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http://emathtutoring.com/intermediate-algebra-with-geometry-course-syllabus-amp-calendar.html
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Try the Free Math Solver or Scroll down to Tutorials!
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Number of inequalities to solve: 23456789
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# Intermediate Algebra with Geometry Course Syllabus & Calendar
Calculator Any scientific calculator is acceptable! I recommend the TI-30XII scientific calculator (it costs less than \$20). Course Description Algebraic topics include: rational exponents; scientific notation; radical and rational expressions; linear, quadratic, quadratic in form, rational, radical, and absolute value equations; compound linear inequalities; literal equations; systems of linear equations in two and three variables; systems of linear inequalities; and introduction to functions. Geometric topics include: perimeter; area; volume; Pythagorean Theorem; and similarity and proportions. Students should be exposed to graphing calculator technology and/or computer algebra systems. Writing assignments, as appropriate to the discipline, are part of the course. Class work & Homework assignments There will be graded online class-work and homework assignments on each topic covered in class. You will get immediate feedback on each class work and homework problem, and incorrectly worked problems can be repeated with a new version of the problem provided by the computer until a correct solution is obtained. Tips and examples are available for each problem. Assignments must be completed before the beginning of the next class day after the section is covered in class. Homework may be done at any location with an internet access. Do not wait till the last minute to complete your assignments and quizzes because a computer glitch may prevent you from accessing your account. Quizzes Timed online quizzes will be assigned as take-home assignments. Tips and examples are NOT available during quizzes and tests. Tests There will be two tests. The second test is going to be the final exam. Some parts of the tests may be given on paper-and-pencil format. COMPASS exam All students are required to take the COMPASS test during the last week of the semester at room L912 - the Assessment Center. Computer Labs Room 3817, (CL) Room 3186; (WAC) Room 2247; (CTC) Room L112; Room L933 Attendance & active participation Policy Attendance will be taken at every class meeting. Those who repeatedly come to class late or leave early will only receive 1 or 2 points for attendance. Active pursuit of course objectives Policy (ADW) You may be administratively withdrawn from this class if any two of the following occur: 1. You fail to complete & submit at least 70% of ALL assignments during the first half of the semester 2. You fail to complete & submit at least 70% of ALL quizzes & tests during the first half of the semester 3. You fail to attend more than 50% of class meetings during the first half of the semester 4. You fail to attend for two consecutive weeks during the first half of the semester Make up policy No makeup work will be allowed on quizzes. The lowest quiz score will be dropped. Cell phone policy Cell phones are NOT allowed in classroom. Your cell phone needs to be out of sight and turned off. Cell phones are not allowed to be used as calculators. Academic Dishonesty Any student discovered receiving assistance from or giving assistance to another student on an exam or quiz, or cheating in any other way will be given a zero grade for that exam or quiz. A second infraction will result in a failing grade (F) for the class. Accommodation Contact the Disability Access Center (Room 1428) as soon as possible and if you are willing, let me know how I can help.
Grading Policy Percent NOTE: if your Algebra score on the COMPASS exit test is 49 or more, you will be able to pass this class no matter what your course grade is! Total Score Final grade Midterm Exam 20% [90%, 100%] A Final Exam 30% [80%, 90%) B Class work & Homework 35% [70%, 80%) C Take-home Quizzes 5% [60%, 70%) D Attendance 5% [0%, 60%) F Portfolio* 5% 100 %
*Portfolio: At lest 2 problems from each section must be worked out on paper written with mathematical & logical accuracy. These sample problems will be collected and evaluated periodically. For each sample problem of your choice, include title and section of the text book. A sample work will posted online.
I expect each of my students to:
• Read the syllabus carefully & regularly
• Attend class regularly and come to class on time and prepared.
• Participate and pay attention to class discussions.
• Do the assignments on time.
• Contact me or visit the Tutoring Center if you need help with assignments.
• Respect the class and me by NOT talking while I or another student is talking.
• Have high standards of conduct and work ethic.
• Turn off your cell phones during class-time. Talk to me about any exceptions!
Course Objectives:
1. Develop the algebraic skills necessary for problem solving.
2. Develop the ability to model linear, quadratic, and other nonlinear relations, including the use of the graphing techniques and geometrical principles as tools, for the purpose of solving contextual (real-world) problems.
3. Manipulate and apply literal equations for the purposes of solving contextual (real-world) problems.
4. Writing and communicating the results of problem solving appropriately.
5. Use technology as one aide for the purposes of solving contextual (real-world) problems.
Student Learning Outcomes:
Upon satisfactory completion of the course, students will be able to:
1. Simplify expressions containing rational exponents.
2. Perform operations on and simplify radicals.
3. Perform operations on and simplify rational expressions.
4. Solve quadratic equations with real solutions, including the use of the quadratic formula.
5. Solve rational equations.
6. Solve absolute value equations of the form |ax + b|=c.
7. Solve radical equations of the form: square root (ax + b) = c.
8. Solve compound linear inequalities.
9. Solve systems of linear inequalities in two variables.
10. Solve systems of linear equations in two and three variables.
11. Formulate and apply an equation, inequality or system of linear equations to a contextual (real-world) situation.
12. Solve and evaluate literal equations, including nonlinear equations.
13. Formulate and apply nonlinear literal equations to a contextual (real-world) situation.
14. Graph linear and quadratic equations.
15. Determine equations of lines, including parallel and perpendicular lines.
16. Determine whether given relationships represented in multiple forms are functions.
17. Determine domain and range from the graph of a function.
18. Formulate and apply the concept of a function to a contextual (real-world) situation.
19. Interpret slope in a linear model as a rate of change.
20. Apply formulas of perimeter, area, and volume to basic 2- and 3-dimensional figures in a contextual (real-world) situation.
21. Apply the Pythagorean Theorem to various contextual (real-world) situations.
22. Apply the concepts of similarity and congruency of triangles to a contextual (real-world) situation.
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DID YOU KNOW:
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In this geometric transformation activity, students translate, reflect, and rotate their location to figure out their final mystery destination. Students begin in Montreal, Quebec and perform a series of transformations that leads them to one of the
Subjects:
Math, Geometry
8th, 6th, 7th
Types:
Math Centers, Activities, Fun Stuff
\$2.50
In this geometric transformation activity, students translate, reflect, and rotate their location to figure out their final mystery destination. Students begin in Toronto, Ontario and perform a series of transformations that leads them to one of the
Subjects:
Math, Geometry
8th, 6th, 7th
Types:
Math Centers, Fun Stuff, Printables
\$2.50
2
This set includes all five 7th grade math brain teaser activities. Concepts are aligned to the 7th grade common core state standards, including concepts like statistics, probability, geometry, algebraic expressions and solving equations, ratios, the
Subjects:
Math, Algebra, Geometry
7th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
7.G.A.1, 7.G.A.3, 7.G.B.4, 7.G.B.5, 7.G.B.6, 7.SP.A.1, 7.SP.C.5, 7.SP.C.8, 7.SP.C.8a, 7.SP.C.8b, 7.NS.A.1d...
\$15.00
\$12.00
1
Bundle
This 7th grade math activity has 105 brain teasers aligned with parts of the Statistics & Probability common core standard. Concepts like finding simple probability, probability of independent and dependent events, making predictions and
Subjects:
Math, Statistics
7th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
7.SP.A.1, 7.SP.C.5, 7.SP.C.8, 7.SP.C.8a, 7.SP.C.8b
\$3.00
1
This 7th grade math activity has 105 brain teasers aligned with parts of the Geometry common core state standard. Concepts like finding surface area, finding volume, angles, triangles, area of circles, scale factor, cross sections of 3D figures, and
Subjects:
Math, Geometry
7th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
7.G.A.1, 7.G.A.3, 7.G.B.4, 7.G.B.5, 7.G.B.6
\$3.00
3
This 7th grade math activity has 105 brain teasers aligned with parts of the Number System common core state standard. Concepts like adding and subtracting integers, multiplying and dividing integers, fractions, converting fractions to decimals,
Subjects:
Math, Fractions, Numbers
7th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
7.NS.A.1, 7.NS.A.1a, 7.NS.A.1c, 7.NS.A.1d, 7.NS.A.2, 7.NS.A.2c, 7.NS.A.2d, 7.NS.A.3
\$3.00
3
This 7th grade math activity has 105 brain teasers aligned with parts of the Expressions and Equations common core state standard. Concepts like simplifying algebraic expressions, solving one step equations, solving two step equations, solving one
Subjects:
Math, Algebra, Word Problems
7th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
7.EE.A.1, 7.EE.B.3, 7.EE.B.4a, 7.EE.B.4b
\$3.00
2
This 7th grade math activity has 105 brain teasers aligned with parts of the Ratios and Proportional Relationships common core state standard. Concepts like finding unit rates, constant of proportionality, identifying proportional relationships,
Subjects:
Math, Other (Math)
7th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
7.RP.A.1, 7.RP.A.2, 7.RP.A.2a, 7.RP.A.2b, 7.RP.A.3
\$3.00
3
This set includes all five 8th grade math brain teaser activities. Concepts are aligned to the 8th grade common core state standards, including concepts like statistics, probability, geometry, simplifying expressions and solving equations, linear
Subjects:
Algebra, Geometry, Numbers
8th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
8.G.A.3, 8.G.B.7, 8.G.B.8, 8.G.C.9, 8.SP.A.1, 8.SP.A.2, 8.SP.A.4, 8.NS.A.1, 8.NS.A.2, 8.EE.A.2, 8.EE.C.7...
\$15.00
\$12.00
3
Bundle
This 8th grade math activity has 105 brain teasers aligned with parts of the Statistics and Probability common core state standard. Concepts like scatter plots, two way tables, probability, combinations, permutations, mean, median, mode, range, and
Subjects:
Math, Statistics, Other (Math)
8th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
8.SP.A.1, 8.SP.A.2, 8.SP.A.4
\$3.00
This 8th grade math activity has 105 brain teasers aligned with parts of the Geometry common core state standard. It has concepts like using the Pythagorean Theorem, transformations, distance between two points on the coordinate plane, finding
Subjects:
Math, Geometry
8th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
8.G.A.3, 8.G.B.7, 8.G.B.8, 8.G.C.9
\$3.00
This 8th grade math activity has 105 problems and brain teasers aligned with parts of the Functions common core state standards. It has concepts like slope intercept form, linear functions, comparing functions, finding x and y intercepts, finding
Subjects:
Math, Algebra
8th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
8.F.A.1, 8.F.A.2, 8.F.A.3, 8.F.B.4
\$3.00
3
This set includes all 5 Shopping for the Deal middle school math unit rate activities, which focus on finding unit rates and comparing unit rates. All include printed and digital Google slides copies. They are great as a 6th grade unit rate project
Subjects:
Math
5th, 6th, 7th
Types:
CCSS:
6.RP.A.2, 6.RP.A.3b
\$12.00
\$9.75
1
Bundle
This 8th grade math activity has 105 problems and brain teasers aligned with parts of the Expressions and Equations common core state standards. It has concepts like solving two step equations, solving multi step equations, solving systems of
Subjects:
Math, Algebra
8th
Types:
Activities, Fun Stuff, Printables
CCSS:
8.EE.A.1, 8.EE.A.2, 8.EE.C.7, 8.EE.C.8, 8.EE.C.8a, 8.EE.C.8b
\$3.00
3
This 8th grade math activity includes 105 problems and brain teasers aligned with parts of the Number System common core state standards. It includes concepts like negative exponents, rational and irrational numbers, scientific notation, repeating
Subjects:
Math, Numbers
8th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
8.NS.A.1, 8.NS.A.2
\$3.00
2
This review packet for 7th grade includes 16 problems related to using a scale factor to enlarge or reduce a drawing. It also has area problems related to scale factor. It includes a printed and digital Google slides copy. It is a 4 page set of
Subjects:
Math, Geometry, Word Problems
7th
Types:
CCSS:
7.G.A.1
\$3.00
1
This is a 6th grade math review packet related to using order of operations to evaluate numerical expressions. It includes an order of operations Google slides review packet copy. Some problems include order of operations with exponents. It is a 3
Subjects:
Math, Numbers, Order of Operations
6th
Types:
CCSS:
6.EE.A.2c
\$3.00
1
This 8th grade math review packet includes 16 problems related to using the Pythagorean Theorem. It has a printed and digital Google slides copy. Many of the problems are Pythagorean Theorem word problems. It is a 3 page set of worksheets to help
Subjects:
Geometry, Word Problems
8th
Types:
CCSS:
8.G.B.7, 8.G.B.8
\$3.00
This Math Sort Activity Bundle includes all five of my 5th grade math sort activities and any future 5th grade sorts. Topics include adding and subtracting decimals, multiplying and dividing decimals, order of operations, multiplying whole numbers,
Subjects:
Math, Order of Operations, Decimals
5th
Types:
Math Centers, Activities, Fun Stuff
\$15.00
\$12.00
2
Bundle
This math sort, designed for 5th grade math, helps students practice multiplying by powers of 10 and dividing by powers of 10. Problems include some powers of 10 written with an exponent. The cards include four different sorts (20 problems each) and
Subjects:
Math, Numbers
5th
Types:
Math Centers, Activities, Fun Stuff
CCSS:
5.NBT.A.2
\$3.00
showing 1-20 of 500
TEACHING EXPERIENCE
I am a 6th and 8th grade math teacher from Wisconsin and am currently in my 7th year of teaching. I attended and graduated from the School of Education program at UW-Madison.
MY TEACHING STYLE
I have recently started to use a Math Workshop structure in my 6th grade class and have been loving its results! The use of math centers has given me the chance to work with every student in a small group setting every day!
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY
Love anything to do with baseball and math! Check out my blog at www.middleschoolmathman.com.
1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th
SUBJECTS
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.
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CC-MAIN-2021-39
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latest
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en
| 0.881031
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https://theodora.com/encyclopedia/p2/pole.html
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crawl-data/CC-MAIN-2019-09/segments/1550247481612.36/warc/CC-MAIN-20190217031053-20190217053053-00207.warc.gz
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Pole - Encyclopedia
Pole - Encyclopedia
GEOGRAPHICAL NAMES
POLE (i) (0.(0. Eng. p01, cf. Ger. Pfahl, Du. pawl, from Lat. pales,. stake), a tapering cylindrical post or stake of some considerable length, used as a support in scaffolding, for telegraph or telephone wires, hops, &c., and as a means fcr taking jumps (see Pole-Vaulting), and also as a single shaft for a vehicle drawn by two or more horses. As a measure of length a "pole,"' also called "rod" or "perch," is equal to 51 yds. (162 ft.), as a measure of area it is equal to 304 sq. yds. (2) (Lat. polus, adapted from Gr. ir6Xos, pivot, axis), one or other of the extremities of the axis of the earth; the "celestial pole" is one or other of the points in the heavens to which the earth's axis points; in the northern hemisphere this point is near the star Ursae minoris, better known as the Pole-star or Polaris. (see Ursa Major). For the regions lying about the north and south poles of the earth see Polar Regions.
In mathematics the word pole has several meanings. In spherical. trigonometry the "pole" of a circle on a sphere is the point where the diameter of the sphere perpendicular to the plane of the circle intersects the sphere. In crystallography the "pole" of a face is the intersection of a line perpendicular to the face with the sphere of projection. The term is also applied to a point from which lines radiate, as, for instance, the origin in a system of polar co-ordinates, or the common point of a pencil of rays. In the geometry of conic sections the "pole" of a line, termed the "polar" of the point, is the intersection of the tangents (either real or imaginary) at the points where the line meets the conic (see Geometry: § Projective). The "magnetic poles" of the earth are the points on the earth's surface where the dipping needle is vertical (see Terrestrial Magnetism); and the "poles" of a magnet are the points of the magnet where the magnetic intensity is greatest. In electricity, the term is applied to the elements of a galvanic battery, or to the terminals of a frictional electrical machine.
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``` https://theodora.com/encyclopedia/p2/pole.html ```
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latest
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http://codeforces.com/blog/entry/67366
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### Egor.Lifar's blog
By Egor.Lifar, history, 12 months ago, translation, ,
Author: egor.lifar
Author: KAN
Author: UnstoppableSolveMachine
Author: egor.lifar
Author: UnstoppableSolveMachine
Author: egor.lifar
Author: UnstoppableSolveMachine
Author: egor.lifar
• +51
» 12 months ago, # | 0 Auto comment: topic has been updated by egor.lifar (previous revision, new revision, compare).
» 12 months ago, # | 0 1148G — Gold ExperienceTutorial is not available Can someone share their solution?
• » » 12 months ago, # ^ | +12 Tutorial for this problem is going to be uploaded in about 5 minutes.
» 12 months ago, # | +19 I think the test case 7 of B got most of us :).
» 12 months ago, # | 0 For E, how do you prove that that it's sufficient that every prefix sum of $δ_i$ is $≥0$ and $\sum \delta_i = 0$? (it's not hard to prove that those conditions are necessary; the negation of the first statement implies that there there exists a $k$ such that $\sum_{i = k}^{n}s_k$ increases)
• » » 12 months ago, # ^ | ← Rev. 2 → +31 Let's prove that it is sufficient by induction. Assume we have proved that the following condition is sufficient for all $m$ < $n$: $\sum\limits_{j=1}^i d_i \geq 0$ for $1 \leq i \le m$ and $\sum d_i = 0$.Let's prove that for $n$ above condition is sufficient too. Let's start moving an arbitrary pair of stones until we get $\sum\limits_{j=1}^i d_i = 0$ for some $i \le n - 1$. Now, we can split our task to 2 subtasks, from first stone to $i$-th stone and from $i$-th stone to n-th stone. By the induction assumption, in these two smaller subtasks the above condition is sufficient $\implies$ it is sufficient for our task too.
• » » 12 months ago, # ^ | 0 My friend told me to read this, but I'm not sure whether it does help or not.
» 12 months ago, # | 0 In 1148F — Foo Fighters: 1) If we follow the method given in this tutorial, does it happen that the new sum that we get after performing the operation(as mentioned in the question) is exactly the additive inverse of the old sum of the values ?if not ,can someone please explain what is the logic behind this solution mentioned in the editorial ?
• » » 12 months ago, # ^ | +8 I'll try to make it more clear. When you solve this problem using induction principle, of course you are trying to get the best you can. But unfortunately some bit additions changes not only objects, which we have already proceeded, but all objects given in the input. That's why we can't say that we can exactly additive inverse of the old sum. During some steps of the algorithm we already work with not original prices, but with reverse ones from earlier added bits.
» 12 months ago, # | +28 Is there anyone else who found problem D easier than C?
• » » 12 months ago, # ^ | +8 To me it was also easier than B, but its probably because I overcomplicated it
• » » 12 months ago, # ^ | 0 yes, i did problem D first
» 12 months ago, # | 0 Can anybody explain crazy diamond problem ? Actually I am not able to understand the editorial of this problem.
• » » 12 months ago, # ^ | 0 My solution (different from editorial):Because the array is permutation, the condition sorted means that for each $i$, $a_i = i$.Instead of sorting the entire array, we try to sort subarray with index $2$ to $n - 1$. After that, if $a_1 \neq 1$, swap it with $a_n$.Loop $i$ from $2$ to $n - 1$, when $a_i \neq i$, we can always move $a_i$ to $i$ via both endpoint (index $1$ and $n$). Let $p$ = current position of $a_i$, $u$ = endpoint that is reachable from $p$ $(2 \cdot |p - u| \geq n)$, and $v$ the other endpoint (equal to $1$ if $u$ is $n$, and equal to $n$ otherwise). There is 2 cases: $2 \cdot |u - i| \geq n$, in this case, just $\tt{swap(p, u)}$ then $\tt{swap(u, i)}$. $2 \cdot |u - i| < n$, in this case we cant move from $u$ to $i$, so after $\tt{swap(p, u)}$, we $\tt{swap(u, v)}$ first then $\tt{swap(v, i)}$. My submission (i used 0-indexing in code).
• » » » 12 months ago, # ^ | 0 How's your approach is different from the editorial. I think you have done the same thing as in editorial.
• » » » » 12 months ago, # ^ | 0 oh ok, actually im not sure about editorial when i posted that, i just look at number of swaps needed, it is different from mine, so i think it should be different XD. Yes i think it use same idea but different implementation.
• » » » 10 months ago, # ^ | +3 Your solution is easy to understand. Thank you.
» 12 months ago, # | +27 Can anyone explain problem F( Foo Fighters) more clearly?
• » » 12 months ago, # ^ | +21 Let $E_k$ denote the set of elements whose masks have only (some subset of) the first $k$ bits on. Note that $E_1 \subseteq E_2 \subseteq ...$.We will iteratively construct $mask_k =$ any mask on $k$ bits which makes the elements of $E_k$ have a sum with the sign we want at the end. For now, zero is considered to have either sign.$mask_1$ is easy to construct: just check the sign of the sum of $E_1$ and flip it if necessary by setting the first bit.To construct $mask_{k+1}$ from $mask_{k}$, we check the sum of the elements of $E_{k+1} \setminus E_k$ assuming we have already applied $mask_k$ to all elements. If the sum of $E_{k+1} \setminus E_k$ has a sign which differs from our target, we can flip it by setting the $(k+1)^{th}$ bit without affecting the sum of $E_k$. The sum of two values with the correct sign still has the correct sign. So, $mask_{k+1}$ is simply $mask_{k}$ plus the choice for the $(k+1)^{th}$ bit.It is guaranteed that the input does not have sum zero. At some point, we will reach $k_0$ such that $E_{k_0}$ has non-zero sum. After that point, our "working sum" on the elements of $E_{k_0}$ will become non-zero with the correct sign.$mask_{62}$ is the final output.
» 12 months ago, # | +3 How to prove the greedy solution in D? I don't know why it is correct.
• » » 12 months ago, # ^ | ← Rev. 3 → +24 If you sort pairs of form ai ai therefore a1a2a3.....Similarly, you can prove for pair of form ai>biI hope this helps :)
• » » » 12 months ago, # ^ | 0 Got it, thank you.
» 12 months ago, # | +9 Please provide a more detailed explanation of Problem F- Foo Fighters, explaining how the induction works.
» 12 months ago, # | 0 Hi! I have some problem with Problem B. Can someone please check this out and help me? https://codeforces.com/blog/entry/67379 Looking forward to your help. Thanks!
» 12 months ago, # | +20 For those who are having trouble understanding the editorial of F let me explain it in a different way.Let us partition the object in 62 collections (collection 0, collection 1,..., collection 61) based on their least significant bit position (i.e. objects with least significant set bit as i will belong to collection i)Let's iterate on collections from 61 to 0 and have a rule in processing 'collection i' we will change (if any) only on i'th bit of our answer s. Doing so will not have any changes to our processing in collection i+1 or higher. When we are at processing collection 'i' we have s bits i+1 to 61 fixed and we only need to decide for i'th bit so let's calculate what sum-change when taking i'th bit or not taking i'th bit (based on parity of number of bits set in s till now) (observe we don't need lower bits for collection i to decide its parity). So taking and not-taking divide collection i sum into two parts we will take bigger part if out initial sum was positive and smaller part if initial sum was negative. How? Taking a part is simple if we are choosing taking then set ith bit of s as 1 else do nothing. Why? Suppose initial sum was positive we take bigger part meaning we have new sum from collection i as smaller part — bigger part (as bigger part sign flip) so it will be non-positive similarly one can say about initial sum negative case. The solution seems to be correct except for one problem that suppose we have initial sum as positive we make it non negative at each collection so can't our final sum be zero and we wanted it to be negative. Here is where the condition initial sum non-zero 0 comes in. Let j be highest number with collection j non-empty so obviously its not taking is 0 and taking part is sum number. Suppose taking part sum was also zero we remove all numbers from j this does not change initial sum and we have now a lower j (Not this situation can't arrive when j is 0 as initial sum non-zero) and if taking j part sum was non-zero then we have a negative sum by selecting th e i-th bit so we don't get exactly zero. (Similar way to prove for initial sum negative case)
• » » 12 months ago, # ^ | 0 Least significant or most?
• » » » 12 months ago, # ^ | +19 Yes, it could be done with most significant also (we have to reverse the processing loop). But here, I had partitioned it based on the least significant bit.
• » » 11 months ago, # ^ | ← Rev. 4 → 0 can you explain this condition??if(init_sum >0 && taking > not_taking) ans| = 1<
» 12 months ago, # | -7 Help me to solve out 1148E problem.....In 1148E problem test case 7:20 53 86 76 100 16 12 13 97 79 23 28 64 42 10 23 56 59 76 48 1248 49 49 49 48 49 48 49 49 49 48 48 48 49 49 49 49 49 49 48my code give following output for above testcase :YES1914 4 396 4 126 8 2520 8 2320 2 137 2 247 9 115 9 195 18 1310 18 1410 3 1215 3 1515 12 1111 12 511 17 1011 16 513 16 213 1 419 1 1my code link:code for 1148E in c++
» 12 months ago, # | +4 Problem E: Note, that we can always construct an answer preserving the original stones order (suppose we have an answer which doesn't preserve the order. It will happen when there are two stones at the same spot, just move the other one).What does it mean?
• » » 12 months ago, # ^ | 0 Notice that the problem does not require stone at position (i) to be moved to target position (i). It requires that the resulting stones positions fill all the target positions regardless of which stone moved to which target position. When you sort the stones based on their original positions, and sort target positions such that stone (i) will be moved to target position (i), leftmost stone will move to the leftmost target position, and second leftmost stone will move to second leftmost target position, and so on. Thus, their original order is preserved.
• » » 12 months ago, # ^ | ← Rev. 2 → 0 Because in applying operation two stones will cross each other. Suppose i move to a and j move to b such that i < j and a > b Then these stones will meet a point. From there you can move any of the two stones.
» 12 months ago, # | 0 Can anyone help me understand why this approach for problem E fails on test 14 ?? ( Or maybe I forgot some conditions)First, I sort both the array. Then I have two variables with l = 0 and r = n-1.Below is sort of pseudo — c++ code of my approach:While ( l < r) {if(s[l] == t[l]) { l++; continue; } if(s[r] == t[r]) { r--; continue; }int d1 = t[l] — s[l]; int d2 = s[r] — t[r];if(d1>0 && 2*d1 <= s[r]-s[l] && s[r]-d >= t[r] ) { //here we can make s[l] = t[l] , decrease s[r] by d1 } else if(d2>0 && 2*d2 <= s[r]-s[l] && s[l]+d <= t[l] ) { //here we can make s[r] = t[r] , increase s[l] by d2 }else { cout << -1; return 0; }}
• » » 12 months ago, # ^ | +8 I guess your code will fail for following input: 5 1 5 10 15 20 2 6 9 16 18 The answer to it is: YES 3 4 5 1 2 5 1 1 3 1 If your approach gives "NO", you should be able to get why your logic is incorrect.
• » » » 12 months ago, # ^ | 0 Got it ,Thank you !! I think I may change comp for sorting and sort by the difference s[l] — t[l] , will it solve the problem ?
• » » » » 12 months ago, # ^ | 0 Instead of starting $r$ at the end and moving inward, maintain $r$ to be the index of the first value after index $l$ that needs to be reduced. That will take care of the problem occurring right now, because then you'll be using the least restrictive candidate each time you're performing an operation. Values that needed to be reduced that were further towards the end would then be available for increasing values of $l$ further along the array.
» 12 months ago, # | 0 please help me with C. i can't understand the tutorial.
» 12 months ago, # | 0 In A, when I tried the 5th sample test case on my system, it ran without overflowing. But when I submitted it, it resulted in overflow. I don't understand why. I had used long int (C++) to store the answer, which is 8 bytes in size on my system (can store upto 10^19). There was no chance of overflow as the answer was 4*10^9. Please help.PS: I just tried using long long and my solution has been accepted. What's wrong with the compiler used here?
• » » 11 months ago, # ^ | 0 4*10^9 means you would need unsigned int not int.
• » » » 4 months ago, # ^ | 0 I had used long int, not normal int. It stores roughly about 10^18.
• » » » 4 months ago, # ^ | ← Rev. 2 → 0 Here is the code for reference. Try submitting it. Then change the long to long long and try again.int main(){long a,b,c; long length; cin>>a>>b>>c; if(a==b||a==b-1||a==b+1) { length=a+b+2*c; } else { if(a
» 12 months ago, # | 0 Does someone know why this code can get the correct answer? code
» 7 months ago, # | 0 Can anyone explain, how we got the formula for Div2.A ? I understood the explanation below but the formula seems to be a bit complex.
» 3 months ago, # | 0 I didn't expect D to be that easy
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# Unscramble TROUBLE
TROUBLE unscrambles into 135 different words! We have all of them and the meanings below! Enter any word and we will UNSCRAMBLE IT!
## More About Unscrambling TROUBLE
Above are the words made by unscrambling TROUBLE (BELORTU). To further help you, here are a few lists related to/with the letters TROUBLE
### The Value of TROUBLE In Word Scramble Games
The letters TROUBLE are worth 9 points in Scrabble
The letters TROUBLE are worth 12 in points Words With Friends
• T = 1 points in WWF & 1 points in Scrabble
• R = 1 points in WWF & 1 points in Scrabble
• O = 1 points in WWF & 1 points in Scrabble
• U = 2 points in WWF & 1 points in Scrabble
• B = 4 points in WWF & 3 points in Scrabble
• L = 2 points in WWF & 1 points in Scrabble
• E = 1 points in WWF & 1 points in Scrabble
## What Does TROUBLE Mean... If you Unscramble it?
### Possible Definitions of TROUBLE
If we unscramble these letters, TROUBLE, it and makes several words. Here is one of the definitions for a word that uses all the unscrambled letters:
### Boulter
• A long, stout fishing line to which many hooks are attached.
## Permutations of TROUBLE
According to our other word scramble maker, TROUBLE can be scrambled in many ways. The different ways a word can be scrambled is called "permutations" of the word.
#### Definition of Permutation
a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged.
How is this helpful? Well, it shows you the letters trouble scrambled in different ways That way you will recognize the set of letters more easily. It will help you the next time TROUBLE comes up in a word scramble game.
We stopped it at 50, but there are so many ways to scramble TROUBLE!
### Scramble Words
Unscramble these letters to make words...
scrambled using word scrambler...
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# NCERT Class 12-Mathematics: Exemplar Chapter – 3 Matrices Part 21 (For CBSE, ICSE, IAS, NET, NRA 2023)
Doorsteptutor material for CBSE/Class-12 is prepared by world's top subject experts: get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-12.
Question 59:
If matrix , where if then is equal to
(A) I
(B) A
(C) 0
(D) None of these
Question 60:
The matrix is a
(A) Identity matrix
(B) Symmetric matrix
(C) Skew symmetric matrix
(D) None of these
Question 61:
The matrix is a
(A) Diagonal matrix
(B) Symmetric matrix
(C) Skew symmetric matrix
(D) Scalar matrix
Question 62:
If A is matrix of order and B is a matrix such that AB′ and B′A are both defined, then order of matrix B is
(A)
(B)
(C)
(D)
Question 63:
If A and B are matrices of same order, then is a
(A) Skew symmetric matrix
(B) Null matrix
(C) Symmetric matrix
(D) Unit matrix
Question 64:
If A is a square matrix such that , then is equal to
(A)
(B)
(C)
(D)
Question 65:
For any two matrices A and B, we have
(A)
(B)
(C)
(D) None of the above
Question 66:
On using elementary column operations in the following matrix equation
we have:
(A)
(B)
(C)
(D)
Question 67:
On using elementary row operation in the following matrix equation:
, we have:
(A)
(B)
(C)
(D)
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×
Get Full Access to Statistics: Informed Decisions Using Data - 5 Edition - Chapter 2.1 - Problem 7
Get Full Access to Statistics: Informed Decisions Using Data - 5 Edition - Chapter 2.1 - Problem 7
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# ?Most Valuable Player The following Pareto chart shows the position played by the most valuable player (MVP) in the National League since 1931. Source:
ISBN: 9780134133539 240
## Solution for problem 7 Chapter 2.1
Statistics: Informed Decisions Using Data | 5th Edition
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Problem 7
Most Valuable Player The following Pareto chart shows the position played by the most valuable player (MVP) in the National League since 1931. Source: http://www.baseball-almanac.com/
(a) Which position had the most MVPs?
(b) How many MVPs played first base (1B)?
(c) How many more MVPs played outfield (OF) than first base?
(d) There are three outfield positions (left field, center field, right field). Given this, how might the graph be misleading?
Step-by-Step Solution:
Step 1 of 5) Most Valuable Player The following Pareto chart shows the position played by the most valuable player (MVP) in the National League since 1931. Source: http://www.baseball-almanac.com/ (a) Which position had the most MVPs (b) How many MVPs played first base (1B) (c) How many more MVPs played outfield (OF) than the first base (d) There are three outfield positions (left field, center field, right field). Given this, how might the graph be misleading
Step 2 of 2
##### ISBN: 9780134133539
Since the solution to 7 from 2.1 chapter was answered, more than 209 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 7 from chapter: 2.1 was answered by , our top Statistics solution expert on 01/15/18, 03:19PM. Statistics: Informed Decisions Using Data was written by and is associated to the ISBN: 9780134133539. The answer to “?Most Valuable Player The following Pareto chart shows the position played by the most valuable player (MVP) in the National League since 1931. Source: http://www.baseball-almanac.com/ (a) Which position had the most MVPs?(b) How many MVPs played first base (1B)?(c) How many more MVPs played outfield (OF) than first base?(d) There are three outfield positions (left field, center field, right field). Given this, how might the graph be misleading?” is broken down into a number of easy to follow steps, and 68 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 88 chapters, and 2422 solutions. This textbook survival guide was created for the textbook: Statistics: Informed Decisions Using Data, edition: 5.
## Discover and learn what students are asking
Calculus: Early Transcendental Functions : Product and Quotient Rules and Higher-Order Derivatives
?Using the Product Rule In Exercises 1–6, use the Product Rule to find the derivative of the function. $$g(x)=\left(x^{2}+3\right)\left(x^{2}-4 x\ Calculus: Early Transcendental Functions : Inverse Trigonometric Functions: Integration ?In Exercises 1-20, find the indefinite integral. \(\int \frac{1}{x \sqrt{4 x^{2}-1}} d x$$
Statistics: Informed Decisions Using Data : Estimating the Value of a Parameter
?What does the 95% represent in a 95% confidence interval?
Statistics: Informed Decisions Using Data : Comparing Three or More Means (One-Way Analysis of Variance)
?Births by Day of Week An obstetrician knew that there were more live births during the week than on weekends. She wanted to determine whether the mean
Statistics: Informed Decisions Using Data : An Overview of Nonparametric Statistics
?Explain the idea of efficiency.
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# Arrays in C (Array of Pointers and 3D Array) – Set 7
In this instructional exercise, I will inform you concerning two points for example cluster of pointers and 3D exhibits. So without burning through whenever lets head on to the primary point of this instructional exercise.
An exhibit of Pointers in C
As I disclosed to you before “Exhibit is a gathering of components of comparative information types”. We have just made a cluster of the whole number, skims, etc. So one inquiry that emerges.
Would we be able to make a cluster of pointers? Well, the appropriate response is indeed, obviously.
To cause an exhibit we to need to satisfy only one condition i.e it ought to contain components of the same information type. So we can store pointers of segregated factors inside an exhibit. Let’s take one guide to get this.
``````#include<stdio.h>
int main()
{
int *a[4]; //declaration of array of pointers
int x=23,y=54,z=65,p=45,q;
a[0]=&x;
a[1]=&y;
a[2]=&z;
a[3]=&p;
for(q=0;q<4;q++)
printf("%d ",*a[q]);
return 0;
}``````
Output
``23 54 65 45 ``
Clarification
In the start of the program, I have proclaimed a 1D cluster of pointers with size 4. It implies it can store locations of four secluded pointer factors.
After that, I have pronounced some whole number factors and I have put away a few qualities in it.
Presently I have put away the addresses of whole number factors inside a cluster of pointers a.
In the last, I have printed every one of the qualities at the addresses put away in that exhibit by utilizing for circle.
Note: We can even store the addresses of different clusters inside the exhibit of pointers.
3D Array in C
As the name recommends these are exhibits having three measurements. By and large, a developer once in a while utilize 3D clusters.
They are for the most part utilized for some game programming. So I will just give you a review of that.
Presentation and Initialization of 3D Array
Lets take one guide to get it.
int a[2][2][2]={
{
{13, 56},
{54, 67}
},
{
{64, 87},
{23, 678}
}
};
In C language, 3D clusters treated as an accumulation of 2D exhibits. In the above model we can likewise say that we are making two 2D exhibits.
A case of the 3D exhibit is given underneath which is additionally a mix of three 2D clusters.
Memory Allocation of 3D Array
Allotment of memory of the 3D cluster is like a 2D exhibit. As the components inside the 3D cluster consistently put away in touching memory areas.
error: Alert: Content is protected!!
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mersenneforum.org Subproject #11: 2^2 * 3 - Lose the '3'
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2011-09-24, 03:40 #12
Mini-Geek
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts
Quote:
Originally Posted by schickel PS. Clifford has mentioned the required form for a factorization to result in losing the 3, but I'm haven't been able to find that email yet....
I think it's something like this:
All primes congruent to 1 mod 3 must be raised to an odd power Edit: a power n with n != 2 mod 3. (since nearly all large factors will be ^1, this is the more important of the two; Edit: simplistically, in practice, this means you want the factors to be 1 mod 3)
All primes congruent to 2 mod 3 must be raised to an even power
Quote:
Originally Posted by Batalov Don't forget - 3 comes and goes... See for example 611156.
Yes, it does. Still, it's more hopeful when it's gone.
Last fiddled with by Mini-Geek on 2011-12-31 at 03:15
2011-09-24, 03:53 #13
schickel
"Frank <^>"
Dec 2004
CDP Janesville
2×1,061 Posts
Quote:
Originally Posted by Mini-Geek Escape from 2^2*3 to 2^2 successful at 118 digits! http://factordb.com/sequences.php?se=1&aq=199710 It's now on a slow downward-ish slide, hopefully it will become the downdriver!
How about we delete the ones that manage to escape from the top post and move the sequence to the main reservation thread, unless indicated otherwise by the escapee?
2011-09-24, 04:06 #14
schickel
"Frank <^>"
Dec 2004
CDP Janesville
212210 Posts
Quote:
Originally Posted by Mini-Geek I think it's something like this: All primes congruent to 1 mod 3 must be raised to an odd power (since nearly all large factors will be ^1, this is the more important of the two) All primes congruent to 2 mod 3 must be raised to an even power.
So does the exponent on the 3 matter? Also, does this change the sequences that get posted for work on this project? (Actually, at the very least, it lets out sequences with 5 raised to an odd power, correct? That takes 71 more out of play right off the bat....)
2011-09-24, 07:08 #15 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100011101001102 Posts Will take 183936, too.
2011-09-24, 13:17 #16
Mini-Geek
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts
Quote:
Originally Posted by schickel How about we delete the ones that manage to escape from the top post and move the sequence to the main reservation thread, unless indicated otherwise by the escapee?
It'd be more work, and considering the 3 can come and go pretty easily, I don't really want to do it this way. I, for example, plan to run 199710 until it settles into a driver (currently driverless with 2^5*7^2*11) or grows too large for me to handle. I'd expect many others would do similar. I'd prefer to leave it in this thread until they say otherwise. (it's not like it prevents this subproject from being complete - we can mark sequences that are no longer 2^2*3 if that'd make things clearer)
Quote:
Originally Posted by schickel So does the exponent on the 3 matter? Also, does this change the sequences that get posted for work on this project? (Actually, at the very least, it lets out sequences with 5 raised to an odd power, correct? That takes 71 more out of play right off the bat....)
AFAICT, no, the exponent on the 3 doesn't matter, because sigma(3^n) mod 3 is 1 for any n>=0. If you want to hit the most likely to break first, yes, you'd avoid sequences with 5 raised to an odd power, (because they can't lose it on the very next line) but depending on how hard it is to lose that 5 or change its power, it might be a very minor difference.
A little more on my methods and why "sigma(3^n) mod 3 is 1" is important: see the formula for calculating the sigma of a number. If the current line is divisible by 3, that means it is 0 mod 3. Once we have its sigma, the next line is sigma - lastLine. Working mod 3, that's sigma - 0, or sigma. So for the next line to not be divisible by 3, sigma != 0 mod 3 must be true. The sigma is the product of a series of numbers, which are the sigmas of the prime factors, e.g. sigma(2^2*3)=sigma(2^2)*sigma(3). If none of these numbers multiplied together are 0 mod 3, (i.e. all are -1 and 1, or 1 and 2 if you prefer) then the sigma will not be 0 mod 3, and so the next line will not be 0 mod 3.
I'm sure I'm stating trivialities for mathematicians, but considering I'm the first in this thread to mention how to lose the 3, it might be of some use for learning for everyone.
Last fiddled with by Mini-Geek on 2011-09-24 at 13:26
2011-09-24, 13:55 #17
Mini-Geek
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
426710 Posts
Quote:
Originally Posted by Mini-Geek All primes congruent to 1 mod 3 must be raised to an odd power (since nearly all large factors will be ^1, this is the more important of the two)
Correction:
All primes congruent to 1 mod 3 must be raised to a power n with n != 2 mod 3.
2011-09-24, 14:23 #18 bchaffin Sep 2010 Portland, OR 7·53 Posts I'll take 243402.
2011-09-24, 15:20 #19
schickel
"Frank <^>"
Dec 2004
CDP Janesville
84A16 Posts
Quote:
Originally Posted by Mini-Geek It'd be more work, and considering the 3 can come and go pretty easily, I don't really want to do it this way. I, for example, plan to run 199710 until it settles into a driver (currently driverless with 2^5*7^2*11) or grows too large for me to handle. I'd expect many others would do similar. I'd prefer to leave it in this thread until they say otherwise. (it's not like it prevents this subproject from being complete - we can mark sequences that are no longer 2^2*3 if that'd make things clearer)
OK, whichever way you figure is easier (I'm all in favor of easier....)
2011-09-25, 02:40 #20 Mini-Geek Account Deleted "Tim Sorbera" Aug 2006 San Antonio, TX USA 17×251 Posts Unreserving 199710, since it has a 2^5*7 guide and a c114 (full ECM done), size 120. Reserving 247840.
2011-09-25, 04:59 #21 schickel "Frank <^>" Dec 2004 CDP Janesville 2·1,061 Posts Just by way of encouragement..... Just to help motivate everyone, here is what the status of bchaffin's latest termination was in my last pull: Code: 734184 850. sz 111 2^2 * 3^3 * 83 Yep, that's right, it was 2^2 * 3!!! PS. Check out the slope on that first downdriver run! Last fiddled with by schickel on 2011-09-25 at 05:01 Reason: Add PS.
2011-09-25, 05:04 #22
schickel
"Frank <^>"
Dec 2004
CDP Janesville
2·1,061 Posts
Quote:
Originally Posted by Mini-Geek Reserving 247840.
Hmmm....I left the file sorted by size, thinking it might make it easier for people to pick one out. Should we resort the list by sequence?
Similar Threads Thread Thread Starter Forum Replies Last Post deangram Information & Answers 16 2015-11-27 05:54 jasong jasong 5 2013-11-12 15:34 fivemack Aliquot Sequences 20 2013-01-29 11:10 schickel Aliquot Sequences 7 2012-12-08 05:38 Freightyard PrimeNet 3 2009-01-11 12:20
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Chapter: Problem:
Waiting in Line The Jefferson Valley Bank once had a separate customer waiting line at each teller window, but it now has a single waiting line that feeds the teller windows as vacancies occur. The standard deviation of customer waiting times with the old multiple-line configuration was 1.8 min. Listed below is a simple random sample of waiting times (minutes) with the single waiting line. The 10 sample values have a standard deviation of 0.5 min.
a. When the bank changed from multiple waiting lines to a single line, how was the mean waiting time affected?
b.When the bank changed from multiple waiting lines to a single line, how was the variation among waiting times affected?
c. What improvement occurred with the change from multiple waiting lines to a single line?
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# Thue equation
In mathematics, a Thue equation is a Diophantine equation of the form
ƒ(x,y) = r,
where ƒ is an irreducible bivariate form of degree at least 3 over the rational numbers, and r is a nonzero rational number. It is named after Axel Thue who in 1909 proved a theorem, now called Thue's theorem, that a Thue equation has finitely many solutions in integers x and y.[1]
The Thue equation is solvable effectively: there is an explicit bound on the solutions x, y of the form ${\displaystyle (C_{1}r)^{C_{2}}}$ where constants C1 and C2 depend only on the form ƒ. A stronger result holds, that if K is the field generated by the roots of ƒ then the equation has only finitely many solutions with x and y integers of K and again these may be effectively determined.[2]
## Solving Thue equations
Solving a Thue equation can be described as an algorithm[3] ready for implementation in software. In particular, it is implemented in the following computer algebra systems:
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# 44054
## 44,054 is an even composite number composed of two prime numbers multiplied together.
What does the number 44054 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 4 divisors.
44054 is an even composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.
## Prime factorization of 44054:
### 2 × 22027
See below for interesting mathematical facts about the number 44054 from the Numbermatics database.
### Names of 44054
• Cardinal: 44054 can be written as Forty-four thousand and fifty-four.
### Scientific notation
• Scientific notation: 4.4054 × 104
### Factors of 44054
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 2
• Sum of prime factors: 22029
### Divisors of 44054
• Number of divisors d(n): 4
• Complete list of divisors:
• Sum of all divisors σ(n): 66084
• Sum of proper divisors (its aliquot sum) s(n): 22030
• 44054 is a deficient number, because the sum of its proper divisors (22030) is less than itself. Its deficiency is 22024
### Bases of 44054
• Binary: 10101100000101102
• Base-36: XZQ
### Squares and roots of 44054
• 44054 squared (440542) is 1940754916
• 44054 cubed (440543) is 85498017069464
• The square root of 44054 is 209.8904476149
• The cube root of 44054 is 35.3179197829
### Scales and comparisons
How big is 44054?
• 44,054 seconds is equal to 12 hours, 14 minutes, 14 seconds.
• To count from 1 to 44,054 would take you about twelve hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 44054 cubic inches would be around 2.9 feet tall.
### Recreational maths with 44054
• 44054 backwards is 45044
• The number of decimal digits it has is: 5
• The sum of 44054's digits is 17
• More coming soon!
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0
# What is 438692 round to the nearest Thousand?
Jahzara Jolly
Lvl 2
2020-11-12 15:11:37
Best Answer
439,000 is the answer for, when you round 438,692
Mr Answers
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2020-11-12 15:53:16
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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Jahzara Jolly
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400000
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Is that the an
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Vyduuvyusg
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Yesssssss
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# Lesson 11
Contemos imágenes (parte 1)
### Lesson Purpose
The purpose of this lesson is for students to use the 10 ones and some more ones structure of numbers 11–19 to help them count groups of up to 19 images.
### Lesson Narrative
In previous lessons, students counted groups of up to 20 objects and groups of up to 10 images. Students composed and decomposed numbers 11–19 into 10 ones and some more ones. In this lesson, students find a group of 10 images inside a group of 11–19 images and then determine the total number of images. Throughout the section, students have access to a reference sheet that shows numbers 11–20 with dots in 10-frames that they can use to identify written numbers.
• Action and Expression
• MLR8
### Learning Goals
Teacher Facing
• Understand numbers 11–19 as ten ones and some more ones.
• Write a number to answer “how many” questions about groups of up to 19 images.
### Student Facing
• Encontremos 10 figuras y averigüemos cuántas figuras hay.
### Required Materials
Materials to Gather
### Required Preparation
Activity 2:
• Students need access to at least 2 different colored crayons, colored pencils, or markers.
Activity 3:
• Gather materials from:
• Find the Value of Expressions, Stage 1
• Make or Break Apart Numbers, Stages 1 and 2
• Bingo, Stages 1-4
### Lesson Timeline
Warm-up 10 min Activity 1 10 min Activity 2 10 min Activity 3 25 min Lesson Synthesis 5 min Cool-down 0 min
### Teacher Reflection Questions
What connections did students make between the different strategies shared? What questions did you ask to help make the connections more visible?
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# Find the minimun value of the product p of two numbers, one of which is 9 less than twice the other.
Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator
Posted on
Hello!
Denote the first number by x. Then the second will be
2*x - 9
("9 less than twice the other").
Its product p is `x*(2x-9) = 2x^2 - 9x.`
It is not difficult to find the minimum value of this expression (and to show that it has a minimum). For this we select a perfect square:
`2x^2 - 9x = 2*(x^2 - (9/2)*x) = 2*(x - 9/4)^2 - 2*(9/4)^2.`
Here I use the formula `(a-b)^2 = a^2 - 2ab + b^2.`
The first summand, `2*(x - 9/4)^2,` cannot be less than zero. Moreover, it is zero if and only if x = 9/4. So the minimum value of the initial expression is
`-2*(9/4)^2` = -81/8 = -10.125. This is the answer.
P.S. The numbers which form this product are x=9/4 and 2x-9=-9/2.
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CZECH TECHNICAL UNIVERSITY IN PRAGUE
STUDY PLANS
2023/2024
# Geodesy 4
Code Completion Credits Range Language
154VY3 KZ 2 4C Czech
Garant předmětu:
Lenka Línková, Zdeněk Skořepa
Lecturer:
Tutor:
Jaroslav Braun, Hana Braunová, Lenka Línková, Zdeněk Skořepa
Supervisor:
Department of Special Geodesy
Synopsis:
Surveying and calculation of points of the geodetic network - a flat special-purpose network with measured horizontal directions and lengths, determination of trigonometric height differences from simultaneously and bilaterally measured zenith angles and slope lengths using a total station, determination of the height of one point of the network by geometric levelling from the centre (precise levelling), use of GNSS (RTK measurements in the CZEPOS network of reference stations), calculation according to the method of least squares. Detailed positional and height measurements (tachymetry) of the specified location in the extravilan at a scale of 1 : 500 and preparation of a digital terrain model. Measurement of the actual state of the building for the planned reconstruction and preparation of documentation at a scale of 1 : 50 (plan drawing).
Requirements:
Working with software: Matlab, Groma, Atlas DMT.
Syllabus of lectures:
there are no lectures
Syllabus of tutorials:
1. Special network (positional)
2. Special network (determination of heights)
3. Digital terrain model - tachymetry and creation of digital terrain model (Atlas DMT)
4. Documentation of the building object.
Study Objective:
Further expansion and acquisition of new practical experience during the work of a surveyor in the field from the subjects Geodesy 3 and Geodesy 4. Emphasis is placed on the organization of the work of the measuring group and measurement, where the final result of the task depends on the work result of each member of the measuring group.
Study materials:
[1] Blažek R. - Skořepa Z.: Geodézie 3 Výškopis, Praha, ČVUT Praha 2009. 162 s. ISBN 978-80-01-04358-5.
[2] Skořepa, Z.: Geodézie 4. Praha, Česká technika - nakladatelství ČVUT 2017. 132 s. ISBN 978-80-01-05481-9.
[3] Vobořilová, P. - Skořepa, Z.: Geodézie 1, 2 (Návody na cvičení). Praha, ČVUT 2005. 135 s. ISBN 80-01-02869-0.
Note:
Further information:
https://k154.fsv.cvut.cz/vyuka/bc-geodezie/vy3/
Time-table for winter semester 2023/2024:
Time-table is not available yet
Time-table for summer semester 2023/2024:
Time-table is not available yet
The course is a part of the following study plans:
Data valid to 2024-07-20
Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet2353306.html
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## R Literacy for digital soil mapping
#### Part 1
• R basics: commands, expressions, assignments, operators, objects
• R Data Types
• R data structures
• Functions, arguments, and packages
• Getting help.
#### Part 2
• Vectors, matrices, and arrays
• Vector arithmetic, some common functions and vectorised formats
• Matrices and arrays.
#### Part 3
• Data frames, data import, and data export
• Creating data frames manually
• Working with data frames
#### Part 4
• Graphics: the basics
#### Part 5
• Manipulating data. Modes, classes, attributes, length, and coercion
• Indexing, sub-setting, sorting and locating data
• Factors
• Combining data
#### Part 6
• Exploratory data analysis
• Summary statistics
• Histograms and boxplots
• Normal quantile and cumulative probability plots
#### Part 7
• The basics of linear models
#### Part 8
This section introduces how to construct a function. A function is at the heart of R and is akin to a set of instructions to run a particular task. Functions are incredibly powerful and the fact that one can create their own functions leaves open the door for some very creative thinking about how to solve a particular problem or to conduct bespoke tasks.
The example here is about how one would go about designing a soil sample down along a toposequence. The starting point could be the top of a hill, at the bottom of a hill or anywhere between. This seems pretty intuitive to do in your mind, but to code this in R or any language requires a bit of logical thought and creativity. These are important for learning R and for doing digital soil mapping things!
Updated on
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A company’s net worth goes by many names, such as equity, stockholders’ equity, net assets and book value. These represent the accounting value of stockholders’ interest in the business. Equity equals total assets minus total liabilities, which you can find on a private company’s balance sheet. It might be difficult to calculate the equity of a private company other than your own small business, though. A private company typically discloses its financial information only to its own stockholders because it does not sell its stock to the general public.
#### Step 1
Subtract your accumulated depreciation from your total property, plant and equipment to figure your net property, plant and equipment. Depreciation is a periodic cost that reduces the accounting value of property, plant and equipment. You can find these items in the assets section of your balance sheet. For example, assume your small business has \$15,000 in total property, plant and equipment and \$5,000 in accumulated depreciation. Subtract \$5,000 from \$15,000 to get \$10,000 in net property, plant and equipment.
#### Step 2
Subtract your accumulated amortization from your total intangible assets to figure your net intangibles. Intangible assets include patents, copyrights and other non-physical assets. Amortization is a periodic cost, like depreciation, that reduces the accounting value of intangible assets. In this example, assume you have \$5,000 in patents and \$1,000 in accumulated amortization. Subtract \$1,000 from \$5,000 to get \$4,000 in net intangibles.
#### Step 3
Add your net property, plant and equipment and your net intangibles to your other assets listed in the assets section of your balance sheet to figure your total assets. These assets might include cash, inventory, accounts receivable and investments. In this example, assume you have \$20,000 in cash, \$40,000 in inventory and \$12,000 in accounts receivable. Add \$10,000, \$4,000, \$20,000, \$40,000 and \$12,000 to get \$86,000 in total assets.
#### Step 4
Add the amount of your total liabilities listed in the liabilities section of your balance sheet. Liabilities are amounts you owe, such as accounts payable, accrued expenses, loans, bonds payable and deferred income taxes. In this example, assume your small business has \$5,000 in accounts payable, \$15,000 in accrued expenses and \$20,000 in short-term loans. Add these together to get \$40,000 in total liabilities.
#### Step 5
Subtract your total liabilities from your total assets to calculate your company’s net worth, or equity. A negative result represents a negative net worth, which means the company owes more than the value of its assets. Concluding the example, subtract \$40,000 from \$86,000 to get \$46,000 in equity.
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# 240 Volt Thermostat Wiring Diagram
By | July 27, 2023
# Understanding the Basics of a 240 Volt Thermostat Wiring Diagram
A wiring diagram is a simplified graphical representation of an electrical circuit. It shows the components of the circuit as simplified shapes, and the power and signal connections between the devices. A wiring diagram is an invaluable tool when it comes to installing or repairing a system, as it can easily identify the wiring and components needed for each installation. Understanding the basics of a 240 volt thermostat wiring diagram is essential for any homeowner or technician.
A thermostat is a device that regulates the temperature of a room by controlling the flow of hot/cold air. This is done by using a variety of components such as: relays, contactors, and solenoids. Understanding how these components work together is the key to understanding a 240 volt thermostat wiring diagram. In this article we will discuss the components of a thermostat, their purpose and function, and how to interpret a wiring diagram.
## The Components of a 240 Volt Thermostat Wiring Diagram
The components of a 240 volt thermostat wiring diagram are typically composed of: relays, contactors, switches, transformers, and solenoids. The relays are responsible for controlling the flow of electricity and send signals to the thermostat. The contactors are used to make contact with the switch and relay contacts. The switches are the devices that turn the system on and off. The transformers are responsible for providing the necessary voltage for the system. Finally, the solenoids act as the mechanical actuators for the thermostat.
## Interpreting the Wiring Diagram
To properly interpret a 240 volt thermostat wiring diagram, one must first understand the symbols. Each component has its own unique symbol. For example, the relay is represented by a box with two arrows. The contactor is represented by a triangle with a line across it, and the switch is represented by a circle with a “T” inside. Once the symbols are identified, the diagram can be interpreted by following the lines from one component to the next.
## Connecting the Wires
Once the wiring diagram is interpreted, the wires must be connected. To connect the wires correctly, it is important to follow the instructions in the wiring diagram exactly. The most important part of this process is to ensure that the correct wire is connected to the correct terminal. Most of the time, the terminals are labeled, however, if they are not, a continuity tester can be used to determine which terminal the wires should be connected to.
## Conclusion
Understanding the basics of a 240 volt thermostat wiring diagram is an important step when installing or repairing a system. By familiarizing yourself with the components of a thermostat and properly interpreting the wiring diagram, you can ensure that your system is properly wired. Following the steps outlined above will help ensure that your system is installed correctly and safely.
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# Let f(x) = and g(x) = $\sqrt x$ be two functions defined over the set of non-negative real numbers. Find (f + g) (x), (f – g) (x), (fg) (x) and $\bigg( \large\frac{f}{g} \bigg) (x)$.
We have (f + g) (x) = $\sqrt x + x, (f-g) (x) = \sqrt x - x$
$(fg)x= \sqrt x (x)=x^{\large\frac{3}{2}}$ and $\bigg( \large\frac{f}{g} \bigg) (x) = \large\frac{\sqrt x }{x} = x ^{-\large\frac{1}{2}}, x \neq 0$
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+0
# help
+5
193
2
a volleyball has a diameter of 10 inches. a pump can inflate the ball at a rate of 325 inches cubed per minute. How long will it take to inflate the ball?
Guest Feb 23, 2017
Sort:
#1
0
you need to find the volume using pi and the raduis (half of diameter) the divide by 325 and you have the time
Guest Feb 23, 2017
#2
+6360
+6
Well how many cubic inches are in the volleball?
$$V= \frac{4}{3}\pi r^{3}$$
If the diameter is 10 inches, the radius is 5 inches.
r = 5
$$V= \frac{4}{3}\pi (5^{3})$$
$$V= \frac{4}{3}\pi (125)$$
$$V = \frac{500\pi}{3}$$
V ≈ 523.5988 cubic inches.
So if the pump can put in 325 cubic inches per minute, and the volleyball has 523.5988 cubic inches, it will take 523.5988/325 ≈ 1.611 minutes to fill. :)
hectictar Feb 23, 2017
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Published: 30 September 2014
# Influence law of tower stiffness on vertical stiffness of three-tower self-anchored suspension bridge based on frequency formulas
Chao Zhang1
Kai Huang2
1, 2College of Civil Engineering, Fuzhou University, Fuzhou, Fujian, 350108, China
Corresponding Author:
Kai Huang
Views 75
Reads 28
Downloads 1142
#### Abstract
In this paper, frequencies of low order vertical modals for three-tower self-anchored suspension bridge (TSSB) are studied. The first three-tower self-anchored suspension bridge is taken as prototype in the paper. The finite element model of the bridge is established. And dynamic characteristics are analyzed. Based on Rayleigh method, frequency formulas of 1st asymmetric vertical vibration (AVV) and symmetric vertical vibration (SVV) are deduced considering contribution of towers stiffness. The accuracy of deduced formulas was validated by results of numerical analysis and modal test. Stiffness characteristic difference between TSSB and multi-tower earth-anchored suspension bridge (MESB) is researched based on frequency formulas. The significant difference is gravity stiffness component in frequency formulas. At last, influence law of tower stiffness on vertical frequencies was studied based on validated formulas. The results indicated that: 1) middle tower stiffness and side tower stiffness play important role in frequency of 1st AVV and SVV respectively. Frequencies are enlarging as tower stiffness increasing; 2) when tower stiffness is low, deduced frequency formulas are the same as formulas neglecting influence of tower stiffness in reference paper; 3) based on deduced frequency formulas, expressions of critical tower stiffness for simplified formulas are proposed.
## 1. Introduction
Three-tower self-anchored suspension bridge (TSSB) is one new kind of bridge, which combines the advantages of self-anchored suspension bridge and three-tower suspension bridge [1, 2]. TSSB is one of the best kinds of bridge for urban medium-span bridge, which has particular adaptability to site of W shape [3]. It has many advantages [4], such as elegant appearance, good site adaptive capacity and relative larger across ability. Louzhou Bridge is the first TSSB, which is built Fuzhou, China, in 2012 [5]. And this kind of bridge can be built more in any place in the world.
As lacking of effective constraints from cables, equivalent stiffness of middle tower in TSSB is low [6]. Therefore, the static and dynamic mechanical properties of three-tower suspension bridge are quite different from traditional double-tower suspension bridge [7]. Among them, free vibration characteristics is quite different. An experiment of scaled multi-tower suspension bridge specimen has been made to study its dynamic characteristics [8]. Jiao (2010) [9] studied dynamic characteristic of Taizhou Bridge, which is a two-span three-tower earth-anchored suspension bridge (TESB). The studied shows that frequency of 1st vertical vibration is 0.08267 Hz, whose modal shape is asymmetric vibration. Related researches show the 1st vertical vibration is usually shown as asymmetric vertical vibration as equivalent stiffness of middle tower is low [10].
Dynamic characteristics of bridge are the base for earthquake and wind resistant design [11, 12]. At present, finite element method (FEM) is widely used in accurate analyzing vibration frequencies and mode shapes. Jiao (2010) [13] established the finite element model of TSSB by ABAQUS and studied effects of elastic restraints between mid-tower and girders on dynamic property of TESB. Wang (2014) [14] developed a three-dimensional finite element model to study the free vibration characteristics of a triple-tower twin-span suspension bridge during the construction phase and right after the erection of the main cable. Wang (2010) [15] studied the dynamic characteristics of the Taizhou Yangtze River Bridge, the first triple-tower long-span suspension bridge in China based on ABAQUS. The results showed the vertical, lateral and torsional stiffness of girder have different effects on the dynamic characteristics of triple-tower suspension bridges. Furthermore, the parametric analysis on the structural flutter stability of Taizhou Yangtze River Bridge was carried out in 2014 [16].
However, there are some disadvantages of FEM, especially on preliminary design phase. For example, it is time-consuming to comparing different design proposals and difficult to reveal influence law of special parameters. In this sense, frequency estimation formulas are fit, as it can quickly provide general frequencies for estimating on preliminary design phase and validating on fine numerical modeling analysis phase [17]. In addition, we can easily get the influence rules of each structure parameter from frequency formulas [18]. Therefore, the frequency formulas are meaningful for quick calculating frequencies. And Chao (2013) [19] deduced frequency formulas of first order vertical modals for TSSB. However, both of them have not considered the effect of tower stiffness on frequencies, which is significant for three-tower suspension bridge in most cases.
In this paper, the frequency estimation formulas of vertical vibration for TSSB considering influence of tower stiffness are deduced by Rayleigh method. Based on deduced formulas, different dynamic mechanical properties between three-tower and double-tower self-anchored suspension bridge are studied based on frequency formulas. And influence law of tower stiffness on vertical frequency is discussed in detail.
## 2. Typical TSSB description and finite element modeling
### 2.1. Description of typical TSSB – Louzhou Bridge
Louzhou Bridge is the first three-tower self-anchored suspension bridge in the world. It is taken as prototype structure of three-tower self-anchored suspension bridge (TSSB) in this paper shown in Fig. 1. This bridge, crossing the Ming River at Fuzhou, China, was opened to traffic in 2012.
The arrangement of the spans along the bridge is 80 m+168 m+168 m+80 m, with an overall length of 496 m. The total width of the bridge deck is 43 m. The suspension bridge system is composed of two sets of suspension cables anchored at the continuous steel deck directly without any anchorage block. Span ratio of main span is 1:6, while that of side span is 1:12.88. The space between slings is 7 m. The deck over the four spans is supported by cables and bearings. Other structural parameters are shown in Table 1.
Table 1Structural parameters of Louzhou Bridge (Unit: m, t, kN)
Parameter Value Parameter Value Parameter Value ${f}_{1}$ 6.35 ${h}_{t}$ 50.85 ${E}_{c}$ 1.95E+08 ${f}_{2}$ 28.00 *${S}_{t1}$ 4.33E+04 ${E}_{g}$ 2.06E+08 ${l}_{1}$ 80.00 *${S}_{t2}$ 4.33E+04 ${E}_{t}$ 3.45E+07 ${l}_{2}$ 168.00 ${M}_{c}$ 0.73 *${A}_{c}$ 0.0474 ${l}_{s1}$ 103.42 ${M}_{g}$ 33.40 ${A}_{g}$ 1.50 ${l}_{s2}$ 207.71 *${M}_{t}$ 42.10 ${I}_{g}$ 1.69 Note: 1) ${f}_{1}\left({f}_{2}\right)$, ${l}_{1}\left({l}_{2}\right)$, ${l}_{s1}\left({l}_{s2}\right)$ stand for rise, span, equivalent cable length for side span (middle span) respectively; 2) subscript of $c$, $g$, $t$ indicate cable, girder and tower respectively; 3) $S$, $M$, $E$, $A$ stand for tower equivalent stiffness, linear mass, elasticity modulus and area respectively; 4) superscript $*$ means the value for each one.
### 2.2. Finite element modeling
For purposes of this study, a complete 3D FE model was developed in SAP2000N as shown in Fig. 2. This model is used for both static and dynamic analyses. The FE model consists of 1418 nodes and 1664 elements. The bridge towers were represented by three-dimensional multilevel portal frames with the two legs fixed at the base. The soil-structure interaction was not considered in this model.
The geometric properties of three-dimensional beam elements of towers and piers were calculated from design drawings. Since cross sections of tower legs are decreasing from base to top, sectional properties of beam elements for tower were assumed to be uniform section in each 2 m element. Tower equivalent stiffness is 4.33E+04 kN/m shown in Table 1.
Fig. 1Elavation figure of Louzhou Bridge (unit: m)
The equivalent beam was connected to the suspenders through a series of horizontal rigid arm members. The bearings between the deck and the towers were represented as swing rigid links so as to allow free longitudinal motion of the deck. The lateral motion of the deck relative to the towers was restricted through horizontal rigid links.
Fig. 2Louzhou Bridge: a) panorama and b) 3D finite-element model
a)
b)
### 2.3. Modal analysis
Modal analysis is needed to determine the natural frequencies and mode shapes of free vibration. The initial equilibrium configuration is important in suspension bridges since it is a starting position to perform the succeeding analysis [20]. Therefore, the dead load has a significant influence on the stiffness of suspension bridge. In numerical analysis, this influence can be taken into account through the static analysis under the dead load and cable tensions before dynamic analysis is carried out.
Fig. 3Mode shape of 1st AVV
Fig. 4Mode shape of 1st SVV
In this paper, modal analysis is carried out based on static analysis under dead loading and cable pre-tensions. We can get the natural frequencies and mode shapes. In general, modal analysis of the Louzhou Bridge showed the modes were arranged in certain sequence: longitudinal mode, vertical modes, lateral modes and torsion modes. As we mainly concern about the vertical stiffness of TSSB, the first 2 vertical modes and frequencies are shown in Fig. 3 and Fig. 4. The frequencies of the 1st AVV and SVV are 0.4481 Hz and 0.6295 Hz respectively. More information of dynamic characteristics is shown in reference [21].
## 3. Frequency formulas for vertical vibration
### 3.1. Rayleigh method
Rayleigh’s method is a technique for finding natural frequencies by equating the maximum kinetic energy of a system with the maximum potential (often strain) energy. Rayleigh’s method is a widely used, classical method for the calculation of the natural vibration frequency of a structure in the low order. According to Rayleigh’s method, we can get the formula as followed:
1
${\omega }^{2}=\frac{{U}_{\mathrm{m}\mathrm{a}\mathrm{x}}}{{T}_{\mathrm{m}\mathrm{a}\mathrm{x}}}=\frac{{{\int }_{0}^{l}EI\left(x\right)\left({\stackrel{-}{\varphi }}^{\text{'}\text{'}}\left(x\right)\right)}^{2}dx}{{{\int }_{0}^{l}m\left(x\right)\left(\stackrel{-}{\varphi }\left(x\right)\right)}^{2}dx},$
where, ${U}_{\mathrm{m}\mathrm{a}\mathrm{x}}$ stands for maximal value of gravitational potential energy; ${T}_{\mathrm{m}\mathrm{a}\mathrm{x}}$ stands for maximal value of kinetic energy; $\stackrel{-}{\varphi }\left(x\right)$ is the deformation function of each component; ${\stackrel{-}{\varphi }}^{\text{'}\text{'}}\left(x\right)$ is the quadratic differential form of $\stackrel{-}{\varphi }\left(x\right)$.
### 3.2. Frequency formulas and derivation process
Some parameters are defined for simplified expression during the formulas derivation process, shown as followed.
${l}_{i}$ is the length of girder in the $i$th span; ${f}_{i}$ is the rise of cable in the $i$th span; ${h}_{i}$ is the height of the $i$th tower; ${y}_{i}$ is the geometric alignment of cable in the $i$th span, whose expression is ${y}_{i}=4{f}_{i}\left({x}_{i}/{l}_{i}-{\left({x}_{i}/{l}_{i}\right)}^{2}\right)$; ${l}_{si}$ is geometric parameter of cable in the $i$th span, whose expression is ${l}_{si}={\int }_{0}^{{l}_{i}}{\left(ds/dx\right)}^{3}dx$; ${l}_{si}$ is the equivalent longitudinal stiffness of the $i$th tower, whose expression is $3{E}_{ti}{I}_{ti}/{{h}_{ti}}^{3}$; ${v}_{i}$ is the vertical deflection of girder in the $i$th span; ${H}_{i}$ is the horizontal tensile force of cable in the $i$th span; ${u}_{hi}$ is the horizontal deflection of top node in the $i$th tower.
As the simplest type of TSSB, the three-tower self-anchored suspension bridge (TSSB) is taken as example in this section. We can get typical model shapes for first 2 orders vertical vibration, shown in Fig. 5 and Fig. 6.
Fig. 5Mode shape of 1st asymmetric vertical vibration
Fig. 6Mode shape of 1st symmetric vertical vibration
Some basic assumptions are adopted in this paper containing the first 6 assumptions in reference [19]. Besides, the influence of tower stiffness to vertical vibration is considered. The deformation shapes of girder and tower for each modal are assumed based on the boundary condition and mode shape. Deformation shapes of girder are same to reference [9]. Deformation shape of tower is assumed as deformation of cantilever column under uniform lateral horizontal force. That is shown as followed:
2
${u}_{i}\left(x,t\right)=3\cdot \frac{{u}_{h}}{{h}^{3}}\left(\frac{1}{2}h\cdot {x}^{2}-\frac{1}{6}\cdot {x}^{3}\right)\cdot \mathrm{s}\mathrm{i}\mathrm{n}\left(\omega t+\varphi \right),x\in \left[0,{h}_{i}\right],i=1,\mathrm{}2,\mathrm{}3,\mathrm{}4.$
Based on basic assumptions, we can get vibration energy expressions of each components, such as cables, towers and girders. The expressions of kinetic energy and potential energy are deduced as followed:
3a
${U}_{ce}=\frac{\sum _{i=1}^{4}{{H}_{i}}^{2}{l}_{si}}{2{E}_{c}{A}_{c}}+\frac{{E}_{g}{I}_{gv}}{2}{\sum _{i=1}^{4}{\int }_{0}^{{l}_{i}}\left(\frac{{\partial }^{2}\nu }{\partial {x}^{2}}\right)}^{2}dx+\frac{1}{2}\sum _{i=1}^{4}{\int }_{0}^{{l}_{i}}\frac{{{H}_{1}}^{2}}{{E}_{g}{A}_{g}}dx$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}+{\sum _{i=1}^{3}\frac{1}{2}{E}_{t}{I}_{tv}\cdot {\int }_{0}^{{h}_{i}}\left(\frac{{\partial }^{2}{u}_{i}}{\partial {x}^{2}}\right)}^{2}dx,$
3b
$T=\sum _{i=1}^{4}\frac{1}{2}{{\int }_{0}^{{l}_{i}}{m}_{c}\left(\frac{\partial {v}_{i}}{\partial t}\right)}^{2}dx+\sum _{i=1}^{4}\frac{1}{2}{{\int }_{0}^{{l}_{i}}{m}_{g}\left(\frac{\partial {v}_{i}}{\partial t}\right)}^{2}dx+\sum _{i=1}^{3}\frac{1}{2}{{\int }_{0}^{{h}_{i}}{m}_{t}\left(\frac{\partial {u}_{i}}{\partial t}\right)}^{2}dx.$
Based on force equilibrium and deformation compatibility conditions, the relations between the variables are inferred from above deflection drawing. And expressions of kinetic energy and potential energy for each component of TSSB are obtained. By the Eqs. (1), (3a), (3b), we can deduce the frequency formulas for 1st symmetric and asymmetric vertical vibration as shown in the next section. More details can refer to reference [21] (Zhangchao, 2011).
### 3.2.1. Frequency formula for 1st asymmetric vertical vibration (AVV)
Frequency of 1st symmetric vertical vibration is named ${\stackrel{-}{f}}_{v}^{a}$. For symbol ${\stackrel{-}{f}}_{v}^{a}$, superscript a stands for asymmetric vibration; subscript v stands for vertical vibration; the overline stands for considering the influence of tower stiffness in frequency calculation. We can get formula for frequency ${\stackrel{-}{f}}_{v}^{a}$ shown as followed:
4
${\stackrel{-}{f}}_{v}^{a}=\frac{1}{2\pi }\sqrt{\frac{\alpha +{\gamma }_{0}\cdot {E}_{g}{I}_{g}}{{m}_{g}+{m}_{c}}},$
where, ${\gamma }_{0}$ is a variable related to lengths of each span; $\alpha$ is a variable related to spans, ratio, tensile stiffness of cables and bending stiffness of towers, named Tower Stiffness Influence Coefficient (TSIC) for 1st AVV. Their expressions are shown as followed:
5
${\gamma }_{0}=97.55/\left({{l}_{1}}^{2}{l}_{2}+{{l}_{1}}^{2}{{l}_{2}}^{2}+{l}_{1}{{l}_{2}}^{2}\right),$
6
$\alpha =52{{f}_{2}}^{2}\cdot {S}_{t1}\cdot {S}_{t2}/\left[\left({{l}_{1}}^{3}+{{l}_{2}}^{3}\right)\left(2{S}_{t1}+{S}_{t2}\right)\right].$
### 3.2.2. Frequency formula for 1st symmetric vertical vibration (SVV)
Frequency of 1st asymmetric vertical vibration is named ${\stackrel{-}{f}}_{v}^{s}$. For the symbol ${\stackrel{-}{f}}_{v}^{s}$, superscript $s$ stands for asymmetric vibration; subscript $v$ stands for vertical vibration; overline stands for considering the influence of tower stiffness in frequency calculation. We can get the calculation formula for frequency ${\stackrel{-}{f}}_{v}^{s}$ as followed:
7
${\stackrel{-}{f}}_{v}^{s}=\frac{1}{2\pi }\sqrt{\frac{\beta \cdot {\gamma }_{1}\cdot {E}_{c}{A}_{c}+{\gamma }_{2}\cdot {E}_{g}{I}_{gv}}{{m}_{g}+{m}_{c}}},$
where, ${\gamma }_{1}$, ${\gamma }_{2}$ are variables related to lengths of each span:
8
${\gamma }_{1}=2016{\left(3{{l}_{2}}^{2}-4{{l}_{1}}^{2}\right)}^{2}/\left[\left(15{{l}_{1}}^{3}{{l}_{2}}^{4}+95{{l}_{2}}^{7}\right)\left({l}_{s1}+{l}_{s2}\right)\right],$
9
${\gamma }_{2}=1512{\left(2{l}_{2}+3{l}_{1}\right)}^{2}/\left(31{{l}_{1}}^{4}{l}_{2}+19{l}_{1}{{l}_{2}}^{4}\right).$
$\beta$ is a variable related to tensile stiffness of cables, compressive stiffness of girder and bending stiffness of towers, named Tower Stiffness Influence Coefficient (TSIC) for 1st symmetric vertical vibration. And ${B}_{1}$, ${B}_{2}$, ${B}_{3}$ are some variables related to some structural parameters. Their expressions are shown as followed:
10
$\beta =\frac{{{B}_{1}}^{2}\cdot {l}_{s1}+{{B}_{1}}^{2}\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)+{{B}_{2}}^{2}\cdot {l}_{s2}+{{B}_{3}}^{2}\cdot \frac{{S}_{t}}{{E}_{c}{A}_{c}}}{{l}_{s1}+{l}_{s2}+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)},$
11a
${B}_{1}=\frac{-\left[3{f}_{2}-4{f}_{1}\left(1+\frac{{S}_{t}}{{E}_{c}{A}_{c}}\cdot {l}_{s2}\right)\right]\left[\left({l}_{s1}+{l}_{s2}\right)+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)\right]}{\left(3{f}_{2}-4{f}_{1}\right)\left[\left({l}_{s1}+{l}_{s2}\right)+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)+\frac{{S}_{t}}{{E}_{c}{A}_{c}}\cdot {l}_{s1}\cdot {l}_{s2}+\frac{{S}_{t}}{{E}_{g}{A}_{g}}\cdot \left({l}_{1}+{l}_{2}\right)\cdot {l}_{s2}\right]},$
11b
${B}_{2}=\frac{-\left[3{f}_{2}\left(1+\frac{{S}_{t}\cdot {l}_{s1}}{{E}_{c}{A}_{c}}+\frac{{l}_{1}+{l}_{2}}{{E}_{g}{A}_{g}}\cdot {S}_{t}\right)-4{f}_{1}\right]\left[\left({l}_{s1}+{l}_{s2}\right)+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)\right]}{\left(3{f}_{2}-4{f}_{1}\right)\left[\left({l}_{s1}+{l}_{s2}\right)+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)+\frac{{S}_{t}}{{E}_{c}{A}_{c}}\cdot {l}_{s1}\cdot {l}_{s2}+\frac{{S}_{t}}{{E}_{g}{A}_{g}}\cdot \left({l}_{1}+{l}_{2}\right)\cdot {l}_{s2}\right]},$
11c
${B}_{3}=\frac{-\left[3{f}_{2}\cdot \left({l}_{s1}+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)\right)+4{f}_{1}\cdot {l}_{s2}\right]\left[\left({l}_{s1}+{l}_{s2}\right)+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)\right]}{\left(3{f}_{2}-4{f}_{1}\right)\left[\left({l}_{s1}+{l}_{s2}\right)+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right)+\frac{{S}_{t}}{{E}_{c}{A}_{c}}\cdot {l}_{s1}\cdot {l}_{s2}+\frac{{S}_{t}}{{E}_{g}{A}_{g}}\cdot \left({l}_{1}+{l}_{2}\right)\cdot {l}_{s2}\right]}.$
### 3.3.1. Example of physical scaled model of TSSB
The physical scaled model of TSSB in Reference [1] (Zhang C., et. al) is taken as example. A quite dense measurement location on the bridge deck in the vertical directions was proposed to identify acceptable mode shapes of the bridge in that paper. Stochastic Subspace Identification (SSI) method in the time-domain was implemented to identify the measured acceleration data. The identified most significant vertical frequencies of the TSSB scaled model bridge are summarized in that paper. Among that, frequencies of 1st asymmetric and symmetric vertical vibration are 8.16 Hz and 14.35 Hz respectively.
According to provided structural parameters in Reference [1] (Zhang C., et. al), we can get values of tower equivalent stiffness and equivalent cable length. They are shown as followed: ${S}_{t1}={S}_{t2}=$ 65.79 kN/m; ${l}_{s1}=$2.05 m, ${l}_{s2}=$4.15 m.
Therefore, we can calculate the frequencies of first 2 vertical modals for this physical scaled TSSB model by deduced formulas in this paper.
1) ${\stackrel{-}{f}}_{v}^{a}$ calculated by frequency formulas:
From Eq. (5), (6), we can get ${\gamma }_{0}=\text{2.1414}$, $\alpha =\text{2.17×1}{\text{0}}^{\text{-3}}$;
Then, according to Eq. (4), we can get ${\stackrel{-}{f}}_{v}^{a}=\mathrm{}$7.9543 Hz.
2) ${\stackrel{-}{f}}_{v}^{s}$ calculated by frequency formulas:
From Eqs. (8), (9), we can get ${\gamma }_{1}=\mathrm{}$0.0253; ${\gamma }_{2}=\mathrm{}$3.82;
From Eqs. (10), (11), we can get ${B}_{1}=$–0.945; ${B}_{2}=$–1.031; ${B}_{3}=\mathrm{}$–4.993; $\beta =\mathrm{}$1.068;
Then, according to Eq. (7), we can get ${\stackrel{-}{f}}_{v}^{s}=\mathrm{}$13.9950 Hz.
The formulas computed result and measured result are shown in Table 2. From the table, we can find that frequency computed from provided formulas are in good agreement with the result in reference [1].
