fermions are the matter-particles, bosons are the force-carriers, and our ‘Boltzmann particles’ are nothing but a classical approximation of the real world. Hence, even if we can’t see them in the actual world, the Fermi-Dirac and Bose-Einstein distributions are the real-world distributions. 🙂 Let me jot down the equations once again: Fermi-Dirac (for fermions): f(E) = 1/[Ae(E − EF)/kT+ 1] Bose-Einstein (for bosons): f(E) = 1/[AeE/kT− 1] We’ve got some other normalization constant here (A), which we shouldn’t be too worried about—for the time being, that is. Now, to see how these distributions are different from the Maxwell-Boltzmann distribution (which we should re-write as f(E) = C·e−E/kT = 1/[A·eE/kT] so as to make all formulas directly comparable), we should just make a graph. Please go online to find a graph tool (I found a new one recently—really easy to use), and just do it. You’ll see they are all like that exponential decay function. However, in order to make a proper comparison, we would actually need to calculate the normalization coefficients and, for the Fermi energy, we would also need the Fermi energy EF(note that, for simplicity, we did equate E0 with zero). Now, we could give it a try, but it’s much easier to googleand find an example online. The HyperPhysicswebsite of Georgia State University gives us one: the example assumes 6 particles and 9 energy levels, and the table and graph below compare the Maxwell-Boltzmann and Bose-Einstein distributions for the model. Now thatis an interesting example, isn’t it? In this example (but all depends on its assumptions, of course), the Maxwell-Boltzmann and Bose-Einstein distributions are almost identical. Having said that, we can clearly see that the lower energy states are, indeed, more probable with Bose-Einstein statistics than with the Maxwell-Boltzmann statistics. While the difference is not dramatic at all in this example, the difference does become very dramatic, in reality, with large numbers (i.e. high matter density) and, more importantly, at very low temperatures, at which bosons can condense into the lowest energy state. This phenomenon is referred to as Bose-Einstein condensation: it causes superfluidity and superconductivity, and it’s real indeed: it has been observed with supercooled He-4, which is notan everyday substance, but real nevertheless! What about the Fermi-Dirac distribution for this example? The Fermi-Dirac distribution is given below: the lowest energy state is now less probable, the mid-range energies much more, and none of the six particles occupy any of the four highest energy levels. Again, while the difference is not dramatic in this example, it can become very dramatic, in reality, with large numbers (read: high matter density) and very low temperatures: at absolute zero, all of the possible energy states up to the Fermi energy level will be occupied, and all the levels above the Fermi energy will be vacant. What can we make out of all of this? First, you may wonder why we actually have more than one particle in one state above: doesn’t that contradict the Fermi exclusion principle? No. We need to distinguish micro- and macro-states. In fact, the example assumes we’re talking electrons here, and so we can have two particles in each energy state—with opposite spin, however. At the same time, it’s true we cannot have three, or more, in any state. That results, in the example we’re looking at here, in five possible distributions only, as shown below. The diagram is an interesting one: if the particles were to be classical particles, or bosons, then 26 combinations are possible, including the five Fermi-Dirac combinations, as shown above. Note the little numbers above the 26 possible combinations (e.g. 6, 20, 30,… 180): they are proportional to the likelihood of occurring under the Maxwell-Boltzmann assumption (so if we assume the particles are ‘classical’ particles). Let me introduce you to the math behind the example by using the diagram below, which shows threepossible distributions/combinations (I know the terminology is quite confusing—sorry for that!). If we could distinguish the particles, then we’d have 2002 micro-states, which is the total of all those little numbers on top of the combinations that are shown (6+60+180+…). However, the assumption is that we cannot distinguish the particles. Therefore, the first combination in the diagram above, with five particles in the zero energy state and one particle in state 9, occurs 6 times into 2002 and, hence, it has a probability of 6/2002 ≈ 0.003 only. In contrast, the second combination is 10 times more likely, and the third one is 30 times more likely! In any case, the point is, in the classical situation (and in the Bose-Einstein hypothesis as well), we have 26 possible macro-states, as opposed to 5 only for fermions, and so that leads to a very different density function. Capito? No? Well, this blog is not a textbook on physics and, therefore, I should refer you to the mentioned site once again, which references a 1992 textbook on physics (Frank Blatt, Modern Physics, 1992) as the source of this example. However, I won’t do that: you’ll find the details in the Post Scriptumto this post. 🙂 Let’s first focus on the fundamental stuff, however. The most burning question is: if the real world consists of fermions and bosons, why is that that we only see the Maxwell-Boltzmann distribution in our actual(non-real?) world? 🙂 The answer is that both the Fermi-Dirac and Bose-Einstein distribution approach the Maxwell–Boltzmann distribution if higher temperatures and lower particle densities are involved. In other words, we cannot seethe Fermi-Dirac distributions (all matter is fermionic, except for weird stuff like superfluid helium-4 at 1 or 2 degrees Kelvin), but they are there! Let’s approach it mathematically: the most general formula, encompassing both Fermi-Dirac and Bose-Einstein statistics, is: Ni(Ei) ∝ 1/[e(Ei − μ)/kT± 1] If you’d google, you’d find a formula involving an additional coefficient, gi, which is the so-called degeneracyof the energy level Ei. I included it in the formula I used in the above-mentioned post of mine. However, I don’t want to make it any more complicated than it already is and, therefore, I omitted it this time. What you need to look at are the two terms in the denominator: e(Ei − μ)/kTand ± 1. From a math point of view, it is obvious that the values of e(Ei − μ)/kT+ 1 (Fermi-Dirac) and e(Ei − μ)/kT− 1 (Bose-Einstein) will approach each other if e(Ei − μ)/kTis muchlarger than ±1, so if e(Ei − μ)/kT>> 1. That’s the case, obviously, if the (Ei − μ)/kT ratio is large, so if (Ei − μ) >> kT. In fact, (Ei − μ) should, obviously, be much larger than kT for the lowestenergy levels too! Now, the conditions under which that is the case are associated with the classical situation (such as a cylinder filled with gas, for example). Why? Well… […] Again, I have to say that this blog can’t substitute for a proper textbook. Hence, I am afraid I have to leave it to you to do the necessary research to see why. 🙂 The non-mathematical approach is to simple note that quantum effects, i.e. the ±1 term, only apply if the concentration of particles is high enough. Indeed, quantum effects appear if the concentration of particles is higher than the so-called quantum concentration. Only when the quantum concentration is reached, particles will start interacting according to what they are, i.e. as bosons or as fermions. At higher temperature, that concentration will notbe reached, except in massive objects such as a white dwarf (white dwarfs are stellar remnants with the mass like that of the Sun but a volume like that of the Earth). So, in general, we can say that at higher temperatures and at low concentration we will not have any quantum effects. That should settle the matter—as for now, at least. You’ll have one last question: we derived Boltzmann’s Law from the kinetic theory of gases, but how do we derive that Ni(Ei) = 1/[Ae(Ei − μ)/kT± 1] expression? Good question but, again, we’d need more than a few pages to explain that! The answer is: quantum mechanics, of course! Go check it out in Feynman’s third Volume of Lectures! 🙂 Post scriptum: combinations, permutations and multiplicity The mentioned example from HyperPhysicsis really interesting, if only because it shows you also need to master a bit of combinatorics to get into quantum mechanics. Let’s go through the basics. If we have n distinctobjects, we can order hem in n! ways, with n! (read: n factorial) equal to n·(n–1)·(n–2)·…·3·2·1. Note that 0! is equal to 1, per definition. We’ll need that definition. For example, a red, blue and green ball can be ordered in 3·2·1 = 6 ways. Each way is referred to as a permutation. Besides permutations, we also have the concept of a k-permutation, which we can denote in a number of ways but let’s choose P(n, k). [The P stands for permutation here, not for probability.] P(n, k) is the number of ways to pick k objects out of a set of n objects. Again, the objects are supposed to be distinguishable. The formula is P(n, k) = n·(n–1)·(n–2)·…·(n–k+1) = n!/(n–k)!. That’s easy to understand intuitively: on your first pick you have n choices; on your second, n–1; on your third, n–2, etcetera. When n = k, we obviously get n! again. There is a third concept: the k-combination (as opposed to the k-permutation), which we’ll denote by C(n, k). That’s when the order withinour subset doesn’t matter: an ace, a queen and a jack taken out of some card deck are a queen, a jack, and an ace: we don’t care about the order. If we have k objects, there are k! ways of ordering them and, hence, we just have to divide P(n, k) by k! to get C(n, k). So we write: C(n, k) = P(n, k)/k! = n!/[(n–k)!k!]. You recognize C(n, k): it’s the binomial coeficient. Now, the HyperPhysicsexample illustrating the three mentioned distributions (Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac) is a bit more complicated: we need to associate q energy levels with N particles. Every possible configuration is referred to as a micro-state, and the total number of possible micro-states is referred to as the multiplicity of the system, denoted by Ω(N, q). The formula for Ω(N, q) is another binomial coefficient: Ω(N, q) = (q+N–1)!/[q!(N–1)!]. Ω(N, q) = Ω(6, 9) = (9+6–1)!/[9!(6–1)!] = 2002. In our example, however, we do not have distinct particles and, therefore, we only have 26 macro-states (as opposed to 2002 micro-states), which are also referred to, confusingly, as distributionsor combinations. Now, the number of micro-states associated with the same macro-state is given by yet another formula: it is equal to N!/[n1!·n2!·n3!·…·nq!], with ni! the number of particles in level i. [See why we need the 0! = 1 definition? It ensures unoccupied states do not affect the calculation.] So that’s how we get those numbers 6, 60 and 180 for those three macro-states. But how do we calculate those average numbers of particles for each energy level? In other words, how do we calculate the probability densities under the Maxwell-Boltzmann, Fermi-Dirac and Bose-Einstein hypothesis respectively? For the Maxwell-Boltzmann distribution, we proceed as follows: for each energy level j (or Ej, I should say), we calculate nj= ∑nij·Pi over all macro-states i. In this summation, we have nij, which is the number of particles in energy level j in micro-state i, while Pi is the probability of macro-state i as calculated by the ratio of (i) the number of micro-states associated with macro-state i and (ii) the total number of micro-states. For Pi, we gave the example of 3/2002 ≈ 0.3%. For 60 and 180, we get 60/2002 ≈ 3% and 180/2002 ≈ 9%. Calculating all the nj‘s for j ranging from 1 to 9 should yield the numbers and the graph below indeed. OK. That’s how it works for Maxwell-Boltzmann. Now, it is obvious that the Fermi-Dirac and the Bose-Einstein distribution should notbe calculated in the same way because, if they were, they would not be different from the Maxwell-Boltzmann distribution! The trick is as follows. For the Bose-Einstein distribution, we give all macro-states equal weight—so that’s a weight of one, as shown below. Hence, the probability Pi is, quite simply, 1/26 ≈ 3.85% for all 26 macro-states. So we use the same nj= ∑nij·Piformula but with Pi = 1/26. Finally, I already explained how we get the Fermi-Dirac distribution: we can only have (i) one, (ii) two, or (iii) zero fermions for each energy level—not more than two! Hence, out of the 26 macro-states, only five are actually possible under the Fermi-Dirac hypothesis, as illustrated below once more. So it’s a very different distribution indeed! Now, you’ll probably still have questions. For example, why does the assumption, for the Bose-Einstein analysis, that macro-states have equal probability favor the lower energy states? The answer is that the model also integrates other constraints: first, when associating a particle with an energy level, we do not favor one energy level over another, so all energy levels have equal probability. However, at the same time, the whole system has some fixedenergy level, and so we cannot put the particles in the higher energy levels only! At the same time, we know that, if we have q particles, and the probability of a particle having some energy level j is the same for all j, then they are likely notto be all at the same energy level: they’ll be distributed, effectively, as evidenced by the very low chance (0.3% only) of having 5 particles in the ground state and 1 particle at a higher level, as opposed to the 3% and 9% chance of the other two combinations shown in that diagram with three possible Maxwell-Boltzmann (MB) combinations. So what happens when assigning an equalprobability to all 26 possible combinations (with value 1/26) is that the combinations that were previously rather unlikely – because they did have a rather heavy concentration of particles in the ground state only – are now much morelikely. So that’s why the Bose-Einstein distribution, in this example at least, is skewed towards the lowest energy level—as compared to the Maxwell-Boltzmann distribution, that is. So that’s what’s behind, and that should also answer the other question you surely have when looking at those five acceptable Fermi-Dirac configurations: why don’t we have the same five configurations starting from the top down, rather than from the bottom up? Now you know: such configuration would have much higher energy overall, and so that’s not allowed under this particular model. There’s also this other question: we said the particles were indistinguishable, but so then we suddenly say there can be two at any energy level, because their spin is opposite. It’s obvious this is rather ad hocas well. However, if we’d allow only one particle at any energy level, we’d have noallowable combinations and, hence, we’d have no Fermi-Dirac distribution at all in this example. In short, the example is rather intuitive, which is actually why I like it so much: it shows how bosonic and fermionic behavior appear rather gradually, as a consequence of variables that are defined at the systemlevel, such as density, or temperature. So, yes, you’re right if you think the HyperPhysics example lacks rigor. That’s why I think it’s such wonderful pedagogic device. 🙂 Share this: Click to share on X (Opens in new window) X Click to share on Facebook (Opens in new window) Facebook Like Loading... Related Tagged Boltzmann distribution Boltzmann's Law Bose-Einstein distribution Bose-Einstein statistics Fermi-Dirac distribution Fermi-Dirac statistics Maxwell-Boltzmann statistics Published by Jean Louis Van Belle View all posts by Jean Louis Van Belle Published 3 thoughts on “Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics” Pingback: The blackbody radiation problem revisited \| Reading Feynman Pingback: New horizons – Reading Feynman Pingback: [A]Symmetries in Nature – Reading Feynman Leave a comment Cancel reply Comment Reblog Subscribe Subscribed Reading Feynman Already have a WordPress.com account? Log in now. Reading Feynman Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Loading Comments...
10831	https://artofproblemsolving.com/wiki/index.php/Parity?srsltid=AfmBOorwRpKMLWPHh7kJBXlAR92sMUMnKl_qb0IIq-nrt7czpxZi8O9s	Art of Problem Solving Parity - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Parity Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Parity Parity refers to whether a number is even or odd. While this may seem highly basic, checking the parity of numbers is often an useful tactic for solving problems, especially with proof by contradictions and casework. This concept begins with integers. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without remainder; an odd number is an integer that is not evenly divisible by 2. (The old-fashioned term "evenly divisible" is now almost always shortened to "divisible".) A formal definition of an even number is that it is an integer of the form n = 2k, where k is an integer; it can then be shown that an odd number is an integer of the form n = 2k + 1. This only applies to integers, not fractions or decimals. Contents 1 Problems 2 Solution 1 3 Solution 2 3.1 Introductory 3.2 Intermediate 3.2.1 Problem 3.2.2 Solution 3.2.3 Problem 2 3.2.4 Solution Problems Example from 1997 AJHSME: Problem: Ten balls numbered to are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. The probability that the sum of the two numbers on the balls removed is even is Solution 1 For the sum of the two numbers removed to be even, they must be of the same parity. There are five even values and five odd values. No matter what Jack chooses, the number of numbers with the same parity is four. There are nine numbers total, so the probability Jill chooses a number with the same parity as Jack's is . Solution 2 We find that it is only possible for the sum to be even if the numbers added are both even or odd. We will get an odd number when we add an even and odd. We can use complementary counting to help solve the problem. There are a total of possibilities since Jack can choose numbers and Jill can pick . There are possibilities for the two numbers to be different since Jack can pick any of the numbers and Jill has to pick from numbers in the set with a different parity than the one that Jack picks. So the probability that the sum will be odd is . Subtracting this by one gets the answer (edited by qkddud) Introductory Many AMC 8 problems fit this category, help us out by putting problems here! Intermediate 2000 AIME II Problem 2: Problem A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola ? Solution Note that and have the same parities, so both must be even. We first give a factor of to both and . We have left. Since there are factors of , and since both and can be negative, this gives us lattice points. 2008 AIME I Problem 3 Problem 2 There exist unique positive integers and that satisfy the equation . Find . Solution Completing the square, . Thus by difference of squares. Since is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since , the factors must be and . Since , we have and ; the latter equation implies that . Indeed, by solving, we find is the unique solution. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
10832	https://calculat.io/en/number/fraction-greater-compare/7--10--3--10	Send You can also email us on infocalculat.io Compare 7/10 and 3/10 Compare / and / Compare Is 7/10 greater than 3/10? Steps to compare fractions 7/10 and 3/10 Comparing fractions like 7/10 and 3/10 involves determining which fraction is greater or if they are equal. Here are the steps to compare these fractions: Step 1: Find a Common Denominator The first step is to find a common denominator for the fractions. The denominator is the bottom number in a fraction. A common denominator is a number that both denominators can divide into without a remainder. In this case, the denominators are 10 and 10. The smallest common multiple of 10 and 10 is 10. Therefore, 10 will be our common denominator. Step 2: Convert the Fractions Next, convert each fraction to an equivalent fraction with the common denominator. To convert 7/10 to a fraction with a denominator of 10, multiply both the numerator (top number) and the denominator by 1. This gives: To convert our second fraction 3/10 to a fraction with a denominator of 10, multiply both the numerator and the denominator by 1. This gives: Step 3: Compare the Numerators Now that the fractions have the same denominator, you can directly compare their numerators. The numerators are 7 for 7/10 and 3 for 3/10. Step 4: Determine the Larger Fraction The fraction with the larger numerator is the greater fraction. Since 7 (numerator of 1/10) is greater than 3 (numerator of 1/10): Conclusion Therefore, Another way to figure out which fraction is larger is to convert them into decimal numbers. To convert a fraction to a decimal, you divide the numerator (top number) by the denominator (bottom number). 7/10 as a decimal = 7/10 = 0.7 3/10 as a decimal = 3/10 = 0.3 Since 0.7 is greater than than 0.3, 7/10 is greater than 3/10. See Also Fraction Comparison Table \| Fractions \| Comparison \| Decimal Form \| --- \| 7/9 and 5/6 \| 7/9 is not greater than 5/6 (it is less) \| 0.778 < 0.833 \| \| 7/9 and 5/7 \| 7/9 is greater than 5/7 \| 0.778 > 0.714 \| \| 7/9 and 5/8 \| 7/9 is greater than 5/8 \| 0.778 > 0.625 \| \| 7/9 and 5/9 \| 7/9 is greater than 5/9 \| 0.778 > 0.556 \| \| 7/9 and 5/12 \| 7/9 is greater than 5/12 \| 0.778 > 0.417 \| \| 7/9 and 7/8 \| 7/9 is not greater than 7/8 (it is less) \| 0.778 < 0.875 \| \| 7/9 and 7/11 \| 7/9 is greater than 7/11 \| 0.778 > 0.636 \| \| 7/9 and 7/12 \| 7/9 is greater than 7/12 \| 0.778 > 0.583 \| \| 7/9 and 8/11 \| 7/9 is greater than 8/11 \| 0.778 > 0.727 \| \| 7/10 and 1/2 \| 7/10 is greater than 1/2 \| 0.7 > 0.5 \| \| 7/10 and 2/3 \| 7/10 is greater than 2/3 \| 0.7 > 0.667 \| \| 7/10 and 2/5 \| 7/10 is greater than 2/5 \| 0.7 > 0.4 \| \| 7/10 and 3/4 \| 7/10 is not greater than 3/4 (it is less) \| 0.7 < 0.75 \| \| 7/10 and 3/5 \| 7/10 is greater than 3/5 \| 0.7 > 0.6 \| \| 7/10 and 3/8 \| 7/10 is greater than 3/8 \| 0.7 > 0.375 \| \| 7/10 and 3/10 \| 7/10 is greater than 3/10 \| 0.7 > 0.3 \| \| 7/10 and 4/5 \| 7/10 is not greater than 4/5 (it is less) \| 0.7 < 0.8 \| \| 7/10 and 4/10 \| 7/10 is greater than 4/10 \| 0.7 > 0.4 \| \| 7/10 and 4/15 \| 7/10 is greater than 4/15 \| 0.7 > 0.267 \| \| 7/10 and 5/8 \| 7/10 is greater than 5/8 \| 0.7 > 0.625 \| \| 7/10 and 5/10 \| 7/10 is greater than 5/10 \| 0.7 > 0.5 \| \| 7/10 and 6/10 \| 7/10 is greater than 6/10 \| 0.7 > 0.6 \| \| 7/10 and 7/8 \| 7/10 is not greater than 7/8 (it is less) \| 0.7 < 0.875 \| \| 7/10 and 7/12 \| 7/10 is greater than 7/12 \| 0.7 > 0.583 \| \| 7/10 and 8/11 \| 7/10 is not greater than 8/11 (it is less) \| 0.7 < 0.727 \| \| 7/10 and 8/15 \| 7/10 is greater than 8/15 \| 0.7 > 0.533 \| \| 7/10 and 9/10 \| 7/10 is not greater than 9/10 (it is less) \| 0.7 < 0.9 \| \| 7/11 and 1/2 \| 7/11 is greater than 1/2 \| 0.636 > 0.5 \| \| 7/11 and 3/5 \| 7/11 is greater than 3/5 \| 0.636 > 0.6 \| \| 7/11 and 3/11 \| 7/11 is greater than 3/11 \| 0.636 > 0.273 \| \| 7/11 and 4/11 \| 7/11 is greater than 4/11 \| 0.636 > 0.364 \| Fraction Comparison Calculator The Fraction Comparison Calculator is a highly intuitive and user-friendly online tool designed to compare two fractions and determine which one is greater. It serves as an invaluable resource for students, teachers, and professionals alike, simplifying the often challenging task of fraction comparison. For example, it can help you find out is 7/10 greater than 3/10? (The answer is: Yes, 7/10 is greater than 3/10). Users can easily input two fractions in separate fields. The calculator accepts fractions in the form of numerators and denominators (e.g., '7/10', '3/10') The calculator simplifies both fractions to their lowest terms before comparison, ensuring accuracy in the results. Upon entering the fractions, the calculator instantly compares them and displays which fraction is greater, or if they are equal. Alongside the numerical result, the calculator provides a graphical representation of the two fractions, helping users to visually grasp the difference in their size. For educational purposes, the tool offers a step-by-step explanation of how the comparison was made, including the process of finding a common denominator (if applicable) and comparing the numerators. This calculator is not only a practical tool for quick comparisons but also an educational resource that enhances the understanding of fractions. Whether it’s for homework, teaching, or daily calculations, the Fraction Comparison Calculator is designed to make fraction comparison straightforward and accessible for everyone. Fraction Comparison Calculator Compare / and / Compare FAQ Is 7/10 greater than 3/10? See Also
10833	https://www.youtube.com/watch?v=kaiek7uo3x8	Organic Chemistry, Chapter 5, McMurry, Stereochemistry Paul Young (ChemistryOnline) 5220 subscribers 68 likes Description 11417 views Posted: 12 Jun 2014 This is the lecture recording for Chapter 5, Stereochemistry, from John McMurry's Organic Chemistry. 8 comments Transcript: Chapter 5 "Stereochemistry" or you'll sit and you'll stare at it and it just won't make any sense at all it has to do with the way your particular brain is organized so as we go through this chapter um we will I will try to present lots of different visual aids to help you see the kind of things we're talking about this is going to require you to be able to take a molecule a carbon something like one of these guys and Visually rotate it in your mind okay now on the second exam you are actually welcome to bring one of these guys okay uh people often ask GE can I bring models to the exams yeah sure you could but in reality nobody has ever had time to put one together okay so but if you bring a single carbon you can quite often just use that to understand the Symmetry rotations and make it simple so all this should make more sense we'll even get to play with them a little bit later um as we go any questions before we Draw the structure of bromocyclopentane. start all right let's start off with a very simple thing here draw the structure of brocycle pentan everybody sits back and says to themselves G I wish those questions like that on the first exam right very very simple you simply draw the cyclohexane you would or cyclopentane and attach a bro something like that would be perfectly acceptable however we all realize that this carbon here the one with the bromine attached is tetrahedral so that means in real life this thing is either coming out towards us or going back into the screen here so tetrahedral carbon is either going to be going front or back from this position but it was okay because I said it doesn't matter these two are the same compound so let's prove this to ourselves um yeah we're going to have to go to the smaller screen I'm sorry but is just going to be too much cut off this is a program called jol um jol is a very versatile free public domain sort of thingy that allows you to take molecules you can actually even build molecules in it and then display them in 3D also the neat thing about it is you can grab these in real time and you can rotate them and do just all sorts of neat stuff with it so this is a brocycle hexane with the bromine going back this is the bromo cyclohexene with the bromine coming forward now as we talked about this I said doesn't matter these are the same molecule and we can prove that let's just take this guy and I'm going to do some rotations here like this like this and by gosh that's pretty good you can see that in both of these cases now the bromine is coming towards us so we have the same molecule in both frames any questions all right let's do the same exercise here with Cy Draw the structure of cis-1-bromo-3-chlorocyclopentane. one bromo three chloro cyop so go ahead and just draw the structure it might have been on the test I don't know again it's very simple we know that if the stereochemistry is sis that means that both and the bromine are going to be on the same side of the wing flame now we can draw this a couple ways this is drawn using wedge diagram we're going to use a lot of wedges in this chapter so let's just get used to looking at them and what this tells us is that this is our ring plane and that the bromine and the chorine are both coming towards us and because of that they are sick now as we did this we also learned that you could take and you could draw the wedges as little dashed wedges showing that they were going back but once again this is our ring plane and both of these are going back into the screen here so they're on the same side this is also Cy roral cycle penting any questions below I have um a rendering this guy is actually a movie and as this thing spins here in real time I'd like you to mentally compare it with this one this is the one that we drew with the two groups coming towards this is the one corresponding to both groups going back again you can see that the Cy back to the full CIS and now we're going to rotate around so that these are now coming towards us when we did this in the chapter on CYO I said for that chapter for that exam it was acceptable to draw either of these structures but I said that was actually wrong the reason it's wrong is that these are not the same compound now we're going to prove that again by going back to J mall and here I have two chloro bromo cyclopentanes one with the two groups coming towards us and one with the two groups going back what I'm going to do is take these guys take one of these and I'm going to rotate it just like we did with the first one so that we try to line it up and show that they are identical I guess I gave it away because no they're not let's just do the simple thing and rotate around this way so now they are both coming towards us so this looks like they are Cy just like these guys but if you were sitting on this carbon this molecule you would say well to my left I have a chlorine and to my right I have a bromine over here Romine is on the left chorine is on the right in fact if you think about it if this border between the two was a mirror this molecule would be the mirror image of this molecule think about what you would see in the mirror chlorine is the closest bromine is the furthest and they're both on the same side these two molecules are their images of each other that makes them different molecules just like a right hand and a left hand they all have fingers and stuff like that but they are M images they're different right and left hands cannot be superimposed they are different same thing with organic molecules the reason that this happens is that carbon is The spatial arrangement of groups around a tetrahedral carbon (the stereochemistry) can be shown tetrahedral here we have two molecules shown this is a bromine a bromo and a chlorine methane so we have two hydrogens of little broin going back this is going back this is the amino acid alanine more complicated but the hydrogen is back the methyl is back the nitrogen here is coming towards us and the carboxilic acid is also coming towards us I can take each of these well I thought I could and set them in motion so that you can see again as this comes around hydrogens are going to be towards us and the bromine and the chlorine are going to be away this guy over here let's do this again hyrogen and methyl are now towards us and as they rotate around methyl and hydrogen are going to be back now we're going to use wedge diagrams like this meaning front and back a lot uh make sure that when you see these you understand exactly what we're talking about It is important to be able to visualize this stereochemistry in order to test molecules for internal planes of symmetry. here one way to describe the fact that compounds can be superimposable on their mirror image or not is to Simply say that if a compound has an internal plane of symmetry then it can be superimposed on his miror image if it does not then it cannot so let's look at these two renderings and see what we mean by a plane of symmetry if we take a mirror and we dropped it straight down in the middle of this molecule it would look something like that the hydrogens would reflect on each other we have split the chlorine and the bromine they reflect on each other this is an internal plane of symmetry this molecule is fully superimposable on a m image just like we saw for the bromo cyclopentane this molecule however if we put a mirror plane right down the middle here we could split the hydrogen and methyl group but the amino group would never be the reflection of the carboxilic acid this will AR whenever you have four different groups on a tetrahedral Center again we can split these two no problem but the meal group it's not the reflection of the carboxilic acid this molecule is not superimposable on his miror image now let's look at some simple The net effect of this asymmetry is to generate a molecule which is not superimposible on it's mirror image. molecules here this looks like the little models I have and this is a movie what we're going to do this represents a mirror going to move it down and then slide these mirror images over each other you see we could line up these two no problem but these two always don't match let's do this with a model these little guys so well my colors are different so doesn't quite but if I take these guys and I set them like this if there was a mirror in between them these guys would reflect so would this orange would be furthest back on both now if I take and I try to line them up I can line up the purple and the yellow but the orange and the green are always reversed no matter how I turn it no matter what I do I cannot fix that why don't you grab one of these little finger toys to play with pass the thing around and again let's go back to jall and let's look at that here we have two molecules they are n images I'm sure you can look at it and see that this is the mirror plane the red guy the broy we reflect in the mirror you would see the IOD all the way in the back and these two are lined up if I take them take one of them and I try to line it up as best I can so now they look identical at least with respect to the um iodin and the bromine but chlorine is back back here and it's front here hydrogen front hydrogen back if I fix the hydrogen and the chlorine like this well that's sort of okay but now we have our iodin and our bromine reversed and see that that's kind of neat isn't Bottom Line: One consequence of tetrahedral geometry is an internal asymmetry which occurs whenever there are four different substituents arranged around a tetrahedral center it this is what we just looked at that's a mirror in between them hydrogens reflect romanes reflect these guys even though they perfectly reflect if we lined them up they would be opposite of each when we have four different substituents we generate what are called A carbon which is attached to four different substituents is called a chiral carbon (chiral for handedness), and a pair of non-superimposible mirror images are called enantiomers. a pair of enantiomers enantiomers is the word for molecules that are mirror images of each other when we have a carbon with four things attached we will call that a chyro har hyro comes from a Greek word for handedness right hand left hand this is both of these are chyo carbons the fact that they are M images means that they are in an any questions you said Kyro was what Kyro Kyro refers to any carbon that has four different substituents attached if you have a carbon with four different substituents it will not be superimposable on its miror image it and its M image will be called in aners all right let's just prove this to ourselves yet again this is bromomethane this is a little mirror in between it as this thing rotates we get around here very clearly these are identical molecules as it rotates back we'll reach the point where we would see what we would see if we looked in the mirror that is the Romine would be closest and the carbon with the three hydrogens would be away if we have two substituents on our carbon so here we have a bromine and a chlorine this is mirror plane Romine is closest Florine is furthest away if I take this molecule and I rotate it now I recorded this from jall so it's a little ragged as it goes around but here I've rotated it that way now I'm going to kind of turn it now I got to fix it a little bit fix it a little more and a little more pretty much now you can see that this and this are in fact the same molecule two things attached or three things attached I suppose counting a hydrogen we have superimposable however as we saw if we go to four things attached this has started off its mirror image and again as it rotates so that the hydrogen is going to be back going forward these two are in opposite positions four things attached we have a pyal center any questions on the concept yeah so for this though like do you need a mirror to to understand it's like if I look at hers I can make mine just the same well okay there's a trick are your two exactly the same borrow hers yeah they are they're exactly the same okay now take any two groups swap pull them out no no the now look at are they the same Mo no they're not that's another thing about a fireal center or a with four things if we take the mirror image and we simply swap any two groups we go to the opposite stereo is swap any two others and you're back where you started kind of a fun exercise um the origin of what's called The Exchange method um it's a very nice way to deal with stereochemistry but MC Murray unfortunately moved it to life chapter 28 so whatever all right our first exercise there let's look at some There must be four different substituents attached to a carbon in order for it to be chiral. molecules and and let's pick out Cyro centers what we do to show a cyal center is we use a little asess so a little star of some sort on the carbon that is in fact chyal now the only candidate in this molecule is going to be this guy remember you're looking for four different substituents with this we have a hydrogen and a bromine they're obviously different here we have a but group attached four carbons on this side we have a pental group attached therefore this is a tyal syndome this molecule would not be superimposable on its miror image so you you'd have to count the hydrogen to oh yeah four things hydrogen counts look at this molecule identify any caral centers in the molecule well this is an SP squ carbon that doesn't count does it all of these have ch2s this carbon here we have a hydrogen and we have a methyl going around the ring this way we have three ch2s before we get our carbonal going around the ring this way we have one ch2 before we hit our carbonal this is different than this therefore this is a chyal center yeah right so I understand why that's carbon Center but isn't it true also that the carbons on the ring itself also have fourr attached to it this guy no just any Carbon on it because you the ch2 and you have the two h on one side four different two are hydrogens okay you have two of anything it's not Caron four different four different things look at this guy here we have a hydrogen and a methyl going around the ring towards the carbonal we have two ch2s going around the ring this way we have two ch2s this is not a caral center this molecule has an internal flame of symmetry that means remember that this is really this hydrogen really is going straight back this really is coming straight towards us so if we drop our miror plane like that this molecule is referred to as being ay AAL simply means it is superimposable on its a image because it has an internal plane of symmetry now let's try another way to see that as this molecule rotates this is the same one we had I'm going to try to stop it at an appropriate time right about here this spot you'll notice we dropped the mirrors straight down the middle here these ch2s would reflect on each other we would split the hydrogen split the methyl split the carbonal that is a plane of symmetry let this can oops well it's not going to do it for me is that there it goes let it continue all the way around and as it gets to the other side where the original drawing was again the hydrogen is back methyls in the front and our plane of symmetry wies right down the middle everyone see that so when that has a PL of symmetry it's always going to be ay it has a symmetry it is a par if it has four different things it For each of the molecules shown below, indicate each of the chiral centers with an asterisk () is go ahead look at these guys these are actually off an old exam and the question was to look at these and put an asterisk on every Power I see nobody has uh luxure handout I did put that up on blackboard yesterday so I really wish you had her especially for this one because this one is a tough one our first compound here if we look at this CH here on the end we have two metal groups attached don't we if you have two of anything it's an ayro carbon this is a ch2 but this guy has a hydrogen that we don't show it has this isopr group going around the ring this way we hit a double bond this way we have all these carbons before we hit there this is a caral center this one this carbon attached to the bromine has two hydrogens it is therefore a chyro ch2 ch2 but this guy has a hydrogen a chlorine a methyl and the rest of the molecule that is a chyal syence so only that carbon only that carbon but this entire molecule now is not superimposable on its M image yeah sorry I'm a little confused I thought um attach to they do we have a hydrogen chlorine a methyl and this thing hyrogen is there this next guy see this very interesting polycyclic uh Center uh that's referred to as an adamantine all of the angles here are 10° um if this was attached to other carbons these are all ch2s and CHS but if it was attached to other carbons this would be the basic structure of a diamond yeah you said that mod is not imposible on its image this molecule is not superimposed doesn't have because it has a TI C all right so this guy if we look at all of these These are ch2s aren't they we can forget those this is a CH but if you trace the path back around to where our meals attached it's always exactly the same this molecule has no chyal centers that means it must have an internal plane of symmetry the internal plane of symmetry we can imagine a mirror dropped through here looking something like that this half of the molecule is a reflection of this half once again let's look at this in a movie and as this thing rotates oops well I almost stopped it in time you can see once we rotate the methal so it's coming straight towards us our plane of symmetry would go straight up and down everything on this side reflecting everything on this side as it goes around in the back you can see this again com around in front once again perfect plane of symmetry right down the middle all right look at these go ahead and assign thereal chemistry can I have more than one or just one can there be more car car oh yes CER yes there can be this is a ch2 we have two hydrogens this guy has a methyl has a hydrogen has this ch2 ch2 attached to the ring ch2 they're different all ch2s these are all s SP squ is one kyol Center C2 with a bromine this is a CH with a chlorine bromomethyl and this guy is Carl this guy five member bring with a methyl group attached remember we started off the lecture with a five m r with a bromine attached same thing there are no chyal centers here our plane of symmetry would run right through the methyl and straight across the rain you can see that if you draw it flat very easily so we'll just flatten this out like this don't put in any stereochemistry and there is our internal plane of symmetry yeah so it doesn't just refer to what it's directly connected to yep to look at the entire L of the this entire molecule is chyal because it contains a chyal center so that means in terms of stereochemistry here we would have of two versions of this molecule it and its mer image for every chyal Center that you have you're going to increase that by two to the N where n is the number of stereo centers the chyro center is also referred to as a stereo Center let's show that with cholesterol For the molecule shown below, indicate each of the chiral centers with an asterisk () go through cholesterol and identify well let's start here this is the alcohol part of cholesterol we have an O we have a hydrogen as you go around the ring this way we have two ch2s here we have one ch2 this is a chyal center now working our way around that's a ch2 that's a ch2 isn't it but this guy has a methyl group this is attached back to a ch2 on this side this is attached to a tertiary carbon here and attached down to an SP squ carbon there that means this is a chyo center next guy have a hydrogen goes back to this one up to this terer up to this ch2 it is also a CH Center ch2 ch2 this is methyl group ch2 tertiary carbon and out here attached to chain here we have a hydrogen going back to this way ch2 here's our chain on our chain we have a hydrogen a methyl all the rest of this stuff all the rest of this stuff this guy has two methyls so it's not a carl center continuing around the ring ch2 CH bonded back here tertiary carbon ch2 same thing here hydrogen sticking up tertiary carbon different tertiary carbon ch2 a total of eight stereo centers eight Hy carbons that means if you built cholesterol with no regard to the stereochemistry around each of these garbet you would have two to the eight stereo isomers that's 256 so potentially there are 256 different isomer stereoisomers of cholesterol in the body there is only one and it looks exactly like this when cholesterol is synthesized in the body the stero biochemistry at every one of these centers is exactly controlled that's the amazing thing about biochemistry yeah so when you chemically um isolate um um cholesterol or chemically make cholesterol yourself in the lab if it's different stereochemistry body process that that's it if we had bad stereo chemistry at one any c your body would not recognize that would still be cholesterol according to is in a Triumph many years ago um cholesterol was actually synthesized from scratch by very very good organic chemist R Woodward and his group very large group at Harvard and they got the stereochemistry right on every carbon an amazing achievement all right in Enantiomers are identical in every physical and chemical property (except in their interactions with other chiral molecules) except for the fact that they rotate the plane of plane polarized light in opposite directions, and hence chiral compounds are often termed "optically active". antimers are actually identical in every physical and chemical wave they are different molecules but in terms of physical properties and chemical properties they are exactly the same they differ in their interaction with other chyro molecules that's why your body only recognizes one cholesterol and they interact differently with polarized light now that might sound really odd this is a light source when light comes out of a light source its polarization is totally random if you take that and run it through a polarized filter polarized filter basically imagine little tiny lines only allows one um Vector associated with the polariz light to pass so we start off with all of these only these guys come through now if we take that polarized light this is just like polarized sunglasses same concept if we take that polarized light and we pass it through a sample tube containing your stuff and an anti of some sort pure in an in solution what happens is it will rotate This Plane starts off like this it rotates it as it goes through this is an analyzer basically what you do is you rotate it until you see the light coming through this angle that you had to rotate it is the specific rotation chyro compounds are therefore called optically active because they interact with plain polarized light differently yeah so an anth act as Annihilator of polarized glass or something right because it changed the way that the light hit some and then just turn to the light right right have different confirmation so here's our compound sitting here and here's our polarized light let's look at it this is kind of a baby physics version of it okay the polarized lightweight can actually be described as two vectors one for a magnetic and one for an electric perform now if you remember when you add vectors basically just take these two arrows combine them like this and you get a resultant so let's pretend that we have a magnetic and electric Vector this is actually the plane of the resulting light now I'm going to take that and shine it on this chyro molecule I'm going to pretend that the iodin and the chlorine here absorb one of these vectors differently so let's just say that the iodin absorbs a red guide here the chlorine does not therefore when you add these two B together you get something that's off of the plane that you started with now let's look at its in antima here now the iodin is on this side that means that our resultant is tipped this way the angle that they're tipped is exactly the same but they're opposite one in an will rotate light one way the other one will rotate at exactly the same amount the other way we call this plus rotation or minus rotation we will actually do a lb where we use a polarimeter polarimeter will allow us to look at the angle that a chyro molecule rotates um plain polarized light one an anima rotates in One Direction other one rotates it exactly the same amount in the opposite direction is that the only thing that makes them difference I'm sorry is that the only thing that really makes that's the only physical difference other physical properties melting point boiling point Etc are the same they just interact differently with plain whole rise light it just seems so specific if we happen to have a solution with equal concentrations of both of them what would happen it would hit this guy rotate R it this way then it would hit one of these rotate it this way you wind up with no n rotation so if you have a solution a mixture of equal concentrations of both there is no rotation of light this is called AIC mixture two anbers equal concentrations does not rotate light each one does individually but they always cancel that of the polarity it's because as we interact with one of these guys this is absorbed we react with the other one if this one's absorbed so our two angles cancel we wind up we straight out down again if you're trying to purify a ky compound you can use a polarimeter as you purify this more more and more you will reach the perfect angle of that enan that's how you can tell me couldn't you also say equilibrium no it's well foric mixture it's an equilibrium yes you rotate One Way rotate the other kind of a physical equilibrium not a SPECIFIC ROTATION (Q). The Specific Rotation is equal to the observed rotation (a) divided by the the pathlength of the cell Iin dm, multiplied by the concentration (C) in g/mL chemical all right there's an equation for this of course we will actually use this in lab Alpha is our rotation this is actually the observed rotation here it's going to depend on the length of the tube that's a little L and it's going to depend on the concentration of the stuff now the units in this aren't units that we typically use these days um the path length is given in decimeters that's 10 concentration is in G per M we refer to this as the specific rotation Alpha D where the D refers to the fact that we're using a light source and we're isolating out what's called the sodium D Line This is obviously a very very old technique this is how in antier were discovered that they rotate one way or rotate the other again in lab we will take a sample do different concentrations and determine the specific rotation of aound how was that thought we're going to shine this SL through polarized LS well we do it really differently these days I just so weird y yeah we do it differently these days but this is the guts of what a polarimeter looks like all right so let's put ourselves way back in time 100 years The direction in which an optically active molecule rotates light is specific for a given molecule, but is not related to the absolute orientation of groups in that molecule around the chiral center. ago we figured out that there are in animers out there they they exist we really want to know in space what the arrange of these atoms are really I mean in this molecule as we draw it here it was sitting on the molecule I have I on my left chlorine on my right how do you know that today you would simply take X-ray defraction and you would actually prove exactly what the molecule looks like back in those days they had no clue so they guessed they guessed they just said well we either right or we're wrong and that was fine when you speak of Amo acids amino acids are D or L aren't they that was the original guess they said okay fine we're going to call this D period turns out they were right but there's a better way it has nothing to do with the rotation here you have a molecule there's no way to look at this molecule and know which way it's going to rotate the light what we do today instead of using things like DNL is we're going to look at the In order to signify the absolute configuration, a system of nomenclature has been established in which groups around the chiral center are assigned "priorities". The lowest priority group is placed towards the back, and the direction (clockwise or counterclockwise) of a line connecting the remaining groups is determined. molecule and we're going to assign priorities to the groups that are attached to the trial Center the rule say we're going to take the lowest priority group and stick it towards the back we will call that number four because there're four things then we're going to go around and we're going to assign priorities to the other groups now we will look at what we have with the lowest priority in the back we will simply draw an ark connecting all these guys if that Ark is counterclockwise we will call this absolute configuration s s stands for sinister which is the word from left they never liked left-handed people did they sinister s if we have counterclockwise on our priorities if we look at the en animer here again we put B in the back we connect this way going around this is our configuration our stand for rectus or right again clockwise rotation all right very simple method it tells us the absolute configuration in space lowest priority to the back assign priorities it's either r or S but what are priorities this is s for sinister this is R for rectus these are our two in animers the rules for priority we'll actually use these rules The Cahn-Ingold-Prelog Rules several times during the uh semester so the worth knowing these are the con inold prog rules three different guys that all work together and these are for uh ranking things that have attached to pyal centers rule number one you look at the four different things that are attached and you say which one has the highest atomic number that will be the highest priority if there's a tie you look out and see what the tide atoms are attached to if we have another carbon attached or something um we just keep track of that if we have multiple Bond so we have a double bond that counts as two carbons or two oxygens or whatever and finally once you've assigned all your priorities we take the lowest priority put it to the back rank everything draw our little Arc try to remember remember how the clot goes and R or S 1. The substituent below with the highest ranking according to the R, S rules is so let's start off with this look at these guys of this set of four which would have the highest priority according to the rules now this means that they're all attached to a cyal center okay so as we do this we look here this chyal Center is attached to a carbon carbon a carbon a carbon this carbon so it's a tie this carbon is attached oops to three hydrogens this carbon is attached to two hydrogens and a carbon but these two are each attached to two carbons we have hydrogen hydrogen and two carbons two carbons is more than one carbon and more than hydrogens so these two we can rule out so let's look at this guy this carbon is attached to two carbons and a hydrogen which we don't show this carbon here is attached two carbons as our double bond and this hydrogen so again it's a tie so we go out one more and we say what are these carbons attached to well here we have three hydrogens here we only have two three is bigger than two therefore C is our highest priority you just work your way down the chain until you get a difference so all four of those are connected to chyro we're all connected to a cyal center right so it's a tie at the first move our next one same thing here which of these is the lowest ranking well that ought to be simple they're all attached to carbons but this guy is attached only to three hydrogens these are attached to other carbons so what would make the double bond more favorable than if our double bond here say had a carbon attached up here that would be higher priority but the fact that we only have two hydrogen versus three three WIS so if that had three with a double bond that would be a higher but that would be a pint veent carbon and organic chemist all over the world would have 3. In the molecule shown below, indicate the substituent with the highest ranking according to the R.S rules. palpitations look at this which of these substituents around this chyro Center has the highest priority well it's easy to tell who's the lowest isn't it this is poor little hydrogen all the rest of these are attached to carbons so go to each of them and list what the carbon is attached to here we have two carbons and a hydrogen here we have have two hydrogens and this oxygen here we have three hydrogens now clearly this guy is going to beat him right but remember rule number one is atomic number even though we have two carbons here oxygen has a higher atomic number and therefore this is our highest priority oxygen has an atomic number of eight carbon's only six therefore even though there's only one oxygen winds um when you doing the atomic number for that one I don't know if this is a stupid question but it's like do you count all of them on on the dimethyl group Noe NOP all we do is look we see look at this say oh my goodness it's an oxygen atomic number eight it wins period it's over doesn't matter how many carbons are attached or whatever it's over because it has a molecule that has a higher carbon and a substituent not of all but a substituent yeah that's do it again going around this is our CYO Center we have a chlorine we have a alen we have a ch2 and a ch3 this is two carbons because remember we count the double bond twice this is two hydrogens and another carbon this is a chlorine and finally this is a carbon with three hydrogens it's the lowest priority and that's the one we would put to the back this is a very simple one because chlorine is going to have the highest atomic number it's 17 versus carbon that is six I will hush for a second quickly run around this molecule sign the priorities remember a carboxilic acid has a doubly bonded oxygen doesn't it has a carbonal bonded to it O So as we run around this molecule we have three hydrogens we have three oxygens here the double bond for the carbonal and the other one for the O we have a carbon and two hydrogens bonded to an oxygen and finally our lowest priority is the hydrogen so the um carox grou is it carboxilic acidic acid is one because it has more oxygens than the other one that's right so when we look at this here we have three oxygens here we have one this is our highest priority so you should you when you see that you should just assume that there's a double even though it's not shown that's right you you know that we're going to do one right now well maybe not now but soon that has the the carbonal shown just in case you forgot look at this one yeah where did you get we'll see a carox acid it'll be here we have a carbon nitrogen triple bond that counts as three nitrogens doesn't it how do you know it's Triple B the car you remember that's a nitr functional group and a nitr has a carbon nitrogen triple bond that's why we did functional groups right would could you remember because carbon has to have four bonds has to have four bonds right nitrogen has to have three all right there's a carbon with two hydrogens bonded to another carbon carbon with two hydrogens bonded to an oxygen carbon bonded to three nitrogens our lowest priority group here is going to be our methyl and as we see one of the great challenges of stereochemistry is that you are going to have to be able to rraw this with the methyl back all right what's our highest priority group everyone wants to say the three nitrogens but of course not oxygen that's a higher atomic number even though there here is our c oxyc acid again double bonded oxygen and then an O for that last one you said that like if we were to redraw this you want to see we do that in just a minute okay yes keep your excitement High all right as we go around we're going to have a carbon bonded to three oxygen a carbonal in the O we have a chlorine two hydrogens in an oxygen and a carbon with three hydrogens that will be our lowest priority and once again to do RNs we have to move that to the back [Music] floring is going to be our highest priority now if we went through and assign the other ones here we have three oxygen here we only have one so this would be number two that would be number three and of course that is number four so this isn't tough to do is it let's just do one last one here ch3 fo group and a carbon carbon double bond three hydrogens two carbons and a hydrogen two hydrogens and a carbon hydrogen is our lowest priority highest two carbons versus one this is one this would be two that would be three and that's four okay we saw this earlier this is the amino acid alanine what I'd like you to do is to look at this and tell me if this molecule is r or S what is the absolute configuration of this molecule step one we want to assign our priorities which group is going to be the highest three oxygen three hydrogens a nitrogen oxygen is going to win isn't it these are our atomic numbers I'm sorry the nitrogen WIS not our carbon this is number one hydrogen is number four here we have three hydrogens here we have three oxygens this is going to be number two and that's going to be number three why is that number one our atomic number here is more remember what's attached directly these are carbons they're tied nitrogen is a higher atomic number right over here atomic number is seven carbon is six so that counts if it's directly attached to the directly attached to the chyro center all right so here are these are our numbers next we have to somehow adjust this molecule so that the lowest priority group is pointing backward now you could take your little thingy here and you could look and say all right well the way this is drawn this hydrogen here is pointing back isn't it so the operation I want is to just rotate this guy this way so the hydrogen is straight back that's the operation I want this is this hydrogen right here and the operation simply looks like this we rotate it so it heads into the back start off with it down rotate it into the back this is also called the steering wheel view our lowest priority is buried at the back and these guys are all around my spokes now let's convert this to a drawing hydrogen is buried in the back nitrogen methyl carboxilic acid our priorities are one two and three so what is our stereochemistry this is going this way look at the clock it goes the other way this is s configuration all of the naturally occurring amino acids are s configuration isn't that worth going to now you see this is why having something like this one of them on an exam can be useful because again you can look at this say the hydrogen is pointing back I just need to rotate this way methal is my orange it comes forward and these guys are basically unchanged same operation on this guy go ahead and take and do it in your head step one assign your priorities which is the highest our double bond we have two carbons this is one carbon that's three hydrogens hydrogen is number four two carbons meets out one carbon this is number one that's number two that's number three now we want we draw it we want to tip it so the hydrogen is going away our methyl comes towards us when we do the tip it looks like this so should the third one always be up well we'll see things can get really interesting here's are numbers 1 2 3 going this way this is what configuration R this is an R now let's look for a shortcut everyone loves a shortcut right let's go back here to this if we didn't rotate this at all if we simply said going here one two three what direction is that clockwise that's an R isn't it we could tell that without even rotating it this one is going to be a little bit different I have my lowest priority group here on the side this one coming out my priorities well this guy is an oxygen this has two carbons has three hydrogens so this one's going to win in terms of priorities but we're going to have to rotate so our lowest group goes to the back now the operation we're going to do is going to look like this look at this purple thingy here relative to the hydrogen when I take it I rotate the hydrogen to the back this purple guy swings all the way around to the other side doesn't it the other two really really don't change position much at all this stays up this stays down they're just rotating this one this way so oops there's our rotation again note whatever's on this side swings over to here these two are pretty much unchanged but whatever was on the other side swings all the way to this side the operation is just rotating around this axis like this and this is what we get we have rotated this guy all the way to this side these two are about the same now our priorities this is one that is two that is three hydrogen of course is four what's our configuration this is our configuration yes everyone see this all right this was a problem directly off of the exam you were asked to asside the absolute configuration of each of these guys so let's do it quickly just like it was in ex for our first one the hydrogen's back that's good whenever you see that you say bromine's going to be number one the eth group's going to be number two this is going to be number three hydrogen the back all we have to do is assign our priorities connect the dots we're going this way this is our configuration our next guy here hydrogen is going to be our lowest priority isn't it so just like the previous problem we're going to have to spin around this axis and move it to the back that's going to take this guy and swing it all the way over to this side take your little thingy look at it you don't see that going to rotate around this axis hydrogen to the back this means the Amal group is going to wind up over here these two will be slightly changed but not badly and we look at something like this once again the amino group is going to swing over to this position now our priorities bromine's going to win we atomic number nitrogen is directly attached that's better than a carbon we draw our Arc what is this this is s configuration this guy I have a quick question so you have a carbon and a nitrogen attached to that you have nitrogen and a hydrogen attached to that right wouldn't the carbon with nitrogen have a higher atomic number than NR attach directly to the chyo center nitrogen carbon okay that makes sense I'm going to do this in two steps I'm going to rotate around this axis first which is going to swing the co to this side I have another question want the carbon have um three bonds nitrogen right there right but again what is ATT directly to the center Chine nitrogen carbon's our Center attach the car all right so I'm going to rotate this this way when I do we're going to swing this over this is going to wind up in the middle now we could actually assign our groups directly using our shortcut or we could complete this by just taking and tipping this so this is back when we do this is going to come forward so are these did you miss that I'm just why don't you just it backwards well we could tip it backwards we still would have to do two rotations though two operations why is that um just the way the thing is drawn you play with your model you can see that but the two the simplest two I think are to rotate first this way and now we're going to take tip hydrogen back when we do that methyl is going to come forward these two guys will stay forward now our priorities attached here we have a nitrogen that's going to win this is a carbon that's a carbon REO or two yeah three oxyg three hydrogens and the answer is s configuration why oh methyl group here is our lowest priority and again we are relieved to see that it's pointing back right so what are our groups Ro to win then chlorine then carbon and this is s configuration these are also exam problems these were the easy ones because these require no rotation at all do they hydrogens are to the back so very quickly look at these and assign their priorities Romine is going to win this is an ethyl group that's going to beat out the methyl we're going this way and that's our chlorine is going to win here eth group Double B this is going to be number two we're going this way and that's our configuration don't you have to use the RS and names RS and s's yep name the good timing on your question though how would you name this compound once again exam number one why won't they ever this easy right we look at here we have a four carbon chain so it's a butane isn't it we have a bromine and carbon number two so it's a two bromo mutane and what is the configuration of that carbon it is R you show the R as an uppercase in parenthesis no space again are two bromo butane yeah okay I have a question on the priorities again why is the second one an R it this is highest this carbon is attached to two carbons and a hydrogen one two well um I took the other numbers off that's the numbering for the alkane chain one last one go ahead and do this again it was an exam problem our hydrogen's poting to the front that just really upsetting isn't it but no problem all we have to do is rotate around the axis until it goes that way when we rotate to get this pointing back this guy is going to swing to this side this one will swing to that side this will swing in front our priorities let's do that again because it's so much fun we're going to move this to the back this is going to swing over this is going to swing over to this side and our methyl Group which was back is now going to swing front you see why if you happen to have this gift in your brain this can be so trivial people that get this can't understand can't badle why people that don't get it can't it's just you know right and left brains s all right so here we are we moved it to the back we have a carbon bonded to two carbons carbon bonded to three carbons and a carbon bonded to three hydrogens and our configuration is any questions we're going to readdress this on a slide in a bit but let's just look at this what if we didn't wrot rotate this at all what if we assigned our priorities this would be number one number two number three right one two three that would be counterclockwise right if we connected these dots one 2 three we're going this way LDS our lowest priority group is pointing forward here's the shortcut if your lowest priority group is pointing up toward you you take whatever configurations you get it's only for high right as long as it's a lowest if the lowest is pointing in the front just do your priorities and write the the wrong answer that's the right answer which is the right answer all right let's do this backwards we know how to name simple Hollow alanes draw the structure of R to bromo butane for me you're going to draw your VI carbon two you can put a bromine you want to draw this with the hydrogen pointing back then you look and you see how everything is arranged if you screwed it up you just swap two groups the molecule would look like this we have a two bromo butane I have drawn my hydrogen going back my priorities are going to be one two three whoops one two three so as I go I'm going this away and that's all now if you would have happen to draw it backwards again all you do is you look at it and say oops erase two of them swap their positions and now you will have r two bromo two chlorobutane draw butane on carbon two you're going to put a chlorine and a bromine that's going to leave us a methyl and an ethyl isn't it the methyl is going to be our lowest priority so we want to put it B draw it do the configuration if it's wrong simply swap any two groups you should get something like this that's back this is going to be priority 1 2 3 and that's going in the S Direction Why is there an there is a one two 3 four aan supposed to be they are one longest chain right looking at a nothing hereit can you explain that one more time what's your longest chain here one 2 3 4 it's a butane now I put the methyl to the back because it's going to be lowest priority then our priorities are one two breathe and that's s we have ethane we have one bromo one chloro that means we have a hydrogen also we're going to put that to the back going to attach a bromine and a chlorine two carbons or one more carbon out and we would get something like this here's our two carbon chain hydrogen back our priorities bromine chlorine methyl going this away and there an S yeah sorry for the second one how did you determine which one is the priority again Romine highest atomic number then chlorine then carbon this carbon we have a hexane so we need six carbons in a row a carbon three we need a methyl on a chlorine and wind up with something like this six carbon chain we have a methyl and carbon 3 and a chlorine at Carbon 3 methyl is our lowest priority I draw it going back and my priorities are the chlorine is number one this guy three carbon chain two carbon chain so three wins this is number two that's number three we're going this way and that's R okay so for the next one what does the one R and the 3s stand for well here we we have two parel centers don't we we have a cyclohexane and in carbon one we going have a methyl in carbon 3 we're going to have another methyl the carbon or the methyl group on carbon one is going to be R the methyl group on carbon three is going to be S so we get to do it all twice well those are our priorities here we go let's call this carbon one I've drawn this with wedgies nice thing about drawing it with wedgies is I can very easily put my hydrogens in the back right our priorities around here this a ch3 both of these are ch2s aren't they so that's a tie but going this way we hit a tertiary carbon going this way we hit a secondary therefore this is going to be carbon one two and if we connect our doots here this is clockwise for our configuration now what if you would mess it up what would you do I would simply erase my wedgie here and put it as a dashed wedgie if I do that I reverse my configuration that is the lowest priority group instead of going back it's going to be coming towards us whoops for a second one here okay for our second one same configuration this is going to be highest second lowest hydrogen back we are going one two three this way counterclockwise and is s can you explain to me why one is one instead of two being one because I would think that two have moreti attached to it than one what do you mean this this one on top oh okay this is a top these are both ch2s right right but this ch2 is attached to a tertiary carbon isn't that attach this attached to another ch2 um let's do our cyc pentane we have a chlorine in one we have a methyl in three and there R and yes so draw your cyclopentane put in your wedges we'll put our hydrogen's going back carbon one is going to have the chlorine on it as goes alphabetically hydrogen is back therefore Florine is highest priority these are both ch2s but this is tertiary that's another ch2 our configuration here is going clockwise so carbon one is our did I draw a carbon 2 or carbon 3 correctly our highest prior priority is going to be this carbon once again because this is attached to a tertiary our ch2 this is another ch2 so we have one two and three I'm going counterclockwise and that's an S so you always want to start with wedge bars that's the simplest that's the simplest yeah I think because in a case like this it's easy to put the hydrogens to the back all right you can probably understand now why on the final exam I've decided to make a stereochemistry part A T on right so you can sit with your little toy and you can do this and this and this all day if you have to until it makes sense exam two however will have these things in real time all right simple problem here we have two caral centers don't we use an Asis to label them now our top one here we have hydrogen methyl o bottom half of the molecule chlorine methyl hydrogen two chyo this guy over here same thing o ch3 h chlorine methyl hydrogen this guy so they each have two chyal centers what is the relationship between these two molecules they're isomers aren't they they have the same molecular formula let's ask the question are they miror images proper magic mirror closest here is the O closest is the O ch3 is far this is going straight up these two are miror images but what about our bottom carbons here chlorine is closest methyl's closest in fact if you just look at the bottom carbon this is identical with this isn't it so we have identical at this power center and in anic at this par Center these are examples of what are called diers unlike an aners that have the same physical properties bi arov are totally different chemical compounds different melting point boiling point different everything they are identical at at least one Center and an antimic at at least one more so if you have more than one you look to see are they identical at all of them in which case they're the same molecule are they entiers at all of them in which case they're ananti but if they differ like this one did they are dier let's do this again methyl hydroxy hydrogen this methyl hydroxy hydrogen this methyl hydroxy hydrogen methyl hydroxy hydrogen they each have two c are they in antiverse drop our mirror Oh's are both closest to the mirror these are both away hydrogen's all in the plane they look like they would be mirror images so the simple answer would be oh yes they're an antiverse but they're not these are identical molecules let's just do this change them into little balls clearly these are identical right or images what I'm going to do is take this guy and I'm going to rotate 180° and oh my goodness they are identical one why in the world is this look at this molecule this has an internal plane of symmetry an internal f it has two chiro carbons but they're the reflection of each other here's our molecule right the top half is the reflection of the bottom half this has an internal of symmetry and it's calledo miso compounds they contain chyo centers but the molecule itself is a chyal it has chyro centers but because it has an internal plane it is not look at another one this is one two dimethyl cyclopentane with Cy stereochemistry we look here we have a methyl a hydrogen ch2 a tertiary carbon methyl hydrogen ch2 tertiary carbon this molecule however is a miso compound because you can put a mirror plane right through the middle and this half reflects this half any questions let's just do one more mindboggling thing here look at these two molecules how many caryo Centers do they have they are all SP s carbons by definition you can have no caral centers and yet this molecule is the mirror image of this room this is a helix it's a staircase here we're going around counterclockwise here we go around clockwise this is stereochemistry on a macroscopic l as opposed to microscopic proteins in the body are this way they are pyal as a glob and they have a mirror image and they're not superimposable on it to let this thing rotate again you can see the spiral to it as it goes around really kind of a cute moment any questions well like I said this is a little bit of a challenging chapter isn't it let's take a very short break here and then let's work through these problems [Applause] clean that's to is all right let's go ahead and do these These are verbatim off of former exam number two for our first one here all you have to do is convert the name into an Alan then go through and place a little asteris on the kyal tarbin we have a pentane in carbons two and four we have methyl groups when we look at this is this guy a caryo has a hydrogen two methyls and a ch2 is this guy same thing two methyls ch2 hydrogen there are no chyal centers here we have a hepan at seven carbons have an ethyl in five we have two methyls in number three this gbet has two methyls therefore it is a chyro this guy here has a hydrogen but it has two ethels it is also a chyal there are Noy centers we have an octane that's eight carbons in carbons four and five we have methyl groups this carbon has a hydrogen a methyl one one two three carbon branch and then the rest of this Sky they are different in the sky this one has a hydrogen a methyl three carbon Branch the rest of this there are two chyal centers in this molecule any questions I have a question um so why is the last one have a pyro Center and first one does this carbon here has a hydrogen and two methyls two meths you have two of anything it's notable what about the second we have two eths two methyls all right look at this guy and assign this configuration R our lowest priority is going to be the hydrogen it's in the back that's good as we go around here our highest priority is going to be the branch coming up then our twocc carbon branch and then our meal all we do is connect the dots and this is s configuration yeahor why is um the priority the priority why is the priority the priority in this well carbon that are attached to our carl center that's a carbon that's a carbon that's a carbon it's a tie right so then you look well this is attached to a ch2 three hydrogens that's a ch2 so it's a tie between these two so you go to the next carbonal this is attached to a let's see this is attached to a ch2 this is attached to a ch3 this guy is only attached to three hyrogen so one 2 3 S configuration this is an epoxide this kind of tests your ability to look at these drawings and say oh I see it our hydrogen is kind of pointing back which means that all these guys this guy is pointing towards us this is back these two are more or less in the plane oxygen is going to be our highest priority here highest atomic number this is a ch2 oxygen this is a ch3 so we have one two and three now because our our hydrogen is pointing back if we kind of looked at this molecule from the top here we would see down this axis that we're going one two three that way let's do that again one two three that way more or less from the top but again because it's pointing back you actually don't have to fiddle with this this is going to be S even the W it's SP hydrogen is our lowest priority is pointing forward right I'm going to take this and rotate it even though I know I don't have to but I will when I rotate it my isopropyl is going to stay more or less still the cyclo is going to swing to this side this going to swing to this side it's going to look like that again isopropyl doesn't move these two will simply swing to opposite sides test would you be looking for us to redraw it or just to tell you the configuration well you can do either whatever works for you that's what's what happens with stereochemistry whatever works in your mind all right our lowest priority is back which is the highest priority carbon well they're carbons and all of them aren't there this is ch2 so that loses this is going to be a CH with two carbons CH with two carbons but these carbons are only attached to hydrogens these guys are attached back to another carbon therefore we're win our configuration is there our configuration once again hydrogen is forward go ahead and mentally perform your flip assign your priorities when we flip this guy is going to be on this side this one's going to be on the [Music] other and it's going to hold like that triple bond here winds up on this side ch2 triple bond winds up over here which is our highest priority carbon directly attached to here we have a triple bond that's a ch2 that's a ch2 so that's a tie but this ch2 is attached to a triple bond this is only attached to a ch2 and our configur duration is our we're going to have to rotate again what do we do the five numbered ring winds up on this side this guy up on this side now our priorities they're all carbons this is a ch2 this is attached to two carbons so is this this is the lowest of the three these ch2s are attached to other carbons these ch2s are attached to ch3s so our highest priority is there and lowest is our ethyl group and again we are R next semester we will learn what a toine sulfinate or tosilate is that's what this is though we'll call it OTS eventually this is our caral Center hydrogen front methyl and apple again we want to rotate this thing what we do the tosilate winds up on this side methyl front there's our Apple oxygen is going to be highest this is a ch2 to a carbon this is a ch3 so this is our configuration now for this whole last group where we had the lowest priority group facing forward we could have used our trick so let's take this structure and simply write our priorities in going to be 1 2 3 and by gosh this would be what counterclockwise it would be S but our lowest priority is pointing forward therefore the answer is R all in that last set with the lowest priority was forward you could have just simply directly assigned and then written the opposite answer whatever works for you all right assign each of these both hydrogens are back that's good we have three oxygens here we have one oxygen there we have a carbon bonded to an oxygen for this carbon the priorities would be whoops I wrote that wrong I'm so sorry oops this of course is our highest priority then this one no no I'm sorry I'm right I'm right I'm so relieved I'm right oxygen is directly attached isn't it so that makes it number one directly attached here is a carbon so that's number two sometimes you can stand too close to these things you know and of course this is number three so this configuration is R but this other one here again our oxygen is going to be highest then our carbon eal and then our CH HOH this is going this way it is also a is this compound now here's a trick this wasn't on the test but here's a trick if you look at this this has aryc acid hydrogen oh right so does this so this could be a meil compound could it it could be a miso but it's not how do you know chro is AAL compound must have power centers but the top half must be the reflection of the bottom half so if this is R this one would have to be S if this was RNs it would be musas well this is just a chiro compound you'd have to compare it to something else two compounds our hydrogens are both pointing back aren't they that's good our priorities are going to be this is ch2 ch2 this is a tertiary carbon it's going to win there's our hydrogen in the back tertiary carbon ch2 methyl group what's our configuration it's an AR what's the configuration on the other one hydrogen is back that's going to be one that's going to be two that's going to be three and this is also R once again this could potentially be a miso compound but it's not it's a meal compound they must be opposite all right so name this thing what is it what's our longest shade one 2 three four five six 78 it is a four five dimethyl Octane and they're both r four r five R you have to four five four five dimethyl octane with this one let's look at this Center first hydrogen's pointing back that's the easy one we'll do that first we have a chlorine this is a ch2 that's attached to an oxygen our configuration here is 1 2 3 is an S now our top one here we could assign it directly here and just reverse it our priorities would be one 2 three that away and that's AR let's prove it here I can take this and flip it hydrogen is back highest second and third 1 2 3 and it's s so the carbon isn't uh directly connected to the chlorine no this carbon is directly attached to the chlorine now once again if we had just looked at it this way we would say 1 2 3 that's going clockwise isn't it R but because the lowest priority is pointing forward it is in fact s last one the oxygen are going to be the highest priority these are ch3s this is a ch2 to a carbon our methyl group is lowest and it's pointing back so our configuration at this carbon we're going this way one two three I always had to think about that but that's s what's our next carbon our methyl is back this is one that's two that's three going this way and that's an R what can you tell me about this molecule is this a misoc compound yes absolutely this is s this is R let's go back get rid of all this if you look at it if we put a mirror plane right down the middle this half is a reflection of this this is r that is s any questions all right well this was on the exam I certain hope everybody the pace continues on Monday we will do chapter six good news is chapter six is trival a
10834	https://en.wikipedia.org/wiki/%CE%92-Lactam	Jump to content β-Lactam العربية বাংলা Bosanski Català Čeština Dansk Eesti Español فارسی Français Galego 한국어 Bahasa Indonesia Italiano Lombard 日本語 Português Romnă Русский Slovenčina Slovenščina Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Türkçe 中文 Edit links From Wikipedia, the free encyclopedia Family of chemical compounds This article is about the class of chemical compounds. For the related antibiotics, see β-Lactam antibiotic. A β-lactam (beta-lactam) ring is a four-membered lactam. A lactam is a cyclic amide, and beta-lactams are named so because the nitrogen atom is attached to the β-carbon atom relative to the carbonyl. The simplest β-lactam possible is 2-azetidinone. β-lactams are significant structural units of medicines as manifested in many β-lactam antibiotics. Up to 1970, most β-lactam research was concerned with the penicillin and cephalosporin groups, but since then, a wide variety of structures have been described. Clinical significance [edit] Main article: β-Lactam antibiotic The β-lactam ring is part of the core structure of several antibiotic families, the principal ones being the penicillins, cephalosporins, carbapenems, and monobactams, which are, therefore, also called β-lactam antibiotics. Nearly all of these antibiotics work by inhibiting bacterial cell wall biosynthesis. This has a lethal effect on bacteria, although any given bacteria population will typically contain a subgroup that is resistant to β-lactam antibiotics. Bacterial resistance occurs as a result of the expression of one of many genes for the production of β-lactamases, a class of enzymes that break open the β-lactam ring. More than 1,800 different β-lactamase enzymes have been documented in various species of bacteria. These enzymes vary widely in their chemical structure and catalytic efficiencies. When bacterial populations have these resistant subgroups, treatment with β-lactam can result in the resistant strain becoming more prevalent and therefore more virulent. β-lactam derived antibiotics can be considered one of the most important antibiotic classes but prone to clinical resistance. β-lactam exhibits its antibiotic properties by imitating the naturally occurring d-Ala-d-Ala substrate for the group of enzymes known as penicillin binding proteins (PBP), which have as function to cross-link the peptidoglycan part of the cell wall of the bacteria. The β-lactam ring is also found in some other drugs such as the cholesterol absorption inhibitor drug ezetimibe. Synthesis [edit] The first synthetic β-lactam was prepared by Hermann Staudinger in 1907 by reaction of the Schiff base of aniline and benzaldehyde with diphenylketene in a [2+2] cycloaddition (Ph indicates a phenyl functional group): Many methods have been developed for the synthesis of β-lactams. The Breckpot β-lactam synthesis produces substituted β-lactams by the cyclization of beta amino acid esters by use of a Grignard reagent. Mukaiyama's reagent is also used in modified Breckpot synthesis. Reactions [edit] Due to ring strain, β-lactams are more readily hydrolyzed than linear amides or larger lactams. This strain is further increased by fusion to a second ring, as found in most β-lactam antibiotics. This trend is due to the amide character of the β-lactam being reduced by the aplanarity of the system. The nitrogen atom of an ideal amide is sp2-hybridized due to resonance, and sp2-hybridized atoms have trigonal planar bond geometry. As a pyramidal bond geometry is forced upon the nitrogen atom by the ring strain, the resonance of the amide bond is reduced, and the carbonyl becomes more ketone-like. Nobel laureate Robert Burns Woodward described a parameter h as a measure of the height of the trigonal pyramid defined by the nitrogen (as the apex) and its three adjacent atoms. h corresponds to the strength of the β-lactam bond with lower numbers (more planar; more like ideal amides) being stronger and less reactive. Monobactams have h values between 0.05 and 0.10 angstroms (Å). Cephems have h values in of 0.20–0.25 Å. Penams have values in the range 0.40–0.50 Å, while carbapenems and clavams have values of 0.50–0.60 Å, being the most reactive of the β-lactams toward hydrolysis. See also [edit] Azetidine Lactone References [edit] ^ Gilchrist T (1987). Heterocyclic Chemistry. Harlow: Longman Scientific. ISBN 978-0-582-01421-3. ^ Flynn EH (1972). Cephalosporins and Penicillins : Chemistry and Biology. New York and London: Academic Press. ^ Fisher, J. F.; Meroueh, S. O.; Mobashery, S. (2005). "Bacterial resistance to β-lactam antibiotics: compelling opportunism, compelling opportunity". Chemical Reviews. 105 (2): 395–424. doi:10.1021/cr030102i. PMID 15700950. ^ Hosseyni S, Jarrahpour A (October 2018). "Recent advances in β-lactam synthesis". Organic & Biomolecular Chemistry. 16 (38): 6840–6852. doi:10.1039/c8ob01833b. PMID 30209477. ^ Brandt C, Braun SD, Stein C, Slickers P, Ehricht R, Pletz MW, Makarewicz O (February 2017). "In silico serine β-lactamases analysis reveals a huge potential resistome in environmental and pathogenic species". Scientific Reports. 7: 43232. Bibcode:2017NatSR...743232B. doi:10.1038/srep43232. PMC 5324141. PMID 28233789.{{cite journal}}: CS1 maint: article number as page number (link) ^ Ehmann DE, Jahić H, Ross PL, Gu RF, Hu J, Kern G, Walkup GK, Fisher SL (July 2012). "Avibactam is a covalent, reversible, non-β-lactam β-lactamase inhibitor". Proceedings of the National Academy of Sciences of the United States of America. 109 (29): 11663–8. Bibcode:2012PNAS..10911663E. doi:10.1073/pnas.1205073109. PMC 3406822. PMID 22753474. ^ Tipper DJ, Strominger JL (October 1965). "Mechanism of action of penicillins: a proposal based on their structural similarity to acyl-D-alanyl-D-alanine". Proceedings of the National Academy of Sciences of the United States of America. 54 (4): 1133–41. Bibcode:1965PNAS...54.1133T. doi:10.1073/pnas.54.4.1133. PMC 219812. PMID 5219821. ^ Tidwell TT (2008). "Hugo (Ugo) Schiff, Schiff bases, and a century of beta-lactam synthesis". Angewandte Chemie. 47 (6): 1016–20. doi:10.1002/anie.200702965. PMID 18022986. ^ Staudinger H (1907). "Zur Kenntniss der Ketene. Diphenylketen". Justus Liebigs Ann. Chem. 356 (1–2): 51–123. doi:10.1002/jlac.19073560106. Archived from the original on 2020-08-02. Retrieved 2019-06-27. ^ Alcaide, Benito; Almendros, Pedro; Aragoncillo, Cristina (2007). "Β-Lactams: Versatile Building Blocks for the Stereoselective Synthesis of Non-β-Lactam Products". Chemical Reviews. 107 (11): 4437–4492. doi:10.1021/cr0307300. PMID 17649981. ^ Hosseyni, Seyedmorteza; Jarrahpour, Aliasghar (2018). "Recent advances in β-lactam synthesis". Organic & Biomolecular Chemistry. 16 (38): 6840–6852. doi:10.1039/C8OB01833B. ISSN 1477-0520. PMID 30209477. ^ Pitts, Cody Ross; Lectka, Thomas (2014-08-27). "Chemical Synthesis of β-Lactams: Asymmetric Catalysis and Other Recent Advances". Chemical Reviews. 114 (16): 7930–7953. doi:10.1021/cr4005549. ISSN 0009-2665. PMID 24555548. Archived from the original on 2022-07-21. Retrieved 2020-12-17. ^ Jump up to: a b "Breckpot β-Lactam Synthesis", Comprehensive Organic Name Reactions and Reagents, Hoboken, NJ, USA: John Wiley & Sons, Inc., pp. 521–524, 2010-09-15, doi:10.1002/9780470638859.conrr115, ISBN 978-0-470-63885-9, archived from the original on 2024-01-16, retrieved 2021-02-04 ^ Bogdanov B, Zdravkovski Z, Hristovski K. "Breckpot Synthesis". Institute of Chemistry Skopje. Archived from the original on 2015-11-06. Retrieved 2014-12-30. ^ Woodward RB (May 1980). "Penems and related substances". Philosophical Transactions of the Royal Society of London. Series B, Biological Sciences. 289 (1036): 239–50. Bibcode:1980RSPTB.289..239W. doi:10.1098/rstb.1980.0042. PMID 6109320. ^ Nangia A, Biradha K, Desiraju GR (1996). "Correlation of biological activity in β-lactam antibiotics with Woodward and Cohen structural parameters: A Cambridge database study". J. Chem. Soc. Perkin Trans. 2 (5): 943–53. doi:10.1039/p29960000943. External links [edit] Synthesis of β-lactams Retrieved from " Category: Beta-lactams Hidden categories: CS1 maint: article number as page number Articles with short description Short description matches Wikidata
10835	https://www.youtube.com/watch?v=pxHd8tLI65Q	This question looks impossibly hard at first. But work carefully and the solution is beautiful. MindYourDecisions 3230000 subscribers 7601 likes Description 268211 views Posted: 18 Oct 2022 This puzzle is adapted from a British Mathematical Olympiad question, but it makes for a very interesting technical coding interview question too. Part one is to find all solutions in non-negative integers so the sum of two square roots is the square root of 2023. The second is to generalize and write Python code to compute all solutions for the square root of a given natural number n. Thanks to Rohit for reviewing the computer code! Based on BMO 2008/2009 Round 2 problem 1 Python Tutor code visualizer References Subscribe: Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas! If you purchase through these links, I may be compensated for purchases made on Amazon. 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My Blog Twitter Merch Patreon Press 343 comments Transcript: hey this is presh talwalkar find all solutions in non-negative integers a and b such that the square root of a plus the square root of b is equal to the square root of two thousand twenty-three that's part 1 of the question part 2 is a coding challenge and i will present a solution in python let's say you are given a non-negative integer n write a program to find all solutions in non-negative integers a and b such that the square root of a plus the square root of b is equal to the square root of n pause the video if you'd like to give this problem a try and when you're ready keep watching to learn how to solve this problem [Music] we begin by tackling the first part of the question whenever you have square roots it's a good idea to try to eliminate the square roots we will first solve for the square root of a then we will square both sides to get that a is equal to the square of the square root of 2023 minus the square root of b we then expand the binomial to get a is equal to 2023 plus b minus 2 times the square root of 2023 b now let's analyze this equation a is a non-negative integer 2023 is an integer b is an integer the only part of the equation that might not be an integer is the square root of 2023 multiplied by b so we'll notice that if and only if this square root evaluates to be an integer we will have a solution so we have a condition that will find a solution if and only if the square root of 2023 multiplied by b is a natural number including 0. then the prime factorization of 2023 is 7 multiplied by 17 squared so this is equivalent to 17 multiplied by the square root of 7b now this will evaluate to be a natural number if the square root of 7b is a natural number so b is equal to 7 multiplied by x squared where x is a non-negative integer so that's the condition for a solution here by very similar logic we can solve for the square root of b and figure out a condition in this case so that a is equal to 7 multiplied by y squared where y is a non-negative integer so we figured out some conditions for b and a we'll substitute that into our original equation from here we can simplify because x and y are non-negative integers so we have y multiplied by the square root of 7 plus x multiplied by the square root of 7 is equal to 17 multiplied by the square root of 7. we divide both sides by the square root of 7 so we have the simple equation x plus y is equal to 17 for non-negative integers x and y this is very easy to solve we have 0 17 1 16 to 15 and so on we have a total of 18 pairs we can then substitute in for x and y to get the pairs a and b so we'll end up with 18 pairs of solutions as shown on the screen and that's the answer to the first part of the question so now we need to take this analysis and generalize it let's say we are given a non-negative integer n and we want to find all solutions in non-negative integers a and b such that the square root of a plus the square root of b is equal to the square root of n how do we do that well let's go over a critical part of the proof in this analysis we saw the critical condition was that we had a solution if and only if the square root of 2023 multiplied by a was a natural number so in our general case we're going to have a solution if and only if the square root of n multiplied by a is a whole number so let's put this idea into pseudo code we're going to write a function solution n we need to have a go over the values of 0 to n we'll evaluate if the square root of a multiplied by n is an integer then we can compute the value of b and we will return an array of all ordered pairs a and b that are solutions so let's implement this in python this is not going to be the most efficient or best code it's just going to be useful for explaining the solution so we'll import the math library we're going to define a solution n we initialize a list for the result for a in the range 0 to n plus 1 a will go over the values of 0 to n we compute the value of b then if b is an integer we're going to add the pair a and the integer b to the solution list and we continue this loop eventually we return the result of all these ordered pairs so let's try n is equal to 16. we're going to print all the solutions here to get a better sense of the code we can use python tutors code visualization which visualizes the code line by line we import the math library we define the solution we have the value n is equal to 16 and we will print all the solutions so we go to the function solution 16 we first initialize an empty list we start with a is equal to zero we evaluate b we check if it's an integer if it is we add the pair a comma b we do this for a is equal to 1 we get another solution and we continue this loop for a is equal to 2 and so on and it will continue until a is equal to 16. in the meanwhile we keep adding ordered pairs a b if we find out that a and b are integer solutions this stops at a is equal to 16 and thus we are going to return the list result and we get the following pairs of 0 16 1 9 4 4 9 1 and 16 0 and those are all the solutions for n is equal to 16. what's amazing about putting this into computer code is that we can input other values of n we can then visualize all of these solutions just at the click of a button this is what happens as n goes over all these other integer values and we just get the results of all these ordered pairs it's so amazing to commute these things so quickly we don't have to go through each case by case now there is a hitch as we incorporate computer code into our solution there are all sorts of difficulties when you start coding so in this vein i had a computer programmer evaluate my code and he gave me some feedback i want to thank rohit for his help he suggested the following python code we import the math library then we define calc b as a function of a and n this will return the value of b based on a and n then we can define the solution n in one line so there are all sorts of ways that you can do this so i think it's wonderful if we can incorporate coding into mathematics thanks for making us one of the best communities on youtube see you next episode of mind your decisions where we solve the world's problems one video at a time
10836	https://www.unitconverters.net/electric-field-strength/millivolt-meter-to-volt-centimeter.htm	Convert Millivolt/meter to Volt/centimeter Home / Electric Field Strength Conversion / Convert Millivolt/meter to Volt/centimeter Convert Millivolt/meter to Volt/centimeter Please provide values below to convert millivolt/meter [mV/m] to volt/centimeter [V/cm], or vice versa. From:millivolt/meter To:volt/centimeter Millivolt/meter to Volt/centimeter Conversion Table \| Millivolt/meter [mV/m] \| Volt/centimeter [V/cm] \| --- \| \| 0.01 mV/m \| 1.0E-7 V/cm \| \| 0.1 mV/m \| 1.0E-6 V/cm \| \| 1 mV/m \| 1.0E-5 V/cm \| \| 2 mV/m \| 2.0E-5 V/cm \| \| 3 mV/m \| 3.0E-5 V/cm \| \| 5 mV/m \| 5.0E-5 V/cm \| \| 10 mV/m \| 0.0001 V/cm \| \| 20 mV/m \| 0.0002 V/cm \| \| 50 mV/m \| 0.0005 V/cm \| \| 100 mV/m \| 0.001 V/cm \| \| 1000 mV/m \| 0.01 V/cm \| How to Convert Millivolt/meter to Volt/centimeter 1 mV/m = 1.0E-5 V/cm 1 V/cm = 100000 mV/m Example: convert 15 mV/m to V/cm: 15 mV/m = 15 × 1.0E-5 V/cm = 0.00015 V/cm Convert Millivolt/meter to Other Electric Field Strength Units Millivolt/meter to Volt/meter Millivolt/meter to Kilovolt/meter Millivolt/meter to Kilovolt/centimeter Millivolt/meter to Microvolt/meter Millivolt/meter to Kilovolt/inch Millivolt/meter to Volt/inch Millivolt/meter to Volt/mil Millivolt/meter to Abvolt/centimeter Millivolt/meter to Statvolt/centimeter Millivolt/meter to Statvolt/inch Millivolt/meter to Newton/coulomb All Converters Common ConvertersEngineering ConvertersHeat ConvertersFluids ConvertersLight ConvertersElectricity Converters ChargeLinear Charge DensitySurface Charge DensityVolume Charge DensityCurrentLinear Current DensitySurface Current DensityElectric Field StrengthElectric PotentialElectric ResistanceElectric ResistivityElectric ConductanceElectric ConductivityElectrostatic CapacitanceInductance Magnetism ConvertersRadiology ConvertersCommon Unit Systems about us \| terms of use \| privacy policy \| sitemap © 2008 - 2025 unitconverters.net
10837	https://stackoverflow.com/questions/29548744/creating-a-truth-table-for-any-expression-in-python	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Creating a truth table for any expression in Python Ask Question Asked Modified 3 years, 1 month ago Viewed 43k times 10 I am attempting to create a program that when run will ask for the boolean expression, the variables and then create a truth table for whatever is entered. I need to use a class and this is what I have so far. I am not sure where to go from here. from itertools import product class Boolean(object): def __init__(self, statement, vars): self.exp = statement self.vars = vars def __call__(self, statement, vars): def main(): expression = raw_input('Give an expression:') vars = raw_input('Give names of variables:') variables = vars.split(' ') b = Boolean(expression, variables) if __name__ == "__main__": main() python boolean Share Improve this question edited Apr 9, 2015 at 20:56 DNA 42.7k1212 gold badges114114 silver badges153153 bronze badges asked Apr 9, 2015 at 20:52 say786say786 10111 gold badge11 silver badge33 bronze badges 5 share with us what might the input / output would look like taesu – taesu 2015-04-09 20:55:19 +00:00 Commented Apr 9, 2015 at 20:55 I don't really understand what you are asking... moreover, I highly doubt you need to use a class, except if this is some homework and that's a requirement for the exercise. A function is probably enough. Bakuriu – Bakuriu 2015-04-09 20:55:58 +00:00 Commented Apr 9, 2015 at 20:55 why are you importing itertools? Daniel – Daniel 2015-04-09 20:56:12 +00:00 Commented Apr 9, 2015 at 20:56 I am required to use a class for this exercise. say786 – say786 2015-04-09 21:08:25 +00:00 Commented Apr 9, 2015 at 21:08 The output for this would be asking for an expression, asking for the variables and then outputting a truth table say786 – say786 2015-04-09 21:09:05 +00:00 Commented Apr 9, 2015 at 21:09 Add a comment \| 4 Answers 4 Reset to default 14 I have a library that does exactly what you want! Check out the github repo or find it here on pypi. The readme describes how everything works, but here's a quick example: from truths import Truths print Truths(['a', 'b', 'x', 'd'], ['(a and b)', 'a and b or x', 'a and (b or x) or d']) +---+---+---+---+-----------+--------------+---------------------+ \| a \| b \| x \| d \| (a and b) \| a and b or x \| a and (b or x) or d \| +---+---+---+---+-----------+--------------+---------------------+ \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| \| 0 \| 0 \| 0 \| 1 \| 0 \| 0 \| 1 \| \| 0 \| 0 \| 1 \| 0 \| 0 \| 1 \| 0 \| \| 0 \| 0 \| 1 \| 1 \| 0 \| 1 \| 1 \| \| 0 \| 1 \| 0 \| 0 \| 0 \| 0 \| 0 \| \| 0 \| 1 \| 0 \| 1 \| 0 \| 0 \| 1 \| \| 0 \| 1 \| 1 \| 0 \| 0 \| 1 \| 0 \| \| 0 \| 1 \| 1 \| 1 \| 0 \| 1 \| 1 \| \| 1 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| \| 1 \| 0 \| 0 \| 1 \| 0 \| 0 \| 1 \| \| 1 \| 0 \| 1 \| 0 \| 0 \| 1 \| 1 \| \| 1 \| 0 \| 1 \| 1 \| 0 \| 1 \| 1 \| \| 1 \| 1 \| 0 \| 0 \| 1 \| 1 \| 1 \| \| 1 \| 1 \| 0 \| 1 \| 1 \| 1 \| 1 \| \| 1 \| 1 \| 1 \| 0 \| 1 \| 1 \| 1 \| \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| 1 \| +---+---+---+---+-----------+--------------+---------------------+ Hope this helps! Share Improve this answer answered Jul 12, 2016 at 23:41 penchantpenchant 2,36333 gold badges1717 silver badges2020 bronze badges Comments 10 You could simply define any boolean function right in python. consider the following example: def f(w,x,y,z): return (x and y) and (w or z) I've wrote a snippet that takes any function f, and returns its truth table: import pandas as pd from itertools import product def truth_table(f): values = [list(x) + [f(x)] for x in product([False,True], repeat=f.func_code.co_argcount)] return pd.DataFrame(values,columns=(list(f.func_code.co_varnames) + [f.func_name])) Using this will yield (in a nicely formatted html if you're using IPython Notebook): truth_table(f) w x y z f 0 False False False False False 1 False False False True False 2 False False True False False 3 False False True True False 4 False True False False False 5 False True False True False 6 False True True False False 7 False True True True True 8 True False False False False 9 True False False True False 10 True False True False False 11 True False True True False 12 True True False False False 13 True True False True False 14 True True True False True 15 True True True True True Cheers. Share Improve this answer answered May 16, 2015 at 15:39 OmerBAOmerBA 84299 silver badges1414 bronze badges 1 Comment Duncan WP Duncan WP With Python 3 you need to use __name__ and __code__ instead of func_name and func_code respectively 3 You probably want to do something like this: from itertools import product for p in product((True, False), repeat=len(variables)): # Map variable in variables to value in p # Apply boolean operators to variables that now have values # add result of each application to column in truth table pass But the inside of the for loop is the hardest part, so good luck. This is an example of what you would be iterating over in the case of three variables: ``` list(product((True, False), repeat=3)) [(True, True, True), (True, True, False), (True, False, True), (True, False, False), (False, True, True), (False, True, False), (False, False, True), (False, False, False)] ``` Share Improve this answer answered Apr 9, 2015 at 21:08 ShashankShashank 13.9k55 gold badges3939 silver badges6363 bronze badges 4 Comments say786 say786 Would that be placed below what I have currently say786 say786 What about building the truth table? And do I need to define product anywhere? Shashank Shashank The import statement defines product. The truth table can be just about anything but I'd recommend using a dictionary where the keys are strings (expressions) and the values are bools. say786 say786 Thanks. I was trying to figure out how to create a truth table that knows to use the function that's the input. 0 If you don't mind providing the number of variables of the function (I think it is possible to get but I don't know how at this moment). I have the following: from itertools import product B = {0,1} varNames= ["r","s","t","u","v","w","x","y","z"] def booltable(f,n): vars = varNames[-n:] header = vars + ["f"] table = [reversed([map(lambda input: [input,f(input)], product(B,repeat=len(vars)))])] return [header] + table If you want to have your 1's at the top (like I do), just reverse the answer. return [reversed([map(lambda input: [input,f(input)],product(B,repeat=len(vars)))])] Here is an example of how to use it, functions are defined using bitwise operations. x \| y - bitwise or of x and y x ^ y - bitwise exclusive or of x and y x & y - bitwise and of x and y ~x - the bits of x inverted ``` Example function def aBooleanFunction(x,y,z): return (x \| y) ^ ~(x ^ z) % 2 # Run display(booltable(aBooleanFunction,3)) ``` The output is: You can then parse this to a table in whatever format you want using, for example, tabulate. Share Improve this answer edited Aug 15, 2022 at 7:25 answered Aug 15, 2022 at 7:24 MetalMathicianMetalMathician 15799 bronze badges Comments Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions python boolean See similar questions with these tags. The Overflow Blog The history and future of software development (part 1) Getting Backstage in front of a shifting dev experience Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure New and improved coding challenges New comment UI experiment graduation Policy: Generative AI (e.g., ChatGPT) is banned Linked 0 print truth table in to odt file Related Making truthtables in python Generate Truth Tables with some variables already set in python 0 Is it possible in Python to write a sort of truth table to simplify the writing of if statements? 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10838	https://www.youtube.com/watch?v=4UdV0pFsf9U	How To Combine Two Ratios Maths With Melv 2870 subscribers 390 likes Description 37073 views Posted: 1 Aug 2017 Combining ratio is now a common type of question for higher and foundation GCSE maths! Turning two ratios into a single ratio can help us with many different types of problems in GCSE maths. The truck to combining ratios is the part that is common to both ratios. Multiply each ratio so the common part is the same, and then you can express as a single ratio! 65 comments Transcript: [Music] hey guys so in this video we're going to look at how to combine two ratios into a single ratio so if we look at this chap here on the front uh we got two ratios so imagine that we're dealing with apples bananas and coconuts then the left ratio tells us uh the ratio of apples to bananas whereas the one on the right tells us the ratio of bananas to coconuts so we want to combine those two ratios into a single ratio apples to bananas to coconuts so let's have a look at how to do it it's quite nice it's not too tricky so the uh the trick to it is the part uh the latter that is common to both ratios so here we've got B's in both of our ratios so what we need need to do is Alter each ratio so that the B part is the same number so at the moment in the left ratio we've got a to B is 3 to 5 but in the right hand ratio we've got B to C is 3 to four so what we need to do is find the lowest common multiple of five and three the two B components so the lowest common multiple of five and three is 15 so then we're going to need to alter each ratio to make the B PA 15 now we can multiply ratios up exactly the same as we can with fractions as long as we do this to one side then we're absolutely fine if we do it on the other side as well so to turn uh this five in the left hand ratio uh into 15 then we're going to need to multiply the whole ratio by three so this would become 99 to 15 and then in the right hand ratio and 3 into 15 we're going to need to multiply by five so that will give us 15 to 20 so now we know that A to B is 9 to 15 and B to C is 15 to 20 so because both of the B components are the same we can now write this as a single ratio 9 to 15 to 20 so A to B to C is 9 to 15 to 20 cool let's check out another one so in this one we've got exactly the same situation just different numbers so the letter that pops up in both ratios is B so we need to check out the b components of each ratio which is three and seven then ask ourselves what's the lowest common multiple of 3 and 7 which is 21 so turn three into 21 on the left hand side we need to multiply it by seven so if we multiply this left hand ratio by seven left and right then we're going to get 14 to 21 and then in the right hand ratio to 107 into 21 we're going to want to multiply by three so we get 21 uh to 27 so now we know that A to B is 14 to 21 and B to C is 21 to 27 so as a single ratio A to B to C is 14 to 21 to 27 jobs are good again let's check out another one okay guys so we're spicing it up here we're using different lers uh exactly the same sort of approach though except for the common component component is in a slightly different position this time so we've got M to n is 4 to7 and M to O is 3 to four now this time as I say the common component is the m part of each ratio so we need to work out then what this lowest common multiple four and three which would be 12 so we need to turn the M component of each ratio into 12 so to do that with the left hand one we're going to multiply by three so we're going to get 12 to 21 and on the right hand side we're going to need to multiply 3 by four and that will give us 12 to 16 so we now know that M to n is 12 to 21 and M to O is 12 to 16 now even though m is at the start of the rtio and O will be at the end of the ratio we can still just write this as M which is 12 to n which is 21 and then to O which would be 16 so that's M to n to oh Kraken let's check out another one okay so this time we've got a a three component ratio and a two component so we play with this exact the same way that we have dealt with the others so we find our common component which here is y so four and five we need to think of the lowest common multiple which would be 20 so we're going to need to multiply four by five in the left hand ratio so everything needs to be multiplied by five so we' get 15 to 5 to 20 and then on the left hand side ratio to turn five into 20 we're going to want to multiply it by four so that's going to give us 20 to 36 so our ratio x h sorry W to X to Y to zed can be written as 15 to 5 to 20 to 36 okay guys so that's how we can combine two ratios into a single ratio I hope it's useful I'll do another one as well GCC exam paper style questions
10839	https://www.youtube.com/watch?v=yagjgIqEAA0	Multiplying Complex Conjugates (Example) James Elliott 11600 subscribers 18 likes Description 1058 views Posted: 18 Sep 2020 This video walks through an example of finding a complex conjugate and then multiplying a complex number by its complex conjugate. For more math help and resources, visit www.hsmathsolutions.com. Transcript: in this example we have the complex number negative 3 minus the square root of 2 i and so i'd like to do a couple of things here the first one is let's go ahead and find the complex conjugate for this number and then let's go ahead and multiply this number by its complex conjugate okay so remember for any complex number written in a plus bi form which is standard form its complex conjugate will be a minus bi we're just going to change that sign right in the middle or you can say we're going to change the sign of the imaginary term okay so when we do that we'll get negative 3 plus the square root of 2 i so this is the complex conjugate so now that we have it let's go ahead and multiply these two together so the negative three let's go ahead and distribute so negative three times negative three will be nine negative 3 times radical 2 i will be minus 3 times the square root of 2 i so let's go ahead and take the negative square root of 2 i and distribute them through there so negative and a negative is a positive 3 radical 2 i and then a negative and a positive will make a negative square root of 2 times the square root of 2 makes a real 2 and then i times i is i squared okay first thing you notice here these middle terms cancel and again that's kind of by design since what we've done here is we have created a difference of squares factorization so nine minus two and then i squared which is really just negative one so nine plus two and that'll equal eleven okay so we started with negative 3 minus the square root of 2 i we found the complex conjugate and we multiplied them together and again worth noting is when you take a complex number and multiply it by its complex conjugate you will always get a real number
10840	https://theretinainstitute.org/toxoplasmic-choroiditis	Treatment of Macular Degeneration and Diabetes is more effective than ever. 2701 North Causeway Boulevard, Metairie, Louisiana 70002, United States (504) 455-0500 THE RETINA INSTITUTE NEW PATIENTS / REVIEWS DR. EBRAHIM CONTACT STORE MORE PORTAL & PAYMENT AMSLER GRID VIDEO GALLERY RESEARCH GENETICS CAREERS ACUTE RETINAL NECROSIS AUTOIMMUNE RETINOPATHY BIRDSHOT CHORIORETINITIS CANCER ASSOC. RETINOPATHY CENTRAL SEROUS/POLYPOIDAL COATS' DISEASE CONE ROD DYSTROPHY CYTOMEGALOVIRUS RETINITIS DIABETIC RETINOPATHY ENDOPHTHALMITIS EPIRETINAL MEMBRANE FLOATERS HIGH BLOOD PRESSURE HISTOPLASMOSIS HIV RETINOPATHY LATTICE DEGENERATION MACULAR DEGENERATION MACULAR HOLE MELANOMA METASTASES MYELINATED RETINAL NFL OCULAR ISCHEMIC SYNDROME OPEN GLOBE INJURY OPTIC NERVE HEAD DRUSEN PROLIF. VITREORETINOPATHY RETINAL ARTERY OCCLUSION RETINAL DETACHMENT RETINAL DYSTROPHIES RETINAL VEIN OCCLUSION RETINITIS PIGMENTOSA RETINOSCHISIS SARCOIDOSIS SICKLE CELL RETINOPATHY STARGARDT DISEASE TOXOPLASMIC CHOROIDITIS UVEITIS VITREOUS DETACHMENT VITREOUS HEMORRHAGE WHITE DOT SYNDROMES PRIVACY POLICY WELLNESS CENTER THE RETINA INSTITUTE LA PRIMARY CARE LA COSMEDIC GEAUX STORE (504) 455-0500 BOARD CERTIFIED EYE CARE IN NEW ORLEANS, LA (504) 455-0500 BOARD CERTIFIED EYE CARE IN NEW ORLEANS, LA(504) 455-0500 BOARD CERTIFIED EYE CARE IN NEW ORLEANS, LA(504) 455-0500 BOARD CERTIFIED EYE CARE IN NEW ORLEANS, LA(504) 455-0500 BOARD CERTIFIED EYE CARE IN NEW ORLEANS, LA CONTACT US TOXOPLASMIC CHORIORETINITIS Darin R. Goldman, in Atlas of Retinal OCT: Optical Coherence Tomography, 2018 Summary Toxoplasmic chorioretinitis, caused by infection of the parasite Toxoplasma gondii, is the most common identifiable cause for posterior uveitis and focal retinitis. The clinical appearance is that of focal yellow or white retinitis with overlying vitreous inflammation. The active area of retinitis is typically adjacent to a darkly pigmented chorioretinal scar, indicative of old disease. The diagnosis of toxoplasmic chorioretinitis is usually determined based on the clinical appearance alone. Unusual cases may be difficult to differentiate from other causes of retinitis. In these cases, additional diagnostic testing, such as serology or aqueous sampling for polymerase chain reaction, may be helpful. OCT, although not definitive in its diagnostic specificity for toxoplasmic chorioretinitis, can elucidate unique findings that support the diagnosis (Figs. 1–4). FIG. 1A–C. (A) Diffuse toxoplasmic chorioretinitis with a large area of focal retinitis (circle). (B) Infrared image highlighting spheroid plaques located on both arterioles and venules. (C) OCT in an area not involved with focal retinitis. Secondary features of active inflammation are present, including visible white blood cells in the vitreous cavity (arrows) and a typical perivascular spheroid deposit in cross section (circle). FIG. 2A and B. (A and B) Diffuse retinal hyperreflectivity is present in a focal area of active retinitis due to toxoplasmosis (arrows). The margins of involved and uninvolved retina are distinct. FIG. 3. Peripapillary toxoplasmic chorioretinitis. There is a highly reflective, thickened vitreous membrane due to localized inflammation. A subclinical retinal detachment is present, but difficult to appreciate. The involved area of focal retinitis is partially obscured by overlying inflammation, which causes shadowing. FIG. 4A–D. (A) Color photograph and (B) fluorescein angiogram of typical toxoplasmic chorioretinitis showing an active lesion adjacent to a scar. (C and D) OCT through the area of active retinitis shows full-thickness involvement with diffuse hyperreflectivity.(Courtesy Lana Rifkin, MD.) Toxoplasmosis Jack S. Remington, ... George Desmonts, in Infectious Diseases of the Fetus and Newborn (Seventh Edition), 2011 Differential Diagnosis of Eye Lesions Congenital Anomalies The healed foci of toxoplasmic chorioretinitis may resemble a colobomatous defect (Fig. 31–11) . The associated ocular, systemic, and serologic changes make toxoplasmosis the most likely diagnosis. Abnormal retinal morphology has been described in one fetal eye , and similar findings have been described in a variety of animal models of the congenital infection. Chromosome analysis was not available for the fetus, however. FIGURE 31–11. Macular scars (pseudocoloboma) of the retina in congenital toxoplasmosis. Other Inflammatory Lesions The differential diagnosis of eye lesions includes many of the inflammatory lesions described in chapters 16, 23, 26, and 28. The lymphochoriomeningitis virus also can cause similar lesions. Birth Injury Intraocular hemorrhage may go unrecognized and may cause retinal damage with gliosis and fibrosis, potentially resulting in retinal detachment. The lesion usually is unilateral, associated cerebral damage is absent, and no serologic evidence is present to support a diagnosis of toxoplasmosis. Retinopathy of prematurity may occur in conjunction with toxoplasmic chorioretinitis. Circulatory Disturbances Congenital aneurysms and telangiectasia of retinal vessels may result in extensive retinal fibrosis, with pigmentation and detachment. The disease usually is unilateral and is not associated with cerebral involvement or other changes. Neoplasms Retinoblastoma rarely may have an appearance similar to that described for ocular toxoplasmosis. It most often is unilateral and is unassociated with visceral or cerebral damage unless an advanced stage has been reached. Pseudoglioma may be difficult to distinguish from a healed chorioretinitis lesion but usually is single and unilateral. Gliomas may be bilateral, progressing from a small nodule to a large polypoid mass protruding into the vitreous. Toxoplasma gondii José G. Montoya, ... Joseph A. Kovacs, in Mandell, Douglas, and Bennett's Principles and Practice of Infectious Diseases (Eighth Edition), 2015 Ocular Toxoplasmosis The decision to treat active toxoplasmic chorioretinitis should be made based on a complete ophthalmologic evaluation. A recent report from the American Academy of Ophthalmology highlighted the limited data that are available from randomized controlled trials with well-defined end points demonstrating the benefits of therapy.459 Treatment is most likely indicated in the following settings: any decrease in visual acuity, macular or peripapillary lesions, lesions greater than one optic disk diameter, lesions associated with a moderate-to-severe vitreous inflammatory reaction, the presence of multiple active lesions, the persistence of active disease for more than 1 month, and any ocular lesions associated with recently acquired infection. Because the disease can be self-limited in immunocompetent patients, many clinicians may not treat small, peripheral retinal lesions that are not immediately vision threatening.9,265,460-462 The reported benefits of medical therapy are related primarily to the clinical presentation.265,460 Because there is so much variation in the clinical manifestations of the retinal disease, and because the disease may be self-limited even without treatment, the response to therapy is difficult to interpret. The combination of pyrimethamine (100-mg loading dose given over 24 hours for 2 days, followed by 25 to 50 mg daily) and sulfadiazine (1 g given four times daily for 4 to 6 weeks), depending on the clinical response, which is considered “classic” therapy for ocular toxoplasmosis, is the most common drug combination used (see Table 280-2).460 TMP-SMX showed responses similar to pyrimethamine-sulfadiazine in a recent randomized, single-blind trial, although the latter regimen was used at lower-than-standard doses.463 Two recent, open, randomized trials found no differences in response rates to intravitreal clindamycin plus dexamethasone compared with an oral regimen combining pyrimethamine, sulfadiazine, leucovorin, and prednisone/prednisolone.464,465 Clindamycin (300 mg orally every 6 hours for a minimum of 3 weeks) has also been used with favorable clinical results.460 Other drugs that may have activity but have been inadequately studied include atovaquone and pyrimethamine plus azithromycin.466,467 Systemic corticosteroids are indicated when lesions involve the macula, optic nerve head, or papillomacular bundle. Photocoagulation has been used both for the treatment of active lesions and for prophylaxis against the spread of lesions because new lesions appear contiguous to old lesions.460 In some patients, vitrectomy and lentectomy may be necessary. Given the high relapse rate seen in some patients with ocular toxoplasmosis, prevention of recurrences would be highly desirable. A randomized, open-label trial of 124 patients in Brazil found that TMP-SMX (1 double-strength tablet every 3 days) was effective in decreasing the frequency of recurrences from 24% to 7% in a population at high risk for recurrences.180 If confirmed in other populations, such a regimen could be beneficial in patients with frequent or severe recurrences. For the approach to ocular toxoplasmosis during pregnancy see “Acute Acquired Toxoplasma Infection in Pregnant Women.” Toxoplasmosis Jose G. Montoya, in Goldman's Cecil Medicine (Twenty Fourth Edition), 2012 Ocular Disease Serologic and PCR testing can be helpful in the diagnosis of toxoplasmic chorioretinitis. An IgG-negative/IgM-negative patient is unlikely to have ocular disease due to toxoplasmosis. However, patients should be tested at reference laboratories (e.g., PAMF-TSL) because their T. gondii–specific IgG can be present but at very low levels such that only a gold standard method like the dye test can detect it. In patients with eye lesions typical of toxoplasmic chorioretinitis (see Fig. 357-2), a positive IgG test result at a relatively low titer (e.g., a dye test at PAMF-TSL ≤512) and a negative IgM test result are diagnostic of ocular disease due to the parasite reactivation. If the serologic test reveals a positive IgM result and confirmatory testing at PAMF-TSL establishes the diagnosis of an acute infection in patients 1 year of age or older, the eye disease is most likely the result of eye involvement in the setting of a recent and postnatally acquired infection. In patients with atypically appearing lesions or in whom the response to anti-Toxoplasma drugs is slow or absent, a T. gondii–specific immune load (aqueous humor) or PCR in ocular fluids (vitreous fluid is preferable to aqueous humor because of probable higher sensitivity, but it is riskier to obtain) should be considered. Toxoplasmosis Jose G. Montoya, in Goldman's Cecil Medicine (Twenty Fourth Edition), 2012 Ocular Disease Serologic and PCR testing can be helpful in the diagnosis of toxoplasmic chorioretinitis. An IgG-negative/IgM-negative patient is unlikely to have ocular disease due to toxoplasmosis. However, patients should be tested at reference laboratories (e.g., PAMF-TSL) because their T. gondii–specific IgG can be present but at very low levels such that only a “gold standard” method like the dye test can detect it. In patients with eye lesions typical of toxoplasmic chorioretinitis (see Fig. 357-2), a positive IgG test result at a relatively low titer (e.g., a dye test at PAMF-TSL ≤512) and a negative IgM test result are diagnostic of ocular disease due to the parasite reactivation. If the serologic test reveals a positive IgM result and confirmatory testing at PAMF-TSL establishes the diagnosis of an acute infection in patients 1 year of age or older, the eye disease is most likely the result of eye involvement in the setting of a recent and postnatally acquired infection. In patients with atypically appearing lesions or in whom the response to anti-Toxoplasma drugs is slow or absent, a T. gondii–specific immune load (aqueous humor) or PCR in ocular fluids (vitreous fluid is preferable to aqueous humor because of probable higher sensitivity, but it is riskier to obtain) should be considered. The immune load (Goldmann-Witmer coefficient analysis of aqueous humor) is calculated as (anti-Toxoplasma IgG in aqueous humor/total IgG in aqueous humor)/(anti-Toxoplasma IgG in serum/total IgG in serum). A value of 2 or greater has been considered by some investigators as evidence of synthesis of intraocular T. gondii–specific IgG in response to the presence of the replicating tachyzoite. Side Effects of Drugs Annual 32 Natascia Corti, ... Christa Wenger, in Side Effects of Drugs Annual,2010 Lincosamides (SED-15, 2063; SEDA-29, 263; SEDA-30, 302; SEDA-31, 437) Clindamycin Observational studies In a retrospective review of the records of 50 consecutive patients with active toxoplasmic chorioretinitis treated with clindamycin, five had gastrointestinal adverse events and six had rashes (129c) Nervous system A 14-year-old girl with autism developed hiccup-like movements after receiving clindamycin 300 mg bd and risperidone 5.5 mg/day for 1 day; 3 days after withdrawal of clindamycin the abnormal movements resolved (130A). Sensory systems Topical clindamycin has been associated with taste disorders. In the adverse drug reactions database of the Netherlands Pharmacovigilance Centre, seven patients were identified with taste disorders (131c). In five cases an oral formulation was involved, in one intravenous administration and in one both formulations were used. Latency was less than 1 day after exposure and in one case taste disorders occurred repeatedly at 10 minutes after every intravenous dose. The adjusted reporting odds ratio was 7.0 (95% CI = 2.8, 17), supporting a causal relationship. Liver Acute hepatotoxicity occurred in a 42-year-old woman after administration of clindamycin for a dental infection (132A). Skin Acute generalized exanthematous pustulosis (AGEP) is a rare skin eruption most commonly caused by medications. AGEP occurred in an 82-year-old Caucasian woman after she had taken clindamycin for 2 days. She was treated with intravenous methylprednisolone, hydrocortisone cream 1%, hydroxyzine, doxepin, and paracetamol. The redness and pustulosis stopped spreading in 1 day and resolved in 5 days (133A). Immunologic A 47-year-old woman developed acute febrile neutrophilic dermatosis(Sweet’s syndrome) after receiving oral and intravenous clindamycin for a tooth infection; after the clindamycin was withdrawn her symptoms resolved over several days (134A). Infection risk In 836 patients aged 65 years or older, clindamycin exposure was associated with the highest rate ratio of infection with C. difficile(RR = 32, 95% CI = 18, 58) (135r). Toxoplasma gondii (Toxoplasmosis) Despina Contopoulos-Ioannidis, José G. Montoya, in Principles and Practice of Pediatric Infectious Diseases (Fifth Edition), 2018 Chronic Infection in Immunocompetent Children During the chronic stage of infection, T. gondii does not appear to cause overt symptoms in most immunocompetent people. However, toxoplasmic chorioretinitis can develop in previously infected, immunocompetent people, most likely because of reactivation of the parasite and subsequent host immune response. Reactivation can occur in congenitally or postnatally infected people. Reactivation of congenital infection appears to occur more commonly between 10 and 30 years of age and is more likely to involve the macula bilaterally. In contrast, ocular disease resulting from reactivation of a postnatally acquired and latent infection appears to occur in people >50 years of age and is more likely to cause peripheral and unilateral retinal lesions. Toxoplasmic Chorio-Retinitis Contact Us Please send us a message or call us for an appointment. We are located conveniently at I-10 and N. Causeway Blvd. near Lakeside Mall. THE RETINA INSTITUTE, New Orleans, LA 2701 N. Causeway Boulevard, Metairie, Louisiana 70002, United States (504) 455-0500 Hours \| \| \| --- \| \| Open today \| 08:00 am – 04:30 pm \| 8am - 4:30pm Monday through Friday Closed Saturday and Sunday Copyright © 2005-2025 The Retina Institute — All Rights Reserved. Most insurance accepted. Care Credit available. 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10841	https://www.emjreviews.com/diabetes/news/reduced-risk-of-infection-with-well-controlled-hba1c-levels/	Subscribe This site is intended for healthcare professionals Subscribe Reduced Risk of Infection with Well-Controlled HbA1c Levels DIABETIC patients who can effectively control their HbA1c levels have been found to have a significantly reduced risk of hospitalisation as a result of infection, according to research conducted at the St. George’s, University of London, London, UK. Since diabetes is known to increase a patient’s risk of infection due to their weakened immune system, researchers from St. George’s, University of London assessed the impact of well-managed blood sugar levels on health outcomes. Records detailing infection rates from hospitals and general practitioner surgeries from 2010–2015 were analysed and 85,000 patients with diabetes aged 40–89 years were included in the study. The results showed that patients with HbA1c levels ≥97 mmol/mol (11%) had an increased risk of hospitalisation as a result of infection compared to those with HbA1c levels between 43 and 53 mmol/mol (6–7%). It was found that those with worse HbA1c control were nearly 3-times more likely to need hospital treatment for infection than those with well-controlled blood sugar levels. This risk was even more pronounced, rising to an 8.5-fold higher risk, for Type 1 diabetes mellitus patients with poorly controlled HbA1c. Further analysis of the records also highlighted that there was a link between high blood sugar levels and the risk of pneumonia, which bears particular significance for the elderly population. Prof Julia Critchley, Population Health Research Institute, St George’s, University of London, commented on the nationwide effect of these results: “Across England as a whole, we found that poor diabetes control accounted for about 20–46% of some of the most serious types of infections (sepsis, bone and joint infections, tuberculosis, and endocarditis) seen in diabetes patients.” This study highlights just how important the control of blood sugar levels is in diabetic patients, not only for managing hypo and hyperglycaemic events but also for reducing the risk of more systemic effects. Rate this content's potential impact on patient outcomes Average rating / 5. Vote count: No votes so far! Be the first to rate this content. Related To This Subject Tirzepatide May Boost Diet-Related Quality of Life in Diabetes Global Study Revealed Persistent Gaps in Diabetes Care Worldwide More articles Innovation and Impact in Diabetes: Interview with Lorenzo Piemonti Fighting Against Diabetes and its Hidden Complications: Interview with Stephan Herzig Nanomedicine in Diabetes Treatment: Novel Drug Delivery Systems Featured journals EMJ Diabetes 12.1 2024 EMJ Diabetes 11.1 2023 Elevating the quality of healthcare globally Elevating the quality of healthcare globally Therapy Area About Us Copyright © 2025 European Medical Group LTD trading as European Medical Journal. All rights reserved. European Medical Journal is for informational purposes and should not be considered medical advice, diagnosis or treatment recommendations. Ts & Cs Privacy Policy Cookie Policy
