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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://everything2.com/user/cjeris/writeups/poset | 1,408,573,941,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500812867.24/warc/CC-MAIN-20140820021332-00220-ip-10-180-136-8.ec2.internal.warc.gz | 66,009,851 | 6,731 | Contraction of "partially ordered set", although in the decades since their introduction, posets have proven to be fundamental enough to deserve their own word, and not have to be a partial anything.
A poset is a set E equipped with a partial order, that is, a binary relation L (usually given the symbol "less than or equal to") which is
reflexive
x L x for every x ∈ E.
transitive
if x L y and y L z, then x L z.
(weakly) antisymmetric
if x L y and y L x, then x = y.
As Einar Hille put it in his nice little book Ordinary differential equations in the complex domain, "the fowl in a hen-yard are partially ordered under the pecking order."
Posets are important in several areas of mathematics and computer science, including logic, set theory, functional analysis, combinatorics, semantics and type theory, and the study of algorithms. | 202 | 841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-35 | latest | en | 0.944883 |
https://www.sarthaks.com/6740/prove-that-sin-2-6x-sin-2-4x-sin-2x-sin-10x | 1,611,568,332,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00023.warc.gz | 966,171,753 | 13,130 | # Prove that sin^2 6x – sin^2 4x = sin 2x sin 10x
1.9k views
edited
Prove that sin26x – sin24x = sin 2x sin10x
by (64.3k points)
selected
We have L.H.S
+1 vote
+1 vote | 78 | 173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-04 | longest | en | 0.89578 |
https://docs.snowflake.com/en/user-guide/querying-approximate-similarity | 1,721,236,412,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00091.warc.gz | 185,960,096 | 48,915 | # Estimating Similarity of Two or More Sets¶
Snowflake uses MinHash for estimating the approximate similarity between two or more data sets. The MinHash scheme compares sets without computing the intersection or union of the sets, which enables efficient and effective estimation.
## Overview¶
Typically, the Jaccard similarity coefficient (or index) is used to compare the similarity between two sets. For two sets, `A` and `B`, the Jaccard index is defined to be the ratio of the size of their intersection and the size of their union:
`J(A,B) = (A ∩ B) / (A ∪ B)`
However, this calculation can consume significant resources and time and, therefore, is not ideal for large data sets.
In contrast, the goal of the MinHash scheme is to estimate `J(A,B)` quickly, without computing the intersection or union.
## SQL Functions¶
The following Aggregate functions are provided for estimating approximate similarity using MinHash:
## Implementation Details¶
As detailed in MinHash (in Wikipedia):
“Let `H` be a hash function that maps the members of `A` and `B` to distinct integer values and, for any set `S`, define `H_min(S)` to be the minimal member of `S` with respect to `H`, i.e. the member `s` of `S` with the minimum value of `H(s)`, as expressed in the following equation:
`H_min(S) = argmin_{s in S} (H(s))`
If we apply `H_min` to both `A` and `B`, we will get the same value exactly when the element of the union `A ∪ B` with minimum hash value lies in the intersection `A ∩ B`. The probability of this being true is the above ratio, therefore:
`Pr[H_min(A) = H_min(B)] = J(A,B)`
Namely, assuming randomly chosen sets `A` and `B`, the probability that `H_min(A) = H_min(B)` holds is equal to `J(A,B)`. In other words, if `X` is the random variable that is 1 when `H_min(A) = H_min(B)` and 0 otherwise, then `X` is an unbiased estimator of `J(A,B)`. Note that `X` has a too large variance to be a good estimator for the Jaccard index on its own (since it is always 0 or 1).
The MinHash scheme reduces this variance by averaging together several variables constructed in the same way using `k` number of different hash functions.”
In order to achieve this, the MINHASH function initially creates `k` number of different hash functions and applies them to every element of each input set, retaining the minimum of each one, to produce a MinHash array (also called a MinHash state) for each set. More specifically, for `i = 0 to k-1`, the entry `i` of the MinHash array for set `A` (shown by `MinHash_A`) corresponds to the minimum value of hash function `H_i` applied to every element of set `A`.
Finally, an approximation for the similarity of the two sets `A` and `B` is calculated as:
`J_apprx(A,B) = (# of entries MinHash_A and MinHash_B agree on) / k`
## Examples¶
In the following example, we show how this scheme and the corresponding functions can be used in order to approximate the similarity of two sets of elements.
First, create two sample tables and insert some sample data:
```CREATE OR REPLACE TABLE mhtab1(c1 NUMBER,c2 DOUBLE,c3 TEXT,c4 DATE);
CREATE OR REPLACE TABLE mhtab2(c1 NUMBER,c2 DOUBLE,c3 TEXT,c4 DATE);
INSERT INTO mhtab1 VALUES
(1, 1.1, 'item 1', to_date('2016-11-30')),
(2, 2.31, 'item 2', to_date('2016-11-30')),
(3, 1.1, 'item 3', to_date('2016-11-29')),
(4, 44.4, 'item 4', to_date('2016-11-30'));
INSERT INTO mhtab2 VALUES
(1, 1.1, 'item 1', to_date('2016-11-30')),
(2, 2.31, 'item 2', to_date('2016-11-30')),
(3, 1.1, 'item 3', to_date('2016-11-29')),
(4, 44.4, 'item 4', to_date('2016-11-30')),
(6, 34.23, 'item 6', to_date('2016-11-29'));
```
Then, approximate the similarity of the two sets (tables `mhtab1` and `mhtab2`) using their MinHash states:
```SELECT APPROXIMATE_SIMILARITY(mh) FROM
((SELECT MINHASH(100, *) AS mh FROM mhtab1)
UNION ALL
(SELECT MINHASH(100, *) AS mh FROM mhtab2));
+----------------------------+
| APPROXIMATE_SIMILARITY(MH) |
|----------------------------|
| 0.79 |
+----------------------------+
```
The similarity index of these two tables is approximated as 0.79, as opposed to the exact value 0.8 (i.e., 4/5).