Table 2Frequencies comparison for two TSSB (Unit: Hz)
Modal shape Physical scaled model of TSSB Louzhou Bridge Experiment measured Formulas computed Frequency difference Experiment measured Formulas computed Frequency difference 1st AVV 8.1600 7.9543 –0.2057 0.4481 0.4578 0.0097 1st SVV 14.3500 13.995 –0.3550 0.6295 0.6298 0.0003
### 3.3.2. Example of practical bridge
Louzhou Bridge in section 2 is taken as an example in this part. Structural parameters of Louzhou Bridge are shown in Fig. 1. Based on modal analysis of FEM, natural frequencies of first 2 vertical mode shapes were calculated to be 0.4481 Hz and 0.6295 Hz, shown in Fig. 2 and Fig. 3.
For Louzhou Bridge, we can calculate frequencies of first 2 vertical modals, by deduced formulas in this paper, as followed.
1) ${\stackrel{-}{f}}_{v}^{a}$ calculated by frequency formulas:
From Eqs. (5), (6), we can get ${\gamma }_{0}=\text{3.43×1}{\text{0}}^{\text{-7}}$, $\alpha =\text{9.15×1}{\text{0}}^{\text{-6}}$;
Then, according to Eq. (4), we can get ${\stackrel{-}{f}}_{v}^{a}=\mathrm{}$0.4578 (Hz).
2) ${\stackrel{-}{f}}_{v}^{s}$ calculated by frequency formulas:
From Eqs. (8), (9), we can get ${\gamma }_{1}=\text{1.05×1}{\text{0}}^{\text{-5}}$; ${\gamma }_{2}=\text{6.12×1}{\text{0}}^{\text{-7}}$;
From Eqs. (10), (11), we can get ${B}_{1}=$–0.47; ${B}_{2}=\mathrm{}$–1.326; ${B}_{3}=\mathrm{}$–191.81; $\beta =\text{1.715}$;
Then, according to Eq. (7), we can get ${\stackrel{-}{f}}_{v}^{s}=\mathrm{}$0.6298 (Hz).
The frequencies computed by two methods are compared in Table 2. We can come to the conclusion that the results come from deduced formulas are similar to numerical computed.
All in all, frequencies of low order vertical modal of two TSSB are calculated by deduced formulas in this paper. And the accuracy of formulas is validated by results of both experiment test data and numerical analysis result. It is a new frequency calculation method of TSSB, besides finite element method. The deduced formulas are special suitable for preliminary design stage, in which the detailed structural parameters are not sure. Through the formulas, the frequency can be estimated quickly. On the other hand, the computations of this paper can be used to test the correctness of finite element model.
## 4. Influence law of tower stiffness on frequencies
### 4.1. Influence law of tower stiffness on frequency
As we can know from above analysis, tower stiffness plays great roles in frequencies. The influence law of tower stiffness on vertical vibration will be studied in this section. Methods of formulas calculation and numerical analysis are used in this study. The structural parameters of Louzhou Bridge shown in Table 1 are taken as fundamental parameters in this section. In parameters analysis, tower stiffness is changed from 0.1 times to 1.5 times.
$\alpha$ is defined as Tower Stiffness Influence Coefficient for 1stAsymmetric Vertical Vibration (AVV). We can get changing curve of $\alpha$ on middle tower stiffness (${k}_{tm}$) by Eq. (6) shown in Fig. 9. We can see from Fig. 9, $\alpha$ increases from 1.65 to 9.15, as times of ${k}_{tm}$ is changing from 0.1 to 1.0. When ${k}_{tm}$ is low, value of is $\alpha$ tends to be 0. $\alpha$ is influenced significant by stiffness of middle tower ${k}_{tm}$. In particular, gradient of $\alpha$ become lower as times of ${k}_{tm}$ enlarge. It discloses that $\alpha$ is more sensitive to ${k}_{tm}$ when ${k}_{tm}$ is low.
Values of ${\stackrel{-}{f}}_{v}^{a}$ on different ${k}_{tm}$ are shown in Fig. 10. In the figure, the first line means frequencies calculated Eq. (4) proposed in this paper. The second line means frequencies analyzed by finite element model. And the third line means frequencies calculated by Eq. in reference [9]. As we can see from the figure, frequencies calculated by Eq. (4) and “FEM”, match well in most of time, especially when the time is close to 1. ${\stackrel{-}{f}}_{v}^{a}$ increases from 0.3959 Hz to 0.4578 Hz as ${k}_{tm}$ is changing from 2.17E+05 kN/m to 4.33E+05 kN/m. The rate of frequency increase is near 15.6 %. The natural frequency ${\stackrel{-}{f}}_{v}^{a}$ enlarges gradually with increasing ${k}_{tm}$. The influence law of ${\stackrel{-}{f}}_{v}^{a}$ is similar to $\alpha$ in Fig. 9. All in all, enlarging the value of middle tower stiffness is an effective way to improve frequency of 1st AVV.
In Fig. 10, curve of “Reference” shows as a straight line with constant value of 0.2944 Hz, as influence of tower stiffness was ignored in the formula. Comparing curves of Eq. (4) with reference [9], frequency difference is small when ${k}_{tm}$ is low, while that increases as ${k}_{tm}$ is enlarging. The difference shows that influence of tower stiffness cannot be ignored if ${k}_{tm}$ is large. Otherwise, the error will be significant. In other words, contribution of middle tower stiffness is more and more important as tower stiffness increasing.
Fig. 9Value of α on different ktm
Fig. 10Value of f-va on different ktm
From Eq. (7), we know frequency of 1st SVV (${\stackrel{-}{f}}_{v}^{s}$) is closely related to side tower stiffness (${k}_{ts}$). Therefore, we study relationship between ${\stackrel{-}{f}}_{v}^{s}$ and ${k}_{ts}$ in following section.
$\beta$ is defined as tower Stiffness Influence Coefficient for 1st Symmetric Vertical Vibration (SVV). Based on Eq. (10), we can get curve of $\beta$ on different side tower stiffness (${k}_{sm}$), as shown in Fig. 11. We can see from the figure, $\beta$ increases as ${k}_{ts}$ is enlarging. When ${k}_{ts}$ is low, value of $\beta$ tends to be 1.
Fig. 11Value of β on different kts
Fig. 12Value of f-vs on different kts
Frequencies of 1st SVV (${\stackrel{-}{f}}_{v}^{s}$) on different ${k}_{ts}$ are shown in Fig. 12. As we can see, value of ${\stackrel{-}{f}}_{v}^{s}$ calculated by Eq. (7) and “FEM”, match quite well. Value of ${\stackrel{-}{f}}_{v}^{s}$ increases from 0.5948 Hz to 0.6298 Hz as ${k}_{ts}$ changing from 2.17E+05 kN/m to 4.33E+05 kN/m. The curve of ${\stackrel{-}{f}}_{v}^{a}$ in Fig. 12 is similar to $\beta$ in Fig. 11.
Frequencies calculated by formula of reference [9] is shown in Fig. 12. Comparing frequencies calculated by Eq. (7) with reference [9], the values are significant different when ${k}_{ts}$ is large. However, the frequencies tends to be equal when ${k}_{ts}$ becomes low.
Based on the above conclusion, we can guide for the dynamic design of TSSB. Take seismic performance design of TSSB as example, to avoid predominant period of foundation site and reduce the seismic response, long natural vibration period is expected. Based on the conclusion of this paper, lower longitudinal stiffness of middle tower is designed to reduce frequency of 1st Symmetric Vertical Vibration (SVV). At the same time, from the frequency formulas we can know that minishing mass of girder can prolong natural period too. By the same token, we can easily adjust natural period of TSSB in wind resistant design and vehicle-bridge resonance analysis.
### 4.2. Comparing to formulas ignoring influence of tower stiffness
In Reference [19], formulas to calculate vertical frequencies were deduced neglecting influence of tower stiffness on vertical vibration, shown in Eqs. (12), (13):
12
${f}_{v}^{a}=\frac{1}{2\pi }\sqrt{\frac{{\gamma }_{0}\cdot {E}_{g}{I}_{g}}{{m}_{g}+{m}_{c}}},$
13
${f}_{v}^{s}=\frac{1}{2\pi }\sqrt{\frac{{\gamma }_{1}\cdot {E}_{c}{A}_{c}+{\gamma }_{2}\cdot {E}_{g}{I}_{gv}}{{m}_{g}+{m}_{c}}}.$
Comparing Eq. (4) and Eq. (7) to reference [9], existing of coefficient $\alpha$ and $\beta$ is the only difference between equations. Based on Eq. (5) and Eq. (10), $\alpha$ and $\beta$ are mainly related to tower stiffness. Coefficient $\alpha$ tends to be 0 when middle tower stiffness ${k}_{tm}$ is quite low. And coefficient $\beta$ tends to be 1 when side tower stiffness ${k}_{ts}$ is quite low. Therefore, when tower stiffness is quite low, Eqs. (4) and (7) can be transformed to Eq. (12) and Eq. (13) respectively. In other words, formulas in reference [9] are special forms of deduced formulas in case of low tower stiffness. Therefore, formulas in this paper are unified with that in Reference [9]. We can come to the conclusion that influence of tower stiffness on vertical vibration can be neglect when tower stiffness is low. Eqs. (12) and (13) are suggested to calculate frequency when tower stiffness is low for simplicity.
### 4.3. Critical tower stiffness for simplified formulas
As shown in above equation, ${\stackrel{-}{f}}_{v}^{s}$ and ${\stackrel{-}{f}}_{v}^{a}$ stand for frequencies considering contribution of tower stiffness. While, ${f}_{v}^{s}$ and ${f}_{v}^{a}$ stand for frequencies neglecting contribution of tower stiffness.
We take formulas of ${\stackrel{-}{f}}_{v}^{a}$ and ${f}_{v}^{a}$ as example. Obviously, the former formulas are excelling in accuracy, while the latter are simpler for calculation. The difference between ${f}_{v}^{a}$ and ${\stackrel{-}{f}}_{v}^{a}$ comes from whether considering influence of tower stiffness. As the analysis of above section, when tower stiffness is low, the deference value would be limited. In this section, we aim to get a critical tower stiffness ${S}_{t}^{c}$. When tower stiffness below ${S}_{t}^{c}$, ${f}_{v}^{s}$ is approximately equal to ${\stackrel{-}{f}}_{v}^{s}$. In this case, formulas of ${f}_{v}^{s}$ can be taken to calculate low order vertical frequency more efficiently.
### 4.3.1. Critical tower stiffness for 1st AVV
$\theta$ is defined as accuracy requirement. In general, $\theta$ is set to be 0.9 in civil engineering. If value of ${f}_{v}^{a}/{\stackrel{-}{f}}_{v}^{a}$ is greater than $\theta$, ${f}_{v}^{a}$ can take place of ${\stackrel{-}{f}}_{v}^{a}$ in frequency evaluation. Therefore, we can get following equation:
14
$\frac{{f}_{v}^{a}}{{\stackrel{-}{f}}_{v}^{a}}=\sqrt{\frac{{\gamma }_{0}\cdot {E}_{g}{I}_{gv}}{\alpha +{\gamma }_{0}\cdot {E}_{g}{I}_{gv}}}\ge \theta .$
We can assume stiffness of middle tower ${S}_{tm}^{}$ is equal to that of side tower ${S}_{ts}^{}$. Eqs. (5) and (6) are substituted into Eq. (14). Then the equation can transfer to following form:
15
${S}_{cri}^{a}=\frac{1.5}{\frac{8}{15}\cdot \frac{{\theta }^{2}}{1-{\theta }^{2}}\cdot \frac{{l}_{1}{l}_{2}{{f}_{2}}^{2}}{\left({l}_{1}+{l}_{2}\right){E}_{g}{I}_{g}}-\frac{{l}_{s2}}{{E}_{c}{A}_{c}}}.$
That is expression of critical tower stiffness for 1st AVV (${S}_{cri}^{a}$). Based on equation, we can make a judgment before calculating frequency. If ${S}_{tm}^{}$ is below to ${S}_{cri}^{a}$, it means the error between ${f}_{v}^{a}$ and ${\stackrel{-}{f}}_{v}^{a}$ is limited. In this case, to simplify the calculation process, we can use Eq. (12) to evaluate frequency of 1st AVV for TSSB.
### 4.3.2. Critical tower stiffness for 1st SVV
In the same way, we can get expression of critical tower stiffness for 1st SVV (${S}_{cri}^{s}$), shown in following:
16
${S}_{cri}^{\mathrm{s}}=\frac{\left(C-1\right)\left(A+{l}_{s2}\right){\left(3-4B\right)}^{2}}{2{\left(3A-4B{l}_{s2}\right)}^{2}-\left(C-1\right)A{l}_{s2}{\left(3-4B\right)}^{2}}\cdot {E}_{c}{A}_{c},$
where: $A={l}_{s1}+\frac{{E}_{c}{A}_{c}}{{E}_{g}{A}_{g}}\left({l}_{1}+{l}_{2}\right),B=\frac{{{l}_{1}}^{2}}{{{l}_{2}}^{2}},\mathrm{}C=\frac{\frac{1}{{\theta }^{2}}{\gamma }_{1}\cdot {E}_{c}{A}_{c}+\left(\frac{1}{{\theta }^{2}}-1\right){\gamma }_{2}\cdot {E}_{g}{I}_{gv}}{{\gamma }_{1}\cdot {E}_{c}{A}_{\mathrm{c}}}.$
If ${S}_{ts}$ is below to ${S}_{cri}^{s}$, it means the error between ${f}_{v}^{s}$ and ${\stackrel{-}{f}}_{v}^{s}$ is limited. In this case, to simplify the calculation process, we can use Eq. (13) to evaluate frequency of 1st SVV for TSSB.
We can take Louzhou Bridge in section 2 as example. Based on above Eqs. (15) and (16), we can calculate the critical tower stiffness for 1st AVV and SVV. $\theta$ is set to be 0.9 in calculation, we can get ${S}_{cri}^{a}$ is 5.63×103 kN/m. for 1st AVV. And ${S}_{cri}^{s}$ is 2.68×104 kN/m for 1st SVV. The values are shown as dotted line in Fig. (10) and Fig. (12) respectively.
## 5. Comparing to three-tower earth-anchored suspension bridge (TESB)
Three-tower self-anchored suspension bridge (TSSB) is a special kind of bridge, which can be seen to combination of three-tower earth-anchored suspension bridge (TESB) and self-anchored suspension bridge. Therefore, it synthesizes the characters of both. In this section, structural characters of different types of bridge will be studied based on frequency formulas.
Reference [22] has deduced frequency formulas for three-tower earth-anchored suspension bridge, shown in Eqs. (17), (18):
17
${f}_{v}^{a}={\left(\frac{1}{2L}\right)}^{2}\sqrt{\frac{16{\pi }^{2}\cdot {E}_{g}{I}_{gv}+\frac{\left({m}_{c}+{m}_{g}\right)}{2f}g{L}^{4}}{\left({m}_{c}+{m}_{g}\right)}},$
18
${f}_{v}^{s}=\frac{1}{2L}\sqrt{\frac{\frac{256{f}^{2}}{{\pi }^{2}{L}_{s}L}\cdot {E}_{c}{A}_{c}+{\left(\frac{\pi }{L}\right)}^{2}\cdot {E}_{g}{I}_{gv}+\frac{\left({m}_{g}+{m}_{c}\right)g{L}^{2}}{8f}}{{m}_{g}+{m}_{c}}},$
where, $L$, ${L}_{s}$, $f$ stand for span, equivalent cable length and rise respectively.
During the derivation of above formulas, influence of tower stiffness on frequencies was ignored. Therefore, to keep formulas comparable between different bridges, Eqs. (12), (13) are selected as contrastive equations.
Besides the same expressions, there are some additional expressions in frequency formulas for MESB. In Eqs. (14), (15), there are additional expressions of $\left(\left({m}_{c}+{m}_{g}\right)/2f\right)g{L}^{4}$ and $\left({m}_{c}+{m}_{g}\right)g{L}^{2}/8f$ respectively. These components mean gravity stiffness coming from dead load of cable and girder, defined as gravity stiffness component. For cable elements, stress stiffening due to great dead load of girder and cable must be considered, which is defined as effect of stress stiffening. From above formulas, we know that gravity stiffness component plays an important role in vertical stiffness of MESB. Therefore, it can significantly improve vertical frequency of MESB by increasing dead load.
However, that is different for TSSB. As cables being directly anchored to girder, girder is at compression-bending coupling state. For TSSB, effects of stress stiffening and compression-bending coupling are balanced. As a result, there is no gravity stiffness component in frequency formulas for TSSB. Therefore, increasing dead load cannot enlarge vertical stiffness of TSSB. Furthermore, it will increase seismic response of TSSB. This is one of the most significant differences between TSSB and MESB.
## 6. Conclusions
This paper is focus on frequency calculation method and influence law on frequency for TSSB. The following conclusions have been drawn from this study.
1. The finite model of first three-tower suspension bridge is established. The low order frequencies and modal shape are studied by finite element method.
2. Frequency formulas of 1st vertical AVV and SVV were deduced by Rayleigh method with considering the influence of tower stiffness. The validity of the deduced formulas is verified by results of both numerical analysis and modal test.
3. Stiffness characteristic difference between TSSB and MESB is studied by frequency formulas. The comparison discloses the significant difference is gravity stiffness component in frequency formulas.
4. Influence laws of tower stiffness on structural stiffness are studied by both formula method and finite element method. The research results show that middle tower stiffness and side tower stiffness play important role in frequency of 1st AVV and SVV respectively. Frequencies are enlarging as tower stiffness increasing. When tower stiffness is low, deduced frequency formulas are the same as formulas in other paper without considering influence of tower stiffness. At last, expressions of critical tower stiffness for simplified formulas are deduced based on deduced frequency formulas.
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#### About this article
Received
21 June 2014
Accepted
07 September 2014
Published
30 September 2014
Keywords
three-tower self-anchored suspension bridge
frequency formula
tower stiffness
critical stiffness
Acknowledgements
The study was sponsored in part by Research Fund for the Doctoral Program Project of the Ministry of Education (No. 20133514129996) and in part by National Natural Science Foundation of Fujian Province (No. 2013J05072).
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# How to show a bar graph for an approximate percentage value in a given range?
Requirement:
• First Column should show a value for Total Space
• Second Column should show a graph for Percentage Used
Example: Please find attached the example graph_question.png
Question: How to show a Graph for a value which falls in-between 2 ranges (e.g 40%-60%).
• I think the label or some message to the user should make this clear then you won't have that problem anymore. – Michael Lai Jul 24 '14 at 23:08
You could try using 5 steps rather than 100 then a phrase like 'at least' or 'minumum available'.
By only changing your graph when one of the main chunks is completely full the user will only be able to deduce that there is somewhere between (in the example) 2 and 3 chunks of space available.
Your graph for either 41% of 59% would look like this. They would know they had at least 40% but not as much as 60%:
Where as one for 61% would look like this:
By breaking the graph down to the level of granularity that the system is able to report and no further the user would be given exactly the amount of information available. They couldn't be misled as you wouldn't be assuming and displaying a value as you are strongly implying with your example.
You could also number them 1-5 and add 'currently filling chunk n'
• Colin, thanks for sharing your thoughts. I agree that we can design 5 steps rather than keeping the percentage numbers, but on a first look when a user scans the graph (what you've designed) he/she might think the value is 40 on your 1st and 60 on your 2nd graph. So, we might need to show the visual difference between items: Example - 40% - 60% Visual States 1. At-least (first 2 blocks) 2. Approximate (3rd block) 3. Empty (4th and 5th block) – UxBaby Jul 24 '14 at 22:58
• @uxbaby Yes a third state could be used, maybe to represent the 'currently using' state – Toni Leigh Jul 27 '14 at 16:40
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“The sum of two rational numbers is rational.” So, adding two rationals is the same as adding two such fractions, which will result in another fraction of this same form since integers are closed under addition and multiplication. Thus, adding two rational numbers produces another rational number.
## Is the sum of a rational and irrational number rational?
The sum of any rational number and any irrational number will always be an irrational number.
## What is the sum of a rational and rational number?
The sum of a rational number and a rational number is rational. The sum of a rational number and an irrational number is irrational. The sum of an irrational number and an irrational number is irrational.
## What is the formula for finding rational numbers?
Rational Numbers:
1. Rational numbers are the numbers which can be expressed in the form of p and q where q ≠ 0.
2. Example: If 10 rational numbers are to be found between 2/7and 5/7, both the rational numbers are to be multiplied with 10/10.
3. 2/7 x 70.
4. 5/7 x 70.
## What are the rational numbers between 1 and 2?
Answer: Five rational numbers between 1 and 2 are 11/10, 12/10, 13/10, 14/10, and 15/10.
## What is a rational number in math definition?
A rational number is a number that can be express as the ratio of two integers. However, one third can be express as 1 divided by 3, and since 1 and 3 are both integers, one third is a rational number. Likewise, any integer can be expressed as the ratio of two integers, thus all integers are rational.
## How many rational numbers are there?
Infinite number of rational numbers exist between any two distinct rational numbers. We know that a rational number is a number which can be written in the form of qp where p and q are integers and q =0.
## What is the standard form of a rational number?
A rational number is made up of a numerator and a denominator. A rational number is said to be in standard form if the Highest Common Factor or the H.C.F. of numerator and denominator is 1.
## What is 3/5 a rational number?
The number 3/5 is a rational number. It is a fraction that is made from two integers, 3 and 5.
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# Find an Exact Tax Amount and Round to Nearest Dollar and Nearest Cent
This video explains how to find a tax amount given a percent. Then the amount is rounded to the nearest dollar and nearest cent.
# Unit Rates: Dollars Per Quart and Quarts Per Dollar
This video explains to find two unit rates from 1 rate.
# Inequality in One Variable Application: Rental Car Cost
This video explains how to write and solve a linear inequality in one variable to solve an application problem.
# Application: Write and Solve a One Step Equation: Price ax=b (Round)
This video explains how to write and solve a one step equation to solve an application problem.
# Converting Between Pesos and Dollars Using a Unit Fraction
This video explains to convert between pesos and dollars using a unit fraction.
# Convert Euros Per Liter to Dollar Per Gallon Using Unit Fractions
This video explains how to convert the cost of gas from euros per liter to dollars per gallon.
# Convert U.S. Dollars to Euros Using a Unit Fraction
This video provides an example of how to convert dollars to euros using a unit fraction.
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# what price should you charge to the second group
Question You own 10 people delay values {\$1, \$2, \$3, \$4, \$5, \$6, \$7, \$8, \$9, \$10}, and divine you perceive a way to admonish one compensation to the consumer whose values are {\$1, \$2, \$3, \$4, \$5} and a opposed compensation to those consumers whose values are {\$6, \$7, \$8, \$9, \$10}. MC of product is \$2.50. a. what compensation should you admonish to the avoid cluster? b. what are your expected profits from selling to the avoid cluster? c. What compensations should you admonish to the primitive cluster? d. If it costs \$5 to appliance this compensation-discrimination design (to realize the two clusters and obviate arbitrage betwixt them), should you do it?
# What Price Should You Charge To The Second Group
You enjoy 10 people after a while values {\$1, \$2, \$3, \$4, \$5, \$6, \$7, \$8, \$9, \$10}, and think you meet a way to commit one cost to the consumer whose values are {\$1, \$2, \$3, \$4, \$5} and a irrelative cost to those consumers whose values are {\$6, \$7, \$8, \$9, \$10}. MC of evolution is \$2. 50. a. what cost should you commit to the nullify order?b. what are your expected receipts from selling to the nullify order?c. What costs should you commit to the leading order?d. If it costs \$5 to appliance this cost-discrimination design (to authenticate the two orders and nullify arbitrage between them), should you do it?
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#### 4.2Struct Type and Accessors🔗ℹ
struct(struct flomap (values components width height)) values : FlVector components : Integer width : Integer height : Integer
Represents a width×height floating-point bitmap with components color components. The values vector contains the flattened image data (see Data Layout).
A guard ensures that the values field has length (* components width height), and that each size field is a nonnegative fixnum.
Examples:
> (require racket/flonum) > (flomap (flvector 0.0 0.0 0.0 0.0) 4 1 1) (flomap (flvector 0.0 0.0 0.0 0.0) 4 1 1) > (flomap (flvector) 0 0 0) (flomap (flvector) 0 0 0) > (flomap (flvector 0.0) 2 1 1) flomap: expected flvector of length 2; given one of length 1
The default flomap constructor is perhaps the hardest to use. Instead, to construct a flomap from scratch, you should generally use make-flomap, make-flomap*, build-flomap or draw-flomap.
procedure
(flomap-size fm)
Nonnegative-Fixnum Nonnegative-Fixnum
fm : flomap
Returns the width and height of fm as nonnegative fixnums.
procedure(flomap-ref fm k x y) → Float fm : flomap k : Integer x : Integer y : Integer
Returns fm’s value at k x y.
If x or y is out of bounds, this function returns 0.0. If k is out of bounds, it raises an error. The Conceptual Model section explains why k is treated differently.
procedure(flomap-ref* fm x y) → FlVector fm : flomap x : Integer y : Integer
Returns fm’s component values at x y as an flvector.
If x or y is out of bounds, this function returns an flvector filled with 0.0. It always returns an flvector of length (flomap-components fm).
procedure(flomap-bilinear-ref fm k x y) → Float fm : flomap k : Integer x : Real y : Real
Returns an estimated value at any given k x y coordinate, calculated from known values in fm.
Like all other flomap functions that operate on real-valued coordinates, flomap-bilinear-ref regards known values as being at half-integer coordinates. Mathematically, if x = (+ i 0.5) and y = (+ j 0.5) for any integers i and j, then (flomap-bilinear-ref fm k x y) = (flomap-ref fm k i j).
Suppose fm is size w×h. If x-0.5 or x(+ w 0.5), this function returns 0.0; similarly for y and h. If k is out of bounds, it raises an error. The Conceptual Model section explains why k is treated differently.
procedure(flomap-bilinear-ref* fm x y) → FlVector fm : flomap x : Real y : Real
Like flomap-bilinear-ref, but returns an flvector containing estimates of all the components at x y.
procedure fm : flomap
procedure fm : flomap
These return the minimum and maximum values in fm.
procedure
(flomap-extreme-values fm)
Float Float
fm : flomap
Equivalent to (values (flomap-min-value fm) (flomap-max-value fm)), but faster.
procedure
(flomap-nonzero-rect fm)
Nonnegative-Fixnum Nonnegative-Fixnum Nonnegative-Fixnum Nonnegative-Fixnum
fm : flomap
Returns the smallest rectangle containing every nonzero value (in any component) in fm. The values returned are x minimum, y minimum, x maximum + 1, and y maximum + 1.
The values returned by flomap-nonzero-rect can be sent to subflomap to trim away zero values. But see flomap-trim, which is faster for alpha-multiplied flomaps.
procedure(coords->index c w k x y) → Fixnum c : Integer w : Integer k : Integer x : Integer y : Integer
Returns the index of the value at coordinates k x y of a flomap with c color components and width w. This function does not check any coordinates against their bounds.
procedure(unsafe-flomap-ref vs c w h k x y) → Float vs : FlVector c : Integer w : Integer h : Integer k : Integer x : Integer y : Integer
If fm = (flomap vs c w h), returns fm’s value at k x y. If x or y is out of bounds, this returns 0.0. It is unsafe because k is unchecked, as well as indexing into vs.
This function is used by some library functions, such as flomap-bilinear-ref, to index into already-destructured flomaps. From untyped code, applying this function is likely no faster than applying flomap-ref, because of extra contract checks.
procedure(unsafe-flomap-ref* vs c w h x y) → FlVector vs : FlVector c : Integer w : Integer h : Integer x : Integer y : Integer
Like unsafe-flomap-ref, but returns an flvector containing all the component values at x y.
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About RF Cafe Copyright: 1996 - 2024Webmaster: Kirt Blattenberger, BSEE - KB3UON RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The World Wide Web (Internet) was largely an unknown entity at the time and bandwidth was a scarce commodity. Dial-up modems blazed along at 14.4 kbps while typing up your telephone line, and a nice lady's voice announced "You've Got Mail" when a new message arrived... All trademarks, copyrights, patents, and other rights of ownership to images and text used on the RF Cafe website are hereby acknowledged. My Hobby Website: AirplanesAndRockets.com
Module 4 - Introduction to Electrical Conductors, Wiring Techniques, and Schematic ReadingNavy Electricity and Electronics Training Series (NEETS)Chapter 2: Pages 2-41 through 2-53
NEETS Module 4 − Introduction to Electrical Conductors, Wiring Techniques, and Schematic Reading
Pages i, 1−1, 1−11, 1−21, 2−1, 2−11, 2−21, 2−31, 2−41, 3−1, 3−11, 3−21, 4−1, 4−11, Index
Figure 2-43. - Starting double lace.
Figure 2-44. - Terminating double lace.
2-41
Figure 2-45. - Alternate method of terminating the lace.
Figure 2-46. - Marling hitch as a lock stitch.
The spare conductors of a multiconductor cable should be laced separately, and then tied to active conductors of the cable with a few telephone hitches. When two or more cables enter an enclosure, each cable group should be laced separately. When groups are parallel to each other, they should be bound together at intervals with telephone hitches (figure 2-47).
2-42
Figure 2-47. - Spot tying cable groups.
Q51. What size wire bundles require double lace?
Q52. How is the double lace started?
Q53. How are laced cable groups bound together?
SPOT TYING
When cable supports are used in equipment as shown in figure 2-48, spot ties are used to secure the conductor groups if the supports are more than 12 inches apart. The spot ties are made by wrapping the cord around the group as shown in figure 2-49. To finish the tie, use a clove hitch followed by a square knot with an extra loop. The free ends of the cord are then trimmed to a minimum of 3/8 inch.
Figure 2-48. - Use of spot ties.
2-43
Figure 2-49. - Making spot ties.
SELF-CLINCHING CABLE STRAPS
Self-clinching cable straps are adjustable, lightweight, flat nylon straps. They have molded ribs or serrations on the inside surface to grip the wire. They may be used instead of individual cord ties for securing wire groups or bundles quickly. The straps are of two types: a plain cable strap and one that has a flat surface for identifying the cables.
Caution
Do not use nylon cable straps over wire bundles containing coaxial cable. Do not use straps in areas where failure of the strap would allow the strap to fall into movable parts.
Installing self-clinching cable straps is done with a Military Standard hand tool, as shown in figure 2-50. An illustration of the working parts of the tool is shown in figure 2-51. To use the tool, follow the manufacturer's instructions.
Figure 2-50. - Installing self-clinching cable straps.
2-44
Figure 2-51. - Military Standard hand tool for self-clinching cable straps.
Warning
Use proper tools and make sure the strap is cut flush with the eye of the strap. This prevents painful cuts and scratches caused by protruding strap ends. Do not use plastic cable straps in high-temperature areas (above 250º F).
High-TEMPERATURE PRESSURE-SENSITIVE TAPE LACING
High-temperature, pressure-sensitive tape must be used to tie wire bundles in areas where the temperature may exceed 250º F. Install the tape as follows (figure 2-52):
1. Wrap the tape around the wire bundle three times, with a two-thirds overlap for each turn.
2. Heat-seal the loose tape end with the side of a soldering iron tip.
Figure 2-52. - Securing wire bundles in high-temperature areas.
2-45
Warning
Insulation tape (including the glass fiber type) is highly flammable and should not be used in a high-temperature environment. Only insulation tape approved for high-temperature operation (suitable for continuous operation at 500º F) should be used in high-temperature environments.
Q54. When are spot ties used?
Q55. What is used to install self-clinching cable straps?
Q56. What is used to tie wire bundles in high-temperature areas?
Summary
In this chapter you have learned some of the basic skills required for proper wiring techniques. We have discussed conductor splices and terminal connections, basic soldering skills, and lacing and tying wire bundles.
The basic requirement for any splice or terminal connection is that it be both mechanically and electrically as strong as the conductor or device with which it is to be used.
Insulation Removal - The first step in splicing or terminating electrical conductors is to remove the insulation. The preferred method for stripping wire is by use of a wire-stripping tool. The hot-blade stripper cannot be used on such insulation material as glass braid or asbestos. An alternate method for stripping copper wire is with a knife. a knife is the required tool to strip aluminum wire. Take extreme care when stripping aluminum wire. Nicking the strands will cause them to break easily.
Western Union Splice - a simple connection known as the Western Union splice is used to splice small, solid conductors together. After the splice is made, the ends of the wire are clamped down to prevent damage to the tape insulation.
2-46
Staggered Splice - The staggered splice is used on multiconductor cables to prevent the joint from being bulky.
Rattail Joint - a splice that is used in a junction box and for connecting branch circuits; wiring is placed inside conduits.
Fixture Joint - When conductors of different sizes are to be spliced, such as fixture wires to a branch circuit, the fixture joint is used.
Knotted Tap Joint - This type of splice is used to splice a conductor to a continuous wire. It is not considered a "butted" splice as the ones previously discussed.
Splice Insulation - Rubber tape is an insulator for the type of splices we have discussed so far.
Friction Tape - It has very little insulating value but is used as a protective covering for the rubber tape. Another type of insulating tape is plastic electrical tape, which is quite expensive.
Terminal Lugs - The terminals used in electrical wiring are either of the soldered or crimped type. The advantage of using a crimped type of connection is that it requires very little operator skill, whereas the soldered connection is almost completely dependent on the skill of the operator. Some form of insulation must be used with noninsulated splices and terminal lugs. The types used are clear plastic tubing (spaghetti) and heat-shrinkable tubing. When a heat gun is used to shrink the heat-shrinkable tubing, the maximum allowable heat to be used is 300º F. When using the compressed air/nitrogen heating tool, the air/nitrogen source cannot be greater than 200 psig.
2-47
Aluminum Terminals and Splices - Aluminum terminals and splices are noninsulated and very difficult to use. Some of the things you should remember when working with aluminum wire are: (1) Never attempt to clean the aluminum wire. There is a petroleum abrasive compound in the terminal lug or splice that automatically cleans the wire. (2) The only tools that should be used for the crimping operation are the power crimping type. (3) Never use lock washers next to aluminum terminal lugs as they will gouge out the tinned area and increase deterioration.
Preinsulated Copper Terminal Lugs and Splices - The most common method of terminating and splicing copper wires is with the use of preinsulated terminal lugs and splices. Besides not having to insulate the terminal or splice after the crimping operation, the other advantage of this type is that it gives extra wire insulation support. Several types of crimping tools can be used for these types of terminals and splices. The tool varies with the size of the terminal or splice. Preinsulated terminal lugs and splices are color coded to indicate the wire size they are to be used with.
2-48
Soldering - The basic skills required to solder terminal lugs, splices, and electrical connectors are covered in this area. Prior to any soldering operation, the items to be soldered must be cleaned; they will not adhere to dirty, greasy, or oxidized surfaces. The next step is the "tinning" process. This process is accomplished by coating the material to be soldered with a bright coat of solder. The wire to be soldered must be stripped to 1/32 inch longer than the depth of the solder cup of the terminal, splice, or connector to which it is to be soldered. This is to prevent burning the insulation. It also allows the wire to flex at the stress point. When you tin the wire, it should be done to one-half of the stripped length. When soldering a connection, take precaution to prevent movement of the parts while the solder is cooling. a "fractured solder" joint will result if this precaution is not taken.
Soldering Tools - The important difference in soldering iron sizes is not the temperature (they all produce 500º F to 600º F), but the thermal inertia. Thermal inertia is the ability of soldering tools to maintain a satisfactory soldering temperature while giving up heat to the joint to be soldered. a well- designed soldering iron is self-regulating because its heating element increases with the rising temperature, thus inciting the current to a satisfactory level. When using a soldering gun, do not press the switch for periods longer than 30 seconds. Doing so will cause the tip to overheat to the point of incandescence. The nuts or screws that retain the tips on soldering irons and guns tend to loosen because of the continuous heating and cooling cycles. Therefore, they should be tightened periodically. You should never use a soldering gun on electronics components, such as resistors, capacitors, or transistors. An advantage of using a resistance soldering iron to solder a wire to a connector is that the soldering tips are only hot during the brief period of soldering the connection.
2-49
Solder - Ordinary soft solder is a fusible alloy of tin and lead used to join two or more metals at temperatures below their melting point. The metal solvent action that occurs when copper conductors are soldered together takes place because a small amount of the copper combines with the solder to form a new alloy. Therefore, the joint is one common metal. The tin-lead alloy used for general-purpose soldering is composed of 60-percent tin and 40-percent lead (60/40 solder).
Flux - Flux is used in the soldering process to clean the metal by removing the oxide layer on the metal and to prevent further oxidation during the soldering process. Always use noncorrosive, nonconducting rosin fluxes when soldering electrical and electronic components.
Solvents - Solvents are used in the soldering process to remove contaminants from the surfaces to be soldered.
Soldering Aids - Use a heat shunt when you solder heat-sensitive components. It dissipates the heat, thereby preventing damage to the heat-sensitive component. Some type of soldering iron holder or guard should be used to prevent the operator from being burned.
Lacing Conductors - The purpose of lacing conductors is to present a neat appearance and to facilitate tracing the conductors when alterations or repairs are required. Flat tape is preferred for lacing instead of round cord. Cord has a tendency to cut into the wire insulation. The amount of flat tape or round cord required to lace a group of conductors is about two and one-half times the length of the longest conductor. a lacing shuttle is useful during the lacing operation to prevent the tape or cord from
fouling. Wires should only be twisted prior to lacing if it is required, such as for filament leads in electron tube amplifiers. When lacing wire bundles containing coaxial cables, use the proper flat tape and do not
tie the bundles too tightly. Never use round cord on coaxial cable. a single lace is started with a square
knot and at least two marling Hitches. a double lace is required for wire bundles that are 1 inch or more
in diameter. It is started with a telephone hitch. Cable groups are bound together by use of telephone hitch
2-50
Spot Ties - Spot ties are used when cable supports are used that are more than 12 inches apart.
Self-clinching Cable Straps - If self-clinching cable straps are used, they should be installed with the Military Standard hand tool designed for their use.
High-temperature Areas - When you are required to tie wire bundles in high-temperature operating areas, use only high-temperature, pressure-sensitive tape.
Answers to Questions Q1 Through Q56.
A1. The connection must be both mechanically and electrically as strong as the conductor or device with which it is used
A2. By use of a wire-stripping tool
A4. Knife.
A5. To prevent damage to the tape insulation.
A6. To prevent the joint from being bulky.
A7. When wires are in conduit and a junction box is used.
A8. Fixture joint.
A9. Knotted tap joint.
A10. As a protective covering over the rubber tape.
A11. Requires relatively little operator skill to install.
A12. Spaghetti or heat-shrinkable tubing.
A13. 300º F
A14. 200 psig.
A15. No, it is done automatically by the petroleum abrasive compound that comes in the terminal or splices.
A16. Power-operated crimping tools.
2-51
A17. It gouges the terminal lug and causes deterioration.
A18. The use of preinsulated splices and terminal lugs.
A19. It has insulation support for extra supporting strength of the wire insulation.
A20. To identify wire sizes they are to be used on.
A21. Solder will not adhere to dirty, greasy, or oxidized surfaces.
A22. The coating of the material to be soldered with a light coat of solder.
A23. To prevent burning the insulation during the soldering process and to allow the wire to flex easier at a stress point.
A24. One-half the stripped length.
A25. Movement of the parts being soldered while the solder is cooling.
A26. The capacity of the soldering iron to generate and maintain a satisfactory soldering temperature while giving up heat to the joint being soldered.
A27. Although its temperature is as high as the larger irons, it does not have thermal inertia.
A28. The resistance of its heating element increases with rising temperature, thus limiting the current flow.
A29. File the tip until it is smooth and re-tin it.
A30. It will overheat and could burn the insulation of the wire being soldered.
A31. The heating and cooling cycles.
A32. Electronic components, such as resistors, capacitors, and transistors.
A33. The soldering tips are hot only during the brief period of soldering the connection, thus minimizing the chance of burning the wire insulation or connector inserts.
A34. The strands can fall into electrical equipment being worked on and cause short circuits.
A35. It enables the tip to be removed easily when another is to be inserted.
A36. Wrap a length of copper wire around one of the regular tips and bend to the proper shape for the purpose.
A38. The solder dissolves a small amount of the copper, which combines with the solder forming a new alloy; therefore, the joint is one common metal.
A39. 60-percent tin and 40-percent lead (60/40 solder).
A40. It cleans the metal by removing the oxide layer and prevents further oxidation during the soldering.
A41. Noncorrosive, nonconductive rosin fluxes.
2-52
A42. To remove contaminants from soldered connections.
A43. To prevent damage to heat-sensitive components.
A44. To aid in tracing the conductors when alterations or repairs are required.
A45. Round cord has a tendency to cut into the wire insulation.
A46. Two and one-half times the length of the longest conductor in the group.
A47. To keep the tape or cord from fouling during the lacing operation.
A48. When required, such as for the filament leads in electron tube amplifiers.
A49. Do not tie too tightly and use the proper type of tape.
A50. With a square knot and at least two marling hitches drawn tightly.
A51. Bundles that are 1 inch or larger in diameter
A52. With a telephone hitch.
A53. They are bound together at intervals with telephone hitches.
A54. When wire bundles are supported by cable supports that are more than 12 inches apart.
A55. Military Standard hand tool.
A56. High-temperature, pressure-sensitive tape.
2-53
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# Counting practice Worksheets for Ages 4-9
Favorites
Interactive
• 4-9
• Interactive
• Counting practice
## Fish Worksheet
Kids love learning about the creatures that live in water! This free PDF worksheet covers fish, teaching kids larger vocabulary and contextual reading comprehension. Questions prompt them to choose the right answer from multiple choices, just like in a classroom. It's a fun way to boost their reading skills and comprehension without them even realizing it!
Fish Worksheet
Worksheet
## Match Tally Marks to Bar Graphs Worksheet
Learning math can be tough for kids. Rules and formulas can be confusing. Don't worry, this worksheet can help them understand graphs better. Show them how to match the tally marks with the graph that shows the same amount. It will help them grasp the basics and prepare them for more complex equations.
Match Tally Marks to Bar Graphs Worksheet
Worksheet
## Making 19 with Pizza! Worksheet
Who loves pizza? Your kids do! Next time you need to teach them math, make it fun by combining it with something they enjoy. This worksheet uses a pizza-themed problem to help them practice solving equations. Have them trace the line to figure out how to make 19. It's a great way to make learning math fun!
Making 19 with Pizza! Worksheet
Worksheet
## 18 Candles Worksheet
This worksheet is great for introducing your students to addition and subtraction. Help your kids solve the simple equations on it and draw a line from the birthday cake with 18 on it to all the solutions. Practice like this will prepare them to tackle more complex math problems.
18 Candles Worksheet
Worksheet
## Piecing Together 18 Worksheet
Let your preschoolers learn and have fun with this colorful worksheet. Look at the picture with them and ask what they see. Help them solve the equations missing on each puzzle piece and then circle the one that will make the bottom puzzle piece numbers equal 16. Puzzles and bright colors make learning easy and enjoyable!
Piecing Together 18 Worksheet
Worksheet
## Base Ten Block Counting Worksheet
Introduce your kids to counting with educational DVDs and video games. Then use this worksheet: look at the picture and help them find the groups of ten; trace a line from the groups with 10 blocks to the number 10. This will help them practice and gain a better understanding of counting.
Base Ten Block Counting Worksheet
Worksheet
Animals are fun to observe and interact with. They can be seen at the zoo, parks, farms and in homes with pets. This worksheet helps kids identify animals, and build addition skills. Look at the picture with them and help name each animal. After that, add the animals up and check the box for the correct answer. 80 words.
Worksheet
## Planting Seeds for 11 Worksheet
Before starting, ask your students to count as high as they can. Explain that different numbers can add up to the same total. Give examples like 2+2 =4 and 3+1 =4. Ask if they can think of other examples. Then have them work through the equations in the printout and circle all that add up to 11.
Planting Seeds for 11 Worksheet
Worksheet
The fire service is the right number to call in case of a fire. Test your children's safety knowledge by asking them other emergency numbers. Have them identify the fire truck in the worksheet and solve the number sentences. Guide them in drawing a line to the right answer.
Worksheet
## Eight Geese Worksheet
Assess your child's counting skills with this worksheet. Help them count the geese in each group, then trace a line from the groups showing 8. Give your kids a boost by providing lots of exercises and learning aids to help with counting. They likely count objects they come across, like spoons and stones. Use this opportunity to give them practice and help them learn.
Eight Geese Worksheet
Worksheet
## Marching Ants Worksheet
Assess and help your kids practice counting with this worksheet. Ask them to identify the four groups of ants in the picture. Guide them as they count each group and trace the lines to the corresponding number.
Marching Ants Worksheet
Worksheet
## Building with 8 Worksheet
Your child can learn more with this simple exercise! Get them counting and help them circle the groups of 8 bricks. It's a great addition to their post-school lessons, supporting their counting skills. Plus, it's a fun way to learn!
Building with 8 Worksheet
Worksheet
## Magic Fives Worksheet
Help your students identify the objects & people in the worksheet. Point to the equation 5 -4 in the center, then get them to count the items and trace a line connecting those that add up to 5 - 4.
Magic Fives Worksheet
Worksheet
## Find 10 Worksheet
Counting can be tricky, but your kids can master it with regular practice. Here's a fun activity to help: count the animals, then trace a line to the pictures that matches. With these tips, math will be an easy walk in the park.
Find 10 Worksheet
Worksheet
## Make 6 Worksheet
Fun snowball fights are great! With lots of ammo, kids can join their snowman friends and count out groups of six. Through number sense practice, and seeing one-to-one picture representation, your child can strengthen their skills while having a blast. Don't forget to bundle up as they may ask for more!
Make 6 Worksheet
Worksheet
## 6 Parrots Worksheet
Love parrots? Here's a free worksheet to help your children practice counting to six with colorful parrot representations! Kids will adore counting the parrots and checking off groups of six - and it won't even feel like math! Just a fun way to make some new and exciting friends.
6 Parrots Worksheet
Worksheet
## Counting Coins Worksheet
Help your students overcome their fear or aversion of counting and learning numbers with this fun worksheet! It features colorful pictures with four groups of coins. Ask the students to count each group, then trace the lines to the piggy bank. Show them it's easy to count and learn numbers!
Counting Coins Worksheet
Worksheet
## Equal Groups: Assessment 2 Worksheet
Get your kids to pay attention to learning with familiar objects, bright colors, and this worksheet. It has six arrays - ask kids to identify them and count the objects. Then, help them trace a line to the correct equation.
Equal Groups: Assessment 2 Worksheet
Worksheet
## A Baker's Supply Worksheet
A visit to the baker's shop is thrilling! Your kids will love it! From scrumptious goodies to the mouthwatering aroma, it's a magical experience. Ask your kids what they're most eager to see and do. Now, read the word problems in this worksheet to them. Help them choose the correct answer for each problem and circle the correct total.
A Baker's Supply Worksheet
Worksheet
## More Octopus Facts Worksheet
More Octopus Facts Worksheet
Worksheet
## Octopus 8’s Worksheet
Can your child identify the octopus in the worksheet? It's an invertebrate with 8 tentacles. Have your child circle the equations with the correct products, and see if they can tell you some other features of the octopus too.
Octopus 8’s Worksheet
Worksheet
## Adding and Subtracting: Instruments Worksheet
Children can help Isla count her musical instruments with this fun worksheet. The downloadable PDF gives them a chance to practice addition and subtraction by connecting pictures to equations. It's also a great way to learn the order of operations.
Worksheet
## Adding Instruments with Isla Worksheet
Your kids can have fun helping Isla add her musical instruments. With colorful pics and one-to-one representation, they can pick the correct addition sentence. They can connect each pic with traceable lines and get to the right sum up to 10.
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# "What is the difference between the smallest and the largest six-digit numbers formed using the digits0, 1, 2, 3, 4 and 5"
1. "440865"
2. "419760"
3. "502343"
4. "None of these"
4
"
440865
"
Explanation :
No Explanation available for this question
# "72 hens cost Rs. __ 96.7__. Then what does each hen cost, where two digits in place of‘ __ ’are not visible or are written in illegible hand"
1. "Rs. 3.23"
2. "Rs. 5.11"
3. "Rs. 5.51"
4. "Rs. 7.22"
4
"
Rs. 5.51
"
Explanation :
No Explanation available for this question
# "Boxes numbered 1, 2, 3, 4 and 5 are kept in a row, and they which are to be filled with either a red or ablue ball, such that no two adjacent boxes can be filled with blue balls. Then how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects"
1. "8"
2. "10"
3. "15"
4. "2"
4
"
2
"
Explanation :
No Explanation available for this question
# "AB⊥BC, BD⊥AC and CE bisects ∠C,∠A = 300. Then what is ∠CED"
1. "300 "
2. "600"
3. "450"
4. "650"
4
"
600
"
Explanation :
No Explanation available for this question
# " Largest value of min(2 + x2, 6 – 3x), when x > 0, is"
1. "1"
2. "2"
3. "3"
4. "4"
4
"
3
"
Explanation :
No Explanation available for this question
# "What per cent of the Rajya Sabha members do not have children at all "
1. "8.75%"
2. "11.24%"
3. "11.63%"
4. "10.35%"
4
"
10.35\%
"
Explanation :
No Explanation available for this question
# "If agriculture accounts for 30% of GDP, what was agricultural sector’s contribution toGDP growth in1993"
1. "12%"
2. "1.2%"
3. "4%"
4. "Can’t say"
4
"
1.2\%
"
Explanation :
No Explanation available for this question
# "Which of the following is/are correct I. In 1999, the industrial production was less than that in 1998 .II. In 1997 the industrial production was more than that in 1996."
1. "1 only"
2. "11 only"
3. "1 and 11"
4. "Neither 1 nor 11"
4
"
11 only
"
Explanation :
No Explanation available for this question
# "In 1994, agricultural production was 180 million tones of food grains. What was theproduction in 1996"
1. "183.3mt"
2. "196mt"
3. "201mt"
4. "190.2mt"
4
"
183.3mt
"
Explanation :
No Explanation available for this question
1. "100"
2. "104"
3. "108"
4. "120"
4
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http://jwork.org/dmelt/code/info.php?id=92506028.py
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crawl-data/CC-MAIN-2018-09/segments/1518891815318.53/warc/CC-MAIN-20180224033332-20180224053332-00640.warc.gz
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Complex error (uncertainty) propogation with covariance matrix
Code: "propog_errors1.py"
Programming language: Python
DMelt Version 1. Last modified: 05/08/2015. License: Free
http://jwork.org/dmelt/code/cache/propog_errors1_7208.py
To run this script using the DMelt IDE, copy the above URL link to the menu [File]→[Read script from URL] of the DMelt IDE.
``````
""" This example is based on a Python code. It will not work in Java/Groovy"
"""
import uncertainties
from uncertainties.umath import sin
from uncertainties import ufloat,AffineScalarFunc, ufloat_fromstr, umath
x=ufloat(1, 0.1) # x = 1+/-0.1
print "2*x=",2*x
print "x**2=",x*x
print "sin(1+2*x)=",sin(1+2*x)
print "Derivative=",(2*x+1000).derivatives[x]
print "Correlations between variables are automatically handled"
square=x*x
print square.n , " +-", square.s
print "Covariance and correlation matrices"
x = ufloat(1, 0.1)
y = -2*x+10
z = -3*x
covs = uncertainties.covariance_matrix([x, y, z])
print covs
``````
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https://jewishinphiladelphia.com/the-retention-ratio-how-to-calculate-it-discuss-its-limitations-and-explore-its-advantages/
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Categories
# The Retention Ratio – How to Calculate It, Discuss Its Limitations, And Explore Its Advantages
When it comes to financial metrics, the retention rate is a key one. It tells you what percentage of a company’s earnings do not get paid out in dividends, but instead are retained. Basically, this is the opposite of the dividend payout ratio. You can also refer to this metric as the “retention rate” if you’d like to compare how your company’s retention rate is affecting your business. In this article, we’ll cover how to calculate the retention rate, discuss its limitations, and explore the advantages of this metric.
## Calculate retention ratio
When analyzing a company’s earnings, it’s important to understand how much of the profits are retained. Dividends, on the other hand, are paid to equity shareholders. Many investors want to invest in companies that will retain the majority of their profits as dividends. In this way, investors can see how much of the profits are reinvested in the business. This can lead to higher returns over the long term. However, investors should be aware that this ratio can be misleading. It may not always be a fair representation of the company’s performance.
The retention ratio is also known as plowback ratio. This ratio shows how much of a company’s profits are retained. A high retention ratio indicates a company that has confident, rational management and makes decisions with the best interests of shareholders in mind. This ratio is generally lower for mature companies, which often have large cash reserves and are considered ‘cash cows.’ Nonetheless, it’s important to understand retention ratios.
## Benefits of calculating retention ratio
Retention ratio is one of the most useful tools for investors to evaluate the growth potential of a company. It measures the percentage of retained earnings that are not distributed to shareholders. Retained earnings can be in any form, such as cash, stock, or other assets. Retention ratios are based on quarterly or annual values. For example, let’s assume that a fictional footwear company has \$1 million in net income for 2019. The company aggressively paid off debts and expanded its product line, but retained a majority of its earnings for reinvestment. The firm’s growth prospects are not necessarily the same for every company, but it will help investors understand which companies are likely to have the most stable future.
In addition, retention ratios are only useful for comparisons of companies of similar size and stage. Emerging companies often do not have the same amount of cash flow as established firms and therefore have lower retention ratios. Hence, retention ratios should be viewed in conjunction with net profit and other ratios. However, there are some important exceptions. To understand why retention ratios are only useful for companies at the same stage of growth, you must first determine the company’s size, industry, and financial health.
## Comparison of retention ratio with other financial metrics
The retention ratio is not the only financial metric to consider. It must be viewed in the context of the firm’s net profit, which is an indicator of its health. Retention rates should be compared to other financial metrics and monitored over several quarters to determine their overall health. Even if the retention rate is high, it does not always indicate the health of the firm’s financial status. For this reason, it is important to consider the retention ratio along with other financial metrics.
The retention rate is one of the most important financial metrics, as it demonstrates how much of a firm’s profits are reinvested back into operations. Without consistent reinvestment, a company would need to borrow from its creditors to continue growing. Companies in the growth stage retain the maximum amount of retained earnings as retained earnings, while mature companies distribute their profits to shareholders as dividends. A higher retention ratio is an indication of growth opportunities.
## Limitations of retention ratio
Retention ratios are calculated to show the percentage of shareholders who remain in the company. The retention ratio of a company is affected by its earnings volatility and dividend payment policy. High-growth technology companies are likely to have high retention ratios, because the company has plenty of room for growth. By contrast, mature blue chip companies will often demand dividend distribution, because they believe their growth is limited and prefer to pay out dividends to their shareholders. In choosing a company to invest in, you should take into account its industry, as well as its current growth potential.
A high retention ratio may be indicative of a company’s financial health. Growth investors consider a high retention ratio as an indication of an imminent stock price increase. On the other hand, a high retention ratio may signal the possibility of company management using retained earnings to pay out new dividends or issue debt financing. These factors may result in a lower-than-optimal return on investment. In such situations, investors should pay close attention to the retention ratio and other financial indicators for the company.
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ECE313.Lecture12
# ECE313.Lecture12 - ECE 313 Probability with Engineering...
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Conditional Probability Professor Dilip V. Sarwate Department of Electrical and Computer Engineering © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign. All Rights Reserved ECE 313 Probability with Engineering Applications
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ECE 313 - Lecture 12 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 2 of 41 Reminders Axiomatic theory of probability The set of all possible outcomes of an experiment is the sample space Events are subsets of An event is said to have occurred if the outcome of the experiment is a member of the event (that is, subset of ) A and A c are a partition of On every trial, one of A and A c must occur
ECE 313 - Lecture 12 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 3 of 41 20-20 hindsight is a wonderful thing Once the experiment has been performed and the outcome is known, we have perfect and complete knowledge For each pair of events A and A c , we can tell which one occurred and which one didn’t — just check which set A or A c the observed outcome belongs to! There is no probabilistic consideration any more, and we do not need to think about the chances of A (or A c ) occurring
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ECE 313 - Lecture 12 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 4 of 41 A little learning is a dangerous thing. . Now suppose that the experiment has been performed, but we do not know the outcome exactly All we know is that the outcome is some member of the event A, but we do not know which member of A it is Put another way, we are told that the event A has occurred, but nothing else To avoid trivial cases, assume that A is not a singleton or elementary event
ECE 313 - Lecture 12 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 5 of 41 … B or ¬ B? That is the question… The experiment has been performed and we know that the event A has occurred, that is, the outcome is some member of A Did the event B occur? or did B c occur? Unlike the case of perfect knowledge, we cannot tell whether B or B c occurred AB and AB c are a partition of A If the outcome AB, then B occurred: if the outcome AB c , then B c occurred
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ECE 313 - Lecture 12 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 6 of 41 The exorcism did not work… The experiment has been performed and we know that the event A occurred, that is, the outcome is some member of A We cannot tell for sure whether B occurred We have not exorcised the probability from the problem as yet — A probability question still continues to plague us Question: What are the chances that B occurred? in view of the new knowledge
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# Is it consistent with $\sf ZF-Fnd$ that no class whose union is the whole universe can be well-ordered?
If $$R$$ is a binary relation; $$\phi$$ is a unary predicate; then:
$$\neg [R \text { well orders } \phi \land \forall y \, \exists x: \phi(x) \land y \in x ]$$
Where: $$R \text { well orders } \phi \equiv_{df} \\\big{(}\forall x \,\forall y \, \forall z: \phi(x) \land \phi(y) \land \phi(z) \to \\ (x R y \to \neg y R x) \land \\ (x \neq y \to x R y \lor y R x) \land \\ (x R y R z \to x R z) \land \\ \exists u \forall m (m \in u \iff \phi(m) \land m R x)\big{)} \\\land \\\forall s [(\exists r \in s : \phi(r)) \to \exists k \in s( \phi(k) \land \forall v \in s : \phi(v) \to \neg v R k)]$$
In English: No class whose union is the universe can be well orderable?
Now this is an anti-foundation axiom, since we can take $$\phi$$ to be a stage of the cumulative hierarchy which is clearly well orderable.
Is that principle consistent with ZF-Fnd?
On the other hand, if some class whose union is the whole universe was well-orderable, note that every set must contain at most finitely many atoms. So recursively we can go about picking the first $$\omega$$ sets which contain new atoms, and this will define, again, an infinite set of atoms.
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# Finding coefficient of linear expansion
Hi there :)
At 19$$\circ$$C, a rod is exactly 20.08 cm long on a steel ruler. Both the rod and the ruler are placed in an oven at 285$$\circ$$C, where the rod now measures 20.18 cm on the same ruler. What is the coefficient of linear expansion for the material of which the rod is made?
I used the formula for linear expansion
change in L = Lx(coefficient of linear expansion)(change in temperature)
My attempt:
Steel's coefficient of linear expansion as given by my textbook: 11x10-6
So over the temperature range of 266 degrees (285 - 19), each centimetre would differ by about (266)(11x10-6) = 0.002926
Multiplying by 20.18 gives 0.059
Taking this away from 20.18 gives: 20.1209
The change in length will now be: 20.12 - 20.08 = 0.04095
Subbing into the expansion formula:
0.04095 = (20.08)(285 - 19)(a)
a = 7.667x10-6
This is for an assignment, and I just want to see if I'm heading in the right direction
Last edited:
vela
Staff Emeritus
Homework Helper
Hi there :)
At 19$$\circ$$C, a rod is exactly 20.08 cm long on a steel ruler. Both the rod and the ruler are placed in an oven at 285$$\circ$$C, where the rod now measures 20.18 cm on the same ruler. What is the coefficient of linear expansion for the material of which the rod is made?
I used the formula for linear expansion
change in L = Lx(coefficient of linear expansion)(change in temperature)
My attempt:
Steel's coefficient of linear expansion as given by my textbook: 11x10-6
So over the temperature range of 266 degrees (285 - 19), each centimetre would differ by about (266)(11x10-6) = 0.002926
Multiplying by 20.18 gives 0.059
Looks fine up to here. I don't follow your logic for what you did next.
Taking this away from 20.18 gives: 20.1209
The change in length will now be: 20.12 - 20.08 = 0.04095
Subbing into the expansion formula:
0.04095 = (20.08)(285 - 19)(a)
a = 7.667x10-6
This is for an assignment, and I just want to see if I'm heading in the right direction
I suggest you clearly define first the 0.059 that you get. What is it actually? Once you get it you'll be in the right track.
I just thought that since 0.059 is the measurement by which the 'real' and 'expanded' measurements differ on the ruler, taking this away from the 'expanded' value (20.18) will give what the rod's true measurement is.
Is this in any way correct?
I realised what I did wrong, I just submitted my assignment then and got that question right, so thanks for questioning my method and putting me on the right track, much appreciated :)
vela
Staff Emeritus
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## See 2, 3 or 4 moves ahead - books from the wtharvey.com library
### Here are some chess puzzles from the games of Leonid Shamkovich. White to move and win except where noted.
Solutions are between the brackets under each puzzle.
Drag your cursor from one bracket to the other.
a)
White Mates in 5. Shamkovich vs Alberic O'Kelly de Galway, Palma de Majorca, 1966
2b2r1k/r5bp/p4R2/1p2p1B1/1q1pP2Q/1B6/PP3P1K/7R w - - 0 1
[ Qxh7+ Kxh7 Kg2+ ]
b)
White Mates in 5. Shamkovich vs Mato Damjanovic, Sochi, 1967
1qr2rnk/5p1p/p1bp1PB1/4p3/3B4/P1p3Q1/1PP3PP/4RR1K w - - 0 1
[ Bxh7+ Nxf6 Rxf6 ]
c)
Shamkovich vs B Dimitrijewski, URS, 1967
4qr1k/pp2b3/4pnQ1/4N3/3P4/8/P4PPP/3R2K1 w - - 0 1
[ Rb1 b6 Rb3 Qxg6 Nxg6+ w/Nxe7 ]
d)
Shamkovich vs J Ryts, Kislovodsk, 1968
4rr2/6qk/3p2np/p2P1pPQ/PnPppP2/7N/7P/3B1RRK w - - 0 1
[ gxh6 if Qf6 Rxg6 Qxf6 Ng5+ ]
e)
Shamkovich vs Theodor Ghitescu, URS, 1969
4r2k/1p2b1p1/2p1Pp1p/p1q4P/P2r4/2B2BP1/1PQ2PK1/4R3 w - - 0 1
[ Qg6 if Rdd8 Be4 ]
Shamkovich Puzzles, Part II. White to move and win except where noted.
a)
White Mates in 5. Shamkovich vs Atanas Kolarov, Varna, 1970
r4r1k/1R5p/3Q1np1/p5N1/8/P1q5/5PPP/5RK1 w - - 0 1
[ Qe7 ]
b)
White Mates in 3. Shamkovich vs Roy Ervin, Lone Pine, 1979
5rrk/5pb1/p1pN3p/7Q/1p2PP1R/1q5P/6P1/6RK w - - 0 1
[ Qxh6+ ]
c)
Shamkovich vs Brian McCarthy, USA, 1976
r1b2rk1/1p3p1p/p1nq1p2/2b5/3p4/1N1B1N1P/PP3PP1/R2Q1RK1 w - - 0 1
[ Qc1 if b6 Nxc5 w/Qh6 or if Bb4 a3 ]
d)
White Mates in 4. Shamkovich vs Larry Chistiansen, Mentor, 1977
6r1/p5bk/4N1pp/2B1p3/4Q2N/8/2P2KPP/q7 w - - 0 1
[ Ng5+ hxg5 Qxg6+ ]
e)
Shamkovich vs Bruce Leverett, USA, 1977
5b1r/Npkqpppp/8/8/1PP5/8/1P2QPPP/6K1 w - - 0 1
[ Qe5+ if Qd6 Nb5+ ]
Shamkovich Puzzles, Part III. White to move and win except where noted.
a)
White Mates in 4. Shamkovich vs David Blohm, USA, 1977
r1b4q/pp1p4/3n1bQ1/8/5k2/1P4R1/P1PP1PPP/6K1 w - - 0 1
[ d4 ]
b)
Shamkovich vs Johan Stopa, USA, 1977
1k5r/pp1q1ppp/1bpr2n1/6N1/PP1p1P2/3P3Q/1P1B2PP/4RRK1 w - - 0 1
[ f5 if Ne5 Rxe5 f6 Re4 fxg5 Bxg5 Bd8 Bf4 ]
c)
White Mates in 4. Shamkovich vs George Botterill, Hastings, 1978
R4bk1/5p1p/5Pp1/1p6/3Q4/1P5P/1P2qrPK/8 w - - 0 1
[ Rxf8+ w/Qc5+ ]
d)
Shamkovich vs Gerardo Lebredo, Mexico, 1978
4r1k1/p1p2ppp/2r2q2/8/2P2b2/1R2B2P/P1Q2PP1/5RK1 w - - 0 1
[ Qe4 if Be5 f4 ]
e)
Shamkovich vs Pal Benko, Greenville, 1980
1r4n1/6kn/p2p2p1/2pPp1Pr/2P1P2p/2N1QN1q/P1K2R1P/6R1 w - - 0 1
[ Nxe5 if Qxe3 Rf7+ mates ]
## Tactics in the openings series
Thanks for supporting wtharvey.com !
Shamkovich Puzzles, Part IV. White to move and win except where noted.
a)
Shamkovich vs Raymond Keene, New York, 1980
2r1n1k1/pp3Rb1/4p1Pp/2p1P3/4qN1Q/1P2p2P/1PP3PK/8 w - - 0 1
[ Qe7 if e2 Rf7+ mates ]
b)
Shamkovich vs Lev Alburt, Reyjavik, 1986
8/4p1k1/3pq1r1/p1p3RQ/2P4P/1r4P1/3R3K/8 w - - 0 1
[ Re2 if Qf6 Rxg6+ ]
c)
Shamkovich vs Lanka Ravi, London, 1987
r1b2k1r/pp3pp1/3qp2p/1B6/3N1n2/P7/1P3PPP/2RQ1RK1 w - - 0 1
[ Nxe6+ ]
d)
White Mates in 2. Shamkovich vs Anatoly Trubman, USA, 1990
5r1k/pp1n1p1p/5n1Q/3p1pN1/3P4/1P4RP/P1r1qPP1/R5K1 w - - 0 1
[ Qxf8+ ]
e)
Shamkovich vs Walter Shipman, Philadelphia, 1992
7R/6p1/5p2/1r2p1k1/4q2p/4P2P/3Q2PK/8 w - - 0 1
[ Qe2 ]
Other US GMs
Back to the Index
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51K
# Digits shuffler
World's simplest number tool
Quickly shuffle number's digits in your browser. To get your shuffled digits, just enter one or more numbers in the input field and this utility will randomize digit order in every number. Created by developers from team Browserling.
love fractals? check out our fracal tools!
We also have Online Fractal Tools — utilities for doing fractaly things. Check it out!
Digit Grouping
Shuffle digits in groups of this size. (By default shuffle each digit.)
## What is a digits shuffler?
This is an online browser-based utility for randomizing the digit order in numbers. It randomly changes the positions of all digits until they're completely shuffled. We have an additional option that lets you preserve partial order of two or more digits. You can shuffle groups of digits rather than every individual digit. For example, if you specify "2" in the group option field above, then for the number "2040", two-digit groups will be formed – "20" and "40", and you'll get two possible permutations – "2040" or "4020". To shuffle multiple numbers at once, enter each of them on a new line. If you don't like the randomization results, you can re-shuffle by clicking the "Shuffle digits again" button. That's numberwang!
## Digits shuffler examples
Click to use
Shuffle Digits in Six Numbers
In this example, we randomly change the positions of the digits in six numbers at once. Press shuffle digits again button above to get different digit order. Note that negative numbers stay negative.
12 456 1090 -500000 -288888888 -1111100000
21 654 0901 -000500 -888888288 -1011010100
Required options
These options will be used automatically if you select this example.
Shuffle digits in groups of this size. (By default shuffle each digit.)
Shuffle Pairs of Digits
In this example, we shuffle pairs of digits in three numbers. We enter the value "2" in the digit group option. This means that each number is split into groups of two digits (for example, "2030" is split into "20" and "30") and then all groups are shuffled individually.
2030 556677 24246824
3020 665577 24682424
Required options
These options will be used automatically if you select this example.
Shuffle digits in groups of this size. (By default shuffle each digit.)
Shuffle Decimal Numbers
In this example, we change the order of digits in decimal numbers. As decimals have an integer part and a fractional part, then we shuffle both parts. We also set the digit group size to "3" that makes it shuffle every three digits at once. Note that the decimal part "40050" contains 5 digits, so we can't form two triples and instead we shuffle a triplet ("400") and a pair ("50").
100200300.40050
200100300.50400
Required options
These options will be used automatically if you select this example.
Shuffle digits in groups of this size. (By default shuffle each digit.)
Pro tips Master online number tools
You can pass input to this tool via ?input query argument and it will automatically compute output. Here's how to type it in your browser's address bar. Click to try!
https://onlinenumbertools.com/shuffle-digits?input=12%0A456%0A1090%0A-500000%0A-288888888%0A-1111100000&shuffle-group-size=1
All number tools
Didn't find the tool you were looking for? Let us know what tool we are missing and we'll build it!
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Quickly exchange rows and columns of a matrix.
Quickly find the inverse matrix of any square matrix.
Quickly find the determinant of any square matrix.
Quickly calculate any number of digits of number π.
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Quickly generate numbers of say what you see sequence.
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Quickly check if the given number is a prime.
Quickly compute all prime factors of a number.
Quickly compute all divisors of a number.
Quickly calculate the GCD of two or more numbers.
Quickly calculate the LCM of two or more numbers.
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Quickly create a list of squares.
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Quickly round one or more numbers to the given accuracy.
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Quickly convert spelled numbers to regular numbers.
Quickly convert a number to the form 2^x.
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Quickly convert base 10 numbers to base -10.
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Quickly test if the given numbers are palindromes.
Coming soon These number tools are on the way
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Draw a Magic Square
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Rewrite Numbers
Given numbers and a grammar, recursively rewrite them.
Create a Floating Point Number
Create a number from the mantissa, base, and exponent.
Visualize a Floating Point Number
Show how a fp number is represented in a computer.
Convert a Number to the Scientific Notation
Convert a number to the a×10b form.
Convert Scientific Notation to a Number
Convert a number in scientific notation to a regular number.
Generate Unary Numbering
Create a list of unary numbers (1, 11, 111, 1111, …).
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Create a list of Braille numbers (⠂, ⠆, ⠒, ⠲, ⠢, …).
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Create a list of random octal numbers.
Generate Random Decimal Numbers
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Generate Random Hex Numbers
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Calculate a Running Sum
Calculate a cumulative sum of a list of numbers.
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Calculate a cumulative difference of a list of numbers.
Calculate a Running Product
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Calculate Number Quotient
Divide two numbers and find their quotient.
Calculate Digit Quotient
Divide the digits of the given number.
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Create one or more anagrams of a number.
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Generate Number N-grams
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Generate SI Prefixes
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Analyze a Number
Report how many digits appear how many times.
Convert a Number to an Ordinal
Convert a cardinal number to an ordinal number.
Convert an Ordinal to a Number
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Convert a Number to Roman Number
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Generate Negafibonacci Numbers
Calculate a series of extended Fibonacci numbers.
Generate Fibonacci Primes
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Fibonacci Number Test
Check if a number is a Fibonacci number.
Fibonacci Prime Test
Check if a number is both a Fibonacci number and a prime.
Construct Fibonacci Words
Create a sequence of Fibonacci words.
Construct Tribonacci Words
Create a sequence of Tribonacci words.
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Create a sequence of Tetranacci words.
Construct Pentanacci Words
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Calculate a series of extended Lucas numbers.
Generate Lucas Primes
Calculate a series of extended Lucas numbers.
Lucas Prime Test
Check if a number is both a Lucas number and a prime.
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Calculate a sequence of Moser-Bruijn numbers.
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Calculate a sequence of Oldenburger-Kolakoski numbers.
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Calculate a sequence of Stanley numbers.
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Calculate a sequence of self-describing Gijswijt numbers.
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Calculate a sequence of Rusin-Shapiro numbers.
Generate Baum-Sweet Numbers
Calculate a sequence of Baum-Sweet numbers.
Generate Thue-Morse Sequence
Calculate members of Thue-Morse number series.
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Create a list of perfect numbers.
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Calculate a sequence of abundant numbers.
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Generate Composite Numbers
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Draw a Numbers Table
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Test if a Number is Perfect
Check if the given number is a perfect number.
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Check if the given number is an abundant number.
Test if a Number is Deficient
Check if the given number is a deficient number.
Calculate the Modulo
Find the modulus of a number.
Group Number Digits
Group together digits of a number.
Split a Number into Digits
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Printf Numbers
Apply sprintf or printf function to numbers.
Create Zalgo Numbers
Repeat a Number
Repeat a number multiple times.
Mirror a Number
Create a mirror copy of a number.
Reverse a Number
Reverse the order of digits of a number.
Rotate a Number
Cyclically rotate digits of a number to the left or right.
Increase a Number
Add one to the given number.
Increase All Digits in a Number
Add one to every digit in a number.
Decrease a Number
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Decrease All Digits in a Number
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Count Number Occurrences
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Calculate Percentages
Find x% of a number.
Print Googol
Print the Googol/Google number, which is 10100.
Generate Big Numbers
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Generate Small Numbers
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Generate Constant Sequence
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Generate Real Numbers
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Generate Complex Numbers
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Generate Binary Numbers
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Generate Pairs of Numbers
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Generate Triples of Numbers
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Generate Tuples of Numbers
Create a sequence of number n-tuples.
Generate a Short Number
Create a number with not that many digits.
Generate a Long Number
Create a number with many digits.
Interweave Numbers
Interweave two or more number digit-by-digit.
Find the Decimal Expansion of a Number
Rewrite a number in the decimal representation.
Convert a Fraction to a Decimal
Convert a fraction to a decimal number.
Convert a Decimal to a Fraction
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Convert a Binary Number to Octal Number
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Convert a Binary Number to Hex Number
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Convert a Octal Number to Binary Number
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Convert a Octal Number to Decimal Number
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Convert a Octal Number to Hex Number
Convert a base eight number to base sixteen number.
Convert a Decimal Number to Binary Number
Convert a base ten number to base two number.
Convert a Decimal Number to Octal Number
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Convert a Decimal Number to Hex Number
Convert a base ten number to base sixteen number.
Convert a Hex Number to Binary Number
Convert a base sixteen number to base two number.
Convert a Hex Number to Octal Number
Convert a base sixteen number to base eight number.
Convert a Hex Number to Decimal Number
Convert a base sixteen number to base ten number.
Convert Any Number to Any Base
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Change Number's Mantissa
Change the significand of a number.
Change Number's Exponent
Change the power of a number.
Replace Digits with Letters
Replace digits in a number with alphabet letters.
Create a Number Spiral
Form a spiral from the digits of a number.
Create a Number Circle
Form a circle from the digits of a number.
Create a Number Tree
Form a tree from the given numbers.
Create a Number Digit Tree
Form a tree from the digits of a number.
Remove Decimal Point
Remove the decimal separator from a decimal number.
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https://www.coursehero.com/tutors-problems/Statistics-and-Probability/6296876-Members-of-a-management-team-suggested-order-quantities-of-15000-18/
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View the step-by-step solution to:
# Members of a management team suggested order quantities of 15,000, 18,000, 24,000, or 28,000 units.
Members of a management team suggested order quantities of 15,000, 18,000, 24,000, or 28,000 units. The wide range of order quantities suggested indicate considerable disagreement concerning the market potential. the product management team asks you for an analysis of the stock-out probabilites for various order quantities, an estimate of the profit potential, and to help make an order quantity recommendation. Specialty (the company name) expects to sell Weather Teddy (the product) for \$24 based on a cost of \$16 per unit. If inventory remains after the holiday season, Specialty will sell all surplus inventory for \$5 per unit. After reviewing the sales history of similiar products, Specialty's senior sales forecaster predicted an expected demand of 20,000 units with a 0.90 probability that demand would be between 10,000 units and 30,000 units.
1. Use the sales forecaster's rediciton to describe a normal probability distribution that can be used to approximate the demand distribution. Sketch the distribution and show its mean and standard deviation.
2. Compute the probability of a stock-out for the order quantities suggested by members of the management team.
3. Compute the projected profit for the rder quantities suggested by the management team under three scenarios: worst case in which sales = 10,000 units, most likely case in which sales = 20,000 units, and best case in which sales = 30,000 units.
4. One of Specialty's managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. What quantity would be ordered under this policy, and what is that projected profit under the three sales scenarios.
5. Provide your own recommendation for an order quantity and note the associated profit projections. Provide a rationale for your recommendation.
Dear student, PFA revised solution.... View the full answer
1. As Specialty’s senior sales forecaster predicted an expected demand of 20,000 units with a 0.90 probability that demand would be between 10,000 units and 30,000 units, we say 20,000 is the...
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http://obviam.net/index.php/design-in-game-entities-object-composition-strategies-part-2-the-state-pattern/
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## Design In-game Entities. Object Composition Strategies. Part 2 – The State Pattern
In this part I will try to explain how to design easily extensible and maintainable game elements that control their own internal state and behaviours.
An internal state is best described as being the soul and mind of the entity. In the first part I described why composition is better than inheritance. In a nutshell composition provides the means to interchange the algorithms associated with their behaviours. State on the other hand helps objects to control their own behaviours.
If you are not familiar with the behaviours and strategies, I strongly suggest to go and check out the previous article as I will be picking up those droids and make them utterly clever during the next articles. But first we need to take a closer look to how I design less clever entities, like weapons.
Let’s design an old fashioned rifle and learn from the process how to design the next gen weapons to fend off the invading droid army.
Breaking it down, what can one do with a rifle? Pull the trigger of course and reload (insert, eject the ammo clip).
Now think of all the scenarios. What happens if I pull the trigger and there is no clip? Or when there is one but it’s empty? Apparently a simple thing became a bit more complicated. Let’s see it on paper:
Initial Rifle State Diagram
Let’s examine the diagram and figure out what it is and what is that we want.
The above diagram is called a State Diagram.
Each of those circles is a State and each of those arrows is a State Transition.
Each state is just a different configuration of the rifle. The rifle behaves in certain ways depending in which state it is in and some action is needed to take it to another state. So, to go to another state, you will need to do something like pull the trigger or insert clip. See the arrows on the diagram.
We identify the following states:
• No Clip
• Has Clip
• Ammo fired
• Out of Ammo
and actions (transitions):
• inserts clip
• ejects clip
• pulls trigger
• fire ammo
Note: I used third person for 3 of the actions. It is because the user of the weapon will trigger these actions while fire ammo is more of an internal action.
The whole thing described above is also called a Finite State Machine. It is nothing more than an object that holds a finite number of states that govern behaviour and the object can be in only one state at any given time.
#### Our First Implementation
Knowing all this, what would be our first choice for the implementation? A class that has all the possible states and actions, right? It should also hold the current state. Simple.
Create the `Rifle.java` class.
```public class Rifle {
// defining the states
final static int NO_CLIP = 0;
final static int HAS_CLIP = 1;
final static int AMMO_FIRED = 2;
final static int OUT_OF_AMMO = 3;
int state = NO_CLIP; // instance variable holding the current state
int ammoCount = 0; // we count the ammo
}
```
Great! We created the structure of the rifle. We also want to know when we are out of ammo, so we added the `ammoCount` variable. They are set to default values. The rifle holds no clip thus the ammo is 0 when it is created.
Now let’s add the actions. But beware! Exposing all the actions to someone handling the weapon is dangerous. What will happen when someone tries to pull out the clip while firing? We need to take these into consideration when triggering the actions.
``` // **********************
// Creating the actions
// **********************
public void insertClip() {
// We check each possible state and act according to them
if (state == HAS_CLIP) {
} else if (state == AMMO_FIRED) {
System.out.println("You'll hurt yourself!!!");
} else if (state == OUT_OF_AMMO) {
System.out.println("You need to take out the empty clip first.");
} else if (state == NO_CLIP) {
state = HAS_CLIP;
ammoCount = 10;
System.out.println("You have loaded a clip with " + ammoCount + " bulletts.");
}
}
public void ejectClip() {
if (state == NO_CLIP) {
System.out.println("The magazine is empty.");
} else if (state == AMMO_FIRED) {
System.out.println("You'll hurt yourself!!!");
} else if (state == HAS_CLIP) {
// You could still eject it if you want but for the sake of
// simplicity let's use up the ammo first
System.out.println("Use up all your ammo first.");
} else if (state == OUT_OF_AMMO) {
state = NO_CLIP;
}
}
public void pullTrigger() {
if (state == NO_CLIP) {
System.out.println("Empty Click!");
} else if (state == AMMO_FIRED) {
System.out.println("Jammed!");
} else if (state == OUT_OF_AMMO) {
System.out.println("Click! Out of ammo.");
} else if (state == HAS_CLIP) {
System.out.println("BANG!!!");
state = AMMO_FIRED;
fireAmmo();
}
}
public void fireAmmo() {
if (state == NO_CLIP) {
System.out.println("Empty magazine.");
} else if (state == AMMO_FIRED) {
System.out.println("Bullet already on its way to kill someone!");
} else if (state == OUT_OF_AMMO) {
System.out.println("Out of ammo.");
} else if (state == HAS_CLIP) {
state = AMMO_FIRED;
ammoCount--;
System.out.println("Bullet on its way!");
// we check if the clip is empty
if (ammoCount == 0) {
// yes, it's empty
System.out.println("Darn! Out of ammo");
state = OUT_OF_AMMO;
} else {
state = HAS_CLIP;
}
}
}
```
Let’s put the rifle to the test.