10842	https://obgyn.onlinelibrary.wiley.com/doi/full/10.1002/ijgo.12614	Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Volume 143, Issue S2 pp. 59-78 FIGO CANCER REPORT 2018 Open Access Cancer of the ovary, fallopian tube, and peritoneum This article relates to: ###### Cancer of the ovary, fallopian tube, and peritoneum: 2021 update Jonathan S. Berek, Malte Renz, Sean Kehoe, Lalit Kumar, Michael Friedlander, Volume 155Issue S1International Journal of Gynecology & Obstetrics pages: 61-85 First Published online: October 20, 2021 Jonathan S. Berek, Corresponding Author Jonathan S. Berek jberek@stanford.edu Stanford Women's Cancer Center, Stanford Cancer Institute, Stanford University School of Medicine, Stanford, CA, USA Jonathan S. Berek, Stanford Women's Cancer Center, Stanford Cancer Institute, Stanford University School of Medicine, Stanford, CA, USA. Email: jberek@stanford.edu Search for more papers by this author Sean T. Kehoe, Sean T. Kehoe Institute of Cancer and Genomics, University of Birmingham, Birmingham, UK Search for more papers by this author Lalit Kumar, Lalit Kumar Department of Medical Oncology, All India Institute of Medical Sciences, New Delhi, India Search for more papers by this author Michael Friedlander, Michael Friedlander Royal Hospital for Women, Randwick, Sydney, NSW, Australia University of New South Wales Clinical School, Sydney, NSW, Australia Search for more papers by this author Jonathan S. Berek, Corresponding Author Jonathan S. Berek jberek@stanford.edu Stanford Women's Cancer Center, Stanford Cancer Institute, Stanford University School of Medicine, Stanford, CA, USA Correspondence Jonathan S. Berek, Stanford Women's Cancer Center, Stanford Cancer Institute, Stanford University School of Medicine, Stanford, CA, USA. Email: jberek@stanford.edu Search for more papers by this author Sean T. Kehoe, Sean T. Kehoe Institute of Cancer and Genomics, University of Birmingham, Birmingham, UK Search for more papers by this author Lalit Kumar, Lalit Kumar Department of Medical Oncology, All India Institute of Medical Sciences, New Delhi, India Search for more papers by this author Michael Friedlander, Michael Friedlander Royal Hospital for Women, Randwick, Sydney, NSW, Australia University of New South Wales Clinical School, Sydney, NSW, Australia Search for more papers by this author First published: 11 October 2018 Citations: 231 Abstract The Gynecologic Oncology Committee of FIGO in 2014 revised the staging of ovarian cancer, incorporating ovarian, fallopian tube, and peritoneal cancer into the same system. Most of these malignancies are high-grade serous carcinomas (HGSC). Stage IC is now divided into three categories: IC1 (surgical spill); IC2 (capsule ruptured before surgery or tumor on ovarian or fallopian tube surface); and IC3 (malignant cells in the ascites or peritoneal washings). The updated staging includes a revision of Stage IIIC based on spread to the retroperitoneal lymph nodes alone without intraperitoneal dissemination. This category is now subdivided into IIIA1(i) (metastasis ≤10 mm in greatest dimension), and IIIA1(ii) (metastasis >10 mm in greatest dimension). Stage IIIA2 is now “microscopic extrapelvic peritoneal involvement with or without positive retroperitoneal lymph node” metastasis. This review summarizes the genetics, surgical management, chemotherapy, and targeted therapies for epithelial cancers, and the treatment of ovarian germ cell and stromal malignancies. 1 INTRODUCTION 1.1 Primary sites: ovarian, fallopian tube, and peritoneal cancer In 2014, the Gynecologic Oncology Committee of FIGO revised the staging to incorporate ovarian, fallopian tube, and peritoneal cancer in the same system. Changing the staging system required extensive international consultation. The primary site (i.e. ovary, fallopian tube, or peritoneum) is designated, where possible. When it is not possible to clearly delineate the primary site, these should be listed as “undesignated”.1, 2 It has been presumed that fallopian tube malignancies were rare.2 However, histologic, molecular, and genetic evidence shows that as many as 80% of tumors that were classified as high-grade serous carcinomas of the ovary or peritoneum may have originated in the fimbrial end of the fallopian tube.3-8 Therefore, the incidence of fallopian tube cancers may have been substantially underestimated. These new data support the view that high-grade serous ovarian, fallopian tube, and peritoneal cancers should be considered collectively, and that the convention of designating malignancies as having an ovarian origin should no longer be used, unless that is clearly the origination site. It has been suggested that extrauterine tumors of serous histology arising in the ovary, fallopian tube, or peritoneum might be described collectively as “Müllerian carcinomas”1, 2 or “pelvic serous carcinomas”.9 The latter tumor designation is controversial because some peritoneal tumors might arise in extrapelvic peritoneum. Therefore, the simple term “serous carcinoma” is preferred, and most of these are high-grade serous carcinomas (HGSC). Although there has been no formal staging for peritoneal cancers, the FIGO staging system is used with the understanding that it is not possible to have a Stage I peritoneal cancer. 1.1.1 Primary site Ovarian epithelial tumors may arise within endometriosis or cortical inclusions of Müllerian epithelium, likely a form of endosalpingiosis. These include low-grade endometrioid carcinomas, clear cell carcinomas, borderline and low-grade serous carcinomas, and mucinous carcinomas. These tumors are thought to evolve slowly from lower-grade precursor conditions (endometriotic cysts, cystadenomas, etc.) and are classified as type I tumors.5 Fallopian tube carcinomas arise in the distal fallopian tube and the majority of these are high-grade serous carcinomas. These are thought to evolve rapidly from more obscure precursors and are designated as type II tumors.5, 6 This latter group encompasses high-grade endometrioid carcinomas and carcinosarcomas. All of these high-grade carcinomas are nearly always associated with mutations in the TP53 gene.5 1.1.2 Lymphatic and lymph node drainage The lymphatic drainage of the ovaries and fallopian tubes is via the utero-ovarian, infundibulopelvic, and round ligament pathways and an external iliac accessory route into the following regional lymph nodes: external iliac, common iliac, hypogastric, lateral sacral, para-aortic lymph nodes and, occasionally, to the inguinal nodes.1, 10-12 The peritoneal surfaces can drain through the diaphragmatic lymphatics and hence to the major venous vessels above the diaphragm. 1.1.3 Other metastatic sites The peritoneum, including the omentum and pelvic and abdominal viscera, is the most common site for dissemination of ovarian and fallopian tube cancers. This includes the diaphragmatic and liver surfaces. Pleural involvement is also seen. Other extraperitoneal or extrapleural sites are relatively uncommon, but can occur.1, 10-12 After systematic pathologic analysis has excluded a tubal or ovarian site of origin, malignancies that appear to arise primarily on the peritoneum have an identical spread pattern, and frequently may involve the ovaries and fallopian tubes secondarily. These “peritoneal” tumors are thought to arise in endosalpingiosis. 1.2 Classification rules Although CT scans can delineate the intra-abdominal spread of disease to a certain extent, ovarian, fallopian tube, and peritoneal cancers should be staged surgically. Operative findings determine the precise histologic diagnosis, stage, and therefore the prognosis, of the patient.1, 9, 10, 12-14 In selected patients with advanced-stage disease, it may be appropriate to initiate chemotherapy prior to surgical intervention, and in these cases, there should be histologic or cytologic confirmation of the diagnosis prior to starting neoadjuvant chemotherapy (see 5.2.2. below). Chest radiograms may serve as a screen for pleural effusions. As distant metastases are infrequent, there is no requirement for other radiological evaluation unless symptomatic. Serum CA125 levels may be useful in determining response to chemotherapy, but they do not contribute to staging. 1.2.1 Fallopian tube involvement Fallopian tube involvement can be divided into three categories. In the first, an obvious intraluminal and grossly apparent fallopian tube mass is seen with tubal intraepithelial carcinoma (carcinoma in situ) that is presumed to have arisen in the fallopian tube. These cases should be staged surgically with a histologic confirmation of disease. Tumor extension into the submucosa or muscularis and to and beyond the serosa can therefore be defined. These features, together with the laterality and the presence or absence of ascites, should all be taken into consideration.1, 3, 6, 7 In the second scenario, a widespread serous carcinoma is associated with a tubal intraepithelial carcinoma. A visible mass in the endosalpinx may not be seen but the histologic findings should be noted in the pathology report since they may indicate a fallopian tube primary. Tumors obliterating both fallopian tube and ovary may belong to this group but whether a presumptive assignment of a tubal origin can be made in such cases is controversial given that tubal intraepithelial carcinoma cannot be confirmed. In the third scenario— risk-reducing salpingo-oophorectomy—tubal intraepithelial carcinoma may be the only finding. It should be reported as originating in the fallopian tube and managed accordingly. The majority of early serous cancers detected are found in the fallopian tube, irrespective of genetic risk.15, 16 1.2.2 FIGO staging The updated, revised FIGO staging system combines the classification for ovarian, fallopian tube, and peritoneum cancer. It is based on findings made mainly through surgical exploration (as outlined above). Table 1 presents the 2014 FIGO staging classification for cancer of the ovary, fallopian tube, and peritoneum. The equivalents within the Union for International Cancer Control (UICC) TNM classification are presented in Table 2. Table 1. FIGO staging classification for cancer of the ovary, fallopian tube, and peritoneum \| \| \| --- \| \| Stage I: Tumor confined to ovaries or fallopian tube(s) \| \| \| \| T1-N0-M0 \| \| IA: Tumor limited to 1 ovary (capsule intact) or fallopian tube; no tumor on ovarian or fallopian tube surface; no malignant cells in the ascites or peritoneal washings \| \| \| \| T1a-N0-M0 \| \| IB: Tumor limited to both ovaries (capsules intact) or fallopian tubes; no tumor on ovarian or fallopian tube surface; no malignant cells in the ascites or peritoneal washings \| \| \| \| T1b-N0-M0 \| \| IC: Tumor limited to 1 or both ovaries or fallopian tubes, with any of the following: \| \| \| IC1: Surgical spill \| \| \| \| T1c1-N0-M0 \| \| IC2: Capsule ruptured before surgery or tumor on ovarian or fallopian tube surface \| \| \| \| T1c2-N0-M0 \| \| IC3: Malignant cells in the ascites or peritoneal washings \| \| \| \| T1c3-N0-M0 \| \| Stage II: Tumor involves 1 or both ovaries or fallopian tubes with pelvic extension (below pelvic brim) or peritoneal cancer \| \| \| \| T2-N0-M0 \| \| IIA: Extension and/or implants on uterus and/or fallopian tubes and/or ovaries \| \| \| \| T2a-N0-M0 \| \| IIB: Extension to other pelvic intraperitoneal tissues \| \| \| \| T2b-N0-M0 \| \| Stage III: Tumor involves 1 or both ovaries or fallopian tubes, or peritoneal cancer, with cytologically or histologically confirmed spread to the peritoneum outside the pelvis and/or metastasis to the retroperitoneal lymph nodes \| \| \| \| T1/T2-N1-M0 \| \| IIIA1: Positive retroperitoneal lymph nodes only (cytologically or histologically proven): \| \| \| IIIA1(i) Metastasis up to 10 mm in greatest dimension \| \| \| IIIA1(ii) Metastasis more than 10 mm in greatest dimension \| \| \| IIIA2: Microscopic extrapelvic (above the pelvic brim) peritoneal involvement with or without positive retroperitoneal lymph nodes \| \| \| \| T3a2-N0/N1-M0 \| \| IIIB: Macroscopic peritoneal metastasis beyond the pelvis up to 2 cm in greatest dimension, with or without metastasis to the retroperitoneal lymph nodes \| \| \| \| T3b-N0/N1-M0 \| \| IIIC: Macroscopic peritoneal metastasis beyond the pelvis more than 2 cm in greatest dimension, with or without metastasis to the retroperitoneal lymph nodes (includes extension of tumor to capsule of liver and spleen without parenchymal involvement of either organ) \| \| \| \| T3c-N0/N1-M0 \| \| Stage IV: Distant metastasis excluding peritoneal metastases \| \| \| Stage IVA: Pleural effusion with positive cytology \| \| \| Stage IVB: Parenchymal metastases and metastases to extra-abdominal organs (including inguinal lymph nodes and lymph nodes outside of the abdominal cavity) \| \| \| \| Any T, any N, M1 \| Table 2. Cancer of the ovary, fallopian tube and peritoneum: FIGO staging (2014) compared with TNM classification.a \| FIGO (designate primary: Tov, Tft, Tp, or Tx) \| UICC \| \| \| --- --- \| \| T \| N \| M \| \| Stage \| \| \| \| \| IA \| T1a \| N0 \| M0 \| \| IB \| T1b \| N0 \| M0 \| \| IC \| T1c \| N0 \| M0 \| \| IIA \| T2a \| N0 \| M0 \| \| IIB \| T2b \| N0 \| M0 \| \| IIIA \| T3a \| N0 \| M0 \| \| \| T3a \| N1 \| M0 \| \| IIIB \| T3b \| N0 \| M0 \| \| \| T3b \| N1 \| M0 \| \| IIIC \| T3c \| N0–1 \| M0 \| \| \| T3c \| N1 \| M0 \| \| IV \| Any T \| Any N \| M1 \| \| Regional nodes (N) \| \| \| \| \| Nx \| Regional lymph nodes cannot be assessed \| \| \| \| N0 \| No regional lymph node metastasis \| \| \| \| N1 \| Regional lymph node metastasis \| \| \| \| Distant metastasis (M) \| \| \| \| \| Mx \| Distant metastasis cannot be assessed \| \| \| \| M0 \| No distant metastasis \| \| \| \| M1 \| Distant metastasis (excluding peritoneal metastasis) \| \| \| Notes The primary site—that is, ovary, fallopian tube, or peritoneum—should be designated where possible. In some cases, it may not be possible to clearly delineate the primary site, and these should be listed as “undesignated.” 2. The histologic type should be recorded. 3. The staging includes a revision of the Stage III patients and allotment to Stage IIIA1 is based on spread to the retroperitoneal lymph nodes without intraperitoneal dissemination, because an analysis of these patients indicates that their survival is significantly better than those who have intraperitoneal dissemination. 4. Involvement of retroperitoneal lymph nodes must be proven cytologically or histologically. 5. Extension of tumor from omentum to spleen or liver (Stage IIIC) should be differentiated from isolated parenchymal splenic or liver metastases (Stage IVB). a Source: Prat J; FIGO Committee on Gynecologic Oncology. Staging classification for cancer of the ovary, fallopian tube, and peritoneum. Int J Gynecol Obstet. 2014;124:1–5. In addition to these changes, several other modifications of the former staging system have been made to better prospectively capture the data. Stage IC is now divided into three categories: IC1 (surgical spill); IC2 (capsule ruptured before surgery or tumor on ovarian or fallopian tube surface); and IC3 (malignant cells in the ascites or peritoneal washings). Stage IIC has been eliminated. The updated staging includes a revision of the Stage IIIC based on spread to the retroperitoneal lymph nodes alone without intraperitoneal dissemination, because an analysis of these patients indicates that their survival is significantly better than those who have intraperitoneal dissemination.17 This category is now subdivided into IIIA1(i) (metastasis ≤10 mm in greatest dimension), and IIIA1(ii) (metastasis >10 mm in greatest dimension). Stage IIIA2 is now “microscopic extrapelvic peritoneal involvement with or without positive retroperitoneal lymph node” metastasis. The wording of Stage IIIB has been modified to reflect the lymph node status. Stage IVB now includes metastases to the inguinal lymph nodes. 1.2.2.1 Regional lymph nodes (N) NX: Regional lymph nodes cannot be assessed. N0: No regional lymph node metastasis. N1: Regional lymph node metastasis. 1.2.2.2 Distant metastasis (M) MX: Distant metastasis cannot be assessed. M0: No distant metastasis. M1: Distant metastasis (excluding peritoneal metastasis). 1.3 Histopathologic classification The majority of cases of ovarian cancer are of epithelial origin. FIGO endorses the WHO histologic typing of epithelial ovarian tumors. It is recommended that all ovarian epithelial tumors be subdivided according to the classification given below.18 The histologic classification of ovarian, fallopian tube, and peritoneal neoplasia is as follows: Serous tumors. Mucinous tumors. Endometrioid tumors. Clear cell tumors. Brenner tumors. Undifferentiated carcinomas (this group of malignant tumors is of epithelial structure, but they are too poorly differentiated to be placed in any other group). Mixed epithelial tumors (these tumors are composed of two or more of the five major cell types of common epithelial tumors. The types are usually specified). Cases with high-grade serous carcinoma in which the ovaries and fallopian tubes appear to be incidentally involved and not the primary origin can be labeled as peritoneal carcinoma or serous carcinoma of undesignated site, at the discretion of the pathologist. Epithelial tumors of the ovary and fallopian tube are further subclassified by histologic grading, which can be correlated with prognosis. This grading system does not apply to nonepithelial tumors.19 Two grading systems are applied. For non-serous carcinomas (most endometrioid and mucinous), grading is identical to that used in the uterus, based on architecture with a one-step upgrade if there is prominent nuclear atypia, as follows: GX: Grade cannot be assessed. G1: Well differentiated. G2: Moderately differentiated. G3: Poorly differentiated. Serous carcinomas are the most common in both the ovary and tube. More than 90% of fallopian tube carcinomas are serous or high-grade endometrioid adenocarcinoma. Other cell types have been reported, but are rare.1, 2, 20 Serous carcinomas are graded in a two-grade system befitting their biology. High-grade serous carcinomas, including both classic appearing and those with SET features (solid, endometrioid-like, and transitional) carry a high frequency of mutations in TP53.21-23 Low-grade serous carcinomas are often associated with borderline or atypical proliferative serous tumors, often contain mutations in BRAF and KRAS and contain wild-type TP53. Most “moderately differentiated” serous carcinomas carry mutations in TP53 and should be combined with the high-grade tumors.19, 22-24 Nonepithelial cancers, although uncommon, are extremely important. These include granulosa cell tumors, germ cell tumors, sarcomas, and lymphomas. They are discussed below as separate entities. Metastatic neoplasms to the ovary, such as tumors arising in the breast, lower reproductive tract sites (cervix or uterine carcinomas) and gastrointestinal tract (signet ring cell [Krukenberg] carcinomas, low grade appendiceal or pancreaticobiliary mucinous tumors and other neoplasms) are graded and staged in accordance with their respective sites of origin.1, 2 2 EPIDEMIOLOGY Malignant tumors of the ovaries occur at all ages with variation in histologic subtype by age. For example, in women younger than 20 years of age, germ cell tumors predominate, while borderline tumors typically occur in women in their 30s and 40s—10 or more years younger than in women with invasive epithelial ovarian cancers, which mostly occur after the age of 50 years. The lifetime risk of a woman in the USA developing ovarian cancer is approximately 1 in 70. Approximately 23% of gynecologic cancers are ovarian in origin, but 47% of all deaths from cancer of the female genital tract occur in women with ovarian cancer. Overall, epithelial ovarian cancer accounts for 4% of all new cancer diagnoses in women and 5% of all cancer-related deaths.1, 2, 25 The overall incidence of epithelial tumors varies from 9 to 17 per 100 000 and is highest in high-income countries, with the exception of Japan.26 However, this incidence rate increases proportionately with age. The largest number of patients with epithelial ovarian cancer is found in the 60–64 years age group. The median age is about a decade earlier in low-income countries. Established risk factors for epithelial ovarian tumors include reproductive risk factors. Women who have never had children are twice as likely to develop this disease. First pregnancy at an early age, early menopause, and the use of oral contraceptives have been associated with lower risks of ovarian cancer.27 The relationship of these variables to fallopian tube cancer is unclear. As noted above, it has been previously presumed that fallopian tube malignancies were rare; however, this has been challenged by evidence to show that many tumors that were classified as serous carcinomas of the ovary or peritoneal cancers appear to have their origin in the fallopian tube.3-7 When the origin is uncertain, the convention of designating all serous cancers, as originating in the ovary should no longer be used and the term “undesignated origin” may be applied at the discretion of the pathologist.18 2.1 Genetics Hereditary factors are implicated in approximately 20% of ovarian, fallopian tube, and peritoneal cancers28-32: Most hereditary ovarian cancers are due to pathogenic mutations in either the BRCA1 or BRCA2 genes. At least 15% of women with high-grade nonmucinous ovarian cancers have germline mutations in BRCA1/2 and, importantly, almost 40% of these women do not have a family history of breast/ovarian cancer. All women with high-grade nonmucinous invasive ovarian cancers should be offered genetic testing even if they do not have a family history of breast/ovarian cancer. Inherited deleterious mutations in BRCA1 and BRCA2 are the major genetic risk factors. Women who carry germline mutations in BRCA1 and BRCA2 have a substantially increased risk of ovarian, tubal, and peritoneal cancer—about 20%–50% with BRCA1 and 10%–20% with BRCA2.29-32 Typically, these cancers occur at an earlier age than sporadic cancers, particularly in BRCA1 mutation carriers, with a median age of diagnosis in the mid-40s. There are a number of other low- to moderate-penetrance genes that can also predispose to ovarian, fallopian tube, or peritoneal cancer. A recent study of next generation sequencing of constitutional DNA samples from 1915 women with ovarian cancer was carried out to identify germline mutations using a panel of 20 genes including BRCA1 and BRCA2, DNA mismatch repair genes, double stranded DNA break repair genes such as CHEK2 and ATM, as well as the BRCA1-associated complex or the BRCA2/Fanconi Anemia pathway genes (including BRIP1, BARD1, PALB2, RAD50, RAD51C, and RAD51D, among others). About 80% of mutations were in BRCA1 or BRCA2. About 3% of patients carried mutations in the Fanconi Anemia pathway genes, while only 0.4% had mutations in mismatch repair genes.33 In an earlier similar study that included 360 patients, 24% carried germline loss-of-function mutations including 18% in BRCA1 or BRCA2 and 6% in BARD1, BRIP1, CHEK2, MRE11A, MSH6, NBN, PALB2, RAD50, RAD51C, or TP53.34, 35 Inherited mutations in the mismatch repair genes associated with Lynch syndrome type II. Women carrying these mutations have an increased risk of a number of cancers including colon, endometrial, and ovarian cancer. Typically, the ovarian cancers that occur are endometrioid or clear cell histologically and are usually Stage I.35 Women with a strong family history of epithelial ovarian, fallopian tube, or peritoneal cancers, particularly if there is a documented germline BRCA mutation, are advised to have a risk-reducing bilateral salpingo-oophorectomy after appropriate counseling and at the completion of childbearing. All women who are suspected of carrying a BRCA germline mutation, based on family history or young age of diagnosis and a high-grade serous or high-grade endometrioid cancer, should be offered genetic testing. BRCA mutations may also occur in women without a family history of breast/ovarian cancer, and genetic testing should be considered in patients from ethnic groups where there is a high incidence of founder mutations (e.g. Ashkenazi Jewish ancestry), and in women with high-grade serous cancers under the age of 70 years.26-30 Australian guidelines advise that all women with invasive epithelial ovarian cancer apart from mucinous cancers diagnosed under the age of 70 should be offered BRCA mutation testing independent of family history and histologic subtype.36 In contrast, the Society of Gynecologic Oncology (SGO) and National Comprehensive Cancer Network (NCCN) guidelines recommend that all women diagnosed with ovarian, fallopian tube, or peritoneal carcinoma, regardless of age or family history, should receive genetic counseling and be offered genetic testing.37 Women whose family history suggests Lynch syndrome type II should undergo appropriate genetic counseling and testing. 3 SCREENING To date, there are no documented effective screening methods that reduce the mortality of ovarian, fallopian tube, or peritoneal cancers. Studies using CA125, ultrasonography of the pelvis, and pelvic examination do not have an acceptable level of sensitivity and specificity, based on trials carried out in women in the general population and those in the high-risk population. The US Preventive Services Task Force recommends against screening asymptomatic women for ovarian cancer with pelvic examination, pelvic ultrasound, or serum tumor marker measurements.38 The low prevalence of disease and lack of high-quality screening methods make it more likely to obtain false-positive results leading to unnecessary interventions. A recent study of multimodal screening using CA125 based on a risk of ovarian cancer algorithm (ROCA) every 4 months and transvaginal ultrasound annually or earlier where indicated by the ROCA in women at high risk of ovarian cancer reported that screening was associated with a low rate of high-volume disease at primary surgery and very high rates of no residual disease after surgery.38 Given that the majority of women with advanced stage ovarian cancer, even with complete resection, will relapse after chemotherapy, this does not seem to be a good alternative to risk-reducing surgery. The authors of the screening study concluded that risk-reducing salpingectomy-oophorectomy remains the treatment choice for women at high risk of ovarian/fallopian tube cancer.38 Women at increased genetic risk should be encouraged to consider risk-reducing bilateral salpingo-oophorectomy, as this is the most effective way to reduce mortality in this population of women.39, 40 An ACOG bulletin has recommended that opportunistic (at the time of a clinically indicated hysterectomy) bilateral salpingectomy be considered in women not at genetic risk who wish to retain their ovaries as a way to reduce their risk of later developing high-grade serous carcinomas.41 4 DIAGNOSIS Patients with epithelial ovarian cancers confined to the ovary or fallopian tube at initial diagnosis have a very good prognosis.42-45 The symptoms are often very insidious and the duration of symptoms not very different between patients with early stage or advanced stage disease.13, 14 This may reflect the different biological behavior of the various histologic subtypes; for example, grade 1 serous, clear cell, mucinous, and endometrioid cancers are commonly early stage at presentation, whereas high-grade serous cancers are most often Stage III because of early dissemination by a more aggressive cancer. Tumor markers such as human gonadotropin (hCG) and alpha-fetoprotein (AFP) are mandatory to exclude germ cell tumors in younger patients with a pelvic mass or suspicious enlargement of an ovary. Approximately two-thirds of all epithelial “ovarian” cancers are Stage III or Stage IV at diagnosis. Presenting symptoms include vague abdominal pain or discomfort, menstrual irregularities, dyspepsia, and other mild digestive disturbances, which may have been present for only a few weeks.13, 14, 46 As the disease progresses, abdominal distention and discomfort from ascites generally worsen, and may be associated with respiratory symptoms from increased intra-abdominal pressure or from the transudation of fluid into the pleural cavities. Abnormal vaginal bleeding is an uncommon symptom. Serous fallopian tube and peritoneal cancers present the same as ovarian cancer. Past analyses have been biased because many fallopian tube cancers have been presumed to arise in the ovaries. A detailed medical history must be taken to ascertain possible risk factors, history of other cancers, and history of cancer in the family. Then a complete physical examination, including general, breast, pelvic, and rectal examination, must be performed.1 Prior to surgery a chest radiograph should be taken to screen for a pleural effusion and a CT scan of the abdomen and pelvis should be performed to delineate the extent of intra-abdominal disease. However, in the absence of extra-abdominopelvic disease, radiological scanning does not replace surgical staging with laparotomy. Tumor markers including CA125, and carcinoembryonic antigen (CEA) should be considered.1 With a high CA125 level, the most common diagnosis would be epithelial ovarian, fallopian tube, or peritoneal cancer. A gastric or colonic primary with metastases to the ovaries may mimic ovarian cancer, and if the CEA is elevated, this should be considered. A ratio of more than 25:1 (CA-125 and CEA) favors an ovarian primary though it does not completely rule out a primary in the gastrointestinal tract.47 A current mammogram should be considered as patients are frequently in the age group where breast cancer is prevalent. A colonoscopy is indicated when symptoms suggest possible bowel cancer.1 The following factors point to the presence of a malignancy, and are useful in the clinical assessment of masses: Age of the patient (young for germ cell, older for epithelial malignancies). Bilaterality. Tumor fixation clinically. Ascites. Ultrasonographically complex, especially if solid areas. CT finding of metastatic nodules. Elevated tumor markers. 5 PRIMARY SURGERY In general, the prognosis of epithelial ovarian, fallopian, and peritoneal malignancies is independently affected by the following1, 48, 49: Stage of the cancer at diagnosis. Histologic type and grade. Maximum diameter of residual disease after cytoreductive surgery. 5.1 Staging laparotomy A thorough staging laparotomy is an important part of early management. If the preoperative suspicion is malignancy, a laparotomy should be performed. If there is no visible or palpable evidence of metastasis, the following should be performed for adequate staging1, 10, 11, 13, 14: Careful evaluation of all peritoneal surfaces. Retrieval of any peritoneal fluid or ascites. If there is none, washings of the peritoneal cavity should be performed. Infracolic omentectomy. Selective lymphadenectomy of the pelvic and para-aortic lymph nodes, at least ipsilateral if the malignancy is unilateral. Biopsy or resection of any suspicious lesions, masses, or adhesions. Random peritoneal biopsies of normal surfaces, including from the undersurface of the right hemidiaphragm, bladder reflection, cul-de-sac, right and left paracolic recesses, and both pelvic sidewalls. Total abdominal hysterectomy and bilateral salpingo-oophorectomy in most cases. Appendectomy for mucinous tumors. Upon opening the abdominopelvic cavity, the peritoneal fluid should be sent for cytology. In the absence of ascites, irrigation should be performed and washings sent for cytology. The laparotomy should proceed with a detailed examination of the contents, including all of the peritoneal surfaces. In addition to the suspicious sites, biopsies from the peritoneal reflection of the bladder, the posterior cul-de-sac, both paracolic gutters, subdiaphragmatic surfaces, and both pelvic sidewalls should be taken. The primary tumor, if limited to the ovary, should be examined to look for capsular rupture. All obvious sites of tumor must be removed wherever possible in addition to total hysterectomy and bilateral salpingo-oophorectomy. The omentum, pelvic, and para-aortic lymph nodes should be removed for histologic examination. In younger women, fertility may be an issue. In these patients, conservative surgery, with preservation of the uterus and contralateral ovary, should be considered after informed consent.43 Clinical judgment is important in the approach to a pelvic mass in the young, reproductive-aged woman. If the suspicion is strong for malignancy, open laparotomy is generally indicated. Laparoscopy may be more appropriate if the suspicion is more for benign disease, where tumor markers (including hCG and AFP) are normal. A biopsy of any suspicious lesion can be performed and frozen section obtained in order to proceed expeditiously with definitive surgery. Ovaries and fallopian tubes should be evaluated as thoroughly as possible to establish the site of origin. If visible, the entire tube, particularly the distal portion, should be submitted for pathology and examined using the SEE-FIM protocol.32 Ovaries should be scrutinized for coexisting endometriotic cysts, adenofibromas, or other benign conditions that could serve as a nidus of tumor development. 5.2 Cytoreductive (debulking) surgery for advanced stage disease 5.2.1 Primary debulking At least two-thirds of patients with ovarian cancer present with Stage III or IV disease. This may affect the performance status and fitness for surgery. However, the most important prognostic indicator in patients with advanced stage ovarian cancer is the volume of residual disease after surgical debulking. Therefore, patients whose medical condition permits should generally undergo a primary laparotomy with total abdominal hysterectomy, bilateral salpingo-oophorectomy, omentectomy, and maximal attempt at optimal cytoreduction.1, 48-50 This may necessitate bowel resection, and occasionally, partial or complete resection of other organs. Systematic pelvic and para-aortic lymphadenectomy of non-enlarged nodes does not improve overall survival, when compared with removal of bulky nodes only, although there is a modest improvement in progression-free survival.51 Level of Evidence A 5.2.2 Interval debulking In selected patients with cytologically proven Stage IIIC and IV disease who may not be good surgical candidates, 3–4 cycles of neoadjuvant chemotherapy (NACT) may be given initially, followed by interval debulking surgery (IDS) and additional chemotherapy as demonstrated in the EROTC and CHORUS Trials.52, 53 These two randomized prospective trials showed that in selected patients, interval debulking surgery after neoadjuvant chemotherapy showed equivalent survival with less morbidity compared with primary cytoreductive surgery. NACT followed by IDS may be particularly useful in patients with a poor performance status, significant medical co-morbidities, visceral metastases, and those who have large pleural effusions and/or gross ascites.54 In selected patients whose primary cytoreduction is considered suboptimal, particularly if a gynecologic oncologist did not perform the initial operation, interval debulking may be considered after 2–3 cycles of systemic chemotherapy.1, 52, 53, 55 Pathologic assessment for residual tumor following neoadjuvant therapy will enable an estimate of residual disease and pathological response.56 There are recent data to indicate that patients who have a good pathological response have a better outcome. A histopathologic scoring system for measuring response to neoadjuvant chemotherapy has been developed and validated by Bohm et al.57 who reported criteria for defining a chemotherapy response score (CRS) based on a three-tier system. A CRS 3 (complete or near complete pathological response) was associated with a better prognosis. Recently, these results have been validated in an independent West Australian cohort.58 6 CHEMOTHERAPY 6.1 Chemotherapy for early stage cancer The prognosis of patients with adequately staged tumors with Stage IA and Stage IB grade 1–2 epithelial cancers of the ovary is very good; adjuvant chemotherapy does not provide additional benefits and is not indicated. For higher-grade tumors and for patients with Stage IC disease, adjuvant platinum-based chemotherapy is given to most patients, although there has been debate about the absolute survival benefit in women with Stage IA and IB cancers who have had thorough surgical staging.42 All patients with Stage II disease should receive adjuvant chemotherapy. The optimal number of cycles in patients with Stage I disease has not been definitively established, but typically between 3 and 6 cycles are administered. The Gynecologic Oncology Group (GOG) 157 study suggested that 3 cycles of carboplatin and paclitaxel was equivalent to 6 cycles, but in subgroup analysis, 6 cycles appeared superior in patients with high-grade serous cancers.50 There is no evidence to support adjuvant therapy for carcinoma in situ of the fallopian tube and it is not recommended.1, 2, 44 Level of Evidence A 6.2 Chemotherapy for advanced stage ovarian cancer Patients who have had primary cytoreduction should receive chemotherapy following surgery1, 59 (Table 3). The accepted standard is 6 cycles of platinum-based combination chemotherapy, with a platinum (carboplatin or cisplatin) and a taxane (paclitaxel or docetaxel).60-64 Docetaxel is an option in patients who have had a significant allergic reaction to paclitaxel or who develop early sensory neuropathy as it has less neurotoxicity, but it is more myelosuppressive than paclitaxel.60 The SCOT-ROC (Scottish Gynecological Cancer Trials Group) study randomly assigned 1077 women with Stages IC–IV epithelial ovarian cancer to carboplatin paclitaxel or docetaxel.60 The efficacy of docetaxel was similar to paclitaxel. The median progression-free survival was 15.1 versus 15.4 months. The MITO 2 trial randomized over 800 patients to receive either carboplatin and liposomal doxorubicin (PLD) or carboplatin and paclitaxel. The median progression-free survival was 19.0 and 16.8 months with carboplatin/PLD and carboplatin/paclitaxel, respectively.65 The median overall survival times were 61.6 and 53.2 months with carboplatin/PLD and carboplatin/paclitaxel, respectively (hazard ratio [HR] 0.89; 95% CI 0.72–1.12; P=0.32). Carboplatin/PLD produced a similar response rate but different toxicity (less neurotoxicity and alopecia but more hematologic adverse effects) and could also be considered as an option in patients where paclitaxel cannot be used. Table 3. Chemotherapy for advanced epithelial ovarian cancer: recommended regimens.a \| Drugs Standard regimens \| Dose \| Administration (h) \| Interval \| No. of treatments \| --- --- \| Carboplatin \| AUC=5–6 \| 3 \| Every 3 wk \| 6–8 cycles \| \| Paclitaxel \| 175 mg/m2 \| \| \| \| \| Carboplatin \| AUC=5–6 \| 3 \| Every 3 wk \| 6 cycles \| \| Paclitaxel \| 80 mg/m2 \| \| Every week \| 18 wk \| \| Carboplatin \| AUC=5 \| 3 \| Every week \| 6 cycles \| \| Docetaxel \| 75 mg/m2 \| \| Every 3 wk \| \| \| Cisplatin \| 75 mg/m2 \| 3 \| Every 3 wk \| 6 cycles \| \| Paclitaxel \| 135 mg/m2 \| \| \| \| \| Carboplatin (single agent)b \| AUC=5 \| 3 \| Every 3 wk \| 6 cycles, as tolerated \| Abbreviation: AUC, area under the curve dose by the methods of Calvert et al. 75 and Nagao et al.76 a Reproduced with permission from Berek et al.,1 p.510. b In patients who are elderly, frail, or poor performance status. Although intraperitoneal chemotherapy has been shown to be associated with improved progression-free survival and overall survival in selected patients with optimally debulked Stage III ovarian cancer, it is not widely used outside the USA because of concerns regarding increased toxicity and catheter-related problems, and the benefits are still debated.66-71 The GOG 172 trial compared intravenous paclitaxel plus cisplatin with intravenous paclitaxel plus intraperitoneal cisplatin and paclitaxel in patients with Stage III ovarian or peritoneal carcinoma, with no residual disease greater than 1 cm in diameter.68 Only 42% of patients in the intraperitoneal group completed 6 cycles of the assigned therapy, but the intraperitoneal group had an improvement in progression-free survival of 5.5 months (23.8 vs 18.3 months; P=0.05) and an improvement in overall survival of 15.9 months (65.6 vs 49.7 months; P=0.03). Level of Evidence A More recently, the GOG 252 trial reported a median progression-free survival of approximately 27–29 months in over 1500 patients with optimal Stage II–III disease treated with regimens consisting of different combinations of intravenous and intraperitoneal cisplatin, carboplatin, and paclitaxel, in combination with bevacizumab.69 The treatment arms included intravenous carboplatin AUC 6/intravenous weekly paclitaxel at 80 mg/m2; intraperitoneal carboplatin AUC 6/intravenous weekly paclitaxel at 80 mg/m; and intravenous paclitaxel at 135 mg/m2 on day one/intraperitoneal cisplatin at 75 mg/m2 on day two/intraperitoneal paclitaxel at 60 mg/m2 on day eight. In addition, each arm received intravenous bevacizumab at 15 mg/kg with cycles 2 through 6 of chemotherapy and then alone for cycles 7 through 22. The median progression-free survival by intent-to-treat analysis was 24.9 (intravenous carboplatin), 27.3 (intraperitoneal carboplatin), and 26.0 months (intraperitoneal cisplatin). An analysis limited to patients with optimal Stage III tumors and no gross residual disease found a median progression-free survival of 31–34 months in all three arms. By comparison, the GOG 172 trial comparing intraperitoneal and intravenous chemotherapy regimens in ovarian cancer had a median progression-free survival of 23.8 months with intraperitoneal cisplatin (vs 18.3 months with intravenous) with an improvement in overall survival in favor of intraperitoneal injection.68 In addition, the median progression-free survival was 60 months in the patients with no residual disease in GOG 172. Differences in the cisplatin arm from the GOG 172 study include a dose reduction from 100 mg to 75 mg and a shorter infusion time from 24 hours to 3 hours.68 If intraperitoneal treatment is used it would be appropriate to follow the GOG 172 protocol rather than the modified protocol with a lower dose of cisplatin accepting the increased toxicity. Combination chemotherapy with either intravenous carboplatin and paclitaxel or intraperitoneal cisplatin and paclitaxel (using the GOG 172 protocol) are the standard treatment options for patients with advanced disease, with evidence to support the addition of bevacizumab in selected patients. The advantages and disadvantages of the intravenous versus intraperitoneal routes of administration of these drugs should be discussed with the patient. Intraperitoneal chemotherapy is applicable only to patients with advanced disease who have had optimal debulking and have less than 1 cm residual disease. It should be used only in centers that have experience with intraperitoneal chemotherapy. The recommended doses and schedule for intravenous chemotherapy are: carboplatin (starting dose AUC 5–6), and paclitaxel (175 mg/m2), every 3 weeks for 6 cycles,51 or the dose-dense regimen of carboplatin AUC 6 every 3 weeks for 6 cycles and weekly paclitaxel 80 mg/m2.70 The Japanese GOG (JGOG) reported the findings of the latter regimen and showed improved progression-free survival and overall survival.71 An Italian trial (MITO-7) investigated a different schedule of weekly carboplatin (AUC 2 mg/mL per min) plus weekly paclitaxel (60 mg/m2) compared with carboplatin (AUC 6 mg/mL per min, administered every 3 weeks) and paclitaxel (175 mg/m2).72 The weekly regimen did not significantly improve progression-free survival compared with the conventional regimen (18.8 vs 16.5 months; P=0.18), but was associated with better quality of life and fewer toxic effects. The results of the ICON 8 trial investigating dose-dense paclitaxel in a non-Japanese population have been recently presented.73 Over 1500 predominantly European patients were randomized to receive one of three regimens. Arm 1: carboplatin AUC 5/6 and paclitaxel 175 mg/m2 every 3 weeks; Arm 2: carboplatin AUC 5/6 every 3 weeks and paclitaxel 80 mg/m2 weekly; and Arm 3: carboplatin AUC 2 and paclitaxel 80 mg/m2 weekly. All patients had received neoadjuvant chemotherapy with planned interval debulking or received chemotherapy after initial primary cytoreductive surgery. There was no benefit found for the dose-dense regimens. The progression-free survival was 24.4 months with every 3-week dosing, compared with 24.9 and 25.3 months in arms 2 and 3, respectively.73 These results are very different to the JGOG trial and it seems that the likely explanation is due to pharmacogenomic differences between these two ethnic groups.74 The recommended doses and schedule for intraperitoneal chemotherapy are paclitaxel 135 mg/m2 intravenously on day one, followed by cisplatin 100 mg/m2 intraperitoneally on day two, followed by paclitaxel 60 mg/m2 intraperitoneally on day eight, every 3 weeks for 6 cycles, as tolerated.68, 69 Many centers modify the dose of cisplatin to 75 mg/m2 rather than 100 mg/m2 that was used in GOG 172 to reduce toxicity, but this could be questioned based on GOG 262 results discussed above.69 Others substitute carboplatin (AUC 5–6) for cisplatin in the regimen and the same caveats regarding lack of evidence apply.69 The role of intraperitoneal carboplatin is being evaluated in JGOG and the results should be available in the near future. Bevacizumab 7.5–15 mg/kg every 3 weeks may be added to these regimens.77, 78 Two studies have reported a modest, but statistically significant increase in progression-free survival in patients receiving maintenance bevacizumab following carboplatin, paclitaxel, and concurrent bevacizumab.77, 78 There is no evidence as yet to demonstrate an overall survival benefit, but a subgroup analysis of the International Collaboration on Ovarian Neoplasms 7 (ICON7) trial reported an improved median survival (30.3 vs 39.4 months) in patients with suboptimal Stage III and Stage IV.77 The role, optimal dose (7.5 mg/kg vs 15 mg/kg), timing (primary vs recurrent disease), and duration of treatment of bevacizumab are still debatable. van Driel et al.79 recently reported results of a randomized trial in which 245 patients with Stage III epithelial ovarian cancer who had received 3 cycles of neoadjuvant chemotherapy underwent interval debulking surgery. These patients were then randomized to receive either 3 more cycles of paclitaxel plus carboplatin with or without hyperthermic intraperitoneal chemotherapy (HIPEC). The addition of HIPEC to interval cytoreductive surgery resulted in longer recurrence-free survival (14.2 vs 10.7 months) and overall survival (45.7 vs 33.9 months) and did not result in higher rates of adverse effects. These findings are provocative and raise important questions. Unfortunately, the study did not have an arm with intraperitoneal cisplatin alone without HIPEC, therefore it is not possible to know whether the improved survival was due to the addition of intraperitoneal cisplatin alone or HIPEC. In patients who may not tolerate combination chemotherapy because of medical comorbidities or advanced age, single-agent, intravenously administered carboplatin (AUC 5–6) can be given. For patients who have a significant hypersensitivity reaction to paclitaxel, an alternative active drug can be substituted (e.g. docetaxel, nanoparticle paclitaxel, or liposomal doxorubicin). Carboplatin hypersensitivity is very uncommon in the first-line setting, but is seen in 10%–20% of patients with recurrent disease who have multiple lines of platinum-based chemotherapy.80 In patients with carboplatin hypersensitivity, desensitization could be attempted, depending on the severity of the reaction, or alternatively cisplatin (50–75 mg/m2) may be an option, but there still may be a risk of a severe allergic reaction. The treatment of all patients with advanced stage disease is approached in a similar manner, with dose modifications based on the toxicity of therapy. Care should be taken when considering combination chemotherapy in patients with a very poor performance status or with compromised renal function. 6.3 Maintenance chemotherapy Almost 80% of women with advanced-stage disease who respond to first-line chemotherapy relapse. There have been several trials conducted to determine if there is a benefit of maintenance therapy in these patients immediately following their primary treatment in an effort to decrease the relapse rate. These were all negative and there is no evidence to support maintenance chemotherapy after completion of first-line therapy. 6.4 PARP inhibitors There is good evidence to support the role of PARP inhibitors as maintenance therapy following response to chemotherapy in patients with platinum-sensitive recurrent ovarian cancer, as well as monotherapy in selected patients with recurrent ovarian cancer.81-85 Patients with BRCA mutations (both germline and somatic) have the greatest benefit, but a subset of patients with tumors with homologous recombination deficiency (HRD) also derive benefit from treatment with PARP inhibitors; the ongoing challenge is how best to identify these patients. The results of these trials are summarized in Table 4.83-85 Readers are directed to the chapter on targeted therapy in this Supplement by Basu et al.86 for further discussion of PARP inhibitors. Table 4. Progression-free survival endpoint in the three phase trials of maintenance PARP inhibitors \| Study \| PARP inhibitor progression-free survival (months) \| Placebo progression-free survival (months) \| Hazard ratio \| --- --- \| \| SOLO 283 \| 19.1 \| 5.5 \| 0.3 \| \| NOVA84 \| \| \| \| \| gBRCA \| 21 \| 5.5 \| 0.27 \| \| Non-BRCA \| 9.3 \| 3.9 \| 0.45 \| \| Non-BRCA HRD+ \| 12.9 \| 3.8 \| 0.38 \| \| ARIEL 3 85 \| \| \| \| \| gBRCA \| 16.6 \| 5.4 \| 0.23 \| \| HRD+ (includes WT/gBRCA) \| 13.6 \| 5.4 \| 0.32 \| 7 SECONDARY SURGERY 7.1 Second-look laparotomy A second-look laparotomy (or laparoscopy) was previously performed in patients who have no clinical evidence of disease after completion of first-line chemotherapy to determine response to treatment. Although of prognostic value, it has not been shown to influence survival, and is no longer recommended as part of the standard of care.87 Level of Evidence C 7.2 Secondary cytoreduction Secondary cytoreduction may be defined as an attempt at cytoreductive surgery at some stage following completion of first-line chemotherapy. Retrospective studies suggest that patients benefit if all macroscopic disease can be removed, which usually means patients with a solitary recurrence. Patients with a disease-free interval longer than 12–24 months and those with only 1–2 sites of disease appear to derive most benefit.88, 89 The role of secondary cytoreductive surgery is being evaluated in randomized clinical trials. The role of secondary debulking surgery has been addressed in the DESKTOP III trial and the results recently presented by Dubois on behalf of the AGO.90 This study included patients with a progression-free survival of greater than 6 months after first-line chemotherapy and who were considered to be good candidates for surgery based on a positive AGO Study Group score, defined as an ECOG performance status score of zero, ascites of 500 mL or less, and complete resection at initial surgery. Du Bois et al.90 reported that the median progression-free survival in 204 women who met this criteria and who were randomized to undergo surgery followed by chemotherapy was 19.6 months, compared with 14 months in 203 women who were randomized to receive only second-line chemotherapy. The primary endpoint of the study is overall survival, which will only be available in a few years. Level of Evidence C 8 FOLLOW-UP FOR MALIGNANT EPITHELIAL TUMORS There is no evidence to show that intensive clinical monitoring during follow-up after completion of primary surgery and chemotherapy with early initiation of chemotherapy in asymptomatic women with recurrent disease improves overall survival or quality of life. In asymptomatic patients with CA125 progression and small volume disease or no radiological evidence of recurrence, it is appropriate to delay starting chemotherapy. However, there may be a subset of patients who are suitable for secondary debulking surgery at the time of recurrence. The objectives of follow-up include: Early recognition and prompt management of treatment-related complications, including provision of psychological support. Early detection of symptoms or signs of recurrent disease. Collection of data regarding the efficacy of any treatment and the complications associated with those treatments in patients treated in clinical trials. Promotion of healthy behavior, including screening for breast cancer in patients with early stage disease, and screening for cervical cancer in patients having conservative surgery. There are no evidence-based guidelines regarding the appropriate follow-up schedule. During the first year following treatment, patients are seen every 3 months with a gradual increase in intervals to every 4–6 months after 2 years and then annually after the fifth year. At each follow-up, the patient should have her history retaken, including any change in family history of cancers and attention to any symptoms that could suggest recurrence; a physical and pelvic examination should be performed. This is an opportunity to refer appropriate patients for genetic testing if it was not done at diagnosis or during treatment. The CA125 has traditionally been checked at regular intervals, but there has been debate regarding the clinical benefit of using CA125 progression alone as a trigger for initiating second-line chemotherapy. A large MRC OV05-EORTC 55955 study showed that treating asymptomatic patients with recurrent ovarian cancer with chemotherapy on the basis of CA125 progression alone did not improve survival and early treatment in asymptomatic patients had a negative impact on quality of life.91 This study has generated considerable debate regarding the use of CA125 for follow-up, but most agree that it is reasonable not to immediately initiate treatment unless there is a clear clinical indication to do so. The timing of treatment should be based on symptoms as well as clinical and radiological findings. Imaging tests such as ultrasonography of the pelvis, CT, MRI, and/or positron emission tomography (PET) scans should be performed only when the clinical findings or the tumor markers suggest possible recurrence. There appears to be no benefit to initiating chemotherapy in an asymptomatic patient with recurrent disease based only on rising CA125 levels in the absence of clinical symptoms or radiological evidence of recurrence. In asymptomatic patients with small volume disease and no radiological evidence of recurrence, close observation is a reasonable option, as well as entry into an appropriate clinical trial or possibly a trial of tamoxifen may be considered. A Cochrane database systematic review of tamoxifen in unselected women with recurrent ovarian cancer reported a 10% objective response and a 32% disease stabilization rate.92 The patients treated were heterogeneous and included asymptomatic patients with rising CA125 levels, and symptomatic patients with chemotherapy-resistant disease who had been heavily pretreated and had a poor performance status. GOG 198 compared tamoxifen and thalidomide in women with recurrent FIGO Stage III or IV epithelial ovarian, tubal, or peritoneal cancer who had completed first-line chemotherapy, and who subsequently had Gynecologic Cancer InterGroup (GCIG) documented CA125 progression. The study reported that women who received thalidomide had a 31% increased risk of disease progression (HR 1.31), compared with those who were given tamoxifen.93 The median progression-free survival was 3.2 months in the thalidomide group versus 4.5 months in the tamoxifen group. This suggests that tamoxifen may have a role in selected patients with a rising CA125 level, and the relationship between estrogen receptor positivity and benefit of tamoxifen in this patient population is being evaluated in current studies. 9 CHEMOTHERAPY FOR RECURRENT EPITHELIAL MALIGNANCIES The majority of patients who present with advanced epithelial cancers of the ovary/fallopian tube/peritoneum will relapse with a median time to recurrence of 16 months. Patients with recurrent ovarian cancer constitute a heterogeneous group with a variable prognosis, and a variable response to further treatment. The most widely used clinical surrogate for predicting response to subsequent chemotherapy and prognosis has been the progression-free interval or the “platinum-free interval,” which is defined as the time from cessation of primary platinum-based chemotherapy to disease recurrence or progression.94, 95 This has been useful to define specific patient populations, but it has a number of limitations and depends on how patients are followed. In particular, it depends on how recurrence is detected and defined. Patients with a treatment-free interval of less than 6 months are classified as platinum resistant and generally treated with nonplatinum-based chemotherapy, while those with a treatment-free interval of more than 6 months are considered to be platinum sensitive and commonly treated with platinum-based chemotherapy. Patients who progress while on treatment or within 4 weeks of stopping chemotherapy are classified as platinum refractory.94, 95 There have been modifications to these definitions, and time to progression or recurrence rather than treatment-free interval or platinum-free interval has been used to define specific patient populations. There has been significant change in practice over the last 20 years and patients have been routinely followed with regular CA125 testing after completion of chemotherapy. For example, the “platinum-resistant” subgroup may include asymptomatic patients with CA125 progression alone at 3 months post chemotherapy or radiological evidence of recurrence as well as those who are symptomatic with clinical recurrence. The Fourth Ovarian Cancer Consensus Conference reached agreement that distinct patient populations should be based on the interval from last platinum therapy and the time to progression. The progression-free interval is defined from the last date of platinum dose until progressive disease is documented.94, 95 For patients whose disease is considered platinum-sensitive, the ICON 4 study showed advantage in terms of overall survival and progression-free survival for a combination of carboplatin and paclitaxel versus single-agent carboplatin.96 Level of Evidence A For patients with neurotoxicity, gemcitabine97 or liposomal doxorubicin98 may be substituted for paclitaxel. A large GCIG study (CALYPSO) compared carboplatin and liposomal doxorubicin (CD) with carboplatin and paclitaxel (CP) in 976 patients.99 The CD arm had statistically superior progression-free survival compared with the CP arm, with a median progression-free survival of 11.3 versus 9.4 months, respectively. There was no significant difference in the overall survival between the treatment groups. Median overall survival was 33 versus 30.7 months for the CP and CD arms, respectively. The CD arm was better tolerated with less severe toxicities, and this combination is now widely used. Level of Evidence A There is evidence that the addition of bevacizumab to the regimen of carboplatin and gemcitabine improves progression-free survival compared with carboplatin and gemcitabine in platinum-sensitive disease. In the OCEANS study,100 484 patients with platinum-sensitive disease were randomly assigned to carboplatin (AUC 4 on day 1) and gemcitabine 1000 mg/m2 on days 1 and 8) with or without bevacizumab (15 mg/kg on day 1) with every 21 days cycles. Bevacizumab could be given concurrently with chemotherapy for a maximum of 10 cycles followed by bevacizumab alone until progression of disease or toxicity. The addition of bevacizumab to carboplatin and gemcitabine resulted in an improvement in progression-free survival (12 vs 8 months; HR 0.48; 95% CI 0.39–0.61); however, there was no difference in overall survival between the two arms. Treatment with bevacizumab was associated with higher rates of serious hypertension (17% vs <1%), proteinuria grade 3 or higher (9% vs 1%), and noncentral nervous system bleeding (6% vs 1%).100 For patients with definite platinum-resistant disease, enrollment on available clinical trials or treatment with nonplatinum chemotherapy should be considered. There are a number of chemotherapy options including liposomal doxorubicin,101 topotecan,101 etoposide,102, 103 and gemcitabine.104, 105 The reported response rates are low, about 10%, with a median time to progression of 3–4 months and a median survival of 9–12 months. Over the last 5 years there have been a number of trials carried out with new agents in patients with platinum-resistant ovarian cancer, including epothilones, trabectedin106 and permetrexed107 with no significant increase in response rates or progression-free survival. No new cytotoxic agent has been approved to treat recurrent ovarian cancer for many years. The role of angiogenesis inhibitors in platinum-resistant ovarian cancer is discussed below. The optimal management of a patient with platinum-resistant or refractory disease is complex and requires a careful assessment of the patient's performance status, symptoms, and extent of disease. Attention to symptom control and good palliative care is an essential component of management. With very few exceptions, recurrent disease is not curable and the aim of treatment is to maintain quality of life and palliate symptoms particularly in patients with platinum-resistant ovarian cancer.108 There are many potential treatment options, including chemotherapy, angiogenesis inhibitors, radiation therapy, or surgery in selected patients and inclusion in clinical trials.89 There is a subset of patients who may benefit from secondary surgical debulking, but they constitute a minority. The role of secondary surgical debulking is being addressed in prospective randomized clinical trials. Level of Evidence C 9.1 PARP inhibitors as monotherapy in patients with recurrent ovarian cancer Olaparib is FDA approved for the treatment of patients with gBRCA-mutated recurrent ovarian cancer who have received three or more prior lines of chemotherapy.109, 110 The FDA granted approval on the basis of the response rate in a single-arm study of olaparib in patients with BRCA mutations and with a wide range of different cancers. The response rate was 34% in heavily pretreated BRCA-positive patients with platinum-resistant recurrent ovarian cancer and the median progression-free survival was 7.9 months.110 Rucaparib is also approved for treatment of BRCA-mutation-associated advanced ovarian cancer after completion of treatment with two or more chemotherapy regimens regardless of whether patients are platinum-sensitive or resistant.111 Rucaparib's approval was based primarily on efficacy data from 106 patients with BRCA-associated recurrent ovarian cancer who had prior treatment with two or more chemotherapy regimens and safety data from 377 patients with ovarian cancer treated with rucaparib 600 mg orally twice daily on two open-label, single-arm trials.112 Investigator-assessed objective response rate was 54% and the median duration of response was 9.2 months.112 10 MANAGEMENT OF EPITHELIAL TUMORS OF LOW-GRADE SEROUS CANCERS Low-grade serous cancers (LGSCs) comprise 5% to 10% of serous ovarian cancers and up to 8% of all ovarian cancers.113 They are typically diagnosed at a younger age than in women with high-grade serous ovarian cancer (HGSOC), with a median age of 47–54 years at diagnosis, and are characterized by a relatively indolent behavior and resistance to cytotoxic chemotherapy.114 In contrast to HGSOC they do not have TP53 mutations, but may have KRAS or BRAF mutations, and activation of the Ras-Raf-MEK-ERK signaling pathway.114-116 Most patients with low-grade serous ovarian cancer (LGSOC) have advanced-stage disease at initial diagnosis and the surgical management is similar to patients with high-grade cancers, with attempts at total resection of tumor—with the exception of fertility-sparing surgery in younger women with tumors confined to the ovary. Neoadjuvant platinum-based chemotherapy for advanced-stage LGSOC or peritoneum was associated with a radiological response rate of 4%, which is much lower than response rates of up to 80% in patients with HGSOC.117 Similarly, the response rates to chemotherapy have been reported to be low in a number of studies and the rate was only 3.7 (4.9% in patients with platinum-sensitive disease and 2.1% in those with platinum-resistant disease) in a report of patients with recurrent LGSC.114 A recent retrospective, exploratory, case-control analysis of over 5000 patients receiving adjuvant chemotherapy in clinical trials included 145 patients (2.8%) with LGSOC, of whom 37 had suboptimal debulking and were evaluable for response evaluation.118 The response rate was higher than other studies at 23.1% in this small subset of patients with LGSOC compared with 90.1% in patients with HGSOC. The majority of patients with LGSOC will relapse despite treatment and have a relatively long survival (median overall survival of 82 months). These patients are often treated with multiple agents over many years for recurrent disease with variable degrees of benefit and the impact of treatment on survival is unclear.118 10.1 Management of low malignant potential (borderline) tumors Compared with invasive epithelial cancers, borderline tumors tend to affect a younger population and constitute 15% of all epithelial tumors of the ovary.119 Nearly 75% of these are Stage I at the time of diagnosis. The following can be said for these tumors120: The diagnosis must be based on the pathology of the primary tumor. Extensive sectioning of the tumor is necessary to rule out invasive cancer. The prognosis of these tumors is extremely good, with a 10-year survival of about 95%. Invasive cancers that arise in borderline tumors are often indolent and generally have a low response to platinum-based chemotherapy. Spontaneous regression of peritoneal implants has been observed. Early stage, serous histology, and younger age at diagnosis are associated with a more favorable prognosis. Although gross residual disease after primary laparotomy is associated with poorer prognosis, mortality from the disease remains low. Those patients who have invasive implants in the omentum or other distant sites are more likely to recur earlier. The role of cytotoxic chemotherapy is questionable as the response rates are low. The causes of death include complications of disease (e.g. small bowel obstruction) or complications of therapy, and only rarely malignant transformation. The mainstay of treatment is primary surgical staging and cytoreduction. For patients with Stage I disease who want to preserve fertility, conservative surgery with unilateral salpingo-oophorectomy can be considered after intraoperative inspection of the contralateral ovary to exclude involvement.121 For patients with only one ovary, or bilateral cystic ovaries, a partial oophorectomy or cystectomy can be considered for fertility preservation. For all other patients, total hysterectomy and bilateral salpingo-oophorectomy are recommended, with maximal cytoreduction if the disease is metastatic. Patients with borderline tumors in all stages of disease should be treated with surgery. A small percentage of patients with invasive implants may respond to chemotherapy but the response to chemotherapy is low. Uncommonly, some patients recur early and have higher-grade invasive cancers and may benefit from chemotherapy.122 In patients with late recurrence of the disease, secondary cytoreduction should be considered, and chemotherapy given only if invasive disease is present histologically. Hormonal therapy has been reported to be associated with clinical benefit in recurrent and metastatic borderline ovarian tumors as well as LGSC. Hormonal therapy was reported to have a response rate of 9% in a retrospective analysis of 64 patients with recurrent LGSC.123 In 26 patients with LGSC of the ovary or peritoneum, adjuvant hormone therapy following debulking surgery was associated with a median progression-free survival of 22 months and recurrence rate of 14.8%.124 In this small study, survival of the patients treated with adjuvant hormonal therapy was not significantly different to an age- and stage-matched control group of patients with LGSC treated with surgery and adjuvant chemotherapy. A recent retrospective analysis was reported of 203 patients with LGSC of the ovary or peritoneum who received either maintenance/adjuvant hormonal treatment or observation, based on physician discretion, following primary cytoreductive surgery and platinum-based chemotherapy.125 Patients who received adjuvant hormonal therapy had significantly longer median progression-free survival (64.9 vs 26.4 months) compared with the patients in the observation group, without significant prolongation of overall survival (115.7 vs 102.7 months). The role of maintenance/adjuvant hormonal therapy in patients with LGSC will soon be tested in a large NRG trial. Follow-up of patients with no evidence of disease is the same as for those with malignant epithelial carcinomas, but at less frequent intervals. If the contralateral ovary has been retained, it should be followed by transvaginal ultrasonography, at least on an annual basis.1, 120, 126 Level of Evidence C 11 MANAGEMENT OF GRANULOSA CELL TUMORS Granulosa cell tumors account for about 70% of sex-cord stromal tumors and 3%–5% of all ovarian neoplasms.2 There are two types of granulosa cell tumors: the juvenile and the adult types. Because of the high estrogen production, the juvenile type typically presents with sexual precocity, while the adult type may present with postmenopausal bleeding. The majority of patients are diagnosed with Stage I tumors. The peak incidence is in the first postmenopausal decade.2, 127 Granulosa cell tumors are generally indolent (i.e. with a tendency to late recurrence). Stage at diagnosis is the most important prognostic factor. Other prognostic factors include age at diagnosis, tumor size, and histologic features. If metastatic, adequate cytoreduction is the mainstay of treatment. If the patient is young and the disease is confined to one ovary, conservative surgery should be performed.128, 129 The infrequency of the disease, and its protracted course, has resulted in a lack of prospective studies. There is no evidence that adjuvant chemotherapy or radiotherapy improves the results of surgery alone for Stage I disease. The value of postoperative adjuvant chemotherapy for higher-risk Stage I disease (tumor size >10 cm, capsule rupture, high mitotic count) is uncertain, and has not been tested in randomized studies. Platinum-based chemotherapy is used for patients with advanced or recurrent disease, with an overall response rate of 63%–80%.129-131 Follow-up is clinical. For patients with elevated levels of inhibin B and/or AMH at initial diagnosis of granulosa cell tumors, inhibin B and/or AMH appear to be reliable markers during follow-up for early detection of residual or recurrent disease.132 There is no evidence-based preference for inhibin B or AMH as a tumor marker.133 Serum inhibin is a useful tumor marker in postmenopausal women. Level of Evidence C 12 MANAGEMENT OF GERM CELL MALIGNANCIES This group of ovarian tumors consists of a variety of histologically different subtypes that are all derived from the primitive germ cells of the embryonic gonad. Malignant germ cell tumors represent a relatively small proportion of all ovarian tumors. Prior to advances in chemotherapy, the prognosis for these aggressive tumors was poor. The use of platinum-based chemotherapeutic regimes has made germ cell malignancies among the most highly curable cancers.127 12.1 Presentation These are most common ovarian tumors in the second and third decades of life. They are frequently diagnosed by finding a palpable abdominal mass in a young woman who complains of abdominal pain. The following are the symptoms of germ cell tumors in order of frequency127: Acute abdominal pain. Chronic abdominal pain. Asymptomatic abdominal mass. Abnormal vaginal bleeding. Abdominal distention. 12.2 Histologic classification The classification of germ cell tumors of the ovary is important to determine prognosis and for treatment with chemotherapy. Germ cell tumors are classified as follows2, 127: Dysgerminoma. Embryonal carcinoma. Polyembryoma. Teratoma (immature; mature; mature with carcinoma [squamous cell, carcinoid, neuroectodermal, malignant struma, etc.]). Extraembryonal differentiation (choriocarcinoma; endodermal sinus tumor [yolk sac tumor]). 12.3 Diagnosis, staging, and surgical management Ovarian germ cell tumors are staged similarly to epithelial carcinomas, although the staging system used for male germ cell tumors is probably more useful. The approach to treatment is based on the principles of management of metastatic germ cell tumors of the testis (i.e. low, intermediate, and poor risk). Dysgerminoma is the equivalent of seminoma in testicular cancer.134 It is exquisitely sensitive to platinum-based chemotherapy and is radiosensitive. The cure rate is high irrespective of the stage. The other histologic subtypes are equivalent to nonseminomatous testicular cancer. The aggressiveness of the disease is dependent on the type, the most aggressive being endodermal sinus and choriocarcinoma, but with combination chemotherapy, they are highly curable.135-139 As chemotherapy can cure the majority of patients, even with advanced disease, conservative surgery is standard in all stages of all germ cell tumors. Conservative surgery means laparotomy with careful examination and biopsy of all suspicious areas, with limited cytoreduction, thereby avoiding major morbidity. The uterus and the contralateral ovary should be left intact. Wedge biopsy of a normal ovary is not recommended as it defeats the purpose of conservative therapy by potentially causing infertility. Patients with advanced disease may benefit from 3 to 4 cycles of neoadjuvant chemotherapy using BEP (bleomycin, etoposide, cisplatin [platinum]) regimen with preservation of fertility.140 Patients who receive conservative surgery with the preservation of one ovary retain acceptable fertility rates despite adjuvant treatment with chemotherapy. There has been no report of higher adverse obstetric outcome or long-term unfavorable sequelae in the offspring.141-144 Secondary surgery is of no proven benefit, except in those patients whose tumor was not completely resected at the initial operation and who had teratomatous elements in their primary tumor. Surgical resection of residual masses may be beneficial in such patients, as there may be mature teratomatous nodules that can continue to increase in size (growing teratoma syndrome), and more rarely can undergo malignant transformation over time to an incurable malignancy, e.g. squamous cell carcinoma.145 12.4 Postoperative management and follow-up of dysgerminoma Patients with Stage IA disease may be observed after surgery. A small proportion of patients may recur, but they can be treated successfully at the time of recurrence with a high rate of cure. Patients with disease beyond the ovary should receive adjuvant chemotherapy. Although radiation therapy is effective, ovarian failure makes it undesirable for patients with an intact ovary. A follow-up surveillance regime for patients with Stage 1A dysgerminoma is outlined in Table 5. This schedule is based on the experience managing seminomas in males and the reports by Patterson et al.146 and Dark et al.147 This pragmatic follow-up schedule and has not been tested in randomized trials. Table 5. Follow-up regime for Stage I germ cell malignancies.a \| Regimen \| Description \| --- \| \| Surveillance \| Baseline CT chest, abdomen, and pelvis, if not performed preoperatively \| \| Repeat CT or MRI, abdomen and pelvis at 3 months after surgery \| \| Repeat CT or MRI abdomen plus pelvis at 12 months \| \| Pelvic ultrasound alternate visits (not when having CT scan) for 2 years if non-dysgerminoma and for 3 years if dysgerminoma \| \| Chest X-ray at alternate visits \| \| Clinical examination \| \| \| 1 year \| Monthly \| \| 2nd year \| 2 monthly \| \| 3rd year \| 3 monthly \| \| 4th year \| 4 monthly \| \| Years 5–10 \| 6 monthly \| \| Tumor marker followup \| Samples: serum AFP and hCG, LDH and CA 125 (regardless of initial value) \| \| 0–6 mo \| 2 weekly \| \| 7–12 mo \| 4 weekly \| \| 12–24 mo \| 8 weekly \| \| 24–36 mo \| 12 weekly \| \| 36–48 mo \| 16 weekly \| \| 48+ mo \| 6 monthly until year 10 \| Abbreviations: AFP, alpha-fetoprotein; hCG, human chorionic gonadotropin; LDH, lactate dehydrogenase. a Adapted from Patterson et al.146 12.4.1 Chemotherapy for dysgerminoma Dysgerminoma is extremely sensitive to chemotherapy, and treatment with chemotherapy cures the majority of patients, even with advanced disease.127, 148 The recommended chemotherapy regimen is as follows: Etoposide (E) 100 mg/m2 IV per day for 5 days every 3 weeks for 3 cycles. Cisplatin (P) 20 mg/m2 IV per day for 5 days every 3 weeks for 3 cycles. Bleomycin (B) 30 000 IU IV/IM on days 1/8/15 for 12 weeks (Optional) (Note: bleomycin is dosed in International Units). If bleomycin is omitted, then 4 cycles of EP are commonly used. Note that various schedules of bleomycin have been used. When there is bulky residual disease, it is common to give 3–4 courses of BEP chemotherapy.148 Level of Evidence B The optimal follow-up schedule has not been clinically investigated in ovarian germ cancers and the frequency of visits and investigations is controversial. Patients who have Stage I tumors and are offered surveillance need to be seen regularly and one option is to utilize the follow-up regimen presented above.147 Patients who have had chemotherapy have a lower risk of recurrence and the frequency of CT scans can be reduced, which is similar to the approach for testicular germ cell tumors.146 Each follow-up visit should involve taking a medical history, physical examination, and tumor marker determination. Although tumor markers are important, radiological imaging is also pertinent, especially for patients whose tumor markers were not raised at diagnosis. CT or MRI scans should be performed as clinically indicated.147 Patients who have not received chemotherapy should be followed closely. Ninety percent of relapses in these patients occur within the first 2 years. At relapse, with few exceptions, these patients can be successfully treated.147 Level of Evidence D 12.5 Postoperative management and follow-up of nondysgerminoma germ cell malignancies These tumors are highly curable with chemotherapy, even with advanced disease. Patients with Stage IA grade 1–2 immature teratoma have a very good prognosis and should be only observed after primary conservative surgery. Adjuvant chemotherapy does not appear to add any survival benefit in this subgroup of patients. All other patients with nondysgerminomas, and higher-stage and higher-grade immature teratomas, should receive postoperative adjuvant chemotherapy.127 The recommended chemotherapy regimen is etoposide 100 mg/m2 per day for 5 days with cisplatin 20 mg/m2 per day for 5 days, and bleomycin at 30 000 IU IM/IV on days 1, 8, and 15 for a total of 12 weeks of treatment. For patients with good prognosis disease, 3 cycles of BEP are recommended, while patients with intermediate/poor risk disease should receive 4 cycles of BEP.127 Patients who relapse after BEP may still attain a durable remission and cure with second-line chemotherapy regimens such as paclitaxel–ifosfamide–cisplatin (TIP).137 High-dose chemotherapy and autologous marrow rescue may be considered in selected patients. These patients should be managed in specialized units. After chemotherapy, patients with metastatic immature teratomas can sometimes have residual masses, which are composed entirely of mature elements. These masses can grow, and should be resected after the completion of chemotherapy.149 Level of Evidence B All patients should have lactate dehydrogenase (LDH), alpha-fetoprotein (AFP), and human gonadotropin (beta hCG) to monitor response to treatment. All patients treated with chemotherapy should be followed-up with medical history, physical examination, and appropriate tumor markers in the same way as dysgerminomas. CT or MRI scans should be performed as clinically indicated.122 Relapses in patients usually occur within the first 2 years after diagnosis127, 137 Level of Evidence D 13 SARCOMA OF THE OVARY Ovarian sarcomas are rare and occur primarily in postmenopausal patients.127, 150 Nevertheless, accurate diagnosis and differentiation from other types of primary ovarian cancer are important, as the prognosis is generally poor. There are two types of sarcoma. Malignant mixed Müllerian tumors (MMMTs), the more common of the two, are biphasic tumors composed of both carcinomatous and sarcomatous elements.150, 151 Most authors agree that most MMMTs are monoclonal in origin and should be thought of and managed as a high-grade epithelial cancer. The sarcomatous component is derived from the carcinoma or from a stem cell that undergoes divergent differentiation. Thus, ovarian carcinosarcomas are best regarded as metaplastic carcinomas. Pure sarcomas are very rare and should be treated according to the specific histologic subtype. These rare sarcomas include fibrosarcomas, leiomyosarcomas, neurofibrosarcomas, rhabdomyosarcomas, chondrosarcomas, angiosarcomas, and liposarcomas. Their management is not discussed here. Patients with early stage MMMTs have a better outcome than those with advanced stage disease, but the overall prognosis is poor. They should be managed similarly to high-grade pelvic serous cancers. Their rarity prohibits any prospective randomized trials. 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Google Scholar Citing Literature Volume143, IssueS2 Special Issue:FIGO Cancer Report 2018 October 2018 Pages 59-78 FIGO Cancer Staging Articles FIGO World Congress – FIGO Committee & Working Group publications ## References ## Related ## Information Recommended Cancer of the ovary, fallopian tube, and peritoneum: 2021 update Jonathan S. Berek, Malte Renz, Sean Kehoe, Lalit Kumar, Michael Friedlander, International Journal of Gynecology & Obstetrics Cancer of the ovary, fallopian tube, and peritoneum: 2025 update Malte Renz, Michael Friedlander, Jonathan S. Berek, International Journal of Gynecology & Obstetrics Cancer of the ovary, fallopian tube, and peritoneum Jonathan S. Berek, Christopher Crum, Michael Friedlander, International Journal of Gynecology & Obstetrics Cancer of the ovary, fallopian tube, and peritoneum Jonathan S. Berek, Christopher Crum, Michael Friedlander, International Journal of Gynecology & Obstetrics FIGO staging of endometrial cancer: 2023 Jonathan S. Berek, Xavier Matias-Guiu, Carien Creutzberg, Christina Fotopoulou, David Gaffney, Sean Kehoe, Kristina Lindemann, David Mutch, Nicole Concin, Endometrial Cancer Staging Subcommittee, FIGO Women's Cancer Committee, International Journal of Gynecology & Obstetrics Metrics 231 Details © 2018 The Authors. International Journal of Gynecology & Obstetrics published by John Wiley & Sons Ltd on behalf of International Federation of Gynecology and Obstetrics This is an open access article under the terms of the Creative Commons Attribution License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited. Check for updates Keywords Cancer staging Chemotherapy Fallopian tube FIGO Cancer Report Ovarian Ovary Peritoneum Publication History 11 October 2018 11 October 2018 Close Figure Viewer Previous FigureNext Figure Caption Download PDF Posted by 8 X users Referenced in 1 patents On 1 Facebook pages 496 readers on Mendeley See more details
10843	https://artofproblemsolving.com/wiki/index.php/Circular_Inversion?srsltid=AfmBOopXpsJ7b7YdJFgQvrkcwTJAjjxnx3XLJlCYZkwZIfliHfdLg-qt	Art of Problem Solving Circular Inversion - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Circular Inversion Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Circular Inversion Circular Inversion, sometimes called Geometric Inversion or simply Inversion, is a transformation where point in the Cartesian plane is transformed based on a circle with radius and center such that , where is the transformed point on the ray extending from through . Note that , when inverted, transforms back to . All points outside of are transformed inside , and vice versa. Points on transform to themselves, meaning . Finally, the transformation of is debated on its existence. Some call the transformation the ideal point, which is infinitely far away and in every direction. Others claim that this point does not have an inverse. Geometric Inversion technically refers to many different types of inversions, however, if Geometric Inversion is used without clarification, Circular Inversion is usually assumed. Circular Inversion can be a very useful tool in solving problems involving many tangent circles and/or lines. Contents 1 Basics of Circular Inversion 1.1 Inversion of a Circle intersecting O 1.2 Inversion of a Circle not intersecting O 1.3 General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle 1.4 Problems 1.4.1 Problem 1 Basics of Circular Inversion Inversion of a Circle intersecting O The first thing that we must learn about inversion is what happens when a circle which intersects the center of the inversion, , is inverted. Let us have circle , with diameter . is chosen arbitrarily on circle . Points and represent the inversions of and , respectively. is the radius of . We seek to show that circle inverts to a line perpendicular to through . By the definition of inversion, we have and . We can combine the two equations to get . Rewriting this gives: Also, since is a diameter of circle , must be right. Now, we consider and . They share an angle - , and we know that Therefore, we have SAS similarity. Therefore, must be right. From there, it follows that all points on circle will be inverted onto the line perpendicular to at . Therefore, the inversion of circle becomes a line. Note that, if circle extends beyond , the argument still holds. All one needs to do is shuffle things around. Inversion of a Circle not intersecting O Now, we study the inversion of a circle not intersecting the center of inversion. Let us have circle not intersecting , the center of , the circle which we invert around. The points where intersect circle are points and , respectively. Point is arbitrary and on circle . We invert points , , and , producing , , and , respectively. We draw and . Because is a diameter, must be right. We wish to show that circle inverts to another circle. The definition of inversion tells us that . From here, we obtain that and By SAS symmetry (exploiting ), the ratios tell us that: Therefore, we have and . Note that , which must equal . Therefore, . But . Therefore, . As this holds for any , all points on circle will invert to a point on a circle with diameter . General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle This is how circular inversion is useful in the first place - we find the radius of an inverted circle to find the radius of the original circle. Let the original circle be and the inverted circle be , with radii of and , respectively. The radius of the circle of inversion is . We draw the tangent line of circle intersecting O. We know that this is also a tangent line to circle from the result from part 2 - the tangent line, by definition, intersects circle at exactly one point, and for every intersection point, part 2 says that there will be another intersection point. Therefore, the tangent line to circle intersects circle at exactly one point, necessitating this line to be a tangent line. Call the intersections and , respectively. We have . We have and = . We can write an equation for by dividing: From the definition of inversion, we have . Subsituting yields: From Power of a Point, we know that , which equals . Subsistuting gives , and solving for gives: Alternately, Problems Problem 1 In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ? (Source) This article is a stub. Help us out by expanding it. 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10844	https://www.facebook.com/SmithsonianNMNH/posts/look-at-the-prominent-ridge-along-the-top-and-sides-of-this-male-gorilla-skull-g/10155899018398230/	Look at... - Smithsonian National Museum of Natural History \| Facebook Log In Log In Forgot Account? Smithsonian National Museum of Natural History's Post Smithsonian National Museum of Natural History August 28, 2018 · Look at the prominent ridge along the top and sides of this male gorilla skull (Gorilla beringei)! It’s called the sagittal crest, and it serves as an anchor point for a gorilla’s immensely powerful chewing muscles. Like in humans, the muscles responsible for closing a gorilla’s jaw run up the side of their face and anchor on the skull, but the gorilla has a much more substantial attachment and a more powerful bite. Want to feel your own temporalis muscles at work? Put your fingers on your temples the next time you eat! All reactions: 82 1 comment 8 shares Like Comment Most relevant Henry Sieverling Alas poor Yorick, I knew him , Horatio. ... 7y
10845	https://mathworld.wolfram.com/Point-PointDistance2-Dimensional.html	Point-Point Distance--2-Dimensional In the case of a general surface, the distance between two points measured along the surface is known as a geodesic. For example, the shortest distance between two points on a sphere is an arc of a great circle. In the Euclidean plane , the curve that minimizes the distance between two points is clearly a straight line segment. This can be shown mathematically as follows using calculus of variations and the so-called Euler-Lagrange differential equation. The line element in is given by \| \| \| (1) \| so the arc length between two points and is \| \| \| (2) \| where and the quantity we are minimizing is \| \| \| (3) \| Finding the derivatives gives \| \| \| (4) \| \| (5) \| so the Euler-Lagrange differential equation becomes \| \| \| (6) \| Integrating and rearranging, \| \| \| (7) \| \| \| \| (8) \| \| \| \| (9) \| \| \| \| (10) \| The solution is therefore \| \| \| (11) \| which is a straight line. Now verify that the arc length is indeed the straight-line distance between the points. and are determined from \| \| \| (12) \| \| (13) \| Solving for and gives \| \| \| (14) \| \| (15) \| so the distance is \| \| \| (16) \| \| (17) \| \| (18) \| \| (19) \| as expected. For two points with exact trilinear coordinates and , the distance between them is \| \| \| (20) \| \| (21) \| where is the area of the triangle (Scott 1894; Carr 1970; Kimberling 1998, p. 31). The shortest distance between two points on a sphere is the so-called great circle distance. See also Geodesic, Great Circle, Line Line Picking, Point-Point Distance--3-Dimensional Explore with Wolfram\|Alpha More things to try: point-point distance—2-dimensional asymptotes (2x^3 + 4x^2 - 9)/(3 - x^2) express 4.8675 through pi and e References Carr, G. S. Formulas and Theorems in Pure Mathematics, 2nd ed. New York: Chelsea, 1970.Kimberling, C. "Triangle Centers and Central Triangles." Congr. Numer. 129, 1-295, 1998.Scott, C. A. Projective Methods in Plane Analytical Geometry, 3rd ed. New York: Chelsea, 1894. Referenced on Wolfram\|Alpha Point-Point Distance--2-Dimensional Cite this as: Weisstein, Eric W. "Point-Point Distance--2-Dimensional." From MathWorld--A Wolfram Resource. Subject classifications
10846	https://www.cuemath.com/commercial-math/marginal-cost/	Marginal Cost Marginal cost is the additional cost that an entity incurs to produce one extra unit of output. In other words, it is the change in the total production cost with the change in producing one extra unit of output. Let us learn more about the marginal cost along with its formula in this article. \| \| \| --- \| \| 1. \| What is Marginal Cost? \| \| 2. \| Marginal Cost Formula \| \| 3. \| How to Calculate Marginal Cost? \| \| 4. \| Marginal Cost Curve \| \| 5. \| Marginal Cost Vs Marginal Benefit \| \| 6. \| FAQs on Marginal Cost \| What is Marginal Cost? Marginal cost is the change in the total cost of production by producing one additional unit of output. It is useful for the firms to find the marginal cost to understand the impact of the production of an additional unit on the overall cost of production and thus to make relevant decisions related to production in their firm. Marginal Cost Definition: Marginal cost is defined as the cost of producing an additional unit of output. It is the ratio of the change in the total production cost to the change in the number of units produced. At zero level of production, i.e when the quantity produces is 0, then the marginal cost is not defined. Learn about the marginal cost formula in the section below. Marginal Cost Formula The formula to calculate the marginal cost of production is given as ΔC/ΔQ, where Δ means change. Here, ΔC represents the change in the total cost of production and ΔQ represents the change in quantity. When the quantity is increased by 1 unit, then the marginal cost of the nth unit of production can also be calculated using the following formula: MCn = TCn - TCn-1, where MC represents marginal cost and TC represents the total cost. Here, it is important to note that the marginal cost is different from the average cost of production, as average cost means the average cost of producing 1 output, while the marginal cost means the change in the cost by producing an additional unit of output. When the marginal cost formula is ΔC/ΔQ, the formula for average cost is TC/TQ, where TC = total cost of production and TQ = total quantity. How to Calculate Marginal Cost? For calculating marginal cost, we just need to learn how to identify the change in the total cost and the change in quantity by looking at the given data. For example, if a company had produced 2 packs of juice earlier with a total cost of $12, and now it produces one extra unit, i.e. 3 packs of juice at a total cost of $15. So, the marginal cost of producing that 1 additional unit of juice pack can be calculated as ΔC/ΔQ, where ΔC = $15 - $12 = $3, and ΔQ = 3 - 2 = 1. Thus, the marginal cost is $3/1 = $3. The steps to calculate the marginal cost of production are given below: Let us take an example of calculating marginal cost. Given below is the data of the total cost of production of a firm producing school uniforms. We will be finding the marginal cost by observing the changes in the total cost and in the output produced. \| Number of Uniforms (Output) \| Total Cost (in $) \| Marginal Cost (ΔC/ΔQ) \| --- \| 5 \| 68 \| 68/5 = $13.6 \| \| 6 \| 76 \| (76-68)/(6-5) = 8/1 = $8 \| \| 8 \| 88 \| (88-76)/(8-6) = 12/2 = $6 \| \| 9 \| 96 \| (96-88)/(9-8) = 8/1 = $8 \| \| 10 \| 110 \| (110-96)/(10-9) = 14/1 = $14 \| Marginal Cost Curve The marginal cost curve is a U-shaped curve. It indicates that initially when the production starts, the marginal cost is comparatively high as it reflects the total cost including fixed and variable costs. In the initial stage, the cost of production is high as it includes the cost of machines, setting up a factory, and other expenses. That is why the marginal cost curve (MC curve) starts with a higher value. Then it shows a decline as with the same fixed cost, many units are produced, keeping the cost of production low. After it reaches the minimum level or point, it again starts rising to show a rise in the cost of production. It is because of the exhaustion of resources or the overuse of resources. The marginal cost curve is given below for your better understanding. Marginal Cost Vs Marginal Benefit Marginal benefit and marginal cost are two important measures of the change in the value or price of a product. We have already discussed the meaning of marginal cost. Now let us understand what is marginal benefit. The marginal benefit describes what consumers are willing to pay to get an additional unit or the item. Marginal cost is a factor or measurement from the producer's side, but the marginal benefit is a measure from the consumer's side. It generally decreases with the increase in consumption by the consumer, as customer satisfaction tends to decrease with the increase in the consumption of the same commodity. The difference between marginal cost and marginal benefit is shown in the table below: \| Marginal Cost \| Marginal Benefit \| --- \| \| Cost of producing an additional unit of output. \| Price of consuming an additional unit of the product. \| \| It is a measure from the producer's side. \| It is a measure from the consumer's side. \| \| Declines with the increase in production initially, but then increases gradually. \| Declines with the increase in the consumption of a good by a customer. \| ► Related Topics Check these interesting articles related to the concept of marginal cost definition. Marginal Cost Examples Example 1: Find the marginal cost of production if a company spent $20 on producing 2 units of output. Solution: Given, the cost of producing 2 units = $20. It implies, ΔC = $20 and ΔQ = 2. So, by using the formula, we get, MC = ΔC/ΔQ = $20/2 = $10. Therefore, the marginal cost of production is $10. Example 2: If a company makes 8 tablets, its total cost is $23,000. If it makes 9 tablets, its total cost is $24,800. Calculate the MC of the 9th tablet. Solution: For producing the 9th tablet, the company is spending $24800 - $23000 = $1800. Therefore, the marginal cost of the 9th tablet is $1800/1 = $1800. Example 3: Complete the table given below by finding the missing values. \| Quantity \| Total Cost (in $) \| Marginal Cost (in $) \| --- \| 0 \| 20 \| 1 \| 35 \| ? \| \| 2 \| ? \| 10 \| \| 3 \| 53 \| ? \| \| 4 \| 65 \| ? \| Solution: By using the marginal cost formula, the above table can be completed as, \| Quantity \| Total Cost (in $) \| Marginal Cost (in $) \| --- \| 0 \| 20 \| 1 \| 35 \| 15 \| \| 2 \| 45 \| 10 \| \| 3 \| 53 \| 8 \| \| 4 \| 65 \| 12 \| go to slidego to slidego to slide Book a Free Trial Class Marginal Cost Practice Questions go to slidego to slide FAQs on Marginal Cost What is the Best Definition of Marginal Cost? The definition of marginal cost states that it is the cost borne by the company to produce an additional unit of output. In other words, it is the change in the total production cost with the change in the quantity produced. What is Marginal Cost in Economics? In economics, marginal cost is a very important concept affecting the supply of the output of any company. It helps the firms in decision-making related to the effectiveness of the production of additional units of output. What are Marginal Cost and Marginal Revenue? When marginal cost is defined as the change in the cost of production by producing an additional unit of output, the marginal revenue states the change in the total revenue by selling an additional unit of output. Both marginal cost and marginal revenue are important factors determining the cost and selling price of the commodities to maximize profits. What is the Formula of Marginal Cost? The marginal cost formula is defined as the ratio of change in production cost to the change in quantity. Mathematically it can be expressed as ΔC/ΔQ, where ΔC denotes the change in the total cost and ΔQ denotes the change in the output or quantity produced. How to Find Marginal Cost? In calculus, marginal cost can be defined as the first derivative of the cost function with respect to the quantity/output. Or, to find marginal cost we can use the formula: MC = ΔC/ΔQ, where ΔC = change in production cost and ΔQ = change in quantity. How to Find Total Cost from Marginal Cost? To find the missing total cost from the marginal cost, we can use the following formula: MCn = (Cn - Cn-1)/(Qn - Qn-1). For example, if the total cost of producing 2 units is $4, and the MC of the 3rd unit is $2.5, then to find the total cost after producing the 3rd unit, we can apply these values in the above formula as: MC = (C - $4)/(3 - 2) 2.5 = (C - $4)/1 C = $2.5 + $4 C = $6.5 Therefore, the total cost of production for 3 units is $6.5. This is how we can find the total cost from the marginal cost. How Marginal Cost Helps in Decision Making? Increasing marginal cost is an alarming indicator for companies to take relevant action to minimize it, as high marginal costs mean high costs of productions for every additional unit which will increase the expenses for a company. So, low marginal cost is an optimal stage of production. When the MC curve reaches its minimum level, it indicates that the company has reached its optimal level of production, and every additional unit after that could be a reason for an increase in the losses. This is how marginal cost helps in decision-making. How to Find Variable Cost from Marginal Cost? Marginal cost is the change in the total cost which is the sum of fixed costs and the variable costs. Fixed costs do not contribute to the change in the production level of the company and they are constant, so marginal cost depicts a change in the variable cost only. So, by subtracting fixed cost from the total cost, we can find the variable cost of production.
10847	https://pubchem.ncbi.nlm.nih.gov/compound/14_N_Nitrogen	(~14~N)Nitrogen \| N2 \| CID 71309542 - PubChem An official website of the United States government Here is how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. NIH National Library of Medicine NCBI PubChem About Docs Submit Contact Search PubChem compound Summary (~14~N)Nitrogen PubChem CID 71309542 Structure Primary Hazards Laboratory Chemical Safety Summary (LCSS) Datasheet Molecular Formula N 2 Synonyms (~14~N)Nitrogen DTXSID80745922 1173020-41-1 RefChem:211351 DTXCID20696666 View More... Molecular Weight 28.010 g/mol Computed by PubChem 2.2 (PubChem release 2025.09.15) Dates Create: 2013-05-17 Modify: 2025-09-27 1 Structures 1.1 2D Structure Structure Search Get Image Download Coordinates Chemical Structure Depiction Full screen Zoom in Zoom out PubChem 1.2 3D Conformer Structure Search Get Image Download Coordinates Interactive Chemical Structure Model Ball and Stick Sticks Wire-Frame Space-Filling Show Hydrogens Animate Full screen Zoom in Zoom out PubChem 2 Names and Identifiers 2.1 Computed Descriptors 2.1.1 InChI InChI=1S/N2/c1-2/i1+0 Computed by InChI 1.07.4 (PubChem release 2025.09.15) PubChem 2.1.2 InChIKey IJGRMHOSHXDMSA-IGMARMGPSA-N Computed by InChI 1.07.4 (PubChem release 2025.09.15) PubChem 2.1.3 SMILES N#[14N] Computed by OEChem 4.2.0 (PubChem release 2025.09.15) PubChem 2.2 Molecular Formula N 2 Computed by PubChem 2.2 (PubChem release 2025.09.15) PubChem 2.3 Other Identifiers 2.3.1 CAS 1173020-41-1 CAS Common Chemistry; EPA DSSTox; European Chemicals Agency (ECHA) 2.3.2 European Community (EC) Number 694-222-0 European Chemicals Agency (ECHA) 2.3.3 DSSTox Substance ID DTXSID80745922 EPA DSSTox 2.3.4 Wikidata Q82694182 Wikidata 2.4 Synonyms 2.4.1 Depositor-Supplied Synonyms (~14~N)Nitrogen DTXSID80745922 1173020-41-1 RefChem:211351 DTXCID20696666 Nitrogen-14N2, 99.99 atom % 14N PubChem 3 Chemical and Physical Properties 3.1 Computed Properties Property Name Property Value Reference Property Name Molecular Weight Property Value 28.010 g/mol Reference Computed by PubChem 2.2 (PubChem release 2025.09.15) Property Name XLogP3-AA Property Value 0.1 Reference Computed by XLogP3 3.0 (PubChem release 2025.09.15) Property Name Hydrogen Bond Donor Count Property Value 0 Reference Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15) Property Name Hydrogen Bond Acceptor Count Property Value 2 Reference Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15) Property Name Rotatable Bond Count Property Value 0 Reference Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15) Property Name Exact Mass Property Value 28.006148008 Da Reference Computed by PubChem 2.2 (PubChem release 2025.09.15) Property Name Monoisotopic Mass Property Value 28.006148008 Da Reference Computed by PubChem 2.2 (PubChem release 2025.09.15) Property Name Topological Polar Surface Area Property Value 47.6 Å² Reference Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15) Property Name Heavy Atom Count Property Value 2 Reference Computed by PubChem Property Name Formal Charge Property Value 0 Reference Computed by PubChem Property Name Complexity Property Value 8 Reference Computed by Cactvs 3.4.8.24 (PubChem release 2025.09.15) Property Name Isotope Atom Count Property Value 1 Reference Computed by PubChem Property Name Defined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Defined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Covalently-Bonded Unit Count Property Value 1 Reference Computed by PubChem Property Name Compound Is Canonicalized Property Value Yes Reference Computed by PubChem (release 2025.09.15) PubChem 4 Related Records 4.1 Related Compounds with Annotation Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here. PubChem 4.2 Related Compounds Same Connectivity Count 7 Similar Compounds (2D) View in PubChem Search Similar Conformers (3D) View in PubChem Search PubChem 4.3 Substances 4.3.1 PubChem Reference Collection SID 504509792 PubChem 4.3.2 Related Substances Same Count 6 PubChem 4.3.3 Substances by Category PubChem 5 Chemical Vendors PubChem 6 Safety and Hazards 6.1 Hazards Identification 6.1.1 GHS Classification Pictogram(s) Signal Warning GHS Hazard Statements H280 (100%): Contains gas under pressure; may explode if heated [Warning Gases under pressure] Precautionary Statement Codes P410+P403P410+P403 ECHA C&L Notifications Summary Aggregated GHS information provided per 39 reports by companies from 1 notifications to the ECHA C&L Inventory. Information may vary between notifications depending on impurities, additives, and other factors. The percentage value in parenthesis indicates the notified classification ratio from companies that provide hazard codes. Only hazard codes with percentage values above 10% are shown. For more detailed information, please visit ECHA C&L website. European Chemicals Agency (ECHA) 6.1.2 Hazard Classes and Categories Press. Gas (Comp.) (100%) European Chemicals Agency (ECHA) 7 Patents 7.1 Depositor-Supplied Patent Identifiers PubChem Link to all deposited patent identifiers PubChem 8 Classification 8.1 UN GHS Classification GHS Classification (UNECE) 9 Information Sources Filter by Source CAS Common ChemistryLICENSE The data from CAS Common Chemistry is provided under a CC-BY-NC 4.0 license, unless otherwise stated. EPA DSSToxLICENSE (~14~N)Nitrogen European Chemicals Agency (ECHA)LICENSE Use of the information, documents and data from the ECHA website is subject to the terms and conditions of this Legal Notice, and subject to other binding limitations provided for under applicable law, the information, documents and data made available on the ECHA website may be reproduced, distributed and/or used, totally or in part, for non-commercial purposes provided that ECHA is acknowledged as the source: "Source: European Chemicals Agency, Such acknowledgement must be included in each copy of the material. ECHA permits and encourages organisations and individuals to create links to the ECHA website under the following cumulative conditions: Links can only be made to webpages that provide a link to the Legal Notice page. EC:694-222-0 EC:694-222-0 (EC: 694-222-0) WikidataLICENSE CCZero (¹⁴N)Nitrogen PubChem GHS Classification (UNECE)GHS Classification Cite Download CONTENTS Title and Summary 1 Structures Expand this menu 2 Names and Identifiers Expand this menu 3 Chemical and Physical Properties Expand this menu 4 Related Records Expand this menu 5 Chemical Vendors 6 Safety and Hazards Expand this menu 7 Patents Expand this menu 8 Classification Expand this menu 9 Information Sources Connect with NLM Twitter Facebook YouTube National Library of Medicine 8600 Rockville Pike, Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov
10848	https://brainly.com/question/43679462	[FREE] What is the minimum angle in degrees that any rectangle that is not a square should be rotated about its - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +24k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +47,6k Ace exams faster, with practice that adapts to you Practice Worksheets +5,2k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified What is the minimum angle in degrees that any rectangle that is not a square should be rotated about its center so that the rectangle maps onto itself? 1 See answer Explain with Learning Companion NEW Asked by isabellamarisol90 • 11/29/2023 0:07 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 4157920 people 4M 0.0 0 Upload your school material for a more relevant answer The minimum angle that any rectangle (that is not a square) should be rotated about its center so that it maps onto itself is 90 degrees. Explanation In order for a rectangle to map onto itself, it must be rotated by a multiple of 90 degrees. This is because a rectangle has four right angles, so any rotation that is a multiple of 90 degrees will preserve the angles and the shape of the rectangle. Therefore, the minimum angle in degrees that any rectangle (that is not a square) should be rotated about its center so that it maps onto itself is 90 degrees. Answered by MehekGupta •29.7K answers•4.2M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 4157920 people 4M 0.0 0 Classical Mechanics - Jeremy Tatum Statistical Mechanics - Daniel F. Styer Non-Euclidean Geometry - Henry Manning Upload your school material for a more relevant answer The minimum angle that any rectangle (that is not a square) should be rotated about its center so that it maps onto itself is 90 degrees. This is due to the rectangle's properties of symmetry which allow it to align back with its original shape after this rotation. Thus, the answer is 90 degrees. Explanation To determine the minimum angle in degrees that any rectangle that is not a square should be rotated about its center to map onto itself, we need to understand the properties of rectangles and their symmetries. Understanding Rotational Symmetry: A rectangle has a specific symmetry. When rotated around its center, the rectangle can align with its original shape through certain angles of rotation. Angles of Rotation: The angles at which a rectangle (not a square) can map onto itself are 90 degrees, 180 degrees, and 270 degrees. This is because: 90 degrees rotation: Rotating a rectangle 90 degrees results in the shorter side aligning with the longer side and vice versa. 180 degrees rotation: Here, the rectangle flips upside down, returning to its original configuration. 270 degrees rotation: This is equivalent to a -90 degree rotation, where again the orientation changes but returns to a recognizable shape. Finding the Minimum Angle: Out of these possible angles (90, 180, 270 degrees), the smallest angle that achieves this symmetry is 90 degrees. Therefore, to maintain its appearance and align correctly, the rectangle must be rotated by the minimum angle of 90 degrees. In conclusion, no matter the dimensions of a rectangle, as long as it is not a square, it will require a rotation of 90 degrees to map onto itself effectively. Examples & Evidence For instance, if you take a rectangle with length 4 units and width 2 units, rotating it 90 degrees will change its orientation but keep the dimensions intact, showing that it aligns correctly after the rotation. In geometric studies, it is established that the rotation of rectangles leads to congruent shapes at these specific angles (90, 180, and 270 degrees), with 90 degrees being the minimum required to achieve self-mapping. Thanks 0 0.0 (0 votes) Advertisement isabellamarisol90 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 5.0 What is the minimum angle of rotation, in degrees, about point p that would carry rectangle abcd onto itself Community Answer What is the minimum angle of rotation, in degrees, about point P that would carry rectangle ABCD onto itself? Respond in the space provided. Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 New questions in Mathematics [13. Associate each number from the first line with the interval it belongs to, from the second line.-2 6 14. Determine which of the following statements are true (A) and which are false (F):a) 1∈(0,2); b) −3∈[−4,−3); c) 3 1∈(−2,1); d) 2 5∈(1,3); e) 2∈(−∞,2]. 15. Determine which of the following statements are true (A) and which are false (F):a) −2∈(−7;−2); b) 2 2∈(1,3;2,82]; c) 3,(6)∈[3,1;3 11]; d) 0∈(−1;0,1]; e) −5∈(−3;2). 16. Write in interval form or by enumerating the elements:а) A={x∈R∣2≤x<4}; f) F={x∈Rx>4 3}; b) B={x∈R∣−1 g)G=\left{x \in R \left\lvert, x \leq-\frac{5}{9}\right.\right}$; c) C={x∈R∣x≤2}; h) H=\left{x \in N \left\lvert\,-3 d)D={x \in Z \mid-2 \leq x<2}$; i) I={x∈Z∣−3≤x≤2}; e) E={x∈R−5 2≤x<5 2}; j) J={x∈R∣x≤17}. 17. Represent the following sets on the number line:а) {x∈R∣−2,5 b){x \in R \mid 4 \leq x \leq 7}$; c) {x∈Z∣−2≤x≤3}; d) {x∈R∣x>7}. 18. Consider the set $A={x \in R \mid-3 a) Write set A as an interval.b) Represent set A on the number line. c) Write an irrational number from set A. d) Specify the smallest natural number from set A. 19. Figure 8 shows the sketch for the back cover of a notebook and the meaning of the inscriptions on this cover. Knowing that this notebook contains 100 sheets, calculate, using a calculator, in what interval the mass of this notebook is, expressed in grams.]( "13. Associate each number from the first line with the interval it belongs to, from the second line. -2 6 Determine which of the following statements are true (A) and which are false (F): a) $1 \in(0,2)$; b) $-3 \in[-4,-3)$; c) $\frac{1}{3} \in(-2,1)$; d) $\frac{5}{2} \in(1,3)$; e) $2 \in(-\infty, 2]$. Determine which of the following statements are true (A) and which are false (F): a) $-2 \in(-7 ;-2)$; b) $2 \sqrt{2} \in(1,3 ; 2,82]$; c) $3,(6) \in\left[3,1 ; \frac{11}{3}\right]$; d) $0 \in(-1 ; 0,1]$; e) $-\sqrt{5} \in(-3 ; 2)$. Write in interval form or by enumerating the elements: а) $A={x \in R \mid 2 \leq x<4}$; f) $F=\left{x \in R \left\lvert\, x>\frac{3}{4}\right.\right}$; b) $B={x \in R \mid-1 g) $G=\left{x \in R \left\lvert\, x \leq-\frac{5}{9}\right.\right}$; c) $C={x \in R \mid x \leq 2}$; h) $H=\left{x \in N \left\lvert\,-3 d) $D={x \in Z \mid-2 \leq x<2}$; i) $I={x \in Z \mid-3 \leq x \leq 2}$; e) $E=\left{x \in R \left\lvert\,-\frac{2}{5} \leq x<\frac{2}{5}\right.\right}$; j) $J={x \in R \mid x \leq 17}$. Represent the following sets on the number line: а) ${x \in R \mid-2,5 b) ${x \in R \mid 4 \leq x \leq 7}$; c) ${x \in Z \mid-2 \leq x \leq 3}$; d) ${x \in R \mid x>\sqrt{7}}$. Consider the set $A={x \in R \mid-3 a) Write set A as an interval. b) Represent set A on the number line. c) Write an irrational number from set A. d) Specify the smallest natural number from set A. Figure 8 shows the sketch for the back cover of a notebook and the meaning of the inscriptions on this cover. Knowing that this notebook contains 100 sheets, calculate, using a calculator, in what interval the mass of this notebook is, expressed in grams.") Which function's graph has asymptotes located at the values x=2 π±nπ?I. y=sec x II. y=cos x III. y=tan x IV. y=cot x A. I onlyB. II onlyC. II and IV onlyD. I and III only The operation is defined by a∗b=a+b+2 ab in arithmetic modulo 4. Draw a table for the operation on the set R={0,1,2,3}. From your table, evaluate: I. 4∗3. II. 2∗(4∗3) Which of the following is a sinusoid? A. x 2+y 2=1 B. y=sin x C. y=[x] D. y=3 x ⋯cos b=2 1[sin(a+b)+sin(a−b)] A. sin a B. cos b C. cos θ D. sin b Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
10849	https://pmc.ncbi.nlm.nih.gov/articles/PMC10752894/	Structural basis for DNA proofreading - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Nat Commun . 2023 Dec 27;14:8501. doi: 10.1038/s41467-023-44198-8 Search in PMC Search in PubMed View in NLM Catalog Add to search Structural basis for DNA proofreading Gina Buchel Gina Buchel 1 Department of Biochemistry and Molecular Biology, Thomas Jefferson University, 1020 Locust St, Philadelphia, PA 19107 USA Find articles by Gina Buchel 1,#, Ashok R Nayak Ashok R Nayak 1 Department of Biochemistry and Molecular Biology, Thomas Jefferson University, 1020 Locust St, Philadelphia, PA 19107 USA Find articles by Ashok R Nayak 1,#, Karl Herbine Karl Herbine 1 Department of Biochemistry and Molecular Biology, Thomas Jefferson University, 1020 Locust St, Philadelphia, PA 19107 USA Find articles by Karl Herbine 1, Azadeh Sarfallah Azadeh Sarfallah 1 Department of Biochemistry and Molecular Biology, Thomas Jefferson University, 1020 Locust St, Philadelphia, PA 19107 USA Find articles by Azadeh Sarfallah 1, Viktoriia O Sokolova Viktoriia O Sokolova 1 Department of Biochemistry and Molecular Biology, Thomas Jefferson University, 1020 Locust St, Philadelphia, PA 19107 USA Find articles by Viktoriia O Sokolova 1, Angelica Zamudio-Ochoa Angelica Zamudio-Ochoa 1 Department of Biochemistry and Molecular Biology, Thomas Jefferson University, 1020 Locust St, Philadelphia, PA 19107 USA Find articles by Angelica Zamudio-Ochoa 1, Dmitry Temiakov Dmitry Temiakov 1 Department of Biochemistry and Molecular Biology, Thomas Jefferson University, 1020 Locust St, Philadelphia, PA 19107 USA Find articles by Dmitry Temiakov 1,✉ Author information Article notes Copyright and License information 1 Department of Biochemistry and Molecular Biology, Thomas Jefferson University, 1020 Locust St, Philadelphia, PA 19107 USA ✉ Corresponding author. Contributed equally. Received 2023 Aug 3; Accepted 2023 Dec 4; Collection date 2023. © The Author(s) 2023 Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit PMC Copyright notice PMCID: PMC10752894 PMID: 38151585 Abstract DNA polymerase (DNAP) can correct errors in DNA during replication by proofreading, a process critical for cell viability. However, the mechanism by which an erroneously incorporated base translocates from the polymerase to the exonuclease site and the corrected DNA terminus returns has remained elusive. Here, we present an ensemble of nine high-resolution structures representing human mitochondrial DNA polymerase Gamma, Polγ, captured during consecutive proofreading steps. The structures reveal key events, including mismatched base recognition, its dissociation from the polymerase site, forward translocation of DNAP, alterations in DNA trajectory, repositioning and refolding of elements for primer separation, DNAP backtracking, and displacement of the mismatched base into the exonuclease site. Altogether, our findings suggest a conserved ‘bolt-action’ mechanism of proofreading based on iterative cycles of DNAP translocation without dissociation from the DNA, facilitating primer transfer between catalytic sites. Functional assays and mutagenesis corroborate this mechanism, connecting pathogenic mutations to crucial structural elements in proofreading steps. Subject terms: Cryoelectron microscopy, DNA synthesis, DNA Here, the authors use cryo-EM to capture nine intermediates along the DNA proofreading pathway using human mitochondrial DNA Polymerase Gamma. The results provide a step-by-step view of the DNA proofreading at single-nucleotide resolution. Introduction Maintaining fidelity of replication of genetic information is among the most critical functions of living organisms. Errors arise as a result of DNA damage but also owing to the occasional incorporation of incorrect (non-cognate) substrates, resulting in mismatched base pairs and potentially deleterious mutations1. Cells have evolved sophisticated mechanisms to fix these errors2; among them is the ability of DNAP to correct the mismatched bases during DNA replication by a proofreading activity3–5. Proofreading employs an intrinsic exonucleolytic activity present in DNAP, or the exonucleolytic activity of auxiliary factors6,7. The former is observed in the Pol A family of DNAPs, which includes bacterial DNA Polymerase I, bacteriophage T7 DNAP, and human mitochondrial DNAP Polγ4,8–12. The N-terminus of these polymerases harbors an exonuclease domain capable of excision of a terminal nucleotide in a canonical metal ion-dependent reaction13. The exonuclease (exo) site is located ~35 Å away from the polymerase (pol) site, with no direct path between them. Therefore, it remains unclear how the misincorporated nucleotide in the nascent DNA can be transferred into the exo site, what triggers the primer separation from the template strand, and how the corrected terminus returns to the pol site after the cleavage6,14,15. One of the hypotheses is based on an intermolecular model, which postulates that upon incorporating a non-cognate base, DNAP dissociates and rebinds the mismatched primer in the exo site16,17. Alternative models suggest an intramolecular mode of proofreading, during which polymerase stays associated with the DNA, but the primer terminus shifts from the pol site to the exo site and returns18–21 or a combination of the two modes14. While there is a general agreement in the field that Pol A enzymes are processive and can proofread DNA without dissociating, the lack of understanding of how editing can be achieved without the engagement of dedicated structural elements in DNAP has persisted until now6,14,15. In this study, we provide a structural basis for the mechanism of proofreading, termed here as “bolt-action,” by human mitochondrial DNAP Polγ and describe the major steps of this process. Results and discussion Capturing Polγ during the multi-step proofreading process To study the proofreading mechanism, we hypothesized that if a processive DNAP, such as Polγ, does not dissociate from the DNA template, it must traverse the misincorporated base from the pol to the exo site through intermediate steps. Two strategies were used to capture these intermediates by preventing the exonucleolytic activity of the enzyme. In the first, we used Wild Type (WT) Polγ (Fig.1a) and a synthetic DNA scaffold with a primer having a terminal mismatched base connected via a nonhydrolyzable phosphorothioate bond (Complex ONE, Fig.1b). In the second, we used a variant of Polγ, in which the catalytic residues in the exo site have been mutated12 (D198A/E200A, Exo- Polγ), and an RNA-DNA scaffold (Complex TWO, Fig.1c). The primer in this complex has been extended by incorporating three cognate dGMP nucleotides and misincorporating one dGMP nucleotide against dTMP, generating a terminal mismatched base pair (Fig.1c). Both complexes showed no significant exonuclease primer degradation within the time frame of the experiment (Supplementary Fig.1a, b). In addition, to validate the data obtained using complex ONE, we prepared WT Polγ complex, in which ~60% of the mismatched DNA primer, containing hydrolyzable phosphodiester bond, has undergone proofreading (Complex THREE, Supplementary Fig.1c). Fig. 1. CryoEM structure of Polγ. Open in a new tab a A schematic showing the domain organization of the Polγ holoenzyme. Polγ_A subunit is in white, Polγ_B1 in royal blue, and Polγ_B2 in sky blue. b, c Assembly of the editing complexes. DNA template strand (here and throughout) is shown in blue, the primer is in red, and the mismatched base is in yellow. d CryoEM density map of Polγ Mismatch Sensing complex (Structure I) at 2.8 Å. The Polγ_B homodimer is shown as a ribbon model (right panel) to reveal the Thumb subdomain (orange). The polymerase (pol) site is colored pink, and the exonuclease (exo) site is yellow. Major structural elements involved in proofreading - Wedge (teal), Guide loop (G-loop, smudge), and Sensor loop (purple)—are shown. The complexes were subjected to single-particle analysis using cryogenic-electron microscopy (CryoEM) (Supplementary Fig.2–6, Supplementary Tables1, 2). A series of high-resolution structures ranging from 2.6 to 3.1 Å provides a detailed account of the proofreading process (Figs.1d and 2a, Supplementary Figs.7 and 8). The CryoEM data revealed that Complex ONE and THREE were represented by five major 3D classes, while Complex TWO—by four major 3D classes, each with a conformation different from the one found in the catalytic Polγ complexes published previously8,22 (Supplementary Figs.2–6). Complexes ONE and THREE data sets produced a structure with a mismatched base in the pol site (Structure I, “Mismatch Sensing” complex), a structure with the mismatched base uncoupled from the pol site (Structure II, “Mismatch Uncoupling” complex), a structure with Polγ during initial backward translocation toward the exo site (Structure VII, “Backtracking Initiation” complex), a structure with a mismatched base at the entrance of the exonucleolytic channel (Structure VIII, “Wedge Alignment” complex), and a structure with two single-stranded nucleotides of the primer located in the exonucleolytic channel (Structure IX, “Primer Separation” complex) (Fig.2a, Supplementary Figs.2, 4, and 7a). Analysis of the Complex TWO dataset resulted in structures representing the consequent forward translocation of Polγ relative to the conformation observed in Structure II by one base pair (bp) (Structure III, “Mismatch Locking” complex) and by two bp (Structures IV, V and VI, “Guide Loop Engagement” complex) (Fig.2a, Supplementary Figs.3 and 7b). Structures IV-VI reveal the same location of the mismatched base of the primer relative to the catalytic sites but show notable changes in protein conformation (Supplementary Fig.7b). Overall, the ensemble of structures represents a stepwise progression of the proofreading process with a single-nucleotide resolution (Fig.2a, Supplementary Figs.7 and 8, Supplementary Videos1 and 2). At the beginning of the proofreading process, the mismatched base is located in the pol site (Structure I); upon completion of the primer’s translocation, this base is found ~35 Å away in the exo site (Structure IX) (Fig.2b). The remaining structures represent the consecutive steps along the primer translocation pathway, which were assigned based on the proximity of the 3′ end of the primer to the pol or exo site (Fig.2b). The non-overlapping conformational states of the Polγ-DNA complexes reveal that upon recognition of the mismatched base and its removal from the pol site (Structures I and II), Polγ translocates forward until the 3′ end of the primer is positioned at the entrance to the channel that leads to the exo site (Structures III, IV, V, and VI, Fig.2a). The transition from the Mismatch Uncoupling to the Mismatch Locking complex requires forward translocation of Polγ by one bp without major conformational changes (Supplementary Videos3, 4). This suggests that Structures III-VI, obtained from analysis of the Complex TWO dataset, represent proofreading steps that follow the Mismatch Uncoupling but precede conformations captured in Structures VII-IX from Complex ONE (Fig.2a, Supplementary Videos1, 2). Upon positioning the 3′ end of the primer at the entrance of the exonuclease channel, Polγ translocates backward (Structure VII) to juxtapose the Wedge helix in the exonuclease domain next to the mismatched base pair (Structure VIII) and then to separate the 3′ end of the primer from the template and divert it towards the exo site in Structure IX (Fig.2). Fig. 2. CryoEM captures multiple steps along the DNA proofreading pathway. Open in a new tab a Polγ structures representing the major steps of the proofreading process. The complexes (Structures I–IV, VII, VIII, and IX, surface representation) are shown in the same orientation of their catalytic subunits. The Mismatch Removal complex is modeled using Structure IX to illustrate the final step of the proofreading. b The trajectory of the mismatched base during proofreading. The structures were aligned using the Polγ_A subunits. The mismatched base of the primer is shown. c The change of DNA trajectory during proofreading. Structures are shown in the same orientation of their catalytic subunit. Change in DNA trajectory (degrees) is indicated relative to the DNA axis in the Mismatch-Sensing complex. The final step of proofreading (Fig.2a, “Mismatch Removal”) and the corresponding structure have not been captured in our data sets due to experimental design since either the exo site (Complex TWO) or the 3′ terminus of the primer (Complex ONE) has been modified, affecting the affinity of the 3′ end of the primer towards the exo site. Nevertheless, the position of the primer termini poised for endonucleolytic reaction can be modeled based on the existing structure of the Klenow fragment of DNAP4. The progression of Polγ along the proofreading pathway is accompanied by ~40° change in the DNA axis relative to DNA in the catalytic structure (Fig.2c) and by the relative motion of its subunits discussed below. Altogether, the observed conformations of Polγ represent consecutive steps in the process of translocation of the 3′ end of the primer from the pol site to the exo site (Fig.2a, b), consistent with a non-dissociative (intramolecular) model of proofreading. Mismatch sensing and uncoupling from the pol site The Complex ONE dataset revealed the initial stages of mismatch recognition (Fig.3a), most likely because the phosphorothioate in the primer slows down the kinetics of DNAP translocation13. Structure I (Mismatch-Sensing complex, Fig.3a) represents the open post-translocated conformation of the primer-template complex, in which the fingers subdomain (res 942–983) is disengaged from DNA. The mismatched base is found in the pol site at about the same distance from the catalytic aspartate residues as the cognate base22 but tilted as compared to the canonical Watson-Crick base-pairing (Fig.3c, d). The mismatched base pair is sensed via minor groove interactions with R853 and Q1102 residues, which are the functional analogs of R615 and Q797 in Bacillus DNAP I23, and R429 and Q615 in T7 DNAP9. The interaction with the mismatch-sensing R853 residue is lost, but new hydrogen bonds are formed with the Q1102 residue of Polγ_A, altering the geometry of the terminal base pair (Fig.3c). In addition, the Y955 residue of the O helix, which is also implicated in correct base-paring sensing24, partially occupies the substrate insertion site, impairing the extension of the mismatched primer (Fig.3c). Fig. 3. Mismatched base sensing and uncoupling from the pol site. Open in a new tab a, b Structures of the Mismatch-Sensing and Uncoupling complexes. The Polγ structures (surface representation) are shown in the same orientation of the Polγ_B homodimer. The thumb domain (res 439–476, orange) is shown as a ribbon representation. Parts of the N-terminal domain of Polγ_A (476–495, 592–623) and thumb (res 797–811) are omitted for clarity. c, d Close-up views of the pol active sites of the mismatched sensing (c) and catalytic (PDB ID 5C51, d) complexes. The complexes were shown in the same orientation of their conserved palm subdomains, the catalytic pol site residues (D890/D1135) are in pink. e A close-up view of DNA in the Mismatch-Sensing and Mismatch Uncoupling complexes. The structures were aligned using their Polγ_B homodimers. The mismatched bases are indicated in yellow, and the bulging part of the Sensor loop (res 861–864) is shown in purple. f, g Close-up views of the DNA binding cavity in Mismatched Sensing (a) and Mismatch Uncoupling (b) complexes. The complexes are shown in the same orientation of the conserved palm subdomain. h Polγ transition from Mismatch Uncoupling to Mismatch Locking complex. The structures were aligned using their Polγ_B homodimers. DNA strands and mismatched bases (yellow) are shown. Following mismatch recognition, the mismatched base is relocated ~15 Å away from the catalytic aspartate residues in the pol site in Structure II (Mismatch Uncoupling complex, Fig.3b). Alignment of Structure I and II using the Polγ_B homodimer suggests that the upstream contacts of Polγ with the DNA are preserved and reveal the backward translocation of DNAP (Fig.3e). This translocation is accompanied by a rigid body 9° rotation of the palm subdomain of the Polγ_A subunit relative to the thumb subdomain of Polγ_A, which remains bound to the upstream DNA (Supplementary Video3). Movement of the Polγ_A palm subdomain around the axis nearly perpendicular to the DNA axis, accompanied by translocation of Polγ along the DNA, positions the 3′ end of the primer 15 Å away from the catalytic aspartate residues in the pol site (Fig.3e). Alignment of Structure I and II using the conserved palm subdomain (res 815–910 and 1095–1236) suggests the movement of the “Sensor loop” (res 851–870), which harbors the mismatch-recognition residue R853 (Fig.3f, g). Residues 861–864 of the Sensor loop are bulging away from their position in the pol site of the Mismatch-Sensing complex, making its position incompatible with the primer trajectory in Structure I and contributing to the uncoupling of the mismatched base from the pol site (Fig.3f, g). Forward translocation of Polγ during proofreading Alignment of the Polγ_B homodimers in Structures II and III reveals the next step of proofreading (Mismatch Locking complex, Structure III), where forward translocation of Polγ by one base pair along the DNA axis is observed (Fig.3h, Supplementary Video4). Polγ moves as a rigid body, not affecting the relative trajectory of the DNA (RMSD 0.8 Å for 1335 Cα atoms). Six bases of the template strand of DNA are seen to neatly fit within the binding cavity of Polγ_A between the fingers and thumb subdomains in the Mismatch Locking complex (Fig.4a, b). The first unpaired template base (n + 1) is flipped out and partially stacks the mismatched base, assuming a “locking” conformation (Fig.4a, b). The stacking interaction of the locking base with the mismatched base of the primer may prevent the premature fraying of the latter and its separation from the DNA template, ensuring timely entry to the exonuclease channel. The conformation of primer/template in Polγ observed in the Mismatch Locking complex is additionally stabilized by interactions with two conserved arginine residues (R337, R338) in the Wedge helix extension region, which becomes ordered in Structure III, and N803 and R807 residues in the thumb subdomain (Fig.4a, b). Fig. 4. Forward translocation of Polγ during proofreading. Open in a new tab a Structure of the Mismatch Locking complex. The Wedge helix (teal) and the G-loop (smudge) are shown. b Close-up views of the 3′-end of the DNA in the Mismatch Locking complex. The n + 1 “locking” DNA base is shown in magenta. c Structures of the Guide Loop Engagement complex (Structures IV–VI). The structures are shown in the same orientation of the Polγ_B homodimer as in the Mismatch Locking complex. Rotation of Polγ_A (degrees) is indicated relative to the Polγ_B homodimer in Structure III by the black arrow. d Close-up views of the 3′ end of the DNA in the Guide Loop Engagement complex. The complex is shown in the same orientation of the palm subdomain of Polγ_A as in b. e Relative motion of Polγ_A and the distal Polγ_B2 subunits of Polγ during proofreading. Conformations in Structure I (closed) and Structure VI (open) are shown. f Close-up view of the long helix of the thumb subdomain in Mismatch Locking (III) and G-loop Engagement (IV–VI) structures showing the bending of the thumb during forward translocation. g Close-up view of the G-loop in Mismatch Locking (III) and G-loop Engagement (IV–VI) structures. h Primer extension assay using the G-loop deletion Polγ in the presence of dGTP. i The G-loop deletion variant of Polγ is inactive in the exonuclease assay. Gels in h, i are representative results from triplicate experiments. The subsequent step in the proofreading pathway involves forward translocation of Polγ along the DNA axis, rotation of the Polγ_A subunit relative Polγ_B homodimer, and refolding of the structural element termed the “Guide loop” (or G-loop, res 757–784), resulting in another distinct Polγ conformation (G-loop Engagement complex, Structure IV, Fig.4c, d, Supplementary Video5). As in Structure III, the DNA binding cavity of Polγ_A in the G-loop Engagement complex accommodates 6 bp of the single-stranded template DNA, which now runs nearly parallel to the Wedge helix (Fig.4c). The n + 1 base completes its rotation and is in a nearly perfect stacking conformation with the mismatched base (Fig.4d), which likely limits any further forward translocation of Polγ. Compared to Structure III, the palm of Polγ_A gradually rotates relative to its thumb subdomain and Polγ_B2 by 13°−26° (Fig.4c), assuming its most “open” conformation and changing the trajectory of DNA by 15° within the catalytic subunit in Structure VI (Fig.2c). The distance between Polγ_B2 and the catalytic subunit increases to ~17 Å as measured by the position of the Arg 232 residue, which sits within the interacting distance with E394 residue in Polγ_B2 in Structures I–III (Fig.4e). In contrast, the later steps (Structures IV-IX) show a significantly enlarged gap between them in the “open” conformation of Polγ (Fig.4e). This relative movement of Polγ subunits appears critical for proofreading activity as it opens up the path for the primer toward the exo site, as has been proposed earlier17. The palm rotation of the Polγ_A subunit is accompanied by alterations in the thumb subdomain, which changes from a bent to a straight conformation (Fig.4f, Supplementary Video5). Alignment of the palm subdomains of Polγ_A in Structures IV and III shows the G-loop translation of ~14 Å and the fingers subdomain rotation by 15°, which pushes the 3′ terminus of the primer toward the exo site (Fig.4b, d, g). A conserved residue in the G-loop, K768, is observed making a hydrogen bond with the phosphate backbone of the primer, moving ~13 Å from its position in Structure III. To probe the functional importance of the G-loop for proofreading activity of Polγ, we generated a deletion variant lacking residues 761–769 (ΔG-loop Polγ). We found that the enzyme’s binding affinity (Supplementary Fig.9a–c) and catalytic activity (Supplementary Fig.9d) were not affected by this deletion. Because the base of the G-loop contributes to Polγ interactions with the DNA primer during primer extension, deletion of the G-loop residues results in a notable decrease in the rate of translocation (Fig.4h). Importantly, ΔG-loop Polγ showed dramatically impaired exonucleolytic activity on the mismatched scaffold, confirming the key role of the G-loop in proofreading (Fig.4i). As mentioned above, Structures IV–VI show similar overall conformations (Fig.4c). The most notable difference between these conformations is the gradual motion of the G-loop (Fig.4g), accompanied by the straightening of the thumb and rotation of the palm subdomain of the Polγ_A subunit, suggesting an essential role of these elements in the proofreading process. Polγ backtracking and primer separation The transition to the G-loop Engagement complex aligns the 3′ end of the mismatched primer with the entrance to the narrow channel leading to the exo site (Fig.4c). Backward translocation of Polγ is now needed to separate the primer from the template strand and to deliver the mismatched base into the exo site (Fig.5a–c). In the Backtracking Initiation complex, Polγ has translocated 1 bp backward as compared to the G-loop Engagement complex observed in Structure VI (Fig.5a, d). The subsequent backward translocation places the Wedge helix atop the mismatched base pair, as observed in the next captured intermediate (Wedge Alignment complex, Structure VIII, Fig.5b, d, e). Further backtracking by 2 bp is observed in the subsequent step of proofreading, which separates the primer from the template strand and positions the 3′ end in the exo site (The Primer Separation complex, Structure IX, Fig.5c, f). Backtracking is defined as the process of backward translocation of polymerase along the DNA that results in a separation of the nascent RNA 3′ terminus from the catalytic site and is associated with the proofreading activity of RNA polymerase25. The process of backtracking is essential for many physiologically relevant processes in bacteria and eukaryotes, such as transcription elongation, pausing, termination, fidelity, and genome instability26. The observed backward translocation of Polγ during proofreading suggests that backtracking may be a universal phenomenon for DNA and RNA polymerases. Fig. 5. Backward translocation of Polγ during proofreading. Open in a new tab a Structure of the Backtracking Initiation complex. b Structure of the Wedge Alignment complex. c Structure of the Primer Separation complex. The structure is shown in the same orientation of the Polγ_B homodimer as the Wedge Alignment complex in b. d Polγ transition from the Guide Loop Engagement (VI) to the Backtracking Initiation (VII) and Wedge Alignment complex (VIII). The structures were aligned using their Polγ_B homodimers. e Close-up view of the mismatched base pair at the entrance to the exonuclease channel in the Wedge Alignment complex (b). The exo catalytic residues (D198/E200) are in pink. f Close-up view of the 3′ end of the primer positioned in the exo site in the Primer Separation complex (c). g, h R309A Polγ variant is deficient in proofreading (g) but active in primer extension (h). i, j R807P Polγ variant is deficient in proofreading (i) but active in primer extension (j). Gels in g–j are representative results from triplicate experiments. In the Wedge Alignment complex, the invariant arginine residue (R309) is seen invading the space between the primer and template strand of DNA (Fig.5e). Substitution of the R309 residue with alanine (R309A) significantly decreases Polγ exonuclease activity (Fig.5g) while not affecting its translocation (Fig.5h). The G-loop assumes its most extended conformation, while the N803, K806, and R807 residues of the thumb subdomain additionally stabilize the DNA primer (Fig.5e). Indeed, substituting the R807 residue with proline (R807P), a mutation found in patients with mitochondrial diseases27–29, dramatically affects Polγ proofreading activity (Fig.5i) while not affecting its translocation (Fig.5j). In the Primer Separation complex, the backtracking of Polγ pushes the Wedge helix against the mismatched base pair causing the primer to peel away from the duplex DNA and into the exonuclease channel (Fig.5c, f). The backward motion is accompanied by a 12° rotation of the Polγ_A at a pivot located at the C-terminal part of the thumb domain (res 476) (Supplementary Video6). In the Primer Separation structure, two terminal nucleotides in the 3′ end of the primer are found in the exonuclease channel (Fig.5c, f). The terminal base is inserted into the exo site and stabilized by hydrogen bonds with N270, D274, D198, and D399 residues (Fig.5f). The proximity of the catalytic residues to the 3′ end of the primer suggests that the exo site of Polγ, similar to E.coli DNAP I4, accommodates three nucleotides of a single-stranded DNA. Therefore, Polγ must backtrack one additional bp to position the phosphodiester bond between two terminal residues of the primer in the exo site in the Mismatch Removal complex (Fig.2a). The “bolt-action” mechanism of proofreading The intermediate steps of the proofreading process are represented by distinct conformations of Polγ, which are stabilized by structural elements involved in the proofreading activity (Fig.2a). These elements—the Sensor loop, the Guide loop, and the Wedge—contain highly conserved residues in mammalian species and are the hot spots for mutations that are associated with debilitating mitochondrial diseases (Supplementary Fig.10). The prevailing point of view in the field of mitochondrial biology is that spontaneous replication errors are indeed responsible for the manifestation of mitochondrial diseases and premature aging1,30,31. Importantly, the link between Polγ proofreading deficiency and mitochondrial dysfunction was further established by the generation of a homozygous knock-in mouse model, termed “mtDNA-mutator” mice, and a genomically engineered fruit fly model, both expressing an exonuclease-deficient mutant of Polγ_A32,33. These studies have shown that Polγ proofreading deficiency led to the accumulation of mtDNA point mutations and deletions, displaying reduced lifespan and premature aging-related phenotypes in the mice and embryonic lethality in the fruit flies32–34. Additionally, others have identified and characterized exonuclease-deficient Polγ_A mutants that are implicated in disease24. Within the scope of this study, we have characterized two additional disease mutations, R309A and R807P, and have provided evidence that defects in structural elements required for Polγ proofreading activity result in proofreading errors, which in turn may be responsible for mtDNA mutations and, therefore, the onset of disease and aging. The captured states of Polγ also suggest a “bolt-action” mechanism of the editing process. The “bolt-action” term refers to the iterative cycles of traversing the 3′ end of the DNA primer between the two catalytic sites, first from the pol site to exo site to eliminate the mismatched base, and then from the exo to pol site to return the corrected 3′ end of DNA (Fig.2a, Supplementary Videos1, 2). To demonstrate the importance of Polγ translocation for proofreading, we prevented the enzyme’s ability to advance towards the downstream DNA region by introducing a non-template strand, as a barrier, into the scaffold constructs (Fig.6a). The presence of the non-template strand did not affect Polγ binding affinity to DNA (Supplementary Fig.11). In the absence of the replicative helicase TWINKLE, Polγ was not able to unwind the DNA and translocate forward on a nick-containing scaffold but incorporated dNTPs when loaded on a gap-10 scaffold (Fig.6b). When scaffolds with the same topology but a mismatched primer were used, proofreading was observed only on the gap-10 scaffold (Fig.6c), confirming that forward translocation is required for proofreading. Fig. 6. The bolt-action mechanism of proofreading. Open in a new tab a Primer-template scaffolds with a nick or a 10-nt gap between the primer and downstream DNA duplex barrier used in (b, c). b Primer extension assays performed with WT Polγ in the presence of all dNTPs on scaffolds shown in (a). c Exonuclease assay performed with WT Polγ on scaffolds having a nick or 10-nt gap and a 3′ end mismatched (MM) base. d Exonuclease assays performed with WT Polγ and Polγ_A. e–g Structural elements involved in proofreading in Polγ are conserved in the Pol A family of DNAPs. Close-up views of the major structural elements in Klenow (PDB 1KLN) (e), T7 DNAP (PDB 2AJQ) (f), and TAQ DNAP (PDB 4KTQ) (g) are shown. Structures are aligned with respect to their conserved palm subdomains. Primer-template DNA in the exo site (f, g) was modeled from the structure of the Klenow fragment (1KLN). Note that the existing T7 and TAQ structures do not reflect the proofreading step at which engagement of the G-loop is expected. Gels in b–d are representative results from triplicate experiments. Since mismatch removal has been prevented by the scaffold design (Complex ONE) or the use of Exo- Polγ (Complex TWO), we could not trap the return pathway of the corrected primer from the exo site to the pol site. Kinetic experiments suggest that this reverse pathway is much faster than delivering the mismatched base into the exo site16. Therefore, it is unclear if it involves the intermediate complexes described in this study. However, based on the observed changes in DNA trajectory, forward and backward translocations of Polγ are expected in order to reanneal the corrected primer to the DNA template and deliver its 3′ end back to the pol site (Supplementary Videos1, 2). As previously proposed, proofreading activity is associated with an equilibrium between DNAP’s polymerization and proofreading modes15,20. This equilibrium can be shifted by altering catalytic properties of the pol or exo sites35 and mutations affecting interactions of Polγ_A with Polγ_B homodimer36–38. Our findings suggest that during proofreading Polγ_B homodimer affects the motion of the thumb subdomain of Polγ, a structural element also involved in the enzyme’s translocation along DNA during replication22. Indeed, in the absence of the Polγ_B homodimer, the catalytic subunit of Polγ exhibits strong exonucleolytic activity (Fig.6d). However, it cannot efficiently discriminate between correct and mismatched bases, resulting in near-complete primer degradation, as evidenced by the accumulation of small cleavage products. In contrast, the holoenzyme accurately removes the mismatched base leaving the primer intact (Fig.6d), in agreement with the proposed role of Polγ_B in proofreading and replication activities19,39–43. Why doesn’t Polγ immediately cleave the next nucleotide upon removal of the mismatched base from the primer? Analysis of structures VIII, IX, and the modeled Mismatch Removal complex suggest that upon mismatch excision, Polγ must completely withdraw the primer from the exo site, allowing the NMP product to diffuse out of the narrow exonuclease channel. We speculate that at this point - the conformation observed in the Wedge Alignment Structure - the pathway for the corrected primer to the pol site would be favored over its return to the exo site. We propose that the Polγ_B homodimer stabilizes the Polγ_A exonuclease domain, thereby controlling the NMP diffusion mechanism and progressive exonucleolytic cleavage. In contrast, the exonuclease domain in Polγ_A might assume more relaxed conformation and allow faster NMP escape, promoting the processive primer degradation (Fig.6d). Our data suggest that the “bolt-action” mechanism of proofreading is likely conserved in Pol A family of DNAPs, which also includes proofreading E.coli DNA polymerase I, bacteriophage T7 DNAP, and proofreading-deficient TAQ DNAP, and human DNAPs Theta and Nu. All of these DNAPs, contain structural elements that our study demonstrates are critical for proofreading activity. Thus, in both the Klenow fragment and T7 DNAP, the Wedge helix harbors bulky amino acid residue—R455 and K102, respectively. These residues are found in a similar orientation to Polγ_A R309 residue and may serve an analogous function in primer separation (Fig.6e, f). Indeed, mutations of these residues result in dramatic changes in the proofreading activity of these enzymes44,45, in agreement with the Polγ_A R309 mutation phenotype demonstrated in this study. In contrast, the Wedge element in TAQ polymerase46 does not contain any residues with the bulky side chain in the Wedge element (Fig.6g), which likely contributes to the lack of proofreading in this enzyme47. The G-loop element is also present in DNAP I and T7 DNAP (Fig.6e, g); however, its role in proofreading in these polymerases has not been verified by mutagenesis or structural methods. Our finding of distinct intermediates along the proofreading pathways in a member of the Pol A family of polymerases raises the question of whether polymerases of other families can also employ the intramolecular mechanism of proofreading21,48. Since some of these polymerases have different domain organizations (such as Pol III49) or use distinct structural elements for primer separation (e.g., the beta-hairpin in T4 DNA polymerase of the Pol B family16), the proofreading pathway likely involves different, as compared to Polγ, intermediate complexes. Nevertheless, the “bolt-action” mechanism of proofreading described in this study may be a common strategy used by all processive DNA polymerases with proofreading activity. Methods Protein expression and purification N-terminal histidine-tagged human Polγ_B (residues 26–485) was expressed and purified as described previously50. To express human Polγ_B, BL21-CodonPlus (DE3)-RIPL (Agilent) were transformed with the respective plasmid and grown at 37 °C in LB media until OD 600 reached 0.5 units. The proteins were induced by the addition of 0.15 mM IPTG for 18 h at 16 °C. Polγ_B was purified by affinity chromatography using Ni-NTA beads (Thermo Fisher Scientific), followed by affinity chromatography using a HiTrap heparin HP column (GE Healthcare). The heparin column was equilibrated in buffer A (40 mM Tris, pH 8.0, 300 mM NaCl, 5% Glycerol, 5 mM β-mercaptoethanol), and the protein was eluted by linear gradient 0–70% of buffer B (40 mM Tris, pH 8.0, 1.5 M NaCl, 5% Glycerol, 5 mM β-mercaptoethanol). Peak fractions were pooled, concentrated, and stored at −80 °C. Variants of N-terminal histidine-tagged human Polγ_A (res 26–1239) were obtained by site-directed mutagenesis (QuikChange, Agilent). Polγ_A variants (WT, G-loop deletion res 761–769, R807P, and R309A) were expressed using SF9 cells and purified as previously described for Exo- (D198A, E200A) with modifications50. Briefly, Polγ_A was purified by affinity chromatography using Ni-NTA beads (Thermo Fisher Scientific), followed by affinity chromatography using a TSKgel Heparin-5PW column (Tosoh Bioscience). The heparin column was equilibrated in buffer A (40 mM Tris-HCl pH 7.9, 150 mM NaCl, 5% Glycerol, 5 mM 2-mercaptoethanol), and Polγ_A was eluted by 0–70% linear gradient of buffer B (40 mM Tris-HCl pH 7.9, 1.5 M NaCl, 5% Glycerol, 5 mM 2-mercaptoethanol). Peak fractions eluted at 33 mS/cm were collected and analyzed using SDS-PAGE, concentrated, and stored at −80 °C. Polγ holoenzyme complex was reconstituted by incubating Polγ_A fractions after Ni-NTA chromatography with a twofold molar excess of purified Polγ_B for 10 min at room temperature. Polγ was purified by affinity chromatography on a TSKgel Heparin-5PW column (Tosoh Bioscience), equilibrated in the buffer described above. Peak fractions eluted at 40 mS/cm were collected and analyzed using SDS-PAGE, concentrated, and stored at −80 °C. DNA and RNA oligonucleotides and scaffold preparation Synthetic DNA oligonucleotides (IDT DNA) and synthetic RNA oligonucleotides (Dharmacon) were used. Phosphorothioate oligonucleotides were purchased as a racemic mixture of the two di-astereomers. Primers sequences (all 5′ to 3’): GAAGACAGTCTGCGGCGCGA (DNA20sA, the asterisk denotes the position of phosphorothioate, Complex ONE), GAAGACAGUCUGCGGCGCGC (RNA20, Complex TWO), GAAGACAGTCTGCGGCGCGC (DNA20), CCAAGTCAGAAGACAGTCTGCGGCGCGC (DNA28), CCAAGTCAGAAGACAGTCTGCGGCGCGA (DNA28A), GAAGACAGTCTGCGGCGCGA (DNA20A), GGTACAACTTGACGACATAGCGTG (DNA24). Template strand sequences (5′ to 3’): ACACACGCGCGCCGCAGACTGTCTTC (DNA20TS, Complex ONE), GGTAGATCCCGCGCGCCGCAGACTGTCTTC (DNA20_3C_TS, Complex TWO), GGTAGATCCCACGCGCCGCAGACTGTCTTC (DNA20_3C_MM_TS), CGGTCGAGTCACGACTCCGATTATGCGCGCCGCAGACTGTCTTCTGACTTGG (DNA28TS), CGGTCGAGTCACGACTCCGATTATCACGCTATGTCGTCAAGTTGTACC (DNA24TS). Non-template strand sequences: TCGTGACTCGACCG (10nt gap_NT), ATAATCGGAGTCGTGACTCGACCG (Nick_NT). To anneal, the scaffolds were diluted in water, heated for 7 min at 95 °C, and cooled down (1 °C/min) for 1 h to 25 °C in a thermocycler. Primer extension and exonuclease assays The primers were 5’-[32 P]-labeled using [γ-32 P]ATP (3000 Ci/mmol) and T4 Polynucleotide Kinase (NEB). The complexes of Polγ or Polγ_A (50 nM) with labeled primer-template scaffolds were assembled in a buffer containing 40 mM Tris-HCl pH 7.9, 60 mM NaCl, 10 mM MgCl 2, and 20 mM 2-mercaptoethanol in the presence of BSA (0.1 mg/ml) for 5 minutes at room temperature. Primer extension was performed using DNA20/DNA30_3C_TS (Figs.3h, 4g, i), “Nick” (DNA28/DNA28TS/Nick_NT, Fig.6b), “Gap 10” (DNA28/DNA28TS/10nt gap_NT, Fig.6b), or DNA20/DNA20TS (Supplementary Fig.9d) in the presence of 0.1 mM dNTPs or 1 mM dGTP, as indicated. Exonuclease assays were performed with DNA20/DNA20_MM_3C_TS (Figs.3i, 4f, h, 6d), “MM/Nick” (DNA28A/DNA28TS/Nick_NT, Fig.6c), or “MM/Gap 10” (DNA28A/DNA28TS/10nt gap_NT, Fig.6c), all of which contained a single mismatch at the 3′ terminal end of the primer. All reactions were carried out for the indicated times at room temperature and stopped by the addition of an equal volume of 95% formamide/0.05 M EDTA. The products of the reaction were resolved by 20% PAGE containing 6 M Urea and visualized by autoradiography using PhosphorImager (GE Healthcare). Electrophoretic mobility shift assay (EMSA) The 5’-Cy3-labeled variants of DNA24 and DNA28TS were obtained from IDT DNA (Supplementary Fig.9a, 11a, b). To perform EMSA, complexes of Polγ/DNA were assembled in a buffer containing 40 mM Tris-HCl pH 7.9, 60 mM NaCl, 10 mM MgCl 2, 5% glycerol, 20 mM 2-mercaptoethanol in the presence of BSA (0.1 mg/ml), and incubated with 150 nM of 5’-Cy3-labeled scaffolds for 10 min at room temperature. The reactions were resolved in 0.5% agarose gels run in 0.5X TBE buffer for 30 minutes at 100 V at 4 °C. The products of the reactions were visualized using Bio-Rad ChemiDoc TM imager and quantified using Bio-Rad Image Lab TM. For each reaction, the fraction of bound DNA was determined as the intensity of the free DNA divided by that in the 0 nM Polγ control (n = 3 independent experiments). Preparation of Polγ Complexes for CryoEM To assemble Complex ONE, 2 μM WT Polγ was mixed with the DNA20A/DNA20_TS scaffold at a 1:1.1 molar ratio in a buffer containing 10 mM Tris-HCl pH 7.9, 100 mM NaCl, 10 mM DTT, and 2 mM MgCl 2 and incubated for 5 minutes at room temperature prior to overnight dialysis at 4 °C in the same buffer. Complex TWO (2 μM) was assembled using Exo- Polγ (D198A, E200A) and the RNA20/DNA20 3C_TS scaffold at a 1:1 molar ratio in a buffer described above. Following incubation at room temperature for 5 minutes, dGTP was added to a final concentration of 0.1 mM, and the primer was extended for 2 minutes at room temperature, followed by dialysis. To assemble Complex THREE, 2 μM WT Polγ was mixed with the DNA20A/DNA20_TS scaffold at a 1:1 molar ratio in a buffer containing 10 mM Tris-HCl pH 7.9, 100 mM NaCl, 10 mM DTT, and 2 mM MgCl 2 and incubated for 20 minutes at room temperature. The progression of mismatch cleavage was monitored using exonuclease assays described above. The complexes were applied to negatively glow-discharged 300 mesh UltraAufoil −1.2/1.3 holey-gold grids (Quantifoil). Grids were blotted with ash-free Whatman® Grade 540 filter paper in a Vitrobot Mark IV (ThermoFisher Scientific) for 4–5 seconds at 4 °C and 95–100% humidity, then vitrified in liquid ethane. The sample quality and distribution were assessed using Glacios Transmission Electron Microscope equipped with a Falcon 4 direct electron detector. Single-particle data acquisition and image processing Polγ data was collected at the Pacific Northwest Center for CryoEM (PNCC) using a Titan Krios transmission electron microscope (ThermoFisher Scientific), operated at 300 kV and equipped with a Bioquantum Energy Filter with a 20 eV slit width. Movies were collected using a Gatan K3 direct electron detector in super-resolution mode with a magnification of 105,000, corresponding to a pixel size of 0.413. A dose rate of 15.9–19.3 e−/s/physical pixel resulted in a total electron dose of 60–70 e−/Å 2, which was applied over 60–70 frames. Data was collected in SerialEM software with defocus values ranging from −0.5 to −2.0 µm. Workflows for image processing of Complex ONE, Complex TWO, and Complex THREE are shown in Supplementary Figs.2–4, respectively. The movie stacks collected for Complex ONE and Complex TWO were processed in CryoSPARC51. The super-resolution movies were frame aligned, motion corrected, gain normalized, dose-weighted, and binned twice with the patch motion correction module. Contrast transfer function (CTF) values were estimated using the patch CTF (CryoSPARC) or CTFFIND452. Micrographs with ice, ethane contamination, and/or poor CTF fit resolution were discarded. A circular blob picker with dimensions of 80–130 Å was used to pick Polγ particles. Local resolution plots were obtained in CryoSPARC. Resolution values for the Polγ_A and Polγ_B subunits were computed in RELION3.053 with focused masks (Supplementary Table1). The reported resolutions of the CryoEM maps are based on FSC 0.143 criterion54. The isotropy of the 3D reconstruction of Structures I-IX was estimated by 3DFSC server55, as shown in Supplementary Figs.5 and 6. The CryoEM density for the 3′ end of the primer in the exonuclease channel of Structure IX was improved with 3D classification in CryoSPARC 4.0. Model building and structure refinement Polγ_A, Polγ_B1, and Polγ_B2 from the Polγ catalytic complex (PDB ID: 4ZTZ [10.2210/pdb4ZTZ/pdb]) were docked into the respective CryoEM maps from Complex ONE and TWO. The template, primer, and mismatched base were placed into the CryoEM density maps. The terminal mismatched guanine bases in Complex ONE structures were modeled as nonhydrolyzable phosphorothioate bases, with 50% occupancy assigned to each stereoisomer. DNA-B and RNA-A restraints from Coot were used to fit the polynucleotide chains. Additional density in Structure II resolved up to 2.6 Å resolution allowed modeling of the loop regions in Polγ_B1 (res 137–161, 169–179), Polγ_B2 (137–179), and Polγ_A (336–340). A polyalanine model for Polγ_B2 docked in a low-pass filtered map was used for Structures IV and IX. The local density fit of the modeled sequence was improved over an iterative process of amino acid fitting in Coot 0.9.8.556, which alternated with real-space refinement in PHENIX57. Real-space refinement was carried out with secondary structure and Ramachandran restraints. Comprehensive model validation was carried out with PHENIX and the PDB validation server ( and is summarized in Supplementary Figs.5 and 6 and Supplementary Table2. Map-to-model Fourier Shell Correlation plots for the nine structures in Complexes ONE and TWO were obtained in PHENIX (Supplementary Figs.5 and 6). Figures and Supplementary Videos were generated with PYMOL and ChimeraX58. Statistics and reproducibility Experiments presented in Figs.4h, i, 5g–j, and 6b–d and Supplementary Figs.1a–c, 9b, d, and 11a, b were repeated at least three times. The representative gel images are shown. Reporting summary Further information on research design is available in theNature Portfolio Reporting Summary linked to this article. Supplementary information Supplemental Information (22.1MB, pdf) Peer Review File (148.3KB, pdf) 41467_2023_44198_MOESM3_ESM.docx (13.9KB, docx) Description of Additional Supplementary files Video 1 (39MB, mp4) Video 2 (27.1MB, mp4) Video 3 (7.2MB, mp4) Video 4 (4.8MB, mp4) Video 5 (9.9MB, mp4) Video 6 (6.6MB, mp4) Reporting Summary (187.1KB, pdf) Acknowledgements We thank current and former members of the Temiakov laboratory. We are indebted to Sean Mulligan, Theo Humphreys, and Marcelo De Farias from PNCC and Tara Fox (NCI) for their expert technical assistance during data collection. We also thank Drs. William McAllister, Michael Anikin, Gino Cingolani, and Hauke Hillen for the critical reading of the manuscript and fruitful discussion. This work was supported by NIH Grant R35 GM131832 (D.T.). A portion of this research was supported by NIH grant U24GM129547 and performed at the PNCC at OHSU and accessed through EMSL (grid.436923.9), a DOE Office of Science User Facility sponsored by the Office of Biological and Environmental Research, and NIH grant S10 OD030457. This research was partly supported by the NCI’s NationalCryo-EM Facility at the Frederick National Laboratory for Cancer Research under contract HSSN261200800001E. Author contributions Experimental design and conceptualization: G.B., A.R.N., D.T. Protein preparation and biochemical experiments: G.B., A.S., V.O.S., A.Z.O. CryoEM Data acquisition: A.R.N., K.H. Single-particle CryoEM analysis and model building: A.R.N. Writing—original draft: D.T. Writing—review & editing: D.T., G.B., A.R.N., A.Z.O. Peer review Peer review information Nature Communications thanks Dong Wang and the other, anonymous, reviewer(s) for their contribution to the peer review of this work. A peer review file is available. Data availability The CryoEM maps and atomic coordinates were deposited in the Electron Microscopy Data Bank ( under accession codes EMD-29745, EMD-29746, EMD-41091, EMD-29747, EMD-29748, EMD-29749, EMD-29751, EMD-29752, and EMD-29750, and in the Protein Data Bank under accession codes 8G5I, 8G5J, 8T7E, 8G5K, 8G5L, 8G5M, 8G5O, 8G5P, and 8G5N. Previously published protein structure data used for analysis in this study are available in the Protein Data Bank (www.rcsb.org) under PDB ID: 5C51 and 4ZTZ (human mitochondrial DNAP Gamma), 1KLN (Klenow fragment of E.coli DNAP I), 2AJQ (T7 DNAP), 4KTQ (Thermus aquaticus DNAP). Source data are provided with this paper. Competing interests The authors declare no competing interests. Footnotes Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. These authors contributed equally: Gina Buchel, Ashok R. Nayak. 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Supplementary Materials Supplemental Information (22.1MB, pdf) Peer Review File (148.3KB, pdf) 41467_2023_44198_MOESM3_ESM.docx (13.9KB, docx) Description of Additional Supplementary files Video 1 (39MB, mp4) Video 2 (27.1MB, mp4) Video 3 (7.2MB, mp4) Video 4 (4.8MB, mp4) Video 5 (9.9MB, mp4) Video 6 (6.6MB, mp4) Reporting Summary (187.1KB, pdf) Data Availability Statement The CryoEM maps and atomic coordinates were deposited in the Electron Microscopy Data Bank ( under accession codes EMD-29745, EMD-29746, EMD-41091, EMD-29747, EMD-29748, EMD-29749, EMD-29751, EMD-29752, and EMD-29750, and in the Protein Data Bank under accession codes 8G5I, 8G5J, 8T7E, 8G5K, 8G5L, 8G5M, 8G5O, 8G5P, and 8G5N. Previously published protein structure data used for analysis in this study are available in the Protein Data Bank (www.rcsb.org) under PDB ID: 5C51 and 4ZTZ (human mitochondrial DNAP Gamma), 1KLN (Klenow fragment of E.coli DNAP I), 2AJQ (T7 DNAP), 4KTQ (Thermus aquaticus DNAP). Source data are provided with this paper. Articles from Nature Communications are provided here courtesy of Nature Publishing Group ACTIONS View on publisher site PDF (5.7 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Results and discussion Methods Supplementary information Acknowledgements Author contributions Peer review Data availability Competing interests Footnotes Supplementary information References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
10850	https://mirtitles.org/2024/06/11/geometric-transformations-1-by-i-m-yaglom/	Mir Books Books from the Soviet Era Skip to content Home About How To Help? Going To Kindergarten by Nadezhda Kalinina ← Geometric Transformations Volume 1 Euclidean And Affine Transformations by P.S. Modenov; A.S. Parkhomenko Geometric Transformations 2 by I. M. Yaglom → Geometric Transformations 1 by I. M. Yaglom Posted on June 11, 2024 by The Mitr In this post, we will see the book Geometric Transformations 1 by I. M. Yaglom. About the book This book, part of the New Mathematical Library series, aims to make key mathematical concepts accessible to high school students and general readers. These volumes often cover topics outside the standard high school curriculum, varying in difficulty. Readers may need to focus more on certain sections but generally require minimal technical knowledge. Mathematical books can’t be read quickly or fully understood on the first pass. Readers should feel free to skip and return to complex parts. Learning mathematics effectively involves solving problems, so readers are encouraged to engage actively with the material using paper and pencil. This work focuses on elementary geometry, exploring significant ideas like the deductive method and geometric transformations, which lead to non-Euclidean geometry. The book prioritizes simplicity and clarity over rigor, avoiding complex definitions and terms to aid understanding. Problems in the book test comprehension and encourage practice. Readers should attempt at least one problem from each group, consulting provided solutions. The book emphasizes methods over results, using problems to illustrate concepts rather than focusing on three-dimensional propositions, keeping the primary ideas intact. You can get the book here. Follow us on The Internet Archive: Follow Us On Twitter: Write to us: mirtitles@gmail.com Fork us at GitLab: Add new entries to the detailed book catalog here. Share this: Share Click to share on Reddit (Opens in new window) Reddit Click to share on X (Opens in new window) X Click to share on Facebook (Opens in new window) Facebook Like Loading... Related About The Mitr I am The Mitr, The Friend View all posts by The Mitr → This entry was posted in books and tagged 1962, displacements, geometry, glide reflection, isometries of the plane, problem book, rotations, symmetry, translations. Bookmark the permalink. ← Geometric Transformations Volume 1 Euclidean And Affine Transformations by P.S. Modenov; A.S. Parkhomenko Geometric Transformations 2 by I. M. Yaglom → Leave a comment Cancel reply This site uses Akismet to reduce spam. Learn how your comment data is processed. Recent Posts गाता जाये नन्हा पंख – वासीली सुखोम्लीन्स्की (The Singing Feather In Hindi by Vasily Sukhomlisnky) Discrete Mathematics Without Formulas – Lectures Without A Board by Alexander Solovyov सतरंगा फुल – वेलेन्टिन कटायेव (Rainbow Flower In Hindi by Valentin Katayev) अनूठे जहाज़ – ख्यातोस्लाव साखर्नीव ( Wondorous Ships In Hindi by Svyatoslav Sakharnov) यह सभी कुत्तें है – ईगर अकीमुश्किन(The Dog Family Stories In Hindi by Igor Akimushkin) Join 2,237 other subscribers Archives Aerodynamics astronomy books chemistry children's books children's stories Editorial Mir electronics engineering folk tales foreign languages publishing french geology history life sciences little mathematics library malysh publishers mathematics mir books mir publishers physics popular science problem books progress publishers raduga publishers science science for everyone soviet technology Éditions Mir Meta Create account Log in Entries feed Comments feed WordPress.com Trackers on the site… There are trackers on the site apart from Wordpress Stats. If you don't know what trackers are, or wish not to be tracked go here to get tools which allow you to do so. ### Blogroll I Collect Soviet Books Soviet Books in Bengali- Arun Som Soviet books in Bengali. Soviet books in Malayalam-Rajaram Vasudevan Soviet books in Marathi Soviet books in telungu-Anil Battula Soviet books- Nikhil Rane Mir Books Blog at WordPress.com. Comment Reblog Subscribe Subscribed Mir Books Already have a WordPress.com account? Log in now. Mir Books Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Loading Comments...