Language: English | 1,164 | 4,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-30 | latest | en | 0.882973 |
http://physics.highpoint.edu/~atitus/mandi/index.php?dcsid=2000&qid=2g30002&view=solution | 1,526,961,376,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864624.7/warc/CC-MAIN-20180522034402-20180522054402-00345.warc.gz | 228,250,537 | 4,795 | Matter & Interactions 2nd ed. Practice Problems Aaron Titus | High Point University home
(N)=# of solutions
2g30002 An electron deflected by charged plates. 2g30002 View Question | View Solution | Download pdf| View Video Question An electron enters a region of uniform electric field between two closely spaced, oppositely charged plates as shown below with an initial speed of m/s. Upon exiting the region, it has been deflected upward. The horizontal displacement of the electron through the plates is 5 cm, and the plates are separated a distance 5 mm. Figure: An electron deflected by oppositely charged plates. Sketch the electric field between the plates. Which plate is positively charged and which plate is negatively charged? Which plate is at a higher electric potential ? Sketch the path of the electron as it travels through the plates. If the vertical deflection of the electron is 1 mm, what is the potential difference across the plates?
Solution (a) Because the plates are closely spaced, the electric field between the plates is nearly uniform and the electric force on the electron will be constant throughout the region between the plates. The electron is deflected upward; therefore, there must be an upward force on the electron. Since and is negative, the electric field must point downward, as shown below. Figure: Electric field and force on an electron between plates. (b) The electric field points away from the positively charged plate and toward the negatively charged plate; therefore, the top plate must be positively charged and the bottom plate negatively charged. Figure: Top plate is positively charged and the bottom plate is negatively charged. (c) Because the electric field between the plates is constant, the electric potential varies linearly with . Electric field points from high potential to low potential. Therefore, the top plate is at a higher potential than the lower plate. (d) Because the force on the electron is in the +y direction, the electron's x-velocity will be constant and its y-velocity will increase in the +y direction. As a result, the electron's path will be a parabola, much like projectile motion. Figure: The path of an electron traveling between the plates is a parabola. (e) The potential difference between the plates for a constant electric field is Since the electric field points in the direction only, where is the plate separation and is the magnitude of the electric field between the plates. Thus, we need to calculate the electric field, using and we can use the motion of the electron and The Momentum Principle to get . So, the next step is to calculate using The Momentum Principle. Because the net force on the electron in the x-direction is zero, the and is constant. Thus, m/s. The time interval during which the electron travels across the region a displacement is Now, apply Newton's second law in the y-direction. The final y-velocity must be obtained by considering the vertical deflection . The average y-velocity is Solving for the final y-velocity gives Thus, the net force in the y-direction on the electron is The force of the electric field on the electron is the ONLY force acting on the electron. (The gravitational force on the electron is negligible.) Thus, the electric field is The potential difference across the plates is | 677 | 3,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.924503 |
https://community.qlik.com/t5/New-to-QlikView/ABC-Analysis-in-Qlikview/td-p/1202496/highlight/true | 1,580,297,991,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00519.warc.gz | 385,103,876 | 49,120 | New to QlikView
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Honored Contributor II
ABC Analysis in Qlikview
Hi All,
I'm trying to build ABC report for Sales analysis. having issue with calculated dimension(Grade) and based on user selection it should show bellow attached expected output screen shot, currently i'm getting wrong results (only the ABC summary report not correct result). if you have any solutions/idea's much appreciated for your help. thanks
Note: used below script for calculated dimension (Grade)
Cumulative%:
If(RangeSum(Peek('run_sum'), TotalNetPrice/TotalSales) < 0.6, 'A Stocks',
If((RangeSum(Peek('run_sum'), TotalNetPrice/TotalSales)) > 0.6 and (RangeSum(Peek('run_sum'), TotalNetPrice/TotalSales)) < 0.9 , 'B Stocks', 'C Stocks')) as Grade,
RangeSum(Peek('run_sum'), TotalNetPrice/TotalSales) as run_sum
Resident Sales
Order By TotalNetPrice desc;
FYI
Current output (refer the ABC summary report)
Tags (4)
1 Solution
Accepted Solutions
MVP
Re: ABC Analysis in Qlikview
Here is the new chart
Dimension
=Aggr(If(
rangesum(above(TOTAL (Sum({<ArticleNo>}TotalNetPrice)/Sum(TOTAL {<ArticleNo>} TotalNetPrice)),0,RowNo(Total)))<0.6,'A Stocks',if(
rangesum(above(TOTAL (Sum({<ArticleNo>}TotalNetPrice)/Sum(TOTAL {<ArticleNo>} TotalNetPrice)),0,RowNo(Total)))>0.6 and rangesum(above(TOTAL (Sum({<ArticleNo>}TotalNetPrice)/Sum(TOTAL {<ArticleNo>} TotalNetPrice)),0,RowNo(Total)))<0.9,'B Stocks',
'C Stocks'
)
), (ArticleNo, (=Sum({<ArticleNo>}TotalNetPrice), DESC)))
Expression
=Count(DISTINCT ArticleNo)
=Count(DISTINCT ArticleNo)/Count(TOTAL DISTINCT ArticleNo)
=Sum(TotalNetPrice)
=Sum(TotalNetPrice)/Sum(TOTAL TotalNetPrice)
28 Replies
MVP & Luminary
Re: ABC Analysis in Qlikview
I'm not quite sure how do you want to classify your articles but if you do it within the script it couldn't be depend on selections anymore - maybe it would be sufficient if you include more dimensions like country and category within your aggregation-load and on which you react in the following peek-load.