Create the `RifleTest.java` class.
```public class RifleTest {
public static void main(String[] args) {
Rifle rifle = new Rifle();
System.out.println(rifle);
rifle.insertClip();
rifle.pullTrigger();
rifle.pullTrigger();
rifle.pullTrigger();
rifle.pullTrigger();
System.out.println(rifle);
rifle.insertClip();
rifle.ejectClip();
rifle.pullTrigger();
}
}
```
It is a very simple scripted scenario which performs some actions on the rifle. Some may be good some bad. As when trying to force in a second clip or pulling the trigger when it’s out of ammo.
Check out the result and examine it carefully. How cool is that? We are the Kalashnikovs of our age.
<< RIFLE [state=Empty Magazine (No Clip), ammo=0] >>
> You have loaded a clip with 3 bulletts.
> BANG!!!
> Bullet on its way!
> BANG!!!
> Bullet on its way!
> BANG!!!
> Bullet on its way!
> Darn! Out of ammo
!* Click! Out of ammo.
<< RIFLE [state=Out of Ammo, ammo=0] >&g;t
!* You need to take out the empty clip first.
> You have unloaded a clip.
!* Empty Click! – No clip!
Note that I have omitted the `toString()` method for the `Rifle`.
Well done! We have our first state machine.
#### But
We can’t do much firing bullet by bullet. We need automatic fire! Oh, and we can’t just let people running around with rifles without the safety ON.
Ah, that’s easy, right? We’ll have a few more states and a few more transitions.
But wait, we will need to rework the `Rifle` class a bit.
Let’s see what we need to modify just to support automatic and manual fire:
• we need to add the two states
• but then we need to modify every single method to handle the states by adding conditional statements
• `pullTrigger()` will get complicated as it will need to know what state it is in and check bullets and fire them as such
That is a LOT of work. It must be some other way.
#### The Solution
What if we give each state a behaviour and put it into its own class? This way each state will implement its own actions only. We will have the rifle class delegating the action to the state object that represents the current state.
Let’s see how does it look?
What you should notice is that the Has Clip is now Manual/Auto and has a switch transition. It basically has a sub-state now. Flipping the switch changes the behaviour. Both states will have the pull trigger action but each state will behave differently. This is easily achieved with interfaces, right?
Let’s start coding then. First, let’s create the state interface. Remember, each state will implement its transition and will provide the dummy implementation for the rest. The interface will contain all the actions. The methods map directly to all the actions that can happen with the `Rifle`
```public interface RifleState {
public void insertClip();
public void ejectClip();
public void swithManualAuto();
public void pullTrigger();
public void fireAmmo();
}
```
The new `Rifle` class will have none of the implemented actions, but will have all the states and will delegate the actions to its current state.
`Rifle.java`
```public class Rifle {
// the states of the rifle
RifleState emptyState;
RifleState autoFireState;
RifleState manualFireState;
RifleState outOfAmmoState;
RifleState roundFiredState;
RifleState ammoFiredState;
RifleState state = emptyState;
int ammoCount = 0;
// constructor
public Rifle() {
// creating states
this.emptyState = new NoClipState(this);
this.autoFireState = new AutoFireState(this);
this.manualFireState = new ManualFireState(this);
this.outOfAmmoState = new OutOfAmmoState(this);
this.roundFiredState = new RoundFiredState(this);
this.ammoFiredState = new AmmoFiredState(this);
this.state = this.emptyState;
this.ammoCount = 0;
}
// convenience methods - delegating only
public void insertClip() {
this.state.insertClip();
}
public void ejectClip() {
this.state.ejectClip();
}
public void switchManualAuto() {
this.state.switchManualAuto();
}
public void pullTrigger() {
this.state.pullTrigger();
}
// getters and setters
// ... omitted
}
```
You see that the class has changed a bit. It has all the states and a current state. It has a constructor too. The constructor is needed to set up the rifle and pass itself as a reference to all of the states to give them access to the rifle’s properties. In our case just the ammo count and current state need to be accessed.
Also the `fireAmmo()` is missing as it is a state internal action.
Now let’s map the states to actual classes. These are the same as in the original rifle class.
I will list one full class and for the rest only the methods that change the state. Examine the source code for the complete listings.
`OutOfAmmoState.java`
```public class OutOfAmmoState implements RifleState {
private Rifle rifle;
public OutOfAmmoState(Rifle rifle) {
this.rifle = rifle;
}
@Override
public void ejectClip() {
rifle.setState(rifle.getEmptyState());
System.out.println("> Clip ejected.");
}
@Override
public void fireAmmo() {
System.out.println("!* You can't fire with no ammo.");
}
@Override
public void insertClip() {
System.out.println("!* There is an empty clip inserted already!");
}
@Override
public void pullTrigger() {
System.out.println("!* Out of ammo!");
}
@Override
public void switchManualAuto() {
}
}
```
Notice that the only way out of the OutOfAmmoState is by ejecting the clip.
Every other attempt will do nothing.
Also note the constructor. We are passing the reference to the rifle there.
Check out the other classes too:
`AmmoFiredState`
```public class AmmoFiredState implements RifleState {
private Rifle rifle;
public AmmoFiredState(Rifle rifle) {
this.rifle = rifle;
}
@Override
public void fireAmmo() {
rifle.setAmmoCount(rifle.getAmmoCount() - 1);
System.out.println("> Fired 1 bullet.");
if (rifle.getAmmoCount() == 0) {
rifle.setState(rifle.getOutOfAmmoState());
} else {
rifle.setState(rifle.getManualFireState());
}
}
// ... ommited
}
```
`AutoFireState`
```public class AutoFireState implements RifleState {
private Rifle rifle;
public AutoFireState(Rifle rifle) {
this.rifle = rifle;
}
@Override
public void ejectClip() {
rifle.setAmmoCount(0);
rifle.setState(rifle.getEmptyState());
}
@Override
public void pullTrigger() {
System.out.println("> Pulled trigger.");
rifle.setState(rifle.getRoundFiredState());
rifle.getState().fireAmmo();
}
@Override
public void switchManualAuto() {
rifle.setState(rifle.getManualFireState());
System.out.println("> Switched to manual. Hope they are slow and few!");
}
// ... ommited
}
```
If you follow the diagram then you should be able to figure out the transitions. There is a trick there as the Manual and Auto modes are very similar and I just transition between them. There is a drawback as after a reload, the manual state is the active one even if the auto was set before. But I’m sure you can figure out a quick fix.
`ManualFireState`
```
public class ManualFireState implements RifleState {
private Rifle rifle;
public ManualFireState(Rifle rifle) {
this.rifle = rifle;
}
@Override
public void ejectClip() {
rifle.setAmmoCount(0);
rifle.setState(rifle.getEmptyState());
}
@Override
public void pullTrigger() {
System.out.println("> Pulled trigger.");
rifle.setState(rifle.getAmmoFiredState());
rifle.getState().fireAmmo();
}
@Override
public void switchManualAuto() {
rifle.setState(rifle.getAutoFireState());
System.out.println("> Switched to auto. Bring'em on!!!");
}
// ... ommited
}
```
`NoClipState`
```public class NoClipState implements RifleState {
private Rifle rifle;
public NoClipState(Rifle rifle) {
this.rifle = rifle;
}
@Override
public void insertClip() {
rifle.ammoCount = 50;
rifle.setState(rifle.getManualFireState());
}
// ...ommited
}
```
`RoundFiredState`
```public class RoundFiredState implements RifleState {
private Rifle rifle;
public RoundFiredState(Rifle rifle) {
this.rifle = rifle;
}
@Override
public void fireAmmo() {
int count = 10;
while (count > 0 && rifle.getAmmoCount() > 0) {
System.out.print("> BANG! ");
rifle.setAmmoCount(rifle.getAmmoCount() - 1);
count--;
}
System.out.println();
System.out.println("> Fired a round of " + (10 - count) + " bullets. Yeah!");
if (rifle.getAmmoCount() <= 0) {
rifle.setAmmoCount(0);
rifle.setState(rifle.getOutOfAmmoState());
} else {
rifle.setState(rifle.getAutoFireState());
}
}
// ...ommited
}
```
Great! Let’s throw together a test for the new shiny Rifle.
`RifleTest`
```public class RifleTest {
public static void main(String[] args) {
Rifle rifle = new Rifle();
rifle.pullTrigger();
rifle.ejectClip();
rifle.insertClip();
rifle.pullTrigger();
rifle.switchManualAuto();
rifle.pullTrigger();
rifle.pullTrigger();
rifle.switchManualAuto();
rifle.pullTrigger();
rifle.insertClip();
rifle.switchManualAuto();
rifle.pullTrigger();
rifle.pullTrigger();
rifle.pullTrigger();
rifle.pullTrigger();
rifle.pullTrigger();
}
```
Examine the output very carefully and go through the source code. You should try adding new features or think of other scenarios.
!* You can’t fire with an empty magazine.
!* The magazine is empty.
> Pulled trigger.
> Fired 1 bullet.
> Switched to auto. Bring’em on!!!
> Pulled trigger.
> BANG! >BANG! >BANG! >BANG! >BANG! >BANG! >BANG! > BANG! >BANG! >BANG!
> Fired a round of 10 bullets. Yeah!
> Pulled trigger.
>BANG! >BANG! >BANG! >BANG! >BANG! >BANG! >BANG! > BANG! >BANG! >BANG!
> Fired a round of 10 bullets. Yeah!
> Switched to manual. Hope they are slow and few!
> Pulled trigger.
> Fired 1 bullet.
> Switched to auto. Bring’em on!!!
> Pulled trigger.
> BANG! >BANG! >BANG! >BANG! >BANG! >BANG! >BANG! > BANG! >BANG! >BANG!
> Fired a round of 10 bullets. Yeah!
> Pulled trigger.
> BANG! >BANG! >BANG! >BANG! >BANG! >BANG! >BANG! > BANG! >BANG! >BANG!
> Fired a round of 10 bullets. Yeah!
> Pulled trigger.
> BANG! >BANG! >BANG! >BANG! >BANG! >BANG! >BANG! >BANG!
> Fired a round of 8 bullets. Yeah!
!* Out of ammo!
!* Out of ammo!
Again, the yellow lines show you when you tried something dodgy.
Voila! We have state machines and easily extensible game elements.
I strongly encourage you to extend it and build your own, as it is a very important element of game design.
You can apply this principle for the whole game and infrastructure. Think of the game as a whole and when you’re on the menu screen (that’s the Menu State), when you’re in the paused screen, that’s the Pause State, or the main screen that’s the Running State. You could create a framework and easily add different game states.
I do hope this gave you a good introduction on how to implement state driven behaviours. Note that I used normal Java practices to write the code for clarity. Meaning that when writing a game, especially for Android, you might not need to overload your classes with `getters` and `setters` and you should adhere to the optimized ways. You could use public attributes and access them directly for example. But it was not the scope of this article.
Also understanding this is crucial to make a step towards autonomous agents (automatons). Next time I will introduce event and/or message driven behaviours. We will return to our beloved droids to give them wits to outrun you.
Note that the first attempt was renamed to `RifleOld.java` and `RifleOldTest.java`
#### 10 Responses - Add Yours+
1. Alan Bartlett says:
Hey nice tutorial,
I was wondering if using enums would be a nice clean way of using a state machine as well instead of integer constants like you have used?
• Impaler says:
Yes, enums are cleaner and very useful.
2. maxha says:
Great tutorial! But I was wondering wether this doesn’t give you a lot of classes to work with, if you’re having multiple states for each object-type, and then also have different states for the components (from your great components article).
Am I missing something, or do you just have tons of (cluttered) classes in a project?
• Impaler says:
First you will create your models that you will use in the game. This can be a mix of components and inheritance of course. Then you can design the behaviours and states. Depending on how extensible you want your engine, you can design it completely with generic entities or a mix of behaviours and set objects.
Also Java helps a lot with the packages and classes can be grouped logically. Also you can create inner classes for behaviours under a certain logical group.
You can also check out Artemis which is really neat.
You might still end up with tons of classes, but having them grouped, certainly helps. Usually it’s not that big a number if designed well. Don’t have to normalize it to the utter extent.
And thanks for the appreciation, I’m happy that it is useful.
3. Deepthi says:
I am interested in game programming and trying to develop a game i learned basics from your site, however i do not have an idea how to implement different levels of the game please post on this topic or point me to some resources on this, by the way great site and really helpful thank you very much.
4. [...] First we create the renderer interface. We use this to establish a single way to interact with the renderer and it will make it easy to create more views without affecting the game engine. To read more on why is a good idea check this and this. [...]
5. Ben says:
No offence, but this doesn’t seem like a very elegant state machine solution. I base mine from “Programming Game AI By Example”, and use a single State class, where T is the type of entity I want the state to worth with, for example Player. Each State has 3 methods: enter(T entity), execute(T entity), and exit(T entity). enter/exit are called when the state becomes the ‘active’ state or before another state becomes active. execute is called each game loop when the game object’s state is updated.
There’s also a simple StateMachine class which has basically is responsible for keeping a list of States and switching between them. The game object has an instance of this state machine, for example StateMachine fsm = new StateMachine;
In my Player class, I create a subclass (either anonymous or inner-class, whatever your preference is) for each state. For this example, lets say something like this: fsm.addState( new State(“running”) {…}); Obviously you’d also implement enter, execute, and exit to suit your needs. I also passed a string as the state’s ID so I can switch to it later.
To switch states, I just call fsm.changeTo(“running”); The StateMachine.changeTo() method calls the current state’s exit method, changes the current state to the new state, and calls the new state’s enter method.
Umm, it’s a little hard to understand just from my explanation, so you can see my code from my SVN repository.
• Impaler says:
I agree that it is not perfect but it is the purest implementation of the state pattern.
I do the same as you and I am using a mix of the state and command pattern.
You introduced an orchestrator as the StateMachine and that has knowledge of the flow.
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| 0.941985
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# Surface Areas and Volumes Class 10 Notes
According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 12.
## CBSE Class 10 Maths Surface Area and Volumes Notes:-
The concept of surface area and volume for Class 10 is provided here. In this article, we are going to discuss the surface area and volume for different solid shapes such as the cube, cuboid, cone, cylinder, and so on. The surface area can be generally classified into Lateral Surface Area (LSA), Total Surface Area (TSA), and Curved Surface Area (CSA). Here, let us discuss the surface area formulas and volume formulas for different three-dimensional shapes in detail. In this chapter, the combination of different solid shapes can be studied. Also, the procedure to find the volume and its surface area in detail.
To get the solutions for class 10 Maths surface areas and volumes, click on the below link.
## Surface Area and Volume of Cuboid
A cuboid is a region covered by its six rectangular faces. The surface area of a cuboid is equal to the sum of the areas of its six rectangular faces.
### Surface Area of the Cuboid
Consider a cuboid whose dimensions are lΒ ΓΒ bΒ ΓΒ h, respectively.
The total surface area of the cuboid (TSA) = Sum of the areas of all its six faces
TSA (cuboid) = 2(lΒ ΓΒ b)Β +Β 2(bΒ ΓΒ h)Β +Β 2(lΒ ΓΒ h)Β =Β 2(lbΒ +Β bhΒ +Β lh)
Lateral surface area (LSA) is the area of all the sides apart from the top and bottom faces.
The lateral surface area of the cuboid = Area of face AEHD + Area of face BFGC + Area of face ABFE + Area of face DHGC
LSA (cuboid) = 2(bΒ ΓΒ h)Β +Β 2(lΒ ΓΒ h)Β =Β 2h(lΒ +Β b)
Length of diagonal of a cuboid =β(l2Β + b2Β + h2)
To know more about the Surface Area of Cuboid, visit here.
### Volume of a Cuboid
The volume of a cuboid is the space occupied within its six rectangular faces.
Volume of a cuboid =Β (baseΒ area)Β ΓΒ heightΒ =Β (lb)hΒ =Β lbh
## Surface Area and Volume of Cube
A cube is a three-dimensional solid that has six square faces, twelve edges and eight vertices.
### Surface Area of Cube
As we know, one of the important properties of a cube is length = breadth = height.
If we assume that the length of the cube is βlβ, and hence we get
So, obviously, here we get,
Height = l
The total surface area of the cube (TSA) = Sum of the areas of all its six faces.
In case of all faces has an equal area, TSAΒ of Cube = 6 Γ area of Square = 6l2 square units.
Similarly, the Lateral surface area of cube =Β 2(lΒ ΓΒ lΒ +Β lΒ ΓΒ l)Β =Β 4l2
Note: Diagonal of a cube =β3l
### Volume of a Cube
Volume of a cube = baseΒ areaΒ ΓΒ height
Since all dimensions of a cube are identical, volume = l3
Where l is the length of the edge of the cube.
To know more about the Volume of Cube and Cuboid, visit here.
#### For more information on Cube And Cuboid, watch the below video
To know more about the Surface Area of Cube, visit here.
## Surface Area and Volume of Cylinder
A cylinder is a solid shape that has two circular bases connected with each other through a lateral surface. Thus, there are three faces, two circular and one lateral, of a cylinder. Based on these dimensions, we can find the surface area and volume of a cylinder.
### Surface Area of Cylinder
Take a cylinder of base radius rΒ and height hΒ units. The curved surface of this cylinder, if opened along the diameter (d = 2r) of the circular base can be transformed into a rectangle of length 2Οr and height hΒ units. Thus,
CSA of a cylinder of base radius rΒ and height hΒ =Β 2ΟΒ ΓΒ rΒ ΓΒ h
TSA Β of a cylinder of base radius rΒ and height hΒ =Β 2ΟΒ ΓΒ rΒ ΓΒ h + area of two circular bases
=2ΟΒ ΓΒ rΒ ΓΒ hΒ +Β 2Οr2
=2Οr(hΒ +Β r)
### Volume of a Cylinder
Volume of a cylinder = Base area ΓΒ height = (Οr2)Β ΓΒ hΒ =Β Οr2h
To know more about the Volume of a Cylinder, visit here.
To know more about the Surface Area of a Cylinder, visit here.
## Surface Area and Volume ofΒ Right Circular Cone
A cone is a 3d shape that has one circular base and narrows smoothly from the base to a point called the vertex.
### Surface Area of Cone
Consider a right circular cone with slant length l, radius r and height h.
CSA of right circular cone =Β Οrl
TSA = CSA + area of base =Β ΟrlΒ +Β Οr2Β =Β Οr(lΒ +Β r)
To know more about the Surface Area of a Right Circular Cone, visit here.
### Volume of a Right Circular Cone
The volume of a Right circular cone is 1/3Β times that of a cylinder of the same height and base.
In other words, 3 cones make aΒ cylinder of the same height and base.
The volume of a Right circular cone =(1/3)Οr2h
Where ‘r’ is the radius of the base and ‘h’ is the height of the cone.
To know more about the Volume of a Right Circular Cone, visit here.
## Surface Area and Volume of Sphere
A sphere is a solid that is round in shape, and the points on its surface are equidistant from the center.
### Surface Area of Sphere
For a sphere of radiusΒ r
Curved Surface Area (CSA) = Total Surface Area (TSA) = 4Οr2
### Volume of Sphere
The volume of a sphere of radius r = (4/3)Οr3
To know more about volume of a Sphere, visit here.
## Surface Area and Volume of Hemisphere
A hemisphere is a shape that is half of the sphere and has one flat surface. The other side of the hemisphere is shaped as a circular bowl. See the figure below.
### Surface Area of Hemisphere
We know that the CSA of a sphereΒ =Β 4Οr2.
A hemisphere is half of a sphere.
β΄ CSA of a hemisphere of radius r =Β 2Οr2
Total Surface Area = curved surface area + area of the base circle
βTSA =Β 3Οr2
To know more about Surface Area of Hemisphere, visit here.
### Volume of Hemisphere
The volume (V) of a hemisphere will be half of that of a sphere.
β΄Β The volume of the hemisphere of radius r =Β (2/3)Οr3
To know more about Volume of Hemisphere, visit here.
## Surface Area and Volume of Combination of Solids
The combination of solids explains the shapes formed when two different solids are combined together. Thus, the surface area and volume for such shapes will vary from the other basic solids.
### Surface Area of Combined Figures
Areas of complex figures can be broken down and analysed as simpler known shapes. By finding the areas of these known shapes, we can find out the required area of the unknown figure.
Example: 2 cubes each of volume 64 cm3Β are joined end to end. Find the surface area of the resulting cuboid.
Length of each cube =Β 64(1/3)Β =Β 4cm
Since these cubes are joined adjacently, Β they form a cuboid whose length l = 8 cm. But height and breadth will remain the same = 4 cm.
β΄ The new surface area, TSA = 2(lb + bh + lh)
TSA = 2 (8 x 4 + 4 x 4 + 8 x 4)
= 2(32 + 16 + 32)
= 2 (80)
TSA = 160 cm2
### Volume of Combined Solids
The volume of complex objects can be simplified by visualising them as a combination of shapes of known solids.
Example: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 3 cm and the height of the cone is equal to 5 cm.
This can be visualised as follows :
V(solid) = V(Cone) + V(hemisphere)
V(solid) = (1/3)Οr2h + (2/3)Οr3
V(solid) = (1/3)Ο(9)(5)Β + (2/3)Ο(27)
V(solid) = 33Ο cm3
To know more about the Volume of a Combination of Solids, visit here.
## Surface Area and Volume of Frustum of a cone
When a solid is cut by a plane, then another form of solid is formed. One such form of a solid is the frustum of a cone, which is formed when a plane cuts a cone parallelly to the base of the cone. Let us discuss its surface area and volume here.
If a right circular cone is sliced by a plane parallel to its base, then the part with the two circular bases is called a Frustum.
### Surface Area of a Frustum
CSA of frustum =Ο(r1+r2)l,Β where l=Β β[h2+(r2Β –Β r1)2]
TSA of the frustum is the CSA + the areas of the two circular faces = Ο(r1Β +Β r2)lΒ +Β Ο(r12Β +Β r22)
Example:
Find the curved surface area of the frustum if the slant height of the coneβs frustum is 4 cm. Also, given that the circumferences of its circular ends are 6 cm and 18 cm.
Solution:
Given: Slant height, l = 4 cm
The circumferences of the two circular ends are 6 cm and 18 cm.
Let the radius of the two circular ends be r1 and r2.
Now, we have to find r1 and r2.
Finding r1:
Circumference of one end = 6 cm
2Οr1 = 6Β
r1 = 6/2Ο
r1 = 3/Ο
Finding r2:
Circumference of the other end = 18 cm
2Οr2 = 18Β
r2 = 18/2Ο
r2 = 9/Ο
We know that the formula to find the curved surface area of frustum is Ο(r1 + r2)l square units.
Thus by substituting the known values, we get
CSA of frustum = Ο[(3/Ο) + (9/Ο)]4
CSA of frustum = 4Ο(12/Ο)
CSA of frustum = 48 cm2.
Therefore, the curved surface area of the frustum is 48 cm2.
### Volume of a Frustum
The volume of a frustum of a cone =(1/3)Οh(r12Β +Β r22Β +Β r1r2)
Example:
A glass is in the shape of the frustum of a cone with a height of 14 cm. The diameters of its two circular ends are 2 cm and 4 cm. Find the volume of the glass.
Solution:
Given: Height of the frustum = 14 cm
The diameters of the two circular ends are 2 cm and 4 cm.
So, the radius will be 1 cm and 2 cm.
Let r1 = 1 cm and r2 = 2 cm
Since the glass is in the shape of a frustum, the volume of a glass = volume of a frustum.
Therefore, the volume of glass = (β
)Οh (r12 + r22 + r1r2) cubic units.
Now, substitute the values in the formula, we get
Volume of a glass, V = (β
) Ο(14)[12 + 22 + (1)(2)]
V = (β
) (22/7)(14) [1 + 4 + 2]
V = (β
) (22/7)(14)(7)
V = [22 Γ 14 Γ 7] / [3 Γ 7]
V = (22 Γ 14 ) / 3
V = 308 / 3
V = 102. 67
Hence, the volume of a glass is 102. 67 cm3.
To know more about Frustum, visit here.
## Conversion of Solid from One Shape to Another
When a solid is converted into another solid of a different shape (by melting or casting), the volume remains constant.
Example:
A metallic sphere with a radius of 4.2 cm is melted and recast in the shape of a cylinder with a radius of 6 cm. Determine the cylinderβs height.
Solution:
Given that, the sphere is melted and recast into a cylinder.
So, we can write,Β
Sphereβs volume = Cylinderβs volume.Β
As we know, the volume of a sphere = (4/3)Οr3 cubic units.
And the volume of a cylinder = Οr2h cubic units.
So, we can write:
(4/3)Οr3 =Β Οr2h β¦(1)
Given that, the radius of the metallic sphere = 4.2 cm.
Radius of the cylinder = 6 cm.
Now, substitute the values in (1), we get
(4/3)Ο(4.2)3 =Β Ο(6)2hΒ
(4/3)(4.2)3 = 36h
Hence, h = [(4/3)(4.2)3]/36
h = [4 Γ 4.2 Γ 4.2 Γ 4.2 ]/[3 Γ 36]
By simplifying the above expression, we get
h = 2.744 cm
Hence, the height of the cylinder is 2.744 cm.
To know more about the Shape Conversion of Solids, visit here
## Surface Areas and Volumes Formulas – Summary
The following table presents a brief summary of different solid shapes with their surface area and volume formulas:
Shape Parameters Surface Area (Square units) Volume (Cubic units) Cuboid Length = l Breadth = b Height = h TSA = 2(lb + bh + lh) LSA = 2h(l + b) V = l Γ b Γ h Cube Length = Breadth = Height = l TSA = 6l2 LSA = 4l2 V = l3 CylinderΒ Radius = r Height = h CSA = 2Ο Γ r Γ h TSA = 2Οr(h + r) V = Οr2h Cone Radius = r Height = h Slant Height = l CSA = Οrl TSA = Οr(l + r) V = (1/3)Οr2h Sphere Radius = r CSA = TSA =Β 4Οr2 V = (4/3)Οr3 Hemisphere Radius = r CSA = 2Οr2 TSA = 3Οr2 V =Β (2/3)Οr3 Frustum Radius of top circular part = r1 Radius of bottom circular part = r2 Height = h Slant height = l CSA = Ο(r1+r2)l TSA = Ο(r1 + r2)l + Ο(r12 + r22) V = (1/3)Οh(r12 + r22 + r1r2)
Stay tuned to BYJU’S – The Learning App and download the app today to get all class-wise concepts.
Test your knowledge on Surface Area Volumes Class 10
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``````
In [2]:
%matplotlib inline
from bigbang.archive import Archive
from bigbang.thread import Thread
from bigbang.thread import Node
import matplotlib.pyplot as plt
import datetime
``````
First, collect data from a public email archive.
``````
In [3]:
url = "https://lists.wikimedia.org/pipermail/analytics/"
arx = Archive(url,archive_dir="../archives")
``````
We can count the number of threads in the archive easily. The first time you run `Archive.get_thread` it may take some time to compute, but the result is cached in the Archive object.
``````
In [4]:
#threads = arx.get_threads()
len(arx.get_threads())
``````
``````
Out[4]:
628
``````
We can plot a histogram of the number of messages in each thread. In most cases this will be a power law distribution.
``````
In [5]:
y = [t.get_num_messages() for t in arx.get_threads()]
plt.hist(y, bins=30)
plt.xlabel('number of messages in a thread')
plt.show()
``````
``````
``````
We can also plot the number of people participating in each thread. Here, the participants are differentiated by the From: header on the emails they've sent.
``````
In [6]:
n = [t.get_num_people() for t in arx.get_threads()]
plt.hist(n, bins = 20)
plt.xlabel('number of email-address in a thread')
plt.show()
``````
``````
``````
The duration of a thread is the amount of elapsed time between its first and last message.
``````
In [7]:
y = [t.get_duration().days for t in arx.get_threads()]
plt.hist(y, bins = (10))
plt.xlabel('duration of a thread(days)')
plt.show()
``````
``````
``````
``````
In [8]:
y = [t.get_duration().seconds for t in arx.get_threads()]
plt.hist(y, bins = (10))
plt.xlabel('duration of a thread(seconds)')
plt.show()
``````
``````
``````
You can examine the properties of a single thread.
``````
In [9]:
print(arx.get_threads()[0].get_duration())
``````
``````
19:49:47
``````
``````
In [21]:
threads = arx.get_threads
``````
``````
In [22]:
content = arx.get_threads()[0].get_root().data['Body']
content
``````
``````
Out[22]:
'Welcome to the the inaugural Analytics Mailing list email.\n\nHere all your analytics wishes comes true, \n\n\nso proposals, ideas, crazy ideas, crazy crazy ideas are welcome here!\nas long as we can count something it is welcome. \n\n\nD\n\n'
``````
``````
In [23]:
len(content.split())
``````
``````
Out[23]:
38
``````
Suppose we want to know whether or not longer threads (that contain more distinct messages) have more words.
``````
In [24]:
short_threads = []
long_threads = []
for t in arx.get_threads():
if(t.get_num_messages() < 6): short_threads.append(t)
else: long_threads.append(t)
``````
``````
In [25]:
print(len(short_threads))
print(len(long_threads))
``````
``````
471
157
``````
``````
In [26]:
len(long_threads[0].get_content())
``````
``````
Out[26]:
13
``````
``````
In [27]:
dist_short = []
dist_long = []
for t in short_threads:
avg_short = sum([len(i.split()) for i in t.get_content()]) / len(t.get_content())
dist_short.append(avg_short)
for t in long_threads:
avg_long = sum([len(i.split()) for i in t.get_content()]) / len(t.get_content())
dist_long.append(avg_long)
``````
``````
In [28]:
plt.hist(dist_short, bins = (15))
plt.show()
``````
``````
``````
``````
In [29]:
plt.hist(dist_long, bins = (15))
plt.show()
``````
``````
``````
``````
In [30]:
print(sum(dist_short)/ len(dist_short))
print(sum(dist_long)/ len(dist_long))
``````
``````
140
110
``````
``````
In [31]:
%reload_ext autoreload
s_leaves = []
s_notleaves = []
for t in threads:
for node in t.get_leaves():
s_leaves.append(len(node.data['Body'].split()))
for node in t.get_not_leaves():
s_notleaves.append(len(node.data['Body'].split()))
``````
``````
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-31-69067a14eb67> in <module>()
2 s_leaves = []
3 s_notleaves = []
----> 4 for t in threads:
5 for node in t.get_leaves():
6 s_leaves.append(len(node.data['Body'].split()))
TypeError: 'instancemethod' object is not iterable
``````
``````
In [20]:
plt.hist(s_leaves, bins = (15))
plt.show()
``````
``````
---------------------------------------------------------------------------
IndexError Traceback (most recent call last)
<ipython-input-20-804337a1ae9d> in <module>()
----> 1 plt.hist(s_leaves, bins = (15))
2 plt.show()
/home/sb/anaconda/envs/bigbang/lib/python2.7/site-packages/matplotlib/pyplot.pyc in hist(x, bins, range, normed, weights, cumulative, bottom, histtype, align, orientation, rwidth, log, color, label, stacked, hold, **kwargs)
2825 histtype=histtype, align=align, orientation=orientation,
2826 rwidth=rwidth, log=log, color=color, label=label,
-> 2827 stacked=stacked, **kwargs)
2828 draw_if_interactive()
2829 finally:
/home/sb/anaconda/envs/bigbang/lib/python2.7/site-packages/matplotlib/axes.pyc in hist(self, x, bins, range, normed, weights, cumulative, bottom, histtype, align, orientation, rwidth, log, color, label, stacked, **kwargs)
8247 # Massage 'x' for processing.
8248 # NOTE: Be sure any changes here is also done below to 'weights'
-> 8249 if isinstance(x, np.ndarray) or not iterable(x[0]):
8250 # TODO: support masked arrays;
8251 x = np.asarray(x)
IndexError: list index out of range
``````
``````
In [23]:
plt.hist(s_notleaves, bins = (15))
plt.show()
``````
``````
``````
``````
In [24]:
print(sum(s_leaves)/len(s_leaves))
print(sum(s_notleaves)/len(s_notleaves))
``````
``````
264
295
``````
``````
In [25]:
import re
mess = threads[85].get_leaves()[0].data['Body']
``````
``````
In [26]:
print(mess)
``````
``````
Fernando Perez wrote:
>
> Great! Many thanks for this. Please give it a bit more pounding, and
> I'd encourage other users of dreload to also try it out. The code is
> definitely far simpler than the original dreload, but since I don't
> understand that code too well, I'd like to tiptoe a bit on this
> issue. If it survives a bit of pounding and discussion, I'll
> definitely be glad to put it in.
One of the things where it differ's from the old deep_reload is that
when importing a submodule, say ipbug.vm, it will not reload
ipbug/__init__.py. I've attached another version, which tries to do just
that and I'm using that version currently without problems.
However I think there must be an even less complicated version.
Clearing sys.modules and 'reimporting' the module like in the following
code seems to work ok.
====
import sys, bbutils.textproc.spellcheck
m=sys.modules.copy()
sys.modules.clear()
sys.modules['sys'] = sys
import bbutils.textproc.spellcheck
====
Currently I don't have the time to investigate this further, but in a
week or two, I'll have another look at this.
- Ralf
-------------- next part --------------
import sys
import __builtin__
builtin_import = None # will be set to __builtin__.__import__ by reload function
old_modules = {} # will be set to sys.modules by reload function
reloaded = {} # names of reloaded modules, uses same keys as sys.modules
def determineParentName(globals):
"""determine name of the module which has called the import statement"""
if not globals or not globals.has_key("__name__"):
return None
pname = globals['__name__']
if globals.has_key("__path__"):
return pname
if '.' in pname:
i = pname.rfind('.')
pname = pname[:i]
return pname
return None
def reloadModuleAndParents(name, globals=None):
"""for name='some.module.bar', reload some, some.module, some.module.bar.
Determines module, which has called the import statement, and prefers
module relative paths, i.e. 'import os' in module m, from m/__init__.py
imports m/os.py if it's there.
"""
global reloaded
def reloadByName(n):
if old_modules.has_key(n):
if reloaded.has_key(n):
return reloaded[n]
reloaded[mname]=1
sys.modules[mname] = old_modules[mname]
if mname != old_modules[mname].__name__:
# module changed it's name. otherwise dreload(xml.sax) fails.
print "Module changed name:", mname, old_modules[mname].__name__
return reloadModuleAndParents(old_modules[mname].__name__)
else:
print "Reloading", mname
return __builtin__.reload(old_modules[mname])
return None
mods = name.split(".")
parent = determineParentName(globals)
retval = None
for i in range(len(mods)):
mname = ".".join(mods[:i+1])
if parent:
relative = "%s.%s" % (parent, mname)
if old_modules.has_key(relative) and old_modules[relative]:
retval=reloadModuleAndParents(relative)
continue
retval=reloadByName(mname)
return retval
def my_import_hook(name, globals=None, locals=None, fromlist=None):
"""replacement for __builtin__.__import__
reloads module 'name' if it hasn't been already reloaded
and then calls original __builtin__.__import__.
"""
## if fromlist:
## print 'Importing', fromlist, 'from module', name
## else:
## print 'Importing module', name
reloadModuleAndParents(name, globals)
return builtin_import(name, globals, locals, fromlist)
# Replacement for reload()
def reload(module, exclude=['sys', '__builtin__', '__main__']):
"""Recursively reload all modules used in the given module. Optionally
takes a list of modules to exclude from reloading. The default exclude
list contains sys, __main__, and __builtin__, to prevent, e.g., resetting
display, exception, and io hooks.
"""
global builtin_import
global old_modules
global reloaded
reloaded = {}
old_modules = sys.modules.copy()
sys.modules.clear()
for ex in exclude+list(sys.builtin_module_names):
if old_modules.has_key(ex) and not sys.modules.has_key(ex):
print "EXCLUDING", ex
sys.modules[ex] = old_modules[ex]
reloaded[ex]=1
builtin_import = __builtin__.__import__
__builtin__.__import__ = my_import_hook
try:
return reloadModuleAndParents(module.__name__)
finally:
# restore old values
__builtin__.__import__ = builtin_import
for m in old_modules:
if not sys.modules.has_key(m):
sys.modules[m] = old_modules[m]
``````
``````
In [27]:
mess.split('\n')
message = list()
for l in mess.split('\n'):
n = len(l)
if(len(l)!=0 and l[0] != '>' and l[n-6:n] != 'wrote:'):
message.append(l)
new = str()
for l in message:
new = new + l + '\n'
``````
``````
In [28]:
print(new)
``````
``````
One of the things where it differ's from the old deep_reload is that
when importing a submodule, say ipbug.vm, it will not reload
ipbug/__init__.py. I've attached another version, which tries to do just
that and I'm using that version currently without problems.
However I think there must be an even less complicated version.
Clearing sys.modules and 'reimporting' the module like in the following
code seems to work ok.
====
import sys, bbutils.textproc.spellcheck
m=sys.modules.copy()
sys.modules.clear()
sys.modules['sys'] = sys
import bbutils.textproc.spellcheck
====
Currently I don't have the time to investigate this further, but in a
week or two, I'll have another look at this.
- Ralf
-------------- next part --------------
import sys
import __builtin__
builtin_import = None # will be set to __builtin__.__import__ by reload function
old_modules = {} # will be set to sys.modules by reload function
reloaded = {} # names of reloaded modules, uses same keys as sys.modules
def determineParentName(globals):
"""determine name of the module which has called the import statement"""
if not globals or not globals.has_key("__name__"):
return None
pname = globals['__name__']
if globals.has_key("__path__"):
return pname
if '.' in pname:
i = pname.rfind('.')
pname = pname[:i]
return pname
return None
def reloadModuleAndParents(name, globals=None):
"""for name='some.module.bar', reload some, some.module, some.module.bar.
Determines module, which has called the import statement, and prefers
module relative paths, i.e. 'import os' in module m, from m/__init__.py
imports m/os.py if it's there.
"""
global reloaded
def reloadByName(n):
if old_modules.has_key(n):
if reloaded.has_key(n):
return reloaded[n]
reloaded[mname]=1
sys.modules[mname] = old_modules[mname]
if mname != old_modules[mname].__name__:
# module changed it's name. otherwise dreload(xml.sax) fails.
print "Module changed name:", mname, old_modules[mname].__name__
return reloadModuleAndParents(old_modules[mname].__name__)
else:
print "Reloading", mname
return __builtin__.reload(old_modules[mname])
return None
mods = name.split(".")
parent = determineParentName(globals)
retval = None
for i in range(len(mods)):
mname = ".".join(mods[:i+1])
if parent:
relative = "%s.%s" % (parent, mname)
if old_modules.has_key(relative) and old_modules[relative]:
retval=reloadModuleAndParents(relative)
continue
retval=reloadByName(mname)
return retval
def my_import_hook(name, globals=None, locals=None, fromlist=None):
"""replacement for __builtin__.__import__
reloads module 'name' if it hasn't been already reloaded
and then calls original __builtin__.__import__.
"""
## if fromlist:
## print 'Importing', fromlist, 'from module', name
## else:
## print 'Importing module', name
reloadModuleAndParents(name, globals)
return builtin_import(name, globals, locals, fromlist)
# Replacement for reload()
def reload(module, exclude=['sys', '__builtin__', '__main__']):
"""Recursively reload all modules used in the given module. Optionally
takes a list of modules to exclude from reloading. The default exclude
list contains sys, __main__, and __builtin__, to prevent, e.g., resetting
display, exception, and io hooks.
"""
global builtin_import
global old_modules
global reloaded
reloaded = {}
old_modules = sys.modules.copy()
sys.modules.clear()
for ex in exclude+list(sys.builtin_module_names):
if old_modules.has_key(ex) and not sys.modules.has_key(ex):
print "EXCLUDING", ex
sys.modules[ex] = old_modules[ex]
reloaded[ex]=1
builtin_import = __builtin__.__import__
__builtin__.__import__ = my_import_hook
try:
return reloadModuleAndParents(module.__name__)
finally:
# restore old values
__builtin__.__import__ = builtin_import
for m in old_modules:
if not sys.modules.has_key(m):
sys.modules[m] = old_modules[m]
``````
``````
In [29]:
print(EmailReplyParser.parse_reply(mess))
``````
``````
---------------------------------------------------------------------------
NameError Traceback (most recent call last)
<ipython-input-29-671024fec275> in <module>()
----> 1 print(EmailReplyParser.parse_reply(mess))
NameError: name 'EmailReplyParser' is not defined
``````
``````
In [ ]:
print(mess)
``````
``````
In [ ]:
threads[85].get_leaves()[0].data
``````
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http://stackoverflow.com/questions/4935541/determining-if-a-leap-day-falls-between-two-days-with-db2-sql
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# Determining if a leap day falls between two days with DB2 SQL
I have a table with two dates, "Start_Date" and "End_Date". In DB2 SQL, is there a way to determine if a leap day falls between these two dates?
Thank you!
-
What version of DB2 and what platform is it running on? – Ian Bjorhovde Feb 8 '11 at 20:08
Sure, you can do this using some date math and the DAYS function, by comparing the number of days between the the start and end date to the number of days between the start date and end date when they've both been shifted by 1 year.
If the number of days between the two dates is the same in both cases, then no leap day has occurred. If the number of days differs, then there has been at least 1 leap day.
This expression will return the number of leap days:
``````select
( DAYS(end_date + 1 year) - DAYS(start_date + 1 year) ) -
( DAYS(end_date) - DAYS(start_date) )
from
sysibm.sysdummy1
``````
This should work as long as end_date >= start_date.
It's trivial to encapsulate this into a scalar User Defined Function.
-
This approach worked great! Thank you very much! – Netbrian Feb 11 '11 at 5:49
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https://www.jiskha.com/questions/85281/who-will-hear-the-voice-of-a-singer-first-a-person-in-the-balcony-50-0-m-away-from-the
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# Physics
Who will hear the voice of a singer first- a person in the balcony 50.0 m away from the stage, or a person 3000 km away at home whose ear is next to the radio? How much sooner? Assume that the microphone is a few centimeters from the singer and the temperature is 20 degree Celcius.
1. 👍
2. 👎
3. 👁
1. Figure the time by using velocity (of light, and then sound).
1. 👍
2. 👎
👤
bobpursley
2. i didn't get u by figuring time by using velocity of sound.
What is the velocity of sound.
1. 👍
2. 👎
## Similar Questions
1. ### Literature
The novel To Kill a Mockingbird features a main character named Scout. Scout is also the voice that tells the story. What point of view is used in the novel? A. first person B. third-person limited
2. ### PE
Multiple Choice The way you communicate with others—how you say things and what you say—plays a role in how messages are received by the other person. Which type of communication uses gestures and tone of voice that can seem
3. ### identify tenses
1. The dou had been given the second best time slot. a. past perfect in the passive voice b. past continuous in the passive voice c. an infinitive in the passive voice Answer a 2. The promotional poster was designed by Zach a.
4. ### ENGLISH - GRAMMAR ????
Do you know how cheese is made? - active voice. A new airport will soon be built near the city. - passive voice. Absolute silence should be maintained. - passive voice.. Are the above are correct?
1. ### Lang Arts
The novel to kill a mockingbird feature a main character named scout. Scout is also the voice that tells the story. What point of view is used in the novel? A.First Person B.Third-Person Limited
2. ### science
If you shout into the Grand Canyon,your voice travels at the speed of sound(340 m/s) to the bottom of the canyon and back,and you hear and echo.How deep is the canyon at a spot where you can hear your echo 5.2 seconds after you
Hello! Can someone please help me out for this question? The loudness, L, of a sound in decibels can be calculated using the formula L= 10 log(I/ F) where I is the intensity of the sound in watts per square metre and F= 10^ -12 .
4. ### language arts 7
language arts 7 A/3.heritage /3.10. Heritage Unit Test Based on the repetition in "Twelfth Song of Thunder," the voice A. coming from above is more important than the voice coming from below. B. coming from below is more important
1. ### LA - Voices
Which statement is true about the passive voice? A verb is in the passive voice when the subject performs the action/ It is generally considered better style for writers to use the passive voice. A passive verb is made from a form
2. ### i need help!!!!
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When is it appropriate to use the passive voice?(1 point) It is never appropriate to use the passive voice. when the person or thing performing the action comes before the verb**** when the person or thing performing the action is
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https://rdrr.io/cran/PracTools/man/nAuditAttr.html
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# nAuditAttr: Sample sizes for an attribute sample in an audit In PracTools: Designing and Weighting Survey Samples
nAuditAttr R Documentation
## Sample sizes for an attribute sample in an audit
### Description
Compute a sample size for an audit where the goal is to control the probability of observing only a small number of errors given an underlying error rate in the population. Auditors refer to this as an attribute sample.
### Usage
nAuditAttr(TolRate=0.05, AccDev, CL, N=5000)
### Arguments
TolRate Proportion of units in the population with an attribute, e.g., errors in an audit. Auditors term this the 'tolerable rate of deviation' in the population to be tested. AccDev Acceptable deviation, which is the number of units with the attribute (i.e., the number of errors) that would be acceptable in the sample. The largest proportion of errors that would be deemed to be acceptable in an audit would be AccDev/N. CL Probability that the sample will contain an acceptable number of errors. Auditors refer to this as 'confidence level'. The probability that the sample will contain AccDev errors or fewer is 1-CL. N Size of the population of records to be audited.
### Details
nAuditAttr computes the minimum sample size required so that the probability, 1-CL of detecting less than or equal to a specified number of errors in the sample, is controlled. The sample is assumed to be selected with equal probabilities. AccDev is the largest number of errors in the sample that will be considered as meeting the audit standards. TolRate is the underlying population error rate, which is typically set to be larger than AccDev/N. The sample size is computed in two ways: (1) using the hypergeometric distribution, which accounts for the size of the population and (2) with the binomial distribution, which will be appropriate if the population is very large. When N is large and the sampling fraction is small, both sample sizes will be approximately the same.
### Value
List object with values:
Pop.Size population size Tol.Dev.Rate proportion of records with errors in population Acceptable.Errors largest number of errors, found in the sample, that will meet audit standards Sample.Size.Hypergeometric minimum sample size needed to detect AccDev errors in the sample computed via the hypergeometric distribution Sample.Size.Binomial minimum sample size needed to detect AccDev errors in the sample computed via the binomial distribution
### Author(s)
George Zipf, Richard Valliant
### References
GAO (2020). Financial Audit Manual, Volume 1, section 450.08. Washington DC; https://www.gao.gov/assets/gao-18-601g.pdf
Stewart, Trevor R. (2012). Technical Notes on the AICPA Audit Guide: Audit Sampling. American Institute of Certified Public Accountants, Inc. New York, NY 10036-8775; https://us.aicpa.org/content/dam/aicpa/publications/accountingauditing/keytopics/downloadabledocuments/sampling_guide_technical_notes.pdf
### Examples
# Examples from the US GAO Financial Audit Manual (2020), Figure 450.1, Table 1
nAuditAttr(AccDev = 0, CL = .90)
nAuditAttr(AccDev = 1, CL = .90)
nAuditAttr(AccDev = 2, CL = .90)
nAuditAttr(AccDev = 3, CL = .90)
nAuditAttr(AccDev = 4, CL = .90)
PracTools documentation built on May 29, 2024, 12:07 p.m.
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# A Brief Summary of Independent Set in Graph Theory
### Graph Basics
Let G be a undirected graph. G=(V,E), where V is a set of vertices and E is a set of edges. Every edge e in E consists of two vertices in V of G. It is said to connect, join, or link the two vertices (or end points).
### Independent Set
An independent set S is a subset of V in G such that no two vertices in S are adjacent. I suppose that its name is meaning that vertices in an independent set S is independent on a set of edges in a graph G. Like other vertex sets in graph theory, independent set has maximal and maximum sets as follows:
The independent set S is maximal if S is not a proper subset of any independent set of G.
The independent set S is maximum if there is no other independent set has more vertices than S.
That is, a largest maximal independent set is called a maximum independent set. The maximum independent set problem is an NP-hard optimization problem.
All graphs has independent sets. For a graph G having a maximum independent set, the independence number α(G) is determined by the cardinality of a maximum independent set.
### Relations to Dominating Sets
• A dominating set in a graph G is a subset D of V such that every vertex not in D is joined to at least one member of D by some edge.
• In other words, a vertex set D is a dominating set in G if and if only every vertex in a graph G is contained in (or is adjacent to) a vertex in D.
• Every maximal independent set S of vertices in a simple graph G has the property that every vertex of the graph either is contained in S or is adjacent to a vertex in S.
• That is, an independent set is a dominating set if and if only it is a maximal independent set.
### Relations to Graph Coloring
• Independent set problem is related to coloring problem since vertices in an independent set can have the same color.
# Hadoop RPC를 이용한 서버/클라이언트 구현
Hadoop은 이미 알려질대로 잘 알려진 분산 컴퓨팅 프레임워크입니다. 많은 사람들이 Hadoop 하면 MapReduce 프로그래밍을 주로 떠올리지만 자체적으로 제공하는 Hadoop RPC와 분산 파일 시스템인 HDFS를 가지고도 재미있는 것을 시도해 볼 수 있을 것 같습니다. 본 포스팅에서는 그 중에서 Hadoop RPC를 이용한 간단한 서버 클라이언트 프로그램의 구현방법을 소개합니다.
Hadoop RPC는 일반적으로 하나의 프로토콜 인터페이스(interface)와 하나의 Server 그리고 하나 이상의 Client(들)로 동작합니다. Hadoop RPC 서버의 인스턴스와 클라이언트 프록시의 인스턴스는 org.apache.hadoop.ipc.RPC 라는 클래스를 통해 얻을 수 있는데 내부적으로는 java reflection을 통해 구현되어 있습니다. 그리고 RPC method의 파라메터와 리턴 값은 오직 자바 primitive type들(예: int, long, String 등등)과 Writable 인터페이스를 구현한 구상클래스만 될 수 있습니다. 또한 Hadoop RPC는 자체적으로 서버와 클라이언트에 대한 기본적인 기능을 제공합니다. 따라서 복잡하게 스레드나 소켓 통신을 직접 구현할 필요가 없으며 개발자는 오로지 RPC 프로토콜 인터페이스와 RPC 메소드들에 대한 내용만 채워 넣으면 됩니다.
### Implementation of RPC Protocol
RPC Protocol은 인터페이스로 정의되어야 하며 이 인터페이스는 org.apache.hadoop.ipc.VersionedProtocol을 상속하여야 합니다. VersionedProtocol은 자체적으로 getProtocolVersion() 메소드를 가지고 있는데 이 메소드는 프로토콜의 버전이 다양할 경우 서버-클라이언트가 다른 버전의 프로토콜로 통신하는 것을 방지하는 역할을 합니다.
RPC 프로토콜은 다음 예제와 같이 간단히 만들 수 있습니다. 아래 예제는 String 값을 반환하는 heartBeat()라는 하나의 RPC 메소드를 가진 RPC 프로토콜 인터페이스입니다.
```import org.apache.hadoop.ipc.VersionedProtocol;
public interface RPCProtocol extends VersionedProtocol {
public long versionID=0;
public String heartBeat() throws IOException;
}
```
### Implementation of RPC Server
위에서 설명한 RPC 프로토콜의 서버 역할을 할 구상 클래스를 구현합니다. 서버 클래스는 간단히 위에서 정의한 RPCProtocol 인터페이스를 implements 하면 됩니다 (아래 예제 참조).
```import java.io.IOException;
public class TestServer implements RPCProtocol {
@Override
public String heartBeat() throws IOException {
return "Hello";
}
@Override
public long getProtocolVersion(String arg0, long arg1) throws IOException {
return 0;
}
/**
* @param args
* @throws IOException
* @throws InterruptedException
*/
public static void main(String[] args) throws IOException, InterruptedException {
TestServer s = new TestServer();
Configuration conf = new Configuration();
Server server = RPC.getServer(s, "localhost", 10000, conf);
server.start();
server.join();
}
}
```
RPCProtocol 인터페이스에서 정의했던 String heartBeat() 메소드 역시 구현되어 있습니다. 반환 값으로 “Hello”가 호출한 RPC 클라이언트에게 전달 될 것입니다.
서버의 시동은 main 메소드에 구현되어 있습니다. 우선 프로토콜의 구상클래스(TestServer)의 인스턴스를 생성하고 RPC.getServer()에 인자로 전달합니다. 또한 getServer 메소드는 추가적으로 서버가 binding할 IP와 port 번호를 인자로 받으며 Server 클래스의 인스턴스를 반환합니다(내부적으로는 TestServer 클래스의 인스턴스에 대한 Listener 스레드를 생성하여 파라메터로 전달된 IP 및 port 번호와 바인딩 시킵니다. 그리고 RPC 콜이 있을 때마다 TestServer의 메소드를 콜하게 됩니다. 처리 결과는 Responder 스레드를 통해 반환하게 됩니다).
RPC.getServer 메소드의 원형은 다음과 같습니다.
`static RPC.Server` `getServer(Object instance, String bindAddress, int port, Configuration conf)`
### Implementation of RPC Client
클라이언트는 RPC.waitForProxy 메소드를 통해서 간단히 얻을 수 있습니다. 그리고 클라이언트는 반환값으로 받은 proxy 인스턴스를 이용해서 손쉽게 RPC method를 콜하고 서버로부터 응답을 받아 올 수 있습니다.
`static VersionedProtocol` `getProxy(Class protocol, long clientVersion, InetSocketAddress addr, UserGroupInformation ticket,Configuration conf, SocketFactory factory)`
```import java.io.IOException;
public class TestClient {
/**
* @param args
* @throws IOException
* @throws InterruptedException
*/
public static void main(String[] args) throws IOException, InterruptedException {
Configuration conf = new Configuration();
RPCProtocol rpc = (RPCProtocol) RPC.waitForProxy(RPCProtocol.class,
String msg = null;
while(true) {
msg = rpc.heartBeat();
System.out.println(msg);
}
}
}
```
위 예제는 프록시 인스턴스 변수인 rpc를 통해 손쉽게 rpc.heartBeat() 메소드를 실행하고 서버로 부터 결과를 얻는 내용을 설명합니다.
### Test
서버를 먼저 실행하고 클라이언트를 실행하면 됩니다. 사실 순서를 바꿔 실행해도 크게 문제 되지 않습니다. Hadoop RPC의 클라이언트는 먼저 실행되었을 경우 RPC 서버에 접속이 될 때까지 1초 단위로 반복하여 접속 시도를 하게 됩니다.
정상적으로 수행되는 경우 다음과 같은 메시지를 확인할 수 있습니다.
```Hello
Hello
Hello
Hello
Hello
...```
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# Taylor Series
Topics: Taylor series, Function, Series Pages: 4 (706 words) Published: March 8, 2012
hp calculators
HP 50g Using Taylor Series
The menu LIMITS AND SERIES Practice using Taylor series
hp calculators HP 50g Using Taylor Series The menu LIMITS AND SERIES The menu LIMITS AND SERIES contains commands related to limits. To access it you press !Ö. You are presented then the calculus menu as a CHOOSE box:
Figure 1
Its second menu item is 2.LIMITS AND SERIES... You can use such CHOOSE boxes much like menus of computer applications. You can move the selection using the arrow keys. You can also jump to a certain menu item by typing the first few letters of the command or the number at the left of the menu item. Pressing the `key or the menu key %OK% will execute the selected item. In this CHOOSE box you press 2 to select 2.LIMITS AND SERIES.. and then ` or %OK% to display the menu:
Figure 2
The command DIVPC needs two polynomials and an integer. It returns the increasing power quotient of the two polynomials up to an order indicated by the integer. The command lim takes an algebraic object and an equation of the form variable=expression. It returns the limit of the algebraic expression when the given variable approaches the expression at the right hand side of the equation. The command SERIES needs an algebraic expression, and equation of the form variable=expression, and an integer. It returns a list at stack level 2 and an equation at stack level 1. The list contains 4 items: The limit of the expression when the given variable approaches the expression at the right hand side of the equation. The equivalent value expression at that point. The power expansion at that point. And finally the order of the residual at that point. The equation on stack level 1 is of the form h=variable-expression, where variable and expression are the same as in the equation variable=expression that we provided as argument for the command. The command TAYLOR0 performs a Maclaurin series expansion of an expression in the default independent variable, VX...
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A154822 Primes p of the form : p+p^2+p^3-+4=prime. 3
2161, 4951, 6421, 8761, 12241, 13411, 19891, 20731, 24631, 27271, 28411, 30091, 34981, 40471, 42331, 42901, 52021, 53731, 58111, 60631, 63361, 65701, 74611, 83641, 90841, 95101, 98251, 103171, 104851, 119671, 120871, 131731, 132661 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 MATHEMATICA lst={}; Do[p=Prime[n]; If[PrimeQ[p+p^2+p^3-4]&&PrimeQ[p+p^2+p^3+4], AppendTo[lst, p]], {n, 2*8!}]; lst Select[Prime[Range[15000]], AllTrue[#+#^2+#^3+{4, -4}, PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Mar 16 2015 *) CROSSREFS Cf. A154821 Sequence in context: A299796 A253715 A205331 * A061335 A159238 A289725 Adjacent sequences: A154819 A154820 A154821 * A154823 A154824 A154825 KEYWORD nonn AUTHOR Vladimir Joseph Stephan Orlovsky, Jan 15 2009 STATUS approved
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https://www.coursehero.com/file/18877/MAE-101-Quiz-2-solution/
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MAE_101-Quiz_2-solution
# MAE_101-Quiz_2-solution - Quiz 2 Solution 1 MAE 101 Statics...
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MAE 101. Statics and Strength of Materials © G. E. Orient 2006 1 Quiz 2 - Solution
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MAE 101. Statics and Strength of Materials © G. E. Orient 2006 2 From x = 0 to x = l 1 , the cross sectional area of the bar is A 1 = 1 in 2 . From x = l 1 to x = l 2 , A=A 2 x, A 2 =0.1 in . The modulus of elasticity of the material is E = 12 × 10 6 psi . There is a gap d = 0.02 in between the right end of the bar and the rigid wall. If an F=40 kip axial force toward the right is applied to the bar at x = l 1 , what is the resulting normal stress in the left half of the bar? l 1 =10 in l 2 =10 in Is the load sufficient to close the gap? Analysis of the two parts Free body diagrams, equilibrium (including the reaction force on the right end) Force-deflection Displacement compatibility Problem 1 in in E A Fl 02 . 0 033 . 0 1 1 = = δ F l 1 l 2 d F B A C R R R R R R R L E A l l R dx x E A R dx E x A R R l l l R l l l R R 2 1 2 2 2 ) / 1 ln( 1 2 1 1 2 1 1 + - = - = - = + + δ E A l R F R L 1 1 ) ( - = δ R L R L x R F R R F R F - = = - + - = 0 d E A l l R E A l R F d R R R L = + - - = + 2 1 2 1 1 ) /
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(b) What volume of chlorine gas (of density 3.17 g/L) will be produced from the reaction of 300 mL of .100M HCl(aq) with ⦠with amines. and volume of the container. The chlorine water indicated a density of 0.992720 g/ml while the salt water indicated a density of 0.99607 g/ml. In addition, the density of the gas is greater than that of air, causing it to remain near ground level and increasing exposure time. All Rights Reserved. Why don't libraries smell like bookstores? Density: 1.42 g/mL Melting point â22 °C (â8 °F; 251 K ... oxidation of vinylidene chloride, or the addition of chlorine to ketene. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool to achieve 1.00 ppm chlorine if the solution contains 6.00 percent chlorine by mass and there are 2.50 × 104 gallons of water in the pool. When did organ music become associated with baseball? It drops to 0.994 g/cm³ at 35 °C at the surface. Krypton: Density at 0° Celsius. How do you put grass into a personification? Is there a way to search all eBay sites for different countries at once? How long will the footprints on the moon last? In the case of Chlorine , density is 0.003214 g/cm³. Chlorine density is 3.16 g/L. Density: 0.567 g/cm 3 (liquid, critical density at 144 °C); 3.21 g/l (gas) at 0°C, 1013 ... 1 ml/m 3 = 2.947 mg/m 3: ORIGIN AND USE. Safety Information. What does struck out mean from the county court? DESCRIPTION (Chlorine): Liquid Chlorine is a clear amber color. Does pumpkin pie need to be refrigerated? It is shipped in steel cylinders under its own vapor pressure of about 85 psig @ 70° F. Common uses for Chlorine are water purification, bleaching and ⦠The density of chlorine at 0 0C and normal atmospheric pressure is 3.2 g/L. Question-1: The volume of chlorine at $$27^{o} C$$ is 564.3 mL. Solved Example for You. The trials for the salt-water pool averaged 11.05 seconds; for the chlorine water pool averaged 11.06 seconds or a difference of 0.01 seconds in favor of the salt water pool. Chlorine (Cl 2). All Rights Reserved. Xenon: Density at 0° Celsius. Calculate the mass of 125 ml of chlorine gas. What is chlorine's density in grams per milliliter or g mL. How long will the footprints on the moon last? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Copyright © 2020 Multiply Media, LLC. Chlorine weighs 0.002898 gram per cubic centimeter or 2.898 kilogram per cubic meter, i.e. Density of ice at 0 degrees Celsius is 916.8 grams per cubic centimeter or milliliter. The density of pure water at 4 °C at its maximum density is 1 g/cm³. Its density depends on pressure, temperature Now calculate the density of chlorine. Inter state form of sales tax income tax? But they were used earlier and we can find them in records. When did organ music become associated with baseball? It may be prepared from chloroacetic acid and ... acyl chloride easily forms esters and amides, while the other end of the molecule is able to form other linkages, e.g. However, there are interesting facts about Chlorine that most don't know about. Asked by Wiki User. Osmium is the densest naturally occurring element, with a density of 22.59 g/cm3. For example, water density at ⦠It has a mass of 1.607 g and torr is 740. Salt or NaCl has a density of 2.16 ish g/ml. Chlorine dioxide, liquid weighs 1.64 gram per cubic centimeter or 1 640 kilogram per cubic meter, i.e. Small amounts of chlorine gas can be generated in the laboratory from the reaction of Manganese(IV) oxide with hydrochloric acid: 4HCl(aq) + MnO2(s) = 2H2O(l) + MnCl2(s) + Cl2(g). Gaseous Chlorine is colorless with a strong odor. Note: Learn more about the density here. What form of id do you need 2 visit rikers island? (a) What mass of Cl2 can be produced from 42.7 g of MnO2 with an excess of HCl(aq)? What is the supporting tissue of the embryo? Copyright © 2020 Multiply Media, LLC. Assuming stp (standard temperature and pressure) the density of Chlorine is 0.0032 g/ml. The odor threshold for chlorine is approximately 0.3â0.5 parts per million (ppm). How do you put grass into a personification? Who is the longest reigning WWE Champion of all time? (Note: The density of air, free of moisture at the same conditions is 1.2929 kg/m 3) Specific Gravity of Cl 2 Liquid: The ratio of the density of saturated liquid chlorine to the density of water at its maximum density - 39°(4°C) 32°F (0°C) 1.467: Specific Heat: The heat required to raise the temperature of a unit weight of chlorine one degree. Is it normal to have the medicine come out your nose after a tonsillectomy? Who is the actress in the saint agur advert? 2 Argon: Density at 0° Celsius. How can creditor collect balance due after auction in Texas? Does pumpkin pie need to be refrigerated? What is chlorine's density in grams per milliliter or g mL? density of chlorine is equal to 2.898 kg/m³. Chlorine: Density given for Cl at 0° Celsius. Chlorine Cl2â¬â¬ P U R E G A S E S Marking ⬠CAS-Number 7782-50-5 Characterization acc. Chlorine gas (Cl2) - Chlorine is the chemical element with the formula Cl2 commonly used as disinfectant. Chlorine is used to disinfect swimming pools. Visit BYJU'S to understand the properties, structure and its uses. Increasing the pressure always increases the water density. The accepted concentration for this purpose is 1 ppm chlorine or 1 g of chlorine per million g of water. Water density changes with the change of the pressure and the temperature. Be the first to answer! What is chlorines density? Experimental Density: 0.842 g/mL / 20 °C Merck Millipore 801550: 0.844 g/mL / 4 °C Merck Millipore 801550: 0.851 g/mL Biosynth Q-200254: 0.846 g/mL Alfa Aesar A13004: 0.851 g/mL SynQuest 1100-5-22: 0.851 g/mL Sigma-Aldrich SIAL-19780: Miscellaneous. mol â1 : Appearance Colorless liquid: Density: 0.89 g/mL Melting point â123.1 °C (â189.6 °F; 150.1 K) Boiling point: 78 °C (172 °F; 351 K) density = mass / volume â mass = volume × density = 100 ml × 0.0032 g/ml = 0.32 g. Log in Ask Question I have got Sigma aldrich's sodium hypochlorite solution (Pdt. Chlorine is eliminated primarily in urine and feces, mainly (81% ⦠As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. ADR UNâ¬1017, Chlorine, 2.3 (8) Class 2, 2â¬TOC ⬠Cylinder Marking shoulder: yellow Essential properties ⬠Yellow-greenish, toxic, corrosive gas with characteristic odor, Who is the longest reigning WWE Champion of all time? Edit: m w's comments here: "start with these 2 equations.. PV = nRT.. n = mass / mw" gives the more general method. In Imperial or US customary measurement system, the density is equal to 0.1809 pound per cubic foot [lb/ft³], or 0.001675 ounce per cubic inch [oz/inch³]. Why don't libraries smell like bookstores? Chlorine is a chemical element with atomic number 17 which means there are 17 protons and 17 electrons in the ... mostly in platinum ores. The molar mass (that is, the mass of one mole) of Cl2 is 70.906 g/mol (IIRC) 70.906 g/mol divided by 22.414 L/mol = the standard density of Cl2 in g/L. CAS Number: Mixture Formula Weight: Mixture Formula: Mixture Hazard Info: Toxic, Corrosive Density (g/mL): 1.02 Boiling Point (°C): Decomposes Freezing Point (°C): 0 Solubility: Water Synonyms: Chlorine Test Solution Shelf Life (months): 36 Storage: WhiteChemicals for science education are available in easy-to-use formats with instructions for students to be introduced to a variety of subjects. $\ce{Hg (l) + Cl2 (g, l, or s) -> HgCl2 (s)}\nonumber$ The following table lists the halogens and their basic properties. Of 0.992720 g/ml while the salt water indicated a density of 1 g/ml mass. Is 740 and volume of the pressure and the temperature density is 0.003214 g/cm³ dioxide liquid... Used as disinfectant and normal atmospheric pressure is 3.2 g/L all eBay sites for countries! Ppm chlorine or 1 186 kilogram per cubic centimeter or 2.898 kilogram per cubic centimeter or milliliter weighs 1.186 per! Reigning WWE Champion of all time o } C\ ) is 564.3 mL region 9 Philippines a! Relay layout for a 1990 vw vanagon or any vw vanagon for matter! G mL do n't know about the only container readily available is a chlorine density g/ml... Question-1: the volume of the container salt water indicated a density ice! ( aq ) common to have the medicine come out your nose a. Structure and its uses are the famous writers in region 9 Philippines 916.8. Chlorine: density given for Cl at 0° Celsius odor threshold for chlorine is approximately 0.3â0.5 parts per (! The matter ) - chlorine is a 150-mL Erlenmeyer flask what mass of mL. Of the pressure and the temperature after auction in Texas saint agur advert medicine come out nose! Given for Cl at 0° Celsius is it normal to have the medicine come out your nose after a?... Dioxide, liquid weighs 1.186 gram per cubic meter, i.e which is g/ml... Liquid weighs 1.186 gram per millimeter ( g/ml ) Besides, there are interesting facts about chlorine most... Of 1.607 g and torr is 740 chlorine density g/ml moon last for chlorine is a 150-mL Erlenmeyer flask a way search... Chemical element with the change of the pressure and the temperature MnO2 with an excess of HCl aq! Visit BYJU 's to understand the properties, structure and its uses salt or NaCl has a mass Cl2! Parts per million ( ppm ) ( Pdt is preparing for an,... For a 1990 vw vanagon for the matter ( chlorine ): liquid chlorine is a 150-mL Erlenmeyer.... Byju 's to understand the properties, structure and its uses Characterization acc, the density of g/ml! Cl2¬¬ P U R E g a S E S Marking ⬠7782-50-5. Common to have the medicine come out your nose after a tonsillectomy or milliliter which we use rarely in.! Rarely in practice density changes with the formula Cl2 commonly used as disinfectant kilogram per cubic centimeter or.. Reigning WWE Champion of all time what is chlorine 's density in grams per or! Will the footprints on the moon last can find them in records of can. To 0.994 g/cm³ at 35 °C at the surface given for Cl at 0° Celsius the chemical element the. Chlorine at 0 0C and normal atmospheric pressure is 3.2 g/L BYJU 's to understand the,... The change of the container for a 1990 vw vanagon or any vanagon. Density at ⦠Calculate the mass of 1.607 g and torr is 740 27^ { o } C\ ) 564.3! 564.3 mL visit rikers island does struck out mean from the county court kilogram! The case of chlorine at \ ( 27^ { o } C\ ) is 564.3 mL amber.. The container from the county court of 0.99607 g/ml 0.3â0.5 parts per million g of MnO2 with an excess HCl! 0.3Â0.5 parts per million ( ppm ) ( ppm ) used earlier and we can find them in.! Of MnO2 with an excess of HCl ( aq ) degrees Celsius is 916.8 grams per cubic or. To understand the properties, structure and its uses, water density changes with the formula Cl2 used... With a density of 22.59 g/cm3 a density of 22.59 g/cm3 is it large enough to contain the acid the. ( ppm ) the density of chlorine gas 0C and normal atmospheric pressure is 3.2 g/L salt. Need to know if i convert mass to liters or density to.. 1 g/ml the case of chlorine, density is 0.003214 g/cm³ 1.607 g and torr is.... Of 125 mL of chlorine gas ( Cl2 ) - chlorine is 0.0032.. To search all eBay sites for different countries at once 0C and normal atmospheric pressure is 3.2 g/L ice! Element, with a density of 22.59 g/cm3 them in records units which use. 150-Ml Erlenmeyer flask due after auction in Texas 27^ { o } ). Saint agur advert are some imperial units which we use rarely in practice from the county court an experiment he! Ppm chlorine or 1 g of chlorine, density is 0.003214 g/cm³ have got Sigma aldrich 's sodium hypochlorite (. Poly bottle Safety & Documentation vw vanagon for the matter salt or NaCl has a density of 22.59 g/cm3 }... This purpose is 1 ppm chlorine or 1 g of water 27^ { o } C\ ) is mL... Who are the famous writers in region 9 Philippines a clear amber color at ⦠Calculate the mass of g... Is 0.0032 g/ml mass of Cl2 can be produced from 42.7 g of water of MnO2 with excess. Liquid chlorine is 0.0032 g/ml salt water indicated a density of which is 1.83 g/ml a what... Instructor is preparing for an experiment, he requires 225 g phosphoric.... Of 0.99607 g/ml available is a clear amber color ( aq ) 's to understand the properties structure. Experiment, he requires 225 g phosphoric acid 1.607 g and torr 740. S Marking ⬠CAS-Number 7782-50-5 Characterization acc excess of HCl ( aq ) the surface the matter 640 per! Chlorine: density given for Cl at 0° Celsius to understand the properties, structure and its.... Is it large enough to contain the acid, the density of at! Chlorine 's chlorine density g/ml in grams per milliliter or g mL 1.607 g and torr 740! 1 186 kilogram per cubic centimeter or 1 g of chlorine at \ ( 27^ { o C\! E S Marking ⬠CAS-Number 7782-50-5 Characterization acc ( chlorine ): liquid is. Is the actress in the case of chlorine gas a S E S Marking ⬠7782-50-5! We use rarely in practice is there a way to search all eBay for! 564.3 mL volume of chlorine at 0 0C and normal atmospheric pressure is g/L... Element, with a density of 0.992720 g/ml while the salt water indicated a density of 2.16 ish g/ml WWE. Champion of all time all eBay sites for different countries at once, i.e at... For different countries at once at \ ( 27^ { o } C\ is! What mass of 125 mL of chlorine at \ ( 27^ { o } C\ ) is 564.3.. Properties, structure and its uses actress in the saint agur advert assuming (. Saint agur advert do n't know about how long will the footprints on moon... 'S to understand the properties, structure and its uses facts about chlorine that most do n't about... 186 kilogram per cubic meter, i.e of HCl ( aq ) how to do everything chlorine density g/ml just need know. 150-Ml Erlenmeyer flask normal atmospheric pressure is 3.2 g/L g/cm³ at 35 °C at the surface chlorine cyanide liquid! Given for Cl at 0° Celsius naturally occurring element, with a density of 22.59 g/cm3 of g/ml... At \ ( 27^ { o } C\ ) is 564.3 mL 186 kilogram per cubic meter i.e... & Documentation produced from 42.7 g of chlorine per million g of is! To liters or density to milliliters 27^ { o } C\ ) is 564.3 mL 186! How to do everything i just need to know if i convert to! We can find them in records 640 kilogram per cubic centimeter or.... It has a mass of 125 mL of chlorine gas ( Cl2 ) - chlorine is approximately 0.3â0.5 parts million! All time is it large enough to contain the acid, the density of g/cm3... Kilogram per cubic meter, i.e a way to search all eBay sites for different at! With the formula Cl2 commonly used as disinfectant of which is 1.83 g/ml we can find them in records is... 0.003214 g/cm³ you need 2 visit rikers island S E S Marking ⬠CAS-Number 7782-50-5 Characterization acc 0.994 at! Element with the change of the pressure and the temperature amber color kilogram per cubic meter, i.e,! Is 0.0032 g/ml for Cl at 0° Celsius millimeter ( g/ml ) Besides, there some! Ish g/ml ⦠Calculate the mass of 1.607 g and torr is 740 per milliliter or g mL Cl 0°. And pressure ) the density of 0.992720 g/ml while the salt water indicated a density 0.99607. Cl2 commonly used as disinfectant the salt water indicated a density of chlorine, density is 0.003214 g/cm³ pressure temperature... Enough to contain the acid, the density of 1 g/ml aldrich 's sodium hypochlorite solution Pdt! However, there are interesting facts about chlorine that most do n't about! Way to search all eBay sites for different countries at once chlorine or 1 640 kilogram per cubic centimeter 2.898... Of MnO2 with an excess of HCl ( aq ) standard temperature and volume of the and. Is 916.8 grams per milliliter or g mL the footprints on the moon last understand the properties, structure its., with a density of 1 g/ml form of id do you need 2 visit rikers?.: the volume of the pressure and the temperature we use rarely in practice g/ml... It large enough to contain the acid, the density of 22.59 g/cm3: liquid chlorine is approximately parts. The salt water indicated a density of chlorine per million g of water S E S Marking ⬠CAS-Number Characterization. The acid, the density of which is 1.83 g/ml will the footprints on the moon last from the court. Visit rikers island long will the footprints on the moon last on moon!
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Công viên STEAM
# Chain Reaction
In this lesson, children will learn about cause and effect by creating chain reactions.
0-30 phút
Nâng cao
Mẫu giáo
## Connect
Show the children the inspiration photo and ask them to describe what they see, then tell them that it shows a model of a ride called Free Fall.
Tell the children that you are going to read a story about a
boy and a girl who were visiting STEAM Park.
Explain that the story will describe a chain reaction, or a sequence of events
that is caused by a trigger.
Read the following story aloud:
Matt and Sienna decided to ride Free Fall, the scariest ride in STEAM Park. They waited in line for only a few minutes and then stepped onto the platform. The machine pulled the rope until they were at the top of the tower.
“Wow! We are way up high!” Matt said.
“I’m so excited to feel my stomach tickle! I wonder when it’s going to drop us,” Sienna said.
They looked at the view of the park as they waited for the fall. Then, the lever holding the rope in place moved and released it. Matt and Sienna screamed and laughed as they dropped. The platform landed on another lever and raised a flag.
“That was the best ride ever!” Sienna said.
“Let’s go again!” Matt said.
Consider asking questions like:
• What caused the platform to drop?
• What happened next?
Explain that the trigger to the sequence of events in the story is that the lever moved and released the rope, which caused the platform to drop. When the platform landed, it caused another event to happen, the raising of the flag. Tell the children that this sequence of events is called a chain reaction.
## Construct
Ask the children to work in pairs to create a chain reaction. Remind them that one event should cause another event to happen.
Show them the inspiration photos for this lesson and ask them to think about how they could make an object move without touching it.
Tell them that they can build separate parts of the chain reaction and then put the model together and test it.
Tip:
You can assign which part each child or pair of children should build. The chain reaction triggers could include throwing the ball, shooting the dart from the cannon, or rolling the car down the ramp. The next part of the chain reaction could include knocking over a line of dominoes, making a gear move, or making the rocking element move.
## Contemplate
Ask the children to share their chain reactions with the rest of the group.
Consider asking questions like:
• What was the first cause or trigger in your chain reaction?
• What was the first event in your chain reaction?
• What was the last event in your chain reaction?
• Did your chain reaction turn out the way you predicted? Why or why not?
## Continue
Ask the children to combine their chain reactions to create one long chain reaction.
Assign a place in the classroom where they can assemble the long chain reaction, then ask them to take turns setting it off and making adjustments until it works.
Tip:
Have the children draw the chain reaction and number the events.
## Did you notice?
Observing the following skills can help you monitor whether the children are developing the necessary competencies in science, technology, engineering, art, and math.
• Identifying cause and effect relationships
• Using technology such as simple gears and wheels in appropriate ways
• Asking questions about science and technology related concepts
• Experimenting/testing “what would happen if” questions
• Observing and describing what happens
## Hỗ trợ giáo viên
Children will:
• Identify cause and effect
• Create their own chain reactions
Children are able to:
• Identify cause and effect relationships
• Use technology such as simple gears and wheels in appropriate ways
• Ask questions about science and technology related concepts
• Experiment/test “what would happen if” questions
• Observe and describe what happens
LEGO, the LEGO logo, the Minifigure, DUPLO, the SPIKE logo, MINDSTORMS and the MINDSTORMS logo are trademarks and/or copyrights of the LEGO Group. ©2023 The LEGO Group. All rights reserved. Use of this site signifies your agreement to the terms of use.
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# Pounds to Kilograms
Convert lbs to kg
Change to Kilograms to Pounds
Share:
## How to convert Pounds to Kilograms
1 [Pounds] = 0.45359237 [Kilograms]
[Kilograms] = [Pounds] / 2.2046226218488
To convert Pounds to Kilograms divide Pounds / 2.2046226218488.
## Example
44 Pounds to Kilograms
44 [lbs] / 2.2046226218488 = 19.95806428 [kg]
## Conversion table
Pounds Kilograms
0.01 lbs0.0045359237 kg
0.1 lbs0.045359237 kg
1 lbs0.45359237 kg
2 lbs0.90718474 kg
3 lbs1.36077711 kg
4 lbs1.81436948 kg
5 lbs2.26796185 kg
10 lbs4.5359237 kg
15 lbs6.80388555 kg
50 lbs22.6796185 kg
100 lbs45.359237 kg
500 lbs226.796185 kg
1000 lbs453.59237 kg
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# Help Needed Understanding Logs
• Oct 10th 2010, 03:29 PM
Help Needed Understanding Logs
I am requesting help understanding natural logs. I understand that if you have lets say
ln a = b
It can be rewritten as
a = e^b
What I would like to know is if there is any way I can simply a function with natural logs in it,
So if I had any function lets say
ln x = 10 - .5 ln A + .75 ln B - 1 ln C
Is there any way I could simplify that into normal terms without a long confusing jumble? If I had inputs, how do I go about solving a function like the above?
My understanding of natural logs is limited.
Thanks for any help.
• Oct 10th 2010, 03:35 PM
skeeter
go to the link and learn the laws of logarithms ... natural and otherwise.
Logarithms - Topics in precalculus
• Oct 10th 2010, 03:45 PM
Jhevon
Quote:
Originally Posted by bradm
I am requesting help understanding natural logs. I understand that if you have lets say
ln a = b
It can be rewritten as
a = e^b
What I would like to know is if there is any way I can simply a function with natural logs in it,
So if I had any function lets say
ln x = 10 - .5 ln A + .75 ln B - 1 ln C
Is there any way I could simplify that into normal terms without a long confusing jumble? If I had inputs, how do I go about solving a function like the above?
My understanding of natural logs is limited.
Thanks for any help.
Yes, you can simplify expressions like these using the laws of logarithms.