10851	http://www.cs.put.poznan.pl/wkotlowski/research/kotlowski_phd.pdf	Poznań University of Technology Institute of Computing Science Wojciech Kotłowski Statistical Approach to Ordinal Classiﬁcation with Monotonicity Constraints Doctoral Dissertation Submitted to the Council of the Faculty of Computer Science and Management of Poznań University of Technology Supervisor: Ph.D., Dr. Habil., Professor Roman Słowiński Poznań, Poland 2008 c ⃝2008 Wojciech Kotłowski Institute of Computing Science Poznan University of Technology Typeset using L A T EX in Computer Modern. BibT EX: @phdthesis{ key, author = "Wojciech Kot{\l}owski", title = "{Statistical Approach to Ordinal Classification with Monotonicity Constraints}", school = "Poznan University of Technology", address = "Pozna{\’n}, Poland", year = "2008", } Contents Preface v 1 Introduction 1 1.1 Problem Setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2.1 Statistical Theory for Machine Learning . . . . . . . . . . . . . . . . . . . . . 2 1.2.2 Ordinal Classiﬁcation and Monotonicity Constraints . . . . . . . . . . . . . . 4 1.2.3 Rank Loss Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Existing Approaches to Ordinal Classiﬁcation with Monotonicity Constraints . . . . 7 1.3.1 Dominance-based Rough Set Approach . . . . . . . . . . . . . . . . . . . . . 8 1.3.2 Utility Functions in MCDA . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.3.3 Isotonic Regression and Monotone Approximation . . . . . . . . . . . . . . . 12 1.3.4 Ordinal Learning Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.3.5 Monotone Neural Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.3.6 Monotone Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.3.7 Ordinal Stochastic Dominance Learner . . . . . . . . . . . . . . . . . . . . . . 15 1.3.8 Isotonic Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.4 Goal and Scope of the Thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2 Probabilistic Model for Ordinal Classiﬁcation with Monotonicity Constraints 19 2.1 Stochastic Dominance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2 Monotonicity of the Bayes classiﬁer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.2.1 Loss functions and monotonicity of the Bayes classiﬁer. . . . . . . . . . . . . 20 2.2.2 Convex loss functions and monotonicity. . . . . . . . . . . . . . . . . . . . . . 21 2.3 Linear Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Appendix: Proofs of the Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3 Nonparametric Methods 31 3.1 Nonparametric Probability Estimation by Isotonic Regression . . . . . . . . . . . . . 31 3.1.1 Maximum Likelihood Estimation . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.1.2 Binary-class Problem and Isotonic Regression . . . . . . . . . . . . . . . . . . 32 3.1.3 Multi-class Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.1.4 Extension Beyond the Training Set and Asymptotic Consistency . . . . . . . 35 3.2 Monotone Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.2.1 Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.2.2 Reduction of the Problem Size . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.2.3 Binary Monotone Approximation . . . . . . . . . . . . . . . . . . . . . . . . . 40 3.2.4 Linear Monotone Approximation . . . . . . . . . . . . . . . . . . . . . . . . . 42 ii 3.2.5 Extension Beyond the Training Set and Asymptotic Consistency . . . . . . . 44 Appendix: Proofs of the Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4 Stochastic Dominance-based Rough Set Approach 55 4.1 Dominance-based Rough Set Approach (DRSA) . . . . . . . . . . . . . . . . . . . . . 55 4.1.1 Classical Rough Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.1.2 Rough Set Theory for Ordinal Classiﬁcation . . . . . . . . . . . . . . . . . . . 58 4.1.3 Generalized Decision in DRSA . . . . . . . . . . . . . . . . . . . . . . . . . . 60 4.1.4 Variable Consistency in DRSA . . . . . . . . . . . . . . . . . . . . . . . . . . 62 4.2 Stochastic extension of DRSA (SDRSA) . . . . . . . . . . . . . . . . . . . . . . . . . 63 4.2.1 DRSA as Most Informative Non-invasive Approach . . . . . . . . . . . . . . . 63 4.2.2 Stochastic Dominance-based Rough Sets . . . . . . . . . . . . . . . . . . . . . 64 4.2.3 Probability Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 4.3 Statistical Learning View: Abstaining Classiﬁers . . . . . . . . . . . . . . . . . . . . 68 4.3.1 Statistical Learning View of Classical Rough Sets . . . . . . . . . . . . . . . . 68 4.3.2 Stochastic DRSA as Monotone Conﬁdence Interval Estimation . . . . . . . . 69 4.3.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 5 Learning Monotone Rule Ensembles 73 5.1 Boosting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 5.1.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 5.1.2 Margin-based Loss Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 5.1.3 Margin Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 5.1.4 LPBoost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 5.2 Monotone Rule Ensembles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 5.2.1 Decision Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 5.2.2 Monotone Rule Ensembles for Binary Classiﬁcation . . . . . . . . . . . . . . . 80 5.2.3 Monotone Rule Ensembles with Linear Loss . . . . . . . . . . . . . . . . . . . 81 5.2.4 Two-phase Procedure of Learning . . . . . . . . . . . . . . . . . . . . . . . . 82 5.2.5 Consistency and Generalization Bounds for the Two-phase Procedure . . . . 82 5.3 Linear Programming Rule Ensembles (LPRules) . . . . . . . . . . . . . . . . . . . . 84 5.3.1 Rule Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 5.3.2 Single Rule Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.4 Sigmoid Loss Monotone Rule Ensembles (MORE) . . . . . . . . . . . . . . . . . . . 88 5.4.1 Combining Binary Classiﬁers . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 5.4.2 Rule induction with Sigmoid Loss . . . . . . . . . . . . . . . . . . . . . . . . 90 5.4.3 Single Rule Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 5.4.4 Analysis of the Step Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Appendix: Proofs of the Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 6 Computational Experiment 97 6.1 Design of the Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 6.1.1 Datasets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 6.1.2 Algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 6.2 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 6.3 Interpretability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 iii Summary of the Contribution 105 List of Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Future Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 iv Preface The role of machine learning is to generalize from a data sample. In order to generalize, ev-ery learning algorithm must be biased. Experience indicates that when this bias comes from the domain knowledge, the algorithm has a chance to generalize better. A typical domain knowledge encountered in applications of machine learning are the statements about orders and about rela-tionships among these orders in data. Indeed, although experts may fail in providing quantitative relationships between attributes of a system under study, they can usually describe their qualita-tive characteristics in terms of orders of value sets of these attributes, and in terms of monotone dependencies between these orders (e.g., “the higher, the better”), which happen to be the relations that are the easiest to express. In the problem of ordinal classiﬁcation (also referred to as ordinal regression), the purpose is to predict for a given object one of the ordered class labels. In this thesis, we require a stronger assumption: a meaningful ordering exists not only between class labels, but also on the value set (domain) of each attribute. Moreover, we assume that we are given a domain knowledge about the problem expressed by the monotonicity constraints: a higher value of an object on an attribute, with other values being ﬁxed, should not decrease its class assignment. Thus, we focus on the problem of ordinal classiﬁcation with monotonicity constraints. The monotonicity constraints are commonly encountered in real-life applications, but rarely taken into account in the theoretical considerations in machine learning. Neglecting the constraints may lead, however, to worse accuracy of the classiﬁers and to inconsistencies in the model. On the other hand, taking the monotonicity constraints into account requires a dedicated, speciﬁc approach. Such an approach, along with in-depth analysis of the problem, is proposed in this thesis. The success of the data analysis was possible due to a continuous improvement of computing resources, but also due to the invention of eﬃcient and accurate algorithms. Those algorithms mainly come from two diﬀerent research ﬁelds – statistics and artiﬁcial intelligence, subﬁeld of computer science. Although these two ﬁelds have very diﬀerent origins, it soon have become clear that they heavily overlap. Statistics has a great impact on the development of artiﬁcial intelligence, but also artiﬁcial intelligence made an inﬂuence on the way of thinking in statistical community. Having in mind the above, we aim in this thesis at providing a statistical framework for ordinal classiﬁcation with monotonicity constraints. Up to our knowledge, such a framework has not been introduced before. We explain existing nonparametric approaches within this framework and propose a generalization of these approaches. We also explain the Dominance-based Rough Set Approach (DRSA) from statistical point of view. DRSA has its roots in logic and was the ﬁrst theory of ordinal classiﬁcation problems with monotonicity constraints. We provide a natural probabilistic extension of this theory, Stochastic DRSA, and show its connections with statistical estimation and Bayesian decision theory. Apart from theoretical foundations of the considered problem, we also propose eﬃcient methods for solving it. The methods are based on decision rules, combined into an ensemble of classiﬁers. This corresponds to a modern approach to rule induction, which employs boosting strategy to vi learning. We provide a theoretical analysis which shows that the monotonicity assumptions allow us to bound the diﬀerence between the performance of a rule ensemble and the performance of an optimal classiﬁer in terms of the empirically measurable value of the so called margin. High performance and good scalability of the proposed rule ensemble methods are veriﬁed in an extensive computational experiment. Acknowledgments. Numerous people contributed to my research in diﬀerent ways. I would like to deeply thank to my supervisor Professor Roman Słowiński for his tremendous support, invaluable help, inspiration and insightful remarks, which made this work deﬁnitely much better. I also express my gratitude to Krzysztof Dembczyński and Salvatore Greco for a very friendly and fruitful cooperation and inspiring ideas during countless discussions. Finally, I wish to thank to my parents, my brothers and Ania for their love, warm support and endless patience. Without them, I could not have brought this work to an end. Chapter 1 Introduction 1.1 Problem Setting Classiﬁcation. Machine learning originated in the ﬁeld of artiﬁcial intelligence but also has deep connections with probability theory and statistics (Duda et al., 2000; Friedman et al., 2003; Vapnik, 1999; Koronacki and Ćwik, 2005). It concerns designing algorithms which have ability to learn. The learning is done by providing the algorithm a set of previously observed training examples (also called objects or observations), described by a set of attributes. Here we deal with a classiﬁcation problem, in which for all the training objects we know a priori the values of a particular attribute with ﬁnite domain, called class (or decision) attribute. The goal of the learning process is to predict the value of the class attribute (class label) on unseen objects as accurate as possible, using the known values of other attributes. Domain knowledge. A very important issue in the learning process is the utilization of the domain (expert) knowledge. In order to generalize, every learning algorithm must be somehow biased (prefer some hypothesis over the others although their similar behavior on the data sample), and when it is biased by the domain knowledge it has a better chance to be biased in the right way (Wolpert, 1996). This results in the improvement of prediction accuracy, but it is not the only advantage: the model consistent with domain knowledge is easier to interpret and easier to be accepted by the domain experts. Notice that although interpretability issues cannot be expressed in a quantitative way, they often play much more important role in real-life applications than a small gain in accuracy. A typical knowledge encountered in real-life applications of machine learning is the knowledge about order and monotonicity. Exploiting this kind of knowledge in the learning process is the main purpose of this thesis. Ordinal classiﬁcation. An ordinal classiﬁcation problem consists in assignment of objects to classes, for which a meaningful order exists. A good example is the classiﬁcation of documents into three groups “irrelevant”, “partly relevant”, “relevant” (Herbrich et al., 1999). As another example, consider classifying patients receiving the chemotherapy with respect to the severity of nausea into the classes “none”, “mild”, “moderate” and “severe” (Anderson, 1984). The distances between class labels are usually meaningless, since the scale is assumed to be purely ordinal. The ordinal classiﬁcation problem shares some characteristics of multi-class classiﬁcation and regression. However, on the contrary to the classiﬁcation, the order between class labels cannot be neglected, and on the contrary to the regression, the scale on the output attribute is not cardinal. 2 1.2. Problem Statement Monotonicity constraints. In the ordinal classiﬁcation with monotonicity constraints we re-quire a meaningful ordering not only between class labels, but also on the value set of each attribute. Moreover, we assume that monotonicity constraints are present in the data: a higher value of an object on an attribute, with values on other attributes being ﬁxed, should not decrease its class assignment; in other words, the expected class label should not decrease with increasing values on condition attributes. As an example, consider the customer satisfaction analysis (Greco et al., 2006), which aims at determining customer preferences in order to optimize decisions about strategies for launching new products, or about improving the image of existing products. The monotonicity constraints are of fundamental importance here. Indeed, consider two customers, A and B, and suppose that the evaluations of a product by customer A on a set of attributes are better than the evaluations by customer B. In this case, it is reasonable to expect that also the comprehensive evaluation of this product (its class label) by customer A is better (or at least not worse) than the comprehensive evaluation made by customer B. As another example, consider the problem of house pricing, i.e. classiﬁcation of houses with respect to their prices, into one of the following classes: “cheap”, “moderate”, “expensive”, “very expensive”. The classiﬁcation is based on the following attributes: lot size, number of bedrooms, bathrooms, garages, whether house contains air conditioning, basement, etc. (Koop, 2000). It is apparent that the price of house A should not be less than that of house B if, for instance, house A has greater number of bedrooms and bathrooms than B, and opposite to B, has basement, and is as good as B on the other attributes. Problems of ordinal classiﬁcation in the presence of monotonicity constraints are commonly encountered in real-life applications, where ordinal and monotone properties follow from the do-main knowledge about the problem. They are encountered in such problems as bankruptcy risk prediction (Słowiński and Zopounidis, 1995; Greco et al., 1998; Ryu and Yue, 2005), breast cancer diagnosis (Ryu et al., 2007), house pricing (Potharst and Feelders, 2002), Internet content ﬁltering (Jacob et al., 2007), credit rating (Doumpos and Pasiouras, 2005), liver disorder diagnosis (Sill and Abu-Mostafa, 1997), credit approval (Feelders and Pardoel, 2003), surveys data (Cao-Van, 2003) and many others. The problem of ordinal classiﬁcation with monotonicity constraints is widely con-sidered under the name multicriteria sorting within multicriteria decision analysis (MCDA) (Roy, 1996; Grabisch, 1996; Greco et al., 2001a, 2002a; Słowiński et al., 2005; Greco et al., 2008). Thus, we should add to the above list numerous applications of multicriteria sorting methods to real-life decision problems. Nevertheless, ordinal classiﬁcation with monotonicity constraints is rarely considered in the machine learning community. There are only few methods which take the monotone nature of data into account. What is even worse, no comprehensive theoretical approach has been established. This thesis aims at creating such approach from the statistical point of view. 1.2 Problem Statement In this section, we overview the statistical theory of machine learning and introduce the concept of ordinal classiﬁcation and dominance relation. This will serve as a basis for the deﬁnition of ordinal classiﬁcation with monotonicity constraints in Chapter 2. 1.2.1 Statistical Theory for Machine Learning Classiﬁcation problem. In the classiﬁcation problem (Friedman et al., 2003), the aim is to predict the unknown class label y ∈Y for a given object (assign object to a class), where Y = Introduction 3 {1, . . . , K} is the set of K class labels. This is usually done by using the available knowledge about the object, expressed in terms of a vector of attributes x = (x1, . . . , xm) ∈X, where X is called attribute space. Here, we assume without loss of generality that the value set of each attribute is a subset of R, so that X ⊆Rm. It is assumed that objects are generated independently according to some ﬁxed but unknown probability distribution P(x, y). Classiﬁcation is performed by constructing a function h(x) (called classiﬁer) which predicts accurately the value of y. The accuracy of the single prediction is measured in terms of the loss function L(y, h(x)), which reﬂects the cost of predicting the class label h(x) when the actual (observed) value is y. The overall accuracy of the classiﬁer h(x) is measured by the expected loss (risk) according the data distribution P(x, y): R(h) = E[L(y, h(x))]. (1.1) Bayes classiﬁer. Thus, the optimal prediction function is the function minimizing the risk (1.1): h∗= arg min h R(h), (1.2) which is called Bayes classiﬁer (Berger, 1993), and R∗:= R(h∗) is called Bayes risk. From the deﬁnition it follows that Bayes risk is the minimal possible risk achievable by any prediction function in a given problem. It follows from statistical decision theory (Berger, 1993) that Bayes classiﬁer is obtained by minimizing the risk according to so called posterior distribution P(y\|x). The derivation of this fact is the following. First, decompose P(x, y) as P(y\|x)P(x). Then: R(h) = Z L(y, h(x))dP(x, y) = Z Z L(y, h(x))dP(y\|x) dP(x) (1.3) which is a decomposition of the form E[·] = E [E[·\|x]]. From (1.3) it follows that the Bayes classiﬁer must minimize the term in parenthesis for any x: h∗(x) = arg min z Z L(y, z)dP(y\|x) = arg min z E[L(y, z)\|x] (1.4) Empirical risk minimization. Since the probability distribution P(x, y) is unknown, we cannot obtain h∗. The learning procedure uses a training set D = {(x1, y1), . . . , (xn, yn)} ⊂X only to construct h to be a good approximation of h∗. Usually, it is performed by minimization of the empirical risk: Remp(h) = 1 n n X i=1 L(yi, h(xi)), (1.5) where function h is chosen from a restricted, predeﬁned class of functions H (linear functions, classiﬁcation trees, etc.), for which we believe that h∗∈H or at least that h∗may be approximated by some function from H. This procedure is usually called empirical risk minimization (ERM) (Vapnik, 1999). Loss functions In the most general case, the loss function takes the matrix form [lyk]K×K, where lyk = L(y, k). An obvious assumption is that lkk = 0 for every k = 1, . . . , K (the correct prediction is not penalized), and lyk > 0 for each y ̸= k (any incorrect prediction is penalized). The most popular loss function for classiﬁcation is zero-one loss, deﬁned as: L0−1(y, h(x)) = ( 0 if y = h(x) 1 if y ̸= h(x) , (1.6) 4 1.2. Problem Statement R1 R2 R3 Figure 1.1: An example of Bayes decision boundary between three classes on R2 so that the misclassiﬁcation is always penalized with unit loss, and for correct classiﬁcation no penalty is imposed. This corresponds to the matrix with zeros on the diagonal and ones anywhere else, lyk = 1y̸=k1. The Bayes classiﬁer has the form: h∗ 0−1(x) = arg maxy∈{1,...,m}P(y\|x) (1.7) so it indicates the most probable class at x. The Bayes classiﬁer divides the attribute space X into K separate regions R1, . . . , RK, such that h∗(x) = k if and only if x ∈Rk. The boundary between those regions is called Bayes decision boundary. It consists of points, for which there are more than one class with the highest probability (mode of the distribution is not unique). A simple example is shown in Figure 1.1, with X = R2. Although in each region there might be nonzero probability for any class, in the region Rk class k is always the most probable. 1.2.2 Ordinal Classiﬁcation and Monotonicity Constraints Ordinal loss matrix. In the ordinal classiﬁcation setting, the loss matrix should be “consistent” with the order between class labels in the sense that the loss should not decrease, as the predicted value “moves away” from the true value. Thus, for any loss matrix [lyk]K×K we assume that each row of the loss matrix is V-shaped (Lin and Li, 2007): ly,k−1 ⩾lyk if k ⩽y, lyk ⩽ly,k+1 if k ⩾y, (1.8) for each y, k = 1, . . . , K. Such matrix will be called ordinal loss matrix. It is straightforward to see that 0-1 loss (1.6) is a proper ordinal loss matrix, however it does not take order into account, because every misclassiﬁcation is given the same penalty. This is diﬀerent for other two popular ordinal loss matrices: absolute and squared error loss. Absolute error loss is deﬁned as: Labs(y, h(x)) = \|y −h(x)\|, (1.9) 11c is the indicator function, i.e. 1c = 1 if c is true, otherwise 1c = 0 Introduction 5 so that the penalty for misclassiﬁcation grows linearly with the diﬀerence between real and predicted class labels. One can show (Berger, 1993; Friedman et al., 2003) that the Bayes classiﬁer for absolute error loss has the form: h∗ abs(x) = med(y\|x), (1.10) the median of the conditional distribution P(y\|x). Notice that the median is a purely ordinal operation: if class labels were encoded by arbitrary real numbers, the Bayes classiﬁer would remain the same. This is not the case of another ordinal loss matrix, squared error loss: Lsqr(y, h(x)) = (y −h(x))2, (1.11) The Bayes classiﬁer is then (Berger, 1993; Friedman et al., 2003): h∗ sqr(x) = E[y\|x], (1.12) the expected value over the conditional distribution P(y\|x). Taking the expected value is not an ordinal operation, because it depends on the particular encoding of class labels and usually does not belong to {1, . . . , K}. Thus, we state that absolute loss is the canonical loss for ordinal classiﬁcation. There is another approach for incorporating the order between classes, in which the ordinal classiﬁcation is treated as a special case of a ranking problem. It is described in the Section 1.2.3. Dominance relation and monotone functions. A dominance relation ⪰is a binary relation on X, deﬁned in the following way: for any x, x′ ∈X we say that x dominates x′, x ⪰x′, if on every attribute, x has evaluation not worse than x′, xj ⩾x′ j, for all j = 1, . . . , m: x ⪰x′ ⇐ ⇒∀j=1,...,m xj ⩾x′ j. The dominance relation ⪰is a partial preorder on X, i.e. it is reﬂexive and transitive. A function h: X →Y is called monotone if the following condition holds for any x, x′ ∈X: x ⪰x′ →h(x) ⩾h(x′). The vector v = (v1, . . . , vn) is called monotone if for every i, j = 1, . . . , n: xi ⪰xj →vi ⩾vj. The diﬀerence between monotone functions and vectors is that the latter corresponds to the training set D only, while the former – to the whole space X. Monotonicity constraints. The concept of monotone function is the core of what we intuitively understand by monotonicity constraints. Therefore, the problem of ordinal classiﬁcation with mono-tonicity constraints is often referred to as the problem of classiﬁcation with monotone functions, i.e. the problem of ﬁnding the accurate classiﬁer within the class of monotone functions. In order to justify imposing monotonicity constraints, one usually assumes that the process generating the data has monotone nature and the aim is then to “discover” (approximate) the process using the available training data D. Such deﬁnition is stated, more or less explicitly, in most of the papers dealing with ordinal classiﬁcation with monotonicity constraints (Ben-David, 1995; Greco et al., 1998, 1999a,b, 2001a; Potharst and Feelders, 2002; Cao-Van, 2003; Popova, 2004; Chandrasekaran et al., 2005; Velikova, 2006). In our probabilistic setting, this corresponds to the statement that the Bayes classiﬁer h∗is a monotone function. An example of decision boundaries with monotone Bayes classiﬁer is shown in Figure 1.2. The problem with such a formulation is that the Bayes classiﬁer is not a primitive concept in classiﬁcation problems, it rather follows from the form of the loss function and from the data distribution. Therefore, we postpone the formal deﬁnition to the Chapter 2, in which we introduce a probabilistic model for monotonicity constraints. 6 1.2. Problem Statement R1 R2 R3 Figure 1.2: An example of monotone Bayes decision boundary between three classes on R2. Compare with non-monotone case in Figure 1.1 Inconsistencies. The monotonicity constraints imply that it should hold: xi ⪰xj = ⇒yi ⩾yj, (1.13) for i, j = 1, . . . , n, i.e. vector y = (y1, . . . , yn) should be monotone. If such property holds, we say that the dataset D is consistent (or monotone). However, for many real datasets, (1.13) is not satisﬁed. We say that an object xi is inconsistent if there exists another object xj, such that xi, xj violate (1.13). In other words, there is xj such that xi dominates xj but yi < yj, or such that xi is dominated by xj but yi > yj. We say that object xi is consistent if it is not inconsistent. Inconsistencies are usually avoided, because none monotone function can approximate accurately inconsistent objects. Some methods for ordinal classiﬁcation with monotonicity constraints work only with consistent data. Others try to remove inconsistencies and operate only on the consistent part of the data. 1.2.3 Rank Loss Function Ranking and scoring. Ranking is deﬁned on a pair of objects: we consider a ranking function r(x, x′) such that r(x, x′) > 0 if x is ranked higher than x′. The ranking loss (Cl´ emen¸ con et al., 2006) is deﬁned as: Lrank(r(x, x′), y −y′) = 1r(x,x′)(y−y′)<0, (1.14) where y, y′ are ranks (class labels). Thus, ranking function is penalized with a unit loss if it makes a ranking error: x is ranked lower (higher) than x′, while y > y′ (y < y′). The aim is to ﬁnd a prediction function minimizing the risk: Rrank(r) = Z Lrank(r(x, x′), y −y′)dP(x, y)dP(x′, y′). (1.15) The Bayes classiﬁer has the simple form: r∗(x, x′) = sign P(y > y′\|x, x′) −P(y′ > y\|x, x′) , (1.16) Introduction 7 In the core of rank loss approach lies the assumption of the existence of the optimal scoring function (Cl´ emen¸ con et al., 2006), i.e. single-argument function s∗(x) such that: r∗(x, x′) > 0 ⇐ ⇒s∗(x) > s∗(x′), (1.17) and then the rank loss takes the form: Lrank(s(x) −s(x′), y −y′) = 1(s(x)−s(x′))(y−y′)<0. (1.18) The assumption (1.17) must be satisﬁed, otherwise there would be no function of a single argument s∗(x) achieving Bayes risk. Herbrich et al. (1999) considered the problem of ordinal classiﬁcation as ﬁnding the scoring function (called “function inducing ordering” in their work) which minimizes the risk (1.15) with the SVM classiﬁcation method. Freund et al. (2003) proposed the extension of AdaBoost in the rank loss formulation (RankBoost). Analysis. The rank loss approach has nonparametric form and (seemingly) avoids imposing any metric on the ordinal scale of ranks, as the error is measured by a number of rank inverses. However, there are two drawbacks. The ﬁrst one is a computational issue: the complexity of the problem grows quadratically with n, making the method computationally infeasible for larger datasets. The second drawback is more fundamental. The scoring function is real-valued, so the output of the ranking procedure is the real number, rather than class label. Therefore, one must somehow estimate the position of the thresholds on the scale of scoring function to change the real values into the class labels. In (Herbrich et al., 1999) the positions of the thresholds were obtained by a margin-based procedure. Generally, the threshold can be obtained only by minimizing (implicitly or explicitly) some loss function. Thus, at the end we anyway do the thing which we tried to avoid: impose some loss function on the ranks’ scale. We believe the ordinal classiﬁcation is diﬀerent to ranking, and our belief is supported by achievements in multicriteria decision analysis (MCDA) (Roy, 1996; Greco et al., 2002a): in the ordinal classiﬁcation (multicriteria sorting in this context) one does not compare objects with each other (as in ranking), one rather assigns them to ordered classes on the basis of their values on considered attributes. The assignment is made by comparing objects with the “class proﬁles”, which are artiﬁcial objects associated with each class: if the object is better than the k-th class proﬁle, its class must be at least k. This approach can be translated to a modiﬁcation of the ranking problem by stating that each object is compared only with the “rank of the class” and considering the “ranking” function r(x, k). Then, the class label predicted for x is the smallest k for which x is still ranked lower: h(x) = min{k: sign(r(x, k)) < 0} (h(x) = K is no such k exists). But if for a given x, we sum the rank loss for each class label, we obtain: K X k=1 Lrank(r(x, k), y −k) = Labs(h(x), y), the absolute error loss. Thus, we conjecture that the absolute error is the canonical loss function for ordinal classiﬁcation. 1.3 Existing Approaches to Ordinal Classiﬁcation with Monotonicity Constraints We overview the existing approaches to ordinal classiﬁcation with monotonicity constraints in the ﬁelds of rough sets, multicriteria decision analysis (MCDA), statistics and machine learn-ing/knowledge discovery. In description of fuzzy integrals, Ordinal Learning Model and monotone decision trees we base on the survey of monotone classiﬁcation methods in (Cao-Van, 2003). 8 1.3. Existing Approaches to Ordinal Classiﬁcation with Monotonicity Constraints object a1 a2 y object a1 a2 y x1 0.15 0.3 1 x4 0.6 0.5 3 x2 0.3 0.15 1 x5 0.7 0.7 2 x3 0.4 0.8 2 x6 0.9 0.8 3 Table 1.1: A set of objects described by 2 attributes a1, a2 and class label y ∈{1, 2, 3}. 1.3.1 Dominance-based Rough Set Approach Dominance-based Rough Set Approach (DRSA) proposed by Greco, Matarazzo, and Słowiński (1999a,b, 2001a) is an extension of the classical rough set approach (Pawlak, 1982) for dealing with multicriteria sorting. By substituting the indiscernibility relation with the dominance relation, it handles inconsistencies coming from violation of the monotonicity constraints. DRSA originated from the fusion of rough set theory and MCDA. It was the ﬁrst approach, which proposed the com-prehensive theory for ordinal classiﬁcation problems with monotonicity constraints in the domain of knowledge discovery. In this section, we only sketch the idea of DRSA, while a deeper insight is postponed to Chapter 4. Extensive surveys of DRSA can be found in (Greco et al., 2001a, 2004b,c; Słowiński et al., 2005). Class unions and approximations. DRSA divides the monotone K-class problem into K −1 binary subproblems. This is done by considering the upward and downward unions of classes which are subsets of data for which class label is at least (or at most) k = 1, . . . , K: Cl⩾ k = {xi : yi ⩾k, i = 1, . . . , n} Cl⩽ k = {xi : yi ⩽k, i = 1, . . . , n}. Then, the k-th problem, k = 2, . . . , K is deﬁned as the problem of discrimination between objects from class union Cl⩾ k and Cl⩽ k−1, i.e. the problem of binary classiﬁcation with two classes Cl⩾ k and Cl⩽ k−1. The core idea of DRSA (and other rough set approaches) is the following: instead of using the whole class unions, which might contain inconsistent objects, use subsets of unions, containing only consistent objects for a given binary problem. Those subsets are called lower approximations, and corresponds to the region of certain knowledge2. Let us deﬁne ﬁrst the dominating and dominated sets for a given x as the subsets of objects which dominate (or are being dominated by) x: D+(x) = {xi : xi ⪰x, i = 1, . . . , n} D−(x) = {xi : xi ⪯x, i = 1, . . . , n}. We remind (see Section 1.2.2) that object xi is inconsistent if there exists another object xj which belongs to a lower class but xj ⪰xi, or which belongs to a higher class but xj ⪯xi. However, in the k-th binary problem we associate objects from Cl⩾ k with one (“positive”) class, while objects from Cl⩽ k−1 with another (“negative”) class. Thus, in the k-th problem object xi ∈Cl⩾ k is inconsistent if there exists xj ∈Cl⩽ k−1 such that xj ⪰xi (and vice-versa). But this is equivalent to D+(xi) ⊈Cl⩾ k and D−(xj) ⊈Cl⩽ k−1. Having above in mind, we are ready to deﬁne lower approximations of upward and downward 2Regions of possible knowledge, upper approximations, are introduced in Chapter 4. Introduction 9 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 Figure 1.3: Example of three-class problem described in Table 1.1. Black points are objects from class 1, dark gray points – from class 2, light gray – from class 3. unions of classes as: Cl⩾ k = {xi : D+(xi) ⊆Cl⩾ k , i = 1, . . . , n}, Cl⩽ k = {xi : D−(xi) ⊆Cl⩽ k , i = 1, . . . , n}. In other words, lower approximations of class unions contain consistent objects, thus can be regarded as certain regions of X for a given binary problem. Example. A simple example of training set is shown in Table 1.1 and in Figure 1.3. Notice that object x4 is inconsistent with x5. We have Cl⩾ 2 = {x3, x4, x5, x6}, Cl⩾ 3 = {x6}, Cl⩽ 1 = {x1, x2}, Cl⩽ 2 = {x1, x2, x3}. Observe that Cl⩾ 1 and Cl⩽ 3 are always trivial (contain all objects). Rule induction (DOMLEM). Having obtained the lower approximations of class unions, we use them to learn a classiﬁer to distinguish between Cl⩾ k and Cl⩽ k−1, for each k = 2, . . . , K. The classiﬁer has the form of the set of decision rules, which generalize description of the information contained in the dataset D and serves as a basis for further classiﬁcation of unseen objects. By decision rule we mean a simple expression of the form “if condition, then decision”. In DRSA, only ordinal types of rules are considered, of the following form: • if xj1 ⩾sj1 and . . . and xjq ⩾sjq, then y ⩾k, • if xj1 ⩽sj1 and . . . and xjq ⩽sjq, then y ⩽k, • if xj1 ⩾sj1 and . . . and xjq ⩾sjq and xj1 ⩽sj1 and . . . and xjq ⩽sjq, then y ∈{k′, k′ + 1, . . . , k −1} where j1, . . . , jp are indices of attributes which appear in the condition part, and sj1, . . . , sjp are some values from the domains of respective attributes. The ﬁrst two types of rules are called certain while the rules of the third type are called approximate3. Certain rules are induced from lower approxi-mations of the appropriate class unions (e.g. rules with decision part y ⩾k are induced from Cl⩾ k ). Approximate rules are induced from the sets of the form: Bk′,k = {x1, . . . , xn}\ Cl⩽ k′−1 ∪Cl⩾ k , k ⩾k′, the boundaries between certain regions. 3There are in fact ﬁve types of rules, since ﬁrst two types can also be induced from upper approximations and are called possible then. 10 1.3. Existing Approaches to Ordinal Classiﬁcation with Monotonicity Constraints Algorithm 1.1: DOMLEM input : Family of sets L = n Cl⩾ k , Cl⩽ k−1, Bk′,k : 2 ⩽k′ ⩽k ⩽K o . output: Set R of decision rules. R := ∅; for each L ∈L do RL := ∅; (rule set covering L) while L ̸= ∅do Start with the rule with empty condition part, Φ = ∅; while Φ covers objects outside L do Evaluate each φ / ∈Φ by \|cov(Φ∪φ)∩L\| \|L\| , where cov(Φ ∪φ) is the set of objects covered by Φ ∪φ; Choose a condition φ / ∈Φ with the highest evaluation; Φ := Φ ∪φ; end Add rule with condition part Φ to RL; Remove from L objects covered by Φ; end for each rule Φ in RL do if Φ is minimal then R := R ∪Φ; end end end return R; There have been several algorithms proposed to induce decision rules within DRSA (Greco et al., 2001c; Pindur and Susmaga, 2003; Stefanowski and Żurawski, 2003; Błaszczyński and Słowiński, 2003; Dembczyński et al., 2003; Pindur et al., 2004). The most popular one, DOMLEM (Greco et al., 2001c), is based on the sequential covering procedure. It tends to generate the so called minimal set of rules (i.e. the non-redundant set of rules covering all objects) with the smallest number of rules. DOMLEM consists of a covering procedure run for each lower approximation and boundary in each subproblem. In the procedure, a set of rules is induced until it covers all objects from the respective approximation (or boundary). Rules are induced one by one, and in each rule the elementary conditions are added one by one (ordered by a speciﬁc evaluation function), until the rule covers only examples from the approximation. The scheme of DOMLEM is shown as Algorithm 1.1. When classifying unseen object, each rule is tested whether it covers the object. Then, some procedure is used to combine the results of all covering rules and classify the object to a class (see e.g. Błaszczyński et al. (2007)). 1.3.2 Utility Functions in MCDA Since the multicriteria sorting, considered within MCDA, coincides with ordinal classiﬁcation with monotonicity constraints, potentially any method dealing with sorting can be regarded as a method for solving ordinal classiﬁcation. However, many methods from MCDA do not suit to the approach used in this thesis, since: • they construct the preference model using diﬀerent forms of prior information (weights of Introduction 11 fj(xj) xj Figure 1.4: An example of piecewise linear function on attribute j with 3 break-points (and 2 boundary points). criteria, preference thresholds, etc.) than the set of training examples, or • they are interactive and demand the presence of decision maker during the learning process, or • they are designed for smaller problems and scale badly with the problem size (both with n and m). In this section we brieﬂy describe methods based on the utility function models, such as the additive functions, Choquet or Sugeno integrals. UTA methods. Originally, the UTA (UTilit´ es Additives) method was ﬁrst proposed by Jacquet-Lagreze and Siskos (1982) for dealing with ranking problem. The UTADIS method (Jacquet-Lagr eze, 1995; Zopounidis and Doumpos, 1997; Siskos et al., 2005) is a variant of UTA for solving the multicriteria sorting. The values of the object on each attribute (criterion) are aggregated into an additive utility function: f(x) = m X j=1 fj(xj), where fj(xj) are marginal utility functions. They are assumed to be piecewise linear consisting of the ﬁxed number of break-points (see Figure 1.4). There are also K + 1 thresholds = −∞= θ0 < θ1 < . . . < θK−1 < θK = ∞and it is assumed that if x is classiﬁed to the class k, i.e. h(x) = k, if θk−1 < f(x) ⩽θk. The construction of all of the marginal utility functions and thresholds is performed by solving a single linear programming problem. For the purpose of robustness in the approach to ranking problems, Greco, Mousseau, and Słowiński (2008) proposed UTAGMS method, which ﬁxes a break-point for every value taken by any of the objects on each attribute. They showed that using such arepresentation, one can model every additive monotone function compatible with training examples. Dembczyński, Kotłowski, and Słowiński (2006b) extended this idea to classiﬁcation and combine the approach with DRSA. They also introduced a speciﬁc penalty term which made the method similar to support vector machines (Vapnik, 1999). Choquet and Sugeno integrals methods. Choquet integral (Choquet, 1953) and its ordinal counterpart, Sugeno integral (Sugeno, 1974), are much more powerful aggregation operators than additive value functions, because they can model interaction between the attributes (Grabisch, 1996). However, they have 2m −2 parameters in general, therefore one usually restricts only to the interaction of the j-th order (for j = 1 we have an additive model). The number of parameters to estimate makes them rather impractical: for small datasets they are overparametrized and thus 12 1.3. Existing Approaches to Ordinal Classiﬁcation with Monotonicity Constraints tend to overﬁt, while for larger datasets the parameters’ estimation becomes computationally very expensive. Choquet integral has been applied in (Verkeyn et al., 2002) to survey data. It constitutes a main part of the sorting procedure implementation TOMASO (Tool for Ordinal Multi-Attribute Sorting and Ordering) (Marichal et al., 2005). 1.3.3 Isotonic Regression and Monotone Approximation The statistical estimation and hypothesis testing in the presence of monotonicity constraints dates back to early 1950s and has been considered much before any other approach mentioned in this survey. The important contribution of this ﬁeld is the algorithm of isotonic regression4 (Brunk, 1955; Ayer et al., 1955). Isotonic regression. Let ⪰be a preorder relation, i.e. reﬂexive and transitive, on X (dominance relation in our case) and let {x1, . . . , xn} be a set of points in X. Let y = (y1, . . . , yn) be a real-valued vector. A vector p∗= (p∗ 1, . . . , p∗ n) is an isotonic regression of y with weight vector w = (w1, . . . , wn) if and only if p∗is the solution of the following optimization problem: minimize n X i=1 wi(yi −pi)2 subject to xi ⪰xj = ⇒pi ⩾pj ∀1 ⩽i, j ⩽n, (1.19) so that p∗minimizes the squared error in the space of all monotone vectors p = (p1, . . . , pn). Isotonic regression is a solution to many order restricted statistical inference problems (Robertson et al., 1998). When ⪰is a simple order (e.g. order between real numbers), an eﬃcient (with complexity O(n log n)) algorithm for solving (1.19) exists (Ayer et al., 1955), PAV (Pool Adjacent Violators). In a general case of any preorder, the problem can be solved in O(n4) (Maxwell and Muchstadt, 1985), which is impractical for larger datasets. However, several heuristics exist, in particular an eﬀective O(n2) algorithm, which very often achieves the results close to optimal, was introduced by Burdakov et al. (2006). Monotone approximation. Let us consider classiﬁcation case with y ∈{1, . . . , K}. We state a problem similar to isotonic regression, but based on the minimization of arbitrary loss matrix: minimize n X i=1 L(yi, di) subject to xi ⪰xj = ⇒di ⩾dj ∀1 ⩽i, j ⩽n di ∈{1, . . . , K} ∀1 ⩽i ⩽n. (1.20) In other words, we would like to minimize a loss matrix within the class of all monotone vectors taking integer values. This is a nonparametric approach to ordinal classiﬁcation, since we do not impose any additional condition, apart from the monotonicity. The problem has the following interpretation: relabel the objects to make the dataset monotone, such that new class labels are as close as possible to the original class labels, where the closeness is measured in terms of the loss function. The new labels in (1.20) are the values of variables di. The set of new labels will be called monotone approximation and we will refer to the problem (1.20) as monotone approximation problem. 4Sometimes monotone regression term is used; the word isotonic means monotone and increasing. Introduction 13 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 Figure 1.5: Example of three-class problem. Dykstra et al. (1999) solved (1.20) for two particular loss functions: absolute error loss L(y, d) = \|y −d\| and squared error loss L(y, d) = (y −d)2. It was shown that the latter problem can be solved by isotonic regression, while the former can be solved by a sequence of K −1 isotonic regression problems. The problem for K = 2 and 0-1 loss appeared in the logical analysis of data, as a problem of ﬁnding monotone Boolean function approximating the data (Boros et al., 1995). It was solved in O(n3) by transformation to the maximum network ﬂow problem, which is an improvement over (Dykstra et al., 1999). Chandrasekaran et al. (2005) considered (1.20) in the most general form, for any K and any loss function, as a part of the isotonic separation procedure (see Section 1.3.8). It was shown that it can be solved either by linear programming or by maximum network ﬂow in O(n3). We were considering the problem of monotone approximation within the context of DRSA (Dembczyński et al., 2006a, 2007a,b). The problem of monotone approximation has a drawback: in most cases the optimal solution is not unique. Thus, the particular solution found by the algorithms mentioned above is accidental (e.g. depends on the parameters of the solver used). We show in Chapter 3 how to cope with this issue. Example. Consider the dataset from Table 1.1, shown in Figure 1.3. Assume the absolute error loss function. Then, there are two optimal solutions: either object x4 will be reassigned to class 2 or object x5 will be reassigned to class 3. Consider one more example presented in Figure 1.5. This example is similar to the previous one, with exception that now y4 = 1 and y1 = 3. There is a unique monotone approximation problem with absolute error loss, which reassigns only a single object x1 to class 1. 1.3.4 Ordinal Learning Model The Ordinal Learning Model (OLM) (BenDavid et al., 1989; Ben-David, 1992) was the ﬁrst method for ordinal classiﬁcation with monotonicity constraints proposed in the machine learning community. It is very simple and consists in choosing the subset D′ ⊆D of training objects (so called “rule base”). Then, classiﬁcation of new objects is done simply by a function of the form: fOLM(x) = max{yi : xi ∈D′, xi ⪯x} (1.21) 14 1.3. Existing Approaches to Ordinal Classiﬁcation with Monotonicity Constraints If the set over which the maximum is chosen is empty (i.e. there is no object from D′ which is dominated by x), then a class label is assigned by a nearest neighbor procedure using the Euclidean distance. The rule base D′ is chosen to be consistent and not to contain redundant objects. An object xi is redundant in D′ if there exists another object xj such that xi ⪰xj and yi = yj, i.e. xi does not have any inﬂuence fOLM(x). Set D′ is constructed in the following way. First, all objects with the same attribute vector are replaced by one vector. The class label of new vector is the average of the class labels of replaced vectors. Then, we start with D′ empty and objects are added to D′ in the order of decreasing class labels. The particular order of objects within a given class is random. An object is added to D′ only if it is consistent with objects already present in D′; otherwise it is rejected. Moreover, redundancy is checked each time: if the new object is redundant, then it is rejected; if the new object makes other objects in D′ redundant, then those objects are rejected. The process is repeated until all the objects are examined. The algorithm has several drawbacks. Firstly, the procedure of building D′ relies on the random order of processing the objects, therefore can return diﬀerent D′ for each run. Moreover, the nearest neighbor rule is not an ordinal and can produce non-monotone results. The averaging of labels is also not ordinal operation. There are procedures for building the consistent subset which are deterministic and much better formally grounded, such as DRSA or monotone approximations, described in previous sections. 1.3.5 Monotone Neural Networks Neural networks have received a great attention and popularity in the machine learning com-munity during the late 1980s and the 1990s. Therefore, it is not surprising that early approaches to ordinal classiﬁcation with monotonicity constraints were also focused on neural network models. Utilizing the domain knowledge about the monotonicity was done in one of two ways: • by adding a second term measuring the “monotonicity error” (the extent to which the model violates monotonicity constraints) to the typical error measure (Sill and Abu-Mostafa, 1997), • by enforcing some constraints on the weights and architecture of the network, which makes the network monotone by construction (Wang, 1994; Wang and Archer, 1994; Sill, 1998; Daniels, 1999). Neural network models have been successfully applied to e.g. bond rating, house pricing (Daniels, 1999) or liver disorders diagnosis (Sill and Abu-Mostafa, 1997). However, their main drawback is the lack of interpretability, which impede the explanation of the model behavior. Moreover, neural networks work only with the cardinal scale on the attributes, so that they may lead to wrong or meaningless results if the attribute scale is purely ordinal. 1.3.6 Monotone Trees Overview. Decision tree algorithms, such as CART (Breiman et al., 1984) and C4.5 (Quinlan, 1993), are very popular in machine learning. They were also considered in the context of ordinal classiﬁcation with monotonicity constraints. The ﬁrst algorithm for tree induction in this context was proposed by Ben-David (1995) and called MID (Monotone Induction of Decision trees). The main idea was to modify the conditional entropy, by adding a term called order-ambiguity-score, which pushed the splitting strategy towards monotonicity. However, this did not guarantee the tree to be a monotone function. Introduction 15 A diﬀerent approach was overtaken by Makino et al. (1999), which presented two algorithms: P-DT (Positive Decision Trees) and QP-DT (Quasi-Positive Decision Trees)5. The former constructs a monotone decision tree (i.e. a tree, which is a monotone function), while the latter – a so called quasi-monotone tree. Both algorithms work only with binary-class problems. They were adapted to handle K-class problems and extended in (Potharst et al., 1997; Potharst, 1999; Potharst and Bioch, 2000; Potharst and Feelders, 2002; Bioch and Popova, 2002), under the names MDT (Monotone Decision Trees) and QMDT (Quasi-Monotone Decision Trees). Decision tree models have also been considered within the DRSA (Giove et al., 2002). In this approach, a cut minimizing the training error for a given class union is chosen for splitting. Three types of trees were considered: single-class (discriminating a single class union only), progressively-ordered (using a single class union, which is, however, progressively changing as the tree is growing) and full range (using all class unions simultaneously). Cao-Van and De Baets (2003) and Cao-Van (2003) considered RT (Ranking Tree) algorithm for growing monotone decision trees. It diﬀers from MDT in using an impurity measure based on the ranking error (number of reversed ranks) and in using a speciﬁc procedure for maintaining the monotonicity of the tree. It can handle inconsistent datasets (while MDT in its original version cannot). MDT. MDT (Potharst and Bioch, 2000) worked originally with consistent (monotone) datasets only, however a modiﬁed version was then proposed to deal with inconsistent data (Bioch and Popova, 2002). It can use any of the popular impurity measures such as entropy or Gini index and the growth of the tree is done in a typical way. The main diﬀerence between MDT and ordinary tree induction method is the so called “cornering technique” for maintaining the monotonicity of a tree. It consists of adding artiﬁcial objects to each node, one in the lower left corner of the node, and another in the upper right corner (notice that each node in the tree represents a subset of X and has the form of hyperrectangle). The lower left object obtains the highest possible class label which does not violate consistency of the dataset, while the upper left object – lowest possible label, respectively. The tree is induced until every node corresponds to the objects from the same class (i.e. until each node is pure). It was shown that a tree generated in such s way is monotone. 1.3.7 Ordinal Stochastic Dominance Learner Ordinal Stochastic Dominance Learner (OSDL) (Cao-Van, 2003; Cao-Van and De Baets, 2004) is an instance-based method for ordinal classiﬁcation with monotonicity constraints. It is based on the dominance relation only. It constructs the estimate of the probability distribution function, ˆ F(x) = ˆ F(x, 1), . . . , ˆ F(x, K) ∈RK, where ˆ F(x, k) is the estimate of P(y ⩽k\|x). First, to each object xi a distribution ˆ f(xi) = ( ˆ f(xi, 1), . . . , ˆ f(xi, K)) is assigned: ˆ f(xi, k) = \|{xj : xj = xi ∧yj ⩽k, j = 1, . . . , n}\| \|{xj : xj = xi, j = 1, . . . , n}\| , i.e. by counting objects with the same values of attributes. Notice that if each object lies in a diﬀerent point in X, each distribution ˆ f(xi) consists of yi ones and K −yi zeros. Next, for each x ∈X, we deﬁne two distribution estimators ˆ Fm(x) and ˆ FM(x) as follows: ˆ Fm(x, k) = min{ ˆ f(xi, k): xi ⪯x, i = 1, . . . , n} ˆ FM(x, k) = max{ ˆ f(xi, k): xi ⪰x, i = 1, . . . , n}. 5In boolean reasoning, the term “positive” is sometimes used instead of “monotone”. 16 1.3. Existing Approaches to Ordinal Classiﬁcation with Monotonicity Constraints If the dataset is consistent then Fm(x, k) ⩾FM(x, k) for each x ∈X, k = 1, . . . , K. In order to handle the inconsistencies, one also deﬁnes: Nm(x, k) = \|{xi : xi ⪯x ∧yi > k}\| NM(x, k) = \|{xi : xi ⪰x ∧yi ⩽k}\|. The ﬁnal estimate ˆ F(x) of the probability distribution depends on whether the situation at x is consistent or not, and is deﬁned as: ˆ F(x, k) = ( (1 −s) ˆ Fm(x, k) + s ˆ FM(x, k) if Fm(x, k) ⩾FM(x, k) (1−s′)Nm(x,k) ˆ Fm(x,k)+s′NM(x,k) ˆ FM(x,k) (1−s′)Nm(x,i)+s′NM(x,k) otherwise. (1.22) where s, s′ ∈[0, 1] are the parameter of the algorithm (chosen by e.g. cross-validation). Having obtained the estimate of the probability distribution ˆ F(x), one can classify the object according to the form of the Bayes classiﬁer for a given loss function, as described in Section 1.2 (for instance, for absolute error loss (1.9) we should choose the median). In (Cao-Van, 2003) the expectation value rounded to the closest integer is suggested (so, squared error loss (1.11) is implicitly used), although it is not a purely ordinal operation. In (1.22), the upper expression is related to the consistent situation. The lower expression is more robust against the inconsistencies (it resembles the variable consistency model used in DRSA, see Chapter 4). For example, suppose there is an object xi from the highest class, yi = K, dominated by every other object, i.e. xi ⪯x for each x ∈X. Then, the upper expression in (1.22) would give ˆ Fm(x, k) = 1 for every k and for every x just because of the single troublesome object xi. The lower expression includes weighting by the number of objects Nm(x, k) and NM(x, k) which reduces the inﬂuence of single, inconsistent objects (see Cao-Van (2003) for more details). Notice that the estimate (1.22) is the extension of the “max” formula (1.21) used in OLM. Although the estimate (1.22) looks sensible, it has a drawback of being based only on the dominance relation. If there are many attributes, the dominance relation becomes sparse. Then, sets {xi : xi ⪰x} and {xi : xi ⪯x} become small, which leads to very poor estimates of probability distribution. 1.3.8 Isotonic Separation Isotonic separation (Chandrasekaran et al., 2005) is a strong, novel tool for solving the problem of ordinal classiﬁcation with monotonicity constraints. It has already been successfully applied to the ﬁrm bankruptcy prediction (Ryu and Yue, 2005), breast cancer diagnosis (Ryu et al., 2007), and Internet content ﬁltering (Jacob et al., 2007). The algorithm works with any loss function and consists of two stages. In the ﬁrst stage, the dataset is “monotonized” (inconsistencies are removed) using the monotone approximation (1.20)6. The second stage is related to the classiﬁcation of unseen objects. It is similar to the nearest-neighbors method, however it maintains monotonicity, i.e. the classiﬁer is monotone. Let (u)+ be a function which returns u if u > 0 and 0 otherwise, i.e. (u)+ = u1u>0. We deﬁne the “distance”: d(x, x′) = m X j=1 (xj −x′ j)+ Notice that if x ⪯x′ then d(x, x′) = 0. Moreover, if x ⪰x′ then d is equivalent to the familiar L1 metric. Let us denote θ(u) = 1u⩾0 and let lyk be a loss of classifying an object as k if the observed 6Of course, Chandrasekaran et al. (2005) do not use the term “monotonize” nor “monotone approximation” in their paper. Introduction 17 value is y. The classiﬁer has the following form: hIS(x) = 1 + K X k=2 θ lk,k−1 min{d(x, xi): y′ i < k, i = 1, . . . , n} −lk−1,k min{d(xi, x): y′ i ⩾k, i = 1, . . . , n} , (1.23) where y′ i are the new, consistent labels of objects (monotone approximation). First, notice that if xi ⪰x then h(x) ⩽y′ i, since then d(x, xi) = 0 and in each term of the sum with k > y′ i the argument of θ is always negative. Similarly, if xi ⪯x then h(x) ⩾y′ i. The particular class label is assigned by comparing the “distances” d to the “nearest neighbors” of x. The function (1.23) has a severe (and very strange) drawback of depending only on the values of the loss function above and below the diagonal, i.e. lk,k−1 and lk−1,k. Notice that, for 0-1 loss, absolute error loss and squared error loss, the expression (1.23) remains the same, which shows a very undesirable behavior in the context of Bayesian analysis considered in Section 1.2. Nevertheless, isotonic separation proved to be very eﬀective in real-life applications. The paper (Chandrasekaran et al., 2005) gives much broader analysis of the problem, including reduction of the problem size, speeding up the classiﬁcation procedure, separation with “doubt regions” (where the classiﬁer can abstain from the answer or indicate more than one class) and continuous outcome (regression) case. 1.4 Goal and Scope of the Thesis This thesis is devoted to the ordinal classiﬁcation with monotonicity constraints. The general goal of the thesis is the following: Provide a comprehensive statistical theory for the problem of ordinal classiﬁcation with monotonicity constraints, as well as an eﬃcient and accurate method for solving the problem. In particular, there are four major objectives associated with this goal. These four objectives are achieved in separate chapters of the thesis. They are characterized brieﬂy below. Probabilistic model for ordinal classiﬁcation with monotonicity constraints. Although ordinal classiﬁcation with monotonicity constraints has been considered in multicriteria decision analysis, rough set approach and machine learning, there is no comprehensive theory which deﬁnes the problem from statistical point of view. We meet these needs in Chapter 2. We show, how monotonicity constraints can be expressed by making general assumptions about the probability distribution. Moreover, we formulate the necessary and suﬃcient conditions for the structure of the loss function which ensures the monotonicity of the optimal Bayes classiﬁer. Nonparametric methods. Having deﬁned the model, we consider the probability estimation. We propose a nonparametric method, based on isotonic regression. Although isotonic regression has already been used to this end in binary-class case with linear preorder relation, our approach for any K and partial preorder is novel. Next, we analyze the problem of monotone approximation from statistical point of view. We also give a general method for reduction of the problem size. We thoroughly analyze the binary-class case and the case of a linear loss function. We show that the monotone approximation can be used as a general method for incorporating monotonicity constraints into the learning process. 18 1.4. Goal and Scope of the Thesis Stochastic framework for Dominance-based Rough Set Approach. DRSA is the approach having its roots in logic, which is probably the reason why no statistical explanation of the approach has ever been proposed. In Chapter 4 we provide such an explanation basing on the statistical theory of ordinal classiﬁcation with monotonicity constraints. The explanation leads us to a natural, stochastic extension of DRSA, which is found to be very useful for data in the presence of noticeable inconsistency. We also show that stochastic DRSA implicitly aims at minimizing a speciﬁc interval loss function. This makes the rough set methods perfectly tailored to the problems where abstaining from classiﬁcation is allowed in some cases. Monotone rule ensembles. Although the statistical theory for ordinal classiﬁcation with mono-tonicity constraints provides an explanation for many concepts and approaches, it does not directly lead to the learning algorithm, which can be used for prediction purposes. In Chapter 5 we intro-duce two such algorithms, which both have the form of ensembles of decision rules. They possess a good prediction performance, low computational costs and maintain simplicity and interpretability. We provide a theoretical analysis which shows that the monotonicity assumptions allow us to bound the diﬀerence between the performance of rule ensemble and the performance of optimal classiﬁer in terms of the empirically measurable value of the so called margin. Computational experiments. The most of the thesis is theoretical. However, the learning algorithms demand empirical evaluation to test how they perform on real data. We address this issue in Chapter 6. We compare our algorithm based on monotone approximation, isotonic regression and rule ensembles with popular existing approaches to ordinal classiﬁcation with monotonicity constraints. To our knowledge, there has not been such an extensive comparison of so many methods on so many datasets for classiﬁcation problem with monotonicity constraints before. Chapter 2 Probabilistic Model for Ordinal Classiﬁcation with Monotonicity Constraints Ordinal classiﬁcation with monotonicity constraints is often referred to as the problem of ﬁnding an accurate classiﬁer within the class of monotone functions. Restricting to such class of functions is justiﬁed either by the assumption that the “target” function is monotone, or by requirement that the constructed model should maintain monotonicity. None of these statements, however, is applicable in the probabilistic setting, in which objects are generated according to some distribution and we do not take into account any semantic information about the data. Therefore, in this section we introduce a general assumption about the probability distribution, expressed in terms of the stochastic dominance in order to capture the concept of monotonicity constraints. Then we show how the monotonicity of the Bayes classiﬁer follows for a speciﬁc class of loss function (Kotłowski and Słowiński, 2008). 2.1 Stochastic Dominance We will formulate the most general assumption about the probability distribution P(x, y) when the monotonicity constraints are present in the ordinal classiﬁcation. The monotonicity constraints require that if x ⪰x′ then x should be assigned a class not lower than x′. In practice, these constraints are not always satisﬁed, leading to the situations referred to as inconsistencies. This suggests that the dominance relation ⪰does not impose “hard” constraints and the constraints should rather be deﬁned in a probabilistic setting. Consider two points x, x′ ∈X, such that x ⪰x′. We believe that the core of the monotonicity concept consist in the observation that the probability that x will get higher class label then x′, should not be smaller than the probability of the opposite event, i.e.: P(y > y′\|x, x′) ⩾P(y < y′\|x, x′) (2.1) In other words, the event that x gets a higher label than x′ is more probable than the even that x′ gets a higher label than x. Moreover, this property should still hold if we merge some of the contiguous classes. For instance, suppose we have four classes in a house pricing dataset: “cheap”, “moderate”, “expensive” and “very expensive”. If we merge classes “cheap” and “moderate” into a single class “not expensive”, obviously, (2.1) should still hold. The intuition behind is that the 20 2.2. Monotonicity of the Bayes classiﬁer problem with merged contiguous classes maintains monotone properties (both, with respect to order on class labels and monotone relationships) and thus can still be regarded as ordinal classiﬁcation with monotonicity constraints. More formally, let Y = {1, . . . , K} be the original set of class labels and let ˜ Y = {1, . . . , ˜ K}, with ˜ K ⩽K, be a set of class labels which results from merging some of the contiguous classes, i.e. ˜ y = 1 ⇐ ⇒y ∈{1, . . . , k1}, ˜ y = 2 ⇐ ⇒y ∈{k1 + 1, . . . , k2}, etc. We expect that for every x ⪰x′ and for every ˜ Y , (2.1) holds. We believe this is the minimal requirement for the probability distribution in the ordinal classiﬁcation with monotonicity constraints. It would be convenient, however, to have more comprehensive conditions than those described above. In particular, we would like to formulate the conditions expressed only by means of the original set of labels Y . Such conditions are introduced by the following theorem: Theorem 2.1. Let x, x′ ∈X be such that x ⪰x′. Then, (2.1) holds for original set of class labels Y and for every set of class labels ˜ Y which results from merging some of the contiguous classes if and only if for the original set Y it holds: P(y ⩽k\|x) ⩽P(y ⩽k\|x′) (2.2) for every k = 1, . . . , K. The proof of the theorem is quite technical and is given in the Appendix at the end of this chapter. Notice that (2.2) is a relation between two probability distributions, conditioned at x and x′, re-spectively. This relation is known as (ﬁrst order) stochastic dominance (Levy, 1998). Therefore, we deﬁne the following property of the probability distribution as the principle of stochastic dominance (Dembczyński et al., 2007b): x ⪰x′ = ⇒P(y ⩽k\|x) ⩽P(y ⩽k\|x′) ∀x, x′ ∈X, k = 1, . . . , K. (2.3) The principle can also be expressed in the following, equivalent way: x ⪰x′ = ⇒P(y ⩾k\|x) ⩾P(y ⩾k\|x′) ∀x, x′ ∈X, k = 1, . . . , K. We will call a probability distribution to be monotonically constrained if it satisﬁes (2.3). The principle of stochastic dominance is the core of what we understand by monotonicity constraints. Notice that in (Cao-Van, 2003; Cao-Van and De Baets, 2004) stochastic dominance was also used, but to deﬁne the properties of the estimator, not the properties of the probability distribution. Now we investigate the consequences of (2.3) for the Bayes classiﬁer. 2.2 Monotonicity of the Bayes classiﬁer 2.2.1 Loss functions and monotonicity of the Bayes classiﬁer. In the classiﬁcation problem, we aim at ﬁnding the classiﬁer which is as close as possible to the Bayes classiﬁer, the best possible classiﬁer. In other words, the Bayes classiﬁer is our “target function” which we try to approximate. Therefore, no surprisingly, we require that in the ordinal classiﬁcation with monotonicity constraints, the Bayes classiﬁer must be monotone. The Bayes classiﬁer may not be unique. First of all, it is deﬁned only up to a zero measure set. To solve this problem, we assume that for every x ∈X, the Bayes classiﬁer returns the class label k with the smallest conditional risk E[L(y, k)\|x]. Secondly, there can be ties between class labels on the conditional risk. Therefore, we assume the lowest label is always chosen. This makes the Bayes classiﬁer unique. Probabilistic Model for Ordinal Classiﬁcation with Monotonicity Constraints 21 Suppose that the probability distribution is monotonically constrained. We will investigate under what assumptions about the loss function, the Bayes classiﬁer is a monotone function. Let lyk = L(y, k) be the loss for predicting class k when the actual class is y, and assume lkk = 0 for each k, and lyk > 0 for each y ̸= k. We remind that the ordinal loss matrix was deﬁned in Section 1.2.2 as: ly,k−1 ⩾lyk if k ⩽y, lyk ⩽ly,k+1 if k ⩾y. Those properties of the loss matrix are not suﬃcient to ensure monotonicity of the Bayes classiﬁer when the probability distribution is monotonically constrained. We will prove this claim by a counter-example. Let us consider two popular loss matrices, 0-1 loss lyk = 1y̸=k and absolute error loss lyk = \|y −k\|. Both are ordinal (see Section 1.2.2), however the Bayes classiﬁer is monotone only for absolute error loss. Indeed, the Bayes classiﬁer for absolute error loss is the median of the distribution. It will be later shown that the median is a monotone function under stochastic dominance assumption. On the other hand, the Bayes classiﬁer for 0-1 loss is the mode of the distribution (most probable class). Consider the following 3-class counter-example: assume that x ⪰x′ and that the probability distribution at x′ is (0.3, 0.3, 0.4) (e.g. P(y = 3\|x′) = 0.4), while the probability distribution at x is (0.1, 0.5, 0.4). Although (2.3) is satisﬁed (distribution is monotonically constrained), the Bayes classiﬁer for x′ indicates class 3, while for x it indicates class 2, which contradicts the monotonicity. Thus, additional constraints must be imposed on the ordinal loss matrix in order to ensure the monotonicity of the Bayes classiﬁer. This issue is solved by the following theorem: Theorem 2.2. Suppose [lyk]K×K is an ordinal loss matrix. Then the Bayes classiﬁer is a monotone function for every monotonically constrained probability distribution P(x, y) if and only if the loss matrix satisﬁes the following constraints: ly,k+1 −lyk ⩾ly+1,k+1 −ly+1,k if k > y, ly,k−1 −lyk ⩾ly−1,k−1 −ly−1,k if k < y. (2.4) We give the proof in the Appendix due to its technical content. From Theorem 2.2 it follows that conditions (2.4) are necessary and suﬃcient for monotonicity of the Bayes classiﬁer. The mono-tonicity property is required in the ordinal classiﬁcation with monotonicity constraints, because otherwise there would be no point in restricting to the class of monotone functions. Hence, we will call the ordinal loss matrix satisfying (2.4) monotone loss matrix. Notice that for K = 2, every loss matrix is monotone. Summarizing, we can deﬁne the ordinal classiﬁcation problem with monotonicity constraints as the problem of minimizing the risk with the monotone loss matrix where the data are generated according to the monotonically constrained distribution. 2.2.2 Convex loss functions and monotonicity. We will investigate the conditions (2.4) for a popular subclass of the loss matrices. Let c: Z →R be a function which assigns to each integer1 a real number. We say that the function c(k) is convex if for each k ∈Z it holds: c(k) ⩽c(k −1) + c(k + 1) 2 . (2.5) 1We denote the set of integers by Z. 22 2.3. Linear Loss This deﬁnitions is equivalent to the more familiar deﬁnition of the form: function c(k) is convex if for every i, j ∈Z and for every λ ∈[0, 1] such that λi + (1 −λ)j ∈Z, we have: c(iλ + (1 −λ)j) ⩽λc(i) + (1 −λ)c(j). (2.6) Here we use (2.5), due to its simplicity. The equivalence of deﬁnitions (2.5) and (2.6) is proved in the Appendix as Lemma 2.2. In multi-class problems, the loss matrix is very often expressed in the following form: lyk = c(y −k), (2.7) with c(0) = 0 and c(k) > 0 if k ̸= 0. The matrices of this form are, for instance, 0-1 loss (with function c(k) = 1k̸=0), mean absolute error loss (c(k) = \|k\|) or squared error loss (c(k) = k2). Moreover, every binary loss matrix is of that form, since it is determined by setting only two parameters, l12 = c(−1) and l21 = c(1). With such a representation of the loss matrices, we can prove the following theorem: Theorem 2.3. Suppose [lyk]K×K is an ordinal loss matrix of the form lyk = c(y −k) for some function c: Z →R, such that c(0) = 0 and c(k) > 0 for every k ̸= 0. Then, the Bayes classiﬁer is monotone if and only if c(k) is convex. Proof. The conditions (2.4) can now be expressed as: c(y −k −1) −c(y −k) ⩾c(y −k) −c(y −k + 1) if k > y, c(y −k + 1) −c(y −k) ⩾c(y −k) −c(y −k −1) if k < y, which are equivalent (along with condition c(0) ⩽c(1)+c(k+1) 2 holding for every loss matrix) to the condition (2.5). Corollary 2.4. Consider loss function of the form lyk = \|y −k\|p, for p ⩾0 and K > 2. Then the Bayes classiﬁer is monotone if and only if p ⩾1. Proof. For any k, expression (2.5) is satisﬁed if and only if p ⩾1. This follows from the well known fact that functions of the form c(x) = \|x\|p are convex if and only if p ⩾1. Corollary 2.4 explains why 0-1 loss (p →0) does not lead to the monotone Bayes classiﬁer under the stochastic dominance assumption, while absolute error loss (p = 1) and squared-error loss (p = 2) do ensure monotonicity. These results suggest that 0-1 is not a proper loss for ordinal classiﬁcation with monotonicity constraints (apart from binary-class case K = 2), since it is not a monotone loss function, according to the deﬁnition (2.4). The absolute error loss is the “boundary” function – it just satisﬁes the minimal requirements for the loss function to be monotone. 2.3 Linear Loss Deﬁnition. Let us consider a speciﬁc loss function known as the linear loss (Berger, 1993): lyk = ( α(k −y) if k > y (1 −α)(y −k) if k ⩽y, (2.8) where 0 < α < 1. For α = 1 2 we have absolute error loss lyk = \|k −y\| (up to proportional constant). The purpose of introducing (2.8) is to model asymmetric costs of misclassiﬁcation: for α > 1 2 the cost of predicting higher class then the actual class y is more penalized than predicting the lower class; for α < 1 2 we have the opposite case. Such a loss function can be useful e.g. in medicine: consider Probabilistic Model for Ordinal Classiﬁcation with Monotonicity Constraints 23 1 2 3 4 5 s( (1) ) s( (2) ) s( (3) ) s( (4) ) s( (5) ) Y s( (Y) ) Figure 2.1: Example of extended linear loss with K = 5. The function s(k) changes the position of each class label on the scale. classifying patients into the classes according to their health condition: “good”, “moderate”, “bad”, “very bad”. Then, classifying the patient’s condition to be better than it really is, is probably more dangerous to her/his health than regarding the patient to be in a worse condition than the real one. As a corollary from Theorem 2.3, we immediately have: Corollary 2.5. The linear loss is a monotone loss matrix. It is also known (Berger, 1993) that such a loss function is minimized by the (1 −α)-quantile of the conditional distribution2 i.e. by such y1−α that P(y ⩽y1−α) ⩾1 −α and P(y ⩾y1−α) ⩾α. For α = 1 2 we have the of median. Extended linear loss. One can even extend the deﬁnition of linear loss by introducing the following deﬁnition: lyk = ( α(s(k) −s(y)) if k > y (1 −α)(s(y) −s(k)) if k ⩽y, (2.9) where s(k): Y →R is a strictly increasing function. It can be interpreted as a “scale changing” function, which positions the class labels on the real axis, thus changing the distances between them. Introducing arbitrary scale may at ﬁrst look like a strong generalization of (2.8). However, it is not hard to show that the arbitrary scale will not change anything on the population level: Theorem 2.6. The Bayes classiﬁer for the extended linear loss (2.9) does not depend on the function s(k). In particular, this implies that the Bayes classiﬁer for extended linear loss is the (1 −α)-quantile of the conditional distribution. Proof. First, notice that since s(k) is strictly increasing, it has an inverse s−1 : s(Y ) →Y . Let y ∈Y be a random variable according to distribution P(y\|x). Let us deﬁne the random variable y′ = s(y). Moreover, for each k ∈Y , let k′ = h(k). Then, arg min k E L(s(y), h(k))\|x = s−1 arg min k′ E L(y′, k′)\|x , (2.10) i.e. minimizing the extended loss (2.9) is equivalent to ﬁrst minimizing the expected loss (2.8) with random variable y′ and then taking the s−1-inverse of the obtained minimizer. The Bayes classiﬁer for (2.8) is y′ 1−α, the (1 −α)-quantile of distribution P(y′\|x). Since P(y′ = k\|x) = P(y = s−1(k)\|x) for every k = 1, . . . , K, then we must have y′ 1−α = s(y1−α). According to (2.10), the Bayes classiﬁer for (2.9) is s−1(y′ 1−α) = s−1(s(y1−α)) = y1−α. 2We remind that in general, p-quantile of probability distribution P(x) is deﬁned as a value xp such that P(x ⩽ xp) ⩾p and P(x ⩾xp) ⩾1 −p. 24 Appendix We conclude that there is no point in using complex scale changing functions and we can stay with the ordinary linear loss. Although parametrized by only a single value α, linear loss is suﬃcient to model most of real-life ordinal classiﬁcation problems. Hence, for the rest of this thesis we will focus on the linear loss function. Appendix: Proofs of the Theorems Proof of Theorem 2.1 Theorem 2.1. Consider x and x′ such that x ⪰x′. Then, (2.1) holds for the original set of class labels Y and for every set of class labels ˜ Y which results from merging some of the contiguous classes if and only if for the original set Y it holds: P(y ⩽k\|x) ⩽P(y ⩽k\|x′) (2.11) for every k = 1, . . . , K. Proof. First we prove the “if” part. Assume (2.11) holds for Y . But then (2.11) also holds for every merged set ˜ Y , since the event {˜ y ⩽k} is equivalent to the event {y ⩽k′} for some k′. Choose any such set ˜ Y and let us denote ∆= P(˜ y ⩾˜ y′\|x, x′)−P(˜ y ⩽˜ y′\|x, x′). We shall prove that ∆⩾0. Let us denote pk = P(˜ y = k\|x) and qk = P(˜ y = k\|x′) for each k = 1, . . . , ˜ K. Due to the independence of y and y′, we have: P(˜ y ⩾˜ y′\|x, x′) = ˜ K X k=2 k−1 X l=1 P(˜ y = k ∧˜ y′ = l\|x, x′) = ˜ K X k=2 k−1 X l=1 pkql, and, similarly, P(˜ y ⩽˜ y′\|x, x′) = P ˜ K k=2 Pk−1 l=1 qkpl, so that: ∆= ˜ K X k=2 k−1 X l=1 (pkql −qkpl). (2.12) We will prove ∆⩾0 by induction on ˜ K. For ˜ K = 1 obviously ∆= 0. Now, ﬁx some ˜ K and assume the theorem holds for ˜ K −1. From (2.11) we have p ˜ K ⩾q ˜ K, because: p ˜ K = 1 − ˜ K−1 X k=1 pk ⩾1 − ˜ K−1 X k=1 qk = q ˜ K. Notice that if we keep p ˜ K +p ˜ K−1 constant, but decrease p ˜ K and increase p ˜ K−1 by the same amount ϵ, ∆decreases; one can easily check by simple analysis of the sums in (2.12) that the total decrease is ϵ(q ˜ K−1 + q ˜ K). We can decrease p ˜ K without violating (2.11) as long as p ˜ K ⩾q ˜ K. Thus, it is enough to show that ∆⩾0 for p ˜ K = q ˜ K. We transform the expression (2.12) to obtain: ∆= ˜ K−1 X k=2 k−1 X l=1 (pkql −qkpl) + ˜ K−1 X l=1 (p ˜ Kql −q ˜ Kpl). (2.13) The second term can be transformed to: ˜ K−1 X l=1 (p ˜ Kql −q ˜ Kpl) = p ˜ K(1 −q ˜ K) −q ˜ K(1 −p ˜ K) = p ˜ K −q ˜ K ⩾0. Probabilistic Model for Ordinal Classiﬁcation with Monotonicity Constraints 25 Thus, it is enough to show that the ﬁrst term is nonnegative. But the ﬁrst term looks almost like the case with ˜ K −1 classes – the only diﬀerence is that probabilities, both qk and pk do not sum to 1. But they both sum to the same number, 1 −p ˜ K, since p ˜ K = q ˜ K. So, if we divide (rescale) the ﬁrst term by 1 −p ˜ K (which will neither change the sign of ∆nor violate (2.11)), we have the case with ˜ K −1, so we can use the induction assumption and ﬁnally we obtain ∆⩾0. Now, we prove the “only if” part. Assume (2.1) holds for every merged set ˜ Y . In particular, it holds when ˜ Y = {1, 2}, such that ˜ y = 1 ⇐ ⇒y ⩽k and ˜ y = 2 ⇐ ⇒y > k; in other words, we merge classes 1, . . . , k into a single class and merge classes k + 1, . . . , K into another class. Let us denote p = P(y ⩽k\|x) and q = P(y ⩽k\|x′). Then, we have 0 ⩽P(˜ y ⩾˜ y′\|x, x′) −P(˜ y ⩽˜ y′\|x, x′) = q(1 −p) −p(1 −q) = q −p, which means: P(y ⩽k\|x) ⩽P(y ⩽k\|x′). Since k was chosen arbitrarily, this ends the proof. Proof of Theorem 2.2 Before we prove the theorem, we ﬁrst need a simple lemma related to the stochastic dominance. The lemma is a basic result in decision theory, but we give the proof for clarity and completeness. Lemma 2.1. Let x ⪰x′ so that P(y\|x) stochastically dominates P(y\|x′). Let z : Y →R be a non-increasing random variable. Then it holds: E[z\|x] ⩽E[z\|x′], (2.14) i.e. the expected value of z according to distribution P(y\|x) is always smaller then the expected value of z according to P(y\|x′). Proof. Let us denote pk = P(y = k\|x), qk = P(yk\|x′) and zk = z(k). Then (2.14) can be rewritten in the following way: K X k=1 pkzk ⩽ K X k=1 qkzk. Let us denote the cumulative distribution by Pk = Pk l=1 pl and Qk = Pk l=1 ql (we assume P0 = Q0 = 0). From the stochastic dominance it follows that Pk ⩽Qk for each k. Then, K X k=1 pkzk = K X k=1 (Pk −Pk−1)zk = K X k=1 Pkzk − K−1 X k=0 Pkzk+1 = = zK + K−1 X k=1 Pk(zk −zk+1) ⩽zK + K−1 X k=1 Qk(zk −zk+1) = K X k=1 qkzk, (2.15) where the inequality follows from the fact that zk −zk+1 ⩾0 for each k. Now, we prove the main theorem: Theorem 2.2. Suppose [lyk]K×K is an ordinal loss matrix. Then the Bayes classiﬁer is a monotone function for every monotonically constrained probability distribution P(x, y) if and only if the loss matrix satisﬁes the following constraints: ly,k+1 −lyk ⩾ly+1,k+1 −ly+1,k if k > y ly,k−1 −lyk ⩾ly−1,k−1 −ly−1,k if k < y (2.16) 26 Appendix Proof. First we prove the “if” part. Suppose conditions (2.16) hold. Let us deﬁne δyk in the following way: δyk = ( lyk −ly,k−1 for k > y, lyk −ly,k+1 for k < y. Let P(x, y) be any monotonically constrained probability distribution and let x, x′ ∈X be any two points such that x ⪰x′. Let us denote pk = P(y = k\|x) and qk = P(y = k\|x′). Let u be a predicted class label. The expected loss for u over the distribution P(y\|x) is as follows: E[L(y, u)\|x] = K X y=1 pylyu = u−1 X y=1 py u X k=y+1 (lyk −ly,k−1) + K X y=u+1 py y−1 X k=u (lyk −ly,k+1), or by using δyk: E[L(y, u)\|x] = u−1 X y=1 py u X k=y+1 δyk + K X y=u+1 py y−1 X k=u δyk. Consider ∆(u\|x) = E[L(y, u + 1)\|x] −E[L(y, u)\|x], the diﬀerence between the expected losses for u + 1 and u: ∆(u\|x) = u X y=1 py u+1 X k=y+1 δyk + K X y=u+2 py y−1 X k=u+1 δyk − u−1 X y=1 py u X k=y+1 δyk − K X y=u+1 py y−1 X k=u δyk = = u X y=1 pyδy,u+1 − K X y=u+1 pyδyu ⩽ u X y=1 qyδy,u+1 − K X y=u+1 qyδyu = ∆(u\|x′), where the inequality comes from Lemma 2.1, since the function z(1) = δ1,u+1, . . . , z(u) = δu,u+1, z(u+1) = −δu+1,u, . . . , z(K) = −δKu is non-increasing according to the assumptions (2.16). This means that the diﬀerence in expected loss for any two contiguous class labels u + 1 and u does not increase as we move from x to x′. But this means that the diﬀerence in expected loss between any class labels v and u does not increase. Now, suppose v is a Bayes classiﬁer for x′, i.e.: v = arg min k∈Y E[L(y, k)\|x′]. Choose some u < v. We have: 0 > E[L(y, v)\|x′] −E[L(y, u)\|x′] ⩾E[L(y, v)\|x] −E[L(y, u)\|x], which means that u cannot be the Bayes classiﬁer for x. Thus, the Bayes classiﬁer must be monotone. Now we prove the “only if” part. Suppose that one of the conditions (2.16) is violated; without loss of generality, we may assume that the ﬁrst one is violated, i.e. ly0k0 −ly0,k0−1 < ly0+1,k0 − ly0+1,k0−1 for some k0 > y0. We shall ﬁnd some probability distribution and objects x ⪰x′ such that the Bayes classiﬁer violates monotonicity condition, i.e. h∗(x) < h∗(x′). First, notice that we can set P(y = k\|x) = 0 for each x ∈X, for every class label k / ∈ {y0, k0, k0 −1}. This will eﬀectively eliminate other classes (they never occur in the problem) so that we end up with three-class problem which is much easier to analyze than a general K-class problem. Therefore, without loss of generality, we assume the following loss matrix: [lij]K×K =    0 l12 l13 l21 0 l23 l31 l32 0   . Probabilistic Model for Ordinal Classiﬁcation with Monotonicity Constraints 27 Without loss of generality we also assume that the violation of (2.16) has the form l23 > l13 − l12 (another possibility is l21 > l31 −l32, but the analysis would be analogical). First, we will construct a probability distribution z = (z1, z2, z3) and later from this distribution we will construct distributions at points x and x′. We will choose distribution z so that the expected loss for predicting class 2 is equal to the loss for predicting 3 and smaller than for class 1, i.e.: Ez[L(y, 3)] = Ez[L(y, 2)], (2.17) Ez[L(y, 3)] < Ez[L(y, 1)], (2.18) where we denoted the expectation over the distribution z by Ez This implies: z1l13 + z2l23 = z1l12 + z3l32, z1l13 + z2l23 < z2l21 + z3l31. Using z3 = 1 −z2 −z1 and knowing that both l32 + l13 −l12 and l13 + l31 are positive, we can transform these expressions to: z1 = A −Bz2, z1 < C −Dz2, (2.19) where: A = l32 l32 + l13 −l12 , B = l23 + l32 l32 + l13 −l12 , C = l31 l31 + l13 , D = l23 + l31 −l21 l31 + l13 . Notice that A, C > 0 and B > 1. We will prove some more inequalities. First, we show that B −D > 0: B −D > 0 ⇐ ⇒(l23 + l32)(l31 + l13) > (l32 + l13 −l12)(l23 + l31 −l21) ⇐ ⇒(l31 −l32)(l23 −l13 + l12) + l12(l23 + l32) +l21(l32 + l13 −l12) > 0, (2.20) which holds, because all the terms in the last equation are positive. Next, we show that BC −AD > 0: BC −AD > 0 ⇐ ⇒(l23 + l32)l31 −l32(l23 + l31 −l21) > 0 ⇐ ⇒l23(l31 −l32) + l32l21 > 0, (2.21) which holds, because, again, all the terms are positive. Finally, we must show that A−C < B −D: A −C < B −D ⇐ ⇒l32(l31 + l13) −l31(l32 + l13 −l12) −(l23 + l32)(l31 + l13) + (l23 + l31 −l21)(l32 + l13 −l12) < 0 ⇐ ⇒ −l23(l31 −k32) −l21(l13 −l12 + l32) −l12l23 < 0, (2.22) which holds because, again, all the terms on the left hand side are negative. Let us substitute the ﬁrst expression in (2.19) into the second expression to obtain: A −z2B < C −z2D ⇐ ⇒z2(B −D) > A −C+ ⇐ ⇒z2 > A −C B −D, because B −D > 0 from (2.20). We now must show that there exist z1, z2 such that 0 < z1, z2 < 1, z1 + z2 < 1, z1 = A −Bz2, and z2 > A−C B−D. Fix z2 = ϵ + max{0, A−C B−D}. From (2.22) it follows that A−C B−D < 1, thus we can always found positive ϵ such that 0 < z2 < 1. Moreover: z1 = A −Bz2 = −Bϵ + min A, A −B A −C B −D = −Bϵ + min A, BC −AD B −D , 28 Appendix and since A > 0 and it follows from (2.21) that BC−AD B−D > 0, we have z1 > 0. Moreover, since A < 1, for suﬃciently small ϵ we have z1 < 1. Finally, notice that: z1 + z2 = A −(B −1)z2 = −ϵ(B −1) + min A, A −(B −1) A −C B −D (we used the fact that B > 1), which means that z1 + z2 < 1 for suﬃciently small ϵ. Thus, all requirements are satisﬁed for z1, z2, z3 to be a probability distribution for which (2.17)-(2.18) hold. Since the inequality in (2.18) is a strict inequality, it will be still satisﬁed for another probability distribution q = (q1, q2, q3) such that q1 = z1 + γ and q2 = z2 −γ and q3 = z3 with suﬃciently small γ. Similarly to the way we got the ﬁrst equation in (2.19) from (2.17), we can show that the following holds: Eq[L(y, 3)] < Eq[L(y, 2)] ⇐ ⇒q1 < A −Bq2, Eq[L(y, 3)] > Eq[L(y, 2)] ⇐ ⇒q1 > A −Bq2. It follows for positive γ that: q1 = z1 + γ = A −Bz2 + γ < A −Bz2 + Bγ = A −B(z2 −γ) = A −Bq2, which means that for distribution q, class label 3 have the lowest cost. Moreover, if we choose another distribution p = (p1, p2, p3) such that p1 = z1 −γ, p2 = z2 + γ, p3 = z3, for the same positive γ, we have: p1 = z1 −γ = A −Bz2 −γ > A −Bz2 −Bγ = A −B(z2 + γ) = A −Bp2, which means that for distribution p class label 2 have the lowest cost. But distribution p stochas-tically dominates distribution q, since p1 = z1 −γ < z1 + γ = q1 and p2 = q2. Thus, we can choose any x, x′ such that x ⪰x′ and assign P(y = k\|x′) := qk, P(y = k\|x) := pk for each k, and from the above analysis it follows that h∗(x) = 2 < 3 = h∗(x′), a contradiction. Proof of Lemma 2.2 In this section we prove the equivalence of two deﬁnitions of convex functions. Lemma 2.2. For any function c: Z →R which assigns to each integer a real number, the two following conditions are equivalent: 1. For every i, j ∈Z and for every λ ∈[0, 1] such that λi + (1 −λ)j ∈Z: c(iλ + (1 −λ)j) ⩽λc(i) + (1 −λ)c(j). (2.23) 2. For every k ∈Z: c(k) ⩽c(k −1) + c(k + 1) 2 . (2.24) Proof. If the function satisﬁes (2.23), than (2.24) holds by choosing i = k −1, j = k + 1 and λ = 1 2. Now, suppose that (2.24) holds. We will show that for all positive integers r, s we have: c(k) ⩽ s s + rc(k −r) + r s + rc(k + s). (2.25) We will prove this by induction on r and s. For r = s = 1 (2.25) holds by assumption. Suppose it holds for every r, s ⩽N and we will prove that it then holds also for r, s ⩽N + 1. Multiple application of (2.25), ﬁrst with some r, s and then with 1, r, leads to: c(k) ⩽ s s + r r r + 1c(k −r −1) + 1 r + 1c(k) + r s + rc(k + s), Probabilistic Model for Ordinal Classiﬁcation with Monotonicity Constraints 29 or equivalently, by rearranging the terms, to: c(k) ⩽ s s + r + 1c(k −r −1) + r + 1 s + r + 1c(k + s), which means that (2.25) holds for r ⩽N + 1 and s ⩽N. Applying once more (2.25) with s, 1, analogously as above, shows that (2.25) holds for r, s ⩽N + 1. This ﬁnishes the induction proof of (2.25). Choose any i, j ∈Z and any λ ∈[0, 1] such that k = λi + (1 −λ)j ∈Z. But then i = k −r and j = k −s for some positive r, s. This means that λ = s s+r and from (2.25) it follows that (2.23) holds, which ends the proof. 30 Chapter 3 Nonparametric Methods In this chapter, we consider the methods of probability estimation (isotonic regression) and clas-siﬁcation (monotone approximation) which are based solely on the assumption about the mononicity constraints of probability distribution (principle of stochastic dominance). No other assumptions are made and hence the only information about the objects being used is obtained through the dominance relation. Those methods are called nonparametric, because we infer about the model using the largest possible class of functions under the monotonicity assumption: the class of all monotone functions. The nonparametric methods are especially useful when the dominance relation is suﬃciently dense (the dimensionality m is not too high) or when the only available information about the objects is given by the dominance relation ⪰(e.g. we have no informations about the attribute vectors). Moreover, as we shall see in Chapter 5, the nonparametric methods can be very useful in combination with some parametric classiﬁcation procedures, constituting a two-phase learning algorithm. 3.1 Nonparametric Probability Estimation by Isotonic Re-gression When the distribution is monotonically constrained, it is enough (under some additional, mild assumption described later) to estimate the probabilities in a nonparametric way. We propose such a nonparametric method, taking into account only the monotonicity constraints expressed by the dominance relation. Our method is based on isotonic regression. Although isotonic regression has already been used to this end in binary-class case with linear preorder relation(Robertson et al., 1998), our approach for any K and partial preorder is new (Dembczyński et al., 2007b; Kotłowski et al., 2008). 3.1.1 Maximum Likelihood Estimation The most popular method of probability estimation in statistics is the maximum likelihood estimation (MLE). Let D be the observed data (training set) and let θ be some vector of parameters which are to be estimated. Then, we deﬁne the MLE estimate of θ, denoted by ˆ θ, as the value of θ, for which the probability of D (likelihood) is maximal: ˆ θ = arg max θ P(D\|θ) = arg max θ L(θ; D), 32 3.1. Nonparametric Probability Estimation by Isotonic Regression where L(θ; D) is the likelihood function. One usually works with the negative log-likelihood ℓ(θ) = −log L(θ; D), since the global minimum of ℓand global maximum of L coincide, i.e. ˆ θ = arg minθ ℓ(θ). In the ordinal classiﬁcation with monotonicity constraints we do not impose any speciﬁc prob-abilistic model apart from the stochastic dominance assumption. Therefore, for each object xi, i = 1, . . . , n, we can regard the probability distribution P(y\|xi) as a set of parameters to esti-mate. By denoting pik := P(y = k\|xi), the likelihood function have the form: L(p\|D) = n Y i=1 pi,yi, so that negative log-likelihood is: ℓ(p) = − n X i=1 log pi,yi. (3.1) The only constraint that we assume is the stochastic dominance principle (2.3): xi ⪰xj = ⇒ k X l=1 pil ⩽ k X l=1 pjl, (3.2) for every i, j = 1, . . . , n, and for each k = 1, . . . , K. Therefore, the nonparametric problem of MLE is deﬁned as the problem of minimizing (3.1) under the constraints (3.2), and the constraints pik ⩾0 and PK k=1 pik = 1, for i = 1, . . . , n, to ensure that the axioms of the probability distribution hold. This is a nonlinear optimization problem, with a convex objective function and linear constraints, so it can be solved quite eﬃciently by general constraint optimization algorithms (Bazaraa et al., 2006). However, the objective function is not strictly convex when K > 2, so that the problem may not have a unique solution. Hence, we will solve the MLE problem only for a binary-class case, while for a general multi-class case we will propose a diﬀerent estimation method, based on the reduction to K −1 binary-class problems. 3.1.2 Binary-class Problem and Isotonic Regression Binary-class MLE. Let us restrict to the case of K = 2 and for the sake of clarity, we will use the set of class labels Y = {0, 1}. Let pi = P(y = 1\|xi) so that P(y = 0\|xi) = 1 −pi. Then, the problem of MLE can be rewritten as: minimize: − n X i=1 yi log pi + (1 −yi) log(1 −pi) subject to: xi ⪰xj = ⇒pi ⩾pj i, j = 1, . . . , n 0 ⩽pi ⩽1 i = 1, . . . , n (3.3) Although the problem (3.3) remains nonlinear, it can easily be shown that the objective function is strictly convex now, i.e. the optimal solution is unique. At this moment, we can use a consistency property (1.13) to signiﬁcantly reduce the size of the problem. We remind that object xi is consistent if for every j = 1, . . . , n, it holds: xi ⪰xj →yi ⩾yj and xi ⪯xj →yi ⩽yj. The reduction procedure is based on the following theorem. Theorem 3.1. Let ˆ p = (ˆ p1, . . . , ˆ pn) be the optimal solution of (3.3). Then, ˆ pi = yi if and only if object xi is consistent. Proof. We consider the case yi = 1 (the case yi = 0 is analogous). If xi is consistent, then from the deﬁnition of consistency (1.13), there is no other object xj, such that xj ⪰xi and yj = 0. Thus, Nonparametric Methods 33 for every xj, such that xj ⪰xi, yj = 1 and yj is also consistent (otherwise, due to transitivity of dominance, xi would not be consistent). Thus, we can set ˆ pj = 1 for xj and ˆ pi = 1 for xi, and these are the values that minimize the objective function in (3.3), while satisfying the constraints. Now, suppose ˆ pi = 1 and assume the contrary, that xi is not consistent, i.e. there exists xj, xj ⪰xi, but yj = 0. Then, due to the monotonicity constraints in (3.3), ˆ pj ⩾ˆ pi = 1, so ˆ pj = 1, and the objective in (3.3) equals to the inﬁnity, which is surely not the optimal solution to the minimization problem (since at least one feasible solution ˆ p ≡1 2 with a ﬁnite objective value exists). Thus, only consistent objects have probability estimates equal to 1 or 0. We can set ˆ pi = yi for each consistent object xi and optimize (3.3) only for inconsistent objects, which usually gives a large reduction of the problem size (number of variables). We have experienced on real data that in most cases at least 80% −90% of variables (objects) are removed. In the next paragraph we show that (3.3) has the same optimal solution as the isotonic regression problem. Isotonic regression. We remind that a vector p∗= (p∗ 1, . . . , p∗ n) is an isotonic regression of y if p∗is the solution of the following problem: minimize: n X i=1 (yi −pi)2 subject to: xi ⪰xj = ⇒pi ⩾pj i, j = 1, . . . , n, (3.4) so that p∗minimizes the squared error in the space of all monotone vectors p = (p1, . . . , pn). Comparing with deﬁnition (1.19) from Chapter 1, we set the weights w = (w1, . . . , wn) to be equal to ones1. Although squared error seems to be arbitrarily chosen, it can be shown that minimizing many other error functions we get the same vector p∗as in the case of (3.4). In particular, we show that this property applies also to the objective function of the MLE problem: Theorem 3.2. (Robertson et al., 1998) Let p∗be an isotonic regression of y. Then, p∗is also the optimal solution to the MLE problem (3.3), i.e. ˆ p = p∗. The proof is based on the analysis from (Robertson et al., 1998) and can be found in the Appendix. To summarize, the problem of MLE for binary-class case (3.3) can be solved by the isotonic regression (3.4), because the optimal solutions of both problems coincide. The isotonic regression is easier to solve due to a simpler objective function (quadratic); e.g. Burdakov et al. (2006) proposed a heuristic algorithm giving results close to optimum in O(n2) (this is in fact the fastest possible time for the order-restricted inference, since the order relation is in general of size O(n2)). Moreover, by using Theorem 3.1 the size of the problem is usually reduced several times before the optimization process has even begun. A simple example of isotonic regression is shown in Figure 3.1. 3.1.3 Multi-class Problem Description of the estimation method. In the multi-class case, the MLE estimation is no longer equivalent to the isotonic regression. Moreover, the MLE estimation is not strictly convex, which results in non-unique solution (Dembczyński et al., 2007b). There is no well-established method for probability estimation for the case of K > 2 ordered classes. Below, we propose such an approach, based on multiple isotonic regression. 1Nevertheless, all the results shown in this chapter are also valid for objects with arbitrary positive weights. 34 3.1. Nonparametric Probability Estimation by Isotonic Regression a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 0 1/3 1/3 1/3 0 1 1/2 1/2 1 1 Figure 3.1: Binary-class example with two attributes. Objects with y = 0 are dark, while with y = 1 – light. The estimate of probability of class 1, ˆ pi, is shown. Notice that for consistent objects (x1, x5, x6, x9, x10) it holds yi = ˆ pi. Let Y = {1, . . . , K} and for a given xi let us deﬁne K −1 dummy variables yik = 1yi⩾k, k = 2, . . . , K. We can think of solving the general K-class problem in terms of solving K −1 binary problems. In the k-th binary problem, dummy variables yik play the role of class labels with Y = {0, 1}, while variables of the problem correspond to estimating the probability P(y ⩾k\|xi). Let us ﬁx k = 2, . . . , K. We deﬁne the vector of estimators ˆ qk = (ˆ q1k, . . . , ˆ qnk) of the probabilities P(y ⩾k\|xi), as the isotonic regression of vector yk = (y1k, . . . , ynk), i.e. the optimal solution to the problem: minimize n X i=1 (yik −pi)2 subject to xi ⪰xj = ⇒pi ⩾pj i, j = 1, . . . , n. (3.5) Having obtained the solution of (3.5) for each k = 2, . . . , K, we construct the estimators ˆ pik of P(y = k\|xi) as: ˆ pik =      ˆ qik if k = K ˆ qik −ˆ qi,k+1 if 2 ⩽k < K 1 −ˆ qi,k+1 if k = 1 (3.6) These estimators are unique because the isotonic regression is unique. They boil down to the previous approach (3.4) in binary-class case. However, as the K −1 problems (3.5) are solved separately, we must guarantee that ˆ qik < ˆ qi,k+1 will never happen, otherwise we would have negative probabilities ˆ pik. Properties of the estimators. We show that for each i = 1, . . . , n, estimators {ˆ pi1, . . . , ˆ piK} form a probability distribution, i.e. they are non-negative and sum to one. However, ﬁrst we prove the following lemma: Lemma 3.1. Let ˆ p be the isotonic regression of the class labels vector y = (y1, . . . , yn). Suppose, we introduce a new vector of class labels y′ = (y′ 1, . . . , y′ n), such that y′ i ⩾yi for all i = 1, . . . , n. Then, ˆ p′, the isotonic regression of y′, has the following property: ˆ p′ i ⩾ˆ pi, for all i = 1, . . . , n. Nonparametric Methods 35 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 (1,0,0) (1/2,1/6,1/3) (1/2,1/6,1/3) (0,1,0) (1,0,0) (0,2/3,1/3) (0,1/2,1/2) (0,1/2,1/2) (0,0,1) (0,0,1) Figure 3.2: 3-class example with two attributes. Objects with y = 1 are black, with y = 2 – dark gray, and with y = 3 – light gray. The vector of probability estimates (ˆ pi1, ˆ pi2, ˆ pi3) is shown below each objects. Notice that for consistent objects (x1, x4, x5, x9, x10) the probability concentrates on a single class yi. The proof can be found in the Appendix. Using Lemma 3.1, the desired properties of the estimators are now easy to show. Theorem 3.3. For each i = 1, . . . , n, estimators {ˆ pi1, . . . , ˆ piK} form a probability distribution, i.e. PK k=1 ˆ pik = 1, and for each k, ˆ pik ⩾0. Proof. It immediately follows from the deﬁnition (3.6) that: K X k=1 ˆ pik = 1 −ˆ qi,2 + K−1 X k=2 ˆ qik −ˆ qi,k+1 + ˆ qiK = 1. Now we prove the non-negativity of ˆ pik. First, notice that the isotonic regression (3.4) is bounded between 0 and 1. This comes from the fact that the problem (3.4) has the same optimal solution as the problem (3.3), which explicitly includes the constraints 0 ⩽pi ⩽1 (see Section 3.1.2). This shows that ˆ pi1 ⩾0 and ˆ piK ⩾0. To show that ˆ pik ⩾0 for k = 2, . . . , K −1, we must show that ˆ qik −ˆ qi,k+1 ⩾0. But this is guaranteed by Lemma 3.1, because class indices yik = 1yi⩾k in the k-th problem are always greater than respective class indices yi,k+1 = 1yi⩾k+1. Summary. The problem of probability estimation for multi-class case is stated as the problem of solving K −1 isotonic regression problems (3.5). The probability estimators are obtained from the optimal solution by (3.6). They always form a proper probability distribution, i.e. they are non-negative and sum to unity. Solving (3.5) in each case k = 2, . . . , K, can be done in exactly the same way as in binary-class case (3.4). For instance, Theorem 3.1 also applies for each k = 2, . . . , K. A simple example of three-class problem is shown in Figure 3.2. 3.1.4 Extension Beyond the Training Set and Asymptotic Consistency Until now, we dealt only with the estimation of the conditional probability distributions for objects from the training set D, i.e. at the points xi, i = 1, . . . , n. However, one can simply extend the estimated probabilities to the whole space X. 36 3.1. Nonparametric Probability Estimation by Isotonic Regression Extension of probability estimates. Consider ﬁrst the binary problem Y = {0, 1} and proba-bility estimate ˆ pi for object xi. Since the estimates were obtained by solving the isotonic regression (3.4), it must hold xi ⪰xj →ˆ pi ⩾ˆ pj. The principle of stochastic dominance (2.3) for K = 2 says that the probability p(x) = P(y = 1\|x) is a monotone function. Therefore, a valid extension ˆ p(x) of the vector of estimators ˆ p = (ˆ p1 . . . , ˆ pn) must satisfy two conditions: 1. ˆ p(x) = ˆ pi (is the extension of the estimators). 2. For every x, x′ ∈X it holds x ⪰x′ →ˆ p(x) ⩾ˆ p(x) (is monotone). Potharst and Feelders (2002) considered the extensions of monotone functions deﬁned on the training set to the whole X. They showed that there is a minimal and a maximal extension, deﬁned as: ˆ pmin(x) = max{ˆ pi : xi ⪯x}, ˆ pmax(x) = min{ˆ pi : xi ⪰x}, and every valid extension ˆ p(x) satisﬁes ˆ pmin(x) ⩽ˆ p(x) ⩽ˆ pmax(x) for every x ∈X. Moreover, every monotone function satisfying the above condition is a valid extension. Therefore, it is worth considering the following parameterized extension: pλ(x) = λˆ pmin(x) + (1 −λ)ˆ pmin(x), (3.7) for λ ∈[0, 1]. The parameter λ can be tuned for a particular problem. In multi-class case, the situation is analogous. Instead of ˆ pi, we have the estimators ˆ qik of P(y ⩾k\|xi), for k = 2, . . . K, obtained from (3.5). The stochastic dominance principle is equivalent to saying that P(y ⩾k\|xi) must be a monotone function for each k. Therefore, the extensions {ˆ q2(x), . . . , ˆ qK(x)} must be monotone. We can proceed analogously as before, deﬁning the minimal, the maximal and λ-parametrized extension ˆ qmin k (x), ˆ qmax k (x) and ˆ qλ k(x). Asymptotic consistency of the estimator. We say that the estimator ˆ θ of parameter θ is strongly consistent if: ˆ θ a.s. − → n→∞θ, (3.8) i.e. the estimator converges to the real value of the parameters almost surely (i.e. with probabil-ity 1), as n goes to inﬁnity. We say that the estimator is weakly consistent, if (3.8) holds with convergence in probability. Strong consistency of isotonic regression. Let us ﬁrst assume Y = {0, 1}. To consider the asymptotic consistency of isotonic regression, let us extend the estimators of probabilities into the whole X using (3.7) for arbitrary λ. There is a large number of papers concerning the consistency of isotonic regression (see, for instance, Robertson and Wright (1975); Christopeit and Tosstorﬀ(1987), or Robertson et al. (1998) for overview). In general, the isotonic regression, extended in the form of (3.7) for any λ, is strongly consistent for every point in the interior of X under the assumption that P(y = 1\|x) is a monotone function and under very mild additional assumption about probability distribution, e.g. if P(x) is absolutely continuous with positive density (Christopeit and Tosstorﬀ, 1987). Now let us assume Y = {1, . . . , K}. Suppose P(x) is such that the binary isotonic regression is strongly consistent and let P(x, y) be monotonically constrained. Then, the strong consistency of the multiple isotonic regression ˆ qλ k(x) immediately follows from the deﬁnition (3.6) and from the fact that P(y ⩾k\|x) is a monotone function for every k = 2, . . . , K. In other words, the estimators Nonparametric Methods 37 ˆ qλ k(x) converge (with probability 1) to the real probabilities as n →∞, as long as the isotonic regression is strongly consistent. Concluding, we are able to state the following theorem: Theorem 3.4. Let P(x, y) be monotonically constraint and let P(x) be absolutely continuous with positive density. Then for each x in the interior of X, we have: ˆ qλ k(x) a.s. − → n→∞P(y ⩾k\|x) . In other words, in practical (non-pathological) cases, the multiple isotonic regression estimators will converge to the true probabilities. 3.2 Monotone Approximation Let us focus again on the ordinal classiﬁcation problem. We will consider the monotone ap-proximation, classiﬁcation method based solely on the mononicity assumption of the probability distribution. The method consist in minimizing the empirical risk in the class of all monotone functions. We will also show that it can be though of as a general method for incorporating the monotonicity constraints into the learning process. Although the problem of monotone approximation has already been considered several times in diﬀerent contexts (see Section 1.3.3 for details), we are ﬁrst to give the detailed analysis of its statistical properties, to show relationship to the isotonic regression and nonparametric maximum likelihood estimation and to summarize the asymptotic convergence issues (Dembczyński et al., 2007b; Kotłowski et al., 2008; Kotłowski and Słowiński, 2008). We also show how to reduce the computational complexity of the problem and how to handle non-uniqueness of the optimal solution. 