But I think you will rather need a calculated dimension within the gui - like described here: Calculated Dimensions and an approach like the following:
Recipe for an ABC Analysis
ABC analisys. As it should be...
Segmentation and custom dimension grouping
ABC Analysis to set and remember classification at runtime
I have adapted a chart from the first blog-posting which is based on a ranking and a second one (the left one) which based on the sales and which should be already quite near to your expected output except for the wrong sorting which will be quite difficult to create with a qlikview release prior to QV 12 - but with them it should be easy: The sortable Aggr function is finally here!
- Marcus
Honored Contributor II
Re: ABC Analysis in Qlikview
Hi Marcus,
Thank you very much for your help, in my requirement is to calculate the grade based on calculated measure and user selected dynamic filter value.
Let's assume we have list box filter as country,Category,Week,size etc, Based on the dimension we need to calculate the Grade (Calculated dimension). would it be possible to derive below summary report by Grade
problem statement:
The problem with calculated dimension is that unable to sort the RangeSum in the descending order and it's creating incorrect Grades.
Best Regards,
Deva
MVP & Luminary
Re: ABC Analysis in Qlikview
Have you tried my suggestion with a sortable aggr-function in QV 12 (maybe at first within a local installation)?
- Marcus
Honored Contributor II
Re: ABC Analysis in Qlikview
Hi Marcus,
Yes, I've tried with the suggested aggregated with sorting option as below,
=Aggr(if(
rangesum(above(SUM((TotalNetPrice)/SUM(TOTAL TotalNetPrice)),0,RowNo(Total)))<0.6,'A Stocks',if(
rangesum(above(TOTAL SUM((TotalNetPrice)/SUM(TOTAL TotalNetPrice)),0,RowNo(Total)))>0.6 and
rangesum(above(TOTAL SUM((TotalNetPrice)/SUM(TOTAL TotalNetPrice)),0,RowNo(Total)))<0.9,'B Stocks',
'C Stocks'
)
),ArticleNo, (TotalNetPrice ,(NUMERIC,DESCENDING)))
But still could not yield the desired output. I have attached the current and expected output. Please share your valuable suggestion/solution for this.
Note:
I would like to have below kind functionality where i would like to sort by expression. Basically a ORDER BY.
Aggr(sum(sales),(sum(sales),(NUMERIC, ASCENDING)))
This will open up lot of possibilities. One good example is Running Total/cumulative total as Calculated Dimension.
Thank you once again.
Best Regards, Deva
MVP & Luminary
Re: ABC Analysis in Qlikview
If I look on the syntax from the new aggr-function I don't think that an expression as sorting-attribute is possible. I haven't a QV 12 available so I couldn't test anything. Perhaps there is a possibility by extending the condition for a ranking-function maybe by nesting them within another aggr-function might work. For this I would like to invite swuehl to this posting who had a lot experience with aggr-functions or he might give useful hints for a different approach.
- Marcus
MVP
Re: ABC Analysis in Qlikview
As I mentioned to you before also, I think troyansky's approach is probably the only approach (AFAIK) available, I have created a sample for you using the approach he mentioned here: So How Many Customers Make Up Most of Your Sales?
The only problem is that you are not including the boundary conditions, but Pareto Select includes it. So when you say Sum(TotalNetPrice) of less than 50%, your chart shows 2 ArticleNo, but pareto includes the anything until it crosses the 50% mark. May be someone (marcus_sommer, Oleg, or someone else) might be able to help you take this even forward.
Oh and forgot to mention that every time you change your selections, you will need to click on Recalculate to make this work. Other method is to add the button actions as OnAnySelect action, but that could drastically slow down your dashboard. But it will definitely help you get rid of the recalculate button
Honored Contributor II
Re: ABC Analysis in Qlikview
Hi Sunny,
Thank you so much for the solution provided. As you said, i'm now facing with one issue. When using Pareto select, the percentage for StockA is 60. Say, 2 stocks are filtered out. The second pareto select's percentage is 90. What now happens is, this time it includes the StockA's 2 article count and also the articles which are less than 90%. When it comes to StockC, the similar thing happens. Is there anyway we can mention the percentage range instead of only the minimum value. Legends of qlikview - marcus_sommer, swuehl. Please pen down your valuable suggestion for this.
Thanks a lot in advance ! Have a wonderful weekend !