(i) $\displaystyle \dsiplaystyle \log_a b = c \implies a^c = b$
(ii) $\displaystyle \dsiplaystyle n \log_a X = \log_a (X^n)$
(iii) $\displaystyle \dsiplaystyle \log_a (XY) = \log_a X + \log_a Y$
(iv) $\displaystyle \dsiplaystyle \log_a \frac XY = \log_a X - \log_a Y$
(v) $\displaystyle \dsiplaystyle \log_a b = \frac {\log_c b}{\log_c a}$
Just a few of the rules that govern logs--the most important ones in my experience. Only some of these are needed to solve the problem you mentioned.
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# Speed up a data-dependent algorithm
Hey Everyone,
I’ve been working to parallelize an algorithm for quite some time now. I am finally done, but my results aren’t impressive - I’m only seeing a 2-3x speed up. I wonder if any of you might have any recommendations.
The algorithm multiplies a column of an NxN matrix by a vector of size N. Each column is dependent on the previous column, so col 2 must wait for col 3 to finish before col 2 can even start. As such, I’ve had to resort to calling my kernel in a for loop from host code. On a whim, I changed the kernel to incorporate a for loop just to see how much faster it would be, and the results were staggering (to me anyway) - 13-16x speedup. This produces incorrect results though, as cuda will execute blocks in any order it sees fit.
The second problem is that I can’t really use shared memory because it takes longer to load it, run the computations then write back to global memory, than just using global memory to begin with. (This is because it must be loaded for each iteration of the for loop in host code).
Anyway, here is the code for the kernel (cuDoubleComplex is a structure similar to the cuComplex structure, but with added position x(px) and position y(py) variables):
``````__global__ void L(cuDoubleComplex *a, cuDoubleComplex* b, cuDoubleComplex *c, int num, int N){
// num refers to the particular column of the matrix
// N = vector size
if(a[tid].px==num && a[tid].py==num){ // handle elements on the diagonal
if(num==0){ // hardcode special case where num==0
c[num] = a[num] * b[num]; // number on the diagonal for num=0
}else{
c[num] = a[tid] * b[num]; // number on the diagonal
}
}
if(a[tid].px==num && a[tid].py > num && a[tid].py < N){ // numbers below the diagonal
b[a[tid].py] = b[a[tid].py] + a[tid] * b[num];
}
}
``````
And I’m calling it like so:
``````for(ctr=0;ctr<N;ctr++) // call L kernel sequentially
L<<<numBlocks,nTL>>>(d_a, d_b, d_c, ctr, N);
``````
Isn’t ‘a’ supposed to be a matrix? What’s the typical size of N? what’s the numBlocks and nTL?
I think you’re probably doing things in a very very wrong manner. Also, I don’t understand why column 1 needs to be dependent on column 0.
If your columns have forward dependency, your rows are independent of each other right?
Maybe you can give a clearer explanation on what it is you are trying to do. A simple matrix multiplication certainly doesn’t have the kind of dependency you mentioned.
A typical size of N is between 6000 and 10000. The matrix is sparse, but we then factorize it to make it a couple times more dense. The calculation isn’t your typical matrix vector multiplication. The matrix is broken up into L and U along the main diagonal. L is column dependent, and then U is row dependent. It is an algorithm used in a power grid simulator to calculate things like line voltage adjustments. I’ve asked the engineers if there is any way around the dependencies and they have said no.
I should go further to explain that the matrix a, is extremely sparse and is in a compacted format. As such, ‘a’ is really just a large vector with x,y coordinates (px, py) and a complex number (vx, vy).
What I’ve done is calculate the number of elements, and then use that in correlation with the number of threads (nTL) to figure out how many blocks (numBlocks) will be needed to have one thread per matrix element. I’ve also found that by using a lower number of threads, I get a slight speed up. e.g. 512 threads vs 1024 (on my 570 fermi based card).
You could be right - I might be doing this in a very very wrong manner, which is why I was hoping for some advice. There are a lot of restrictions plaguing this algorithm, which make it hard to make a good design decision.
Using 512 threads per block gives more registers per thread. I’m surprised the kernel could use more than 32 registers. It could be some problem with the multiplication part.
If I did not understand your code wrongly, your algorithm has these characteristics:
Each element is gone through N times.
c is the only output, c[i] is only dependent on the ith diagonal value and b[i].
b[i] is dependent on all elements in the ith row as well as all the b values before the current b.
If I’m not wrong, then there is very little parallelism that could be used, especially when the matrix has high sparsity. Does your CPU code go through each element N times as well? If that’s the case you’d be better off optimising your CPU code. If you think your CPU code is highly efficient, then I guess the sparsity if the matrix is not as high as I imagined.
EDIT: Just looked at the algorithm again. Hope I’m not wrong:
c[i] = a[i, i] * b[i]
b[i] = sum(r = 0, to i-1) a[r, i]* b[r];
parallelism could be found only when i gets large. That is when the summation has slightly larger number of elements to work on.
Advise from the above analysis: group your elements by rows. Do not use GPU when for the first few hundreds of rows. Go through your elements in rows, not in columns. Each time only go through all elements in the same row. Go through each element only once. At each go, calculate b at the end. Calculate c after all the rows have been gone through.
On another thought, I realized the above advice may not be so good… maybe by expanding b[r] you could get more paralellism… I guess it’s still dependent on the sparsity of your matrix and you’ll have to try various methods to see which one gives better result.
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# Low Dimensional Topology
## October 7, 2016
### A bird’s eye view of topological recursion
Filed under: Quantum topology — dmoskovich @ 8:53 am
Eynard-Orantin Theory (topological recursion) has got to be one of the biggest ideas in quantum topology in recent years (see also HERE). Today I’d like to attempt a bird’s eye explanation of what all the excitement is about from the perspective of low-dimensional topology.
Mirror symmetry is a physical idea that relates two classes of problems:
1. A-Model: Measurement of a “volume” of a moduli space. In particular, counting the number of points of a moduli space that is a finite set of points.
2. B-Model: Computation of matrix integrals.
We may think of the A-model as “combinatorics and geometry” and of the B-model as “complex analysis”. Why might relating these classes of problems be important?
• Mirror symmetry might help us to compute a quantity of interest that we would not otherwise know how to compute. Sometimes enumeration may be simpler (e.g. the Argument Principle) and sometimes complex analysis may be simpler (when integrating by parts is easier than counting bijections).
• An object in one model may readily admit an interpretation, whereas its mirror dual’s meaning may be a mystery. This is the case in quantum topology- quantum invariants, which live on the B-model side, are powerful, but their topological meaning is a mystery. On the other hand, the A-model invariants (hyperbolic volume, A-polynomial) have readily understood geometric/topological meaning.
Mirror symmetry (as currently understood) doesn’t in-fact directly solve either problem, but it does provide heuristics. There is no known formula to compute the mirror dual problem to a given problem- mirror duals in mathematics have tended to be noticed post-facto. Mirror symmetry is also not mathematically rigourous, so each prediction of mirror symmetry must be carefully analyzed and proven. In addition, the mathematical meaning of mirror symmetry is unclear.
Despite this, quantum topology has received a number of Fields medals for work in and around mirror symmetry, including Jones (1990), Witten (1990), Kontsevich (1998), and Mirzakhani (2014). Several of our most celebrated conjectures, such as the AJ conjecture relating a quantum invariant to a classical invariant, stem from it.
Topological recursion observes that all known B-model duals of A-model problems can be framed in a common way (a holomorphic Lagrangian immersion of an open Riemann surface in the contangent bundle $T^\ast \mathbb{C}$ with some extra structure). This was observed first in special cases, and then it was noticed that the picture generalizes. Topological recursion thus reveals a common framework to all known mathematical examples of mirror symmetry. This simplifies B-model duals to A-model problems and places them in a common framework (a-priori they are complex integrals with a lot of variables without much else in common). It also provides tools to prove mirror duality in special cases. Explicitly, all of the information of an a-priori complicated mirror dual can be recovered from an embedded open Riemann surface (plus some extra structure), whose information is again encapsulated via an explicit formula in information in lower genus surfaces. Together with Mariño’s Remodeling Conjecture, we can say that topological recursion “tidies up” the B-model side of mirror symmetry, and elucidates what it means for something to be a “B-model dual” do an “A-model problem”.
One insight which topological recursion provides is that many of the simplest cases of mirror symmetry are Laplace transforms. Perhaps this is a window to understanding mirror symmetry itself? An vague conjecture along the lines of “in some contexts, mirror symmetry and the Laplace transform are the same thing in disguise” is given by Dumitrescu, Mulase, Safnuk, and Sorkin. For quantum topologists, another insight provided by topological recursion is that it suggests ways of reframing our favourite quantum invariants, such as the Jones polynomial, as objects which have more ready topological meaning, such as tau functions of integrable systems.
So, in conclusion, topological recursion provides a common framework for B-model objects such as quantum invariants. The hope is that this will elucidate their meaning and facilitate proving their mirror duality to better-understood mathematical objects. It does this by tidying up the B-model side into something structured which begins to look tractable.
Topological recursion has already led to several breakthroughs, including the simplest known proof of Witten’s conjecture and of Mirzakhani’s recurrence, and the subject is still in its infancy. It fits well with what we know by recovering all the “right” invariants at low orders (hyperbolic volume, analytic torsion) and hitting some heuristically expected keywords (e.g. $K_2$). Topological recursion is white-hot at the moment.
Disclaimer: I’m not an expert and some things I said might be wrong- please correct mistakes, inaccuracies, and omissions in the comments!
## 1 Comment »
1. Thanks for the brevfing, its nicely sintesized and clear. Perhaps you’ll be interested in the newer verzion of the Topological Recursion, “Airy Structures”. arXiv:1701.09137 and arXiv:1703.03307v2 I hope yo like it.
Comment by Helder Larraguivel — March 27, 2017 @ 3:35 am
Blog at WordPress.com.
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note - ch 7
# note - ch 7 - Econ 251 Fall 09 Handout 11 09/29/09 Chapter...
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Econ 251 Fall 09 Handout 11 09/29/09 1 Chapter 7 – Utility and Demand I. Utility = Consumer’s objective : Since our income is limited => ____________________________________ Simplified case : __________________ consumption choices must satisfy __________________________ Def. Budget Line The Budget Line is then: Real Income (in terms of good X) = Relative Price of good X (in terms of good Y) =
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Econ 251 Fall 09 Handout 11 09/29/09 2 Example : Nick has income I=\$20 and faces P D = \$1, P H =\$2 Nick’s budget constraint is: Graphically: Q: What happens to the budget line if the price of donuts increases , ceteris paribus? Suppose P D rises to \$2. New budget line: Graphically: b Any time you change the price of a good, the budget line ______________________
Econ 251 Fall 09 Handout 11 09/29/09 3 Q: What happens to the budget line if income decreases , ceteris paribus? Suppose P
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## This note was uploaded on 10/04/2011 for the course ECONOMICS 251 taught by Professor Kelly during the Spring '11 term at Purdue.
### Page1 / 7
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Ask a homework question - tutors are online
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Need Help Understanding This Code
• Question
• Hello, and thanks. Can someone help me out understanding the bold code? I know it is comparing sequential numbers, but I do not see how this correlates to descending counts.
```namespace CSharp1Exercises.Text
{
public class Strings
{
/// <summary>
/// Write a program and ask the user to enter a few numbers separated by a hyphen. Work out
/// if the numbers are consecutive. For example, if the input is "5-6-7-8-9" or "20-19-18-17-16",
/// display a message: "Consecutive"; otherwise, display "Not Consecutive".
/// </summary>
public void Exercise1()
{
Console.Write("Enter a few numbers (eg 1-2-3-4): ");
var numbers = new List<int>();
foreach (var number in input.Split('-'))
numbers.Sort();
var isConsecutive = true;
for (var i = 1; i < numbers.Count; i++)
{
if (numbers[i] != numbers[i - 1] + 1)
{
isConsecutive = false;
break;
}
}
var message = isConsecutive ? "Consecutive" : "Not Consecutive";
Console.WriteLine(message);
}
}
}```
Sunday, May 13, 2018 3:38 AM
• E.g.input= "5-6-7-9"
then numbers will be a list with numbers[0]=5,numbers[1]=6,numbers[2]=7,numbers[3]=9
list first needs to sorted
Now we have to check whether the list contains consecutive numbers
So logic is that numbers[i] should be equal to numbers[i-1]+1; we should check this for all elements in the list.
initially isConsecutive = true i.e. assume it to be Consecutive
i Condition to check-->numbers[i] != numbers[i - 1] + 1 Description
1 numbers[1]!=numbers[0]+1 6=5+1-->false-->continue
2 numbers[2]!=numbers[1]+1 7=6+1-->false-->continue
3 numbers[3]!=numbers[2]+1 9!=7+1--> true-->break
Here as soon as condition is matched we break the loop & set isConsecutive = false & hence numbers are not consecutive
You can test it step by step as
``` public void Exercise1()
{
Console.Write("Enter a few numbers (eg 1-2-3-4): ");
var numbers = new List<int>();
foreach (var number in input.Split('-'))
numbers.Sort();
var isConsecutive = true;
for (var i = 1; i < numbers.Count; i++)
{
Console.Write("{0} {1} {2}+1",numbers[i], numbers[i] == numbers[i - 1] + 1 ? "Equals" : "Not Equals",numbers[i - 1]);
if (numbers[i] != numbers[i - 1] + 1)
{
isConsecutive = false;
break;
}
}
var message = isConsecutive ? "Consecutive" : "Not Consecutive";
Console.WriteLine(message);
}```
Sunday, May 13, 2018 3:57 AM
• The Line
`Console.Write("{0} {1} {2}+1",numbers[i], numbers[i] == numbers[i - 1] + 1 ? "Equals" : "Not Equals",numbers[i - 1]);`
is used for getting step by step information
Here, string is formatted with arguments as
{0}=numbers[i]
{1}=(numbers[i] == numbers[i - 1] + 1 ? "Equals" : "Not Equals")
{2}=numbers[i - 1]
So we are displaying the info. in format such as 6 Equals 5+1 or like 9 Not Equals 7+1
Here 6, Equals & 5 are 3 arguments
Number of parameters must match with provided arguments i.e. each of {0}, {1}, {2} should be provided with an argument. (argument=actual value passed)
Sunday, May 13, 2018 5:22 AM
• If numbers were descending just change the condition like number[i]!=number[i-1]-1 as next number should be one less than previous
Check by following code:
``` public void Exercise1()
{
Console.Write("Enter a few numbers (eg 1-2-3-4): ");
var numbers = new List<int>();
foreach (var number in input.Split('-'))
//Sort Ascending
numbers.Sort();
//Reverse to get Descending
numbers.Reverse();
Console.Write("Sorted List:");
foreach (int item in numbers)
Console.Write(item + " ");
Console.WriteLine();
var isConsecutive = true;
for (var i = 1; i < numbers.Count; i++)
{
Console.Write("{0} {1} {2}-1\n", numbers[i], numbers[i] == numbers[i - 1] - 1 ? "Equals" : "Not Equals", numbers[i - 1]);
if (numbers[i] != numbers[i - 1] - 1)
{
isConsecutive = false;
break;
}
}
var message = isConsecutive ? "Consecutive" : "Not Consecutive";
Console.WriteLine(message);
}```
Sunday, May 13, 2018 8:08 AM
• Actually, I'm wondering if the intent of the exercise is to actually sort the input the user has entered it, or is it to determine if they have entered the numbers sequentially (so, without sorting)?
Your example shows sorting. Without sorting, you could check it like this:
```var isConsecutive = true;
var isAscending = true;
if (numbers.Length > 1 && numbers[0] > numbers[1])
isAscending = false;
for (var i = 1; i < numbers.Count; i++)
{
if (isAscending && numbers[i] != numbers[i - 1] + 1)
{
isConsecutive = false;
break;
}
else if (!isAscending && numbers[i] != numbers[i - 1] - 1)
{
isConsecutive = false;
break;
}
}
```
~~Bonnie DeWitt [C# MVP]
http://geek-goddess-bonnie.blogspot.com
Sunday, May 13, 2018 4:40 PM
• So, it seems the answer is that it doesn't matter the order of number entry. Only whether the series of numbers are consecutive. At least according to the answer provided.
Oh!! ... wait a minute ... does that mean that the code you initially posted was part of the exercise, and that it showed the answer to the exercise? If so, then forget everything I said. I misunderstood. I thought that the code you posted was your attempt at solving the exercise.
~~Bonnie DeWitt [C# MVP]
http://geek-goddess-bonnie.blogspot.com
Sunday, May 13, 2018 11:25 PM
All replies
• E.g.input= "5-6-7-9"
then numbers will be a list with numbers[0]=5,numbers[1]=6,numbers[2]=7,numbers[3]=9
list first needs to sorted
Now we have to check whether the list contains consecutive numbers
So logic is that numbers[i] should be equal to numbers[i-1]+1; we should check this for all elements in the list.
initially isConsecutive = true i.e. assume it to be Consecutive
i Condition to check-->numbers[i] != numbers[i - 1] + 1 Description
1 numbers[1]!=numbers[0]+1 6=5+1-->false-->continue
2 numbers[2]!=numbers[1]+1 7=6+1-->false-->continue
3 numbers[3]!=numbers[2]+1 9!=7+1--> true-->break
Here as soon as condition is matched we break the loop & set isConsecutive = false & hence numbers are not consecutive
You can test it step by step as
``` public void Exercise1()
{
Console.Write("Enter a few numbers (eg 1-2-3-4): ");
var numbers = new List<int>();
foreach (var number in input.Split('-'))
numbers.Sort();
var isConsecutive = true;
for (var i = 1; i < numbers.Count; i++)
{
Console.Write("{0} {1} {2}+1",numbers[i], numbers[i] == numbers[i - 1] + 1 ? "Equals" : "Not Equals",numbers[i - 1]);
if (numbers[i] != numbers[i - 1] + 1)
{
isConsecutive = false;
break;
}
}
var message = isConsecutive ? "Consecutive" : "Not Consecutive";
Console.WriteLine(message);
}```
Sunday, May 13, 2018 3:57 AM
• Thank you Vikram!
Sunday, May 13, 2018 4:04 AM
• Vikram, why is the last part necessary?
Console.Write("{0} {1} {2}+1", numbers[i], numbers[i] == numbers[i - 1] + 1 ? "Equals" : "Not Equals", numbers[i - 1]);
Sunday, May 13, 2018 4:25 AM
• The Line
`Console.Write("{0} {1} {2}+1",numbers[i], numbers[i] == numbers[i - 1] + 1 ? "Equals" : "Not Equals",numbers[i - 1]);`
is used for getting step by step information
Here, string is formatted with arguments as
{0}=numbers[i]
{1}=(numbers[i] == numbers[i - 1] + 1 ? "Equals" : "Not Equals")
{2}=numbers[i - 1]
So we are displaying the info. in format such as 6 Equals 5+1 or like 9 Not Equals 7+1
Here 6, Equals & 5 are 3 arguments
Number of parameters must match with provided arguments i.e. each of {0}, {1}, {2} should be provided with an argument. (argument=actual value passed)
Sunday, May 13, 2018 5:22 AM
• According to the logic description, if the numbers were descending, rather than ascending, the argument wont work, from what I see? if number[0] = 8, number[1]=7, number[2]=6, number[3] =5
then:
i Condition to check-->numbers[i] != numbers[i - 1] + 1 Description
1 numbers[1]!=numbers[0]+1 7=8+1 would be true right?
What am I missing?
Sunday, May 13, 2018 7:31 AM
• If numbers were descending just change the condition like number[i]!=number[i-1]-1 as next number should be one less than previous
Check by following code:
``` public void Exercise1()
{
Console.Write("Enter a few numbers (eg 1-2-3-4): ");
var numbers = new List<int>();
foreach (var number in input.Split('-'))
//Sort Ascending
numbers.Sort();
//Reverse to get Descending
numbers.Reverse();
Console.Write("Sorted List:");
foreach (int item in numbers)
Console.Write(item + " ");
Console.WriteLine();
var isConsecutive = true;
for (var i = 1; i < numbers.Count; i++)
{
Console.Write("{0} {1} {2}-1\n", numbers[i], numbers[i] == numbers[i - 1] - 1 ? "Equals" : "Not Equals", numbers[i - 1]);
if (numbers[i] != numbers[i - 1] - 1)
{
isConsecutive = false;
break;
}
}
var message = isConsecutive ? "Consecutive" : "Not Consecutive";
Console.WriteLine(message);
}```
Sunday, May 13, 2018 8:08 AM
• That's really kind of you to take the time Vikram. I will go over the iterations part of my beginner course again tonight and review what you provided!!
Sunday, May 13, 2018 10:22 AM
• Actually, I'm wondering if the intent of the exercise is to actually sort the input the user has entered it, or is it to determine if they have entered the numbers sequentially (so, without sorting)?
Your example shows sorting. Without sorting, you could check it like this:
```var isConsecutive = true;
var isAscending = true;
if (numbers.Length > 1 && numbers[0] > numbers[1])
isAscending = false;
for (var i = 1; i < numbers.Count; i++)
{
if (isAscending && numbers[i] != numbers[i - 1] + 1)
{
isConsecutive = false;
break;
}
else if (!isAscending && numbers[i] != numbers[i - 1] - 1)
{
isConsecutive = false;
break;
}
}
```
~~Bonnie DeWitt [C# MVP]
http://geek-goddess-bonnie.blogspot.com
Sunday, May 13, 2018 4:40 PM
• Thanks Bonnie, that's very interesting! I appreciate the alternative explanation. It shows me there are different ways to do the same thing. At first I got thrown by Sort. I missed its function. I couldn't understand why numbers entered in descending order were not causing isConsecutive to be false. In the inclusion of sort it makes only one mathematical parameter numbers[i] = numbers[i-1]+1 necessary right? I guess it is a simplification of "sorts".
Sunday, May 13, 2018 8:43 PM
• Thanks Bonnie, that's very interesting! I appreciate the alternative explanation. It shows me there are different ways to do the same thing.
Well, it's really not the same thing. It all depends on what the intent of the exercise is.
Are you supposed to determine "consecutiveness", based on the numbers *alone*?
-- Or --
Are you supposed to determine "consecutiveness", based on the numbers *as they were entered*?
The code I proposed will actually work for either scenario (although, as you said, with the sort you only need to check in one direction, ascending). But, keep in mind that these *are* two different interpretations of the exercise. I have no idea which is expected ... sorting or not.
At first I got thrown by Sort. I missed its function. I couldn't understand why numbers entered in descending order were not causing isConsecutive to be false. In the inclusion of sort it makes only one mathematical parameter numbers[i] = numbers[i-1]+1 necessary right? I guess it is a simplification of "sorts".
It sounds to me, from the summary comment, that numbers entered in descending order should also be considered for "consecutiveness".
~~Bonnie DeWitt [C# MVP]
http://geek-goddess-bonnie.blogspot.com
Sunday, May 13, 2018 9:16 PM
• The only instruction provided was as follows:
``` /// <summary>
/// Write a program and ask the user to enter a few numbers separated by a hyphen. Work out
/// if the numbers are consecutive. For example, if the input is "5-6-7-8-9" or "20-19-18-17-16",
/// display a message: "Consecutive"; otherwise, display "Not Consecutive".
/// </summary>```
So, it seems the answer is that it doesn't matter the order of number entry. Only whether the series of numbers are consecutive. At least according to the answer provided.
Sunday, May 13, 2018 9:21 PM
• The only instruction provided was as follows:
``` /// <summary>
/// Write a program and ask the user to enter a few numbers separated by a hyphen. Work out
/// if the numbers are consecutive. For example, if the input is "5-6-7-8-9" or "20-19-18-17-16",
/// display a message: "Consecutive"; otherwise, display "Not Consecutive".
/// </summary>```
So, it seems the answer is that it doesn't matter the order of number entry. Only whether the series of numbers are consecutive. At least according to the answer provided.
Right, and I was thinking that the comment *could* mean that "5-7-6-8-9" or "20-18-19-17-16" is *not* consecutive!!
~~Bonnie DeWitt [C# MVP]
http://geek-goddess-bonnie.blogspot.com
Sunday, May 13, 2018 11:20 PM
• So, it seems the answer is that it doesn't matter the order of number entry. Only whether the series of numbers are consecutive. At least according to the answer provided.
Oh!! ... wait a minute ... does that mean that the code you initially posted was part of the exercise, and that it showed the answer to the exercise? If so, then forget everything I said. I misunderstood. I thought that the code you posted was your attempt at solving the exercise.
~~Bonnie DeWitt [C# MVP]
http://geek-goddess-bonnie.blogspot.com
Sunday, May 13, 2018 11:25 PM
• You did point out something very important to understand! I should have stated that more clearly, but thanks a lot for your input. It really helps!!
Monday, May 14, 2018 2:56 AM
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Question:
Solve the simultaneous equation 3x + 4y = 24 4x + 3y = 22
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Jugal V.
here multipy eq 1 by 4 we get ----> 12x + 16y = 96 multiply eq 2 by 3 we get ---> 12x + 9y = 66 now subtract ecq 2 from eq 1 7y = 30 y = 30/7 therefore x = 16/7
Subject:Electrical Engineering
TutorMe
Question:
A 230V , 50 Hz supply is given to the RL circuit with R = 100 ohm and L = 1 H. Calculate the current through the circuit and the power consumed.
Inactive
Jugal V.
R = 100 ohm , $$X_{L}$$ = $$2\pi$$fL = 314.159 ohm The total impedance Z = 100 + 314.159 = 414.159 ohm $$\phi$$ = $$tan^-1(X_{L}$$/R) = 72.34 Current I = V/Z = 100/414.159 = 0.241 A P = VIcos$$\phi$$ = 230*0.241*0.3033 = 16.81 Watt
Subject:Algebra
TutorMe
Question:
If $$x^3 + Y^3 = 9$$ and x + y = 3 , Find the value of $$x^4 + y^4$$.
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Jugal V.
We know that $$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$ Therefore, 9 = 3$$(x^2 - xy + y^2 )$$ 3 = $$(x^2 + 2xy + y^2 - 3xy)$$ 3 = $$(x + y)^2 - 3xy$$ 3 = 9 - 3xy 2 = xy Therefore, substitute $$x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$$ = $$((x + y)^2 - 2xy))^2 - 2(xy)^2$$ = $$(9 -(2*2))^2 - 2(4)$$ = 25 - 8 = 17
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For discussion of other cellular automata.
Hunting
Posts: 3019
Joined: September 11th, 2017, 2:54 am
Denseflakes, glider is T(moves at c/2), have a common sparky sort-of-high-period oscillator, have TWO spiral oscillator(one statorless), BORDERLINE EXPLOSIVE. EDIT: 14o! evolves into 4 gliders + a large SL.
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
obo!``````
EDIT: It actually have TWO common sparky sort-of-high-period oscillator. One(the one I found first) is posted by Moosey below
p20:
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
7bo\$5b5o\$b3ob5ob3o\$b3ob5ob3o\$4ob5ob4o\$b3ob5ob3o\$b3ob5ob3o\$5b5o\$7bo!``````
(That's probably the longest RLE I have ever typed.)
The two spiral oscillators:
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
bobo\$3o\$bo!``````
Code: Select all
``````x = 5, y = 4, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
2bo\$b4o\$3o\$b2o!
``````
Relatives:
(EXPLOSIVE)This one have an almost sierpinski generator(Denseflakes w/o S8)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c
4o!``````
(TOO STABLE)p4 denseflakes with a p12(Denseflakes w/o S6a)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6c7c8
obo10b4o!``````
(WITH NATURAL RAKES!)p3 denseflakes. Bullet's a flipping oscillator. Show good behaviour when running symmetrical patterns. (Denseflakes w/o S4a)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4e5ai6ac7c8
obo!``````
(TOO STABLE)The denseflake dies while there's a p4 sparker. (Denseflakes w/o S3i)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3a4ae5ai6ac7c8
b5o\$7o\$b5o``````
(TOO STABLE)Denseflake dies. guess what the above p4 sparker will evolve into. (Denseflakes w/o S3a)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3i4ae5ai6ac7c8
b5o\$7o\$b5o``````
(TOO STABLE) Guess what the above p4 sparker will evolve into. (Denseflakes w/o S1e)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S2i3ai4ae5ai6ac7c8
b5o\$7o\$b5o``````
It have a sparky oscillator:
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S2i3ai4ae5ai6ac7c8
bbbbo\$ bbbbo\$ bbobo\$ booob\$ ooooo\$ oooob\$ bbo!``````
Last edited by Hunting on June 25th, 2019, 9:22 am, edited 1 time in total.
MLP will live on forever, so does John Horton Conway.
Moosey wrote:
February 5th, 2019, 7:51 pm
“New knightship tagalong!”
“Quick, hide it!”
My TODO list
LeapLife - DirtyLife - LispLife
I could make a rule in ten seconds flat
Moosey
Posts: 3286
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
Contact:
Denseflakes w/0 s8 has many rakes:
Code: Select all
``````x = 21, y = 11, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c
11bo\$9bob2o\$8b2obo\$9bo4\$18b2o\$b2o14b4o\$4o14b2o\$b2o!
``````
Original rule powerful sparker
Code: Select all
``````x = 6, y = 6, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
2bo\$b4o\$5o\$b5o\$b4o\$3bo!
``````
A near-explosive denseflakes relative.
Code: Select all
``````x = 13, y = 7, rule = B2ci3ai/S1e2i3ai4aen5ai6ac7c8
3bo5bo\$b4obob4o\$b4obob4o\$5obob5o\$b4obob4o\$b4obob4o\$3bo5bo!
``````
You can add all B4 transitions except B4a and non-symmetric patterns don’t explode unless they gain symmetry.
Here’s one which emits what appears to be a knightship but is actually not, in B4-aDenseFlakes
Code: Select all
``````x = 21, y = 2, rule = B2ci3ai4-a/S1e2i3ai4ae5ai6ac7c8
21o\$20bo!
``````
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Gamedziner
Posts: 796
Joined: May 30th, 2016, 8:47 pm
Location: Milky Way Galaxy: Planet Earth
Hunting wrote:(TOO STABLE)p4 denseflakes with a p12(Denseflakes w/o S6a)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6c7c8
obo10b4o!``````
(TOO STABLE)Denseflake dies. guess what the above p4 sparker will evolve into. (Denseflakes w/o S3a)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3i4ae5ai6ac7c8
b5o\$7o\$b5o``````
c/16 in former rule:
Code: Select all
``````x = 4, y = 6, rule = B2ci3ai/S1e2i3ai4ae5ai6c7c8
2b2o3\$bo\$3o\$3o!``````
p29 in latter rule:
Code: Select all
``````x=9, y=5, rule=B2ci3ai/S1e2i3i4ae5ai6ac7c8
4bo2\$o3bo3bo2\$4bo!``````
Code: Select all
``````x = 81, y = 96, rule = LifeHistory
58.2A\$58.2A3\$59.2A17.2A\$59.2A17.2A3\$79.2A\$79.2A2\$57.A\$56.A\$56.3A4\$27.
A\$27.A.A\$27.2A21\$3.2A\$3.2A2.2A\$7.2A18\$7.2A\$7.2A2.2A\$11.2A11\$2A\$2A2.2A
\$4.2A18\$4.2A\$4.2A2.2A\$8.2A!
``````
Hunting
Posts: 3019
Joined: September 11th, 2017, 2:54 am
Moosey wrote:Denseflakes w/0 s8 has many rakes:
Code: Select all
``````x = 21, y = 11, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c
11bo\$9bob2o\$8b2obo\$9bo4\$18b2o\$b2o14b4o\$4o14b2o\$b2o!
``````
Original rule powerful sparker
Code: Select all
``````x = 6, y = 6, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
2bo\$b4o\$5o\$b5o\$b4o\$3bo!
``````
A near-explosive denseflakes relative.
Code: Select all
``````x = 13, y = 7, rule = B2ci3ai/S1e2i3ai4aen5ai6ac7c8
3bo5bo\$b4obob4o\$b4obob4o\$5obob5o\$b4obob4o\$b4obob4o\$3bo5bo!
``````
You can add all B4 transitions except B4a and non-symmetric patterns don’t explode unless they gain symmetry.
Here’s one which emits what appears to be a knightship but is actually not, in B4-aDenseFlakes
Code: Select all
``````x = 21, y = 2, rule = B2ci3ai4-a/S1e2i3ai4ae5ai6ac7c8
21o\$20bo!
``````
Thanks! Those are impressive. Also I edited the original post to post another sparky oscillator.
Also, an SL seed:
Code: Select all
``````x = 21, y = 2, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
3ob3o!``````
Hunting
Posts: 3019
Joined: September 11th, 2017, 2:54 am
Gamedziner wrote:
Hunting wrote:(TOO STABLE)p4 denseflakes with a p12(Denseflakes w/o S6a)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6c7c8
obo10b4o!``````
(TOO STABLE)Denseflake dies. guess what the above p4 sparker will evolve into. (Denseflakes w/o S3a)
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3i4ae5ai6ac7c8
b5o\$7o\$b5o``````
c/16 in former rule:
Code: Select all
``````x = 4, y = 6, rule = B2ci3ai/S1e2i3ai4ae5ai6c7c8
2b2o3\$bo\$3o\$3o!``````
p29 in latter rule:
Code: Select all
``````x=9, y=5, rule=B2ci3ai/S1e2i3i4ae5ai6ac7c8
4bo2\$o3bo3bo2\$4bo!``````
Thanks too!
I really wish people to apgsearch Denseflakes, and its relatives.
MLP will live on forever, so does John Horton Conway.
Moosey wrote:
February 5th, 2019, 7:51 pm
“New knightship tagalong!”
“Quick, hide it!”
My TODO list
LeapLife - DirtyLife - LispLife
I could make a rule in ten seconds flat
Moosey
Posts: 3286
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
Contact:
Here’s a rule which may support conduits, orthogonoids, etc.
Code: Select all
``````x = 50, y = 5, rule = B2e3aikr4aet8/S1c2-i34y
15b2obo10b2obo8bo6b2o\$15bobo11bobo11bo5bo\$11bobobo2b2o5bobobo2b2o7b3o
4b2o\$4o8bob4obo6bobo2bobo\$o2bo7b2o12b2o2b2o!
``````
It has many natural SLs, large SLs may be unusually cheap.
Also, it has a natural puffer.
Code: Select all
``````x = 5, y = 5, rule = B2e3aikr4aet8/S1c2-i34y
ob3o\$bo2bo\$2o2bo\$2b2o\$2bobo!
``````
Misc salvo
Code: Select all
``````x = 99, y = 22, rule = B2e3aikr4aet8/S1c2-i34y
6b2o14b2o15b2o\$2o4bo15bo16bo\$2o4b2o9b2o3b2o15b2o\$17b2o16b2o\$35b2o10\$
97b2o\$74b2o21bo\$46b2o26bo22b2o\$2o22b2o4b2o7b2o5bo20b2o5b2o15b2o\$o11b2o
10bo5bo8bo6b2o19bo23bo\$bo10bo12bo4b2o8bo27bo23bo\$o11b2o10bo14bo27bo23b
o\$2o22b2o13b2o26b2o22b2o!
``````
Any reflectors?
Code: Select all
``````x = 10, y = 9, rule = B2e3aikr4aet8/S1c2-i34y
5b2ob2o\$2ob2o2bobo\$obo2bo\$3bo\$2bo\$2b2o2\$6b3o\$6bobo!
``````
EDIT:
There is a G-to-U reflector, no U reflectors or U-to-Gs or G-to-Gs:
(Yet)
Code: Select all
``````x = 10, y = 14, rule = B2e3aikr4aet8/S1c2-i34y
o\$2bo\$3o7\$4bobo\$5bo2b2o\$5b2o2bo\$8bo\$8b2o!
``````
Code: Select all
``````x = 11, y = 13, rule = B2e3aikr4aet8/S1c2-i34y
o\$2bo\$3o7\$4bobo\$5bo2b2o\$5b2o2bo\$8bobo!
``````
Gun that is a wick
Code: Select all
``````x = 23, y = 11, rule = B2e3aikr4aet8/S1c2-i34y
2o4b2o4b2o4b2o\$o5bo5bo5bo\$bob2o2bobo3bobo3bob2o\$obob3obobobobobobobobo
\$10bobo4bo\$4bobo3b3o\$4bobo\$5bo10bobo\$10bobo3b3o\$10bobo\$11bo!
``````
Amusing p18
Code: Select all
``````x = 9, y = 9, rule = B2e3aikr4aet8/S1c2-i34y
2o4bo\$bobo3b2o\$obobobobo\$5bo3\$7bo\$5bo\$5b2o!
``````
P9
Code: Select all
``````x = 9, y = 10, rule = B2e3aikr4aet8/S1c2-i34y
2o4bo\$bobo3b2o\$obobobobo\$5bo4\$6bobo\$7b2o\$6bo!
``````
Stretcher
Code: Select all
``````x = 8, y = 7, rule = B2e3aikr4aet8/S1c2-i34y
5b2o\$7bo\$5b2o\$2obo\$obob2o\$5bo\$4b2o!
``````
Tagalongs, Branching SSes:
Code: Select all
``````x = 61, y = 23, rule = B2e3aikr4aet8/S1c2-i34y
45bobobobobobo\$45b3ob3ob3o\$47bo3bobo2\$45bobo3bobo\$45b3obob3o\$49b2o\$50b
o2\$48bobobobo\$33bobobobo8b3ob3o\$33b3ob3o12bo\$35bobo\$50bobo5bobo\$18bobo
bobo8bobobobo10b3obobob3o\$18b3ob3o8b3ob3o13b2ob2o\$obo17bo14bobo15bo3bo
\$3obobobo\$3b2ob3o9bobo12bobobobo11bobo3bobo\$3bo14b3o12b3ob3o11b3o3b3o
2\$bobo\$b3o!
``````
Code: Select all
``````x = 59, y = 36, rule = B2e3aikr4aet8/S1c2-i34y
36bobobobobobobobobobobobo\$36b3ob3ob3ob3ob3ob3o\$38bo3bo3bo3bo3bobo2\$
38bobobobobobobobobobo\$38b3ob3ob3ob3ob3o\$8bobobobobobobobobobobobo9bo
3bo3bo3bobo\$8b3ob3ob3ob3ob3ob3o\$10bobo5bobo5bobo11bobobobobobobobo\$40b
3ob3ob3ob3o\$10bobobobobobobobobobo13bo3bo3bobo\$10b3ob3ob3ob3ob3o\$12bob
o5bo3bobo15bobobobobobo\$42b3ob3ob3o\$12bobobobobobobobo17bo3bobo\$12b3ob
3ob3ob2o\$14bobo5bobo19bobobobo\$44b3ob3o\$14bobobobobobo21bobo\$14b3ob3ob
3o\$16bo3bobo23bobo\$2bobo41b3o\$2b3obobobo5bobobobo\$5b2ob3o5b3ob3o\$5bo
12bobo\$bobo\$b3obobobo8bobo\$4b2ob3o8b3o\$4bo\$obo\$3obobobo\$3b2ob3o\$3bo2\$b
obo\$b3o!
``````
Code: Select all
``````x = 27, y = 26, rule = B2e3aikr4aet8/S1c2-i34y
obobobobobobobobobobobobobo\$3ob3ob3ob3ob3ob3ob3o\$2bobo3bo3bo5bobo3bo2\$
2bobobobobobobobobobobobo\$2b3ob3ob3ob3ob3ob3o\$4bobo3bo3bo5bobo2\$4bobob
obobobobobobobo\$4b3ob3ob3ob3ob3o\$6bobo3bo3bo3bo2\$6bobobobobobobobo\$6b
3ob3ob3ob3o\$8bobo5bobo2\$8bobobobobobo\$8b3ob3ob3o\$10bobo3bo2\$10bobobobo
\$10b3ob3o\$12bobo2\$12bobo\$12b3o!
``````
Growing ship:
Code: Select all
``````x = 8, y = 13, rule = B2e3aikr4aet8/S1c2-i34y
4b3o\$bo2bobo\$obo\$obo\$3bo\$2bo\$3bo\$2bo2b3o\$3bobobo\$2bo\$3bo\$2bo\$bo!
``````
Related rule with same tech (more or less) & another ship
Code: Select all
``````x = 24, y = 22, rule = B2e3aikr4aet8/S1c2-i34ety
22b2o\$22b2o7\$o\$2bo\$3o\$19bo\$17b5o\$17bobobo4\$4bobo\$5bo2b2o\$5b2o2bo\$8bo\$
8b2o!
``````
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Posts: 571
Joined: May 7th, 2016, 8:53 am
Contact:
Code: Select all
``````x = 8, y = 13, rule = B357/S23
2bo\$bob3o\$2o2b3o\$b4ob2o\$2b3o2bo4\$2b3o2bo\$b4ob2o\$2o2b3o\$bob3o\$2bo!
``````
FWKnightship
Posts: 451
Joined: June 23rd, 2019, 3:10 am
Location: 我不告诉你
Hunting wrote:Denseflakes, glider is T(moves at c/2), have a common sparky sort-of-high-period oscillator, have TWO spiral oscillator(one statorless), BORDERLINE EXPLOSIVE. EDIT: 14o! evolves into 4 gliders + a large SL.
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
obo!``````
EDIT: It actually have TWO common sparky sort-of-high-period oscillator. One(the one I found first) is posted by Moosey below
p20:
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
7bo\$5b5o\$b3ob5ob3o\$b3ob5ob3o\$4ob5ob4o\$b3ob5ob3o\$b3ob5ob3o\$5b5o\$7bo!``````
(That's probably the longest RLE I have ever typed.)
The two spiral oscillators:
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
bobo\$3o\$bo!``````
Code: Select all
``````x = 5, y = 4, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
2bo\$b4o\$3o\$b2o!
``````
p3,p4,p6,p8,p14:
Code: Select all
``````x = 115, y = 18, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
25b2o\$50b2o\$24b4o19b2ob2ob2o\$5b2o17b4o22b2o\$5b2o16b6o21b2o\$4b4o16b4o
16bo5b2o5bo13bo28bo10bo\$5b2o14bo3b2o3bo13bo5b2o5bo13bo6b2o18b5o6b5o\$2b
o6bo9b4o6b4o45b2o18b5o6b5o\$4o4b4o5bob5o4b5obo8b6o4b6o10b6o2b3obo15b7o
4b7o\$4o4b4o5bob5o4b5obo8b6o4b6o10b6o2b3obo16b6o4b6o\$2bo6bo9b4o6b4o45b
2o18b4o3b2o3b4o\$5b2o14bo3b2o3bo13bo5b2o5bo13bo6b2o20b2o3b2o3b2o\$4b4o
16b4o16bo5b2o5bo13bo33b2o\$5b2o16b6o21b2o53b2o\$5b2o17b4o22b2o\$24b4o19b
2ob2ob2o\$50b2o\$25b2o!
``````
c/2,2c/4,4c/8:
Code: Select all
``````x = 56, y = 28, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
5b2o15bo6bo\$4b4o13b3o4b3o8b3o10b3o\$3ob4ob3o31b3o2b3o\$5b2o17b4o12bo2b3o
2b3o2bo\$bo3b2o3bo13b4o14b4o2b4o\$2b2ob2ob2o14b4o14b4o2b4o\$25b2o14bo2bo
4bo2bo\$24b4o18b2o\$44bo4bo7\$21b3o4b3o2\$22bo2b2o2bo8b3o12b3o\$24b4o14b4o
2b4o\$24b4o11bo2b4o2b4o2bo\$25b2o14b12o\$24b4o13b12o\$44b2o2b2o5\$42bo8bo!
``````
EDIT:p2,p6,p17:
Code: Select all
``````x = 82, y = 43, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
34b2o\$73b4o\$4bo27b6o35b4o\$4bo27b6o32b2ob4ob2o\$34b2o34b2ob4ob2o\$3b3o67b
4o\$3b3o62b4o6b4o\$34b2o32b5ob2ob5o\$33b4o31b5ob2ob5o\$3b3o19b2o6b4o6b2o
23b4o6b4o\$3b3o19b2o4b2o4b2o4b2o28b4o\$23bob3o2b3o4b3o2b3obo23b2ob4ob2o\$
4bo18bob3o2b3o4b3o2b3obo23b2ob4ob2o\$4bo20b2o4b2o4b2o4b2o28b4o\$25b2o6b
4o6b2o28b4o\$33b4o\$34b2o2\$6b2o\$6b2obo24b2o\$5b3obo22b6o\$6b2o24b6o\$3bo\$b
4ob2obo24b2o\$b4ob2obo2\$2b2o2b2o5\$8bo\$6b5o\$6b5o\$5b7o\$6b5o\$3bo3b4o\$b4o3b
o\$b5o\$7o\$b5o\$b5o\$3bo!
``````
EDIT2:p4,p6,p8,p10,p14,p18:
Code: Select all
``````x = 153, y = 98, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
77b2o7b2o\$54b2o3b2o16b2o3bo3b2o48b3o\$7bo22b2o3b2o20bo18b5obob5o17bo29b
3o\$7bo25bo23bo19b4obob4o18bo29b3o\$28bo9bo12bo5bo5bo10bo3b3obob3o3bo15b
o\$6b3o19bo4bo4bo12bo11bo8b4o3bo5bo3b4o42b5o\$6b3o63b5o11b5o39b2ob5ob2o\$
29bobo3bobo14b3o5b3o11b4o9b4o10b3o5b3o18b4ob5ob4o\$5bobobo64b3o11b3o39b
4o3bo3b4o\$5bo3bo18bo4bo4bo12bo11bo\$7bo20bo9bo12bo5bo5bo9b4o11b4o14bo
24b2o9b2o\$6b3o24bo23bo48bo\$6b3o21b2o3b2o20bo16b3o11b3o15bo\$54b2o3b2o
13b4o9b4o\$7bo64b5o11b5o\$7bo64b4o3bo5bo3b4o\$74bo3b3obob3o3bo43bo5bo\$77b
4obob4o46bo5bo\$76b5obob5o\$77b2o3bo3b2o45b3o3b3o\$3bo73b2o7b2o45b3o3b3o\$
3bo2b2o2b2o94bo24b4o2bo2b4o\$103b2obob2o19bob4ob3ob4o\$bob2ob3ob2ob2o40b
5o40b2o4bo4b2o16bob2o3b3o3b2ob2o\$bob2ob3ob2o18bo5bo19b3o41b2o3b3o3b2o
24bo\$30bo2bo2bo19b3o46b3o19b3o15b3o\$3b2o5b2obo19bo22b3o40bo6bo6bo13b3o
7bo7b3o\$bob2o5b2obo15b3o3b3o13bo5bo5bo35bo13bo11b4o8bo8b4o\$bob2o5b2o
14b2ob3o3b3ob2o10b4o5b4o37b2o7b2o10b2ob4o17b4ob2o\$29b3o3b3o13b5o3b5o
34b6o5b6o10b2o21b2o\$3b2ob3ob2obo37b4o5b4o37b2o7b2o16b2o15b2o\$2ob2ob3ob
2obo13b2o9b2o11bo5bo5bo35bo13bo13b4ob2o7b2ob4o\$56b3o40bo6bo6bo14b2o15b
2o\$3b2o2b2o2bo17b3o3b3o18b3o46b3o17b2o21b2o\$11bo14b2ob3o3b3ob2o15b3o
41b2o3b3o3b2o9b2ob4o17b4ob2o\$29b3o3b3o17b5o40b2o4bo4b2o12b4o8bo8b4o\$
33bo69b2obob2o17b3o7bo7b3o\$30bo2bo2bo69bo20b3o15b3o\$30bo5bo100bo\$128b
2ob2o3b3o3b2obo\$131b4ob3ob4obo\$131b4o2bo2b4o\$57bo75b3o3b3o\$55b5o46bo
26b3o3b3o\$55b5o44b5o\$54b7o43b5o25bo5bo\$55b5o43b7o24bo5bo\$52bo3b3o3bo
41b5o\$50b4o7b4o\$50b5o5b5o40b3o\$33bo15b6o5b6o39b3o\$30b2obob2o13b5o5b5o
33bo7bo7bo\$27b2o9b2o10b4o7b4o31b4o13b4o\$27b2obo2bo2bob2o12bo3b3o3bo33b
4ob2o7b2ob4o\$32b3o20b5o35b5ob3o5b3ob5o\$26bobo3b3o3bobo13b7o35b4ob2o7b
2ob4o\$26bo13bo14b5o36b4o13b4o\$29b2ob3ob2o17b5o38bo7bo7bo\$25b2ob3ob3ob
3ob2o15bo47b3o\$29b2ob3ob2o67b3o\$26bo13bo\$26bobo3b3o3bobo63b5o\$32b3o68b
7o\$27b2obo2bo2bob2o64b5o\$27b2o9b2o64b5o\$30b2obob2o69bo\$33bo7\$102b2o5b
2o2\$101b4o3b4o\$101b4o3b4o\$102b2o2bo2b2o\$102b2ob3ob2o\$105b3o\$96b2o8bo8b
2o\$94bob4o13b4obo\$94bob4o6bo6b4obo\$96b2o17b2o\$99b2o11b2o\$98b4obo5bob4o
\$99b2o11b2o\$96b2o17b2o\$94bob4o6bo6b4obo\$94bob4o13b4obo\$96b2o8bo8b2o\$
105b3o\$102b2ob3ob2o\$102b2o2bo2b2o\$101b4o3b4o\$101b4o3b4o2\$102b2o5b2o!
``````
EDIT3:p2,p3,p4,p5,p6,p7,p8,p9,p15,p16,p17,p24,p27,p30,p34:
Code: Select all
``````x = 410, y = 121, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
378b2o2\$376b6o20bo6bo\$376b6o\$25bo352b2o\$23b5o21b3o117bo24bo210b2o\$6b2o
12b2ob5ob2o19bo26b2o2b2o25bo56b2o2bo24bo27b2o3bo3b2o23bo68bo23bo3bo\$3b
2o18b5o21b3o53b7o21bobo91bo28bo7b2o27b4o27b4o24b4o48b4o\$2ob2o2b2o9bob
3o5b3obo15b5o23b8o23b3o18b2o4bo4b2o23b3ob3o21b5o24b4o5b4o30b2o21b2o4b
4o4b2o20b5o20bo2b5o18bo8bo19b6o\$3b2o2b2ob2o6bob3o5b3obo16b3o21b2ob8ob
2o20b3o24bo29b3ob3o21b5o21b2ob4ob3ob4ob2o18b3o5b3obo16b2ob2ob2ob4ob2ob
2ob2o18b5o23b5o16bo10bo18b6o\$7b2o11b3o5b3o19bo22b2o2b6o2b2o21bo20b4obo
b4o25bo3bo19b2o3bo3b2o21b2o3b3o3b2o19bob3o6b2obo19b2o4b4o4b2o21b4o22bo
2b3o19bo6bo20b6o\$4b2o14b3o5b3o18b3o18b2ob3o2b4o2b3ob2o39b4o3b4o22b2o7b
2o14bob4o3b4obo19b2o9b2o19bob2o36b4o29bo22b3o2bo18b4ob2ob4o18b6o\$17b2o
b3o5b3ob2o39b2o4b2o4b2o21bo21b3o3b3o20b2ob4obob4ob2o11bob4obob4obo53b
2o6b3o82b5o22bo6bo20b6o\$23b5o45b2o10b2o21bo21b3o3b3o23b4o3b4o16b3o5b3o
62b4obo81b5o2bo16bo10bo18b6o\$20b2ob5ob2o77bo20b4o3b4o23b2o5b2o86b2o2b
4obo81b4o21bo8bo20b4o\$23b5o80bo20b4obob4o23b2ob3ob2o18b3o3b3o57b5o90bo
3bo\$25bo108bo31b3o21b4ob4o57b5o3b2o139b2o\$47b2o59bo19b2o4bo4b2o50b3ob
3o\$5b2o43b3o54b3o23bobo31bo23b3ob3o60bo66b3o50b2o\$5b2o42b3o24bo5bo24b
3o57bo90bo65b5o47b6o20bo6bo\$3bob3o39bob3o24bo5bo22b7o80bo2b2o127b5o47b
6o\$3bob2o40bob3o27bo28bo83bo131b5o\$9b2o42bo21b3obob3o241b3o50b2o\$5b2o
42b2o2bo21b3obob3o\$3bob3o68bo5bo\$3bob2o67bo9bo\$5b2o67bo9bo3\$326bo\$49b
2o28bo244b4o\$108bo215b5o\$73b2o9b2o20b5o212b5o\$8bo38bo4bo23b2obob2o23b
5o213b4o\$4b2o2bo38b6o21b4o3b4o20b7o213bo\$47b6o19bob5ob5obo18b7o\$3b3ob
3o37b6o19bob5ob5obo17bo7bo\$bob3ob3obo35bo4bo21b4o3b4o20bo2bo2bo\$bob2o
3b2obo64b2obob2o24b3o\$73b2o9b2o22bo\$3b7o39b2o\$3b7o69bo28bo\$108bo\$4b2o
2bo99bo\$8bo99bo2\$49bo58bo\$46bob3obo54b3o\$47b5o53bo2bo2bo\$6bo41b3o53bo
7bo\$6bo39bo5bo52b7o\$46bo2bo2bo52b7o\$5b3o41bo56b5o\$5b3o98b5o\$108bo2\$5b
3o\$48b2o\$48b2o\$5b3o39b4o\$48b3obo\$49b2obo\$5b3o\$49b2o\$48b3ob2o\$5b3o41b2o
\$46b2o2\$5b3o\$5b3o2\$6bo\$6bo45bo\$48b2o2bo2\$47b3ob3o\$45bob3ob3obo\$45bob2o
3b2obo2\$47b7o2\$45bob2o3b2obo\$45bob3ob3obo\$47b3ob3o2\$48b2o2bo\$52bo5\$47b
2o2b2o2\$46b8o\$46b8o\$43b2o2b6o2b2o\$43b3o3b2o3b3o\$40b2ob3obo4bob3ob2o\$
43b3o3b2o3b3o\$43b2o2b6o2b2o\$46b8o\$46b8o2\$47b2o2b2o6\$44bo2b2o2b2o2b2o\$
42b17o\$39b2ob17ob2o\$42b17o\$45bo3bo3bo3bo\$41bobo3bo3bo3bo3bo\$41bo3bo3bo
3bo3bobo\$43bo3bo3bo3bo\$42b17o\$39b2ob17ob2o\$42b17o\$44b2o2b2o2b2o2bo!
``````
Last edited by FWKnightship on July 1st, 2019, 2:08 am, edited 3 times in total.
I'm too shy to talk to other members.
But I want to upload my apgsearch results to Catagolue like others.
Code: Select all
``````x = 12, y = 1, rule = JvN29
FWKNIGHTSHIP!``````
Hunting
Posts: 3019
Joined: September 11th, 2017, 2:54 am
FWKnightship wrote:
Hunting wrote:Denseflakes, glider is T(moves at c/2), have a common sparky sort-of-high-period oscillator, have TWO spiral oscillator(one statorless), BORDERLINE EXPLOSIVE. EDIT: 14o! evolves into 4 gliders + a large SL.
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
obo!``````
EDIT: It actually have TWO common sparky sort-of-high-period oscillator. One(the one I found first) is posted by Moosey below
p20:
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
7bo\$5b5o\$b3ob5ob3o\$b3ob5ob3o\$4ob5ob4o\$b3ob5ob3o\$b3ob5ob3o\$5b5o\$7bo!``````
(That's probably the longest RLE I have ever typed.)
The two spiral oscillators:
Code: Select all
``````x=0, y=0, rule=B2ci3ai/S1e2i3ai4ae5ai6ac7c8
bobo\$3o\$bo!``````
Code: Select all
``````x = 5, y = 4, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
2bo\$b4o\$3o\$b2o!
``````
p3,p4,p6,p8,p14:
Code: Select all
``````x = 115, y = 18, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
25b2o\$50b2o\$24b4o19b2ob2ob2o\$5b2o17b4o22b2o\$5b2o16b6o21b2o\$4b4o16b4o
16bo5b2o5bo13bo28bo10bo\$5b2o14bo3b2o3bo13bo5b2o5bo13bo6b2o18b5o6b5o\$2b
o6bo9b4o6b4o45b2o18b5o6b5o\$4o4b4o5bob5o4b5obo8b6o4b6o10b6o2b3obo15b7o
4b7o\$4o4b4o5bob5o4b5obo8b6o4b6o10b6o2b3obo16b6o4b6o\$2bo6bo9b4o6b4o45b
2o18b4o3b2o3b4o\$5b2o14bo3b2o3bo13bo5b2o5bo13bo6b2o20b2o3b2o3b2o\$4b4o
16b4o16bo5b2o5bo13bo33b2o\$5b2o16b6o21b2o53b2o\$5b2o17b4o22b2o\$24b4o19b
2ob2ob2o\$50b2o\$25b2o!
``````
c/2,2c/4,4c/8:
Code: Select all
``````x = 56, y = 28, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
5b2o15bo6bo\$4b4o13b3o4b3o8b3o10b3o\$3ob4ob3o31b3o2b3o\$5b2o17b4o12bo2b3o
2b3o2bo\$bo3b2o3bo13b4o14b4o2b4o\$2b2ob2ob2o14b4o14b4o2b4o\$25b2o14bo2bo
4bo2bo\$24b4o18b2o\$44bo4bo7\$21b3o4b3o2\$22bo2b2o2bo8b3o12b3o\$24b4o14b4o
2b4o\$24b4o11bo2b4o2b4o2bo\$25b2o14b12o\$24b4o13b12o\$44b2o2b2o5\$42bo8bo!
``````
Its nice to see not all people forgot this rule. I have to say, impressive, especially the p14! We have p1, 3, 4, 6, 8, 14, 20, 24.
MLP will live on forever, so does John Horton Conway.
Moosey wrote:
February 5th, 2019, 7:51 pm
“New knightship tagalong!”
“Quick, hide it!”
My TODO list
LeapLife - DirtyLife - LispLife
I could make a rule in ten seconds flat
Moosey
Posts: 3286
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
Contact:
Here's a nice rule with a slope 18 (!) natural spaceship ((1,18)c/48)
Code: Select all
``````x = 3, y = 5, rule = B3-eqy4ceiqtz5cy6c8/S2-i3-a4i5q6a7e
2bo\$bo\$2o\$b2o\$2bo!
``````
Reminds me of some crazy puffer I found a while ago.
EDIT:
this
Code: Select all
``````x = 52, y = 49, rule = B3-ey4kt5e7c/S23-a4ityz
ob2o3bo5b3o2bo10bo3b3o3b5ob4o2bo\$5ob5ob3ob3obobo4b3o7bo5b6o\$bo2bob2ob
2o2b2ob2o8b2o7b4obobo5bo2bo\$3ob4o3b3o5b5o3b8o2b5o2b3ob4o\$b2obob6ob3obo
bo4bobob2o3bo2bo2b2obo3bob2o\$b4o5b4o3b2ob4o2b2o2bo2b2o3bob3o2bo2b2obo\$
bob7obobobo2b2o5b3ob2obob2o4bo3bobo4bo\$2ob3ob2ob4ob5o7bo7bobob2obobobo
bobo\$3ob3o2bob4ob4o6b5ob2ob2ob3o4b2ob2obo\$b3ob3o2b2o4bob3obob2obo2bo2b
o4b4o5b3obo\$4bob6o3b6obobo5bobo6bo2bo3bo4bo\$3o2bo2bo4b2o2b2o3b2o4b3ob
2o2b2o5bo2bobo\$b2obob4o2bob2o3b2o2b2obob6ob2ob7o2bob3o\$3b3o3b2obo3bo3b
5o4bobob4o7b2obo2b2o\$bo2bob3ob2obobo2b3o2bo4bob3o2b4o6bobo\$bo2bo4bo5bo
b3o3bob4ob3o3b2obobo4bo2b3o\$ob4obo3b2o2bobobob3o2bo4b2ob2obo4bobob3o\$o
b2ob5ob2ob3ob2o2bo2bo2b2obo2b3o3bob2ob4o2bo\$bob2ob3o7b3obo8bo3b2o2b3o
2bobo2bobo\$3obo2b3o2b2o2b2o2bo3bo4bo3b2o2b5obobobob2o\$o3b2ob3o2b2ob4o
2b3o4bo3bob2obob3o3b2o3bo\$2o6bo5b5o2b2o4b3o2b2o4b2ob2o2bobo3bo\$3o3b2ob
2ob4obob2obobo2bobo2bo2b2obo2bob3ob5o\$o3bo2bo4bo3b2obobo2bob2o2bo2b7ob
o2bob2o2b2o\$ob2o2bo4bobo2bo4b2obo2b2ob2o2bob3ob5obobob2o\$o4bobobo4b5ob
o3bo3bobobobo2b2ob2o2b2ob2obo\$bo5bob3ob2ob3obob2o6bobo2bob2ob2o3b5obo\$
2bo2bo2b4ob2o2b2ob3o2bob2o5b2o2b3ob3o2b3obo\$2bo2bo10bob2ob2obob3ob2o2b
ob3ob2o4b4o\$2b2o3b2obo3bo2b5obobo2bo6b2o4bo2b2ob5o\$2bo4bo2b3o3bo3b2obo
3bob4o3b2ob5obo2bob2o\$2bob11ob2o3bob3obob4ob3o3b2o2bo3b2obo\$3bobo2bobo
3b2o5b3o2bo4bobob3obo4bobo\$bo4b9ob2ob3ob3o4bobob4o2bo2bo2bob2o\$4o2b5o
2bobo3b2o3bo2b3o6b2obob4obo2bobo\$3b3obobobobobobob3o7bo3b2o4b4o2b2o2b
3o\$4b2o2b2ob2o2b3o5bobo3b3o2bo2bo2bo2b2o2b2o2bo\$b4o2bob3o3bob2o2bob6o
3b4obob2o3bobob4o\$o2bo3bo2bo3bo3b4o3b4ob2ob3obobob2ob3obo2bo\$o2bo2bo2b
2ob2ob4o3bobo3bo8b2obobo3bob3o\$2b4o4bobobob4obobob2o2b3ob2ob3o3bob4o2b
2o\$bobo3bo2b2ob4o2bobob3o5b2ob3obob2ob3o2b2o\$o5b2ob3obo3bob2obob4ob3ob
ob4ob2o3b2obo2bo\$o7bo3bobo2b3obo3bobo3b6ob3ob2ob2o2bobo\$3ob3ob4o4b3o6b
ob2obo2bo4bob2ob3o2bo\$3b2o3bobob3obo2b4obo2bobobobobo3bob4o3b2o\$b10ob
5o2bo2bob4ob2o7b2o3b4ob4o\$bobo2b3o5bo5b3obo2bobobo3b2obob2o2bo2b2o\$3o
2b2obobob5o3b3ob3obobobobo2b3o2bob2ob2obo!``````
EDIT:
this relative to the first rule in this post is cool:
Code: Select all
``````x = 6, y = 10, rule = B3-qy4ceiqtz5cy6c8/S2-i3-a4i5q6a7e
b3o\$bo2bo\$o3bo\$4o\$bo3\$4b2o\$3b3o\$4bo!
``````
The B goes the wrong way
Apgsearching now.
another ship: (normal ship for comparison)
Code: Select all
``````x = 75, y = 48, rule = B3-qy4ceiqtz5cy6c8/S2-i3-a4i5q6a7e
6bo3\$2bo3bo\$2obo\$b3o\$2bo7\$10bo\$5b3o50bo\$5bob2obo\$5bob2o\$54bo3bo\$52b2ob
o\$13b2o38b3o\$16bo37bo\$13bo2bo\$13b2o8\$21bo\$22bo45bo\$20b3o47bo\$19b2o46bo
bobo\$66b2o3bo\$70b2o\$67bobo\$68b2o7\$73bo\$74bo\$72b3o\$71b2o!
``````
There's also a 2c/4 which demands you ea1t your veggies (that was not a typo, that was its apgcode)
Code: Select all
``````x = 4, y = 5, rule = B3-qy4ceiqtz5cy6c8/S2-i3-a4i5q6a7e
2b2o\$2o\$o2bo\$2obo\$3bo!``````
EDIT:
somewhat related rule with at least 4 dirt-common natural ships (particularly the c/7, as the B demonstrates)
Code: Select all
``````x = 155, y = 31, rule = B3-qy4-a5cey6c7/S2-i3-a4it5q6a7e
129b2o\$131bo8bobo7bo2bo\$128bo2bo8bo2bo10bo\$131bo8bo2bo8bobo\$129b2o10b
3o7b4o13\$149bo\$150bo\$148b3o8\$2bo\$b2o\$2o\$b2o!
``````
The c/7's back spark is strong, if fairly inaccessible.
Code: Select all
``````x = 5, y = 14, rule = B3-qy4-a5cey6c7/S2-i3-a4it5q6a7e
b3o\$o3bo\$o3bo7\$b3o\$o2bo\$2b2o\$3bo\$o!
``````
distaff?
EDIT:
unrelated rule with a nice failedrep and a replicator too
Code: Select all
``````x = 90, y = 269, rule = B3-e4e/S2-i3-a4i5k
11bo\$10b3o\$9b2ob2o2\$11bo14\$35b2o19bo27bo\$36bo18b3o24b3o\$36bobo15b2ob2o
22bo\$37b2o42b2o\$56bo7\$42b2o42b2o\$43bo42bobo\$40b3o45bo\$40bo47b2o28\$24bo
\$23bobo\$24bo190\$2b3o\$2bo\$3bo8\$3o\$2bo\$3o2\$3o\$2bo\$3o!
``````
EDIT:
relative:
Code: Select all
``````x = 36, y = 43, rule = B3-e4e/S2-i3-a4i5k6kn
17b3o\$17bobo\$17bo\$18b2ob2o\$22bo\$19bo2bo\$17b2ob2o\$19bo8b2o\$15bo3bo9b2o\$
16bo2bo7bobo\$16b2obo7b2o2b3o\$16b2ob2o10bo2b2o\$17b3o11b2ob2o\$18bo9bo4b
3o\$28bo2bo\$26bo4bo\$26bobo\$26bo3bo\$27b3o\$27b3o11\$3bo7b2o\$bobobo5b3o\$2o
3bo6bo\$bo2b2o\$2b3o\$3bo3\$13bo13b2o\$25bob2o\$13bo10b2o\$25b2o3bobo\$27b4o!
``````
That oscillator is super common BTW; a B becomes a glider + that osc.
Code: Select all
``````x = 3, y = 4, rule = B3-e4e/S2-i3-a4i5k6kn
o\$2o\$b2o\$2o!
``````
Relative:
2 of these come from a B
Code: Select all
``````x = 10, y = 10, rule = B3-e4e7e/S2-i3-a4i5k6kn
b3o\$o2bo\$2obo4\$7b3o\$9bo\$7bobo\$7b2o!
``````
A rule which isn't B2a but behaves just like it:
Code: Select all
``````x = 19, y = 9, rule = B3-e4et7c/S2-i3-a4i5k6kn
16bo\$15b3o\$15bob2o3\$2bo\$b2o\$2o\$b2o!
``````
Imagine if this were a B synth
Code: Select all
``````x = 3, y = 4, rule = B3-e4e/S2-i3-a4i5k6kn8
o\$2o\$b2o\$2o!
``````
SMOG
EDIT:
another nice one
Code: Select all
``````x = 3, y = 4, rule = B3-e4e/S2-i3-a4i5kr6kn8
o\$2o\$b2o\$2o!
``````
Code: Select all
``````x = 3, y = 7, rule = B3-e4e/S2-i3-a4i5kr6kn8
3o\$o\$3o2\$3o\$o\$3o!
``````
Still retains the tub blowing up to the SL I call an "ocean" (if a tub is a smaller version of a pond... of course it is smaller than a lot of lakes, but...)
Code: Select all
``````x = 3, y = 3, rule = B3-e4e/S2-i3-a4i5kr6kn8
bo\$obo\$bo!
``````
Ocean layer:
Code: Select all
``````x = 21, y = 18, rule = B3-e4e/S2-i3-a4i5kr6kn8
18bo\$18b2o\$19b2o\$18b2o12\$2obo\$b3o\$2bo!
``````
EDIT:
I like this rule I found a while ago. (I lost it in my files but found it today)
It's three transitions from life, it's nonexplosive, and it has loads of 9c/20 pi-based technology, including (probably) every size of cordership starting with 2.
Code: Select all
``````x = 56, y = 117, rule = B3/S23-q4ij
53b2o\$53bobo\$55bo\$53bobo\$53b2o30\$11bo\$10bobo\$10bo2bo6b2o\$11b2o7bobo\$
11b2o9bo\$11b2o7bobo\$10bo2bo6b2o\$11b2o2\$40bo\$39bobo\$39bo2bo6b2o\$11b2o
27b2o7bobo\$10bo2bo6b2o18b2o9bo\$11b2o7bobo17b2o7bobo\$2o2b2o5b2o9bo16bo
2bo6b2o\$11b2o7bobo17b2o\$o4bo4bo2bo6b2o\$10bo2bo\$2o2b2o\$10bo2bo\$10bo29b
2o\$10bo28bo2bo6b2o\$10b2o28b2o7bobo\$11b2o27b2o9bo\$18b3o19b2o7bobo\$19b2o
18bo2bo6b2o\$18b3o18bobo\$10b3o27bo\$10b2o\$10bobo22\$6bo\$5bobo\$5bo2bo6b2o\$
6b2o7bobo\$17bo\$6b2o7bobo\$5bo2bo6b2o\$5bo2bo\$5bo4bo\$6bo2b3o\$8bo3bo\$9bo3b
o\$8b2o\$8bo3bo5b2o\$10bo7bobo\$9b2o9bo\$9b2o7bobo\$8bo2bo6b2o\$9b2o5\$9b2o\$8b
o2bo6b2o\$9b2o7bobo\$9b2o9bo\$9b2o7bobo\$8bo2bo6b2o\$8bobo\$9bo!
``````
Bunch of ships, some a là gustavo, but all discovered by hand.
Code: Select all
``````x = 339, y = 1300, rule = B3/S23-q4ij
280bo\$279bobo\$279bo2bo6b2o\$280b2o7bobo\$280b2o9bo\$280b2o7bobo\$279bo2bo
6b2o\$280b2o5\$280b2o\$279bo2bo6b2o\$280b2o7bobo\$269b2o2b2o5b2o9bo\$280b2o
7bobo\$269bo4bo4bo2bo6b2o\$279bo2bo\$269b2o2b2o\$279bo2bo\$279bo\$279bo\$279b
2o\$280b2o\$287b3o\$288b2o\$287b3o\$279b3o\$279b2o\$279bobo23\$273bo\$272bobo\$
271bo2bo6bo\$272bo7bobo\$280bo2bo6b2o\$258bo22b2o7bobo\$258bobo31bo\$258b2o
21b2o7bobo\$280bo2bo6b2o\$280bo2bo\$244bo35bo4bo\$244bobo34bo2b3o\$244b2o
37bo3bo\$284bo3bo\$283b2o\$230bo52bo3bo5b2o\$230bobo52bo7bobo\$230b2o63bo\$
284b2o7bobo\$283bobo7b2o\$216bo\$216bobo64b2o\$216b2o66bo\$267b2o\$232b2o16b
2o\$232bobo15bobo14bobo13bo\$233bobo15bobo15b2o11bobo\$220b2o12bo5bo11bo
5bo17bo5bo2bo6b2o\$220b2o18bo17bo17bo6b2o7bobo\$229b2o9bo17bo10b3o4bo17b
o\$224b2o3bobo36bob2o12bo7bobo\$225bo3bo38bo2bo10bo3bo5b2o\$204b2o16bo46b
3o10b2o\$204b2o16b2o45b2o12bo3bo\$243b2o22b3o12bo3bo\$243bobo20bo13bo2b3o
\$243bo22bobo10bo4bo\$279bo2bo\$279bo2bo6b2o\$257b2o21b2o7bobo\$257bobo31bo
\$257bo22b2o7bobo\$279bo2bo6b2o\$271bo7bobo\$270bo2bo6bo\$271bobo\$272bo31\$
272bo\$271bobo\$270bo2bo6bo\$271bo7bobo\$279bo2bo6b2o\$257bo22b2o7bobo\$257b
obo31bo\$257b2o21b2o7bobo\$279bo2bo6b2o\$279bo2bo\$243bo35bo4bo\$243bobo34b
o2b3o\$243b2o37bo3bo\$283bo3bo\$282b2o\$229bo52bo3bo5b2o\$229bobo52bo7bobo\$
229b2o52b2o9bo\$283b2o7bobo\$282bo2bo6b2o\$215bo67b2o\$215bobo\$216b2o\$216b
2o\$215bobo\$215bo67b2o\$282bo2bo6b2o\$283b2o7bobo\$229b2o52b2o9bo\$229bobo
52bo7bobo\$229bo52bo3bo5b2o\$282b2o\$283bo3bo\$243b2o37bo3bo\$243bobo34bo2b
3o\$243bo35bo4bo\$279bo2bo\$279bo2bo6b2o\$257b2o21b2o7bobo\$257bobo31bo\$
257bo22b2o7bobo\$279bo2bo6b2o\$271bo7bobo\$270bo2bo6bo\$271bobo\$272bo25\$
283bo\$282bobo\$282bo2bo6b2o\$283b2o7bobo\$283b2o9bo\$283b2o7bobo\$282bo2bo
6b2o\$283b2o5\$283b2o\$282bo2bo6b2o\$283b2o7bobo\$283b2o9bo\$283b2o7bobo\$
282bo2bo6b2o\$282bobo\$283bo12\$274bo\$273bobo\$272bo2bo6bo\$273bo7bobo\$281b
o2bo6b2o\$259bo22b2o7bobo\$259bobo31bo\$259b2o21b2o7bobo\$281bo2bo6b2o\$
281bo2bo\$245bo35bo4bo\$245bobo34bo2b3o\$245b2o37bo3bo\$285bo3bo\$284b2o\$
231bo52bo3bo5b2o\$231bobo52bo7bobo\$231b2o52b2o9bo\$285b2o7bobo\$284bo2bo
6b2o\$217bo67b2o\$217bobo\$217b2o3\$203bo81b2o\$203bobo78bo2bo6b2o\$203b2o
80b2o7bobo\$285b2o9bo\$285b2o7bobo\$189bo94bo2bo6b2o\$189bobo92bobo\$189b2o
94bo3\$187bo8bo8bo8bo8bo8bo8bo8bo\$186bobo6bobo6bobo6bobo6bobo6bobo6bobo
6bobo\$186bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo6b2o\$187b2o7b
2o7b2o7b2o7b2o7b2o7b2o7b2o7bobo\$261bo\$187b2o7b2o7b2o7b2o7b2o7b2o7b2o7b
2o7bobo\$186bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo6b2o\$186bobo
6bobo6bobo6bobo6bobo6bobo6bobo6bobo\$187bo8bo8bo8bo8bo8bo8bo8bo3\$189b2o
94bo\$189bobo92bobo\$189bo94bo2bo6b2o\$285b2o7bobo\$285b2o9bo\$203b2o80b2o
7bobo\$203bobo78bo2bo6b2o\$203bo81b2o3\$217b2o\$217bobo\$217bo67b2o\$284bo2b
o6b2o\$285b2o7bobo\$231b2o52b2o9bo\$231bobo52bo7bobo\$231bo52bo3bo5b2o\$
284b2o\$285bo3bo\$245b2o37bo3bo\$245bobo34bo2b3o\$245bo35bo4bo\$281bo2bo\$
281bo2bo6b2o\$259b2o21b2o7bobo\$259bobo31bo\$259bo22b2o7bobo\$281bo2bo6b2o
\$273bo7bobo\$272bo2bo6bo\$273bobo\$274bo34\$273bo\$272bobo\$271bo2bo6bo\$272b
o7bobo\$280bo2bo6b2o\$258bo22b2o7bobo\$258bobo31bo\$258b2o21b2o7bobo\$280bo
2bo6b2o\$280bo2bo\$244bo35bo4bo\$244bobo34bo2b3o\$244b2o37bo3bo\$284bo3bo\$
283b2o\$230bo52bo3bo5b2o\$230bobo52bo7bobo\$230b2o63bo\$285bo7bobo\$283bo3b
o5b2o\$216bo66b2o\$216bobo65bo3bo\$216b2o65bo3bo\$281bo2b3o\$280bo4bo\$202bo
77bo2bo\$202bobo75bo2bo6b2o\$202b2o77b2o7bobo\$292bo\$290bobo\$188bo93bo7b
2o\$188bobo93bo\$188b2o\$282bo\$286bo7bo\$284bobo6bobo\$285b2o6bo2bo6b2o\$
294b2o7bobo\$305bo\$294b2o7bobo\$187b2o104bo2bo6b2o\$187bobo103bo2bo\$187bo
105bo4bo\$294bo2b3o\$296bo3bo\$201b2o94bo3bo\$201bobo92b2o\$201bo94bo3bo5b
2o\$298bo7bobo\$297b2o9bo\$215b2o80b2o7bobo\$215bobo78bo2bo6b2o\$215bo81b2o
3\$229b2o\$229bobo\$229bo67b2o\$296bo2bo6b2o\$297b2o7bobo\$243b2o52b2o9bo\$
243bobo52bo7bobo\$243bo52bo3bo5b2o\$296b2o\$297bo3bo\$257b2o37bo3bo\$257bob
o34bo2b3o\$257bo35bo4bo\$293bo2bo\$293bo2bo6b2o\$271b2o21b2o7bobo\$271bobo
31bo\$271bo22b2o7bobo\$293bo2bo6b2o\$285bo7bobo\$284bo2bo6bo\$285bobo\$286bo
39\$296bo\$295bobo\$294bo2bo6bo\$295bo7bobo\$303bo2bo6b2o\$281bo22b2o7bobo\$
281bobo31bo\$281b2o21b2o7bobo\$303bo2bo6b2o\$303bo2bo\$267bo35bo4bo\$267bob
o34bo2b3o\$267b2o37bo3bo\$307bo3bo\$306b2o\$253bo52bo3bo5b2o\$253bobo52bo7b
obo\$253b2o52b2o9bo\$307b2o7bobo\$306bo2bo6b2o\$239bo67b2o\$239bobo\$239b2o
3\$225bo81b2o\$225bobo78bo2bo6b2o\$225b2o80b2o7bobo\$307b2o9bo\$307b2o7bobo
\$211bo94bo2bo6b2o\$211bobo92bobo\$211b2o94bo5\$211b2o94bo\$211bobo92bobo\$
211bo94bo2bo6b2o\$307b2o7bobo\$307b2o9bo\$225b2o80b2o7bobo\$225bobo78bo2bo
6b2o\$225bo81b2o3\$239b2o\$239bobo\$239bo67b2o\$306bo2bo6b2o\$307b2o7bobo\$
253b2o52b2o9bo\$253bobo52bo7bobo\$253bo52bo3bo5b2o\$306b2o\$307bo3bo\$267b
2o37bo3bo\$267bobo34bo2b3o\$267bo35bo4bo\$303bo2bo\$303bo2bo6b2o\$281b2o21b
2o7bobo\$281bobo31bo\$281bo22b2o7bobo\$303bo2bo6b2o\$295bo7bobo\$294bo2bo6b
o\$295bobo\$296bo46\$294bo\$293bobo\$292bo2bo6bo\$293bo7bobo\$301bo2bo6b2o\$
279bo22b2o7bobo\$279bobo31bo\$279b2o21b2o7bobo\$301bo2bo6b2o\$301bo2bo\$
265bo35bo4bo\$265bobo34bo2b3o\$265b2o37bo3bo\$305bo3bo\$304b2o\$251bo52bo3b
o5b2o\$251bobo52bo7bobo\$251b2o52b2o9bo\$305b2o7bobo\$304bo2bo6b2o\$237bo
67b2o\$237bobo\$237b2o3\$223bo81b2o\$223bobo78bo2bo6b2o\$223b2o80b2o7bobo\$
305b2o9bo\$305b2o7bobo\$209bo94bo2bo6b2o\$209bobo92bobo\$209b2o94bo3\$195bo
\$195bobo\$195b2o3\$64b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o
7b2o\$64b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o97bo\$
279bobo\$68b2o209bo2bo6b2o\$68bobo209b2o7bobo\$68bo211b2o9bo\$198b2o80b2o
7bobo\$198bobo78bo2bo6b2o\$82b2o114bo81b2o\$82bobo\$82bo\$212b2o\$212bobo\$
96b2o114bo67b2o\$96bobo180bo2bo6b2o\$96bo183b2o7bobo\$226b2o52b2o9bo\$226b
obo52bo7bobo\$110b2o114bo52bo3bo5b2o\$110bobo166b2o\$110bo169bo3bo\$240b2o
37bo3bo\$240bobo34bo2b3o\$124b2o114bo35bo4bo\$124bobo149bo2bo\$124bo151bo
2bo6b2o\$254b2o21b2o7bobo\$254bobo31bo\$138b2o114bo22b2o7bobo\$138bobo135b
o2bo6b2o\$138bo129bo7bobo\$267bo2bo6bo\$268bobo\$152b2o115bo\$152bobo\$152bo
3\$166b2o\$166bobo\$166bo3\$180b2o\$180bobo\$180bo3\$194b2o94bo\$194bobo92bobo
\$194bo94bo2bo6b2o\$290b2o7bobo\$290b2o9bo\$208b2o80b2o7bobo\$208bobo78bo2b
o6b2o\$208bo81b2o3\$222b2o\$222bobo\$222bo67b2o\$289bo2bo6b2o\$290b2o7bobo\$
236b2o52b2o9bo\$236bobo52bo7bobo\$236bo52bo3bo5b2o\$289b2o\$290bo3bo\$250b
2o37bo3bo\$250bobo34bo2b3o\$250bo35bo4bo\$286bo2bo\$286bo2bo6b2o\$264b2o21b
2o7bobo\$264bobo31bo\$264bo22b2o7bobo\$286bo2bo6b2o\$278bo7bobo\$277bo2bo6b
o\$278bobo\$279bo35\$289bo\$288bobo\$288bo2bo6b2o\$289b2o7bobo\$289b2o9bo\$
289b2o7bobo\$288bo2bo6b2o\$289b2o5\$289b2o\$276b4o8bo2bo6b2o\$280bo3bo4b2o
7bobo\$276bo3bo2b2o15bo\$278bo4bo14bobo\$276bobo4bo5b3o6b2o\$275bo2bo3bo6b
2obo\$229b2o7b2o7b2o7b2o7b2o8b2ob2obo10bo\$229b2o7b2o7b2o7b2o7b2o7b2o4bo
8bo2bo\$277bo3bo7bob2o\$277bobob2o5bo2b2o\$287b3ob2o\$276bo11bo2bo6b2o\$
298bobo\$235b2o63b2o\$234b2o62bobo\$236bo51bo9b2o\$288b2o\$289bo\$249b2o37bo
2b2o\$248b2o40bo\$250bo34b2o3b2o\$284b3o2bo\$285bo2bo6b2o\$263b2o21b2o7bobo
\$262b2o33b2o\$264bo21b2o7bobo\$285bo2bo6b2o\$285bobo\$276b2o8bo\$277bobo\$
278bo31\$235bo\$234bobo\$233bo2bo6bo\$234bo7bobo\$242bo2bo6b2o\$220bo22b2o7b
obo\$220bobo31bo\$220b2o21b2o7bobo\$242bo2bo6b2o\$242bo2bo\$206bo22bobo10bo
4bo\$206bobo20bo13bo2b3o\$206b2o22b3o12bo3bo\$131bobo15b2o16b2o16b2o45b2o
12bo3bo\$131bo3bo13b2o16b2o16bo46b3o10b2o\$136bo51bo3bo38bo2bo10bo3bo5b
2o\$133b2o52b2o3bobo36bob2o12bo7bobo\$192b2o9bo17bo10b3o4bo17bo\$165b2o
16b2o18bo17bo17bo6b2o7bobo\$165b2o16b2o12bo5bo11bo5bo17bo5bo2bo6b2o\$
196bobo15bobo15b2o11bobo\$195bobo15bobo14bobo13bo\$195b2o16b2o\$230b2o\$
162b2o\$162bobo\$162bo3\$176b2o\$176bobo\$176bo3\$190b2o94bo\$190bobo92bobo\$
190bo94bo2bo6b2o\$286b2o7bobo\$286b2o9bo\$204b2o80b2o7bobo\$204bobo78bo2bo
6b2o\$204bo81b2o3\$218b2o\$218bobo\$218bo67b2o\$285bo2bo6b2o\$286b2o7bobo\$
232b2o52b2o9bo\$232bobo52bo7bobo\$232bo52bo3bo5b2o\$285b2o\$286bo3bo\$246b
2o37bo3bo\$246bobo34bo2b3o\$246bo35bo4bo\$282bo2bo\$282bo2bo6b2o\$260b2o21b
2o7bobo\$260bobo31bo\$260bo22b2o7bobo\$282bo2bo6b2o\$274bo7bobo\$273bo2bo6b
o\$274bobo\$275bo31\$278bo\$277bobo\$276bo2bo6bo\$277bo7bobo\$285bo2bo6b2o\$
263bo22b2o7bobo\$263bobo31bo\$263b2o21b2o7bobo\$285bo2bo6b2o\$285bo2bo\$
249bo35bo4bo\$249bobo34bo2b3o\$249b2o37bo3bo\$289bo3bo\$161bo126b2o\$160bob
o72bo52bo3bo5b2o\$160bo2bo6b2o63bobo52bo7bobo\$161b2o7bobo62b2o52b2o9bo\$
161b2o9bo116b2o7bobo\$161b2o7bobo115bo2bo6b2o\$160bo2bo6b2o49bo67b2o\$
161b2o58bobo\$221b2o3\$207bo81b2o\$161b2o44bobo78bo2bo6b2o\$160bo2bo6b2o
35b2o80b2o7bobo\$161b2o7bobo116b2o9bo\$161b2o9bo116b2o7bobo\$161b2o7bobo
20bo94bo2bo6b2o\$160bo2bo6b2o21bobo92bobo\$160bobo30b2o94bo\$161bo\$176bo
8bo\$176bo8bo\$176bo8bo\$194b2o94bo\$194bobo92bobo\$194bo94bo2bo6b2o\$290b2o
7bobo\$290b2o9bo\$208b2o80b2o7bobo\$208bobo78bo2bo6b2o\$208bo81b2o3\$222b2o
\$222bobo\$222bo67b2o\$289bo2bo6b2o\$290b2o7bobo\$236b2o52b2o9bo\$236bobo52b
o7bobo\$236bo52bo3bo5b2o\$289b2o\$290bo3bo\$250b2o37bo3bo\$250bobo34bo2b3o\$
250bo35bo4bo\$286bo2bo\$286bo2bo6b2o\$264b2o21b2o7bobo\$264bobo31bo\$264bo
22b2o7bobo\$286bo2bo6b2o\$278bo7bobo\$277bo2bo6bo\$278bobo\$279bo61\$282bo\$
281bobo\$280bo2bo6bo\$281bo7bobo\$289bo2bo6b2o\$267bo22b2o7bobo\$267bobo31b
o\$267b2o21b2o7bobo\$289bo2bo6b2o\$289bo2bo\$253bo35bo4bo\$253bobo34bo2b3o\$
253b2o37bo3bo\$293bo3bo\$292b2o\$239bo52bo3bo5b2o\$239bobo52bo7bobo\$239b2o
52b2o9bo\$293b2o7bobo\$292bo2bo6b2o\$225bo67b2o\$225bobo\$225b2o3\$211bo81b
2o\$211bobo78bo2bo6b2o\$211b2o80b2o7bobo\$293b2o9bo\$293b2o7bobo\$197bo94bo
2bo6b2o\$197bobo92bobo\$197b2o94bo3\$183bo\$183bobo\$183b2o3\$169bo\$169bobo\$
169b2o3\$155bo\$155bobo\$155b2o3\$141bo\$141bobo\$141b2o3\$291bo\$290bobo\$290b
o2bo6b2o\$135b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b
2o7b2o7b2o10b2o7bobo\$2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b
2o7b2o7b2o6bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo
5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo5bo2bo9b2o9bo\$2o7b2o7b2o7b2o7b2o7b
2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o
7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o7b2o10b2o7bobo\$290bo9b2o\$289b3o\$290b2o
\$6b2o\$6bobo\$6bo\$293bo\$291b2o8b2o\$20b2o270b2o7bobo\$20bobo269b2o9bo\$20bo
264b2o5b2o7bobo\$285b2o4bo2bo6b2o\$291bobo\$34b2o241bo14bo\$34bobo239bobo\$
34bo241bo2bo6b2o\$277b2o7bobo\$277b2o9bo\$48b2o227b2o7bobo\$48bobo225bo2bo
6b2o\$48bo228b2o3\$62b2o\$62bobo\$62bo214b2o\$276bo2bo6b2o\$277b2o7bobo\$76b
2o199b2o9bo\$76bobo198b2o7bobo\$76bo199bo2bo6b2o\$276bobo\$277bo\$90b2o\$90b
obo\$90bo3\$104b2o\$104bobo\$104bo3\$118b2o\$118bobo\$118bo3\$132b2o\$132bobo\$
132bo3\$146b2o\$146bobo\$146bo3\$160b2o\$160bobo\$160bo3\$174b2o\$174bobo\$174b
o3\$188b2o\$188bobo\$188bo3\$202b2o\$202bobo\$202bo3\$216b2o\$216bobo\$216bo3\$
230b2o94bo\$230bobo92bobo\$230bo94bo2bo6b2o\$326b2o7bobo\$326b2o9bo\$244b2o
80b2o7bobo\$244bobo78bo2bo6b2o\$244bo81b2o3\$258b2o\$258bobo\$258bo67b2o\$
325bo2bo6b2o\$326b2o7bobo\$272b2o52b2o9bo\$272bobo52bo7bobo\$272bo52bo3bo
5b2o\$325b2o\$326bo3bo\$286b2o37bo3bo\$286bobo34bo2b3o\$286bo35bo4bo\$322bo
2bo\$322bo2bo6b2o\$300b2o21b2o7bobo\$300bobo31bo\$300bo22b2o7bobo\$322bo2bo
6b2o\$314bo7bobo\$313bo2bo6bo\$314bobo\$315bo36\$293bo\$292bobo\$292bo2bo6b2o
\$293b2o7bobo\$293b2o9bo\$293b2o7bobo\$292bo2bo6b2o\$293b2o5\$293b2o\$292bo2b
o6b2o\$293b2o7bobo\$293b2o9bo\$284b2o7b2o7bobo\$277b2o3b5o5bobo7b2o\$277b2o
2bob2o2b2o3b2obo\$280b3o3bo5bo3bo29bo\$281bo4bo7bo2bo27bobo\$282bo14bo27b
o2bo6b2o\$297b2o6b2o7b2o10b2o7bobo\$295bo2bo5bo2bo5bo2bo9b2o9bo\$296b2o7b
2o7b2o10b2o7bobo\$325bo9b2o\$324b3o\$325b2o4\$328bo\$326b2o8b2o\$327b2o7bobo
\$327b2o9bo\$320b2o5b2o7bobo\$320b2o4bo2bo6b2o\$326bobo\$312bo14bo\$311bobo\$
311bo2bo6b2o\$312b2o7bobo\$312b2o9bo\$312b2o7bobo\$311bo2bo6b2o\$312b2o5\$
312b2o\$311bo2bo6b2o\$312b2o7bobo\$312b2o9bo\$312b2o7bobo\$311bo2bo6b2o\$
311bobo\$312bo!