3.2.1 Problem Statement Problem formulation. The monotone approximation is based on relabeling objects from the training set in order to remove the inconsistencies and “monotonize” the data. Let us consider the minimization of the empirical risk (1.5) within the class of all monotone functions: minimize n X i=1 L(yi, di) subject to xi ⪰xj →di ⩾dj i, j = 1, . . . , n di ∈{1, . . . , K} i = 1, . . . , n, (3.9) where L(y, k) is any monotone loss function. In this section, we adopt the interpretation that optimal values ˆ di of the problem (3.9) are new labels of the objects. Then, the problem can be stated in the following way: reassign (relabel) the objects to make the dataset consistent (monotone) such that new class labels ˆ di are as close as possible to the original class labels y, where the closeness is measured in terms of the loss function. We believe the new labels ˆ di are closer to the Bayes classiﬁer h∗(xi) than the original, non-monotone labels yi. Therefore, the monotone approximation can be thought of as the data “improvement” towards the Bayes classiﬁer. Algorithms for solving monotone approximation. Monotone approximation can be solved by either linear programming or network ﬂow. In both cases, it must be transformed to a more useful form. 38 3.2. Monotone Approximation Let dik, for k = 2, . . . , K, be a binary variable with the following interpretation “dik = 1 iﬀnew class label of object xi is at least k“. Such interpretation implies that dik ⩾di,k+1; for instance, for K = 5, di = 1 is decoded as [di1 = 0, di2 = 0, di3 = 0, di4 = 0], d2 is decoded as [1, 0, 0, 0], d3 as [1, 1, 0, 0], d4 as [1, 1, 1, 0] and d5 as [1, 1, 1, 1]. Thus, the new label of object xi can be obtained from di = 1+PK k=2 dik. The monotonicity of new labels implies that for any xi ⪰xj we must have dik ⩾djk for each k = 2, . . . , K. Finally, the loss function can be reformulated as: L(yi, di) = yi X k=2 (lyi,k−1 −lyi,k)(1 −dik) + K X k=yi+1 (lyi,k −lyi,k−1)dik = K X k=2 (lyi,k −lyi,k−1)dik + K X k=yi+1 (lyi,k−1 −lyi,k) (3.10) and the second sum on the right-hand side can be dropped, because it is constant. Thus, we transformed problem (3.9) into the following problem: minimize n X i=1 K X k=2 (lyi,k −lyi,k−1)dik subject to xi ⪰xj = ⇒dik ⩾djk i, j = 1, . . . , n; k = 2, . . . , K, di,k ⩾di,k+1 i = 1, . . . , n; k = 2, . . . , K −1, dik ∈{0, 1} i = 1, . . . , n; k = 2, . . . , K −1. (3.11) This is a linear program with integer variables. However, the integer condition (last constraint) can be relaxed to 0 ⩽dik ⩽1 and we end up with an ordinary linear program. The relaxation of this constraint follows from the fact that the matrix of coeﬃcients of the constraints is totally unimodular and the right hand sides of the constraints are integer. In such a case, every basic feasible (hence also optimal) solution is always integer. Therefore, we do need to impose integer constraints, because we will obtain an integer solution anyway (see Papadimitriou and Steiglitz (1998); Chandrasekaran et al. (2005) for more details). Thus,(3.11) can be solved by linear programming. One can also show (Boros et al., 1995; Chandrasekaran et al., 2005) that this problem is a dual of the maximum network ﬂow problem, so it can be solved in O(n3). However, linear program with an eﬃcient solver was found to be a faster way to solve (3.11) on real datasets. 3.2.2 Reduction of the Problem Size We provide in this section a general method for reduction of the size of the monotone approxima-tion problem. A reduction method has been proposed for binary classiﬁcation in (Chandrasekaran et al., 2005). We proposed a reduction method for 0-1 loss in multi-class problem in (Dembczyński et al., 2006a). Here we show how to reduce the problem (3.11) in the most general case. Lower and upper labels. For each xi, let us deﬁne the lower and upper class labels, respectively as: li = min{yj : xj ⪰xi, j = 1, . . . , n} ui = max{yj : xj ⪯xi, j = 1, . . . , n}. (3.12) Lower and upper labels are the indicators of the inconsistencies in the dataset, as the following, simple lemma states: Nonparametric Methods 39 Lemma 3.2. Let li and ui be the lower and upper class labels deﬁned in (3.12). Then, we have: 1. li ⩽yi ⩽ui. 2. ui = li if and only if xi is consistent. 3. For each xi ⪰xj we have li ⩾lj and ui ⩾uj. Proof. We successively prove three parts of the theorem: 1. Since xi ⪰xi, yi is in the set over which the minimum and maximum is taken in (3.12). This immediately implies li ⩽yi ⩽ui. 2. If xi is consistent, then according to the deﬁnition under the equation (1.13), for every object xj ⪰xi it must hold yj ⩾yi. This implies that li ⩾yi and from the property 1 we have yi = li. Similarly, one can show that yi = ui. Assume li = ui. This means that yi = li, so for every object xj ⪰xi it must hold yj ⩾yi. From yi = ui we conclude that for each object xj ⪯xi it must hold yj ⩽yi. Thus, xi is consistent. 3. If xi ⪰xj, then {yt : xt ⪰xi, t = 1, . . . , n} ⊆{yt : xt ⪰xj, t = 1, . . . , n}. This implies li ⩾lj, since the minimum of the subset must be greater than the minimum of the whole set. Analogously, one can show that ui ⩾uj. We will now prove one more lemma which will be needed in the construction of the reduction method: Lemma 3.3. Suppose [lyk]K×K is a monotone loss matrix, as deﬁned in (2.4). Then, the loss function is strictly increasing, i.e.: lyk > ly,k+1 if k < y, ly,k−1 < lyk if k > y. (3.13) Proof. We will prove the ﬁrst inequality in (3.13); the second one can be proved analogously. From (2.4) we have that for k > y, ly,k+1 −lyk ⩾ly+1,k+1 −ly+1,k. Repeating this iteratively, we must ﬁnally get ly,k+1 −lyk ⩾ly+2,k+1 −ly+2,k ⩾. . . ⩾lk,k+1 −lkk > 0, where the last inequality comes from lkk = 0 and lyk > 0 for y ̸= k. Reduction procedure. The problem of monotone approximation, formulated in (3.11) has n × (K −1) variables. Removing any of those variables is desirable. Here we prove the theorem from which we know a priori the optimal values of some of the variables: Theorem 3.5. Let ˆ dik, i = 1, . . . , n, k = 2, . . . , K, be any optimal solution to the problem (3.11). Let ˆ di = 1 + PK k=2 ˆ dik. Then, we have li ⩽ˆ di ⩽ui. Proof. Since (3.11) is equivalent to (3.9), we will refer to the latter. Assume we have any optimal solution ˆ di, i = 1, . . . , n. Let I be a subset of those i for which ˆ di < li . Similarly, let J be a subset of those i for which ˆ di > ui. Let us introduce solution ˜ di such that ˜ di = li for i ∈I, ˜ di = ui for i ∈J and ˜ di = ˆ di otherwise. Since from Lemma 3.2 we know that li ⩽yi ⩽ui and from Lemma 3.3 we know that the loss matrix is strictly increasing, it follows that ˜ di has a lower objective value 40 3.2. Monotone Approximation that ˆ di, if any of the sets I or J is nonempty. Indeed, for every i ∈I and every i ∈J, the label ˜ di is surely “closer” than ˆ di to the real label yi. Therefore, it is enough to prove that the solution ˜ di is feasible. Then, I and J must be empty, because otherwise it would contradict the optimality of ˆ di. To prove the feasibility of ˜ di in the problem (3.9), we must show that: xi ⪰xj = ⇒˜ di ⩾˜ dj i, j = 1, . . . , n. (3.14) Notice that for i ∈I, ˜ di > ˆ di and for i ∈J, ˜ di < ˆ di Choose any xi ⪰xj. First we consider i ∈I, then i ∈J and ﬁnally the case i / ∈I ∪J: 1. Case i ∈I. Then if j ∈I, ˜ di = li ⩾lj = ˜ dj. If j / ∈J, ˜ di > ˆ di ⩾ˆ dj ⩾˜ dj. 2. Case i ∈J. Then ˜ di = ui ⩾uj ⩾˜ dj. 3. Case i / ∈I ∪J. Then if j ∈I, ˜ di ⩾li ⩾lj = ˜ dj. If j / ∈I, ˜ di = ˆ di ⩾ˆ dj ⩾˜ dj. Theorem (3.5) implies that we can remove consistent objects from the optimization process, since we know a priori that ˆ di = yi for such objects. Moreover, for other objects, we can ﬁx some of dik to be 0 or 1 (and hence remove them from the optimization process) so that we ensure that li ⩽ˆ di ⩽ui holds. This dramatically reduces the size of the problem: similarly to the isotonic regression, we have experienced from real data that in most cases more than 80 −90% of variables are removed. 3.2.3 Binary Monotone Approximation Let us consider the simplest problem of monotone approximation, when K = 2 and Y = {0, 1}. The loss function has the form: [lyk]K×K = 0 l01 l10 0 ! , and is always a monotone loss matrix. Let us denote α = l01 l01+l10 . Then, one can easily show that the Bayes classiﬁer h∗(x) has one of the following forms: h∗(x) = 1P (y=1\|x)⩾α h∗(x) = 1P (y=1\|x)>α, or can be any monotone function between those two. The monotone approximation problem (3.11) can be presented in the simpliﬁed form: since we have K = 2, we can omit the index k for the variables. Moreover, the objective function for a given i is l01di if yi = 0 or l10(1 −di) if yi = 1. It can be written more concisely as wyi\|yi −di\|, where: w0 = l01 l01 + l10 = α w1 = l10 l01 + l10 = 1 −α, (3.15) and such weights follow from dividing the loss function by l01 + l10. Then, we can write (3.11) as: minimize n X i=1 wyi\|yi −di\| subject to xi ⪰xj = ⇒di ⩾dj i, j = 1, . . . , n. (3.16) Nonparametric Methods 41 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 0 0 0 0 0 1 0−1 0−1 1 1 Figure 3.3: Binary-class monotone approximation with α = 1 2; the dataset is the same as in Figure 3.1. The new class label is shown below on the right of each object. The optimal solution is not unique; two objects have class labels 0−1, which means that they are assigned label 0 in the lowest optimal solution and label 1 in the greatest optimal solution. Notice that the relaxed constraint 0 ⩽di ⩽1 was dropped, because if there were any di ⩾1 (or di ⩽0) in any feasible solution, we could decrease their values down to 1 (or increase up to 0), obtaining a new feasible solution with smaller value of the objective function of (3.16). We transformed the problem into (3.16) to show that it closely resembles isotonic regression (3.4). In the isotonic regression problem, we minimize L2-norm (sum of squares) between vectors y = (y1, . . . , yn) and p = (p1, . . . , pn), while in (3.16) we minimize L1-norm (sum of absolute values). In fact, both problems are closely connected, which is shown by the following theorem: Theorem 3.6. Suppose ˆ p = (ˆ p1, . . . , ˆ pn) is the optimal solution to the problem of isotonic regression (3.4). Then, the solution ˆ d∗= ( ˆ d∗1, . . . , ˆ d∗n) given by ˆ d∗i = 1ˆ pi>α for each i = 1, . . . , n, and the solution ˆ d∗= ( ˆ d∗ 1, . . . , ˆ d∗ n) given by ˆ d∗ i = 1ˆ pi⩾α for each i = 1, . . . , n, are the optimal solutions to the problem of binary monotone approximation (3.16) with weights (3.15). Moreover, if ˆ d = ( ˆ d1, . . . , ˆ dn) is the optimal integer solution to the problem of binary monotone approximation, it must hold ˆ d∗i ⩽ˆ di ⩽ˆ d∗ i , for all i = 1, . . . , n. In particular, if ˆ d∗= ˆ d∗, then the solution to the binary monotone approximation problem is unique. The proof can be found in the Appendix. Let us call ˆ d∗the greatest, and ˆ d∗the smallest optimal solutions. Theorem 3.6 states that if the MLE estimator (isotonic regression) ˆ pi is greater (or smaller) than α, then the optimal value for the corresponding variable ˆ di in the binary monotone approximation problem (3.16) is 1 (or 0). The interest of this result lies in the fact that the functions 1ˆ pi⩾1 and 1ˆ pi>1 minimize the loss function on the training set D, while the functions 1P (y=1\|xi)⩾1 and 1P (y=1\|xi)>1 (Bayes classiﬁers) minimizes the expected loss (risk). Thus, the correspondence between probability estimates and the Bayes classiﬁer estimate on D is the same as the correspondence between real probabilities and the real Bayes classiﬁer. We will often write α-binary monotone approximation, if we want to stress that the weights have the form w0 = α and w1 = 1 −α. 42 3.2. Monotone Approximation Handling non-uniqueness of the solution. It follows from Theorem 3.6 that the monotone approximation may have non-unique solution. On the other hand, the isotonic regression is unique; moreover, if ˆ pi > 1 2 then ˆ di = 1, and if ˆ pi < 1 2 then ˆ di = 0. Therefore, the only non-unique variables in the monotone approximation are for those i, for which ˆ pi = 1 2. To investigate this issue in a greater detail, let us present a useful property of the isotonic regression. Suppose A is a subset of {1, . . . , n} and f = (f1, . . . , fn) is a real-valued vector. We deﬁne Av(f, A) = 1 \|A\| P i∈A fi to be the average value of f on set A. Now suppose ˆ p is the isotonic regression of y. By a level set of ˆ p, denoted by [ˆ p = a], we mean the subset of {1, . . . , n} on which ˆ p has constant value a, i.e. [ˆ p = a] = {i: ˆ pi = a}. The following theorem holds: Theorem 3.7. (Robertson et al., 1998) Suppose ˆ p is the isotonic regression of y. If a is any real number such that the level set [ˆ p = a] is not empty, then a = Av(y, [ˆ p = a]). Theorem 3.7 states that for a given xi, ˆ pi equals to the average of yj over all the objects xj having the same value ˆ pj = ˆ pi. Since there is a ﬁnite number of divisions of the dataset {x1, . . . , xn} into level sets, we conclude that there is a ﬁnite number of values that ˆ p can possibly take. In our case, since yi ∈{0, 1}, all the values ˆ pi must be of the form r r+s, where r is the number of objects from class y = 1 in the level set, while s is the number of objects from class y = 0. Using Theorem 3.7, we can construct the procedure of ﬁnding the greatest and the smallest optimal solutions in the binary case. First of all, notice that when α is not of the form r r+s for some integers r, s ⩽n, then the binary monotone approximation is unique (there will be no ˆ pi = α). On the other hand, if α is of the form r r+s, let us increase α by suﬃciently small ϵ, such that α+ϵ is not of the form r r+s and there is no other number γ = r r+s for some r, s ⩽n such that α < γ < α + ϵ. Then, the solution to the (α + ϵ)-binary monotone approximation is unique and is the same as the greatest solution to the α-binary monotone approximation, ˆ d∗, because there is no ˆ pi such that α < ˆ pi ⩽α + ϵ. One can show that ϵ ⩽n−2 is suﬃcient. Similarly, decreasing α by ϵ will lead us to the smallest solution ˆ d∗. Thus, we have proved the following theorem. Theorem 3.8. If α is not of the form r r+s for some r, s ⩽n, the α-binary monotone approximation is unique. Otherwise, the greatest α-binary monotone approximation ˆ d∗can be found by increasing the value of α by ϵ ⩽n−2 and solving the problem (3.16). Similarly, the smallest α-binary monotone approximation ˆ d∗can be found by decreasing the value of α by ϵ ⩽n−2 and solving again the problem (3.16). Summarizing, we see that ﬁnding the greatest and the lowest optimal solutions corresponds to solving the monotone approximation at most twice, with α slightly perturbed by ±ϵ, ϵ ⩽n−2. Example. A simple example of binary monotone approximation can be found in Figure 3.3. Comparison with Figure 3.1 shows the relationships between new labels and probability estimates, as stated in Theorem 3.6. 3.2.4 Linear Monotone Approximation Monotone approximation with linear loss. Now we will investigate the problem of monotone approximation (3.11) with the extended linear loss matrix (2.9). We have already shown in Theorem 3.6 that there exists a correspondence between binary isotonic regression and binary monotone approximation. In the forthcoming theorem we will show that an analogous correspondence between multiple isotonic regression and monotone approximation with linear loss takes place. We will also show that both monotone approximation with extended linear loss and with the ordinary linear loss (2.8) lead to the same solution, similarly as it is on the population level (c.f. Theorem 2.6). Nonparametric Methods 43 Theorem 3.9. Let ˆ qk be the isotonic regression of yk, k = 2, . . . , K, as deﬁned in (3.5). Then, the solutions ˆ d∗= ( ˆ d∗1, . . . , ˆ d∗n) and ˆ d∗= ( ˆ d∗ 1, . . . , ˆ d∗ n) deﬁned as: ˆ d∗i = 1 + PK k=2 1ˆ qik>α, ˆ d∗ i = 1 + PK k=2 1ˆ qik⩾α, (3.17) are the optimal solutions to the monotone approximation problem with extended linear-loss (2.9). Moreover, every other optimal solution ˆ d = ( ˆ d1, . . . , ˆ dn) satisﬁes ˆ d∗i ⩽ˆ di ⩽ˆ d∗ i , for each i = 1, . . . , n. Proof. Let us transform the objective function of the monotone approximation (3.11) for a given i, denoted by Li: Li = K X k=2 (lyi,k −lyi,k−1)dik = = K X k=2 −(1 −α)yik(s(k) −s(k −1))dik + α(1 −yik)(s(k) −s(k −1))dik = K X k=2 (s(k) −s(k −1)) −(1 −α)yikdik + α(1 −yik)dik , where yik = 1yi⩾k, as usual. By adding the constant value (which does not change the optimization process) PK k=2(s(k) −s(k −1))yik(1 −α) we obtain: Li = K X k=2 (s(k) −s(k −1))(1 −α)yik(1 −dik) + α(1 −yik)dik = K X k=2 (s(k) −s(k −1))wyik\|yik −dik\|, where w0 = α and w1 = 1 −α. Thus, the total loss has the form: K X k=2 (s(k) −s(k −1)) n X i=1 wyik\|yik −dik\| ! . (3.18) For each k, the loss function looks exactly like the loss in the binary monotone approximation (3.16) (except the term s(k)−s(k−1)), where yik now plays the role of the binary class label. Unfortunately, those K −1 binary problems are not independent due to constraint dik ⩾di,k+1 in (3.11), which involves the variables for diﬀerent k. However, we will show that constraint dik ⩾di,k+1 is not needed for linear loss and can be removed. Indeed, we will show that the greatest optimal solutions to the K −1 binary problems solved independently still satisfy the constraint dik ⩾di,k+1. Consider the optimal solution to the binary problems for some k and k + 1. It follows that yik ⩾yi,k+1 for each i (because 1yi⩾k ⩾1yi⩾k+1). Then, according to Lemma 3.1, the isotonic regression of yk = (y1k, . . . , ynk), denoted by ˆ qik, is greater than or equal to the solution to the isotonic regression of yk+1, denoted by ˆ qi,k+1. But then, from Theorem 3.6 it follows that the greatest optimal solution to the binary monotone approximation problem with yik, ˆ d∗ ik = 1ˆ qik⩾α is greater than or equal to the greatest optimal solution to the binary monotone approximation problem with yi,k+1, ˆ d∗ i,k+1 = 1ˆ qi,k+1⩾α. Thus, the constraint dik ⩾di,k+1 is satisﬁed for the greatest optimal solution. One can similarly show the same for the smallest solution. What we have proved above is that the solutions (3.17) composed of the greatest and the smallest optimal solutions to K −1 binary problems are feasible solutions of the original problem (i.e. linear monotone approximation (3.16)); but since the minimum of a more constrained problem cannot decrease, they are also the greatest and the smallest optimal solutions to the original problem. 44 3.2. Monotone Approximation There are several important conclusions following from Theorem 3.9. First of all, the problem of monotone approximation (3.11) for extended linear loss can be solved by solving a sequence of K −1 simple weighted binary problems. Although it seems that we now have K −1 problem instead of one problem, from the computational point of view it is a great gain, because we reduced the problem with (K −1) × n variables to K −1 subproblems with n variables each, i.e. we decomposed a big problem into a sequence of smaller ones. This decomposition will be especially useful in Chapter 5, for dealing with rule ensemble models. Moreover, a closer look at (3.17) reveals an interesting fact: for each i = 1, . . . , n, every ˆ di such that ˆ d∗i ⩽ˆ di ⩽ˆ d∗ i , is the (1 −α)-quantile of the probability distribution {ˆ pi1, . . . , ˆ piK}, obtained in (3.6) from the multiple isotonic regression. This corresponds exactly to the relationship between real probability distribution {pi1, . . . , piK} and the Bayes classiﬁer h∗(xi). In other words, the relationship between the estimates on D is the same as the relationship between corresponding quantities on the population level. The theorem also shows that there is no point in using sophisticated extended linear loss, because for every scale function s(k), the solutions are identical. At ﬁrst look, this seems to be quite counter-intuitive, since in medicine we could position class 1 (“ill”) far away from class 2 (“healthy”) and class 3 (“very healthy”), e.g. s(1) = 0, s(2) = 0.8, s(3) = 1. This would be done due to the fact that we do not care so much about the condition of the patient provided her or she is healthy. However, as the above analysis shows, this is pointless: we would obtain exactly the same results for e.g. s(1) = 0, s(2) = 0.01 and s(3) = 1. This shows the unusual property of the linear loss of being independent of a particular scale. Finally, notice that similarly to the case of binary monotone approximation, we can give a simple procedure for ﬁnding the greatest and the smallest solutions. If α is not of the form r r+s for some r, s ⩽n, then the linear monotone approximation is unique. Otherwise, we can increase α by ϵ ⩽n−2 and solve the K −1 binary problems (3.16) to obtain the greatest linear monotone approximation. Similarly, we obtain the smallest linear monotone approximation if we decrease α by ϵ. Example. Consider the example shown in Figure 3.4, illustrating how three-class problem is transformed to two binary problems. For simplicity we assume that α = 1 2, i.e. the loss function is an ordinary (symmetric) absolute error. Notice that for any object, the new class label in the top ﬁgure can be obtained from ˆ di = 1 + PK k=2 ˆ dik, i.e. by summing up new class labels on middle and lower ﬁgures and adding 1. Moreover, new class labels on the middle ﬁgure are always greater or equal then those on the lower ﬁgure. This is exactly the conclusion of Theorem 3.9. Remark. The monotone approximation with linear loss plays an important role in the subsequent chapters, therefore from now on we will often omit the term “linear loss” and call it simply monotone approximation or monotone α-approximation, if we want to stress that we are using particular value of asymmetry constant in (2.8). 3.2.5 Extension Beyond the Training Set and Asymptotic Consistency Extensions of the monotone approximation. The monotone approximation (with linear loss) ˆ d is deﬁned only at training points xi, i = 1, . . . , n. Since we are dealing with a class of all monotone functions, we can extend monotone approximation to X by using any monotone function d: X →Y , such that d(xi) = ˆ di, for each i = 1, . . . , n. Similarly as in Section 3.1.4, we deﬁne the following Nonparametric Methods 45 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 1 1−2 1−2 2 1 2 2−3 2−3 3 3 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 0 0−1 0−1 1 0 1 1 1 1 1 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 0 0 0 0 0 0 0−1 0−1 1 1 Figure 3.4: Monotone approximation with linear loss for α = 1 2. In the top ﬁgure the 3-class dataset is shown with new labels assigned; in the middle and bottom ﬁgures are shown the datasets used in 2 binary subproblems (with new labels). Label e.g. 2 −3 means that the object is assigned label 2 in the lowest optimal solution and label 3 in the greatest optimal solution. 46 3.2. Monotone Approximation minimal and maximal extensions: dmin(x) = max{ ˆ d∗i : xi ⪯x}, dmax(x) = min{ ˆ d∗ i : xi ⪰x}, (3.19) We will now prove that every valid extension is capped between dmin(x) and dmax(x): Theorem 3.10. Let d: X →Y be monotone and let there exists an optimal solution to the mono-tone approximation ˆ d such that d(xi) = ˆ di, for each i = 1, . . . , n (i.e. d(x) is a valid extension of the monotone approximation). Then, for each x ∈X, dmin(x) ⩽d(x) ⩽dmax(x). Proof. We know that for each i, ˆ d∗i ⩽ˆ di ⩽ˆ d∗ i . Choose any x ∈X. Then, since d(x) is monotone, we must have that if xi ⪯x then d(x) ⩾d(xi) = ˆ di. But it means that d(x) ⩾max{ ˆ di : xi ⪯x} ⩾ max{ ˆ d∗i : xi ⪯x} = dmin(x). Analogously, we can show that d(x) ⩽dmax(x). Unfortunately, the converse of Theorem (3.10) is not true: there may exist a monotone function d: X →Y such that for each x ∈X, dmin(x) ⩽d(x) ⩽dmax(x), but d(x) is not a valid extension of monotone approximation. Strong consistency of monotone approximation. We will consider the problem of consistency of monotone approximation with linear loss. Let h: X →Y be a classiﬁer. Let us denote Rn(h) the risk of h when it is trained on the dataset D = {(x1, y1), . . . , (xn, yn)}. We say that h is weakly consistent (Devroye et al., 1996) if: E[Rn(h)] n→∞ − →R∗, where the expectation is taken over the random training sets D of particular size n. In other words, as the size of the training set increases, the risk of classiﬁer averaged over the random choice of training data approaches the Bayes risk, i.e. h approaches the best possible classiﬁer h∗. We say that h is strongly consistent if: Rn(h) n→∞ − →R∗ holds with probability one, i.e. for almost every sequence of data (x1, y1), (x2, y2), . . ., h approaches the Bayes classiﬁer h∗. We will now prove that under mild assumption about the probability distribution, monotone approximation with linear loss is strongly consistent. To this end suppose X = V × Rm0, where V = V1×. . .×Vm−m0 is a ﬁnite set. This corresponds to a very general situation encountered in real-life problems, where some of the attributes V1, . . . , Vm−m0 are ordered and have ﬁnite domains while the rest of the attributes are continuous. Every vector x ∈X we can be divided into x = xV ⊕xR, where xV ∈V and xR ∈Rm0. Now we can state the following theorem: Theorem 3.11. Assume P(x, y) is monotonically constrained and let X = V × Rm0, where V is ﬁnite. Assume P(x) has density2. Let d(x) be any valid extension of the linear monotone approximation. The d(x) is strongly consistent, i.e. Rn(d) n→∞ − →R∗ with probability one. 2i.e. P(x) is absolutely continuous with respect to the product measure N m−m0 × λm0, where N is a counting measure and λ is a Lebesgue measure. Nonparametric Methods 47 Proof. We ﬁrst prove the theorem when Y = {0, 1}. Let A be a collection of subsets of X. Let NA(x1, . . . , xn) be the number of diﬀerent sets in: {{x1, . . . , xn} ∩A: A ∈A} (Devroye et al., 1996). Let us deﬁne the monotone layer as the set A such that if x ∈A then for every x′ ⪰x also x′ ∈A. Then, notice that the set of all possible monotone functions f : X →Y , denoted F, corresponds to the set of all monotone layers A in the sense that for any layer A there is a corresponding function f such that x ∈A if and only if f(x) = 1. Let x = xV ⊕xR, where xV ∈V and xR ∈Rm0. Let \|V \| be the cardinality of V . The number of all possible monotone layers cannot exceed 2\|V \| × N, where N is the number of all possible monotone layers on Rm0. Therefore, we can bound: NA(x1, . . . , xn) ⩽2\|V \|NA(xV 1, . . . , xV n), and similarly: E[NA(x1, . . . , xn)] ⩽2\|V \|E[NA(xR1, . . . , xRn)], where the expectation is taken over all possible sets {x1, . . . , xn} according to P(x). Notice that from the assumption, P(xR) has density. Devroye et al. (1996) showed (Theorem 13.13 and remark below Corollary 13.3) that for xR ∈Rm0 if P(xR) has density, then: E[NA(xR1, . . . , xRn)] = 2o(n). But this means that also E[NA(x1, . . . , xn)] = 2o(n). Then, we proceed analogously as in the proof of Corollary 13.3 in (Devroye et al., 1996), where the Vapnik-Chervonenkis inequality (Theorem 12.5) has been used. The main diﬀerence is that we have asymmetric costs of misclassiﬁcation: the linear loss for binary problem can be written as L(y, k) = α1k>y + (1 −α)1k<y. However Vapnik-Chervonenkis inequality can be slightly modiﬁed to handle the uneven costs, c.f. Lemma 29.1 in (Devroye et al., 1996). Thus, similarly as in (Devroye et al., 1996)) we can conclude that: Rn(d) n→∞ − →inf f∈F R(f) with probability one, where d(x) is the classiﬁer which minimizes empirical risk in the class of all monotone functions F. But d(x) can be every valid extension of the binary monotone approxi-mation, because every valid extension achieves the minimum of the empirical risk on the dataset. Moreover, since P(x, y) is monotonically constrained, from Corollary 2.4 we have that the Bayes classiﬁer h∗∈F and therefore the inﬁmum is achieved by h∗. Thus, we obtain: Rn(d) n→∞ − →R∗= R(h∗) with probability one. Now consider the general case Y = {1, . . . , K}. We will ﬁrst prove the theorem for dmin(x). Let yk = 1y⩾k and dk(x) = 1dmin(x)⩾k, for k = 2, . . . , K. If we denote the linear loss (2.8) by L(y, k), then it is easy to see that: L(y, dmin(x)) = K X k=2 L(yk, dk(x)). (3.20) For each k, consider the random variable yk = 1y⩾k and let P(x, yk) denote the distribution induced from P(x, y). Notice that P(x, yk) is monotonically constrained and has density. Moreover, h∗ k(x) = 1h∗(x)⩾k is the Bayes classiﬁer for P(x, yk) with linear loss. From Theorem 3.9 it follows that dk(xi) = 1ˆ qik⩾α for each xi. From Theorem 3.6 1ˆ qik, i = 1, . . . , n is a monotone approximation 48 Appendix for k-th binary problem (with vector yk), so dk(xi) is the extension of k-th monotone approximation. Therefore, we can apply the theorem for binary-class case and conclude that: Rn(dk) n→∞ − →R(h∗ k), (3.21) with probability one, i.e. for some set Ωk such that P(Ωk) = 1. This means that for Ω= TK k=2 Ωk (3.21) holds simultaneously for each k. But since P(Ω) = 1 and from (3.20) we obtain: Rn(dmin) n→∞ − →R∗ with probability one. Similar conclusion can be drawn for dmax. Let d: X →Y be any valid extension of monotone approximation. Then dmin(x) ⩽d(x) ⩽dmax(x) and we conclude that: Rn(d) n→∞ − →R∗ The distribution assumption in Theorem 3.11 is very weak and will be satisﬁed in almost every real problem of ordinal classiﬁcation with monotonicity constraints. Therefore, monotone approx-imation is almost universally consistent under the monotonicity assumptions. Unfortunately, the theorem tells nothing about rates of convergence. Those rates may in fact be very slow, especially when the dimensionality of the space m is high. Therefore, monotone approximation will not be used directly as a classiﬁcation method, except for the low-dimensional problems. Instead, it will be used as a preprocessing method for “monotonizing” the data. Then, the data will be passed to the proper learning method, which will train the monotone ensemble of weak classiﬁers. This issues will be discussed in Chapter 5. Appendix: Proofs of the Theorems Proof of Theorem 3.2 Before proving Theorem 3.2, we sketch the assumptions and the content of the theorem, which leads to the so called generalized isotonic regression. Details can be found in (Robertson et al., 1998). Suppose that Φ is a convex function ﬁnite on an interval I containing the values of the coordinates of y (i.e. for each i = 1, . . . , n, yi ∈I) and Φ has value +∞elsewhere. Let φ be a nondecreasing function on I such that for each u ∈I φ(u) is a subgradient of Φ, i.e. φ(u) is a number between the left derivative of Φ at u and the right derivative of Φ at u. For each u, v ∈I deﬁne the function ∆Φ(u, v) by: ∆Φ(u, v) = Φ(u) −Φ(v) −(u −v)φ(v). (3.22) Then the following theorem holds: Theorem 3.12. (Robertson et al., 1998) Let p∗be an isotonic regression of y i.e. p∗solves (3.4). Then it holds: n X i=1 ∆Φ(yi, pi)) ⩾ n X i=1 ∆Φ(yi, y∗ i ) + n X i=1 ∆Φ(y∗ i , pi) for any monotone vector p with the values of the coordinates in I, so that p∗minimizes n X i=1 ∆Φ(yi, pi) in the space of all monotone vectors p with values of the coordinates in I. The minimizing function is unique if Φ is strictly convex. Nonparametric Methods 49 u Φ Φ( (u) ) 0.0 0.2 0.4 0.6 0.8 1.0 −1.0 −0.8 −0.6 −0.4 −0.2 0.0 Figure 3.5: Function Φ(u) = u ln u + (1 −u) ln(1 −u). Theorem 3.12 states that for any convex function Φ satisfying the assumptions, the isotonic regression minimizes also the function ∆Φ. Thus, Theorem 3.12 can be used to show that the isotonic regression provides a solution for a wide variety of restricted estimation problems in which the objective function does not look at all like least squares (Robertson et al., 1998). Here, this property will be used to solve the MLE problem (3.3). Theorem 3.2. (Robertson et al., 1998) Let p∗be an isotonic regression of y. Then, p∗is also the optimal solution to the MLE problem (3.3). Proof. Let I = [0, 1] and deﬁne Φ to be (Robertson et al., 1998): Φ(u) = ( u ln u + (1 −u) ln(1 −u) for u ∈(0, 1) 0 for u ∈{0, 1} (see Fig. 3.5). One can show that Φ is indeed convex on I. The ﬁrst derivative φ is given by: φ(u) =      −∞ for u = 0 ln u −ln(1 −u) for u ∈(0, 1) +∞ for u = 1. Then, ∆Φ(u, v) for u, v ∈(0, 1) is given by: ∆Φ(u, v) = u ln u + (1 −u) ln(1 −u) −u ln v −(1 −u) ln(1 −v). (3.23) It is easy to check that ∆Φ(u, v) = 0 if u = v = 1 or u = v = 0, and that ∆Φ(u, v) = +∞for u = 0, v = 1 or u = 1, v = 0. Now, suppose that we want to minimize the function Pn i=1 ∆Φ(yi, pi) between all monotone vectors p with the coordinates in the range I = [0, 1]. Then, the ﬁrst two terms in (3.23) depend only on yi, so they can be removed from the objective function, thus leading to the problem of minimizing: − n X i=1 yi ln pi + (1 −yi) ln(1 −pi) between all monotone vectors p with coordinates in the range I, which is exactly the MLE problem (3.3). 50 Appendix Proof of Lemma 3.1 Lemma 3.1. Let ˆ p be the isotonic regression of the class indices vector y = (y1, . . . , yn). Suppose, we introduce a new vector of class indices y′ = (y′ 1, . . . , y′ n), such that y′ i ⩾yi for all i = 1, . . . , n. Then, ˆ p′, the isotonic regression of y′ has the following property: ˆ p′ i ⩾ˆ pi, for all i = 1, . . . , n. Proof. Assume the contrary, let ˆ p′ be the isotonic regression of y′, and suppose there exists i, such that ˆ p′ i < ˆ pi. Deﬁne two other solutions, ˆ p+ and ˆ p−in the following way: ˆ p+ i = max{ˆ pi, ˆ p′ i}, (3.24) ˆ p− i = min{ˆ pi, ˆ p′ i}. (3.25) Notice that ˆ p+ ̸= ˆ p′ and ˆ p−̸= ˆ p, since for some i, ˆ p′ i < ˆ pi. We show that ˆ p+, ˆ p−are feasible solutions, i.e. they satisfy constraints of (3.4). Suppose xi ⪰xj. Then, since ˆ p, ˆ p′ are feasible, it follows that ˆ pi ⩾ˆ pj and ˆ p′ i ⩾ˆ p′ j. But from deﬁnition of ˆ p+ i we have that ˆ p+ i ⩾ˆ pi and ˆ p+ i ⩾ˆ p′ i, so it also holds that ˆ p+ i ⩾ˆ pj and ˆ p+ i ⩾ˆ p′ j. Then, ˆ p+ i ⩾max{ˆ pj, ˆ p′ j} = ˆ p+ j . Similarly, from the deﬁnition of ˆ p− j we have that ˆ p− j ⩽ˆ pj and ˆ p− j ⩽ˆ p′ j, so it also holds that ˆ p− j ⩽ˆ pi and ˆ p− j ⩽ˆ p′ i. But then ˆ p− j ⩽min{ˆ pi, ˆ p′ i} = ˆ p− i . Thus, both ˆ p+, ˆ p−are feasible. Let us denote the objective function of (3.4) by F(y, p) = Pn i=1(yi −pi)2. Then, we have: F(y′, ˆ p+) −F(y′, ˆ p′) = n X i=1 ˆ p+2 i −ˆ p′2 i −2y′ iˆ p+ i −2y′ iˆ p′ i = = n X i=1 (ˆ p+ i −ˆ p′ i)(ˆ p+ i + ˆ p′ i) −2y′ i(ˆ p+ i −ˆ p′ i) . (3.26) Since from the deﬁnition (3.24) it holds that ˆ p+ i −ˆ p′ i ⩾0 and from the assumption of the theorem it holds y′ i ⩾yi, we have: n X i=1 2y′ i(ˆ p+ i −ˆ p′ i) ⩾ n X i=1 2yi(ˆ p+ i −ˆ p′ i), (3.27) so that: F(y′, ˆ p+) −F(y′, ˆ p′) ⩽ n X i=1 (ˆ p+ i −ˆ p′ i)(ˆ p+ i + ˆ p′ i) −2yi(ˆ p+ i −ˆ p′ i) . (3.28) Moreover, from (3.24)-(3.25) it holds that ˆ p+ i + ˆ p− i = ˆ p′ i + ˆ pi, so that: ˆ p+ i −ˆ p′ i = ˆ pi −ˆ p− i , (3.29) and by adding 2ˆ p′ i to both sides of (3.29): ˆ p+ i + ˆ p′ i = 2(ˆ p′ i −ˆ p− i ) + (ˆ pi + ˆ p− i ). (3.30) Nonparametric Methods 51 u 1u> >α α 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 u 1u≥ ≥α α 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Figure 3.6: Functions 1u>α and 1u⩾α used in the Theorem 3.6, shown for value α = 1 2. They diﬀer only in point u = α. Putting (3.29)-(3.30) into (3.28), we ﬁnally obtain: F(y′, ˆ p+) −F(y′, ˆ p′) ⩽ n X i=1 (2(ˆ p′ i −ˆ p− i ) + (ˆ pi + ˆ p− i ))(ˆ pi −ˆ p− i ) −2yi(ˆ pi −ˆ p− i ) = n X i=1 2(ˆ pi −ˆ p− i )(ˆ p′ i −ˆ p− i ) + (ˆ pi −ˆ p− i )(ˆ pi + ˆ p− i ) −2yi(ˆ pi −ˆ p− i ) = n X i=1 2(ˆ pi −ˆ p− i )(ˆ p′ i −ˆ p− i ) + ˆ p2 i −2yiˆ pi −ˆ p−2 i + 2yiˆ p− i = n X i=1 2(ˆ pi −ˆ p− i )(ˆ p′ i −ˆ p− i ) + F(y, ˆ p) −F(y, ˆ p−) < n X i=1 2(ˆ pi −ˆ p− i )(ˆ p′ i −ˆ p− i ), (3.31) where the last inequality comes from the assumption that ˆ p is the isotonic regression of y and ˆ p ̸= ˆ p−. In the last sum, however, for each i, either ˆ pi = ˆ p− i or ˆ p′ i = ˆ p− i , so the sum vanishes. Thus, we have: F(y′, ˆ p+) −F(y′, ˆ p′) < 0, (3.32) which is a contradiction, because as ˆ p′ is the isotonic regression of y′, it is the unique optimal solution for class indices y′. Proof of Theorem 3.6 Theorem 3.6. Suppose ˆ p = (ˆ p1, . . . , ˆ pn) is the optimal solution to the problem of isotonic regression (3.4). Choose some value α ∈[0, 1]. Then the solution ˆ d∗= ( ˆ d∗1, . . . , ˆ d∗n) given by ˆ d∗i = 1ˆ pi>α for each i = 1, . . . , n and the solution ˆ d∗= ( ˆ d∗ 1, . . . , ˆ d∗ n) given by ˆ d∗ i = 1ˆ pi⩾α for each i = 1, . . . , n (see Figure 3.6) are the optimal solutions to the problem of binary monotone approximation (3.16) with weights (3.15). Moreover, if ˆ d = ( ˆ d1, . . . , ˆ dn) is the optimal integer solution to the problem of binary monotone approximation, it must hold ˆ d∗i ⩽ˆ di ⩽ˆ d∗ i , for all i = 1, . . . , n. In particular, if ˆ d∗= ˆ d∗, then the solution to the binary monotone approximation problem is unique. Proof. Let us deﬁne a function Φ(u) on the interval I = [0, 1] in the following way (see Figure 3.7): Φ(u) = ( α(u −α) for u ⩾α, (1 −α)(α −u) for u < α. (3.33) 52 Appendix u Φ Φ( (u) ) 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.1 0.2 0.3 0.4 0.5 Figure 3.7: Function Φ(u)) deﬁned in (3.33) for α = 0.6. It is easy to check that Φ(u) is a convex function, but not a strictly convex function. Φ has derivative φ(u) = α −1 for u ∈[0, α), and φ(u) = α for u ∈(α, 1]. At point u = α, Φ(u) is not diﬀerentiable, but each value in the range [α −1, α] is a subgradient of Φ(u). First, suppose we set φ(α) = α −1. We remind from (3.22) that: ∆Φ(u, v) = Φ(u) −Φ(v) −(u −v)φ(v). Now, assume u ∈{0, 1}. To calculate ∆Φ(u, v), we need to consider four cases, depending what are the values of u and v: 1. u = 0, v > α; then Φ(u) = α(1 −α), Φ(v) = α(v −α), φ(v) = α, so that ∆Φ(u, v) = α. 2. u = 0, v ⩽α; then Φ(u) = α(1−α), Φ(v) = (1−α)(α−v), φ(v) = α−1, so that ∆Φ(u, v) = 0. 3. u = 1, v > α; then Φ(u) = α(1 −α), Φ(v) = α(v −α), φ(v) = α, so that ∆Φ(u, v) = 0. 4. u = 1, v ⩽α; then Φ(u) = α(1 −α), Φ(v) = (1 −α)(α −v), φ(v) = α −1 so that ∆Φ(u, v) = 1 −α. However, we can comprehensively write those results as: ∆Φ(u, v) = wu\|1v>α −u\|, for u ∈{0, 1}, where wu are given by (3.15). Thus, according to Theorem 3.12, ˆ p is the optimal solution to the problem: minimize n X i=1 wyi\|1pi>α −yi\| subject to xi ⪰xj = ⇒pi ⩾pj i, j = 1, . . . , n. (3.34) Notice that ˆ d∗is also the optimal solution to the problem (3.34), because 1u>α is nondecreasing on u, so if ˆ p satisﬁes constraints of (3.34), then so does ˆ d∗. Moreover, 11u>α = 1u>α, so the value of the objective function in (3.34) is the same for both ˆ p and ˆ d∗. But ˆ d∗is integer and, for integer solutions, problems (3.34) and (3.16) are the same, so ˆ d∗is the solution to the problem (3.16) with the lowest objective value among all the integer solutions to this problem. But, from the analysis of the unimodularity of constraints matrix of (3.16) we know that if ˆ d∗is the solution to (3.16) with the lowest objective value among the integer solutions, it is also the optimal solution, since there exists an optimal solution to (3.16), which is integer. Nonparametric Methods 53 Now, setting φ(α) = α, we repeat the above analysis, which leads to the function 1u⩾α instead of 1u>α and shows that also ˆ d∗is the optimal solution to the problem (3.16). We now prove the second part of the theorem. Assume v ∈{0, 1} and ﬁx again φ(α) = α −1. To calculate ∆Φ(u, v), we consider again four cases, depending what are the values of u and v: 1. u > α, v = 0; then Φ(u) = α(u −α), Φ(v) = α(1 −α), φ(v) = α −1, so that ∆Φ(u, v) = u −α > 0. 2. u ⩾α, v = 1; then Φ(u) = α(u −α), Φ(v) = α(1 −α), φ(v) = α, so that ∆Φ(u, v) = 0. 3. u ⩽α, v = 0; then Φ(u) = (1−α)(α−u), Φ(v) = α(1−α), φ(v) = α−1, so that ∆Φ(u, v) = 0. 4. u < α, v = 1; then Φ(u) = (1 −α)(α −u), Φ(v) = α(1 −α), φ(v) = α, so that ∆Φ(u, v) = α −u > 0. From Theorem 3.12 it follows that: n X i=1 ∆Φ(yi, pi) ⩾ n X i=1 ∆Φ(yi, ˆ pi) + n X i=1 ∆Φ(ˆ pi, pi) (3.35) for any monotone vector p with coordinates in the range [0, 1]. Notice that if the last term in (3.35) is nonzero, then p cannot be optimal to the problem (3.34) (since then ˆ p has strictly lower cost than p). Suppose now that ˆ d is the optimal integer solution to the binary monotone approximation problem (3.16). But then it is also the solution to the problem (3.34) with the lowest objective value between all the integer solutions (since both problems are exactly the same for integer solutions). Since ˆ d∗is an optimal solution to the problem (3.34) and it is integer (so that there exists integer solution which is optimal), ˆ d is also an optimal solution to this problem. Then, however, the last term in (3.35) must be zero, so for each i = 1, . . . , n it must hold ∆Φ(ˆ pi, ˆ di) = 0 (since all those terms are nonnegative). As ˆ d is integer, it is clear from the above analysis of ∆Φ(u, v) for v being integer that it may only happen, if the following conditions hold: ˆ pi > α = ⇒ˆ di = 1, ˆ pi < α = ⇒ˆ di = 0, (3.36) for all i = 1, . . . , n. From the deﬁnitions of ˆ d∗and ˆ d∗it follows that for ˆ pi = α it holds that ˆ d∗i = 0 and ˆ d∗ i = 1, for ˆ pi > α it holds ˆ d∗i = ˆ d∗ i = 1 and for ˆ pi < α it holds ˆ d∗i = ˆ d∗ i = 0. From this and from (3.36) we conclude that: ˆ d∗i ⩽ˆ di ⩽ˆ d∗ i , for all i = 1, . . . , n, for any optimal integer solution ˆ d to the problem (3.16). 54 Chapter 4 Stochastic Dominance-based Rough Set Approach In this chapter, we explain the Dominance-based Rough Set Approach (DRSA) from the statis-tical point of view and introduce the extension of DRSA, based on the probabilistic model ordinal classiﬁcation with monotonicity constraints. We start with the overview of classical rough set the-ory and DRSA. Then we introduce a statistical framework and method for probability estimation, which leads to the stochastic extension of DRSA (Dembczyński et al., 2007b; Kotłowski and Słow-iński, 2008). Finally, we show the equivalence of the extension to the monotone conﬁdence interval estimation. This Chapter heavily relies on the results from Chapter 3. It should be useful to the researchers working on the rough set theory; some of the results apply also to the “classical” rough set approach, based on the indiscernibility relation. 4.1 Dominance-based Rough Set Approach (DRSA) The rough sets approach, proposed by Pawlak (1982), has been used to deal with the classiﬁ-cation problems. It is not able, however, to handle inconsistencies coming from consideration of order relation and monotone relationships in the data, therefore it cannot be applied to the or-dinal classiﬁcation with monotonicity constraints. That is why Greco, Matarazzo, and Słowiński (1999a,b, 2001a) have proposed a new rough set approach that is able to deal with such kind of inconsistencies, called Dominance-based Rough Set Approach (DRSA). The origins of DRSA can be, however, trace back to the papers of Słowiński (1993, 1994); Pawlak and Słowiński (1994). Extensive surveys of DRSA can be found in (Greco et al., 2001a, 2004b,c; Słowiński et al., 2005). More information about the axiomatization of the DRSA can be found in (Słowiński et al., 2002a; Greco et al., 2004a). Before describing DRSA, we overview the classical rough set approach. 4.1.1 Classical Rough Set Theory Indiscernibility relation. The classical rough set approach (Pawlak, 1982, 1991; Pawlak et al., 1995; Pawlak, 2002; Pawlak and Skowron, 2007) neither takes into account monotonicity constraints nor classes and attributes are ordered. It is based on the assumption that objects having the same description (values on attributes) are indiscernible (similar) with respect to the available information. The indiscernibility relation thus generated induces a partition of the universe into 56 4.1. Dominance-based Rough Set Approach (DRSA) object a1 a2 y object a1 a2 y object a1 a2 y x1 1 1 1 x9 2 2 1 x17 3 2 2 x2 1 1 1 x10 2 2 1 x18 1 3 1 x3 2 1 1 x11 2 2 1 x19 1 3 3 x4 2 1 3 x12 2 2 1 x20 2 3 3 x5 2 1 2 x13 2 2 3 x21 2 3 3 x6 3 1 2 x14 2 2 3 x22 2 3 1 x7 3 1 2 x15 3 2 2 x23 3 3 3 x8 1 2 1 x16 3 2 2 x24 3 3 3 Table 4.1: Example: 24 objects described by 2 attributes a1, a2 and class label y ∈{1, 2, 3}. blocks of indiscernible objects, called granules. The indiscernibility relation I is deﬁned as: I = {(xi, xj): xil = xjl ∀i, j = 1, . . . , n, ∀; l = 1, . . . , m}, where xd is the value of object x on attribute d. The equivalence classes of I are called granules. The equivalence class for an object xi is denoted I(xi). Lower and upper approximations. Any subset of the universe may be expressed in terms of the granules either precisely (as a union of granules) or approximately only. In the latter case, the subset may be characterized by two ordinary sets, called lower and upper approximations. The subsets approximated in the classiﬁcation problems are classes Clk, k ∈Y , deﬁned as the subsets of objects from the training set having class value k: Clk = {xi : yi = k}. The lower and upper approximations of class Clk are deﬁned, respectively, as: Clk = {xi : I(xi) ⊆Clk, i = 1, . . . , n}, Clk = {xi : I(xi) ∩Clk ̸= ∅, i = 1, . . . , n}. It always holds that: Clk ⊆Clk ⊆Clk. (4.1) Therefore, if an object xi belongs to Clk, it is certainly also an element of Clk, while if xi belongs to Clk, it may belong to class Clk. Variable precision. For application to the real-life data, some less restrictive deﬁnitions were introduced under the name of variable precision rough sets (VPRS) (Ziarko, 2001, 2005). The new deﬁnitions of approximations (where lower approximation is usually replaced by the term positive region) are expressed in the probabilistic terms in the following way. Let Pr(y = k\|I(xi)) be a probability that any object from granule I(xi) belongs to the class Clk. Notice that in this deﬁnition we assume the probabilities are the same for each object within the same granule. The probabilities are unknown, but are estimated by frequencies Pr(y = k\|I(xi)) = \|Clk ∩I(xi)\| \|I(xi)\| . Then, the lower approximation of class Clk is deﬁned as: Clk = {xi : Pr(y = k\|I(xi)) ⩾u, i = 1, . . . , n}, (4.2) Stochastic Dominance-based Rough Set Approach 57 a1 a2 G1 G2 G3 G4 G5 G6 G7 G8 G9 1 2 3 1 2 3 a1 a2 G1 G2 G3 G4 G5 G6 G7 G8 G9 1 2 3 1 2 3 Figure 4.1: Example of three-class problem. Black points are objects from class 1, dark gray points – from class 2, light gray – from class 3. The value sets of two attributes a1, a2 are {1, 2, 3}. On the upper picture lower approximations are presented as the shaded area: Cl1 = G1 ∪G4, Cl2 = G3 ∪G6, Cl3 = G9. On the lower picture, variable precision lower approximations are presented with precision threshold u = 2/3; we have Cl1 = G1 ∪G4 ∪G5, Cl2 = G3 ∪G6, Cl3 = G8 ∪G9 so it is the sum of all granules, for which the probability of class Clk is at least equal to some threshold u ⩾1 2. Similarly, the upper approximation of class Clk is deﬁned as: Clk = {xi : Pr(y = k\|I(xi)) ⩾l, i = 1, . . . , n}, where l ⩽ 1 2 is usually set to 1 −u for the complementarity reasons. An example of lower ap-proximations for three-class toy problem is shown in Table 4.1.1 and in Figure 4.1.1. Recently, the VPRS model has been investigated from the viewpoint of some desirable properties of monotonic-ity in (Błaszczyński et al., 2008), and some new deﬁnitions of this model, ensuring the desirable properties, were given. Notice that concepts of rough approximations are related to the training set only so they con-tribute to the description of the data and therefore are part of the knowledge discovery process. However rough sets can be used for prediction (classiﬁcation) purposes as well. Firstly, one can search for the minimal description of objects (set of attributes) which does not increase the incon-sistency of the data, called reduct; this process can be considered as feature selection. Moreover, one can use lower and upper approximations as a basis for induction of certain and possible rules, respectively (Stefanowski, 1998). 58 4.1. Dominance-based Rough Set Approach (DRSA) Variable precision as maximum likelihood estimation (MLE). It can be shown that fre-quencies used for estimating probabilities in are the MLE estimators under assumption of common class probability distribution within a given granule (notice that the assumption that each object in a given granule has the same probability distribution, can be thought of as a probabilistic version of principle of indiscernibility). Let us choose some granule G = I(x). Let nG be the number of objects in G, and for each class Clk, let nk G be the number of objects from this class in G. Then the class label y has a multinomial distribution when conditioned on granule G. Let us denote those probabilities Pr(y = k\|G) by pk G. Then the conditional probability of observing the n1 G, . . . nK G objects in G, given p1 G, . . . , pK G (conditional likelihood) is the following: L(p; G) = K Y k=1 (pk G)nk G so that the negative log-likelihood is: ℓ(p) = −ln L(p; G) = − K X k=1 nk G ln pk G (4.3) The minimization of ℓwith additional constraint PK k=1 pk G = 1 leads to the well-known fomula for MLE estimators ˆ pk G in multinomial distribution: ˆ pk G = nk G nG (4.4) which are exactly the frequencies used in VPRS. This observation will lead later to the stochastic generalization of DRSA. 4.1.2 Rough Set Theory for Ordinal Classiﬁcation The classical rough set approach is not able to handle inconsistencies coming from consideration of order relations and monotonicity constraints (Greco et al., 2000, 2001a, 2002b). Consider for example assigning companies into one of the bankruptcy risk classes, e.g. “low risk”, “medium risk” and “high risk”; the classes are clearly (preference) ordered. Let the set of attributes be: product quality, market share and debt ratio; all those attributes are monotonically correlated with the risk classes: as the product quality and market share increase, and debt ratio decreases, the companies are assigned to lower risk classes. Consider two ﬁrms A and B. If ﬁrm A has a low value while ﬁrm B has a high value of the debt ratio, and evaluations of these ﬁrms on other attributes are equal, then, from bankruptcy risk point of view, ﬁrm A is at least as good as ﬁrm B. Suppose, however, that ﬁrm A has been assigned to a class of higher risk than ﬁrm B. This is obviously inconsistent with the monotone structure of the problems. Nevertheless, in the classical rough set approach, the two ﬁrms will be considered as just discernible and no inconsistency will be stated (Greco et al., 2001a; Słowiński et al., 2002b; Greco et al., 2007a). For this reason, Greco, Matarazzo, and Słowiński (1999a,b, 2001a) have proposed a new rough set approach that is able to deal with this kind of inconsistencies. This innovation is based on substitution of the indiscernibility relation by a dominance relation in the rough approximations of classes, therefore the new theory has been named Dominance-based Rough Set Approach (DRSA). In the rough set theory, the term decision attribute is often used for the class attribute, the term decision value is used for a class label and the term condition attribute is used for the attributes other than the decision attribute. We remind about this fact, because the meaning of the names appearing in this chapter (e.g. condition and decision granules, generalized decision, etc.) becomes then clear. Stochastic Dominance-based Rough Set Approach 59 object a1 a2 y object a1 a2 y x1 0.1 0.12 1 x6 0.64 0.88 2 x2 0.2 0.54 3 x7 0.73 0.28 3 x3 0.28 0.73 1 x8 0.82 0.34 2 x4 0.41 0.45 2 x9 0.88 0.64 3 x5 0.5 0.22 1 x10 0.96 0.93 3 Table 4.2: A set of 10 objects described by 2 attributes a1, a2 and class label y ∈{1, 2, 3}. Dominance relation and dominance principle. The dominance relation ⪰is deﬁned as a binary relation on X in the following way (cf. Section 1.2): for any x, x′ ∈X we say that x dominates x′, x ⪰x′, if on every attribute, x has evaluation not worse than x, xj ⩾x′ j, for all j = 1, . . . , m. The dominance relation ⪰is a partial pre-order on X, i.e. it is reﬂexive and transitive. The dominance principle can be expressed as follows. For every xi, xj, where i, j = 1, . . . , n, it holds: xi ⪰xj = ⇒yi ⩾yj. (4.5) Notice that the dominance principle is related to the training set D only. The dominance principle follows from the monotone relationship between class labels and attribute values. However, in many real-life applications, the dominance principle is not satisﬁed, i.e. there exists at least one pair of objects violating (4.5). We say that an object xi is inconsistent if there exists another object xj, such that xi, xj violates (4.5). Otherwise, we say that object xi is consistent. We will sometimes also use the following expression: object xi is consistent with xj, if a pair xi, xj satisﬁes (4.5). Granules as dominance cones. The rough approximations concern granules resulting from information carried out by the class indices. The approximation is made using granules resulting from information carried out by (condition) attributes. These granules are called decision and condition granules, respectively. The decision granules can be expressed by unions of classes: Cl⩾ k = {xi : yi ⩾k, i = 1, . . . , n} Cl⩽ k = {xi : yi ⩽k, i = 1, . . . , n}. The condition granules are dominating and dominated sets deﬁned, respectively, as: D+(x) = {xi : xi ⪰x, i = 1, . . . , n} D−(x) = {xi : x ⪰xi, i = 1, . . . , n}. Remark that decision granules and condition granules are orthogonal cones in decision and condi-tions space, respectively. Lower and upper approximations. Lower approximations of Cl⩾ k and Cl⩽ k are deﬁned as follows: Cl⩾ k = {xi : D+(xi) ⊆Cl⩾ k , i = 1, . . . , n}, (4.6) Cl⩽ k = {xi : D−(xi) ⊆Cl⩽ k , i = 1, . . . , n}. (4.7) They include the objects which certainly belong to class Cl⩾ k (or Cl⩽ k ), i.e. without any ambiguity caused by inconsistency. Indeed, the certainty comes from the fact that object xi belongs to the 60 4.1. Dominance-based Rough Set Approach (DRSA) lower approximation of class union Cl⩾ k (respectively Cl⩽ k ), if no other object in the training set contradicts it, i.e. xi is consistent with every other object outside of Cl⩾ k (respectively Cl⩽ k ). Otherwise, if there exists an object outside Cl⩾ k , which dominates xi, then due to the dominance principle (following from the monotonicity constraints) we cannot say that xi belongs to Cl⩾ k with certainty. Upper approximations of Cl⩾ k and Cl⩽ k are deﬁned as: Cl ⩾ k = {xi : D−(xi) ∩Cl⩾ k ̸= ∅, i = 1, . . . , n}, Cl ⩽ k = {xi : D+(xi) ∩Cl⩽ k ̸= ∅, i = 1, . . . , n}. They include the objects which possibly belong to class Cl⩾ t (or Cl⩽ t ), i.e. with a possible ambiguity caused by inconsistency. Notice that for any k ∈Y , we have Cl⩾ k ∪Cl⩽ k−1 = {x1, . . . , xn}. This is not the case with the lower approximations. Therefore, we deﬁne the boundary (doubtful) region for class unions Cl⩾ k and Cl⩽ k−1 as: Bk = {x1, . . . , xn}(Cl⩾ k ∪Cl⩽ k−1), (4.8) which includes the area which does not belong to lower approximations of class unions Cl⩾ k and Cl⩽ k−1. Notice that DRSA decomposes the analysis of inconsistencies into K −1 binary problems: for each k = 2, . . . , K we have lower approximations Cl⩾ k , Cl⩽ k−1 and boundary Bk which together form the whole set {x1, . . . , xn}. Such a decomposition will be also used in the stochastic extension of DRSA. A simple example of training set is presented in Table 4.2 and lower approximations of class unions are shown in Figure 4.2. The quality of approximation is deﬁned as a ratio of the number of objects from the dataset that are consistent with respect to the dominance principle, to the number of all objects from the dataset: γ = 1 − SK k=2 Bk n . (4.9) This is a measure of inconsistency present in the dataset. 4.1.3 Generalized Decision in DRSA For the purpose of this paper, we will focus our attention on another concept from DRSA (as we shall shortly see, equivalent to the notion of approximations), the generalized decision. Suppose, object xi ∈Cl⩾ k . Since the lower approximation of class union Cl⩾ k is a region in which objects certainly belong to Cl⩾ k , we can state that class index of xi should be at least k. Choosing the greatest k, for which xi ∈Cl⩾ k holds (denoted by li), we know that the class index of xi must be at least li and we cannot give more precise statement, since we are not certain that the class index of xi is at least li + 1, because xi / ∈Cl⩾ li+1. On the other hand, if xi ∈Cl⩽ k , we known that the class index of xi must be at most k. Similarly, choosing the lowest k for which xi ∈Cl⩽ k (denoted by ui), we end up with the interval of classes [li, ui], for which we known that object xi must belong to. This interval is often denoted by δi, and is called generalized decision: δi = [li, ui], (4.10) where: li = max n k: xi ∈Cl⩾ k , k = 1, . . . , K o , ui = min n k: xi ∈Cl⩽ k , k = 1, . . . , K o . (4.11) Stochastic Dominance-based Rough Set Approach 61 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 Figure 4.2: Example of three-class problem described in Table 4.2. Black points are objects from class 1, dark gray points – from class 2, light gray – from class 3. On the upper picture lower approximations Cl⩽ 1 = {x1, x5} and Cl⩾ 2 = {x4, x6, x7, x8, x9, x10} are shown, on the lower picture – Cl⩽ 2 = {x1, x4, x5} and Cl⩾ 3 = {x9, x10}. Approximations Cl⩾ 1 and Cl⩽ 3 are not shown, since they are trivially {x1, . . . , xn}. The generalized decision determines an interval of decision classes to which an object may belong due to the inconsistencies with the dominance principle. Investigating the deﬁnitions of lower approximations (4.6)-(4.7) one can show that generalized decision can be easily computed without reference to the lower approximation: li = min{yj : xj ⪰xi, j = 1, . . . , n}, ui = max{yj : xi ⪰xj, j = 1, . . . , n}. (4.12) Thus, li is the lowest class, to which objects dominating xi belong; ui is the highest class, to which objects dominated by xi belong. Obviously, li ⩽yi ⩽ui and if li = ui, then object xi is consistent with respect to the dominance principle with every other object xj, for each i, j = 1, . . . , n. The wider the generalized decision, the less precise knowledge about the object we have. One can show (it follows directly from deﬁnition) that if xi ⪰xj, then li ⩾lj and ui ⩾uj, i.e. functions li and ui are monotone; in other words, if we replace the original class labels yi by li (or ui), then we will obtain consistent dataset. Notice that generalized decision is equivalent to lower and upper class labels (3.12) deﬁned in Section 3.2.2. Let us also remark that the description with generalized decisions is fully equivalent to the 62 4.1. Dominance-based Rough Set Approach (DRSA) a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 [1,1] [1,3] [1,3] [2,2] [1,1] [2,3] [2,3] [2,3] [3,3] [3,3] Figure 4.3: Generalized decisions δi = [li, ui] (shown in brackets). description with rough approximations. Namely, dominance-based lower approximations may be expressed using the generalized decision: Cl⩾ k = {xi : li ⩾k, i = 1, . . . , n}, Cl⩽ k = {xi : ui ⩽k, i = 1, . . . , n}. Generalized decisions are calculated in Figure 4.3 for the simple dataset from Table 4.2. 4.1.4 Variable Consistency in DRSA The deﬁnitions of lower approximations and generalized decisions for DRSA are quite restrictive: an object may be excluded from the lower approximation due to a single inconsistent object. Indeed, from the deﬁnition, object xi belong to lower approximation Cl⩾ k if D+(xi) ⊆Cl⩾ k . It is enough that there is one object xj ∈D+(xi) (so that xj ⪰xi) with yj < k to exclude xi from Cl⩾ k . Moreover, suppose there exist one object dominating all the other objects from dataset, but its class index is the lowest one. Then, all the objects from the dataset, apart from those belonging to the lowest class, will we excluded from any lower approximations (they all will fall into the boundary region). That is why relaxed deﬁnitions of lower approximations have been introduced under the name variable consistency DRSA (VC-DRSA) (Greco et al., 2001b), which allow object xi to be incor-porated into lower approximations, if a high fraction of objects dominating xi (or being dominated by xi) is consistent with xi. Thus, lower approximations (4.6)-(4.7) are redeﬁned as: Cl⩾ k =   xi : D+(xi) ∩Cl⩾ k \|D+(xi)\| ⩾l, i = 1, . . . , n   , (4.13) Cl⩽ k =   xi : D−(xi) ∩Cl⩽ k \|D−(xi)\| ⩾l, i = 1, . . . , n   , (4.14) The threshold l is called consistency level and is a parameter controlling the acceptable range of ambiguity. The upper approximations are deﬁned by complementarity: Cl ⩾ k = {x1, . . . , xn}\Cl⩽ k−1 Cl ⩽ k−1 = {x1, . . . , xn}\Cl⩾ k . Stochastic Dominance-based Rough Set Approach 63 There is, however, a problem with deﬁnitions (4.13)-(4.14), since they may lead to the non-monotone assignments to the lower approximations. Consider the dataset from Table 4.2 shown in Figure 4.2. If we set l = 4/5 then x6 ∈Cl⩽ 2 since \|D−(x6)∩Cl⩽ 2 \| \|D−(x6)\| = 5 6. On the other hand, x3 / ∈Cl⩽ 2 , because \|D−(x3)∩Cl⩽ 2 \| \|D−(x3)\| = 2 3; but it is counter-intuitive, since x3 ⪯x6 and, moreover, y3 < y6. This problem has been addressed in (Błaszczyński et al., 2007, 2008), where some modiﬁed deﬁnitions have been proposed which do not lead to non-monotone assignments. 4.2 Stochastic extension of DRSA (SDRSA) 4.2.1 DRSA as Most Informative Non-invasive Approach Let us start with proving an interesting fact about DRSA, using the concept of generalized decision (see Section 4.1.3). The proven fact will show that if our approach should not make any additional assumptions that cannot be backed and if it should not invade into the training set by changing or removing the objects (such an approach is called non-invasive (D¨ untsch and Gediga, 2000; Cao-Van, 2003)), the DRSA is the best we can do. In other words, if we want to extend DRSA, we will need to modify the dataset to some extent. Consider the family of class intervals {αi = [ai, bi]: i = 1, . . . , n}. Let us deﬁne the following relation between such families: the family of class intervals {αi = [ai, bi]: i = 1, . . . , n} is more informative than the family {βi = [ci, di]: i = 1, . . . , n} if for each i, αi ⊆βi. We show now that the generalized decision (thus, also DRSA rough approximations) is in fact the unique optimal non-invasive approach that holds the maximum amount of information which can be obtained from the data (Dembczyński et al., 2007a): Theorem 4.1. The family of generalized decisions {δi = [li, ui]: i = 1, . . . , n} is the most infor-mative family of class intervals, among families of class intervals of the form {αi = [ai, bi]: i = 1, . . . , n} with the following properties: 1. The sets {(xi, ai): i = 1, . . . , n} and {(xi, bi): i = 1, . . . , n}, composed of objects with, respec-tively, class labels ai and bi assigned instead of yi, are consistent with the dominance principle, i.e. if xi ⪰xj, then ai ⩾aj and bi ⩾bj (monotonicity). 2. For each i = 1, . . . , n, it holds ai ⩽yi ⩽bi (non-invasiveness). Proof. It was already noticed in Section 4.1.3 that condition 2 holds for the generalized decisions δi. Condition 1 also holds since if xi ⪰xj then {xr : xr ⪰xi} ⊆{xr : xr ⪰xj}, so li = min{yr : xr ⪰ xi} ⩾min{yr : xr ⪰xj} = lj. Analogously, one can show that ui ⩾uj. We now prove the minimality (or maximal informativeness). For any object xi and class in-terval αi = [ai, bi] satisfying the conditions 1-2, it must hold that ai ⩽li. This is because li = min{yr : xr ⪰xi}, so there exists object xj such that xj ⪰xi and li = yj. But due to conditions 1 and 2 it must hold that ai ⩽aj ⩽yj = li. Analogously, one can show that it must hold that bi ⩾ui. Thus, we conclude that for every family of class intervals {αi : i = 1, . . . , n} satisfying the conditions 1-2, it must hold that δi ⊆αi for each i = 1, . . . , n. Hence, it follows that {δi : i = 1, . . . , n} is the most informative family. Thus, if we want to obtain more informative monotone class intervals, we need to be invasive. This constitutes the motivation of stochastic DRSA. 64 4.2. Stochastic extension of DRSA (SDRSA) 4.2.2 Stochastic Dominance-based Rough Sets In this section, we extend the deﬁnitions of dominance-based lower approximations and gener-alized decision to the stochastic case. Probabilistic model. In Section 4.1.1, when we showed that VPRS comes from the MLE, we have made the assumption that in a single granule, each object has the same conditional class probability distribution. This is due to the property of indiscrenibility of objects within a granule. In case of DRSA, indiscernibility is replaced by a dominance relation, so that a diﬀerent relation between the probabilities must hold. However, we have already found in Section 2.1 a rudimentary property of the probability distribution for ordinal classiﬁcation with monotonicity constraints, the stochastic dominance principle: x ⪰x′ = ⇒P(y ⩾k\|x) ⩾P(y ⩾k\|x′) ∀x, x′ ∈X, k = 1, . . . , K. (4.15) In other words, if object xi dominates object xj, probability distribution conditioned at point xi stochastically dominates probability distribution conditioned at xj. This principle has the advantage of being a natural and intuitive extension of dominance principle (4.5) to the stochastic case. Stochastic lower approximations. Having stated the probabilistic model, we introduce the stochastic DRSA by relaxing the deﬁnitions of lower approximations of classes: Cl⩾ k = {xi : P(y ⩾k\|xi) ⩾α, i = 1, . . . , n}, (4.16) Cl⩽ k = {xi : P(y ⩽k\|xi) ⩾α, i = 1, . . . , n} = = {xi : P(y ⩾k + 1\|xi) ⩽1 −α, i = 1, . . . , n}, (4.17) where α ∈( 1 2, 1] is a ﬁxed threshold. Thus, lower approximation of class union, say Cl⩾ k , is a region, in which objects are assigned to Cl⩾ k with a high probability (at least α). The boundary region Bk = X(Cl⩾ k ∪Cl⩽ k−1) is the region in which objects belong to any of unions Cl⩾ k and Cl⩽ k−1 with probability in the range (1 −α, α). Two special cases are important. When α = 1, lower approximation corresponds to the certain region for a given class union and, as we shall shortly see, the stochastic deﬁnition boils down to the classical deﬁnition of the dominance-based lower approximation. When α becomes close to 1 2, only objects for which P(y ⩽k −1\|xi) = P(y ⩾ k\|xi) = 1 2 are in the boundary Bk, which corresponds to the Bayes boundary between classes (Duda et al., 2000). Notice that we excluded values α ⩽1 2, because otherwise the lower approximations of complementary class unions would overlap. Notice that it may happen for some object xi that although it does not belong to the class union Cl⩾ k , it belongs to Cl⩾ k (because its class probability satisﬁes Pr(y ⩾k\|xi) ⩾α). The interpretation of this fact is the following: although the class label of xi observed in the dataset is smaller than k, i.e. yi < k, such event is less likely than the event yi ⩾k; hence we should change its class union to the more probable one. Therefore, stochastic approximations lead to reassigning the objects. This is not a surprise, because from Theorem 4.1 it follows that in non-invasive way we cannot do better than DRSA. Stochastic decision. Having deﬁned the lower approximation, we can obtain generalized decision through the relations (4.11), similarly to the non-stochastic case (notice, however, that the formula (4.12) no longer holds in stochastic case): li = max{k: P(y ⩾k\|xi) ⩾α, k = 1, . . . , K} ui = min{k: P(y ⩾k + 1\|xi) ⩽1 −α, k = 1, . . . , K}. (4.18) Stochastic Dominance-based Rough Set Approach 65 To distinguish between non-stochastic and stochastic deﬁnitions, the class intervals deﬁned in (4.18) will be called stochastic decision. By carefully looking at (4.18) and reminding the deﬁnition of quantiles (see e.g. Section 3.2.4), one can notice that li is the greatest (1−α)-quantile and ui is the smallest α-quantile of the conditional class distribution P(y\|xi). Therefore, the stochastic decision [li, ui] is a conﬁdence class interval such that y ∈[li, ui] with probability at least 2α −1. Moreover, we can prove that two important properties of generalized decision also hold in the stochastic case: Theorem 4.2. Let δi be a stochastic decision, deﬁned by (4.18). Then, for every i, j = 1, . . . , n, it holds: 1. li ⩽ui, 2. xi ⪰xj →li ⩾lj and ui ⩾uj. Proof. From the deﬁnition of li, it follows that for each k such that P(y ⩾k) < α, it holds li < k. Therefore, it must hold li < min{k: P(y ⩾k) < α}. But since α ∈( 1 2, 1], we have α > 1 −α and therefore {k: P(y ⩾k) ⩽1 −α} ⊆{k: P(y ⩾k) < α}. This implies that li < min{k: P(y ⩾k) ⩽1 −α}, or equivalently li ⩽min{k: P(y ⩾k + 1) ⩽1 −α} = ui. This proves the ﬁrst property. The second property follows directly from the consideration of linear loss function in Section 3.2.4. The Bayes classiﬁer for such loss function is (1−α)-quantile, where α is the parameter which reﬂects the strength of asymmetry. Due to the fact that linear loss is a monotone loss matrix, it follows from Theorem 2.2 that Bayes classiﬁer (i.e. (1 −α)-quantile function) is monotone under the stochastic dominance assumption for any value α ∈[0, 1]. This proves the second property. Notice that we do not longer have li ⩽yi ⩽ui, because it only holds with probability 2α −1. The above relation holds with certainty only for α = 1. We will shortly see that this case correspond to the non-stochastic DRSA. This is actually in the spirit of Theorem 4.1, where we showed that if both li ⩽yi ⩽ui and monotonicity hold, the best solution is DRSA. Moreover, notice that as α increases, the stochastic decisions broaden (so they become less informative), but on the other hand the probability of catching the observed class value in the class interval increases. The stochastic decision can be interpreted in the following way: although the original label of the object xi is yi, the object is associated with class interval [li, ui], which contains the most probable class labels. In a special case, when li = ui ̸= yi, we may say that object xi is reassigned from class yi to li = ui. This resembles the monotone approximation problem and, as we will shortly see, the resemblance is not accidental. 4.2.3 Probability Estimation The stochastic lower approximations and stochastic decision are both deﬁned with respect to the probabilities. However, the probability distribution is unknown. Therefore, we need to estimate the probabilities from data. Then, to obtain the lower approximations, we will plug the estimators into (4.17)-(4.16) instead of real probabilities. In Section 4.3.2, we will show how we can omit the step of probability estimation and directly obtain stochastic lower approximations. To estimate the probabilities, we will use the method described in detail in Section 3.1, which we remind brieﬂy here. This method is equivalent to the MLE for the binary class case. For the general K −1-class case, it is based on solving K −1 problems of isotonic regression, to obtain in the k-th problem the estimators of probabilities P(y ⩾k\|xi). For a given xi, let us deﬁne K −1 dummy variables yik = 1yi⩾k for k = 2, . . . , K. In the k-th binary problem, dummy variables yik play the role of class labels with Y = {0, 1}, while variables of 66 4.2. Stochastic extension of DRSA (SDRSA) the problem correspond to estimating the probabilities P(y ⩾k\|xi). Let ˆ P(y ⩾k\|xi), the estimator of corresponding probability, be deﬁned as the optimal solution to the following isotonic regression problem: minimize n X i=1 (yik −pi)2 subject to xi ⪰xj = ⇒pi ⩾pj i, j = 1, . . . , n. (4.19) This is the least squares estimation with the monotonicity constraints, which ensures that the probability estimators satisfy the stochastic dominance principle. In Section 3.1, we showed that although the K −1 problems are solved separately, it holds ˆ P(y ⩾k\|xi) ⩾ˆ P(y ⩾k + 1\|xi), i.e. the estimators form a proper probability distribution. Example. Consider the problem shown in Figure 4.4. On the top chart, the probability estima-tors ˆ P(y ⩾2\|xi), ˆ P(y ⩾3\|xi) are shown (notice that ˆ P(y ⩾1\|xi ≡1, so it is never shown). Plugging those estimators into the deﬁnitions (4.16)-(4.16) and setting α = 0.6, we obtain the lower approximations and stochastic decisions shown in the middle and lower pictures. It is instructive to compare this with Figures 4.2 and 4.3 to see, what has changed comparing with non-stochastic case. First, notice that the lower approximation Cl⩽ 2 expanded in stochastic case, by including the objects x2, x3 and x6, which were on the boundary in the non-stochastic case. The reason behind this phenomenon is the following: although x2 belongs to class Cl3, it is dominated by objects x3 and x6, which belongs to lower classes. Consider estimating the probability P(y ⩾3\|xi) in the second binary problem, when class Cl3 forms “positive” class y = 1 and classes Cl1 and Cl2 form negative class y = 0. Then, the object x2 is in minority (one against two objects from “negative” class); therefore, object x2 will get relatively low probability ( 1 3) of belonging to class Cl3 and hence for threshold α = 0.6 (rather unrestrictive) it can be incorporated into lower approximation Cl⩽ 2 . Notice that for α > 2 3, it would not be incorporated into Cl⩽ 2 and would stay on the boundary. Statistical explanation of DRSA. Consider the k-th binary problem of probability estimation. According to Theorem 3.1, if object xi is consistent in the k-th binary problem, i.e. there exists no other object such that xi ⪰xj and yik < yjk, or such that xi ⪯xj and yik > yjk, then the probability estimate ˆ P(y ⩾k\|xi) will be equal to 1. Moreover, only such objects possess the property ˆ P(y ⩾k\|xi) = 1. This implies that if we set α = 1, the requirement ˆ P(y ⩾k\|xi) ⩾α = 1 in the deﬁnition of stochastic lower approximations (4.16)-(4.17) will be satisﬁed only by objects consistent in the k-th binary problem. On the other hand, only such objects are incorporated to the non-stochastic lower approxima-tions (4.7)-(4.6), either Cl⩽ k−1 or Cl⩾ k . Indeed, one can simply show that condition D+(xi) ⊆Cl⩾ k is equivalent to xj ⪰xi →yjk = 1, while condition D−(xi) ⊆Cl⩽ k to xj ⪯xi →yjk = 0. Thus, we conclude the stochastic approximations boil down to the non-stochastic ones in the limit α = 1. From the above analysis, we have two statistical interpretations of the non-stochastic DRSA. From the point of view of lower approximations, DRSA estimates subsets of the training set D, for which either P(y ⩾k\|xi) = 1 or P(y ⩽k −1\|xi) = 1; those are the objects which certainly belong to one of the complementary class unions and they form respective lower approximations. From the point of view of generalized decision, DRSA estimates the conﬁdence intervals which cover the whole probability distribution, i.e. such intervals [li, ui] that P(y ∈[li, ui]\|xi) = 1. In other words, those intervals are broad enough but not broader) to ensure that the observed class labels falls inside the intervals. Stochastic Dominance-based Rough Set Approach 67 a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 (0,0) (1/2,1/3) (1/2,1/3) (1,0) (0,0) (1,1/3) (1,1/2) (1,1/2) (1,1) (1,1) a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 [1,1] [1,2] [1,2] [2,2] [1,1] [2,2] [2,3] [2,3] [3,3] [3,3] a1 a2 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 [1,1] [1,2] [1,2] [2,2] [1,1] [2,2] [2,3] [2,3] [3,3] [3,3] Figure 4.4: Estimation of the probabilities and stochastic approximations for dataset from Table 4.2. In the top ﬁgure, estimators ˆ P(y ⩾2\|xi), ˆ P(y ⩾3\|xi) are shown for each object. In the middle ﬁgure there are marked stochastic lower approximations Cl⩾ 2 and Cl⩽ 1 , and in the lower ﬁgure – Cl⩾ 3 and Cl⩽ 2 . In both ﬁgures, stochastic decision is shown in brackets. We set α = 0.6. 68 4.3. Statistical Learning View: Abstaining Classiﬁers Inﬂuence of α. The parameter α can be chosen arbitrary and it is a sort of consistency level, similarly as in VC-DRSA. We have already discussed the limit α = 1. As α decreases, we are able to tighten the stochastic decision but, on the other hand, some of original class labels yi will fall outside those intervals. This is especially useful when the dataset is highly inconsistent. For instance, consider the case when α = 1 (non-stochastic case) and there exists one object x0 with y0 = 1, which dominates 100 objects xi, i = 1, . . . , 100, with class labels yi = 2. Then, since every object will be inconsistent, we have li = 1 and ui = 2 for i = 0, . . . , 100; moreover, none of the objects will enter any lower approximation. However, there is probably something wrong with x0, rather than with other 100 objects (if we removed x0, every other object would be consistent). Choosing level α < 0.99 will lead to assigning li = 2 and ui = 2 for every i = 0, . . . , 100. In other words, object x0 will be eﬀectively reassigned to class Cl2 and will enter to the lower approximation Cl⩾ 2 . As α →1 2, all the stochastic decisions shrink to single points li = ui, with exception of those objects, for which ˆ P(y ⩾k\|xi) = 1 2 for some k. Thus, most of the objects will be reassigned and obtain class labels li = ui. This is very similar to the monotone approximation problem encountered in Chapter 3. In the next section, we will focus on the relationship between the monotone approximation and stochastic decision. 4.3 Statistical Learning View: Abstaining Classiﬁers In this section, we will look at the problem of stochastic DRSA from the point of view of statistical learning and Bayesian decision theory, described in the Section 1.2.1. This will not only provide us with a comprehensive view o statistical DRSA, but will also computationally improve estimation of the lower approximations: we will show that they can be obtained without estimating the probabilities at all (Dembczyński et al., 2008a). We will extend the theory considered in Chapter 1 to the case where the classiﬁer is allowed to abstain from any answer. Such functions are known as abstaining classiﬁers (Chow, 1970; Pietraszek, 2005) and are analogous to a domain expert, who is able to say “I don’t know”, when his knowledge is not suﬃcient to draw any conclusion. Such an approach is much safer than making an uncertain decision and hence can be preferred in some domains (e.g. in medical diagnosis). We will show that rough set (both classical and dominance-based) are perfectly tailored for considering the abstaining classiﬁers. 4.3.1 Statistical Learning View of Classical Rough Sets Before we present the statistical learning view of stochastic DRSA, we will ﬁrst consider the VPRS model. A Bayesian-decision approach has already been proposed for VPRS in (Yao and Wong, 1992; Yao, 2007); the theory presented here for VPRS diﬀers slightly from that described in (Yao and Wong, 1992). Let us consider the classiﬁcation problem from the statistical learning point of view, as described in Section 1.2.1. The aim is to ﬁnd a classiﬁer h(x) which minimizes the expected loss (1.1) over the data distribution. Since the probability distribution is unknown, we minimize instead the empirical risk (1.5) within some restricted class of functions. However, now we allow the classiﬁer to refrain from the answer, which is denoted by h(x) =?. The loss function suitable for the problem is the following: Lβ(y, h(x)) =      0 if h(x) = y 1 if h(x) ̸= y β if h(x) =? (4.20) Stochastic Dominance-based Rough Set Approach 69 As we see, there is a penalty β for refraining from the answer. To be consistent with the classical rough set theory, we assume that every function must be constant within each granule, i.e. for each G = I(xi) for some object xi, we have: xi, xj ∈G = ⇒h(xi) = h(xj) i, j = 1, . . . , n, (4.21) which is in fact the principle of indiscernibility. Notice that this is the only restriction imposed on the class of functions. We now state: Theorem 4.3. The function ˆ h, minimizing the empirical risk with loss function (4.20) between all functions satisfying (4.21) is equivalent to the VPRS in the sense that ˆ h(G) = k if and only if granule G belongs to the lower approximation of class k with the precision threshold u = 1 −β, otherwise ˆ h(G) =?. Proof. As (4.21) is the only restriction imposed on the class of functions, we can analyze the value of any given function h independently for each granule. Let us choose then a granule G = I(xi) for an object xi. Let us also denote the number of objects in G as nG, and for each class index k ∈Y , let nk G be the number of objects from class k in G. The total loss of a function h in the granule G is the following: L(h(G)) = ( nG −nk G if h(G) = k βnG if h(G) =? . This follows from the fact that if h(G) = k, then for each xi ∈G such that yi ̸= k, function h suﬀers loss 1. On the other hand, if h(G) =?, for each xi ∈G, function h suﬀers loss β. The best strategy is then to choose the majority class in G or abstain from answer, depending on what loss is lower. The preferred strategy is to choose the majority class, if for some k it holds nG −nk G ⩽βnG, or if: nk G nG ⩾1 −β. (4.22) If no k satisﬁes this relation, the preferred strategy is to choose ˆ h(G) =?. Comparing this result with Section 4.1.1, one can show that the decision ˆ h(G) = k is chosen if and only if the granule G belongs to the lower approximation of class k with the precision threshold u = 1 −β. If there is no lower approximation of any class to which G belongs, the function ˆ h abstains from answer. Concluding, the variable precision rough sets can be derived by considering the class of functions constant in each granule and choosing the function ˆ h, which minimizes the empirical risk for loss function (4.20) with parameter β = 1 −u. For each granule G, if G ⊆Clk for some k ∈Y , then ˆ h(xi) = k for each xi ∈G. Otherwise ˆ h(xi) =? (abstaining from the answer). As we see, classical rough set theory suits well for considering the problems, when the classiﬁcation procedure is allowed to abstain from predictions for some xi. 4.3.2 Stochastic DRSA as Monotone Conﬁdence Interval Estimation The formulation of stochastic DRSA presented in this section, considered from the point of view of statistical learning, is completely new and ﬁrst appeared in our paper (Dembczyński et al., 2007a). However, a Bayesian-decision view of VC-DRSA has been also proposed in (Greco et al., 2007b). Interval functions. The case with stochastic DRSA is more sophisticated, because we assume the classiﬁers are monotone. We propose the following concept of abstaining classiﬁers in the ordinal classiﬁcation with monotonicity constraints. Let us consider the classiﬁer h(x), which assigns to 70 4.3. Statistical Learning View: Abstaining Classiﬁers 1 2 3 4 5 6 7 8 y h l( (x) ) h u( (x) ) h( (x) ) Figure 4.5: The problem with K = 8. The actual class label y = 2, while the predicted interval is h(x) = [hl(x), hu(x)] = [5, 7]. The total penalty is thus 2β (for imprecision) + 3 (for misclassiﬁca-tion). each point x the interval of classes, denoted [hl(x), hu(x)]. The width of the interval reﬂects the imprecision of the classiﬁer’s response, i.e. to what degree it refrains from the precise answer. The lower and upper ends of each interval are supposed to be monotone functions: xi ⪰xj = ⇒hl(xi) ⩾hl(xj) i, j = 1, . . . , n, xi ⪰xj = ⇒hu(xi) ⩾hu(xj) i, j = 1, . . . , n. (4.23) The interval loss function Lint(y, h(x)) is composed of two terms. First term is a penalty for the size of the interval (degree of imprecision) and equals to β(hu(x) −hl(x)). Second term measures the accuracy of the classiﬁcation and is zero, if y ∈[hl(x), hu(x)], otherwise h(x) suﬀers additional loss equal to the distance of y from the closer interval end: Lint(y, h(x)) = β(hu(x) −hl(x)) + 1y / ∈[hl(x),hu(x)] min{\|y −hl(x)\|, \|y −hu(x)\|} (4.24) This model incorporates the abstaining classiﬁers into the monotone framework in a very intuitive and consistent way. The interval of classes reﬂects the imprecision of the classiﬁer. Parameter β establishes the trade-oﬀbetween the precision of the response and its reliability. This is the ﬁrst model of abstaining classiﬁers proposed for the ordinal classiﬁcation with monotonicity constraints. Interval loss minimization. Let us consider the problem of minimizing (4.24) on the dataset D, subject to constraint that hu and hl are monotone: minimize n X i=1 Lint(yi, h(xi)) subject to xi ⪰xj = ⇒hl(xi) ⩾hl(xj) i, j = 1, . . . , n xi ⪰xj = ⇒hu(xi) ⩾hu(xj) i, j = 1, . . . , n hu(xi) ⩾hl(xi) i = 1, . . . , n hu(xi), hl(xi) ∈{1, . . . , K} i = 1, . . . , n (4.25) Let Lα(yi, k) be the linear loss, as deﬁned in Section 3.2.4: Lα(y, k) = ( α(k −y) if k > y (1 −α)(y −k) if k ⩽y, . One can show that: Stochastic Dominance-based Rough Set Approach 71 Lemma 4.1. It holds Lint(y, h(x)) = L1−β(y, hl(x)) + Lβ(y, hu(x)). Proof. We prove the lemma by considering three cases y > hu(x), hl(x) ⩽y ⩽hu(x) and y < hl(x). For y > hu(x), we have Lint(y, h(x)) = y −hu(x) + β(hu(x) −hl(x)), while L1−β(y, hl(x)) = β(y −hl(x)) and Lβ(y, hu(x)) = (1 −β)(y −hu(x)). Similarly, for hl(x) ⩽y ⩽hu(x) we have Lint(y, h(x)) = β(hu(x)−hl(x)), while L1−β(y, hl(x)) = β(y −hl(x)) and Lβ(y, hu(x)) = β(hu(x) −y). Finally, for y < hl(x) we have Lint(y, h(x)) = hl(x)−y+β(hu(x)−hl(x)) , while L1−β(y, hl(x)) = (1 −β)(hl(x) −y) and Lβ(y, hu(x)) = β(hu(x) −y). Notice that the solution to the problem (4.25) can be non-unique. The question arises, whether we can solve both linear loss problems separately. We have for each i the constraint hu(xi) ⩾hl(x). However, one can prove that if we remove the constraint from the optimization process (which is equivalent to solving problems with hu and hl separately), the constraint is still satisﬁed at optimality: Theorem 4.4. Consider the interval loss minimization (4.25), relaxed by removing constraint hu(xi) ⩾hl(x) and assume β > 1 2. Then, the optimal solution to this problem, which consists of the linear monotone approximation ˆ hu with α = β and the linear monotone approximation ˆ hl with α = 1 −β, is also the optimal solution to the original problem of interval loss minimization (4.25). Proof. We showed in Section 3.2.4 that one obtains linear monotone approximation by solving K−1 independent binary monotone approximation problems with weights w0 = α and w1 = 1 −α. Let ˆ huk(xi), for i = 1, . . . , n be the optimal solution in the k-th binary problem with α = β. Similarly, let ˆ hlk(xi), for i = 1, . . . , n be the optimal solution in the k-th binary problem with α = 1 −β. Since β > 1 2, we have β ⩾1 −β. From Theorem 3.6, we know that ˆ huk(xi) ⩾1ˆ pi>β and that ˆ hlk(xi) ⩽1ˆ pi⩾1−β, where ˆ p = (ˆ p1, . . . , ˆ pn) is the isotonic regression of yk = (y1k, . . . , ynk). This implies that for each i = 1, . . . , n, ˆ huk ⩾ˆ hlk. Since k was arbitrary, we have ˆ hu(xi) ⩾ˆ hl(xi). Thus, the relaxed condition hu(xi) ⩾ hl(x) is satisﬁed anyway and the pair (ˆ hu, ˆ hl) is the optimal solutions to the original problem (4.25). Stochastic DRSA is equivalent to interval loss minimization. Let us focus on the stochastic DRSA expressed in terms of the stochastic decision. Then, the following equivalence holds: Theorem 4.5. The function ˆ h(xi) = [ˆ hl(xi), ˆ hu(xi)], the solution to the interval loss minimization in the class of monotone intervals functions (4.25), such that ˆ hl is the greatest (1 −β)-linear monotone approximation and ˆ hu is the smallest β-linear monotone approximation, is equivalent to the stochastic decision with threshold α = 1 −β: for i = 1, . . . , n, ˆ hl(xi) = li and ˆ hu(xi) = ui. Proof. Since the interval loss minimization separates into two linear monotone approximation prob-lem, we will also consider functions ˆ hl(xi) and ˆ hu(xi) separately. Let us start with ˆ hl(xi). This is the (1 −β)-linear monotone approximation, which can be obtained by solving K −1 binary monotone approximations. From the deﬁnition (substituting α = 1−β) we have that li = max{k: ˆ P(y ⩾k\|xi) ⩾1−β}, where ˆ P(y ⩾k\|xi) is the probability estimator based on the isotonic regression of the vector yk = (y1k, . . . , ynk), as described in Section 4.2.3. From Theorem 3.6 we know that ˆ hlk(xi) = 1 ˆ P (y⩾k\|xi)⩾1−β; hence we have li = max{k: ˆ hlk(xi) = 1} = ˆ hl(xi). Similarly, for ˆ hu(xi), we have ui = min{k: ˆ P(y ⩾k + 1\|xi) ⩽β}. However, we know that ˆ huk(xi) = 1 ˆ P (y>k\|xi)>β; hence ui = min{k: ˆ hu,k+1(xi) = ˆ hl(xi)}. 72 4.3. Statistical Learning View: Abstaining Classiﬁers Let us notice that the optimal solution which appears in Theorem 4.5 (i.e. the “greatest-smallest” solution) is the solution of the minimal width of the intervals. Moreover, notice that in the context of what was written in Section 3.2.4, such solution is actually obtained by choosing value α decreased by a suﬃciently small positive ϵ, for which the unique linear monotone approximation exists. Concluding, the stochastic DRSA can be derived by considering the class of interval functions, for which the lower and upper ends of intervals are monotone vectors, and choosing the function ˆ h, which minimizes the empirical risk with loss function (4.24) with parameter β = 1 −α. For each xi, i = 1, . . . , n, ˆ h(xi) is then a stochastic decision [li, ui]. 4.3.3 Summary We have presented the stochastic extension of DRSA, based the appropriate probabilistic model, satisfying stochastic dominance principle. The lower approximations were deﬁned as the subsets of the dataset D, for which the class unions Cl⩾ k or Cl⩽ k are observed with probability at least α. Moreover, the equivalent formulation of stochastic approximation is given by introducing the stochastic decision, which assigns to each object class interval between (1 −α)-quantile and α-quantile of the class conditional distribution. Since the probability distribution is unknown, the stochastic approximations (or, equivalently, stochastic decisions) must be estimated from data. One can follow two approaches, based either on lower approximations or stochastic decision, and both lead to exactly the same results: The approach based on lower approximations. For each class union Cl⩾ k and for each object xi, i = 1, . . . , n, the probability P(y ⩾i\|xi) is estimated by solving K −1 isotonic regressions. Having obtained the probability estimates, one can calculate the lower approximations. Notice that isotonic regression has the pessimistic complexity O(n4); however there exists strong method of reduction of the problem size, described in Section 3.2.2; moreover, there are fast and reliable heuristic algorithms for isotonic regression, working in O(n2). The approach based on stochastic decisions. For each object xi, i = 1, . . . , n, stochastic decision [li, ui] is obtained by minimizing the interval loss on the dataset D within the space of all monotone vectors. This corresponds to solving 2(K −1) binary monotone approximations; this reduces the complexity to O(n3), which is the improvement in comparison with O(n4) for isotonic regression. This improvement follows from the fact that we directly estimate the stochastic decisions (hence also lower approximations), without estimating the probabilities. Chapter 5 Learning Monotone Rule Ensembles In the previous chapters, we derived a theoretical basis for dealing with ordinal classiﬁcation problems in the presence of monotonicity constraints. We proposed the monotone approximation, a nonparametric method of classiﬁcation based on considering the class of all monotone functions. We also explained and extended DRSA from the stochastic point of view and proved it to be a generalization of monotone approximation to the interval loss functions. In this chapter, we show how nonparametric methods are used in practice to solve the ordinal classiﬁcation problems with monotonicity constraints. This is accomplished by the following two-phase procedure. In the ﬁrst phase, we apply monotone approximation to the training data in order to get rid of the inconsistencies. In other words, we “monotonize” the training data. The second phase consists in building the model on the monotonized data. The model has the form of the monotone rule ensemble, i.e. the set of decision rules, in which rules are combined in an additive way to form a monotone function. The process of rule induction is based on the boosting strategy of learning. The model makes no errors on the training set, i.e. we say the model separates the monotonized data. Notice that such a two-phase procedure is in the spirit of the rough set approach to data analysis: ﬁrst the rough approximations are induced from the training set, and then a model is constructed from the approximations, which is consistent with the dataset (i.e. separates the data). In fact, monotonization of the data corresponds exactly to using stochastic dominance-based rough approximations with consistency level 1 2 (see Chapter 4). First, we give the overview of boosting, introduce the idea of margin and margin-based loss functions, and present linear programming formulation of boosting (LPBoost). Then, we formulate a general scheme of learning the monotone rule ensembles and describe the two-phase procedure. We examine the asymptotic consistency and generalization bounds of the procedure, which suggests the use of a linear programming boosting framework to generate an ensemble with the maximal value of the margin. This leads to the algorithm called linear programming monotone rule ensembles (LPRules). At the end of this chapter, we describe another algorithm generating a monotone rule ensemble, based on the sigmoid approximation to the linear loss function, called sigmoid loss monotone rule ensemble (MORE). To make our notation consistent with the notation used in the majority of boosting papers, we assume in this chapter that Y = {−1, 1} in case of binary classiﬁcation problems. Let us also deﬁne function sgn (x) to be equal to 1 if x ⩾0 and −1 for x < 0 (sign function). 74 5.1. Boosting Algorithm 5.1: Boosting procedure. input : set of training examples D, T – number of base learners to be generated. output: ensemble of base learners {b1, . . . , bT }. f0(x) := 0 for t = 1 to T do (at, bt) := arg mina∈R,b∈B P (yi,xi)∈D L(yi, ft−1(xi) + ab(xi)); fm(x) = fm−1(x) + atbt(xi); end 5.1 Boosting 5.1.1 Overview Boosting appeared in the ﬁeld of computation learning theory (Kearns and Vazirani, 1994), as the answer to the question, whether a “weak” learning algorithm which, performs just slightly better than random guessing, can be “boosted” into an arbitrarily accurate learning algorithm (Freund and Schapire, 1997). The ﬁrst provable polynomial-time boosting algorithm was proposed by Schapire (1990). A fast development and great popularity of boosting started, however, from introducing the AdaBoost algorithm by Freund and Schapire (1997), which proved to be very eﬃcient and surprisingly resistant to overﬁtting on real datasets (Quinlan, 1996; Breiman, 1998). Here, we follow the formulation of boosting introduced by Breiman (1999), Friedman et al. (1998) and Mason et al. (1999), who showed that boosting greedily minimizes a speciﬁc loss function on dataset. Note, however, that a dual point of view exists, explaining boosting as minimization of a particular Bregman distance (Kivinen and Warmuth, 1999; Collins et al., 2000; Warmuth et al., 2006). Let us consider the binary-class case Y = {−1, 1}. Let f(x) be some real-valued function, such that object x is classiﬁed according to the sign of f(x), h(x) = sgn (f(x)), i.e. x is classiﬁed to the “positive” class (y = 1) if f(x) ⩾0, and x is classiﬁed to the “negative” class (y = −1) otherwise. In boosting, we assume that f(x) has the form: f(x) = T X t=1 atbt(x), i.e. it is the linear combination (ensemble) of T functions bt ∈B, called the base classiﬁers; it is usually assumed that bt(x) ∈{−1, 1}; the set of base classiﬁers B can be, e.g., the space of all classiﬁcation trees, decision rules, perceptrons, etc. Let L(y, f(x)) be a loss function, which penalizes the prediction of the real-valued function f(x) ∈R; notice that this is a diﬀerent loss function then the loss L(y, h(x)) deﬁned before, which penalized the classiﬁer h(x) ∈{1, . . . , K}. The function f(x) is learned in the iterative way, by greedily minimizing the loss L(y, f(x)) on the training set. We start with f(x) = 0. Let ft−1(x) be the ensemble at the end of iteration t; a new base classiﬁer bt(x) and its weight at are added to the ensemble without adjusting classiﬁers which have already been added, by minimizing the loss L(y, ft−1(x) + ab(x)) on the dataset, with respect to a ∈R and b ∈B. The method is formally presented as Algorithm 5.1. There is no single way to extend the idea of boosting for an arbitrary number of classes. The most popular approach is to transform K-class problem into the sequence of binary problems. In “one-against-all” strategy, K binary class problems are constructed and in the k-th problem, the k-th class forms positive class y = 1, while K −1 other classes form negative class y = −1. In “one-against-one” strategy, one constructs K(K −1)/2 binary problems and for each problem the objects Learning Monotone Rule Ensembles 75 from one class are discriminated from the objects from another class. In both methods, one uses the binary class boosting several times and then combines the results. The less popular strategy, but well-motivated theoretically, is to solve a single problem with the vector margin function and vector loss function (Friedman et al., 1998; Zou et al., 2005). In this thesis, we use a diﬀerent procedure for handling multi-class case, tailored to the problem of ordinal classiﬁcation. 5.1.2 Margin-based Loss Functions The aim of the classiﬁcation problem is to minimize the expected loss function, which will be called target loss. Although the loss function used in the boosting procedure (which will be called training loss) can be the same as the target loss, it leads to a poor performance of the classiﬁers trained in this way. This is caused by the fact that the target loss is usually discontinuous and insensitive to the changes of f(x). For instance, consider 0-1 loss: any change of f(x) with no change in sign does not aﬀect the value of 0-1 loss. This makes greedy procedures working poorly, because there is no local improvement possible. Moreover, it is often the case that apart from choosing the class label minimizing the target loss, one needs to obtain the class probabilities, being the measure the 23- of prediction reliability. This also suggests using diﬀerent loss function for training the classiﬁer, which is capable of probability estimation. Let us focus again on the binary class case with Y = {−1, 1}. Let us call yf(x) a margin for x; notice that x is correctly classiﬁed if the margin is positive; moreover, the higher the margin, the more certain the prediction. Many binary loss functions can be expressed as single argument functions, depending only on the margin, i.e. L(y, f(x)) = L(yf(x)). For instance, 0-1 loss can be written as L(u) = 1u>0, where u = yf(x). Thus, 0-1 loss is insensitive to the value of the margin. Below we present the margin-sensitive loss functions, which can be used within the boosting procedure. Log-likelihood and hinge loss. The negative log-likelihood loss function can be derived by considering the MLE approach (Friedman et al., 1998; Friedman, 2001) and thus it estimates the probabilities of class labels. It has the following form: Llog(u) = ln 1 + e−2u , (5.1) where u = yf(x). The Bayes function for the log-likelihood loss, i.e. function f ∗(x) minimizing the expected loss, has the form: f ∗(x) = 1 2 ln P(y = 1\|x) P(y = −1\|x). (5.2) This shows that log-likelihood loss estimates the (logit of) probabilities at x. Log-likelihood loss is a convex function, yet it is nonlinear. Therefore, it is sometimes approxi-mated by a piecewise linear function: Lh(u) = (1 −u)+ (5.3) (where u+ = u1u⩾0 means the positive part), called hinge loss. Such a function can be incorporated into the mathematical programming problem in terms of the linear constraints. It was used in the linear programming formulation of boosting (Demiriz et al., 2001). Exponential loss. Exponential loss is another convex function estimating the probabilities. It appeared (implicitly) in the ﬁrst boosting algorithm, AdaBoost (Freund and Schapire, 1997). It has the following, simple form: Lexp(u) = e−u (5.4) 76 5.1. Boosting −2 −1 0 1 2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 yf(x) L(y,f(x)) 0−1 loss sigmoid loss log−likelihood loss hinge−loss exponential loss Figure 5.1: Margin-sensitive loss functions. 0-1 loss is shown for comparison. where, as usual, u = yf(x). Minimizing the expected value of (5.4) with respect to P(y\|x) leads to the Bayes function: f ∗(x) = 1 2 ln P(y = 1\|x) P(y = −1\|x) (5.5) which is the same as in case of the log-likelihood loss (5.2). Exponential loss is easy to cope with due to its multiplicative nature. It leads to a very simple reweighting scheme when used with boosting (in fact, this is how the boosting was originally formulated). Sigmoid loss. Sigmoid loss function is the continuous approximation of the 0-1 loss: Lsigm(u) = 1 1 + eu . (5.6) Although not convex, it is diﬀerentiable. It does not estimate the probabilities, because the Bayes function is rather aberrant: f ∗(x) =      +∞Pr(y = 1\|x) > 1 2, −∞Pr(y = −1\|x) < 1 2, 0 otherwise. There are some theoretical justiﬁcations for using this loss function (Mason et al., 1999, 2000). The practical premise is that this loss function is the most similar to the 0-1 loss. Moreover, contrary to the exponential and the log-likelihood loss, this loss function is bounded (its range is from 0 to 1), therefore it is less sensitive to outliers (it does not grow to inﬁnity for misclassiﬁed objects). The margin-sensitive loss functions, along with 0-1 loss, are shown in Figure 5.1. Learning Monotone Rule Ensembles 77 5.1.3 Margin Theory It has been observed in experiments that boosting methods achieve very good prediction accu-racy and usually do not overﬁt, even when the number of base classiﬁers is very large (Drucker and Cortes, 1996; Quinlan, 1996; Breiman, 1998). Moreover, even when boosting drives the training error down to zero, the test error still tends to decrease. This counterintuitive phenomenon has been explained in terms of the ability of boosting to produce combinations of classiﬁers with large margins. Consider y ∈{−1, 1}, let f(x) = PT t=1 atbt(x), as before, and assume that weights are non-negative and normalized, PT t=1 at = 1, i.e. f(x) is a convex combination of classiﬁers. We refer to the margin distribution as to the set of values {yif(xi), i = 1, . . . , n}. The margin theorems (Schapire et al., 1998; Schapire and Singer, 1999; Breiman, 1999; Koltchinskii and Panchenko, 2002, 2006) bound the expected risk of f(x) in terms of its margin distribution. We cite the result obtained by Koltchinskii and Panchenko (2006), which states that for every distribution P(y, x), with probability 1 −δ, the following inequality holds: P(yf(x) ⩽0) ⩽ inf γ∈(0,1]  PD(yf(x) ⩽γ) + M   s d nγ2 + s log 1 δ n )    , (5.7) where d is the Vapnik-Chervonenkis dimension (Vapnik, 1998) of the base classiﬁer, M is a universal constant and PD(·) is the empirical probability distribution, i.e.: PD(yf(x) ⩽γ) = 1 n n X i=1 1yif(xi)⩽γ. Thus, with high probability the expected 0-1 loss (generalization error) is bounded by the fraction of objects from the training set with margin yf(x) greater than γ plus some complexity term which increases with decreasing value of γ. The margin bound shows that ensembles with high margin on the training set generalize well, independently of their sizes. This led to the analysis of ensemble methods from the point of view of their margin distributions (Schapire et al., 1998; Breiman, 1999; Demiriz et al., 2001; R¨ atsch and Warmuth, 2005; Rudin et al., 2007a,b; Warmuth et al., 2008a). 5.1.4 LPBoost Demiriz et al. (2001) introduced a new boosting algorithm formulated as a linear program, LPBoost (Linear Programming Boosting), which was later extended by Leskovec and Shawe-Taylor (2003) and then by Warmuth et al. (2008b). LPBoost has the property of directly maximizing the minimal margin on the dataset. Let us consider the set of all possible base classiﬁers B = {bj : j = 1, . . . , J} (we can assume that B is ﬁnite because we have a ﬁnite training set and there are at most 2n classiﬁers distinguishable on the training set). Let us denote the classiﬁcation function by f(x) = PJ j=1 ajbj(x), with aj ⩾0 and PJ j=1 aj = 1. The problem of maximizing the minimal margin on the dataset can be formulated as a linear program (Breiman, 1999): max : ρ subject to: yi PJ j=1 ajbj(xi) ⩾ρ i = 1, . . . , n, PJ j=1 aj = 1, aj ⩾0 . (5.8) Indeed, a closer look at the constraints of (5.8 reveals that ρ corresponds to the minimal margin on the training set. Let us refer to ρ as to the hard margin. Unfortunately, maximizing the minimal 78 5.2. Monotone Rule Ensembles margin corresponds to optimizing the “worst-case examples” and thus it may put too much attention to the noise or outliers, sacriﬁcing the overall accuracy. Hence, the optimization goal is relaxed, leading to the soft margin case: max : ρ −C Pn i=1 ξi, subject to: yi PJ j=1 ajbj(xi) ⩾ρ −ξi i = 1, . . . , n, PJ j=1 aj = 1, xi, aj ⩾0. (5.9) It can be shown that the parameter C corresponds to the number of objects violating the margin; strictly speaking, 1 C upper-bounds the number of margin errors, Pn i=1 1yif(xi)⩽ρ (R¨ atsch et al., 2000). To solve (5.9), it is useful to consider the dual problem (Demiriz et al., 2001): min : β subject to: Pn i=1 qiyibj(xi) ⩽β j = 1, . . . , J, PJ j=1 qi = 1, 0 ⩽qi ⩽C i = 1, . . . , n. (5.10) There are n dual variables qi and J constraints, where each constraint corresponds to one base classiﬁer. However, instead of taking into account all possible base classiﬁers at once, we use the column generation technique and add base classiﬁers iteratively to the problem. We start with the empty set of classiﬁers, denoted by J0. In each iteration, we choose the classiﬁer which maximally violates the corresponding constraint in (5.10), i.e. the one with the highest value Pn i=1 qiyibj(xi). We add it to J0 and solve the problem (5.10) restricted only to the classiﬁers from J0 (so called restricted dual). We proceed in this way until no classiﬁer violates the constraints. Then this means we are at optimum of (5.10). The scheme of the LPBoost algorithm is presented as Algorithm 5.2 (Demiriz et al., 2001). Notice that the minimum is always reached in a ﬁnite number of steps (as opposed to e.g. AdaBoost). The classiﬁers are chosen according to the value Pn i=1 qiyibj(xi), which is the weighted classiﬁcation error scaled from −1 to 1. In other words, in each iteration we choose the classiﬁer which performs the best according to the current weights. From the dual point of view, we are at minimum if we ﬁnd the least favorable distribution (Breiman, 1999) qi, i = 1, . . . , n. Notice that the regularization constant C prevents us from putting too much weight on any of the objects. Finally, notice that it follows from the duality slackness conditions (Papadimitriou and Steiglitz, 1998) that for each i = 1, . . . , n we have: qi  yi X j∈J0 ajbj(xi) + ξi −ρ  = 0. In other words, only the “hardest” objects, with margins lower then the soft margin ρ (i.e. those making margin errors) receive non-zero weights. The LPBoost formulation is simple and elegant. Moreover, it tends to produce a very sparse solution, with only few non-zero coeﬃcients. This improves the interpretability of the model, which will appear to be especially useful for monotone rule ensembles. 5.2 Monotone Rule Ensembles We now introduce an ensemble of speciﬁc base classiﬁers, known as decision rules. Since we deal with the ordinal classiﬁcation with monotonicity constraints, we will restrict to the ensembles which are monotone functions. Learning Monotone Rule Ensembles 79 Algorithm 5.2: Linear programming boosting (LPBoost). input : set of training examples D, C – regularization penalty. output: ensemble of base learners {b1, . . . , bT }. optimum := false; J0 := ∅; qi := 1 n, i = 1, . . . , n; repeat jmax := arg maxj=1,...,J Pn i=1 qiyibj(xi); if Pn i=1 qiyibjmax(xi) ⩽β then optimum := true; end else J0 := J0 ∪{jmax}; Solve the problem: min : Pn i=1 β subject to: Pn i=1 qiyibj(xi) ⩽β j ∈J0, PJ j=1 qi = 1, 0 ⩽qi ⩽C i = 1, . . . , n. to obtain new weight vector qi, i = 1, . . . , n and β; end until optimum = true; Obtain aj, j ∈J0 from the dual variables. 5.2.1 Decision Rules We will consider a learning algorithm involving decision rules. Such decision rules are simple and interpretable logical statements of the form: “if conditions then decision”. They can be treated as simple classiﬁers that give a constant response to examples satisfying the condition part, and abstain from the response for other examples. The problem of induction of decision rules has been widely considered in machine learning (Michalski, 1983; Clark and Niblett, 1989; Cohen, 1995; F¨ urnkranz, 1996), logical analysis of data (Boros et al., 2000) and rough set approaches to knowledge discovery (Pawlak, 1991; Słow-iński, 1992; Grzymala-Busse, 1992; Stefanowski, 1998; Greco et al., 2001c, 2005). The most popular algorithms were based on a sequential covering procedure (also known as separate-and-conquer ap-proach). In this technique, a rule is learned which covers a part of the training examples, then examples are removed from the training set and the process is repeated until no examples remain. The interest in decision rule models is still growing in machine learning – let us mention such algo-rithms as RuleFit (Friedman and Popescu, 2005), SLIPPER (Cohen and Singer, 1999), Lightweight Rule Induction, (LRI) (Weiss and Indurkhya, 2000), ENDER (Błaszczyński et al., 2006b,a; Dem-bczyński et al., 2008c,e), MLRules (Dembczyński et al., 2008d). All these algorithms follow a speciﬁc iterative approach to decision rule generation by treating each decision rule as a subsidiary base classiﬁer in the ensemble. This approach can be seen as a generalization of the sequential covering, because it approximates the solution of the prediction task by sequentially adding new rules to the ensemble without adjusting those that have already been added (RuleFit is an exception since it generates the trees ﬁrst and then transforms them into rules). Each rule is ﬁtted by concentrating 80 5.2. Monotone Rule Ensembles on objects which were hardest to classify correctly by rules already present in the ensemble. All these algorithms can be explained within the framework of boosting. However, all the algorithms mentioned above were designed to deal with regular classiﬁcation problems, where neither the order relation nor monotonicity constraints present in the data are taken into account. Here, we propose a methodology for monotone rule induction, i.e. generating rules that are adapted to the ordinal classiﬁcation problem and do not violate monotonicity constraints. 