Regards, Deva
Honored Contributor II
Re: ABC Analysis in Qlikview
Can you help me to advise on this Pareto Percentage (%) range selection issue. Thanks
need to set like below,
StockA: <=60%
StockB: >60 and <=90%
StockC: >90%
Regards,
Deva
MVP
Re: ABC Analysis in Qlikview
I am out of ideas on how we can take this forward, lets get some other experts involved may be johnw, marcowedel | 1,767 | 7,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-05 | latest | en | 0.765948 |
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% greater than
Author Message
Manager
Status: GMAT Preperation
Joined: 04 Feb 2010
Posts: 103
Concentration: Social Entrepreneurship, Social Entrepreneurship
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WE: Consulting (Insurance)
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17 Sep 2010, 04:00
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Question Stats:
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A group of tourists needed to walk 45 miles. During the first day the group walked 19 miles. To complete the requires distence during the secound day. the group had to cover the distance which was approximatly what percent greater than the distance walked during the first day?
a)11%
b)16%
c)27%
d)37%
e)58%
[Reveal] Spoiler: OA
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Joined: 18 Jul 2010
Posts: 46
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17 Sep 2010, 04:04
45 miles
19 miles the first day so 26 miles left
the equation the solve is the following: 26 = 19*(1+x/100)
which lead to x = 700/19 = (approximately) 37 so ANS : D
Manager
Status: GMAT Preperation
Joined: 04 Feb 2010
Posts: 103
Concentration: Social Entrepreneurship, Social Entrepreneurship
GPA: 3
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Followers: 2
Kudos [?]: 127 [0], given: 15
Show Tags
17 Sep 2010, 04:08
I had the same explaination in the book ? whats the logic for "the equation the solve is the following: 26 = 19*(1+x/100)"?
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Joined: 18 Jul 2010
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17 Sep 2010, 04:15
1
KUDOS
you want to know: 'How much "GREATER" ?"
In other words: what is the percentage of 19 you have to add to 19 to get 26!
Is is clearer?
Be careful, the question is not "By how much percent do you have to multiply 19 to get 26"
If it was the question: you would have the following equation: 19*x/100 = 26.
And, without doing the calculation, I am pretty sure x = 137 !!
Why? Because 19*(137/100) = 19*(100/100 + 37/100) = 19*(1 + 37/100) and 37 is the answer of the previous question !!
Is it better ?
Re: % greater than [#permalink] 17 Sep 2010, 04:15
Display posts from previous: Sort by | 822 | 2,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2016-44 | latest | en | 0.90135 |
http://mathhelpforum.com/discrete-math/197045-finding-binomial-coefficient.html | 1,503,393,908,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110573.77/warc/CC-MAIN-20170822085147-20170822105147-00171.warc.gz | 261,152,490 | 10,746 | # Thread: Finding a Binomial Coefficient
1. ## Finding a Binomial Coefficient
What is the coefficient of x^5 in the binomial expression (x - 2x^-2)^20 when multiplied by (1+3x^3)
So far I have:
(20 over k)(-1)^k (2)^k x^(20-3k)
so then 20-3k = 5 such that k = 5
So then (20 over 5 )(-1)^5 (2)^5
But then I don't know how to handle multiplying it by ( 1 + 3x^3)
2. ## Re: Finding a Binomial Coefficient
$(x - 2x^{-2})^{20}=a_{20}x^{20}+a_{17}x^{17}+\dots+a_5x^5+a_ 2x^2+\dots+a_{-37}x^{37}+a_{-40}x^{-40}$ for some coefficients $a_k$. Also, $1+3x^3=b_0x^0+b_3x^3$ for $b_0=1$, $b_3=3$. The coefficient of $x^5$ in $(x - 2x^{-2})^{20}(1+3x^3)$ is $\sum_{i+k=5}a_ib_k$, i.e., $a_2b_3+a_5b_0$.
3. ## Re: Finding a Binomial Coefficient
Originally Posted by ThatPinkSock
What is the coefficient of x^5 in the binomial expression (x - 2x^-2)^20 when multiplied by (1+3x^3)
So far I have: (20 over k)(-1)^k (2)^k x^(20-3k)
so then 20-3k = 5 such that k = 5
So then (20 over 5 )(-1)^5 (2)^5
but then I don't know how to handle multiplying it by ( 1 + 3x^3)
I have a different take on this question from you and reply #2.
Each term in $\left(x-2x^{-2}\right)^{20}$looks like $\binom{20}{k}(-2)^{20-k}x^{3k-40}$
Now you want $x^5$ from $[x^{3k-40}]~\&~3x^3{x^{3k-40}}$
So $k=14~\&~k=15$ WHY? | 556 | 1,290 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-34 | longest | en | 0.665984 |
https://www.convertunits.com/from/stick/to/mile | 1,685,691,552,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00126.warc.gz | 767,349,798 | 12,885 | ## Convert stick to mile
stick mile
## More information from the unit converter
How many stick in 1 mile? The answer is 528. We assume you are converting between stick and mile. You can view more details on each measurement unit: stick or mile The SI base unit for length is the metre. 1 metre is equal to 0.32808398950131 stick, or 0.00062137119223733 mile. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between sticks and miles. Type in your own numbers in the form to convert the units!
## Quick conversion chart of stick to mile
1 stick to mile = 0.00189 mile
10 stick to mile = 0.01894 mile
50 stick to mile = 0.0947 mile
100 stick to mile = 0.18939 mile
200 stick to mile = 0.37879 mile
500 stick to mile = 0.94697 mile
1000 stick to mile = 1.89394 mile
## Want other units?
You can do the reverse unit conversion from mile to stick, or enter any two units below:
## Enter two units to convert
From: To:
## Definition: Mile
A mile is any of several units of distance, or, in physics terminology, of length. Today, one mile is mainly equal to about 1609 m on land and 1852 m at sea and in the air, but see below for the details. The abbreviation for mile is 'mi'. There are more specific definitions of 'mile' such as the metric mile, statute mile, nautical mile, and survey mile. On this site, we assume that if you only specify 'mile' you want the statute mile.