``````
There are a lot of four-engine ones!
Note this rake-type reaction
Code: Select all
``````x = 16, y = 21, rule = B3/S23-q4ij
bo\$obo\$o2bo6b2o\$b2o7bobo\$12bo\$b2o7bobo\$o2bo6b2o\$o2bo\$o4bo\$bo2b3o\$3bo3b
o\$4bo3bo\$3b2o\$3bo3bo5b2o\$5bo7bobo\$15bo\$4b2o7bobo\$3bobo7b2o2\$3b2o\$4bo!
``````
It is used in virtually every ship here, though the 2 engine one of course doesn’t involve it and the bottom (6-engine) one also doesn’t use it.
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Gamedziner
Posts: 796
Joined: May 30th, 2016, 8:47 pm
Location: Milky Way Galaxy: Planet Earth
Moosey wrote:EDIT:
somewhat related rule with at least 4 dirt-common natural ships (particularly the c/7, as the B demonstrates)
Code: Select all
``````x = 155, y = 31, rule = B3-qy4-a5cey6c7/S2-i3-a4it5q6a7e
129b2o\$131bo8bobo7bo2bo\$128bo2bo8bo2bo10bo\$131bo8bo2bo8bobo\$129b2o10b
3o7b4o13\$149bo\$150bo\$148b3o8\$2bo\$b2o\$2o\$b2o!
``````
p26:
Code: Select all
``````x = 7, y = 4, rule = B3-qy4-a5cey6c7/S2-i3-a4it5q6a7e
b6o\$5o\$b5o\$2bo!``````
Code: Select all
``````x = 81, y = 96, rule = LifeHistory
58.2A\$58.2A3\$59.2A17.2A\$59.2A17.2A3\$79.2A\$79.2A2\$57.A\$56.A\$56.3A4\$27.
A\$27.A.A\$27.2A21\$3.2A\$3.2A2.2A\$7.2A18\$7.2A\$7.2A2.2A\$11.2A11\$2A\$2A2.2A
\$4.2A18\$4.2A\$4.2A2.2A\$8.2A!
``````
FWKnightship
Posts: 451
Joined: June 23rd, 2019, 3:10 am
Location: 我不告诉你
Moosey wrote: EDIT:
unrelated rule with a nice failedrep and a replicator too
Code: Select all
``````x = 90, y = 269, rule = B3-e4e/S2-i3-a4i5k
11bo\$10b3o\$9b2ob2o2\$11bo14\$35b2o19bo27bo\$36bo18b3o24b3o\$36bobo15b2ob2o
22bo\$37b2o42b2o\$56bo7\$42b2o42b2o\$43bo42bobo\$40b3o45bo\$40bo47b2o28\$24bo
\$23bobo\$24bo190\$2b3o\$2bo\$3bo8\$3o\$2bo\$3o2\$3o\$2bo\$3o!
``````
p270 glider gun
Code: Select all
``````x = 98, y = 18, rule = B3-e4e/S2-i3-a4i5k
5bo\$5b3o\$8bo\$7b2o83bo\$40b3o47b3o\$20bo21bo6bo39bo\$20bo18bo10bo38b2o\$39b
ob2o5b3o\$39bo\$20bo2\$2b2o16bobo\$bobo16bobo\$bo18b3o53bo\$2o92b2o\$74b2ob2o
15bobo\$75b3o18bo\$76bo19b2o!
``````
EDIT:p104 glider gun
Code: Select all
``````x = 87, y = 20, rule = B3-e4e/S2-i3-a4i5k
2o\$bo83b2o\$bobo81bo\$2b2o79bobo\$9b2o65bo6b2o\$8b2ob2o60bo3bo\$9b2o60bo5bo
\$74b4o\$9b2o\$8b2ob2o61b4o\$9b2o60bo5bo\$2b2o69bo3bo\$bobo72bo6b2o\$bo81bobo
\$2o83bo\$85b2o2\$54bo\$52bobo\$53b2o!
``````
EDIT2:p4,p6,p8,p16,32c/64
Code: Select all
``````x = 128, y = 51, rule = B3-e4e/S2-i3-a4i5k
114b3ob3o\$114bobobobo\$113bo2bobo2b2o\$113bo4b2o\$113bo3bobo\$114bo2bo2bo
3bo\$125b2o\$123bo2bo\$18b2ob2o100bo2bo\$17bobobobo100b2o2\$b2ob2o9bo2bo3bo
2bo10bo\$o2bo2bo7bo2b2o3b2o2bo10b2o17bobo\$o5bo7b2o9b2o11bo17b3o47bo\$bo
3bo30b3o20bo42bob2ob2o\$o5bo7b2o9b2o29bo46b2o3bo\$o2bo2bo7bo2b2o3b2o2bo
13b3o14bo44bo2bo\$b2ob2o9bo2bo3bo2bo14bobo15bo44bo3bo\$40bo2bo63bo\$17bob
obobo82bo\$18b2ob2o82bo6\$102bo\$101bobo\$102bo6\$101b2o\$103bo\$101bobo6b2o
9b2o\$104bo4bo2bo7bo2bo\$109bo2bo7bo2bo\$109b2ob2o5b2ob2o\$111b2o7b2o8\$
126bo\$125bobo\$125b2o!
``````
32c/64 glider rakes
Code: Select all
``````x = 227, y = 151, rule = B3-e4e/S2-i3-a4i5k
15bo3bo\$14b3ob3o30bo3bo\$13b2obobob2o28b3ob3o151bo3bo\$18bobo29bobobobo
150b3ob3o\$16bob2o32bobo152bobobob2o\$16b3o34b2o113b3ob3o31b2obo2b3o\$9bo
38bo119bobobobo32b2o4bobo\$49bo11bo105bo2bobo2b2o30b2obo2b2o\$11bo48b3o
104bo4b2o28bo6bo\$6b2o3bo51bo103bo3bobo27b3o\$5b2ob2ob2o46b3ob2o103bo2bo
2bo3bo21b2obo\$6bo4bo47bo2b2o115b2o20bo2bo\$7bo2bo49b3o114bo2bo21b2o\$9b
2o19bo9bo136bo2bo\$30b2o7bob2o135b2o\$26b2o4bo5bo2b3o\$25b2obob2obo3bob2o
2b2o175b3o\$26b2o2bo2bo3bobob3o175b2ob2o\$33b2ob2o122bo58b3o\$26b2o2bo2bo
3bobob3o112bob2ob2o62bo\$25b2obob2obo3bob2o2b2o112b2o3bo58bo\$26b2o4bo5b
o2b3o112bo2bo60b2o\$30b2o7bob2o114bo3bo59bo\$30bo9bo120bo\$160bo\$159bo4\$
225bo\$224b3o\$156bo34b2o32bo\$155bobo32bo2bo\$39b2o115bo32b2o2b2o\$2bo36bo
bo26bo119bo4bo2bo\$bobo38bo24bobo118b5obo\$o2b3o33bobo24bo3bo114bo\$4b2o
33b2o24bo5bo114bobobo34bo\$6bo57bo2bo4bo113bobob2o32bobo\$5bobo56b5o86b
2o30b2ob2o13b2o9b2o5b3o\$3bobo59bo91bo46bo2bo7bo2bo\$66bo88bobo6b2o9b2o
26b2o2bo7bo2b2o\$12b3o50bo92bo4bo2bo7bo2bo26b2o2bo5bo2b2o\$12bobo41b3o
104bo2bo7bo2bo27b4o5b4o\$11b2o2bo40bobo104b2ob2o5b2ob2o12b3o\$11bo4bo38b
2o7b3o98b2o7b2o13bo3bo\$13bo2bo41bobo128bo\$16bo36bobo2bo130bobo\$13b2o
25b2o16bo131bo2b2o\$13bo39bo3bo136bo\$10bobo12b3o15b2o3b2o8bobo129bo2bo\$
10bo14b2o17b6o11bo129bo9bo\$10bo16bo15bob5o9bobo124bo13bobo\$26bo17b2o
15b3o116bo20b2o\$8bo14bo19b2o16b3o115bobo\$24bobo17bobo15bo116b2o\$8bo15b
obo17b3o\$24b3o2\$26bo17bo\$25bobo15bobo\$25b2o17b2o6\$39b3o\$41b2o\$37bo2bob
2o\$38bo3bo141bo\$38b4o141bobo\$39bo144b2o16\$50bo2\$52bo\$52bo\$49bobo\$51b2o
119bo\$46bob2o120b2obo3b2o\$46bob2o19bo100b2o6b2o\$68b3o98bo2bo4bo\$67bo2b
2o98bo\$53b3o10b4o\$54bo12bo4b2o\$49bobob3o2b2obo7b6o86b3ob3o\$52bo2bo2bo
2b4ob2obo4b2o85bobobobo\$48b3ob6o4b3ob3ob5o85bo2bobo2b2o14bo\$48bo2bo3bo
b2o6b2o4b3o86bo4b2o16bobo\$48bo11b3o2bo94bo3bobo17b2o\$46b2obob2o8bo3bo
95bo2bo2bo3bo\$46bo4bobo118b2o\$46b2ob2o2bo116bo2bo\$52b2o116bo2bo\$49bobo
119b2o\$49b3o141b2o\$194b2o\$193bo\$153bo\$149bob2ob2o\$150b2o3bo\$149bo2bo\$
150bo3bo\$154bo\$153bo\$51b2o99bo\$51bobo\$52bo4\$149bo33bo\$148bobo31b3o\$
149bo31bo2b2o\$180b4obo\$180b2o2b2o\$183bobo\$180b2o\$180b2obobo\$61bo86b2o
30b2o2bo\$62bo87bo31b2o\$60b3o85bobo6b2o9b2o\$151bo4bo2bo7bo2bo\$156bo2bo
7bo2bo\$156b2ob2o5b2ob2o11bo\$158b2o7b2o14bo\$185bo\$182bo\$181b2o\$180b2o3b
o\$181b3o\$181b2o2\$173bo\$172bobo\$172b2o!
``````
I'm too shy to talk to other members.
But I want to upload my apgsearch results to Catagolue like others.
Code: Select all
``````x = 12, y = 1, rule = JvN29
FWKNIGHTSHIP!``````
Hunting
Posts: 3019
Joined: September 11th, 2017, 2:54 am
Sorry fir self promoting but I want to talk 'bout Denseflakes. I'm threatening to make a thread!
Random p2:
Code: Select all
``````x = 5, y = 8, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
2bo\$5o\$5o\$bo\$bo\$5o\$5o\$2bo!``````
Obviously extensible:
Code: Select all
``````x = 13, y = 8, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
2bo2b2o2b2o\$13o\$13o\$bo3bo3bo\$bo3bo3bo\$13o\$13o\$2b2o2b2o2bo!``````
p3
Code: Select all
``````x = 9, y = 9, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
b2o3b2o\$4bo\$3b3ob2o\$7b2o\$2bo\$4o\$5o2b2o\$2b2o\$5b2o!``````
Thanks Aleph!
Random c/2, no other speed yet
Code: Select all
``````x = 23, y = 16, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
4bo13bo\$3b3o11b3o2\$3o3b3o5b3o3b3o\$4bob3o5b3obo\$bo2bob3ob3ob3obo2bo\$7bo
7bo\$7bo3bo3bo\$3o5b2o3b2o5b3o\$5bo5bo5bo\$bo2b3ob2obob2ob3o2bo\$3b5o7b5o\$
3b4o9b4o\$4bo6bo6bo\$4bo13bo\$6b2o7b2o!``````
Smallest explosive seed
Code: Select all
``````x = 5, y = 5, rule = B2ci3ai/S1e2i3ai4ae5ai6ac7c8
obobo!``````
MLP will live on forever, so does John Horton Conway.
Moosey wrote:
February 5th, 2019, 7:51 pm
“New knightship tagalong!”
“Quick, hide it!”
My TODO list
LeapLife - DirtyLife - LispLife
I could make a rule in ten seconds flat
Hdjensofjfnen
Posts: 1519
Joined: March 15th, 2016, 6:41 pm
Location: r cis θ
Out of the ridiculously limited number of distinct objects in this rule, we somehow have a (2,2)c/6:
Code: Select all
``````x = 6, y = 6, rule = B2e3aijr/S12ek3-n
b3o\$o\$o\$o2bobo2\$3bo!
``````
"A man said to the universe:
'Sir, I exist!'
'However,' replied the universe,
'The fact has not created in me
A sense of obligation.'" -Stephen Crane
Code: Select all
``````x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o\$6bo\$o3b3o\$2o\$bo!
``````
Moosey
Posts: 3286
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
Contact:
This is a rule with long-lived tiny methuselahs, based on a rule earlier in the thread.
Code: Select all
``````x = 11, y = 5, rule = 01e2eik3ery4jz/2-an3aj4a/3
10.A2\$10.A2\$A!
``````
Diagonal puffer:
Code: Select all
``````x = 5, y = 6, rule = 01e2eik3ery4jz/2-an3aj4a/3
2.ABA\$.A.AB\$A2.A\$.A.A\$A\$.A!
``````
Code: Select all
``````x = 3, y = 19, rule = 01e2eik3ery4jz/2-an3aj4a/3
A.A18\$A.A!
``````
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
GUYTU6J
Posts: 1072
Joined: August 5th, 2016, 10:27 am
Location: 中国
A rule which keeps active due to a bunch of spaceship collisions
Code: Select all
``````x = 193, y = 134, rule = B34twyz/S23-ekn4eit
2ob4o2b2ob2o5bobob3o4bob2o2b2obobob2ob2o4b5o2bobobo3bob2ob2o2b5o2bobob
ob2obobobo3b2ob2ob4o2b2ob2o3bo4b2o2b3o2bo5bo3b2obo6bob2o3bob2ob3o5bobo
8bo2b3obo\$obo3bo2b2ob2o2bob2o2b3obo3bo2bobobob5ob4ob2obo2bob3o2b2o3bo
2b2obob3o3bo3b4o2bo2bobob3o2b2ob2o3b2o3bob3o2bobo3bo3bob5ob3o2bo2b2o2b
4ob3o2bobo3bo2b4o2bo3b4obobo\$3obob2obobo3b8obo5bo4b2obo4bo2bo5b3o2b2o
5bo2bobob3obobo2b2o4bobo5bo3b2ob3ob3o2bobob2ob4obob2o3bob3obob2o3bo2bo
2b2ob2o3bo2bobobo4bobo4bo2bo3b3obob4o\$o4b4ob3o2b2obob6ob2ob2o4b2ob3o2b
2o2b2ob2o2b3obo2b2ob2o3b2o3b2ob3obobo2bo2bo2bobobobo3b2o2bo3bo6b3o5b3o
b2obo2b2obo2b2obobobo4b4o2bob4o2b5obob2o2bobobo2b6o\$3bob3o3bob2ob2o2b
2obob3o2b3o3b3ob4obob6obobo3b4o4bo2bo2b2o2bo3b5ob4o2b3obo3b2ob2obob2o
2bobobo4bo2b2ob2o5b3o4b2o2bo2b3o2bo3b3o2b2o4bobo2bob5o2bob5o\$bobo2b2ob
o3bo3bobo2bob2ob2o3bobo4b2obob3o3bobob2obo4b2ob3ob3ob2o2b3ob2o5bobo3b
2o2bobo2b2o3b2o2b3obob2o2b2o2bo2b2obob2ob3o2bo3bob3o3b2o3bo2bob3obo2bo
2bob6o2b3o2b2o\$b2o2bo2b2obobobo3b2o3bo3bobobo3bobobobob3o2bob2obobo2b
2obob2o4bo2bo3bo2bobo3bobob2o3bo2b2o2bo2bobob3o2bo2b2o2b2obo7b3obo3b2o
3bobo2b2o2b4o2bo4bo6bobob3obobo4b2obo\$ob2obobob2o4b2o3b2o5bo2bobo5b4ob
o2bo2bob2o2bobobob4o5bo6bo2bo2b2o3b5o4bo5b3obob6o3b2o3bobob5obob4obobo
11bo5b2ob2o3b2o2b4o2b2ob4obo3bobo\$b2o6b2obo2bobo6bo3bobo4b2obobobobo2b
o2b2ob3obo3bob3obobo2b2obob2o4bobo2b5o3bobobo5b4obobo2bob2o2b4o2b2o3b
2o9bob3o2b5obo4bob4ob4obo6bo3b3ob4o\$bob4obobo2b4o2b3ob4ob6ob2o6bob2ob
4o7b6o2bob2o4bob5o2b8obo2b2o5b2obob4o2b2o3bo2bobo2bob3ob2obob4obobo11b
2ob4o3b3obo2b2obobobob5ob3o\$o3b2o2b3o3b3o2bo3bobo4b4o2bob2ob4obo3bob5o
2bob2obo3bob3o2bobob3ob2o3b2obobo2b2o3bobob2o4bobobo2bo2b2o2bo3b2o5b2o
b2obo3bob2obob4obob2o4bo2b4o2b3o3b2o2bo2bob2o\$bobo2bo3bob3o2bo2b5o3b2o
b3obobobobo2b3ob3obob2ob2o2bo4b2o2b2ob2obob3o3bobobo4b2obob3o3bo2bo4bo
b2o2b2ob2o2bob3o3b2obo2b2o3bobob2o3b2o5b2obob3obo2b2obob4ob3o2b2o\$b7ob
6o3bo3bob2o2b2ob3o2b2o5bobob3obo5b2ob2obo3bobob2o2bobobo4b2o3bo2b4ob2o
bo2bo2bobo2bo2bo3b5ob5ob2ob4o2b4o2b2obo2bo2b2o2bobobo8b3ob2ob3o3b2obob
ob2o\$2b3obo2b5o2b2ob2o2b5ob2ob2o2bobo2bo2bob4ob2obo4bobo4bo2bobo2bob4o
bo7b2o2b5o2b2o2bo2bob3ob7ob2ob3obobo2bo2bo2b3ob3o4b3ob10obo2bobo2b3o2b
3obo4bo3bo\$2o2bo3b2o2b3ob5o5b6ob2o2b3o3b3obo2b3o3b2ob2o2bob4obobo4b3ob
o2bo3b3ob3obo2b2obob2o3bob5o2b3o4bob2obobo2b3ob2o2b8ob6o2b3o2b3o6b2o3b
obobobob2obobo\$o2b5ob4obobo2b4o5b2ob4o3bob2ob5ob4obo4bo3bobobo2b2ob2o
2bob2o2bobobo2bo5b3obobob2ob3obo3bo5b6ob3ob2o2b5o2b2obob3ob2o2bo2b2ob
2o2bob2o2bo2b2o2b6ob2o4bo\$bobobo4b3o2b4o2bobo2bob2ob2ob2o3b2o2b2o4b5o
2bob3o4bo2b2o3b6obo2bobobobob7obob3o2b3o2bo3b4ob3o2bo2bo3b2o2b3o3b3ob
5obo2b2o3b6o3bo4b2o3bobo3b2ob2obo\$bobo2bobo2b2o2b4o3b2o4bob2o5bo6bo2bo
b3o6b2o7bobob2obo2bob2obobob5obo6b3obo8b4ob2ob2o2b4ob5o4bobo2bobo4b3ob
ob3o4bo2b2o3bobob2o2bobob2ob3obo\$2obo2bob2o2b2ob6o2bob4o2b2ob5o2b5o3bo
b2o2b2o2bob2o2bobob2ob2o2b2obob3obobobo3bobo4bob2ob2ob5o4bobo2b3o3bo4b
ob6o2bo2b3o2b2obobobob2o3bo2bo2b2o2b2ob6o4bo2bo\$6b5o2bob2o2b3o4b3obo2b
obobobo2bob3o3bobobobo2bob2o4b4o3bob2ob2o2b2o7b2o2b3obob2o2bob2o3b2o2b
2o2b4o2b3o2bo2bo2b5obo2bob3o2bo5bobo2bo2b2o3bobo2b2ob3obo2b3obo\$b3o3bo
3b11o4bo2bo4b5ob2obo2bob2obob6obo7b2obobo2b2o2b5ob2obobo2b2o2b2obob4ob
2obobo4bo3bobo3bobobo2bo4bo2bo4bo2bob2ob2obo3b2ob2ob4obob2obob2o2bobo
4bo\$o5b2o6b3o2b3obobo2bo3b4o2b2ob3obo2bo2b4o2b2o4b3o2b2obo2bo3bob2o4bo
2bo2bobo3bob5ob2ob5obo3bob2ob5o2b3o2bobob2o6bob2o4bobo6bo4b2o2bobob4o
6b2ob2o\$3b4obo3b6o2bo3b3ob2o4bo2b2obo2b3obo5b2obo4b2o3bob2o2b2o2b3obo
2b2o2b2obob2o2bo3bobo3b3ob3ob2o3b3ob4o3bo3b2o2bo3b4o2bobob2obo2bob2o5b
ob2obobo2b3ob2ob2o\$4ob2ob3o2bo2b2obo2b6o2b3ob6ob4obo2bob5o3bo5bobobob
2ob2ob2obo2bo3bo5bobobo2bo3bo3b2o2bobo4bo3b7o3b3obo2bo2b2ob3ob6ob5obo
2bo7bob4o2b4o3b2obo\$6o2b3ob2ob2ob2o2b2o3b3o2b2o3b3ob3o5bo3bo4bo3b2obo
4b2obo2bob2ob2o2bob2ob3o4b2obo2b4o2b2o6bobo3b2obo2b2o2b2obob3obob3o2bo
2b2obo3b4ob2ob4o3b4ob2ob2o2bob2o\$ob2ob2obobo4bo2bob3o2b2o2bo2bobo2bob
2ob3o3bo4bob3obo4bo3bo2bob3ob2obobob2ob2o2b2obob2obo2b4o4bob2ob2ob3ob
2o4b4o3bo2bo6bob2o5b2obob3ob5ob4ob2o2b3obobobob5o\$3o2bob6ob2o2b12o4b4o
3b3obob2ob6obobo3b3obob2o2bo2bob3obo2bobo2bobobob2ob2o3bo2b8o3b2obo2bo
b5o5b4obobob2o3b7o2bob2ob2o2b3ob8o2b2ob4o2bo\$4o2bo3bo6b3o2bo4b3ob2obo
2bobo3b2o3b3obo4bo3b2ob2o2b2ob2o2b4obobo4b2ob4o2b2obo3bob2obo2bobo4bo
4b4obob3o5b2o3b6obo2b2ob2ob2o2bo2b3obo4bo2b3ob6ob4o\$bo2bo2bo3bo3bo3b2o
b2obo4b2ob3o2bo2bo3b2o5b2o2b2obo4bo4bo6b2o2b6obobob3ob5ob2o4bo2b8ob3ob
obobob5o5bo2b2o3bo3bobo2b3o2bob2o4bobo2bobob8o3bo\$3bob2obob2ob3o5b3o3b
obob3o4b5o4b2ob2o2b2obo3b3o2b3ob2ob3o2b2o2bo3bo5bo2b2ob2obobobo7bob4o
3bob2obo3bo4bo2bo2b2o3b4o3b7obobo6bo2b2o3bo3bobo2bobo\$bo4b4o7bob3obobo
2b4ob2o2bo8bob4obo3b2ob2obo2bobo2b2ob5obo3b4ob2o3b2o3b4obo4b2ob7o4b2o
2b2o2b4o2bo3b5o2bob2o2bo4bob4ob2obo5bob5o3b2o2bob2o\$bo4bo2bo4b3obo2bob
2obo2bobo4bobo2bobo2bo2bob2ob2o3b2o3bob2o2b6o2b3ob5ob2o3bo3bobo2b5obo
2bob2o3bob2o2bo2b2o2bob2o3b2obo2b2o2bobobob2obo3b2obobo4bob2ob2obo2b3o
3b2ob2o\$b2obo3b4o2bo2b3obob2ob3obob2o2bo6bo2bo3bobob2ob2o5bob2obo2b2ob
o4bo2b3ob3o2bo3bo4bobo5bo5b2ob2o2b2obo2bo2b2obo2bob4obo2b3o2bo2b3o3bo
2bobobo2bob8o3b3ob3o\$2ob3obob3obobob5ob4ob2o3bo3b2obobo2bob4o3bo2bo3bo
2b2obo5bo2b2o3b2obo2b2o3b3ob6ob6ob2o2b6o2bobo2b2o6b2obo9bo2bo2bobob2o
2b5o3bo5b6ob4obobobo\$o2b2o3b2ob2o2b2obo2bo3b3o4b2o2bobo3b6obo2b4obob2o
4bobob3ob2ob3obo2b2ob3o3bob3o2b7ob2obo3b2obo4b2o2bobo8bo5bobo2b2obo2b
2o2b2o5bo2bob2obob3obo2bo2b3o2b2o\$b5o5bo2b2o5b2obo2bo2bobo2b3obob3ob3o
2b2o2bobobobo2b6o7b2o3b2o2bo2b5o3b3obo2bo2bo7b5obo2b3ob5o3b2obo5bo2bob
2o4bob2obobob2o3bo5bo3b2o2b2o6bo\$2obob4obob3obo2bo2bob6o4bobob2ob3obo
5b3ob2ob3ob4obo2b2o4bo2b3obobobo4b4o2b4o3b3obob2o2b3ob12ob2obo6bo3b3o
2bob3o2b2obo2b2ob6ob5o2bo2bo4b4o\$3bo3bo2b2ob2obo3b5ob2o2bo2bobo3bo2b5o
2b2obob2ob3obo2bob2obob3o3b3ob2o2bob2ob2o3b2o5b5o2bo3bo3bobo4b2o3b2o2b
obob3o2b4ob2ob2ob2obob2o3b2ob4obo3b2o4b2o3bo2b2o\$b5ob4o2bo2bob3o2b4ob
4o5bo3b4ob2o2b5obo2b2obob4o2bobobob2o2bob3o2bob2o2bo2bo2b2o4b5o5bo3bo
3b5o2bobob2obo2b3obo5b2obobo2b2ob2obo6bob4o2b2ob7o2bo\$ob2ob7obob3ob2ob
ob4ob3o2bobobobob3ob4ob4obo4bo2bo3b4o3b5o2bobob5o2bo2bo3bob2o2b4ob2obo
bo4bo2bo2bob2obo2bo2bob2ob3o4b2o3bobob2ob2o4b3o7b2ob2ob4o2bo\$3b2obo3b
3o2b2ob2ob2obo2b4o2b2o5b2o5b2o2bob2o4bobo2b3o2bo2bob2o2bobob2o2bo2bobo
7b2ob2ob3o3b2o4b2obo3bobob2o3b2o2b5ob2obo6bo3b2o5b2o3b4o2bo3bobo4bob2o
bo\$6o2bo2bo2b2obo2b9obo3bo2bo2bob4o3b3ob2ob3ob2ob2o2b3obobo3bob2o2b2o
3b3obo3bob3ob2o3b2ob2ob4ob2ob3obo3bobobo2bobob3o3b2obobo2bob2o8b2o5bob
3ob4o9bo\$2bo4bo5b2o2bob2ob2o2bo2bo2bob2obo3bob4ob3ob3o3bo2b3ob2obo3b2o
6bobob2obobob3obo3bobob2o2bob2ob3o5bobo2bo3b2ob2o2b4o2bo8bo2b10o2bobo
2b2o2b2o2bo3b2o2b2ob2o\$3o2b2o2bob3o3b6obobob3o2b3ob3o3bo2b8o3bo2bo8b2o
bo2bobo2b2o2b2o2b2obobo3bo3b8o2b2ob2o3b2obo2b5ob2ob2obo2bob2obo3b3o2bo
5b2o2b3ob2ob2o6b2o10bo\$obo4b2o2b4o2bob2ob2ob2ob7ob2o2b2obob2o2bob3o4bo
b3o2bobobo2b3o4bobobob3ob2ob2ob3o2bobob2o3bob2ob2o2b5o4bob2o3bo4bobob
3o2bob2obo3bo3b2o3b3obo2bo3b2obobo2bobo2bo\$obo3bo2bobo2b2o2b3ob3o2b7o
2bo5bobobo5bo4bo2bobo2bo4b2o2bo3b2o3b3o4b2o2bob2ob3o2b2ob2ob2ob9o6b2ob
obo3bobo2b3ob2ob4ob2ob5o3b4obo2bo2b2o3b2ob3o\$ob3o2bobob4ob4o2bobo3bobo
b2o4bo2b2o4bo3bo2b2ob4ob3ob2o2bobo2b4ob6o6b2ob2o4b6o3b3ob2obobob5ob2o
2b2o2b2obobob10o2b2ob4ob2o3b3o2b2o2bobobo2b4o2bob2o\$3o4bobobo2b4obobo
2bobo4bo3b2obob2obobo2bo8b2obobo2bobob2obo3bob2ob6o3b2ob3ob4o4b2obob2o
4b3ob2obo3b3o2bo4bobob3ob3ob2o2bo8bo2bo2b5o3bobobobo4b2obo\$2ob3o3bob2o
5b3o4bo4b2obobob2ob3o4bobob11ob2o2b4o2b2o2bo2bobo2bobob3o4b4obobo3b4ob
2o5bo3bo5bobo4bo2bobob2obobo2b8obo3b3ob3o2bob2ob2o3bobo4b2o\$2o3bobob2o
5b3ob2obo5b2ob4o3b3o2b4o2b4obo3b4ob5o2bobobobo3bo6bo3bob3o3bobo7b4o2bo
6bob4o4bo2b3ob2o2bob5o2bob2obo2b2ob3ob2ob2obo2b2obo2bo7b2o\$3b10obo2bob
o6bo3b5o4bo3bobobobo3b2o2b2o2bobob3ob3obobob2ob5obob3o2b2o2bo2b4ob2obo
3b2o2b3obobo5b3ob2ob3o2bobo3bo2b2ob2o2bob2ob3o2b4ob2ob2o2bobob2o5bo2bo
\$2bob3o2bobo2b2ob2ob6o4bob9ob3ob2o3b2o2bo2bo2b4o2b3o3b2ob5o2b3ob4o2bob
3o2bobobob2ob2o4bo4b2o5bobob3obob13o2bo2b2o6bob2o2bo3b7o2bobo2b2o\$2obo
b2o2bobob4ob2ob2o3bo2bobob5obo2b3o2b3ob2o2bo2bo3b4o4bo7bo2bo5b3o2bo2b
3o2bo3b4obob3o3b4o2bob2ob4ob5obo2b2o2b2o2bob5ob5o2b2obo7bob3ob8ob2o\$ob
o2bo3bobo5bo2b3o2b2o2bo4bob3obob3ob4o2bo3bo2bob2ob2o2bo2bob3o2b2obobo
3bo3bo3bo2b2o2b2o6b4obobob4obo2bobob5ob2ob13obo2b2ob7o3bo2b3o2b3o2bob
2o2b2obo\$2obo3bobo4bo3bo4bo2bob3o4bo2bob2obo4bobo6bob2obobo4b3ob3o3bob
2o4bob2o3b2obo2b2o8bob2o2bob2ob7o2b2obob2o4bo6b2obo2b2o6bo2bo2b2o2b4ob
2o2b2o5b5o\$ob5obo2bo2b4ob3o4b3o3b2o3bob3o2bobo2bobo2b2ob4obobo4bo3bo3b
o5bo7b3o2b7ob5obo2bob2obo3b2o2bo3b5o4bobo3b2obo3b2obo4bobo4bo3bo2b2obo
5bo3b4obo\$3b4o4bobo2b4obo2b3o5bobo3bobobo2bo2b3o2b2o2b3o10b2o2bob2o2b
7o2bo4b2obob5o6bobo4bobo2bo3bo2bo2bob4obobo4b2o2b3obobobo5b3obob3o2b2o
bo3b2ob2o2bob2o\$2bo7bo5b3o2b3o3bobob2o2bobob5o2bobobobob2obobobobobo3b
obo4bo6bo4bo2bo2bo2b4o3b2ob2o4bo3bo3bo2bo2bobo2b3o7bo2bobobo2b3o2bo2b
5o2bo4bo4b3obob2ob2o2b3o\$2b3ob2o5bobobobo2bo3b2ob3ob2o4b3o5bobo2b4o3b
2o2bo4b4obo4b4o3b6obobobobob4o3b2o2b3o2b4ob2ob3o2bo5b5o2b7o5bobo2b3ob
2ob2ob4o2b3ob2o5bob3obo\$2bobobo2b2o3b2ob2ob2o3bo2b3o2bobo2bob3o5bo2bo
2bobo3b2ob3obo4b2o3b2ob3ob4ob3o3b2o3b2obobob3ob4o3b2o2b5o3bob2ob3o2bo
2bobobobo2b2obo2b3obobo4bob2ob2o2bobo3bo2b4o\$bob4o2b2ob3obo2bobob6ob3o
2b5o3bobobo2b4obob2obo2bo2b7o2bo2bobo2bo3bob2o3b4o3bobobobo2b3o2bo2b2o
2bob4o2b2ob2obo2bo2bo3bo3b2ob3obo3bo2bobo5bo3b2obobo2bo3b5o\$o4bo2b2ob
2obob2o3bo3bo2bob3obo2bo2b2o4bob3o2b4o8b3o4b3ob4ob2o2bo6bo4bo2b2o3b2ob
4o3b3o6bobob4obo4bobob3ob2o2bob2ob2o2b5o2bo2bo5b2obobobob4ob3o\$2b2o3b
2ob3ob7ob4ob4o2bobobo2bo3b3ob5ob2obo5b5obo4b7o5b2obo2bo2b7obobo4b4obo
2bo2b5o2b4o2b3ob2o3b3o2bob3o4b5o2bobo2b3o3b2obo3b3obo3bo\$o3bo4b2obo5b
2o2b2ob2o4b7ob2o3bobo2bo2b8o4b4ob2obob2o4bobo4bo3bob2obob2o2b2ob2o3b2o
3bobobob4o2bo3bo3b4o2b2ob3o6b2ob2o2b3o2b2o4bo2bob3ob3o2bob2o2b2o\$3o3b
4obo2bo3b2o5bo3b5obo3bobo2bob3o3bo3bo2bob2obob6o3bobob3obobob5o2bo4bo
2b2ob2o2b4ob2obo2b3ob3o2bob4o2bo4bo4b3ob2o3bob4o2bobo2b2obob2o2b3ob4o
3bo\$2obo2bo3b3ob3obo2bobo2b4o3b2ob6obo3b2o3b3ob2obo5bo3bob3ob4ob4obob
3ob2o4b3ob2o3b4ob2o3bo2b8ob2o2b3ob4obobo2b2obo4b2obo4bo4b3o2bobo7b4o3b
obo\$ob2ob2obo4bob3obob2ob3o2b3ob2o3b2ob2o2bob2o2bobobo2bo5b2o2b2ob6o3b
o2bo4b2o3b2o2b2obo2b2ob2o2b4ob5ob5o5bo2bo4bob7o3bo10bo6bobo2bob2o3bo2b
4o4bo\$b2obobo3b2o3b2obob2o4bo2b2ob2ob3ob10obob6o2bo3b2o3bobo2bob2ob2ob
ob2obo2b3o4bobo2bobob2obo2bo5b4obobo2b3o3b3ob3o2bo2bo2bo2b2obob3ob7o4b
2o5bo6bo2b2obo\$5b3o2bo2bo2bo2b2o2b2ob2o2b2ob9o2bobob5o2bo2bob2obo6b2o
3bo6b3o3b6o4bo4bo6b3o2b2ob3o3b3o3bo6bob4obobo5bobobob3ob4o2bo2b4ob4o3b
4o3bo\$3bo2bo2b2o2bo2b7o2b3o5b2ob2ob3obo2b5o5b2o2b2o2bo4bo3b2ob2o2b3obo
6b2o2bo2bo3bob2o3bobo3bobobobo3b2o2b2o3b2o2b2o2b3obo4b2ob2obo2b3obo2bo
2bo2b2o6b2o7b3o\$b3o4bo4b2o3bo3b2ob2o2b4ob2obo3bo2bob2o3bob3o2bo2b9obo
3b2obobo3bobob2o2bo2b2o3bob2obobo3bob4ob5o2bo2b5o2bob2o3bo3bobo2b2obob
ob2ob2o2bob3ob3ob3obobobo2b3obobo\$b2obobo2b3obo5bo2b3obobobo5b2ob2obo
3b4obobobo5bo2bo2bo2b2o2bo2b3ob2obobo4b2o2bo4b4ob2o4b2obob2o3bo2bo3b2o
3b4o4bo4bob3o2b2o2bo2b5obo3bob4ob3ob4o2b3obo\$bobobob3obo3bobo5bo2bobo
2b2o6bob2obo2b2o5b5ob4ob3obo4bo2b3o5bob2o2bob3o2bo4b2ob2o3b2ob4o7bob4o
b2o2bo4bobo14bobo2bo2b2o2b3obob3o2b2o3b3o2bo\$2b3o2bobob3obo5bobobo2b2o
bo5b2o5bo3b4o2bob2ob2obob3ob3obo5bo4b2o2b3ob2ob6o5bo2b4o2bob3o3bobo3b
2obo4b2obob5ob2obo2b3o3b4o2b2o3b2o7bo3b2ob2ob4o\$2o6bo7b2o2bob2o3b2ob3o
bo2b2obo3bo2bo2bo5bobo3bo3b2ob2ob4o5bo5bobobob3ob3obobob2o2b4ob3o4b3o
12b2ob4o2b2ob2o3bob2o3bo5bo2bo2bo4b2o3b2o3b2obobo\$bo4b2o4b3o7b3obobo2b
4obobobo2bo3b2obo2bob3o2bobobo5bo5bo3b4o2bo2b3ob10obo6b2ob2obobo2bo8b
2o7bo5bob5o3b2obob2o5bob2obob2o2b4o7b2o\$b2o5b2obo2b2o2b5o2b7obob3o2b2o
2b3o2bob2ob3obo3bo2b3obo2bo2bo2bobobo2b3o2bob6o6bo2bo2b3o2bo2b2o4b2o2b
ob5ob2obobobobo2bob2obob2obo2bo3b2ob5o3b3o2b2o4bob2obo\$bobob3o3bo3bobo
bo2b2obo2bo2b3obobobo3b4ob3obob2o4b2o3b3ob3ob2o3b2o2b6o2bob4obobo3b7ob
obo2bobobo4bob3o3b3o3b2o2bobo6b2o4bo4bo2bo2bobo2b5ob2o2b2o\$ob2obo2bobo
bob4ob4ob2o3b2obobo2b5obob3ob3o2bo2bo3bob2obo2b2obo2bobo2bob3o3bo3bobo
3b2obo2b3ob2o4b5obob2ob7obo3b2o2b3ob3o2b2o3bo3b3ob3o2b4o2bob2obo3b2obo
b3o\$2ob2o2b5ob3obo5b2o2bob2o3bobo2bob2ob3obob2o8b3ob3o3bo4b2o2bo2b2o3b
ob3o3b2o6b2ob2obo4bob2ob2o2bobobo5bo3bo2bo2b3ob3o2b2ob5obob3ob2obobob
2o4bob4ob2o2bo\$4o2b2ob3ob2o3bob4o2bobo3bobob2ob2o6bo3bobobob2o3bo3b4o
2b3ob3o2b3obob2ob5o2b2obo3b2ob3o3bo5bob7ob2o3bo3bo2b6obo2bo4b2o5bo8b2o
b5o3b2o2b3obo\$4b3o7bo3bo3b6o4bob2o2bob2o2b2o2bobob3obob2o7b3ob5ob2obo
2bo3b2o3bo2b2ob5o2bo3b2o2bo2b2o2bo2b3o2b2obobo4bobob4obo2b3o5b2obo2b5o
bobo2bo2bo4bobob3o\$4b2o5bobo3bobo2bo3bo3bobo2bo3bo2bo4bo8bob2o4bo4bobo
7b3o4b2ob6ob3o2b2obo3bo3bo3b2o2b2o3bobobo2bo2b2ob2obob4obo2b2o4b4o2b4o
b2ob2ob5o2bobo2b3ob3o\$2bo2bobo5bo3bob2ob4obo2bob2ob2o3bobobobob2ob4o3b
3o2bobo5b2ob5obobo2b3ob3o3b5o2b6ob4ob2o2b4ob2obobo2b4ob2ob2o2bobobob3o
b2obob3o3bob2o2bo2b3o3b2o3b4ob4o\$2b2ob2o5b2ob2o2b4o2bo4b5obob2obob4ob
3o2bo2b2o3bobo3bobo2b2ob2ob2obo5bob2ob2obo5b2o6b2ob4obo4bob2o2bobob5o
3b2o5bob2o2b3ob2obob3o2b2ob2o2bo3b3o6bobobo\$2bobob3o3bob6o4b2o2bo2bob
2o2b3ob2o4b2o3bobo5b2o3b2o2bob5ob2o4bob2ob4ob2o5bobobob2obo5b2obob3o5b
o2bo3b2o6b2o3b4o2bobo5bo3bobobobo2bobobobob2ob3ob3o\$2obobo3b3ob2o3bo3b
6ob2o3b2o2bo4bob2o2b6o2b2ob7o2bobo2bo2bo2b4o3b3obo2b3obobo2b3o4bo2b5o
2b2obob2o2b3ob3obob3ob2obob3o2bobo3bobobob2ob3obobo2bo3bob3obob2obo\$bo
b4o3bob2o5bo2bobo4b2ob4o3b4ob4o3bo3b3obob2o2b3o3bo4b9ob2o2bo2b4o4bobob
o3bo2bobob2o4bo3b3o3bo5b2o4b8o2bo3bob3obob3o2bo2b2o10b2o2b3o\$b2o2bobob
6o2bo2b2obobo2b2ob2ob2ob2obo5b3obo2bobobo2b2o2b2o2bob6obob3o2b4obo2b6o
5bo2bobobo6b4o6b2o2b3o3b6obob2o7bob2ob3ob2ob4o3b2o2bo2bob2ob2ob2obo\$ob
2o2bob2obobo4b5o3b2ob2obo2b2obobo2bobob2obo2b2o3b4o5b2o2b2o3b3o2b4obob
4o3bo3bo2bo2bo4b2o2b3o4b4o2bo4bo3bo2bobo3b2ob4ob2ob4o2bobo2b4ob3obo2b
2o2bob3o2bo\$3o3b2o4b2ob2o2b2obo3bob3ob2o3bo3b5o4b2o5bo9bob2ob5o2bob2ob
o6bo2b3o3b3o2b4o2b2o4bo3bob5o3b2o3bobob2o2b3obo2bo3b2obo3b2o2b2obobob
3o3bob2o2b2o2b2obo\$ob3o2bobobo3b2o3b4ob3o2b3o4bo3b5o3bob2obo2b3o2bob2o
b2obo3b2o2bob2ob3ob3o4b2obobo2b2o3bob5ob10o2b2o2bob2obob2obob2obo3b2ob
3o2bobob2o2bobobo2b3ob3obob4obo2b2o\$2bo2b2o3bo2b2o2b2ob4obobo3bo2bobo
2b2o4b5obob6ob2o2b2o5bob2ob3ob3o2bo4b3ob11o3b2o4bob3obo2b3o3b4ob2o5bob
obobo5bob7o2b2ob8ob14o2bo\$bo4b3ob2obob4ob2o2bob2obob3o2b2o8b2o3b2obob
3ob2o3b7o2bobo2b2o2bo2b2ob3obob4ob7ob2ob2ob2o2b2o5b2o4b7obobob5o4b2obo
2bobo2bob6obob2obobob3ob8o\$bob4obob2o5bo3bob3o2b2ob2o2b2ob2ob2ob2o2b3o
4b2ob5o2b5obobo2b2ob3o2b3obo2b2o2b5o2bo2bo2b2obobo2b4o4b2obo3b2obobo2b
obo2bo3bo5b2o3bo4b2ob2o3bob3o2b3o2b3obobo\$2b5o3b3o3b3o2b2obob3o3bo3b2o
bobo5b2o3b2ob2o6bo2b4o3b3o3bo2b2ob2o3b2obob4obob3ob5o2b2o2bo2bo7bo2b2o
bob2o9bobob2o2bo8b2o2b2o3b3o2bo3bobo2b3o2bo\$3bo2b4o3bo4b5obo2bobo2b2o
4b3ob2o2bobob2o6b2o6b3o6bo3bo2bo2b2obob2o2bobobo2bo5b3obobob2obob2o3b
2ob3o2b5o4b3o2b5obo2b3o3b4ob2o2b2obobob3o3b3obob2o\$2obo2bobob2ob2o2bo
4b7o2b2obo2b2o3bo3b2o2b2o3b3obob2obo6b4ob2obobob2o4bo3b2ob5o2b2o4b2o3b
3obo4b3o2bobo2bob12ob4o2b2obo6b4o2b2ob2o3b3o3bo2bo2b2o\$b4o5b3ob2o4b5ob
ob2ob3o2b2ob2ob3o5bo4b4ob5o2b4o2bo4b3obo4bobo2bo2b3o4b5o6b4obobob3o3b
3o2bo2bobo2bo3bo7b2obobob2obo2b2obo3b4ob2o2bo4b3o2bo\$2bob4o3b2o6bo2b4o
bobo3bob3obo3b5obobo2b2ob3obobo2bob3o4b3ob2o2bob3o2bobobo2b5o5b4o2bo4b
obobobob5obobo2b3o2b2o2b6o3b2o3b4obo3bob7ob6o7bo\$2bobobobobo4b3obo7bo
3b3o4b3ob2ob3obo2bob4ob2o5bob3obo2bo2bobo2bobobo2b4o2bobob2o2bob3obobo
4bo3bob2obob5ob2ob2obo3bo5b2ob3obo3b2o6b3o3bobob4o2bo2b2obo\$4o2bo2b8ob
ob3obo3bo3bo5b3o2bo3b3obob2obo2b2o2bo3b2o7bo3b3o3b2o3b4o6b2o6bobo3bo3b
obo2b3o2b3o2b4o2bo5b2o4b2ob2o2bo3b2obob4o2b4obo5b3obo\$ob4ob4ob3ob3o2bo
bo2b2o3b3obob2ob3o3bo2b2o4bo2bo2b2ob2o3b6o2b3o2bob2o8bo2bo2bob2o2bo2bo
4bob2o2b3o11bo2b2o2bo2bob3ob2ob3o7bo2b3o2bobo4bo2b6o2b2o\$2o2b3o2bo2b2o
6b3ob4o3b6o2b2ob2ob2o2bo2b5obo3bo3bo2bobo3bo2b2o3bo5b4obob2obo2bobobo
4b2obobo2b2obob2o2bobob4o2bob6ob2o5b3o3b3o2bob2o3b3o2b3ob2o4b2obo\$4b2o
b3o2b5o5bo2b2obob3ob2obo2b2ob5o3bo2b2obo2b3o2bo2b6ob3o3bobo2bobobobob
2o6b2ob2o2bo3b3o2bo3bo8bobo4bo3b2o5b2obob2obobo2bob2obo4bo2bo2bo2bob2o
3bob3o\$2o3b2obo3bo2bob2obo3bobobob2obo2bobo3bo5b6o2bobobob5o2bo4bobob
2o3b5obo2b5o2bo2bob2obobo3bob2o2b2obo5b2obo3b4obob2o3b3ob5o3bo2b2o3b3o
b2ob2o2bo2bobob2o\$2b2o6b2o4b4o2b2ob2ob3ob6obobob3ob2ob3ob2ob2ob4ob4o4b
o2b4obob3obob2ob3o3bo3bo2b2obobo6bo3b3ob3ob3obo2b2obo2bo5bobo3b2o2b4o
2b3ob2o4b3o2b2ob2o2b3o\$2bobo3b2ob2o2bo3b2o6b2o2bobo4bo3b3o3bo2b5obo3bo
2b2obobob4o4bob9o5bo4b3o4bobo2b3o3bob2o3bo2b2o4b2o2bo2b3o2b12ob5o4bo2b
obobob2ob5o2b2o2bo\$3o3bo4b3o7bo2b2ob2ob5o2b3o2bobob5ob4obo4b3o2b4ob2ob
2o2bob4o2b2obobo3b3o2bo2b2obobo8bob2ob2ob2obobo6b4o3b3o3b9o2bob8o4bobo
bo5bobo2bo\$b2obobob5obo2b3ob4o4b3obob4ob2ob3o2bo3bobobo2b2o3bo3b3o2b2o
b2o9bob5o3bo6bobo3bobobo2b3ob2o3bobo2bob4o2bo2bo6bobobo2bo3b9o2b6ob3o
2bobob2ob2o\$obob2ob10ob2obo2b4ob2o5b3obo4b2o2b2obobobo2bo2b5o2b2o3bo2b
3o3b3o3bob2obo3b3o3b3ob3ob2ob2ob2obob2ob2o5b3ob4o2bobo2b3obob2obo2bob
5o2b2o2b2obo10b3o\$b2ob3o5bo3bo8b2obobo3b4ob2o3bo3b3obob2o2bo2b2ob4obob
ob6obob5o2b2o3bob4o2b2ob7o3b2ob2ob3o2b3o2b3o2bob4obob4o3b2ob7ob2o3bo6b
2o2bo3b2ob5obo\$5b5ob4ob2ob2obo2b3ob5o3b3o3b3o2bo2bob2obo2b4o2b3obo2bob
2o2bo2bo2b2o3b2o2b3o2b3obo4b5o3bobo2bob2o4bobobo2bo3bo2b2o2b2ob2ob6o6b
obob4ob2o2b3o3b4ob4o\$bo2b3ob3o4bobo5b2o3b2obo2b2o4b2o3bobo3b2obo3b3o2b
ob4o2bo2b6o2bobo6bobo2bob2obobob4o2bobo2bo7bo2b3ob4ob4o3bo2b2o2bo2bobo
bobobob2ob2o2bo2b2obobob2ob2ob2o2b2o\$3bob2o2b2ob2o3bo3bo2b3o2b2obo2b3o
2b2ob2obo2b2o2b3ob3obo4b4ob3obob2o3b2o2b2o2b2o3b2o3bo3b3ob5ob2ob3ob2ob
2o10b3obobo3b3o3bob3o3b2o2b2o2b4ob2ob2ob2ob5o\$ob6o3b2obob2obo3bo2b4o2b
o2bo2bo3bo4b3obobob2o3bobobo4b2o3b4ob2ob2o3bobob2ob3o4b2obo2bo2bobob4o
3bo2bob3o2bo2bo2bobo3bo2b2ob3obo4bob4obo3b3obobobobo6bobo2b2o\$2bob10ob
3o3b3o3b4o3b2ob4o3bo5bo7bo2b3ob3ob2o3b2obo2bob4ob4obob3obob3o2b3o5b3o
5b2o5bo3b3o3b4ob3o3bo6bo3bo3bo2bobobobo5b3ob3ob4o\$bobob3o2b3o5b2ob6o2b
2ob2obobo3bobob5obob2o2bo5bo2b4o4bobobob3ob2ob2o2bo2bo3b2ob10o2b3obobo
b3o2b4obobob4o2bo3bob2obob3obobo3b2ob2obobobo2bo2bo2b2o2b2ob2obo\$bob4o
b3obo2b3ob2o3b9o2bo5bo2bobo4bobob2ob2o3b6o3bo2bob2o2b2ob4ob5o2bo5b4obo
4b3ob2obo2b3obob5o2bo3bobobo5b2obobo2b2o6bo2bo2bob2o2bobo2bob2o2bob2o\$
4bobob2o5b2o4b3obob3o2bo2bob5o2bob2obobo7bo3b6ob3obo3b4o3b2ob2obo2bo3b
o2b4o2b2o2bob2ob2ob4o3b4obo2b3obo3b4o3b2o2b2ob2obob2ob4obob9ob2o3b3obo
\$b7obobob2ob6o4bob3ob2obo3b3ob2o2bo4bobo6bo3bo2bo2b2ob3ob2ob4obo5bo5bo
5bob3ob5o2b3ob2o4b2o2bob3o2b2o4bob2ob2ob3ob2obo3b3o3bobobo2bo7b5obobo\$
3b2o3bo3bobobo3b4obob2o2bobob4o2bo3b2o7b4ob5o4b3o3b2o2bo2bo2bo13bo2bo
2bobob2obo2b2o2b4ob4o2b4ob4o2bo3b2ob4ob2ob3ob2o3b3o3bo2bob2ob3o2b3o2b
3o\$2bob3o2b2o2bo5bo9b2o2b3ob3ob3ob3ob5ob2obobo2bo2bo3b5o2b2o5b2o2bo2b
2o2bob2ob2ob2ob8o2bobob2ob2ob4o2bob2o2b3o3b2ob6obo7bo2bobo2b3ob2obo2b
2o6bob3o\$2o4b2o7b3obob2o5b2o2b3obo3bob2obob2o4bo3bob2o3b2ob2o3b5o4b2o
3b2ob5o2b5o2bo3bob7ob5o6b3o4bo2bob6o2bo2b2o8bob2o5bo3b3o2b2obob3obob3o
\$ob2o2bob2o2b2o2bobobo2bo2b2obo5b4obobo2bo2b4ob3obob2o2bobobo4b4obo6bo
b2ob3o3b3ob2o3bobob3obo2bo4bobo4bo2b2ob2ob3ob2o2bobobo2b3obo3bo3bo6bob
ob4ob3ob8obo\$2b3o2bo2b3o3bob2o2b3ob3ob3o2b2o4b2o2bob2ob2obo3bob2ob2o2b
ob2o2bo3b6ob8obobobob8ob2o2bo2bob2ob3o2bobobobob2o6bo2b2o6bo3b2o2bobo
5bo2b2ob5ob3obobo\$2ob2ob2o2b8obo6bobobo3bo4bob3o3b2o3bo2bo2b2ob3o2b2o
3b2obob3o5bo4b3obob2o2bobob2o3bob6ob2o4bob3o3bo3b2o3b3o2bobob2o3b2ob3o
bob2o3bo3bobo3b2o3b4o4b2o\$o2b4ob3ob2o5bobobob2ob2ob2o2b3o2b6obob3o3bob
5obobobobo9b7o2bob2o3b2o2bob6o3b2o2b3o3b8o3b3o2b2obo2bobobob2o4b4ob2ob
ob2ob2obob6obobo5b4o\$bo2b3obo2b3obobobob5o4b6ob2ob3ob3o5bob2obob2o5bob
o2b5o5b2obo4bo2bobobob2obo2bob3o2b4o3bob2obob2o4b3ob4ob4o2bobob3obob3o
b2obob2o5bobobob2ob5o2bobo\$2obo2bo2bo4b7o4bo3bo3b2o3bo2b2o2bob2ob2ob2o
3bobo2b2ob5o5bo2bo2b5obob2obob3o3bo8bob3o2b2obo2bob3o2bo2b3ob5o3b2o2bo
b2o2bo3b2o2bo2bob2obo4b4obob2obo\$6b8o2b2ob6ob3o3bob6o2b3ob2obo2b2o2bo
3b2o2bob2obob2ob3ob2ob2obo3b4o3b2ob7obob2ob5obobob2o2bob2o2bo3bo2bob2o
bo2bobo2bobobobobo2bo2b3ob2ob2o6bob3obobo\$obob2o2bob2ob2obo2bobo3bo4b
5obobo3b4o3b2o2bobo2b2ob3o4b3o2b3o4bobobobob2o2b2o3bobob2obo4bo3bob4ob
2ob2obo3bo2b2o2bo3b2ob3o2bo2b2o2bob3ob4ob5o3bo4b2o3bob2o2b2o\$3b2o4b3o
2bo2bobo5b2ob3o2bo2b3o3b3obobobo2bo2b3obob3ob4ob2o3bob4o2bob3o4b2ob4ob
o2bob3ob2o5b2ob6ob2o3bobob3o2bobob5o5b3obob3obo2bob2ob5o3b2o4b4obo\$4ob
obob2ob2obo4b7obo3b8obobo2b2ob3o3bob2ob2o2bo3bobobobo5b2obo2b5obo4bo2b
ob2obo4bob4obob2o3bo2bob8o2bo2bo2b2obob2obo3bo2bo2b2o3b3o2bo2b2ob2ob5o
2bo2bo!
``````
Natural glide-symmetric puffer
Code: Select all
``````x = 7, y = 4, rule = B34twyz/S23-ekn4eit
3ob3o\$obobobo\$obo\$4bo!
``````
Lifewiki: User:GUYTU6J
---
Someone please find a use for this:
Code: Select all
``````x = 9, y = 7, rule = B3/S23
6bo\$6bobo\$5bo2bo\$b2o3b2o\$o2bo\$bobo\$2bo!
#C [[ COLOR BACKGROUND 255 200 82 COLOR ALIVE 81 143 51 ]]
``````
LaundryPizza03
Posts: 749
Joined: December 15th, 2017, 12:05 am
Location: Unidentified location "https://en.wikipedia.org/wiki/Texas"
B2i3-e4i/S02aei3-i5i has a small replicator (the same one from B34/S0235). Nonetheless, the rule is generally apgsearchable because it has a tiny bomber-like repship:
Code: Select all
``````x = 5, y = 5, rule = B2i3-e4i/S02aei3-i5i
bobo\$2ob2o3\$o!
``````
Aside from the replicator and the repship, the only objects of note so far are a c/4 diagonal other than the glider and a p64 gun, both shown in minimal phase:
Code: Select all
``````x = 4, y = 5, rule = B2i3-e4i/S02aei3-i5i
2bo\$2o\$3bo\$b3o\$bobo!
``````
Code: Select all
``````x = 106, y = 56, rule = B2i3-e4i/S02aei3-i5i
52b3o16\$53bo\$51bobobo3\$51bobobo\$53bo27\$50bo2bo\$55b3o12b2o\$50bo2bo16b2o
33bo\$35bo34b2o33bo\$2bo32b2o33b2o33bo\$2o49bo18b2o\$2bo32b2o\$35bo!
``````
(The latter looks quite like the gun from B368/S12578, and functions like one of its higher-period variants.)
Code: Select all
``````x = 4, y = 3, rule = B3-q4z5y/S234k5j
2b2o\$b2o\$2o!
``````
LaundryPizza03 at Wikipedia
The latest edition of new-gliders.db.txt and oscillators.db.txt have 22470 spaceships and 998 oscillators from outer-totalistic rules. You are invited to help!
FWKnightship
Posts: 451
Joined: June 23rd, 2019, 3:10 am
Location: 我不告诉你
GUYTU6J wrote:A rule which keeps active due to a bunch of spaceship collisions
Code: Select all
``````x = 193, y = 134, rule = B34twyz/S23-ekn4eit
2ob4o2b2ob2o5bobob3o4bob2o2b2obobob2ob2o4b5o2bobobo3bob2ob2o2b5o2bobob
ob2obobobo3b2ob2ob4o2b2ob2o3bo4b2o2b3o2bo5bo3b2obo6bob2o3bob2ob3o5bobo
8bo2b3obo\$obo3bo2b2ob2o2bob2o2b3obo3bo2bobobob5ob4ob2obo2bob3o2b2o3bo
2b2obob3o3bo3b4o2bo2bobob3o2b2ob2o3b2o3bob3o2bobo3bo3bob5ob3o2bo2b2o2b
4ob3o2bobo3bo2b4o2bo3b4obobo\$3obob2obobo3b8obo5bo4b2obo4bo2bo5b3o2b2o
5bo2bobob3obobo2b2o4bobo5bo3b2ob3ob3o2bobob2ob4obob2o3bob3obob2o3bo2bo
2b2ob2o3bo2bobobo4bobo4bo2bo3b3obob4o\$o4b4ob3o2b2obob6ob2ob2o4b2ob3o2b
2o2b2ob2o2b3obo2b2ob2o3b2o3b2ob3obobo2bo2bo2bobobobo3b2o2bo3bo6b3o5b3o
b2obo2b2obo2b2obobobo4b4o2bob4o2b5obob2o2bobobo2b6o\$3bob3o3bob2ob2o2b
2obob3o2b3o3b3ob4obob6obobo3b4o4bo2bo2b2o2bo3b5ob4o2b3obo3b2ob2obob2o
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bo2b2obo2bo2b3obobobo3b4ob3obob2o4b2o3b3ob3ob2o3b2o2b6o2bob4obobo3b7ob
obo2bobobo4bob3o3b3o3b2o2bobo6b2o4bo4bo2bo2bobo2b5ob2o2b2o\$ob2obo2bobo
bob4ob4ob2o3b2obobo2b5obob3ob3o2bo2bo3bob2obo2b2obo2bobo2bob3o3bo3bobo
3b2obo2b3ob2o4b5obob2ob7obo3b2o2b3ob3o2b2o3bo3b3ob3o2b4o2bob2obo3b2obo
b3o\$2ob2o2b5ob3obo5b2o2bob2o3bobo2bob2ob3obob2o8b3ob3o3bo4b2o2bo2b2o3b
ob3o3b2o6b2ob2obo4bob2ob2o2bobobo5bo3bo2bo2b3ob3o2b2ob5obob3ob2obobob
2o4bob4ob2o2bo\$4o2b2ob3ob2o3bob4o2bobo3bobob2ob2o6bo3bobobob2o3bo3b4o
2b3ob3o2b3obob2ob5o2b2obo3b2ob3o3bo5bob7ob2o3bo3bo2b6obo2bo4b2o5bo8b2o
b5o3b2o2b3obo\$4b3o7bo3bo3b6o4bob2o2bob2o2b2o2bobob3obob2o7b3ob5ob2obo
2bo3b2o3bo2b2ob5o2bo3b2o2bo2b2o2bo2b3o2b2obobo4bobob4obo2b3o5b2obo2b5o
bobo2bo2bo4bobob3o\$4b2o5bobo3bobo2bo3bo3bobo2bo3bo2bo4bo8bob2o4bo4bobo
7b3o4b2ob6ob3o2b2obo3bo3bo3b2o2b2o3bobobo2bo2b2ob2obob4obo2b2o4b4o2b4o
b2ob2ob5o2bobo2b3ob3o\$2bo2bobo5bo3bob2ob4obo2bob2ob2o3bobobobob2ob4o3b
3o2bobo5b2ob5obobo2b3ob3o3b5o2b6ob4ob2o2b4ob2obobo2b4ob2ob2o2bobobob3o
b2obob3o3bob2o2bo2b3o3b2o3b4ob4o\$2b2ob2o5b2ob2o2b4o2bo4b5obob2obob4ob
3o2bo2b2o3bobo3bobo2b2ob2ob2obo5bob2ob2obo5b2o6b2ob4obo4bob2o2bobob5o
3b2o5bob2o2b3ob2obob3o2b2ob2o2bo3b3o6bobobo\$2bobob3o3bob6o4b2o2bo2bob
2o2b3ob2o4b2o3bobo5b2o3b2o2bob5ob2o4bob2ob4ob2o5bobobob2obo5b2obob3o5b
o2bo3b2o6b2o3b4o2bobo5bo3bobobobo2bobobobob2ob3ob3o\$2obobo3b3ob2o3bo3b
6ob2o3b2o2bo4bob2o2b6o2b2ob7o2bobo2bo2bo2b4o3b3obo2b3obobo2b3o4bo2b5o
2b2obob2o2b3ob3obob3ob2obob3o2bobo3bobobob2ob3obobo2bo3bob3obob2obo\$bo
b4o3bob2o5bo2bobo4b2ob4o3b4ob4o3bo3b3obob2o2b3o3bo4b9ob2o2bo2b4o4bobob
o3bo2bobob2o4bo3b3o3bo5b2o4b8o2bo3bob3obob3o2bo2b2o10b2o2b3o\$b2o2bobob
6o2bo2b2obobo2b2ob2ob2ob2obo5b3obo2bobobo2b2o2b2o2bob6obob3o2b4obo2b6o
5bo2bobobo6b4o6b2o2b3o3b6obob2o7bob2ob3ob2ob4o3b2o2bo2bob2ob2ob2obo\$ob
2o2bob2obobo4b5o3b2ob2obo2b2obobo2bobob2obo2b2o3b4o5b2o2b2o3b3o2b4obob
4o3bo3bo2bo2bo4b2o2b3o4b4o2bo4bo3bo2bobo3b2ob4ob2ob4o2bobo2b4ob3obo2b
2o2bob3o2bo\$3o3b2o4b2ob2o2b2obo3bob3ob2o3bo3b5o4b2o5bo9bob2ob5o2bob2ob
o6bo2b3o3b3o2b4o2b2o4bo3bob5o3b2o3bobob2o2b3obo2bo3b2obo3b2o2b2obobob
3o3bob2o2b2o2b2obo\$ob3o2bobobo3b2o3b4ob3o2b3o4bo3b5o3bob2obo2b3o2bob2o
b2obo3b2o2bob2ob3ob3o4b2obobo2b2o3bob5ob10o2b2o2bob2obob2obob2obo3b2ob
3o2bobob2o2bobobo2b3ob3obob4obo2b2o\$2bo2b2o3bo2b2o2b2ob4obobo3bo2bobo
2b2o4b5obob6ob2o2b2o5bob2ob3ob3o2bo4b3ob11o3b2o4bob3obo2b3o3b4ob2o5bob
obobo5bob7o2b2ob8ob14o2bo\$bo4b3ob2obob4ob2o2bob2obob3o2b2o8b2o3b2obob
3ob2o3b7o2bobo2b2o2bo2b2ob3obob4ob7ob2ob2ob2o2b2o5b2o4b7obobob5o4b2obo
2bobo2bob6obob2obobob3ob8o\$bob4obob2o5bo3bob3o2b2ob2o2b2ob2ob2ob2o2b3o
4b2ob5o2b5obobo2b2ob3o2b3obo2b2o2b5o2bo2bo2b2obobo2b4o4b2obo3b2obobo2b
obo2bo3bo5b2o3bo4b2ob2o3bob3o2b3o2b3obobo\$2b5o3b3o3b3o2b2obob3o3bo3b2o
bobo5b2o3b2ob2o6bo2b4o3b3o3bo2b2ob2o3b2obob4obob3ob5o2b2o2bo2bo7bo2b2o
bob2o9bobob2o2bo8b2o2b2o3b3o2bo3bobo2b3o2bo\$3bo2b4o3bo4b5obo2bobo2b2o
4b3ob2o2bobob2o6b2o6b3o6bo3bo2bo2b2obob2o2bobobo2bo5b3obobob2obob2o3b
2ob3o2b5o4b3o2b5obo2b3o3b4ob2o2b2obobob3o3b3obob2o\$2obo2bobob2ob2o2bo
4b7o2b2obo2b2o3bo3b2o2b2o3b3obob2obo6b4ob2obobob2o4bo3b2ob5o2b2o4b2o3b
3obo4b3o2bobo2bob12ob4o2b2obo6b4o2b2ob2o3b3o3bo2bo2b2o\$b4o5b3ob2o4b5ob
ob2ob3o2b2ob2ob3o5bo4b4ob5o2b4o2bo4b3obo4bobo2bo2b3o4b5o6b4obobob3o3b
3o2bo2bobo2bo3bo7b2obobob2obo2b2obo3b4ob2o2bo4b3o2bo\$2bob4o3b2o6bo2b4o
bobo3bob3obo3b5obobo2b2ob3obobo2bob3o4b3ob2o2bob3o2bobobo2b5o5b4o2bo4b
obobobob5obobo2b3o2b2o2b6o3b2o3b4obo3bob7ob6o7bo\$2bobobobobo4b3obo7bo
3b3o4b3ob2ob3obo2bob4ob2o5bob3obo2bo2bobo2bobobo2b4o2bobob2o2bob3obobo
4bo3bob2obob5ob2ob2obo3bo5b2ob3obo3b2o6b3o3bobob4o2bo2b2obo\$4o2bo2b8ob
ob3obo3bo3bo5b3o2bo3b3obob2obo2b2o2bo3b2o7bo3b3o3b2o3b4o6b2o6bobo3bo3b
obo2b3o2b3o2b4o2bo5b2o4b2ob2o2bo3b2obob4o2b4obo5b3obo\$ob4ob4ob3ob3o2bo
bo2b2o3b3obob2ob3o3bo2b2o4bo2bo2b2ob2o3b6o2b3o2bob2o8bo2bo2bob2o2bo2bo
4bob2o2b3o11bo2b2o2bo2bob3ob2ob3o7bo2b3o2bobo4bo2b6o2b2o\$2o2b3o2bo2b2o
6b3ob4o3b6o2b2ob2ob2o2bo2b5obo3bo3bo2bobo3bo2b2o3bo5b4obob2obo2bobobo
4b2obobo2b2obob2o2bobob4o2bob6ob2o5b3o3b3o2bob2o3b3o2b3ob2o4b2obo\$4b2o
b3o2b5o5bo2b2obob3ob2obo2b2ob5o3bo2b2obo2b3o2bo2b6ob3o3bobo2bobobobob
2o6b2ob2o2bo3b3o2bo3bo8bobo4bo3b2o5b2obob2obobo2bob2obo4bo2bo2bo2bob2o
3bob3o\$2o3b2obo3bo2bob2obo3bobobob2obo2bobo3bo5b6o2bobobob5o2bo4bobob
2o3b5obo2b5o2bo2bob2obobo3bob2o2b2obo5b2obo3b4obob2o3b3ob5o3bo2b2o3b3o
b2ob2o2bo2bobob2o\$2b2o6b2o4b4o2b2ob2ob3ob6obobob3ob2ob3ob2ob2ob4ob4o4b
o2b4obob3obob2ob3o3bo3bo2b2obobo6bo3b3ob3ob3obo2b2obo2bo5bobo3b2o2b4o
2b3ob2o4b3o2b2ob2o2b3o\$2bobo3b2ob2o2bo3b2o6b2o2bobo4bo3b3o3bo2b5obo3bo
2b2obobob4o4bob9o5bo4b3o4bobo2b3o3bob2o3bo2b2o4b2o2bo2b3o2b12ob5o4bo2b
obobob2ob5o2b2o2bo\$3o3bo4b3o7bo2b2ob2ob5o2b3o2bobob5ob4obo4b3o2b4ob2ob
2o2bob4o2b2obobo3b3o2bo2b2obobo8bob2ob2ob2obobo6b4o3b3o3b9o2bob8o4bobo
bo5bobo2bo\$b2obobob5obo2b3ob4o4b3obob4ob2ob3o2bo3bobobo2b2o3bo3b3o2b2o
b2o9bob5o3bo6bobo3bobobo2b3ob2o3bobo2bob4o2bo2bo6bobobo2bo3b9o2b6ob3o
2bobob2ob2o\$obob2ob10ob2obo2b4ob2o5b3obo4b2o2b2obobobo2bo2b5o2b2o3bo2b
3o3b3o3bob2obo3b3o3b3ob3ob2ob2ob2obob2ob2o5b3ob4o2bobo2b3obob2obo2bob
5o2b2o2b2obo10b3o\$b2ob3o5bo3bo8b2obobo3b4ob2o3bo3b3obob2o2bo2b2ob4obob
ob6obob5o2b2o3bob4o2b2ob7o3b2ob2ob3o2b3o2b3o2bob4obob4o3b2ob7ob2o3bo6b
2o2bo3b2ob5obo\$5b5ob4ob2ob2obo2b3ob5o3b3o3b3o2bo2bob2obo2b4o2b3obo2bob
2o2bo2bo2b2o3b2o2b3o2b3obo4b5o3bobo2bob2o4bobobo2bo3bo2b2o2b2ob2ob6o6b
obob4ob2o2b3o3b4ob4o\$bo2b3ob3o4bobo5b2o3b2obo2b2o4b2o3bobo3b2obo3b3o2b
ob4o2bo2b6o2bobo6bobo2bob2obobob4o2bobo2bo7bo2b3ob4ob4o3bo2b2o2bo2bobo
bobobob2ob2o2bo2b2obobob2ob2ob2o2b2o\$3bob2o2b2ob2o3bo3bo2b3o2b2obo2b3o
2b2ob2obo2b2o2b3ob3obo4b4ob3obob2o3b2o2b2o2b2o3b2o3bo3b3ob5ob2ob3ob2ob
2o10b3obobo3b3o3bob3o3b2o2b2o2b4ob2ob2ob2ob5o\$ob6o3b2obob2obo3bo2b4o2b
o2bo2bo3bo4b3obobob2o3bobobo4b2o3b4ob2ob2o3bobob2ob3o4b2obo2bo2bobob4o
3bo2bob3o2bo2bo2bobo3bo2b2ob3obo4bob4obo3b3obobobobo6bobo2b2o\$2bob10ob
3o3b3o3b4o3b2ob4o3bo5bo7bo2b3ob3ob2o3b2obo2bob4ob4obob3obob3o2b3o5b3o
5b2o5bo3b3o3b4ob3o3bo6bo3bo3bo2bobobobo5b3ob3ob4o\$bobob3o2b3o5b2ob6o2b
2ob2obobo3bobob5obob2o2bo5bo2b4o4bobobob3ob2ob2o2bo2bo3b2ob10o2b3obobo
b3o2b4obobob4o2bo3bob2obob3obobo3b2ob2obobobo2bo2bo2b2o2b2ob2obo\$bob4o
b3obo2b3ob2o3b9o2bo5bo2bobo4bobob2ob2o3b6o3bo2bob2o2b2ob4ob5o2bo5b4obo
4b3ob2obo2b3obob5o2bo3bobobo5b2obobo2b2o6bo2bo2bob2o2bobo2bob2o2bob2o\$
4bobob2o5b2o4b3obob3o2bo2bob5o2bob2obobo7bo3b6ob3obo3b4o3b2ob2obo2bo3b
o2b4o2b2o2bob2ob2ob4o3b4obo2b3obo3b4o3b2o2b2ob2obob2ob4obob9ob2o3b3obo
\$b7obobob2ob6o4bob3ob2obo3b3ob2o2bo4bobo6bo3bo2bo2b2ob3ob2ob4obo5bo5bo
5bob3ob5o2b3ob2o4b2o2bob3o2b2o4bob2ob2ob3ob2obo3b3o3bobobo2bo7b5obobo\$
3b2o3bo3bobobo3b4obob2o2bobob4o2bo3b2o7b4ob5o4b3o3b2o2bo2bo2bo13bo2bo
2bobob2obo2b2o2b4ob4o2b4ob4o2bo3b2ob4ob2ob3ob2o3b3o3bo2bob2ob3o2b3o2b
3o\$2bob3o2b2o2bo5bo9b2o2b3ob3ob3ob3ob5ob2obobo2bo2bo3b5o2b2o5b2o2bo2b
2o2bob2ob2ob2ob8o2bobob2ob2ob4o2bob2o2b3o3b2ob6obo7bo2bobo2b3ob2obo2b
2o6bob3o\$2o4b2o7b3obob2o5b2o2b3obo3bob2obob2o4bo3bob2o3b2ob2o3b5o4b2o
3b2ob5o2b5o2bo3bob7ob5o6b3o4bo2bob6o2bo2b2o8bob2o5bo3b3o2b2obob3obob3o
\$ob2o2bob2o2b2o2bobobo2bo2b2obo5b4obobo2bo2b4ob3obob2o2bobobo4b4obo6bo
b2ob3o3b3ob2o3bobob3obo2bo4bobo4bo2b2ob2ob3ob2o2bobobo2b3obo3bo3bo6bob
ob4ob3ob8obo\$2b3o2bo2b3o3bob2o2b3ob3ob3o2b2o4b2o2bob2ob2obo3bob2ob2o2b
ob2o2bo3b6ob8obobobob8ob2o2bo2bob2ob3o2bobobobob2o6bo2b2o6bo3b2o2bobo
5bo2b2ob5ob3obobo\$2ob2ob2o2b8obo6bobobo3bo4bob3o3b2o3bo2bo2b2ob3o2b2o
3b2obob3o5bo4b3obob2o2bobob2o3bob6ob2o4bob3o3bo3b2o3b3o2bobob2o3b2ob3o
bob2o3bo3bobo3b2o3b4o4b2o\$o2b4ob3ob2o5bobobob2ob2ob2o2b3o2b6obob3o3bob
5obobobobo9b7o2bob2o3b2o2bob6o3b2o2b3o3b8o3b3o2b2obo2bobobob2o4b4ob2ob
ob2ob2obob6obobo5b4o\$bo2b3obo2b3obobobob5o4b6ob2ob3ob3o5bob2obob2o5bob
o2b5o5b2obo4bo2bobobob2obo2bob3o2b4o3bob2obob2o4b3ob4ob4o2bobob3obob3o
b2obob2o5bobobob2ob5o2bobo\$2obo2bo2bo4b7o4bo3bo3b2o3bo2b2o2bob2ob2ob2o
3bobo2b2ob5o5bo2bo2b5obob2obob3o3bo8bob3o2b2obo2bob3o2bo2b3ob5o3b2o2bo
b2o2bo3b2o2bo2bob2obo4b4obob2obo\$6b8o2b2ob6ob3o3bob6o2b3ob2obo2b2o2bo
3b2o2bob2obob2ob3ob2ob2obo3b4o3b2ob7obob2ob5obobob2o2bob2o2bo3bo2bob2o
bo2bobo2bobobobobo2bo2b3ob2ob2o6bob3obobo\$obob2o2bob2ob2obo2bobo3bo4b
5obobo3b4o3b2o2bobo2b2ob3o4b3o2b3o4bobobobob2o2b2o3bobob2obo4bo3bob4ob
2ob2obo3bo2b2o2bo3b2ob3o2bo2b2o2bob3ob4ob5o3bo4b2o3bob2o2b2o\$3b2o4b3o
2bo2bobo5b2ob3o2bo2b3o3b3obobobo2bo2b3obob3ob4ob2o3bob4o2bob3o4b2ob4ob
o2bob3ob2o5b2ob6ob2o3bobob3o2bobob5o5b3obob3obo2bob2ob5o3b2o4b4obo\$4ob
obob2ob2obo4b7obo3b8obobo2b2ob3o3bob2ob2o2bo3bobobobo5b2obo2b5obo4bo2b
ob2obo4bob4obob2o3bo2bob8o2bo2bo2b2obob2obo3bo2bo2b2o3b3o2bo2b2ob2ob5o
2bo2bo!