5.2.2 Monotone Rule Ensembles for Binary Classiﬁcation We assume the binary-class case Y = {−1, 1}, so that “positive” class is labeled with +1, while “negative” class is labeled with −1. Deﬁnition. Let Xj be the value set of the j-th attribute, i.e. the set of all possible values for the j-th attribute. Condition part of the rule consist of elementary conditions of the form xj ⩾sj or xj ⩽sj for some sj ∈Xj. Let Φ denote the set of elementary conditions constituting the condition part of the rule. We can deﬁne rule as a function Φ(x) which is equal to 1 or −1 if an objects x satisﬁes all the conditions of the rule (i.e. object is covered by the rule), otherwise Φ(x) = 0. As we will see, rules with Φ(x) = 1 votes for a higher class, while rules with Φ(x) = −1 votes for a lower class. We assume that either Φ(x) ⩾0 for all x ∈X, or Φ(x) ⩽0 for all x ∈X. In other words, for all covered objects, function Φ(x) consequently returns the same value, either −1 or +1. Here we consider a real-valued classiﬁcation function which is a linear combinations of T decision rules: f(x) = T X t=1 atΦt(x), (5.11) where at, t = 1, . . . , T are positive coeﬃcients which will be called weights of the rules. Object x is classiﬁed to the class indicated by the sign of f(x), i.e. h(x) = sgn (f(x)). The combination (5.11) has a very simple interpretation as a voting procedure: rules with Φ(x) ⩾0 vote for the positive class, while rules with Φ(x) ⩽0 – for the negative class. Object x is classiﬁed to the class with a higher vote (which is equivalent to classiﬁcation according to the sign of f(x)). Monotonicity of rule ensemble. We assume the monotonicity constraints are present in the data and thus we require that function f(x) must be monotone. The following theorem establishes the suﬃcient conditions for the monotonicity of rule ensemble: Theorem 5.1. Let f(x) be a rule ensemble, i.e. a function of the form (5.11). Then, in order to maintain monotonicity of f(x), it is suﬃcient that for each rule t = 1, . . . , T, Φt(x) ⩾0 and all elementary conditions in Φt are of the form xj ⩾sj, or Φt(x) ⩽0 and all elementary conditions are of the form xj ⩽sj. Proof. Assume that all the elementary conditions are of the form xj ⩾sj if Φt(x) ⩾0 and are of the form xj ⩽sj if Φt(x) ⩽0. In each case, a single rule rt(x) = atΦt(x) is a monotone function. Indeed, suppose x ⪰x′ and consider Φt(x) ⩾0. Then, since xj ⩾x′ j, if x′ j ⩾sj, then also xj ⩾sj, so that Φt(x) ⩾Φt(x′) and thus rt(x) ⩾rt(x′). Similarly, consider Φt(x) ⩽0. Then, since xj ⩾x′ j, if xj ⩽sj then also x′ j ⩽sj so that Φt(x) ⩾Φt(x′) and thus rt(x) ⩾rt(x′). Hence, f(x) is the sum of monotone functions, so is a monotone function. Rules, which satisfy the conditions of Theorem 5.1 are called monotone. Moreover, monotone rules with Φt(x) ⩾0 will be called upward, while those with Φt(x) ⩽0 will be called downward. The function f(x) of the form (5.11), consisting of monotone rules is called monotone rule ensemble. Learning Monotone Rule Ensembles 81 5.2.3 Monotone Rule Ensembles with Linear Loss Let us go back to the general K-class case and consider the rule ensemble algorithm for mini-mizing the linear loss function (2.8): L(y, k) = ( α(k −y) if k > y (1 −α)(y −k) if k ⩽y, (5.12) We will show how to solve the general problem by decomposition into K −1 binary problems. Decomposition procedure. Suppose we have an access to the learning algorithm for binary classiﬁcation problems, which can produce classiﬁer h(x) ∈{−1, 1} with small weighted 0-1 loss: L(h(x), y) = wy1yh(x)<0, (5.13) for y ∈{−1, 1}, where w−1 = α and w1 = 1 −α. It is not hard to verify that the weighted 0-1 loss is exactly the linear loss (5.12) for binary-class case. Let y be the class label and let us deﬁne yk = sgn (y −k) for k = 2, . . . , K, similarly as in Chapter 3 (with the exception that now y ∈{−1, 1}). Therefore, yk = 1 corresponds to the class union “at least k”, while yk = −1 corresponds to the class union “at most k −1”. Suppose we train K −1 binary classiﬁers described above, k = 2, . . . , K, where the k-th classiﬁer hk(x) is trained on the dataset with original class labels y substituted by labels yk. We combine binary classiﬁers into a single K-class classiﬁer in the following way: h(x) = 1 + K X k=2 1hk(x)=1. (5.14) Notice that we do not assume classiﬁers are consistent, i.e. we do not assume hk(x) ⩾hk′(x) when k < k′. We will show that the ﬁnal classiﬁer makes error not greater then the sum of errors of base classiﬁers: Theorem 5.2. Let L(y, hk(x)) be the linear loss, given by (5.13), suﬀered by the k-th binary classiﬁer and let L(y, h(x)) be the linear loss suﬀered by a composite classiﬁer deﬁned by (5.14). Then we have: L(y, h(x)) ⩽ K X k=2 L(yk, hk(x)) Proof. For simplicity, we denote hk(x) by hk and h(x) by h. We have: K X k=2 L(yk, hk) = K X k=2 ((1 −α)1yk=11hk=−1 + α1yk=−11hk=1) = y X k=2 (1 −α)1hk=−1 + K X k=y+1 α1hk=1 ⩾(1 −α)(y −h)1hy = L(y, h) Thus, we see that all we need is an accurate learning algorithm for binary problems with weighted 0-1 loss. 82 5.2. Monotone Rule Ensembles Monotonicity of the composite classiﬁer. In the ordinal classiﬁcation with monotonicity constraints we require that the classiﬁer h(x) must be monotone. One can simply show that if binary classiﬁers hk(x) are all monotone, then so is h(x). Indeed, assume x ⪰x′. Then, h(x) = K X k=2 hk(x) ⩾ K X k=2 hk(x′) = h(x′). Hence, if for each k, hk(x) is a rule ensemble composed of monotone rules, then h(x) is a monotone function. 5.2.4 Two-phase Procedure of Learning We now give more details about how the binary classiﬁers hk(x) are learned. Let us ﬁx some k = {2, . . . , K}. We ﬁrst transform the dataset D = {(xi, yi), i = 1, . . . , n} into the dataset Dk = {(xi, yik), i = 1, . . . , n}, such that yik = sgn (yi −k). Next, we perform monotone approximation of Dk and obtain the monotonized dataset D′ k = {(xi, y′ ik), i = 1, . . . , n}, where y′ ik are new labels, corresponding to the monotone approximation ˆ dik. Notice that it follows from the deﬁnition that y′ ik are monotone, i.e. xi ⪰xj →y′ ik ⩾y′ jk. As soon as the data are monotonized, we train monotone rule ensemble fk(x) on D′ k such that it makes no errors on the dataset. In other words, y′ ikfk(xi) > 0 for all i = 1, . . . , n, and we say that such rule ensemble separates D′ k. This constitutes a two-phase procedure of learning the monotone rule ensemble: ﬁrst, the dataset is monotonized, and then, the monotone rule ensemble separating the dataset is constructed. To prove the validity of our procedure, we must show that a monotone rule ensemble separating the data always exists. It appears that the existence of a separating rule ensemble is strictly related to the monotonicity (consistency) of the dataset. Theorem 5.3. There exists a monotone rule ensemble f(x) separating a dataset D = {(x1, y1), . . . , (xn, yn)}, yi ∈{−1, 1}, if and only if D is monotone, i.e. for each i, j = 1, . . . , n, we have xi ⪰xj →yi ⩾yj. Proof. From the monotonicity of f(x) it follows that xi ⪰xj →f(xi) ⩾f(xj) →sgn (f(xi)) ⩾ sgn (f(xj)). Moreover, f(x) separates D so that we have sgn (yif(xi)) = sgn (yjf(xj)) = 1. Since yi ∈{−1, 1}, it follows that yi ⩾yj, so D is monotone. Assume D is monotone. Then, for each xi such that yi = 1 we construct a rule Φ(x) = 1x⪰xi with ai = 1; this is a valid decision rule. Similarly, for each xi with yi = −1 we construct a rule Φ(x) = −1x⪯xi with ai = 1. Since D is monotone, rules with Φi(x) ⩾0 cover only the objects with yj = 1, while rules with Φi(x) ⩽0 cover only the objects with yj = −1. Hence, rule ensemble makes no errors on D. Since D′ k is monotone, we are guaranteed that a separating monotone rule ensemble exists. Moreover, to create separating rule ensemble we must ﬁrst monotonize the data. 5.2.5 Consistency and Generalization Bounds for the Two-phase Proce-dure Strong consistency. This section brings theoretical justiﬁcation for the two-phase procedure of learning monotone ensembles. We start with showing that for a very large class of monotoni-cally constrained probability distributions, including attributes with both discrete and continuous domains, the two-phase procedure is strongly consistent. Learning Monotone Rule Ensembles 83 Theorem 5.4. Let Y = {1, . . . , K} and consider minimizing the linear loss. Assume P(x, y) is monotonically constrained and let X = V × Rm0, where V is ﬁnite and m0 ⩽m. Assume P(x) has density. Then the classiﬁer h(x) returned by the two-phase procedure is strongly consistent, i.e. Rn(h) n→∞ − →R∗= R(h∗) with probability one, where Rn(h) is the risk of h when trained on the dataset of size n and h∗is a Bayes classiﬁer. Proof. Fix k = 2, . . . , K. The monotone rule ensemble hk(x) separates set D′ k and is monotone, which implies that hk(x) is a valid monotone extension of the monotone approximation y′ ik, i = 1, . . . , n. Thus, from Theorem 3.11 we have Rn(hk) n→∞ − →R(h∗ k) with probability one. This event holds for all k = 2, . . . , K simultaneously also with probability one (because K is ﬁnite). From Theorem 5.2 we have: R∗⩽Rn(h) ⩽ K X k=2 Rn(hk) n→∞ − → K X k=2 R(h∗ k) = R∗, which ends the proof. Theorem 5.4 shows that the two-phase method achieves asymptotically the smallest possible risk, but it does not tell anything about the rate of convergence or about any non-asymptotic error bounds. The rest of this section is devoted to bounding the deviations of the two-phase procedure from the Bayes risk in terms of the margin achieved on the monotonized dataset. Margin theorem for uneven misclassiﬁcation costs. We start our analysis with extending the margin bound (5.7) into the case of weighted 0-1 loss. Consider the binary class case Y = {−1, 1} and real-valued function f(x). Fix some margin value γ and deﬁne the violation of the margin for object xi as event {yif(xi) ⩽γwyi}, where w−1 = α and w1 = 1 −α, as usual. Notice that this deﬁnition incorporates the uneven costs of misclassiﬁcation into the margin: it eﬀectively increases the margin for more penalized class and hence prompts the classiﬁer to predict this class more often. Similar procedure has been used by Karakoulas and Shawe-Taylor (1999) to derive AdaUBoost, boosting algorithms for uneven costs of misclassiﬁcation. Let us also denote: Rγ emp(f) = 1 n n X i=1 wi1yif(xi)⩽γwyi (5.15) for the fraction of weighted margin errors and: R(f) = E[wy1yf(x)<0] for the expected weighted 0-1 loss (risk). The following theorem holds: Theorem 5.5. Let f(x) = P j ajbj(x) be the convex combination of classiﬁers bj(x) ∈{−1, 1}, i.e. aj are non-negative and P j aj = 1. Then, with probability 1−δ, for every distribution P(y, x), the following inequality holds: R(f) ⩽ inf γ∈(0,1]  Rγ emp(f) + M   s d nγ2 + s log 1 δ n    , (5.16) where d is the Vapnik-Chervonenkis dimension of the base classiﬁer and M is some universal con-stant. 84 5.3. Linear Programming Rule Ensembles (LPRules) The proof closely follows the proof for the ordinary (symmetric) 0-1 loss and is given in the appendix. The main diﬀerence between the asymmetric and symmetric case is in the choice of the contraction functions. Notice that the asymmetric margin leads to the contractions γφ1(x) and γφ−1(x) which lead, in turn, to the bound 1/γ. If we used symmetric margin, we would obtain contractions αγφ1(x) and (1 −α)γφ−1(x) which would lead to the bound 1/(min{α, 1 −α}γ). If α were close to 0 or 1, such a bound would become very loose. This motivates our choice of the asymmetric margin in the algorithm. Generalization bounds for the two-phase method. We are now ready to give our main theorem for the two-phase method. The theorem shows that the accuracy of the ensemble with large margin remains close to the accuracy of the Bayes classiﬁer. There are four problems which are to be addressed: 1. Single rule returns one of three values {−1, 0, 1}, contrary to the ordinary classiﬁer which returns one of two values {−1, 1}. To cope with such three-valued classiﬁers, we will show that there always exists an ensemble of ordinary base classiﬁers having their outputs in the set {−1, 1}, which is equivalent to the rule ensemble and has at least the same value of the margin. 2. We must extend the theorem to the multi-class problem with linear loss. We will do it by bounding the linear loss of the composite classiﬁer in terms of sum of the losses of binary classiﬁers. 3. The ensemble is learned on the monotonized data, which are not i.i.d. anymore. We will by-pass this problem by noticing that the ensemble learned on monotonized data without misclassiﬁcation errors makes on the original data no more errors than the number of relabeled objects. 4. In margin theorems, one compares the fraction of margin errors to the real accuracy of the ensemble. Here we compare the accuracy of the ensemble with the Bayes risk by noticing that with high probability the fraction of relabeled objects does not exceed much the Bayes risk. Let L(y, k) be the linear loss (5.12) and deﬁne R(h) = E[L(y, h(x)] to be the expected linear loss of the composite classiﬁer (returned by the two-phase method). The following theorem holds: Theorem 5.6. Assume P(x, y) is monotonically constrained. Let h(x) be the classiﬁer returned by the two-phase method. Let fk(x), k = 2, . . . , K, be the k-th rule ensemble trained on the monotonized dataset D′ k, such that there exists γk > 0 for which y′ ikfk(xi) ⩾γkwy′ ik for all i = 1, . . . , n (i.e. fk(x) achieves hard margin γk on the dataset D′ k). Then, with probability at least 1 −δ: R(h) ⩽R∗+ M  2(K −1) s log 2(K−1) δ n + rm n K X k=2 1 γk  . for some universal constant M. Theorem 5.6 states that an ensemble with a high value of the hard margin on the monotonized data performs not much worse than the Bayes classiﬁer. 5.3 Linear Programming Rule Ensembles (LPRules) In the previous section, we presented a two-phase procedure of learning rule ensembles. First, the K-class problem is converted into K −1 binary problems, such that in the k-th problem, objects Learning Monotone Rule Ensembles 85 with class labels at least k are discriminated from objects with class labels at most k −1, leading to the training set Dk. Next, we monotonize Dk to obtain consistent dataset D′ k and rules are trained on D′ k such that all objects from the training set are correctly classiﬁed and the ensemble achieves hard margin γk on the dataset. We bounded the diﬀerence between the accuracy of the classiﬁer made in such a way and the accuracy of the Bayes classiﬁer in terms of the margin γk. This suggests that ensembles achieving large margin on the dataset have better generalization power. In this section, we show how to obtain an ensemble with a high margin value. Our algorithm is based on the LPBoost framework, which is known to directly maximize the margin. As pointed out in (Demiriz et al., 2001), LPBoost ends at optimal solution with the highest possible value of the margin. On the other hand, solutions produced by LPBoost are very sparse. It is a very desirable property in our case, because a compact rule ensemble is much easier to interpret. 5.3.1 Rule Induction Fix k and consider the k-th binary classiﬁcation problem. As the dataset D′ k is monotonized, it follows from Theorem 5.3 that it is separable by the set of monotone rules. Hence, we can use the “hard margin” linear program (5.8). After translating the program into the rule framework and adapting it to the problem of searching for the asymmetric margin (with uneven costs of misclassiﬁcation), we have: max : ρ subject to: yi PJ j=1 ajΦj(xi) ⩾wyiρ i = 1, . . . , n, PJ j=1 aj = 1, aj ⩾0, (5.17) where w−1 = α, w1 = 1 −α and the sums are over all possible rules. As there are at most n possible values that objects from D′ k can take on each attribute, the number of rules J can reach 2nm in the worst case. Therefore, we are not able to solve the problem directly but we can use a column generation method in the dual program, similarly as described in Section 5.1.4, but with one exception. Demiriz et al. (2001) noticed that pushing the weights away from zero increases the stability of the algorithms. Following this observation, we add a set of constraints wyiqi ⩾κ n parametrized by a nonnegative value κ. Notice that κ ⩽1 in order to keep the dual feasible. Hence, the “hard margin” dual takes the form: min : β subject to: Pn i=1 qiyiΦj(xi) ⩽β j = 1, . . . , J, PJ j=1 wyiqi = 1, wyiqi ⩾κ n i = 1, . . . , n. (5.18) The procedure of rule induction is described as Algorithm 5.3. Notice that we need a linear pro-gramming solver to obtain a new weight vector in each iteration. We also need a procedure for ﬁnding the rule conditions Φj with a high value Pn i=1 qiyiΦj(xi). This value will be called edge (Breiman, 1998) and will be denoted by Eq(j). The edge corresponds to the rescaled weighted zero-one error on the training set. Thus, in each iteration we try to ﬁnd a rule with the highest possible accuracy according to the distribution with weights qi. This is very similar to the way in which base classiﬁers are generated in other boosting algorithms. Notice that the minimum is always reached in a ﬁnite number of steps when for all rules we have Eq(j) ⩽β. 86 5.3. Linear Programming Rule Ensembles (LPRules) Algorithm 5.3: Linear Programming Rule Ensembles (LPBoost). input : separable set of training examples D. output: rule ensemble {Φ1, . . . , ΦT } with weights {a1, . . . , aT }. optimum := false; J0 := ∅; qi := 1 n, i = 1, . . . , n; repeat jmax := arg maxj=1,...,J Pn i=1 qiyiΦj(xi); if Pn i=1 qiyiΦjmax(xi) ⩽β then optimum := true; end else J0 := J0 ∪{jmax}; Solve the problem: min : Pn i=1 β subject to: Pn i=1 qiyiΦj(xi) ⩽β j ∈J0, PJ j=1 wyiqi = 1, wyiqi ⩾κ n i = 1, . . . , n. to obtain new weight vector qi, i = 1, . . . , n and β; end until optimum = true; Obtain aj, j ∈J0 from the dual variables, choose from J0 a subset T of rules with nonzero weights and return as the ﬁnal rule ensemble. 5.3.2 Single Rule Generation Let us now focus on the main aspect of Algorithm 5.3 which is eﬃciently executing the operation: arg max j=1,...,J n X i=1 qiyiΦj = arg max j=1,...,J Eq(j). Since the number of rules is very large, one cannot do arg max by checking the edges of all rules, thus, it is reasonable to use a heuristic procedure instead. The procedure aims at ﬁnding the rule with a high value of the edge. In each iteration of the algorithm, one runs the procedure twice, ﬁrst generating an upward rule and next a downward rule. Then, rule with a higher edge is chosen, while the other rule is discarded. The procedure for generating upward and downward rules is very similar. The only diﬀerence is the set of allowed conditions (xj ⩾sj for upward, and xj ⩽sj for downward rules) and the class which we name to be positive. When upward rule is generated, objects with y = 1 are positive, while for downward rule, positive objects belong to the class y = −1. Description of the procedure. The procedure is presented as Algorithm 5.4. We start with the most general rule, covering the whole X. Then, conditions are added subsequently until the rule covers only positive examples. The quality of each candidate condition φ is deﬁned as the ratio of the sum of weights of negative examples to the sum of weights of positive examples removed by adding the condition. In other words, let P ⊆{1, . . . , n} be a set of positive examples, let Φ be a set of conditions before adding φ, while Φ′ is a set of conditions after adding φ. Then, the quality Learning Monotone Rule Ensembles 87 Algorithm 5.4: Rule Generation. input : weight vector {q1, . . . , qn}, set of positive examples P ⊆{1, . . . , n}. output: Rule conditions Φ. t := 0; Φt := ∅; repeat choose condition φ with the highest quality Q(φ) deﬁned by (5.19); Φt+1 := Φt ∪φ; t := t + 1; until ∀i/ ∈PΦt(xi) = 0. Choose from {Φ0, . . . , Φt} a rule with the highest edge, Φmax; Remove from Φmax those conditions, which removal does not decrease the edge; return Φmax; of candidate condition φ is deﬁned as: Q(φ) = P i/ ∈P qi1Φ(xi)̸=Φ′(xi) P i∈P qi1Φ(xi)̸=Φ′(xi) . (5.19) If the denominator is zero, the quality is inﬁnite. We assume that when we compare two conditions with inﬁnite qualities, the one with higher numerator is chosen (“more inﬁnite” one). This assump-tion essentially simpliﬁes the notation, as we do not need to handle the case with zero denominator separately. The quality (5.19) has a very intuitive meaning: we always choose a condition which decreases the number of covered positive examples as little as possible and decreases the number of covered negative examples as much as possible. Notice that the edge of the rule increases only if Q(φ) > 1. Nevertheless, we keep adding conditions even when Q(φ) < 1, until the rule covers positive examples only. The reason for this is that we keep in the memory a rule with the highest edge encountered so far, so we can always roll back the procedure and return to this rule. On the other hand, adding conditions until the rule covers no negative examples helps to avoid local minima of the edge. Notice that we use a ratio measure (5.19) rather than the diﬀerence between positive and negative examples (which precisely corresponds to the edge of the rule). This is due to the fact that a ratio measure prefers more prudent steps, i.e. conditions chosen using Q(φ) remove smaller numbers of examples. To make this clear, suppose we choose between two conditions: φ1 removes 2 positive and 3 negative examples, while φ2 removes 8 positive and 6 negative examples from the rule. Although φ2 increases the edge by 2, while φ1 by 1, the ratio measure Q(φ) prefers φ1, because it corresponds to a more gentle step, with possibility of improvement in later steps. Having chosen the condition with the highest edge, denoted by Φmax, we enter a “pruning” phase which consists in removing spare conditions, i.e. conditions which do not decrease the edge. This makes the rule simpler (having less conditions) which, in turn, increases its generalization ability. Finally, notice that the rule generation procedure described above can be substituted by any other reasonable procedure which is capable of ﬁnding a rule which minimizes a weighted error on the training set. The main Algorithm 5.3 remains unchanged. Complexity issues. The computational complexity of the two-phase procedure is rather poor in the worst case, because solving the linear program may take a long time (yet still polynomial 88 5.4. Sigmoid Loss Monotone Rule Ensembles (MORE) in n and T). Moreover, the monotone approximation has time complexity O(n3). Nevertheless, both procedures work very fast on the real datasets, much below the worst-case time. An addi-tional speed-up is achieved by using all the methods for reducing the complexity of the monotone approximation, described in Chapter 3. The rule generation procedure is linear in n, therefore it scales very well with the size of the dataset. The scaling behavior is theoretically quadratic in m in the worst case, when the number of conditions is comparable to the number of attributes. In real cases, however, the number of conditions is much smaller than the number of attributes, which makes the procedure scalable also with m. 5.4 Sigmoid Loss Monotone Rule Ensembles (MORE) We present in this section another approach to rule induction. The algorithm has been in-troduced in (Dembczyński et al., 2008b) and was the ﬁrst boosting-based algorithm for ordinal classiﬁcation with monotonicity constraints. Therefore, it was simply named monotone rule ensem-bles and abbreviated MORE. Here, we will refer to the algorithm as sigmoid loss monotone rule ensembles, however we keep the original abbreviation (MORE). The algorithm was not motivated by the margin theorem presented in Section 5.2.5. It is rather motivated by statistical considerations described by Friedman et al. (1998) and developed by Friedman and Popescu (2004). Moreover, it uses a speciﬁc, non-convex approximation of the zero-one loss function – the sigmoid loss. The algorithm also works by transforming the general K-class problem into K −1 binary subproblems, similarly as it was presented in Section 5.2. The main diﬀerence is the method of combining binary classiﬁers into the multi-class classiﬁer, which will be described below. 5.4.1 Combining Binary Classiﬁers We start with presenting a diﬀerent method of combining K −1 binary classiﬁers into a single K-class classiﬁer, originally used in MORE (Dembczyński et al., 2008b). Let us assume that each binary classiﬁer has the form hk(x) = sgn (fk(x)), where fk(x) is a real-valued function (e.g. ensemble of classiﬁers) such that the magnitude \|fk(x)\| corresponds to the conﬁdence of prediction – the higher the magnitude, the more we are certain about predicting the class sgn (fk(x)). Then, we could take advantage of additional information by using the value of fk(x) instead of merely the sign. Since fk(x) is an increasing function of the conﬁdence of classiﬁcation to the class union {k, . . . , K}, then it should hold: fk+1(x) ⩽fk(x), (5.20) because we are always more certain about classifying object to the set {k, . . . , K} than to its subset {k + 1, . . . , K}. If (5.20) holds, then the classiﬁcation procedure h(x) is doubtless: we seek for k, for which the sequence fk(x), k = 2, . . . , K, changes the sign and classify x to the class k. This is equivalent to comprehensively writing h(x) = 1 + PK k=2 1fk(x)⩾0. However, binary problems are solved independently, so we cannot guarantee that such con-straints hold in each case. We deal with the violation of the constraints (5.20) using the isotonic regression in the following way. Fix x and notice that from (5.20) it follows that fk(x) must be a monotonically decreasing function of k. If fk(x) is not monotonically decreasing, we search for another function gk(x) which is monotonically decreasing and is as close as possible to fk(x) in the Learning Monotone Rule Ensembles 89 sense of squared error: min K X k=2 (fk(x) −gk(x))2. This is exactly the problem of isotonic regression, as described in Section 1.3.3. It can be thought of as “monotonizing” the function fk(x) violating the constraints (5.20). What is surprising, we do not even need to solve the isotonic regression. Let us consider the following algorithm of combining K −1 classiﬁers fk(x), k = 2, . . . , K, to obtain a class label h(x) ∈{1, . . . , K}. The algorithm calculates votes for each class and the class with the highest vote is chosen as the prediction h(x). Let us denote the vote for class k as votek. The vote is calculated in the following way: votek(x) = k X l=2 fl(x) (5.21) The following theorem holds: Theorem 5.7. Consider the classiﬁer: b(x) = 1 + K X k=2 1gk(x)⩾0, where for each k = 2, . . . , K, gk(x) is the isotonic regression of fk(x). Let votek = Pk l=2 fl(x) and let us deﬁne the classiﬁer h(x) = arg maxk votek (in case of ties, we choose the highest label). Then, h(x) = b(x). Proof. Let us ﬁx x; we will omit the dependency on x and write fk, gk, h, b, etc. Notice that h is such that for any k > h it holds voteh > votek, while for any k ⩽h it holds voteh ⩾votek. To show that b = h, it is enough to show that if k > b then voteb > votek and if k ⩽b then voteb ⩾votek. Let us deﬁne G+ = {k: gk ⩾0} and similarly G−= {k: gk < 0}. Notice that G+ = {2, . . . , b} and G−= {b + 1, . . . , K}. Moreover, let us also denote f(A) = P k∈A fk. We will use Theorem 1.4.3 from (Dykstra et al., 1999), which states that for every k = 2, . . . , K it holds: f({2, . . . , k} ∩G−) < 0 f({k + 1, . . . , K} ∩G+) ⩾0. Suppose k ⩽b. Then: voteb −votek = b X l=k+1 fl = f({2, . . . , b} ∩{k + 1, . . . , K}) = f({k + 1, . . . , K} ∩G+) ⩾0, so that voteb ⩾votek. Similarly, if k > b then: votek −voteb = k X l=b+1 fl = f({2, . . . , k} ∩{b + 1, . . . , K}) = f({2, . . . , k} ∩G−) < 0, so that voteb > votek, which ends the proof. We would like to have the monotonicity property of the classiﬁer created according to Theo-rem 5.7, i.e. if x ⪰x′ then h(x) ⩾h(x′). The following theorem gives suﬃcient conditions for monotonicity: Theorem 5.8. For each k = 2, . . . , K, let fk(x) be a monotone function, i.e.: x ⪰x′ →fk(x) ⩾ fk(x′). Then, the classiﬁer h(x), obtained by choosing the class label with the highest vote (5.21), is a monotone function. 90 5.4. Sigmoid Loss Monotone Rule Ensembles (MORE) Proof. Choose any r, s ∈{2, . . . , K} such that r ⩾s. Then: voter(x) −votes(x) = r X k=s+1 fk(x) ⩾ r X k=s+1 fk(x′) = voter(x′) −votes(x′) It follows from the deﬁnition that if k ⩽h(x′) then voteh(x′)(x′) −votek(x′) ⩾0. But this also means that voteh(x′)(x) −votek(x) ⩾0 which, in turn, means that h(x) ⩾h(x′). We presented an alternative method of combining binary classiﬁers which takes into account the prediction conﬁdence of each classiﬁer. The main problem of this method is that one cannot bound the loss of the composed classiﬁers by means of the votes of binary classiﬁers, as it was done in Theorem 5.2. This convinced us to abandon this method in case of using with LPRules. Nevertheless, although the loss of a composite classiﬁer cannot be simply bounded, nothing prevents this method to work well in practice on real-life problems. Therefore, this method was ﬁnally used with MORE. 5.4.2 Rule induction with Sigmoid Loss In each of the binary subproblems, rule induction is performed by minimizing the weighted 0-1 loss (linear loss) function on the dataset. For Y = {−1, 1}, 0-1 loss can be expressed as L0−1(yf(x)) = wy1yf(x)<0. This function, however, is neither smooth nor diﬀerentiable. Therefore, following the arguments raised in Section 5.1.2, we approximate 0-1 loss with the sigmoid function (5.6): Lsigm(yf(x)) = wy 1 1 + eyf(x) . (5.22) Thus, we minimize the following empirical risk: Remp(f) = n X i=1 Lsigm(yif(xi)) (5.23) However, ﬁnding a set of rules minimizing (5.23) is computationally hard. That is why we follow the boosting strategy, i.e. the rules are added one by one, greedily minimizing (5.23). We start with the “default” rule deﬁned as: a0 = arg min a Remp(a) = arg min a n X i=1 Lsigm(ayi). (5.24) The default rule is a real value but can be thought of as a rule covering the whole space X. In each subsequent iteration, a new rule is added by taking into account previously generated rules. Let ft−1(x) be a classiﬁcation function after t −1 iterations, consisting of the ﬁrst t −1 rules and the default rule. In the t-th iteration, a decision rule can be obtained by solving: (at, Φt(x)) = arg min Φ,a Rt(Φ, a) (5.25) where we deﬁned: Rt(Φ, a) = n X i=1 Lsigm(yi(ft−1(xi) + aΦ(xi))). (5.26) Monotonization of the data. Similarly as in LPRules, we are able to do the monotone approx-imation of the k-th dataset Dk and feed the k-th learning algorithm with the monotonized data D′ k. Now, however, the monotonization process is not necessary as we do not aim at separating the dataset with rules. Nevertheless, the process of rule induction is a greedy optimization procedure Learning Monotone Rule Ensembles 91 Algorithm 5.5: Monotone Rule Ensemble (MORE). input : set of training examples D, T – number of rules to be generated. output: monotone rule ensemble {r1, . . . , rT }. f0(x) := arg min±a R0(1, ±a); for t = 1 to T do Φt = arg minΦ Rt(Φ, a); rt(x) = aΦt(x); ft(x) = ft−1(x) + rt(x); end and monotonization may be helpful for the process to converge faster. This results in a smaller number of rules and better interpretability of the model. What is more, we did not observe any signiﬁcant deﬁciency in accuracy, when monotonization is applied. Since the monotone approximation usually takes much less time than the process of rule in-duction, using it as a preprocessing step does not cost much computational eﬀort, therefore it was incorporated into the rule induction procedure. 5.4.3 Single Rule Generation The exact solution of (5.25) is still computationally hard, because we need to determine the optimal Φt(x) and at simultaneously. Therefore, we restrict our algorithm to the case, in which all rules have weights of the same magnitude, equal to a, i.e. we only allow at = a. The magnitude a is a ﬁxed parameter of the algorithm. With such a restriction, (5.25) becomes: Φt(x) = arg min Φ Rt(Φ, a), (5.27) and it requires calculating only two loss values at points ft−1(xi) and ft−1(xi) ± a (+a for upward rules and −a for downward rules) for every object xi. The default rule is solved in a similar way by restricting the minimization of (5.24) to the values a0 ∈{−a, a}. In each subsequent iteration, problem (5.27) can be solved via a heuristic procedure for rule generation, deﬁned as follows. The procedure generates ﬁrst the upward rule and next the downward rule. Then, both rules are compared and the one with lower empirical risk is chosen, while another one is discarded. The procedures for generating the upward and downward rules are almost identical, therefore they will be presented simultaneously: • At the beginning, Φt is empty (no elementary conditions are speciﬁed), i.e. Φt(x) ≡1. • In each step, an elementary condition xj ⩾sj (for upward rule) or xj ⩽sj (for downward rule) is added to Φ that minimizes Rt(Φ, a). Such expression is searched by consecutive testing of elementary conditions, attribute by attribute. Let x(1) j , x(2) j , . . . , x(N) j be a sequence of ordered values of the j-th attribute, such that x(i−1) j ⩾x(i) j , for i = 2, . . . , n. Each elementary condition of the form xj ⩾sj (for upward rule) or xj ⩽sj (for downward rule) for each sj = x(i−1) j +x(i) j 2 is tested. • The previous step is repeated until Rt(Φ, a) cannot be decreased. The above procedure is very fast and proved to be eﬃcient in computational experiments. The attributes can be sorted once before generating any rule. The procedure for ﬁnding Φt resembles 92 5.4. Sigmoid Loss Monotone Rule Ensembles (MORE) the way the decision trees are generated. Here, we look only for one path from the root to the leaf. Moreover, let us notice that minimal value of Rt(Φ, a) is a natural stop criterion in building a single rule. The whole procedure is presented as Algorithm 5.5. 5.4.4 Analysis of the Step Length We now analyze the behavior of the rule induction algorithm depending on the value of the parameter a, i.e. the magnitude of the weight for each rule. Notice that this parameter corresponds to the scale of the loss function, since: f(x) = ±a + T X t=1 aΦt(x) = a ±1 + T X t=1 Φt(x) ! = a ˜ f(x), where ˜ f(x) = ±1 + PT t=1 Φt(x). Then: Lsigm(yf(x)) = 1 1 + exp(ay ˜ f(x)) . Thus, small values of a cause the loss function to broaden and the changes in the slope of the function are smaller (the loss becomes similar to the linear function). On the other hand, large values of a cause the sigmoid loss to become similar to the 0-1 loss. In general, large values of a correspond to a larger complexity (Mason et al., 1999), because we are able to decrease the error signiﬁcantly in a smaller number of steps. For a better insight into the problem, we state the following general theorem: Theorem 5.9. Minimization of (5.27) for any twice diﬀerentiable loss function L(yf(x)) and any a is equivalent to the minimization of: Rt(Φ, a) = X i∈R− wt i + 1 2 X Φ(xi)=0 wt i −avt i . (5.28) where we deﬁne: R−= {i: Φ(xi)yi < 0} (5.29) wt i = −∂ ∂uL(u) u=yift−1(xi) (5.30) vt i = 1 2 ∂2 ∂u2 L(u) u=yift−1(xi)±γiayi . (5.31) for some γi ∈[0, 1], where the “plus” sign in (5.31) is for rules Φ(xi) ⩾0, while the “minus” sign is for rules Φ(xi) ⩽0. Proof. From Taylor expansion it follows that for a twice diﬀerentiable loss function we have: L(u + z) = L(u) + z ∂ ∂uL(u) + z2 2 ∂2 ∂u2 L(u + γz) for some γ ∈[0, 1]. By denoting Li = L(yi(ft−1(xi))), using (5.30)-(5.31) and substituting u = yift−1(xi) and z = Φ(xi)ayi, we have for every xi such that Φ(xi) ̸= 0: L(yi(ft−1(xi) + aΦ(xi))) = Li −ayiΦ(xi)wt i + a2Φ2(xi)vt i. Thus, the empirical risk becomes: Rt(Φ, a) = X Φ(x)i̸=0 Li −ayiΦ(xi)wt i + a2vt i + X Φ(x)i=0 Li. Learning Monotone Rule Ensembles 93 The term Pn i=1 Li is constant, so it can be dropped from the optimization process. Thus, we equivalently minimize: Rt(Φ, a) = X Φ(x)i̸=0 −ayiΦ(xi)wt i + a2vt i = X i∈R− awt i − X i∈R+ awt i + X Φ(x)i̸=0 a2vt i, where R+ = {i: Φ(xi)yi > 0} and R−is deﬁned in (5.29). We now use the fact that P i∈R+ = Pn i=1 −P i∈R−−P Φ(xi)=0 and that P Φ(xi)̸=0 = Pn i=1 −P Φ(xi)=0 to obtain: Rt(Φ, a) = n X i=1 (a2vt i −awt i) + 2a X i∈R− wt i + a X Φ(xi)=0 (wt i −avt i), and by dropping the ﬁrst constant term and dividing by constant value 2a, we prove the theorem. Thus, a establishes a trade-oﬀbetween misclassiﬁed and unclassiﬁed examples. Values vt i are always positive, because the loss function is decreasing. Sigmoid loss is convex for yf(x) > 0 and concave for yf(x) < 0, therefore, as a increases, uncovered examples satisfying yift−1(xi) > 0 (“correctly classiﬁed”) are penalized less, while the penalty for uncovered “misclassiﬁed” examples (yift−1(xi) < 0) increases. This leads to the following conclusion: although the rule covers only a part of the examples, with respect to uncovered examples it still tries to make a small error; remark that the weights of the uncovered examples depend on the curvature of the function (second derivative) rather than on the slope (ﬁrst derivative). Appendix: Proofs of the Theorems Proof of Theorem 5.5 Theorem 5.6. Let f(x) = P j ajbj(x) be the convex combination of classiﬁers bj(x) ∈{−1, 1}, i.e. aj are non-negative and P j aj = 1. Then, with probability 1 −δ, for every distribution P(y, x) the following inequality holds: R(f) ⩽ inf γ∈(0,1]  Rγ emp(f) + M   s d nγ2 + s log 1 δ n    , (5.32) where d is the Vapnik-Chervonenkis dimension of the base classiﬁer and M is some universal con-stant. Proof. Our proof is very similar to the proof of Koltchinskii and Panchenko (2002). We follow more accessible version of the proof given by Lugosi (2002)1 and describe here only the diﬀerences from the unweighted loss case. Let us deﬁne two functions φ1(x) and φ−1(x) as follows: φ1(x) =      1 −α if x ⩽0 0 if x ⩾γ(1 −α) 1 −α −x/γ if x ∈(0, γ(1 −α)) φ−1(x) =      α if x ⩽0 0 if x ⩾γα α −x/γ if x ∈(0, γα)) . 1Lugosi (2002) proves the theorem for a ﬁxed value of γ, but it can be extended to all values of γ in the same way as in (Koltchinskii and Panchenko, 2002). 94 Appendix Observe that wy1yf(x)<0 ⩽φy(yf(x)) ⩽wy1yf(x)⩽γwy. Thus, we have: sup f∈F (R(f) −Rγ emp(f)) ⩽sup f∈F Eφy(yf(x)) −1 n n X i=1 φyi(yif(xi)) ! , where F is the class of all ensembles. Using the bounded diﬀerence inequality (McDiarmid, 1989; Devroye et al., 1996) with ci = n−1 max{α, 1 −α} ⩽n−1 we bound the distance of the right-hand side from its expectation exactly as in (Lugosi, 2002). Deﬁne z = yf(x) and zi = yif(xi). By symmetrization argument we have: E sup f∈F Eφy(z) −1 n n X i=1 φyi(zi) ! ⩽E sup f∈F 1 n n X i=1 σi(φy′ i(z′ i) −φyi(zi)) ! where σi, i = 1, . . . , n, are i.i.d. symmetric sign variables, and (x′ i, y′ i) forms a sample of size n, being independent of (xi, yi) and having the same distribution. Let us deﬁne φ(x) = φ1(x)+φ−1(x) 2 . Then, using: φy′ i(z′ i) −φyi(zi) = (φy′ i(z′ i) −φ(z′ i)) −(φy′ i(0) −φ(0)) + (φ(z′ i) −φ(zi)) + (φyi(0) −φy′ i(0)) −(φyi(zi) −φ(zi)) −(φyi(0) −φ(0)) we bound: E sup f∈F 1 n n X i=1 σi(φy′ i(z′ i)−φyi(zi))⩽2E sup f∈F 1 n n X i=1 σi(φyi(zi)−φ(zi)−φyi(0)+φ(0)) + E sup f∈F 1 n n X i=1 σi(φ(z′ i) −φ(zi)) + E sup f∈F 1 n n X i=1 σi(φyi(0) −φy′ i(0)). (5.33) The last term on the right-hand side is equal to zero. The middle term can be bounded as in (Lugosi, 2002) by noticing that ψ(x) = γ(φ(x) −φ(0)) is a contraction. By noticing the symmetry φyi(zi) − φ(zi) = −(φ−yi(zi) −φ(zi)) and exploiting the fact that σi are i.i.d. symmetric sign variables, the expression under sup in the ﬁrst term on the right-hand side equals to φ1(zi)−φ(zi)−φ1(0)+φ(0), which is also the contraction after multiplying by γ. Thus, in summary we bound: E sup f∈F 1 n n X i=1 σi(φy′ i(z′ i) −φyi(zi)) ! ⩽4 γ E sup f∈F 1 n n X i=1 σif(xi). Notice that the bound is twice the bound of the symmetric case which, in turn, doubles universal constant M in comparison to the symmetric case. The rest of the proof proceeds exactly in the same way as in (Lugosi, 2002). Proof of Theorem 5.6 Theorem 5.7. Assume P(x, y) is monotonically constrained. Let h(x) be a classiﬁer output by the two-phase method. Let fk(x), k = 2, . . . , K, be k-th rule ensemble trained on the monotonized dataset D′ k such that there exists γk > 0 for which y′ ikfk(xi) ⩾γkwy′ ik for all i = 1, . . . , n (i.e. fk(x) achieves hard margin γk on the dataset D′ k). Then, with probability at least 1 −γ: R(h) ⩽R∗+ M  2(K −1) s log 2(K−1) δ n + rm n K X k=2 1 γk  . for some universal constant M. Learning Monotone Rule Ensembles 95 Proof. We start with a transformation of rules (which can abstain from the response) to classiﬁers with output in {−1, 1} and with well-deﬁned VC dimension. Let f(x) = PT t=1 atΦt(x). With each rule Φt(x), t = 1, . . . , T, we associate a cone gt(x) = 2Φt(x) −1 for upward rules (Φ(x) ⩾0) and gt(x) = 2Φt(x) + 1 for downward rules (Φ(x) ⩽0). Notice that gt(x) ∈{−1, 1} and either the set {gt(x) = 1} or the set {gt(x) = −1} is an axis-parallel cone, i.e. set of the form {x: x ⪰x0} or of the form {x: x ⪯x0} for some x0. The class of axis-parallel cones has VC dimension m (Devroye et al., 1996). We add to the ensemble one more cone g0(x) = sgn P t∈T + at −P t∈T −at which covers the whole X and the subsets T + and T −correspond to the upward and downward rules, respectively. Then, one can easily show that: T X t=0 ctgt(x) = T X t=1 atΦt(x), when ct = at 2 for t ⩾1 and c0 = 1 2\| P t∈T + at−P t∈T −at\|. Hence, we see that each rule ensemble has an equivalent cone ensemble. Now, suppose rule ensemble is L1-normalized, i.e. PT t=1 at = 1. The corresponding cone ensemble must be divided by the sum of weights ℓ= PT t=0 ct to be normalized; let us denote such ensemble by f ′(x) = f(x)/ℓ. Notice that: ℓ= 1 2 X t∈T + at − X t∈T − at + T X t=1 αt 2 ⩽ T X t=1 at = 1, so that if yf ′(x) ⩽γ, then also yf(x) ⩽γ. In other words, the fraction of margin errors for the normalized cone ensemble upperbounds the fraction of margin errors for the normalized rule ensemble. Thus, we can prove the theorem for an ensemble of cones, which are ordinary classiﬁers with well-deﬁned VC dimension, and the theorem will also hold for a rule ensemble. Let Rγ emp(fk) be deﬁned as before in (5.15), as the weighted fraction of margin violations for the k-th monotone ensemble with normalized weights. Consider the k-th ensemble fk(x) trained on the monotonized data D′ k. The ensemble makes no margin errors on D′ k, i.e. for all i, it holds y′ ikfk(xi) ⩾γkwy′ ik. But since the sets Dk and D′ k diﬀer only on the relabeled objects, the ensemble makes on Dk a margin error Rγk emp not greater than: Rγk emp(f) ⩽Γ = 1 n n X i=1 wyik1yik̸=y′ ik. But the upper bound Γ equals to the objective function of the binary monotone approximation problem (3.16), hence it is minimized by values y′ ik in the class of all monotone functions. From the assumption about the monotonically constrained distribution it follows, however, that the Bayes classiﬁer h∗ k(x) is monotone, which means that: Rγk emp(f) ⩽Γ ⩽1 n n X i=1 wyik1yik̸=h∗ k(xi) = 1 n n X i=1 L(yik, h∗ k(xi)), where L stands for the linear loss. Notice that E[L(yk, h∗ k(x))] = R∗ k, the Bayes risk in the k-th binary problem, so using Hoeﬀding’s bound (Devroye et al., 1996) we can state that for every δ′ ∈(0, 1): P  1 n n X i=1 L(yik, h∗ k(xi)) −R∗ k ⩾ s log 1 δ′ 2n  ⩽δ′. From Theorem 5.5 we have with probability at most δ′′: R(fk) ⩾ inf γ∈(0,1]  Rγ emp(fk) + M   r m nγ2 + s log 1 δ′′ n    . 96 Appendix By setting δ′ = δ′′ = δ 2(K−1), we have with probability at most δ′ + δ′′ = δ K−1: R(fk) ⩾R∗ k + M   r m nγ2 k + 2 s log 2(K−1) δ n  . (5.34) To bound the risk in a general K-class case, notice that with probability at most δ there exists k ∈{2, . . . , K} such that (5.34) holds. But this means that with probability at most δ: K X k=2 R(fk) ⩾ K X k=2 R∗ k + M K X k=2   r m nγ2 k + 2 s log 2(K−1) δ n  . We take advantage of the fact that: R∗= E[L(y, h∗(x)] = K X k=2 E[L(yk, h∗ k(x))] = K X k=2 R∗ k, and from Theorem 5.2 we have: R(h) = E[L(y, h(x)] ⩽ K X k=2 E[L(yk, hk(x))] = K X k=2 R(fk). Thus, we conclude that with probability at most δ: R(h) ⩾R∗+ M  2(K −1) s log 2(K−1) δ n + rm n K X k=2 1 γk  . Chapter 6 Computational Experiment In this chapter, we verify the eﬃciency of our methods in the computational experiment. We also compare our methods with already existing approaches to ordinal classiﬁcation with monotonicity constraints and analyze the results with use of the nonparametric statistical tests. The experiment is conducted on several real datasets, for which it is known that there exist monotone relationships between attribute values and class labels. As the accuracy is not the only criterion of assessing the quality of the learning method, we also consider the interpretability of the models. The interpretabilty was actually the major point for which we deal with decision rules, therefore an analysis of this issue is conducted in this chapter. 6.1 Design of the Experiment 6.1.1 Datasets We found 12 datasets, for which it is known that monotonicity constraints are present. We did not search for monotonicity directions by calculating any particular statistics, rather we obtained the directions using the domain knowledge about the problem. Four datasets, which are the results of surveys, were taken from Ben-David (1992, 1995) and were accompanied by the following descriptions: • ESL (employee selection) dataset contains proﬁles of applicants for certain industrial jobs. • SWD (social workers decisions) dataset contains real-world assessments of qualiﬁed social work-ers regarding the risk facing children if they stayed with their families at home. • LEV (lecturers evaluation) dataset contains examples of anonymous lecturer evaluations, taken at the end of MBA courses. • ERA (Employee Rejection/Acceptance) dataset was originally gathered during an academic decision-making experiment aiming at determining which are the most important qualities of candidates for a certain type of jobs. Three datasets are related to the problem of house pricing: • Housing dataset comes from the UCI repository (Asuncion and Newman, 2007) and concerns housing values in suburbs of Boston. • Windsor dataset ﬁrst appeared in (Koop, 2000) and concerns housing values in Windsor, Canada. 98 6.1. Design of the Experiment Table 6.1: Description of data sets used in experiments. Data set #attributes #objects #classes ESL 4 488 8 SWD 10 1000 4 LEV 4 1000 5 ERA 4 1000 8 Housing 8 506 4 Windsor 11 546 4 DenBosch 9 119 2 Wisconsin 9 699 2 Ljubljana 8 286 2 Car 6 1728 4 CPU 6 209 4 Balance 4 625 3 • DenBosch dataset was taken from (Daniels, 1999) and concerns housing values in small Dutch city Den Bosch; see (Daniels and Feelders, 2000) for overview. From the three house pricing datasets only DenBosch dataset contained a discrete output variable (price discretized into two levels). We decided to discretize the price variable in Housing and Windsor into four levels containing equal number of objects (i.e. quartiles of the price distribution), similarly as in (Feelders and Pardoel, 2003). There are ﬁve other datasets taken from the UCI repository: • Wisconsin breast cancer dataset. • Ljubljana breast cancer dataset, in which some of the non-monotone attributes and attributes containing most of missing values, have been removed. • Car evaluation data. • CPU performance data, for which the class attribute was discretized into four levels, containing equal number of objects. • Balance scale dataset. For all datasets, objects with missing values were removed, since not every method is able to deal with missing values (rule ensembles have a very natural way to handle the missing values, which is included in our implementation, however, its description is beyond the scope of this thesis). The quantitative characteristics of the datasets are shown in Table 6.1. 6.1.2 Algorithms In the experiment, we used ten algorithms in total, among which three were invented by us and introduced in this thesis. Four other methods are well known in the ﬁeld of ordinal classiﬁcation with monotonicity constraints and can be though of as the state of the art in this domain. The last three methods are ordinary classiﬁers which are not suited to the ordinal classiﬁcation and do not take monotonicity constraints into account. We include these methods to assess whether incorporating domain knowledge about order and monotonicity gives any improvement in accuracy. Up to our knowledge, there was no such an extensive comparison of so many algorithms ever before in this ﬁeld. Computational Experiment 99 State-of-the-art methods. We considered four already existing methods to ordinal classiﬁcation with monotonicity constraints: • Ordinal Learning Model (OLM) with implementation obtained from Weka (Witten and Frank, 2005). • Ordinal Stochastic Dominance Learner (OSDL), implementation was obtained from Weka, bal-anced version was used (as it was recommended by Cao-Van (2003) to give better results). The internal cross-validation for tuning the interpolation parameter was turned on, because algorithm was quite fast anyway. • Isotonic Separation (IsoSep) was implemented by us, using Weka and linear programming solver lp solve (Berkelaar, 2005), according to (Chandrasekaran et al., 2005), with absolute error cost matrix. As suggested, all the attributes were normalized before calculating the distance in the classiﬁcation procedure. • Rule induction with VCDomLEM algorithm obtained from J. Błaszczyński and M. Szeląg, as a part of the Java Rough Sets (JRS) library. The classiﬁcation procedure used in the algorithm was the one described in (Błaszczyński et al., 2007). A variable consistency version was used, with consistency level 0.9. The methods were described in detail in Section 1.3. Our methods. Apart from already existing approaches, we tested the following three methods, introduced in this thesis: • Multiple isotonic regression (IsoReg), using the heuristic algorithm of Burdakov et al. (2006). In order to estimate the probabilities outside the training set, we must use some extension of isotonic regresion (see Section 3.1.4). We used the extension deﬁned by (3.7) with λ = 0.5. The classiﬁcation was performed by taking the median of the distribution. • Linear Programming Rule Ensemble (LPRules), using our implementation in Weka and linear programming solver lp solve. LPRules does not have any parameters to set apart from a technical parameter κ (the smallest value of object’s weight), which maintains the stability of the optimization steps. We observed that the performance of the algorithm remains the same when we change κ provided we keep it small but nonzero. Thus, we set κ = 02. • Sigmoid-loss Monotone Rule Ensembles (MORE), using our implementation in Weka. Since we did not optimize other algorithms, we decided to choose standard parameters, setting the number of rules M to 50 (per each binary subproblem) and the scale (step length) a to 0.5. Notice that in the real applications, those values would be obtained by cross-validation, probably leading to better results. Ordinary methods. We decided to use three ordinary classiﬁers in order to check whether incorporating domain knowledge about order and monotonicity gives any improvement in accuracy: • j48 is the Weka’s implementation of famous tree induction algorithm C4.5 (Quinlan, 1993). Unfortunately, j48 is designed to minimize 0-1 error and does not handle the order between class labels; using it directly on multi-class problems led to very poor results in terms of the mean absolute error. Therefore, we decided to improve its performance and use it in the ordinal setting by combining it with a simple approach to ordinal classiﬁcation proposed by Frank and Hall (2001). 100 6.2. Experimental Results This approach divides K-class ordinal problem into K −1 binary problems in a similar way as it was done in this thesis. Then, for each binary problem, a base classiﬁer (j48 in this case) is learned and the conditional probability of upward class union P(y ⩾k\|x) is estimated. Plugging those estimates (instead of real probabilities) into the expression deﬁning the Bayes classiﬁer allows us to minimize any ordinal loss function. For instance, if we use the absolute error loss function, then the Bayes classiﬁer is the median of conditional distribution, so the output of our classiﬁer will be the median of the estimated distribution. • SVM (Support Vector Machine) classiﬁer proposed by Boser et al. (1992); Vapnik (1998). We used its implementation in Weka. A standard linear kernel was used and the complexity parameter was set to default value. Similarly as in the case of j48, we used a simple approach to ordinal classiﬁcation, because SVM in its basic version is not suited to capture the order between the labels and minimizes 0-1 error rather than the mean absolute error. • NB (Na¨ ıve Bayes), which estimates the probabilities P(x\|y) and uses Bayes’ rule to obtain P(y\|x). We plug those probabilities into the expression deﬁning the Bayes classiﬁer to out-put the predicted label. For instance, when dealing with absolute error loss function, our prediction is the median of the estimated distribution. 6.2 Experimental Results Error estimation. The measure of error was chosen to be the mean absolute error (MAE). It reﬂects the ordinal nature of the problem by penalizing the classiﬁer according to the diﬀerence between predicted and observed class labels. The error of each classiﬁer was estimated by a 10-fold cross validation, repeated 10 times to improve the replicability of the experiment (so that the results of the experiment do not depend on particular train/test folds splits). Apart from the average error, its standard deviation was estimated from the measurements on each fold and each trial (100 measurements in total). To avoid underestimation, the standard deviation of error was estimated conservatively, by taking into account the dependence between the subsequent testing samples in the repeated cross-validation, as described in (Nadeau and Bengio, 2003). The results of both average error and standard deviation are given in Table 6.2. Testing statistical signiﬁcance. It was stressed (Bengio and Grandvalet, 2004; Nadeau and Bengio, 2003) that there is no good (unbiased) way of estimating the standard error of accuracy and using such estimates in signiﬁcance test may give misleading, unreliable results. Therefore, we perform nonparametric signiﬁcant test. To compare multiple classiﬁers on the multiple datasets, we follow (Demˇ sar, 2006), and apply the Friedman test, which uses ranks of each algorithm to check whether all the algorithms perform equally well (null hypothesis). The ranks are calculated for each dataset and the lowest rank (1) means that an algorithm performed the best on this dataset, while the highest rank (10) means that an algorithm performed the worst. Then, the ranks are averaged for each classiﬁer and Friedman statistics is computed: χ2 F = 12d c(c + 1)   c X j=1 ¯ r2 j −c(c + 1)2 4  , where d is the number of datasets, c is the number of compared classiﬁers and ¯ rj is the average rank of j-th classiﬁer; χ2 F is distributed approximately according to χ2 with c −1 degrees of freedom. Notice that Friedman statistics depends on results of the experiment only through the average ranks of the classiﬁers – the particular values of the mean absolute error do not matter. Computational Experiment 101 Table 6.2: Results of the experiment. For each dataset (row) and each classiﬁer (column) two values are given: average mean absolute error (above) and standard deviation of error (below, starting with ±). Dataset OLM OSDL IsoSep DOMLEM LPRules MORE IsoReg J48 SVM NB DenBosch 0.282 0.157 0.183 0.125 0.168 0.133 0.165 0.172 0.202 0.126 ±0.039 ±0.039 ±0.037 ±0.034 ±0.034 ±0.03 ±0.038 ±0.032 ±0.036 ±0.031 Wisconsin 0.17 0.039 0.03 0.038 0.041 0.031 0.04 0.046 0.03 0.037 ±0.014 ±0.008 ±0.007 ±0.008 ±0.009 ±0.007 ±0.008 ±0.009 ±0.007 ±0.007 ESL 0.371 0.353 0.328 0.432 0.323 0.344 0.328 0.369 0.355 0.333 ±0.024 ±0.025 ±0.023 ±0.024 ±0.024 ±0.023 ±0.023 ±0.022 ±0.023 ±0.024 SWD 0.452 0.44 0.442 0.449 0.435 0.441 0.44 0.442 0.435 0.457 ±0.017 ±0.017 ±0.018 ±0.017 ±0.017 ±0.016 ±0.017 ±0.016 ±0.016 ±0.016 LEV 0.427 0.405 0.398 0.513 0.396 0.413 0.393 0.415 0.444 0.441 ±0.018 ±0.016 ±0.017 ±0.015 ±0.016 ±0.016 ±0.017 ±0.018 ±0.016 ±0.017 ERA 1.256 1.271 1.271 1.393 1.263 1.269 1.261 1.217 1.271 1.227 ±0.031 ±0.033 ±0.034 ±0.04 ±0.033 ±0.03 ±0.033 ±0.032 ±0.029 ±0.031 Housing 0.527 0.775 0.286 0.337 0.274 0.288 1.187 0.332 0.314 0.506 ±0.032 ±0.021 ±0.02 ±0.025 ±0.021 ±0.023 ±0.032 ±0.023 ±0.025 ±0.033 CPU 0.29 0.215 0.099 0.086 0.073 0.065 0.232 0.1 0.371 0.18 ±0.035 ±0.028 ±0.02 ±0.019 ±0.018 ±0.017 ±0.033 ±0.019 ±0.03 ±0.033 Balance 0.224 0.111 0.19 0.221 0.063 0.126 0.207 0.271 0.137 0.085 ±0.02 ±0.014 ±0.017 ±0.015 ±0.009 ±0.015 ±0.016 ±0.021 ±0.017 ±0.013 Ljubljana 0.325 0.274 0.241 0.289 0.25 0.251 0.24 0.259 0.299 0.252 ±0.033 ±0.027 ±0.024 ±0.029 ±0.026 ±0.022 ±0.021 ±0.021 ±0.023 ±0.025 Windsor 0.576 0.512 0.52 0.519 0.516 0.538 0.534 0.565 0.491 0.509 ±0.028 ±0.025 ±0.028 ±0.028 ±0.026 ±0.025 ±0.026 ±0.025 ±0.026 ±0.029 Car 0.084 0.031 0.045 0.023 0.03 0.061 0.038 0.09 0.078 0.177 ±0.01 ±0.005 ±0.006 ±0.005 ±0.004 ±0.006 ±0.005 ±0.008 ±0.007 ±0.008 Friedman statistics gives 26.73 which exceeds the critical value 15.51 (for conﬁdence level 05), so we can reject the null hypothesis and state that classiﬁers are not equally good. Next, we proceed to a post-hoc analysis and calculate the critical diﬀerence (CD) according to the Nemeneyi procedure: CD = qα r c(c + 1) 6d , where critical value qα is based on Studentized range statistics divided by √ 2 (see e.g. Demˇ sar (2006) for the table of critical values). We obtain CD = 3.91 which means that algorithms with diﬀerence in average ranks more than 3.91 are signiﬁcantly diﬀerent. In Figure 6.2, average ranks were marked on a line, and groups of classiﬁers that are not signiﬁcantly diﬀerent were connected. This shows that algorithms such as LPRules and MORE are signiﬁcantly diﬀerent only to OLM. No other signiﬁcant diﬀerences were obtained. This is mostly due to the fact that we used only 12 datasets and applied the weak and conservative nonparametric test. The diﬀerence in ranks must be very high in order to state that one algorithm outperforms another. Using some parametric test (such as ANOVA) would probably lead to stronger results, but we prefer to be as prudent as possible in drawing conclusions about the statistical signiﬁcance. 102 6.2. Experimental Results Table 6.3: Ranks of each classiﬁer for each dataset. Rank 1 corresponds to the best average error for a given dataset, while rank 10 corresponds to the worst. In the last row, the average ranks across the datasets are given for each classiﬁers. Dataset OLM OSDL IsoSep VCDomLEM LPRules MORE IsoReg J48 SVM NB DenBosch 10 4 8 1 6 3 5 7 9 2 Wisconsin 10 6 1 5 8 3 7 9 2 4 ESL 9 6 3 10 1 5 2 8 7 4 SWD 9 3 7 8 1 5 4 6 2 10 LEV 7 4 3 10 2 5 1 6 9 8 ERA 3 8 7 10 5 6 4 1 9 2 Housing 8 9 2 6 1 3 10 5 4 7 CPU 9 7 4 3 2 1 8 5 10 6 Balance 9 3 6 8 1 4 7 10 5 2 Ljubljana 10 7 2 8 3 4 1 6 9 5 Windsor 10 3 6 5 4 8 7 9 1 2 Car 8 3 5 1 2 6 4 9 7 10 avg. rank 8.5 5.25 4.5 6.25 3.0 4.42 5.0 6.75 6.17 5.17 10 9 8 7 6 5 4 3 2 1 CD = 3.91 OLM DOMLEM OSDL J48 IsoSep IsoReg LPRules MORE SVM NB Figure 6.1: Critical diﬀerence diagram Conclusions. Although statistical signiﬁcance could not be conﬁrmed, some important remarks can still be given: • On the “top” of the algorithms’ list there are two methods introduced in this thesis (LPRules and MORE), followed by IsoSep. Isotonic Separation is known to behave very well in practice (Chandrasekaran et al., 2005; Jacob et al., 2007; Ryu and Yue, 2005) and our methods perform not worse. This means that LPRules and MORE can probably be used successfully in practice and make predictions with high accuracy. • OSDL and IsoReg perform very similar to each other and worse than the top-of-the-list meth-ods. Notice that both OSDL and IsoReg are purely based on the dominance relation ⪰, i.e. they classify by comparing the test object with training objects using ⪰. The classiﬁcation procedure is practically the same in both cases, the main diﬀerence is how those algorithms deal with inconsistencies: IsoReg uses statistically justiﬁed multiple isotonic regression, while OSDL uses simpler, but also faster procedure. Since the amount of inconsistencies is usually not very high, their total accuracy is thus very similar. The performance of IsoReg and OSDL is poor for some datasets, which shows that using only the dominance relation is not enough – one must use some “parametric” approach, e.g. by Computational Experiment 103 restricting to the class of decision rules. • OLM perform very poorly in almost every case. Note, however, that OLM was the ﬁrst algorithm, proposed already in 1992, to deal with monotone problems. • VCDomLEM has a very high variance in classiﬁcation error: on some datasets the error is low, while on the other datasets the algorithm works best among all the algorithms. We do not know how to explain this behavior, but we noticed that VCDomLEM usually works worse on the data with high level of inconsistency. • The ordinary classiﬁers which do not take monotonicity constraints into account, perform worse than our methods. This was anticipated in the previous chapters, as there is a theoretical justiﬁcation for using monotone classiﬁers for such problems. However, the more important fact is that the ordinary classiﬁers can be inconsistent with the domain knowledge about the monotonicity. This leads to the problems with interpretation of the model; moreover, decision maker may want the model to be consistent with the knowledge she or he possesses about the problem. Those issues are usually much more substantial than a small gain in accuracy. Although neglecting the monotonicity does not deteriorate much the accuracy of the classiﬁer, neglecting the order between the class labels do. When we ﬁrst used J48 and SVM directly, without ordinal setting, the results were very bad. The only exception was Na¨ ıve Bayes, which works well without taking the order into account, but its good performance follows from the fact that it is a generative method (in contrary to J48 and SVM, which are discriminative), i.e. it estimates the joint distribution P(x, y). 6.3 Interpretability Let us consider the intelligibility of the considered methods, i.e. how easy or diﬃcult is the interpretation of resulting models. Since it is not possible to strictly measure and compare the intepretability, we will brieﬂy comment on each algorithm, indicating its pros and cons with respect to this aspect. Moreover, we will estimate the size of the model (description length), deﬁned as the amount of memory needed to save the model. This rather corresponds to the compactness of the model, but there is no other measurable property which is closer to interpretability. Instance-based methods. Our main competitor, IsoSep, is an instance-based method of classi-ﬁcation and “nearest neighbor” (with respect to the asymmetric “distance”) is found each time an object is to be classiﬁed. Although there exists a reduction method (Chandrasekaran et al., 2005) which allows us to store only the Pareto and the anti-Pareto frontier of each class (the set of objects not dominated by other objects from the class and the set of objects which do not dominate other objects from the class), those sets can be very large, especially when the dimensionality m is high. In general, they grow linearly with n. This makes Isotonic Separation quite slow in the classiﬁcation phase. Moreover, when classifying new objects, the nearest neighbor chosen with respect to the speciﬁc, asymmetric distance function does not explain clearly why a given prediction was made. OLM stores the model by saving a subset of the training set D called “rule base”. OSDL and Isotonic Regression (IsoReg) keep all of the objects, but a similar reduction as for IsoSep can be used to store only a subset of D. Additionally, IsoReg stores probabilities instead of class labels. All those models have a very similar size and interpretability properties to IsoSep. The main diﬀerence is that, contrary to IsoSep, they do not use nearest-neighbor procedure to classify new objects, they use only the dominance relation to this end, which is a much clearer and comprehensible way of classiﬁcation. 104 6.3. Interpretability Table 6.4: Average numbers of rules (per class union) for LPRules, MORE and VCDomLEM. Data set # objects # rules LPRules MORE VCDomLEM ESL 448 3 25 5.0 SWD 1000 6.7 24.9 16.8 LEV 1000 3.2 25 10.7 ERA 1000 2.6 25 10.2 Housing 506 26.6 25 32.2 Windsor 546 19.1 25 22.3 DenBosch 119 4.7 25 6.1 Wisconsin 699 9.7 25 9.3 Ljubljana 286 8.6 25 11.2 Car 1728 5.2 24.8 10.2 CPU 209 4.1 25 5.0 Balance 625 32.2 25 44.2 Rule-based methods. MORE and LPRules build a model which consists of a set of rules with assigned weights (respones). VCDomLEM is similar but it does not assign weights to the rules. It is well known that decision rules are one of the most interpretable forms of knowledge representation. Rule weight expresses the importance and strength of the rule vote, hence its meaning is very clear. Thus, the intepretability of the whole model mostly depend on the number of rules in the ensemble. Table 6.4 contains average numbers of rules per class union for the datasets used in the experiment. MORE produces M rules in each binary subproblem, which is M/2 rules per class union (there are upward and downward class unions in each subproblem). In our case, it was 25 rules per class union which is usually much smaller than the number of objects. It may sometimes (but rarely) happen that less rules are produced when no rule decreasing the current loss function can be found. LPRules generates rules until the optimum (the largest margin) is reached. Sometimes it can take even up to 100 iteration, but at the optimum most of the generated rules have zero weights, because the solution of the linear program is usually very sparse. This means that the ensembles produced by LPRules are usually very small, much smaller then those produced by MORE, especially for the problems in which attributes have purely ordinal scale and ﬁnite domains. We believe the sparsity of the ensemble is the most important advantage of LPRules. VCDomLEM generates rule sets larger than LPRules, but signiﬁcantly smaller than MORE. Summary. The algorithms developed by us proved to be eﬃcient and competitive with the best existing approaches to ordinal classiﬁcation with monotonicity constraints (they obtained the best results actually, but the diﬀerence was not found to be statistically signiﬁcant due to a small number of datasets). What distinguishes our approaches from the main competitor, isotonic separation, is the simplicity and interpretability of the decision rule model. Moreover, isotonic separation is a “lazy learning” algorithms, therefore it needs to keep a large part of the dataset to classify unseen objects. This, in turn, slows down the classiﬁcation procedure. In case of monotone rule ensembles, all we need is a small set of rules combined by a weighted sum in an ensemble, which captures the information contained in the dataset. Summary of the Contribution In this thesis, we dealt with the problem of ordinal classiﬁcation with monotonicity constraints. As stated in Section 1.4, our goal was to provide a comprehensive and consistent statistical theory for this problem, along with an eﬃcient and accurate method for solving it. In our opinion, the goal has been achieved. To support this claim, we provide below a summary of our contributions to the ﬁeld of ordinal classiﬁcation with monotonicity constraints. List of Main Results Our results can be considered from both theoretical and practical point of view. The theoretical results of the thesis can be found mainly in Chapters 2, 3 and 4, and they are related to the develop-ment of a statistical learning methodology for ordinal classiﬁcation with monotonicity constraints. The practical results were presented in Chapters 5 and 6, and they are related to introducing novel monotone classiﬁcation methods, which are veriﬁed in course of a computational experiment. Below, we list our contributions in the chronological order: Statistical framework for ordinal classiﬁcation with monotonicity constraints. We pro-posed a deﬁnition of the problem from probabilistic point of view. This was done by introducing monotonically constraint (with respect to the stochastic dominance) probability distribution. Then, we also showed that some more constraints must be imposed on the loss function in order to assure that the Bayes classiﬁer is monotone. Multiple isotonic regression. We considered the problem of probability estimation. We pro-posed a nonparametric method, taking into account only the monotonicity constraints expressed by the dominance relation. Our method is based on isotonic regression and is equivalent to MLE in binary-class case. Although isotonic regression has already been used to this end in binary-class case with linear preorder relation, our approach for any number of classes and partial preorder is new. Monotone approximation. We proposed a general method for removing the inconsistencies and “monotonizing” the data in a statistically safe way. The method is called monotone approximation and is based on relabeling the training objects. This is done in a nonparametric way, by empirical risk minimization within the class of all monotone functions. Although the problem of relabeling has already been used, we are ﬁrst who derived this problem from the statistical perspective and who showed its connection with isotonic regression. We also ﬁrst raised and solved the problem of non-unique optimal solution and we gave a detailed analysis of possible problem reduction to speed up the computations. Finally, we proved the convergence of our method to the Bayes classiﬁer for a wide class of probability distributions. 106 Future Research Stochastic Dominance-based Rough Sets. We extended DRSA into the stochastic setting by deﬁning the lower approximations (or, equivalently, generalized decision) in the probabilistic way. Notice that this was done in a diﬀerent manner than in the previous variable consistency approaches, which were also based on probabilistic reasoning. After deﬁning the model, we proposed to estimate the probabilities (which are unknown) by multiple isotonic regression. Our extension has classical DRSA in the deterministic limit. Then, we considered the problem of monotone conﬁdence interval estimation, which can be though of as a generalization of monotone approximation into the abstaining classiﬁers. We intro-duced to this end a speciﬁc, interval loss function and showed that stochastic DRSA is equivalent to minimizing this loss within the class of all monotone functions. Since the problem is linear, we obtain a fast algorithm for calculating the lower approximations without estimation of probabilities (i.e. isotonic regression does not need to be solved). Our theory extends well beyond DRSA and can be though of as a statistical base for explanation of all rough set approaches. Monotone Rules Ensembles. We introduced new algorithms based on decision rules for ordinal classiﬁcation with monotonicity constraints. Our algorithms are based on the so called two-phase procedure: in the ﬁrst phase, the monotone approximation (or, equivalently, Stochastic DRSA) is applied to the training data in order to get rid of the inconsistencies. In the second phase, the rule ensemble is built on the monotonized data. We explained why monotonization is made and how it is connected with separability of the data by a set of rules. We proved that the accuracy of the ensemble learned on the monotonized data without errors, and having a large margin, remains close to the accuracy of the optimal classiﬁer. We proposed a method of reducing the main K-class problem into K −1 binary subproblems and showed the necessary conditions for monotonicity of the classiﬁer. This constitutes to a general theory of monotone rule ensembles. Next, we proposed two particular rule induction algorithms for dealing with a linear loss func-tion. The ﬁrst algorithm, LPRules (Linear Programming Rule Ensemble), is based on the linear programming boosting which is known to directly maximize the minimal margin on the dataset. LPRules produces very small and accurate ensembles. The second algorithm, MORE (Sigmoid Loss Monotone Rule Ensemble) is based on greedily minimizing a diﬀerentiable approximation of the linear loss (sigmoid loss), using the boosting strategy to learning. MORE also achieves high accuracy on the real datasets. Our rule induction methods produce compact and interpretable sets of rules, consistent with the domain knowledge about the order and monotonicity, maintaining very good prediction perfor-mance. Future Research There still remain many unsolved questions in the investigated research area. We mention below some problems which we would like to study in the future: Mining monotonicity constraints. It is not always clear whether the monotone relationship holds for our data, although the expert may say so. Moreover, even if no domain knowledge is available, it can be worth considering automatic way of mining the monotonicity constraints from data. This improves our knowledge about the problem and thus mining the monotonicity constraints can be regarded as a part of the knowledge discovery process. Summary of the Contribution 107 Reduction of dimensionality. The nonparametric methods (isotonic regression, monotone ap-proximation) are based only on dominance relation. As the dimensionality of the problem (number of attributes m) increases, the dominance relation becomes sparse and the nonparametric methods become less reliable, since less objects are comparable by a dominance relation. This is one of the manifestations of the famous curse of dimensionality (Bellman, 1961). Generally, if m is too high, one should decrease the dimension of the space by removing some of the attributes. 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10852	https://math.stackexchange.com/questions/640460/in-the-equation-y-ax2-bx-c-of-a-parabola-what-do-a-b-c-r	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams In the equation $y = ax^2 + bx + c$ of a parabola, what do "$a$", "$b$", "$c$" represent? Ask Question Asked Modified 6 years, 1 month ago Viewed 5k times 4 $\begingroup$ I have trouble grasping some basic things about parabolas. (This should be easily found on Google, but for some reason I couldn't find an answer that helped me). I know one simple standard equation for a parabola: $$y = ax^2 + bx + c$$ My problem is: I'm not sure what the following letters represent: $a$, $b$, and $c$. Please try to explain to me what each of these letters represent in the equation, in a simple manner so I will understand, since I have very basic knowledge in math. Thank you geometry conic-sections Share edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Jan 16, 2014 at 13:35 user3150201user3150201 55322 gold badges55 silver badges1515 bronze badges $\endgroup$ 1 $\begingroup$ Just a few extra points not obvious from the answers below: $c$ will be the y-intercept of the parabola (where it crosses the y-axis, i.e. when $x=0$). $a$ will changes the width of the parabola. Also if $a$ is positive the parabola will have a minimum value (like a smiley face) and if it's negative then it will have a maximum value (like a sad face). $b$ is more complex, it both changes the width and the position of the turning point. Without $b$, the parabola would be symmetrical about the y-axis. $\endgroup$ Dan – Dan 2014-01-16 14:26:46 +00:00 Commented Jan 16, 2014 at 14:26 Add a comment \| 6 Answers 6 Reset to default 2 $\begingroup$ It would be worth your while to learn another standard form of the equation of a parabola, and you can complete the square, given $y = ax^2 + bx + c$, to obtain this form: $$4p(y - k) = (x-h)^2$$ The vertex of the parabola is given by $(h, k)$. $$h = \frac{-b}{2a};\quad k = \frac{4ac - b^2}{4a}$$ $$4p = \frac 1a$$ Share answered Jan 16, 2014 at 13:55 amWhyamWhy 211k198198 gold badges283283 silver badges505505 bronze badges $\endgroup$ 0 Add a comment \| 1 $\begingroup$ You can use an DGS to get an idea. I created a worksheet for you: Use your mouse to change the numbers on the slider and see what happen. To get smaller steps use the left mouse button to active the slider and then use left- or right-arrow on you keyboard. Share edited Jan 16, 2014 at 14:28 answered Jan 16, 2014 at 13:42 ulead86ulead86 3,4612929 silver badges4545 bronze badges $\endgroup$ 4 1 $\begingroup$ It should flip when $a$ becomes negative... $\endgroup$ Dan – Dan 2014-01-16 14:21:53 +00:00 Commented Jan 16, 2014 at 14:21 1 $\begingroup$ Yeah, they've written the function wrong, taking $a$ inside square (formula is given on the left). $\endgroup$ Ruslan – Ruslan 2014-01-16 14:24:19 +00:00 Commented Jan 16, 2014 at 14:24 $\begingroup$ Yes, thats true. I didn't enough attention the first time but now everything should be fine. $\endgroup$ ulead86 – ulead86 2014-01-16 14:29:07 +00:00 Commented Jan 16, 2014 at 14:29 $\begingroup$ As an aside, Desmos seems to handle this more smoothly and it's slightly easier for the user to create their own version for other families of equations. $\endgroup$ Mark S. – Mark S. 2014-02-16 09:04:40 +00:00 Commented Feb 16, 2014 at 9:04 Add a comment \| 0 $\begingroup$ Complete the square and get $$ y = ax^2 + bx + c = a(x-h)^2 + k, $$ where $$ h = -\frac{b}{2a} \quad\text{and}\quad k= -\frac{b^2-4ac}{4a} $$ So, $ y = ax^2 + bx + c $ is a translated and scaled version of $y=x^2$. Share answered Jan 16, 2014 at 13:56 lhflhf 222k2020 gold badges254254 silver badges585585 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ The "Cartesian connection" The most usual context is where $a,b,c$ are real numbers. For this to be a parabola, we would usually require $a\neq 0$. This equation describes a parabola whose axis is parallel to the $y$-axis in the Cartesian plane $\Bbb R\times \Bbb R$. How does it describe the shape of the parabola itself? Here it is: The points of the parabola are exactly the solutions to the given equation. That means that the points on the parabola, when plugged into the equation, make a true statement, and conversely, the only points that can be plugged in to make the equation true are points on the parabola. This is a very important connection to grasp when learning about graphs and equations. Oddly, many students "know" how to graph an equation without realizing this relationship between the equation and the graph. The individual coefficents I can understand why you might seek for a meaning for each individual coefficient, but actually the truth is a little more complicated. One thing you can say is that $c$ i the $y$ intercept of the graph. Another thing is that if $a>0$, the graph is opening up like $\cup$, and when $a<0$ the graph is opening down like $\cap$. Beyond those two facts, the rest of the most important information is a mixture of $a,b$ and $c$. Through various algebraic manipulations, you can show that the vertex of the parabola has coordinates $(\frac{-b}{2a},\frac{4ac - b^2}{4a})$, and that its $x$ intercepts, if they exist, are at $(\frac{-b\pm\sqrt{b^2-4ac}}{2a},0)$ (the $x$ intercepts exist when this evaluates to real values.) Notice how the vertex's $x$ coordinate sits halfway between the $x$ intercepts on the real line (when the intercepts exist.) The focus is another major feature. The focus turns out to be at $(\frac{-b}{2a},\frac{1-b^2}{4a}+c)$. Notice how the vertex and focus are lying on the same line. That's to be expected with the equation for "vertical" parabolas that you described. Anyhow, this illustrates that the individual coefficients do not each dictate one thing about the parabola, but rather that their mixtures control the various features. To derive all of this and really understand it, you'll have to spend some time patiently with basic algebra and the geometric definition of a parabola. Some examples courtesy of Desmos online grapher: Here's an example plotting two points and a parabola. One is on the parabola and one is not You can check manually that the one that is on the parabola satisfies the equation, and the one that is not does not satisfy the equation. Share edited Jan 16, 2014 at 14:35 answered Jan 16, 2014 at 13:39 rschwiebrschwieb 161k1616 gold badges184184 silver badges445445 bronze badges $\endgroup$ 2 $\begingroup$ Thanks for answering. I still don't understand. I know that this equation represents a parabola whose axis is parallel to the y axis, and that a,b and c are real numbers. But what do they represent in the parabola? $\endgroup$ user3150201 – user3150201 2014-01-16 13:43:59 +00:00 Commented Jan 16, 2014 at 13:43 $\begingroup$ @user3150201 K, I added elaboration. How's it look now? $\endgroup$ rschwieb – rschwieb 2014-01-16 14:41:30 +00:00 Commented Jan 16, 2014 at 14:41 Add a comment \| 0 $\begingroup$ They are just numerical coefficients; try to plot the fuction with any values (except $a=0$ which would reduce the parabola to a straight line). Wolfram Alpha or Excel would be convenient. For $x=0$, the value of y will just be equal to c. If ($b^2 - 4 a c$) is positive, you will have two values of x for which y will be zero (these are the roots of the quadratic equation). The parabola will go to an extremum for $x = - b / (2 a)$; this extremum will be a minimum if $a > 0$ and a maximum if $a < 0$. Can I do more for you ? Share edited Aug 18, 2019 at 13:39 user689538 answered Jan 16, 2014 at 13:47 Claude LeiboviciClaude Leibovici 294k5555 gold badges130130 silver badges316316 bronze badges $\endgroup$ Add a comment \| -3 $\begingroup$ So, we have f(x) = ax^2 + bx + c, where a is distinct from 0. 'A' is related with growth. If 'a' is negative, function decrease, if 'a' is positive, function increase. 'C' is of course y-argument of point (0, f(x)) (let's imagine this as a place where function cross OY axis). Share answered Jan 29, 2015 at 15:32 Nelliusz FrÄcekNelliusz FrÄcek 3766 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry conic-sections See similar questions with these tags. 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10853	https://math.stackexchange.com/questions/3648675/extremal-graph-theory-problems	Extremal graph theory problems - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Extremal graph theory problems Ask Question Asked 5 years, 5 months ago Modified5 years, 3 months ago Viewed 449 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I'm currently working on some assignments, but I'm not getting anywhere. Maybe you can give me some ideas. 1) Show for a graph G:=(V,E)G:=(V,E) with n n vertices and minimum graph degree δ=⌊(r−2)n r−1⌋+1 δ=⌊(r−2)n r−1⌋+1 contains a subgraph K r K r. I don't have any clue. 2) Show for a graph G:=(V,E)G:=(V,E) with n=\|V(G)\|≥5 n=\|V(G)\|≥5 without triangles and who is not bipartite the following holds: \|E(G)\|≤⌊(n−1)2 4⌋+1\|E(G)\|≤⌊(n−1)2 4⌋+1 Here I think that a vertex v v with degree d(v)=1 d(v)=1 should exist. Then G′:=G−v G′:=G−v is a graph with \|V(G′)\|=n−1\|V(G′)\|=n−1 vertices and without triangles. According to Turan the graph G′G′ has maximal \|E(G′)\|≤⌊\|V(G′)\|2 4⌋\|E(G′)\|≤⌊\|V(G′)\|2 4⌋ edges. And therefore G G, due to the additional edge, has maximal the amount of edges as stated in the claim. However, I cannot cleanly prove why such a vertex v v have to exist. 3) G=(V,E)G=(V,E) is a graph with n=\|V(G)\|≥2 n=\|V(G)\|≥2 without a hamiltonian path. Show that: \|E(G)\|≤(n 2)−(n−1)\|E(G)\|≤(n 2)−(n−1) Determine the extremal graph and show the uniqueness. I have no ideas here either. :( I would be grateful for any help! Thank you very much graph-theory extremal-graph-theory Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Apr 28, 2020 at 18:33 Tri SoTri So 21 1 1 bronze badge 6 What have you tried? These are starter questions, and should be almost obvious if you start writing things out. E.g. for 1) Fix a vertex, how many vertices is it connected to? Pick one of those vertices, how many common vertices do they have? Iterate, can we force out a K r K r in this way?Calvin Lin –Calvin Lin 2020-04-28 19:24:49 +00:00 Commented Apr 28, 2020 at 19:24 The first one i solved yet by induction. For the second i have written my idea, but im not sure why v v have to exist. For the third i dont have a clue. :S Tri So –Tri So 2020-04-28 19:34:49 +00:00 Commented Apr 28, 2020 at 19:34 Can you add how you did 1 1? I would love to see the induction process for it. 3) Can you find the extremal graph? Hint: It has nearly all (n 2)(n 2) edges, so which n−1 n−1 would you want to exclude?Calvin Lin –Calvin Lin 2020-04-28 22:47:20 +00:00 Commented Apr 28, 2020 at 22:47 1 Note that for question 2, the cycle graph on 5 vertices verifies \|V(C 5)\|≥5\|V(C 5)\|≥5, it is triangle-free, and is not bipartite. However each vertex has degree 2, there is no such v v Thomas Lesgourgues –Thomas Lesgourgues 2020-04-28 23:42:28 +00:00 Commented Apr 28, 2020 at 23:42 Right Thomas, so my assumption is wrong. Calvin, i post the solution for 1 tonight, after i finished the other questions. For the third one is my guess G:=K n G:=K n with one isolated vertex. But it's only extremal if the enequality is shown.Tri So –Tri So 2020-04-29 07:15:18 +00:00 Commented Apr 29, 2020 at 7:15 \|Show 1 more comment 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. 3)) Suppose to the contrary that G G has at least (n 2)−(n−1)+1(n 2)−(n−1)+1 edges. If n=2 n=2 then G G has an edge and so a Hamiltonian path. So let’s assume n≥3 n≥3. We can try to show that G G has a Hamiltonian path directly, but a teaching way is to apply Ore's theorem. If G G is complete then it is Hamiltonian. Otherwise we add to it an arbitrary edge e e and obtain a graph G′G′ with at least (n 2)−n+3(n 2)−n+3 edges. Let v v and w w be any pair of distinct non-adjacent vertices of G′G′. Since in a complete graph on n n vertices each vertex has a degree n−1 n−1 and G′G′ lacks at most n−3 n−3 edges to a complete graph, deg v+deg w≥n−1+n−1−(n−3)−1=n deg⁡v+deg⁡w≥n−1+n−1−(n−3)−1=n (the last −1−1 is taken into account for an edge v w v w is a complete graph). By Ore’s theorem G′G′ has a Hamiltonian cycle, so G G, which is a G′G′ with a removed edge, has a Hamiltonian path. Now suppose that G G is an extremal graph, that is a graph with (n 2)−(n−1)(n 2)−(n−1) edges and without a Hamiltonian path. An instance of such graph is a complete graph with all edges incident to a fixed vertex removed. We claim that there are no other extremal graphs. Indeed, suppose to the contrary that G G is such a graph. If G G has at least three vertices of degree one then n≥4 n≥4 and G G has at most (n−3 2)+3(n−3 2)+3 edges and, which follows (n−3 2)+3≥(n 2)−(n−1)(n−3 2)+3≥(n 2)−(n−1), and n≤2 n≤2, a contradiction. Thus we can add an edge to G G such that the graph G′G′ has no vertices of degree one. An analysis of the Hamiltonicity proof of G′G′ shows that it can fail only if there exist two non-adjacent vertices v v and w w of G′G′ such that each edge which G′G′ lacks to a complete graph is incident either to v v or to w w and, moreover, deg v=deg w=2 deg⁡v=deg⁡w=2, n>4 n>4, and v v and w w have the same sets of neighbors. Thus G′G′ lacks 2(n−3)−1 2(n−3)−1 edges to a complete graph. Thus 2(n−3)−1≤n−2 2(n−3)−1≤n−2, which follows n≤5 n≤5, so n=5 n=5 and G′G′ is isomorphic to a graph K 5 K 5 without a triangle. But then it is easy to check that the graph G G, which is a graph G′G′ with one edge removed has a Hamiltonian path, a contradiction. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jun 28, 2020 at 8:05 Alex RavskyAlex Ravsky 108k 5 5 gold badges 65 65 silver badges 203 203 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions graph-theory extremal-graph-theory See similar questions with these tags. 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10854	https://www.tes.com/teaching-resource/rounding-to-significant-figures-12204810	Rounding to significant figures \| Teaching Resources International Resources Education Jobs Schools directory News Courses Store Search Tes for schoolsLog inRegister for free ResourcesEducation JobsSchools directoryNewsMagazineCoursesLog outHelp Home feed My list CoursesMy JobsJob alertsMy CVCareer preferencesResourcesAuthor dashboard Settings Edit accountLog out HomeResourcesJobsSchools directoryNewsMagazineCoursesRegister for freeLog inHelp Home feedEdit accountAbout usMy productsTes for schoolsWork for TesLog out Education jobs My jobsJob alertsMy CVCareer preferences Resources DownloadsSaved resourcesAuthor dashboardAdd resource Courses Access courses News MagazineSubscriptionsPayments Rounding to significant figures Subject: Mathematics Age range: 11-14 Resource type: Lesson (complete) Richardtock's Shop 4.62 494 reviews Maths resources. Working on Project-A-Lesson. A full lesson in a PowerPoint. For busy teachers who still want outstanding engaging tasks and learning checks Last updated 20 September 2022 Share this Share through email Share through twitter Share through linkedin Share through facebook Share through pinterest File previews docx, 412.06 KB pdf, 357.32 KB pptx, 250.29 KB CHANGELOG : Massively updated 23/09/2021 to clean up a lot of stuff and add a section on identifying how many significant figures are in a number. I also collated the activities into a worksheet. 20/9/22 : Changed the worked examples to avoid duplicate digits. Full lesson. Starts with a pattern spot (you could use mini whiteboards), moves on to some increasingly difficult problems, then a lovely pixel puzzle. Creative Commons "Sharealike" Review 4 Something went wrong, please try again later. mah57 2 years ago report 4 Thank you for this very helpful. 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10855	https://mate.unipv.it/mora/ctf2013/Lezione09.pdf	Esercizi Esercizio 4. Un materiale radioattivo e caratterizzato da un tempo di dimezzamento pari a 800 anni. Dopo quanto tempo un campione di tale materiale si sar a ridotto del 15%? Qual e il tempo di di-mezzamento di un secondo campione che si riduce del 15% in 800 anni? Soluzione: 1. Indicando con P il peso iniziale del materiale, si deve avere 1 2nP = 85 100P. Dunque 2n = 100 85 , da cui n = log2 100 85 . Il tempo richiesto e 800 · log2 100 85 . 2. Il tempo di dimezzamento del secondo campione e 800 log2 100 85 . Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Esercizi Esercizio 5. Un materiale radioattivo e caratterizzato da un tempo di dimezzamento pari a 1000 anni. Dopo quanto tempo un campione di 1 Kg di tale materiale si sara ridotto del 20%? Dopo quanto tempo un altro campione di 2 Kg di tale materiale si sar a ridotto al 20%? Soluzione: Si deve avere 1 2n · K0 = 80 100K0 avendo indicato con K0 il peso iniziale. Dunque 2n = 5 4 e n = log2 5 4 e il tempo e 1000 · log2 5 4. Nel secondo caso si deve avere 1 2n · K0 = 20 100K0. Dunque 2n = 5 e n = log2 5 e il tempo e 1000 · log2 5. Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Esercizi Esercizio 6. Una sostanza radioattiva ha un tempo di dimezzamento T pari a 100 anni. Quanto tempo deve trascorrere aﬃnch´ e di un campione della sostanza rimanga il 50% del quantitativo iniziale? Soluzione: Dalla deﬁnizione di tempo di dimezzamento segue immediatamente che deve trascorrere un tempo di dimezzamento, cioe 100 anni. Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Esercizi Esercizio 7. Una popolazione cellulare e formata all’istante t = 0 da N0 cellule aventi tempo di raddoppio T = 10 giorni. Dopo quanti giorni la popolazione e pari a 8N0? Qual e il tempo di raddoppio di una seconda popolazione cellulare che passa da N0 cellule a 8N0 cellule in 10 giorni? Soluzione: 1. 30 giorni 2. 10 3 giorni Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Esercizi Esercizio 8. Una popolazione cellulare e formata ad un certo istante da N0 individui ed e caratterizzata da un tempo di raddoppio pari a 14 giorni. Dopo quanto tempo la popolazione risultera composta da 10N0 individui? Qual e il tempo di raddoppio di un secondo campione che passa da N0 a 10N0 individui in 14 giorni? Soluzione: 1. 14 · log2 10 giorni 2. 14 log2 10 giorni Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Studio Qualitativo di Funzione Reperire un certo numero di informazioni per descrivere a livello qualitativo l’andamento del graﬁco di una funzione f 1. campo di esistenza (cioe, l’insieme di deﬁnizione) 2. segno: per quali x si ha f(x) ≥0 ? 3. intersezioni con gli assi: (0, f(0)); per quali x si ha f(x) = 0 4. comportamento agli estremi del campo di esistenza 5. continuit a 6. monotonia 7. massimi e minimi 8. graﬁco qualitativo Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Campo di Esistenza Il campo di esistenza e l’insieme di tutti i punti nei quali la funzione e deﬁnita. Nel caso di una funzione composta si determina, caso per caso, te-nendo conto degli insiemi di deﬁnizione delle funzioni base con le quali la funzione e stata costruita. Esempio: data la funzione f(x) = 1 ln(4 −x2) • il logaritmo e deﬁnito per 4 −x2 > 0 ⇔x ∈(−2, 2) • il denominatore deve essere diverso da zero ln(4 −x2) ̸= 0 ⇔4 −x2 ̸= 1 ⇔x ̸= ± √ 3 Il campo di esistenza di f e (−2, − √ 3) ∪(− √ 3, √ 3) ∪( √ 3, 2). Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Comportamento agli Estremi Se il campo di esistenza D e costituito dall’unione di piu intervalli (li-mitati o illimitati), occorre prendere in considerazione separatamente gli estremi di ognuno di questi intervalli. • Se gli estremi appartengono a D, si calcola semplicemente il valore della funzione in tali punti. Esempi: f(x) = √x, D = [0, +∞), f(0) = √ 0 = 0 f(x) = q x(1 −x), D = [0, 1], f(0) = 0, f(1) = 0 • Se gli estremi non appartengono a D, si introduce il concetto di limite. Esempi: lim x→0 1 x2 lim x→0+ 1 √x lim x→+∞ 1 x2 lim x→−∞ 1 x2 Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Limite Destro Finito Quando la variabile x assume valori “vicini” ad a (sempre maggiori di a), i corri-spondenti valori di f(x) si avvicinano sempre pi u al valore L. ε ε L -L + ε L - ε L + y = f (x) O x y O x y L L a scelta di y = f (x) a a + δ δ scelta di ε ε ε limite destro ﬁnito lim x→a+ f(x) = L Si dice che f(x) tende al limite L per x che tende ad a da destra se: per ogni ε > 0 esiste un δε > 0 tale che \|f(x) −L\| < ε per ogni x ∈(a, a + δε). Esempi: (1) lim x→1+ √ x −1 = 0 , (2) lim x→0+ \|x\| x = 1. Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Limite Sinistro Finito Quando la variabile x assume valori “vicini” a b (sempre minori di b), i corrispondenti valori di f(x) si avvicinano sempre piu al valore L. ε ε L -L + ε L - ε L + y = f (x) x y ε O x y L L O scelta di y = f (x) b scelta di ε δ ε b δ b -limite sinistro ﬁnito lim x→b−f(x) = L Si dice che f(x) tende al limite L per x che tende a b da sinistra se: per ogni ε > 0 esiste un δε > 0 tale che \|f(x) −L\| < ε per ogni x ∈(b −δε, b). Esempi: (1) lim x→1− √ 1 −x = 0 , (2) lim x→0− \|x\| x = −1. Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Limite Finito per x →x0 Se la funzione possiede sia il limite destro che il limite sinistro nel punto x0 e se entrambi sono uguali al valore L, si dice che lim x→x0 f(x) = L (limite ﬁnito) Quando la variabile x assume valori “vicini” a x0 (diversi da x0), i corrispondenti valori di f(x) sono “vicini” al valore L. Si dice che f(x) tende al limite L per x che tende ad x0 se: per ogni ε > 0 esiste un δε > 0 tale che \|f(x) −L\| < ε per ogni x ∈(x0 −δε, x0 + δε) con x ̸= x0. Esempi: (1) lim x→1 (2x + 1) = 3 , (2) lim x→1 x2 −1 x −1 = 2 Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Limite Inﬁnito Quando la variabile x assume valori “vicini” ad x0 (diversi da x0), i corrispondenti valori di f(x) crescono arbitrariamente. x 0 x x + 0 δ M x -scelta di 0 M 0 δ M δ x y = f (x) y = f (x) x scelta di M M O y O M y limite inﬁnito lim x→x0 f(x) = +∞ Si dice che f(x) tende a +∞per x che tende ad x0 se: per ogni M > 0 esiste un δM > 0 tale che f(x) > M per ogni x ∈(x0 −δM, x0 + δM) con x ̸= x0. Esempi: (1) lim x→0 1 x2 = +∞. Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Osservazioni sui Limiti per x →x0 Poich´ e nella deﬁnizione di limite si richiede x ̸= x0, non ha alcuna importanza l’eventuale valore assunto dalla funzione nel punto x0: f(x) =    x2 per x ̸= 0 1 per x = 0 f(0) = 1 , ma lim x→0 f(x) = 0 g(x) =      1 x2 per x ̸= 0 0 per x = 0 g(0) = 0 , ma lim x→0 g(x) = +∞ Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Limite Finito per x →+∞ Quando la variabile x cresce arbitrariamente, i corrispondenti valori di f(x) sono sempre pi u “vicini” al valore L. L + ε L - ε L - ε ε ε scelta di x ε scelta di ε L + x x y = f (x) O y L O x y L y = f (x) limite ﬁnito lim x→+∞f(x) = L Si dice che f(x) tende al limite L per x che tende ad +∞se: per ogni ε > 0 esiste un xε > 0 tale che \|f(x) −L\| < ε per ogni x ∈(xε, +∞). Esempi: (1) lim x→+∞ x + 1 x = 1 , (2) lim x→+∞e−x = 0. Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Il Limite Puo Non Esistere Il limite di una funzione pu o non esistere: • f(x) = \|x\| x , x ̸= 0. Non esiste il limite per x →0. Infatti, il limite destro e limite sinistro esistono, ma sono diversi: lim x→0+ f(x) = 1 , lim x→0−f(x) = −1 . • f(x) = 1 x, x ̸= 0. Non esiste il limite per x →0. Infatti, i limiti destro e sinistro sono inﬁniti di segno opposto: lim x→0+ f(x) = +∞, lim x→0−f(x) = −∞. • f(x) = sin x. Non esiste il limite per x →+∞. Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14 Alcuni Limiti da Ricordare • lim x→+∞xn = +∞ per ogni n ∈N, n ̸= 0 • lim x→−∞xn =    +∞ se n e pari −∞ se n e dispari • lim x→+∞ax =    0 se 0 < a < 1 +∞ se a > 1 lim x→−∞ax =    +∞ se 0 < a < 1 0 se a > 1 • lim x→+∞loga x =    −∞ se 0 < a < 1 +∞ se a > 1 • lim x→0+ loga x =    +∞ se 0 < a < 1 −∞ se a > 1 Matematica con Elementi di Statistica, Prof.ssa M.G. Mora – a.a. 2013/14
10856	https://www.youtube.com/watch?v=aHvHKv9BTZg	Signed and Unsigned Numbers Made Easy! – Bits, Bytes & Binary Numbers HuwsTube 66100 subscribers 368 likes Description 17071 views Posted: 18 Jul 2022 Numbers that are signed (can have positive or negative values) have different ranges of values from numbers that are unsigned (positive only values) and it’s all down to a simple 1 or 0 – the vaklue of the leftmost bit in a byte. Why is that? In this lesson, I explain exactly how each bit contributes a value to a given number and how that value can be switched from plus to minus. I also return to a problem I looked at some time ago: why do big positive numbers sometimes get changed to negative? MORE LESSONS To follow all this series on bits, bytes and binary numbers, go to the “Down To The Metal” playlist: Lessons on specific topics: Why Do Big Numbers Turn Negative? What is a Binary Number? Bits, Bytes and Binary Numbers: Be sure to subscribe to the Code With Huw channel so that you get a notification when I upload more lessons in this series. PROGRAMMING BOOKS If you want to learn programming in more depth (and also support this channel!) you might think of buying one of my books. I have written books on C programming, Using Pointers in C, Recursion and other programming topics. You can buy my books (paperback or Kindle) on Amazon. “CODE WITH HUW” ON TWITTER: “CODE WITH HUW” ON FACEBOOK: Good luck! And good programming! 26 comments Transcript: as I showed in my last video the maximum possible value of a number is determined by the number of bits available to the data type each BTE has eight bits a two bite data type can store numbers from 0 to 65535 a one bite data type can store numbers from 0 to 255 bigger data types that use more bytes can store bigger numbers for Simplicity here I'm going to stick with one by data types what I want to show in this lesson is how signed numbers that's those that can have both positive and negative values are different from unassigned numbers which can only be positive and why sometimes big positive numbers become negative numbers with eight bits of available an unsigned data type can represent numbers from 0 to 255 as I showed in another video which is linked down below signed data types are either positive or negative depending on the value of the leftmost bit if assigned data type has a negative value this bit is set to one if it's positive it's zero now let's look a bit more closely at that leftmost bit and find out exactly the effect it has with both signed and with unsigned numbers let's assume this is a one bite data type one bite is eight bits and the leftmost bit is set to one with all the other bits set to zero as I explained in my last video the value of a bit that is set to one in this leftmost column is 128 that is it adds the value 128 to the decimal number represented by the whole bite the unsigned data type in this case in binary is 1 followed by seven zeros so in decimal that's 128 but what happens when that leftmost bit is set in assigned data type well in the last lesson we saw that it switches the value represented by the bite to a negative number so if this is a signed number its value would be minus 128 now that seems reasonable because according to Microsoft's document ation assigned data type can have a maximum value of 127 and a minimum of minus 128 in my spreadsheet I've written a couple of simple formulas to help to show what's going on here when a bit is set the value shown in the column header is added to this row recall that in a bite the decimal value of each column is twice that of the column to its right so when this this bit is set one is added to the decimal total when this bit is set two is added and so on if you don't understand this go back and watch my previous video which explains binary numbers in more detail to work out the unsigned decimal value represented by a binary value that is the eight zeros and ones shown here I just add all the values calculated for each column that is for each bit in the bite that's what this sum formula does but if this binary value represents assigned number then the maximum positive value will be 127 if the sum of the columns is greater than 127 the decimal value switches to the lowest possible value for this data type which is minus 128 and then it adds the sum of the other cells to that value that's what this formula does what this says is if the sum of all eight cells here that is of all eight bits in a bite is greater than 127 then add the sum of these seven cells to minus 128 but why is the maximum 127 well remember with assigned value I need some way of representing both a positive and a negative number as we've seen when this leftmost bit is set to one that switches the value of the entire bite to negative in fact it does a bit more than that it not only sets the sign plus or minus it also changes the value as we've seen for an unsigned data type the leftmost bit adds 128 to the entire decimal number represented by the B when all bits are one the maximum possible value is 255 which in my spreadsheet is the sum of all the column headers but with a signed number the leftmost bit is used to switch between positive and negative which means it's not available to add the positive value 128 as it is for an unassigned number so to get the maximum positive value for a signed number I must ignore that leftmost bit I do that by setting it to zero and now I can see that the maximum positive number for a signed one by number is 127 so what happens if I set the leftmost bit to one with all the others set to zero I can't go beyond the maximum positive value for assed data type so this now switches to the the lowest possible value that is minus 128 so clearly since all these seven bits are zero and contribute nothing to the value of the resulting number this leftmost bit must have done more than just switch the sign from plus to minus it's contributed the value minus 128 so you can think of this leftmost bit as adding its value to an unsigned number and subtracting its value from assigned number the other seven bits always add positive values for example when the first bit and the last bit are set for an unsigned value one is added to 128 giving the decimal value for the bite of 129 but when the first bit and the last bit are set for assigned value one is added to minus 128 giving the decimal value minus 127 now I set this bit and three is added to the value of the leftmost bit 128 + 3 for an unsigned number gives 131 - 128 + 3 for a signed number gives - 125 we can see that in the code too here C is a Char which is assigned data type and we can see from the Microsoft documentation a standard Char can have values within the range negative 128 to positive 127 I sign 127 and my code here shows the binary equivalent all these seven bits are set to one and this leftmost bit is set to zero now I try to set C to 128 and I run the program again this time it shows that only the leftmost bit is now set to one remember for an unsigned number that bit represents 128 but for a signed number it becomes the lowest possible value which uh is the negative number minus 128 right so now let me try 129 run my program again and the binary number is now one followed by six zeros then another one the resulting decimal value isus 127 which is exactly what my spreadsheet shows for this binary number for a signed number like this Char the bit pattern which for an unsigned number gives positive 129 while for a signed number it gives minus 127 why is that because for a signed number the leftmost bit gives the value minus 128 this rightmost bit then adds one to that - 128 + one results in- 127 what we are seeing here is the behavior that I first showed a couple of lessons ago when we saw how big numbers can wrap around into negative values and that's what's happened here a Char has a maximum positive value of 127 so when I assign a bigger number the bits in The Bite are set just as they would be for an unsigned data type but now the value added by the leftmost bit is negative rather than positive and that's why big numbers that go beyond the upper range of possible values end up starting all over again at the lowest value of their range for sign data types the lowest possible value is a negative number for unsigned data types the lowest possible value is zero now this may take a bit of thinking about so if you don't get it right away go back and watch my other videos about bits bytes and binary numbers the links as always are down below then take some time to experiment with my code which is shown here but wait a minute how does this code actually work if you've been staring at it and can't figure out exactly what's going on well that's not really too surprising because even though this program is really short there's well there's quite a lot going on in it I'll explain how this code works in the next lesson thanks for watching if you like this video give it a thumbs up and subscribe to my channel and click the Bell to be notified whenever I upload new videos
10857	http://physics.bu.edu/~redner/211-sp06/class04/atwood.html	Atwood's Machine Atwood's machine is a device where two masses, M & m, with M > m, are connected by a massless string passing over a massless pulley. What is the acceleration of each mass? Use DID TASC Diagram and coordinate system s Isolate the system s Draw all forces acting Take components Apply F=maand apply constraints Solve Check! Steps 1-4: See diagram and transparency. Step 5: Apply Newton's 2nd law and constraints Constraints: non-stretchy rope implies T 1 = T 2 = T and a M=-a m. Newton's Second Law: ma m = T-mg and Ma M = T-Mg Step 6: Solve Newton's Second Law +Ma M = T-Mg -ma M = T-mg Subtracting the two equations gives: (M + m) a M = (m-M) g a M=m - M M + m g Step 7: Check your answer! Look at simple limiting cases If m = M then a M = 0. The system is balanced. If m = 0 then a M = -g. M is in freefall. If m > M then a M is positive. M accelerates up.