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 486 | 1,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-23 | latest | en | 0.867055 |
https://www.mrexcel.com/archive/vba/macro-to-find-missing-numbers/ | 1,550,904,960,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249490870.89/warc/CC-MAIN-20190223061816-20190223083816-00487.warc.gz | 882,467,196 | 8,934 | # Macro to find missing numbers
Posted by Homero on June 26, 2001 6:54 PM
Please if anybody can helpme on this: I have numbers on column A, 122,123,125,130, and I have in column D: 126,127,129, and I need to put in any cell a macro, to know what numbers are missing? In this formula I can put ( because I know) the small number (122) and the high number (130) and then click on a macro button and when the macro runs showme the numbers what are missing (in this example : 124 , 128)
thanks for all
Posted by Kevin James on June 26, 2001 9:48 PM
Working smart, not hard
Hi Homero,
Yes, that can be written in a macro, but why work so hard.
You can write a formula that merely says:
If Number is greater than Above+1, then "skipped number" else ""
For example
A1 contains 125
A2 contains 127
In B2 enter the formula:
=if(A2>(A1+1),"skipped number","")
Copy that down for however many numbers in have in column A.
Kevin
Posted by Homero on June 27, 2001 10:24 AM
Re: Working smart, not hard
Posted by Homero on June 27, 2001 10:27 AM
Re: Working smart, not hard
ok but I have this two columns with a lot of entrys ( I have in just only one column 950 numbers of invoices, and in the other column 560 entrys) and I cant check one by one if I miss to put any number of invoice.
thanks
Homer | 373 | 1,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-09 | latest | en | 0.935825 |
https://hellothinkster.com/blog/making-math-word-problems-accessible-for-fourth-graders/ | 1,726,119,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00407.warc.gz | 263,627,763 | 28,727 | # Making Math Word Problems Accessible for Fourth Graders
Last Updated on May 26, 2022 by Thinkster
Math word problems are so frustrating that even parents dread seeing the mathematical paragraphs appear on their fourth grader’s homework assignments. If you hate doing word problems as an adult, then you know they’re no picnic for your 9-year-old child. Approach the task with enthusiasm so that your student will follow suit. Adopting a winning attitude is half the battle in math. You need to remember that math is all about logic—and numbers. Follow the numbers, and the word problem isn’t quite as confusing any longer.
1. Teach a Logical Process
If your child is struggling with 4th grade math word problems, teach him a logical process to go through to determine what needs to be done. These steps should be:
• Question – Read the problem to determine what the question is.
• Information – Determine what information you have.
• Clue words – What words tell you the math process to use.
• Equation – Use the information, question and clue words to write an equation.
• Check your work – Does your answer make sense compared to the given information and the question?
Once your child can learn to use this process on a regular basis, you will find that he has much more success with word problems.
2. Teach Common Clue Words
Many fourth graders read a word problem and have no idea what to do with it. Yet most word problems on this grade level have clue words in them. Teach your child to identify those clue words, to help ease the struggle. Here are the clue words:
• Addition – Combined, increased, total of, sum, added to, together, plus
• Subtraction – Minus, less than, less, fewer than, difference, decreased, take away, more than
• Multiplication – Multiplied, product of, times, of
• Division – Divided by, into, per, quotient of, percent, out of, ratio of
Students who can pair these clue words with the math process will be much more successful with 4th grade math word problems.
3. Provide Practice
In order for a child to gain mastery of word problems, he needs practice. Simply throwing a sheet of word problems at him is not going to do the job, however. He needs a way to practice that is engaging and enticing.
Thinkster Math, an iPad-based math tutoring program, is an excellent option for this. Thinkster provides personalized math tutoring covering all math topics, including word problems, and designated tutors provide instruction to ensure that students make adjustments as needed when practicing. Students who use Thinkster are engaged in their learning and enjoy excellent gains in difficult math concepts, such as 4th grade math word problems.
4. Use Manipulates or Diagrams
Sometimes visualizing the problem can give the student the tools needed to solve it. For problems with small amounts, you can use math manipulates to help your child picture what is happening. For larger amounts or measurements, draw a diagram. This action gets additional learning processes involved and helps make the word problem a visual concept for the child to consider.
The key to making fourth grade math word problems solvable for your child is to make them understandable and then provide the right practice and support. It can also be worthwhile to get your child a math tutor. By doing so, you will be able to help your child succeed in mathematics.
Summary
Article Name
Making Math Word Problems Accessible for Fourth Graders
Description
Is your child struggling with 4th grade math word problems? Here is what you can do to help.
Author
Publisher Name
Thinkster Math
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Kyle's Converter > Flow > Acre-inches Per Hour > Acre-inches Per Hour to Million Acre-feet Per Year
Acre-inches Per Hour (ac in/h) Million Acre-feet Per Year (MAF/yr) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Million Acre-feet Per Year to Acre-inches Per Hour
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Acre-Inch per Hour:
Volume flow rate of 1 acre-inch per hour. Acre-inch being a volume of 660 feet by 66 feet by 1 inch; assuming an international foot of exactly 0.3048 meters. 3600 seconds per hour. 1 ac in/h ≈ 0.0285528203136 m3/s.
1 Million Acre-Feet per Year:
River flow rate of 1 000 000 acre-feet per year. Acre-foot being 660 feet by 66 feet by 1 foot; assuming an international foot of exactly 0.3048 meters. Assuming a civic year of 365 days. 1 MAF/yr ≈ 39.1134524843836 m3/s.