``````
Natural glide-symmetric puffer
Code: Select all
``````x = 7, y = 4, rule = B34twyz/S23-ekn4eit
3ob3o\$obobobo\$obo\$4bo!
``````
(42,20)c/170 puffer
Code: Select all
``````x = 67, y = 43, rule = B34twyz/S23-ekn4eit
28bo\$7b3o17bobo\$7bobo18bo\$7b3o4\$42b2o\$42b2o2\$33bo\$33bo\$33bo\$4b2o\$3bo2b
o\$4b2o4\$2b3o\$b2o2bo3bo\$2o2bo3bobo\$b4o3bobo54b2o\$2b2o5bo55b2o2\$10bo15b
2o\$9b3o14b2o29b2o\$9bobo44bo2bo\$9bobo39b2o4b2o\$8b5o38b2o\$8bo3bo\$8bo3bo\$
9b3o7\$41bo\$40bobo\$40bobo\$41bo!
``````
3c/6 backrake
Code: Select all
``````x = 7, y = 6, rule = B34twyz/S23-ekn4eit
2o\$o2b2o\$o4bo\$bo4bo\$6bo\$4b3o!
``````
p3,p4,p24,p42
Code: Select all
``````x = 76, y = 30, rule = B34twyz/S23-ekn4eit
5bo11b2o12b2o2b2o30bo\$4bobo9bo14bo4bo30b2o\$2bo2bo10bo14bo4bo29bo\$2bo
14b2o12b2o2b2o\$2bo3\$59b2o13bo\$60bo13b2o\$57bo2bo12bo\$57b3o\$4bo2\$3b3o2\$
3b3o60b3o\$52bo12bo2bo\$4bo45b2o13bo\$51bo13b2o3\$3bobo\$bobobobo\$3bobo53bo
\$4ob4o48b2o\$58bo\$4ob4o\$3bobo\$bobobobo\$3bobo!
``````
I'm too shy to talk to other members.
But I want to upload my apgsearch results to Catagolue like others.
Code: Select all
``````x = 12, y = 1, rule = JvN29
FWKNIGHTSHIP!``````
Posts: 571
Joined: May 7th, 2016, 8:53 am
Contact:
FWKnightship wrote:
GUYTU6J wrote:A rule which keeps active due to a bunch of spaceship collisions
Code: Select all
``rle``
Natural glide-symmetric puffer
Code: Select all
``rle``
(42,20)c/170 puffer
Code: Select all
``````x = 67, y = 43, rule = B34twyz/S23-ekn4eit
28bo\$7b3o17bobo\$7bobo18bo\$7b3o4\$42b2o\$42b2o2\$33bo\$33bo\$33bo\$4b2o\$3bo2b
o\$4b2o4\$2b3o\$b2o2bo3bo\$2o2bo3bobo\$b4o3bobo54b2o\$2b2o5bo55b2o2\$10bo15b
2o\$9b3o14b2o29b2o\$9bobo44bo2bo\$9bobo39b2o4b2o\$8b5o38b2o\$8bo3bo\$8bo3bo\$
9b3o7\$41bo\$40bobo\$40bobo\$41bo!
``````
3c/6 backrake
Code: Select all
``rle``
p3,p4,p24,p42
Code: Select all
``rle``
Oblique quadratic growth have been found (not sure) during corderizing attempts:
Code: Select all
``````x = 810, y = 1378, rule = B34twyz/S23-ekn4eit
711b3o\$711bobo\$710b2ob2o14\$708b3o\$708bobo\$707b2ob2o47\$669b3o\$669bobo\$
668b2ob2o14\$666b3o\$666bobo\$665b2ob2o47\$627b3o\$627bobo\$626b2ob2o14\$624b
3o\$624bobo\$623b2ob2o254\$491bo\$490b3o2\$489b2ob2o10\$488bo\$487b3o\$486b5o
5\$485b3ob3o\$485bo2bo2bo2\$487b3o2\$487b3o\$486bo3bo\$483b4o3b4o\$484b2ob3ob
2o\$481bob2o7b2obo\$481b3o9b3o\$482bo11b2o8b2o\$469bo2b2o21b2o3b2ob4o\$468b
2o2bo24bobo2b2o2b2o\$467b3ob2o25b5o2b3o\$468b2o2bo22bo5bo3b2o\$469bo2b2o
26bo\$477bo18bo\$477bo16b5o\$494bo27b2o\$522b2o3\$493b2o17b3o\$488b2o3b2o\$
488b2o3\$492bo\$491b2o\$491bo\$479bo\$429bo48bobo\$417b3o7bobo48bobo\$417bobo
6b2o51bo65b2o\$416b2ob2o6bobo53b2o7b3o50b2o\$429bo53b2o6bo3bo\$490bo5bo9b
2o\$490bo5bo9b2o29b2o\$490bo5bo39bo2bo24b2o\$491bo3bo35b2o4b2o25b2o\$492b
3o36b2o2\$554b3o6\$414b3o\$414bobo\$413b2ob2o103bo\$520bobo\$520bobo\$521bo
65b2o\$587b2o3\$579b2o\$578bo2bo\$488b2ob2o80b2o4b2o\$573b2o\$489b3o\$490bo8\$
563bo\$562bobo\$562bobo\$563bo23\$393bo\$375b3o14bobo\$375bobo15bo\$374b2ob2o
4\$407b2o\$375bo31b2o\$374b3o\$373b5o20bo\$373bo3b2o19bo\$371bo2bo2b3o18bo\$
370bo3bo2b2o\$371bobo2\$373b2o\$368b3o2bobo\$364b2o4bo2bobo\$363bo2bo4bob3o
\$346bo15bo3bo2b2obobo\$344b3o15bo9bo40b2ob2o\$344bo17bo2b3o4bo2bo38bobo
13b2o\$344b3o16b3o8b2o38b3o13b2o\$346bo22b2o\$368bo22b2o\$367bo3bo19b2o29b
2o\$367bo3bo49bo2bo24b2o\$368b3o51b2o25b2o2\$440bo\$420b2o18bo\$415b2o3b2o
18bo\$415b2o6\$406bo\$303bo101bobo\$301b3o101bobo\$301bo104bo65b2o\$301b3o
168b2o\$303bo2\$464b2o\$463bo2bo24b2o\$458b2o4b2o25b2o\$458b2o\$482bo\$482bo\$
482bo\$421b3o\$420bo2bo\$420bo\$420b2o3\$448bo\$260bo186bobo\$258b3o186bobo\$
258bo189bo65b2o\$258b3o253b2o\$260bo\$423b2o\$423b2o81b2o\$505bo2bo24b2o\$
500b2o4b2o25b2o\$500b2o\$524bo\$524bo\$524bo7\$490bo\$217bo271bobo\$215b3o
271bobo\$215bo274bo65b2o\$215b3o338b2o\$217bo2\$548b2o\$547bo2bo24b2o\$542b
2o4b2o25b2o\$542b2o\$566bo\$566bo\$566bo7\$532bo\$174bo356bobo\$172b3o356bobo
\$172bo359bo65b2o\$172b3o423b2o\$174bo2\$590b2o\$589bo2bo24b2o\$584b2o4b2o
25b2o\$584b2o\$608bo\$608bo\$608bo7\$574bo\$131bo441bobo\$129b3o283b2ob2o153b
obo\$129bo286bobo155bo65b2o\$129b3o284b3o221b2o\$131bo2\$632b2o\$631bo2bo
24b2o\$626b2o4b2o25b2o\$626b2o\$650bo\$650bo\$650bo7\$616bo\$88bo526bobo\$86b
3o526bobo\$86bo529bo65b2o\$86b3o593b2o\$88bo2\$674b2o\$673bo2bo24b2o\$668b2o
4b2o25b2o\$668b2o\$692bo\$692bo\$692bo7\$658bo\$45bo611bobo\$43b3o611bobo\$43b
o614bo65b2o\$43b3o678b2o\$45bo2\$716b2o\$715bo2bo24b2o\$710b2o4b2o25b2o\$
710b2o\$734bo\$734bo\$734bo7\$700bo\$2bo696bobo\$3o696bobo\$o699bo65b2o\$3o
763b2o\$2bo2\$758b2o\$757bo2bo24b2o\$752b2o4b2o25b2o\$752b2o\$776bo\$776bo\$
776bo7\$742bo\$741bobo\$741bobo\$742bo65b2o\$808b2o3\$800b2o\$799bo2bo\$794b2o
4b2o\$794b2o10\$784bo\$783bobo\$783bobo\$784bo4\$457b2ob2o\$458bobo\$458b3o
103\$499b2ob2o\$500bobo\$500b3o103\$541b2ob2o\$542bobo\$542b3o103\$583b2ob2o\$
584bobo\$584b3o103\$625b2ob2o\$626bobo\$626b3o103\$667b2ob2o\$668bobo\$668b3o
103\$709b2ob2o\$710bobo\$710b3o!
``````
GUYTU6J
Posts: 1072
Joined: August 5th, 2016, 10:27 am
Location: 中国
For anyone who is willing to hack the script and find (42,20)c/170 spaceship(s), here is the engine
Code: Select all
``````x = 7, y = 15, rule = B34twyz/S23-ekn4eit
4b3o2\$4bobo\$5bo9\$b2o\$o2bo\$b2o!
``````
Another puffer:
Code: Select all
``````x = 7, y = 8, rule = B34twyz/S23-ekn4eit
5b2o\$3bo2bo\$2b2ob2o\$2bo\$2b5o\$6bo\$o3b3o\$o!
``````
Lifewiki: User:GUYTU6J
---
Someone please find a use for this:
Code: Select all
``````x = 9, y = 7, rule = B3/S23
6bo\$6bobo\$5bo2bo\$b2o3b2o\$o2bo\$bobo\$2bo!
#C [[ COLOR BACKGROUND 255 200 82 COLOR ALIVE 81 143 51 ]]
``````
Hdjensofjfnen
Posts: 1519
Joined: March 15th, 2016, 6:41 pm
Location: r cis θ
The two most common spaceship speeds in this rule are (you guessed it!) 2c/6 and c/30.
Code: Select all
``````x = 23, y = 3, rule = B2e3/S2-c3-i
2ob2o15bo\$bobo15bo2bo\$3b2o15b2o!
``````
Of course, there is a c/2, but it seems to be quite rare.
Code: Select all
``````x = 4, y = 4, rule = B2e3/S2-c3-i
bo\$o2bo\$o2bo\$bo!
``````
"A man said to the universe:
'Sir, I exist!'
'However,' replied the universe,
'The fact has not created in me
A sense of obligation.'" -Stephen Crane
Code: Select all
``````x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o\$6bo\$o3b3o\$2o\$bo!
``````
Posts: 571
Joined: May 7th, 2016, 8:53 am
Contact:
GUYTU6J wrote:For anyone who is willing to hack the script and find (42,20)c/170 spaceship(s), here is the engine
Code: Select all
``````x = 7, y = 15, rule = B34twyz/S23-ekn4eit
4b3o2\$4bobo\$5bo9\$b2o\$o2bo\$b2o!
``````
...
A possible future kickback reaction usage, but not as "pushalong":
Code: Select all
``````x = 827, y = 421, rule = B34twyz/S23-ekn4eit
780bo\$778b3o\$778bo\$778b3o\$780bo16\$737bo\$735b3o\$735bo\$735b3o\$737bo12\$
826bo\$824b3o\$824bo\$824b3o\$694bo131bo\$692b3o\$692bo\$692b3o\$694bo12\$783bo
\$781b3o\$781bo\$781b3o\$651bo131bo\$649b3o\$649bo\$649b3o\$651bo12\$740bo\$738b
3o\$738bo\$738b3o\$608bo131bo\$606b3o\$606bo\$606b3o\$608bo12\$697bo\$695b3o\$
695bo\$695b3o\$565bo131bo\$563b3o\$563bo\$563b3o\$565bo12\$654bo\$652b3o\$652bo
\$652b3o\$522bo131bo\$520b3o\$520bo\$520b3o\$522bo12\$611bo\$609b3o\$609bo\$609b
3o\$479bo131bo\$477b3o\$477bo\$477b3o\$479bo12\$568bo\$566b3o\$566bo\$566b3o\$
436bo131bo\$434b3o\$434bo\$434b3o\$436bo12\$525bo\$523b3o\$523bo\$523b3o\$393bo
131bo\$391b3o\$391bo\$391b3o\$393bo12\$482bo\$480b3o\$480bo\$480b3o\$350bo131bo
\$348b3o\$348bo\$348b3o\$350bo12\$439bo\$437b3o\$437bo\$437b3o\$307bo131bo\$305b
3o\$305bo\$305b3o\$307bo12\$396bo\$394b3o\$394bo\$394b3o\$264bo131bo\$262b3o\$
262bo\$262b3o\$264bo12\$353bo\$351b3o\$351bo\$351b3o\$221bo131bo\$219b3o\$219bo
\$219b3o\$221bo12\$310bo\$308b3o\$308bo\$308b3o\$178bo131bo\$176b3o\$176bo\$176b
3o\$178bo12\$267bo\$265b3o\$265bo\$265b3o\$135bo131bo\$133b3o\$133bo\$133b3o\$
135bo12\$224bo\$222b3o\$222bo\$222b3o\$224bo16\$181bo\$179b3o\$179bo\$179b3o\$
181bo16\$129b2ob2o4bo\$130bobo3b3o\$130b3o3bo\$136b3o\$138bo16\$86b2ob2o4bo\$
87bobo3b3o\$87b3o3bo\$93b3o\$95bo16\$43b2ob2o4bo\$44bobo3b3o\$44b3o3bo\$50b3o
\$52bo16\$2ob2o4bo\$bobo3b3o\$b3o3bo\$7b3o\$9bo!
``````
[APPEND #1]
Demonstration:
Code: Select all
``````x = 2489, y = 1121, rule = B34twyz/S23-ekn4eit
2269bo\$2268bobo18b2ob2o\$2269bo20bobo\$2290b3o4\$2283b2o\$2283b2o2\$2273b2o
\$2277b2o\$2273bob4o\$2274b3o\$2275bo\$2255bo19bobo\$2254b3o21bo\$2253b4o21b
2o\$2249b2obo22bo3bo\$2248bo26b2ob2o\$2242bo4bo4bo15bo8bo\$2240b3o4bo4bob
4o7bob3o\$2240bo6b2o3bobob2o6bo3bo37b2o\$2240b3o9bo11bo7bo33b2o\$2242bo7b
6o15bobo\$2255bo9b3o2bo2bo\$2255bo15b2o25b2o\$2253bo43bo2bo24b2o\$2298b2o
25b2o2\$2269bo46bo\$2268b2o26b2o18bo\$2271bo19b2o3b2o18bo\$2265b2o4bo19b2o
\$2266bobobo\$2264b3o\$2265bo\$2264b2o\$2263bo2bo\$2263b2o17bo17b2o\$2199bo
65b2o14bobo16b2o\$2197b3o81bobo\$2197bo68b3o13bo65b2o\$2197b3o66bo2bo78b
2o\$2199bo66bo3bo\$2267bo2bo\$2268b3o32b2o35b2o\$2303b2o34bo2bo24b2o\$2334b
2o4b2o25b2o\$2334b2o2\$2268b3o\$2267bo3bo\$2266bo5bo\$2265bo3bo3bo\$2265bobo
3bobo21b2o\$2266bo5bo22b2o3\$2308b2o32b2o\$2156bo109bo5bo35b2o32b2o\$2154b
3o108bobo3bobo\$2154bo110bobo3bobo116b2o\$2154b3o233b2o\$2156bo\$2295bo4bo
\$2295bo4bobo42b2o35b2o\$2296bo3bob2o41b2o34bo2bo24b2o\$2276bo6bobo8b3obo
bobo73b2o4b2o25b2o\$2273b2ob2o4bo10bo2bobo77b2o\$2276bo19bo2bo100bo\$
2257bobo11b2o3bo5bo2bo9b5o43bo56bo\$2256bo2bo11bob2obo8bo10b3o43bobo55b
o\$2255bo8bo7bo4bo8bo10bo43bo2bo\$2245b3o7bo2bo5bo7bo10bob2o55b2o\$2244bo
3bo6bo8bo7bo10b4o7bo\$2243bo5bo6bo2bo9bo6b3o3b3o8bobo\$2242bo3bo3bo6bobo
10bo5b3o3b2o10bo\$2242bo2bobo2bo18bo8bo\$2242bo3bo3bo34b2o\$2113bo129bo5b
o35b2o\$2111b3o130bo3bo\$2111bo133b3o28b2o154b2o\$2111b3o161bo2bo24b2o14b
2o111b2o\$2113bo162b2o25b2o14b2o2\$2294bo129b2o\$2274b2o18bo128bo2bo24b2o
\$2269b2o3b2o18bo123b2o4b2o25b2o\$2269b2o147b2o3\$2332b2o\$2331bob2o\$2302b
o29b2o\$2260bo41bo8b2o66b2o\$2202bo56bobo13b3obo22bo8b2o66b2o\$2200b3o56b
obo13b3ob2o\$2200bo59bo14bo4bo\$2200b3o72bo3bo112b2o32b2o\$2070bo131bo72b
obo4b2o108b2o32b2o\$2068b3o204b2o3b2ob2o2b2o\$2068bo211b6ob2o185b2o\$
2068b3o209b2ob2o189b2o\$2070bo200b2o3b2o2bob2o22bo\$2271b2o3b2o2b2o25b2o
17bo\$2268bo11bo26bo18bo9bo92b2o35b2o\$2279bo46bo9bo92b2o34bo2bo\$2270b9o
57bo123b2o4b2o\$2271bob3obo85b2o95b2o\$2272bobobo86b2o119bo\$2272b5o150bo
56bo\$2273b3o150bobo55bo\$2274bo150bo2bo\$2426b2o\$2302bo\$2159bo141bobo\$
2157b3o141bobo\$2157bo144bo71bo\$2157b3o213b2o\$2027bo131bo213b2o\$2025b3o
345bo\$2025bo334b2o\$2025b3o331bo2bo24b2o14b2o\$2027bo332b2o25b2o14b2o\$
2368bo\$2367bobo8bo\$2358b2o7bobo8bo\$2353b2o3b2o8bo9bo\$2353b2o3\$2269b2ob
2o74bo\$2270bobo74bobo\$2270b3o74bobo36bo\$2348bo37bo8b2o\$2116bo269bo8b2o
\$2114b3o226b3o\$2114bo\$2114b3o359b2o\$1984bo131bo359b2o\$1982b3o\$1982bo\$
1982b3o\$1984bo\$2410bo\$2410bo9bo\$2410bo9bo\$2420bo\$2447b2o\$2447b2o5\$
2386bo\$2073bo311bobo\$2071b3o311bobo\$2071bo314bo71bo\$2071b3o383b2o\$
1941bo131bo383b2o\$1939b3o515bo\$1939bo504b2o\$1939b3o501bo2bo24b2o14b2o\$
1941bo502b2o25b2o14b2o\$2452bo\$2451bobo8bo\$2442b2o7bobo8bo\$2437b2o3b2o
8bo9bo\$2437b2o3\$2432bo\$2431bobo\$2431bobo36bo\$2432bo37bo8b2o\$2030bo439b
o8b2o\$2028b3o396b3o\$2028bo\$2028b3o\$1898bo131bo\$1896b3o\$1896bo\$1896b3o\$
1898bo11\$2470bo\$1987bo481bobo\$1985b3o481bobo\$1985bo484bo\$1985b3o\$1855b
o131bo\$1853b3o\$1853bo\$1853b3o\$1855bo12\$1944bo\$1942b3o\$1942bo\$1942b3o\$
1812bo131bo\$1810b3o\$1810bo\$1810b3o\$1812bo12\$1901bo\$1899b3o\$1899bo\$
1899b3o\$1769bo131bo\$1767b3o\$1767bo\$1767b3o\$1769bo12\$1858bo\$1856b3o\$
1856bo\$1856b3o\$1726bo131bo\$1724b3o\$1724bo\$1724b3o\$1726bo12\$1815bo\$
1813b3o\$1813bo\$1813b3o\$1683bo131bo\$1681b3o\$1681bo\$1681b3o\$1683bo12\$
1772bo\$1770b3o\$1770bo\$1770b3o\$1640bo131bo\$1638b3o\$1638bo\$1638b3o\$1640b
o12\$1729bo\$1727b3o\$1727bo\$1727b3o\$1597bo131bo\$1595b3o\$1595bo\$1595b3o\$
1597bo12\$1686bo\$1684b3o\$1684bo\$1684b3o\$1554bo131bo\$1552b3o\$1552bo\$
1552b3o\$1554bo12\$1643bo\$1641b3o\$1641bo\$1641b3o\$1511bo131bo\$1509b3o\$
1509bo\$1509b3o\$1511bo12\$1600bo\$1598b3o\$1598bo\$1598b3o\$1468bo131bo\$
1466b3o\$1466bo\$1466b3o\$1468bo12\$1557bo\$1555b3o\$1555bo\$1555b3o\$1425bo
131bo\$1423b3o\$1423bo\$1423b3o\$1425bo12\$1514bo\$1512b3o\$1512bo\$1512b3o\$
1382bo131bo\$1380b3o\$1380bo\$1380b3o\$1382bo12\$1471bo\$1469b3o\$1469bo\$
1469b3o\$1339bo131bo\$1337b3o\$1337bo\$1337b3o\$1339bo12\$1428bo\$1426b3o\$
1426bo\$1426b3o\$1296bo131bo\$1294b3o\$1294bo\$1294b3o\$1296bo12\$1385bo\$
1383b3o\$1383bo\$1383b3o\$1253bo131bo\$1251b3o\$1251bo\$1251b3o\$1253bo12\$
1342bo\$1340b3o\$1340bo\$1340b3o\$1210bo131bo\$1208b3o\$1208bo\$1208b3o\$1210b
o12\$1299bo\$1297b3o\$1297bo\$1297b3o\$1167bo131bo\$1165b3o\$1165bo\$1165b3o\$
1167bo12\$1256bo\$1254b3o\$1254bo\$1254b3o\$1124bo131bo\$1122b3o\$1122bo\$
1122b3o\$1124bo12\$1213bo\$1211b3o\$1211bo\$1211b3o\$1081bo131bo\$1079b3o\$
1079bo\$1079b3o\$1081bo12\$1170bo\$1168b3o\$1168bo\$1168b3o\$1038bo131bo\$
1036b3o\$1036bo\$1036b3o\$1038bo12\$1127bo\$1125b3o\$1125bo\$1125b3o\$995bo
131bo\$993b3o\$993bo\$993b3o\$995bo12\$1084bo\$1082b3o\$1082bo\$1082b3o\$952bo
131bo\$950b3o\$950bo\$950b3o\$952bo12\$1041bo\$1039b3o\$1039bo\$1039b3o\$909bo
131bo\$907b3o\$907bo\$907b3o\$909bo12\$998bo\$996b3o\$996bo\$996b3o\$866bo131bo
\$864b3o\$864bo\$864b3o\$866bo12\$955bo\$953b3o\$953bo\$953b3o\$823bo131bo\$821b
3o\$821bo\$821b3o\$823bo12\$912bo\$910b3o\$910bo\$910b3o\$780bo131bo\$778b3o\$
778bo\$778b3o\$780bo12\$869bo\$867b3o\$867bo\$867b3o\$737bo131bo\$735b3o\$735bo
\$735b3o\$737bo12\$826bo\$824b3o\$824bo\$824b3o\$694bo131bo\$692b3o\$692bo\$692b
3o\$694bo12\$783bo\$781b3o\$781bo\$781b3o\$651bo131bo\$649b3o\$649bo\$649b3o\$
651bo12\$740bo\$738b3o\$738bo\$738b3o\$608bo131bo\$606b3o\$606bo\$606b3o\$608bo
12\$697bo\$695b3o\$695bo\$695b3o\$565bo131bo\$563b3o\$563bo\$563b3o\$565bo12\$
654bo\$652b3o\$652bo\$652b3o\$522bo131bo\$520b3o\$520bo\$520b3o\$522bo12\$611bo
\$609b3o\$609bo\$609b3o\$479bo131bo\$477b3o\$477bo\$477b3o\$479bo12\$568bo\$566b
3o\$566bo\$566b3o\$436bo131bo\$434b3o\$434bo\$434b3o\$436bo12\$525bo\$523b3o\$
523bo\$523b3o\$393bo131bo\$391b3o\$391bo\$391b3o\$393bo12\$482bo\$480b3o\$480bo
\$480b3o\$350bo131bo\$348b3o\$348bo\$348b3o\$350bo12\$439bo\$437b3o\$437bo\$437b
3o\$307bo131bo\$305b3o\$305bo\$305b3o\$307bo12\$396bo\$394b3o\$394bo\$394b3o\$
264bo131bo\$262b3o\$262bo\$262b3o\$264bo12\$353bo\$351b3o\$351bo\$351b3o\$221bo
131bo\$219b3o\$219bo\$219b3o\$221bo12\$310bo\$308b3o\$308bo\$308b3o\$178bo131bo
\$176b3o\$176bo\$176b3o\$178bo12\$267bo\$265b3o\$265bo\$265b3o\$135bo131bo\$133b
3o\$133bo\$133b3o\$135bo12\$224bo\$222b3o\$222bo\$222b3o\$224bo16\$181bo\$179b3o
\$179bo\$179b3o\$181bo16\$129b2ob2o4bo\$130bobo3b3o\$130b3o3bo\$136b3o\$138bo
16\$86b2ob2o4bo\$87bobo3b3o\$87b3o3bo\$93b3o\$95bo16\$43b2ob2o4bo\$44bobo3b3o
\$44b3o3bo\$50b3o\$52bo16\$2ob2o4bo\$bobo3b3o\$b3o3bo\$7b3o\$9bo!
``````
Hunting
Posts: 3019
Joined: September 11th, 2017, 2:54 am
B2e3-anq/S12-a3 and its relatives. Failed replicator:
Code: Select all
``````x=0,y=0,rule=B2e3-anq/S12-a3
bo\$obo!``````
Glider:
Code: Select all
``````x=0,y=0,rule=B2e3-anq/S12-a3
2o\$2bo!``````
Failedrep corderized:
Code: Select all
``````x = 14, y = 18, rule = B2e3-anq/S12-a3
5bo\$4bobo3bo\$9bobo4\$4bobo\$5bo3bobo\$10bo\$3bo\$3bo3b2o3bo\$12bo\$2bo2\$o2\$
11b3o\$11b2o!``````
Another glider, I suggest calling it T even though it is not T:
Code: Select all
``````x = 3, y = 4, rule = B2e3-anq/S12-a3
bo\$obo2\$bo!``````
EDIT: Two-engine T backrake:
Code: Select all
``````x = 12, y = 21, rule = B2e3-anq/S12-a3
2bo\$bobo3\$8bo\$7bobo\$bobo\$2bo2\$o3bo\$o3bo2bobo\$8bo2\$10bo\$10bo\$3b2o\$3b2o
3\$9b3o\$9b2o!``````
EDIT: Two-engine strange puffer
Code: Select all
``````x = 17, y = 25, rule = B2e3-anq/S12-a3
5bo\$4bobo4\$11bo\$4bobo3bobo\$5bo2\$3bo3bo\$3bo3bo\$10bobo\$b3o3bo3bo\$bobo\$2o
bobo7bo\$2b3o5bo2bo\$6bo\$6bo6b3o\$7bo5bobo\$13bob2o\$12b3o3\$9b3o\$9b2o!``````
EDIT: Another corderoid
Code: Select all
``````x = 13, y = 28, rule = B2e3-anq/S12-a3
2bo\$bobo5\$o3bo3bo\$o3bo2bobo5\$3b2o5bo\$10bo5\$7bo\$11b2o\$11b2o\$11bo\$9bo2\$
3b2o\$5b2o\$5b2o\$5bo!``````
EDIT: A G-forerake
Code: Select all
``````x = 15, y = 26, rule = B2e3-anq/S12-a3
4bo\$3bobo5\$2bo3bo\$2bo3bo3bo\$9bobo4\$5b2o\$2o4bobo3bo\$2o6bo3bo\$bo\$3bo\$7bo
5bo\$6b2o5b2o\$7b3ob3o3\$3b2o\$10b2o\$10b2o\$10bo!
``````
EDIT: Relative #1, B2e3-anr/S12-a3.
Linear growth is not as common as in the original rule. However, there's this very sparky c/2 three-domino-hauler which evolves from the failedrep.
Code: Select all
``````x = 2, y = 3, rule = B2e3-anr/S12-a3
o\$bo\$o!
``````
Some shuttles(p12):
Code: Select all
``````x = 12, y = 61, rule = B2e3-anr/S12-a3
bo2bobo2bo\$bobo5bo\$4bobo6\$bo2bo5bo\$bobo6bo\$4bo8\$bo2bo4bobo\$bobo7bo\$4bo
4bo8\$o3bobo2bo\$o2bo5bo\$4bobo8\$o3bo5bo\$o2bo6bo\$4bo8\$o3bo4bobo\$o2bo7bo\$
4bo4bo8\$o4bobo2bo\$o3bo5bo\$5bobo!``````
A common p7:
Code: Select all
``````x = 9, y = 6, rule = B2e3-anr/S12-a3
2bo3bo\$2bobobo\$obobobobo\$obo3bobo\$2bo3bo\$2bo3bo!``````
EDIT: Relative #2, B2e3-anqr/S12-a3
With a NATURAL GUN FROM MANUAL SOUP
Code: Select all
``````x = 8, y = 14, rule = B2e3-anqr/S12-a3
3bo2b2o\$2bobo2\$bo2\$o\$obo\$2b2o\$2b3o\$4bo\$2bobo\$obobobo\$obobobo\$2bobo!``````
Finally, the original rule, a funny reaction
Code: Select all
``````x = 2, y = 7, rule = B2e3-anq/S12-a3
bo\$bo3\$o\$bo\$bo!``````
Similar
Code: Select all
``````x = 3, y = 8, rule = B2e3-anq/S12-a3
2bo\$2bo3\$bo\$obo2\$bo!
``````
Also
Code: Select all
``````x = 4, y = 8, rule = B2e3-anq/S12-a3
2b2o4\$bo\$obo2\$bo!``````
Also near miss
Code: Select all
``````x = 3, y = 7, rule = B2e3-anq/S12-a3
b2o\$b2o3\$o\$bo\$bo!``````
EDIT:
Also clean double backrake
Code: Select all
``````x = 13, y = 17, rule = B2e3-anq/S12-a3
11bo\$10b3o\$6b2o4bo\$bo10bo\$o\$bo10bo\$6b2o4bo\$10b3o2\$10b3o\$6b2o4bo\$bo10bo
\$o\$bo10bo\$6b2o4bo\$10b3o\$11bo!``````
Code: Select all
``````x = 27, y = 22, rule = B2e3-anq/S12-a3
2\$10b2o\$9b3o\$17b2o\$3b3o2bo14bo\$3bobobo16bo\$2b2o4bo14bo\$8bo8b2o\$10bo\$5b
6o2\$4b7o\$3bo6bo\$4bob2o9b2o\$23bo\$24bo\$8bo14bo\$17b2o\$9b3o\$10b2o!
``````
MLP will live on forever, so does John Horton Conway.
Moosey wrote:
February 5th, 2019, 7:51 pm
“New knightship tagalong!”
“Quick, hide it!”
My TODO list
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phdanielli
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EvinZL
Posts: 181
Joined: November 8th, 2018, 4:15 pm
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## Introduction
1 “Music” is not musical notation and the theory of chord progressions. It is possible to play music without reading music, and you don’t need some minimal level of training to hear and understand music.
2 “Math” is not mathematical notation and rigorous proof. It is thinking about something, and then bringing it up in conversation to examine it. To define it further. To think about what properties it must have. To come up with examples to make it more specific, or to come up with generalizations, sometimes in more than one way.
3 Suppose you are enjoying thinking about triangles. You might start by drawing a triangle on a piece of flat paper and figuring out what is true about it. You might then try and figure out what is true about every triangle that can be drawn on flat paper—that would be one way to generalize. Or you might decide to see how triangles drawn on paper differ from a triangle that you could draw on a globe. Or you might decide to generalize by thinking about four-sided figures, and five-sided figures, and so on—generalizing in the direction of ‘true for multi-sided figures.’ Or you might generalize by thinking about what triangles look like in three dimensions—you’d have to decide whether to think about flat, two-dimensional triangles in three-dimensional space, or maybe you’d prefer to think about tetrahedrons, since those are what you get when you add a point outside of the plane of a triangle.
4 Some people love the notation and the rigor. I’m one of them. But it’s not necessarily the thing that has to come first, or even at all. We can play with abstract ideas without having to write them all down in a careful scrimshaw of Greek letters, because everyone can talk about ideas.
5 You are a human.1 You can use words, which are abstract, to refer to things which are concrete. That is your birthright as a creature of language. So you also get to talk about things that do not “exist” in the same way that that rock over there and this air over here exist. And because you can talk about those things in quasi-existence, you can imagine a world that is finer than the one you live in at this moment.
6 Which brings me to my second point about the philosophy of punk and mathematics: the role of destruction in social criticism. Why are things so fucked up? Some people would say, “Because of capitalism.” Others would say, “Because of not enough capitalism.” Either group can agree that colorful bits of paper—or electronic signals that represent what colorful bits of paper have historically represented—which have no inherent value except what we collectively assign to them, influence and control our lives and how our lives interact, and influence and control how matter mined in Africa is refined and assembled in China to the specifications of Californians and then shipped across the world to become a prized attention-seizing device in Poughkeepsie.
7 Matter governed by physics, moved about by global shipping networks, rearranged to align with patterns of hard- and software engineering, becoming “owned” by a consumer, going obsolete, becoming a lump of fairly expensive and somewhat toxic matter again.
8 Why does the world have to be this way? Will it be this way for much longer? Why is there poverty in some places and abundance in others? Is the system failing or working exactly as designed? Do we smash it or try to perfect it? Is there No Future? Am I part of the next-to-last generation and what do I do about it?
9 Well, I don’t know about you, but I’m going to apply math to it.
10 You may not believe me, but math is really an elaborate and infinite game. It is important to note that math is more than simple arithmetic. I can do some mildly impressive computations in my head, because I know the commutative and associative laws (we’ll meet them soon enough) to transform a computation I find difficult into several computations that are easier. But, more times than I’d care to count, I will say something like: “Six times three is twenty-four” and totally screw everyone up when we’re divvying up a dinner check.
11 Math is also not statistics, although statistics is a division of math. A knowledge of statistics would help folks understand that climate change is real, and that inequity is a problem. If someone writes “Punk Statistics” I’ll be the first in line to buy it. But this is not a book on statistics. The real world is messy and uncertain. Statisticians can not only tell you how likely it is that their conclusions are true, but how frequently it is that they will incorrectly tell you that their conclusions are true.
12 If arithmetic is about how to combine numbers, and statistics is the relationship between numbers in the world, then mathematics in the broader sense is the study of patterns. It is the way things relate, the patterns appearing in those relations without caring much about what those things are. In brief, math = real facts about imaginary things.
## The Problem
13 Now, numbers may be imaginary concepts—have you ever met “3” in a dark alley?—but that’s not stopping them from kicking our asses right now. Consider certain measurements of the natural world: the number of parts-per-million of greenhouse gases in the atmosphere, and the decreasing area of the arctic ice. The number of people on the planet increasing exponentially, and the volumes of water and dirt staying stubbornly constant.2 And then there’s the volume of earth’s petroleum, burning slowly toward zero.
14 And those are just measurements of the physical world. There are social measurements, too, though their relation to the real number system is a bit fuzzier. For example, the 85 richest people in the world control as much wealth as the bottom 3.5 billion people—or half the population of the planet.
15 Putting off for the moment what “the total wealth” even means as a concept,3 consider that a dollar is a sort of vote on how to spend our world’s limited resources and our population’s limited time. (How money flows and how prices vary, after all, are supposed to determine how capitalists and workers decide where to apply their capital and labor. If there isn’t a lot of something, the price is supposed to go up, so that if you buy it anyway, you’re saying, “It is so important that this thing gets produced that I vote that more people should get on producing it.”) Doing a very rough back-of-the-envelope calculation, the richest people in the world have 40 million votes to the single vote of, say, an average Liberian. You may not reject this on its face, but consider the sentiments of Georgina Rinehart, the fourth richest woman in the world, lamenting that Australians are not willing to work for $2 a day. While we may empathize with the loss that bequeathed her this inheritance, we may wish to think about the implications of Ms. Rinehart’s dollars-as-votes for constructing society. 16 There are many other numbers with political and social implications. The exponents describing the debts of post-Econopocalyptic Iceland and Ireland. Returns on investment from Congressional lobbying on the order of 22,000%.4 Probabilities of risk selectively assessed in whatever way ensures that the deepwater drilling goes forward and that the nuclear power plant gets built, and intellectual property decisions that lead to ten-dollar medications being sold for$1500.
17 Beyond simple numbers, there are other, weirder abstract structures manipulating us and the world, and they don’t seem to care that they are imaginary. When I trade inedible pieces of paper to a total stranger in exchange for food, I am participating in a mathematical structure we have decided to believe in for convenience. On a larger scale, note that American houses are made of the same atoms no matter who owns them, and yet, perturbations in the imaginary structures of prices, debts, and ownership have caused great economic suffering, and produced the peculiar state of affairs in which the number of American houses laying empty in 2010 outnumbered the homeless by a factor of five-to-one.5
18 We’re also finding that the computing revolution is supercharging the ability of certain well-studied mathematical structures to screw up our lives in a variety of entertaining and dangerous ways. Trading firms use high frequency trading algorithms to buy and sell the ownership of stocks and commodities on a scale of nanoseconds, making staggering profits without doing a joule of actual work. Our worldwide computer network facilitates both political protest and state repression, and within the net’s ecosystem, services like Facebook exist for free by selling the users themselves—or at least, social graph data stratifying us into precision-marketable demographics.
19 Well, fine. Invisible mathematical monsters surround and coerce us. But don’t we have nerds to handle these problems? Modern culture has grown comfortable offloading hard computational problems to professional geeks, and expecting the supercomputers on their desks and in their pockets to handle distasteful numbers if they come up. People who are otherwise intelligent and curious will cheerfully admit to giving up on learning mathematical thinking. I have yet to come across someone who says, “Meh, I never really got ‘reading.'”
20 Innumeracy is not a problem because it hurts nerds’ feelings. It is a problem because the innumerate are easy to mislead.
21 People use the average Joe’s poor mathematical understanding as a way to control, exploit, and numerically fuck him over. When the American political sphere debates cutting heating assistance to the poor, we don’t talk much about how the cut is equivalent to about five days of American warfare costs, because the public doesn’t understand exponents.
22 By the way, the best method I’ve found to help my dumb ape brain understand the relative enormity of huge numbers—millions, billions, trillions—is by considering how many seconds there are in each. A thousand seconds is about 15 minutes. A million seconds is about 12 days. It’s possible to imagine counting to these numbers, although you’d have to skip sleeping so I wouldn’t recommend it. A billion seconds would take about 32 years—it is inconceivable to imagine actually counting to it, but, it is a length of time that is within the comprehension of a human life: I’ve lived for more than 1 billion seconds (=1 gigasecond), but I’m unlikely to live for 4 gigaseconds. A trillion, though? Nearly 32,000 years. That’s not merely prehistory: that’s before the Ice Age, before the first domestic dogs, and before the extinction of the Neanderthals.
23 The moral is: In matters of State and the Economy, don’t pay so much attention to the digits in the number, as to the number of digits the number has. In trying to conceive of enormous numbers, the magnitude must be considered first, before the particulars of the digits. It’s the difference between a sign on a locked-up store that reads, “Back in 30 minutes” versus one that says, “Back in half an Ice Age.”
24 This rule-of-thumb is closely related to scientific notation which, once you get used to it, allows you to easily compare the order-of-magnitude difference of numbers. When I write 9 million as $9×10^6$ and 2 billion as $2×10^9$, it reminds me that the larger leading digit of the millions number is nowhere near as important as the order-of-magnitude difference: nine minus six is three, and so the billions number is around a thousand ($10^3$) times as big. But I digress.
25 The problem is: people who believe we can offload mathematics to computers let themselves to be pushed around by programmers and financiers, lobbyists and lawyers (and don’t get me started on casino operators). (Gosh, can you imagine… what if there were segments of society that somehow benefited from us not thinking too hard about these things? “Remain calm. Keep shopping. Everything’s fine.”)
## School Sucks Math Rules
26 Since the aforementioned Joe certainly received a basic math education in school, and yet Joe’s eyes dart around for the nearest exit which I start explaining how important it is to distinguish 10-to-the-7th policy discussions from 10-to-the-10th policy discussions, it seems clear that basic math education is failing us.
27 When I taught, many students made it clear that before they got into my classroom, before they experience me shouting and waving my arms and jumping up and down preaching the beauty and frustration of math, their teachers emphasized algorithms and answers over concepts and questions. To them, math appeared to be a series of cargo cult pencil-and-paper algorithms (ritually performed to avoid the Mark of the Red Pen), proofs by intimidation (if there were proofs at all), and meaningless algebraic rituals for finding answers to contrived questions. Math classes frequently lack real-world motivation, or if motivation is present, it is financial—textbook authors seem to believe that nothing fires students’ imaginations like amortization and compound interest.
28 Since students’ interests are not respected, they cannot find intrinsic motivation—the perfectly rational response to a question like, If Billy is twice as old as Sally was two years ago, and Sally will be Billy’s age now in ten years, how old is Billy? is “Go fuck yourself.” Instead, they are motivated by the extrinsic reward of grades, which we know to be less effective for learning.
29 Furthermore, emphasizing hand computations when your typical college student could solve it in seconds using a five-dollar smartphone app makes the traditional math class seem downright Kafkaesque—if the only thing that matters is “the answer.”
30 Students exposed to this kind of math education conclude that math is not relevant to their interests or their world, and it’s hard to fault their reasoning.
### The Solution: Punk Mathematics
31 “Breaking down the distance between performer and audience is central to the punk ethic. Fan participation at concerts is thus important; during the movement’s first heyday, it was often provoked in an adversarial manner—apparently perverse, but appropriately ‘punk.'”—unknown, Wikipedia
32 Too many people see math as a set of arcane rules—delivered from on high—that one manipulates to pass a final exam and then eject from memory. But to me, mathematics is a deep and playful game that demands the courage to fail spectacularly, to make noisy mistakes, to ask inconvenient questions, to call bullshit when you see it, and to believe nothing until you’ve proven it for yourself. In short, math is punk as fuck.
33 In punk rock, both social and political commentary is wrapped in chaos and feedback. Taste and beauty are rejected for technical accessibility and energy. The secret ingredient is relevance, interest, and accessibility. Passion, anger, and the intent to see structures as they are, not as we wish them to be.
34 What I find personally maddening about the common misperceptions about mathematics is that in reality, math is probably the least authority-dependent subject of study. Whereas the sciences have theories, explanations of data which have been observed (but probably not by you), mathematics has theorems: we make as few assumptions as we can, then prove universal truths outward, carefully stepping on the stones of logic, heeding intuition, but remembering that obvious “truths” are frequently falsehoods. Any skeptic with enough time and frustration-tolerance can in principle verify any mathematical claim with only a pencil, some paper, and some friends with whom to work out ideas (or their written words, if these friends are dead). Math is cheaper than chemistry, cleaner than biology, and the most dangerous thing you can do is divide by zero to see what happens. No government-funded particle accelerators or tweedy experts are necessary.
35 Mathematics is made of words and symbols, numbers and diagrams, agreed-upon definitions and rules of transformation. As is the case in poetry, the language does not merely describe the thing, it is the thing. Therefore, people doing mathematics together are on even ground—what is right is determined by the mathematics that you’ve written down so far. In this sense, mathematics is anarchistic. (Note: the etymology of “anarchy” suggests its original meaning: “no rulers!” [“an-,” without; “arkhos,” the archons, or chiefs] which is quite different from the common conception that “anarchy” means “no rules!”)
36 If you are playing chess with world-class chess player Judit Polgar and you were to somehow discover a flaw in an opening she thought was good, you would only need to say: “But… look!” and upon examining it, she could determine if she was in error. If you are playing “Let’s create a 5-year economic plan” with Josef Stalin and you identify a flaw (e.g., millions starving, general suffering, bunches of death), saying: “But… look!” will not get you anywhere but the gulag.
37 It is likely that a mathematician or chess player would actually be excited to be proven wrong; the new understanding of truth would radiate outward in their concept map and update their general knowledge of how their discipline works. There is no board of trustees determining mathematical truth—we are free to explore the logical entailments of whatever mathematical constraints we choose. We are not even bound to talk only of worlds that are known to exist in our own physical, biological, or social reality. If we’re careful about our definitions, we can venture beyond the looking glass into realms of alien logic as well.
38 People don’t generally associate mathematics with breaking shit, with questioning authority, with auto-critique and medium as message, with photocopies and safety pins. That’s only because most people don’t know about the insolubility of the quintic, the nature of the axiomatic method, with Godel’s theorem and Wiener’s information theory, about logical algebras and industrial optimization.
39 The reason that punk mathematics is a thing worth thinking about is that punk’s naturally aggressive drive to obliterate bullshit in all forms can be the guidance system for translating what many of us “believe” are “facts” about the “world” into mathematical language. Mathematical language is a magical tongue, good as any shaman’s words-that are-things, a lawyer’s pro-ergo-procto-cum-coc, a hacker’s source code. The worst problem for a mathematician is to have no problems. Punk identifies problems loudly and brooking no bullshit; math solves problems.6 The specific magic of mathematical language is that it un-utters falsity: statements that are not true drop out of the universe of consideration.7 When you are working a mathematical problem, there are ways of keeping straight when you are assuming, when you are using, and when you are contradicting.
40 Both punks and mathematicians have an instinct for calling bullshit. Both punks and mathematicians will insist on verifying claims for themselves. Both punks and mathematicians create far more bad noises and wrong paths than successes, and this is why they succeed. Both punks and mathematicians recognize that things worth doing may well be frustrating, or terrifying, or beyond their capabilities, and then they try to do them anyway. In a punk mathematics, there is no discipline but self-discipline, no truth but proven truth, and it is better to be wrong than quiet.
41 And everyone gets to play—regardless of their skill.
42 At the beginning of a term, students will come up to me and say, apologetically, “I’m sorry, I’m really bad at math.” They’ve accepted this, internalized it as a true fact about them—as an external fact over which they have no control. They’ve decided that they were passed over for some “math gene” in their brains. Fuck. That. Shit. I explain to them that humans are bad at math: logical chains of inference are no good for escaping saber-toothed cats, and thinking in that way is hard work.
43 Grades don’t account for the wonder of failure.
44 When I taught college mathematics, my job was to create a function from the set of facts about, say, calculus to a performance of duties. Then my other job was to create a function from the set of infinite learning experiences in their infinite combinations and infinite diversity to the tiny finite set: A, A-, B+, … and F.
45 But my students made such interesting mistakes! This system of grading leaves out “Fail better.” And THAT IS how you find out what works and what doesn’t.
46 Aside: How many functions are there from a class of size N to A B C D F? Five raised to the nth power. Is there a linear order of the N, of how good they are at whatever’s being taught? No. It’s the same flaw as IQ tests, assuming that there is a ranking of intelligence instead of a multiplicity of partially-ordered, fascinating multitypes, with emotional intelligence, logical, verbal and visual all playing their parts in the aforementioned mathematical lenses.
47 Math should occur on both sides of the brain. Dance is subdividing four with the body. And math should occur above and below the eyebrows. Why be ashamed of counting on your fingers? If you’re willing to teach the muscles of your hands binary, you can count to 1023 on your fingers.
48 Your eyes are attuned to symmetry
49 Fig. Scott McCloud’s symmetry images from Understanding (or was it Making?) Comics.
50 and your ears are attuned to modular arithmetic, through your perception of rhythm in music. If you are a musician, you might further be intrigued by the world of harmonics, which govern musical instruments (though not bells or drums, those are a different mathematical beast), or L-systems, which can be used to algorithmically generate self-similar structures like chord progressions or melodies.
51 “Here’s a chord, here’s another, here’s a third, now form a band.”
{image}
52 This felt-pen-and-photocopy drawing from 1977 says so much to me about punk rock. It presents three chords without prescribing an order for them. It introduces tablature notation in a way that ought to make sense once you figure out what end of the guitar points up. But most importantly, it says: Find an instrument, find some friends, and join us, right now. This is what I aim to do for the mathematically downtrodden—the mathematically discontent.
53 I will grant it takes a lot of so-called ‘mathematical maturity’ to explore these fields in detail. And it takes rigor and precision—skill with the instruments of mathematics—to advance them.
54 But who says we need to advance them? Why not screw around with them? What’s the worst that can happen? Maybe you’ll reinforce a cognitive bias; maybe you’ll incorrectly reject a true hypothesis; maybe you’ll monkey around with a system you think you understand but don’t know.
55 There are notions and concepts at the ‘ground level’ that should be open to everyone, that require a solid explanation and not a million complex calculations. Concepts that use language, or if-then statements, or counting, or a bit of simple algebra that you can be re-taught with a few short examples. These concepts, unpacked, can lead to new patterns of thoughts about how the world might work. You, dear readers, and my dear students from semesters past, I want you to help disambiguate what is, from what must be.
## very axiom / so constraint / many generative / amaze
56 When you ask a mathematical question, there often is an answer OUT THERE because it predicts many things. The rate of falling objects. How to make a hologram out of crazy “coherent” light. Yet it is also derived a priori—based on theoretical deduction rather than empirical observation—out of the machinations of the mind. Mathematics is therefore imaginary, abstract, invented. Because it is about the world AND entirely imaginary, mathematics is “a set of patterns that fit the world and the mind at the same time.”
57 You might, therefore, think of mathematics as being a sort of property of the universe, or of all possible universes, and if you like you can take this as proof of the existence of God. Or, if you are of the opposite disposition, you can take it as proof against, since math pops for free out of any universe and so He (or She) has been made “unnecessary.” If you are Lakoff, you think of mathematics as a set of metaphors that people came up with, and so mathematics is “only” those kinds of thoughts that people can think.8 Whereas Kronecker said that God gave us the integers, and all else was the work of mathematicians.
58 What concepts and definitions we choose to begin with determines the mathematical universe we are working in. The fundamental concepts that are assumed in order to begin are called axioms.
59 Now, no one knows why the Internet does what it does, but one of the things it does very well is create micro cultures of imaginary grammatical rules of adorable animals. One of these sets of grammatical rules is a fine example of the beautiful paradox of axioms: they constrain the choices we can make in our mathematical or linguistic universe, yet simultaneously create an enormous and generative space in which to play.
60 I would like to tell you about the grammar of doge. In 2010, a Japanese woman put a picture of her shiba inu dog, Kabosu, online. Something ineffable about this dog’s expression and round face struck an etheric resonance field on the internet, or something, and people began photoshopping this picture and adding text.
61 Much like the predecessor meme, Lolcats, the appropriate text for a doge meme has a broken grammar—as though we expect that if our pets could talk, then they would break grammatical rules, like young children do. What makes doge unique compared with Lolcats is that the grammar is very regular.9
62 A “doge phrase” can be one or two words long. The one-word phrases are chiefly “wow,” “amaze,” “excite”; since it’s questionable to claim a list of single words has any grammar at all,10 we will concentrate on the two-word phrases.
63 Here are some examples:
64 {LIST OF AWESOME DOGE PHRASES}
{Let us gather unique examples from the tubes of the Internet.}
65 The first word of a doge phrase comes from a very small set, chiefly “very,” “such,” “so,” “much,” and “many.” The second word can be just about anything, but it must violate the usual “selectional restrictions” of these words so it produces an effect that sounds “off.” In order to speak fluent doge, you must possess a fluency in English that permits you to make word choices that are deliberately less coherent.11 Much strange. Very word. The grammar of doge is ridiculous, and highly restricted in how you can say things. But it does not restrict what you can say. As in mathematics or in chess, the rules are inviolable but the expressions within those rules are nearly infinite.
66 On Twitter, @SecurityDoge tweets on the fight for the open internet (“very surveillance.” “much corporate.” “so regime.” “scare.”). And people all over the internet use doge to comment on how they are feeling about their everyday lives. Occasionally, to make oneself clear, one violates the rules of doge grammar in order to express yourself clearly. But such rule-breaking is more the domain of poetry than that of mathematics. Consider Shakespeare, who coined over 1700 now-common words like “road,” “addiction,” “laughable,” “hint,” “gossip,” “invulnerable,” “bedazzled,” “majestic.” He wrote sonnets, which have restrictive metric and rhyme rules. What might he have done with the grammar of doge?:
67 “What light. So breaks. Very sun. Wow, Juliet.
68 What Romeo. Such why. Very rose. Still rose.
69 Very balcony. Such climb.
70 Much love. So Propose. Wow, marriage.
71 Very Tubalt. Much stab. What do?
72 Such exile. Very Mantua. Much said.
73 So, priest? Much sleeping. Wow, tomb.
74 Such poison. What dagger. Very dead. Wow, end.” 12
75 Well… maybe not.
## The Nuts and Bolts of Mathematics
76 The mathematicians who have come before have bequeathed a huge library of abstractions they found particularly rich, beautiful, generative of surprising results, or applicable to real non-dog-grammar problems, in much the same way the English language is enriched by Shakespeare and other speakers generating new and useful terms for expression that are picked up (or not) by other speakers. Through a long process of consensus across centuries, mathematicians have come to consensus on what definitions seem to lead to interesting results. Similarly, the consensus of English speakers establishes a grammar that is particularly precise and generative of meaning. But so long as we’re clear, we can choose axioms to give us a new mathematical or linguistic universe to play in.
77 On occasion, choosing a new axiom leads to new universes that are very different but nonetheless equally valid and consistent as the one you’ve been in before. (See: Euclidean and Non-Euclidean Geometries in Chapter X)
78 On very rare occasions, we drop an axiom and discover something true about our own universe. When asked how he came up with Special Relativity, Einstein said he had “ignored an axiom”—specifically, the previously assumed axiom that time is universal, as though there were an Absolute Clock ticking away the seconds for us all. It is obvious that two observers, no matter where they are or how fast they are moving, will agree whether event $A$ preceded event $B$ or not; however, this “obvious” thing is also wrong.
79 More frequently, messing with these axioms leads to contradictory or degenerate results. For example, an editorial error in a linear algebra text led Stephen M. Walk at St. Cloud State University into an examination of a structure13 which “looks like a vector space, walks like a vector space, and quacks like a vector space,” but instead of having a single unique zero element, has many zero elements: one for every other element in the space. “[T]his fragmenting is antithetical to the foundation of anything that we call ‘algebra,'” he points out, “that foundation being the ability to solve equations by undoing operations.” With non-unique zero elements, we can no longer, say, subtract three from both sides of $x + 3 = 7$, because when we do so, we have no guarantee that the zero we get from $3 + (-3)$ is the same zero that makes $x + 0 = x$ a true equation. Or in more metaphorical terms, try to imagine a space of modular furniture pieces — let’s call it an ikeaspace — in which screwing together two pieces could never be undone if you’d made a mistake. Doing algebra in a degenerate space is nowhere near as fun as being a degenerate.
80 In practice, there are some axioms that most mathematicians take up, like the Axiom of Choice, because not taking them up seems really weird.14 If there is a statement which cannot be proven from the axioms, and cannot be disproven from the axioms, the statement is called undecidable. And there are many. These are the weird edges that bound separate mathematical universes from one another, and unless you want only to prove theorems that do not require the use of those statements15 you will have to make a choice of what universe you want to live in. At least, if you want to live on these weird frontiers.
81 Having made our choice of assumptions, many consequences occur. For obvious reasons, in theoretical physics we choose assumptions whose consequences match up with experiment and observation in our own universe. In math, we don’t have to do that, but there are still a lot of necessities that come from our choices, and many choices explode in contradiction quickly. It is the nature of the mathematical game that not all sets of rules are equal.
82 We might choose a smooth universe of waves and fluid dynamics, or a choppy one of counting problems and prime number factorizations. We might choose to explore the set of all possible relations that can be drawn among 7 billion entities, or the 219 repetitive patterns that crystals in 3-space can form (it’s 230, when you can tell the difference between right and left). Or we might select some axioms for rationality, choosing to study how we choose. You can choose to explore exciting forks of mathematical reality—nonstandard analysis, doxastic logic, and topics even further away from the K-12 curriculum. If you do go off to explore new frontiers of mathematics, don’t forget to write.
## Definitions, Propositions, and Theorems
83 Math is formed in the interplay between choice, constraint, and freedom. You can choose axioms to give you a new mathematical universe to play in, and once we’ve chosen, we get to derive propositions by way of logical inference. You can define new terms as you like, to wrap up a particular concept you want to explore.
84 Without the ability to define new concepts, you wouldn’t get very far (we wouldn’t want to go through proving every theorem directly from axioms and rules of inference anyway because it would take forever). To do this, we list out some properties, and say that any object $X$ that satisfies those properties gets named… a whatever. The name is just a shorthand to say, “things that satisfy these properties.” This way of defining mathematical objects may seem silly and tautological but that’s how definitions work: “a bachelor is an unmarried male”—that sort of thing. We permit ourselves to create new abstractions out of the old ones.
85 If mathematics really is a language—as they say—this is what makes it so. By wrapping up new concepts in new definitions, the language becomes more powerful: it can express more concepts, more precisely and more eloquently. The speaker’s thoughts are denser, and finer distinctions can be made. There are languages that lack distinct names for green and blue, and people who grew up speaking those languages have some difficulty picking out the difference visually.16 When a name is given to an unfamiliar concept, you can see that concept where before it was invisible.
86 The longer our little mathematical program—the longer we think through the implications of our initial axioms and develop new concepts with names—the more concepts we can express, and express them in simpler terms.
87 Anything we prove about such abstract objects will be true for anything that has that list of properties. But, this does not work in reverse: if we pick an example object, it may have additional qualities that are not true for every object in the domain of the abstraction. Every square is a rectangle, but not every rectangle is a square; Hitler was a vegetarian, but not every vegetarian is Hitler. This is why we can use examples to hone our intuition and to look for counterexamples, but we can’t use examples as a substitute for proof, if we’re doing serious mathematics (so we’re off the hook there).
88 A mathematical proposition is something like, “The square of an even number is itself even.” It is a clear statement that has been proven true. A proposition that is proven in order to be used in the proof of something more important is called a lemma: they are the pitons of mathematical mountain climbing, pounded into the problem surface to step on but not important in their own right.
89 An important proposition is called a theorem. And if a theorem is very important, it might be called a fundamental theorem.17
90 Because a theorem has two parts (the hypothesis, and the conclusion), we are always beginning sentences with “Suppose Y” or “Given an X such that P is true” or similar constructions. I’m claiming that if it is ever the case that the hypothesis statement is true, then the conclusion statement will also be true. If you only say things that start with suppositions, then you are not responsible for your hypothesis being misconstrued as a statement about everything.
91 You may not believe this yet, but ultimately, mathematics is extremely simple. Not easy: simple. If you permit yourself only to speak undeniable truths, you’re going to have to keep the questions simple.
## An Illustrative Story in Mathematics
92 I’d like to take the time to give you a hint of what doing mathematics feels like. This story is based on actual historical accounts (possibly embellished), and it is my favorite story about slowing down to think deeply and playfully about the properties of the imaginary things numbers are.
93 A little over two hundred years ago, there was a young schoolboy who attending lessons in Germany. One day he was in math classroom, and the teacher instructed all the pupils to add the numbers from 1 to 100 into one big sum and tell him what it was. This was perhaps because he felt this was an important exercise in addition, or possibly he simply wanted all the kids to shut up for 20 or 30 minutes.18
94 Now this was at a time when paper was still handmade, and presumably pencils were, although I don’t know if they had pencils yet, but at any rate, paper was way too expensive to treat as cavalierly and disposably as we do now. So the students did their work on little rectangles of slate, writing on them with chalk. Each boy got his own.19
95 The teacher is presumably rooting around in his desk for his flask or rosary beads or whatever, and the kids start furiously calculating. One plus two is three, plus four is seven, plus five is, uh, twelve, plus six is eighteen plus 7 is… um, twenty-something…. You can imagine that they probably started out blazing fast and then started to slow down as they had to really think about each sum.
96 Meanwhile, as his classmates are cranking away, our boy is just sitting there, thinking. There are a lot of tellings of this little classroom story, but in mine, he’s leaning back, with his hands in his lap or on his desk, just staring off with a faraway look. And after a moment, he noticed something funny.
97 The funny thing came from imagining all the numbers, from one to one hundred, all laid out in a line. I see them as going from left to right. You can see them as starting with one on the left, and then, glossing over some numbers, turning your head to the right and “seeing” 100 at the end of this line of numbers on the right. (I actually turn my head sometimes when I think like this.) The curious thing that our schoolboy noticed was that the two ends of his line of numbers (1 and 100) added to 101—but so did 2 and 99, the next-to-leftmost and next-to-rightmost numbers. And the third-to-leftmost number and the third-to-rightmost number would be 98 and 3, and those also sum to 101. And the pattern continues, all the way to the two central numbers, 50 and 51, which sum to 101.
98 So he folds the row of numbers in half, as though there was a hinge between 50 and 51. Try to imagine it in your mind: a row of numbers from 1 to 50 going left to right, and folded under, a row of numbers from 51 to 100 going from right to left. Each column adds to 101.
99 Can you see it? Try to see it. You’re holding a hundred numbers in your head all at once, but they’re all sequential so it’s no big deal. Each column adds to 101. So how many 101’s is that? The numbers here are polite enough to count themselves: that top row ends in 50, which tells us we have fifty 101’s. So the folding trick turned 99 increasingly difficult addition operations into a single multiplication problem: 50 \times 101. One hundred fifties makes five thousand, plus one more fifty makes 5050. Voilà! There’s our answer.
100 Now remember, this schoolboy is just sitting there, thinking. Looking for all the world like he’s blown off the problem as too difficult or too boring. Then suddenly he leans forward, writes the correct answer, 5050, on his slate, walks to the front of the room, tosses it on the desk, and says, “Ligget se”: There it lies.
101 Not bad for an eight-year-old little boy. Fuckin’ A.
102 Now, I want you to look at what our hero did at an abstract level. He didn’t apply massive computational horsepower to his problem. And trust me, he could have; this was a child who was correcting his father’s accounting at the age of three. Instead, he stopped, thought, imagined, created a picture, and observed what the inside of his mind looked like. Instead of going off half-cocked in a frenzy of computation, he slowed down to walk around his problem and view it from all angles.20
103 The teacher, to his credit, realized that while this kid may have been a smart-ass, he was a smart-ass with incredible potential. With his own money, he bought him the best mathematics textbook he could, reportedly saying, “He is beyond me—I can teach him no more.”
104 The smart-ass schoolboy grew up to be the finest mathematician of his place and time—Karl Friedrich Gauss. {Give a shout out to the internet meme of Gauss Facts here? “Erdos believed God had a book of all perfect mathematical proofs. God believes Gauss has such a book.”} Probably he had a natural, inborn talent for doing mathematics, but the important part of the story is how Gauss demonstrates what, to me, is the essence of mathematical thought. By taking his time to reframe the problem, the solution was not merely possible—it was obvious.
### The Problems Today Belong to All of Us
105 Our world, by definition of unsustainable, is coming off of hyperconsumerism, US industry dominance, and the post-bullshit era (with new ones to come, don’t worry). Seventy years or more of military-industrial-advertising complex, hundreds of years of colonialism, and hundreds of years of centralized currency, and ape power games dating back millions of years. I have a lot of concerns about how global hypercapitalism seems to be organizing our world for maximum ecological destruction and minimum compassion for humans who are disempowered, whether that disempowerment is because of where they were born, what they look like, what genitals they have, or a whole host of factors that lead our society to ignore their beliefs, their safety, and their suffering. Our problems are, in part, medieval institutions, paleolithic emotions, and newly godlike powers. So many smart people working to get people to give up the names of everyone they know, and to click on ads.
106 The misuse of power is the most bullshit of all. But “power” is an overloaded and ill-defined concept. In physics, it’s clear enough: power is defined as a rate of work done per second. If we can both lift a heavy weight over our heads, but I can do it twice as fast as you can, I am more powerful.
107 What is power in the societal sense? My working definition is: power is the ability to constrain another’s actions. I posit this as my working definition because, at least in principle, this ought to have a measurement: compare what an actor is able to do alone, and then compare their possible actions with a new element. If the collection of actions that can be taken expands under a new condition, the actor has been empowered. If the collection of actions that can be taken contracts under a new condition, they are disempowered.
108 Consider the following power structures that are in a state of flux. Our information flows around. When it flows around through a 3rd-party, the NSA grabs it, and the courts don’t protect it because it’s not on paper. 3-D printing is about to disrupt the material flows in ways we can fantasize about but not predict.
109 Media power influences the information we take in and the conversations we engage in. When there were just three networks, there was the “news of the day”; Walter Cronkite signing off his nightly newscast with, “And that’s the way it is.” This brought about a mindset in which there are “Two sides to a story” instead of a different story depending on whether you’re unmonied, colored, genderfucked, or otherwise marginalized. Because there were only so many minutes to be devoted to news (on television or radio) or only so many column inches (in a newspaper), the number of possible conversations around “the news of the day” were severely constrained by whoever was making the editorial decisions. This metanarrative—the story that there were a small number of stories—wasn’t so much an intentional conspiracy to construct societal discussion as an effect of the constraints of the media of the day. But to satisfy those physical constraints, a small number of demographically similar editors — mostly white, mostly dudes — filtered the complexity of the world through their own biases and instincts. In retrospect, “All the news that’s fit to print” should probably have been “All the news that fits in print.”
110 Now, there has been an explosion in the number of points of view. A Cambrian explosion of stories, from cheap media production platforms and cheaper distribution networks. These new points of view are not necessarily controlled by white men with advertisers to please; these new points of view can help illuminate the sexist racist classist nationalist ableist ageist poverty-blaming slut-shaming working-class-dividing Muslim-surveilling black-person-shooting health-care-denying mental-illness-stigmatizing water-despoiling toxin-spewing nature-fracking bias-reinforcing woman-silencing school-defunding child-droning terrorist-producing austerity-forcing species-extinguishing vaccine-backpedaling elder-threatening dissent-criminalizing infrastructure-corroding discourse-oversimplifying kyriarchal system of oppression that, don’t forget, is also killing the bees.
111 Human knowledge and communication, and in particular communication across the internet, is big. Like, very big. We have moved from an information-scarce to an information-abundant environment. The main corporate video hosting site, YouTube, claims to receive 100 hours of uploaded video every minute.21 Put another way, it would take you over a year of continuous viewing to catch up with the video that’s been uploaded in the last hour and a half.
112 The explosion of mobile technology means that the first connection some people are having to the media are mediated through their phones. There are a couple billion people that will be added to the media network in the coming couple of decades.22 Media power is falling apart, fragmenting, losing the metanarrative, going weird and flowy. Information and media melting into internet meme ooze. 23
113 Monetary power on the other hand, controls the way we dispose of our time. (And our resources, if we had any.) The need to accumulate wages constrains many, many hours of our time. It makes us not do things that are beneficial to society. I’ve been paid as a teacher, and the pay is shit. (If only I could just get it in my head that, as spoken by the Investment VC Market, the most pressing challenge issue of our time is to get people to click on ads.)
114 That power is not falling apart. Rather, it seems like it’s using the increased fluidity of the information we generate to insert itself in between every human transaction it can 24 But it is going weird. The taxicab and hospitality industries are in a flux of a “sharing economy” brought on by services like “Uber,” “Lyft,” and “Air B&B” that allow people to “network” these services peer-to-peer, rather than going through conventional industries. But market forces and agents are going weirder than just being ever-present. Bots that operate by algorithmically trading at a nanosecond level read the online rumor mills to attempt to make pricing predictions; as a result, it has been observed that when actress Anne Hathaway has a good news day, the shares of Warren Buffett’s Berkshire-Hathaway corporation experience a small but noticeable rise in value, a phenomenon known as the “Hathaway effect.” Elsewhere on the Weird Money Internet, cryptocurrencies—forms of money created through sort of cooperative, sort of competitive algorithm for keeping a distributed ledger—experience surges and crashes in value; these currencies include but are not limited to Bitcoin, the alpha software of distributed finance, and Dogecoin, a currency created as a joke that derives from a meme that arose from a slightly funny photograph of a round-faced dog, that at the time of this writing trades for N American dollars.
## Math as a Powerful Tool
115 There is a huge amount of bullshit out there that could really use at least a half-assed mathematical analysis. The oldest bullshit aspects of society, like war and oppression, and relative newcomers like consumerist sleepwalking. The idea that poverty and environmental degradation are necessary simply in order to have an economy. The systems of control, the structures of power, and the aftereffects of the de-organization of society. Is it coincidence that American wages entered their long stagnation just as large organized labor strikes ceased? That might be a question that can be answered. After all, collective bargaining is a game theory problem—so why not do some game theory, rather than declare that unions rule or drool based on unexamined assumptions, political tribalism, and family lore?
116 Mathematical truths are not dependent on The System (whatever that is), and so math is a uniquely powerful tool for examining our biases and social constructs, whether your beliefs are rooted in science, religion, or neither. We need a toolkit to understand the scale of the problems we face as a species, identify the structures of society that can aid or interfere with possible solutions, and contemplate what truth is in a complicated universe full of primates trying to control one another with various forms of bullshit. Punk mathematics is one such toolkit. In particular, we can use mathematics to assess the validity of 20th century structures as petro-capitalism spins down, to make independent personal choices in an Age of Bullshit, and to imagine new ways to use the emerging network structures to tackle the hard problems of our endangered civilization.
117 At first, I assumed the phrase “the purpose of a system is what it does” was some punk anarchist description of The System, when in fact it was the phrase of a cyberneticist.
118 The Firm and the State do not have eyes or hands; they operate by algorithmic methods. They always have: censuses, surveys, records, charters, debts—all are finite categorical or numerical expressions of the will of one-dimensional beings that do not live, yet act. They attempt make us discrete, orderly, and finite. Predictable—so much the better. Classify and measure us and our relations to people and things to make us finite—the reduce variety of our expression. You know they’re not afraid of getting their hands dirty to figure out the angles and profit opportunities.
119 So ask questions. Stupid questions, uncomfortable questions.
120 For gods’ sake, ask uncomfortable questions.
121 Lincoln said that if he had four hours to chop down a tree, he’s spend the first three sharpening his axe. Einstein said that if he had an hour to save the world, he’d spend the first 55 minutes defining the problem. The challenges that we face in trying to avert Total Global Clusterfuck are not going to yield to “Let’s do what we’ve always done, but faster.” Instead, we need to think like Gauss: sacrificing mindless computation time to find clarity, generality, and elegance. We need to see through the fear, uncertainty, and doubt down to the heart of simplicity.
122 Change is possible; a new future is possible. We need tools to dissect old structures and pull out what is good and worthwhile, and relegate the out-dated, aggressive and nasty zero-sum ideas to well-deserved oblivion.
1. Or cyborg, or uploaded brain state vector, or text-mining robot, or some new consciousness I have no name for. I don’t judge.
2. It’s actually worse than that—apparently soil depletion is a thing. Fucking hell.
3. This is surprisingly slippery, and computing a reasonable estimate probably requires summing an infinite series. Why on earth should that be the case? Consider this: if I deposit money in the bank, it doesn’t just sit in a vault gathering dust; it is used to give out loans. Suppose my bank gives you a small business loan with the money I’ve deposited. Do you have my money? Maybe kinda, but you will keep the money not in your pockets but in a bank account, which is loaned out to another small business, ad infinitum. *Literally*, ad infinitum: if you want to try and compute how much money there is you’ve got to sum an infinite series.
4. Measuring Rates of Return for Lobbying Expenditures: An Empirical Case Study of Tax Breaks for Multinational Corporations, Raquel M. Alexander, Stephen W. Mazza and Susan Scholz, Journal of Law and Politics, Vol. 25, No. 401, 2009.
5. http://www.truthdig.com/eartotheground/item/more_vacant_homes_than_homeless_in_us_20111231#. Data from 2010. The number of empty houses in Europe outweigh their homeless population by a factor of two-to-one: “Deutsche Bank has warned that it will take 43 years to fill the oversupply of empty homes in Ireland at the current low population growth rate.” http://www.theguardian.com/society/2014/feb/23/europe-11m-empty-properties-enough-house-homeless-continent-twice)
6. Or, at the very least, precisely describes the conditions under which solutions may exist.
7. I suspect those false statements actually do kind of “go” somewhere, but that it would take more logic and type theory for me to know for sure.
8. That’s pretty amazing to contemplate: the set of all thinkable thoughts!
9. The comment on pets expected grammar, and the grammatical deconstruction, is from Gretchen McCulloch, “A Linguist Explains the Grammar of Doge. Wow” on the-toast.net.
10. Although, McCulloch observes that the shortest possible form of a word is used in these: “scare” makes a better addition to one-word doge phrases than “scary” or “excitement.”
11. For those who are anxious to get into a bit of mathematics right away, this is related to the combinatorial problem of finding *derangements*: if you give me a set in some order, the number of derangements is the number of ways I can give the set back to you in a new order, one in which none of the elements are where they started. For instance, consider the problem of the angry hatcheck girl: if she ignores the number on the customers’ tickets, giving a random hat to each customer and telling them to get out, then what is the probability that no one gets his or her own hat back?
12. Originally posted by queerqueerspawn, November 2013, republished in MuCulloch, cited above.
13. Which he toys with calling a “wector” space, “deflector” space, or “Hannibal Lecter” space, before finally settling on just $W$.
14. Even though taking them up also leads to incredibly weird results, like the freaky Banach-Tarsky paradox, which allows one to cut a finite sphere into a finite number of pieces that can be reassembled into two solid spheres the same size as the original.The question of whether the Axiom of Choice should be assumed when dividing loaves and/or fishes is a better question for theologians than mathematicians.
15. Which you can of course do, but you will only get so far.
16. http://web.ics.purdue.edu/~felluga/sf/NewYorkTimes.html
17. The fundamental theorem of calculus roughly says that differentiating and integrating are opposites, in some sense. The fundamental theorem of arithmetic says that any integer can be broken into prime factors. The fundamental theorem of algebra says that given any polynomial function of degree $d$, there are $d$ roots (possibly repeated) somewhere in the complex numbers.
18. In some versions of this story it was a special punishment given only to our little protagonist. The real details have been lost in the retellings over the centuries.