10858	https://www.forourschool.org/math-guides/compensation	Using Compensation for Quick Mental Math \| Math Guide ← Use ForOurSchool.org to host your school's Math-A-Thon ← Use ForOurSchool.org to host your school's Math-A-Thon Math Guides / Compensation All Guides Area Model Array Model Arrow Method Box Model Break Apart Compensation Cross Multiply Distributive Property Dot Cards Equal Groups Finger Counting Matrix Model Number Bond Partial Products PEMDAS Place-Value Chart Prime Factorization Tape Diagram How To Use Compensation Method For Quick Mental Math The Compensation method is an approach to addition and subtraction equations using mental calculations. It involves rounding numbers to convenient values to ease the arithmetic process. To use the Compensation method for addition or subtraction: Identify one number to adjust by rounding it to a nearby, more convenient value. Compensate for the adjustment by making an opposite adjustment to the other number in the equation. Add or subtract the adjusted numbers to find the solution. For example, let's consider the addition equation 38 + 45: Begin with 38 because it is closer to its rounded value. Adjust 38 by adding 2 to make it a round number of 40. To compensate for the adjustment, subtract 2 from 45, resulting in 43. Add the adjusted numbers: 40 + 43 = 83, which is the final sum. Related Math Games 1 1st Grade Addition metric distance word problem within 20 Addition of 3 numbers within 20 Addition of 3 numbers word problem within 20 Addition with 2-digit numbers within 100 Subtraction metric distance word problem within 20 Subtraction of 3 numbers word problem within 20 2 2nd Grade Addition money word problem up to 50 cents Addition with 3 numbers within 100 Addition within 100 Addition within 1,000 Addition within 200 Addition word problem time Addition word problem within 100 Counting money up to 1 dollar Counting money word problem up to 1 dollar Subtraction money word problem up to 50 cents Subtraction within 100 Subtraction within 1,000 Subtraction within 200 Subtraction word problem within 100 Subtraction word problem with time 3 3rd Grade Addition and subtraction of 3 numbers within 100 Addition number names within 100 Addition, subtraction, multiplication, division of 4 numbers within 100 Addition, subtraction, multiplication, division word problem of 4 numbers within 100 Addition with 3 digit numbers within 1,000 Addition with time Addition word problem with time Counting money up to 2 dollars Subtraction number names within 100 Subtraction with 3 digit numbers within 1,000 Subtraction with time Subtraction word problem with time 4 4th Grade Addition within 100,000 Addition word problem within 10,000 Addition word problem within 100,000 Subtraction within 100,000 Subtraction word problem within 100,000 5 5th Grade Addition and subtraction of 3 2-digit numbers Addition and subtraction word problem of 3 2-digit numbers Addition, subtraction, and multiplication of 3 numbers within 200 6 6th Grade Division of decimals Division of decimals word problem 7 7th Grade Addition and subtraction of 3 2-digit numbers Discounts, Taxes, and Tips © 2025, made with ❤️ by A PS107 Parent Terms of ServicePrivacyChildren's PrivacyPricingSupportContact Us
10859	https://www.scribd.com/document/839202797/A-11-A13-134a	Table A-11: Specific Volume, Internal Energy, Enthalpy, Entropy \| PDF \| Enthalpy \| Scientific Phenomena Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 35 views 5 pages Table A-11: Specific Volume, Internal Energy, Enthalpy, Entropy The document contains a temperature table for saturated refrigerant-134a, detailing specific volume, internal energy, enthalpy, and entropy at various temperatures and pressures. It includes… Full description Uploaded by lemanhdung2122005 AI-enhanced title and description Go to previous items Go to next items Download Save Save A-11-A13 (134a) For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save A-11-A13 (134a) For Later You are on page 1/ 5 Search Fullscreen 916 PROPERTY TABLES AND CHARTS TABLE A–11 Saturated refrigerant-134a—Temperature table Temp., T 8 C Sat. press., P sat kPa Specific volume, m 3 /kg Internal energy, kJ/kg Enthalpy, kJ/kg Entropy, kJ/kg∙K Sat. liquid, v f Sat. vapor, v g Sat.liquid, u f Evap., u fg Sat.vapor, u g Sat.liquid, h f Evap., h fg Sat.vapor, h g Sat.liquid, s f Evap., s fg Sat.vapor, s g 2 4 0 51.2 5 0.0 00 70 5 3 0.36 0 64 2 0.03 6 20 7.42 20 7.38 0.00 22 5.86 22 5.86 0.0 00 00 0.9 68 69 0.96 86 9 2 38 56.8 6 0.00 07 08 2 0.32 71 8 2.47 2 20 6.06 20 8.53 2.51 2 22 4.62 22 7.13 0.01 07 1 0.9 55 16 0.96 58 8 2 36 62.9 5 0.00 07 11 1 0.29 74 0 4.98 7 20 4.69 20 9.68 5.03 2 22 3.37 22 8.40 0.02 13 7 0.9 41 82 0.96 31 9 2 34 69.5 6 0.00 07 14 1 0.27 08 2 7.50 9 20 3.32 21 0.83 7.55 9 22 2.10 22 9.66 0.03 19 6 0.9 28 67 0.96 06 3 2 32 76.7 1 0.00 07 17 1 0.24 70 6 10.0 4 20 1.94 21 1.97 10.0 9 22 0.83 23 0.93 0.04 24 9 0.9 15 69 0.95 81 9 2 30 84.4 3 0.00 07 20 1 0.22 57 7 12.5 8 20 0.55 21 3.12 12.6 4 21 9.55 23 2.19 0.05 29 7 0.9 02 89 0.95 58 6 2 28 92.7 6 0.00 07 23 2 0.20 66 6 15.1 2 19 9.15 21 4.27 15.1 9 21 8.25 23 3.44 0.06 33 9 0.8 90 24 0.95 36 4 2 26 10 1.73 0.00 07 26 4 0.18 94 7 17.6 7 19 7.75 21 5.42 17.7 5 21 6.95 23 4.70 0.07 37 6 0.87 77 6 0.9 51 52 2 24 11 1.37 0.00 07 29 6 0.17 39 8 20.2 3 19 6.34 21 6.57 20.3 1 21 5.63 23 5.94 0.08 40 8 0.86 54 2 0.9 49 50 2 22 12 1.72 0.00 07 32 8 0.15 99 9 22.8 0 19 4.92 21 7.71 22.8 9 21 4.30 23 7.19 0.09 43 5 0.85 32 3 0.9 47 58 2 20 13 2.82 0.00 07 36 1 0.14 73 5 25.3 7 19 3.49 21 8.86 25.4 7 21 2.96 23 8.43 0.10 45 6 0.84 11 9 0.9 45 75 2 18 14 4.69 0.00 07 39 4 0.13 58 9 27.9 6 19 2.05 22 0.00 28.0 7 21 1.60 23 9.67 0.11 47 3 0.82 92 7 0.9 44 01 2 16 15 7.38 0.00 07 42 8 0.12 55 0 30.5 5 19 0.60 22 1.15 30.6 7 21 0.23 24 0.90 0.12 48 6 0.81 74 9 0.9 42 34 2 14 17 0.93 0.00 07 46 3 0.11 60 5 33.1 5 18 9.14 22 2.29 33.2 8 20 8.84 24 2.12 0.13 49 3 0.80 58 3 0.9 40 76 2 12 18 5.37 0.00 07 49 8 0.10 74 4 35.7 6 18 7.66 22 3.42 35.9 0 20 7.44 24 3.34 0.14 49 7 0.79 42 9 0.9 39 25 2 10 20 0.74 0.00 07 53 3 0.09 96 00 38.3 8 18 6.18 22 4.56 38.5 3 20 6.02 24 4.55 0.1 54 96 0.7 82 86 0.9 37 82 2 8 21 7.08 0.00 07 57 0 0.09 24 38 41.0 1 18 4.69 22 5.69 41.1 7 20 4.59 24 5.7 6 0.16 49 1 0.7 71 54 0.9 36 45 2 6 23 4.44 0.00 07 60 7 0.08 58 88 43.6 4 18 3.18 22 6.82 43.8 2 20 3.14 24 6.9 5 0.17 48 2 0.7 60 33 0.9 35 14 2 4 25 2.85 0.00 07 64 4 0.07 98 89 46.2 9 18 1.66 22 7.94 46.4 8 20 1.66 24 8.1 4 0.18 46 9 0.7 49 21 0.9 33 90 2 2 27 2.36 0.00 07 68 3 0.07 43 88 48.9 4 18 0.12 22 9.07 49.1 5 20 0.17 24 9.3 3 0.19 45 2 0.7 38 19 0.9 32 71 0 29 3.01 0.00 07 72 2 0.06 93 35 51.6 1 17 8.58 23 0.18 51.8 3 19 8.67 25 0.5 0 0.20 43 2 0.7 27 26 0.9 31 58 2 31 4.84 0.00 07 76 1 0.06 46 90 54.2 8 17 7.01 23 1.30 54.5 3 19 7.14 25 1.6 6 0.21 40 8 0.7 16 41 0.9 30 50 4 33 7.90 0.00 07 80 2 0.06 04 12 56.9 7 17 5.44 23 2.40 57.2 3 19 5.58 25 2.8 2 0.22 38 1 0.7 05 65 0.9 29 46 6 36 2.23 0.00 07 84 3 0.05 64 69 59.6 6 17 3.84 23 3.51 59.9 5 19 4.01 25 3.9 6 0.23 35 1 0.6 94 96 0.9 28 47 8 38 7.88 0.00 07 88 6 0.05 28 29 62.3 7 17 2.23 23 4.60 62.6 8 19 2.42 25 5.0 9 0.24 31 8 0.6 84 35 0.9 27 52 10 41 4.89 0.00 07 92 9 0.04 94 66 65.0 9 17 0.6 1 23 5.69 65.4 2 19 0.80 25 6.22 0.2 52 82 0.6 73 80 0.9 26 6 1 12 44 3.31 0.00 07 97 3 0.04 63 54 67.8 2 16 8.9 6 23 6.78 68.1 7 18 9.16 25 7.33 0.2 62 43 0.6 63 31 0.9 25 7 4 14 47 3.19 0.00 08 01 8 0.04 34 71 70.5 6 16 7.3 0 23 7.86 70.9 4 18 7.49 25 8.43 0.2 72 01 0.6 52 89 0.9 24 9 0 16 50 4.58 0.00 08 06 4 0.04 07 98 73.3 1 16 5.6 2 23 8.93 73.7 2 18 5.80 25 9.51 0.2 81 57 0.6 42 52 0.9 24 0 9 18 53 7.52 0.00 08 11 2 0.03 83 17 76.0 7 16 3.9 2 23 9.99 76.5 1 18 4.08 26 0.59 0.2 91 11 0.6 32 19 0.9 23 3 0 20 57 2.07 0.00 08 16 0 0.03 60 12 78.8 5 16 2.1 9 24 1.04 79.3 2 18 2.33 26 1.64 0.3 00 62 0.6 21 92 0.9 22 5 4 22 60 8.27 0.00 08 20 9 0.03 38 67 81.6 4 16 0.4 5 24 2.09 82.1 4 18 0.55 26 2.69 0.3 10 12 0.6 11 68 0.9 21 8 0 24 64 6.18 0.00 08 26 0 0.03 18 69 84.4 4 15 8.6 8 24 3.13 84.9 8 17 8.74 26 3.72 0.3 19 59 0.6 01 48 0.9 21 0 7 26 68 5.84 0.00 08 31 2 0.03 00 08 87.2 6 15 6.8 9 24 4.15 87.8 3 17 6.90 26 4.73 0.3 29 05 0.5 91 31 0.9 20 3 6 28 72 7.31 0.00 08 36 6 0.02 82 71 90.0 9 15 5.0 8 24 5.17 90.7 0 17 5.03 26 5.73 0.3 38 49 0.5 81 17 0.9 19 6 7 30 77 0.64 0.00 08 42 1 0.02 66 48 92.9 3 15 3.2 4 24 6.17 93.5 8 17 3.13 26 6.71 0.3 47 92 0.5 71 05 0.9 18 9 7 32 81 5.89 0.00 08 47 7 0.02 51 31 95.7 9 15 1.3 7 24 7.17 96.4 9 17 1.19 26 7.67 0.3 57 34 0.5 60 95 0.9 18 2 9 34 86 3.11 0.00 08 53 5 0.02 37 12 98.6 7 14 9.4 8 24 8.15 99.4 1 16 9.21 26 8.61 0.3 66 75 0.5 50 86 0.9 17 6 0 36 91 2.35 0.00 08 59 5 0.02 23 83 10 1.56 14 7.55 24 9.11 10 2.34 16 7.19 26 9.53 0.37 61 5 0.54 07 7 0.9 16 92 38 96 3.68 0.00 08 65 7 0.02 11 37 10 4.47 14 5.60 25 0.07 10 5.30 16 5.13 27 0.44 0.38 55 4 0.53 06 8 0.9 16 22 40 10 17.1 0.00 08 72 0 0.01 99 68 10 7.39 14 3.61 25 1.00 10 8.28 16 3.03 27 1.31 0.39 49 3 0.52 05 9 0.9 15 52 42 10 72.8 0.00 08 78 6 0.01 88 70 11 0.34 14 1.59 25 1.92 11 1.28 16 0.89 27 2.17 0.40 43 2 0.51 04 8 0.9 14 80 44 11 30.7 0.00 08 85 4 0.01 78 37 11 3.30 13 9.53 25 2.83 11 4.30 15 8.70 27 3.00 0.41 37 1 0.50 03 6 0.9 14 07 ( i' )( r )( i" )( s' )( s" ) adDownload to read ad-free 917 APPENDIX 1 TABLE A–11 Saturated refrigerant-134a—Temperature table ( Concluded ) Temp., T 8 C Sat. press., P sat kPa Specific volume, m 3 /kg Internal energy, kJ/kg Enthalpy, kJ/kg Entropy, kJ/kg∙K Sat. liquid, v f Sat. vapor, v g Sat.liquid, u f Evap., u fg Sat.vapor, u g Sat.liquid, h f Evap., h fg Sat.vapor, h g Sat.liquid, s f Evap., s fg Sat.vapor, s g 46 11 91.0 0.00 08 92 4 0.01 68 66 11 6.28 13 7.43 25 3.71 11 7.34 15 6.46 27 3.80 0.4 23 11 0.49 02 0 0.91 33 1 48 12 53.6 0.00 08 99 7 0.01 59 51 11 9.28 13 5.30 25 4.58 12 0.41 15 4.17 27 4.57 0.4 32 51 0.48 00 1 0.91 25 2 52 13 86.2 0.00 09 15 1 0.01 42 76 12 5.35 13 0.89 25 6.24 12 6.62 14 9.41 27 6.03 0.4 51 36 0.45 94 8 0.91 08 4 56 15 29.1 0.00 09 31 7 0.01 27 82 13 1.52 12 6.29 25 7.81 13 2.94 14 4.41 27 7.35 0.4 70 28 0.43 87 0 0.90 89 8 60 16 82.8 0.00 09 49 8 0.01 14 34 13 7.79 12 1.45 25 9.23 13 9.38 13 9.09 27 8.47 0.4 89 30 0.41 74 6 0.90 67 6 65 18 91.0 0.00 09 75 1 0.00 99 59 14 5.80 11 5.06 26 0.86 14 7.64 13 2.05 27 9.69 0.5 13 30 0.39 04 8 0.90 37 9 70 21 18.2 0.00 10 03 7 0.00 86 50 15 4.03 10 8.17 26 2.20 15 6.15 12 4.37 28 0.52 0.5 37 63 0.36 23 9 0.90 00 2 75 23 65.8 0.00 10 37 3 0.00 74 86 16 2.55 10 0.62 26 3.17 16 5.01 11 5.87 28 0.88 0.5 62 52 0.33 27 9 0.89 53 1 80 26 35.3 0.00 10 77 4 0.00 64 39 1 71.4 3 92.2 2 26 3.66 17 4.27 10 6.35 28 0.63 0.58 81 2 0.30 11 3 0.8 89 25 85 29 28.2 0.00 11 27 3 0.00 54 84 1 80.8 1 82.64 26 3.45 18 4.11 95.3 9 27 9.51 0.6 14 87 0.2 66 32 0.88 1 20 90 32 46.9 0.00 11 93 8 0.00 45 91 1 90.9 4 71.19 26 2.13 19 4.82 82.2 2 27 7.04 0.6 43 54 0.2 26 38 0.86 9 91 95 35 94.1 0.00 12 94 5 0.00 37 13 2 02.4 9 56.25 25 8.73 20 7.14 64.9 4 27 2.08 0.6 76 05 0.1 76 38 0.85 2 43 10 0 39 75.1 0.00 15 26 9 0.00 26 57 21 8.7 3 29.7 2 24 8.46 22 4.80 34.2 2 25 9.02 0.7 22 24 0.09 16 9 0.81 39 3 Source of Data: Tables A 2 11 through A 2 13 are generated using the Engineering Equation Solver (EES) software developed b y S. A. Klein and F. L. Alvarado. The routine used in calculations is the R134a, which is base d on the fundamental equation of state developed by R. Tillner 2 Roth and H.D. Baehr, “An International Standard Formulation for the Thermodynamic Properties of 1,1,1,2-Tet rafluoroethane (HFC-134a) for temperatures from 170 K to 455 K a nd pressures up to 70 MPa,” J. Phys. Chem, Ref. Data , Vol. 23, No. 5, 1994. The enthalpy and entropy values of saturated liquid are set to zero at 2 40 8 C (and 2 40 8 F). adDownload to read ad-free 918 PROPERTY TABLES AND CHARTS TABLE A–12 Saturated refrigerant-134a—Pressure table Press., P kPa Sat. temp., T sat 8 C Specific volume, m 3 /kg Internal energy, kJ/kg Enthalpy, kJ/kg Entropy, kJ/kg∙K Sat. liquid, v f Sat. vapor, v g Sat.liquid, u f Evap., u fg Sat.vapor, u g Sat.liquid, h f Evap., h fg Sat.vapor, h g Sat.liquid, s f Evap., s fg Sat.vapor, s g 60 2 36.9 5 0.00 07 09 7 0.31 10 8 3.79 5 20 5.34 20 9.13 3.83 7 22 3.96 22 7.80 0.01 63 3 0.94 81 2 0.96 44 5 70 2 33.8 7 0.00 07 14 3 0.26 92 1 7.67 2 20 3.23 21 0.90 7.72 2 22 2.02 22 9.74 0.03 26 4 0.92 78 3 0.96 04 7 80 2 31.1 3 0.00 07 18 4 0.23 74 9 11.1 4 20 1.33 21 2.48 11.2 0 22 0.27 23 1.47 0.04 70 7 0.91 00 9 0.95 71 6 90 2 28.6 5 0.00 07 22 2 0.21 26 1 14.3 0 19 9.60 21 3.90 14.3 6 21 8.67 23 3.04 0.06 00 3 0.89 43 1 0.95 43 4 100 2 26.3 7 0.00 07 25 8 0.19 25 5 17.1 9 19 8.01 21 5.21 17.2 7 21 7.19 23 4.46 0.07 18 2 0.88 00 8 0.95 19 1 120 2 22.3 2 0.00 07 32 3 0.16 21 6 22.3 8 19 5.15 21 7.53 22.4 7 21 4.52 23 6.99 0.09 26 9 0.85 52 0 0.94 78 9 140 2 18.7 7 0.00 07 38 1 0.14 02 0 26.9 6 19 2.60 21 9.56 27.0 6 21 2.13 23 9.19 0.11 08 0 0.83 38 7 0.94 46 7 160 2 15.6 0 0.00 07 43 5 0.12 35 5 31.0 6 19 0.31 22 1.37 31.1 8 20 9.96 24 1.14 0.12 68 6 0.81 51 7 0.94 20 2 180 2 12.7 3 0.00 07 48 5 0.11 04 9 34.8 1 18 8.20 22 3.01 34.9 4 20 7.95 24 2.90 0.14 13 1 0.79 84 8 0.93 97 9 200 2 10.0 9 0.00 07 53 2 0.09 99 51 38.2 6 18 6.25 224.5 1 38.4 1 20 6.09 24 4.50 0.15 44 9 0.78 33 9 0.93 78 8 240 2 5.38 0.00 07 61 8 0.08 39 83 44.4 6 18 2.71 227.1 7 44.6 4 20 2.68 24 7.32 0.17 78 6 0.75 68 9 0.93 47 5 280 2 1.25 0.00 07 69 7 0.07 24 34 49.9 5 17 9.54 229.4 9 50.1 6 19 9.61 24 9.77 0.19 82 2 0.73 40 6 0.93 22 8 32 0 2.46 0.00 07 7 71 0.06 36 81 54.9 0 17 6.65 23 1.55 55.1 4 19 6.7 8 25 1.93 0.21 63 1 0.71 39 5 0.9 30 26 36 0 5.82 0.00 07 8 40 0.05 68 09 59.4 2 17 3.99 23 3.41 59.7 0 19 4.1 5 25 3.86 0.23 26 5 0.69 59 1 0.9 28 56 40 0 8.91 0.00 07 9 05 0.05 12 66 63.6 1 17 1.49 23 5.10 63.9 2 19 1.6 8 25 5.61 0.24 75 7 0.67 95 4 0.9 27 11 45 0 12.4 6 0.00 07 9 83 0.04 56 77 68.4 4 16 8.58 23 7.03 68.8 0 18 8.7 8 25 7.58 0.26 46 2 0.66 09 3 0.92 55 5 50 0 15.7 1 0.00 08 0 58 0.04 11 68 72.9 2 16 5.86 23 8.77 73.3 2 18 6.0 4 25 9.36 0.28 02 1 0.64 39 9 0.92 42 0 55 0 18.7 3 0.00 08 1 29 0.03 74 52 77.0 9 16 3.29 24 0.38 77.5 4 18 3.4 4 26 0.98 0.29 46 0 0.62 84 2 0.92 30 2 60 0 21.5 5 0.00 08 1 98 0.03 43 35 81.0 1 16 0.84 24 1.86 81.5 0 18 0.9 5 26 2.46 0.30 79 9 0.61 39 8 0.92 19 6 65 0 24.2 0 0.00 08 2 65 0.03 16 80 84.7 2 15 8.51 24 3.23 85.2 6 17 8.5 6 26 3.82 0.32 05 2 0.60 04 8 0.92 10 0 70 0 26.6 9 0.00 08 3 31 0.02 93 92 88.2 4 15 6.27 24 4.51 88.8 2 17 6.2 6 26 5.08 0.33 23 2 0.58 78 0 0.92 01 2 75 0 29.0 6 0.00 08 3 95 0.02 73 98 91.5 9 15 4.11 24 5.70 92.2 2 17 4.0 3 26 6.25 0.34 34 8 0.57 58 2 0.91 93 0 80 0 31.3 1 0.00 08 4 57 0.02 56 45 94.8 0 15 2.02 24 6.82 95.4 8 17 1.8 6 26 7.34 0.35 40 8 0.56 44 5 0.91 85 3 85 0 33.4 5 0.00 08 5 19 0.02 40 91 97.8 8 15 0.00 24 7.88 98.6 1 16 9.7 5 26 8.36 0.36 41 7 0.55 36 2 0.91 77 9 90 0 35.5 1 0.00 08 58 0 0.02 27 03 10 0.84 14 8.03 24 8.88 10 1.62 16 7.69 26 9.31 0.37 38 3 0.54 32 6 0.91 70 9 95 0 37.4 8 0.00 08 64 0 0.02 14 56 10 3.70 14 6.11 24 9.82 10 4.52 16 5.68 27 0.20 0.38 30 7 0.53 33 3 0.91 64 1 10 00 39.3 7 0.0 00 87 00 0.02 03 29 10 6.47 14 4.24 25 0.71 10 7.34 16 3.70 27 1.04 0.39 19 6 0.52 37 8 0.91 57 4 12 00 46.2 9 0.0 00 89 35 0.01 67 28 11 6.72 13 7.12 25 3.84 11 7.79 15 6.12 27 3.92 0.42 44 9 0.48 87 0 0.91 32 0 14 00 52.4 0 0.0 00 91 67 0.01 41 19 12 5.96 13 0.44 25 6.40 12 7.25 14 8.92 27 6.17 0.45 32 5 0.45 74 2 0.91 06 7 16 00 57.8 8 0.0 00 94 00 0.01 21 34 13 4.45 12 4.05 25 8.50 13 5.96 14 1.96 27 7.92 0.47 92 1 0.42 88 1 0.90 80 2 18 00 62.8 7 0.0 00 96 39 0.01 05 68 14 2.36 11 7.85 26 0.21 14 4.09 13 5.14 27 9.23 0.50 30 4 0.40 21 3 0.90 51 7 20 00 67.4 5 0.0 00 98 87 0.00 92 97 14 9.81 11 1.75 26 1.56 15 1.78 12 8.36 28 0.15 0.52 51 9 0.37 68 4 0.90 20 4 25 00 77.5 4 0.0 01 05 67 0.00 69 41 16 7.02 96.4 7 26 3.49 16 9.66 11 1.18 28 0.84 0.57 54 2 0.31 70 1 0.89 24 3 30 00 86.1 6 0.0 01 14 10 0.00 52 72 18 3.09 80.1 7 26 3.26 18 6.51 92.5 7 27 9.08 0.62 13 3 0.25 75 9 0.87 89 3 ( i' )( i" )( s') ( s" ) adDownload to read ad-free 919 APPENDIX 1 TABLE A–13 Superheated refrige rant-134a T 8 C v m 3 /kg u kJ/kg h kJ/kg s kJ/kg·K P 5 0.06 MPa ( T sat 5 2 36.95 8 C)Sat. 0.31 10 8 20 9.13 22 7.80 0.9 64 5 2 20 0.33 60 8 22 0.62 24 0.78 1.01 75 2 10 0.35 04 8 22 7.57 24 8.60 1.04 78 0 0.36 47 6 23 4.67 25 6.56 1.07 75 10 0.37 89 3 24 1.94 26 4.68 1.10 67 20 0.39 30 2 24 9.37 27 2.95 1.13 54 30 0.40 70 5 25 6.97 28 1.39 1.16 37 40 0.42 10 2 26 4.73 28 9.99 1.19 16 50 0.43 49 5 27 2.66 29 8.75 1.21 92 60 0.44 88 3 28 0.75 30 7.68 1.24 64 70 0.46 26 9 28 9.01 31 6.77 1.27 32 80 0.47 65 1 29 7.43 32 6.02 1.29 98 90 0.49 03 2 30 6.02 33 5.43 1.32 61 10 0 0.50 41 0 31 4.76 34 5.01 1.35 21 P 5 0.18 MPa ( T sat 5 2 12.73 8 C)Sa t. 0.11 04 9 22 3.01 24 2.90 0.93 98 2 10 0.11 18 9 22 5.04 24 5.18 0.94 85 0 0.1 17 22 23 2.49 25 3.59 0.97 99 10 0.12 24 0 24 0.02 26 2.05 1.01 03 20 0.12 74 8 24 7.66 27 0.60 1.04 00 30 0.13 24 8 25 5.43 27 9.27 1.06 91 40 0.13 74 1 26 3.33 28 8.07 1.09 76 50 0.14 23 0 27 1.38 29 7.00 1.12 57 60 0.14 71 5 27 9.58 30 6.07 1.15 33 70 0.15 19 6 28 7.93 31 5.28 1.18 06 80 0.15 67 3 29 6.43 32 4.65 1.20 75 90 0.16 14 9 30 5.09 33 4.16 1.23 40 10 0 0.16 62 2 31 3.90 34 3.82 1.26 03 P 5 0.28 MPa ( T sat 5 2 1.25 8 C)Sa t. 0.07 24 3 22 9.49 24 9.77 0.93 23 0 0.07 28 2 23 0.46 25 0.85 0.93 62 10 0.07 64 6 23 8.29 25 9.70 0.96 81 20 0.07 99 7 24 6.15 26 8.54 0.99 87 30 0.08 33 8 25 4.08 27 7.42 1.02 85 40 0.08 67 2 26 2.12 28 6.40 1.05 77 50 0.09 00 0 27 0.28 29 5.48 1.08 62 60 0.09 32 4 27 8.58 30 4.69 1.11 43 70 0.09 64 4 28 7.01 31 4.01 1.14 19 80 0.09 96 1 29 5.59 32 3.48 1.16 90 90 0.10 27 5 30 4.30 33 3.07 1.19 58 10 0 0.10 58 7 31 3.17 34 2.81 1.22 23 11 0 0.10 89 7 32 2.18 35 2.69 1.24 84 12 0 0.11 20 5 33 1.34 36 2.72 1.27 42 13 0 0.11 51 2 34 0.65 37 2.88 1.29 98 14 0 0.11 81 8 35 0.11 38 3.20 1.32 51 v m 3 /kg u kJ/kg h kJ/kg s kJ/kg·K P 5 0.10 MPa ( T sat 5 2 26.37 8 C)0.19 25 5 21 5.21 23 4.46 0.95 19 0.19 84 1 21 9.68 23 9.52 0.97 21 0.20 74 3 22 6.77 24 7.51 1.00 31 0.21 63 0 23 3.97 25 5.60 1.03 33 0.22 50 6 24 1.32 26 3.82 1.06 28 0.23 37 3 24 8.81 27 2.18 1.09 19 0.24 23 3 25 6.46 28 0.69 1.12 04 0.25 08 8 26 4.27 28 9.36 1.14 85 0.25 93 7 27 2.24 29 8.17 1.17 62 0.26 78 3 28 0.36 30 7.15 1.20 36 0.27 62 6 28 8.65 31 6.28 1.23 06 0.28 46 5 29 7.10 32 5.57 1.25 73 0.29 30 3 30 5.71 33 5.01 1.28 36 0.30 13 8 31 4.48 34 4.61 1.30 97 P 5 0.20 MPa ( T sat 5 2 10.09 8 C)0.09 99 5 22 4.51 24 4.50 0.93 79 0.09 99 1 22 4.57 24 4.56 0.93 81 0.10 48 1 23 2.11 25 3.07 0.96 99 0.10 95 5 23 9.69 26 1.60 1.00 05 0.11 41 8 24 7.36 27 0.20 1.03 04 0.11 87 4 25 5.16 27 8.91 1.05 96 0.12 32 2 26 3.09 28 7.74 1.08 82 0.12 76 6 27 1.16 29 6.70 1.11 64 0.13 20 6 27 9.38 30 5.79 1.14 41 0.13 64 1 28 7.75 31 5.03 1.17 14 0.14 07 4 29 6.27 32 4.41 1.19 84 0.14 50 4 30 4.93 33 3.94 1.22 50 0.14 93 3 31 3.75 34 3.62 1.25 13 P 5 0.32 MPa ( T sat 5 2.46 8 C)0.06 36 8 23 1.55 25 1.93 0.93 03 0.06 60 9 23 7.56 25 8.70 0.95 45 0.06 92 5 24 5.51 26 7.67 0.98 56 0.07 23 1 25 3.52 27 6.66 1.01 58 0.07 53 0 26 1.62 28 5.72 1.04 52 0.07 82 3 26 9.83 29 4.87 1.07 39 0.08 11 1 27 8.17 30 4.12 1.10 22 0.08 39 5 28 6.64 31 3.50 1.12 99 0.08 67 5 29 5.24 32 3.00 1.15 72 0.08 95 3 30 3.99 33 2.64 1.18 41 0.09 22 9 31 2.87 34 2.41 1.21 06 0.09 50 3 32 1.91 35 2.31 1.23 68 0.09 77 5 33 1.08 36 2.36 1.26 27 0.10 04 5 34 0.41 37 2.55 1.28 83 0.10 31 4 34 9.88 38 2.89 1.31 36 v m 3 /kg u kJ/kg h kJ/kg s kJ/kg·K P 5 0.14 MPa ( T sat 5 2 18.77 8 C)0.14 02 0 21 9.56 23 9.19 0.94 47 0.14 60 5 22 5.93 24 6.37 0.97 24 0.15 26 3 23 3.25 25 4.61 1.00 32 0.15 90 8 24 0.68 26 2.95 1.03 31 0.16 54 4 24 8.24 27 1.40 1.06 25 0.17 17 2 25 5.95 27 9.99 1.09 13 0.17 79 4 26 3.80 28 8.72 1.11 96 0.18 41 2 27 1.81 29 7.59 1.14 75 0.19 02 5 27 9.97 30 6.61 1.17 50 0.19 63 5 28 8.29 31 5.78 1.20 21 0.20 24 2 29 6.77 32 5.11 1.22 89 0.20 84 7 30 5.40 33 4.59 1.25 54 0.21 44 9 31 4.19 34 4.22 1.28 15 P 5 0.24 MPa ( T sat 5 2 5.38 8 C)0.08 39 8 22 7.17 24 7.32 0.93 48 0.08 61 7 23 1.30 25 1.98 0.95 20 0.09 02 6 23 9.00 26 0.66 0.98 32 0.09 42 3 24 6.76 26 9.38 1.01 34 0.09 81 2 25 4.63 27 8.17 1.04 29 0.10 19 3 26 2.61 28 7.07 1.07 18 0.10 57 0 27 0.73 29 6.09 1.10 02 0.10 94 2 27 8.98 30 5.24 1.12 81 0.11 31 0 28 7.38 31 4.53 1.15 55 0.11 67 5 29 5.93 32 3.95 1.18 26 0.12 03 8 30 4.62 33 3.51 1.20 93 0.12 39 8 31 3.46 34 3.22 1.23 56 P 5 0.40 MPa ( T sat 5 8.91 8 C)0.0 512 66 2 35.10 255.61 0.9 27 1 0.0 515 06 2 35.99 256.59 0.9 30 6 0.0 54 213 244.19 265.88 0.9 62 8 0.0 56 796 252.37 275.09 0.9 93 7 0.0 59 292 260.60 284.32 1.0 23 7 0.0 61 724 268.92 293.61 1.0 52 9 0.0 64 104 277.34 302.98 1.0 81 4 0.0 66 443 285.88 312.45 1.1 09 5 0.0 68 747 294.54 322.04 1.1 37 0 0.0 71 023 303.34 331.75 1.1 64 1 0.0 73 274 312.28 341.59 1.1 90 8 0.0 75 504 321.35 351.55 1.2 17 2 0.0 77 717 330.56 361.65 1.2 43 2 0.0 79 913 339.92 371.89 1.2 68 9 0.0 82 096 349.42 382.26 1.2 94 3 ( i ) adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Saturation Table R 410A No ratings yet Saturation Table R 410A 5 pages Table A-4E: Saturated Water-Temperature Table 50% (2) Table A-4E: Saturated Water-Temperature Table 8 pages Mechanics of Materials SI Edition 7th - Solution Manual No ratings yet Mechanics of Materials SI Edition 7th - Solution Manual 10 pages Steam Table No ratings yet Steam Table 6 pages Principles of Engineering Thermodynamics - SI Version 8th Edition No ratings yet Principles of 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10860	https://fiveable.me/analytical-chemistry/unit-2/significant-figures-measurement-uncertainty/study-guide/yCndln7wwHHT2mCN	Significant figures and measurement uncertainty \| Analytical Chemistry Class Notes \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade ⚗️Analytical Chemistry Unit 2 Review 2.1 Significant figures and measurement uncertainty All Study Guides Analytical Chemistry Unit 2 – Chemical Analysis: Measurements & Data Topic: 2.1 ⚗️Analytical Chemistry Unit 2 Review 2.1 Significant figures and measurement uncertainty Written by the Fiveable Content Team • Last updated September 2025 Written by the Fiveable Content Team • Last updated September 2025 print study guide copy citation APA ⚗️Analytical Chemistry Unit & Topic Study Guides Introduction to Analytical Chemistry Chemical Analysis: Measurements & Data 2.1 Significant figures and measurement uncertainty 2.2 Accuracy, precision, and error analysis 2.3 Statistical methods for data analysis and interpretation 2.4 Calibration methods and standard addition Sample Preparation and Handling Gravimetric and Titrimetric Analysis Spectroscopic Methods Separation Techniques Electroanalytical Methods Mass Spectrometry Thermal Analysis Automated and Computerized Systems Quality Assurance and Control in Chemistry Applications of Analytical Chemistry print guide report error Significant figures and measurement uncertainty are crucial concepts in analytical chemistry. They help scientists communicate the precision and reliability of their measurements, ensuring data integrity and reproducibility. Understanding these concepts allows chemists to report results accurately and interpret data correctly. By mastering significant figures and measurement uncertainty, you'll be better equipped to analyze and present chemical measurements effectively. Significant Figures in Measurement Importance of Significant Figures Significant figures are the meaningful digits in a measured or calculated quantity that indicate the precision and uncertainty of the value The number of significant figures in a reported value reflects the precision of the measuring instrument and the certainty of the measurement More significant figures imply higher precision and less uncertainty Fewer significant figures suggest lower precision and greater uncertainty Reporting the appropriate number of significant figures is crucial for maintaining the integrity and reproducibility of scientific data Overreporting can imply a false sense of precision Underreporting can lead to a loss of valuable information Significant figures are essential for communicating the limitations and reliability of experimental results They allow other researchers to assess the quality and reproducibility of the data They help in determining the appropriate level of precision for subsequent calculations and analyses Rules for Determining Significant Figures All non-zero digits (1-9) are always significant Example: In the number 123.45, all five digits are significant Zeros between non-zero digits are always significant Example: In the number 1002.05, all six digits are significant Leading zeros (to the left of the first non-zero digit) are never significant, as they merely indicate the position of the decimal point Example: In the number 0.0123, only the last three digits are significant Trailing zeros (to the right of the last non-zero digit) are significant only if the decimal point is explicitly shown Example: In the number 12.3000, all six digits are significant because the decimal point is shown Example: In the number 12300 (without a decimal point), only the first three digits are significant Determining Significant Figures in Calculations Addition and Subtraction In addition and subtraction, the result should have the same number of decimal places as the least precise measurement Example: 12.3 + 1.456 = 13.8 (rounded to one decimal place, as 12.3 has only one decimal place) Example: 10.1 - 9.78 = 0.3 (rounded to one decimal place, as 10.1 has only one decimal place) When adding or subtracting measurements with different units, convert them to the same unit before determining the number of significant figures Example: 5.2 cm + 12.34 mm = 5.2 cm + 1.234 cm = 6.4 cm (rounded to one decimal place) Multiplication and Division In multiplication and division, the result should have the same number of significant figures as the quantity with the least number of significant figures Example: 2.3 × 1.456 = 3.3 (rounded to two significant figures, as 2.3 has only two significant figures) Example: 12.34 ÷ 2.1 = 5.9 (rounded to two significant figures, as 2.1 has only two significant figures) When multiplying or dividing measurements with different units, the result should have the appropriate unit derived from the input units Example: 5.2 cm × 3.1 cm = 16 cm² (rounded to two significant figures) Measurement Uncertainty and Its Sources Concept of Measurement Uncertainty Measurement uncertainty is the doubt that exists about the result of any measurement, which is an estimate of the range of values within which the true value is expected to lie No measurement is perfect, and there is always some level of uncertainty associated with the measured value Measurement uncertainty arises from various sources, including instrument limitations, environmental factors, operator error, and the inherent variability of the measured quantity Understanding and quantifying measurement uncertainty is essential for assessing the reliability and comparability of experimental results Sources of Measurement Uncertainty Instrument limitations, such as the resolution of the measuring device and its calibration, contribute to systematic errors in measurements Example: A ruler with a resolution of 1 mm cannot accurately measure lengths smaller than 1 mm Example: An improperly calibrated balance may consistently give readings that are higher or lower than the true value Environmental factors, like temperature, humidity, and pressure, can affect the performance of measuring instruments and introduce uncertainties Example: Changes in temperature can cause thermal expansion or contraction of materials, affecting length measurements Example: Variations in humidity can influence the mass of hygroscopic substances during weighing Operator errors, including parallax errors and inconsistencies in measurement techniques, can lead to random errors and increased uncertainty Example: Parallax errors occur when the observer's eye is not aligned properly with the measuring scale, leading to inaccurate readings Example: Inconsistent placement of the measuring instrument or variations in the applied force during measurements can introduce random errors Estimating and Reporting Uncertainty Expressing Uncertainty Uncertainty in measured values is typically reported as an absolute or relative uncertainty, depending on the context and the desired level of precision Absolute uncertainty is expressed in the same units as the measured value and indicates the range within which the true value is expected to lie Example: A length measurement of 5.2 cm with an absolute uncertainty of ±0.1 cm would be reported as 5.2 ± 0.1 cm Relative uncertainty is the ratio of the absolute uncertainty to the measured value, often expressed as a percentage Example: A length measurement of 5.2 cm with an absolute uncertainty of ±0.1 cm has a relative uncertainty of (0.1 cm / 5.2 cm) × 100% ≈ 2% When reporting uncertainty, the number of significant figures in the uncertainty should match the number of significant figures in the measured value Example: 5.2 ± 0.1 cm (both the measured value and the uncertainty have two significant figures) Propagation of Uncertainty Propagation of uncertainty rules are used to estimate the uncertainty in calculated values based on the uncertainties of the input measurements For addition and subtraction, the absolute uncertainties are added in quadrature (square root of the sum of squares) Example: If $A = 5.2 \pm 0.1$ cm and $B = 3.4 \pm 0.2$ cm, then for $C = A + B$, the absolute uncertainty in $C$ is $\sqrt{(0.1)^2 + (0.2)^2}$ cm $\approx 0.2$ cm For multiplication and division, the relative uncertainties are added in quadrature Example: If $A = 5.2 \pm 0.1$ cm and $B = 3.4 \pm 0.2$ cm, then for $D = A \times B$, the relative uncertainty in $D$ is $\sqrt{(0.1/5.2)^2 + (0.2/3.4)^2} \times 100%$ $\approx 3%$ These propagation rules allow for the estimation of uncertainty in derived quantities based on the uncertainties of the input measurements BackNext 2.2 Study Content & Tools Study GuidesPractice QuestionsGlossaryScore Calculators Company Get $$ for referralsPricingTestimonialsFAQsEmail us Resources AP ClassesAP Classroom every AP exam is fiveable history 🌎 ap world history🇺🇸 ap us history🇪🇺 ap 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10861	https://search.library.ucla.edu/discovery/fulldisplay?docid=alma9915804443606533&context=L&vid=01UCS_LAL:UCLA&lang=en&adaptor=Local%20Search%20Engine&tab=Articles_books_more_slot&query=sub%2Cexact%2C%20Quantitative%20genetics%20%2CAND&mode=advanced&offset=0	Full display result ; Principles of population genetics / Daniel L. Hartl, Andrew G. Clark. ; Hartl, Daniel L.; ; Sunderland, Mass. : Sinauer Associates; ; ©1989; Send to How to get it Please sign in to check if there are any request options. 0" flex="none" class="flex-none"> For Special Collections, Film & TV Archive, and Clark Library, select the location below to display the Request link. MaintwoRequest FormRequest reply MaintwoRequest FormRequest reply Locations Powell Library Available , Powell Library ; QH455 .H37 1989 Biomed Library Out of library , Biomed Library ; QH455 .H331p 1989 LOCATION ITEMS Institutions ListAlma Member Getit Institutions ListAlma Member Getit Links MARC Record Details Author / Contributor Hartl, Daniel L. Hartl, Daniel L. Clark, Andrew G., 1954- Clark, Andrew G., 1954- Title Principles of population genetics / Daniel L. Hartl, Andrew G. Clark. Principles of population genetics / Daniel L. Hartl, Andrew G. Clark. Publication Information Sunderland, Mass. : Sinauer Associates, ©1989. Sunderland, Mass. : Sinauer Associates, ©1989. Edition 2nd ed. 2nd ed. Type Book Book Physical Description xii, 682 pages : illustrations ; 25 cm xii, 682 pages : illustrations ; 25 cm Contents Contents: Darwinian evolution in Mendelian populations. -- Random genetic drift. -- Mutation and the neutral theory. -- Natural selection. -- Inbreeding and other forms of nonrandom mating. -- Population subdivision and migration. -- Molecular population genetics. -- Evolutionary genetics of quantitative characters. -- Ecological genetics and speciation. Contents: Darwinian evolution in Mendelian populations. -- Random genetic drift. -- Mutation and the neutral theory. -- Natural selection. -- Inbreeding and other forms of nonrandom mating. -- Population subdivision and migration. -- Molecular population genetics. -- Evolutionary genetics of quantitative characters. -- Ecological genetics and speciation. Notes Includes indexes. Includes indexes. Includes bibliographical references (pages 628-664). Includes bibliographical references (pages 628-664). Language English English Subject Population genetics Population genetics Quantitative genetics Quantitative genetics Populationsgenetik Populationsgenetik Lehrbuch Lehrbuch Populatiegenetica Populatiegenetica Einführung Einführung populatiegenetica populatiegenetica population genetics population genetics kwantitatieve genetica kwantitatieve genetica quantitative genetics quantitative genetics Population and Quantitative Genetics Population and Quantitative Genetics Kwantitatieve- en populatiegenetica Kwantitatieve- en populatiegenetica Genre Einführung Einführung Einführung Einführung Lehrbuch Lehrbuch MESH Subjects Genetics, Population Genetics, Population Identifier LC : 88030818 LC : 88030818 ISBN : 0878933026 ISBN : 0878933026 ISBN : 9780878933020 ISBN : 9780878933020 OCLC : (OCoLC)18558351 OCLC : (OCoLC)18558351 OCLC : (OCoLC)ocm18558351 OCLC : (OCoLC)ocm18558351 MMS ID 9915804443606533 Library Catalog Former System Number 1580444-ucladb We are committed to updating our catalog records and finding aids whenever feasible to revise problematic descriptions and subjects, including the addition of relevant context. To report harmful language, please use this form. Related Resources A primer of population genetics A primer of population genetics Principles of population genetics Principles of population genetics Principles of population genetics Principles of population genetics Genetics of populations Genetics of populations Genetics of populations Molecular genetic ecology Human population genetics : the Pittsburgh symposium Approche pluri-disciplinaire des isolats humains Populations and genetics : legal and socio-ethical perspectives
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10863	https://www.khanacademy.org/math/cc-fifth-grade-math/imp-algebraic-thinking/imp-writing-expressions/v/translating-expressions-with-parentheses	Translating expressions with parentheses (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Understand the importance of parentheses in expressions, and how they can change the meaning of calculations. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted CodeLoader 10 years ago Posted 10 years ago. Direct link to CodeLoader's post “Does "divide in half" act...” more Does "divide in half" actually mean "multiply by half" or "take a half of.."? Answer Button navigates to signup page •13 comments Comment on CodeLoader's post “Does "divide in half" act...” (149 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Miss H 10 years ago Posted 10 years ago. Direct link to Miss H's post “All three of those expres...” more All three of those expressions: "divide in half", "multiply by half", or "take a half of" are different ways of saying the same thing, which is dividing a quantity into 2 equal parts, or halves. Half is a really common concept in everyday life so there are many ways that people express it. Remember that division is the same as multiplying by the reciprocal, which is why "multiply by 1/2" is the same as "divide by 2". 4 comments Comment on Miss H's post “All three of those expres...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Basketballgirl2010 2 months ago Posted 2 months ago. Direct link to Basketballgirl2010's post “He says "this is fun let'...” more He says "this is fun let's keep going" Me muting the video cuz I watched it five times and still dont get it "uh huh you do that" Answer Button navigates to signup page •3 comments Comment on Basketballgirl2010's post “He says "this is fun let'...” (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Phoenix 4 months ago Posted 4 months ago. Direct link to Phoenix's post “I've never seen a problem...” more I've never seen a problem with multiplication that doesn't have an x Answer Button navigates to signup page •1 comment Comment on Phoenix's post “I've never seen a problem...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer TheReal3A 4 months ago Posted 4 months ago. Direct link to TheReal3A's post “Multiplication can be rep...” more Multiplication can be represented in other ways! There's the middle dot "·" which can be used so that it's not confused with the letter (you'll learn why). E.g. 5·3 = 5x3 = 15. And also if you have 2 things next to each other without any operations, there's an implied multiplication there. E.g. 3(5 + 2) = 3x(5 + 2), and also 4a = 4xa. These alternative representations of multiplication are important later on when you get to Algebra! 2 comments Comment on TheReal3A's post “Multiplication can be rep...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Lord skorupi 3 years ago Posted 3 years ago. Direct link to Lord skorupi's post “When he put the coma in t...” more When he put the coma in the wrong spot I said in my deepest voice (Not very deep) "HE SAID IT WRONG" and then burst out in laughter and then he fixed his mistake. LOL Answer Button navigates to signup page •5 comments Comment on Lord skorupi's post “When he put the coma in t...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer :o 2 years ago Posted 2 years ago. Direct link to :o's post “for the pink question he ...” more for the pink question he could of just done 43-16-11 Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Dennis/Intern 2 years ago Posted 2 years ago. Direct link to Dennis/Intern's post “No, he can't it says on t...” more No, he can't it says on the pink line it says:"43 minus the sum of 16 and 11 "the sum is the total of 16 and 11 so you have to do that first if you do it as (43-16-11) you get 43-16-11 while the answer is 38. 1 comment Comment on Dennis/Intern's post “No, he can't it says on t...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Pizza dude 5 months ago Posted 5 months ago. Direct link to Pizza dude's post “I am prepared now for any...” more I am prepared now for anyone who walks up to me on the street to ask me math problems :) Answer Button navigates to signup page •3 comments Comment on Pizza dude's post “I am prepared now for any...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer christophmanoj a year ago Posted a year ago. Direct link to christophmanoj's post “Is it called PEMDAS or BO...” more Is it called PEMDAS or BODMAS Answer Button navigates to signup page •4 comments Comment on christophmanoj's post “Is it called PEMDAS or BO...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer joshua a year ago Posted a year ago. Direct link to joshua's post “Both are correct, and the...” more Both are correct, and they describe the same thing. Just pick one that you find more comfortable with. 1 comment Comment on joshua's post “Both are correct, and the...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... JonathanZ 3 months ago Posted 3 months ago. Direct link to JonathanZ's post “I love your video” more I love your video Answer Button navigates to signup page •1 comment Comment on JonathanZ's post “I love your video” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer lildave/caleb 2 years ago Posted 2 years ago. Direct link to lildave/caleb's post “how do you translate expr...” more how do you translate expressions with parenthesis the right way? Is there also an easier way? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Lreed 2 years ago Posted 2 years ago. Direct link to Lreed's post “If they're are numbers in...” more If they're are numbers in parenthesis you are suppose to do what it says first 2 comments Comment on Lreed's post “If they're are numbers in...” (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... riaanmittal1 21 days ago Posted 21 days ago. Direct link to riaanmittal1's post “how to make a video” more how to make a video Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Video transcript [Voiceover] What I hope to do in this video, is give ourselves some practice interpreting statements and writing them as mathematical expressions, possibly using parentheses. So let's get started. And for any of these statements, if you get so inspired, and I encourage you to get so inspired, pause the video and see if you can write them as mathematical expressions. So this first one says 700 minus 19, divided in half. So we could say, another way to think about divided in half is divided by two, so we could write this as 709 minus 19, and we're going to do that first, so that's why I put the parentheses around it, divided by two, or divided in half. That's one way that we could write this. Now the next one, and once again, pause it if you get inspired, and I encourage you to. Three times the sum of 56 and seven. So it's gonna be three times the sum of 56 and seven. So the sum of 56 and seven, we want to take that first, so it's going to be 56 plus the seven, that's the sum of 56 and seven, and then we want to do three times that. We want to do three times this sum. So we could write it like that. Another way we could write it, when you're dealing with parentheses, and you're going to see this more and more as you get into more and more fancy algebra, I guess you could say, but what I'm about to show you isn't so fancy, is, you don't have to write the multiplication sign here. You could just write three, and then open parentheses, 56 plus seven, and this, too, is three times the sum of 56 and seven. And you want to be very careful, because you might be tempted to maybe do it without the parentheses, so you might be tempted to do something like this, three times 56 plus seven, but this one isn't, obviously, three times the sum of 56 and seven. In fact, the standard way to interpret this is that you would do the multiplication first. You would do three times 56, and then add seven, which is going to give you a different value, and you could try it out, than if you were to add the 56 and the seven first. So, to make sure that you do the 56 and the seven first, you want to put this parentheses around it. So let's keep going. The sum of three times 56 and seven. So we're gonna take the sum of two things. The first thing that we're gonna take the sum of is three times 56. So, three times 56, and seven. Let me do that in a different color. And seven. So this right over here is the sum of three times 56, and seven. Now it's always good to write the parentheses. It makes it a little bit cleaner, a little bit more obvious. Look, I'm gonna take the three times 56, I'm gonna do that first, and then I'm gonna add seven, but based on what I just told you, the standard way, if someone were to just write three times 56 plus seven, this actually can still be interpreted as the sum of three times 56, and seven, because as I just said, the standard, the convention, so to speak, is to do your multiplication first. Order of operations, which you may or may not, if you're not familiar, you will be familiar with it soon, is to do the multiplication first, and then add the seven, or then do the addition. But just to make it clear, it doesn't hurt to put the parentheses there. Three times 56, plus seven. Now we have 43 minus the sum of 16 and 11. So, 43 minus, so we're gonna have 43 minus, minus the sum of 16 and 11. So, minus the sum of 16 and 11. So, from 43, we're gonna take the sum of 16 and 11, and so, once again, the parentheses make it clear that we're going to take the sum of 16 and 11, and we're gonna take that from 43. The parentheses are very, very, very important here, because if we just did 43 minus 16 plus 11, the standard way of interpreting this would be 43 minus 16, and then adding 11, which would give you a different value than 43 minus the sum of 16 and 11. So once again, the parentheses are very, very, very important here to make it clear that you're gonna add the 16 and 11 first, and then subtract that sum from 43. This is fun, let's keep going. 10 times the quotient of 104 and eight. So, we're gonna do 10 times something. 10 times the quotient of 104 and eight, and so the quotient of 104 and 8 we could write like this, 104 divided by eight, or, based on what we told you a little earlier, you could write this as 10 times the quotient of 104 and eight, or 104 divided by eight. Now let's just do this last one. Four times as large as the expression 175 minus 58. So I'm gonna do four times as large as something, so I'm gonna multiply something times four. It's four times as large as the expression 175 minus 58. And once again, I could write it as four times as large as the expression, let me do that in that purple color, as the expression 175 minus 58. Either way, and once again, if you were to do it like this, if you didn't write the parentheses, then, it wouldn't be the same thing, 'cause if the parentheses weren't here, then you would want to do the four times 175 first, and then subtract the 58, which isn't what this statement is telling us. And this last one, I think, brings up an interesting thing for us to think about, because if someone were to walk up to you on the street, and they were to show you-- Whoops, what's going on with my computer? And they were to show you two different expressions. Well, the first expression said two-- Let's write it this way, actually, I'm not gonna even speak 'em out. I'm just gonna write it down. I'm just gonna write some crazy number here. Some crazy numbers here. So that's one expression that someone were to write, and let's say another one is this one, and I'm intentionally-- What, I put the commas in the wrong place. Let me make sure I get this right. Alright, that's 183,576. This is 37,399. So that's one expression, and then another expression is this. And I'm intentionally not reading it out. Well, I'll read it out a little bit, 37,399. And someone said, "Quick! "Which expression is larger?" And you might be tempted, or you might not be tempted, but you might be tempted, "Oh, let me calculate this thing. "Gee, I'm gonna have to write this thing down, "or use a calculator or something, "or whatever else to add "183,576 "plus 37,399, "and then I'm gonna have to multiply that by two, "and figure out what that number is equal to, "and then I would have to take "183,576 plus 37,399 "and figure out what that is, "multiply that by seven, "and figure out what that's going to be. "That's hard! "That's gonna take--" Not hard, it's just gonna take you some time, you might make some careless mistakes. But the big realization to say, "Well, which one is larger? "Well I don't have to even calculate these things!" 'Cause this is two times this craziness right over here, this thing that's gonna be 200 something thousand, and this is seven times that thing that is going to be 200 and something thousand. So seven times that thing is going to be larger than two times that thing, and so, one way to-- Before you dive deep, and start computing things, it's always good to take a step back and say, "Hey, look, can I look at how the expressions are formed, "the structure of these expressions?" And say, "Look, this is two times this thing, "and this is seven times this thing." Well, the seven is going to be, this one right over here is going to be a larger expression. Anyway, hopefully you enjoyed that as much as I did. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: exercise Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. 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10864	https://www.youtube.com/watch?v=IpW5BRLYXkI	Circumcircle of Isosceles Triangle with GXWeb Saltire Software 887 subscribers 4 likes Description 365 views Posted: 30 Dec 2020 Let ABC be a triangle with AC=AB. E is a point on BC. Line AE meets the circumcircle of ABC at F. Show that AB²=AE.AF 3 comments Transcript: I'm going to prove this theorem if ABC um is an isoceles triangle with AC equals a and if e is a point on b c uh line a e meets the circum circle of ABC at F show that a^ 2 = a e AF um okay well that will uh appear obvious this once we get going so first of all we're going to start with this triangle ABC and it's going to be isoceles then we can do that let's make this length B and this length B and this length a so here's our isoceles triangle and we're going to need the circum Circle which we can do using our Circle Construction now we're going to need um another point on BC well can make it external here um so here's my external point and make it lie on BC using the incidence constraint and now I'm going to join that that point up to a and create a point uh draw point where it intersects the circum Circle here F and so now what I want to look at is um the product of a e time a so distance from a e times the distance from a to F and we see that's b^ s the square of the length AB uh which is what we wanted to show
10865	http://mrsgalgebra.pbworks.com/w/page/28354850/Factoring%20Polynomials	Mrs. Grieser's Algebra Wiki: WikiGrieser / Factoring Polynomials Mrs. Grieser's Algebra Wiki: WikiGrieser log inhelp Get a free wiki\|Try our free business product Wiki Pages & Files View Edit To edit this page, request access to the workspace. Already have an account? Log in! Factoring Polynomials ===================== Page history last edited by Andrea Grieser;)15 years, 1 month ago There are a variety of way to factorpolynomials. First, we need to recall what it means to factor. If we a factor a number, such as 6, want to find all integers that divide evenly (that is, leave no remainder) into 6. The factors of 6 are 1, 2, 3, and 6, since these all divide evenly into 6. However, 4 is not a factor of 6, since it does not divide evenly into 6. With polynomials, we look to find other polynomials, all in their simplest form, thatdivide evenly into the polynomial. We will rewrite the polynomial as a product of lower-degree polynomials. For example, the factored form of x 2 - x - 2 is (x -2)(x + 1). If youmultiply the polynomials that make up the factors, you will get your original polynomial as the product. We will consider factoring types to help us classify different factoring methods. Type I Factoring This type of factoring involves "factoring out" (performing reversedistribution) of common monomial factors from a polynomial. For example, factor 6x 2 + 3x We need to find the greatest common monomial. First we will look at the numbers: is there a greatest common factor (GCF)for 6 and 3? Yes, the GCF of 6 and 3 is 3. Next we look at the variables. Is there a GCF for x 2 and x? Yes, the GCF of x 2 and x is x. So my greatest common monomial is 3x. Now we will do the opposite of what we would do duringdistribution. We will divide each term by our GCF of 3x, positioning the GCF outside the parentheses, with the remaining quotients inside the parentheses: 3x(2x + 1) We can check our factoring by distributing back: 3x(2x + 1) =6x 2 + 3x, the same polynomial we started with. Type II Factoring This type of factoring involves recognizing when a polynomial made up of two terms where both terms areperfect squares. This kind of polynomial is the result of multiplying twobinomials where one adds and one subtracts the second term (this is a case of a sum and difference product, such as (x + a)(x - a) = x 2 - a 2). For example, if we are given the polynomial x 2 -9, we see that it is the difference of two terms, both of which are perfect squares (the square root of x 2 is x, and the square root of 9 is 3). We can factor this using the sum and difference pattern: (x + 3)(x - 3) Again, always multiply the factors back to verify. Type III Factoring This type of factoring involves factoring polynomials of the form x 2 + bx + c (note that the leadingcoefficient is 1). When factoring these types of polynomials, we ask the question "what two numbers multiply to c yet also add to b?" For example, if we are given x 2 - x - 2 and wish to factor it, we first want to find two numbers that multiply to -2 (the c value)and add to -1 (the b value). These two numbers are -2 and 1:-21 = -2 and-2 + 1 = -1. This means that the polynomial will factor to (x - 2)(x + 1) (as always,multiply these factorsto verify we get the polynomial we started with). Note that if you are stumped as to what factors to use, list all the factors of the c term, and add each one until you get the b term's value. This method is the "mental math" method described in factoring second degree polynomials. Here is a video demonstrating Type III factoring using mental math: Fliqz has shut down their service. To access this video, email support with this video id: fae77024a89943c39d3934770c31f70e Type IV Factoring This type of factoring is called "factoring by grouping." If we are presented with a 4 terms polynomial, we split the polynomial into 2 groups.We then factor each group individually. If we notice a common factor between the two groups, we then factor that common factor out using type I factoring. For example, if we are asked to factor n 3 + 6n 2 + 5n + 30, we would split the polynomial into two groups: (n 3 + 6n 2) + (5n + 30). Now we factor each group individually. Using type I factoring, the first group,n 3 + 6n 2, factors to n 2(n + 6) (we factor out the common factor of n 2). Then we factor the second group, 5n + 30. These terms have a common factor of 5, so it factors to 5(n + 6). So... n 3 + 6n 2 + 5n + 30 = (n 3 + 6n 2) + (5n + 30) = n 2(n + 6) + 5(n + 6) Now we notice that these two groups have a common polynomial factor of (n + 6), so we will factor it out: n 2(n + 6) + 5(n + 6) = (n + 6)(n 2 + 5) Again, multiply these factors to verify it is the same as the original polynomial. We will use factoring by grouping when factoring type V polynomials. Type V Factoring This type of factoring involves factoring polynomials of the form ax 2 + bx + c. Note that this is similar to type III polynomials, except that the leading coefficient may be a number different than 1 (but it will not be 0). There are several different methods forfactoring these types of polynomials. This video demonstrates using factoring by grouping for these types of polynomials: Fliqz has shut down their service. To access this video, email support with this video id: 2252ce3a23024794a12a980b577c39d0 Theseclass notes provide further explanation and examples for type I factoring. Theseclass notes provide further explanation and examples for type II and III factoring. Theseclass notes provide further explanation and examples for type IV and V factoring. Thisflowchart walks you through the steps to determine what kind of factoring to do, and provides a summary of different methods for factoring type V polynomials. Many thanks to Mrs. Donohue fromThomas Jefferson High School of Science and Technology for her ideas about factoring types Factoring Polynomials ===================== #### Page Tools ### Insert links Insert links to other pages or uploaded files. PagesImages and files Insert a link to a new page 1. Loading... 1. No images or files uploaded yet. Insert image from URL Tip: To turn text into a link, highlight the text, then click on a page or file from the list above. ### Comments (0) You don't have permission to comment on this page. 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Log in! ### Navigator Algebra Integers Inverse Operations Inverse Variation Laws of Exponents Laws of Radicals Legs of a Right Triangle Linear Equations Linear Inequalities Logic Mathematical Conventions Mean Absolute Deviation Measures of Central Tendency Mental Math Method for Factoring Second Degree Polynomials Monomials Multiplicative Inverse (Reciprocal) Multiplying a Polynomial by a Monomial Multiplying Polynomials Multiplying Polynomials by Repeated Distribution (Horizontal Multiplication) Multiplying Polynomials Using a Table Multiplying Polynomials Vertically Nets Numerical Expressions Odds Order of Operations (PEMDAS) Percents Perfect Square Perimeter Point-Slope Form of a Line Polygons Polynomial Blog Question 1 Polynomial Blog Question 2 Polynomials Prisms Probability Probability: Tree Diagrams Properties of Equality Properties of Inequality Proportions Pyramids Pythagorean Theorem Quadratic Equations (Parabolas) Quadratic Formula Quadrilaterals Radicals Range of a Relation or Function Rational Equations Rationalizing the Denominator Rational Numbers Ratios Real Number Subsets optionsPagesFiles ### SideBar Mrs. G's Algebra Wiki. 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10866	https://prep.yocket.com/gmat/gmat-data-insights-practice-questions	Published Time: 2025-04-18 GMAT Data Insights Practice Questions with Answers Yocket Prep Home Prep Tools Prep Courses Free Resources Enquire Now Pricing More GRE Diagnostic Test Find Your Starting Line in Just 30 Minutes! Take a quick diagnostic test to identify your strengths and weaknesses. Get personalized insights to boost your GRE prep today! Take the Test Now Login / Sign Up Get GMAT Prepared with Yocket Prep Prepare smarter, not harder, with Yocket Prep Signup on Yocket Prep Now Table of Content Data Insights Section: Overview Types of GMAT Data Insights Questions GMAT Data Insights Practice Questions with Solution Where to Find GMAT Data Insights Sample Papers? GMAT Data Insights Preparation Strategies From the Desk of Yocket FAQs on GMAT Data Insights Practice Questions Articles you might like GMAT Permutation & Combination Practice Questions with Answers Free GMAT Integrated Reasoning Practice Questions with Answers GMAT Ratio and Proportion: Tips, Practice Questions & Answers GMAT Quantitative Probability Questions With Answers GMAT Reading Comprehension: Strategies, Practice Questions & Answe... GMAT Data Insights Practice Questions with Answers The data insights section was introduced in GMAT in 2023. Combining the erstwhile integrated reasoning and data sufficiency sections, the data insights portion of GMAT tests your ability to analyze and synthesize data from multiple sources and make informed decisions from that data. But how do you solve them? What are the data insights question strategies? Read this blog by Yocket to learn more about Data Insights questions. Key highlights: GMAT has 20 data insight questions to be covered within a 45-minute time frame. The score range for the data insights section is 60-90, with a score interval of 1 and a standard error of measurement of 3 points. The syllabus for the Data Insights section of GMAT Focus Edition consists of data sufficiency, table analysis, graphics interpretation, multi-source reasoning, and two-part analysis. Data Insights Section: Overview The official GMAT focus edition has a total of 64 questions, of which 20 are from the Data Insights section. The total time limit for this section is 45 minutes. The score range for this section is 60-90, with a score interval of 1 and a standard error of measurement of 3 points. The Data Insights section of the GMAT Focus edition is meant to test your ability in data analysis, problem-solving, interpreting data, and making informed decisions from them. Some parts included in the data insights section for GMAT are: Data Sufficiency: You get a question and 2 statements, and you have to decide whether they are sufficient to answer the question. Graphics Interpretation: You get data presented in different graphical formats, interpret them, and answer the question. Table Analysis: You get data in tabular formats, where you analyse it and answer the questions. Two-Part Analysis: You get a prompt and a question-answer table, and you have to answer 2 interdependent questions. Multi-Source Reasoning: You get statements, tables, or graphs that might or might not be interrelated, and interpret their relationship to answer the question. Types of GMAT Data Insights Questions The GMAT Data Insights section’s syllabus comprises 5 different subtopics. You will have to answer 20 questions under those subtopics. All of these questions will measure your skills in interpreting relations between various types of data provided, calculation skills, correlation between different forms of data, and such. Here is a breakdown of all the GMAT subtopics: 1. Data Sufficiency There are around 6-7 Data Sufficiency or DS questions in the Data Insights section. You can easily identify these questions by reading them as these questions will have two parts to them, there will be a problem for you to solve and then you will have statements following it. Based on the problem and statements you will have to interpret whether the statements provide sufficient information or not. 2. Graphics Interpretation In the Data Insights section, you will find questions with problems in the form of graphs, or parts of the question will have graphs. Apart from basic x-y graphs, the questions could also be in the form of pie charts, statistical curve distributions, and such. The key to answering these questions is to understand the relevance of the graph to the question as well as pick only the necessary information in the question. 3. Table Analysis The table analysis questions evaluate your ability to make correlations and interpret results. Within these questions, you will be provided with some information regarding the table, a table with definite information and then there will be statements. In front of each statement, there are the options for true or false, based on the information and the table you will have to select true or false for the statements. 4. Two-Part Analysis Another part of the Data Insights section of the GMAT focus edition is the two-part analysis question. These questions can be of any type, verbal or mathematical. Your task is to understand the provided question and statements and then interpret the answer. You will be provided with at least two statements that might or might not be interconnected. Make sure that you read the question carefully, and interpret the answer after assessing all the possibilities. 5. Multi-Source Reasoning As suggested by the name, this section is based on information from different sources like statements and graphs. Based on the different information presented, you have to find the correct answer. The type of questions can vary from asking true or false to providing the right answer. GMAT Data Insights Practice Questions with Solution These questions might come out as difficult at first, but once you start practicing them, you will understand just what you need to look for. Here are some practice D Insights Questions to help you example to help you understand it a little more: 1. Data Sufficiency: Under the Data Sufficiency questions, you will be presented with two statements and a theory relating to the statements. Your work is to interpret the relation between them. Here is a practice question to help you understand the concept: Question:Is the average of four distinct positive integers 2, 7, w, and x greater than 10? Given Statements: A) w = 2x + 3 B) w + x = 25 Choose the correct answer based on the statements provided. A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient. B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient. C) Both statements (1) and (2) together are sufficient, but neither statement alone is sufficient. D) Each statement alone is sufficient. E) Statements (1) and (2) together are not sufficient. Correct answer: B) Statement (2) alone is sufficient, but Statement (1) alone is not sufficient. Explanation: We are asked whether the average of 2, 7, w, and x > 10. The average = (2 + 7 + w + x) / 4 = (9 + w + x)/4 We need: (9 + w + x)/4 > 10 => 9 + w + x > 40 => w + x > 31 So we need to determine if w + x > 31. Statement A: w = 2x + 3 Let's test with values: Suppose x = 5 → w = 2(5) + 3 = 13 → w + x = 18 < 31 Suppose x = 12 → w = 2(12) + 3 = 27 → w + x = 39 > 31 → Different results. Not sufficient. Statement B: w + x = 25 This directly tells us w + x = 25 → 25 is less than 31, so average is less than 10. Sufficient. 2. Graph Interpretation: For the questions related to the Graphics Interpretation, you will have different types of pictograms ranging from bars to pie charts and based on them you will have to provide the correct interpretation. See this practice question below to understand the concept. The graph below shows the sales revenue of two companies, A and B, over five years: Question: When did Company A's sales revenue exceed Company B's sales revenue by the greatest amount? A) 2018 B) 2019 C) 2020 D) 2021 E) 2022 Correct answer: B) 2019 Explanation: Since the graph is not shown here, we assume the chart shows both companies’ revenues over 5 years. You calculate the difference (A – B) for each year and find the maximum. Let’s say: 2018: A = 10M, B = 9M .. Difference = 1M 2019: A = 12M, B = 8M .. Difference = 4M 2020: A = 11M, B = 10M .. Difference = 1M 2021: A = 14M, B = 13M .. Difference = 1M 2022: A = 13M, B = 12M .. Difference = 1M 3. Table Analysis: Under the questions relating to table analysis, you can get a mix of tables and statements. You have to carefully understand the provided data and then choose the correct answer. Use the practice question below to understand the concept. Here is a company’s different department’s revenue report for the year 2024: \| Department \| Q1 Revenue (USD) \| Q2 Revenue (USD) \| Q3 Revenue (USD) \| Q4 Revenue (USD) \| Total Annual Revenue (USD) \| Employee Count \| --- --- --- \| Sales \| 1,250,000 \| 1,450,000 \| 1,600,000 \| 1,750,000 \| 6,050,000 \| 45 \| \| Marketing \| 800,000 \| 950,000 \| 1,000,000 \| 1,200,000 \| 3,950,000 \| 30 \| \| Operations \| 600,000 \| 700,000 \| 750,000 \| 850,000 \| 2,900,000 \| 20 \| \| Research \| 500,000 \| 550,000 \| 600,000 \| 650,000 \| 2,300,000 \| 18 \| \| IT Support \| 400,000 \| 450,000 \| 500,000 \| 600,000 \| 1,950,000 \| 12 \| True/False Questions True/False: The Sales department experienced the highest revenue growth between Q1 and Q4. True/False: The IT Support department’s total revenue was less than half of the Sales department’s total revenue. True/False: The Operations department had the lowest revenue in every quarter. True/False: The Marketing department's revenue increased by more than 30% from Q1 to Q4. True/False: The Research department’s Q4 revenue was 30% higher than its Q1 revenue. Solution: Statement 1: Sales department experienced the highest revenue growth between Q1 and Q4. Sales: 1.75M – 1.25M = 0.5M Marketing: 1.2M – 0.8M = 0.4M Operations: 0.85M – 0.6M = 0.25M Research: 0.65M – 0.5M = 0.15M IT: 0.6M – 0.4M = 0.2M True Statement 2: IT Support’s total revenue was less than half of Sales. IT Support: 1.95M Half of Sales: 6.05M / 2 = 3.025M → 1.95M < 3.025M True Statement 3: Operations had the lowest revenue in every quarter. Check Q1: IT = 0.4M, lower than Ops (0.6M) Same in other quarters. False Statement 4: Marketing revenue increased by more than 30% from Q1 to Q4. From 0.8M to 1.2M % Increase = (1.2 – 0.8)/0.8 = 0.4 / 0.8 = 0.5 = 50% True Statement 5: Research Q4 was 30% higher than Q1. From 0.5M to 0.65M % Increase = (0.65 – 0.5)/0.5 = 0.15 / 0.5 = 30% True Answers: True True False True True 4. Two-part Analysis: Alex, Betty, and Cherry are working together on a single task. Working individually, Alex can complete the job in 8 hours, Betty in 24 hours, and Cherry in 48 hours. They decide to work in rotations of 1 hour each, with each person working alone for one hour at a time, in a repeating cycle of three hours. That is, the first person works for 1 hour, then the second, then the third, and then the cycle repeats in the same order. The total time taken to complete the task depends on who starts first. Question: If all three continue working in this fixed sequence until the job is finished, what is the minimum and maximum amount of time required to complete the job based on the starting order? Select one option in each column: \| Minimum hours \| Maximum hours \| --- \| \| A) 15 \| A) 15 \| \| B) 15.5 \| B) 15.5 \| \| C) 16.17 \| C) 16.17 \| \| D) 17 \| D) 17 \| \| E) 17.33 \| E) 17.33 \| Answer: Minimum hours: B) 15.5 Maximum hours: E) 17.33 Explanation: Let’s first calculate their work rates (job per hour): Alex: 1 job in 8 hrs → 1/8 per hour Betty: 1 job in 24 hrs → 1/24 per hour Cherry: 1 job in 48 hrs → 1/48 per hour Every 3-hour cycle includes all three working one hour each. So total work done in one cycle: = (1/8 + 1/24 + 1/48) = (6 + 2 + 1) / 48 = 9/48 = 3/16 of the job in 3 hours To complete 1 job: (1 job) ÷ (3/16 per 3 hrs) = 16 hours (approx) But depending on who starts, the final partial hour might differ. Minimum Time (Fastest Person Starts First): Let Alex (fastest) start, followed by Betty, then Cherry. After 15 hours (5 full cycles), total work = 5 × 3/16 = 15/16 Remaining work = 1 – 15/16 = 1/16 Alex works next at 16th hour with speed = 1/8 per hour Time to do 1/16 work = (1/16) ÷ (1/8) = 0.5 hour Total time = 15 + 0.5 = 15.5 hours Minimum = 15.5 Maximum Time (Slowest Person Starts First): Let Cherry (slowest) start, followed by Betty, then Alex. After 15 hours = 5 full cycles = 15/16 work done Cherry works next at 16th hour (6th cycle starts) Cherry’s speed = 1/48 per hour Remaining work = 1/16 → Time = (1/16) ÷ (1/48) = 3 hours → Total time = 15 + 3 = 18, but the job finishes sometime during Cherry's 3 hours Actual total time ~ 17.33 hours Thus, Maximum = 17.33 5. Multi-source Reasoning: The multi-source reasoning questions are as suggested by the name, based on different resources. These resources can be in the form of charts, statements, and tables. Make sure that you use only the related information to answer the question. Use this practice question to understand the concept. A company is considering implementing a new employee wellness program. The program would include on-site fitness classes, healthy snack options, and stress management workshops. Email from HR Manager I am excited to propose a new employee wellness program to promote a healthier and happier work environment. According to our recent employee survey, 75% of employees are interested in participating in on-site fitness classes, and 60% would like to have healthy snack options available. Best, [HR Manager] Company Financial Report Category Budget Employee Benefits$100,000 Employee Wellness Programs$20,000 Total$120,000 Article from Wellness Industry Magazine "The average ROI for employee wellness programs is 3:1. For every dollar invested, companies will save three dollars in healthcare and even gain increased productivity." Question 1 What is the primary advantage of an employee wellness program? A) Increased employee satisfaction B) Improved employee productivity C) Reduced healthcare costs and increased productivity D) Enhanced company reputation E) Increased employee retention Answer: C) Reduced healthcare costs and increased productivity Explanation: From the article: “average ROI = 3:1... save on healthcare and gain productivity” Thus, only C mentions both cost savings + productivity Question 2 If the company decides to allocate 20% of its employee benefits budget to the employee wellness program, how much money will be available for the program? A) $10,000 B) $15,000 C) $20,000 D) $25,000 E) $30,000 Answer: C) $20,000 Explanation: 20% of 100,000 = $20,000 Where to Find GMAT Data Insights Sample Papers? Since Data Insights has been recently introduced into GMAT, there isn’t a huge repository of questions and papers. However, it is essential that you understand the exam pattern, syllabus, and question distribution. You can check the official GMAT website, mba.com, which offers a GMAT starter kit and official practice questions, and test prep platforms like Yocket Prep and Kaplan. Also, keep an official preparation book handy, like the GMAT Official Guide: Data Insights Review, 2023-2025, and others. GMAT Data Insights Preparation Strategies The Data Insights questions make up 31.25% of the GMAT Focus edition, and to gain an edge over the competition it is important that you not only solve these questions correctly but also efficiently. Here are some tips to help you with the Data Insights GMAT Strategies: Use only the necessary information: Many questions will provide you with extra information that can throw off your calculation. Make sure that you only use relevant information. No assumptions: The geometric figures might not be drawn to scale; only use specific numbers when specified in the question. Read the question: Your question will have various elements, and the verbal questions will mostly build up concepts sentence by sentence. Make sure you do not miss any point and put the concept together in the right sequence. Check the condition: Before answering any true/false question, carefully check the condition and match your results with the statements and tables. Check all the possibilities: Before answering any question, check out all the options and rule out the wrong ones. From the Desk of Yocket Did you know that almost 2 lakh students take the GMAT exam each year? With such high competition and limited resources, how do you plan to succeed? If you really want to get the right score for your dream B-school, join Yocket Prep and start your preparation the right way. We will not only provide you with question papers but also with solutions, counselors, and doubt clearance. Don’t wait, start your prep with us today! Continue Reading FAQs on GMAT Data Insights Practice Questions How many questions are in the Data Insights section of the GMAT focus edition? There are 20 data insight questions in the GMAT focus edition exam. What is the score range of the Data Insights section of the GMAT focus edition? The score range for this section is 60-90. Can I skip the Data Insights section in my GMAT exam? Though you do have the option, it is absolutely not recommended as it will significantly affect your test score. Can I prepare for GMAT without coaching? Yes, there are various resources available online to help you prepare for GMAT without any coaching. What is the syllabus for the Data Insights section of the GMAT focus edition? The syllabus for the Data Insights section of the GMAT focus edition consists of data sufficiency, table analysis, graphics interpretation, multi-source reasoning, and two-part analysis. Is the Data Insights section of GMAT focus edition tough? Yes, comparatively the Data Insights section of GMAT focus edition is tough. What is the time limit for the Data Insights section of GMAT focus edition? The time limit for the Data Insights section of the GMAT focus edition is 45 minutes. A test-taking platform that helps with exam preparation for studying abroad. COMPANY About us Privacy Policy Help Center Careers GUIDES GMAT Guide GRE Guide IELTS Guide TOEFL Guide BLOGS GMAT Blogs GRE Blogs IELTS Blogs TOEFL Blogs GET IN TOUCH 18002706088 support@yocket.in © 2025 Yocket Prep. All rights reserved. GRE® is a registered trademark of Educational Testing Service (ETS). This website is not endorsed or approved by ETS.
10867	https://www.otoscape.com/staging.html	Published Time: Tue, 04 Feb 2025 02:32:33 UTC AJCC 8th Edition Staging Guidelines AJCC 8th Edition Staging Guidelines Otoscape Sections Basic Science 0 chapters General Otolaryngology 0 chapters Facial Plastics 0 chapters Pediatric Otolarygology 0 chapters Otology & Neurotology 0 chapters Rhinology 0 chapters Head & Neck Oncology 0 chapters Laryngology 0 chapters Cancer Staging Summary of AJCC 8th Edition Staging Guidelines Introduction Shared Nodal Categorization & Group Staging Oral Cavity Oropharynx (p16-) Oropharynx (p16+) Hypopharynx Larynx Esophagus and Esophagogastric Junction Nasal Cavity & Paranasal Sinuses Nasopharynx Major Salivary Glands Sarcomas of the Head & Neck Cutaneous Carcinoma of the Head & Neck Cutaneous Melanoma Mucosal Melanoma of the Head & Neck Merkel Cell Carcinoma Thyroid (Differentiated) Thyroid (Medullary) Thyroid (Anaplastic) Parathyroid Images Glossary About Otoscape Disease Site Introduction Shared Nodal Categorization & Group Staging Oral Cavity Oropharynx (p16-) Oropharynx (p16+) Hypopharynx Larynx Esophagus and Esophagogastric Junction Nasal Cavity & Paranasal Sinuses Nasopharynx Major Salivary Glands Sarcomas of the Head & Neck Cutaneous Carcinoma of the Head & Neck Cutaneous Melanoma Mucosal Melanoma of the Head & Neck Merkel Cell Carcinoma Thyroid (Differentiated) Thyroid (Medullary) Thyroid (Anaplastic) Parathyroid AJCC 8 th Edition Head & Neck Staging Guidelines Table of Contents Introduction Shared Nodal Categorization & Group Staging Oral Cavity Oropharynx (p16-) Oropharynx (p16+) Hypopharynx Larynx Esophagus and Esophagogastric Junction Nasal Cavity & Paranasal Sinuses Nasopharynx Major Salivary Glands Sarcomas of the Head & Neck Cutaneous Carcinoma of the Head & Neck Cutaneous Melanoma Mucosal Melanoma of the Head & Neck Merkel Cell Carcinoma Thyroid (Differentiated) Thyroid (Medullary) Thyroid (Anaplastic) Parathyroid
10868	https://artofproblemsolving.com/wiki/index.php/Divisibility_rules/Rule_for_11_proof?srsltid=AfmBOorjWik7kE8vPd_8ILkOe1r3vR2f1x6620d2qNd-xa7u7HZTTY-T	Art of Problem Solving Divisibility rules/Rule for 11 proof - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Divisibility rules/Rule for 11 proof Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Divisibility rules/Rule for 11 proof A number is divisible by 11 if the alternating sum of the digits is divisible by 11. Proof An understanding of basic modular arithmetic is necessary for this proof. Let where the are base-ten numbers. Then Note that . Thus This is the alternating sum of the digits of , which is what we wanted. Here is another way that doesn't require knowledge of modular arithmetic. Suppose we have a 3-digit number that is expressed in the form: we then can transpose this into: and that equals: which equals Since the first addend, will always be divisible by 11, we just need to make sure that is divisible by 11. You can use this for any number. Here it is again, with an even-numbered digit number: So you just need to check for divisibility with 11. See also Back to divisibility rules Retrieved from " Category: Divisibility Rules Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
10869	https://askfilo.com/user-question-answers-smart-solutions/find-the-range-of-3335353139373330	Question asked by Filo student Find the range of y=∣sinx∣+∣cosx∣. Views: 5,262 students Updated on: Jun 18, 2025 Text SolutionText solutionverified iconVerified Concepts Range of a Function, Absolute Value, Trigonometric Functions (Sine and Cosine), AM-GM Inequality Explanation We are asked to find all possible values (the range) of the function y=∣sinx∣+∣cosx∣ as x varies over all real numbers. Since sinx and cosx each range from −1 to 1, their absolute values will range from 0 to 1. The sum can potentially vary from 0 (minimum) to a certain maximum value. Let's analyze the function systematically. Step-By-Step Solution Step 1: Express in terms of a single variable Let ∣sinx∣=a and ∣cosx∣=b. For any x, a,b≥0 and a2+b2=sin2x+cos2x=1. So, y=a+b, where a,b≥0 and a2+b2=1. Step 2: Find the minimum value of y Since a,b≥0, their sum y will be minimized when one of them is 0 and the other is 1. Step 3: Find the maximum value of y Let us maximize y=a+b subject to a2+b2=1, a≥0, b≥0. Let us use Cauchy-Schwarz (AM-GM is not tight here), or solve via calculus by writing: Let a=sinθ≥0, b=cosθ≥0 Consider x in [0,π/2] only (as both are positive), then ∣sinx∣+∣cosx∣=sinx+cosx We know that sinx+cosx=2sin(x+π/4) So, maximum value of sinx+cosx=2 when x+π/4=π/2⟹x=π/4. For any other quadrant, one of sinx, cosx could be negative, but their absolute values will mirror this situation by symmetry, so the minimum and maximum do not change. So, ymax=2 and ymin=1. Step 4: Write the range Therefore, the range is [1,2]. Final Answer The range of y=∣sinx∣+∣cosx∣ is: [1,2] Students who ask this question also asked Views: 5,359 Topic: Smart Solutions View solution Views: 5,021 Topic: Smart Solutions View solution Views: 5,757 Topic: Smart Solutions View solution Views: 5,179 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| Find the range of y=∣sinx∣+∣cosx∣. \| \| Updated On \| Jun 18, 2025 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 12 Passed \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
10870	https://www.clinicalkey.com/dura/browse/bookChapter/3-s2.0-C20211025500	Page loading - ClinicalKey Skip to search Skip to navigation Skip to content Skip to help and feedback ClinicalKey Choose search scope All Types This Book All Types Books Journals Clinical Overviews Clinical Trials Drug Monographs Guidelines Patient Education Multimedia Procedure Videos Clinical Calculators Clinical Focus Search English CME/MOC Login Register Help Call Us Email Us Help Search for diagnoses, conditions, drugs and more... Main Navigation Anonymous (Login or Register) Browse Books Journals Clinical Overviews Drug Monographs Guidelines Patient Education Multimedia Procedure Videos Tools Clinical Calculators CME/MOC Change Language Store Remote Access Help & Feedback Call Us Email Us Help About ClinicalKey Learn more about other offerings and editions Browse Books Journals Clinical Overviews Drug Monographs Guidelines Patient Education Multimedia Procedure Videos Tools Clinical Calculators 246711 Google Play - Go to our Google Play page Apple App Store - Go to our Apple App Store page Facebook - Go to our Facebook page LinkedIn - Go to our LinkedIn page X - Go to our X page Elsevier (opens in a new window) Contact Us (opens in a new window) Resource Center (opens in a new window) Registered User Agreement (opens in a new window) Help (opens in a new window) Terms & Conditions (opens in a new window) Privacy Policy (opens in a new window) Accessibility (opens in a new window) We use cookies to help provide and enhance our service and tailor content. By continuing you agree to the Cookie Settings. All content on this site: Copyright © 2025 Elsevier Inc. or its licensors and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply. (opens in a new window)
10871	https://www.pearsonhighered.com/assets/samplechapter/0/1/3/4/0134703693.pdf	128 5 UNIT 5-1 Define ratio and proportion 5-2 Locate and decipher ratios in common health applications such as drug labels 5-3 Determine if two ratios are a proportion 5-4 Simplify ratios and complex ratios 5-5 Apply ratio definitions to express unit rates 5-6 Solve for x or an unknown in a proportion 5-7 Apply proportions to nutrition labels 5-8 Apply proportion to solve dental stone mixtures Ratio and Proportion Student Learning Outcomes After completing the tasks in this unit, you will be able to: Pre-Test 1. One kilogram equals 2.2 pounds. How many pounds equal 24.5 kilo grams? 2. Solve for x. 48 : 64 = x : 124 3. One cup contains 8 ounces. How many full cups are in 138 ounces? 4. Solve for x. 1 8 : 3 = 1 4 : x 5. Solve for x. x : 225 = 2 : 5 6. Write the ratio that represents three registered nurses to twelve certified nursing assistants. UNIT 5 Ratio and Proportion 129 7. The patient’s pulse is documented as 73 beats per minute. How many beats will be documented for 5 minutes if the rate remains the same? 8. Two tablespoons of peanut butter are used for each sandwich. One tablespoon has three teaspoons. If 49 sandwiches are made, how many teaspoons of peanut butter are needed to make these sandwiches? 9. Write the medication dosage for the amount on this label. Write it as a ratio. Include the unit of measure. 10. The tapioca recipe calls for 2 1 2 cups of milk to make six servings. How many cups of milk are needed for 15 servings? 11. The dietitian ordered 4 ounces of chicken per patient for a dinner meal. The kitchen served 496 ounces of chicken this evening. How many patients received a 4-ounce portion of chicken? 12. The patient went to the laboratory to have blood drawn by a phlebotomist. Each tube of blood holds 2.5 milliliters of blood. Four tubes of blood were drawn. How many milliliters of blood were drawn? 13. On average, one registered nurse at Valley View uses 62 gloves a day to examine patients. In 5 days, how many gloves will that particular nurse use? 14. Solve for x. 4 75 : 15 :: x : 20 15. How many minutes are in 320 seconds? Your answer should be in minutes and seconds. 1. 2. 120 mL ORAL SUSPENSION 2418 Each teaspoonful (5 mL) contains hydroxyzine pamoate equivalent to 25 mg hydroxyzine hydrochloride. USUAL DAILY DOSAGE: Adults: 1 to 4 teaspoonfuls 3-4 times daily. Children: 6 years and over– 2 to 4 teaspoonfuls daily in divided doses. Under 6 years–2 teaspoonfuls daily in divided doses. READ ACCOMPANYING PROFESSIONAL INFORMATION. PROTECT FROM MOISTURE AND HUMIDITY. Store below 77°F (25°C) SHAKE VIGOROUSLY UNTIL PRODUCT IS COMPLETELY RESUSPENDED. DYE FREE FORMULA PL Pharmaceuticals 25 mg/5 mL 120 ml \| NDC 0069-5440-97 Rx only For Oral Use Only 0069-5440-97 05-2255-32-6 Practice Label 130 UNIT 5 Ratio and Proportion Practice 1 Write the following relationships as ratios using a colon. Reduce to lowest terms, if necessary. 1. 6 days out of 7 days _______ 2. twelve teeth out of thirty-two teeth _____ Overview Ratio is a way to show a relationship between two items. We are always counting and comparing items in our daily lives: hours at work versus hours away from work, number of yogurts we have eaten versus the number of yogurts still in the refrigerator, and so on. Ratios simply help us compare two items, objects, or amounts. Proportion compares two equal ratios in a mathematical equation. We use proportion to either increase or decrease one part of the ratio in the equation so that the unit expressed or found is in the same relationship with the other part of the specific ratio and so that this ratio, when completed, shows the same relationship as the other ratio. Ratio REVIEW A ratio is used to show a relationship between two numbers or a comparison of two items. The numbers are separated by a colon (:) as in x : y. For example, three nurses and four medical assistants working a clinic shift can form a ratio. Ratios may be presented in three formats that provide the set-up for solving proportions. a. 3 : 4 (three nurses to four medical assistants) b. 3 4 (three nurses to four medical assistants) c. 3 is to 4 (three nurses to four medical assistants) The relationship can represent something as simple as the 1 : 3 ratio commonly used to mix frozen juices. We use one can of frozen juice concentrate to three cans of water. Ratios are fractions that represent a part-to-whole relationship. Often when we work with ratios, we use the fraction format to reduce the ratio to its simplest form. Ratios are always reduced to their lowest form. For example, 8 hours of sleep to 24 hours in a day would be expressed as 8 : 24 S 8 24 = 8 1 8 3 = 1 3 so the ratio is 1 : 3. SET-UP HINT Writing your ratios in the proper order is important. Follow the order of each number of the ratio in the sentence or problem. It is a good idea to include the units. In that way, you can be sure that you have followed the same pattern in each ratio. UNIT 5 Ratio and Proportion 131 Simplifying ratios is an important skill. To simplify a ratio, divide the first number by the second. For example, simplify the following ratio: 4 1 2 : 6. 4 1 2 , 6 S 9 2 , 6 1 S 9 2 1 6 = 9 12 S 3 3 3 4 = 3 4 , The answer is 3 : 4. As another example, simplify the following ratio: 11 1 4. Convert the mixed number into an improper fraction, and then reduce, if necessary. 11 1 4 S 11 4 + 1 = 45 S 45 4 = 45 : 4 The answer is 45 : 4. Practice 2 Simplify the following ratios. Write each answer as a ratio. 1. 24 : 3 1 4 = __________ 2. 7 8 : 14 = ________ 3. 25 : 5 6 = ________ 4. 1 3 : 45 = ________ 5. 0.8 : 2 5 = ________ 6. 1 2 : 1 8 = _________ 7. 4 1 3 : 7 = ________ 3. 8 students out of 15 students _______ 4. 16 scalpels to 45 syringes ______ 5. 7 inlays to 14 crowns ________ which becomes 3 : 4 as a simplified ratio 132 UNIT 5 Ratio and Proportion Ratios in Health Care Drug labels are another place that ratios may be seen in health care. Careful reading of the drug label will help locate the dosage of medication per tablet or per amount of solution. Notice that each label uses a specific language indicating a ratio. These formats are milligrams/milliliters (mg/mL), micrograms/milligrams (mcg/mg), mg per tablet, milligrams in milliliters (mg in mL), etc. Careful reading will help identify what is in each tablet, each milliliter of medication, etc. 8. 0.875 : 1 4 = _________ 9. 2 1 2 = ________ 10. 2 3 : 0.33 = _________ lorazepam 4 mg/mL 1mL Vial NDC 6077-112-81 For IM USE REFRIGERATE FOR IV ROUTE, SEE DIRECTIONS. For IV use, additional dilution is required; see accompanying information. Usual Dosage: See accompanying information. Do not use if solution is discolored or contains a precipitate. PROTECT FROM LIGHT Lot: Exp.: 462-145-00 PL Pharmaceuticals Each milliliter contains 4 milligrams of lorazepam. Practice Label Practice 3 Read the labels and write the ratio of the medication indicated in each label. Write the ratio in simplified form. 1. _________ NDC 0025-6589-82 Derived from porcine intestinal mucosa For intravenous for subcutaneous use Sterile, nonpyrogenic Each mL contains: 1000 USP units heparin sodium: 9 mg sodium chloride; 0.15% methylparaben; 0.015% propylparaben; water for injection q.s. Made isotonic with sodium chloride. Hydrochloric acid and/or sodium hydroxide may have been added for pH adjustment. Use only if solution is clear and seal is intact. Store at 25°C (77°F); excursions permitted to 15° to 30°C (59° to 86°F). For educational Purposes Only 354588–23 heparin sodium 10,000 units/mL Injection, USP Rx only Multiple Dose Vial Lot. Exp. Date PL Pharmaceuticals Practice Label UNIT 5 Ratio and Proportion 133 2. _________ 3. _________ 4. _________ 5. _________ lincomycin hydrochloride 10 mL multidose vial NDC 1123-4567-89 300 mg/mL PL Pharmaceuticals Practice Label NDC 68094-767-59 LITHIUM Citrate Syrup, USP 8 mEq/5 mL Store at 25 ºC (77 ºF) Delivers 5 mL Each 5 mL contains: 8 mEq of Lithium lon (Equivalent to 300 mg of Lithium Carbonate per 5 mL) Alcohol 0.3% Sugar-Free 768 RO 6498 Rx only PL Pharmaceuticals Practice Label morphine Sulfate Injection, USP FOR SUBCUTANEOUS, INTRAMUSCULAR OR SLOW INTRAVENOUS USE NOT FOR EPIDURAL OR INTRATHECAL USE WARNING: May be habit forming PROTECT FROM LIGHT DO NOT USE IF PRECIPITATED Each mL contains morphine sulfate 10 mg. sodium dihydrogen phosphate 10 mg. disodium hydrogen phosphate 2.8 mg. sodium formaldehyde sulfoxy-late 3 mg and phenol 2.5 mg in Water for Injection. Sulfuric acid used to adjust pH to 2.5–6.0. Sealed under nitrogen. Usual Dose: See package insert for complete prescribing information. Note: Slight discoloration will not alter efcacy. Discard if markedly discolored. Caution: Federal law prohibits dispensing without prescription. NDC 0641-0180-25 Rx only PL Pharmaceuticals 10 mg/mL 1 mL Vial Practice Label 029158 D074377113 SHAKE WELL BEFORE USING. RECONSTITUTE WITH 63 mL WATER Keep bottle tightly closed. Any unused portion must be discarded 10 days after mixing. (when reconstituted) 100 mL ORAL SUSPENSION 02–9158 Each 5 mL contains 125 mg cefdinir after reconstitution. DIRECTIONS FOR RECONSTITUTION Prepare suspension at time of dispensing by adding a total of 63 mL water to the bottle. Tap bottle to loosen the powder, then add about half the water, and shake. Add the remaining water and shake to complete suspension. This provides 100 mL of suspension. USUAL DOSAGE: Children-14 mg/kg/day in a single dose or in two divided doses, depending on age, weight, and type of infection. See package enclosure for full prescribing information. This bottle contains 2.5 g cefdinir. Do not accept if seal over bottle opening is broken or missing. Keep this and all drugs out of the reach of children. Store dry powder and reconstituted suspension at 25°C (77°F); excursions permitted to 15°–30°C (59°–86°F) [see USP Controlled Room Temperature]. Use within 10 days. SHAKE WELL BEFORE EACH USE. Keep bottle tightly closed. Rx only PL Pharmaceuticals 125 mg per 5 mL Lot Expiration date of powder. 100 mL \| NDC 0074-3771-13 Practice Label 134 UNIT 5 Ratio and Proportion Proportion REVIEW Proportions can be applied to almost every health care profession in one way or another. In addition to on-the-job applications, proportions provide a simple and quick method for solving many everyday math problems such as measurement conversions, recipe conversions for increasing or decreasing the amounts of ingre-dients, and map mileage. Unit Rates Another example of ratio in health care comes from dealing with insurance cover-age. There is something called a medical loss ratio, which is used in managed care to measure medical costs as a percentage of premium revenues or income. It is a type of loss ratio used to measure the percentage of premiums paid out in claims rather than expenses. A desirable ratio is 17 : 20 or 85% or less. In addition, ratios can compare two items. For example, ratios are often used to find the unit rate, which is a rate having 1 as its denominator. Rate is a ratio of two measurements having different units. Example : $4.10 (a package of gauze strips) 24 (individual number of strips) To get the unit rate, divide the bottom number (the denominator) into the top number (the numerator). $4.10 , 24 = 0.170833 or 17 cents per sheet of gauze 17 cents : 1 sheet or 17 : 1 Practice 4 Express each as a unit rate. 1. $4182 room bill for a hospital stay 3 days in the hospital _____ 2. 120 pound goal to lose 52 weeks in a year _______ 3. 24 ounces 3 cups ________ 4. 14 cups of sugar 252 cookies _______ 5. 14235 patients 365 days ________ UNIT 5 Ratio and Proportion 135 Once you become well versed in the set-up, you may drop the units. However, labeling the units is very helpful to ensure the proper set-up. 3 4 = 15 20 or 3 : 4 :: 15 : 20 Proportions are two or more equivalent ratios or fractions in which the terms of the first ratio/fraction have the same part-to-whole relationship as the second ratio/fraction. SET-UP HINT Notice how the terms for each category (boxes and gloves) are across from each other. This is the proper set-up: boxes across from boxes and gloves across from gloves. Example If one box of gloves contains 100 gloves, then 4 1 2 boxes will contain how many gloves? 1 box 100 gloves = 41 2 boxes x number of gloves 4 1 2 = 4.5 S 4.5 100 = 450 MATH SENSE : : means “is” or = Test the two ratios/fractions to see whether they are equivalent by multiplying diagonally (cross multiply). 3 4 = 15 20 4 15 = 60 and 3 20 = 60. This is a proportion. If the two numbers that are diagonal result in the same answer when they are multiplied, you are working with a proportion. Proportions are powerful tools in health care. You can rely on them for solving a majority of your math conversions and problems. Practice 5 Check to see if the following ratios are proportions. 1. 5 : 2 = 4 : 1 Yes No 2. 16 : 15 = 8 : 7 Yes No 3. 40 : 30 = 4 : 3 Yes No 4. 10 : 16 = 5 : 8 Yes No 5. 100 : 1 = 50 : 2 Yes No Solving for x REVIEW The ratio and proportion method is used to solve for x. Solving for x uses the known or given information to find what is not known or given. Since a proportion con-sists of two equal ratios, the relationship between the numerator and denominator 136 UNIT 5 Ratio and Proportion of each fraction is the same for each ratio of the proportion. This is important to know when one must increase or decrease a solution or mixture because the ratio of ingredients must remain constant. Solving for x is done in two steps. STEP 1: Set the problems up like fractions. If units of measure such as inches and feet are given, place inches across from inches and feet across from feet. Then cross multiply (diagonally) the two numbers. Set the ratios up like fractions using a vertical line. 3 4 = ? 16 3 16 = 48 STEP 2: Divide the answer from step 1 by the remaining number. 12 4)48 -4T 8 -8 0 The quotient 12 is the answer to ? or x. This method is an easy way to find the answers for measurement conversions, dosage conversions, and math questions that provide part but not all of the information. Practice 6 Solve for x or ? 1. 12 : 45 = x : 15 _______ 2. x : 6 = 15 : 60 ________ 3. 25 : 45 = 75 : x _______ 4. 7 : x = 21 : 24 ________ 5. 3 : 9 = ? : 81 ________ 6. 13 : 39 = 1 : ? ________ 7. 2 : 11 = ? : 77 ________ 8. x : 125 = 5 : 25 ________ 9. 2 : 26 = 4 : ? ________ 10. 1 : x = 5 : 200 ________ UNIT 5 Ratio and Proportion 137 Using ratios is often the simplest method of solving other health care math problems, such as dosage calculations and measurement problems. Example Zoe weighs 35 pounds. Her doctor prescribed a drug that relates milligrams of medication to kilograms of body weight. The pharmacy technician will need to convert pounds to kilograms. By using the ratio of 1 kilogram to 2.2 pounds, the answer is quickly computed. known unknown 1 kilogram 2.2 pounds = ? kilograms 35 pounds STEP 1: Multiply the numbers diagonally. 1 35 = 35 STEP 2: Divide 35 by 2.2. The answer is 15.9 kilograms. So, 35 pounds equals 15.9 kilograms. Converting between kilograms and pounds is a common procedure in health care. Practice 7 Set up and solve these conversions using ratio and proportions. Round to the nearest tenth, if necessary. Use the conversion 1 kilogram = 2.2 pounds. 1. Convert 16.4 kilograms to pounds. ______ 2. Convert 125.8 kilograms to pounds. _____ 3. Convert 35 kilograms to pounds. _____ 4. Convert 75 kilograms to pounds. _____ 5. Convert 83.5 kilograms to pounds. ______ 6. Convert 16 pounds to kilograms. _____ 7. Convert 25 1 2 pounds to kilograms. _____ 8. Convert 108 pounds to kilograms. _____ 9. Convert 215.6 pounds to kilograms. _____ 10. Convert 165 pounds to kilograms. ______ 138 UNIT 5 Ratio and Proportion Example How many pounds are in 24 ounces? Set the problem up by placing what you know on the left side of the equation and what you do not know on the right side. If you set up all your problems with the known on the left and the unknown on the right, there will be less information for your brain to process because the pattern will be familiar to you. known unknown 1 pound 16 ounces = ? pounds 24 ounces STEP 1: 1 24 = 24 STEP 2: 24 , 16 = 1.5 The answer is 1 1 2 pounds or 1.5 pounds. The answer in a ratio may have a decimal or a fraction in it. Example Bob is 176 centimeters (cm) tall. How tall is he in inches? Round the answer to the nearest tenth. known unknown 1 inch 2.54 cm = ? inches 176 cm STEP 1: 1 176 = 176 STEP 2: 176 , 2.54 = 69.29 69.29 254)17600 - 1524T 2360 - 2286T 740 - 508T 2320 - 2286 34 So, after the division: 176 , 2.54 = 69.29. Rounded to the nearest tenth, the answer is 69.3 inches. Notice that the units in the metric system are often rounded to the nearest tenth. This is why the division problem above adds a decimal and zeros: to work out the dividend to the hundredths place so that the final answer is rounded to the tenth. UNIT 5 Ratio and Proportion 139 Notice that the conversions are set up so that the unit (1) elements are all on the left and that these will be placed on the top of the known part of the ratio and propor-tion equation. This simplifies the learning process, expedites learning, and helps recall of these conversions. APPROXIMATE EQUIVALENTS 1 inch = 2.54 centimeters 1 cup = 8 ounces 1 foot = 12 inches 1 pint = 480 milliliters 1 yard = 3 feet 1 quart = 32 ounces 1 pound = 16 ounces 1 quart = 960 milliliters 1 kilogram = 2.2 pounds 1 pint = 2 cups 1 tablespoon = 3 teaspoons 1 fluid ounce = 30 milliliters 1 quart = 2 pints 1 teaspoon = 5 milliliters 1 gallon = 4 quarts 1 fluid ounce = 2 tablespoons Certain fields use rounded measures; for example, instead of 480 mL and 960 mL, they use 500 mL and 1000 mL. Check with your instructor. SET-UP HINT Some basic guidelines need to be followed when formatting answers in measurement conversions: If the answer is in feet, yards, cups, pints, quarts, gallons, teaspoons, tablespoons, or pounds, use fractions if there is a remainder. If the answer is in kilograms, milliliters, or money amounts, use decimals. The correct format ensures correct answers, which are often rounded to either the tenth or the hundredth place. SET-UP HINT Because inches are rounded to the nearest tenth, go to the hundredths place and stop dividing. At that point, you will have enough information to round to the nearest tenth. Practice 8 Using the ratio and proportion set-up below, solve the following conversions. known unknown = Set up these conversions using ratios and proportions. 1. 23 feet = yards S 1 yd 3 ft = ? 23 ft 2. 18 quarts = gallons 3. 3 quarts = pints 4. 2 1 4 pints = cups 140 UNIT 5 Ratio and Proportion 5. 3 tablespoons = teaspoons 6. 2 1 2 quarts = milliliters 7. 1 2 cup = ounces 8. 1 injection at $29.50 = 3 injections at 9. 3 1 2 pounds = ounces 10. 3 medicine cups = milliliters (One medicine cup equals 1 fluid ounce.) 11. 12.5 mL = teaspoons 12. 5 fluid ounces = tablespoons 13. tablespoons = 15 teaspoons 14. 64 ounces = cups 15. 750 milliliters = pints 16. 48 inches = feet 17. 5 pounds = ounces 18. quarts = 5000 milliliters 19. kilograms = 11 pounds 20. 15 cups = ounces Using Proportions and Metric Units to Measure Dental Stone Sometimes dental assistants need to mix dental stone material for dental molds. The amount of stone and water varies according to the size of the mold needed. A dental assistant learns the importance of mixing a uniform and consistent mate-rial for the mold. This task uses the metric system: Dental stone is measured by weight in grams, and room temperature water is measured by volume in milliliters. Knowledge of the metric system coupled with the use of ratio and proportion help maintain the correct ratio of dental stone to water. UNIT 5 Ratio and Proportion 141 The typical ratio of dental stone to water is: 263 grams of dental stone powder 80 milliliters of room temperature water Dental assistants use this ratio as the standard to solve variances in either stone or water to create the proper amount of material for the mold. On occasion, they may be asked to use other ratios of dental stone to water, depending on the type of material or the mold being created. Example We will set up the problems as proportions, cross multiply, and then divide by the leftover number. This will solve for the unknown. If the dentist requests a smaller mold using 35 mL of water, how much stone should be used? The set-up is: 263 g 80 mL = ? g 35 mL STEP 1: 263 35 = 9205 STEP 2: 9205 , 80 = 115.0625. Round to the nearest whole number. The final answer is that 115 grams of stone to 35 milliliters of water should be used. 100 mL 80 mL 60 mL 40 mL 20 mL Practice 9 Solve the following using this ratio: 263 grams dental stone powder 80 milliliters of water Round to the nearest whole number. 1. If you use 75 grams of stone, how many milliliters of room temperature water are needed? 142 UNIT 5 Ratio and Proportion 2. If you use 125 milliliters of water, how many grams of dental stone are needed? 3. If you use 95 grams of stone, how many milliliters of room temperature water are needed? 4. If you use 75 milliliters of water, how many grams of dental stone are needed? 5. If you use 65 grams of stone, how many milliliters of room temperature water are needed? 6. If you use 55 grams of stone, how many milliliters of room temperature water are needed? 7. If you use 40 milliliters of water, how many grams of dental stone are needed? 8. If you use 35 grams of stone, how many milliliters of room temperature water are needed? 9. If you use 50 grams of stone, how many milliliters of room temperature water are needed? 10. If you use 60 milliliters of water, how many grams of dental stone are needed? Word Problems Using Proportions REVIEW When solving word problems involving proportions, follow these three basic steps. UNIT 5 Ratio and Proportion 143 Example If 12 eggs cost $1.49, how much do 18 eggs cost? Eggs Cost = Eggs Cost S S 12 eggs $1.49 = 18 eggs $? STEP 1: Set the problem up so that the same types of elements are directly across from one another. STEP 2: Multiply the diagonal numbers. $1.49 18 = 26.82 STEP 3: Divide the answer from step 2 by the remaining number in the problem. 26.82 , 12 = $2.235 or $2.24 So, 18 eggs cost $2.24. SET-UP HINT Ensure that you understand the story problem, and then place the known information on the left side of the proportion and the unknown information on the right. By doing so, you will not switch the ratio relationships, but instead will rely on the known part-to-whole relationships. Example How many milligrams of medication would a nurse administer to a 95-pound child if the prescribed dose was 30 milligrams for every 10 pounds? STEP 1: 30 milligrams 10 pounds = x milligrams 95 pounds STEP 2: 30 95 = 2850 STEP 3: 2850 , 10 = 285 milligrams So, the child would receive 285 milligrams of the medication. Example The lab mixes a 12% solution for the physician. A 12% solution has a 3 : 25 ratio. This ratio includes 3 grams of powder in 25 milliliters of solution. How many grams of powder will be added to 65 milliliters of solution? STEP 1: 3 grams 25 milliliters = x grams 65 milliliters STEP 2: 3 65 = 195 STEP 3: 195 , 25 = 7.8 grams So, 7.8 grams of powder will be used to mix 65 milliliters of the 3 : 25 solution. 144 UNIT 5 Ratio and Proportion Solving for x in More Complex Problems Using Proportion REVIEW Decimals and fractions may appear in your proportion problems. Although the numbers may be visually distracting, the very same principles apply when solving these proportions. Practice 10 Solve the following word problems. 1. A caplet contains 325 milligrams of medication. How many caplets contain 975 milligrams of medication? 2. If a dose of 100 milligrams is contained in 4 milliliters, how many milliliters are in 40 milligrams? 3. If 35 grams of a pure drug are contained in 150 milliliters, how many grams are contained in 75 milliliters? 4. Three tablets of ulcer medication contain 375 milligrams of medication. How many milligrams are in 12 tablets? 5. If 1 kilogram equals 2.2 pounds, how many kilograms are in 61.6 pounds? Example 0.25 mg : 0.8 mL = 0.125 mg : x mL STEP 1: Place mg across from mg and mL across from mL. Place the known information on the left side of the equation and the unknown informa-tion on the right side. known unknown 0.25 mg 0.8 mL = 0.125 mg x mL Cross multiply 0.8 0.125 mg = 0.1. STEP 2: 0.1 , 0.25 = 0.4 mL UNIT 5 Ratio and Proportion 145 Example 1 8 : 1 2 :: 1 : x 1 8 1 2 = 1 x STEP 1: Set up and cross multiply. Multiply 1 2 1 = 1 2. STEP 2: Divide 1 2 by 1 8. 1 2 , 1 8 S 1 2 8 1 = 8 2 , which is reduced to 4. Sometimes you will find that medical dosages have both fractions and decimals. Analyze the situation and convert the numbers to the same system format. As a general rule, fractions are always more accurate for calculating than decimals because some decimal numbers have repeating digits, which create variable answers. Example 1 16 : 1.6 :: 1 8 : x This looks like this as a fraction: 1 16 1 6 10 = 1 8 x STEP 1: Convert 1.6 into a fraction. So, 1.6 = 1 6 10. Then multiply 1 6 10 1 8 = 2 10. 1 16 1 6 10 = 1 8 x 1 6 10 1 8 = 16 10 1 8 = 16 80 or 2 10 STEP 2: Divide 2 10 by 1 16. 2 10 , 1 16 S 2 10 16 1 = 32 10. Reduced to 3 2 10 S 3 1 5. 146 UNIT 5 Ratio and Proportion Nutritional Application of Proportions REVIEW Carbohydrates, fats, and proteins provide fuel factors for our bodies. These factors are easily applied by using proportions to solve for the unknown. Practice 11 Include a unit of measure in your answer. Round any partial unit to the nearest tenth. 1. 2.5 mg : 2 mL = 4.5 mg : x mL ______ 2. 12 mg : 2.5 mL = 4 mg : x mL _____ 3. 7.5 mg : 5 mL = 24 mg : x mL _____ 4. 0.2 mg : 1 tab = 6 mg : x tabs _______ 5. 1 4 grains : 15 mg = ? grains : 60 mg _____ 6. x mg : 1 2 tab = 6 mg : 4 tabs ______ 7. 1 100 grains : 2 mL = 1 150 grains : x mL ____ 8. 600 mg : 1 cap = x mg : 2 caps _____ 9. 1000 units : 1 mL = 2400 units : x mL _____ 10. 1 tab : 0.1 mg = x tabs : 0.15 mg ______ 11. A drug comes in 100 milligram tablets. If the doctor orders 150 milligrams daily, how many tablets should the patient receive daily? 12. A medical chart states that the patient weighs 78.4 kilograms. What is the patient’s weight in pounds? Round to the nearest tenth. SET-UP HINT Tablets can be divided if they are scored; use 1 2 , not 0.5. SET-UP HINT Carbohydrates S 4 calories per 1 gram Fats S 9 calories per 1 gram Proteins S 4 calories per 1 gram UNIT 5 Ratio and Proportion 147 Example 400 carbohydrate calories = grams known unknown 1 gram 4 calories = ? grams 400 calories STEP 1: Multiply diagonally. 1 400 = 400 STEP 2: Divide answer from step 1 (400) by the remaining number in the equation (4). 100 4)400 - 4 00 So, 400 carbohydrate calories are available in 100 grams of carbohydrates. Practice 12 Use proportion to solve the following problems. 1. 81 calories of fat = grams 2. 120 calories of protein = grams 3. 36 calories of carbohydrate = grams 4. 145 calories of carbohydrate = grams 5. calories are in 12 grams of protein. 6. calories are in 99 grams of fat. 7. calories are in 328 grams of carbohydrate. 8. calories are in 2450 grams of protein. Proportion is also useful in solving nutritional problems that involve amounts of sodium, calories, fat, and protein in food or an amount in a drug dosage. The pro-portion will use the information in a scenario to solve for the unknown quantities of a specific amount. 148 UNIT 5 Ratio and Proportion Practice with Food Labels Knowing how to read food labels is important because patients often need to limit their salt, sugar, and fat intake to help ensure good health. Proportion is useful in figuring out the amounts of these ingredients when portioning—increasing or decreasing portions. Example If one glass of milk contains 280 milligrams of calcium, how much calcium is in 1 1 2 glasses of milk? 1 glass 280 milligrams = 11 2 glasses ? milligrams 280 1 1 2 = 420 mg of calcium Practice 13 Solve these nutritional problems using ratio and proportion. 1. One-half cup of baked beans contains 430 milligrams of sodium. How many milligrams of sodium are there in 3 4 cup of baked beans? 2. Baked beans contain 33 grams of carbohydrates in a 1 2 cup serving. How many milligrams of carbohydrates are in three 1 2 cup servings? 3. A 1 2 cup serving of fruit cocktail contains 55 milligrams of potassium. How many milligrams of potassium are in 2 cups of fruit cocktail? 4. If 1 2 cup of fruit cocktail contains 13 grams of sugar, then 1 1 4 cup of fruit cocktail contains how many grams of sugar? 5. Old-fashioned oatmeal contains 27 grams of carbohydrates per 1 2 cup of dry oats. How many grams of carbohydrates are available in 2 1 4 cups of the dry oats? UNIT 5 Ratio and Proportion 149 Practice 14 Carefully read the label and then use its information to solve each question. 1. If 1 2 cup of soup equals 120 milliliters, then how many milliliters (mL) are in 3 1 3 cups of soup? 2. If a can has 2.5 servings, how many cans are needed to serve 10 people? 3. One serving contains 90 calories; how many calories are in 4 1 2 servings? 4. One gram of fiber constitutes 4% of a daily dietary value. How many grams of fiber would be present in 25% of the daily value? 5. How many grams of carbohydrates are present if the portion meets 15% of the daily value of carbohydrates? Round to the nearest tenth. Use the information from the label to complete these proportions. Albert’s Tomato Soup Nutrition facts Amount/serving %DV Amount/serving %DV Total fat 0 g 0% Total carbohydrates 20 g 7% Saturated fat 0 g 0% Fiber 1 g 4% Cholesterol 0 mg 0% Sugars 15 g Sodium 710 mg 30% Protein 2 g Serving size ½ cup (120 mL) Condensed soup Servings about 2.5 Calories 90 Fat calories 0 Percent daily values (%DV) are based on a 2000 calorie diet. Vitamin A 12% Vitamin C 12% Calcium 0% Iron 0% Big Al’s Organic Sweet and Juicy Dried Plums Nutrition facts Serving size 1½ oz (40 g in about 5 dried plums) Servings per container about 30 Amount per serving Calories 100 Calories from fat 0 %DV %DV Potassium 290 mg 8% Total carbohydrates 24 g 8% Dietar y fber 3 g 11% Total fat 0 g 0% Saturated fat 0 g 0% Cholesterol 0 mg 0% Sodium 5 mg 0% Soluble fber 1 g Vitamin A 10% (100% as beta carotene) Insoluble fber 1 g Vitamin C 0% Sugars 12 g Calcium 2% Protein 1 g Iron 2% Percent daily values (%DV) are based on a 2000 calorie diet. Your daily values may be higher or lower depending on your calorie needs. Big Al’s Organic Sweet and Juicy Dried Plums/Prunes 150 UNIT 5 Ratio and Proportion 6. How many total grams (g) of weight are present in 34 prunes? 7. If 100 calories are consumed with 5 prunes, how many calories are consumed with 12 prunes? 8. If 5 prunes have 290 milligrams (mg) of potassium and that accounts for 8% of percent daily value, how many prunes are needed to equal 15% of the percent daily value? Round to the nearest whole number. 9. If 5 prunes provide 10% of the Vitamin A needed daily, what percent of the daily percent of Vitamin A is present in 20 prunes? 10. If a serving size is 1 1 2 ounces (oz), how many ounces are five servings? Use the information from the label to complete these proportions. 11. If 1 cup of soy milk contains 4 grams (g) total fat, then how many grams of total fat are in 2 3 4 cups of Jade’s Soy Milk? 12. If a cup of soy milk contains 85 milligrams of sodium, and an individual consumes 2 2 3 cups of soy milk per day, what is the sodium intake from soy milk? Round to the nearest whole number. 13. If one serving of Jade’s Soy Milk provides 1% of the amount of daily carbohydrates needed, how many milliliters make 5% of the daily carbohydrate intake? 14. One serving contains 80 calories; how many calories are in 3 1 4 servings? 15. If 1 cup of Jade’s Soy Milk provides 1 gram of fiber and 4% of the recommended daily fiber intake, how many cups of this soy milk are needed to make up 25% of the dietary fiber? Jade’s Soy Milk Nutrition facts Amount/serving %DV Amount/serving %DV Total fat 4 g 6% Total Carbohydrates 4 g 1% Saturated fat 0.5 g 3% Fiber 1 g 4% Trans fat 0 g Cholesterol 0 mg 0% Polyunsaturated fat 2.5 g Sugars 12 g Monounsaturated fat 1 g Protein 7 g Sodium 85 mg 4% Potassium 300 mg 8% Vitamin A 10% Vitamin C 0% Calcium 30% Iron 6% Serving size 1 cup (240 mL) Servings about 8 per 1.89 L Calories 80 Fat calories 35 Percent daily values (%DV) are based on a 2000 calorie diet. Vitamin D 10% Folate 6% Selenium 8% Magnesium 10% UNIT 5 Ratio and Proportion 151 Unit Review Critical Thinking with Ratio and Proportion 1. Use ratio and proportion to find x. 1 8 3 4 = 12 x 2. The physician ordered Valium 3.5 milligrams intramuscularly every 6 hours as needed for anxiety. The pharmacist has Valium in his pharmacy in the following supply: 10 milligrams per 2 milliliters. How many milliliters will the patient receive in each dose for this drug order? 3. Using the information from problem 2, what would be a full dosage of medication, in milliliters, given on schedule for one day? 4. The average human heart beats 4320 times every 60 minutes. Express this as a unit rate. 5. Write 450 calories in three servings as a unit rate. 6. Mixing infant formula is an important task. For every 2 ounces of instant infant formula, Mary needs 6 ounces of tepid, sterilized water. If the can of instant infant formula contains 8 ounces, how many ounces of sterilized water are needed? 7. At Valley View Center, there are three part-time employees for every two full-time employees. If there are 48 full-time employees, how many part-time employees are there? 8. The exercise pool at Care Vista has a pump that can drain 5500 gallons of treated water from the pool in 90 minutes. How many gallons can it drain in two and a half hours? Round to the nearest whole number. 9. Greek yogurt has 140 calories in 6 ounces. What is the unit rate of calories in each ounce? Round to the nearest whole number. 10. There are 14 grams of protein in a 6-ounce container of Greek yogurt. What is the per-unit rate of grams of protein per ounce? Round to the nearest whole number. UNIT 5 Ratio and Proportion 151 152 UNIT 5 Ratio and Proportion 11. In 12.5 hours, the patient had an even distribution of 750 milligrams of medication via intravenous solution. How many milligrams of medication did the patient receive per hour? 12. Write the following ratio as a decimal: 1 : 25. 13. The dental assistant is to mix the modeling mix for a dental model. She has learned that she should use 50 milliliters of water for 100 milligrams of the modeling mix. How many milligrams of modeling mix are used with 75 milliliters of water? 14. The lab technician is working on a special research project. She can prepare 45 pipettes every hour for the project. If she works consistently at this rate, how many pipettes will she have ready in 45 minutes? Round to the nearest whole number. 15. A disinfectant solution has 10 grams of disinfectant powder to 100 milliliters of hot water. Keeping the ratio at 1 : 10, how many grams of disinfectant powder are used for 330 milliliters of hot water? Professional Expertise • Always reduce ratios to their simplest form. • Similar units in ratios should be across from each other when setting up two fractions as a proportion. • Read the problem carefully to find out if you are solving for a ratio, a unit rate, or a proportion. Ratio and Proportion Post-Test Show all your work. 1. Write a definition for proportion. Provide one health profession application or example. 2. 30 : 120 = ? : 12 3. One glass contains 8 ounces. How many full glasses are in 78 ounces? 4. 1 2 : 4 = 1 3 : x UNIT 5 Ratio and Proportion 153 5. x : 625 = 1 : 5 6. If 10 milligrams are contained in 2 milliliters, how many milligrams are contained in 28 milliliters? 7. A tablet contains 30 milligrams of medication. How many tablets will be needed to provide Ms. Smith with 240 milligrams of medication? 8. 100 micrograms of a drug are contained in 2 milliliters. How many milliliters are contained in 15 micrograms? 9. 1 100 : 6 = ? : 8 10. 0.04 : 0.5 = 0.12 : ? 11. How many minutes are in 130 seconds? Your answer will have both minutes and seconds. Show your set-up. 12. Four out of every six dental patients request fluoride treatment after their dental cleaning treatments. If 120 patients have dental cleanings this week, how many will choose to have fluoride treatments as well? 13. If the doctor’s office uses 128 disposal thermometer covers each day, how many covers will be used in a 5-day workweek? 14. Solve: 1 125 : 3 :: : 12 15. If the doctor ordered 6 ounces of cranberry juice four times a day for 4 days, how many total ounces would be served to the patient?
10872	https://artofproblemsolving.com/wiki/index.php/Searching_the_community?srsltid=AfmBOoo5VfpQoe9BkGjPKwJhmOrOf0rJmNU_vboa-Y0wOvC446m_uImj	Art of Problem Solving Searching the community - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Searching the community Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Searching the community This article is an introduction to searching the AoPS forums. All searching takes place from one of three pages: the main webpage, any of the specific forums, or from the dedicated webpage. Contents [hide] 1 Searching methods 1.1 Searching from the community 1.2 Searching from a forum 1.3 Searching from the main webpage 2 Getting the most out of search 2.1 Searching for a specific post 2.2 General searching tips 2.3 Customizing search queries 3 Searching for LaTeX and Asymptote 3.1 Searching for LaTeX 3.2 Searching for Asymptote Searching methods Searching from the community Searching from the community page The first of these two options is shown in the image to the right. It is primarily useful for quickly searching, especially when the search is of a general nature and without a specific post in mind. The method is simple: simply type your query into the "Search Community" field, click the magnifying glass (or press the Enter key), and posts containing terms matching your query will be shown to you. The parts of the posts that match your query will be highlighted in yellow. For example, when searching for problems involving a circle, one would simply type "circle" into the indicated field, getting a result similar to the following image: Result of searching "circle" Clicking on any of these results will open the full topic in which the post was made, starting from the post you clicked on. For example, if the post you clicked on was the fourth post in its topic, you would originally see that post and could scroll up (or down) to see the rest of the topic. If these are not the results you were looking for, you can click the "Edit search settings" link in the top right corner to adjust your search parameters. Searching from a forum It is possible to initiate a search from a specific forum, using the toolbar on its header. This will limit the search results to posts in that forum. The header of a forum To search from a forum, click the magnifying glass on its header (see image to the right). This will redirect you to the main search webpage, with the forum information already filled in. Searching from the main webpage Searching from the webpage This page can be reached in three ways: by clicking on the "Advanced search" button under the quick-search option shown above, through the Community drop-down menu, or directly through the link Compared to the other options, the webpage gives you significantly more ways to customize your search. There are five different fields to be filled in, but any of them can be left blank. For example, to search for all posts by rrusczyk made within the last year, the "search term" and "Posted In forum" fields should be left blank, but the other fields should be filled out with the appropriate information. It is even possible to leave every field blank, in which case the search results will mimic the global feed. The five fields are: Search fields \| Field \| Function \| --- \| \| Search term \| Fairly self-explanatory: enter the term(s) that you are searching for. Note that titles, tags, posts, and sources are all simultaneously searched. To search these separately, click the indicated option. It is possible to search different places for different terms simultaneously; for example, to search for AMC problems using Simon's Favorite Factoring Trick, search for posts containing "SFFT" in topics with source containing "AMC". The indicated option will also allow you the choice of restricting your search to opening posts. This field is blank by default, indicating that all posts are searched regardless of content. Searching for multiple terms will return posts matching at least one of those terms, prioritizing posts "closer" to the query as a whole. See the following section for ways to further customize these queries. Note that it is also possible to search for multiple tags, which will return posts under at least one of those tags. \| \| Posted By User \| Enter the name(s) of user(s) to restrict your search to. The search results will only contain posts posted by the user(s) you entered. Note that the posters of the original topics may be different from the users you enter here. You can also search by user ID instead of username, if you happen to know the ID of the user you are searching for (you can find this in their profile URL, which is of the form artofproblemsolving.com/community/user/). Users entered must be valid users and will be authenticated prior to searching. This field is blank by default, indicating that posts by all users are searched. \| \| Posted In Forum \| Enter the name(s) of forum(s) to restrict your search to. The search results will only contain posts posted in the forum(s) you entered. If you reached this webpage through clicking on the magnifying glass in one of the forums, this field will be pre-populated with that forum. Forums entered must be valid forums and will be authenticated prior to searching. This field is blank by default, indicating that all forums are searched. Note: On the previous website, forums were associated with a unique ID that could be used in searching, similar to user ID numbers, but this is no longer the case; type in the name of the forum instead. \| \| Dates \| Select the timeframe in which posts you're searching for must have been made. The options are "Any" (default), "During the last 24 hours", "During the last week", "During the last month", "During the last year", or any manually entered date range. \| \| Sorting \| Select whether to sort results by "Relevance" (default), "Newest first", or "Oldest first". If "relevance" is selected, the posts will be sorted according to a scoring algorithm that approximates how well a post correlates to the search query. The other two options, "Newest first" and "Oldest first", are self-explanatory. \| Posts by rrusczyk in the last year For example, to search for the aforementioned posts made by rrusczyk within the last year, the "Search term" field should be blank, the "Posted by User" field should contain rrusczyk, the "Posted in Forum" field should be blank, the "Dates" field should be set to "During the last year", and the "Sorting" field can be set according to how the searcher wishes the results to be presented. The image to the right shows how this search looks on the webpage. Topics started by rrusczyk in the last year To search instead for topics that rrusczyk has started within the last year, the steps are the same, except that the "search only the first post of each topic" option should be selected. To reach that option, click first on the "Click here to search titles, posts, sources, and tags separately" text, then check the box indicated. The image on the left shows how this search looks on the webpage. Of course, in both searches it is possible to further narrow the results by including the forum it was posted in, text that should be present in returned posts, and so on. It is also possible to expand the search to posts by rrusczyk and copeland, or even postsbycopelandaboutrrusczyk. Getting the most out of search Searching for a specific post One of the most common uses of search is to find a specific post that you remember some details about, but cannot find. For example, you might recall having once read a particularly well-written solution, a very nice question, or an important announcement, but you can't remember quite where you saw it before. How frustrating! Obviously, the more details you remember about the post, the more likely you are to find it. If you remember that, for example, the post was written sometime in the summer of 2013, you can set the date range from (for example) May to September of that year. If you remember the author, that narrows down the possibilities significantly, but even remembering that it was one of several authors is a good way to whittle the possibilities down. Finally, remembering the forum that it was posted in would be excellent, but even if you only vaguely remember the post, you can make some educated guesses. For example, if the post was about a new and exciting technique for the USAMO, it was probably posted in either the Contests & Programs forum or the High School Olympiads forum. If the post was a collection of MATHCOUNTS strategies, it's almost certainly in Middle School Math. General searching tips Of course, the above strategies are helpful, but still leave a lot of posts to sift through - unless you remember some of the language used in the post. You might remember, for example, a particularly memorable phrase in the post, or perhaps they used a math problem as reference that you remember a bit about. In these cases, Include, in your search query, unusual or uncommon words. For example, searching for just the word "circumcenter", along with the tips in the previous section, narrows down the possibilities significantly Avoid common words such as "a", "the", "of", and so on, even if you remember a word-for-word phrase from the post containing these words. You'll get bogged down in posts matching those words - which there are a lot of! Avoid searching words that are contained within another common word, as they will get matched to words you certainly didn't intend. For example, search queries containing the word "in" might return results for "logging", "Inequality", and so on. The word "a" is particularly guilty of this, as it often returns words containing the letter "a". Be sure to avoid searching for words that are special modifiers - see the below section. In particular, avoid searches containing the words "and" and "not". Similarly, do not include quotes ("), asterisks (), question marks (?), plus signs (+), minus signs (-), or parentheses as these all represent special commands. Search engines in general are not very good at searching for mathematical symbols or numbers, so you are generally better off searching for text instead. For example, searching the text of a problem rather than equations or expressions is likely to produce better results. For example, if you remember a post contained the phrase "the three perpendicular bisectors of a triangle intersect at the circumcenter", your search query should be something similar to "perpendicular bisectors intersect circumcenter", as these are all specific words that limit the number of posts returned (unlike words like "of", "a", "the", and "at"). Customizing search queries We've already mentioned that posts are matched to the search query using a scoring algorithm, which approximates how "close" a post is to the given query. However, especially when searching for multiple terms, this algorithm often weights certain factors in different ways than the user intended. As such, there are several ways to customize your search queries. Customizing search queries Operator Example Result [[(no adjustment)incenter circumcenter Returns posts containing the word "incenter" and/or the word "circumcenter", with higher weight given to posts containing both. ++incenter circumcenter Returned posts must contain the word "incenter", but may or may not contain "circumcenter". Again, higher weight is given to posts containing both words. --incenter circumcenter excenter Returned posts must not contain the word "incenter". Returned posts will contain the word "circumcenter" and/or the word "excenter", with higher weight given to posts containing both. AND incenter AND circumcenter Returned posts must contain both "incenter" and "circumcenter". This is equivalent to the query "+incenter +circumcenter". NOT incenter NOT circumcenter Returned posts will contain "incenter", but will not contain "circumcenter". Equivalent to the "-" and "!" operators. "" (quotes)"incenter circumcenter"Returns posts containing the phrase "incentercircumcenter". Punctuation is usually ignored in results, so posts containing "incenter, circumcenter" or "incenter-circumcenter" will also be returned. ? and te?t, inc, ine Wildcard symbols. The? symbol allows any character to replace it, so posts containing the words "test" or "text" will match the query "te?t". The symbol allows any number of characters to replace it, so posts containing the words "incenter", "incircle", "inclusive", etc. will match the query "inc". The symbol can also be used in the middle of a word, so posts containing the words "interface", "incircle", "intermediate", "infinite", etc. will all match the query "ine". () (parentheses)incenter AND (circumcenter OR excenter)Grouping symbols to allow one boolean command (e.g. AND, NOT, +, -,!) to modify multiple elements. The example returns posts that contain both "incenter" and at least one of "circumcenter" or "excenter". Equivalent to the query (incenter AND circumcenter) OR (incenter AND excenter) (Boolean logic applies to search strings). \AND An "escape" character that allows you to search for reserved keywords and symbols. For example, the above query will return posts containing the word "and". Without the escape character, the search engine would parse the query "AND" as a command linking two nonexistent terms, and would thus throw an error. Similarly, the search term "incenter \AND circumcenter" would include results containing the word "incenter" and the word "and", but not the word "circumcenter". Note: in order to search for terms containing backslashes, such as LaTeX commands, the somewhat non-intuitive \ is necessary (the first backslash "escapes" the second one). Note that, when using search modifiers, the yellow background indicating words that match your query will not generally be entirely accurate; for example, searching for "incenter AND circumcenter" will return posts containing both words as expected, but the word "and" will also be highlighted. Don't get confused by this - the search engine is still searching according to your query (and is not searching for the word "and"), the highlighting is simply slightly misleading. Searching for LaTeX and Asymptote Searching for LaTeX or Asymptote (Vector Graphics Language) can be complicated, as they are their own languages, but this can actually be a big advantage! Predicting what code they've used is usually quite easy, and since the keywords are generally quite unique, searching for them will often narrow results down to posts using LaTeX/Asymptote themselves. Below are some common commands that are often used, to help you search for them. Searching for LaTeX LaTeX is a programming language for rendering mathematical statements, and is very popular on AoPS (and other mathematical sources). LaTeX commands are enclosed in dollar signs for rendering, but you should not generally include those while searching. If you are searching for "full" commands (e.g. \frac{1}{2}), it is highly recommended that you enclose the entire command with quotation marks (e.g. "\frac{1}{2}"; recall double backslash is needed), as otherwise it may parse as three different search terms (\frac, 1, and 2). Common LaTeX commands Command Use Example Rendered \frac Creates fractions\frac{1}{2} \sqrt Creates square (or, more generally, th) roots\sqrt{3}, \sqrt{3}, \leq, \neq, \geq Less than or equal to, not equal to, greater than or equal to (respectively)a \leq b, a \neq b, a \geq b \alpha Renders greek letters (most often alpha, beta, epsilon, pi, theta, phi, and omega)\alpha+\beta=\pi-\epsilon \rightarrow, \implies, \iff Used for implication, algorithms, etc.a>b \implies a^2>b^2, a>b \iff a-b>0, A\rightarrow A+1 \sum, \prod, \int Used for summation, product, and integration symbols\sum_1^5 x=15, \prod_1^5x=120, \int_1^5x=12 See LaTeX:Symbols for a more comprehensive list of LaTeX commands, and use the TeXeR to test how commands look when rendered. Searching for Asymptote Asymptote is a programming language for creating diagrams. Unlike LaTeX, dollar signs are not necessary, and Asymptote code is enclosed in [asy] tags. Asymptote commands take the form of most modern programming languages; lines are generally of the form command(param1, param2, ...) (this is different from TeX, where each parameter is enclosed in separate brackets). "Unfortunately" (for our purposes), most Asymptote commands are simply the word-for-word descriptors of their function; for example, the command that returns the midpoint of a path is simply "midpoint", and the command that returns the circumcenter of a triangle is "circumcenter". This is further complicated by the usage of variables, the names of which are entirely up to the posters (unlike TeX, in which variables are generally not used). Below are commonly used Asymptote commands that are not actual words, so they are less likely to be confused with other posts during searching. Common Asymptote commands Command Purpose Command Purpose defaultpen Adjusts the default settings for the pen. Very likely for this command to be in an Asymptote drawing.orthographic Adjust the "camera position" in 3-D drawings. Very likely for this command to be in a 3-D Asymptote drawing. filldraw Colors a given shape in some manner. Useful for shading in parts of diagrams; used reasonably foten.unitsize Sets the default size of a "unit" to a specified value. Default is very small. Not used particularly often because defaultpen is more general. intersectionpoint Determines the intersection (if there is exactly one) of two paths. Very useful command, and thus in many drawings.intersectionpoints Determine all the intersections of two paths. Obviously very similar to the previous command, but slightly less used as we generally want to know the intersection point of lines (rather than arbitrary paths). rightanglemark Marks a given angle as right. Used, obviously, in diagrams containing right angles; there are many of these.anglemark Marks a given angle with a given value. Used surprisingly little, since directly marking angles isn't generally that important, but angle chase solutions will make heavy use of it. linewidth Sets the (visual -- lines have no real width!) width of a line. Also little used because defaultpen is more general.currentpicture A reference to the current picture. Used as an auxiliary command in other commands. This is the default value, so it is used only in very complicated drawings (or as a formality/copied and pasted sections). Note that simply searching "asy" will generally turn up Asymptote drawings or discussions of such; especially when combined with a username or date range, simply remembering a post had any Asymptote in it whatsoever is usually a good way to find it. As Asymptote becomes more and more popular, the previous statement will become less true (LaTeX, for example, is in a huge number of posts these days), but for now the numbers are small enough that you can sift through them pretty quickly. 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10873	https://simple.wikipedia.org/wiki/Calf	Calf - Simple English Wikipedia, the free encyclopedia Jump to content [x] Main menu Main menu move to sidebar hide Getting around Main page Simple start Simple talk New changes Show any page Help Contact us About Wikipedia Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Give to Wikipedia Create account Log in [x] Personal tools Give to Wikipedia Create account Log in Calf [x] 6 languages Cebuano Deutsch English Français Plattdüütsch Tagalog Change links Page Talk [x] English Read Change Change source View history [x] Tools Tools move to sidebar hide Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Sandbox Edit interlanguage links Print/export Make a book Download as PDF Page for printing In other projects Wikimedia Commons Wikidata item From Simple English Wikipedia, the free encyclopedia A calf can be: A word for the young of some animals, usually cows, but also whales, dolphins, giraffes, bison, hippopotamuses, rhinoceroses, yaks and elephants. A muscle in the lower leg. This disambiguation page lists articles associated with the title Calf. If an internal link led you here, you may wish to change the link to point directly to the intended article. Retrieved from " Category: Disambiguation pages Hidden categories: All article disambiguation pages All disambiguation pages This page was last changed on 17 June 2025, at 05:49. Text is available under the Creative Commons Attribution-ShareAlike License and the GFDL; additional terms may apply. See Terms of Use for details. Privacy policy About Wikipedia Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search Calf 6 languagesAdd topic
10874	https://stackoverflow.com/questions/63825681/finding-similarities-and-contradictor-statements-in-strings-with-python	nlp - Finding similarities and contradictor statements in strings with Python - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding similarities and contradictor statements in strings with Python Ask Question Asked 5 years ago Modified5 years ago Viewed 334 times Part of NLP Collective This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I want to write code that will find similarities between strings. I found this script on StackOverflow, and it words good, but the problem is that I want to find contradictor statements. I have no idea how to do this, so im asking for help ```python from difflib import SequenceMatcher def similar(a, b): return SequenceMatcher(None, a, b).ratio() x = similar("sky is blue","color of the sky is blue") print(x) 0.6285714285714286 x = similar("sky is blue","color of the sky is not blue") print(x) 0.5641025641025641 x = similar("people are telling that sky is blue","people are telling that is not blue") print(x) 0.8857142857142857 ``` You see that I get 0.88 score for two contradictor statements, so that score should be much less, or it should return some kind of label (true or fake, contradict or something like that) My inputs are text inputs, sentances (3-15 words). Something like fact - checking NLP Collective python nlp nltk Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Sep 10, 2020 at 11:19 tagataga asked Sep 10, 2020 at 8:25 tagataga 3,913 16 16 gold badges 73 73 silver badges 153 153 bronze badges 5 1 It will be good if you add expected output.Olvin Roght –Olvin Roght 2020-09-10 08:29:51 +00:00 Commented Sep 10, 2020 at 8:29 Does this help? stackoverflow.com/questions/55162668/…DirtyBit –DirtyBit 2020-09-10 08:32:32 +00:00 Commented Sep 10, 2020 at 8:32 @OlvinRoght I have added expected output taga –taga 2020-09-10 08:37:17 +00:00 Commented Sep 10, 2020 at 8:37 This is not as trivial as it seems. Most similarity measures are statistical ones based on vectors and bag of word models. Therefore, they are insensitive to small negation markers like "not". The solution to this task highly depends on the structural variability of your input data. Could you give more precise info on your input and the exact text type? On a free text corpus, e.g. Wikipedia, this might be impossible without deeper analysis of syntax.CLpragmatics –CLpragmatics 2020-09-10 09:27:16 +00:00 Commented Sep 10, 2020 at 9:27 @CLpragmatics My inputs are lowercase text inputs, sentences (3-15 words). Something like fact - checking taga –taga 2020-09-10 11:20:21 +00:00 Commented Sep 10, 2020 at 11:20 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. try to use Sentence Embedding , it convert the sentences to vectors [20 65 32 ...] then use them to calculate the similarity. Sentence embedding depend on word embedding which convert each word in the sentence to vector then take the SUM or AVG as you need -> how to do word embedding ? you have 2 choices , first of them is to use pre-trained model of word2vec like Google's model but it's too much space for u 1.5GB!!! but it will give you incredible results the second choice is to build your own model here you will find the code to use Google's pre-trained model and the sentence embedding function Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 20, 2020 at 0:28 Esraa IbrahimEsraa Ibrahim 26 1 1 bronze badge Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions python nlp nltk See similar questions with these tags. NLP Collective See more This question is in a collective: a subcommunity defined by tags with relevant content and experts. The Overflow Blog The history and future of software development (part 1) Getting Backstage in front of a shifting dev experience Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure New and improved coding challenges New comment UI experiment graduation Policy: Generative AI (e.g., ChatGPT) is banned Report this ad Report this ad Linked 5Calculate similarity between list of words Related 2Finding locations where two strings differ 10Comparing sentences according to their meaning 6Compare similarity of terms/expressions using NLTK? 4check if two words are related to each other 8search similar meaning phrases with nltk 8Check the similarity between two words with NLTK with Python 3An Algorithm to Determine How Similar Two Sentences Are 1Sentence meaning Similarity and frequency 3finding semantic similarity between 2 statements 1How can I know through similarity how specific string is included in the sentence? 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10875	https://www.mededportal.org/doi/10.15766/mep_2374-8265.572	Delirium, an Interactive Learning Experience \| MedEdPORTAL Search Publications Search Advanced search Skip main navigation Close Drawer Menu Open Drawer MenuHome AUTHORS Submission Instructions Author Center Author Development REVIEWERS Reviewer Center Reviewer Recognition ABOUT Journal Information Editorial Oversight CALLS FOR SUBMISSION Artificial Intelligence Community Engagement Language-Appropriate Health Care OPEN ACCESS March 14, 2007 Delirium, an Interactive Learning Experience Ugochi Ohuabunwa, MD 1 , Nurcan Ilksoy, MD 2 , Erica Brownfield, MD 3 , Birju Patel, MD 4 , Jonathan Flacker, MD 5 Ugochi Ohuabunwa, MD 1 Emory University School of Medicine Google Scholar More articles by this author , Nurcan Ilksoy, MD 2 Emory University School of Medicine Google Scholar More articles by this author , Erica Brownfield, MD 3 Emory University School of Medicine Google Scholar More articles by this author , Birju Patel, MD 4 Emory University School of Medicine Google Scholar More articles by this author , Jonathan Flacker, MD 5 Emory University School of Medicine Google Scholar More articles by this author Author Information Ugochi Ohuabunwa, MD 1,Nurcan Ilksoy, MD 2,Erica Brownfield, MD 3,Birju Patel, MD 4,Jonathan Flacker, MD 5 1 Emory University School of Medicine 2 Emory University School of Medicine 3 Emory University School of Medicine 4 Emory University School of Medicine 5 Emory University School of Medicine Sections Abstract Educational Objectives About Tools Download Citations Share Facebook X Email Abstract It is important to expose medical students to a geriatrics curriculum. This resource is a case about delirium designed to create active-learning groups among third-year medical students. At our institution, it is delivered as part of a 2-hour workshop, but it can also be offered on its own. The total time required for this case is about 40 minutes. Presentation of the case is followed by discussion within small groups of learners (about 10 students each). Each group is given 15 minutes to review the case and identify salient aspects of the history, physical examination, risk factors, and management options. Then, the whole group of learners (about 30 students at our institution) reviews the information and conclusions of each small group through a discussion facilitated by a faculty member. The facilitator, equipped with the standard answer key, checks off the correct answers from each group on the key. The facilitator also highlights key teaching points and fills in unanswered questions, thereby providing the students with a rewarding learning experience. At the end of the session, the group with the highest number of correct answers receives a gift. The competitive nature of the interaction encourages active participation by the groups. Assessment of the achievement of the learning objectives is done in a subsequent OSCE session where the students are required to evaluate a patient presenting with delirium. This approach has been well received by our students. Evaluations obtained from them in the last 3 years indicate that they find this method of education very valuable. This resource forms a part of the Reynolds Program on medical education in geriatrics at the Emory University School of Medicine. The target population of third-year medical students constitutes a prime entry point for creating and mobilizing interest in geriatrics. This exercise offers them a learning opportunity that can facilitate growth in the knowledge of geriatric medicine required to build a professional expertise bank catering to the growing US geriatric population, as well as to pass on such knowledge and skills to future generations. We have developed and included a Likert-scale questionnaire to allow future learners to evaluate the utility of the resource. Educational Objectives By the end of this workshop, learners will be able to: Identify risk factors for delirium in older adults. Describe diagnostic criteria for delirium in older adults. Describe appropriate management options for delirium in older adults. Sign up for the latest publications from MedEdPORTAL Add your email below FILES INCLUDED References Related Details FILES INCLUDEDIncluded in this publication: Delirium Facilitator's Manual.doc Delirium Case Guide.doc Delirium Evaluations Summary.doc Delirium Feedback Form.doc To view all publication components, extract (i.e., unzip) them from the downloaded .zip file. Downloadeditor’s note This publication may contain technology or a display format that is no longer in use. Citation Ohuabunwa U, Ilksoy N, Brownfield E, Patel B, Flacker J. Delirium, an Interactive Learning Experience. MedEdPORTAL. 2007;3:572. None Cited byJackson J, Strowd L and Peters T(2020)The Simulated Virology Clinic: A Standardized Patient Exercise for Preclinical Medical Students Supporting Basic and Clinical Science Integration, MedEdPORTAL, 16, Online publication date: 1-Jan-2020. Article Metrics View all metrics Views Downloads No data available. 413 117 Total 6 Months 12 Months Total number of views and downloads Copyright & Permissions © 2007 Ohuabunwa et al. This is an open-access publication distributed under the terms of the Creative Commons Attribution-NonCommercial-Share Alike license. Keywords Elderly Delirium Altered Mental Status Aging Older Adult Disclosures None to report. Funding/Support None to report. tabs.loading Close Figure Viewer Browse All FiguresReturn to FigureChange zoom level Zoom in Zoom out Previous FigureNext Figure Caption back Back to top Connect With Us CALLS FOR SUBMISSION Artificial Intelligence Community Engagement Language-Appropriate Health Care QUICK LINKS Journal Information Editorial Oversight Submission Instructions Author Center Reviewer Center ###### TERM & CONDITIONS###### PRIVACY STATEMENT###### DISCLAIMER Copyright © 2025 AAMC 655 K Street, NW, Suite 100, Washington, DC 20001-2399 ✓ Thanks for sharing! 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10876	http://mathrefresher.blogspot.com/2005/05/exponents.html	Math Refresher: Exponents Math Refresher Review of fundamental math concepts in a straight-forward, accessible way. Monday, May 02, 2005 Exponents 1. Introduction to Exponents An exponent is an elegant shorthand for multiplication. Instead of 5 5 5, you can write 5 3 Instead of 3 3 3 3 3 3 3, you can write 3 7 The number that gets multiplied is called the base. The number of multiplications that occur is called the power. So, in the above example, 3 is the base and 7 is the power. Of course, this method only applies when the power is a positive integer. Later on, I will discuss what it means when a power is 0, positive, or even a fraction. So 4 2 = 4 4 = 16 And 4 3 = 4 4 4 = 64 And 4 1 = 4 = 4 x and y notation In mathematics, when we want to talk about "any", we use a letter such as x or y or z. For example, if we wanted to say that 1 to any power equals 1, we could write this as follows: 1 x = 1 Using x-and-y notation, we can create a definition for the positive exponents. Definition 1: Positive Exponents x ymeans xmultiplied with itself ytimes. xis called the base yis called the power 3. Multiplication of Exponents Multiplying exponents of the same base can be determined based on the above definition. 4 2 4 3 = = (4 4) (4 4 4) = 4 4 4 4 4 = 4 5 So, when exponents get multiplied, if they have the same base, you can add the powers and create a new exponent. Here are some more examples: 5 5 5 10 = 5 15 2 10 2 1000 = 2 1010 Of course, this does not work if two exponents have a different base. In mathematics, a method such as this can be presented as a theorem. A theorem is any statement that can be derived from previous results. In this case, we are able to prove a theorem regarding the method of adding the powers of the same base. Here's the theorem Theorem 1: x y x z = x(y+z) (1) We know that xy= x multipled to itself y times and that xz= x multipled to itself z times. (Definition of Positive Exponents). (2) Multiplying all those x's, we have (y + z) x's multiplied together. (3) Now x multiplied to itself (x + z) times = x(y + z)by the Definition of Positive Exponents. QED QED is put at the end of a proof to show it is done. It is an abbreviation for a latin phrase that means basically that the proof is finished. It serves the same purpose in a proof as a period does in a sentence. Division of Exponents To talk about division, it is useful to introduce the following definition: Definition 2: Division a = b / cmeans a is equal to bdivided byc. ais refered to as the quotient. bis refered to as the dividend. cis refered to as the divisor. Division with exponents of the same base can also be determined based on the definition for positive exponents: 4 2 / 4 1 = = ( 4 4 ) / ( 4 ) = = 16 / 4 = 4 = 4 1 To divide two exponents of the same base, you simply subtract the two powers. Here are some examples: 5 3 / 5 1 = 5 2 4 10 / 4 5 = 4 5 Now, what happens if we are dividing by a number greater than the top (in other words, where the divisor is greater than the dividend)? In this case, we are left with a fraction. 5 1 / 5 3 = 1 / 5 2 4 5 / 4 10 = 1 / 5 5 This leads us to a third definition: Definition 3: Negative Exponents x(-y)means that we have a fraction of 1over xmultiplied by itself ytimes. Here are some examples. 5-1 = 1 / 5 4-3 = 1 / 4 3 And what happens if the subtraction results in 0? We can answer this with the following theorem: Theorem 2: x 0 = 1 (1) By basic arithemitic, we know that x 0 = x(1 - 1) (2) Since 1 - 1 = 1 + (-1), we can rewrite this as: x(1 + -1) (3) Now x(1 + -1)= x1 x(-1)by Theorem 1. (4) Now, x(-1) = 1/x, by Definition 3. (5) So, we are left with x (1/x) = 1 QED We can also introduce a corollary to this theorem. A corollary is a small proof that is derived directly from the logic of a theorem. Corollary 2.1: x 0 = 1 implies that x ≠ 0 (1) Now x 0 = x(1 - 1) (2) Which means that x 0 = x / x (3) But this implies that x ≠ 0since division by 0is not allowed. QED Another way of saying this result is that 0 0 just like 0/0 or even 1/0 is undefined. We can summarize division of exponents with the following theorem. Theorem 3: x y / x z = x(y - z) Case I: y = z In this case x y / x z = 1 = x 0 = x(y - z). Case II: y > z In division, we are able to cancel out all the common factors. Since y > z, we cancel out z factors from both dividend and divisor and we are left with x(y-z). Case III: y < z Again, we cancel out common factors. Since z > y, we are left with a fraction of 1 / [x(z-y)]which, by definition 3, equals x(-(z-y)) = x(y-z) QED 5. Fractional Exponents There is more that we can talk about. What about fractional exponents such as x(1/2)? It turns out that based on our definitions, corrolaries, and theorems, we are now ready to take on fractional exponent. Let's start with 1/2. We know that x 1/2 x 1/2 = x(1/2 + 1/2) by Theorem 1. Now x(1/2 + 1/2) = x(1) = x. So x 1/2 is none other than the square root of x. Let's start out by looking at a definition for what a root is. Definition 4: an nth root of x is a number that multiplied n times equals x. Sometimes, nth roots are whole numbers. The cube root of 27 is 3 since 3 3 3 = 27. Likewise, the 4th root of 16 is 2. 1 is its own 5th root since 1 1 1 1 1 = 1. This gives us our last theorem: Theorem 4: x 1/n = the nth root of x (1) x1/nmultiplied by itself ntimes equals x1/n + 1/n + 1/n + etc.. (2) Now 1/n + 1/n + etc. ntimes equals n/n which equals1. (3) Therefore x1/nmultipled by itself ntimes equals x1 (4) And this is the very definition of an nthroot. QED Posted by Larry Freeman at 8:00 PM Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest 4 comments : Anonymous said... Excellent Nov 6, 2007, 9:40:00 AM stuvxz said... indeed excellent! Feb 28, 2008, 9:23:00 AM Anonymous said... Four to the fifth power divided by four to the tenth power is one over four, (with the four to the fifth power)- right? Jul 15, 2008, 5:15:00 PM Larry Freeman said... Four to the fifth power = 4^5 = 44444 Four to the tenth power = 4^10 = 4444444444 Four to the fifth power divided by four to the tenth power = 4^5/4^10 = 4^(5-10) = 4^(-5) = 1/(4^5) = 1/(44444) Four to the fifth power divided by four to the sixth power = 4^5/4^6 = 4^(5-6) = 4^(-1) = 1/(4^1) = 1/4. I hope that helps. Jul 15, 2008, 5:59:00 PM Post a Comment Newer PostHome Subscribe to: Post Comments ( Atom ) Followers About Me Larry FreemanView my complete profileBlog Archive ►2009 ( 32 ) ►11/15 - 11/22 ( 4 ) ►10/04 - 10/11 ( 1 ) ►09/27 - 10/04 ( 5 ) ►09/20 - 09/27 ( 5 ) ►09/13 - 09/20 ( 1 ) ►09/06 - 09/13 ( 1 ) ►08/30 - 09/06 ( 3 ) ►02/01 - 02/08 ( 5 ) ►01/25 - 02/01 ( 6 ) ►01/11 - 01/18 ( 1 ) ►2008 ( 11 ) ►10/19 - 10/26 ( 2 ) ►02/24 - 03/02 ( 3 ) ►01/27 - 02/03 ( 2 ) ►01/20 - 01/27 ( 2 ) ►01/06 - 01/13 ( 2 ) ►2007 ( 41 ) ►12/30 - 01/06 ( 2 ) ►12/02 - 12/09 ( 4 ) ►11/25 - 12/02 ( 2 ) ►10/14 - 10/21 ( 1 ) ►09/16 - 09/23 ( 2 ) ►07/15 - 07/22 ( 1 ) ►07/01 - 07/08 ( 3 ) ►06/24 - 07/01 ( 3 ) ►06/17 - 06/24 ( 1 ) ►06/10 - 06/17 ( 3 ) ►06/03 - 06/10 ( 2 ) ►05/27 - 06/03 ( 1 ) ►05/13 - 05/20 ( 2 ) ►05/06 - 05/13 ( 1 ) ►04/29 - 05/06 ( 3 ) ►04/22 - 04/29 ( 1 ) ►04/15 - 04/22 ( 3 ) ►03/18 - 03/25 ( 2 ) ►02/25 - 03/04 ( 1 ) ►01/28 - 02/04 ( 1 ) ►01/07 - 01/14 ( 2 ) ►2006 ( 75 ) ►12/31 - 01/07 ( 1 ) ►11/05 - 11/12 ( 1 ) ►10/29 - 11/05 ( 1 ) ►10/22 - 10/29 ( 1 ) ►09/24 - 10/01 ( 5 ) ►09/17 - 09/24 ( 4 ) ►09/10 - 09/17 ( 8 ) ►09/03 - 09/10 ( 2 ) ►08/27 - 09/03 ( 4 ) ►08/13 - 08/20 ( 1 ) ►07/16 - 07/23 ( 1 ) ►06/18 - 06/25 ( 1 ) ►06/04 - 06/11 ( 1 ) ►05/28 - 06/04 ( 1 ) ►05/21 - 05/28 ( 2 ) ►05/14 - 05/21 ( 4 ) ►05/07 - 05/14 ( 2 ) ►04/30 - 05/07 ( 3 ) ►04/23 - 04/30 ( 3 ) ►04/16 - 04/23 ( 5 ) ►04/09 - 04/16 ( 1 ) ►04/02 - 04/09 ( 1 ) ►03/26 - 04/02 ( 3 ) ►03/19 - 03/26 ( 2 ) ►03/12 - 03/19 ( 5 ) ►03/05 - 03/12 ( 6 ) ►02/19 - 02/26 ( 1 ) ►02/05 - 02/12 ( 2 ) ►01/08 - 01/15 ( 2 ) ►01/01 - 01/08 ( 1 ) ▼2005 ( 16 ) ►12/18 - 12/25 ( 1 ) ►10/09 - 10/16 ( 2 ) ►09/18 - 09/25 ( 1 ) ►08/28 - 09/04 ( 2 ) ►07/31 - 08/07 ( 1 ) ►07/24 - 07/31 ( 2 ) ►07/17 - 07/24 ( 1 ) ►06/12 - 06/19 ( 1 ) ►05/15 - 05/22 ( 1 ) ►05/08 - 05/15 ( 1 ) ▼05/01 - 05/08 ( 3 ) Greatest Common Divisor CoPrime Numbers: xn + yn = zn Exponents Powered by Blogger.