Conversions Table
1 Acre-inches Per Hour to Million Acre-feet Per Year = 0.000770 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0511
2 Acre-inches Per Hour to Million Acre-feet Per Year = 0.001580 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0584
3 Acre-inches Per Hour to Million Acre-feet Per Year = 0.002290 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0657
4 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0029100 Acre-inches Per Hour to Million Acre-feet Per Year = 0.073
5 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0037200 Acre-inches Per Hour to Million Acre-feet Per Year = 0.146
6 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0044300 Acre-inches Per Hour to Million Acre-feet Per Year = 0.219
7 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0051400 Acre-inches Per Hour to Million Acre-feet Per Year = 0.292
8 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0058500 Acre-inches Per Hour to Million Acre-feet Per Year = 0.365
9 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0066600 Acre-inches Per Hour to Million Acre-feet Per Year = 0.438
10 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0073800 Acre-inches Per Hour to Million Acre-feet Per Year = 0.584
20 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0146900 Acre-inches Per Hour to Million Acre-feet Per Year = 0.657
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40 Acre-inches Per Hour to Million Acre-feet Per Year = 0.029210,000 Acre-inches Per Hour to Million Acre-feet Per Year = 7.3
50 Acre-inches Per Hour to Million Acre-feet Per Year = 0.0365100,000 Acre-inches Per Hour to Million Acre-feet Per Year = 73
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Topic: Meander
Meander Valley Council - Home - Northern Tasmania Meander Valley Council - Home - Northern Tasmania Meander Valley is a large and diverse area in Northern Tasmania. It includes the mountains of the Great Western Tiers, extensive forests and productive land, the historic towns and villages of Deloraine and Westbury, and Prospect Vale, an urban suburb of Launceston and is home to the Country Club Resort. www.meander.tas.gov.au (97 words)
Meander - Wikipedia, the free encyclopedia A meander is a bend in a river, also known as an oxbow loop. When a meander gets cut off from the main stream body, an oxbow lake is formed. Due to the way a meander is formed, the river flows faster on the outside edges of the meander and slower along the inside edge.We see deposition on the inner edge because the river, now moving slowly, cannot hold the weight of the sediment it is carrying. en.wikipedia.org /wiki/Meander (351 words)
Meander (architecture) - Wikipedia, the free encyclopedia In art and architecture, a meander is a decorative border constructed from a continuous line, shaped into a repeated motif. Meanders are common decorative elements in Greek art and Roman art. The symbol of the meander is considered to be a Greek trademark and is commonly used as a symbol of Greek nationalism because it represents a link to an ancient past. en.wikipedia.org /wiki/Meander_(architecture) (234 words)
Meander (mathematics) - Wikipedia, the free encyclopedia In mathematics, a meander or closed meander is a self-avoiding closed curve which intersects a line a number of times. Two meanders are said to be equivalent if they are homeomorphic in the plane. Two open meanders are said to be equivalent if they are homeomorphic in the plane. en.wikipedia.org /wiki/Meander_(mathematics) (400 words)
Meander C Evidence Area C is distinguished from either A or B by the presence of a number of fairly obvious meander scars along the west side of the main river channel, suggesting previous positions of the river while the meander loop was still open. Between this old section of the main channel and the western arm of the meander loop is an elevated ridge of land that is roughly the same height as the land in the interior of the loop (ox-bow). This may suggest that the east arm of the meander loop was migrating westward at the neck prior to cut-off, thus producing a contracting, not expanding, neck. www.nysm.nysed.gov /research_collections/research/history/neck/reportc.html (1267 words)
CNF ARTICLES : ISSUE 1 - JONES - MEANDER In what we do on foot, meandering implies an aimless wandering, with the pleasant connotation that the very aimlessness of the wander is something freely, even happily, chosen. The meanders of water seem equally aimless, but are, it turns out, very regular in their irregularity -- although if you were walking along the bank of a meandering river, you might find that hard to believe. Meanders then may be the norm, not the exception. www.creativenonfiction.org /thejournal/articles/issue01/01jones_meander.htm (831 words)
Meander C Because the canal itself was about to cut through a large meander of the original river to the north, a new channel had to be cut to keep the river connected. It is supposed that by the beginning of the 18th century, this eastern arm of the meander had migrated westward to where it nearly joined the western arm, and at that time, it presented to navigators an obvious candidate for an easy cut-through channel. It is further suggested that after the meander was cut off, the active part of the river continued to migrate, with the channel moving easterly toward the ox-bow from the west. www.nysm.nysed.gov /research_collections/research/history/neck/meanderc.html (415 words)
incised meander - Hutchinson encyclopedia article about incised meander (Site not responding. Last check: 2007-10-08) The cathedral city of Durham is situated inside an incised meander of the River Wear. In a river, a deep steep-sided meander (bend) formed by the severe downwards erosion of an existing meander. Such erosion is usually brought about by the rejuvenation of a river (for example, in response to a fall in sea level). encyclopedia.farlex.com /incised+meander (132 words)