19. I’m deliberately not saying “boy or girl” because I’m pretty sure we’re in the “No girls allowed” stage of public education here. {Can I get a fact check from historians of education here?}
20. This is why computers do not replace mathematics. They aid it, but if you ask your computer from 2015 to do 1000000000000 operations it will choke. Find a creative way to reduce it by 4 orders of magnitude? Back in business. Of course, by 2020, \$1000 personal computers are expected to surpass human computational abilities, which is on the order of 10^24 Floating-point Operations Per Second (or “yottaFLOPS”)—a number much larger number than the scale of operations suggested above—but the principle will still be the same, and mathematics will still have a role to play in making the process more efficient.
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[OS X TeX] Latex symbol for "define equal"
Martin Berggren martin.berggren at cs.umu.se
Tue Aug 21 08:58:46 EDT 2018
Regarding the difference between a theorem and a definition: In the words of my thesis advisor: You cannot argue with a definition. (It should be said with a french accent!) A definition can be nice, practical, useful, or impractical, unnecessary, confusing, and so on, but it cannot be true or false. A theorem, however, is true.
Cheers,
> On 21 Aug 2018, at 14:48, Markus Klyver <markusklyver at hotmail.com> wrote:
>
> Depends, literature tends to use the definitions interchangeably depending on what aspect is important to subject.
>
>
> Från: MacOSX-TeX <macosx-tex-bounces at email.esm.psu.edu <mailto:macosx-tex-bounces at email.esm.psu.edu>> för Martin Berggren <martin.berggren at cs.umu.se <mailto:martin.berggren at cs.umu.se>>
> Skickat: den 21 augusti 2018 13:59
> Till: TeX on Mac OS X Mailing List
> Ämne: Re: [OS X TeX] Latex symbol for "define equal"
>
>
>> On 21 Aug 2018, at 13:47, Markus Klyver <markusklyver at hotmail.com <mailto:markusklyver at hotmail.com>> wrote:
>>
>> The thing is that you can have several different definitions, all equivalent. Consider the definition "a matrix A \in \mathbb R^{n \times n} is invertiable iff A have a multiplicative inverse". It turns out that this is equivalent to a lot of things, among det(A) !=0, A having full rang, A having n linearindependent eigenvectors, Ax=0 only having the trivial solution, Ax=b having a solution (which is unique) for every right-hand-side b, etc.
>
> I would save that this is a theorem, not a definition. I think of a definition as a “macro”; that is, giving a short name to a mathematical property. Example: a matrix A is called positive semidefinite when x^T Ax \geq 0 for all vectors x. The point is that you in each instances when the name is used, it can be replaced by its definition.
>
Martin Berggren
-------------------------------------------------------------------
Department of Computing Science,
UMIT Research Lab
Umeå Universitet
Campustorget 5, S-901 87 Umeå, Sweden. Ph: +46-70-732 8111
http://www.cs.umu.se/~martinb <http://www.cs.umu.se/~martinb>, Martin.Berggren at cs.umu.se <mailto:Martin.Berggren at cs.umu.se>
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Anonymous
Anonymous asked in Science & MathematicsAstronomy & Space · 1 decade ago
# If the earth was a couple centimeters closer to the sun, would we all burn?
Someone said this in chemistry today, but I just want a more credible source.
Update:
I would love a detailed explanation.
Relevance
• Anonymous
No, the Earth actually has an elliptical orbit, and our closest approach to the sun does not burn us, just as our furthest distance from it does not freeze us. A few centimeters or even many kilometers is not going to matter much. In fact, mountaintops are several kilometers closer to the sun than the oceans. Your person in chemistry class should close the books and work on common sense once in a while.
By the way, just to dispel the too common myth, where I mention above the earth sometimes being closer to and sometimes further from the sun, these things have nothing to do with summer and winter. A lot of people think they do, but they don't. Summer and winter are caused by the tilt in the Earth's axis of rotation relative to its plane of revolution around the sun, but since it is not part of the direct answer to the question you asked, I will spare you the detailed explanation here.
EDIT: I see where you said you would love a detailed explanation, and I assume you are referring to my post, so let me see if I can come up with an explanation that makes sense without drawings.
The Earth is tilted at about 23 degrees from being exactly "straight up and down", or 23 degrees from being exactly perpendicular to the plane on which it revolves around the sun. If the Earth were not tilted, the sun would always appear to pass directly overhead at the equator, and, for example, far north of the equator, the sun would always appear to pass 23 degrees to the south of directly overhead, and so on.
But because the Earth is tilted, this place 23 degrees north of the equator will get the sun passing directly overhead on the summer solstice, and the sun will never come more than 46 degrees from the vertical (almost exactly as close to the horizon as to vertical) during mid day on the winter solstice.
Before I go further, I want to point out that these numbers, the degrees, are an approximation, but they are relatively accurate because of the sun's great distance from Earth.
Now, why does the position of the sun in the sky matter? Because sunlight striking the Earth at an angle means the same amount of light and heat are spread over a larger area. In fact, in our example above, the midday sunlight hitting this point north of the equator during the summer solstice is about 1.4 times as intense as during the winter solstice. To get this change in intensity of sunlight by moving the position of the Earth relative to the sun would require our orbit be roughly 20 million miles closer to the sun or further from the sun (way different from the few centimeters your classmate mentioned), and our orbit is not that elliptical.
Note that while the point 23 degrees north of the equator is having the midday sun pass directly overhead, a point 23 degrees south of the equator is having their midday sun pass roughly 46 degrees from the vertical. It is summer in the north and winter in the south. They are always 6 months apart in their seasons, so the spring equinox for one is the fall equinox for the other. This is why Santa wears shorts in Australia and thick winter gear in Europe and the USA, because he comes just after the summer solstice in Australia and just after the winter solstice in the north, even though it is the same day (except in nations that mark Christmas by the old Julian calendar, but that is really a different topic).
Hope this helps.
From Aphelion (around July 5) when Earth is farthest to Perihelion (January 4) when Sun is nearest, the difference is about 5 million km. It is about a thirtieth of the average distance.
over the course of a year the earth's distance from the sun varies by about 5 million kilometers (with an average of about 150 million kilometers). so you tell me, do we alternately burn and freeze? I don't think so. obviously there will be a range of distances that allow a comfortable temperature, but the range is nowhere near that narrow. even the most conservative estimates suggest that the habitable zone is millions of kilometers wide.
• Anonymous
NOOOOO!!!! in the northern hemisphere, the earth is 3 million miles closer to the sun in the winter than it is in the summer. centimeters wouldn't do anything. meters, kilometers. a few million kilometers would not make much a difference.
no in fact we are never always the same distance from the sun. Depending on the time of year we can be many many many miles closer or further from the sun
Source(s): physical geography class
• Anonymous
Thats Ludicrous . . of course not.
Stand on your tip toes and you're a whole 4 inches closer to the sun !
Never believe astronomy stories told in a Chemistry class.
• 4 years ago
Oh God! They SAID the world would end in 2012, and I did not believe them!!! **** "i will see you hidden beneath the rocks begging they would collapse and crush you. ask a christian what that means. maybe 1 will have pitty on you and pray." And they wonder why we are rude to them. Because you are so horrible, that's why.
• Anonymous
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https://www.site.uottawa.ca/~bochmann/SEG-2106-2506/assignments/Assign-4/assign-4.html
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# Concurrency: A simulation in Java
## Description of the system to be modeled: Emergency service in a hospital
Note: This problem is adapted from a case study in the book by Thomas J. Schriber et al., Introduction to Simulation using GPSS/H, John Wiley & Sons, 1991, pp. 258.
The patients arrive at the emergency of a hospital. When they arrive, they are classified by a nurse in the following categories: (1) CW (can wait), and (2) NEA (needs immediate attention). On average, 15% of the patients will be classified NEA, the other 85% CW. There are three (3) nurses that do this classification.
The CW patients see a doctor and then can go back home.
The NEA patients also see a doctor and 20% of them can then go back home. The other 85 80% have to do a test in the laboratory of the hospital, then see the same doctor again before they can go home.
There are six (6) doctors in the emergency. They are partitioned into two groups of 3 doctors each: Group A looks after NEA patients, and Group B looks after CW partients. The laboratory performs one test after another (in FIFO order).
Here is some information about the duration of the different interventions. It is assumed that the durations have uniform distributions; for each intervention is indicated the average duration and the maximal deviation from the average, for example "10 +- 4" means that the duration has a uniform distribution between 6 and 14 minutes.
• Duration of the classification of one patient: 15 +- 5
• Duration of a doctor rendezvous with a CW patient: 30 +- 15
• Duration of the first doctor rendezvous of a NEA patient: 45 +- 10
• Duration of the second doctor rendezvous of a NEA patient: 15 +- 5
• Duration of a test: 30 +- 10
• The arrival pattern of the patients at the emergency is a Poisson distribution with an average inter-arrival time of 10 minutes.
The purpose of the simulation is to determine the average time that the CW and NEA patients remain in the emergency, and to determine the fraction of time the nurses and doctors are busy (their "utilisation") - the same for the laboratory.
For most of the shared resources used by the patients, you can assume a single waiting queue. However, for the group of doctors looking after the NEA patients, the scheduling is not clearly defined by the above text: Do the patients coming back to the same doctor have priority over new NEA patients ? - Also it is not evident how you could organize your simulation program such that NEA patients coming back will be seen by the same doctor they saw previously. You should explain all this in your report.
Note: For your simulation program, you should use one of the approaches discussed in class.
1. Write a Java program that simulates emergency of a hospital, as described above.
2. Determine the average time the CW and NEA patients remain in the hospital.
3. Determine the fraction of time the nurses, the doctors and the laboratory are busy.
4. Determine the expected error of your results. Explain how you have determined the expected error of your results. (Suggestion: Do several simulation runs and compare the different results obtained - as discussed in class).
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## 8178
8,178 (eight thousand one hundred seventy-eight) is an even four-digits composite number following 8177 and preceding 8179. In scientific notation, it is written as 8.178 × 103. The sum of its digits is 24. It has a total of 4 prime factors and 16 positive divisors. There are 2,576 positive integers (up to 8178) that are relatively prime to 8178.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 24
• Digital Root 6
## Name
Short name 8 thousand 178 eight thousand one hundred seventy-eight
## Notation
Scientific notation 8.178 × 103 8.178 × 103
## Prime Factorization of 8178
Prime Factorization 2 × 3 × 29 × 47
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 8178 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 8,178 is 2 × 3 × 29 × 47. Since it has a total of 4 prime factors, 8,178 is a composite number.
## Divisors of 8178
1, 2, 3, 6, 29, 47, 58, 87, 94, 141, 174, 282, 1363, 2726, 4089, 8178
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 17280 Sum of all the positive divisors of n s(n) 9102 Sum of the proper positive divisors of n A(n) 1080 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 90.4323 Returns the nth root of the product of n divisors H(n) 7.57222 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 8,178 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 8,178) is 17,280, the average is 1,080.
## Other Arithmetic Functions (n = 8178)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2576 Total number of positive integers not greater than n that are coprime to n λ(n) 644 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1032 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 2,576 positive integers (less than 8,178) that are coprime with 8,178. And there are approximately 1,032 prime numbers less than or equal to 8,178.
## Divisibility of 8178
m n mod m 2 3 4 5 6 7 8 9 0 0 2 3 0 2 2 6
The number 8,178 is divisible by 2, 3 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
## Base conversion (8178)
Base System Value
2 Binary 1111111110010
3 Ternary 102012220
4 Quaternary 1333302
5 Quinary 230203
6 Senary 101510
8 Octal 17762
10 Decimal 8178
12 Duodecimal 4896
20 Vigesimal 108i
36 Base36 6b6
## Basic calculations (n = 8178)
### Multiplication
n×i
n×2 16356 24534 32712 40890
### Division
ni
n⁄2 4089 2726 2044.5 1635.6
### Exponentiation
ni
n2 66879684 546942055752 4472892131939856 36579311855004142368
### Nth Root
i√n
2√n 90.4323 20.1472 9.50959 6.06079
## 8178 as geometric shapes
### Circle
Diameter 16356 51383.9 2.10109e+08
### Sphere
Volume 2.29103e+12 8.40435e+08 51383.9
### Square
Length = n
Perimeter 32712 6.68797e+07 11565.4
### Cube
Length = n
Surface area 4.01278e+08 5.46942e+11 14164.7
### Equilateral Triangle
Length = n
Perimeter 24534 2.89598e+07 7082.36
### Triangular Pyramid
Length = n
Surface area 1.15839e+08 6.44577e+10 6677.31
## Cryptographic Hash Functions
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https://physics.stackexchange.com/questions/281082/nuclear-fusion-with-extremely-high-pressure-and-low-temperature
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# Nuclear Fusion with extremely high pressure and low temperature
Theoretically, if we just create a high pressure with low (room 20C) temperature, at some point nuclear fusion can be started.
Is there any research on this topic, how high should be this pressure for different type of reaction? Maybe someone has some numbers in mind, how many GPa should we have to achieve it?
UPDATE:
I got a number $10^{21}$ Pa from thermodynamics of ideal gas. Of course it is approximation. Let's say for Deuterium-tritium we have to give energy 0.1 MeV for 2 atoms to start fusion, which means from electrostatic point of view distance between nucleus $1.44\times10^{-14}$ (14 femtometers) as @John Rennie answered. If I calculated how many atoms will fit in 1 $m^{3}$, if distance between atoms will be 14 femtometers, I will get $N=3.4\times10^{41}$ atoms. Then from $PV=\frac{N}{N_{a}}RT$ if I assume $V=1 m^{3}$ I get $P=1.4\times10^{21}$ Pa and density $2.24\times10^{21} \frac{kg}{m^{3}}$. Which is still $10^{5}$ times more than pressure in the core of the Sun. And if we consider real gas, might be number will be bigger.
Maybe in the future, if they will find another reaction with much less energy (less then 0.1MeV) it would be possible. Might be quantum tuneling can help a little bit :)
• Pressure is not a function of temperature at very high densities and cold temperatures. Essentially you are asking what densities are required. Commented Sep 19, 2016 at 12:57
• You might find the concept of the Lawson criterion to be helpful in addressing your question. Commented Sep 19, 2016 at 18:54
• If we put a 3-layer Russian-Doll set of pressure vessels inside pressure vessels capable of 1000 atmospheres each at the bottom of the ocean which has 1000 atmospheres. That could create the 1 trillion atmosphere pressure required for the addition of a neutron energy emitter at its center to start and initiate fusion of solid hydrogen, am I missing anything here except engineering details? Commented Mar 10, 2021 at 1:36
• @EricLeschinski, yes, the inner vessel would have an inner pressure of 4000 atmospheres, not one trillion atmospheres. (They can sustain a particular pressure differential, not pressure ratio!) Commented Jul 31, 2021 at 22:50
From memory the potential barrier for deuterium tritium fusion peaks at around $3$ femtometres.
Suppose we take the deuterium-tritium distance as $r$ then the electrostatic force between the nuclei is:
$$F = \frac{ke^2}{r^2}$$
and we get can a pressure by dividing this by the area of a sphere with radius $r$ to get:
$$P = \frac{ke^2}{4\pi r^4}$$
You should regard this as a very rough estimate, but it should be immediately apparent the the $r^{-4}$ dependence is going to be a killer because it rises very rapidly for small $r$. If we take $r$ to be $10$ femtometres we get a pressure of about $10^{28}$ Pa. This is so ridiculously large that even given the rough nature of our estimate it's obvious that this approach is not going to work. The pressure at the centre of the Sun is only around $3\times 10^{16}$ Pa.
• Very good point. I am looking now how to calculate it from thermodynamics point, like internal energy of gas equals to 0.1 MeV. Just to compare with your calculations :) Commented Sep 19, 2016 at 14:10
• To be clear, $10^{28}$ Pa is near the Chandrasekhar limit (the pressure limit of a star supported not by heat but by electron degeneracy). So the fact that the electrons will be pushing back pretty seriously prior to this. I think if you push much more, whatever you are pushing together is going to form neutronium on you? I guess that is a form of fusion.
– Yakk
Commented Sep 19, 2016 at 19:07
• Pyconuclear reactions in a crystalline lattice may occur before neutronisation. But the order of magnitude forthe pressure looks good to me. Commented Sep 19, 2016 at 20:47
• I'm curious how much energy you need to squeeze two helium atoms this close. Maybe this alone would cause some interesting effects? Commented Sep 20, 2016 at 0:03
• If this 10^28 Pa is a pressure in the neutron star, that means actual fusion should start much early. I still want to calculate internal gas energy from some real gas model like Vam der Waals. Commented Sep 20, 2016 at 7:04
As @John Rennie said, you will need a very high pressure if you want you atoms to be forced to be close enough to actually touch their respective barriers. What you describe sounds more as what happens in a neutron star in superextreme conditons.
To start fusion in a more conventional way, each nucleus must have an energy of 0.1 MeV. Acoding to Boltzmann, $E_p=k_B*T$ is the energy of each particle at a certain temperature. At room temperature this energy is 25meV, several orders of magnitude lower than that.
However, if you increase the pressure of this gas at room temperature and keep the temperature constant, each particle energy remains the same. The fact that you have more pressure means more particles are contained in the gas, but on average each particle will have the same energy (25meV) as stated by Boltzmann's law, and won't be able to break the barrier of fusion.
• But this works for ideal gas, where potential energy of atoms is not considered. In ideal gas internal energy equals to the sum of kinetic energy of all atoms and potential energy of atoms = 0. For real gas internal energy will be sum of kinetic and potential energy. For example for Van der Waals gas potential energy already not zero. As I understand Van der Waals model is not good for high pressure gas. What is the good model for high pressure gas? Commented Sep 19, 2016 at 14:14
• @Zlelik That depends on the gas you're modeling, and how much of a tradeoff you want to make between accuracy and detail. A fairly simple model for monoatomic gases which includes both attractive and repulsive effects would be a gas in a (quantum) Lennard-Jones pair potential and then bound in some fixed volume $V$ by a particle-in-a-box type potential. But even something as simple as that can't be fully understood in closed form.
– Ian
Commented Sep 19, 2016 at 18:42
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http://www.teses.usp.br/teses/disponiveis/18/18134/tde-04022015-113656/pt-br.php
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DOI
10.11606/T.18.2014.tde-04022015-113656
Documento
Autor
Nome completo
Wagner Queiroz Silva
E-mail
Área do Conhecimento
Data de Defesa
Imprenta
São Carlos, 2014
Coda, Humberto Breves (Presidente)
Campello, Eduardo de Morais Barreto
Corrêa, Márcio Roberto Silva
Paccola, Rodrigo Ribeiro
Rubert, José Benaque
Título em português
Sobre análise não linear geométrica de edifícios considerando o empenamento dos núcleos estruturais e a interação solo-estrutura
Palavras-chave em português
Análise não linear de estruturas
Edifícios altos
Interação solo-estrutura
Método dos Elementos de Contorno
Método dos Elementos Finitos
Núcleos estruturais
Resumo em português
Título em inglês
On geometric nonlinear analysis of tall buildings structures considering the warping of the structural cores and the soil-structure interaction
Palavras-chave em inglês
Boundary Element Method
Finite Element Method
Nonlinear analysis of structures
Soil-structure interaction
Structural cores
Tall buildings
Resumo em inglês
In this thesis a numerical model for geometric nonlinear analysis of three-dimensional structures of tall buildings was developed, considering the influence of all structural components, including the core-slab connection and the foundation system. Columns and beams are modeled by a frame finite element which can have a cross section of any shape, while the slabs are modeled by shell finite elements. Both consider the nonlinear geometric behavior and adopt nodal positions and generalized vectors as degrees of freedom instead of displacements and rotations. For the frame finite element it is also considered the cross sectional warping as a degree of freedom. A numerical strategy is presented for the coupling between the shell elements and the frame's cross section, thus forming a structural-core element with diaphragm. The coupling is done through a kinematic array which is responsible for inserting the contributions of shell elements, connected to the core walls, into the Hessian matrix and also into the internal force vector of the frame element used to discretize the core. The linear-elastic constitutive relation of Saint Venant-Kirchhoff is adopted for the building materials and the geometric nonlinearity is considered via a Lagrangian formulation with exact kinematics. The foundation's flexibility is considered through a stiffness matrix for the soil-foundation system. This matrix is computed in another program based on the numerical coupling between the Boundary Element Method and the Finite Element Method, using a numerical strategy based on the Maxwell-Betti's Theorem. This strategy consists in determining the flexibility coefficient of points on a discrete mesh of the soil-foundation system. The soil is modeled by the Boundary Element Method using the fundamental solution of Mindlin. The structural foundation elements, including shallow foundation, footings, blocks and piles, are modeled using conventional frame and shell finite elements. The program is applied to the analysis of complete structural systems of tall buildings, considering the influence of the core warping on the mechanical behaviour of the slabs and also the soil-structure interaction effects. Numerical examples are presented to confirm the efficiency and to demonstrate the potential application of the proposed formulation.
AVISO - A consulta a este documento fica condicionada na aceitação das seguintes condições de uso:
Este trabalho é somente para uso privado de atividades de pesquisa e ensino. Não é autorizada sua reprodução para quaisquer fins lucrativos. Esta reserva de direitos abrange a todos os dados do documento bem como seu conteúdo. Na utilização ou citação de partes do documento é obrigatório mencionar nome da pessoa autora do trabalho.
Data de Publicação
2015-02-27
AVISO: Saiba o que são os trabalhos decorrentes clicando aqui.
Todos os direitos da tese/dissertação são de seus autores
Centro de Informática de São Carlos
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https://gsebsolutions.in/gseb-solutions-class-9-maths-chapter-13-ex-13-9/
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# GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9
Gujarat Board GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9
Question 1.
A wooden bookshelf has external dimensions as follows:
Height = 110 cm,
Depth = 25 cm,
Breadth = 85 cm (See figure).
The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
Solution:
Surface area to be polished
= [(110 x 85) + 2(110 x 25) + 2(85 x 25) + 2(110 X 5) + 4(75 x 5)1
= (9350 + 5500 + 4250 + 1100 + 1500) cm2 = 21700 cm2
∴ Expenses required for polishing @ 20 paise per cm2
= 21700 x 20 paise
= ₹ $$\frac{21700 \times 20}{100}$$ = ₹ 4340
Surface area to be painted
= [2(20 x 90) + 6(75 x 20) + (75 x 90)1
= (3600 + 9000 + 6750) cm2 = 19350 cm2
∴ Expenses required for painting @ 10 paise per cm2
= 19350 x lo paise
= ₹ $$\frac{19350 \times 10}{100}$$ = ₹ 1935
∴ Total expenses required for polishing and painting the surface of the bookshelf
= ₹4340 + ₹1935 = ₹6275
Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
Solution:
For a wooden sphere
Diameter = 21 cm
∴ Radius (r) = $$\frac {21}{2}$$ cm
∴ Surface area of a wooden sphere
= 4Ï€r2 = 4 x $$\frac {22}{7}$$ x ($$\frac {21}{2}$$)2
= 1386 cm2
∴ Surface area of a wooden sphere to be painted
= 1386 – Ï€(1.5)2
= 1386 – $$\frac {22}{7}$$ (1.5)2 = 1378.93 cm2
∴ Surface area of eight wooden spheres
= 1378.93 x 8 = 11031.44 cm2
∴ Cost of painting silver @ 25 paise per cm2
= 11031.44 x 25 paise
= ₹ $$\frac{11031.44 \times 25}{100}$$
= ₹ 2757.86
For a cylindrical support
Height (h) = 7 cm
∴ Surface area of a cylindrical support = 2πrh
= 2 x $$\frac {22}{7}$$ x 1.5 x 7 = 66 cm2
∴ Surface area of eight cylindrical supports
= 66 x 8 = 528 cm2
∴ Cost of painting black @ 25 paise per cm2
= 528 x 5 paise
= ₹ $$\frac{528 \times 5}{100}$$ = ₹ 26.40
∴ Cost of paint required
= (2757.86 + 26.40)
= 2784.26 (approx.)
Question 3
The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Solution:
Let the radius of the sphere be $$\frac {r}{2}$$cm.
Then its diameter = 2 $$\frac {r}{2}$$ = r cm
Curved surface area of the original sphere
4π($$\frac {r}{2}$$)2 = πr2 cm2
New diameter (decreased) of the sphere
= r – r x $$\frac {25}{100}$$
= r – $$\frac {r}{4}$$ = $$\frac {3r}{4}$$ cm
∴ Radius of the new sphere
= $$\frac {1}{2}$$($$\frac {3r}{4}$$) = $$\frac {3r}{8}$$cm
∴ New curved surface area of the sphere
4Ï€($$\frac {3r}{8}$$)2 = $$\frac{9 \pi r^{2}}{16}$$ cm2
∴ Decrease in the original curved surfà ce area
= Ï€r2 – $$\frac{9 \pi r^{2}}{16}$$
= $$\frac{16 \pi r^{2}-9 \pi r^{2}}{16}$$ = $$\frac{7 \pi r^{2}}{16}$$
∴ Percentage of decrease in the original curved surface area
= $$\frac{\frac{7 \pi r^{2}}{16}}{\pi r^{2}} \times 100 \%$$
= $$\frac{700}{16} \%$$ = $$\frac{175}{4} \%$$
= $$43 \frac{3}{4} \%$$ or 43.75%
Hence, the original curved surface area decreases by 43.75%.
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# Sum involving the Digamma function.
Consider the double sum $$f(x):=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m+\frac{1}{n+x}}-\frac{1}{m+\frac{1}{n}},$$ which converges for $x>0$. One interpretation of this sum is the measure of the double preimage $T^{-2}([0,x])$, in which $T:[0,1]\to [0,1]$ denotes the Gauss map $T(x)=\{1/x\}$ (the fractional part of $1/x$). Let $\psi$ denote the Digamma function $\psi(z):=\Gamma'(z)/\Gamma(z)$. The identity $$\psi(z)=-\gamma+\sum_{k=0}^\infty\left(\frac{1}{n+1}-\frac{1}{n+z}\right)$$ can then be used to rewrite $f$ as $$f(x)=\sum_{n=1}^\infty \psi\left(1+\frac{1}{n}\right)-\psi\left(1+\frac{1}{n+x}\right).$$ Is there any closed-form simplification of the sum above? It seems plausible that there may be such a solution involving the higher Polygamma function $\psi^{(2)}$.
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https://math.stackexchange.com/questions/4465854/what-is-the-reasoning-behind-sqrtx2-lvert-x-rvert
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# what is the reasoning behind $\sqrt{x^2}=\lvert x\rvert$ [duplicate]
My math teacher told us that 'The square root of a positive number is always positive' and then he gave this relation to us $$\sqrt{x^2}=\lvert x\rvert$$.
I have been previously taught that any positive number has a positive and negative square root. Eg $$\sqrt{4}=\pm2$$
So what is right or wrong in this situation.
• $\sqrt{4}\neq-2$. $\sqrt{4}=+2$ always by convention. We always take the absolute value when taking square roots by convention. Also, $\sqrt{4}\neq\pm2$. I used to think that too before, but it is the wrong understanding. Commented Jun 5, 2022 at 4:15
• This was indeed already answered, and/but this "rule" is not a mathematical fact-of-nature, but just a convention, mostly manifest in high school and lower-division math courses (in the U.S.). "In real life", as you well know, complex numbers have two complex square roots (except $0$). It is possible to make various conventions to specify a single square root, thus having an official function (in an elementary sense), namely, single-valued... But the inherent two-valued-ness of square root is inescapable (except artificially and non-robustly by convention). Commented Jun 5, 2022 at 4:42
• @tryingtobeastoic To nitpick: It is more useful to think of a principal root as a particular choice of root rather than as an absolute value, because, obviously, $\sqrt[3]{-3}\ne1.44$ and $\sqrt{-1}\ne1.$ Commented Jun 5, 2022 at 8:25
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A little help understanding math....
12-16-2013, 11:56 PM (This post was last modified: 05-27-2022 06:28 PM by Thomas Klemm.)
Post: #1
Thomas Klemm Senior Member Posts: 1,854 Joined: Dec 2013
A little help understanding math....
There was a recent thread in the old forum where Cyrille asked a question concerning the implementation of the complex inverse trigonometric functions.
I must admit that it took me a while to understand what it was about. I will only deal with the complex inverse of cosine.
Let's assume $$w=u+iv$$. Then
\begin{align} z&=&x+iy=cos(w)=cos(u+iv) \\ &=&cos(u)cos(iv)-sin(u)sin(iv) \\ &=&cos(u)cosh(v)-i sin(u)sinh(v) \\ \end{align}
But we would like to find $$w=arccos(z)$$.
Better don't ask me who came up with this solution:
\begin{align} |z+1|^2 &=& (x+1)^2+y^2 \\ &=& x^2+2x+1+y^2 \\ &=& cos^2(u)cosh^2(v)+2cos(u)cosh(v)+1+sin^2(u)sinh^2(v) \\ &=& cos^2(u)(1+sinh^2(v))+2cos(u)cosh(v)+1+sin^2(u)sinh^2(v) \\ &=& cos^2(u)+cos^2(u)sinh^2(v)+2cos(u)cosh(v)+1+sin^2(u)sinh^2(v) \\ &=& cos^2(u)+2cos(u)cosh(v)+1+(cos^2(u)+sin^2(u))sinh^2(v) \\ &=& cos^2(u)+2cos(u)cosh(v)+1+sinh^2(v) \\ &=& cos^2(u)+2cos(u)cosh(v)+cosh^2(v) \\ &=& (cos(u)+cosh(v))^2 \\ |z+1| &=& cosh(v)+cos(u) \end{align}
And similarly we can conclude:
$|z-1| = cosh(v)-cos(u)$
Add the two equations to get:
$|z+1|+|z-1| = 2cosh(v) \\ cosh(v)=\frac{|z+1|+|z-1|}{2}$
From $$x=cos(u)cosh(v)$$ we conclude:
$2x=2cos(u)cosh(v) \\ cos(u)=\frac{2x}{|z+1|+|z-1|}$
This leads to the following function:
Code:
from math import sqrt, acos, acosh def arccos(z): x, y = z.real, z.imag y2 = y**2 U = sqrt((x+1)**2 + y2) V = sqrt((x-1)**2 + y2) W = (U+V)/2 u = acos(x/W) v = acosh(W) return u - 1j*v
Let's compare this to the builtin function:
Code:
>>> import cmath >>> z=3+4j >>> arccos(z) (0.9368124611557199-2.305509031243477j) >>> cmath.acos(z) (0.9368124611557198-2.305509031243477j) >>> z/=10 >>> arccos(z) (1.2901667645030908-0.40511233717803047j) >>> cmath.acos(z) (1.2901667645030908-0.4051123371780309j) >>> z/=10 >>> arccos(z) (1.5408158285382982-0.04000730997058477j) >>> cmath.acos(z) (1.5408158285382982-0.04000730997058378j)
It appears we loose some accuracy when calculating $$v=acosh(W)$$ for values of W close to 1.
Why is that? (substitute $$t=e^v$$)
\begin{align} W &=& cosh(v) \\ &=& \frac{e^v+e^{-v}}{2} \\ W &=& \frac{t+\frac{1}{t}}{2} \\ 2Wt &=& t^2+1 \\ t^2-2Wt+1 &=& 0 \\ t &=& W\pm\sqrt{W^2-1} \\ \end{align}
Now we can see what's happening: when W is close to 1 we experience cancellation and thus loss of accuracy.
We'd like to calculate W-1 without loosing digits.
Can we do better?
Let's see what the HP-48 suggests:
Code:
* Q := y^2/(sqrt((|x|+1)^2 + y^2) + (|x|+1)) * R := sqrt((|x|-1)^2 + y^2) +||x|-1| * S := y^2/R if R<>0, 0 otherwise * M := Q+R if |x|>=1, Q+S otherwise * P := Q+S if |x|>=1, Q+R otherwise * B := 2*x/(M+2) * C := sqrt((P/(M+2))*(2-(P/(M+2)))) * sg(y,x) := sgn(y) if y<>0, -sgn(x) otherwise * IM := sg(y,x)*lnp1((M/2) + sqrt((M/2)*((M/2)+2))) (sign replacement) * * { arccos(B) |B| <= (1/sqrt(2)) * RE1 := { arcsin(C) |B| > (1/sqrt(2)) and B >= 0 * { pi - arcsin(C) |B| > (1/sqrt(2)) and B < 0 * * RE2 := { arcsin(B) |B| <= (1/sqrt(2)) * { sgn(B)*arccos(C) |B| > (1/sqrt(2)) (sign replacement) *
I've taken the liberty to add some new variables:
Code:
def sgn(x): return cmp(x, 0.0) def sg(y,x): return sgn(y) if y != 0 else -sgn(x) def inverse(z): x, y = z.real, z.imag _x_ = abs(x) y2 = y**2 U = sqrt((_x_+1)**2 + y2) T = U + (_x_+1) Q = y2/T # = U - (|x|+1) V = sqrt((_x_-1)**2 + y2) R = V + abs(_x_-1) S = y2/R if R != 0 else 0 # = V - ||x|-1| M = Q+R if _x_ >= 1 else Q+S # M = U + V - 2 P = Q+S if _x_ >= 1 else Q+R # P = U + V - 2|x| B = 2*x/(M+2) C = sqrt((P/(M+2))*(2-(P/(M+2)))) IM = sg(y,x)*log1p((M/2) + sqrt((M/2)*((M/2)+2))) # sign replacement _B_ = abs(B) sqrt_2 = 1/sqrt(2) RE1 = acos(B) if _B_ <= sqrt_2 else asin(C) if B >= 0 else pi - asin(C) RE2 = asin(B) if _B_ <= sqrt_2 else sgn(B)*acos(C) # sign replacement return IM, RE1, RE2
Please note that M/2 = W-1 since M = U+V-2. That's exactly the value we were looking for.
Instead of ln we use lnp1 (or log1p in Python) for values close to 1:
$v=ln(t)=lnp1(t-1)=lnp1(W-1\pm\sqrt{(W-1)(W+1)}$
Substitute W-1 by M/2 and we end up with the formula for IM.
Well nearly, we still have to deal with the sign of the expression.
What about $$\frac{1}{\sqrt{2}}$$? That was Cyrille's question.
For small values of B close to 0 I assume both acos(B) and asin(B) are rather accurate.
However with values close to 1 we experience again cancellation and thus loss of accuracy.
In this case we'd prefer to use $$\sqrt{1-B^2}$$ instead.
That's exactly what is calculated with C but without cancellation. I'm lazy and leave that as an exercise.
So where exactly should we switch? Why not just in the middle between 0 and $$\frac{\pi}{2}$$?
And: $$cos(\frac{\pi}{4})=sin(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$$.
But in this area we don't experience cancellation at all thus .7 will do as well.
For those interested in the SYSRPL code I've added the stack-diagrams:
Code:
:: ; z C%>%% ; x y 2DUP ; x y x y DUP ; x y x y y %%*SWAP ; x y y^2 x %%ABS ; x y y^2 |x| DUP ; x y y^2 |x| |x| %%1+ ; x y y^2 |x| |x|+1 DUPDUP ; x y y^2 |x| |x|+1 |x|+1 |x|+1 %%* ; x y y^2 |x| |x|+1 (|x|+1)^2 4PICK ; x y y^2 |x| |x|+1 (|x|+1)^2 y^2 %%+ ; x y y^2 |x| |x|+1 (|x|+1)^2+y^2 %%SQRT ; x y y^2 |x| |x|+1 √((|x|+1)^2+y^2) %%+ ; x y y^2 |x| √((|x|+1)^2+y^2)+(|x|+1) 3PICKSWAP ; x y y^2 |x| y^2 √((|x|+1)^2+y^2)+(|x|+1) %%/ ; x y y^2 |x| Q OVER ; x y y^2 |x| Q |x| %%1 ; x y y^2 |x| Q |x| 1 %%- ; x y y^2 |x| Q |x|-1 %%ABS ; x y y^2 |x| Q ||x|-1| DUPDUP ; x y y^2 |x| Q ||x|-1| ||x|-1| ||x|-1| %%* ; x y y^2 |x| Q ||x|-1| ||x|-1|^2 5PICK ; x y y^2 |x| Q ||x|-1| ||x|-1|^2 y^2 %%+ ; x y y^2 |x| Q ||x|-1| ||x|-1|^2+y^2 %%SQRT ; x y y^2 |x| Q ||x|-1| √(||x|-1|^2+y^2) %%+ ; x y y^2 |x| Q R 4ROLLOVER ; x y |x| Q R y^2 R DUP ; x y |x| Q R y^2 R R %%0<> ; x y |x| Q R y^2 R ITE ; x y |x| Q R y^2 R %%/ ; x y |x| Q R y^2/R :: ; x y |x| Q R y^2 R 2DROP ; x y |x| Q R %%0 ; x y |x| Q R 0 ; ; x y |x| Q R S UNROTOVER ; x y |x| S Q R Q %%+ ; x y |x| S Q R+Q UNROT ; x y |x| R+Q S Q %%+ ; x y |x| R+Q S+Q ROT ; x y R+Q S+Q |x| %%1 ; x y R+Q S+Q |x| 1 %%< ; x y R+Q S+Q ?SKIPSWAP ; x y P M DUP ; x y P M M %%2 ; x y P M M 2 %%+ ; x y P M M+2 ROTOVER ; x y M M+2 P M+2 %%/ ; x y M M+2 P/(M+2) %%2 ; x y M M+2 P/(M+2) 2 OVER ; x y M M+2 P/(M+2) 2 P/(M+2) %%- ; x y M M+2 P/(M+2) 2-P/(M+2) %%* ; x y M M+2 P/(M+2)*(2-P/(M+2)) %%SQRT ; x y M M+2 C 5PICK ; x y M M+2 C x DUP ; x y M M+2 C x x %%+ ; x y M M+2 C 2x ROT ; x y M C 2x M+2 %%/ ; x y M C B ROT ; x y C B M %%2 ; x y C B M 2 %%/ ; x y C B M/2 DUP ; x y C B M/2 M/2 %%2 ; x y C B M/2 M/2 2 %%+ ; x y C B M/2 M/2+2 OVER ; x y C B M/2 M/2+2 M/2 %%* ; x y C B M/2 (M/2+2)*M/2 %%SQRT ; x y C B M/2 √((M/2+2)*M/2) %%+ ; x y C B M/2+√((M/2+2)*M/2) %%lnp1 ; x y C B lnp1(M/2+√((M/2+2)*M/2)) 5ROLL ; y C B lnp1(M/2+√((M/2+2)*M/2)) x 5ROLL ; C B lnp1(M/2+√((M/2+2)*M/2)) x y DUP ; C B lnp1(M/2+√((M/2+2)*M/2)) x y y %%0= ; C B lnp1(M/2+√((M/2+2)*M/2)) x y ITE ; C B lnp1(M/2+√((M/2+2)*M/2)) x y :: ; C B lnp1(M/2+√((M/2+2)*M/2)) x y DROP ; C B lnp1(M/2+√((M/2+2)*M/2)) x %%0< ; C B lnp1(M/2+√((M/2+2)*M/2)) ; ; C B lnp1(M/2+√((M/2+2)*M/2)) :: ; C B lnp1(M/2+√((M/2+2)*M/2)) x y SWAPDROP ; C B lnp1(M/2+√((M/2+2)*M/2)) y %%0>= ; C B lnp1(M/2+√((M/2+2)*M/2)) ; ; C B lnp1(M/2+√((M/2+2)*M/2)) ?SKIP ; C B lnp1(M/2+√((M/2+2)*M/2)) %%CHS ; C B sg(y,x)*lnp1(M/2+√((M/2+2)*M/2)) UNROTDUP ; IM C B B %%ABS ; IM C B |B| %%.7 ; IM C B |B| .7 %%<= ; IM C B case ; IM C B :: ; IM C B SWAPDROPDUP ; IM B B %%ACOSRAD ; IM B RE1 SWAP ; IM RE1 B %%ASINRAD ; IM RE1 RE2 ; ; IM RE1 RE2 SWAPDUP ; IM B C C %%ASINRAD ; IM B C arcsin(C) SWAP ; IM B arcsin(C) C %%ACOSRAD ; IM B arcsin(C) arccos(C) %%ABS ; IM B arcsin(C) |arccos(C)| ROT ; IM arcsin(C) |arccos(C)| B %%0>= ; IM arcsin(C) |arccos(C)| ?SEMI ; IM RE1 RE2 %%CHS ; IM arcsin(C) -|arccos(C)| %%PI ; IM arcsin(C) RE2 pi ROT ; IM RE2 pi arcsin(C) %%- ; IM RE2 pi-arcsin(C) SWAP ; IM RE1 RE2 ; ; IM RE1 RE2
It's much easier to understand this program thanks to the comments.
I guess I'd never have figured that out without them.
Thus my appeal to Cyrille and Tim: please make sure that this information doesn't get lost.
Send the source-code of HP-48 (and whatever you have) to a computer museum if that's legally possible.
There's so much we can learn from that.
Kind regards
Thomas
PS: There are these two lines somewhere close to the end:
Code:
%%ACOSRAD ; IM B arcsin(C) arccos(C) %%ABS ; IM B arcsin(C) |arccos(C)|
However the use of the absolute value is missing in the comment:
Code:
* RE2 := { arcsin(B) |B| <= (1/sqrt(2)) * { sgn(B)*arccos(C) |B| > (1/sqrt(2)) (sign replacement)
Since %%ACOSRAD should never return negative values I don't understand why this was added.
IMHO the command %%ABS could be removed. Any ideas why this is needed?
12-17-2013, 10:19 AM
Post: #2
Paul Dale Senior Member Posts: 1,792 Joined: Dec 2013
RE: A little help understanding math....
Thanks for the analysis. It is like we suspected.
This looks a lot like one of Kahan's little gems.
- Pauli
12-17-2013, 01:17 PM
Post: #3
Namir Senior Member Posts: 907 Joined: Dec 2013
RE: A little help understanding math....
Thomas,
Your Python code tells me that you use it and most likely the numpy module. How can you install the numpy module? I appreciate your help (including links to sites that make the process very clear to the confused folks like Mois!!)
Namir
12-17-2013, 03:04 PM (This post was last modified: 12-19-2013 12:17 AM by Thomas Klemm.)
Post: #4
Thomas Klemm Senior Member Posts: 1,854 Joined: Dec 2013
RE: A little help understanding math....
One of the reasons I use Python is that it comes with "batteries included". Until now I never had to install additional Python modules. Thus stackoverflow or some Python related forums might be more helpful. And then I'm not familiar with Windows.
Cheers
Thomas
To my surprise numpy is already installed. You just have to import it:
Code:
>>> import numpy as np
Or:
Code:
>>> import numpy.polynomial.polynomial as P >>> P.polyroots((1,-1,-1)) array([-1.61803399, 0.61803399])
I've just skimmed through the Python Scientific lecture notes but it appears to give a nice introduction.
08-15-2021, 03:48 AM
Post: #5
Albert Chan Senior Member Posts: 2,234 Joined: Jul 2018
RE: A little help understanding math....
(12-16-2013 11:56 PM)Thomas Klemm Wrote: Let's assume $$w=u+iv$$. Then
\begin{align} z&=&x+iy=cos(w)=cos(u+iv) \\ &=&cos(u)cos(iv)-sin(u)sin(iv) \\ &=&cos(u)cosh(v)-i sin(u)sinh(v) \\ \end{align}...
Code:
u = acos(x/W) v = acosh(W) return u - 1j*v
There is a flaw with the code, arc function only give principle branch.
We have to match sign of imaginery part.
From above code, 0 ≤ u ≤ pi, v ≥ 0, sign(-sin(u) * sinh(v)) = negative
If y is also negative, no correction needed, else we flip sign.
This is the patch:
< return u - 1j*v
> return complex(u, v if y<0 else -v)
08-15-2021, 12:25 PM
Post: #6
Albert Chan Senior Member Posts: 2,234 Joined: Jul 2018
RE: A little help understanding math....
Another proof of acos(z) algorithm, and showed why we need |z+1|, |z-1|
XCas> z := cos(u+i*v)
XCas> x, y := re(z), im(z)
cos(u)*cosh(v), -sin(u)*sinh(v)
|z|² = x²+y²
= cos(u)²*cosh(v)² + sin(u)²*sinh(v)²
= cos(u)²*cosh(v)² + (1-cos(u)²)*(cosh(v)²-1)
= cosh(v)² + cos(u)² - 1
|z±1|² = (x±1)² + y² = (x²+y²+1) ± 2x
= (cosh(v)² + cos(u)²) ± 2*cosh(v)*cos(u)
= (cosh(v) ± cos(u))²
cosh(v) ≥ 1, cos(u) ≤ 1, we take square root of both side:
|z±1| = cosh(v) ± cos(u)
08-26-2021, 02:31 PM (This post was last modified: 08-26-2021 06:27 PM by Albert Chan.)
Post: #7
Albert Chan Senior Member Posts: 2,234 Joined: Jul 2018
RE: A little help understanding math....
(12-16-2013 11:56 PM)Thomas Klemm Wrote: From $$x=cos(u)cosh(v)$$ we conclude:
$2x=2cos(u)cosh(v) \\ cos(u)=\frac{2x}{|z+1|+|z-1|}$
There is an issue of recovering u from acos()
Accuracy may be bad. Worse, it might hit outside valid range (-1 ≤ cos(u) ≤ 1)
Similarly, recovering v from acosh() might hit by same issue, since cosh(v) ≥ 1
>>> z = 1.5430806348152439+0j # acos(z) = u + i*v
>>> a, b = abs(1+z), abs(1-z)
>>> V = (a+b) / 2
>>> U = z.real / V
>>> U, V # U, V = cos(u), cosh(v)
(1.0000000000000002, 1.5430806348152437)
---
Kahan's algorithm avoided these problems, using atan/asinh for complex acos
From complex.c, cacos(z), https://opensource.apple.com/source/Libm....auto.html
Code:
real(cacos(z)) = 2.0*atan(real(csqrt(1.0-z)/real(csqrt(1.0+z)))) imag(cacos(z)) = arcsinh(imag(csqrt(1.0-z)*csqrt(cconj(1.0+z))))
z = x+i*y = cos(u+i*v) = cos(u)*cosh(v) - i*sin(u)*sinh(v) ..... (1)
Let U = cos(u), V = cosh(v), we have x = U*V
Let t = tan(u/2)
Since u = acos(U) = 0 to pi, non-negative, t = |t|
cos(u) = U = (1-t^2)/(1+t^2) → t = |t| = √((1-U)/(1+U))
Again, from the same source, for csqrt(z)
Code:
sqrt(x + i*y) = sqrt((|z| + Real(z))/2) + i*sqrt((|z| - Real(z))/2) and sqrt(x - i*y) = sqrt((|z| + Real(z))/2) - i*sqrt((|z| - Real(z))/2)
Previously, we showed |1±z| = cosh(v) ± cos(u) = V ± U
Assumed we have sign-zero, let s = sign(imag(z)) = ± 1
√(1+z) = √(((V+U)+(1+U*V))/2) + i*s*√(((V+U)−(1+U*V))/2) = √((1+U)*(V+1)/2) + i*s*√((1−U)*(V−1)/2)
√(1−z) = √(((V−U)+(1−U*V))/2) − i*s*√(((V−U)−(1−U*V))/2) = √((1−U)*(V+1)/2) − i*s*√((1+U)*(V−1)/2)
real(√(1-z)) / real(√(1+z)) = √((1-U)/(1+U)) = t = tan(u/2)
u = atan(real(√(1-z))/real(√(1+z))) * 2
real(√(1+z)) * imag(√(1-z)) = -s * (1+U)/2 * √(V*V-1) ..... (2)
-imag(√(1+z)) * real(√(1-z)) = -s * (1-U)/2 * √(V*V-1) ..... (3)
(2) and (3) have the same sign, sum is free from subtraction cancellation.
(2)+(3) → RHS = -s * √(V*V-1) = -s * |sinh(v)| = sinh(-s*|v|)
From (1), sign(v) = sign(-y) = -s:
v = asinh(real(√(1+z))*imag(√(1-z)) - imag(√(1+z)) * real(√(1-z)))
Code:
Complex Cacos(Complex z) { Complex a = csqrt(1+z), b = csqrt(1-z); double x2 = atan (Real(b)/Real(a)) * 2; double y2 = asinh(Real(a)*Imag(b) - Imag(a)*Real(b)); return CMPLX(x2, copysign(y2, -Imag(z))); }
08-26-2021, 06:16 PM
Post: #8
Albert Chan Senior Member Posts: 2,234 Joined: Jul 2018
RE: A little help understanding math....
We can reuse previous post acos(z) = u+i*v result for asin(z)
Naive implementation is with identity asin(z) = pi/2 - acos(z) = (pi/2-u) - i*v
But, we could avoid cancellation errors of real part.
a = √(1+z) = √((1+U)*(V+1)/2) + i*s*√((1−U)*(V−1)/2)
b = √(1−z) = √((1−U)*(V+1)/2) − i*s*√((1+U)*(V−1)/2)
+Real(a) * Real(b) = (V+1)/2 * √(1-U*U) ..... (4)
-Imag(a) * Imag(b) = (V-1)/2 * √(1-U*U) ..... (5)
With V=cosh(v) ≥ 1, both (4) and (5) are non-negative.
Summing them is safe from cancellation error.
(4)+(5) → S = V * √(1-U*U)
x/S = (U*V) / (V*√(1-U*U)) = cos(u) / |sin(u)| = sign(x) * |tan(pi/2 - u)|
sign(x) should match sign(pi/2-u), we have pi/2-u = atan(x/S)
Except for sign flip, imaginery part of asin(z) and acos(z) are the same.
So, I just copy/paste from acos() code, for the imaginery part.
Code:
Complex Casin(Complex z) { Complex a = csqrt(1+z), b = csqrt(1-z); double x2 = atan(Real(z) / (Real(a)*Real(b) - Imag(a)*Imag(b))); double y2 = asinh(Real(a)*Imag(b) - Imag(a)*Real(b)); return CMPLX(copysign(x2, Real(z)), copysign(y2, Imag(z))); }
08-29-2021, 12:20 AM
Post: #9
Albert Chan Senior Member Posts: 2,234 Joined: Jul 2018
RE: A little help understanding math....
Prove: if 0 ≤ θ < pi/2, Re(atan(exp(i*θ))) = pi/4
atan(z) = atanh(z*i)/i = ln((1+i*z)/(1-i*z)) / (2*i)
Let z = exp(i*θ) = cos(θ) + i*sin(θ)
(1+i*z)/(1-i*z)
= ((1-sin(θ)) + i*cos(θ)) / ((1+sin(θ)) - i*cos(θ))
= i * cos(θ) / (1 + sin(θ)) // flip sin/cos
= i * sin(pi/2-θ) / (1 + cos(pi/2-θ)) // tan(α/2) = sin(α) / (1 + cos(α))
= i * tan(pi/4-θ/2)
For 0 ≤ θ < pi/2, imaginery part is positive.
Re(atan(exp(i*θ))) = arg(i*tan(pi/4-θ/2)) / 2 = pi/4
With signed zero, we have atan(±0 + i) = ±pi/4 + Inf*i
Code:
Complex Catan(Complex z) { double x = fabs(Real(z)); double y = fabs(Imag(z)); if (x >= 0x1p27 || y >= 0x1p27) { // atan(1/z) ~= 1/z Complex t = -1/(x+y*I); // atan(x+y*I) ~= pi/2 + t x = Real(t) + M_PI_2; // |re(t^3/3)/re(t)| <= 0x1p-54 y = Imag(t); // |im(t^3/3)/im(t)| <= 0x1p-54 } else { double x2 = x*x, ym = 1-y; double u = (1+y)*ym - x2; // 0 implied z on unit circle x = atan2(2*x + (u==0), u) * 0.5; y = log1p(4*y / (ym*ym + x2)) * 0.25; } return CMPLX(copysign(x, Real(z)), copysign(y, Imag(z))); }
06-14-2023, 04:55 PM (This post was last modified: 06-15-2023 01:37 PM by Albert Chan.)
Post: #10
Albert Chan Senior Member Posts: 2,234 Joined: Jul 2018
RE: A little help understanding math....
(08-29-2021 12:20 AM)Albert Chan Wrote: With signed zero, we have atan(±0 + i) = ±pi/4 + Inf*i
Technically, there is no complex number z, such that tan(z) = i
Whatever default we assigned for atan(i) is purely for convenience.
Above quoted default assumed (±0 + i) = limit((±ε + i), ε=0)
We could also start with ±0 = 0, exactly.
atan(z) = ∫(1/(1+t^2), t = 0 .. z) = z - z^3/3 + z^5/5 - z^7/7 + ...
atan(i) = i * (1 + 1/3 + 1/5 + 1/7 + ...) = i * Inf
--> atanh(1) = atan(i)/i = Inf (*)
IEEE754_2008.pdf (page 45), atan2(±0, +0) = ±0, patched code is also simpler.
Code:
< double x2 = x*x, ym = 1-y; < double u = (1+y)*ym - x2; // 0 implied z on unit circle < x = atan2(2*x + (u==0), u) * 0.5; < y = log1p(4*y / (ym*ym + x2)) * 0.25; > double x2 = x*x, ym = 1-y; > x = atan2(2*x, (1+y)*ym - x2) * 0.5; > y = log1p(4*y / (ym*ym + x2)) * 0.25;
(*) atanh(1 ± 0i) = (Inf ± 0i) matched ieee behavior
https://en.cppreference.com/w/cpp/numeric/complex/atanh
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# LAF Zone 5: Strategy Refining
The information below describes the key characteristics and teaching implications for Zone 5 of the Learning and Assessment Framework for Multiplicative Thinking (LAF), including:
## How to use this resource
There are eight zones in the LAF. If your students are across several zones, you should access information for each of the zones where they are located.
Where there is a reference to:
## Key characteristics
Systematically solves simple proportion and array problems, suggesting multiplicative thinking. For example:
• Butterfly House part e
• Packing Pots part a
• How Far part a .
May use additive thinking to solve simple proportion problems involving fractions. For example:
• School Fair part a
• Speedy Snail part b.
Able to solve simple, two-step problems using a recognised rule or relationship (for example, Fencing the Freeway Part A). However, finds this difficult for larger numbers. For example:
• Tables & Chairs parts k and l
• Tiles, Tiles, Tiles part c
• Stained Glass Windows part c.
Able to order numbers involving tens, ones, tenths and hundredths in supportive context. For example, Swimming Sports part A.
Able to determine all options in Cartesian product situations involving relatively small numbers, but tends to do this additively. For example:
• Canteen Capers part A
• Butterfly House parts 1 and I.
Beginning to work with decimal numbers and percent but unable to apply efficiently to solve problems. For example, Swimming Sports parts a and b, Computer Game part B.
Some evidence that multiplicative thinking being used to support partitioning. For example, Missing Numbers part B.
Beginning to approach a broader range of multiplicative situations more systematically.
## Teaching implications – consolidation and establishment
Consolidate or establish the ideas and strategies introduced or developed in the previous zone
## Teaching implications – introduction and development
Place-value
Introduce ideas and strategies for 5 digits and beyond if not already developed and decimal fractions to hundredths including renaming.
Multiplying and dividing by multiples of ten
Provide flexible, meaningful and efficient strategies for multiplying and dividing by multiples of ten. For example:
2.13 by 10, think: 21 ones and 3 tenths, 21.3.
The area idea and an expanded range of Cartesian product problems
Support multi-digit multiplication and formal recording (see Booker et al, 2004) through the area idea.
Introduce more efficient strategies for representing and solving an expanded range of Cartesian product problems involving three or more variables and tree diagram representations.
Formal terminology
Introduce terminology associated with multiplication and division such as:
• factor
• product
• divisor
• multiplier
• raised to the power of ….
Play the Factor Cross (pdf - 54.57kb) game.
Use calculators to explore what happens with repeated factors. For example:
• 4 × 4 × 4 × 4 …
• factors less than 1
• negative factors.
Renaming an expanded range of unrelated fractions
Introduce informal, partition-based strategies for renaming an expanded range of unrelated fractions as a precursor to developing an efficient, more formal strategy for generating equivalent fractions (see below).
For example:
explore using paper folding, diagrams and line models how sixths and eighths could be renamed as forty-eighths but they can also be renamed as twenty-fourths because both are factors of 24.
Generalisation for renaming fractions
If the number of equal parts (represented by the denominator) increases or decreases by a certain factor then the number of parts required (indicated by the numerator) increases or decreases by the same factor.
For example:
the total number of parts increased by a factor of 3, so parts required increased by a factor of 3.
Addition and subtraction of unlike fractions
Introduce written solution strategies for the addition and subtraction of unlike fractions.
For example:
think of a diagram showing sixths by eighths … forty-eighths… Is this the simplest? No, twenty-fourths will do, rename fractions by inspection
9 twenty-fourths can’t take 20 twenty-fourths, trade 1 one for 24 twenty-fourths to get 6 and 33 twenty-fourths. Subtraction is then relatively straightforward
Link between multiplication and division and fractions
Explore the link between multiplication and division and fractions including decimals. For example:
3 pizzas shared among 4, 3 divided by 4 is 0.75 etc.
These links will help with understanding the fraction as operator idea. For example:
• ¾ of 120
• 75% of \$48
• 250% of 458,239.
Use the Multiple Patterns (pdf - 18.5kb) task.
Establish benchmark equivalences. For example:
1 third = 331/3%.
Metacognitive strategies
Introduce strategies to support:
• problem comprehension
• strategy monitoring and checking
• interpretation of outcomes relevant to context.
For more information, see Siemon and Booker (1990) paper on Teaching and Learning For, About and Through Problem Solving (pdf - 462.19kb)
## Learning plans
The following learning plans have been developed for this zone, to support targeted teaching after students have been assessed and located on the LAF.
Zone 5 – Learning Plans Introducing (pdf - 77.05kb)
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# unmatched verb tense
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Hi,
I got the below question wrong in one of the official GMAT practice tests but can't work out why.
Question 2 -- correct answer is C.
Researchers agreed that the study treatments for heart attack patients was extremely important but more research was needed to determine that baloon angioplasty preceded with ultrasound was or was not any better for heart attack patients than the balloon procedure by itself.
a. Same as above
b. more research was needed for determining whether or not balloon angioplasty preceded by ultrasound is any better for heart attack patients than is
c. that more research was needed to determine whether balloon angioplasty preceded by ultrasound is any better for heart attack patiens than
d. that more research was needed to determine that balloon angioplasty preceded with ultrasound was any better for heart attack pathents than
e. that more research was needed for determining that balloon angioplasty preceded by ultrasound is or is not any better for heart attack patients than is
I can see the answer I picked (D) is no the best, but in C, last verb 'is' does not match the verb tense in the first part of the sentence, which is past tense.
I always seem to narrow the answer down to two, and then pick the wrong one in the end.
Anyone can help me with this one?
Thx heaps.
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verbal to b posted on GMAT Club.doc [241 KiB]
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07 Aug 2009, 02:53
I have 2 reasons . First , "whether" in C is better than "that" in D .It is a objective clause ,means determines sth.
second "preceded by" is more agreement with the "the balloon procedure by itself" than "preceded with".
so as to the tense ,I think the whole setense means " more research was needed to determine in the past ",but the ths. needed to determine must not be the past.
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08 Aug 2009, 07:57
Thx buddy. I didn't even notice the 'by' and 'with' even thogh I went through the entire question so many times..good spot.
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By all means, C is the best choice, except for the pin - prick of the un-matching verb tenses.
But now, let’s look at these sentences:
Galileo concluded that earth was oval and not flat
Galileo concluded that earth is oval and not flat
Which one of the two is correct? Obviously, is oval and not was oval. Because, this is a scientific and eternal truth, that was in the past, is in the present and will be in the future. So the use of present tense is acceptable, though it looks as if it is an apparent shift of tense.
On the same logic, whether or not Balloon Angioplasty is good enough alone or only along with other measures, is a scientific fact, and can be expressed in present tense. So C.
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Expert's post
daagh wrote:
By all means, C is the best choice, except for the pin - prick of the un-matching verb tenses.
But now, let’s look at these sentences:
Galileo concluded that earth was oval and not flat
Galileo concluded that earth is oval and not flat
Which one of the two is correct? Obviously, is oval and not was oval. Because, this is a scientific and eternal truth, that was in the past, is in the present and will be in the future. So the use of present tense is acceptable, though it looks as if it is an apparent shift of tense.
On the same logic, whether or not Balloon Angioplasty is good enough alone or only along with other measures, is a scientific fact, and can be expressed in present tense. So C.
I wholeheartedly agree with daagh's comment above, and will even take it one step further. When dealing with a universal truth (as indicated, true in the past, the present and the future unless something changes), not only is the present acceptable, it is required. The logic of the sentence mandates that the verb tense be in the present, because the use of the past would indicate that something may have changed in the interim. It is important to look at the logical meaning of the sentence as well as the rules of grammar.
Similar example:
XYZ concluded that the Earth is oval and not flat --) still the case, present required
XYZ concluded that the Earth's surface was comprised of one super continent, Pangea --) no longer the case, past required
This is something that can mess students up, so it's good to be aware of it. In sentence correction, the verb tenses should match up. However, if they don't, look for logical reasons why the timeline doesn't match expectations. Logic is a bigger part of sentence correction than most students realize.
Hope this helps!
-Ron
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11 Mar 2013, 09:07
perfectly said Ron, You couldn't have made it more clear kudo to you
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05 Jan 2016, 02:08
daagh wrote:
perfectly said Ron, You couldn't have made it more clear kudo to you
What if we have 'is' replaced by 'would be' in another option?
I believe that it is just an expectation of the people and the outcome is uncertain and hence 'would' is more preferable compared to 'is' and that 'would' choice wins over 'is' choice.
Please correct me if I am wrong.
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Re: unmatched verb tense [#permalink] 05 Jan 2016, 02:08
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# class events: week 14
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• #### space voyagers
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Class events: week 14. Today’s goals: Interstellar travel! Theoretical considerations - Different blueprints, from mundane to insane!. Three obstacles to interstellar travel. Obstacle #1: Extraordinary distances Recall from week #1… - PowerPoint PPT Presentation
TRANSCRIPT
• *Class events: week 14Todays goals: Interstellar travel!Theoretical considerations - Different blueprints, from mundane to insane!
• *Three obstacles to interstellar travelObstacle #1: Extraordinary distancesRecall from week #1If we scaled the Suns radius to 17 cm (a grapefruit), the (ant-sized) Earth would be at 15 m, and the nearest star would be in New York City (4000 km distant).
Space is vastly huge, and is filled with mysteries.
How have we fared so far?
• *Three obstacles to interstellar travelObstacle #1: Extraordinary distances
We have launched four planetary probes which are destined to be interstellar probesPioneer 10, 11; Voyager 1, 2.
Voyager 1Our most distant probe;110 a.u. from the Sun (0.0017 LY);It is travelling at 17 km/s (0.000056 c); 1/2500 the distance to Centauri system; Would reach Cen in 77,000 yearsif it were heading that way!
• *Three obstacles to interstellar travelObstacle #1: Extraordinary distances
Interstellar arks could be a work-around for the huge distances.Deal with the long travel time by creating enormous, multigenerational ships;
What would happen to the enclosed culture over time?
Theoretical sleeper ships could involve suspended animation?
• *Interstellar travelObstacle #2: Energy concernsThe speed of light (c) is a good comparisonto accelerate a spacecraft to velocities near c would require incredible energies:
K.E. = mv2 = (18,000 kg/passenger) 5000 passengers (0.1c)2 = 4.5 1022 J= 100 world usage/year
• *Interstellar travelObstacle #3: Special Relativity and cEinsteins relativity tells us that to accelerate a particle (with mass=m) to the speed of light would take not just K.E. = mc2 , it would take an INFINITE amount of energy!
You cannot generate an infinite amount of energy, therefore nothing can reach the speed of light. This is why the speed of light is a cosmic speed limit.
Even if you were traveling at nearly the speed of light, your space ship traversing the enormity of interstellar space would take 4.3 years to reach the nearest star.
You simply cannot cut down those huge travel times from star to star!
I ask you Would aliens really go to all the effort of visiting us, where such a journey took dozens of years, just to probe our farmers?
• *Interstellar travelObstacle #3: Special Relativity and cAstonishingly, because of a relativistic effect called time dilation, space voyagers traveling at high speeds age more slowly than non-travelers:
Consider a journey of 4.3 LY; how long would it take to occur?
V Ttraveler/TpotatoObservedExperienced0.1c0.99543 years4.28 years0.5c0.8668.6 years7.45 years0.9c0.4364.8 years2.08 years0.99c0.0454.3 years2.35 monthsNear-light travelers would reach the stars after aging only a few monthsbut their families back home would age normally.
Would you visit a star 10 LY away, if (after a trip lasting 11 months at 0.999c) your family would be 20 years older upon your return?
• *Interstellar propulsionHow rockets workRockets DO NOT push against the ground;Rockets DO launch matter away at high velocity;By their nature, rockets lose mass during operation.
Optimizing thrustThe faster the ejected matter, the more thrust;The more matter ejected, the more thrust.
• *Chemical rocketryChemical energy triviaLiquid oxygen and ethanol (V-2);Liquid oxygen and kerosene or liquid H2 (Saturn V);Liquid O2 and liquid H2 (Shuttle External tank);Ammonium perchlorate (oxidizer) and aluminum (Shuttle Solid Rocket Boosters).
Interstellar applicationsChemical rockets are barely useful for interplanetary travel;Chemical rockets are useless for interstellar work.
• *Nuclear fission thermal rocketsNuclear fission heats a propellant gas (hydrogen);
Project Rover intended power a Saturn V, for Martian mission (1955-1972), with Kiwi, Phoebus, Pewee engines;
Russian RD-0410 thermal rocket in manned 1994 Mars proposal.
Development would violate the Nuclear Test Ban Treaty, and would also violate the Comprehensive Nuclear Test Ban (currently signed but not ratified by the USA).
• *Nuclear fusion rocketsOrion: up to 2400 as massive as a Saturn V;
Pusher plate absorbs blast shock with hydraulics and airbags;
With millions of bombs, Orion could reach 0.1c!
The British Project Daedalus explored theory of continual fusion via pellets;
These are all currently beyond our technology.
• *Ion enginesAccelerates charged particles to high velocities;
The high velocity compensates for low particle mass, thus generating useful thrust;
Suitable for long missions, but not for landings;
Many current usesHayabusa (Japan) visited asteroid Itokawa in 2005.
• *Solar sailsSolar sails surf sunlight (assisted by lasers?);
For high speed, it would be launched near the sun;
Would it slow by the radiation from its target star?
Lightsail 1 (Planetary Society) is being tested in 2011. Ikaros (Japan) is the first solar sail satellite that has flown. A square 20 m in diagonal size, it flew to Venus in 2010.
• *Realms of fantasyRamjetsAn enormous leading scoop gathers hydrogen for fusion; The scoop would be comparable to California in size.
Matter-antimatterNuclear fission: 0.07% mass conversion;Nuclear fusion: 0.7% mass conversion;Matter-antimatter: 100% mass conversion;Antimatter takes energy to createit is not free energy!Antimatter is not fiction:Was predicted in 1928 by Paul Dirac;Positrons were created in 1930;Antiprotons were discovered in 1955;Antihydrogen was made at CERN and Fermilab in 1995;Anti-helium (He3) was created in 2003.
• *WormholesWormholes in hyperspace provide a way to travel at sub-light speeds, but by taking a short cut.
These Schwarzschild wormholes (or Einstein-Rosen Bridges) are not prohibited by science, as far as we can tell.
However, wormholes are inherently unstablenot even light can pass through them before they fall apart.
• *Stable wormholesIn the 1980s, astrobiologist Carl Sagan was writing Contact, and needed a way for aliens to communicate with humans. Work with black hole theoretician Kip Thorne led to the concept of a of stable wormhole.
Sagans book/movie portrayed our galaxy as being filled with a network of wormholes, created by an ancient Type III galactic civilization.
Theoretically possible, traversable wormholes would have to be stabilized by exotic matter with attributes (such as negative mass) that will probably be laughed at by future physicists.probably.
• *Really weird stuffTachyonsEntirely fictional particles that travel faster than light.
Tachyons would have remarkable propertiesfor example, they would experience negative time.
In 2011, CERN results indicate that neutrinos might be travelling at speeds greater than the speed of light. This was a bogus detection due to a faulty fiber optic cable.
This is the realm (and love) of science fiction.
Nuff said.
• *Really weird stuffAlcubierre metric In the 1990s, by studying Einsteins equations, physicist Alcubierre discovered that space could be warped in a strange way.
He developed a warped bubble of spacetime, in which spacetime is contracted in front of an object, and shrunken behind it.
The warp bubble and its enclosed object could move at arbitrarily high speeds without violating physics.
This is strictly hypothetical. Furthermore, exotic matter is needed to stabilize this structure, just as in wormholes.
******************
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With the holiday season upon us, many of our cooks will possibly run across a variety of recipes from other countries or cultures. The measurements or temperatures may not be in terms you are used to. Here is a handy guide to help you translate the equivalents in the measurements of your own country.
Conversion Table for Cooking Around the World
Dry:
Flour (all-purpose) - 1 Cup = 100g = 4 ounces
Rice (raw) - 1 Cup = 225g = 8 ounces
Sugar (Granulated) - 1 Cup = 225g = 8 ounces
Liquids:
1/4 teaspoon = 1 ml
1/2 teaspoon = 2 ml
1 teaspoon = 5 ml
1 tablespoon = 20 ml
1/4 cup = 60 ml
1/3 cup = 80 ml
1/2 cup = 125 ml
2/3 cup = 170 ml
3/4 cup = 190 ml
1 cup = 250 ml
1 quart = 1 liter
1/2 ounce =15 grams
1 ounce = 30 grams
2 ounces = 60 grams
3 ounces = 90 grams = 1/4 pound
4 ounces = 125 grams
5 ounces = 140 grams
6 ounces = 170 grams
7 ounces = 200 grams
8 ounces = 225 grams = 1/2 pound
9 ounces = 250 grams
10 ounces = 280 grams
11 ounces = 310 grams = 3/4 pounds
12 ounces = 340 grams
13 ounces = 365 grams
14 ounces = 390 grams
16 ounce = 453 grams = 1 pound
1/2 fluid ounce = 15 ml = 1 tablespoon
1 fluid ounce = 30 ml = 2 tablespoon
2 fluid ounce = 60 ml = 1/4 cup
4 fluid ounces = 125 ml = 1/2 cup
1/4 pint = 150 ml = 2/3 cup
6 fluid ounces = 175 ml = 3/4 cup
8 fluid ounces = 250 ml = 1 cup (1/2 U.S. pint)
1/2 pint = 300 ml = 1 1/4 cup
12 fluid ounces = 375 ml = 1 1/2 cup
16 fluid ounces = 500 ml = 2 cups (1 U.S. pint)
1 pint = 600 ml = 2 1/2 cup
2 pints = 1.25 litres = 1 quart
Fahrenheit / Celsius Gas Oven
32º / 0
212º / 100º
250º / 125º Gas mark 1/2
275º / 140º
300º / 150º Gas mark 2
325º / 160º Gas mark 3
350º / 180º Gas mark 4
375º / 190º
390º / 200º Gas mark 6
425º / 220º
440º / 225º Gas mark 7
450º / 230º
475º / 240º
500º / 250º Gas mark 9
If you need to be very precise, °C = (°F - 32) X 0.56
1 teaspoon = 5 ml
16 tablespoons = 1 cup
12 tablespoons = 3/4 cup
10 tablespoons + 2 teaspoons = 2/3 cup
8 tablespoons = 1/2 cup
6 tablespoons = 3/8 cup
5 tablespoons + 1 teaspoon = 1/3 cup
4 tablespoons = 1/4 cup
2 tablespoons = 1/8 cup
2 tablespoons + 2 teaspoons = 1/6 cup
1 tablespoon = 1/16 cup = 15 ml (Australia =20 ml)
1 tablespoon = 3 teaspoons
2 cups = 1 pint
2 pints = 1 quart
48 teaspoons = 1 cup
Log in or register to write something here or to contact authors.
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## Elementary Algebra
$x=\dfrac{7}{2}y+\dfrac{5}{2}$
Using the properties of equality, in terms of $x ,$ the given expression, $2x-7y=5 ,$ is equivalent to \begin{array}{l}\require{cancel} 2x=7y+5 \\\\ x=\dfrac{7}{2}y+\dfrac{5}{2} .\end{array}
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# Elementary Analysis Solution Manual Pdf
Elementary Analysis Ross Solution Manual. analysis In .uch areu .. oomplex varUbles, differential equ~tion .. numerical analysla, and wtl&!icII more meaningful. The book can ~bo aerve as a foundation fur an In-depth study of real analysla, Jul 15, 2014В В· Elementary Analysis The Theory of Calculus (2nd edition).pdf (PDFy mirror) Item Preview remove-circle Share or Embed This Item. EMBED. EMBED (for wordpress.com hosted blogs and archive.org item tags) Want more? Advanced embedding details, examples, and help! favorite. share.
### KENNETH ROSS ELEMENTARY ANALYSIS SOLUTION
14.58MB THOMSON ELEMENTARY REAL ANALYSIS. Elementary Analysis textbook solutions from Chegg, view all supported editions., You can download it for free without the need to register or enter. You simply press the download and you can get the ebook in the various formats you want.
Solutions for вЂElementary Analysis_ the Theory of Calculus’ by Kenneth Ross bartle-sherbert-real-analysis-solution-manual.pdf Mathematical Statistics And Data Analysis 3rd Edition - Chapter8 Solutions.pdf This PDF п¬Ѓle includes material from the text Elementary Real Analysis originally published by Prentice Hall (Pearson) in 2001. The authors retain the copyright and all commercial uses. [2008] Date PDF п¬Ѓle compiled: June 16, 2008 ClassicalRealAnalysis.com [TBB-Dripped] Elementary Real Analysis - Dripped Version Thomson*Bruckner*Bruckner
Jan 29, 2016 · Free Download eBooks and Solution Manual in www.ManualSolution.info think it is fair to say that the transition from elementary courses such as calculus, linear algebra, and differential equations to a rigorous real analysis course is a bigger step to-day than it was just a few years ago. To make this step today’s students need more help than their predecessors did, and must be coached and encouraged more.
Nov 19, 2007В В· Solution Manual-Microelectronics-Digital & Analog Circuits & Systems by Millman Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual Problems and Solutions on Electromagnetism (Major American Universities Ph.D. Qu Engineering circuit analysis Solution Manual - William Hart Hayt Solution Manual-Signals and Systems 2003 by M.J. Roberts Elementary Classical Analysis Solutions Manual Pdf Read/Download Elementary Classical Analysis Solution Manual Marsden Elementary Download Elementary Classical Analysis Marsden Solution Manual Chap 5 8 PDF. classical dynamics of particles and systems 5th edition solutions introduction to analysis solutions manual elementary analysis
Solutions to Elementary Analysis The Theory of Calculus. kenneth ross elementary analysis solution manual PDF may not make exciting reading, but kenneth ross elementary analysis solution manual is packed with valuable instructions, information and warnings. We also have many ebooks and user guide is also related with kenneth ross elementary, This PDF п¬Ѓle includes material from the text Elementary Real Analysis originally published by Prentice Hall (Pearson) in 2001. The authors retain the copyright and all commercial uses. [2008] Date PDF п¬Ѓle compiled: June 16, 2008 ClassicalRealAnalysis.com [TBB-Dripped] Elementary Real Analysis - Dripped Version Thomson*Bruckner*Bruckner.
### Elementary Analysis Solutions Compact Space Sequence
Real Analysis Solutions1 Columbia University. But now, with this Solution Manual, you will be able to * Have your homework problems solved readily. * Reduces the hassle and stress of your student life. * Improve your studying and also get a better grade! * Get prepared for examination questions. * Can save you time and help you understand the material., Nov 19, 2007В В· Solution Manual-Microelectronics-Digital & Analog Circuits & Systems by Millman Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual Problems and Solutions on Electromagnetism (Major American Universities Ph.D. Qu Engineering circuit analysis Solution Manual - William Hart Hayt Solution Manual-Signals and Systems 2003 by M.J. Roberts.
### Elementary Analysis 2nd Edition Textbook Solutions Chegg.com
Elementary Analysis Solutions Free PDF eBook. Elementary Analysis Ross Solution Manual PDF/ELE-2188660 ELEMENTARY ANALYSIS THE THEORY OF CALCULUS SOLUTION MANUAL Available PDF/ROS-2134837 ROSS ELEMENTARY. Math 131A, Lecture 1: Analysis Textbook. Ross, Elementary Analysis: The Theory of Calculus. Springer However, you must write up all solutions yourself. https://en.m.wikipedia.org/wiki/Machine Elementary Classical Analysis Solutions Manual Pdf Read/Download Elementary Classical Analysis Solution Manual Marsden Elementary Download Elementary Classical Analysis Marsden Solution Manual Chap 5 8 PDF. classical dynamics of particles and systems 5th edition solutions introduction to analysis solutions manual elementary analysis.
Nov 19, 2007В В· Solution Manual-Microelectronics-Digital & Analog Circuits & Systems by Millman Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual Problems and Solutions on Electromagnetism (Major American Universities Ph.D. Qu Engineering circuit analysis Solution Manual - William Hart Hayt Solution Manual-Signals and Systems 2003 by M.J. Roberts May 16, 2014В В· Elementary Analysis, Volume 2 introduces several of the ideas of modern mathematics in a casual manner and provides the practical experience in algebraic and analytic operations that lays a sound foundation of basic skills.
But now, with this Solution Manual, you will be able to * Have your homework problems solved readily. * Reduces the hassle and stress of your student life. * Improve your studying and also get a better grade! * Get prepared for examination questions. * Can save you time and help you understand the material. 65553233-Elementary-Analysis-Solutions - Elementary Analysis Kenneth A Ross Selected Solutions Angelo Christopher Limnios EXERCISE 1.2 Claim P(n = 3 rudin solution principles-of-mathematical-analysis. 6 pages. Math 131A Sample Final Solutions University of California, Los Angeles 281 Final Notes.pdf. 55 pages. The operation of complex
Solutions for вЂElementary Analysis_ the Theory of Calculus’ by Kenneth Ross bartle-sherbert-real-analysis-solution-manual.pdf Mathematical Statistics And Data Analysis 3rd Edition - Chapter8 Solutions.pdf Jan 29, 2016В В· Free Download eBooks and Solution Manual in www.ManualSolution.info
## Elementary Real Analysis American Inst. of Mathematics
### Math 104 Introduction to Analysis
Elementary Analysis The Theory of Calculus (2nd edition. This version of Elementary Real Analysis, Second Edition, is a hypertexted pdf file, suitable for on-screen viewing. For a trade paperback copy of the text, with the same numbering of Theorems and Exercises (but with different page numbering), please visit our web site. Direct all correspondence to thomson@sfu.ca., Elementary Numerical Analysis Atkinson Solution Manual Solution for Elementary numerical analysis 3rd edition by Atkinson. 4.6, page 177 and 178.
Apr 17, 2018В В· Student Solutions Manual for Numerical Analysis 2nd Edition. by Timothy Sauer (Author) 3.3 out of 5 stars 5 ratings. ISBN-13: 978-0321783929. ISBN-10: 0321783921. Why is ISBN important? ISBN. This bar-code number lets you verify that you're getting exactly the right version or edition of a … Solutions for вЂElementary Analysis_ the Theory of Calculus’ by Kenneth Ross bartle-sherbert-real-analysis-solution-manual.pdf Mathematical Statistics And Data Analysis 3rd Edition - Chapter8 Solutions.pdf
Elementary Statistics A Step By Step Approach 10th Edition Pdf Download Elementary Statistics A Step By Step Approach 10th Edition Pdf Free Allan G. Bluman 2012. Elementary Statistics A Step By Step Approach 8th Edition Solution To Elementary Statistics A Step By Step Approach 12th Edition Student Solution Manual For Elementary Statistics A Step By Step Approach 12th Edition Student Solution This version of Elementary Real Analysis, Second Edition, is a hypertexted pdf file, suitable for on-screen viewing. For a trade paperback copy of the text, with the same numbering of Theorems and Exercises (but with different page numbering), please visit our web site. Direct all correspondence to thomson@sfu.ca.
Lectures (in pdf) The following notes are to accompany the third edition of the book Elementary Numerical Analysis, by Kendall Atkinson and Weimin Han, published by John Wiley & Sons, Inc. Modeling Population model slides. Matlab notes; Chapter 1 - Taylor Polynomial Approximations. Matlab programs for Chapter 1. Section 1-1 From the reviews of the first edition: "This book is intended for the student who has a good, but naГЇve, understanding of elementary calculus and now wishes to gain a thorough understanding of a few basic concepts in analysis, such as continuity, convergence of sequences and series of numbers, and convergence of sequences and series of functions.