10877	https://womenshealth.labcorp.com/tests/004317/progesterone	004317: Progesterone \| Labcorp Women's Health Skip to main content Close Menu Logins Individuals & Patients Find a LabView Test ResultsPay a BillShop for Tests View Individuals & Patients Page Providers Test FinderProvider LoginEducation & ExpertsContact Us View Providers Page Health Systems & Organizations Hospitals & Health SystemsEmployee Wellness & TestingManaged Care & PayorsResources View Organizations Page Biopharma & Investigators Nonclinical ResearchCentral Laboratory ServicesOrder a KitContact Us View Biopharma Section Managing Your Health Shop for Health Tests Explore Women's Health Annual Wellness Guidelines More Diseases & Therapeutic Areas Cancer & Oncology Neurology Rheumatology More Treatment Modalities & Methods Cell & Gene Therapies Precision Medicine Radiopharmaceuticals Vaccines More Scientific Lab Disciplines Genetics Digital Pathology & AI Toxicology More Labs Diagnostic Reference & Specialty Labs Nonclinical Labs Central Labs Industries Healthcare Pharma Medtech Crop & Agricultural Chemical & Environmental Novel Food Safety Testing Education, events & experts About us News Careers Investors Help Open Menu Utility Menu Login Gold Connect Labcorp Link PATHways Search Submit Main Menu Search Submit Main Navigation PatientsPatients ### Patients TestsToggle Women's Health & WellnessHereditary CancerPre-PregnancyPregnancyPediatricSexual Health Reproductive Genetics TestingToggle About Genetic TestingDiseases & ConditionsTalking to Your DoctorGenetic Counseling Patient ResourcesToggle Test ResultsEducational VideosPatient StoriesOnline Genetics Resources Cost & BillingToggle BillingFinancial AssistanceEstimate My Cost ProvidersProviders Providers Products & ServicesToggle Test MenuFeatured TestsFamily StudiesGenetic CounselingVirtual Workflow Testing TypeToggle Advanced DiagnosticsWomen's Wellness TestingCarrier ScreeningPrenatal & Pregnancy ScreeningHereditary Cancer TestingSexual Health TestingCervical Cancer Screening ResourcesToggle Diseases & ConditionsInternationalPosters & PublicationsSample ReportsFormsOnline Clinical Resources Provider HelpToggle Ordering & ResultsBilling & InsuranceCustomer ServiceGold ConnectLabcorp Link Genetic Counseling LoginLogin Login Patient LoginGold ConnectLabcorp Link Estimate My Cost View All Labcorp Sites Close Search Test Menu Test Menu Search Diagnostic Test Menu Breadcrumb Home Physicians Test Menu 004317: Progesterone Progesterone Share Print TEST004317 Test number copied CPT 84144 Test Details Specimen Requirements Test Details Turnaround Time Within 1 day View Requisition FormsLogin or Register for Labcorp Link™ Use Establish the presence of a functioning corpus luteum or luteal cell function; confirm basal body temperature measurements for the occurrence of ovulation; obtain an indication of the day of ovulation; evaluate the functional state of the corpus luteum in infertility patients; assess placental function during pregnancy; ovarian function test Special Instructions This test may exhibit interference when sample is collected from a person who is consuming a supplement with a high dose of biotin (also termed as vitamin B7 or B8, vitamin H, or coenzyme R). It is recommended to ask all patients who may be indicated for this test about biotin supplementation. Patients should be cautioned to stop biotin consumption at least 72 hours prior to the collection of a sample. Limitations As with all tests containing monoclonal mouse antibodies, erroneous findings may be obtained from samples taken from patients who have been treated with monoclonal mouse antibodies or who have received them for diagnostic purposes.1 In rare cases, interference due to extremely high titers of antibodies to streptavidin and ruthenium can occur.1 The test contains additives that minimize these effects. Methodology Electrochemiluminescence immunoassay (ECLIA) References Hilborn S, Krahn J. Effect of time of exposure of serum to gel-barrier tubes on results for progesterone and some other endocrine tests. Clin Chem. 1987 Jan; 33(1):203-204. 3802491 Nippoldt TB, Reame NE, Kelch RP, Marshall JC. The roles of estradiol and progesterone in decreasing luteinizing hormone pulse frequency in the luteal phase of the menstrual cycle. J Clin Endocrinol Metab. 1989 Jul; 69(1):67-76. 2499593 Rebar RW. The ovaries. In: Wyngaarden JB, Smith LH Jr, eds.Cecil Textbook of Medicine. 18th ed. vol 2. Philadelphia, Pa: WB Saunders Co;1988:1425-1446. Romero R, Scoccia B, Mazor M, Wu YK, Benveniste R. Evidence for a local change in the progesterone/estrogen ratio in human parturition at term. Am J Obstet Gynecol. 1988 Sep; 159(3):657-660. 2971319 Stewart MO1, Whittaker PG, Persson B, Hanson U, Lind T. A longitudinal study of circulating progesterone, oestradiol, hCG and hPL during pregnancy in type 1 diabetic mothers. Br J Obstet Gynaecol. 1989 Apr; 96(4):415-423. 2751954 Additional Information Progesterone is a steroid hormone with a molecular weight of 314.5 daltons.2 Progesterone is mainly formed in the cells of the corpus luteum and during pregnancy in the placenta. Progesterone is increased in congenital adrenal hyperplasia due to 21-hydroxylase, 17-hydroxylase, and 11-β-hydroxylase deficiency. Progesterone is decreased in primary or secondary hypogonadism and short luteal phase syndrome. The progesterone concentration correlates with the development and regression of the corpus luteum. Whereas progesterone is barely detectable in the follicular phase of the female cycle, a rise in the progesterone level is observed one day prior to ovulation. Increased progesterone synthesis occurs during the luteal phase. In the second half of the cycle pregnanediol is excreted in urine as the main degradation product of progesterone. Progesterone brings about the conversion of the uterine mucosa into a tissue rich in glands (secretion phase), in order to prepare for the intrauterine implantation of the fertilized ovum. During pregnancy, progesterone inhibits the contraction of the myometrium. In the mammary gland, progesterone (together with estrogens) promotes the proliferation and secretion disposition of the alveoli.2,3 The determination of progesterone is utilized in fertility diagnosis for the detection of ovulation and assessment of the luteal phase.3,4 Specimen Requirements Information on collection, storage, and volume Specimen Serum (preferred) or plasma Volume 1 mL Minimum Volume 0.7 mL (Note: This volume does not allow for repeat testing.) Container Red-top tube, gel-barrier tube, or green-top (lithium heparin) tube. Do not use oxalate, EDTA, or citrate plasma. Storage Instructions Room temperature Collection If a red-top tube or plasma is used, transfer separated serum or plasma to a plastic transport tube. 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10878	https://www.chegg.com/homework-help/questions-and-answers/room-temperature-bulk-modulus-glycerine-435-times-10-9-n-m-2-density-glycerine-1260-kg-m-3-q22769087	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: At room temperature the bulk modulus of glycerine is about 4.35 times 10^9 N/m^2 and the density of glycerine is about 1260 kg/m^3. Calculate the speed of sound in glycerin. (Express your answer to three significant figures.) Solution: The speed … Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & ServicesChegg Products & Services Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
10879	https://math.stackexchange.com/questions/708431/proving-the-inequality-a-b-leq-a-c-c-b-for-real-a-b-c	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Proving the inequality $\|a-b\| \leq \|a-c\| + \|c-b\|$ for real $a,b,c$ Ask Question Asked Modified 11 years, 6 months ago Viewed 14k times 4 $\begingroup$ Let $a,b,c$ real numbers. Prove the inequality $\|a-b\| \leq \|a-c\| + \|c-b\|$. Prove that equality holds if and only if $a \leq c \leq b$ or $b \leq c \leq a$. I've tried starting with just $a \leq c$ and using field properties to reconstruct the inequality, however I haven't been able to make it work. I also tried making the negatives positive and stripping the inequalities and making something happen but again I don't know if that's a proper rule and it didn't seem to get me anywhere. real-analysis inequality proof-writing absolute-value Share edited Mar 11, 2014 at 19:39 TMM 10.1k33 gold badges3838 silver badges5454 bronze badges asked Mar 11, 2014 at 19:24 AchillesAchilles 34511 gold badge88 silver badges2020 bronze badges $\endgroup$ 2 1 $\begingroup$ $c-c=0{}{}{}{}$ $\endgroup$ Najib Idrissi – Najib Idrissi 2014-03-11 19:44:27 +00:00 Commented Mar 11, 2014 at 19:44 $\begingroup$ Similar questions: math.stackexchange.com/questions/675977/… $\endgroup$ Martin Sleziak – Martin Sleziak 2014-03-12 13:04:57 +00:00 Commented Mar 12, 2014 at 13:04 Add a comment \| 4 Answers 4 Reset to default 7 $\begingroup$ You could use the triangle inequality and the fact that you can write $\|a-b\|=\|(a-c)+(c-b)\|$. Share edited Mar 12, 2014 at 13:04 Martin Sleziak 56.3k2020 gold badges211211 silver badges391391 bronze badges answered Mar 11, 2014 at 19:36 homomathematicushomomathematicus 34511 silver badge1010 bronze badges $\endgroup$ 3 $\begingroup$ ahhh I see thanks. $\endgroup$ Achilles – Achilles 2014-03-11 19:39:47 +00:00 Commented Mar 11, 2014 at 19:39 1 $\begingroup$ Ok, I still don't have it solved I think i'm more confused now then I was before. $\endgroup$ Achilles – Achilles 2014-03-11 20:06:03 +00:00 Commented Mar 11, 2014 at 20:06 2 $\begingroup$ Put $x=a-c$ and $y=c-b$. Now apply $\|x+y\| \le \|x\|+\|y\|$. $\endgroup$ TonyK – TonyK 2014-03-11 20:10:18 +00:00 Commented Mar 11, 2014 at 20:10 Add a comment \| 3 $\begingroup$ As others have pointed out, you should prove this by using the triangle inequality. I also think you should try to understand the problem intuitively so I drew a picture: $\|a-b\|$ represents the distance between the points $a$ and $b$ on the number line. If $c$ is between $a$ and $b$, or is equal to either $a$ or $b$, then the distance from $a$ to $c$, which is $\|a-c\|$, plus the distance from $c$ to $b$, which is $\|c-b\|$, will equal the total distance from $a$ to $b$. In other words, the length of the two red line segments in case $2$ equals the length of the green line segment in case $1$. If $c$ does not lie between $a$ and $b$, as in case $3$, then the distance from $c$ to either $a$ or $b$ must be greater than the distance from $a$ to $b$. Note that the above is in not a proof but it can help you understand the problem. Share edited Mar 11, 2014 at 20:39 answered Mar 11, 2014 at 20:14 SidSid 4,43266 gold badges2626 silver badges5252 bronze badges $\endgroup$ 1 $\begingroup$ Thank you , I really appreciate you taking the time to draw a picture. I do understand much better now. Thanks again. $\endgroup$ Achilles – Achilles 2014-03-11 20:21:04 +00:00 Commented Mar 11, 2014 at 20:21 Add a comment \| 3 $\begingroup$ Just for fun, here is a proof that does not use the triangle equality, and instead uses the definition $$ \|x\| \;=\; x\text{ max }-x $$ together with the properties of $\;\text{max}\;$: \begin{align} & \|a-c\| \;+\; \|c-b\| \ = & \qquad \text{"the above definition of $\;\|\quad\|\;$, twice"} \ & ((a-c)\text{ max }(c-a)) \;+\; ((c-b)\text{ max }(b-c)) \ = & \qquad \text{"$\;+\;$ distributes over $\;\text{max}\;$, three times"} \ & (a-c+c-b)\text{ max }(a-c+b-c)\text{ max }(c-a+c-b)\text{ max }(c-a+b-c) \ = & \qquad \text{"arithmetic: simplify"} \ & (a-b)\text{ max }(a+b-2c)\text{ max }(2c-a-b)\text{ max }(b-a) \ \ge & \qquad \text{"basic property of $\;\text{max}\;$: $\;x\text{ max }y \ge x\;$} \ & \qquad \phantom{\text{"}}\text{-- to get rid of the parts that contain $\;c\;$"} \ & (a-b)\text{ max }(b-a) \ = & \qquad \text{"the above definition of $\;\|\quad\|\;$"} \ & \|a - b\| \ \end{align} Share answered Mar 12, 2014 at 14:19 MarnixKlooster ReinstateMonicaMarnixKlooster ReinstateMonica 5,62644 gold badges3535 silver badges6666 bronze badges $\endgroup$ 2 $\begingroup$ Basically this is a variation on this (math.stackexchange.com/a/457759/11994) proof of the triangle equality. $\endgroup$ MarnixKlooster ReinstateMonica – MarnixKlooster ReinstateMonica 2014-03-12 14:21:08 +00:00 Commented Mar 12, 2014 at 14:21 $\begingroup$ Nice job , pretty clever $\endgroup$ Achilles – Achilles 2014-03-13 21:45:41 +00:00 Commented Mar 13, 2014 at 21:45 Add a comment \| 1 $\begingroup$ The inequality is a simple application of the triangle inequality as already mentioned (with $x=a-c$ and $y=c-b$). The equality therefore holds when $xy=(a-c)(c-b)\geq 0$, so either when $a-c\leq 0$ and $c-b\leq 0$ or when $a-c\geq 0$ and $c-b\geq 0$, which eventually means when $a\leq c\leq b$ or $b\leq c\leq a$. Share answered Mar 12, 2014 at 11:57 vlxmvlxm 12111 silver badge44 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis inequality proof-writing absolute-value See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 2 Prove: Use the triangle inequality to prove that for all $x, y, z, \| x z \| ¤ \| x y \| + \| y z \|$ Proof for absolute value inequality of three variables: $\|x-z\| \leq \|x-y\|+\|y-z\|$ 0 Prove this statement (inequality): $\|x-y\| \leq\|x-z\|+\|y-z\|$ 2 Prove the triangle inequality Proof of an inequality with the triangle inequality Related 1 Proving Complex Inequality $\|\|z_1z_2\|-1\| \leq\|z_1z_2 + i\| \leq \|z_1z_2\| +1$. 1 An inequality of symmetric polynomials using AGM inequality? 0 Suppose $0\leq{t}\leq{x}\leq1$ and let $z=\frac{t}{x}$ prove $\|1-z\|\leq1$ 0 Show that $\|1-z\| \leq \|z\|+1$ 2 Inequality of $\vert \text{tanh}(x)-x+\frac{x^3}{3}\vert\leq\frac{1}{8}$ for $x\in[-1/4,1/4]$ 2 Proving the Triangle Inequality using only the Axioms of Inequality on the set of Real Numbers 2 Prove an inequality with absolute values 0 Show $\sum_{k = 1}^\infty \frac{x!}{(x + k)!} \leq \frac{1}{x}$ 0 Proving $2\|c_1\|^2\|c_2\|^2 \leq c_1c_2^ + c_1^c_2$ for $\|c_1\|,\|c_2\|\leq 1$ Hot Network Questions On the Subject of Switches What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? how do I remove a item from the applications menu Change default Firefox open file directory Is it safe to route top layer traces under header pins, SMD IC? 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10880	https://www.tiger-algebra.com/drill/-3k=24/	Copyright Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. Solution - Linear equations with one unknown Other Ways to Solve Step by Step Solution Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : -3k-(24)=0 Step by step solution : Step 1 : Pulling out like terms : 1.1 Pull out like factors : -3k - 24 = -3 • (k + 8) Equation at the end of step 1 : Step 2 : Equations which are never true : 2.1 Solve : -3 = 0 This equation has no solution. A a non-zero constant never equals zero. Solving a Single Variable Equation : 2.2 Solve : k+8 = 0 Subtract 8 from both sides of the equation : k = -8 One solution was found : How did we do? Why learn this Terms and topics Related links Latest Related Drills Solved Copyright Ⓒ 2013-2025 tiger-algebra.com
10881	https://www.mytutor.co.uk/answers/1720/GCSE/Maths/What-are-the-differences-between-decimal-points-and-significant-figures/	What are the differences between decimal points and significant figures? \| MyTutor Find a tutorHow it worksPricesResourcesFor schoolsBecome a tutor +44 (0) 203 773 6024 Log inSign up Answers>Maths>GCSE>Article What are the differences between decimal points and significant figures? The total number of digits to the right of the point are known as 'decimal points'whereassignificant figuresstart from the first non-zero number.Note: As significant figures start from the first non-zero number this could be to the left of the decimal point.We then use the laws of rounding to come up with our final answers! d.p. = decimal points, s.f. = significant figures Examples: a) 704.539018 is 704.53902 (to 5d.p.) and 704.54 (to 5s.f.) b) 0.0091 is 0.01 (to 2d.p.) but 0.0091 (to 2s.f.) c) 0.100400 is 0.10040 (to 5d.p.) and also 0.10040 (to 5s.f.) because,from the first significant figure (non-zero)onwards,all numbers are significant. This includes zeros between non-zeros and all trailing zeros. NL Answered by Niall L. •Maths tutor 76538 Views See similar Maths GCSE tutors Need help with Maths? One-to-one online tuition can be a great way to brush up on your Maths knowledge. Have a Free Meeting with one of our hand picked tutors from the UK's top universities Find a tutor Download MyTutor's free revision handbook? This handbook will help you plan your study time, beat procrastination, memorise the info and get your notes in order. 8 study hacks, 3 revision templates, 6 revision techniques, 10 exam and self-care tips. Download handbook Free weekly group tutorials Join MyTutor Squads for free (and fun) help with Maths, Coding & Study Skills. Sign up Related Maths GCSE answers All answers ▸ #### How can you calculate the distance between 2 points in a grid if they're not on the same horizontal or vertical line? Answered by Tom C. #### Given a graph of y = f(x) how do you sketch a graph of y = f(x)+4, y = f(x-5), y = -f(x+4) or any other similar transformations Answered by Jack E. #### factorise x^2+5x+6=0 and find x Answered by Ashley O. #### Work out the value of (16/81)^(3/4) Answered by Stefano S. We're here to help Contact us+44 (0) 203 773 6020 Company Information CareersBlogSubject answersBecome a tutorSchoolsSafeguarding policyFAQsUsing the Online Lesson SpaceTestimonials & pressSitemapTerms & ConditionsPrivacy Policy Popular Requests Maths tutorChemistry tutorPhysics tutorBiology tutorEnglish tutorGCSE tutorsA level tutorsIB tutorsPhysics & Maths tutorsChemistry & Maths tutorsGCSE Maths tutors ##### CLICK CEOP Internet Safety Payment Security Cyber Essentials MyTutor is part of the IXL family of brands: ##### IXL Comprehensive K-12 personalized learning ##### Rosetta Stone Immersive learning for 25 languages ##### Wyzant Trusted tutors for 300 subjects ##### Education.com 35,000 worksheets, games, and lesson plans ##### Vocabulary.com Adaptive learning for English vocabulary ##### Emmersion Fast and accurate language certification ##### Thesaurus.com Essential reference for synonyms and antonyms ##### Dictionary.com Comprehensive resource for word definitions and usage ##### SpanishDictionary.com Spanish-English dictionary, translator, and learning resources ##### FrenchDictionary.com French-English dictionary, translator, and learning ##### Ingles.com Diccionario ingles-espanol, traductor y sitio de apremdizaje ##### ABCya Fun educational games for kids © 2025 by IXL Learning
10882	https://www.sparkl.me/learn/cambridge-igcse/physics-0625-supplement/equation-for-average-orbital-speed-v-2pr-t/revision-notes/2838	1.1.1 Structure and function of a split-ring commutator in motors 1.2.1 Principle of operation of an iron-core transformer 1.2.2 Equation for transformer efficiency: IpVp = IsVs 1.2.3 Equation for power loss in transmission lines: P = I²R 1.3.1 Magnetic forces due to interactions between magnetic fields 1.3.2 Relative strength of a magnetic field represented by field line spacing 1.4.1 Definition of charge in coulombs 1.4.2 Electric field as a region where a charge experiences a force 1.4.3 Electric field direction based on force on a positive charge 1.4.4 Electric field patterns: point charge, charged spheres, parallel plates 1.5.1 Equation for current: I = Q / t 1.5.2 Conventional current direction (positive to negative) 1.6.1 Equation for e.m.f.: E = W / Q 1.6.2 Equation for potential difference: V = W / Q 1.7.1 Current-voltage graphs for a resistor, filament lamp, and diode 1.7.2 Variation of resistance with length and cross-sectional area of wire 1.8.1 Kirchhoff’s first law: sum of currents at a junction 1.8.2 Kirchhoff’s second law: sum of voltages in a closed loop 1.8.3 Total resistance in parallel circuits 1.9.1 Equation for potential divider: R1/R2 = V1/V2 1.9.2 Use of a potential divider in circuits 1.10.1 Direction of induced e.m.f. opposes change causing it 1.10.2 Relative directions of force, field, and induced current 1.11.1 Graph of e.m.f. variation with time in a generator 1.12.1 Variation of magnetic field strength around wires and solenoids 1.13.1 Direction of force on moving charges in a magnetic field 2. Waves 2.1.1 Definition of monochromatic light as light of a single frequency 2.2.1 Speed of electromagnetic waves in vacuum (3.0 × 10^8 m/s) 2.2.2 Use of microwaves, radio waves, and visible light in communications 2.2.3 Difference between digital and analog signals 2.2.4 Transmission of sound as digital or analog signals 2.2.5 Advantages of digital signals: data transmission speed and accuracy 2.3.1 Definition of compression and rarefaction in sound waves 2.3.2 Speed of sound comparison in solids, liquids, and gases 2.3.3 Uses of ultrasound in non-destructive testing and medical scanning 2.4.1 Effect of wavelength on diffraction at an edge 2.4.2 Effect of wavelength and gap size on diffraction 2.5.1 Using ray diagrams for reflection by plane mirrors 2.6.1 Definition of refractive index: n = sin i / sin r 2.6.2 Equation for critical angle: n = 1 / sin c 2.6.3 Use of optical fibers in telecommunications 2.7.1 Using ray diagrams to show virtual images by converging lenses 2.7.2 Use of a single lens as a magnifying glass 2.7.3 Use of converging and diverging lenses to correct vision defects 3. Space Physics 3.1.1 Definition of a light-year and conversion to meters 3.1.2 Life cycle of a star, including formation, main sequence, red giant, white dwarf, supernova, neutron 3.2.1 Redshift as evidence for an expanding Universe 3.2.2 Cosmic Microwave Background Radiation (CMBR) as evidence for the Big Bang 3.2.3 Expansion of the Universe stretching CMBR into the microwave spectrum 3.2.4 Hubble’s Law: speed of a galaxy moving away is proportional to its distance 3.2.5 Hubble constant (H₀) and estimation of the Universe’s age using 1 / H₀ 3.2.6 Supernova brightness as a method to determine galactic distances 3.2.7 Current estimate for Hubble constant and its implications for cosmic expansion 3.3.1 Equation for average orbital speed: v = 2πr / T 3.4.1 Planets, minor planets, and comets have elliptical orbits 3.4.2 Sun is not at the exact center of elliptical orbits (except in nearly circular ones) 3.4.3 Analyzing planetary data: orbital distance, period, density, surface temperature 3.4.4 Gravitational field strength decreases with distance from the Sun 3.4.5 Planets travel faster when closer to the Sun due to conservation of energy 3.5.1 Nuclear fusion in the Sun as the source of energy release 4. Motion, Forces, and Energy 4.1.1 Equations for power: P = W / t and P = ΔE / t 4.2.1 Understanding scalar and vector quantities 4.2.2 Identifying scalar quantities: distance, speed, time, mass, energy, temperature 4.2.3 Identifying vector quantities: force, weight, velocity, acceleration, momentum, field strengths 4.2.4 Determining the resultant of two perpendicular vectors graphically or by calculation 4.3.1 Equation for pressure in fluids: Δp = ρgΔh 4.4.1 Definition of acceleration as change in velocity per unit time 4.4.2 Equation for acceleration: a = Δv / Δt 4.4.3 Determining acceleration from speed-time graphs 4.4.4 Understanding deceleration as negative acceleration 4.4.5 Describing motion under gravity with and without air resistance 4.5.1 Understanding weight as the effect of a gravitational field on mass 4.6.1 Determining whether one liquid will float on another using density data 4.7.1 Definition and use of spring constant: k = F / x 4.7.2 Understanding the concept of limit of proportionality on a load-extension graph 4.7.3 Equation of motion: F = ma (Force and acceleration in the same direction) 4.7.4 Understanding motion in a circular path due to perpendicular force 4.8.1 Applying the principle of moments to situations with multiple forces 4.8.2 Demonstration that an object in equilibrium has no resultant moment 4.9.1 Equation for momentum: p = mv 4.9.2 Equation for impulse: impulse = FΔt = Δ(mv) 4.9.3 Applying the principle of conservation of momentum in one dimension 4.9.4 Equation for resultant force: F = Δp / Δt 4.10.1 Equation for kinetic energy: Ek = 1/2 mv² 4.10.2 Equation for change in gravitational potential energy: ΔEp = mgΔh 4.10.3 Applying the conservation of energy principle to multi-stage processes 4.11.1 Equation for work done: W = Fd = ΔE 4.12.1 Understanding that almost all energy resources (except nuclear, geothermal, tidal) originate from th 4.12.2 Nuclear fusion as the energy source for the Sun 4.12.3 Current research on large-scale energy production from nuclear fusion 4.12.4 Equation for efficiency: Efficiency = (useful energy output / total energy input) × 100% 4.12.5 Equation for efficiency in terms of power: Efficiency = (useful power output / total power input) × 5. Nuclear Physics 5.1.1 Alpha particle scattering experiment supporting the nuclear model 5.1.2 Evidence from scattering experiment for a small, dense, positively charged nucleus 5.2.1 Processes of nuclear fission and nuclear fusion 5.2.2 Mass and energy changes in fission and fusion reactions 5.2.3 Relationship between proton number and charge on nucleus 5.2.4 Relationship between nucleon number and mass of nucleus 5.3.1 Correcting for background radiation in radioactivity measurements 5.4.1 Deflection of alpha, beta, and gamma radiation in electric and magnetic fields 5.4.2 Explanation of ionizing effects based on charge and kinetic energy 5.5.1 Explanation of unstable isotopes due to neutron excess or heavy nucleus 5.5.2 Nuclear changes occurring in alpha, beta, and gamma emission 5.5.3 Neutron decay equation: neutron → proton + electron 5.5.4 Writing nuclear equations for radioactive decay using nuclide notation 5.6.1 Calculation of half-life from raw data or decay curves 5.6.2 Use of radioisotopes in medical imaging and treatment 5.6.3 Use of radioisotopes in industrial thickness monitoring 5.6.4 Factors influencing choice of radioisotope for specific applications 5.7.1 Methods to minimize radiation exposure: time, distance, shielding 5.7.2 Protective measures for handling radioactive materials 5.7.3 Biological effects of ionizing radiation, including DNA damage and cancer risk 6. Thermal Physics 6.1.1 Forces and distances between particles affect properties of solids, liquids, and gases 6.1.2 Pressure in gases explained in terms of molecular collisions and force per unit area 6.1.3 Motion of microscopic particles due to collisions with smaller molecules (Brownian motion) 6.2.1 Equation for gas pressure-volume relationship: pV = constant (for a fixed mass at constant temperatu 6.3.1 Explanation of thermal expansion in terms of molecular motion and arrangement 6.4.1 Internal energy increase linked to an increase in particle kinetic energy 6.4.2 Definition of specific heat capacity: c = ΔE / (mΔθ) 6.4.3 Experiments to determine specific heat capacity of solids and liquids 6.5.1 Differences between boiling and evaporation 6.5.2 Effects of temperature, surface area, and air movement on evaporation 6.5.3 Cooling effect of evaporation and its applications 6.6.1 Conduction in solids explained through atomic vibrations and electron movement in metals 6.6.2 Why conduction is poor in liquids and gases 6.7.1 Constant temperature requires equal rates of energy absorption and emission 6.7.2 Effect of energy imbalance on object temperature 6.7.3 Earth’s temperature balance between absorbed and emitted radiation 6.7.4 Experiments comparing good and bad emitters of infrared radiation 6.7.5 Experiments comparing good and bad absorbers of infrared radiation 6.7.6 Rate of radiation emission depends on surface temperature and area 6.8.1 Complex applications of conduction, convection, and radiation in real-world scenarios Equation for average orbital speed: v = 2πr / T Topic 2/3 Revision Notes Flashcards Past Paper Analysis Questions Videos Your Flashcards are Ready! 15 Flashcards in this deck. or How would you like to practise? Choose Difficulty Level. Choose Easy, Medium or Hard to match questions to your skill level. Choose Learning Method. Choose Easy, Medium or Hard to match questions to your skill level. 3 Still Learning I know 12 Previous Next Equation for Average Orbital Speed: v=T2πr Introduction Understanding the average orbital speed is fundamental in space physics, particularly when studying celestial bodies like Earth. This concept is pivotal for the Cambridge IGCSE Physics curriculum, specifically under the chapter "The Earth" in the unit "Space Physics." Mastery of the equation v=T2πr not only facilitates the comprehension of Earth's motion but also lays the groundwork for exploring more complex astronomical phenomena. Key Concepts 1. Definition of Average Orbital Speed The average orbital speed (v) of a celestial body is the constant speed at which it would need to travel to complete one full orbit around another body in a given period (T), assuming a perfectly circular orbit with radius (r). This simplifies the complex variations in speed that occur in elliptical orbits, providing a useful approximation for many practical applications. 2. Derivation of the Equation v=T2πr To derive the average orbital speed equation, consider a body moving in a circular orbit. The circumference of the orbit is 2πr, where r is the radius of the orbit. The time taken to complete one full orbit is the orbital period T. Therefore, the average speed is the total distance traveled divided by the time taken: v=T2πr This equation assumes a uniform circular motion, where the speed remains constant along the orbit. 3. Components of the Equation Radius of Orbit (r): The distance from the center of the Earth to the orbiting body. For Earth orbiting the Sun, this is approximately 1.496×1011 meters. Orbital Period (T): The time taken to complete one full orbit. For Earth around the Sun, T is about 365.25 days. Average Orbital Speed (v): The speed at which the Earth travels along its orbital path, averaging about 29.78 km/s. 4. Application of the Equation This equation is essential for calculating the speed of any object in a stable orbit. For example, satellites orbiting Earth must maintain a specific speed to counteract gravitational pull, ensuring they do not spiral into the planet or drift away into space. By adjusting the radius or orbital period, engineers can determine the required speed for various satellite missions. 5. Real-World Examples Earth's Orbit: Using v=T2πr, the Earth's average orbital speed around the Sun is calculated to be approximately 29.78 km/s. Satellites: Geostationary satellites orbit Earth at an altitude of about 35,786 km with an orbital period matching Earth's rotation, requiring a specific speed to remain fixed over a point on the equator. Space Probes: Probes like Voyager 1 rely on achieving sufficient speed to escape Earth's gravity and journey through the solar system. 6. Factors Affecting Average Orbital Speed Orbital Radius (r): Larger orbits require higher speeds to balance gravitational forces. Orbital Period (T): Shorter periods necessitate higher speeds to complete the orbit in less time. Mass of Central Body: While not directly in the equation, the mass influences gravitational force, affecting the required speed for a stable orbit. 7. Importance in Physics Curriculum Mastering the equation for average orbital speed is crucial for students preparing for the Cambridge IGCSE Physics exam. It integrates concepts of circular motion, gravitational forces, and mathematical proficiency, providing a comprehensive understanding necessary for tackling more advanced topics in space physics. 8. Mathematical Examples Example 1: Calculate the average orbital speed of Earth around the Sun. Given: Orbital radius, r=1.496×1011 meters Orbital period, T=365.25 days = 3.156×107 seconds Using the equation: v=3.156×1072π×1.496×1011≈29.78×103m/s- Thus, the average orbital speed of Earth is approximately 29.78 km/s. Example 2: Determine the average orbital speed of a satellite orbiting Earth at an altitude of 2×107 meters with a period of 5×104 seconds. Given: Orbital radius, r=2×107 meters Orbital period, T=5×104 seconds Using the equation: v=5×1042π×2×107=5×1044π×107=54π×103≈2513.274m/s- Therefore, the average orbital speed of the satellite is approximately 2.51 km/s. 9. Graphical Representation Graphing the relationship between orbital radius (r) and average orbital speed (v) can visually demonstrate the inverse relationship; as the radius increases, the average speed decreases for a constant orbital period. This is crucial for understanding satellite deployment strategies and the mechanics of planetary motion. 10. Limitations of the Average Orbital Speed Equation Circular Orbits Only: The equation assumes a perfectly circular orbit, whereas most celestial orbits are elliptical. Constant Speed Assumption: In reality, an object in an elliptical orbit varies its speed, moving faster at periapsis and slower at apoapsis. Neglecting Gravitational Variations: The equation does not account for variations in gravitational forces that can affect orbital speed. Advanced Concepts 1. Derivation from Newtonian Mechanics The average orbital speed can also be derived using Newton's law of universal gravitation and centripetal force. For a body of mass m orbiting a central mass M, the gravitational force provides the necessary centripetal force: r2GMm=rmv2 Simplifying, we get: v2=rGM Taking the square root: v=rGM Comparing this with the average orbital speed equation: v=T2πr Equating and solving for T: T=2πGMr3 This derivation connects the average orbital speed with fundamental gravitational principles, highlighting the interplay between mass, distance, and time in orbital mechanics. 2. Application of Kepler's Third Law Kepler's Third Law states that the square of the orbital period (T2) is directly proportional to the cube of the semi-major axis of the orbit (r3): T2∝r3 For circular orbits, this simplifies to: T=2πGMr3 Combining this with the average orbital speed equation allows for a deeper understanding of celestial mechanics, enabling predictions of orbital periods based on distance and mass. 3. Energy Considerations in Orbital Motion Orbital motion involves both kinetic and potential energy. The total mechanical energy (E) of a body in orbit is the sum of its kinetic energy (KE) and gravitational potential energy (PE): E=KE+PE=21mv2−rGMm Using the average orbital speed equation: KE=21m(T2πr)2=T22π2mr2 Substituting T2=GM4π2r3 from Kepler's Third Law: KE=GM4π2r32π2mr2=2rGMm Thus, the total energy becomes: E=2rGMm−rGMm=−2rGMm This negative energy indicates a bound system, essential for maintaining stable orbits. 4. Perturbations and Orbital Stability In reality, orbits are subject to perturbations from various forces, such as gravitational influences from other celestial bodies, atmospheric drag (for low Earth orbits), and radiation pressure. These perturbations can alter the orbital speed and radius over time, affecting the average orbital speed. Advanced orbital mechanics involves calculating these perturbations to predict and maintain orbital stability. 5. Relativistic Corrections At speeds approaching the speed of light or in strong gravitational fields, Newtonian mechanics give way to Einstein's theory of relativity. Relativistic corrections to the average orbital speed become significant in such scenarios, altering the predictions made by the classical equation v=T2πr. These corrections are essential in high-precision applications like GPS satellite systems and understanding the orbits of objects near massive stars or black holes. 6. Tidal Forces and Orbital Decay Tidal forces between celestial bodies can lead to orbital decay, gradually reducing the orbital radius and altering the average orbital speed. For example, Earth’s tides affect the Moon’s orbit, causing it to slowly recede from Earth and its orbital period to increase. Understanding these long-term changes requires integrating the average orbital speed equation with models of tidal interactions. 7. Interdisciplinary Connections Astronomy: Calculating the orbital speeds of planets and stars to understand galactic dynamics. Engineering: Designing satellite orbits and space missions based on required speeds and periods. Environmental Science: Assessing the impacts of orbital changes on Earth's climate and space weather. Economics: Resource allocation for space missions, considering the costs associated with achieving specific orbital speeds. 8. Complex Problem-Solving Scenarios Problem 1: A satellite orbits Earth at a radius of 7×106 meters with an orbital period of 6000 seconds. Calculate its average orbital speed. Given: Orbital radius, r=7×106 meters Orbital period, T=6000 seconds Using the equation: v=6×1032π×7×106=6×10314π×106=614π×103≈7,330m/s- Thus, the average orbital speed is approximately 7.33 km/s. Problem 2: Determine the orbital period of a planet orbiting a star with mass 2×1030 kg at a distance of 1×1011 meters, given its average orbital speed is 3×104 m/s. Given: Orbital radius, r=1×1011 meters Average orbital speed, v=3×104 m/s Using the equation: T=v2πr=3×1042π×1×1011≈3×1046.2832×1011≈2.0944×107seconds- Converting seconds to years: T≈3.154×1072.0944×107≈0.664years- Thus, the orbital period is approximately 0.664 years. 9. Numerical Simulations and Modeling Modern astrophysics employs numerical simulations to model orbital dynamics beyond the scope of analytical equations. These simulations can incorporate factors like orbital eccentricity, multi-body interactions, and relativistic effects, providing more accurate predictions of average orbital speeds in complex systems. 10. Future Research Directions Exoplanetary Orbits: Studying average orbital speeds to detect and characterize planets beyond our solar system. Space Mission Design: Optimizing orbital parameters for deep space missions and asteroid deflection strategies. Gravitational Wave Astronomy: Understanding the orbital speeds involved in systems that emit detectable gravitational waves. Comparison Table \| \| \| \| --- \| Aspect \| Average Orbital Speed Equation \| Newtonian Derivation \| \| Definition \| v=T2πr \| v=rGM \| \| Primary Variables \| Orbital radius (r), Orbital period (T) \| Gravitational constant (G), Mass of central body (M), Orbital radius (r) \| \| Assumptions \| Circular orbit, Constant speed \| Circular orbit, Gravitational force provides centripetal force \| \| Applications \| Calculating average speed for satellites, planets \| Deriving orbital period from gravitational principles \| \| Limitations \| Only for circular orbits, ignores gravitational variations \| Assumes Newtonian mechanics, not applicable for relativistic speeds \| Summary and Key Takeaways The average orbital speed equation, v=T2πr, is essential for understanding celestial motion. Derived from circular motion principles, it links orbital radius and period to speed. Advanced concepts include Newtonian derivations, energy considerations, and relativistic effects. The equation is foundational for applications in astronomy, engineering, and beyond. Understanding its limitations and interdisciplinary connections enriches its practical utility. Coming Soon! Examiner Tip Tips Remember the Formula: Think of v=T2πr as the circumference over time. Unit Consistency: Always convert all units to SI units to avoid calculation errors. Practice Problems: Regularly solve different orbital speed problems to reinforce your understanding and prepare for exam variations. Did You Know Did You Know The International Space Station (ISS) travels at an average orbital speed of approximately 7.66 km/s, allowing it to circle the Earth roughly every 90 minutes. Additionally, some exoplanets have orbital speeds so high that they complete a full orbit around their star in just a few Earth days, drastically different from Earth's 365-day journey. Common Mistakes Common Mistakes Mistake 1: Confusing orbital radius (r) with orbital diameter. Incorrect: Using diameter in the equation v=T2πr. Correct: Ensure r is the radius, half of the diameter. Mistake 2: Using days instead of seconds for the orbital period (T). Incorrect: Plugging 365 days directly into the equation. Correct: Convert days to seconds before using T. FAQ What is average orbital speed? Average orbital speed is the constant speed needed for a celestial body to complete an orbit around another body in a given period, assuming a circular orbit. How do you derive the equation v=T2πr? By dividing the circumference of the orbit (2πr) by the orbital period (T), assuming uniform circular motion. Why is unit conversion important in orbital speed calculations? Consistent units, preferably SI units, ensure accurate calculations and prevent errors when applying the orbital speed equation. Can this equation be used for elliptical orbits? No, the equation assumes a circular orbit. Elliptical orbits require a different approach as speed varies at different points. What factors can alter the average orbital speed? Orbital radius, orbital period, and the mass of the central body are primary factors affecting average orbital speed. How is average orbital speed applied in satellite deployment? Engineers calculate the necessary speed to maintain a stable orbit, ensuring satellites neither fall back to Earth nor drift away into space. 1. Electricity and Magnetism 1.1.1 Structure and function of a split-ring commutator in motors 1.2.1 Principle of operation of an iron-core transformer 1.2.2 Equation for transformer efficiency: IpVp = IsVs 1.2.3 Equation for power loss in transmission lines: P = I²R 1.3.1 Magnetic forces due to interactions between magnetic fields 1.3.2 Relative strength of a magnetic field represented by field line spacing 1.4.1 Definition of charge in coulombs 1.4.2 Electric field as a region where a charge experiences a force 1.4.3 Electric field direction based on force on a positive charge 1.4.4 Electric field patterns: point charge, charged spheres, parallel plates 1.5.1 Equation for current: I = Q / t 1.5.2 Conventional current direction (positive to negative) 1.6.1 Equation for e.m.f.: E = W / Q 1.6.2 Equation for potential difference: V = W / Q 1.7.1 Current-voltage graphs for a resistor, filament lamp, and diode 1.7.2 Variation of resistance with length and cross-sectional area of wire 1.8.1 Kirchhoff’s first law: sum of currents at a junction 1.8.2 Kirchhoff’s second law: sum of voltages in a closed loop 1.8.3 Total resistance in parallel circuits 1.9.1 Equation for potential divider: R1/R2 = V1/V2 1.9.2 Use of a potential divider in circuits 1.10.1 Direction of induced e.m.f. opposes change causing it 1.10.2 Relative directions of force, field, and induced current 1.11.1 Graph of e.m.f. variation with time in a generator 1.12.1 Variation of magnetic field strength around wires and solenoids 1.13.1 Direction of force on moving charges in a magnetic field 2. Waves 2.1.1 Definition of monochromatic light as light of a single frequency 2.2.1 Speed of electromagnetic waves in vacuum (3.0 × 10^8 m/s) 2.2.2 Use of microwaves, radio waves, and visible light in communications 2.2.3 Difference between digital and analog signals 2.2.4 Transmission of sound as digital or analog signals 2.2.5 Advantages of digital signals: data transmission speed and accuracy 2.3.1 Definition of compression and rarefaction in sound waves 2.3.2 Speed of sound comparison in solids, liquids, and gases 2.3.3 Uses of ultrasound in non-destructive testing and medical scanning 2.4.1 Effect of wavelength on diffraction at an edge 2.4.2 Effect of wavelength and gap size on diffraction 2.5.1 Using ray diagrams for reflection by plane mirrors 2.6.1 Definition of refractive index: n = sin i / sin r 2.6.2 Equation for critical angle: n = 1 / sin c 2.6.3 Use of optical fibers in telecommunications 2.7.1 Using ray diagrams to show virtual images by converging lenses 2.7.2 Use of a single lens as a magnifying glass 2.7.3 Use of converging and diverging lenses to correct vision defects 3. Space Physics 3.1.1 Definition of a light-year and conversion to meters 3.1.2 Life cycle of a star, including formation, main sequence, red giant, white dwarf, supernova, neutron 3.2.1 Redshift as evidence for an expanding Universe 3.2.2 Cosmic Microwave Background Radiation (CMBR) as evidence for the Big Bang 3.2.3 Expansion of the Universe stretching CMBR into the microwave spectrum 3.2.4 Hubble’s Law: speed of a galaxy moving away is proportional to its distance 3.2.5 Hubble constant (H₀) and estimation of the Universe’s age using 1 / H₀ 3.2.6 Supernova brightness as a method to determine galactic distances 3.2.7 Current estimate for Hubble constant and its implications for cosmic expansion 3.3.1 Equation for average orbital speed: v = 2πr / T 3.4.1 Planets, minor planets, and comets have elliptical orbits 3.4.2 Sun is not at the exact center of elliptical orbits (except in nearly circular ones) 3.4.3 Analyzing planetary data: orbital distance, period, density, surface temperature 3.4.4 Gravitational field strength decreases with distance from the Sun 3.4.5 Planets travel faster when closer to the Sun due to conservation of energy 3.5.1 Nuclear fusion in the Sun as the source of energy release 4. Motion, Forces, and Energy 4.1.1 Equations for power: P = W / t and P = ΔE / t 4.2.1 Understanding scalar and vector quantities 4.2.2 Identifying scalar quantities: distance, speed, time, mass, energy, temperature 4.2.3 Identifying vector quantities: force, weight, velocity, acceleration, momentum, field strengths 4.2.4 Determining the resultant of two perpendicular vectors graphically or by calculation 4.3.1 Equation for pressure in fluids: Δp = ρgΔh 4.4.1 Definition of acceleration as change in velocity per unit time 4.4.2 Equation for acceleration: a = Δv / Δt 4.4.3 Determining acceleration from speed-time graphs 4.4.4 Understanding deceleration as negative acceleration 4.4.5 Describing motion under gravity with and without air resistance 4.5.1 Understanding weight as the effect of a gravitational field on mass 4.6.1 Determining whether one liquid will float on another using density data 4.7.1 Definition and use of spring constant: k = F / x 4.7.2 Understanding the concept of limit of proportionality on a load-extension graph 4.7.3 Equation of motion: F = ma (Force and acceleration in the same direction) 4.7.4 Understanding motion in a circular path due to perpendicular force 4.8.1 Applying the principle of moments to situations with multiple forces 4.8.2 Demonstration that an object in equilibrium has no resultant moment 4.9.1 Equation for momentum: p = mv 4.9.2 Equation for impulse: impulse = FΔt = Δ(mv) 4.9.3 Applying the principle of conservation of momentum in one dimension 4.9.4 Equation for resultant force: F = Δp / Δt 4.10.1 Equation for kinetic energy: Ek = 1/2 mv² 4.10.2 Equation for change in gravitational potential energy: ΔEp = mgΔh 4.10.3 Applying the conservation of energy principle to multi-stage processes 4.11.1 Equation for work done: W = Fd = ΔE 4.12.1 Understanding that almost all energy resources (except nuclear, geothermal, tidal) originate from th 4.12.2 Nuclear fusion as the energy source for the Sun 4.12.3 Current research on large-scale energy production from nuclear fusion 4.12.4 Equation for efficiency: Efficiency = (useful energy output / total energy input) × 100% 4.12.5 Equation for efficiency in terms of power: Efficiency = (useful power output / total power input) × 5. Nuclear Physics 5.1.1 Alpha particle scattering experiment supporting the nuclear model 5.1.2 Evidence from scattering experiment for a small, dense, positively charged nucleus 5.2.1 Processes of nuclear fission and nuclear fusion 5.2.2 Mass and energy changes in fission and fusion reactions 5.2.3 Relationship between proton number and charge on nucleus 5.2.4 Relationship between nucleon number and mass of nucleus 5.3.1 Correcting for background radiation in radioactivity measurements 5.4.1 Deflection of alpha, beta, and gamma radiation in electric and magnetic fields 5.4.2 Explanation of ionizing effects based on charge and kinetic energy 5.5.1 Explanation of unstable isotopes due to neutron excess or heavy nucleus 5.5.2 Nuclear changes occurring in alpha, beta, and gamma emission 5.5.3 Neutron decay equation: neutron → proton + electron 5.5.4 Writing nuclear equations for radioactive decay using nuclide notation 5.6.1 Calculation of half-life from raw data or decay curves 5.6.2 Use of radioisotopes in medical imaging and treatment 5.6.3 Use of radioisotopes in industrial thickness monitoring 5.6.4 Factors influencing choice of radioisotope for specific applications 5.7.1 Methods to minimize radiation exposure: time, distance, shielding 5.7.2 Protective measures for handling radioactive materials 5.7.3 Biological effects of ionizing radiation, including DNA damage and cancer risk 6. Thermal Physics 6.1.1 Forces and distances between particles affect properties of solids, liquids, and gases 6.1.2 Pressure in gases explained in terms of molecular collisions and force per unit area 6.1.3 Motion of microscopic particles due to collisions with smaller molecules (Brownian motion) 6.2.1 Equation for gas pressure-volume relationship: pV = constant (for a fixed mass at constant temperatu 6.3.1 Explanation of thermal expansion in terms of molecular motion and arrangement 6.4.1 Internal energy increase linked to an increase in particle kinetic energy 6.4.2 Definition of specific heat capacity: c = ΔE / (mΔθ) 6.4.3 Experiments to determine specific heat capacity of solids and liquids 6.5.1 Differences between boiling and evaporation 6.5.2 Effects of temperature, surface area, and air movement on evaporation 6.5.3 Cooling effect of evaporation and its applications 6.6.1 Conduction in solids explained through atomic vibrations and electron movement in metals 6.6.2 Why conduction is poor in liquids and gases 6.7.1 Constant temperature requires equal rates of energy absorption and emission 6.7.2 Effect of energy imbalance on object temperature 6.7.3 Earth’s temperature balance between absorbed and emitted radiation 6.7.4 Experiments comparing good and bad emitters of infrared radiation 6.7.5 Experiments comparing good and bad absorbers of infrared radiation 6.7.6 Rate of radiation emission depends on surface temperature and area 6.8.1 Complex applications of conduction, convection, and radiation in real-world scenarios Get PDF PDF Share Explore How would you like to practise? 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10883	https://www.omnicalculator.com/law-of-sines-ambiguous-case	Law of Sines: The Ambiguous Case Biology Chemistry Construction Conversion Ecology Everyday life Finance Food Health Math Physics Sports Statistics Other Law of Sines: The Ambiguous Case 1. Introduction: What is the law of sines? This article will teach you how to use the law of sines to solve ambiguous SSA (side-side-angle) cases. But before, let's start by recalling the law of sines. The law of sines applies to any triangle, equating the ratios of the sines of the angles to the lengths of the corresponding sides. a sin⁡α=b sin⁡β=c sin⁡γ\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} sin α a=sin β b=sin γ c If you want to know more about the law of sines, check out our detailed article: What is the Law of Sines? (Sine Rule Explained). In short, we can use the law of sines to find unknown angles and lengths in triangles where we know either two lengths and an opposite angle, or two angles and an opposite side. 🙋 If you want to explore the formula directly, try our law of sines calculator 🇺🇸. 2. When does the ambiguous case of the sine rule happen? Using the law of sines to determine an unknown side or angle can result in an ambiguous answer. The ambiguous case of the sine rule occurs under the following conditions: You only know the angle α\alpha α and sides a a a and c c c; The angle α α α is acute (α<9 0∘α \lt 90^\circ α<9 0∘); a a a is shorter than c c c (a<c a \lt c a<c); and a a a is longer than the altitude h h h from angle β\beta β, where h=c×sin⁡α h = c \times \sin \alpha h=c×sin α (or a<c×sin⁡α a \lt c \times \sin \alpha a<c×sin α). This situation leads to two possible solutions because the same sine value corresponds to two different angles: θ\theta θ and 18 0∘−θ 180^\circ - \theta 18 0∘−θ. It is important to note that the law of sines and the ambiguous case can only occur if we are given the lengths of two sides and an opposite angle; however, depending on the values, there are three possibilities to distinguish in this case: no triangle exists, one triangle exists, or two triangles exist. We have summarized the possible cases when two sides a a a and b b b, and an opposite angle α\alpha α are known in the table below. Law of sine and ambiguous case table. \| Case \| If α\alpha α is acute \| If α\alpha α is obtuse \| --- \| ab a \gt b a>b \| One triangle exists \| One triangle exists \| \| h<a<b h \lt a \lt b h<a<b \| Two triangles exist \| \| \| a=b a = b a=b \| \| No triangle exists \| 💡 Check out our Pythagorean theorem 🇺🇸 to learn more about triangles! 3. Example of the law of sines and the ambiguous case Suppose you have a triangle A B C ABC A BC with a=3 in a = 3 \ \text{in}a=3 in, b=9 in b = 9 \ \text{in}b=9 in, α=1 0∘\alpha = 10^\circ α=1 0∘. Let's determine all possible measures of β\beta β (to the nearest degree). Check how many triangles are possible. Since α\alpha α is acute, we calculate the altitude: h=b×sin⁡α=9 in×sin⁡(1 0∘)≈1.56 in\begin{align} h &= b \times \sin \alpha = 9 \ \text{in} \times \sin(10^\circ) \ &\approx 1.56 \ \text{in}\ \end{align}h=b×sin α=9 in×sin(1 0∘)≈1.56 in Because h<a<b h < a < b h<a<b, it is possible to build two different triangles. Apply the law of sines 🇺🇸: sin⁡β b=sin⁡α a sin⁡β 9 in=sin⁡(1 0∘)3 in sin⁡β=sin⁡(1 0∘)3×9 in sin⁡β≈0.521\begin{align} \frac{\sin \beta}{b} &= \frac{\sin \alpha}{a}\[1.2em] \frac{\sin \beta}{9\ \text{in}} &= \frac{\sin(10^\circ)}{3\ \text{in}}\[1.2em] \sin \beta &= \frac{\sin(10^\circ)}{3} \times 9\ \text{in}\[1.2em] \sin \beta &\approx 0.521 \end{align}b sin β9 in sin βsin β sin β=a sin α=3 in sin(1 0∘)=3 sin(1 0∘)×9 in≈0.521 Solve for β\beta β: β=sin⁡−1(0.521)β≈31.4∘\begin{align} \beta &= \sin^{-1}(0.521)\[.2em] \beta &\approx 31.4^\circ \end{align}β β=sin−1(0.521)≈31.4∘ Since sin⁡(θ)\sin(\theta)sin(θ) has the same value for θ\theta θ and 18 0∘−θ 180^\circ - \theta 18 0∘−θ, there is another possible solution: β=18 0∘−31.4∘=148.6∘ \beta = 180^\circ - 31.4^\circ = 148.6^\circ β=18 0∘−31.4∘=148.6∘ Check if the triangle angle sum is less than 18 0∘180^\circ 18 0∘: α+β=1 0∘+148.6∘=158.6∘<18 0∘\alpha + \beta = 10^\circ + 148.6^\circ = 158.6^\circ < 180^\circ α+β=1 0∘+148.6∘=158.6∘<18 0∘ So this value of β\beta β is also possible. There are two possible measures for β\beta β: β≈31.4∘\beta \approx 31.4^\circ β≈31.4∘ or β≈148.6∘\beta\approx 148.6^\circ β≈148.6∘. Hence, two distinct triangles can be formed. 🙋 Interested in geometry in general? Try our triangle area calculator 🇺🇸. 4. Conclusion: Law of sines and the ambiguous case The ambiguous case of the sine rule occurs in the SSA configuration, when two sides and a non-included angle are given. In this situation, the law of sines may produce two triangles, one triangle, or no triangle at all. Knowing how to test for acute and obtuse angles allows you to interpret the results correctly. FAQs ### Why does the law of sines sometimes give two solutions? Because the sine of an angle and its supplement are equal, i.e., sin(θ) = sin(180° − θ). This means that when solving for an angle with the sine rule, both an acute and an obtuse angle might satisfy the equation. ### In which case does the ambiguous case of the sine rule occur? The ambiguous case of the sine rule occurs only in the SSA configuration when two sides and a non-included angle are known. It does not appear in the ASA, AAS, SAS, or SSS cases. ### Can the ambiguous case happen in right triangles? No. In right triangles, the sine rule gives a unique solution, because the right angle eliminates the possibility of an obtuse solution. The ambiguous case only arises in oblique (non-right) triangles. This article was written by Claudia Herambourg and reviewed by Steven Wooding. We make it count! Calculator Categories Biology Chemistry Construction Conversion Ecology Everyday life Finance Food Health Math Physics Sports Statistics Other Discover Omni Press Editorial policies Partnerships Meet Omni About Resource library Contact We're hiring!
10884	https://sjr.uptodate.com/contents/hand-foot-and-mouth-disease-and-herpangina/print	Hand, foot, and mouth disease and herpangina - UpToDate Subscribe Sign in English Deutsch Español 日本語 Português Why UpToDate? Product Editorial Subscription Options Hand, foot, and mouth disease and herpangina View Topic Loading... Share Font Size Small Normal Large Bookmark Rate Feedback Tools Formulary drug information for this topic No drug references linked in this topic. Find in topic Formulary Print Share Feedback Font Size Small Normal Large Print Options [x] Text Print Official reprint from UpToDate® www.uptodate.com © 2025 UpToDate, Inc. and/or its affiliates. All Rights Reserved. Official reprint from UpToDate® www.uptodate.com © 2025 UpToDate, Inc. and/or its affiliates. All Rights Reserved. Hand, foot, and mouth disease and herpangina Authors:José R Romero, MD, FAAP, FIDSA, FPIDS, FAAASLiset Olarte, MD, MScSection Editor:Morven S Edwards, MDDeputy Editor:Diane Blake, MD Literature review current through:Aug 2025. This topic last updated:Jul 08, 2025. INTRODUCTION Hand, foot, and mouth disease (HFMD) is a clinical syndrome characterized by an oral enanthem and a macular, maculopapular, or vesicular rash of the hands and feet (and possibly other locations) . HFMD is one of the most recognizable viral exanthems in children and adults . HFMD was first described in a summer outbreak that occurred in Toronto, Canada in 1957 and was caused by coxsackievirus A16 . Since then, at least 15 other enterovirus serotypes have been shown to cause HFMD, most commonly Coxsackievirus A serotypes. Herpangina is a benign clinical syndrome characterized by fever and a painful papulo-vesiculo-ulcerative oral enanthem . It can be clinically differentiated from HFMD and primary herpetic gingivostomatitis . Herpangina was first described in the 1920s, but the viral etiology was not established until 1951 [4,6,7]. Herpangina is caused by at least 22 enterovirus serotypes, most commonly Coxsackievirus A serotypes. An overview of HFMD and herpangina will be presented here. Other enterovirus infections are discussed separately. (See "Enterovirus and parechovirus infections: Epidemiology and pathogenesis" and "Enterovirus and parechovirus infections: Clinical features, laboratory diagnosis, treatment, and prevention".) PATHOGENESIS Human enterovirus infection occurs after oral ingestion of virus that is shed from the gastrointestinal or upper respiratory tract of infected individuals (ie, via the ingestion of fecal material, oral secretions, or, for some serotypes, respiratory secretions) . Human enterovirus infection also may occur following contact with vesicle fluid or oral and respiratory secretions [3,8,9]. Virus may be detected in the stool for six weeks and sometimes for several months after infection. The duration of shedding from the oropharynx is generally less than four weeks. Prolonged shedding in the stool and the innate environmental stability of the enteroviruses favors their transmission. To continue reading this article, you must sign in. For more information or to purchase a personal subscription, click below on the option that best describes you: Medical Professional Resident, Fellow or Student Hospital or Institution Group Practice Disclaimer: This generalized information is a limited summary of diagnosis, treatment, and/or medication information. It is not meant to be comprehensive and should be used as a tool to help the user understand and/or assess potential diagnostic and treatment options. It does NOT include all information about conditions, treatments, medications, side effects, or risks that may apply to a specific patient. It is not intended to be medical advice or a substitute for the medical advice, diagnosis, or treatment of a health care provider based on the health care provider's examination and assessment of a patient's specific and unique circumstances. Patients must speak with a health care provider for complete information about their health, medical questions, and treatment options, including any risks or benefits regarding use of medications. This information does not endorse any treatments or medications as safe, effective, or approved for treating a specific patient. UpToDate, Inc. and its affiliates disclaim any warranty or liability relating to this information or the use thereof. The use of this information is governed by the Terms of Use, available at 2025© UpToDate, Inc. and its affiliates and/or licensors. All rights reserved. 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10885	https://clinicalview.gehealthcare.com/quick-guide/maximizing-arrhythmia-detection-ek-pro-arrhythmia-algorithm	Contact us Subscribe to Newsletter Clinical View By clinicians for clinicians Share "Share to Facebook" "Share to X" "Share to LinkedIn" "Share to WhatsApp" "Share to Email" Add to bookmarks Subscribe to newsletter Subscribe to Newsletter See all resources article Maximizing Arrhythmia Detection with the EK-Pro Arrhythmia Algorithm Because a stable cardiac rhythm is essential to effective contractility and adequate cardiac output, fast and accurate arrhythmia detection is critical. The EK-Pro algorithm uses four leads for arrhythmia detection, exceeding the American Heart Association’s minimum recommendation. This Quick Guide from GE Healthcare provides advice on applying the EK-Pro algorithm to help achieve reliable arrhythmia interpretation. It covers these practical aspects of arrhythmia monitoring.Signal quality. Careful skin preparation and the use of high-quality electrodes are keys to ensuring a good ECG signal.Relearning the patient’s QRS pattern. During ECG monitoring, it may be necessary to use the manual Relearn QRS feature when a substantial change in the patient’s ECG pattern has occurred. The manual relearn can correct false arrhythmia alarms and heart rate values. In addition, it restores the ST measurements. Pacemaker detection. Pacemaker detection should be activated when patients with pacemakers are monitored. When using a CARESCAPE™ Patient Data Module (PDM), Tram™ Module or Combination monitoring, pacemaker detection may need to be manually activated.ECG arrhythmia alarms. GE monitors have three arrhythmia analysis modes: Off, Lethal and Full. The Lethal mode is the standard; the Off and Full modes are optional choices. The Lethal mode detects asystole, ventricular fibrillation/ventricular tachycardia and ventricular tachycardia. For the most current arrhythmia calls, please consult the user manual specific to your current monitor for the details. ECG and cardiac care Rhythm disorders Image 5 min read Jun 23, 2020 Download article resources (195.19 KB) 6likes Arrhythmia Detection Arrhythmia means any disturbance or irregularity of the cardiac rhythm. A stable cardiac rhythm is essential to maintain effective contractility and adequate cardiac output. Maintaining an optimal cardiac output leads to adequate organ perfusion and can improve a patient’s clinical outcomes and survival. Therefore, fast and accurate detection of arrhythmias is critical. Continuous ECG monitoring allows unique views of the electrical activity of the heart from two planes, frontal and horizontal. Viewing the electrical activity of the heart from multiple angles can be very useful to the clinician especially when it applies to arrhythmia detection. Our monitoring system allows the clinician to monitor in multi-lead or single-lead analysis. In general, the more leads used in the arrhythmia analysis leads to a more reliable arrhythmia interpretation. Multi-Lead Analysis examines ECG leads l, ll, lll and V lead (whether they are displayed or not) to help eliminate false alarms and improve the ability of the system to: Detect beats that are not seen in all leads. Discriminate artifact that appears in one lead compared to the other leads. Provide a smart-lead fail feature, where the failed lead is identified, and, if available, another lead is provided for analysis. Non-disturbed ECG arrhythmia analysis even after a lead change. Sometimes ventricular beats can be more obvious in some of the leads than in others where the changes in morphology are minor. It is also possible that QRS amplitude can be low in one lead and normal in others. Therefore, the sensitivity of the algorithm may increase when more than one lead is used. The recognition of ventricular beats may be improved by multi-lead monitoring, and the same applies to QRS detection. The decision between normal and ventricular beats may be more reliable when information from more than one lead is available. The EK-Pro arrhythmia algorithm uses I, II, III, and the V/VA lead for arrhythmia detection. The American Heart Association (AHA) has recommended that two or preferably three or more leads should be displayed and monitored simultaneously. Exceeding the AHA’s minimum recommendation, the distinctive EK-Pro algorithm utilizes four simultaneous leads for analysis. Even though the multi-lead ECG analysis is preferred and recommended, the algorithm utilizes single lead arrhythmia analysis automatically when 3 lead wired ECG is used. Single lead analyses may also be used with multi-lead monitoring. This is done using manual selection from the user interface. Single-lead usage can be considered, for example, if majority of the leads are affected with noisy signal or low amplitude QRS resulting in false ECG alarms and incorrect HR readings. In this case, the user should select the best available lead to be displayed on the monitor and select single lead mode from the user interface. EK-Pro algorithm, in single lead mode, uses lead II as a default. As an alternative the user is able to choose lead I, III or Va as a source for single lead arrhythmia analysis. Mirvis. D. M. et al. Instrumentation and practice standards for electrocardiographic monitoring in special care units. A report for health professionals by a Task Force of the Council on Clinical Cardiology. American Heart Association 79, 464-471 (February 1989). How are arrhythmias monitored? The EK-Pro algorithm processing can be represented by three major phases that help determine heart rate, ST segment deviation, and arrhythmia detection: Continuous Correlation Event measurement and Classification Contextual Analysis Continuous Correlation filters the incoming signal before detecting the incoming events as part of the QRS detection process. Each detected QRS complex is compared to previously detected QRS complexes and the templates are updated accordingly, or a new template is created and the beat is added to the beat list. Event measurement and classification is a process by which the multi-lead waveform templates used for beat classification and measurement accurately track subtle, progressive changes in beat shapes. With this technology, automated measurements can be made consistently and accurately since waveform artifact is effectively minimized in the waveform templates by using this updating process. Contextual Analysis allows the algorithm to use information gained from neighboring beats for identifying arrhythmia events. This allows the algorithm to evaluate features and information about the rhythm to make the best possible decision regarding the beat’s origin (e.g., R-R interval, early, late, wide, narrow and run length). A beat is classified as normal, ventricular or artifact and then in conjunction with the decision of the past beat templates, the EK Pro algorithm makes a final determination of the arrhythmia detected and the heart rate. These features, Continuous Correlation, Incremental Template Updating and Contextual Analysis all work in concert to provide accurate arrhythmia recognition. The EK Pro arrhythmia criteria are listed in the table at the end of this document. Practical aspects in arrhythmia monitoring Signal quality Careful skin preparation and the use of high-quality electrodes are key to ensuring a good ECG signal especially when utilizing arrhythmia monitoring. A good signal helps to ensure accurate arrhythmia detection and helps decrease false and nuisance alarms. If there is artifact in all of the analyzed ECG leads, the EK-Pro algorithm provides a noisy ECG message. If the condition continues the algorithm will provide a “Arrhythmia Paused” or “Arrhy Suspend” message depending on the GE monitor in use. At this point all arrhythmia detection will be suspended until the integrity of the ECG signal is restored. Relearning the patient’s QRS pattern Automatic relearning of the patient’s QRS pattern takes place under the following conditions: The measurement mode changes between the 3–lead mode and any other lead mode. The ECG 1 Lead selection is changed in the 3–lead mode. The Va lead selection is changed in the 5– and 6–lead modes. The ECG cable is connected (with PSM and PDM modules). The Lead Analysis setting is changed from Multi lead to Single lead. During ECG monitoring, you may need to use the manual Relearn QRS feature when a substantial change in the patient’s ECG pattern has occurred (e.g. due to change of electrodes or electrode locations). The manual relearn of patient’s QRS pattern might correct false arrhythmia alarms and heart rate values. In addition, it restores the ST measurements. Relearning the QRS morphology can be completed manually in the “Advanced” menu of the ECG parameter window. Relearning typically takes 30 seconds or less. Pacemaker Detection Pacemaker detection should be activated when patients with pacemakers are monitored. Pacemaker detection is always activated with E-Modules (i.e. E-PSM or E-PRESTN). When utilizing a CARESCAPE™ Patient Data Module (PDM), Tram™ Module or Combination monitoring pacemaker detection may need to be manually activated. The Patient Data Module and Tram x51 series modules use multi-vector pace detection. The PDM picks 3 channels to detect pacemaker pulse amplitude. Here are some additional guidelines for monitoring pacemaker patients when using one of these modules. When using the 5- or 6-leadwire patient cables with all the electrodes attached, pace detection occurs on three ECG leads simultaneously. Channels I, II and V1 are used for pace detection as a default, when 5 or 6 lead wire cable is connected. If these leads are noisy or not available, multi-vector pace detection switches to better signal or next available leads. Pace detection switches to single-lead automatically when a 3-leadwire patient cable is used. Pace detection in combination monitoring or telemetry utilizes multi-vector pace detection. The Apex Pro™ telemetry system picks 2 channels to detect pacemaker pulse amplitude. Lead II and Va are used for pace detection as a default, when 5- or 6- lead wire cable is connected to the transmitter. ECG Arrhythmia alarms In GE monitors there are three arrhythmia analysis modes: Off, Lethal and Full. The Lethal mode is the standard, where the Off and Full are optional choices. The Lethal mode detects asystole, ventricular fibrillation/ventricular tachycardia and ventricular tachycardia. Please note, that bradycardia is considered as lethal arrhythmia in neonatal mode only when using Tram module.Different generations of EK-Pro algorithms are used in CARESCAPE Bx50 v2 monitors, Solar™ monitors, Dash™ monitors, ApexPro telemetry and CARESCAPE Telemetry systems: CARESCAPE Bx50 monitors with software v2 in conjunction with PDM and PSM modules are utilizing EK-Pro v13 algorithm. If Tram module (CARESCAPE B850 only) or telemetry in combo mode is used with CARESCAPE Monitors, the ECG algorithm in use is EK-Pro v11. Solar Monitors, Dash monitors, Transport Pro™ and Telemetry systems utilize EK-Pro v11 algorithm. The details of arrhythmia alarms criteria are described in the following table. Please see pdf file below! For the most current arrhythmia calls, please consult the user manual specific toyour current monitor for the details. Additional resources For white papers, guides and other instructive materials about our clinical measurements, technologies and applications, please visit Latest content View all content 1 of Image Infographic Sep 23, 2025 5 min ### Neonatal anesthesia: what you need to know In the United States alone, approximately six million children, including 1.5 million infants, undergo surgical procedures under... 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10886	https://www.youtube.com/watch?v=a5_ApJrEZLI	Evaluating Piecewise Functions [fbt] Fort Bend Tutoring 87600 subscribers 217 likes Description 15083 views Posted: 2 Nov 2014 This video by Fort Bend Tutoring shows the process of evaluating piecewise functions. There are four (4) examples in all. The lesson is instructed by Larry "Mr. Whitt" Whittington. Intro/Outro by Perry "Lelo" Graham. Subscribe to Fort Bend Tutoring [fbt] here: Check out our Fort Bend Tutoring Amazon Affiliate Store for recommendations on products and textbooks to help you in your academic endeavors! Graphing Piecewise Functions Please donate to assist us in bringing the world more free videos through our YouTube Channel using the link: This Donation button is secured using PayPal. 33 comments Transcript: do you feel like installing a concrete swimming pool in your backyard and then drowning your math textbook and your math homework in it that's Grand we can help you avoid getting wet and costly lawyer fees with this video from Fort Ben tutoring and Mr wit hello ladies and gentlemen this is Mr wit with Fort Ben tutoring fbt and today's tutorial is going to be about evaluating peace wise functions yeah yeah that's right functions where they have cobbled together parts of other functions to make one function yeah they pieced it together thus the name piecewise function all right so here we have our first problem so what you're looking at here is a peie wise function in function notation that's right that f ofx means y f ofx g ofx h of X all of those are just names of the functions so basically the says y = x - 2 as long as X is less than 3 and Y = 5 - x as long as X is greater than or equal to three so basically this all depends upon the domain and that's why when you're evaluating pwi's functions all you're going to be concerned about is the conditions at first so for our first problem right here where it says F of5 it's asking you what is the yvalue when the x is is5 so first begin with your conditions it says to use x - 2 whenever X is less than 3 then your function says use 5 - x as long as X is greater than or equal to 3 so what we realize in our first problem is that our x value that they're referring to is5 well5 is definitely less than three so we're going to be using this first part of the function the x - 2 so you'll be replacing the X with5 so you'll write it as --5 - 2 and then simplify that we know that like Signs add so your answer is going to be -7 all right just like that and I don't know why this equal sign looks the way it does but it's ugly I'm going to fix it all right let me let me there you go trim that up there you go nice little haircut for you all right so I just chopped off part of the leg of that equal sign and now it's even now all right so here our answer is7 I'm going to put a red box around this Red Box in it just like that ladies and gentlemen that was part a of this problem the next thing they want to do is they want us to find what f of1 is that means your x value is NE -1 find the yvalue when X is1 so we always start by looking at the condition here our value is -1 well that value ne1 is less than three so once again we'll be using that x - 2 in order to evaluate the function so we'll be replacing x with -1 so you'll rewrite it as -1 - 2 and then you're going to simplify mhm so -1 - 2 is going to give you -3 so then Red Box it because this is the answer there you go now we have part C and in part C our x value that we're plugging in is zero so once again we go to our conditions the condition say to use x - 2 whenever X is less than 3 to use 5 - x whenever X is greater than or equal to 3 so 0 is less than three so once again we're going to the x - 2 here so replace the X with zero so we'll rewrite this as 0 -2 and then simplify that to get -2 then go ahead and give right up your answer there you go red boxed it all right now we have D part D and it says find the result when X is three so back to your conditions all right so here it says to use the first one whenever X is less than three but here our value for x is three so notice that our condition says whenever X is greater than or equal to three to use the 5 - x so what we'll do is we'll replace the X in this second part of your function here so you'll rewrite this as 5 - 3 so the answer when simplified is 2 two all right so there we have it ladies and gentlemen let me just box this up for you there you go that's your answer for Part D finally for part e here our value of x that we're plugging in is five remember based on your conditions whenever X is greater than or equal to three we should use the second part of this function here so we'll be plugging in five into the 5 - x here all right replacing that x with five so we can rewrite this as 5 - five and then simplify this to get zero and that's it I'm boxing this up just like that all right so that answer there is zero yeah that was problem number one let's move on to problem number two now all right problem number two we have f ofx = - 12x^2 + 2 when X is less than or equal to 2 and then we have 12x when X is greater than 2 they want to know the Y value when x = -4 in part A so the first thing you check is which part of the function you use whenever X is -4 so the 12x^2 + 2 should be used whenever X is less than or equal to 2 well -4 is definitely less than or equal to two so that means we're going to be plugging in -4 into this part of the function so rewriting that I'll have -12 -4 2ar + 2 mhm just like that then using your order of operations we have - one2 -42 is 16 thank you very much + 2 then we know that -2 16 is8 and so we have -8 + 2 and this equals -6 and that's the answer all right let's go ahead and box that up readed boxing it so your y value when X is -4 is -6 so our x value in problem B is going to be -2 so that means we have the same situation going on meaning that we use the first part of the pie wise function because our value of x is less than or equal to two so all you got to do is plug it in so plugging this in we have -2 -2 2 + 2 all right we'll go ahead and simplify this knowing that -22 is 4 and then -2 4 is -2 and then -2 + 2 you had zero Okay so zero is the answer let's go ahead and box it up boxing up my answers that's what I do all right on to part C here part C it says X is zero find out the value of the function when x equals 0 so going back to your conditions we'll need to use which part the first part or the second part oh yeah we'll need to use the first part why because our condition says to use the first part whenever X is less than or equal to two and Z is less than or equal to two ladies and gentlemen mhm so so go ahead and plug that in all right I love plugging in zero because it's easy so we have -2 0 sared + 2 all right so that means that 0 squar is 0 and then 0 -2 or you could say 1/2 0 it's still zero it's still zero then you'll add two to that 0 + 2 is two the answer is two red box it that's the answer to C okay so we're now moving on to D ladies and gentlemen to D all right our x value is two okay so since our x value is two let's check out our conditions our conditions state that we should use the first part of the function whenever X is less than or equal to two mm it says to use the second part of the function whenever X is greater than two mhm since our value of x is two ladies and gentlemen we should definitely use the first part of the function because our condition states that X is less than or equal to two when you're using that first part of the function and that's exactly what we have we have a value of two so that satisfies our first condition so plug it in there all right so let's do just that we'll have the - one2 2^ 2 + 2 simplifying this we know that 2^ 2 is 4 and then - 1/2 4 is -2 and then -2 + 2 yeah that's zero the answer is zero boxing it up red boxing it all right finally for E we have a value of x that's pos4 based on your conditions here we know that we should be using the second part of our function why because four is greater than two so once again you always start by looking at the conditions First Once you found out what your value of x is that you're plugging in mhm and we're looking in a positive four here and four is definitely greater than two it's it's greater than two okay four is greater than two don't test me got it thank you now plug that into this second part of the function okay there you go so we're taking half let's go ahead and rewrite this here all right so we'll be taking half of four there you go all right how you like that so half of four is two yes 1 half time 4 is two there you go that's the answer that is the answer I'm going to box this up done and done so that completes problem number two there you go moving along to problem number three all right F ofx = -2X whenever X is less than -3 F ofx = 3x - 1 whenever X is between and including the values -3 and 2 mhm and we'll be using -4x for the value of f ofx whenever X is greater than 2 that's our pie wise function make sure I pay attention to the conditions let's check it out we're going to start out with part A where it's asking us to plug in5 for our value of x we'll start with the conditions first because we need to know which part of the function to use notice that this piecewise function is comprised of three separate parts all right three separate rules based on our value within the domain so5 mhm 5 is less than -3 so I definitely need to use this first part of the function so you'll replace the value of x in -2X with the -5 so let's rewrite that we'll have -2 -5 simplifying this -2 5 is positive1 it is positive 10 there you go I'm going to box this up because it's the answer there you go that's part A I am moving on to Part B now moving on to Part B when X is1 mhm when X is1 is -1 less than -3 no it's not no does negative-1 lie within the interval -3 to 2 yes it does that means I need to use the second part of this function all right we'll be replacing the X here in three x -1 with -1 why because1 lies between -3 and 2 there you go so let's see what happens plugging this in we'll have 3 -1 - 1 simplifying this 3 -1 is -3 and then -3 - 1 gives us -44 all right let's box it up read boxing it then in part C we have a value of x which is -3 that we will be plugging in here okay based on our conditions when x = -3 we'll be using the second part of the function why because here our value says that X has to be greater than or equal to -3 and less than or equal to 2 in order to use the second rule of our pie wise function so replace our value of x with -3 and that second part of the function there all right so here we have it we have 3 -3 - 1 and then simplifying this 3 -3 is going to be 9 -9 - 1 is -10 and this is our answer all right moving on in part D here we have F of two meaning we want to find out the Y value when x equals to 2 so X equaling to two means that we're going to be using which part of our piecewise function the first part the second part the third part uh if it equals two we need to use the second part because notice that this interval here includes the value of two so we'll be plugging in two into the second part of that piecewise function all right so I'll rewrite it as 3 2 - 1 all right and so 3 2 is 6 and then 6 - 1 is 5 so our answer here is positive 5 all right finally we'll be plugging in four for X So based on our conditions here from the original problem we know that positive4 is greater than two all right so that means we have to use the third part of our P wise function so plugging in 4 into the third part of the function we'll rewrite it as -4 4 all right just like that and then we know that -4 4 is just -16 yes there you have it and those are the answers for c d and e and I'mma box them all up that's right I didn't forget so we got a box for you a box for you and a box for you I feel like Oprah there you go so there you have it ladies and gentlemen that's going to be problem number three finally we're going to be hitting up problem number four now all right let's turn to page all right problem number four here we go here we go here we go here we go we're evaluating right okay so here we have problem number four we're asked to find the Y value when X = 6 our conditions are not defined when x = -6 so as far as we know our yalue for when xal -6 does not exist so depending on your teacher's preference you can state that this answer does not exist or you can say that it is undefined or you can say that there is no solution because our piecewise function is not defined for when a value of xal -6 all right it's not part of the original domain it's outside of the domain that's why any of these all right would be acceptable all right so it does not exist it's the null set it's an empty set it's no solution for that problem okay so there you go you just can't do it so we're moving on in Part B it asks us to find the value of y when X is -3 well when X is -3 we need to use the first part of the function so replacing this x value in that first part of the function we'll have 510 time -3 2 all right well using your order of operations here you know that -3 2ar means -3 -3 which is posive 9 then 9 510 is going to give you a result of 4 and 510 all right so this is the answer for Part B 4 and 510 in other words multiplying by 510 is the same as dividing by two or multiplying by 1/2 all right and half of nine is four and 510 now for part C part C we have a value of x that is -2 whenever X is -2 let's check our conditions here we should be using the first part of the function why because this compound inequality has a less than or equal to -2 value here that means that when x equals -2 all right not just when it's greater than -2 when it's equal to -2 we're still using the first part of the function so we'll be plugging in our value of -2 into that first part so I'll be writing 510 -2 2 and then simplifying this -2 2 is pos4 and then 4 a half or 4 5/10 is going to give us 2 and that's it that's the answer for part C next we have part D where we have zero being replaced for X all right so find out the y-value whenever x equals 0 that's the F of0 value that they're looking for so looking at our conditions first we'll find that zero lies in this region right here in this interval right here for X it's between -2 and positive2 so we use the second part of our pie wise function and replacing x with zero yeah you got it the answer is just zero all you had to do is replace x with zero on that one I like that part let's have all the problems look like that right okay so finally we're looking at e part e they want us to find out the yvalue when x equals 3 when x = 3 is going to lie in the interval between 2 and 4 all right so it's right in here it's that third part of the pie wise function that we need to use so replace your X in that x^2 - 4 with three let's go ahead and write that up so we'll rewrite it as 3^ 2 - 4 all right and 3^ 2 - 4 means 9 - 4 and 9 - 4 is always 5 okay there you go there you go you couldn't see that huh there you go so the answer is five let's go ahead and box up all these answers I've been dying to do this here we go so we'll just say that this is the null set all right or you could say it's does not exist or we can say it's no solution so find out which one your teacher prefers there you go that's the one you'll mark down for part A then for Part B we said our result was 4 and 510 part C it's two uh D yeah d d is going to be zero and then finally part e is five all right well ladies and gentlemen that concludes this video on evaluating peace wise functions I had so much fun doing that with you thanks appreciate that and could you please rate comment and subscribe to our YouTube channel and as always if you're able please donate it helps us bring you more free math tutorials from me Mr wit and Fort Ben tutoring peace thanks for watching we feel great knowing that you got some help and you're safe and sound now if you'd be so kind as to like this video subscribe to our YouTube channel Fort B tutoring and like us on Facebook we be much obliged
10887	https://artofproblemsolving.com/wiki/index.php/Geometric_sequence?srsltid=AfmBOorZICCso7vguHw_1s4Seu_sMXDnA2-uGi9PlC46uFAhr3nQ9dR6	Art of Problem Solving Geometric sequence - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Geometric sequence Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Geometric sequence In algebra, a geometric sequence, sometimes called a geometric progression, is a sequence of numbers such that the ratio between any two consecutive terms is constant. This constant is called the common ratio of the sequence. For example, is a geometric sequence with common ratio and is a geometric sequence with common ratio ; however, and are not geometric sequences, as the ratio between consecutive terms varies. More formally, the sequence is a geometric progression if and only if . A similar definition holds for infinite geometric sequences. It appears most frequently in its three-term form: namely, that constants , , and are in geometric progression if and only if . Contents [hide] 1 Properties 2 Sum 2.1 Finite 2.2 Infinite 3 Problems 3.1 Introductory 3.2 Intermediate 4 See also Properties Because each term is a common multiple of the one before it, every term of a geometric sequence can be expressed as the sum of the first term and a multiple of the common ratio. Let be the first term, be the th term, and be the common ratio of any geometric sequence; then, . A common lemma is that a sequence is in geometric progression if and only if is the geometric mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of geometric sequences. Sum A geometric series is the sum of all the terms of a geometric sequence. They come in two varieties, both of which have their own formulas: finitely or infinitely many terms. Finite A finite geometric series with first term , common ratio not equal to one, and total terms has a value equal to . Proof: Let the geometric series have value . Then Factoring out , mulltiplying both sides by , and using the difference of powers factorization yields Dividing both sides by yields , as desired. Infinite An infinite geometric series converges if and only if ; if this condition is satisfied, the series converges to . Proof: The proof that the series convergence if and only if is an easy application of the ratio test from calculus; thus, such a proof is beyond the scope of this article. If one assumes convergence, there is an elementary proof of the formula that uses telescoping. Using the terms defined above, Multiplying both sides by and adding , we find that Thus, , and so . Problems Here are some problems with solutions that utilize geometric sequences and series. Introductory 2025 AMC 8 Problem 20 Intermediate 1965 AHSME Problem 36 2005 AIME II Problem 3 2007 AIME II Problem 12 See also Arithmetic sequence Harmonic sequence Sequence Series Retrieved from " Categories: Algebra Sequences and series Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
10888	https://artofproblemsolving.com/wiki/index.php/Square_root_property?srsltid=AfmBOorMZvoUSH6d75duaDC5a4R5ivSFu6Q2HV5KEDdYwDONU6X7oVUZ	Art of Problem Solving Square root property - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Square root property Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Square root property The square root property is a method for solving quadratic equations. Method First we must convert the quadratic to either the form or . First format take the square root of both sides so the roots are and . Second format take the square root of both sides isolate and those are the roots. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
10889	https://www.studocu.com/en-us/document/indiana-wesleyan-university/cellular-biology/ch-13-objectives-summary-molecular-biology-of-the-cell/2198149	CH 13 Objectives - Summary Molecular Biology of the Cell - CELL BIOLOGY LECTURE OBJECTIVES Chapter - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI CH 13 Objectives - Summary Molecular Biology of the Cell Main material covered in this chapter Course Cellular Biology (BIO 351) 48 documents University Indiana Wesleyan University Academic year:2015/2016 Uploaded by: Anonymous Student Indiana Wesleyan University Recommended for you 6 molecular biology of the cell chapter 3 summary Molecular Biology of the Cell Summaries 100% (25) 8 Summary Molecular Biology of the Cell chapter 4 and 6 Biochemistry Summaries 98% (46) 5 molecular biology of the cell chapter 4 summary Molecular Biology of the Cell Summaries 100% (22) 13 molecular biology of the cell chapter 6 summary Molecular Biology of the Cell Summaries 100% (20) 6 molecular biology of the cell chapter 5 summary Molecular Biology of the Cell Summaries 100% (15) Comments Please sign in or register to post comments. Report Document Students also viewed CH 15 Objectives - Main material covered in this chapter CH 16 Objectives - Main material covered in this chapter CH 17 Objectives - Main material covered in this chapter CH 18 Objectives - Main material covered in this chapter CH 19 Objectives - Main material covered in this chapter CH 20 Objectives - Main material covered in this chapter Related documents CH 02 Objectives 2015 - Summary Molecular Biology of the Cell CH 03 Objectives 2015 - Summary Molecular Biology of the Cell CH 04 Objectives 2015 - Summary Molecular Biology of the Cell CH 05 Objectives 2015 - Summary Molecular Biology of the Cell CH 06 Objectives 2015 - Main material covered in this chapter CH 14 Objectives - Main material covered in this chapter Related Studylists Alberts Advanced Cell Biology - 17/12/2024 Preview text CELL BIOLOGY LECTURE OBJECTIVES Chapter 13: Intracellular Vesicular Traffic After attending this series of lectures and studying the text, lecture presentations, and notes, the student should be able to: State the central theme of vesicular transport.  Vesicular transport is the predominant mechanism for exchange of proteins and lipids between membrane-bound organelles in eukaryotic cells. Describe the 3 major pathways involved in vesicular transport.  Biosynthetic-Secretory o Make and deliver to extracellular o Leads outward from the ER toward the golgi apparatus and cell surface, with a side route leading to lysosomes  Endocytic o Bring outside in o Leads inward from the plasma membrane  Retrieval o Take membrane back to compartment where it came from o Balance the flow of membrane between compartments in the opposite directions, bringing membrane and selected proteins back to the compartment of origin Describe a clathrin-coated vesicle. Include in your description what the donor and target compartments are for these vesicles, and functional descriptions of the following: triskelion, adaptor proteins, cargo receptor, dynamin GTPase, PIP phosphatase, Hsp70, and auxillin.  Clathrin Coated Vesicle: transport from golgi apparatus and plasma membrane o Donor compartment (Golgi and Plasma Membrane)  Buds membrane-bounded transport vesicles that carry cargo destined for acceptor compartments. o Target compartment (cell surface or lysosome)  Target of the vesicle; end location o Triskelion (falls off)  Formed by clathrin  3 legged structure  Determine geometry  Assemble into coated pits on cytosolic membrane surface  3 heavy and 3 light chains o Adaptor proteins (falls off)  Bridge between clathrin and cargo receptor  Discrete inner layer of the coat, between clathrin cage and membrane  Bind clathrin coat to the membrane and trap various transmembrane proteins, including receptors that capture soluble cargo molecules inside the vesicle o Cargo receptor  Carries desired soluble proteins into vesicles o Dynamin GTPase  Required for pinching off  Uncoating: o PIP phosphatase  Depletes PI(4,5)P 2 from membrane  Weakens binding of adaptor proteins  Co packaged into clathrin coated vesicles o Hsp  Peals off clathrin  Acts as an ATPase  Using energy of ATP hydrolysis to peel it off  Activated by Auxillin  Activates ATPase Describe a COPII-coated vesicle. Include in your description what the donor and target compartments are for these vesicles, and functional descriptions of the following: Sar1, GEF, Sec23, Sec24, Sec13, and Sec31. State the quality checkpoint used to determine when proteins can exit the ER.  COPII- coated vesicle- transport of large cargo molecules o Donor Compartment: ER o Target Compartment: Golgi o Sar  Inserts amphipathic helix into membrane and recruits coat proteins o GEF  Activates Sar  Removes GDP and adds GTP  Reveals amphipathic helix and then Sar1 can insert  o GTP bound Sar 1 binds to a complex of two COPII adaptor coat proteins:  Sec  Like twist ties  Alpha helical to twist vesicle down closer o T-SNARE  Target snare  Target membranes have complementary receptors o Rabs  Monomeric GTPases  For every Rab, there is a specific tether  Interact with target effector  Hydrolyzes its GTP after fusion  Rab effector proteins interact with active Rab proteins on target membranes o Tethering Protein  Rab effector on the target membrane  Attaches to Rab GTP on vesicle o NSF  ATPase that is requires to pry apart identical pairs of v- and t-snares  Use to form vesicular tubular clusters  Crucial protein that cycles membranes and the cytosol and catalyzes the disassembly process  Botulism Toxin o Interacts with receptors (gangliosides) o Heavy and light chain  The light chain cuts up the snares  Cant release ACh  flaccid paralysis Define heterotypic and homotypic membrane fusion and give examples of each. Describe the process of homotypic membrane fusion and state what results when it occurs between vesicles coming off the ER.  Heterotypic o Membrane of one compartment fuses to a membrane of a different compartment o Ex: HIV  HIV fusion protein rearranges, releasing energy that helps pull two membranes together  Gp120 (glycoprotein) attaches to CD4 (identifier of T helpers) and cytokine receptor   detach insertion of fusion membrane protein  ratchet/ twist o  Homotypic o Fusion of membranes from same compartment o Ex: tubular clusters from ER  Vesicular Tubular Clusters  When ER-derived vesicles fuse with one another  Proteins leave ER in COPII vesicles  Vesicles coalesce by homotypic fusion to form vesicular tubular clusters  move along microtubules to Golgi Describe how proteins go: from ER to the Golgi apparatus and from Golgi apparatus back to ER (include what signals retrieval and state the name for this type of transport).  Retrograde transport o Uses COPI coats o ER retrieval signals present on ER membrane proteins o Soluble ER proteins rely on their own KDEL (Lys-Asp-Glu-Leu) sequence recognized by KDEL receptor o Proteins can also be retained in their organelle by aggregation  Kin recognition o o In all eukaryotic cells, diverse morphology:  Breakdown intra- and extra-cellular debris  Destroy phagocytized microorganisms  Produce nutrients  Plant vacuole o Related to animal lysosomes  Can act as storage for nutrient or waste  Degradative compartment  Maintain turgor pressure  Homeostatic device  Eg: increase H+ import when external pH decreases  Targeting Proteins to Lysosomes o In cis-golgi, mannose 6-phosphate added to N-linked oligosaccharide of hydrolase:  Hydrolase is recognized by its signal patch  M6P added by N-acetylglucosamine phosphotransferase (GlcNacPT)  Three pathways feed into lysosomes o Endocytosis o Phagocytosis o autophagy   Targeting Proteins o Signal patch of lysosomal hydrolase with an N-linked oligosaccharide with terminal mannose residue binds to GlcNAcPT along with UDP-GlcNAc o Transfers GlcNAc with two phosphates to Mannose in catalytic site  Release of UMP o Release the GlcNAc-P  Remove GlcNAc, leaves mannose 6-phosphate exposed Give a general description of a lysosomal storage disease and specific information on what functional enzyme is lacking in the following lysosomal storage diseases: Gaucher’s, Nieman-Pick, Tay-Sachs.  Lysosomal Storaage Disease o If GlcNAcPT is defective/missing, lysosomal storage disease occurs  Inclusion-cell disease  Hydrolases secreted instead of transported to lysosomes o Gaucher’s  Liver, lung, spleen, bone marrow  Acid hydrolase deficiency: glucocerebrosidase  Spleen packed with glucocerebrosides and cant get rid of it  Lose lung tissue and gas exchange abilities o Nieman-Pick  Acid Hydrolase deficiency: acid sphingomyelinase  Accumulates in spleen, brain, and liver o Tay-Sachs  Acid hydrolase deficiency: hexosaminidase A  Lacking  Ganglioside GM2 accumulation in brain List and completely describe the three types of endocytosis. Include in your description the proteins and vesicles that are involved in these processes. Use LDL reception, uptake, and receptor recycling as an example of one of the types. Describe the disease that results from LDL receptor malfunction.  Phagocytosis o Phagosomes fuse with lysosomes inside the cell o Opsin- protein on surface of cell that allows attachment to pull it in o Microorhanisms, dead cells ingested via endocytic vesicles called phagosomes  Pinocytosis o Small but of plasma membrane in the form of small pinocytic vesicles. Eventually returned to plasma membrane o Often begins at clathrin-coated pit o Can also begin at caveolae (little cavities- static) that are present in lipid rafts  No cytosolic protein  Involves caveolin, integral membrane proteins, and no coat proteins  Protein on plasma membrane that helps facilitate caveolae formation o Not selective  Receptor-Mediated Endocytosis o First seen in cholesterol uptake o Specific molecules brought in o Need: receptor, adaptor protein, and clathrin o Ex: LDL: low density lipoprotein  Is ligand for receptor that recruits clathrin  Contains cholesterol molecules  Receptor recycling  LDL releases cholesterol and the LDL receptors are returned to the plasma membrane (10min)  Low pH in endosomes causes receptors to release cargo  Put with hydrolytic enzymes in lysosome that digest the LDL  Cholesterol released into cytosol  LDL Mutation o Familial Hypercholestrolemia  Dominantly inherited LDL receptor mutation  Xanthomas o Cholesterol deposits  Coronary artery disease  Total cholesterol > 300mg/dl  Serum LDL > 200mg/dl o  Degradation o Ex: opiod and EGF receptors o lysosomes Describe the two secretory pathways and what must occur before exocytosis can occur.  Constitutive Secretory Pathway o All cells require o Make product and release immediately  Secretory Pathway o Specialized cells  Eg: histamine, hormone, digestive enzyme, neurotransmitters  Make product and regulate and hold it until needed o wait by plasma membrane until signaled to release their contacts  Exocytosis o Requires maturation of vesicles  Aggregated secretory proteins become more concentrated as lumen pH drops and membrane is retrieved  Gets smaller and more concentrated  During maturation, protolytic processing occurs  Different cells have different prohormone convertases that give different products from a gene o Initial cleavages are made by proteases that cut next to pairs of positively charged amino acids (Lys-Arg, Lys-Lys, Arg-Lys, or Arg- Arg pairs) o Trimming reactions then produce the final secreted products o Different cell types produce different concentrations of individual processing enzyme  CH 13 Objectives - Summary Molecular Biology of the Cell Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 8 0 Save CH 13 Objectives - Summary Molecular Biology of the Cell Course: Cellular Biology (BIO 351) 48 documents University: Indiana Wesleyan University Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 8 0 Save CELL BIOLOGY LECTU RE OBJECTIVES Chapter 13: Intracellular Vesicular Traffic After attending this series of lectures and studyi ng the text, lecture presentations, and notes, the student should be able to: 1.State the central theme of vesicular transport. Vesicular transport is the predominant mechanism for exchange of proteins and lipids between membrane-bound organelles in eukaryotic cells. 2.Describe the 3 major pathways involved in vesicula r transport. Biosynthetic-Secretory o Make and deliver to extracellular o Leads outward from the ER toward the golgi apparat us and cell surface, with a side route leading to lysosomes Endocytic o Bring outside in o Leads inward from the plasma membrane Retrieval o Take membrane back to compartment where it came from o Balance the flow of membrane between compartm ents in the opposite directions, bringing membrane and selected proteins back to the compartme nt of origin 3.Describe a clathri n-coated vesicle. Include in your description what the donor and target compartments are for these vesicle s, and functional descriptions of the following: triskelion, adaptor proteins, cargo receptor, dynamin GTPase, PIP phosphatase, Hsp70, and auxi llin. Clathrin Coated Vesicl e: transport from golgi apparatus and plasma membrane o Donor compartment (Golgi and Plasma Membrane) Buds membrane-bounded transport vesicles that carry cargo destine d for acceptor compartments. o Target compartment (cell surface or lysosome) Target of the vesicle; end locati on o Triskelion (falls off) Formed by clathrin 3 legged structure Determine geometry Assemble into coated pits on cytosolic mem brane surface 3 heavy and 3 light chains o Adaptor proteins (falls off) Bridge between clathrin and cargo receptor Discrete inner layer of the coat, between clathri n cage and membrane Bind clathrin coat to the membrane and trap various transmembra ne proteins, including receptors that capture soluble cargo molecul es inside the vesicle o Cargo receptor Carries desired soluble proteins into vesicles o Dynamin GTPase Required for pinching off Uncoating: o PIP phosphatase Depletes PI(4,5)P 2 from membrane Weakens binding of adaptor proteins Co packaged into clathri n coated vesicles o Hsp70 Peals off clathrin Acts as an ATPase Using energy of ATP hydrolysis to peel it off Activated by Auxillin Activates ATPase 4.Describe a COPII-coated vesicle. Include in your descript ion what the donor and target compartments are for these vesicle s, and functional descriptions of the following: Sar1, GEF, Sec23, Sec24, Sec13, and Sec31. State the qualit y checkpoint used to determine when proteins can exit the ER. COPII- coated vesicle- transport of large cargo mole cules o Donor Compartment: ER o Target Compartment: Golgi o Sar1 Inserts amphipathic helix into membra ne and recruits coat proteins o GEF Activates Sar1 Removes GDP and adds GTP Reveals amphipathic helix and then Sar1 can insert  o GTP bound Sar 1 binds to a complex of two COPII adaptor coat protei ns: Sec23 Form inner coat Sec24 Form inner coat Several different binding sites for the cytosolic tai ls of cargo receptors Entire complex is curved to match the diamet er of the vesicle o Complex of two additional COPII coat proteins: Sec13 Sec31 Form outer shell of the coat Can assemble on their own into symmetri cal cages o Membrane bound, active Sar1-GTP recruits COPII adaptor proteins to the m embrane They select certain trans membrane proteins and cause the mem brane to deform Adaptor proteins recruit outer coat proteins which help form a bud Membrane fusion  pinching off o One exit signal understood: ERGIC53 Lectin that recognizes clotti ng factors V and VIII 5.State what the donor and target compartm ents are for COPI-coated vesicles. Donor: Golgi or cluster Target: ER 6.Describe how vesicles are marked so that they get to the proper targe t compartment. Include functional descriptions of the following: v-SNARE, t-SNARE, Rabs, te thering protein, and NSF. Describe how botulism toxin interrupts vesic ular targeting and state the results of such interruption. How vesicles are marked so that they go to the proper target compartm ent  o V-SNARE Vesicle snare Like twist ties Too long to read on your phone? Save to read later on your computer Save to a Studylist Alpha helical to twist vesicle down closer o T-SNARE Target snare Target membranes have complementa ry receptors o Rabs Monomeric GTPases For every Rab, there is a specific tethe r Interact with target effector Hydrolyzes its GTP after fusion Rab effector proteins interact with active Rab protei ns on target membranes o Tethering Protein Rab effector on the target membrane Attaches to Rab GTP on vesicle o NSF ATPase that is requires to pry apart ident ical pairs of v- and t-snares Use to form vesicular tubular clusters Crucial protein that cycles membranes and the cytosol and cat alyzes the disassembly process Botulism Toxin o Interacts with receptors (gangliosides) o Heavy and light chain The light chain cuts up the snares Cant release ACh  flaccid paralysis 7.Define heterotypic and homotypic membrane fusion and give example s of each. Describe the process of homotypic membrane fusion and state what results when it occ urs between vesicles coming off the ER. Heterotypic o Membrane of one compartment fuses to a membrane of a different compart ment o Ex: HIV HIV fusion protein rearranges, releasing energy that helps pull two mem branes together Gp120 (glycoprotein) attaches to CD4 (identifier of T helpers) and cytokine receptor  detach insertion of fusion membrane protein  ratchet/ twist o Homotypic o Fusion of membranes from same compart ment o Ex: tubular clusters from ER Vesicular Tubular Clusters When ER-derived vesicles fuse with one another Proteins leave ER in COPII vesicle s Vesicles coalesce by homotypic fusion to form vesicul ar tubular clusters  move along microtubules to Golgi 8.Describe how proteins go: from ER to the Golgi apparat us and from Golgi apparatus back to ER (include what signals retrieval and state the name for this type of transport). Retrograde transport o Uses COPI coats o ER retrieval signals present on ER membrane protei ns o Soluble ER proteins rely on their own KDEL (Lys-Asp-Glu-Leu) sequence recogni zed by KDEL receptor o Proteins can also be retained in their organelle by aggregati on Kin recognition o 9.Describe the Golgi apparatus. Include functional descript ions of the following: cis- and trans- faces. State the 4 target compartment s where vesicles coming from the Golgi apparatus may go. State where the Golgi apparatus may be observed. Golgi Apparatus o Major site of carbohydrate synthesis as well as sorting and dispatc hing station o Carbohydrate added to proteins and lipids from ER o Glycosylation starts in the ER lume n and finishes in the Golgi Anatomy o Collection of membrane-enclosed cisternae o Cis-face through which transported substance s enter Can continue in Golgi or go back to ER o Trans face through which transported substances exit Can move on to lysosomes, secretory vesicles, cel l surface, or return o Especially prominent in the small intestine goblet and pla sma cells H&E does not stain carbs so Golgi shows up as negati ve Humoral immunity  heavily glycosylated (Plasma) 10.In brief, describe the protein processing that can occur in the Golgi apparat us. Define N-linked oligosaccharides and list 2 classes of N-linked oligosaccha rides that may be added to proteins. Define O-linked oligosaccharide s. Describe some functions of protein glycosylation. Asparagine-linked (N-linked) oligosaccharides in glycoproteins (unique to Eukaryotes- aids folding and transport) o Complex Trimming of original N-linked oligosaccha ride with addition sugar addition o High-Mannose Many mannose residues added to original N-linked oligosacc harides o Highly ordered pathway involving mannosidases, sugar nucle otides, and glycosyl transferases in each cisternae O-linked Oligosaccharide s in glycoproteins o Linked to hydroxyl of serine of threonine side chain Sugar Chain Functions o May prevent digestion of glycoprotei n by proteases o Provide protection with some movement o Cell-cell adhesion – lecti ns o Cell surface receptors glycosylated, aids in selecti vity of response to stimuli – signaling enhanced Each cistern has a functional compartm entalization o Determined by biochemical sub fractiona tion and EM using antibody labeling of processing enzyme 11.Give functional descriptions of a lysosome and a plant vacuol e. Describe how the necessary proteins for lysosome function arrive at the lysosome; use a hydrola se as an example. Include in your description the signals involved and the functional descript ions of M6P and GlcNac phosphotransferase. Lysosome o Contain acid hydrolases for intracell ular digestion Document continues below Discover more from: Cellular BiologyBIO 351Indiana Wesleyan University 48 documents Go to course 5 CH 01 Objectives - Summary Molecular Biology of the Cell Cellular Biology Summaries 100% (13) 13 CH 19 Objectives - Main material covered in this chapter Cellular Biology Summaries 100% (6) 9 Cell Bio [Chapter 2 Objectives]Cellular Biology Lecture notes 100% (3) 12 CH 17 Objectives - Main material covered in this chapter Cellular Biology Summaries 100% (3) 12 CH 04 Objectives 2015 - Summary Molecular Biology of the Cell Cellular Biology Summaries 100% (3) 14 CH 03 Objectives 2015 - Summary Molecular Biology of the Cell Cellular Biology Summaries 100% (3) Discover more from: Cellular BiologyBIO 351Indiana Wesleyan University48 documents Go to course 5 CH 01 Objectives - Summary Molecular Biology of the Cell Cellular Biology 100% (13) 13 CH 19 Objectives - Main material covered in this chapter Cellular Biology 100% (6) 9 Cell Bio [Chapter 2 Objectives] Cellular Biology 100% (3) 12 CH 17 Objectives - Main material covered in this chapter Cellular Biology 100% (3) 12 CH 04 Objectives 2015 - Summary Molecular Biology of the Cell Cellular Biology 100% (3) 14 CH 03 Objectives 2015 - Summary Molecular Biology of the Cell Cellular Biology 100% (3) o In all eukaryotic cells, diverse morphology: Breakdown intra- and extra-cellular debris Destroy phagocytized microorgani sms Produce nutrients Plant vacuole o Related to animal lysosomes Can act as storage for nutrient or waste Degradative compartment Maintain turgor pressure Homeostatic device Eg: increase H+ import when external pH decreases Targeting Proteins to Lysosomes o In cis-golgi, mannose 6-phosphate added to N-linked oligosacc haride of hydrolase: Hydrolase is recognized by its signal patch M6P added by N-acetylglucosami ne phosphotransferase (GlcNacPT) Three pathways feed into lysosomes o Endocytosis o Phagocytosis o autophagy  Targeting Proteins o Signal patch of lysosomal hydrolase with an N-linked oligosaccha ride with terminal mannose residue binds to GlcNAcPT along with UDP-GlcNAc o Transfers GlcNAc with two phosphates to Mannose in cata lytic site Release of UMP o Release the GlcNAc-P Remove GlcNAc, leaves mannose 6-phosphate exposed 12.Give a general description of a lysosomal storage disease and speci fic information on what functional enzyme is lacking in the following lysosomal storage disea ses: Gaucher’s, Nieman-Pick, Tay-Sachs. Lysosomal Storaage Disease o If GlcNAcPT is defective/missing, lysosomal storage disea se occurs Inclusion-cell disease Hydrolases secreted instead of transported to lysosomes o Gaucher’s Liver, lung, spleen, bone marrow Acid hydrolase deficiency: glucocerebrosidase 1 out of 11 Share Download Download More from:Cellular Biology(BIO 351) More from: Cellular BiologyBIO 351Indiana Wesleyan University 48 documents Go to course 5 CH 01 Objectives - Summary Molecular Biology of the Cell Cellular Biology Summaries 100% (13) 13 CH 19 Objectives - Main material covered in this chapter Cellular Biology Summaries 100% (6) 9 Cell Bio [Chapter 2 Objectives]Cellular Biology Lecture notes 100% (3) 12 CH 17 Objectives - Main material covered in this chapter Cellular Biology Summaries 100% (3) More from: Cellular BiologyBIO 351Indiana Wesleyan University48 documents Go to course 5 CH 01 Objectives - Summary Molecular Biology of the Cell Cellular Biology 100% (13) 13 CH 19 Objectives - Main material covered in this chapter Cellular Biology 100% (6) 9 Cell Bio [Chapter 2 Objectives] Cellular Biology 100% (3) 12 CH 17 Objectives - Main material covered in this chapter Cellular Biology 100% (3) 12 CH 04 Objectives 2015 - Summary Molecular Biology of the Cell Cellular Biology 100% (3) 14 CH 03 Objectives 2015 - Summary Molecular Biology of the Cell Cellular Biology 100% (3) More from:Albertsby- - More from: Alberts by- - 16 documents Go to Studylist 5 CH 01 Objectives - 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10890	https://physics.stackexchange.com/questions/686441/how-do-we-determine-the-order-of-the-degeneracy-of-silicon-bands	solid state physics - How do we determine the order of the degeneracy of silicon bands? - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How do we determine the order of the degeneracy of silicon bands? Ask Question Asked 3 years, 9 months ago Modified3 years, 8 months ago Viewed 839 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. In an exercise on intrinsic semiconductors, I was asked to compute the effective state density in conduction and valence bands of Silicon (resp. N c(T)N c(T) and P v(T)P v(T)), at T=300 T=300 K. Those can be expressed as: N c(T)=1 4(2 m c k B T π ℏ 2)3/2 N c(T)=1 4(2 m c k B T π ℏ 2)3/2 and P v(T)=1 4(2 m v k B T π ℏ 2)3/2,P v(T)=1 4(2 m v k B T π ℏ 2)3/2, with m c m c (resp. m v m v) the effective mass of conduction (resp. valence) carriers and k B k B Boltzmann's constant. I know that one can account for the degeneracy of the bands by replacing the effective masses in the above formulae by m 3/2 c,v→∑i m 3/2 i m c,v 3/2→∑i m i 3/2 where m i m i is the effective mass corresponding to the i i th extremum in the band. What I do not understand is that in the solution, it is said that the degeneracy in the conduction band is of 6, while the degeneracy in the valence band is of 4=2+2 4=2+2 due to the so called "heavy" and "light" holes, yielding m 3/2 c→6 m 3/2 c,m 3/2 v→2(m 3/2 h h+m 3/2 l h).m c 3/2→6 m c 3/2,m v 3/2→2(m h h 3/2+m l h 3/2). Could someone enlighten me on these numbers? I guess I am also a bit confused with the concept of heavy and light holes in the valence band so any clarification would be greatly appreciated. Thanks! solid-state-physics semiconductor-physics Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications asked Jan 2, 2022 at 8:56 XavierXavier 55 8 8 bronze badges 2 Bands with different curvature have different effective masses. The fact that the energies of the top of the two valence bands in silicon are the same is just a fact of the band structure.Jon Custer –Jon Custer 2022-01-02 15:44:42 +00:00 Commented Jan 2, 2022 at 15:44 @JonCuster Thank you, I've just found this page that gives a comprehensive explanation on the origin of heavy & light holes. This page also demonstrates the 6-fold degeneracy of the conduction band of silicon.Xavier –Xavier 2022-01-03 10:08:33 +00:00 Commented Jan 3, 2022 at 10:08 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Problem solved! This page gives a comprehensive explanation on the origin of heavy & light holes in semiconductors, while this one demonstrates the 6-fold degeneracy of the conduction band of silicon : "The six-fold degeneracy of the valleys arise due to the symmetry of the lattice along the , , and directions". Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jan 3, 2022 at 10:10 XavierXavier 55 8 8 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! 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10891	https://www.researchgate.net/publication/231337965_Disruptions_in_Round_Robin_Tournaments	Published Time: 2012-03-16 (PDF) Disruptions in Round Robin Tournaments Article PDF Available Disruptions in Round Robin Tournaments March 2012 Authors: Matt Wiser Matt Wiser This person is not on ResearchGate, or hasn't claimed this research yet. Sudipta Sarangi Virginia Tech Surajit Borkotokey Dibrugarh University Download full-text PDFRead full-text Download full-text PDF Read full-text Download citation Copy link Link copied Read full-textDownload citation Copy link Link copied Citations (1)References (31) Abstract In this paper we study disruptions in round robin tournaments by considering the addition and removal of contestants at arbitrary stages of an on-going tournament. Our focus is on how such disruptions will alter the tournament schedule. We find that adding contestants is an easy problem, while for removing contestants we can only provide bounds. Discover the world's research 25+ million members 160+ million publication pages 2.3+ billion citations Join for free Public Full-text 1 Content uploaded by Surajit Borkotokey Author content All content in this area was uploaded by Surajit Borkotokey Content may be subject to copyright. Disruptions in Round Robin T ournaments Matt Wiser∗Sudipta Sarangi†Sura jit Borkotok ey ‡§ Abstract In this paper we study disruptions in round robin tournaments by considering the addition and remov al of contestants at arbitrary stages of an on-going tournament. Our focus is on how such disruptions will alter the tournament sc hedule.W e ﬁnd that adding contestants is an easy problem, while for removing con testants we can only provide bounds. Keywords: Games, T eams, Round Robin T ournaments, Scheduling, Adding Contestants, Removing Contestan ts, Edge Coloring. ∗Department of Economics, Louisiana State University,USA-70808, email:mwiser1@tigers.lsu.edu †Department of Economics, Louisiana State University, USA-70808, email:sarangi@lsu.e du ‡Department of Mathematics, Dibrugarh University, India-786004, email:surajitb or@yahoo.com §Surajit Borkotokey acknowledges the Indo-US Science and T echnology Forum for pro viding him a fellowship to visit Louisiana State University, during 2011-12 and also expresses his gratitudes for the hospitality provided by the Department of Economics, Louisiana State University during his stay there. 1 1 Introduction The typical round-robin tournament is a competition in which eac h contestant meets all other contestants in turn.A round robin tournament is the most eﬃcient wa y to ﬁnd the champion among a ﬁnite num b er of contestants since eac h contestant or team has equal chance to perform against all other participants.Usually, the p ool stages within a wider tournament are conducted on a round-robin basis.Examples with pure round-robin scheduling include the FIF A W orld Cup,UEF A European Football Championship and UEF A Cup in football, Sup er Rugby (Rugby Union) in the Southern Hemisphere during its past incarnations, the Cricket W orld Cup, Indian Premier League Tw enty-20 Crick et and the likes. Professional sports are big business and the revenue of a sports league may be aﬀected by the quality of the sc hedule (Rasmussen and Tric k ).Disruption in a tournament may be caused by v arious human and non-human factors, among which tw o very signiﬁcant factors are removing and adding contestan ts (teams) in the middle of the tournament. Players dropping out in the middle of the tournamen t are more often seen in individual events.1 Similarly, there may be cases where contestants join after the tournamen t is b eing started.So the problem of rescheduling a tournament after disruption is an in teresting area of study that inherits ideas from combinatorics, graph theory and operation research as well. In this paper, we ﬁnd the exact number of rounds required after an eﬃcient rescheduling of a round robin tournament when contestan ts join in the middle of the tournament and obtain bounds for rounds that can be saved due to the remov al or drop out of contestants. The benchmark of our study is a theorem by Vizing on edge coloring of graphs which shows that maximum n umber of colors to color the edges of a graph dep ends on its degree 1 The most recent example of dropping in the middle of an international ev ent is that of Maria Sharapova withdrawing with injury from the WT A Championship 2011, Istanbul after losing her second match to Li Na ( The New Y ork Times, October 26, 2011). 2 and multiplicity.Here, we can identify eac h contestant with a v ertex, and each color with a round, and the color of the edge connecting two con testants signiﬁes the round in which the two con testants will meet.How ever, the original result does not deal with that of adding new nodes or removing them.W e assume that the addition or removal is exogenous so that such mov es do not interfere the performances in the tournament.F urthermore, we allow for adding or removing an y number of contestants and such action can tak e place at any stage of the Round Robin tournament. Putting constraints to the scheduling is an in teresting domain of study for a number of researchers viz.Trick , Barone et al. , Ikebe and T amura , Nemhauser, and T rick , Easton, Nemhauser,and Trick .A go od general survey of the scheduling problems in a round robin tournament is given b y Rasmussen and Tric k .They have identiﬁed eight constrain ts that have been so far imposed on a round robin tournament in order to eﬀect the schedule.Interested reader may see[3,5,8,9,13–18,24,29] for details. Howev er, in all these cases, scheduling takes place before the tournament starts and thus no disruption occurs in the middle of it.T o the b est of our knowledge, disruption due to addition and remov al of contestants has not yet been considered by any of the researc hers. The rest of the paper is organized as follows.In Section 2, w e present the Mathemat- ical Preliminaries.Section 2 and Section 3 deal with situations that involv e adding and removing of contestan ts resp ectively.Finally, Section 4 gives some concluding remarks and identiﬁes some areas for future research. 2 Mathematical Preliminaries Let us denote by N, N r, N r the sets of initial contestants, contestan ts to be added and those to be removed after the r th round in a round robin tournament respectively. Denote their cardinalities by the small scripts namely n, n r and n r.When scheduling a tournament, the games must be allocated to a number of rounds (time slots) in such a 3 way that eac h team plays at most one game in each round.When the number of teams n is even at least (n−1) rounds are required and when n is odd at least n rounds are required to schedule a single round robin tournament.The allocation of games to rounds can be presented as a timetable.Eac h row of the timetable corresponds to a round while the columns correspond to the matches.The sequence of home games, awa y games and byes (if there are an y) according to which a team plays during the tournament is known as a home awa y pattern (pattern).If b yes occur in the tournament, a pattern is normally represented by a v ector with an entry for each round containing either an H, an A, or a B. The combination of a pattern set and a corresponding timetable constitutes a schedule for a tournament (Rasmussen and T rick ).Since round robin tournaments hav e a correspondence to graphs, the timetable can b e identiﬁed with a complete simple graph G= (V, E )with V being the num b er of vertices representing n con testants and E the set of edges representing the matches.A matching in G is a set of independent edges (non-adjacent edges) and a matching in whic h all the no des in V are incident to an edge is called a perfect matching.The graph induced by a complete matching is a 1-regular graph (the degree is 1 for all nodes).In order to simplify our problem, here, we have considered a single round robin tournament having sc hedules without a pattern set.Inheriting the idea of “matching” from graph theory, we alternatively call ev ery p ossible game between tw o players a “matc hing”.Similarly w e do not distinguish b etween a “game” and a “match”.2 Deﬁnition 2.1. An edge is multiple if there exists at least one more edge with the same end vertices; otherwise it is simple.The multiplicity of an edge is the num b er of multiple edges sharing the same end vertices; the multiplicity of a graph is the maxim um multiplicity of its edges. Finding an edge coloring which uses the minimal num b er of colors has been a widely studied problem in computer science and graph theory.T echniques have been developed 2 Howev er,there is a signiﬁcant diﬀerence between these two terms in some of the games, viz.in the game of T ennis,a match consists of sets and sets consist of games. 