other labyrinths - meander Also known as the "Greek Key," the meander pattern is a path basic to the Classical labyrinths. To meander means to wander aimlessly, which suggests to some that the Greeks really didn't have the key. Meander Pattern in Albany, California built by Alex Champion. www.labyrinthsociety.org /html/other_labyrinths03.html (127 words)
Jefferson Iris with Meander White Opalescent Jefferson Iris with Meander White Opalescent Master Berry Bowl was produced in 1903-1906 from a mould designed by Jefferson Glass Company. Jefferson Iris with Meander Canary Opalescent Jelly Compote on Pedestal was produced in 1903-1906 from a mould designed by Jefferson Glass Company. Jefferson Iris with Meander Canary Opalescent Swung Vase (made from the spooner) was produced in 1903-1906 from a mould designed by Jefferson Glass Company. www.maxframe.com /GLASS/Jefferson/IrisWithMeander/WhiteOpalescent.htm (416 words)
Essay or Coursework - Water flows faster and is deeper on the outside edge of a meander curve This is the general theory, water tries to go in straight lines and so when it comes across a meander it carries on going straight. When this fastest flow reaches a meander it also travels in a straight line and so flows straight to the outside edge. There is also another reason and this is that the river has a lot of momentum and so when it goes round a corner this momentum drags it wider and to the outside edge of the meander. www.coursework.info /i/14690.html (391 words)
Midlands Meander The Midlands Meander is what weavers, potters, woodcrafters, leather workers, artists, metalworkers, box makers, herb growers, cheese makers, beer brewers and pianos have in common. As you tour the Midlands Meander you will also notice that this area is home to a number of the country’s most famous schools. The geographic boundaries of the Midlands Meander extend from Hilton in the south to Hidcote in the north and span from the Dargle Valley in the west to Currys Post in the east. www.drakensberg.net /midlands-meander.html (649 words)
The Valley Of the Meander Sometimes it turns back, sometimes turns left and right, meandering like a snake, and finally reaches the sea. Even the snakes and fish lose their way in this muddy and aimless flow; finding themselves on the shore and become a prey for storks. This kind of flowing has created the "meandering" motif which then had spread all over the world. www.meandertravel.com /kusadasi/meandervalley.htm (381 words)
ArtLex's Me-Mh page Meanders are among the ten classes of patterns. In the 1980s, Marden developed a fluid and expressive style that he said was influenced by nature, Eastern mysticism, and calligraphy. A detail of Wall Chart exhibits the sort of meandering design Wilkinson drew all over this work -- "what appears to be an obsessivly elaborate improvisational drawing of intestines, Pre-Columbian motifs, a labyrinth of inner and outer space," said a critic on artseensoho.com. www.artlex.com /ArtLex/Me.html (3916 words)
Meander Review Watching the video sample game on the Meander website, the authors are somehow able to (even in rubber gloves) always touch the ball so that it goes directly down the correct channel. Meander rarely seemed to make it back to its box, and several people asked me about acquiring copies. Because it is a bit of a pain to order and expensive to ship directly from the Netherlands, I've sent missives off to a couple of the online retailers to see about stocking a few copies. www.thegamesjournal.com /reviews/Meander.shtml (1119 words)
On the eddy-Kuroshio interaction: Meander formation process The simulation successfully reproduced the four phases of the interaction: (1) westward propagation of the eddy; (2) advection of the eddy by the Kuroshio; (3) meander formation; and (4) detachment of the eddy from the Kuroshio and their repetition. During the growth of the meander, the necessary barotropic kinetic energy is produced through the shallowing of the thermocline of the anticyclonic eddy as it elongates and splits. The growth of the meander ceases when the split anticyclonic eddies merge, the thermocline deepens, and the eddy detaches itself from the Kuroshio as a result of its own westward thrust. www.agu.org /pubs/crossref/2003/2002JC001583.shtml (484 words)
A Musical Meander (Site not responding. Last check: 2007-10-08) A Musical Meander differs from its ancestors; as the name additionally implies, the program usually contains an interlude at some point during which music with origins other than the Western Classical tradition is played. On other occasions the meander may be a long meander, when it might offer a discussion or interview regarding some particular musical issue, involving local or visiting experts. The meander may also provide live music performed in the studio by local or visiting artists. www4.semo.edu /krcu/meander.html (271 words)
The AmaJuba Birding Meander The Amajuba Birding Meander region is renowned for its water birds. One of the advantages of the Meander is that it lies on the crossroads between Gauteng and Durban, Durban and the Lowveld, and the Free State and Zululand. The Meander has been developed to give birdwatchers easier access to most of the wonderful birds to be found in the region. www.antbear.co.za /information/birding-meander.htm (423 words)
Meander Geometry Trends of Urbanized Beal Slough in Lincoln, Nebraska (Site not responding. Last check: 2007-10-08) Meander Geometry Trends of Urbanized Beal Slough in Lincoln, Nebraska We have determined the meander wavelength/bankfull-dominant discharge width ratios and the meander radius/bankfull-dominant discharge width ratios in relation to the longitudinal course on a selected urban reach of Beal Slough in Lincoln, Nebraska. During the geomorphic assessment of this reach, which was an integral part of designing the stream interventions, the specific locations of prominent meander fronts were located. www.pubs.asce.org /WWWdisplay.cgi?0410818 (151 words)