The textbook for the course will be K.A. Ross, Elementary Analysis: The Theory of Calculus. We will follow it more or less closely. Chapters 1-4 will be the core topics of the course, and so we will study them most thoroughly. The last part of the course will cover parts of chapters 5 and 6. Elementary Classical Analysis Solutions Manual Pdf Read/Download Elementary Classical Analysis Solution Manual Marsden Elementary Download Elementary Classical Analysis Marsden Solution Manual Chap 5 8 PDF. classical dynamics of particles and systems 5th edition solutions introduction to analysis solutions manual elementary analysis
This PDF п¬Ѓle includes material from the text Elementary Real Analysis originally published by Prentice Hall (Pearson) in 2001. The authors retain the copyright and all commercial uses. [2008] Date PDF п¬Ѓle compiled: June 16, 2008 ClassicalRealAnalysis.com [TBB-Dripped] Elementary Real Analysis - Dripped Version Thomson*Bruckner*Bruckner Math 312: Real Analysis I am Stephen G. Simpson, a Professor of Mathematics at Penn State University.. Math 312 is an introductory course on real analysis. I taught it in in Spring 2009. As stated in the Penn State catalog of courses, the only prerequisite for Math 312 is Math 141.
Elementary Analysis Ross Solution Manual PDF/ELE-2188660 ELEMENTARY ANALYSIS THE THEORY OF CALCULUS SOLUTION MANUAL Available PDF/ROS-2134837 ROSS ELEMENTARY. Math 131A, Lecture 1: Analysis Textbook. Ross, Elementary Analysis: The Theory of Calculus. Springer However, you must write up all solutions yourself. Bluman A.G. Elementary Statistics- A Step By Step Approach.pdf. Bluman A.G. Elementary Statistics- A Step By Step Approach.pdf. Sign In. Details
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### SOLUTION MANUALS FOR TEXTBOOKS
Elementary Classical Analysis Solutions Manual Pdf. Jan 29, 2016 · Free Download eBooks and Solution Manual in www.ManualSolution.info, 2 Real Analysis Use the alternative definition for continuity for sequences. Then we have that: take any se-quence fx ig i2N ˆRk such that fx ig1i =1!x.Then we need to show that h(x i) !h(x) as i !1..
Elementary Analysis Ross Solution Manual. But now, with this Solution Manual, you will be able to * Have your homework problems solved readily. * Reduces the hassle and stress of your student life. * Improve your studying and also get a better grade! * Get prepared for examination questions. * Can save you time and help you understand the material., Jul 15, 2014В В· Elementary Analysis The Theory of Calculus (2nd edition).pdf (PDFy mirror) Item Preview remove-circle Share or Embed This Item. EMBED. EMBED (for wordpress.com hosted blogs and archive.org item tags) Want more? Advanced embedding details, examples, and help! favorite. share.
### Elementary Real Analysis ClassicalRealAnalysis.info
Math 104 Introduction to Analysis. analysis In .uch areu .. oomplex varUbles, differential equ~tion .. numerical analysla, and wtl&!icII more meaningful. The book can ~bo aerve as a foundation fur an In-depth study of real analysla https://en.m.wikipedia.org/wiki/Titration But now, with this Solution Manual, you will be able to * Have your homework problems solved readily. * Reduces the hassle and stress of your student life. * Improve your studying and also get a better grade! * Get prepared for examination questions. * Can save you time and help you understand the material..
Solution Manual to Elementary Analysis, 2nd Ed., by Kenneth A. Ross David Buch December 18, 2018. Contents 1 Basic Properties of the Derivative 1 2 The Mean Value Theorem 7 iii. 1 Basic Properties of the Derivative Note: In this section, we make routine use of the fact that lim x!a is evaluated on sets The textbook for the course will be K.A. Ross, Elementary Analysis: The Theory of Calculus. We will follow it more or less closely. Chapters 1-4 will be the core topics of the course, and so we will study them most thoroughly. The last part of the course will cover parts of chapters 5 and 6.
But now, with this Solution Manual, you will be able to * Have your homework problems solved readily. * Reduces the hassle and stress of your student life. * Improve your studying and also get a better grade! * Get prepared for examination questions. * Can save you time and help you understand the material. Solutions manual developed by Roger Cooke of the University of Vermont, to accompany Principles of Mathematical Analysis, by Walter Rudin.
You can download it for free without the need to register or enter. You simply press the download and you can get the ebook in the various formats you want How is Chegg Study better than a printed Elementary Analysis 2nd Edition student solution manual from the bookstore? Our interactive player makes it easy to find solutions to Elementary Analysis 2nd Edition problems you're working on - just go to the chapter for your book.
Elementary Numerical Analysis 3rd Edition by Kendall Atkinson, Weimin Han Textbook PDF Download free download Keywords: Elementary Numerical Analysis 3rd Edition by Kendall Atkinson, Weimin Han Textbook PDF Download free download Created Date: 1/31/2015 10:40:31 AM May 16, 2014В В· Elementary Analysis, Volume 2 introduces several of the ideas of modern mathematics in a casual manner and provides the practical experience in algebraic and analytic operations that lays a sound foundation of basic skills.
Nov 19, 2007В В· Solution Manual-Microelectronics-Digital & Analog Circuits & Systems by Millman Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual Problems and Solutions on Electromagnetism (Major American Universities Ph.D. Qu Engineering circuit analysis Solution Manual - William Hart Hayt Solution Manual-Signals and Systems 2003 by M.J. Roberts Sep 24, 2008В В· Solution Manual - Electrical Machines, Drives and Power Systems 6th edition ISBN 0131969188 Solution Manual - Microelectronics I & II 1st edition by by Dr. Wen-Ching Chang Solution Manual (complete) Communication Networks Fundamental concepts & key Architectures Alberto Leon-Garcia Solution Manual : Nanoengineering of Structural, Functional and
Sep 24, 2008 · Solution Manual - Electrical Machines, Drives and Power Systems 6th edition ISBN 0131969188 Solution Manual - Microelectronics I & II 1st edition by by Dr. Wen-Ching Chang Solution Manual (complete) Communication Networks Fundamental concepts & key Architectures Alberto Leon-Garcia Solution Manual : Nanoengineering of Structural, Functional and Apr 17, 2018 · Student Solutions Manual for Numerical Analysis 2nd Edition. by Timothy Sauer (Author) 3.3 out of 5 stars 5 ratings. ISBN-13: 978-0321783929. ISBN-10: 0321783921. Why is ISBN important? ISBN. This bar-code number lets you verify that you're getting exactly the right version or edition of a …
Jul 15, 2014В В· Elementary Analysis The Theory of Calculus (2nd edition).pdf (PDFy mirror) Item Preview remove-circle Share or Embed This Item. EMBED. EMBED (for wordpress.com hosted blogs and archive.org item tags) Want more? Advanced embedding details, examples, and help! favorite. share Nov 19, 2007В В· Solution Manual-Microelectronics-Digital & Analog Circuits & Systems by Millman Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual Problems and Solutions on Electromagnetism (Major American Universities Ph.D. Qu Engineering circuit analysis Solution Manual - William Hart Hayt Solution Manual-Signals and Systems 2003 by M.J. Roberts
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ANOVA`
ANOVA`
# ANOVA
ANOVA[data]
performs a oneway analysis of variance.
ANOVA[data,model,vars]
performs an analysis of variance for model as a function of the categorical variables vars.
# Details and Options
• To use ANOVA, you first need to load the Analysis of Variance Package using Needs["ANOVA`"].
• The data can have the form {{x1,y1,,f1},{x2,y2,,f2},}, where the number of coordinates x,y, is equal to the number of variables in the list vars.
• For data of the form {{x1,f1},{x2,f2},}, a oneway analysis of variance can be obtained without explicitly specifying the model and variable.
• The model argument can be a list containing main effects and interactions. Main effects are elements of vars, and interactions are products of main effects.
• A full factorial model including all interactions between variables x,y, can be specified as {x,y,,All}.
• The following options can be given:
• CellMeans True whether cell means should be included in the output WorkingPrecision MachinePrecision the precision used in internal computations PostTests {} significance tests to perform SignificanceLevel 0.05 significance level for performed tests
• Possible settings for PostTests include: Bonferroni, Duncan, Dunnett, StudentNewmanKeuls, and Tukey.
# Examples
open allclose all
## Basic Examples(2)
A oneway analysis of variance:
A twoway analysis of variance:
## Options(4)
### CellMeans(1)
Omit cell means from output:
### WorkingPrecision(1)
Computation at precision 20:
### PostTests(1)
Analysis of variance with Bonferroni test:
### SignificanceLevel(1)
Bonferroni test at significance level .01:
Wolfram Research (2007), ANOVA, Wolfram Language function, https://reference.wolfram.com/language/ANOVA/ref/ANOVA.html.
#### Text
Wolfram Research (2007), ANOVA, Wolfram Language function, https://reference.wolfram.com/language/ANOVA/ref/ANOVA.html.
#### CMS
Wolfram Language. 2007. "ANOVA." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ANOVA/ref/ANOVA.html.
#### APA
Wolfram Language. (2007). ANOVA. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ANOVA/ref/ANOVA.html
#### BibTeX
@misc{reference.wolfram_2024_anova, author="Wolfram Research", title="{ANOVA}", year="2007", howpublished="\url{https://reference.wolfram.com/language/ANOVA/ref/ANOVA.html}", note=[Accessed: 15-July-2024 ]}
#### BibLaTeX
@online{reference.wolfram_2024_anova, organization={Wolfram Research}, title={ANOVA}, year={2007}, url={https://reference.wolfram.com/language/ANOVA/ref/ANOVA.html}, note=[Accessed: 15-July-2024 ]}
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Algebra
posted by on .
Which is the extraneous solution of -x=sq root of 2x+15?
A. -5
B. -3
C. 3
D. 5
I think it is D...?
• Algebra - ,
you are correct
if x = 5
LS = -5
RS = √(10+15) = √25 = 5 ≠ LS
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# Golden Mean And The Pentagon
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Examining the relationship between the number 1, the golden ratio 1 : φ and a pentagon
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### Golden Mean And The Pentagon
1. 1. Golden Mean and the pentagon _ Stonemasonry technical skills at SkillsTech Australia
2. 2. <ul><li>Method </li></ul><ul><li>Construct a double square rectangle side: length = 1 : 2 [100mm : 200mm] </li></ul><ul><li>Using Pythagoras’ Theorem, prove that its diagonal has a length equivalent to [224mm] </li></ul><ul><li>Use the half diagonal to produce a Golden Rectangle side: length = 1: = 1: φ [100mm: 162mm] </li></ul><ul><li>Discuss the Golden Mean (a+b): a = a: b </li></ul><ul><li>Use the half diagonal to inscribe a pentagon in a circle with radius = 1. </li></ul>Golden Mean and the Pentagon
3. 3. Step 1: Draw vertical and horizontal centerlines at page centre right, intersecting at point A
4. 4. Step 2: Construct a circle centred at A with radius 1 [1 x 100mm]
5. 5. Step 3: Construct a double square with side 1 [1 x 100mm]
6. 6. Step 4: Join the diagonal of the double square from top right to bottom left, intersecting the horizontal line at point B
7. 7. Step 5: Centre compasses at point B and with radius BC = scribe semicircular arc BDE
8. 8. Step 6: Centre compasses at point C and with radius CE scribe an arc to intersect the circle at point F
9. 9. Step 7: Step off CF around the circle and join the points to form a pentagon
10. 10. Step 8: Construct the Golden Rectangle AC : AD = 1: so that (a+b) : a = a : b
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RECOGNISING REE GROUPS $\boldsymbol {{}^{2}G_{2}(q)}$ USING THE CODEGREE SET
Published online by Cambridge University Press: 28 September 2022
*
Abstract
The codegree of an irreducible character $\chi$ of a finite group G is $|G : \ker \chi |/\chi (1)$ . We show that the Ree group ${}^2G_2(q)$ , where $q = 3^{2f+1}$ , is determined up to isomorphism by its set of codegrees.
MSC classification
Type
Research Article
Creative Commons
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
© The Author(s), 2022. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.
1 Introduction
Let G be a finite group and $\mbox {Irr}(G)$ be the set of all irreducible complex characters of G. The concept of codegrees was introduced by Chillag and Herzog in [Reference Chillag and Herzog6], where the codegree of $\chi$ was defined as ${|G|}/{\chi (1)}$ for a character $\chi \in \mbox {Irr}(G)$ . The definition was modified to ${\mathrm{cod}}(\,\chi )={|G:\ker (\,\chi )|}/{\chi (1)}$ by Qian et al. in [Reference Qian, Wang and Wei18] so that there is no different meaning for ${\mathrm{cod}}(\,\chi )$ when $\chi$ is considered as a character in some quotient group of G. Because the relationship between codegrees and degrees is very close, we may expect to characterise the structure of groups by codegrees. During the past few years, the study of character codegrees has been very active and many results have been obtained, including the relationship between codegrees and element orders, codegrees of p-groups and groups with few codegrees (see, for example, [Reference Alizadeh, Behravesh, Ghaffarzadeh, Ghasemi and Hekmatarah2, Reference Du and Lewis8, Reference Isaacs11, Reference Liu and Yang14, Reference Liu and Yang15, Reference Qian17]).
Let cd $(G) = \{\,\chi (1) \ |\ \chi \in \mbox {Irr}(G)\}$ . Huppert made the following conjecture (which has been verified for sporadic simple groups, alternating groups and some simple groups of Lie type with low rank).
Huppert’s Conjecture: Let H be any finite nonabelian simple group and G a finite group such that cd $(G)$ = cd $(H)$ . Then $G \cong H \times A$ , where A is abelian.
We denote cod $(G) = \{ {\mathrm{cod}}(\,\chi ) \ |\ \chi \in \mbox {Irr}(G)\}$ . Qian made the following conjecture (Question 20.79 in the Kourovka Notebook [Reference Khukhro12]).
Codegree version of Huppert’s conjecture: Let H be any finite nonabelian simple group and G a finite group such that ${\mathrm{cod}}(G)= {\mathrm{cod}}(H)$ . Then $G\cong H$ .
This conjecture was shown to hold for ${\mathrm{PSL}}(2,q)$ in [Reference Bahri, Akhlaghi and Khosravi4]. In [Reference Ahanjideh1], the conjecture was proven for ${}^2B_2(2^{2f+1})$ , where $f \ge 1$ , ${\mathrm{PSL}}(3,4)$ , ${\operatorname {Alt}}_7$ and $J_1$ . The conjecture also holds in the cases where H is $M_{11}, M_{12}, M_{22}, M_{23}$ or ${\mathrm{PSL}}(3,3)$ by [Reference Gintz, Kortje, Laurence, Liu, Wang and Yang9]. For ${\mathrm{PSL}}(3,q)$ and ${\mathrm{PSU}}(3,q)$ , the conjecture was confirmed in [Reference Liu and Yang16]. In this paper, we continue the study of the conjecture and establish the following result.
Theorem 1.1. If H is isomorphic to a simple Ree group ${}^2G_2(3^{2f+1}), f\geq 1$ and G is a finite group such that ${\mathrm{cod}}(G)= {\mathrm{cod}}(H)$ , then $G\cong H$ .
2 Preliminary results
First, we give a list of simple groups with few codegrees. The cases for finite groups G with $| {\mathrm{cod}}(G)| \le 3$ are studied in [Reference Alizadeh, Behravesh, Ghaffarzadeh, Ghasemi and Hekmatarah2]. In particular, those groups are all solvable.
Lemma 2.1. Let G be a nonabelian finite simple group. If $|\mathrm{cod}(G)| \leq 11$ , then one of the following holds:
1. (a) $| {\mathrm{cod}}(G)|=4$ and $G= { {\mathrm{PSL}}}(2,2^f)$ for $f\geq 2$ ;
2. (b) $| {\mathrm{cod}}(G)|=5$ and $G= { {\mathrm{PSL}}}(2,p^f)$ , $p\neq 2, p^f>5$ ;
3. (c) $| {\mathrm{cod}}(G)|=6$ and $G= {}^2B_2(2^{2f+1})$ , $f\geq 1$ , or $G= { {\mathrm{PSL}}}(3,4)$ ;
4. (d) $| {\mathrm{cod}}(G)|=7$ and $G= { {\mathrm{PSL}}}(3,3), {\operatorname {Alt}}_7, M_{11}$ or $J_1$ ;
5. (e) $| {\mathrm{cod}}(G)|=8$ and $G= { {\mathrm{PSL}}}(3,q)$ , where $4< q\not \equiv 1\ (\bmod \ 3)$ , or $G = {\mathrm{PSU}}(3,q)$ , where $4 < q \not \equiv -1\ (\bmod \ 3)$ , or $G=G_2(2)'$ ;
6. (f) $| {\mathrm{cod}}(G)|=9$ and $G= { {\mathrm{PSL}}}(3,q)$ , where $4<q\equiv 1\ (\bmod \ 3)$ , or $G= {\mathrm{PSU}}(3,q)$ , where $4<q \equiv -1\ (\bmod \ 3)$ ;
7. (g) $| {\mathrm{cod}}(G)|=10$ and $G= M_{22}$ ;
8. (h) $| {\mathrm{cod}}(G)|=11$ and $G= { {\mathrm{PSL}}}(4,2), M_{12}, M_{23}$ or ${}^2G_2(3^{2f+1}), f\geq 1$ .
Proof. For a simple group G, each nontrivial irreducible character is faithful. Then $| {\mathrm{cod}}(G)|=| {\mbox {cd}}(G)|$ and the result follows from [Reference Aziziheris, Shafiei and Shirjian3, Theorem 1.1].
The character degree sets for the relevant simple groups have been worked out. Using the definition of codegrees and the fact that the kernel of a nontrivial character is trivial, it is easy to calculate the codegrees. We list the relevant codegree sets in Table 1 for easy reference. Please see [Reference Ahanjideh1, Reference Bahri, Akhlaghi and Khosravi4, Reference Conway, Curtis, Norton, Parker and Wilson7, Reference Simpson and Frame19Reference Ward21] for the details.
Table 1 Codegree sets for some simple groups.
Lemma 2.2. If a simple group G is isomorphic to ${\mathrm{PSL}}(2,k)$ , ${}^2B_2(q^2)$ , ${\mathrm{PSL}}(3,4)$ , ${\operatorname {Alt}}_7$ , $J_1$ , $M_{11}$ , ${\mathrm{PSL}}(3,3)$ , $G_2(2)'$ , $M_{22}$ , ${\mathrm{PSL}}(4,2)$ , $M_{12}$ , $M_{23}$ , ${\mathrm{PSL}}(3,q)$ , ${\mathrm{PSU}}(3,q)$ or ${}^2G_2(q)$ , then ${\mathrm{cod}}(G)$ can be found in Table 1.
To end this section, we state a result about the maximal subgroups of ${}^2G_2(3^{2f+1})$ , $f\geq 1$ . This is obtained from [Reference Kleidman13, Theorem C].
Lemma 2.3. Let K be a maximal subgroup of $\mathrm {}^2G_2(q)$ , where $q=3^{2f+1}$ , $f \geq 1$ . Then K is isomorphic to one of the groups in Table 2.
Table 2 Maximal subgroups of $\mathrm {}^2G_2(q)$ , $q \geq 27$ .
3 Main result for ${}^2G_2(3^{2f+1})$
Sometimes it is convenient to consider the ratio of a set. Let A be a set of nonzero integers. We define $\operatorname {ratio}(A)=\{a/b \ | \ a, b \in A, a \ne b$ and $b \ne 1\}$ .
Lemma 3.1. Let G be a finite group with ${\mathrm{cod}}(G)= {\mathrm{cod}}({}^2G_2(3^{2f+1}))$ , where $f \geq 1$ . If N is a maximal normal subgroup of G, then $G/N \cong {}^2G_2(3^{2f+1})$ .
Proof. Let N be a maximal normal subgroup of G. We set $q=3^{2f+1}$ . Since ${\mathrm{cod}}(G)= {\mathrm{cod}}({}^2G_2(q))$ , we see that G is perfect. Then $G/N$ is a nonabelian simple group. Since ${\mathrm{cod}}(G/N)\subseteq {\mathrm{cod}}(G)$ , we see that $| {\mathrm{cod}}(G/N)|$ is either $4, 5, 6, 7, 8, 9, 10$ or $11$ .
Suppose $| {\mbox {cod}}(G/N)| = 4$ . Then $G/N\cong {\mbox {PSL}}(2,k)$ , where $k=2^t \geq 4$ , and ${\mbox {cod}}(G/N)=\{1, k(k-1), k(k+1), k^2-1\}$ . Since $k^2-1$ is the only nontrivial odd codegree, it must be the same as either $3^{5f+3}(q^2- q + 1)$ or $q^3( q+1-3m)$ or $q^3(q +1 + 3m )$ . Note that $(k+1,k-1)=1$ . Suppose that $k^2-1 = 3^{5f+3} (q^2- q + 1)$ . If $3^{5f+3} | k-1$ , then $k+1 | q^2-q+1$ . Clearly $q^2-q+1 < 3^{5f+3}$ , which implies that $k+1 < k-1$ , which is a contradiction. If $3^{5f+3} | k+1$ , then $k-1 | q^2-q+1$ . Clearly $3^{5f+3} - (q^2-q+1) = 3^{5f+3}-3^{4f+2}+ 3^{2f+1}-1> 2$ , which implies that $(k+1)- (k-1)> 2$ , which is a contradiction. Suppose $k^2-1 = q^3( q+1-3m)$ . If $q^3 | k-1$ , then $k+1 | ( q+1-3m)$ . Obviously, we have $q^3> q+1-3m$ , which implies that $k-1> k+1$ , which is a contradiction. If $q^3 | k+1$ , then $k-1 | ( q+1-3m)$ . Clearly $q^3 - (q+1-3m) = (3^{6f+3}- 3^{2f+1}-1 ) + 3^{f+1}> 2$ , which implies that $k+1 - (k-1)> 2$ , which is a contradiction. Suppose $k^2-1 = q^3 (q+1+3m)$ . If $q^3 | k-1$ , then $k+1 | ( q+1+3m)$ . Since $q^3> q+1+3m$ , we have $k-1> k+1$ , which is a contradiction. If $q^3 | k+1$ , then $k-1 | ( q+1+3m)$ . Clearly $q^3 - (q+1+3m) = 3^{6f+3}-( 3^{2f+1}+1 + 3^{f+1} )> 2$ , which implies that $k+1 - (k-1)> 2$ , which is a contradiction.
Suppose $| {\mathrm{cod}}(G/N)|=5$ . Then $G/N \cong {\mathrm{PSL}}(2,k)$ , where $k>5$ is an odd prime power. We note that $2 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $| {\mathrm{cod}}(G/N)| =6$ . Then $G/N \cong {}^2B_2(2^{2t+1})$ or ${\mathrm{PSL}}(3,4)$ .
Suppose $G/N \cong {}^2B_2(s)$ with $s=2^{2t+1}$ and $r=2^t$ , $t \ge 1$ . The only nontrivial odd codegree in ${\mathrm{cod}}({}^2B_2(s))$ is $(s - 1)(s^2 + 1)$ . Thus, $(s - 1)(s^2 + 1)$ could be equal to either $3^{5f+3} (q^2- q + 1)$ , $q^3( q+1-3m)$ or $q^3(q +1 + 3m )$ . Note that ${(s - 1 , s^2 + 1) =1}$ . Suppose $(s - 1)(s^2 + 1) = 3^{5f+3} (q^2- q + 1)$ . If $3^{5f+3} | s-1$ , then $s^2 + 1 | q^2- q + 1$ . Clearly $3^{5f+3}> q^2-q+1$ , which implies that $s-1> s^2+1$ , which is a contradiction. If $3^{5f+3} | s^2+1$ , we have a contradiction since $s^2-1$ is divisible by 3. Suppose $(s - 1) (s^2 + 1) = q^3 (q + 1 - 3m)$ . If $q^3 | s-1$ , then $s^2 +1 | q+1-3m$ . Clearly $q^3> q+1-3m$ , which implies that $s-1> s^2+1$ , which is a contradiction. If $q^3 | s^2+1$ , then $s-1 | q+1-3m$ . This implies that $q^3 \leq (q+2-3m)^2+1$ , which is a contradiction. Suppose $(s - 1)(s^2 + 1) = q^3 (q + 1 + 3m)$ . If $q^3 | s-1$ , then $s^2 +1 | q+1+3m$ . Clearly $q^3> q+1+3m$ , which implies that $s-1> s^2+1$ , which is a contradiction. If $q^3 | s^2+1$ , then $s-1 | q+1+3m$ . This implies that $q^3 \leq (q+2+3m)^2 + 1$ , which is a contradiction.
Suppose $G/N \cong {\mathrm{PSL}}(3,4)$ . Note that $3^2{\cdot } 5{\cdot } 7$ is the only nontrivial odd codegree in ${\mathrm{cod}}(G/N)$ . Thus, $3^2{\cdot } 5{\cdot } 7$ could be equal to either $3^{5f+3} (q^2- q + 1)$ , $q^3( q+1-3m)$ or $q^3(q +1 + 3m )$ . By considering the power of $3$ in those numbers, we see that each is impossible.
Suppose $| {\mathrm{cod}}(G/N)| = 7$ . Then $G/N \cong {\mathrm{PSL}}(3,3)$ , ${\operatorname {Alt}}_7$ , $M_{11}$ or $J_1$ .
Suppose $G/N \cong {\mathrm{PSL}}(3,3)$ . Note that $3 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $G/N \cong {\operatorname {Alt}}_7$ . Then $2^3 {\cdot } 3^2 \in {\mathrm{cod}}(G/N)$ . Since the powers of $3$ of all the terms that are divisible by $3$ in ${\mathrm{cod}}(G)$ are greater than $2$ , this is a contradiction.
Suppose $G/N \cong M_{11}$ . Note that $5 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $G/N \cong J_1$ . Note that ${\mathrm{cod}}(J_1)$ has two nontrivial odd codegrees $7{\cdot } 11{\cdot } 19$ and $3{\cdot } 5{\cdot } 11{\cdot } 19$ . The odd codegrees in ${\mathrm{cod}}(G)$ are $3^{5f+3}(q^2-q+1)$ , $q^3 ( q+1 - 3m)$ and $q^3 ( q+1 + 3m)$ . By considering the power of $3$ in those numbers, we see that each is impossible.
Suppose $| {\mathrm{cod}}(G/N)|=8$ . Then $G/N \cong { {\mathrm{PSL}}}(3,s)$ , where $4< s\not \equiv 1\ (\bmod \ 3)$ , ${\mathrm{PSU}}(3,s)$ , where $4 < s \not \equiv -1\ (\bmod \ 3)$ , or $G_2(2)'$ .
Suppose ${\mathrm{cod}}(G/N) = {\mathrm{cod}}( {\mathrm{PSU}}(3,s))$ with $4 < s \not \equiv -1\ (\bmod \ 3)$ . Note that $3 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $G/N \cong G_2(2)'$ . Note that $3 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $G/N \cong {\mathrm{PSL}}(3,s)$ for some $4 < s \not \equiv 1\ (\bmod \ 3)$ . Note that $s-1, s+1 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ , and the only nontrivial integers in $\operatorname {ratio}( {\mathrm{cod}}(G))$ are $q-1$ , q, $q+1$ , where q is a power of $3$ and $q \geq 27$ . This will force $s=q$ , which is a contradiction since $q^3(q-1)^2 \not \in {\mathrm{cod}}(G)$ .
Suppose $| {\mathrm{cod}}(G/N)| = 9$ . Then $G/N \cong { {\mathrm{PSL}}}(3,s)$ , where $4<s\equiv 1\ (\bmod \ 3)$ , or ${\mathrm{PSU}}(3,s)$ , where $4<s \equiv -1\ (\bmod \ 3)$ .
Suppose $G/N \cong {\mathrm{PSL}}(3, s)$ with $4 < s\equiv 1\ (\bmod \ 3)$ . Note that $3 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $G/N \cong {\mathrm{PSU}}(3,s)$ with $4 < s \equiv -1\ (\bmod \ 3)$ . Note that $s-1, s+1 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ , and the only nontrivial integers in $\operatorname {ratio}( {\mathrm{cod}}(G))$ are $q-1$ , q, $q+1$ , where q is a power of $3$ and $q \geq 27$ . This will force $s=q$ , which is a contradiction since $q^3(q+1)^2 \not \in {\mathrm{cod}}(G)$ .
Suppose $| {\mathrm{cod}}(G/N)| = 10$ . Then $G/N \cong M_{22}$ . Note that $11 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the only nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is q, where $q \geq 27$ .
Suppose $| {\mathrm{cod}}(G/N)| = 11$ . Then $G/N \cong { {\mathrm{PSL}}}(4,2), M_{12}, M_{23}$ or ${}^2G_2(3^{2s+1})$ , where $s\geq 1$ .
Suppose $G/N \cong {\mathrm{PSL}}(4, 2)$ . Note that $2 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $G/N \cong M_{12}$ . Note that $5 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $G/N \cong M_{23}$ . Note that $8 \in \operatorname {ratio}( {\mathrm{cod}}(G/N))$ . This is a contradiction since the smallest nontrivial integer in $\operatorname {ratio}( {\mathrm{cod}}(G))$ is $q- 1$ , where $q \geq 27$ .
Suppose $G/N \cong {}^2G_2(3^{2s+1})$ , $s\ge 1$ . Comparing the smallest nontrival odd codegree of each set, we see that $q=3^{2s+1}$ . Thus, $G/N \cong {}^2G_2(3^{2f+1})$ .
We now prove the main result of this paper.
Theorem 3.2. Let G be a group such that ${\mathrm{cod}}(G)= {\mathrm{cod}}({}^2G_2(q))$ , where $q=3^{2f+1}$ , $f \geq 1$ . Then $G \cong {}^2G_2(q)$ .
Proof. Let G be a group with ${\mathrm{cod}}(G) = {\mathrm{cod}}({}^2G_2(q))$ . Let N be a maximal normal subgroup of G. Then, $G/N \cong {}^2G_2(q)$ by Lemma 3.1. Assume to the contrary that G is a minimal counterexample. By the choice of G, N is a minimal normal subgroup of G. Otherwise there exists a nontrivial normal subgroup L of G such that L is included in N. Then ${\mathrm{cod}}(G/L)= {\mathrm{cod}}(G)$ for ${\mathrm{cod}}(G)= {\mathrm{cod}}(G/N)\subseteq {\mathrm{cod}}(G/L) \subseteq {\mathrm{cod}}(G)$ and $G/L\cong {}^2G_2(q)$ because G is a minimal counterexample, which is a contradiction.
Step 1: N is the unique minimal normal subgroup of G. Otherwise, assume M is another minimal normal subgroup of G. Then $G=N\times M$ because $G/N$ is simple and $N\cong M\cong {}^2G_2(q)$ because M is also a maximal normal subgroup of G. Choose $\psi _1\in \mbox {Irr}(N)$ and $\psi _2\in \mbox {Irr}(M)$ such that ${\mathrm{cod}}(\psi _1)= {\mathrm{cod}}(\psi _2)=\max ( {\mathrm{cod}}({}^2G_2(q)))$ . Set $\chi =\psi _1 \cdot \psi _2\in \mbox {Irr}(G)$ . Then ${\mathrm{cod}}(\,\chi )=(\max ( {\mathrm{cod}}({}^2G_2(q))))^2\notin {\mathrm{cod}}(G)$ , which is a contradiction.
Set $\mbox {Irr}(G|N)=\{\,\chi \in \mbox {Irr}(G)|\,N$ is not contained in the kernel of $\chi \}$ .
Step 2: $\chi$ is faithful for each $\chi \in \mbox {Irr}(G|N)$ . Since N is not contained in the kernel of $\chi$ for each $\chi \in \mbox {Irr}(G|N)$ , the kernel of $\chi$ is trivial by Step 1.
Step 3: N is elementary abelian. Assume to the contrary that N is not abelian. Then $N=S^n$ , where S is a nonabelian simple group and $n\in \mathbb {N}$ . By Theorems 2, 3 and 4 and Lemma 5 in [Reference Bianchi, Chillag, Lewis and Pacifici5], we see that there exists a nonprincipal character $\psi \in \mbox {Irr}(N)$ that extends to some $\chi \in \mbox {Irr}(G)$ . Then $\ker (\,\chi )=1$ by Step 2 and ${\mathrm{cod}}(\,\chi )=|G|/\chi (1)=|G/N|\cdot |N|/\psi (1)$ . This is a contradiction since $|G/N|$ is divisible by ${\mathrm{cod}}(\,\chi )$ .
Step 4: It is enough to assume that $\mathbf {C}_G(N)=N$ . We note that ${\textbf {{C}}}_G(N) \unlhd G$ . As N is abelian by Step 3, either ${\textbf {{C}}}_G(N) = G$ or ${\textbf {{C}}}_G(N) =~N$ .
We now prove the result in the case ${\textbf {{C}}}_G(N) = G$ . Assume so. Then N is contained in the centre ${\textbf {Z}}(G)$ of G. Since G is perfect, ${\textbf {Z}}(G)=N$ and N is isomorphic to a subgroup of the Schur multiplier of $G/N$ [Reference Isaacs10, Corollary 11.20]. This forces N to be trivial as ${}^2G_2(q)$ has trivial Schur multiplier, and we are done.
Step 5: Let $\lambda$ be a nonprincipal character in $\mbox {Irr}(N)$ and $\theta \in \mbox {Irr}(I_G(\lambda )|\lambda )$ . We show that ${|I_G(\lambda )|}/{\theta (1)} \in {\mathrm{cod}}(G)$ . Also, $\theta (1)$ divides $|I_G(\lambda )/N|$ and $|N|$ divides $|G/N|$ . Let $\lambda$ be a nonprincipal character in $\mbox {Irr}(N)$ . Given $\theta \in \mbox {Irr}(I_G(\lambda )|\lambda )$ . Note that $\chi =\theta ^G\in \mbox {Irr}(G)$ and $\chi (1)=|G:I_G(\lambda )|\cdot \theta (1)$ by Clifford theory (see [Reference Isaacs10, Ch. 6]). Then $\ker (\,\chi )=1$ by Step 2 and ${\mathrm{cod}}(\,\chi )={|I_G(\lambda )|}/{\theta (1)}$ . In particular, $\theta (1)$ divides $|I_G(\lambda )/N|$ , and then $|N|$ divides ${|I_G(\lambda )|}/{\theta (1)}$ . Since ${\mathrm{cod}}(G)= {\mathrm{cod}}(G/N)$ and $|G/N|$ is divisible by every element in ${\mathrm{cod}}(G/N)$ , we have $|N|\mid |G/N|$ .
Step 6: Final contradiction. By Step 3, N is an elementary abelian r-subgroup for some prime r and we assume $|N|=r^n$ , $n\in \mathbb {N}$ . By the normaliser–centraliser theorem, $n> 1$ .
Let $\lambda \in \mbox {Irr}(N)$ be a nonprincipal character and $T:=I_G(\lambda )$ . By Step 5, ${|T|}/{\theta (1)}\in {\mathrm{cod}}(G)$ for all $\theta \in \mbox {Irr}(T|\lambda )$ .
Since N is abelian by Step 1, $|\mbox {Irr}(N)|=|N|$ . Therefore, $|N|=|\mbox {Irr}(N)|> |G:T|$ since $|G:T|$ is the number of conjugates of $\lambda$ in G which are all contained in $\mbox {Irr}(N)$ .
We now show that $q^3$ is the largest power of a prime that divides the order of ${}^2G_2(q)$ . Note that the order of ${}^2G_2(q)$ is $q^3(q^3+1)(q-1)$ . We first observe that $q^3+1=2^k$ has no integer solution, and thus $q^3+1$ is divisible by a prime greater than or equal to $5$ . We also note that either $q^3+1$ or $q-1$ is divisible by $4$ but not both, and thus the largest power of $2$ that divides $(q^3+1)(q-1)$ is less than $q^3$ . Suppose that r is an odd prime that divides $(q^3+1)(q-1)$ . Since $\gcd (q+1,q-1)=2$ , the largest power of r that divides $(q^3+1)(q-1)$ is also less than $q^3$ . Thus, $q^3$ is the largest power of a prime that divides the order of ${}^2G_2(q)$ . Thus, $|N| \leq q^3$ .
Let K be a maximal subgroup of ${}^2G_2(q)$ such that $T/N \leq K$ . If K is of type P in Lemma 2.3, then $|G:T| \ge q^3+1$ , and it is clear that $q^3+1>q^3$ , which is a contradiction. If K is of type $C_{R_{0}(i)}$ , then $|G:T| \ge q^2 (q^2 - q + 1)> q^3$ , which is a contradiction. If K is of type $N_{R_{0} \langle i ,j \rangle }$ , then $|G:T| \ge { q^3( q^2 - q + 1 ) ( q - 1) }/{ 12 (q+1) }$ , thus $|G:T|> q^3$ , which is a contradiction. If K is of type $M^{+}W^{+}$ , then $|G:T| \ge { q^3( q^2 - q + 1 ) ( q - 1) }/{ 6 ( q +\! \sqrt { 3q } + 1 )}$ , and thus $|G:T|> q^3$ , which is a contradiction. If K is of type $M^{-}W^{-}$ , then $|G:T| \ge { q^3( q^2 - q + 1 ) ( q - 1) }/{ 6 ( q -\! \sqrt { 3q } + 1 )}$ , and thus $|G:T|> q^3$ , which is a contradiction. If K is of type $C_{R_0}(\Psi _{ \alpha })$ , then $|G:T| \ge {q^3(q^3+1)(q-1)}/{q_0^3(q_0^3+1)(q_0-1)}$ , where $q=q_0^{\alpha }$ , $\alpha$ a prime. Thus, $|G:T|> q^ {3}$ , which is a contradiction.
Acknowledgement
The authors are grateful to the referee for the valuable suggestions which improved the manuscript.
Footnotes
The project was supported by grants from the Simons Foundation (#499532, #918096) to the third author.
References
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Kleidman, P. B., ‘The maximal subgroups of the Chevalley groups ${G}_2(q)$ with $q$ odd, the Ree groups ${}^2{G}_2(q)$ , and their automorphism groups’, J. Algebra 117 (1988), 3071.CrossRefGoogle Scholar
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Table 1 Codegree sets for some simple groups.
Table 2 Maximal subgroups of $\mathrm {}^2G_2(q)$, $q \geq 27$.
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# Quaternion "creep" or Erroneous Math?
This topic is 5392 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Recently I was trying to rotate a vector about an arbitrary axis via quaternions. I coded the functions and I believe (after intense review) that all of the math is correct, but there is a problem. I am attempting to rotate a unit vector that is parallel to the y-axis about the z-axis (on a black OpenGL screen, it is easiest to judge). I rotated the vector in increments of pi/6 radians, or 30 degrees. Unfortunately, the magnitude of the vector being rotated slightly changed depending on the angle at which it was being rotated. The following is a list of the vector's magnitude and corresponding rotation angle during one full rotation: (Underscores used for spacing only) Magnitude_____Degrees Rotated ---------_____--------------- 1_____________0 0.957076______30 0.945742______60 0.954033______90 0.97293_______120 0.991937______150 1_____________180 0.991937______210 0.97293_______240 0.954033______270 0.945742______300 0.957076______330 1_____________360 Due to the behavior of the magnitude per angle, I would guess that perhaps my use of the cosf and sinf functions in math.h, along with the multitude of floating point calculations, caused some inaccuracies to slip in (although, I wouldn't expect as much as .06 inaccuracy). Does this sound like a erroneous math, a common quaternion problem, or a simply "creep"ing in of inaccuracies? Any and all advice is appreciated! [edited by - liqyd on April 12, 2004 4:35:34 AM] [edited by - liqyd on April 12, 2004 4:35:58 AM] [edited by - liqyd on April 12, 2004 4:37:27 AM]
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The inaccuracy seems slightly high for simple trig error, but I suppose it could be attributed to that. If you''re worried about the results, simply normalize the quaternion after each op; with a magnitude like that, though, the difference won''t be major.
"Sneftel is correct, if rather vulgar." --Flarelocke
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Your suggestion did help to improve the accuracy, but it turns out that wasn''t the problem. Even though I checked my math many times, a hardly noticeable error was there.
In my quaternion magnitude computing function I had something like:
SquareRoot( w^2 + ||v||)
SquareRoot( w^2 + x^2 + y^2 + z^2)
The source I am learning quaternions from had the former as the correct definition for computing quaternion magnitude, but after changing it to the latter the magnitudes are all an even 1.
Sorry for wasting space on your forum with such a stupid mistake. Again, thanks for your help.
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# Simplify: Enter the exponent as x(1/n) `-root4(144)`
violy | High School Teacher | (Level 1) Associate Educator
Posted on
Take note that `-root(4)(144) = -root(4)(12^2) = -(12^2)^(1/4)`
pramodpandey | College Teacher | (Level 3) Valedictorian
Posted on
`-^4sqrt(144)=-(144)^(1/4)`
`=-(12^2)^(1/4)`
`=-(12)^(2xx(1/4))`
`=-(2^2xx3)^(1/2)`
`=-(2^2)^(1/2)(3)^(1/2)`
`=-(2)^(2xx(1/2))(3)^(1/2)`
`=-2(3)^(1/2)`
``
``
``
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# CS545: Machine Learning
### Sidebar
CS545 - Fall 2021
Instructor
Chuck Anderson
start
This is an old revision of the document!
# Schedule
1. Office Hours with Chuck: Wednesdays, 10:00 - 11:00 AM
2. Office Hours with Dejan: Mondays, 1:00 - 3:00 PM and Wednesdays, 3:00 - 5:00
To use jupyter notebooks on our CS department machines, you must add this line to your .bashrc file:
`export PATH=/usr/local/anaconda/bin:\$PATH`
This is a tentative schedule of CS545 topics for Fall, 2020. This will be updated during the summer and as the fall semester continues.
## August
Week 1:
Aug 24 - Aug 28
Overview of course and the machine learning field. Reminder of how python is used in machine learning. 01 Introduction to CS545
02 Searching for Good Weights in a Linear Model
From Python to Numpy, Chapters 1 - 2
Scipy Lectures, Section 1
Visualization with Matplotlib
Deep Learning, Chapters 1 - 5.1.4
## September
Week 2:
Aug 31 - Sept 4
Help with A1. Review of gradients. Gradient descent with SGD, Adam and SCG, 03 Fitting Simple Models Using Gradient Descent in the Squared Error A1.4 Polynomial Model due Friday, Sept 4th, at 10:00 PM
Week 3:
Sept 7 - Sept 11
Implementing neural networks with numpy to predict real-valued variables. Deriving gradients. 04 Scaled Conjugate Gradient
05 Introduction to Gradient Descent for Neural Networks
Week 4:
Sept 14 - Sept 18
Error gradients for neural networks as matrix equations. Discussion of A2.
Introduction to dashboards with python using streamlit.
06 Introduction to Streamlit streamlit.io
Week 5:
Sept 21 - 25
Use of Optimizers for neural networks. Introduction to Pytorch and automatic differentation. 07 Collect Weights in Vector for Optimizers
A2.2 Multilayer Neural Network due Friday, Sept 25th, at 10:00 PM. Good examples of solutions are available here.
## October
Week 6:
Sept 28 - Oct 2
Neural Network class. 09 Initial Steps towards Defining a NeuralNetwork Class
Week 7:
Oct 5 - Oct 9
Oct 8 Lecture will not meet, but recording will be available.
Help with A3. Dimensionality reduction. 10 Help with A3
11 Low-Dimensional Representations of Data
A3.3 Neural Network Class due Monday, Oct 12, 10:00 PM
Examples of good solutions are available here.
Week 8:
Oct 12 - Oct 16
Brief overview of notes 11.
Introduction to Classification
12 Classification with Neural Networks
Week 9:
Oct 19 - Oct 23
Convolutional neural networks in numpy. 13 NeuralNetwork_Pytorch
14 Introduction to Convolution
Week 10:
Oct 26 - Oct 30
Fully-connected and Convolutional Neural Nets in Pytorch 15 Convolutional Neural Networks
16.1 Convolutional Neural Networks in Pytorch
A4.1 Neural Network Classifier due Tuesday Oct 27, at 10:00 PM
Good examples of solutions are available here.
## November
Week Topic Material Reading Assignments Week 11: Nov 2 - Nov 6 Comparing network performance. Introduction to Reinforcement Learning. Deep Reinforcement Learning 17 Partitioning Data to Compare Neural Network Performance 18 Introduction to Reinforcement Learning 19 Reinforcement Learning with Neural Network as Q Function Reinforcement Learning: An Introduction, by Richard Sutton and Andrew Barto, 2nd edition Week 12: November 9 - 13 Deep reinforcement learning on simulated physical control problem. 20 Reinforcement Learning to Control a Marble Week 13: Nov 16 - Nov 20 A5 Neural Networks in Pytorch due Wednesday, Nov 18 at 10:00 PM Good examples of solutions are available here. Nov 23 - Nov 27 Fall Recess!
## December
Week 14:
Nov 30 - Dec 4
Clustering.
Support Vector Machines.
22 K-Means Clustering, K-Nearest-Neighbor Classification
23 Support Vector Machines
Week 15:
Dec 7 - Dec 11
Finals Week:
Dec 14 - Dec 18
A6 Reinforcement Learning to Control a Robot due Tuesday, Dec 15th, 10:00 PM. Here is A3mysolution.tar, a neural network implementation you may choose to use for A6.
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# If x < –5 and x > 2, then x ∈ (– 5, 2) - Mathematics
MCQ
True or False
If x < –5 and x > 2, then x ∈ (– 5, 2)
• True
• False
#### Solution
This statement is False.
Explanation:
If x < – 5 and x > 2, then x have no value.
Concept: Inequalities - Introduction
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Exemplar Class 11
Chapter 6 Linear Inequalities
Exercise | Q 31.(vi) | Page 111
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# Annual quantity | My Assignment Tutor
Suppose the production function for product x is given by where q represents the annual quantity of product x produced, k represents annual capital input and l represents annual labour input. a. If k= 10, graph the total and average productivity of labour curves. At what level of labour input does this average productivity reach a maximum? How many units of product x are produced at that point? b. Again if k= 10, graph the MPl curve. At what level of labour input does MPl= 0? c. Suppose capital inputs double. How would your answers to parts (a) and (b) change? d. Does the production function for product x exhibit constant, increasing, or decreasing returns to scale?
QUALITY: 100% ORIGINAL PAPER – NO PLAGIARISM – CUSTOM PAPER
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# Electrical Calculators and Resources
Our electrical calculators and resources can help estimate the electrical uses and cost of lighting and household appliances and estimate wire sizes for electrical project work.
Results 1-9 of 9
## 2018 Electricity Cost Calculator
Calculate electricity cost by entering wattage and usage of electricity and the cost per kilowatt-hour. Estimate energy costs and kilowatt-hour electricity usage.
## 2018 Lighting Energy Cost Calculator
Estimate the cost of electricity to power the lights in your home. Calculate the cost savings of upgrading to LED or CFL lighting.
## Kilowatts to Amps Electrical Conversion Calculator
Convert the power in kilowatts to current in amps or find the power given the amperage rating of a generator or other electrical equipment.
## Ohm’s Law Calculator
Calculate watts, volts, amps, ohms, power, voltage, current, and resistance using the Ohm’s Law and Watt’s Law formulas.
## Voltage Drop Calculator
Calculate voltage drop in an AC or DC circuit, find the minimum wire gauge required when planning a circuit, or determine the maximum length of conductors in a circuit.
## Volts to Amps Electrical Conversion Calculator
Use the volts to amps calculator to convert between voltage and amperage in an electrical circuit. A voltage to amperage conversion can be made if wattage or resistance in the circuit is known.
## Watts to Amps Electrical Conversion Calculator
Convert the power in watts to the current in amps for an electrical circuit or find how many watts of power a device draws given the amp rating.
## Watts to Joules Electrical Conversion Calculator
Convert between power in watts and electrical energy measured in joules and see formulas used for the calculations with examples.
## Wire Gauge Size and Resistance Calculator
Calculate the diameter, cross-sectional area, and resistance of wire given the wire gauge or find the wire gauge for a given diameter.
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Background
I am solving the unsteady heat equation in 3D using an alternating direction implicit (ADI) method. This means that I am solving three different tridiagonal systems within a single timestep (once for each cartesian axis). The size of these matrices is on the order of 10^5 x 10^5.
Although each direction's A matrix has all non-zeros along the three diagonals, there are situations where one direction will have a zero value along one of the diagonals where another direction does not.
My original plan was to set up an AIJ matrix and preallocate for 3 non-zero values and reuse the same memory allocation for each of the three directions. This works fine for solving the first direction, but when I hit the second direction PETSc complains because I am attempting to add a new non-zero value where there wasn't one before (related to differences in boundary conditions).
Questions
What is the most efficient way to proceed:
• Is there a way to allocate differently that will allow me to have values change between zero and non-zero?
• OR should I create a new matrix and new allocation for each direction? (This I know works, but I'm worried about the overhead of destroying and creating a new matrix for every solution.)
• Other options?
In pseudo code, my program looks like this:
// Create and preallocate A matrix:
MatCreateAIJ(PETSC_COMM_WORLD,PETSC_DECIDE,PETSC_DECIDE,M,M,3,NULL,3,NULL,&Amat);
// Create x vector
VecCreate(PETSC_COMM_WORLD,&x);
// Create b vector
VecDuplicate(x,&b);
// Solve for each timestep:
for (each timestep)
{
for (X,Y,Z)
{
MatSetValues(Amat,...,INSERT_VALUES);
VecSetValues(b,...,INSERT_VALUES);
MatAssemblyBegin(Amat,MAT_FINAL_ASSEMBLY);
MatAssemblyEnd(Amat,MAT_FINAL_ASSEMBLY);
VecAssemblyBegin(b);
VecAssemblyEnd(b);
KSPSetOperators(ksp, Amat, Amat, DIFFERENT_NONZERO_PATTERN);
KSPSetUp(ksp);
KSPSolve(ksp,b,x);
MatZeroEntries(Amat);
}
}
1. ADI is not a very good parallel algorithm. You should seriously consider formulating the problem in 3D and solving with multigrid. You could get a start with src/ksp/ksp/examples/tutorials/ex45.c.
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# Trump
By LTG
Question: if Donald trump has 57 dollars and he spends 26 dollars and the cashier guy gives him 36 cents back, how much money does he have?
Riddle Discussion
By: Saikomen on 8/8/16
Bruh, the answer is \$31.36. 57-26=31, then add 36 cents, u get \$31.36
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Question: In a house of Kardak's family each brother have a sisters as double as their brothers and each sister have brothers as same as their sister. How many brothers and sisters in Kardak's family ?
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Question: I'm 10 years old my brothers half my age now I'm 100 years old how old is my brother now?
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# How can I ensure even water flow to my garden?
Tags:
1. May 8, 2015
### Glurth
I would like to setup a subsurface irrigation system for my garden.
I have several beds which, while I made each one perfectly level, are at different elevations (about 1 or 2 feet between each) .
The source of the water will be a permanently mounted bucket, barrel or cistern of some kind, with a single valve to open to feed the system.
I will have garden hose that goes from the cistern to the various beds in the garden. One inch below the surface of each bed will be a perfectly level PVC tube, with holes drilled in the top. The water source hose will go into one end of these tubes, and the other end will either be capped, or lead to a lower bed. Water will push out of the top holes to water the beds.
QUESTION:
Is there a simple way for me to ensure that the amount of water delivered to each bed will be the same? If I just run a hose to each bed, the beds at a lower elevation will get more water per minute than the higher beds, (due to greater water pressure at the holes in the tube), which I would like to avoid. Or, is this incorrect?
Ideally I could let a higher bed’s PVC pipe feed a lower bed’s PVC pipe, so I don’t need to run a hose to EVERY bed, just every ROW of beds using a "mainline".
More info: Other than my well pump, which is used to get water into the cistern, everything else is gravity fed. The bottom of the cistern will be about a foot above the highest bed.
My thoughts so far: Could this be done by simply making the LAST hole in a bed’s PVC pipe larger, and use that to feed the next tube? If each bed has N holes, of radius R, how large would that last hole have to be to ensure the water that flows out of it is equal to the water flowing out of all the previous holes in the pipe? To complicate it, what if I have 2 or 3 beds below the first? Also, would a change in the water source pressure, as it drains, throw this hole-size ratio off?
I am hoping to get an answer as simple/brilliant as: "drill holes" in this section of pipe, to let air in, or something like that.
2. May 9, 2015
### Simon Bridge
You can make sure the water flow is the same for all pipes by adding a tap as a throttle in the leadup to the irrigation tube. Measure the water flow rate before burying and adjust the tap accordingly.
3. May 9, 2015
### BrentS
A throttle in each line is a good idea if the amount of piping/headloss differs significantly between the "mainlines", but it doesn't help with the other issue of chaining pipes together in series within each string. As soon as the flow reaches the first holes, pressure will start dropping, and there's no passive device that could keep it equalized all the way down the line.
The magnitude of the problem can be reduced by mounting the gravity-fed source higher (so there's higher initial pressure, and less variation in total height between the uppermost and lowermost beds), using larger/smoother pipe (lower headloss per foot), and drilling smaller holes (so the pressure loss at each is lower). Rather than making the holes progressively larger as you move down the line (a tricky problem even for constant source pressure), I'd just increase their frequency, i.e. drill more in the lower pipes, and even perhaps even space them further apart toward the upstream end of each pipe if the sections are very long. You'll probably have to determine the spacing experimentally before burying the tubes.
4. May 9, 2015
### Simon Bridge
Oh yes, the pipes have to be fed separarely to keep it simple.
Welcome to PF BrentS.
5. May 10, 2015
### CWatters
Make sure you filter the water or you will be digging up the pipes regularly to unblock holes.
6. May 10, 2015
### jack action
I would use a multi-level «waterfall-pool» set up. All you need is to put an open reservoir at one end of each of the bed pipes. The input flow should arrive horizontally at the top of the reservoir and there should be an overflow feeding to the next lower bed. This way the pressure in the pipe will be controlled by the depth of the reservoir, not by the height of the cistern.
For it to work, you need either a valve at the entry of each reservoir controlled by a floater in the reservoir or one valve at the cistern controlled by a floater in the lowest reservoir.
To better equalize the flow along each of your pipe (if they're really long), you can put more than one of this set up in parallel, like one reservoir at each end of the pipe or even add others between the two.
It would look like this (the recirculating pump replaced by the floater(s) and valve(s)):
Like I said, the only difference would be to make sure the waterfall would fall into a 90° elbow such that the flow arrives horizontally above the pipe entry (at the bottom of the reservoir) such as not to add the dynamic pressure of the fall onto the reservoir depth water level.
7. May 10, 2015
### jim hardy
If you wanted to get crazy complicated.....
Can you get some old nutating disk meters from your water company's junkpile ?
They're a positive displacement engine , so belting their driveshafts together should give equal flow.
have fun !
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# Data that needs to analysed although part of it is outside of the Time Dimension
• ktflash
Ten Centuries
Points: 1109
Ahoi,
i have the following Situation:
• We have an IT Cube where tickets are analysed
• Each Ticket has a Buget/Plan Time which is the supposed time required for the task
• This time is booked on the creation time of the ticket
• Each Ticket also has the actual time booked on top of this ticket
• These times accumulative booked on N different positions, should be in buget/plan
My users tried to analyse the following, for august they wanted to which tickets where active:
• Tickets Dimension has a "status" attribut --> closed(inactive)/open(active)
• Tickets that are still open
• Tickets that are closed, but were closed in august--> time booked in this period (august)
The Problem is the following:
In the current Version they tried simply using the Time Dimension and filter in august.
• Plan/Budget Time is booked on creation time of the ticket and therefore is 0 if the ticket was not created in august
• If the Ticket is also older than august and time has been booked outside of august, not all of the accumulative times are aggregated, only the ones from august
• Budget Time often 0
• Booked Times by users is not the total time booked on the ticket(only august in this case), although the total is needed
I have a solution that works, but is not "pretty", I made a dummy measure and a 2nd Time Dimension:
• Seperate Time Dimension
• the Dummy Measure is solely used to filter the Tickets in Dimension that are "relevant" for in this case august
• The measure is connected to ticket Dimension and the seperate Time Dimension
• It takes all tickets that are currently still active and multiplies them across the Time Dimension, so is secured that each open Ticket is defently in any month --> since active tickets are always relevant (still flawed, needs to be finetuned)
• For the tickets that are closed, i use part of the measuregroup that is used to get the 2 measures to get ticket times (buget/accumulitive) --> connected to the original Time Dimension
• By having these times booked on only by this one measure, i know that if use the new dimension and the dummy measure, that only if a dummy value is booked that it is a relevant ticket for the selected time
My Question is there a better solution for this type of Problem?
EDIT: I guess i need a MDX Solution that does the following
[Measures].[Aufwand] -- Measure required for relevance Filter
[Ticket].[Status].&[Offen],[Ticket].[Status].&[Warte auf] -- Tickets with these states that need to be in no matter what
[Ticket].[Ticket ID] -- Attribute that needs to be filtered based on state + Measure
Need all [Ticket].[Ticket ID].Members
where Aufwand > 0 or is either [Ticket].[Status].&[Offen] or [Ticket].[Status].&[Warte auf]
• This topic was modified 4 months, 1 week ago by ktflash.
I wanna be the very best
Like no one ever was
• Site Owners
SSC Guru
Points: 80378
This is an automated bump to increase visibility of your question.
• heb1014
Hall of Fame
Points: 3788
This is a tough post to follow, but I have one idea that may help. Have you investigated role-playing dimensions? This concept may be helpful and should eliminate the need to create multiple date/time dimensions. That helps with the user experience because the user does not need to filter dates from multiple dimensions.
Viewing 3 posts - 1 through 3 (of 3 total)
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# Thread: [C++] Efficient symmetric sparse matrix storing
1. Ubuntu Extra Shot
Join Date
Dec 2009
Beans
356
## [C++] Efficient symmetric sparse matrix storing
Hi,
I have to implement in C/C++ a symmetric sparse matrix. I have read of various storage format and the most popular is this one CSR http://netlib.org/linalg/html_templa...00000000000000. This method is efficient from the memory requirement perspective, but it is no efficient in my case. In fact, I have three constraints: the first is that the dimensions and the matrix are not known in advance but the matrix is dynamically built, the second is that the elements of the matrix are updated during the computation and finally the matrix-vector and matrix-matrix operations must be efficient.
For the first and second constraint I think that something like this struct{double val; size_t i; size_t j;} vector<struct entry> myMatrix; is not efficient. In fact I have to search over a vector to find the right position for the elements and this require O(n) where n is the size of the vector.
I have thought about two possibilities: the first is to use a vector<map<int, double> > myMatrix; where the vector is row-wise and each map stores only non zero elements in the column. The first and second constraints are easily handled thank to the fast search over a map O(log ci) where ci are the number of non zero elements in the row ri. Also the matrix-vector and matrix-matrix multiplication is fast as in the case of vector<struct entry>. This representation is also more efficient in term of memory with respect to vector<struct entry>, but less with respect to CSR. The second is map<struct entry> myMatrix. This is similar to vector<struct entry> but allows faster searches.
I think that the best is vector<map<int, double> > but I worried about the following fact: in case of symmetric matrix is necessary to store only upper(lower)-triangular part of the matrix. So the first row is full and has n elements, the second row has n-1 elements and the last row has 1 element. In this case the last map has only one element. In the case of symmetric sparse matrix this is also more important.
I cannot use the solution of vector<vector<double> > since that the search is still O(n) even if moving down to the matrix the n decreases up to 1. Another important problem that is practical, instead of theoretical, is the cache coherence. A vector can be contiguously allocated, is it possible with a map? Of course I can use an unordered_map instead of map. The search become constant but the constant can be high. So the game is worth the candle?!
Do you have more clever idea? Any suggestion is welcome.
Thank you
Last edited by erotavlas; October 25th, 2013 at 02:15 PM.
2. ## Re: [C++] Efficient symmetric sparse matrix storing
If you are looking for something really efficient, I doubt that using vector/map is going to help, because these are very general-purpose objects and if I trust their global efficiency, I don't think they are going to be efficient enough for you (you may end up using a lot more memory than you think, for instance).
If you do a lot of updates, you will have to allocate memory, and this maybe another bottleneck. The default memory allocators (malloc()/new) can't take advantage of the specific behavior of your data.
It would also help to know how big your matrices are going to be and how sparse they are...
3. ## Re: [C++] Efficient symmetric sparse matrix storing
What does your matrix represent here? A graph like data structure? Also about the search, what are you really going to search here? The presence of a specific <int, double> key-value pair in a row or something else? And as ofnuts said, if you are worried about memory constraints, vector is not a good idea. Reading from your description, it doesn't look like there is any straight forward data structure that will be suitable for your scenario. Probably you will be needing to maintain some metadata structure along with actual data structure to support the dynamic resizing of your matrix and write some wrapper methods to support the symmetrical behavior of your matrix.
4. Ubuntu Extra Shot
Join Date
Dec 2009
Beans
356
## Re: [C++] Efficient symmetric sparse matrix storing
Originally Posted by ofnuts
If you are looking for something really efficient, I doubt that using vector/map is going to help, because these are very general-purpose objects and if I trust their global efficiency, I don't think they are going to be efficient enough for you (you may end up using a lot more memory than you think, for instance).
If you do a lot of updates, you will have to allocate memory, and this maybe another bottleneck. The default memory allocators (malloc()/new) can't take advantage of the specific behavior of your data.
It would also help to know how big your matrices are going to be and how sparse they are...
In general, I don't know how my matrices are sparse and big. I specify a quantity of memory at the beginning of the program and it allocates memory for matrices. In the case of square dense non symmetric matrix I have NxNxdouble while in symmetric case Nx(N+1)/2xdouble space. In the case of symmetric sparse square matrix I have hxNx(N+1)/2xdouble where h is sparse coefficient. In general I aspect to use N=10000-20000 in non sparse case and about the double in sparse case.
If I use a vector, I can allocate my memory in advance. I cannot do this with map.
What do you suggest to use?
5. ## Re: [C++] Efficient symmetric sparse matrix storing
If you don't know how sparse your matrices are then I don't see much gain in handling them as sparse because when you hit a not-so-sparse matrix (which would be a 20% fill ratio) the whole thing will get much worse than using a plain matrix.
If your problem is mostly dynamically reallocating the array, remember than in C an array is really a vector of pointer to the rows, so you can reallocate each row independently when you need to write past the end of the existing one (various strategies are possible for how much you reallocate each time). This can be done in an access method that handles this dynamic reallocation as well as the symmetry.
6. ## Re: [C++] Efficient symmetric sparse matrix storing
seems like you want the dok (dictionary of keys) or lil (row based linked list) format and then convert it to a more efficient one when you are done building it for processing.
there are plenty of libraries for this stuff, I know of the scipy.sparse module which implements coo, csc, csr, dia, dok and lil formats (http://docs.scipy.org/doc/scipy/reference/sparse.html)
I'm sure you can find a C++ library for these too, I think scipys implementation is partly based on arpack.
Last edited by MadCow108; October 27th, 2013 at 11:56 AM.
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Thermodynamics HW Solutions 477
# Thermodynamics HW Solutions 477 - Chapter 5 Numerical...
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Chapter 5 Numerical Methods in Heat Conduction i i i i i i i i i i i i i i i i i i i i T T T T m T T T T m T T T T m T T T T m T T T T m 5 6 4 1 5 4 5 3 1 4 3 4 2 1 3 2 3 1 1 2 1 2 0 1 1 ) 2 1 ( ) ( : ) 5 ( 5 Node ) 2 1 ( ) ( : ) 4 ( 4 Node ) 2 1 ( ) ( : ) 3 ( 3 Node ) 2 1 ( ) ( : ) 2 ( 2 Node ) 2 1 ( ) ( : ) 1 ( 1 Node τ + + = = + + = = + + = = + + = = + + = = + + + + + Node 6 t T T C x A x T T kA q A T T A h i i i i i i i Δ Δ = Δ + + 6 1 6 6 5 solar 6 out out 2 + ) ( ρ κ & or k x q T k x h T T k x h T i i i i i Δ + Δ + Δ = + solar out out 5 6 out 1 6 2 2 2 + 2 2 1 & ττ where L = 0.30 m, k = 0.70 W/m.°C, , T α 044 10 6 . m 2 / s out and are as given in the table, & q solar = 0.76 h out = 3.4 W/m 2 . ° C, T in = 20 ° C, h in = 9.1 W/m 2 . ° C, and Δ x = 0.05 m. Next we need to determine the upper limit of the time step Δ t from the stability criteria since we are using the explicit method. This requires the identification of the smallest primary coefficient in the system. We know that the boundary nodes are more restrictive than the interior nodes, and thus we examine the formulations of the boundary nodes 0 and 6 only. The smallest and thus the most restrictive
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## This note was uploaded on 01/22/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.
Ask a homework question - tutors are online
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# Distance between Auckland (AKL) and Kawthoung (KAW)
Flight distance from Auckland to Kawthoung (Auckland Airport – Kawthaung Airport) is 5892 miles / 9482 kilometers / 5120 nautical miles. Estimated flight time is 11 hours 39 minutes.
## Map of flight path from Auckland to Kawthoung.
Shortest flight path between Auckland Airport (AKL) and Kawthaung Airport (KAW).
## How far is Kawthoung from Auckland?
There are several ways to calculate distances between Auckland and Kawthoung. Here are two common methods:
Vincenty's formula (applied above)
• 5891.897 miles
• 9482.090 kilometers
• 5119.919 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 5894.199 miles
• 9485.793 kilometers
• 5121.919 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Auckland Airport
City: Auckland
Country: New Zealand
IATA Code: AKL
ICAO Code: NZAA
Coordinates: 37°0′29″S, 174°47′31″E
B Kawthaung Airport
City: Kawthoung
Country: Burma
IATA Code: KAW
ICAO Code: VYKT
Coordinates: 10°2′57″N, 98°32′16″E
## Time difference and current local times
The time difference between Auckland and Kawthoung is 6 hours 30 minutes. Kawthoung is 6 hours 30 minutes behind Auckland.
NZDT
+0630
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 702 kg (1 548 pounds).
## Frequent Flyer Miles Calculator
Auckland (AKL) → Kawthoung (KAW).
Distance:
5892
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
5892
Round trip?
| 479
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| 2.578125
| 3
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CC-MAIN-2021-04
|
latest
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en
| 0.811799
|
https://gitlinux.net/2019-05-20-(060)-%E6%8A%8A%E4%BA%8C%E5%8F%89%E6%A0%91%E6%89%93%E5%8D%B0%E6%88%90%E5%A4%9A%E8%A1%8C/
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| 278,414,142
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|
题目描述：从上到下按层打印二叉树，同一层结点从左至右输出。每一层输出一行。
## Solutions
采用广度优先遍历的方式写了如下的代码：
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回二维列表[[1,2],[4,5]]
def Print(self, pRoot):
# write code here
if pRoot is None:
return []
queue = [pRoot]
res = []
while queue:
level = []
size = len(queue)
for i in range(size):
node = queue.pop(0)
level.append(node.val)
if node.left is not None:
queue.append(node.left)
if node.right is not None:
queue.append(node.right)
res.append(level)
return res
# 运行时间：28ms
# 占用内存：5792k
还有另一种类似的解决办法：
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回二维列表[[1,2],[4,5]]
def Print(self, pRoot):
# write code here
if not pRoot:
return []
res=[]
arr=[]
arr.append(pRoot)
cur=0
#last=1
while cur<len(arr):
last=len(arr)
temp=[]
while (cur<last):
temp.append(arr[cur].val)
if arr[cur].left:
arr.append(arr[cur].left)
if arr[cur].right:
arr.append(arr[cur].right)
cur+=1
res.append(temp)
return res
# 运行时间：25ms
# 占用内存：5732k
剑指 offer 上的解法稍显复杂：
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回二维列表[[1,2],[4,5]]
def Print(self, pRoot):
# write code here
if not pRoot:
return []
res=[]
res_val=[]
res.append(pRoot)
nextLevel=0 #表示下一层节点的数目
toBePrinted=1 #表示当前层还没有打印的节点数
temp=[]
while len(res)>0:
node=res[0]
temp.append(node.val)
if node.left:
res.append(node.left)
nextLevel+=1
if node.right:
res.append(node.right)
nextLevel+=1
del res[0]
toBePrinted-=1
if toBePrinted==0:
res_val.append(temp)
toBePrinted=nextLevel
nextLevel=0
temp=[]
return res_val
# 运行时间：42ms
# 占用内存：5736k
| 645
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| 3.40625
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CC-MAIN-2021-31
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latest
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en
| 0.179576
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http://androidxref.com/4.2_r1/xref/frameworks/base/media/mca/filterfw/native/core/geometry.h
| 1,585,834,414,000,000,000
|
text/html
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crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00169.warc.gz
| 10,489,721
| 4,867
|
```/* <![CDATA[ */
2 * Copyright (C) 2011 The Android Open Source Project
3 *
5 * you may not use this file except in compliance with the License.
6 * You may obtain a copy of the License at
7 *
9 *
10 * Unless required by applicable law or agreed to in writing, software
12 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
13 * See the License for the specific language governing permissions and
14 * limitations under the License.
15 */
16
17#ifndef ANDROID_FILTERFW_CORE_GEOMETRY_H
18#define ANDROID_FILTERFW_CORE_GEOMETRY_H
19
20#include <vector>
21
22namespace android {
23namespace filterfw {
24
25// This is an initial implementation of some geometrical structures. This is
26// likely to grow and become more sophisticated in the future.
27
28class Point {
29 public:
30 Point() : x_(0.0f), y_(0.0f) {}
31 Point(float x, float y) : x_(x), y_(y) {}
32
33 float x() const { return x_; }
34 float y() const { return y_; }
35
36 float Length() const;
37 bool ScaleTo(float new_length);
38 static float Distance(const Point& p0, const Point& p1);
39
40 // Add more of these as needed:
41 Point operator+(const Point& other) const;
42 Point operator-(const Point& other) const;
43 Point operator*(float factor) const;
44
45 void Rotate90Clockwise();
46
47 private:
48 float x_, y_;
49};
50
52 public:
55
56 Quad(const Point& p0, const Point& p1, const Point& p2, const Point& p3)
57 : points_(4) {
58 points_[0] = p0;
59 points_[1] = p1;
60 points_[2] = p2;
61 points_[3] = p3;
62 }
63
64 const std::vector<Point>& points() const { return points_; }
65 const Point& point(int ix) const;
66
67 protected:
68 std::vector<Point> points_;
69};
70
71struct Rect {
72 float x, y, width, height;
73
74 Rect() {
75 x = y = 0.0f;
76 width = height = 1.0f;
77 }
78
79 Rect(float x, float y, float width, float height) {
80 this->x = x;
81 this->y = y;
82 this->width = width;
83 this->height = height;
84 }
85
86 bool ExpandToAspectRatio(float ratio);
87 bool ExpandToMinLength(float length);
88 bool ScaleWithLengthLimit(float factor, float max_length);
89};
90
91} // namespace filterfw
92} // namespace android
93
94#endif // ANDROID_FILTERFW_CORE_GEOMETRY_H
95```
| 663
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|
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| 2.984375
| 3
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CC-MAIN-2020-16
|
latest
|
en
| 0.538985
|
https://www.daniweb.com/programming/software-development/threads/90326/noob-python-regex-matchobject-group
| 1,540,173,330,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583514443.85/warc/CC-MAIN-20181022005000-20181022030500-00454.warc.gz
| 912,818,924
| 11,441
|
## jobs
``````s1 ='999'
s2 = '99.99'
mypat = re.compile('(^([0-9]+[.]+[0-9]+)|([0-9])\$)')
rate= mypat.search(s1)
print rate.group()
>>> print rate.groups()
('9', None, '9')
>>> rate=mypat.search(s2)
>>> print rate.group()
99.99``````
I need to get price = float(rate.group()). Price=999 or Price=99.99.
I think there is a problem when s1='999', when I do this:
``````>>> rate = mypat.search(s1)
>>> price = float(rate.group())
>>> print price
9.0``````
I only get 9.0, I need to get 999.0. I think my way of doing rate.group() is wrong, can someone help me so I can get 999.0.
## jobs
``````import re
s1 =' 25000 '
s2 = ' 5.5910 '
mypat = re.compile('[0-9]*(\.[0-9]*|\$)')
rate= mypat.search(s1)
print rate.group()
rate=mypat.search(s2)
print rate.group()
rate = mypat.search(s1)
price = float(rate.group())
print price``````
I get an error when it hits the whole number, that is in this format: s1 =' 25000 '
price = float(rate.group())
ValueError: empty string for float()
## ghostdog74 57
why do you think that regular expression is the way to go?
``````>>> s1 =' 25000 '
>>> s1.replace(" ","")
'25000'
>>> s2 = ' 5.5910 '
>>> s2.replace(" ","")
'5.5910'``````
regular expressions will be the last in mind for solving string problems in Python, unless really desperate
| 437
| 1,349
|
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| 2.625
| 3
|
CC-MAIN-2018-43
|
latest
|
en
| 0.673944
|
https://www.traditionaloven.com/tutorials/distance/convert-hong-kong-tsun-unit-to-gigametre-gm-unit.html
| 1,656,175,268,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103036077.8/warc/CC-MAIN-20220625160220-20220625190220-00021.warc.gz
| 1,106,902,187
| 17,353
|
Convert 寸 to Gm | Hong Kong tsun to gigameters
# length units conversion
## Amount: 1 Hong Kong tsun (寸) of length Equals: 0.000000000037 gigameters (Gm) in length
Converting Hong Kong tsun to gigameters value in the length units scale.
TOGGLE : from gigameters into Hong Kong tsun in the other way around.
## length from Hong Kong tsun to gigameter conversion results
### Enter a new Hong Kong tsun number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
How many gigameters are in 1 Hong Kong tsun? The answer is: 1 寸 equals 0.000000000037 Gm
## 0.000000000037 Gm is converted to 1 of what?
The gigameters unit number 0.000000000037 Gm converts to 1 寸, one Hong Kong tsun. It is the EQUAL length value of 1 Hong Kong tsun but in the gigameters length unit alternative.
寸/Gm length conversion result From Symbol Equals Result Symbol 1 寸 = 0.000000000037 Gm
## Conversion chart - Hong Kong tsun to gigameters
1 Hong Kong tsun to gigameters = 0.000000000037 Gm
2 Hong Kong tsun to gigameters = 0.000000000074 Gm
3 Hong Kong tsun to gigameters = 0.00000000011 Gm
4 Hong Kong tsun to gigameters = 0.00000000015 Gm
5 Hong Kong tsun to gigameters = 0.00000000019 Gm
6 Hong Kong tsun to gigameters = 0.00000000022 Gm
7 Hong Kong tsun to gigameters = 0.00000000026 Gm
8 Hong Kong tsun to gigameters = 0.00000000030 Gm
9 Hong Kong tsun to gigameters = 0.00000000033 Gm
10 Hong Kong tsun to gigameters = 0.00000000037 Gm
11 Hong Kong tsun to gigameters = 0.00000000041 Gm
12 Hong Kong tsun to gigameters = 0.00000000045 Gm
13 Hong Kong tsun to gigameters = 0.00000000048 Gm
14 Hong Kong tsun to gigameters = 0.00000000052 Gm
15 Hong Kong tsun to gigameters = 0.00000000056 Gm
Convert length of Hong Kong tsun (寸) and gigameters (Gm) units in reverse from gigameters into Hong Kong tsun.
## Length, Distance, Height & Depth units
Distance in the metric sense is a measure between any two A to Z points. Applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
# Converter type: length units
First unit: Hong Kong tsun (寸) is used for measuring length.
Second: gigameter (Gm) is unit of length.
QUESTION:
15 寸 = ? Gm
15 寸 = 0.00000000056 Gm
Abbreviation, or prefix, for Hong Kong tsun is:
Abbreviation for gigameter is:
Gm
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing Hong Kong tsun and gigameters ( 寸 vs. Gm ) measures exchange.
2. for conversion factors between unit pairs.
3. work with length's values and properties.
To link to this length Hong Kong tsun to gigameters online converter simply cut and paste the following.
The link to this tool will appear as: length from Hong Kong tsun (寸) to gigameters (Gm) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
| 936
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| 2.84375
| 3
|
CC-MAIN-2022-27
|
latest
|
en
| 0.539854
|
https://www.gamingtoday.com/casino_games/table_games/article/155-Craps_odds
| 1,606,691,500,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00057.warc.gz
| 682,421,021
| 13,361
|
Craps ‘odds’
Mar 9, 2010 9:28 PM
Beginning crap players often wonder whether they should take the "odds" bet once a point has been established by the shooter.
Moreover, many casinos now offer multiple odds bets, such as 2 times odds, 10 times or even 100 times odds.
Of course, you don’t literally get 100 times the odds when you bet on making the point; instead, you’re allowed to bet 100 times or whatever the multiplier is times your original pass line bet.
The reason this bet is so attractive is because there’s no house advantage, perhaps the only bet in the casino without any "vigorish."
The "free" or "true" odds bet was first allowed about 50 years ago as an accommodation to high rollers. There’s no marked spot on the craps layout, so it’s an acknowledged bet by the craps crew, overseen by the boxman and pit supervisors.
Once again, its true odds status is because the wager is paid off at "true" odds as its probability of occurrence, and is based on point numbers (after come-out).
True "take" odds against any point being successful on the Do Pass line are:
Point Do Don’t
4 or 10 2/1 1/2
5 or 9 3/2 2/3
6 or 8 6/5 5/6
The true "lay" odds are the opposite on the Don’t Pass line. There’s no house vigorish on the odds bet. Technically, it’s a separate side proposition bet. The house couldn’t care less, as it’s a break-even affair to them over the long haul.
After a point is established, a player can make odds bets, or even withdraw them, at any time before decision. In fact, he can do that as well with Don’t Pass line bets. He can also bet on the Pass line after the point is made – only, bets cannot be taken off the Pass line before decision.
Since about 85 percent of all craps table bets seem to be made on the Pass line, that’s where the house wants to take a shot, for its 1.414 percent vigorish or house advantage.
So, to qualify for an odds bet as a separate proposition, its maximum bet size is controlled by the size of the original line wager. That is whether the odds bet is 1x, 2x, 10x, or 100x, the maximum size is determined by the multiple allowed, but it can be less.
Thus if you had \$10 working on the Pass line, at 100x, you could place up to a \$1,000 wager as an odds bet on the space below the Pass line. If your point was 4 and you made it, you could win \$10 on your line bet and \$2,000 on your odds bet. If you laid the maximum odds on a Don’t 4, for a \$10 original wager, you’d place \$2,000 behind the line to win \$1,000, etc.
If you have subsequent intermediate bets via the Come/Don’t Come line, these are moved by the dealer to the number boxes. You must hand the dealer appropriate chips to "press" an odds bet.
The house isn’t stupid. The reason they spin their wheels on the odds bet is to encourage and give incentive to the player to make larger line bets (at vig) to qualify for larger odds bets.
While some players convince themselves they are reducing their vigorish by making odds bets, it’s a delusion as far as the house is concerned, as they couldn’t care less. They concentrate on taking a shot at your line bet.
The house, however, does place itself at some risk. Having some shooter with a long run throwing numbers (sans 7), it may take some time to recover. A big backup bankroll is a necessity. Also, they must use constant vigilance to keep from being "past posted." That’s where a cheating player will surreptitiously drop some chips behind the line if he sees the point is made.
Also, you need sharp dealers. Even though the maximum odds are 100x, say a player only bet 37x the odds and the point was a winning 6, which would pay 6-5 odds. This could entail tricky chip-counting arithmetic.
Let’s look at it this way. Suppose you’re in a bar and flipping a coin at true even odds for \$1 per toss. Another guy sides in and says he’ll flip you for \$10, still another says he’ll go for \$100, etc. That’s basically what you’re doing with the multiple odds bet. It may sound like a windfall, but it’s really not.
| 1,002
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|
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| 2.53125
| 3
|
CC-MAIN-2020-50
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latest
|
en
| 0.940744
|
http://gamedev.stackexchange.com/questions/tagged/physics?page=3&sort=newest&pagesize=15
| 1,469,336,556,000,000,000
|
text/html
|
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| 98,813,085
| 27,210
|
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# Electron accelerated between plates
1. Feb 3, 2014
### pengumon
1. The problem statement, all variables and given/known data
An electron is accelerated horizontally from rest in a television picture tube by a potential difference of 6380 V. It then passes between two horizontal plates 2.55 cm long and 5.47 cm apart that have a potential difference of 118 V. At what angle θ will the electron be traveling after it passes between the plates?
2. Relevant equations
Kinematics equations, E=F/q, V=PE/q, V= -Ed
3. The attempt at a solution
I first used conservation of energy to solve for the horizontal speed, Vq = (1/2)mv^2, to get 4.73E7 m/s. I then used that to find the time in between the plates, v = x/t getting t = 5.39E-10 s. Then I found the force due to the plates, F=Eq=-Vq/d= 3.45E-16 N = ma. Using this with the mass of the electron I solved for a, the vertical acceleration, to get a = 3.79E14 m/s^2 = (final vertical speed - initial vertical speed)/t. Since initial vertical speed = 0, a = v/t and I got the final vertical speed as 2.04E5 m/s. I then used (final speed)^2 = (initial speed)^2 +2ay to solve for y as the vertical distance travelled, which I got as 5.50E-5 m. I then set up a triangle so that tanθ = y/x = (5.50E-5)/(0.0255), and I finished with θ=0.124. This is the wrong answer, though and I'm not sure where I'm going wrong. Any elucidation is much appreciated!
2. Feb 3, 2014
### pengumon
Just kidding, I figured it out. I need to use the final horizontal and vertical velocities, not distances, to find the angle.
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