4 to attempt to ﬁnd such colorings, ideally a coloring consisting of a known minimal n umber of colors.For exam ple one can look into Asratian , Enochs , Liu to name a few. A good amount of work has been done in the round robin scheduling problems using the notion of edge coloring.The ﬁrst,and most famous, result is due to Vizing , can b e stated as, Theorem 2.2. (Vizing,1964) F or a graph with maximum de gre e d and maximum multi- plicity m, an edge c oloring exists with a maximum of d+m colors.In particular, for a complete simple gr aph with n vertices, we have d=n−1 and m= 1 so that the edge coloring exists with a maximum of n c olors. This result is important, as each color can be treated as a round, where the color of the edge connecting two con testants signiﬁes the round in which the tw o contestants will meet. Therefore, Theorem 2.2 applies to the scheduling problem and gives us n 2(alternatively n−1 2) number of matchings in eac h round from among a total of n(n−1) 2 matchings when n is even (odd) such that each round has exactly (at most) one matc hing of a contestant. Deﬁnition 2.3. The load for a round is the number of matched competitors (equivalently matchings) in a given round. Maximal load consists of the maximum possible load, either n 2 or n−1 2 depending on parity, existing for every round. Deﬁnition 2.4. An eﬃcient schedule is one in which ev ery round has a maximal load. Note that, if n is even and it is possible to color the graph in n−1 colors, in particular, then such a coloring is eﬃcient, since eac h edge has a maximal load. Deﬁnition 2.5. The minimum required number of colors for the edges of a given graph is called the chromatic index of the graph. Deﬁnition 2.6. A cubic graph is a 3-regular graph. 5 It can be seen that there are other graphs which cannot be edge-colored with d colors. An interesting result due to Hoyler needs special mentioning here in order to apply it in a subsequent section. Theorem 2.7. (Hoyler,1981, pg 3) It is NP-complete to determine whether the chr omatic index of a cubic graph is 3 or 4. Deﬁnition 2.8.A bye is a ﬁctional contestant (Rasmussen and T rick ) who is assumed to be matched in each round of a tournament.A bye also refers to the practice of allowing a contestant or team to adv ance to the next round of a playoﬀ tournament without pla ying. 3 Adding Con testants In this section, we consider the problem of appending the set N r to the set N of a round robin tournament in progress.Recall that, if n is o dd (respectively even), the initial eﬃcient schedule for N has a total of n(n−1) 2 matchings with n(respectively (n−1)) rounds each having a maximal load of n−1 2(respectively n 2). Proposition 3.1.Suppose n r c ontestants are added after r r ounds in a round robin tour- nament with n contestants initially.Then, Case 1:If n is even with n r= 1, the new tournament wil l re quire a total of n+r rounds. Case 2:If n+n r is odd, and n r 6= 1, the new tournament will requir e a total of n+r+n r rounds.3 Case 3:If n+n r is even, the new tournament wil l re quire a total of n+r+n r−1 rounds. Pro of. Let r≥2.Observe that apart from the case of adding a single contestant to an ev en sized group, the best possible way of rescheduling after round r is equiv alent to making a new schedule with contestan ts from N∪N r.When n r= 1 and n is even (i.e n+ 1 is odd), 3 Case 2 actually covers 2 cases namely : (a) n is o dd but n r is even and (b) n is even, n r(6= 1) is odd. Similarly, Case 3 covers t wo cases, either both n and n r even or both n and n r o dd. 6 then among these odd contestants, one contestan t remains unmatched in every round.So we can assume that the new tournament is one with odd contestan ts where the entering contestant w as unmatched in a previous round. Now, by application of Vizing’s theorem , any eﬃcien t schedule with contestants from N∪N r needs a total of(n+n r)(n+n r−1) 2 matchings.We obtain binding constrain ts on rounds to get the exact number. Let ˆ r be the number of rounds required by the contestan ts from N r all together to complete their matchings among themselves and with con testants from N. Complying with the rules of a round robin tournament, we m ust have ˆ r≥(n+n r−1).This is our ﬁrst binding constraint. Since the maximal load in each round in Case 1 and Case 2 is n+n r−1 2, the total number of rounds for the new schedule will be n+n r.Since r≥2, at least one round has already been completed. Let r∗be the number of additional rounds required to complete the tournament.Thus, we must ha ve r∗≤n+n r−1.This gives a second binding constrain t on the number of rounds.Since ˆ r≤r∗by hypothesis, we get the exact n umber of rounds for the rescheduling as required. F or Case 3, the maximal load in each round will be n+n r 2, and dividing the total number of matches required by this maximal load giv es n+n r−1 rounds as required.This completes the proof. 4 Removing Con testants W e now discuss the case of removing contestants in a round robin tournamen t and ﬁnd the minimum num b er of rounds needed to exhaust the matchings in N\N r.W e formalize this by ﬁrst considering the remov al of a single contestant and then generalize this for n r≥1. Lemma 4.1. Removing a c ontestant from an ongoing r ound robin tournament do es not 7 change the number of r equir ed rounds if n is even.F or odd n, it re duces the number of rounds by at most 2. Pro of. If n is even, removing a single contestan t from N(i.e scheduling for N\N r with n r= 1) does not save any rounds, since this also requires n−1 rounds similar to that for N. If n is odd ( n−1 even), there are n−1 2 matchings in each round, both before and after the remov al.As n−1 contestants require (n−1)(n−2) 2 matchings, this would imply that w e require at least n−2 total rounds, as opposed to the n rounds required originally.Th us we hav e a saving of at most 2 rounds.This completes the pro of. Lemma 4.1 provides an upper bound for saving rounds.Let m be the number of match- ings involving the remo ved contestant up to the r th round.Then m is either r−1 or r, depending on whether the removed contestan t has had an unmatched round.If we remove her after the ﬁrst round, and she is the one who was unmatched in the ﬁrst round (i.e m= 0), we can sav e 2 rounds.If 0< m ≤n−1 2, we can sav e one round,and restructure the remaining rounds so that the total number is one less.Observe that, the feasibility of reducing a round is dependent on precisely which contestan t is removed and the order of the rounds that were originally set.Thus computing the exact number of rounds that can be saved is rather complicated when removing con testants. When adding contestants, all added contestan ts are essentially identical, whereas in removing contestan ts,this is not true, since diﬀerent contestants would ha ve played dif- ferent num b er of matches, and thus will hav e diﬀerent impacts upon removal.Clearly, this problem is equivalen t to the edge coloring problem discussed by Fa vrholdt et al. , where we can deﬁne a graph with the vertices being the remaining contestan ts, and edges connecting only those contestants whose matc hings among themselves would take place after the r th round.The colors in our edge coloring will then correspond to the rounds in the new tournament held after the remov al of the members in N r. 8 Recall that, Hoyler’s theorem (Section 2) suggests that, determining the minimal number of colors to complete an edge coloring of a graph is NP-Complete and so no algo- rithm exists to ﬁnd an exact solution to this problem in polynomial time.4 As a corollary to Theorem 2.7, we hav e the following: Corollary 4.2. A polynomial time algorithm to ﬁnd the exact numb er of rounds re quire d to complete a r ound robin tournament when a c ontestant is removed at some r ound does not exist. Note that this statement holds good even if more than one contestants are remo ved. Much of the research follo wing Hoyler has focused on the approximability of the NP- complete optimization problem (see for example Bodlaender , Petrank , Leven ). There is also some application related work.In, Li et al.show that the neighboring mobile sensors must report signals on diﬀering frequencies, with the sensors moving in real time.Th us they have a graph that is constantly shifting, making speed crucial in ﬁnding a work able coloring.In a very recent paper, Costa et al. obtain an optimal version of simulated annealing algorithms.In this pap er, variations in the selection of edges to recolor and the acceptance or rejection of this new coloring is incorporated.These reﬁnements increase the speed of the algorithm, however the basic tec hnique remains the same. F ormally, let us consider the scheduling problem with mem bers from N\N r, which 4 In , the NP-Completeness of edge coloring is shown on a cubic graphs.A cubic graph can be constructed from subgraphs which are equiv alent to creating an arbitrary num b er of copies of a variable and the negation of the variable, with the v ariable b eing true or false depending on the coloring, true statements being represented by edge pairs with the same coloring and false statemen t by edge pairs with diﬀerent coloring.There are also comp onents which join together three of these pairs such that they are 3-colorable if and only if a three variable clause con tains at least one true variable.Thus, 3-colorablity of these graphs is equivalen t to the 3-satisﬁability problem, which is the canonical NP-Complete problem. As a subset of the problem is not solvable in polynomial time, the problem cannot be generally solvable in polynomial time, so 3-colorability of cubic graphs is NP-Complete, and similarly the n-colorability of a graph with maximal degree n must be NP-Complete. 9 requires a total of(n−n r)(n−n r−1) 2 matchings.In the ﬁrst r rounds we ha ve n 2matchings per round, and thereafter each round will have n−n r 2matchings.Let r s denote the number of rounds to be saved by resc heduling the tournament for N\N r.Given m(the num b er of matchings in the previous rounds inv olving the removed contestants), that depends on the contestants remo ved as well as the initial scheduling, we can obtain the bounds on r s. W e present our results in four diﬀerent propositions depending on the parity of n and n r. First note that a revised schedule of the tournament after the remo val of contestants may diﬀer from the original schedule if w e can move some matchings to diﬀerent rounds to minimize the total number of rounds required. Proposition 4.3.If n and n r are both even, then, Case 1:max{0, n r−rn r n−n r} ≤ r s≤n−r−1 if r>n −n r; Case 2:max{0, n r−rn r n−n r} ≤ r s≤n r if r≤n r−1 and r≤n−n r; Case 3:max{0, n r−rn r n−n r} ≤ r s≤j n 2−nn r−n(n r−1)−rn+2 rn r n−n r k if r > n r−1 and r≤n−n r. F urther, these are the best p ossible bounds of r s in a general r e-scheduling pr oblem without additional condition. Pro of. Given that n and n r are both even, each remov ed contestant m ust have had a match in each round in the initial schedule.W e ﬁrst obtain the common lower bound r s for all the three cases.This is equivalen t to ﬁnding the maximum possible number of rounds that can be saved by the remo val of contestants.Let n r−rn r n−n r≥0 without any loss of generality. If m i denotes the number of matches pla yed by the removed con testants in round i, we have n r 2≤m i≤min{n r,n 2}.Note that if n r 6<n 2, then r s≤n 2.Therefore, we consider the extreme case where all the removed con testants in the original schedule would play the maximum num b er of matches after the r th round.This suggests that the n r removed contestants and the n−n r remaining con testants divide the total rounds of the tournament into the ﬁrst n r−1 and the next n−n r−1 rounds, so that n r 2≥min{n r−1, n −n r−1}. Therefore, all subsequent rounds will hav e the maximum num b er of matches inv olving a 10 removed con testant as desired.Thus we hav e, rn r 2≤m≤rn r for r≤n r−1,and (4.1) (n r−1)n r 2+ (r−(n r−1))n r≤m≤rn r for r > n r−1.(4.2) It follows that, there will be at least rn 2−rn r=r(n−2 n r) 2 valid matc hes in the ﬁrst r rounds among a total of rn 2(since n is even, each round in the original schedule con- sists of n 2 matches).Subtracting this from the required(n−n r)(n−n r−1) 2 matches gives us (n−n r)(n−n r−1)−r(n−2 n r) 2 remaining matches.Dividing this by the n−n r 2 maximal matches per round after the remov al gives us(n−n r)(n−n r−1)−r(n−n r)+rn r n−n r=n−n r−1−r+rn r n−n r remaining rounds as required.As the initial group of n contestan ts would require n−r−1 further rounds after the remov al, we have sav ed n r−r n r n−n r rounds by the remov al of n r contestants after round r.This provides the common lower bound for all the three cases.It can be seen from the construction itself that this is the best possible lower bound, one can have with the most general situation. The upper bounds for r s can b e obtained as follows: Case 1:In the extreme possible situation, the case r>n −n r implies that all matches be- tween the remaining mem b ers are already completed, and hence we can reduce a maximum of n−r−1 rounds in the new schedule. Case 2:If r≤n r−1, we have r n r 2≤m, thus giving us at most rn 2−rn r 2=r(n−n r) 2 still valid matc hes.W e require a total of(n−n r)(n−n r−1) 2 matches so that at least another (n−n r)(n−n r−r−1) 2 matches are needed to complete the resulting tournament sc hedule.Di- viding by the maximum n umber of n−n r 2 matches per round gives us the requirement of n−n r−r−1 further rounds.As we need n−r−1 further rounds without remov al, this leaves us with a reduction of at most n r rounds. Case 3:If r > n r−1, we have (n r−1)n r 2+(r−(n r−1))n r≤m.Subtracting this from the r n 2 matches in the ﬁrst r rounds gives us r n−(n r−1)n r 2−rn r+(n r−1)n r=rn+(n r−1)n r−2 r n r 2 remain- ing valid matc hes.Subtracting this from the (n−n r)(n−n r−1) 2 required total matches gives us 11 n 2−nn r−n(n r−1)−rn+2 rn r 2 required remaining matches.Dividing this by the n−n r 2 maximum matches per round results in j n 2−nn r−n(n r−1)−rn+2 rn r n−n r k remaining rounds required.This completes the proof. F ollowing example provides an illustration of Proposition 4.3. Example 4.4. Let N={1,2,3,4,5,6}b e the set of contestants in the initial tournamen t and N r={1,2,3,4}, the set of contestants to be remov ed.Th us n= 6 and n r= 4. Case 1:Let r= 3 so that r>n −n r= 2.In what follows, we consider a schedule where maximum num b er of matches inv olving the removed contestants is k ept after round r= 3. Rounds Match 1 Match 2 Match 3 Round 1 2 4 3 1 5 6 Round 2 1 5 6 4 2 3 Round 3 3 6 1 2 4 5 Round 4 6 2 5 3 1 4 Round 5 4 3 2 5 6 1 T able 1:The Original Round Robin Sc heduling It is clear from T able 1, that after removing the contestants from the tournament, ex- actly two rounds can be sav ed, which complies with the bounds of r s as by Proposition 4.3, we hav e 0≤r s≤2. Similarly, for Case 2,if we take r= 1, the same schedule as of T able 1 will keep the maximum num b er of matchings with the remov ed contestants after round 1.By Prop osi- tion 4.3,2≤r s≤4 and from T able 1, we can see that exactly four rounds can be saved. Case 3 is trivially true in this case as there does not exist an r such that r > n r−1 and r≤n−n r with n= 6 and n r= 4. 12 Proposition 4.5.If n is odd and n r is even, then Case 1:max{0, n r−rn r n+1−n r} ≤ r s≤n−r if r>n −n r Case 2:max{0, n r−rn r n+1−n r} ≤ r s≤n r if r≤n r−1 and r≤n−n r Case 3:max{0, n r−rn r n+1−n r} ≤ r s≤j(n+1)2−(n+1)n r−(n+1)(n r−1)−r(n+1)+2 rn r n+1−n r k if r > n r−1 and r≤n−n r.Further, these ar e the best possible b ounds of r s in a general r e-scheduling problem without additional c ondition. Pro of. We proceed by adding a bye (Deﬁnition 2.8) to the original con testant set N so that the new set has cardinality n+ 1(even).Th us the pro of follows exactly in the same way as that of Proposition 4.3, with n being replaced by n+ 1. W e illustrate Prop osition 4.5 by the following example: Example 4.6. Let N={1,2,3,4,5,6,7}be the set of contestants in the initial tourna- ment and N r={1,2,3,4}, the set of contestan ts to b e removed.Thus n= 7 and n r= 4. Case 1:Let r= 4 so that r>n −n r= 3.Let us denote by 8, a bye contestant.Thus we obtain a schedule where maximum n umber of matches involving the remov ed contestants is kept after round r= 4. Rounds Slot 1 Slot 2 Slot 3 Slot 4 Round 1 7 3 6 4 5 1 8 2 Round 2 4 5 2 7 6 3 1 8 Round 3 3 5 6 2 4 1 7 8 Round 4 5 2 4 3 1 7 6 8 Round 5 4 7 2 1 6 5 3 8 Round 6 4 2 1 3 6 7 5 8 Round 7 5 7 2 3 6 1 4 8 T able 2:The Original Round Robin Sc heduling 13 It is clear from T able 2,that after removing the contestan ts from the tournament, exactly three rounds can be saved, which complies with the bounds of r s as by Proposi- tion 4.3, we hav e 0≤r s≤3. Similarly, for Case 2, if we take r= 3, the same schedule as of T able 2 keeps the max- imum num b er of matches of the remov ed contestants after round 3.By Prop osition 4.5, 1≤r s≤4 and from T able 2,observe that exactly four rounds can be saved.Case 3 is trivially true in this case as there does not exist an r such that r > n r−1 and r≤n−n r with n= 7 and n r= 4. Proposition 4.7.If both n and n r are o dd, then, Case 1:max{0, n r+ 1 −r(n r+1) n−n r} ≤ r s≤n−r−1 if r>n −n r. Case 2:max{0, n r+ 1 −r(n r+1) n−n r} ≤ r s≤n r+ 1 if r≤n r−1 and r≤n−n r. Case 3:max{0, n r+ 1 −r(n r+1) n−n r} ≤ r s≤j(n+1)2−(n+1)(n r+1)−(n+1)n r−r(n+1)+2 r(n r+1) n−n r k if r > n r−1 and r≤n−n r. F urther, these are the best p ossible bounds of r s in a general r e-scheduling pr oblem without additional condition. Pro of. Here also we append the bye (as in Proposition 4.5) to N, however, since n r is odd, we remov e the bye matches scheduled with the members of N r.Therefore, we actually remove n r+ 1 contestants, which is even, and th us the result follows from Proposition 4.5. Proposition 4.8.If n is even and n r is odd, then, Case 1:max{0, n r−1−r(n r−1) n−n r+1 } ≤ r s≤n−r if r>n −n r+ 1 Case 2:max{0, n r−1−r(n r−1) n−n r+1 } ≤ r s≤n r−1 if r≤n r−2 and r≤n−n r+ 1 Case 3:max{0, n r−1−r(n r−1) n−n r+1 } ≤ r s≤j n 2−n(n r−1)−n(n r−2)−rn+2 r(n r−1) n−n r+1 k if r > n r−2 and r≤n−n r+1.Further, these ar e the best possible b ounds of r s in a general r e-scheduling problem without additional c ondition. Pro of. The proof pro ceeds exactly in the same way as that of Proposition 4.7 with n r being 14 replaced by n r−1.This follows from the fact that in spite of having a match initially, after remov al, every round will have an unmatched con testant.This contestant will be matched with a bye, who is functionally the same as a removed con testant in the extreme cases, so the net eﬀect is equivalen t to removing one contestant less. 5 Conclusion This paper investigates the consequences of adding or removing an arbitrary n umber of contestants at an arbitrary stage in a round robin tournamen t.Though adding additional participants turns out to be relatively uninteresting, the remov al of participants from a competition already in progress is an interesting problem.W e have established bounds on the reduction of rounds that can be achieved by remo val of participants and these bounds are based solely on the eﬀect on the total number of matchings.However, in this paper we do not consider other scheduling issues.These are to a certain degree impossible to consider in a general case.We w ould also have to consider all the possible groups of n r contestants to remove from the original group n.Thus, even for a small case of 10 contestan ts, with 4 being removed, we w ould have 9!×10! 6!×4! = 76204800 separate cases to consider for each round. Howev er, there may be reﬁnements possible that would allow some progress to be made. A natural extension would be to consider multiple points of entry or exit.This would obviously add large computational diﬃculties, howev er some version of this may prove worthwhile.F or example,multiple points of having a single random contestan t removed may be a tractable problem. The eventual goal of this w ork is to move towards w orking on Swiss-type tournaments where contestants or teams need to be paired to face each other for sev eral rounds of competition.These have been used to model promotions, and the ability to add exogenous entry or exit would be of help.This would allo w for handling cases of employees leaving a ﬁrm due to outside oﬀers or movemen t to a diﬀerent city, and the hiring of exp erienced, 15 qualiﬁed people, who would thus not hold entry lev el p ositions. References Da vid R. Appleton,May the Best man win ?Journal of the Royal Statistic al Society. 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In this article, the possibility of adding or removing contestants from a round-robin tournament that is already under way is discussed. ... Having More Fun with Organized Kissing Article Jan 2014 Matthew Wiser Gloria Yeomans-Maldonado Sudipta Sarangi Several current issues in economics are centered on scheduling and matching problems, notably including the 2012 Nobel Prize winning work. Such problems usually lie outside the scope of most undergraduate courses. We present a relatively simple problem that can be used to introduce the graph theory needing to teach these interesting but somewhat difficult topics. View Show abstract An Application of Combinatorial Optimization to Statistical Physics and Circuit Layout Design Article Full-text available May 1988 Francisco Barahona Martin Grötschel Michael Jünger Gerhard Reinelt We study the problem of finding ground states of spin glasses with exterior magnetic field, and the problem of minimizing the number of vias (holes on a printed circuit board, or contacts on a chip) subject to pin preassignments and layer preferences. The former problem comes up in solid-state physics, and the latter in very-large-scale-integrated (VLSI) circuit design and in printed circuit board design. Both problems can be reduced to the max-cut problem in graphs. Based on a partial characterization of the cut polytope, we design a cutting plane algorithm and report on computational experience with it. Our method has been used to solve max-cut problems on graphs with up to 1,600 nodes. View Show abstract Some results on an edge coloring problem of Folkman and Fulkerson Article Full-text available Aug 2000 DISCRETE MATH Armen S. Asratian In 1968, Folkman and Fulkerson posed the following problem: Let G be a graph and let (n1,…,nt) be a sequence of positive integers. Does there exist a proper edge coloring of G with colors 1,2,…,t such that precisely ni edges receive color i, for each i=1,…,t? If such a coloring exists then the sequence (n1,…,nt) is called color-feasible for G. Some sufficient conditions for a sequence to be color-feasible for a bipartite graph where found by Folkman and Fulkerson, and de Werra. In this paper we give a generalization of their results for bipartite graphs. Furthermore, we find a set of color-feasible sequences for an arbitrary simple graph. In particular, we describe the set of all sequences which are color-feasible for a connected simple graph G with Δ(G)⩾3, where every pair of vertices of degree at least 3 are non-adjacent. View Show abstract Fixture-scheduling for the Australian Football League using a Multi-Objective Evolutionary Algorithm Conference Paper Full-text available Jan 2006 Luigi Barone Lyndon While P. Hughes Philip Hingston AFL football is a team sport that entertains millions and contributes a huge amount of money to the Australian economy. Scheduling games in the AFL is difficult, as a number of different, often conflicting, factors must be considered. In this paper, we propose the use of a multi-objective evolutionary algorithm for determining such a schedule. We detail the technical details needed to apply a multi-objective evolutionary algorithm to this problem and report on experiments that show the effectiveness of this approach. Comparison with actual schedules used in the AFL demonstrates that this approach could make a useful contribution. View Show abstract May the Best Man Win? Article Jan 1995 David R. Appleton Simulations of various kinds of sporting tournament have been carried out to assess their relative ability to produce as winner the best of the entrants. A strong contender when it is necessary to play relatively few games is the seeded draw and process. When the players are closely matched and there can be more games the round robin played twice is most effective. View Show abstract Allocating Games for the NHL Using Integer Programming Article Aug 1993 Jacques A. Ferland Charles Fleurent In recent years, the National Hockey League (NHL) expanded from 21 to 24 teams. In order to accommodate its scheduling process, the league had to determine new game allocations, that is, the number of times the teams play against each other. This paper introduces a procedure based on integer linear programming that generates several game allocation scenarios that have been presented to the league managers. The model takes into account the constraints specified by the NHL, and the objective function allows the league managers to specify their preference on the distribution of the games. View Show abstract Forging''optimal solutions to the edge-coloring problem Article Jan 2001 Brandon Enochs R. Wainwright View Polynomial algorithms for graph isomorphism and chromatic index on partial k-trees Conference Paper Jan 2006 Lect Notes Comput Sci Hans L. Bodlaender In this paper we show that Graph Isomorphism and Chromatic Index are solvable in polynomial time when restricted to the class of graphs with treewidth k (k a constant) (or equivalently, the class of partial k-trees). Also, we show that there exist algorithms that find tree-decompositions with treewidth k of graphs with treewidth k, in O(n 3) time, (k constant). View Show abstract Some Topics on Edge-Coloring Chapter Jun 2007 Lect Notes Comput Sci Guizhen Liu Changqing Xu In this paper some new results on edge coloring of graphs are introduced. This paper deals mainly with edge cover coloring, g-edge cover coloring, (g, f)-coloring and equitable edge coloring. Some new problems and conjectures are presented. View Show abstract Scheduling the Italian Football League: An ILP-based approach Article Jul 2006 COMPUT OPER RES Federico Della Croce Dario Oliveri Scheduling the Italian Major Football League (the so-called “Serie A”) consists in finding for that league a double round robin tournament schedule that takes into account both typical requirements such as conditions on home-away matches and specific requests of the Italian Football Association such as twin-schedules for teams belonging to the same home-town. In this paper, we present a solution procedure able to derive feasible schedules that are also balanced with respect to additional cable televisions requirements. This procedure adapts the recent approach by Nemhauser and Trick to schedule a College Basketball Conference that considers however only half of the teams involved here. The proposed procedure is divided into three phases: phase 1 generates a pattern set respecting the cable televisions requirements and several other constraints; phase 2 produces a feasible round robin schedule compatible with the above pattern set; finally, phase 3 generates the actual calendar assigning the teams to the patterns. The procedure allows to generate within short time several different reasonable calendars satisfying the cable television companies requirements and satisfying various other operational constraints while minimizing the total number of violations on the home-away matches conditions. View Show abstract A polynomial-time algorithm to find an equitable home–away assignment Article May 2005 OPER RES LETT Ryuhei Miyashiro Tomomi Matsui We propose a polynomial-time algorithm to find an equitable home–away assignment for a given timetable of a round-robin tournament. Our results give an answer to a problem raised by Elf et al. (Oper. Res. Lett. 31 (2003) 343), which concerns the computational complexity of the break minimization problem in sports timetabling. View Show abstract Show more Recommended publications Discover more Conference Paper Full-text available Selfish Routing and Path Coloring in All-Optical Networks August 2007 Ioannis Milis Aris Pagourtzis Katerina Potika We study routing and path coloring problems in all-optical networks as non-cooperative games. We especially focus on oblivious payment functions, that is, functions that charge a player according to her own strategy only. 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They demonstrate the nature of contemporary entrepreneurship and innovation by drawing analogies from physics and the article closes with a case study of their theory in practice. Read more Article On perfection and imperfection of one-realizations of a given set October 2016 · Discrete Mathematics Kefeng Diao Vitaly Voloshin Bing Xue Ping Zhao For any set S of positive integers, a mixed hypergraph H is a one-realization of S if its feasible set is S and each entry of its chromatic spectrum is either 0 or 1. In this paper, we focus on several special kinds of one-realizations of a given set S and get the following results: each colorable mixed hypergraph can be embedded into a one-realization of S; there exist one-realizations of S ... [Show full abstract] which are C-perfect; furthermore, for any positive integer k, there exists a one-realization H of S with αC(H)-χ(H)>k. 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10892	https://www.quora.com/How-do-I-find-the-maximum-or-minimum-value-of-a-quadratic-function-easily	Something went wrong. Wait a moment and try again. Minimum Value Types of Functions Basic Algebra Functions (mathematics) Basic Functions Maximum Value 5 How do I find the maximum or minimum value of a quadratic function easily? Vishakh Nair Studied at Dougherty Valley High School (Graduated 2019) · Upvoted by Aaron Briseno , B.S Mathematics & Teaching, University of California, Los Angeles (2010) · 6y I believe you meant “quadratic expressions” and not “quadratic equations”. The general formula for a quadratic expression is ax2+bx+c For maximum and minimum values, the derivative should be zero. So, 2ax+b=0 Which means x=−b2a A quadratic expression is a parabola, so it has either a maximum value or a minimum value. This can be found by taking the second derivative, which is 2a. If this is negative, we have a maximum point. If it is positive, we have a minimum point. So, in short, a quadratic expression ax2+bx+c has a maximum value at x=−b2a if a is negative, and h I believe you meant “quadratic expressions” and not “quadratic equations”. The general formula for a quadratic expression is ax2+bx+c For maximum and minimum values, the derivative should be zero. So, 2ax+b=0 Which means x=−b2a A quadratic expression is a parabola, so it has either a maximum value or a minimum value. This can be found by taking the second derivative, which is 2a. If this is negative, we have a maximum point. If it is positive, we have a minimum point. So, in short, a quadratic expression ax2+bx+c has a maximum value at x=−b2a if a is negative, and has a minimum value at x=−b2a if a is positive. Pooja Thakkar Software Devlopment (2018–present) · Author has 228 answers and 975.4K answer views · 7y Originally Answered: What are the maximum and minimum value of a quadratic equation? · In the other case, when the graph extends downwardly, then the vertex determines themaximum of f(x), i.e., when a < 0, then at x = –b/2a, f(x) attains its maximum equal to (–D/4a). These same concepts are applicable in numerical which demand the computation of maximum or minimum values of a function. We have already discussed the concept of quadratic expressions in the previous sections. It is a simple topic that can fetch you some direct questions which are scoring as well.It has been observed that the graph of f(x) = ax2 + bx + c extends upwardly or downwardly in all cases accordingly to a > In the other case, when the graph extends downwardly, then the vertex determines themaximum of f(x), i.e., when a < 0, then at x = –b/2a, f(x) attains its maximum equal to (–D/4a). These same concepts are applicable in numerical which demand the computation of maximum or minimum values of a function. We have already discussed the concept of quadratic expressions in the previous sections. It is a simple topic that can fetch you some direct questions which are scoring as well.It has been observed that the graph of f(x) = ax2 + bx + c extends upwardly or downwardly in all cases accordingly to a > 0 or a < 0. Now, when graph extends upwardly, then the vertex determines the minimum of f(x) i.e. when a > 0, then at x = b/2a, f(x) attains its minimum equal to (–D/4a).In the other case, when the graph extends downwardly, then the vertex determines the maximum of f(x), i.e., when a < 0, then at x = –b/2a, f(x) attains its maximum equal to (–D/4a). These same concepts are applicable in numerical which demand the computation of maximum or minimum values of a function. Some key Points When a > 0, then maximum of f(x) does not exist and when a < 0, then minimum of f(x) does not exist. This happens because of R being the domain of the function f(x) = ax2 + bx + c. The above graphically interpretation of maximum & minimum can be verified algebraically as under:The given quadratic expression isf(x) = ax2 + bx + cWe try to make the first term a perfect square and so we get the following equation= [√ax + b/2√a]2 + (c-b2/4a) (if a > 0)The first term on R.H.S is a perfect square and so it can never be negative. It can at most be equal to zero for any x which happens at x = –b/2a.If a > 0 then f(x)½min = -D/4a at x = -b/2aIf a < 0 then f(x) = a1x2 + bx + c where a = –a1 = - (√(a1x) – b/2√a1)2 + (c + b2/4a1) The first term of R.H.S. can be at the most zero which is for x = –b/2a.If a < 0 then f(x)½max = -D/4a at x = –b/2 Some Useful Results If f(x) is a polynomial and a, b are the real numbers such that f (a). f (b) < 0 i.e., f(a) and f(b) are of opposite signs, then between a and b f(x) has at least one root and if it has more than one root between a and b then the no. of roots is odd. This theorem is basically used for obtaining approximate roots. If f (a). f (b) > 0 then between a and b there are either no roots or even no. of roots. An exception occurs when X axis turns out to be a tangent to the curve. Michael Lamar PhD in Applied Mathematics · Author has 3.7K answers and 17.5M answer views · 7y Originally Answered: What are all the ways that you can think of to find the maximum/minimum of a quadratic function? · Here’s one less common proof that x=−b2a is the value that maximizes or minimizes a quadratic function: Let f(x)=ax2+bx+c. Let’s find the two values of x for which f(x)=c. f(x)=ax2+bx+c=c ⟹x(ax+b)=0 ⟹x=0 & x=−ba We know that quadratic functions are laterally symmetric. It follows that the line of symmetry corresponds to the value of x that is half-way in the middle of the two values of x that give f(x)=c. So we know that the line of symmetry occurs at x=−b2a. By symmetry, every value of f other than the maximum or minimum must occur twice, so it Here’s one less common proof that x=−b2a is the value that maximizes or minimizes a quadratic function: Let f(x)=ax2+bx+c. Let’s find the two values of x for which f(x)=c. f(x)=ax2+bx+c=c ⟹x(ax+b)=0 ⟹x=0 & x=−ba We know that quadratic functions are laterally symmetric. It follows that the line of symmetry corresponds to the value of x that is half-way in the middle of the two values of x that give f(x)=c. So we know that the line of symmetry occurs at x=−b2a. By symmetry, every value of f other than the maximum or minimum must occur twice, so it follows that the value of x that maximizes or minimizes f must be at the axis of symmetry, x=−b2a. Then the actual maximum or minimum value must be f(−b2a)=a(−b2a)2+b⋅−b2a+c=b24a−b22a+c=c−b24a If we find that x=0 and x=−ba are the same value, then there is only solution, x=0. In this case, it is x=0 that maximizes or minimizes f and its maximum or minimum value is just c. Related questions How can you find the maximum or minimum value of a quadratic function without graphing it? How can you determine the maximum or minimum value of a quadratic function? Illustrate using an example. How do I find the maximum value of a quadratic function? What are the maximum and minimum value of a quadratic equation? How do you find the maximum and minimum value of y in a quadratic graph? Assistant Bot · 1y To find the maximum or minimum value of a quadratic function, you can follow these steps: Quadratic Function Form A quadratic function is typically written in the form: f(x)=ax2+bx+c where a, b, and c are constants. Steps to Find Maximum or Minimum Identify the Coefficient a: If a>0, the parabola opens upwards, and the function has a minimum value. If a<0, the parabola opens downwards, and the function has a maximum value. Calculate the Vertex: The x-coordinate of the vertex, where the maximum or minimum occurs, can be found using the formula: x=−b2a Evaluate the Func To find the maximum or minimum value of a quadratic function, you can follow these steps: Quadratic Function Form A quadratic function is typically written in the form: f(x)=ax2+bx+c where a, b, and c are constants. Steps to Find Maximum or Minimum Identify the Coefficient a: If a>0, the parabola opens upwards, and the function has a minimum value. If a<0, the parabola opens downwards, and the function has a maximum value. Calculate the Vertex: The x-coordinate of the vertex, where the maximum or minimum occurs, can be found using the formula: x=−b2a Evaluate the Function at the Vertex: Substitute the x-coordinate back into the function to find the maximum or minimum value: f(−b2a)=a(−b2a)2+b(−b2a)+c Example Consider the function: f(x)=2x2−4x+1 Here, a=2 (positive), so there is a minimum value. Calculate the x-coordinate of the vertex: x=−−42⋅2=44=1 Substitute x=1 back into the function: f(1)=2(1)2−4(1)+1=2−4+1=−1 Thus, the minimum value is −1 at x=1. Summary Use x=−b2a to find the vertex's x-coordinate. Substitute this value back into the function to find the maximum or minimum value. Megan Cook Former Chief Information Officer – CIO · 2y Method 1 of 3:Beginning with the General Form of the Function A quadratic function is one that has an x2{\displaystyle x^{2}} term. It may or may not contain an x{\displaystyle x} term without an exponent. There will be no exponents larger than 2. The general form is f(x)=ax2+bx+c{\displaystyle f(x)=ax^{2}+bx+c}. If necessary, combine similar terms and rearrange to set the function in this general form. For example, suppose you start with f(x)=3x+2x−x2+3x2+4{\displaystyle f(x)=3x+2x-x^{2}+3x^{2}+4}. Combine the x2{\displaystyle x^{2}} terms and the x{\displaystyle x} terms to get the following i Method 1 of 3:Beginning with the General Form of the Function A quadratic function is one that has an x2{\displaystyle x^{2}} term. It may or may not contain an x{\displaystyle x} term without an exponent. There will be no exponents larger than 2. The general form is f(x)=ax2+bx+c{\displaystyle f(x)=ax^{2}+bx+c}. If necessary, combine similar terms and rearrange to set the function in this general form. For example, suppose you start with f(x)=3x+2x−x2+3x2+4{\displaystyle f(x)=3x+2x-x^{2}+3x^{2}+4}. Combine the x2{\displaystyle x^{2}} terms and the x{\displaystyle x} terms to get the following in general form: f(x)=2x2+5x+4{\displaystyle f(x)=2x^{2}+5x+4} Determine the direction of the graph. A quadratic function results in the graph of a parabola. The parabola either opens upward or downward. If a{\displaystyle a}, the coefficient of the x2{\displaystyle x^{2}} term, is positive, then the parabola opens upward. If a{\displaystyle a} is negative, then the parabola opens downward. Look at the following examples: For f(x)=2x2+4x−6{\displaystyle f(x)=2x^{2}+4x-6}, a=2{\displaystyle a=2} so the parabola opens upward. For f(x)=−3x2+2x+8{\displaystyle f(x)=-3x^{2}+2x+8}, a=−3{\displaystyle a=-3} so the parabola opens downward. For f(x)=x2+6{\displaystyle f(x)=x^{2}+6}, a=1{\displaystyle a=1} so the parabola opens upward. If the parabola opens upward, you will be finding its minimum value. If the parabola opens downward, you will find its maximum value. Calculate -b/2a. The value of −b2a{\displaystyle -{\frac {b}{2a}}} tells you the x{\displaystyle x} value of the vertex of the parabola. When the quadratic function is written in its general form of ax2+bx+c{\displaystyle ax^{2}+bx+c}, use the coefficients of the x{\displaystyle x} and x2{\displaystyle x^{2}} terms as follows: For a function f(x)=x2+10x−1{\displaystyle f(x)=x^{2}+10x-1}, a=1{\displaystyle a=1} and b=10{\displaystyle b=10}. Therefore, find the x-value of the vertex as: x=−b2a{\displaystyle x=-{\frac {b}{2a}}} x=−10(2)(1){\displaystyle x=-{\frac {10}{(2)(1)}}} x=−102{\displaystyle x=-{\frac {10}{2}}} x=−5{\displaystyle x=-5} As a second example, consider the function f(x)=−3x2+6x−4{\displaystyle f(x)=-3x^{2}+6x-4}. In this example, a=−3{\displaystyle a=-3} and b=6{\displaystyle b=6}. Therefore, find the x-value of the vertex as: x=−b2a{\displaystyle x=-{\frac {b}{2a}}} x=−6(2)(−3){\displaystyle x=-{\frac {6}{(2)(-3)}}} x=−6−6{\displaystyle x=-{\frac {6}{-6}}} x=−(−1){\displaystyle x=-(-1)} x=1{\displaystyle x=1} Find the corresponding f(x) value. Insert the value of x that you just calculated into the function to find the corresponding value of f(x). This will be the minimum or maximum of the function. For the first example above, f(x)=x2+10x−1{\displaystyle f(x)=x^{2}+10x-1}, you calculated the x-value for the vertex to be x=−5{\displaystyle x=-5}. Enter −5{\displaystyle -5} in place of x{\displaystyle x} in the function to find the maximum value: f(x)=x2+10x−1{\displaystyle f(x)=x^{2}+10x-1} f(−5)=(−5)2+10(−5)−1{\displaystyle f(-5)=(-5)^{2}+10(-5)-1} f(−5)=25−50−1{\displaystyle f(-5)=25-50-1} f(−5)=−26{\displaystyle f(-5)=-26} For the second example above, f(x)=−3x2+6x−4{\displaystyle f(x)=-3x^{2}+6x-4}, you found the vertex to be at x=1{\displaystyle x=1}. Insert 1{\displaystyle 1} in place of x{\displaystyle x} in the function to find the maximum value: f(x)=−3x2+6x−4{\displaystyle f(x)=-3x^{2}+6x-4} f(1)=−3(1)2+6(1)−4{\displaystyle f(1)=-3(1)^{2}+6(1)-4} f(1)=−3+6−4{\displaystyle f(1)=-3+6-4} f(1)=−1{\displaystyle f(1)=-1} Report your results. Review the question you have been asked. If you are asked for the coordinates of the vertex, you need to report both the x{\displaystyle x} and y{\displaystyle y} (or f(x){\displaystyle f(x)}) values. If you are only asked for the maximum or minimum, you only need to report the y{\displaystyle y} (or f(x){\displaystyle f(x)}) value. Refer back to the value of the a{\displaystyle a} coefficient to be sure if you have a maximum or a minimum. For the first example, f(x)=x2+10x−1{\displaystyle f(x)=x^{2}+10x-1}, the value of a{\displaystyle a} is positive, so you will be reporting the minimum value. The vertex is at (−5,−26){\displaystyle (-5,-26)}, and the minimum value is −26{\displaystyle -26}. For the second example, f(x)=−3x2+6x−4{\displaystyle f(x)=-3x^{2}+6x-4}, the value of a{\displaystyle a} is negative, so you will be reporting the maximum value. The vertex is at (1,−1){\displaystyle (1,-1)}, and the maximum value is −1{\displaystyle -1}. Method 2 of 3:Using the Standard or Vertex Form 1. Write your quadratic function in standard or vertex form. The standard form of a general quadratic function, which can also be called the vertex form, looks like this: f(x)=a(x−h)2+k{\displaystyle f(x)=a(x-h)^{2}+k} If your function is already given to you in this form, you just need to recognize the variables a{\displaystyle a}, h{\displaystyle h} and k{\displaystyle k}. If your function begins in the general form f(x)=ax2+bx+c{\displaystyle f(x)=ax^{2}+bx+c}, you will need to complete the square to rewrite it in vertex form. To review how to complete the square, see Complete the Square. Determine the direction of the graph. Just as with a quadratic function written in its general form, you can tell the direction of the parabola by looking at the coefficient a{\displaystyle a}. If a{\displaystyle a} in this standard form is positive, then the parabola opens upward. If a{\displaystyle a} is negative, then the parabola opens downward. Look at the following examples: For f(x)=2(x+1)2−4{\displaystyle f(x)=2(x+1)^{2}-4}, a=2{\displaystyle a=2}, which is positive, so the parabola opens upward. For f(x)=−3(x−2)2+2{\displaystyle f(x)=-3(x-2)^{2}+2}, a=−3{\displaystyle a=-3}, which is negative, so the parabola opens downward. If the parabola opens upward, you will be finding its minimum value. If the parabola opens downward, you will find its maximum value. Identify the minimum or maximum value. When the function is written in standard form, finding the minimum or maximum value is as simple as stating the value of the variable k{\displaystyle k}. For the two example functions given above, these values are: For f(x)=2(x+1)2−4{\displaystyle f(x)=2(x+1)^{2}-4}, k=−4{\displaystyle k=-4}. This is the minimum value of the function because this parabola opens upward. For f(x)=−3(x−2)2+2{\displaystyle f(x)=-3(x-2)^{2}+2}, k=2{\displaystyle k=2}. This is the maximum value of the function, because this parabola opens downward. Find the vertex. If you are asked for the coordinates of the minimum or maximum value, the point will be (h,k){\displaystyle (h,k)}. Note, however, that in the standard form of the equation, the term inside the parentheses is (x−h){\displaystyle (x-h)}, so you need the opposite sign of the number that follows the x{\displaystyle x}. For f(x)=2(x+1)2−4{\displaystyle f(x)=2(x+1)^{2}-4}, the term inside the parentheses is (x+1), which can be rewritten as (x-(-1)). Thus, h=−1{\displaystyle h=-1}. Therefore, the coordinates of the vertex for this function are (−1,−4){\displaystyle (-1,-4)}. For f(x)=−3(x−2)2+2{\displaystyle f(x)=-3(x-2)^{2}+2}, the term inside the parentheses is (x-2). Therefore, h=2{\displaystyle h=2}. The coordinates of the vertex are (2, 2). Method 3 of 3:Using Calculus to Derive the Minimum or Maximum 1. Start with the general form. Write your quadratic function in general form, f(x)=ax2+bx+c{\displaystyle f(x)=ax^{2}+bx+c}. If necessary, you may need to combine like terms and rearrange to get the proper form. Begin with the sample function f(x)=2x2−4x+1{\displaystyle f(x)=2x^{2}-4x+1}. Use the power rule to find the first derivative. Using basic first-year calculus, you can find the first derivative of the general quadratic function to be f′(x)=2ax+b{\displaystyle f^{\prime }(x)=2ax+b}. For the sample function f(x)=2x2−4x+1{\displaystyle f(x)=2x^{2}-4x+1}, find the derivative as: f′(x)=4x−4{\displaystyle f^{\prime }(x)=4x-4} Set the derivative equal to zero. Recall that derivative of a function tells you the slope of the function at that selected point. The minimum or maximum of a function occurs when the slope is zero. Therefore, to find where the minimum or maximum occurs, set the derivative equal to zero. Continue with the sample problem from above: f′(x)=4x−4{\displaystyle f^{\prime }(x)=4x-4} 0=4x−4{\displaystyle 0=4x-4} Solve for x. Use basic rules of algebra to rearrange the function and solve the value for x, when the derivative equals zero. This solution will tell you the x-coordinate of the vertex of the function, which is where the maximum or minimum will occur. 0=4x−4{\displaystyle 0=4x-4} 4=4x{\displaystyle 4=4x} 1=x{\displaystyle 1=x} Insert the solved value of x into the original function. The minimum or maximum value of the function will be the value for f(x){\displaystyle f(x)} at the selected x{\displaystyle x} position. Insert your value of x{\displaystyle x} into the original function and solve to find the minimum or maximum. For the function f(x)=2x2−4x+1{\displaystyle f(x)=2x^{2}-4x+1} at x=1{\displaystyle x=1}, f(1)=2(1)2−4(1)+1{\displaystyle f(1)=2(1)^{2}-4(1)+1} f(1)=2−4+1{\displaystyle f(1)=2-4+1} f(1)=−1{\displaystyle f(1)=-1} Report your solution. The solution gives you the vertex of the maximum or minimum point. For this sample function, f(x)=2x2−4x+1{\displaystyle f(x)=2x^{2}-4x+1}, the vertex occurs at (1,−1){\displaystyle (1,-1)}. The coefficient a{\displaystyle a} is positive, so the function opens upward. Therefore, the minimum value of the function is the y-coordinate of the vertex, which is −1{\displaystyle -1}. Sponsored by Bigin by Zoho CRM Too many tools slowing your business down? Try Bigin—the easiest CRM. Helps manage deals, customers & and business proccess, all in one place. CMA K S Narayanan Cost Accountant having 24 years Telecom Experience · Author has 7K answers and 13.5M answer views · 7y Originally Answered: How do we get the expression for the maximum or minimum value of a quadratic equation? · The general form of Quadratic Expression is : f(x) = ax² + bx + c f’(x) = 2ax + b For maxima or minima, f’(x) = 0 => 2ax + b = 0 => x = -b/2a Further, value of expression is maximum when f’’(x) < 0 => 2a < 0 or a is negative. In this case, the graph ( parabola ) will open downwards. Value of expression is minimum when f’’(x) > 0 => 2a > 0 or a is positive. In this case, the graph ( parabola ) will open upwards. To find minima or maxima, put the value of x = -b/2a in f(x). :-) Related questions How do I find maximum and minimum value of a function? What is it possible for a quadratic function to have both minimum and maximum values? Explain by giving examples. What is the maximum or minimum value of a quadratic function if it has no turning point? What is the method for finding the minimum or maximum of a quadratic function without completing the square? Is it possible to find the maximum/minimum value of a function without using calculus? Ayush Thakur Studied at St. Joseph's Academy, Savita Vihar (Graduated 2018) · 8y Originally Answered: What are the maximum and minimum value of a quadratic equation? · It all depends whether the parabola is opening upwards or downwards. For upwards opening parabola coefficient of x 2 is positive. Therefore its maximum value be infinite and its minimum value will be − d / 4 a where d is discriminant of the equation and a is coefficient of x 2 . For downwards opening parabola coefficient of x 2 is negative. Therefore in this case its minimum will be infinite and maximum value will − d / 4 a where d is discriminant of the equation and a is coefficient of x 2 . 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I should not be answering this question because I can come up with only 3 no 4 no 5 … no 4 ways to find the minimum, which by no means could be all the ways. Definition: Given the focus and directrix, the extreme point is halfway between them. If the focus is above the directrix the quadratic is concave upward and has a minimum. Calculus: Written in standard form, y(x) = a x^2 + b x + c the extreme point has x = -b/(2 a) which I can show from calculus Quadratic functions have this astounding property that if they have a maximum there is no minimum. Yet if they have a minimum there is no maximum! I should not be answering this question because I can come up with only 3 no 4 no 5 … no 4 ways to find the minimum, which by no means could be all the ways. Definition: Given the focus and directrix, the extreme point is halfway between them. If the focus is above the directrix the quadratic is concave upward and has a minimum. Calculus: Written in standard form, y(x) = a x^2 + b x + c the extreme point has x = -b/(2 a) which I can show from calculus and I’m sure there’s an algebraic derivation as well (which I accidentally derive later). If the coefficient `a’ is positive then the quadratic has a minimum. Graph: Thirdly you might be able to graph the function and pick the coordinates off the graph, and having drawn it you’d know what sort of extreme point you had found. Exploit symmetry: Written in standard form the quadratic function crosses the y axis at y(0) = c. Use root finding to find the two values of x at which (a x^2 + b x + c = c) … oh wait this is turning out to be the same method 2. (x = 0 and x = -b/a, averaged) Algorithmic: Finally you could apply an optimization algorithm to find x at the extreme point. Optimizers find minima. If `a’ is negative apply the algorithm to negative y(x). If the domain is limited then the quadratic will have a maximum and minimum. In addition to finding x and y(x) evaluate the quadratic at the ends of the interval as well. Use a bit of creative thinking to figure out what’s what and where is where. Jane Smith Ghost at McDonald's (fast food chain) (2000–present) · Author has 76 answers and 253.4K answer views · 7y Originally Answered: What are all the ways to find the maximum and minimum of a quadratic function? · These were the only ones I could think of: Lets say y = x² - x - 2 Factorise it to get the values of x when y is 0 and find the middle value of the x's, then insert it in the given equation to obtain y. Ex: 0 = (x -2)(x+1) x = 2, x = -1 Mid value of x = (2–1)/2 = 1/2 When x = 1/2, y = -2.25 Differentiate it and equate it to 0 and find x. dy/dx = 2x - 1 0 = 2x - 1 x = 1/2 y = -2.25 Plot the equation and look at the min or max poing Sponsored by CDW Corporation Need AI intuition without compromising compute power? Explore AMD Ryzen™ and Windows 11-powered x86 PCs from CDW to accelerate modern business objectives. Rushikesh Jadhav b tech. from Shri Guru Gobind Singhji Institute of Engineering and Technology, Nanded · 7y Originally Answered: How do we get the expression for the maximum or minimum value of a quadratic equation? · There is not any fix expression for minimum and maximum value. If we take domain from minus infinity to infinity and if we consider the graph of equation then there could only be one value minimum or maximum. For that value we differentiate the equation with respect to X or which is variable term then compare it with zero we get the minimum or maximum value. Then again differentiate and put the value of x from above equation ( X which we got by comparing the differential equation with zero) if it is positive then it is minimun and vice versa. Ex. x²+4x+4=0 For first differential 2x+4=0 this gives X There is not any fix expression for minimum and maximum value. If we take domain from minus infinity to infinity and if we consider the graph of equation then there could only be one value minimum or maximum. For that value we differentiate the equation with respect to X or which is variable term then compare it with zero we get the minimum or maximum value. Then again differentiate and put the value of x from above equation ( X which we got by comparing the differential equation with zero) if it is positive then it is minimun and vice versa. Ex. x²+4x+4=0 For first differential 2x+4=0 this gives X=-2 So at X=-2 it could be minimum or maximum value. For that again differentiate wrt X We get 2 that is positive then it is minimun value. Gaurav Kumar Former Mathematics Learner · Author has 536 answers and 1.7M answer views · 5y Originally Answered: What is the general formula of the minimum and maximum values of quadratic equations? · Dip Bhattacharya M.S + Ed D in Mathematics & Secondary Math Education, Clarion University Of Pennsylvania (Graduated 1982) · Author has 3.1K answers and 2M answer views · 4y The equation does not have any Max or Min value. However , a Quadratic function could have a Max or Min because if you plot it , it will look like a parabola either like U ( in that case we have a min ) or it will look like an inverted u ( in that case we have a Max ) All Quadratic functions can be f(x) = ax^2 + bx + c and Max or min will happen when f’(x) = 0 2ax + b = 0 will give x = -b/2a To find if it is a Max ( there are other ways ) , look at the sign of the 2nd derivative that is the sign of 2a or just a , if a is negative ( it will look like an inverted U and we have a Max , and if the the The equation does not have any Max or Min value. However , a Quadratic function could have a Max or Min because if you plot it , it will look like a parabola either like U ( in that case we have a min ) or it will look like an inverted u ( in that case we have a Max ) All Quadratic functions can be f(x) = ax^2 + bx + c and Max or min will happen when f’(x) = 0 2ax + b = 0 will give x = -b/2a To find if it is a Max ( there are other ways ) , look at the sign of the 2nd derivative that is the sign of 2a or just a , if a is negative ( it will look like an inverted U and we have a Max , and if the the sign is positive , it will look like U and we have a min J Paul Tisdel Jr Former Mathematics Teacher · Author has 108 answers and 27.8K answer views · 3y In a quadratic function graph, the vertex of the graph is always the location of the maximum or minimum depending on the graph orientation. The vertex always has its x value at -b/(2a). Where a is the coefficient of x^2 and b is the coefficient of x. y = ax^2 + bx + c once you know x, substitute it back into the equation to find y, the max or min. Related questions How can you find the maximum or minimum value of a quadratic function without graphing it? How can you determine the maximum or minimum value of a quadratic function? Illustrate using an example. How do I find the maximum value of a quadratic function? What are the maximum and minimum value of a quadratic equation? How do you find the maximum and minimum value of y in a quadratic graph? How do I find maximum and minimum value of a function? What is it possible for a quadratic function to have both minimum and maximum values? Explain by giving examples. What is the maximum or minimum value of a quadratic function if it has no turning point? What is the method for finding the minimum or maximum of a quadratic function without completing the square? Is it possible to find the maximum/minimum value of a function without using calculus? What is the maximum value of a quadratic inequality? What is the range of the quadratic function? What are the requirements for determining the minimum and maximum values of quadratic equations? How do you find the minimum of a quadratic function? A quadratic function is expressed in standard form. (2 marks) a. Explain two different techniques for determining the maximum or minimum value of the function. b. Is it always possible to use both techniques? Explain About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10893	https://people.math.wisc.edu/~angenent/171/sinaddition.pdf	Derivation of the addition formulas for Sine and Cosine Assume the circle has radius 1, and that its center is the origin. Then sin(α+β) = AE = AC+CE. Computation of AC: OAB is a right triangle, with OA=1. Hence AB = 1× sin β = sin β. The angles α and ∠OFE are complementary, so ∠OFE=90-α. The angles ∠OFE and ∠AFB are equal, so ∠AFB=90-α. The angles ∠AFB and ∠BAF are also complementary, so ∠BAF=90-∠AFB=90-(90-α) = α. Now we know one angle and one side of the right triangle ABC, namely, ∠BAF= α, and AB = sin β. Therefore we get AC=cos α sin β. A B C E D O α β F Computation of CE. EC equals OD. OBD is a right triangle. The hypothenuse of OBD is OB, which is also the adjacent side of the right triangle OBA. Hence OB = cos β. The angle ∠OBD equals the angle α since BD is parallel to the x-axis. Now we know one angle and one side of the right triangle OBD, namely, ∠OBD= α, and OB = cos β. Therefore we get EC =OD = OB sin (∠OBD) =cos β sin α. Math 171--Fall 1997 Angenent Add the results of these two computations together, and you get sin(α+β) = sin α cos β + sin β cos α As an exercise, try getting the formula for cos (α+β) in the same way. Hint: cos (α+β) = OE = CD= BD-BC; now use the same two right triangles ABC and OBD. How to make this drawing: First draw the circle, then angles α and β. Then draw AB so that OBA is a right triangle. Next, draw BD parallel to the x-axis and AE parallel to the y-axis.
10894	https://www.ck12.org/flexi/geometry/triangle-proofs/how-to-calculate-the-area-of-an-isosceles-triangle/	Flexi answers - How to calculate the area of an isosceles triangle? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Geometry Theorems about Triangles Question How to calculate the area of an isosceles triangle? Flexi Says: An isosceles triangle has two sides of the same length and a third side of different lengths. i The formula for the perimeter and area of an isosceles triangle is i Perimeter=2 a+b Altitude=a 2−b 2 4 Area of isosceles triangle=1 2×Base×Height=1 2×b×a 2−b 2 4 To learn more about the area of triangles, click here! Analogy / Example Try Asking: If △ABC is isosceles with vertex angle ∠ABC, and 𝐴𝐸 ≅ 𝐹𝐶 , which theorem or postulate can be used to prove △AEB ≅ △CFB?Prove that a triangle must have atleast two acute angles.In a triangle ABD, AB = 13, BC = 16 and AC = 8. Find the perimeter of the triangle formed by joining the midpoints of the sides of the triangle. How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy × Image Attribution Credit: Source: License:
10895	https://web.njit.edu/~vitaly/121/notes121M.pdf	ELECTRICITY & MAGNETISM Lecture notes for Phys 121. Part II. Dr. Vitaly A. Shneidman Department of Physics, New Jersey Institute of Technology, Newark, NJ 07102 (Dated: April 26, 2025) Abstract This part of Lecture Notes is mostly on magnetism. It continues Part I on electrostatics and current. Please refer to Part I for the general information (Abstract) and for the Introduction on vectors and vector ﬁelds. 1 Contents I. Magnetic ﬁeld and magnetic force. 2 1. Properties of magnetic ﬁeld lines 4 II. Magnetic force 5 A. Circular motion of a particle 7 B. 3D motion of charged particles 9 C. Magnetic force on a wire 11 III. Magnetic Fields from Currents 13 A. Straight wire 13 B. Magnetic force between two parallel wires 14 C. The Biot-Savarat law 15 1. Applications of Biot-Savarat 16 D. The Ampere’s theorem 19 1. Applications of AT H ⃗ B · d⃗ s = µ0Ienc 20 IV. The Faraday’s Law 25 A. Experimental 25 B. Theory 25 1. Magnetic ﬂux 25 C. The Law 26 D. Lenz’s rule 29 E. Induced electric ﬁeld 30 F. Advanced: relativity of electric and magnetic ﬁelds 30 V. Self-induction 32 A. Inductance - deﬁnition 32 B. RL circits 33 1. Decay of current 33 2. Build-up of current 34 C. Magnetic Energy of an Inductor 36 D. Advanced: Energy stored in magnetic ﬁeld 37 0 VI. Electromagnetic oscillations 38 A. math 38 B. LC-circuit, free oscillations 39 1. Diﬀerential equation 40 2. Advanced: Damped oscillations 41 C. Driven oscillations and resonance 43 1. Negligible R 43 2. R ̸= 0 44 3. Phasors 45 D. Power 48 E. Transformers 49 VII. Maxwell Equations and Electromagnetic Waves 51 A. The Maxwell equations 52 B. And then there was Light! 53 1 Dr. Vitaly A. Shneidman, Phys 121h, 9th Lecture I. MAGNETIC FIELD AND MAGNETIC FORCE. We now start studying a new vector ﬁeld, the magnetic ﬁeld ⃗ B: Notations: ⃗ B , Units: [B] = T (tesla) Earth B < 10−4 T (depends on location, magnetic ”South pole” is close to geographic ”North”) a refrigerator magnet: B ∼10−2 T , MRI: B ∼1.5 T , Lab magnet: B > 20 T , . . . Safety: time-independent magnetic ﬁeld of several teslas or less is mostly safe, but if ﬁeld is time-dependent or if you move in or out of magnetic ﬁeld, it can become dangerous even for modest values of the ﬁeld (will discuss later in connection with Faraday’s eﬀect). How does the magnetic ﬁeld originate (in other words, What are the sources of this ﬁeld)? Unlike the electrostatic ﬁeld E which is due to electric charges, there are no magnetic charges (or, more cautiously, such charges have not been discovered despite intense search during the last 100 years - see the cover of Physics Today from a few years ago). In the absence of ”elementary” sources of magnetic ﬁeld, one can identify currents, including the microscopic current as electrons in orbital motion around a nucleus, and permanent magnets. Many elementary particles, including electrons, have their own magnetic moments. Competition of the inherent magnetic moment of the electrons with magnetic eﬀects due to orbital motion determines magnetic properties of a material; usually, orbital motion prevails but on occasions (iron, nickel) the inherent magnetic moment takes the upper hand. 2 ”Elementary” sources of magnetic ﬁeld. Left -there are NO known ”magnetic charges” (particles with only North or only South poles). Right - a microscopic ring current, as electrons around a nucleus (note that direction of motion is opposite for electrons, due to a negative charge). The North pole is on top, and the South pole is on the bottom. Many elementary particles have their own magnetic dipole moments, with a similar structure of ﬁeld. Conventions for pictures Problems with magnetism are always 3D. To simplify graphics, will use a 2D convention, as in Fig. 1: if ﬁeld goes into the page you see ”feathers-of-an-arrow” which ﬂies away from you; if ﬁeld goes out of the page, you see ”head-of-an-arrow” ﬂying towards you. (A similar convention will be used later when drawing currents which ﬂow in or out of the page). FIG. 1: Convention for 2D pictures of magnetic ﬁeld. Left - ﬁeld goes into the page; right - ﬁeld goes out of the page. 3 1. Properties of magnetic ﬁeld lines Similar to electric ﬁeld: • tangent - direction of ⃗ B • density - proportional to ⃗ B • do not cross Diﬀerent from electric ﬁeld: • magnetic lines never start or end (i.e. there are NO magnetic charges) • magnetic lines can loop (they loop around currents). This does not contradict conser-vation of energy since the static magnetic ﬁeld does not do work. 4 II. MAGNETIC FORCE The force acting on a charge moving with velocity ⃗ v in ﬁeld ⃗ B is given by ⃗ F = q⃗ v × ⃗ B (1) Similarly to electric force, the magnetic force is proportional to charge and ﬁeld magnitudes. But, this force acts only on moving charges and is perpendicular (not parallel) to direction of the ﬁeld ⃗ B. v B F v B F Magnetic force on a positive particle, ⃗ F = q⃗ v × ⃗ B. Note that ⃗ F (blue) is perpendicular to both ⃗ v (red) and ⃗ B (green) , and reaches maximum when ⃗ v and ⃗ B are perpendicular to each other. For parallel (or antiparallel) ⃗ v and ⃗ B the force would be zero. Power: P = ⃗ F · ⃗ v = 0 ⇒no work by static magnetic ﬁeld (2) 5 i  j  k  i x j =k  j xk = i  k x i = j  i xk =- j , etc. Example. Find acceleration of an electron if ⃗ v = V (⃗ i+⃗ j) , ⃗ B = b(⃗ j +⃗ k) with V = 106 m/s and b = 1 T. Solution. Use the ”ring diagram” above (or, look up the components of the cross product): ⃗ a = 1 me ⃗ F = 1 me (−\|e\|)⃗ v × ⃗ B = 1 me (−\|e\|)V b(⃗ i +⃗ j) × (⃗ j + ⃗ k) = = 1 me (−\|e\|)V b(⃗ i×⃗ j +⃗ i×⃗ k+⃗ j ×⃗ k) = 1 me (−\|e\|)V b(⃗ k−⃗ j +⃗ i) = 1 me \|e\|V b(−⃗ k+⃗ j −⃗ i) ∼1017m s2 Example: separation of particles by charges and masses - see Fig. 2. FIG. 2: Separation of particles of diﬀerent signs in the magnetic ﬁeld, as observed in radioactive decay. The positive (red) particles deviate to the right (they were called ”alpha-particles” and turned out to be nuclei of Helium). The negative (blue) particles deviate left. (they were called ”beta-particles” and turned out to be electrons). Neutral particles (green) do not deviate. (they were called ”gamma-particles”, and turned out to be quanta of electromagnetic radiation). Note that the path of negative particles is more curved, due to smaller mass. 6 A. Circular motion of a particle Since the magnetic force ⃗ F is perpendicular to the velocity, the latter changes only the direction while the magnitude (”speed”) remains constant. The same is true for ⃗ F. Thus, the magnetic force leads to circular motion. Circular motion of a positive particle in magnetic ﬁeld. The velocity of a particle is shown in red, and the magnetic force (blue) provides a centripetal force for the circular motion. The period of revolution is independent of the velocity (!) - a faster particle will make a larger circle, completing the revolution in the same time. For the actual calculations we use the 2nd Law mv2 r = qvB , or r = mv qB (3) This is the experimental way to ﬁnd q/m for elementary particles, and eventually to ﬁnd their masses (which otherwise are very small)- see the example on next page. We now use the above equation to ﬁnd the period of revolution T = 2πr/v: T = 2π m qB (4) Doest not depend on the velocity (energy) of a particle (!) 7 Example (Advanced). A particle is accelerated through a potential diﬀerence ∆V and the enters magnetic ﬁeld B at a right angle, where it revolves in a circle of radius r. Find q/m and the period of revolution. from work-energy theorem 1 2mv2 = q∆V ⇒v = r q m2∆V we substitute this into r = mv qB ⇒r = m qB r q m2∆V = 1 B rm q 2∆V ⇒ q m = 2∆V (rB)2 The period of revolution : T = 2πm q 1 B , is independent of ∆V Here are a few typical q/m ratios proton : q m ≃1.6 · 10−19 C 1.67 · 10−27 kg ∼108 C kg electron : q m ≃1.6 · 10−19 C 9.11 · 10−31 kg ∼1011 C kg alpha-particle : q m ≃2 × 1.6 · 10−19 C 4 × 1.67 · 10−27 kg ≈0.5 · 108 C kg The cyclotrone (advanced topic, not on exam) Since magnetic ﬁeld alone cannot change energy, need electric ﬁeld to accelerate particles. The cyclotrone is placed in strong magnetic ﬁeld which is perpendicular to the plane of the picture and which makes a particle to move in a circular ark. When the particle reaches the space between the half-rings an electric ﬁeld parallel to ⃗ v accelerates the particle, increasing its energy and radius of the ark (but not changing the period of revolution!). When the particle, after completing a half circle, again enters the space between the half-rings, the electric ﬁeld changes to the opposite direction again being parallel to velocity vector and further accelerating the particle, etc. 8 B. 3D motion of charged particles In a uniform magnetic ﬁeld, circular motion is observed only if ⃗ v of a particle is strictly perpendicular to ⃗ B. More generally, the velocity can be broken into 2 components, ⃗ v\|\| parallel to ⃗ B and v⊥perpendicular to B. Since the magnetic force contains a cross product, the parallel component of the velocity will not matter and force will be only in direction perpendicular to ⃗ B with magnitude determined only by v⊥. This will determine the motion in a plane perpendicular to ⃗ B while motion along ⃗ B will proceed with a constant speed v\|\|. A combination of linear and circular motion will produce a helix - see the left picture below. The actual calculations, however, are on advanced side and can be skipped in the ﬁrst reading. Left: uniform ﬁeld. Brake ⃗ v into components parallel and perpendicular to ⃗ B ⃗ v = ⃗ v\|\| + ⃗ v⊥, ⃗ F = q⃗ v⊥× ⃗ B r = mv⊥ qB , T = 2π m qB v\|\| = const ⇒step of the helix = v\|\|T = 2πmv\|\| qB Right: non-uniform ﬁeld. Paths of charged particles still approximately wound around the ﬁeld lines. Examples: Aurora Borealis and ’magnetic bottle’ - in class. 9 Motion in combined ⃗ B and ⃗ E ﬁelds When both ⃗ B and ⃗ E are present, one has to consider the total force which is the sum of electric force ⃗ Fe = q ⃗ E and magnetic force ⃗ Fm = q⃗ v × ⃗ B ⃗ F = q ⃗ E + ⃗ v × ⃗ B (5) For example, the condition that force is zero (i.e. particle moves in a straight line) results in a condition ⃗ E + ⃗ v × ⃗ B = 0 regardless of the sign of the charge. Note that in order to have a zero force, ⃗ E cannot be parallel to ⃗ B and in the simplest case when it is perpendicular, E = vB In a general case, eq.(5) should be combined with the Newton equation m⃗ a = ⃗ F Solution is hard, and can be obtained analytically only when both ﬁelds are uniform and perpendicular to each other. Otherwise, it is the work for a computer. An example of a cycloid is given below. E Motion of a charged particle in magnetic ﬁeld, ⃗ B which points in the vertical z-direction; the electric ﬁeld ⃗ E points in the y-direction, into the page. With vz = 0 the particle stays in the plane but drifts in the x-direction. [The latter is important in understanding the rather advanced Hall eﬀect]. 10 C. Magnetic force on a wire Conventional current inside a wire can be represented as motion of positive charges q with drift velocity v = vd. Each charge experiences a force q⃗ v × ⃗ B (shown for a single charge in the picture). v + + + + + + + + + + + + + + + + + + + + + B F The forces add up to give the total force on the segment of wire of length L. Transition from formulas for a single charge which moves with velocity ⃗ v to formulas for a small wire which carries current i and has length ⃗ L (vector in direction of current), is given by q⃗ v →i⃗ L (6) Thus Fw = iLB , or in vector form ⃗ F = i⃗ L × ⃗ B (7) Example: current (blue) up in the plane of the page; ﬁeld out of the page. Find the direction of force (red). Note: picture can be rotated. Now, derivation of eq. (7). Force on a single charge q: ⃗ Fi = q⃗ vd × ⃗ B (⃗ vd is the drift velocity). Force on a unit volume (with n the number of charges in unit volume and ⃗ J = nq⃗ vd the density of current): n · ⃗ Fi = nq⃗ vd × ⃗ B = ⃗ J × ⃗ B Multiply by volume V = LA (with L length of wire and A the cross-sectional area) and use i = JA to get eq. (7). 11 FIG. 3: Torque on a frame with current (left) and brush contacts(right) used to ensure spinning in one direction. A loop with current will experience a torque when placed in magnetic ﬁeld. Major application - electric motor. If ⃗ A is the vector-area of the loop (in the direction of the normal) and θ is the angle between ⃗ A and ⃗ B, then torque τ = AIB sin θ (8) In vector form: magnetic moment of a loop ⃗ µ = I ⃗ A (9) where the r.h. rule is used to ﬁnd direction of µ and it should be multiplied by the number of turns N for a coil. Then, ⃗ τ = ⃗ µ × ⃗ B (10) Here is the math derivation. Let the frame in ﬁg. 3 be composed of the following vectors (starting from axis, along currents): ⃗ a, ⃗ b, −2⃗ a , −⃗ b, ⃗ a. Note, ⃗ A = 2⃗ a ×⃗ b, while forces on the ”b” sides equal ±I⃗ b × ⃗ B = ±⃗ Fb . Thus, ⃗ τ = 2⃗ a × ⃗ Fb = I ⃗ A × ⃗ B, which proves the above equations. 12 Dr. Vitaly A. Shneidman, Phys 121h, 10th Lecture III. MAGNETIC FIELDS FROM CURRENTS A. Straight wire Magnetic ﬁeld (red circular lines) due to a long straight current (blue). Note the direction of the ﬁeld determined by the r.h. rule. Right: same in a 2D view. Direction: r.h. rule. Thumb of r.h. goes with the current in the wire; curled ﬁngers show direction of ⃗ B. Field lines circle around the current. The formula for the magnitude B(r) can be derived from various fundamental principles (whatever we agree to treat as ”fundamental”). At the moment we use it as ”empirical result”: B = µ0i 2πr , µ0 = 4π · 10−7 m · T A (11) µ0 −magnetic permeability constant. 13 B. Magnetic force between two parallel wires Consider two parallel wires, a and b with currents ia and ib in the same direction. Wires are separated by a distance d and have length L each. We ﬁnd the force on wire b in two steps: 1) ﬁnd the ﬁeld due to a at the location of b and 2) ﬁnd the force on b We will do steps 1) and 2) ﬁrst for the direction and then for the magnitude. From direc-tion we get that parallel currents attract (and opposite -antiparallel- currents repel!). For magnitude, step 1) gives from eq. (11) Ba = µ0ia 2πd Then, from formula for force on a wire F = iLB: Fb = ibLBa = ibLµ0ia 2πd Fb = µ0iaibL 2πd (12) Checkpoint: ﬁnd the force on wire a. Should satisfy the 3rd Law. 14 C. The Biot-Savarat law How to ﬁnd a ﬁeld due to a wire of arbitrary complex shape? Contribution of a small wire with current i is d ⃗ B = µ0 4πid⃗ s × ⃗ r r3 (13) Here d⃗ s is the vector which indicates the elementary wire (in the direction of the current), while ⃗ r goes from the wire to the observation point. The superposition principle. Fields of diﬀerent elementary wires add up as vectors. Can be used, in principle, to ﬁnd a ﬁeld from a wire of any complex shape or from several wires. Note: there will be no ﬁeld in the direction of the wire. 15 1. Applications of Biot-Savarat Straight wire. Can derive the above formula 11, but will get it more eﬃciently from the Ampere’s theorem. However, will discuss an example of ﬁeld due to 2 parallel wires to illustrate the superposition. Superposition of ﬁelds due to two currents, Left: one into the page (red) and one out of the page (blue); resultant ﬁeld at a point equidistant from both currents will be up. Right (for which illustrative calculations are done below): both currents out of the page. Let 2a - distance between wires, r - distance from each wire to the observation point; h = √ r2 −a2. Full ﬁeld ⃗ B - black (horizontal) sum of red and blue. Note triangle formed by red and blue ﬁelds is similar to the one formed by the two currents and observation point. Thus, B/2 B1 = h r , B = 2B1 h r = µ0I πr h r 16 Ring (center only). The simpliﬁcation here is that d⃗ s is always perpendicular to ⃗ r, and that r = const = R (R is the radius of the ring) for every element of the ring. Thus, dB = µ0i 4πr2ds. The direction is determined by the r.h. rule (it is more convenient to curl ﬁngers with the current; then the thumb points in the direction of the ﬁeld). Contributions of individual elements just add up. Thus Bring = µ0i 4πr2 Z 2πR 0 ds = µ0i 2R For an arc with angle φ < 2π the ﬁeld is reduced proportionally Barc = µ0i 2R · φ 2π (14) FIG. 4: Example. Superposition of ﬁeld from 2 half-loops with R > r: B1 = µ0I/(2r) × 0.5 (out of the page) and B2 = µ0I/(2R) × 0.5 < B1 (into the page). B = B1 −B2 = µ0I 4 1 r −1 R 17 Example. Consider a semi-inﬁnite wire with current I to the right. Find ﬁeld B at a distance D from the end, perpendicular to the direction of current. dB (into) I D 0 -∞ r  dx dB = µ0 4πI dx r sin α r3 , r = √ x2 + D2 , sin α = D r B = Z 0 −∞ dB(x) = µ0 4πID Z 0 −∞ dx (x2 + D2)3/2 = µ0 4πI x D (x2 + D2)1/2 0 −∞ B = µ0I 4πD (which is half of the ﬁeld of an inﬁnite wire). 18 D. The Ampere’s theorem n  Preliminaries: Geometry. A closed contour (geometric line) is considered, and a positive direction along this contour is selected. A current is said to be enclosed if it passes inside the contour, piercing the imaginary surface stretched on this contour. The contribution of such a current is positive if it ﬂows in the positive direction determined by the r.h. rule. The Ampere’s theorem: I ⃗ B · d⃗ s = µ0ienc = µ0 (I1 −I2 + . . .) (15) 19 1. Applications of AT H ⃗ B · d⃗ s = µ0Ienc The AT is always valid, but to make it useful for calculations in each speciﬁc case one needs to select a good contour which is consistent with the symmetry of the problem (when it exists). Then the circulation can be evaluated even before ⃗ B is known, and then used in the AT to ﬁnd ⃗ B. FIG. 5: Calculating B outside and inside a long wire. Current - into the page (”X”); ﬁeld lines - CW. Outside, r > R - red dashed Ampere’s loop; inside r < R - black dashed. In both cases H ⃗ B · d⃗ l = 2πrB, but only current inside the loop contributes to AT - full current I for the outside loop, and I(r/R)2 for the inside loop. Long wire: outside Good contour - circle around the wire, in the perpendicular plane. Radius r, center at the wire. Then I ⃗ B · d⃗ s = B I ds = 2πrB = (from AT) µ0i which gives eq. (11) = µ0i/(2πr) (and R of the wire does not matter as long as r > R). Long wire: inside Assume that the current is uniformly distributed (has a constant density) inside the wire with radius R. Contour - same, with the center on the axes of the cylindrical wire. Circulation - same as above, 2πrB. Enclosed current is now diﬀerent ienc = i πr2 πR2 ⇒B(r) = µ0i 2πR r R , r ≤R Note that ﬁeld is the same when the surface of the wire is approached either from the inside or from the outside. 20 R r Bmax B Magnetic ﬁeld as a function of the distance r from the axis of the wire of radius R, both inside the wire (r < R) and outside the wire (r > R). The maximum Bmax = µ0I/(2πR) is achieved on the surface r = R. Solenoid. Consider a long, tightly wound coil with current i. The coil has n turns of wire per meter, and the radius turns out to be unimportant. First, we need to understand the structure of ﬁeld lines in order to get a feeling for the symmetry.One has practically straight ﬁeld lines inside the solenoid, while outside they curve and are at a very large distance from each other. Direction of lines is given by the r.h. rule. The cross section is shown on right, with the upper currents (red) going into the page and the lower currents (blue) going out of the page. The magnetic ﬁeld lines are shown by horizontal red arrows.(North pole is on the left). The selected contour is shown by a dashed rectangle. Only the lower side gives a non-zero contribution to circulation, and only those currents which fall inside the rectangle contribute to the Ampere’s theorem. We now select a ”good” contour. Consider a 2D cross section of the coil. For a contour select a long rectangle (with length L) which is partly inside the solenoid and which contains many turns, N = nL ≫1. Two sides are parallel to the axes of the solenoid, and only one of those is inside. Note that only the latter contributes to circulation, so that one has I ⃗ B · ds = BL = (from AT) = µ0Ienc = µ0NI = µ0nLI ⇒BL = µ0nLI B = µ0nI (16) Note that B does not depend on the distance from the axes of the solenoid, as long as one remains inside. 21 x x x x x x x x x x x x x x x x x x x x x x x x Magnetic ﬁeld inside a toroidal coil with inner and outer radii R1 and R2 , with current I and N turns of the wire. In the cross section on right the in/out currents are shown by red/blue circles. The magnetic ﬁeld lines (red) are CCW inside the toroid. The Ampere’s loop (dashed) has a radius r and is also inside the toroid. Note that only the out-of-the-page (blue) currents contribute to AT. I ⃗ B · d⃗ l = 2πrB , Ienc = NI ⇒from AT: B = µ0IN/(2πr) , R1 < r < R2 For R2 ≈R1 ≈r one has N/(2πr) = n, the density of turns and one recovers eq.(16) for a straight solenoid. 22 Example. (Combination of ﬁelds and forces). A rectangular a × b frame with current I2 is placed at a distance c from a straight wire with current I1. Find the forces on each of the 4 sides: ⃗ F1 - upper, ⃗ F2 - lower, ⃗ F3 -left side, ⃗ F4 - right side. I1 x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x a b I2 c Solution (2 step): First, ﬁnd ﬁeld (blue), then ﬁnd forces - see picture. F1 = µ0 I1I2 b 2πc , F2 = µ0 I1I2 b 2π(c + a) dF3 = I2B(x)dx , B(x) = µ0I1/(2πx) F3 = µ0I1I2 2π Z c+a c dx x = µ0I1I2 2π ln c + a c = F4 Fnet = F1 −F2 = µ0I1I2 2π ab c(c + a) 23 Advanced. Field and magnetic interaction of point charges. In Biot-Savarat Law Id⃗ s →q⃗ v : ⃗ B = µ0 4πq⃗ v × ⃗ r r3 Find the ratio of Fm/Fe for 2 identical charges with velocities ⃗ v separated by a distance d. v v Fm Fm Fe Fe + + B = µ0 4πq v d2 ⇒Fm = qvB = µ0 4πq2v2 d2 Fe = 1 4πǫ0 q2 d2 ⇒Fm Fe = µ0ǫ0v2 What is µ0ǫ0? Note: [µ0ǫ0] = s2 m2 , 1 √µ0ǫ0 = 1 4π × 10−74π × 9 × 109 1/2 = 9 × 10161/2 = 3 × 108 m s ⇒ Fm Fe = v2 c2 , c - speed of light 24 Dr. Vitaly A. Shneidman, Phys 121, 11th Lecture IV. THE FARADAY’S LAW FIG. 6: An old circus show ”Lady-on-a-Horse”. The bulb lights up without any apparent reason?? A. Experimental Consider a contour. If the magnetic ﬂux (the number of lines which penetrate through the contour) changes with time, there will be an EMF induced in the contour, as in Fig. 7. If the contour is conducting, there also will be current. The faster is the change in the ﬂux, the larger is the EMF. B. Theory 1. Magnetic ﬂux Deﬁnition: ΦB = Z ⃗ B · d ⃗ A (17) Integration can go over ANY surface limited by a given contour. Units: Wb (webers) 25 FIG. 7: Illustration of the law of induction. If the magnetic ﬂux through the blue contour changes, there will be an induced EMF in the contour (and there will be current if the contour is conducting). The magnitude of EMF is given by the Faraday’s law, and the direction is determined by the Lenz’s rule. In the illustration, since the ﬂux tends to decrease, the induced current will attempt to support the changing ﬂux, being directed as shown. C. The Law E = −dΦB dt (18) Note that the law is valid, and has the same form, regardless of the physical reason why ΦB is changing. Several typical situations can be identiﬁed for Φ(t) ≃⃗ A · ⃗ B 26 • B = B(t) ⇒E = −⃗ A · d ⃗ B dt e.g. when a bar magnet approaches the contour (or the contour approaches the mag-net), or the ﬁeld through a given contour is generated by another coil with a time-dependent current. Mutual induction: 27 • ΦB ≈BA cos [φ(t)] ⇒E = −AB d cos φ dt as in frame revolving between the poles of a bar magnet (prototype of an electric generator). π ω 2 π ω t ℰmax ℰmax ℰ For φ(t) = ωt equation (18) will give E = Emax sin ωt , Emax = ωBA 28 • A = A(t) ⇒E = −B dA dt E.g., for a rod which moves with velocity v - see Fig. 8: A = Lx(t) ⇒dA dt = Lv ⇒ E = vLB FIG. 8: Induction in a contour formed by rails and a sliding rod. The magnetic ﬁeld B (which goes into the page) is constant, but the ﬂux ΦB = BA is changing since the area A is increasing. The magnitude of EMF is given by E = −BdA/dt = −vLB. The direction follows from the Lenz’s rule. The current will be given by I ≈E/R with R being the resistance of rails plus rod. For non-conduction (or, imaginary) rails there will be no current, but the EMF still will be induced • if the contour is a conducting loop, the ﬂux can also change due to change in current in the loop itself. This leads to self-induction and will be discussed separately. D. Lenz’s rule The magnetic ﬂux due to an induced current opposes the changes in the original magnetic ﬂux. This allows one to ﬁnd the direction of induced current. This is consistent with conservation of energy. If current is impossible (non-conducting loop), Lenz’s rule gives the direction of EMF. Example: ﬁg. 8. (1) Flux Φ = AB into the page and grows ⇒(2) induced ﬁeld Bind out of the page ⇒(3) induced current - CCW 29 E. Induced electric ﬁeld Consider a charge q being moved around a contour with EMF induced by a changing ﬂux. One has E = W q = 1 q I ⃗ F · d⃗ s = I ⃗ E · d⃗ s Thus, the Fraday’s Law, eq. (18) can be written as I ⃗ E · d⃗ s = −dΦB dt (19) Note that ⃗ E is very diﬀerent from the electrostatic ﬁeld we studied before. Similarities between induced and electrostatic electric ﬁeld: • both ﬁelds act on charges with forces ⃗ F = q ⃗ E • both ﬁelds can be represented by ﬁeld lines Diﬀerences: • lines of induced electric ﬁeld can loop (and this does not contradict the conservation of energy!) • lines of induced ﬁeld do not start or end (i.e. this ﬁeld is produced not by charges, but by changing magnetic ﬂux). F. Advanced: relativity of electric and magnetic ﬁelds will be discussed in class 30 31 Dr. Vitaly A. Shneidman, Phys 121, 12th Lecture V. SELF-INDUCTION Slider on rails: E = lvB , Iind = E/R , What if R = 0 ?? A. Inductance - deﬁnition Start with static current i, then Φtot = Li Units: [L] = H (henry) = T · m2/A Example: Solenoid, N turns, length l, area A B = µ0iN/l , Φtot = NAB = NAµ0iN/l ⇒ L/l = µ0An2 , n = N/l (20) Now consider i = i(t) ⇒ Φtot = Φtot(t). From Faraday’s Law: EL = −dΦtot/dt, thus Self-induced EMF: EL = −Ldi dt (21) 32 B. RL circits 1. Decay of current loop rule: EL −IR = 0 , ⇒LdI dt + IR = 0 (22) Solution: I(t) = I0e−t/τL , τL = L/R −inductive time constant (23) [Checkpoint: verify that [τL] = s.] Note, if R →0 (superconductor) the current never decays. 33 2. Build-up of current loop rule: E + EL −iR = 0 ⇒Ldi dt + iR = E switch to new current I = i−E/R , LdI dt+IR = 0 with I(0) = −E/R This is the same diﬀerential equation as for RL I(t) = I(0)e−t/τL ⇒i(t) = E R + I(0)e−t/τL = E R −E Re−t/τL i(t) = E R 1 −e−t/τL (24) Note: at t = 0 the inductor has ”inﬁnite resistance”; at t →∞ (constant current) it is just a piece of wire. 34 Example: t →0 : ”clip oﬀ” the inductor I1 = I2 = E R1 + R2 t →∞: replace inductor by a wire I2 = 0 , I1 = E/R1 35 C. Magnetic Energy of an Inductor UL = − Z ∞ 0 ELi dt = L Z ∞ 0 idi dtdt = L Z I 0 i di or UL = 1 2LI2 (25) Example An L = 4.0 mH inductor with some initial current I0 is discharged through a 0.25 Ωresistor. How long does it take to lose half of the initial current? Half of the initial energy? Solution τ = L R = 16 · 10−3 s ; i(t) I0 = e−t/τ = 0.5 t/τ = −ln 0.5 = ln 2 , t = τ ln 2 ≈1.1 · 10−2 s U(t) ∼i(t)2 ⇒U(t) U(0) = e−2t/τ = 0.5 2t/τ = −ln 0.5 = ln 2 , t = 1 2τ ln 2 ≈5.5 · 10−3 s 36 D. Advanced: Energy stored in magnetic ﬁeld From B = µ0In , L = µ0An2l and volume V = lA U = 1 2LI2 = 1 2µ0An2l(B/µ0n)2 = B2V 2µ0 U/V = B2 2µ0 (26) 37 Dr. Vitaly A. Shneidman, Phys 121, 13th Lecture VI. ELECTROMAGNETIC OSCILLATIONS A. math i i i = √ −1 e±i i iα = cos α ± i i i sin α (27) cos α = ei i iα + e−i i iα 2 , sin α = ei i iα −e−i i iα 2i i i (28) d dte±i i iωt = ±i i iωe±i i iωt , d2 dt2e±i i iωt = −ω2e±i i iωt (29) 38 B. LC-circuit, free oscillations The capacitor is originally charged with Q = Qmax. 0.5 1 1.5 2 reduced time -1 -0.5 0.5 1 reduced charge and current 0.5 1 1.5 2 reduced time 0.2 0.4 0.6 0.8 1 energies Free oscillations. Reduced time is t/T, with T = 2π √ LC, the period of oscillations. Left -charge on capacitor q(t)/Qmax (red) and current in the inductor I(t)/Imax (blue). Note that when charge is maximum (or minimum) current is zero, and vice versa. Right - electric (red) and magnetic (blue) energies. Their sum (green line) is the total energy which remains constant (and all energies are reduced by this value). 39 1. Diﬀerential equation From the loop rule: EL + VC = 0 EL = −LdI dt = Ld2q dt2 and VC = q/C Ld2q dt2 + q/C = 0 , a diﬀerential equation for q(t) Look for a solution q(t) = Q cos(ωt) , ω −? (or, q(t) = Qei i iωt) d2q/dt2 = −ω2q , ⇒q(t) −Lω2 + 1/C = 0, ⇒ ω = 1/ √ LC (30) Checkpoint. Calculate explicitly the electric and magnetic energy, and make sure their sum remains constant. 40 2. Advanced: Damped oscillations If the resistance of the LC circuit is non-negligible, the electro-magnetic energy will not be conserved, and the amplitude will get smaller with time - see Fig. 9. One can get the description math-ematically by adding a term Rdq/dt into the l.h.s. of the above diﬀerential equation. 1 2 3 4 5 6 reduced time -0.5 0.5 1 current FIG. 9: Free oscillations in an RLC-circuit. The upper (green) line shows the decaying amplitude due to dissipation of energy in the resistance R. For a suﬃciently large R (”overdamped case”) there would be no oscillations at all. Loop equation. q/C + EL −I(t)R = 0 , I(t) ∼ei i iωt q(t) = − Z I dt ∼−1 i i iωei i iωt , EL = −LdI dt ∼i i iωLei i iωt −1 i i iωC −i i iωL −R = 0 ω2LC −i i iωRC −1 = 0 41 ω2 0 ≡ 1 LC , x ≡ω ω0 : x2 −ixR r C L −1 = 0 x = 1 2i i iR r C L ± p 1 −R2C/4L ω = ±Ω+ i i iγ Ω= ω0 p 1 −R2C/4L , γ = ω0 1 2R p C/L I(t) ∼ei i iωt = e±i i iΩte−γt 42 C. Driven oscillations and resonance E = Em sin (ωdt) FIG. 10: LC-circuit. Driven oscillations. 1. Negligible R Consider Fig. 10 with the driving EMF given by E = Em sin (ωdt) with ωd diﬀerent from ω0 = 1 √ LC This should be added into the r.h.s. of the diﬀerential equation: Ld2q dt2 + q/C = Em sin (ωdt) or d2q dt2 + ω2 0q = Em L sin (ωdt) look for a solution q(t) = Qm sin (ωdt) Now the frequency is known, but Qm is the maximum, yet unknown charge to appear on the capacitor. Similarly to free oscillations d2q/dt2 = −ω2 dq ⇒−ω2 dq + ω2 0q = Em L sin (ωdt) or Qm sin (ωdt) −ω2 d + ω2 0 = Em/L · sin (ωdt) and 43 Qm = Em L 1 ω2 0 −ω2 d The amplitude for current is similar, and is given by Im = ωdQm. Note a dramatic increase of the amplitude when ωd →ω0. This is the resonance - see Fig. 11 0.5 1 1.5 2 2.5 3 reduced frequency 10 20 30 40 scaled amplitude FIG. 11: Resonance in driven oscillations. When the driving frequency ωd is close to the natural frequency ω0 = 1/ √ LC there is an enormous increase in the amplitude. 2. R ̸= 0 Loop: E −q C + EL −I(t)R = 0 E = Emei i iωdt , I(t) = Imei i iωdt q = Z I(t) dt = 1 i i iωd Imei i iωdt , EL = −LdI dt = −Li i iωdImei i iωdt 44 Em − 1 i i iωdCIm −i i iωdLIm −ImR = 0 Im R + i i i ωdL − 1 ωdC = Em \|Im\| Z = \|Em\| , Z = s R2 + ωdL − 1 ωdC 2 3. Phasors i i i = ei i i π 2 ⇒i i iei i iωt = ei i i(ωt+ π 2 ) 1 i i i = e−i i i π 2 ⇒ 1 i i i ei i iωt = ei i i(ωt−π 2 ) FIG. 12: A phasor. It spins in the counter-clockwise direction with the angular frequency ωd and its vertical projection determines the physical quantity, e.g. voltage or current. Phasors for voltage (red) and current (blue) for 3 diﬀerent elements: resistor (left), capacitor (middle) and inductor (right). Note that for a resistor the voltage and current have the same 45 phase, for the capacitor current leads the voltage by 90o, and for the inductor the current lags by 90o. FIG. 13: Solving a driven RLC circuit using phasors. The current (blue) is identical since all ele-ments are in series. Voltages (red) are to be added as vectors, which can be done using Pythagorean theorem. The resultant (dashed) should correspond to the driving EMF. In case of a resonance the voltage on the inductor (leading voltage - upper left) and the voltage on capacitor (lagging voltage - lower right) would completely cancel each other. Resistor: VR = ImR Capacitive reactance: XC = 1 ωdC , VC = \|Im\| XC Inductive reactance: XL = ωdL , VL = \|Im\| XL V 2 R + (VL −VC)2 = \|Em\|2 and Impedance: Z = q R2 + (XL −XC)2 Amplitude of current and phase angle: I = Em Z , tan φ = XL −XC R 46 resonance: ωd = ω0 , XL = XC Z = R = min , I = Em/R = max R = 1 ohm R = 5 ohm 950 1000 1050 1100 fd , Hz 20 40 60 80 100 120 Im , A Amplitude of current for a variable driving frequency fd with ﬁxed L = 0.1 H, C = 0.253 µF (with fres = 1000 Hz), E = 120 V and R = 1 Ω(blue) or R = 5 Ω(red-dashed). 47 D. Power 5 10 15 t -1.0 -0.5 0.5 1.0 sin(t) 5 10 15 t 0.2 0.4 0.6 0.8 1.0 sin2(t) I(t) = Im sin (ωdt) ⇒¯ I = 0 ??? I2(t) = I2 m sin2 (ωdt) = I2 m 1 2 (1 −cos(2ωdt)) ⇒¯ I2 = 1 2I2 m IRMS = p ¯ I2 = 1 √ 2Im . Note: IRMS = VRMS Z ¯ P = I2(t)R = I2 RMSR (31) or ¯ P = IRMSVRMS cos φ , cos φ = R/Z 48 E. Transformers Φ1 = N1BA , Φ2 = N2BA V1 = −dΦ1 dt , V2 = −dΦ2 dt V2 = V1 N2 N1 N2 > N1 - ”step up”, N2 < N1 - ”step down” Current: Consider power P1 = I1V1 cos φ1 , P2 = I2V2 cos φ2 Ideal: matched φ1 and φ2: I1V1 = I2V2 , I2 = I1 N1 N2 49 Transmission of energy. Power plant (step up): V = max , I = min - minimal losses I2R. Consumer (step down): V = 110 volt ... 50 Dr. Vitaly A. Shneidman, Phys 121, 14th Lecture VII. MAXWELL EQUATIONS AND ELECTROMAGNETIC WAVES + -The idea of Maxwell’s construction. Upper left: a standard wire with current (a 2-dimensional cross-section); magnetic ﬁeld is determined from the Ampere’s circulation theorem, I ⃗ B · ⃗ s = µ0i Upper right: the wire is cut and a parallel plate capacitor with plate area A is inserted. The current, however, keeps coming in the original direction, charging the upper plate positively and the lower negatively. There is an electric ﬁeld between the plates E(t) = Q(t)/Aǫ0, with a time-dependent electric ﬂux, dΦE/dt = d (AE) /dt = d (Q/ǫ0) /dt = i/ǫ0 The magnetic ﬁeld is expected not to change. Lower ﬁgure: the wire and the capacitor are removed (as scaﬀolding). The magnetic ﬁeld is due to the changing electric ﬂux: I ⃗ B · d⃗ s = µ0ǫ0dΦE/dt ǫ0 = 1 4πk ≃ 1 4π · 9 · 109 , µ0 = 4π10−7 , µ0ǫ0 = . . . 51 A. The Maxwell equations I ⃗ E · d ⃗ A = qenc/ǫ0 (32) I ⃗ B · d ⃗ A = 0 (33) I ⃗ E · d⃗ s = −dΦB dt (34) I ⃗ B · d⃗ s = µ0ienc + 1 c2 dΦE dt (35) with c = 1/√µ0ǫ0 ≃3 · 108 m/s, the speed of light. 52 B. And then there was Light! Maxwell’s equations in empty space I ⃗ E · d ⃗ A = 0 I ⃗ B · d ⃗ A = 0 I ⃗ E · d⃗ s = −dΦB dt I ⃗ B · d⃗ s = 1 c2 dΦE dt x y z E B FIG. 14: Structure of an electromagnetic wave. The electric ﬁeld (red) is in the ±z-direction, the magnetic ﬁeld (blue) is in the ±x-direction and the wave propagates in the y-direction. (This is known as a plane, polarized wave). The distance in the y-direction in which the oscillations start to repeat themselves is the wavelength. 53 For example, the wave in Fig. 14 is described by ⃗ E = ⃗ E0 sin n 2π y λ −ft o ⃗ B = ⃗ B0 sin n 2π y λ −ft o with ⃗ E0 = c ⃗ B0 f is known as frequency and λ as wavelength. One has f = c λ for any wavelength in empty space. 54 Generation of EM waves Consider R →0: Im = Em Z , Z = lim R→0 p R2 + (XL −XC)2 = \|XL −XC\| XL = ωdL , XC = 1 ωdC , ω0 = 1 √ LC ⇒Im ∝ 1 \|ω2 d −ω2 0\| Radiation power ∝ω4 power = max ⇒ω = max ω ≃ 1 √ LC ⇒LC = min 55 L/l = µ0A N l 2 , L/l = min ⇒l = max ∼ Linear antenna. Red line shows a snapshot of the current density. (Antenna works as a time-dependent dipole). If length l is adjusted to correspond to half a wavelength at the driving fre-quency, l = πc/ωd the antenna is ”in resonance” and will radiate intensely. 56
10896	https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/21%3A_Carboxylic_Acid_Derivatives-_Nucleophilic_Acyl_Substitution_Reactions/21.05%3A_Chemistry_of_Acid_Anhydrides	Skip to main content 21.5: Chemistry of Acid Anhydrides Last updated : Sep 30, 2024 Save as PDF 21.4: Chemistry of Acid Halides 21.6: Chemistry of Esters Page ID : 448783 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Preparation of Acid Anhydrides Acid anhydrides are typically prepared by nucleophilic acyl substitution reaction of an acid chloride with a carboxylate anion, as we saw in the previous section. Both symmetrical and unsymmetrical acid anhydrides can be prepared in this way. Reactions of Acid Anhydrides The chemistry of acid anhydrides is similar to that of acid chlorides, although anhydrides react more slowly. Thus, acid anhydrides react with water to form acids, with alcohols to form esters, with amines to form amides, and with LiAlH4 to form primary alcohols. Only the ester- and amide-forming reactions are commonly used, however. Conversion of Acid Anhydrides into Esters Acetic anhydride is often used to prepare acetate esters from alcohols. For example, aspirin (acetylsalicylic acid) is prepared commercially by the acetylation of o-hydroxybenzoic acid (salicylic acid) with acetic anhydride. Conversion of Acid Anhydrides into Amides Acetic anhydride is also commonly used to prepare N-substituted acetamides from amines. For example, acetaminophen, a drug used in over-the-counter analgesics such as Tylenol, is prepared by reaction of p-hydroxyaniline with acetic anhydride. Only the more nucleophilic –NH2 group reacts rather than the less nucleophilic –OH group. Notice in both of the previous reactions that only “half” of the anhydride molecule is used, while the other half acts as the leaving group during the nucleophilic acyl substitution step and produces acetate ion as a by-product. Thus, anhydrides are inefficient, and acid chlorides are normally preferred for introducing acyl substituents other than acetyl groups. Exercise Write the mechanism of the reaction between p-hydroxyaniline and acetic anhydride to prepare acetaminophen. Answer : This is a typical nucleophilic acyl substitution reaction, with p-hydroxyaniline as the nucleophile and acetate ion as the leaving group. Exercise What product would you expect from reaction of one equivalent of methanol with a cyclic anhydride, such as phthalic anhydride (1,2-benzenedicarboxylic anhydride)? What is the fate of the second “half” of the anhydride? Answer : Monomethyl ester of benzene-1,2-dicarboxylic acid 21.4: Chemistry of Acid Halides 21.6: Chemistry of Esters
10897	https://www.youtube.com/watch?v=ebdB_HdAdHE	Using Parenthesis and Brackets in Math & Algebra Math and Science 1650000 subscribers 177 likes Description 10031 views Posted: 2 Mar 2024 This tutorial demystifies the use of parentheses in mathematics, an essential concept for students, educators, or anyone looking to enhance their understanding of mathematical operations and expressions. Parentheses play a crucial role in determining the order in which operations are performed, directly influencing the outcome of mathematical expressions. This video aims to clarify how and when to use parentheses, making complex equations more manageable and understandable. The video begins with an introduction to the basic rules of arithmetic operations and the significance of the order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). It then delves into the specific role of parentheses in altering the standard order, ensuring operations within them are prioritized. Through clear, step-by-step examples, viewers will learn how to correctly insert and solve expressions within parentheses, including nested parentheses and their impact on calculation outcomes. The tutorial covers a range of scenarios, from simple arithmetic to more complex algebraic expressions, demonstrating the power of parentheses in simplifying and structuring calculations. Additionally, the video provides tips for avoiding common mistakes and emphasizes the importance of careful placement and resolution of parentheses in ensuring mathematical accuracy. By the end of this tutorial, viewers will have a solid grasp of using parentheses effectively in math, boosting their problem-solving skills and confidence in tackling diverse mathematical challenges. Join us to master the art of using parentheses and elevate your mathematical proficiency. More Lessons: Twitter: Transcript: let's take a look at the problem 13 plus 3 close parentheses minus five what do we do by now you should know always do what's inside the parentheses and inside the parentheses is 13 plus 3 that's what we do first that's going to be 16. so we can kind of write that with the parentheses around there we still have to do the minus 5. now the parentheses here around the 16 they don't mean anything anymore because we've already done the operation so 16 minus 5. 16 minus 5 is 11 and that is the final answer all of these problems will be worked the same way let's take a look at problem number two let's say we have 16 times 2. and we're going to then close the parentheses out and add 4. what did we do first we always do what's inside the parentheses 16 times 2. now you may not remember what 16 times 2 is so go over here and figure it out 16 times 2. 6 times 2 is 12. carry the one one times two is two one more is three so the answer that we get here is 32. now we can put parentheses around it but really it they disappear once you do the operation so it's 32 plus 4. the next thing we have to do is to add 32 plus 4 which is 36 and that's the final answer 36. all right so you see the numbers are getting slightly bigger but that doesn't change anything about what we're doing now let's crank up the complexity just a little bit by taking a look at five uh times six minus one close parentheses minus two so we have a nested a double set of parentheses remember in the last lesson we said if you ever have one set of parentheses inside of another set then all you do is you go to the innermost one first so the way you kind of think about it is you're like well okay I have to do the parentheses first I'm going to do what's inside of here first but what is inside of here is yet some other junk with another set of parentheses so I must do this first so the six minus 1 is really going to be the first thing I'm going to do what is 6 minus one it's five I can drop this set of parentheses here but I still have to multiply by 5 and I still have the outermost set of parentheses basically the six minus 1 became five I still have to multiply by this and I still have the other set of parentheses so now I need to do this set of parentheses 5 times 5. that's 25 and again I can drop the parentheses once I do the calculation I still have the minus 2. 25 minus 2 going down is 23. 23 is the final answer all right making good progress let's take a look at 10 times parentheses 8 minus 3. what do we do always what's in the parentheses inside of the parentheses first so what do we have 8 minus 3 that's going to be 5 right so we put a 5 here but we and we can drop the parentheses at that point after we calculate the answer but we still have to multiply by 10 because all we did was this first we still have the multiplication times 10 to do 10 times 5 is 50. that's the only thing left and so 50 is the final answer to that problem all right let's take a look at a bigger problem let's take a look at 22 and we'll add to that open parentheses 3 times open another set of parentheses 5 minus one close close this parenthesis goes with this one and this pair goes together as well so we have to do the innermost parentheses first this one because we look at the whole thing we say well we must do what's inside here first but what is inside of here is yet another set of parentheses we must then do this one first five minus 1 is 4. I can drop this parentheses around the 4 after I do the calculation but I still have to do the multiplication by the 3 and I still have the multi the addition of the 22. all I've done is add these together I'm sorry done this subtraction here and then this Remains the Same inside the parentheses so now what do I do I have the 22 still left to do I have to do what's inside of here first 3 times 4 is 12. again I could put the parentheses here but I'm going to drop them anyway so as soon as you do the calculation you can drop the parentheses now what is 22 plus 12 maybe I don't remember that so go off to the side twenty two twelve 4 right here 3 right here and so the answer is 34. and 34 is the final answer all right that was the halfway mark of these problems let's take a look at 70. plus parentheses 3 plus parentheses 3 times 5 close parentheses close parentheses this is the innermost set of parentheses I must do this one first so I say what is 3 times 5 anyway well it's going to be 15. but I still have the plus 3 to do all of that is still encapsulated in the outer parentheses and then I still have the Plus 70. so now at this step what do I do I do the parentheses again 3 plus 15 is going to be 18 and then I still have to add the 70. I can drop the parentheses after I do the calculation so what is 70 plus 18. we have a 8 here and we have an 8 here and so the answer is 88. 88 is the final answer all right making good progress one step at a time that's how we do this do not try to do too many things in your mind students will say oh I'll do this I'll do this I'll do and try to do it all in your head and you're just going to make a mistake so don't do that let's take a look at 25 multiplied by parentheses 8 plus 1. close times 2. well we look at the whole thing we say well I have to do what's inside of this parentheses but what is inside of there is yet another set of parentheses so I have to do this one first the 8 plus 1 becomes a nine I still have to multiply by the 25 and then I still have the outer parentheses here all I did was this first that's it I still have to do this what is 25 times 9. I don't actually know so I go off to the side and try to figure out what is 25 times 9. 9 9 times 5 is 45 I carry the 4. 9 times 2 is 18 19 20 21 22 225. so inside of here is 225 I can drop these parentheses after I've calculated what's inside I have to multiply times two so I go off to the side and I say well what is 225 times 2. 5 times 2 is 10 carrying the one two times two is four one more is five two times two is four answer is four hundred fifty 450 that's the correct answer that's the final answer all right just a couple more problems and we are done with this let's take a look at 123 minus open parentheses open another parentheses 8 times 2 close minus 10 Close how do we handle this well we ignore everything except the parentheses and inside of this set we have another set so we have to do the eight times two first eight times two is what 16. and then we still have to subtract 10 and all of that is still wrapped in the outer parentheses and then this 123 minus whatever is over here is still there all we've done is done this operation first now we have to do this set of parentheses what is 16 minus 10. it's just 6. we can drop the parentheses after we calculate and now we have 123 minus 6. we go down 122 121 120 119 118 117 117 is the final answer you could set it up and subtract over there if you want but the answer is 117. all right only two more problems here we go I'm going to throw a little bit of a curveball at you here let me show you this let me get the whole thing down then I'll explain here's a curly brace and then we're going to have a 6 Plus uh double parentheses three plus one close times three close closing curly brace times four now what does this mean so what we have here is you can have parentheses inside of each other but if you have like a bunch of parentheses three or four levels deep inside of each other it gets very hard to keep track of it because all the parentheses look the same so what you might see is Curly braces this is just another set of parentheses with a special shape this curly brace goes with this one this parenthesis goes with this one this one goes with this one you can also instead of curly braces you might see like a bracket three plus two plus one and it's very easy to see the brackets go together and the parentheses go together here the curly braces go together you just treat them like parentheses that's all you do so what do we have we have a set of parentheses here we go inside but now we have a set of parentheses here so we go inside we have a set of parentheses here we must do the three plus one first which is a four we can drop this set of parentheses and it's going to be now 4 times 3 from here the curly brace will be here the plus six will be here and then we still have times four now what do we do we go on the inside innermost parentheses 4 times 3 is 12. we still have this plus six we have this curly brace right here times four now what do we have we have to do what's inside the parentheses which are just they're just fancy parentheses or curly braces what is 12 plus 6 or 6 plus 12. it's going to be 18. we can drop the curly braces after we calculate we still have to multiply by four what is 18 times 4 I don't know let me go over here and figure it out 18 times 4. 8 times 4 is 32. 4 times 1 is 4 and then three more is seven and so the answer that we get from all of this stuff is 72 that's the final answer so if you ever see weird looking parentheses don't freak out don't worry about it it's just enough another another set of parentheses if you start stacking three or four levels of parentheses up it gets very hard to figure out where the partner parenthesis goes with so we can use these curly braces we can also use brackets so let me show you that in the next problem all right what if we have three times now I'll use a bracket here five plus double parentheses 26 minus 9 then I'll close one off then I'll have multiply by four I'll close another off and then I'll have another bracket here so how do we read this it means this parenthesis goes with this one this parenthesis goes with this one and then these brackets go together and and the reason we use brackets instead of another set of parentheses is because if I put parentheses it gets hard to read so we can easily read the brackets I could have used curly braces if I wanted to so I look at the brackets I have to do what's in here first inside is another set of parentheses but inside here is yet another set of parentheses so I have to do 26 minus 9. what is 26 minus 9. you can count backwards or you can just think 26 minus 10 would be 16 but it's actually minus 9. so it has to be 17 in there so 26 minus 9 is 17. so I'll put a 17 down there all right so 26 minus 9 is just the 17 and now I have this parenthesis here times 4 this parentheses here the closing bracket is here I have the plus 5 with the opening bracket here and then the times 3. all I have done is I have the 5 is from here I've done this I've made it 17 the times 4 is the same as it was before now what is 17 times 4. I have to go off to the side and do that 17 times 4. 7 times 4 is 28 and 2 times 4 times 1 is 4 5 6. so the answer I get is 68 for this number so put a 68 right here I can drop those parentheses now and then I have plus five I still have to multiply by 3 after I'm done with all that so I have to do what's inside of the brackets first what does 68 plus 5 just go up 69 70 71 72 73. so what I have here is 73 I can drop the brackets now and I have to multiply by 3. so what is 73 times 3 73 times 3 3 times 3 is 9 and 9 times I'm sorry 7 times 3 is 21. so I get an answer of 219 that is the final answer so at the beginning of this lesson if I had shown you something like this in the beginning or this or even something like this in the beginning at the very beginning of the lesson you would have been like whoa that's too hard because it looks hard but once you start with simple problems very simple problems problems that everybody can understand problems like this do what's inside of the parentheses first then whatever you get you just follow the rest of the problem and then we slowly make it a little more complicated with the double parentheses and a little more complicated with the double parentheses until finally we end up at the end of it after all of doing all this practice getting to problems like this and showing you step by step how to think about it then you can conquer it the other advice I'll give you write your work down just the way I have done it a lot of students will say oh three plus one is four and then four times three is twelve and then 12 plus 6 is this but eventually you will make a mistake you have to show your work so please understand this practice these and follow me on to the next lesson we're going to conquer in a little more detail the concept of the order of operations learn anything at mathandscience.com
10898	https://math.stackexchange.com/questions/4654383/proof-of-youngs-inequality-for-increasing-functions	real analysis - Proof of Young's inequality for increasing functions - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Proof of Young's inequality for increasing functions Ask Question Asked 2 years, 6 months ago Modified8 months ago Viewed 613 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I am looking at a proof of the converse Young's inequality which is stated as follows: Let f f and g g be two continuous, one-to-one and increasing functions such that f(0)=g(0)=0,g−1(x)≥f(x)f(0)=g(0)=0,g−1(x)≥f(x) for all non-negative x x. If for every positive real numbers a and b we have a b≤∫a 0 f(x)d x+∫b 0 g(x)d x a b≤∫0 a f(x)d x+∫0 b g(x)d x then f f and g g are inverse. I look at a proof given by Tatsuo Takahashi here: We prove this in the case that f(x)≠g−1(x)f(x)≠g−1(x) for all non-negative real x, by showing that there exists a pair a,b a,b such that a b>∫a 0 f(x)d x+∫b 0 g(x)d x.a b>∫0 a f(x)d x+∫0 b g(x)d x. f(x)≠g−1(x)f(x)≠g−1(x) for all non-negative real x implies that there exists a positive real c, for which f(c)≠g−1(c).f(c)≠g−1(c). Therefore, g−1(x)≥f(x),g−1(x)≥f(x), can be rewritten as g−1(c)>f(c).g−1(c)>f(c). As f,g are continuous, there exists a δ δ such that g−1(x)>f(x),g−1(x)>f(x), when x is between (c−δ,c].(c−δ,c]. Consider the function ∫c 0 f(x)d x+∫g−1(c)0 g(x)d x+∫c 0 g−1(x)−f(x)d x=∫g−1(c)0 g(x)d x+∫c 0 g−1(x)d x.∫0 c f(x)d x+∫0 g−1(c)g(x)d x+∫0 c g−1(x)−f(x)d x=∫0 g−1(c)g(x)d x+∫0 c g−1(x)d x. I follow everything up until this point, then we have the line "Using the substitution g−1(x)=x,g−1(x)=x, this is equal to ∫c 0 x d g−1(x)d x d x+∫c 0 g−1(x)d x=∫c 0 d(x g−1(x))=c g−1(c)∫0 c x d g−1(x)d x d x+∫0 c g−1(x)d x=∫0 c d(x g−1(x))=c g−1(c) " Why exactly does this follow? Especially, the second result of the three results. real-analysis calculus integration Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Mar 8, 2023 at 19:49 Mittens 46.6k 6 6 gold badges 60 60 silver badges 115 115 bronze badges asked Mar 7, 2023 at 22:00 KeynesianSpacemanKeynesianSpaceman 81 7 7 bronze badges 3 Is g−1 g−1 differentiable? it is not stated in your assumptions nor in the assumptions in the paper you quoted.Mittens –Mittens 2023-03-08 00:22:10 +00:00 Commented Mar 8, 2023 at 0:22 @OliverDíaz I presume so KeynesianSpaceman –KeynesianSpaceman 2023-03-08 00:22:50 +00:00 Commented Mar 8, 2023 at 0:22 Then state that as an assumption. My impression is that it is valid without differentiability (the functions are Lebesgue a.s. differentiable but not absolutely continuous so an appropriate version of integration by parts is needed.)Mittens –Mittens 2023-03-08 00:30:38 +00:00 Commented Mar 8, 2023 at 0:30 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. The argument is clearer if we write the substitution in the form t:=g(x)t:=g(x), rather than reusing x x. Then x=g−1(t)x=g−1(t) and d x=d g−1(t)d t d t d x=d g−1(t)d t d t so we can write ∫g−1(c)x=0 g(x)d x=∫c t=0 t d g−1(t)d t d t.(1)(1)∫x=0 g−1(c)g(x)d x=∫t=0 c t d g−1(t)d t d t. At this point you can change the dummy variable of integration on the RHS of (1) from t t back to x x. As for the equality ∫c 0 x d g−1(x)d x d x+∫c 0 g−1(x)d x=∫c 0 d(x g−1(x)),∫0 c x d g−1(x)d x d x+∫0 c g−1(x)d x=∫0 c d(x g−1(x)), this is simply the product rule applied to the function x↦x g−1(x)x↦x g−1(x): d d x[x g−1(x)]=x⋅d g−1(x)d x+1⋅g−1(x)d d x[x g−1(x)]=x⋅d g−1(x)d x+1⋅g−1(x) and therefore ∫c 0(x d g−1(x)d x+g−1(x))d x=∫c 0(d d x[x g−1(x)])d x=x g−1(x)∣∣c 0=c g−1(c).∫0 c(x d g−1(x)d x+g−1(x))d x=∫0 c(d d x[x g−1(x)])d x=x g−1(x)\|0 c=c g−1(c). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 7, 2023 at 23:03 grand_chatgrand_chat 41k 1 1 gold badge 46 46 silver badges 86 86 bronze badges 3 Differentiability is not assumed.Mittens –Mittens 2023-03-08 00:22:25 +00:00 Commented Mar 8, 2023 at 0:22 @OliverDíaz Right. The function g−1 g−1 is continuous and strictly increasing, so in the non-differentiable case a Riemann-Stieltjes integral can handle the change of variables.grand_chat –grand_chat 2023-03-08 18:04:27 +00:00 Commented Mar 8, 2023 at 18:04 1 @gran_chat: I understand that. It is just that in the OP there is no such assumption and it is not too difficult to deal with the general case (ie. integration by parts via Lebesgue-Stieltjes or Riemann-Stieltjes).Mittens –Mittens 2023-03-08 18:42:06 +00:00 Commented Mar 8, 2023 at 18:42 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. In general, if g g is continuous, g(0)=0 g(0)=0, and strictly monotone increasing (g(x)<g(y)g(x)<g(y) whenever x<y x<y) on an interval [0,T][0,T], then g−1 g−1 exists, is continuous and and is strictly monotone increasing; furthermore, then for all 0≤a≤T 0≤a≤T and 0≤b≤g(T)0≤b≤g(T) ∫a 0 g+∫b 0 g−1≥a b(0)(0)∫0 a g+∫0 b g−1≥a b with equality iff b=f(a)b=f(a). There are several proofs of this fact, and this has been considered many times in MSE in the past (postings 1 and 2, and 3 for example). Some proofs are based on splitting a bounded region in the plane in two easy to estimate pieces and others are based on Riemann-Stieltjes-Lebsgue integration by parts. Taking (0)(0) as a known, suppose f,g f,g satisfy the assumptions in the OP and that there is 0f(c)g−1(c)>f(c). Then ∫c 0 g−1>∫c 0 f∫0 c g−1>∫0 c f and so c g−1(c)=∫c 0 g−1+∫g−1(c)0 g>∫c 0 f+∫g−1(c)0 g c g−1(c)=∫0 c g−1+∫0 g−1(c)g>∫0 c f+∫0 g−1(c)g contradicting the assumption that a b≤∫a 0 f+∫b 0 g a b≤∫0 a f+∫0 b g for all 0≤a≤T 0≤a≤T and 0≤b≤g(T)0≤b≤g(T). Therefore, g−1≡f g−1≡f. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jan 10 at 23:03 answered Mar 8, 2023 at 20:38 MittensMittens 46.6k 6 6 gold badges 60 60 silver badges 115 115 bronze badges Add a comment\| You must log in to answer this question. 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10899	https://testbook.com/maths/complete-graph	Get Started Exams SuperCoaching Live Classes FREE Test Series Previous Year Papers Skill Academy Pass Pass Pass Pro Pass Elite Pass Pass Pro Pass Elite Rank Predictor IAS Preparation More Free Live Classes Free Live Tests & Quizzes Free Quizzes Previous Year Papers Doubts Practice Refer and Earn All Exams Our Selections Careers IAS Preparation Current Affairs Practice GK & Current Affairs Blog Refer & Earn Our Selections Test Series A Complete Graph, denoted as (K_{n}), is a fundamental concept in graph theory where an edge connects every pair of vertices. It represents the highest level of connectivity among vertices and plays a crucial role in various mathematical and real-world applications. In this mathematics article, we will explore the complete graph in graph theory along with its properties, applications, and complete graph examples. What is a Complete Graph? A complete graph, denoted as (K_{n}), is a fundamental concept in graph theory, which is a branch of mathematics that deals with the study of networks or structures composed of nodes (vertices) and edges connecting these nodes. In a complete graph, every pair of distinct nodes is directly connected by an edge. In other words, a complete graph is a graph in which there is an edge between every pair of nodes. UGC NET/SET Course Online by SuperTeachers: Complete Study Material, Live Classes & More Get UGC NET/SET SuperCoaching @ just ₹25999₹7583 Your Total Savings ₹18416 Explore SuperCoaching Want to know more about this Super Coaching ? People also like Prev Next Assistant Professor / Lectureship (UGC) ₹26999 (66% OFF) ₹9332 (Valid for 3 Months) Explore this Supercoaching IB Junior Intelligence Officer ₹6999 (80% OFF) ₹1419 (Vaild for 12 Months) Explore this Supercoaching RBI Grade B ₹21999 (76% OFF) ₹5499 (Valid for 9 Months !) Explore this Supercoaching Characteristics of Complete Graph The main characteristics of a complete graph include: Connectivity: In a complete graph, connectivity prevails, ensuring the existence of a path between any pair of vertices within the graph. Edge Count: Each vertex within a complete graph boasts a degree of ((n - 1)), where '(n)' signifies the number of vertices in the graph. Consequently, the total number of edges equals (\frac{n\times(n - 1)}{2}). Symmetry: Every edge within a complete graph exhibits symmetry, being undirected and connecting two vertices in a consistent manner. Transitivity: Complete graphs adhere to transitivity, meaning that if vertex A connects to vertex B and vertex B connects to vertex C, then there is also a direct connection between vertex A and vertex C. Regularity: Complete graphs demonstrate regularity, signifying that each vertex shares the same degree, ensuring uniformity throughout the graph. How to Identify a Complete Graph? To identify whether a graph is a complete graph, it is essential to verify if each vertex is linked to every other vertex. Two techniques for confirming a complete graph are as follows: Inspect Vertex Degrees: In a complete graph containing (n) vertices, every vertex exhibits a degree of (n - 1). Therefore, if it is evident that each vertex in the graph possesses a degree of (n - 1), then the graph qualifies as a complete graph. Count Edges: A complete graph comprising (n) vertices encompasses (\frac{n\times(n - 1)}{2}) edges. Hence, by counting the edges within the graph and confirming that they total (\frac{n\times(n - 1)}{2}), you can conclude that the graph is a complete graph. Please note that these approaches are valid when it is ensured that the graph is acyclic. To learn more about the different types of graphs in graph theory. Test Series 139.8k Users NCERT XI-XII Physics Foundation Pack Mock Test 323 Total Tests \| 3 Free Tests English,Hindi 3 Live Test 163 Class XI Chapter Tests 157 Class XII Chapter Tests View Test Series 94.1k Users Physics for Medical Exams Mock Test 74 Total Tests \| 2 Free Tests English 74 Previous Year Chapter Test View Test Series 65.1k Users NCERT XI-XII Math Foundation Pack Mock Test 117 Total Tests \| 2 Free Tests English,Hindi 3 AIM IIT 🎯 49 Class XI 65 Class XII View Test Series Special Types of Complete Graph Complete graphs are a fundamental concept in graph theory, and there are several types of complete graphs based on different characteristics and properties. Here are some of the notable types: Bipartite Complete Graph ((K_m,n)): In a bipartite complete graph, the vertices are divided into two disjoint sets, let's say '(m)' and '(n)', where each vertex in set '(m)' is connected to every vertex in set '(n)'. This type of complete graph is often used in matching problems and combinatorial optimisation. Crown Graph: A crown graph is formed by adding a single vertex to a classic complete graph. This added vertex is connected to every vertex in the original complete graph, creating a star-like structure. Platonic Graphs: Platonic graphs are complete graphs with additional constraints, typically based on geometric shapes. For example, the tetrahedral graph is a complete graph with four vertices, and the edges represent the edges of a tetrahedron. Complete Bipartite Graph ((K_n,n)): In a complete bipartite graph, there are two disjoint sets of '(n)' vertices each, and every vertex in one set is connected to every vertex in the other set, but no edges exist within the same set. Applications of Complete Graph Here are some of the key applications of complete graphs: Networking and Communication: Complete graphs model fully connected networks, ensuring direct communication between all devices or nodes. Social Network Analysis: Used to study social networks where each person is connected to every other person, providing insights into information flow and influence. Internet Topology: Helps in understanding the interconnectedness of the internet, studying how websites or routers are linked. Bioinformatics: Used to represent interactions in biological systems, such as protein-protein interaction networks. Clique Analysis: Fundamental in identifying and analysing cliques (subsets of vertices where every pair is adjacent) in various domains, including computer science and social sciences. Optimisation Problems: Applied in various optimisation problems, including scheduling, circuit design, and network flow analysis. We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams. More Articles for Maths Quarter in Maths Weighted Graph Planar Graph Regular Graph Father of Polynomials Truth Tables and Logical Statements - Comprehensive Guide Venn Diagram in Set Theory - Representation, Examples & FAQs Handshaking Theorem Explained Decision Trees Difference Between Kite and Rhombus Complete Graph FAQs What is the complete graph definition? A complete graph is a graph in which a unique edge connects every pair of distinct vertices. What are the number of edges in a complete graph? In a complete graph with '(n)' vertices, there are (\frac{n\times(n - 1)}{2}) edges. What is a complete bipartite graph? A complete bipartite graph is a type of graph in which every vertex in one set is connected to every vertex in another set, with no edges within the same set. How many spanning trees are possible from complete graph? In a complete graph with '(n)' vertices, there are '(n^{(n - 2)})' spanning trees possible. What is the difference between bipartite and complete bipartite graph? A bipartite graph is one where the vertices can be divided into two disjoint sets, while a complete bipartite graph is a specific type of bipartite graph where every vertex in one set is connected to every vertex in the other set. What is the notation for a complete graph with 'n' vertices? The notation is (K_{n}), where '(n)' represents the number of vertices. 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