Meander Inn Bed & Breakfast - Nellysford, Virginia - BBOnline.com / Introduction This 85 years old Victorian farmhouse country inn is situated on 40 acres of horses grazed pasture and woods skirted by hiking trails and traversed by the... Nestled in the foothills of Virginia's Blue Ridge Mountains, in the peaceful surroundings of the Rockfish Valley, The Meander Inn opens its doors to guests looking to relax, refresh and regroup in the comfort of country farm-style living. The Meander Inn offers beautiful grounds with river, in a lush setting perfect for any wedding or your next private event. www.bbonline.com /va/meanderinn (404 words)
Canadian Geographic Magazine: Maps, Travel, Photography, Geography Contests, and Canadian Geographic Magazine ... The bends, known as meanders, reflect the way in which a river minimizes resistance to flow and spreads energy as evenly as possible along its course. Velocity is lowest along the bed and walls of streams or rivers because it is there that water encounters the most friction, and therefore the flow is reduced. Thus, a meandering pattern is created along the course of the river, with shallow water and point bars on the inside bends and steep banks on the outside. www.canadiangeographic.ca /landforms/meanders.asp (373 words)
Issue of August 12, 2003 A person who is "meandering" is strolling along in a very leisurely fashion, gawking at the scenery, stopping or changing course frequently, and, most importantly, showing absolutely no sign of urgency about getting wherever it is that they are supposedly going. Commonly applied to a person, the verb "meander" means to wander aimlessly, or, if there is a destination in mind, to get there by a very circuitous route. By the 17th century, we were using "meander" as a verb to describe the action of a person who takes a long, long time to get there. www.word-detective.com /081203.html (6025 words)
Jefferson Iris with Meander Canary Opalescent Jefferson Iris with Meander Canary Opalescent Master Berry Bowl was produced in 1903-1906 from a mould designed by Jefferson Glass Company. Jefferson Iris with Meander Canary Opalescent Plate (made from 4.5" berry bowl and flattened out) was thought to be produced in 1903-1906 from a mould designed by Jefferson Glass Company. Jefferson Iris with Meander Canary Opalescent Swung Vase (whimsey made from the Spooner mould) was produced in 1903-1906 from a mould designed by Jefferson Glass Company. www.maxframe.com /GLASS/Jefferson/IrisWithMeander/CanaryOpalescent.htm (492 words)
North Dakota Streams: Meander (Site not responding. Last check: 2007-10-08) The waters of the Sheyenne River are shown flowing through a stream meander and toward the bottom of the image. Higher water velocities on the outside (left side) of this stream meander have induced bank erosion, forming an oversteepened slope (cutbank). The net consequence of erosion on the outside of the meander and deposition on in the inside is the shift or migration of the meander loop in the direction of the cutbank. www.ndsu.nodak.edu /nd_geology/nd_streams/meander1.htm (126 words)
Meander Inn - Specials - Virginia B&B - Shenandoah Valley Lodging Virginia Packages must be paid in full on arrival at The Meander Inn. A non refundable deposit of 50% shall be made at the time of the reservation. The Meander Inn offers this and more, take advantage of our specials and prepared to be pampered. www.meanderinn.com /specials.htm (416 words)
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http://au.metamath.org/mpegif/limsupval2.html | 1,508,843,952,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00204.warc.gz | 30,865,957 | 13,319 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > limsupval2 Unicode version
Theorem limsupval2 11954
Description: The superior limit, relativized to an unbounded set. (Contributed by Mario Carneiro, 7-Sep-2014.) (Revised by Mario Carneiro, 8-May-2016.)
Hypotheses
Ref Expression
limsupval.1
limsupval2.1
limsupval2.2
limsupval2.3
Assertion
Ref Expression
limsupval2
Distinct variable group: ,
Allowed substitution hints: () () () ()
Proof of Theorem limsupval2
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 limsupval2.1 . . 3
2 limsupval.1 . . . 4
32limsupval 11948 . . 3
41, 3syl 15 . 2
5 imassrn 5025 . . . . 5
62limsupgf 11949 . . . . . . 7
7 frn 5395 . . . . . . 7
86, 7ax-mp 8 . . . . . 6
9 infmxrlb 10652 . . . . . . 7
109ralrimiva 2626 . . . . . 6
118, 10mp1i 11 . . . . 5
12 ssralv 3237 . . . . 5
135, 11, 12mpsyl 59 . . . 4
145, 8sstri 3188 . . . . 5
15 infmxrcl 10635 . . . . . 6
168, 15ax-mp 8 . . . . 5
17 infmxrgelb 10653 . . . . 5
1814, 16, 17mp2an 653 . . . 4
1913, 18sylibr 203 . . 3
20 limsupval2.3 . . . . . . 7
21 limsupval2.2 . . . . . . . . 9
22 ressxr 8876 . . . . . . . . 9
2321, 22syl6ss 3191 . . . . . . . 8
24 supxrunb1 10638 . . . . . . . 8
2523, 24syl 15 . . . . . . 7
2620, 25mpbird 223 . . . . . 6
27 infmxrcl 10635 . . . . . . . . . . 11
2814, 27mp1i 11 . . . . . . . . . 10
2921sselda 3180 . . . . . . . . . . . 12
3029ad2ant2r 727 . . . . . . . . . . 11
316ffvelrni 5664 . . . . . . . . . . 11
3230, 31syl 15 . . . . . . . . . 10
336ffvelrni 5664 . . . . . . . . . . 11
3433ad2antlr 707 . . . . . . . . . 10
35 ffn 5389 . . . . . . . . . . . . 13
366, 35mp1i 11 . . . . . . . . . . . 12
3721ad2antrr 706 . . . . . . . . . . . 12
38 simprl 732 . . . . . . . . . . . 12
39 fnfvima 5756 . . . . . . . . . . . 12
4036, 37, 38, 39syl3anc 1182 . . . . . . . . . . 11
41 infmxrlb 10652 . . . . . . . . . . 11
4214, 40, 41sylancr 644 . . . . . . . . . 10
43 simplr 731 . . . . . . . . . . . 12
44 simprr 733 . . . . . . . . . . . 12
45 limsupgord 11946 . . . . . . . . . . . 12
4643, 30, 44, 45syl3anc 1182 . . . . . . . . . . 11
472limsupgval 11950 . . . . . . . . . . . 12
4830, 47syl 15 . . . . . . . . . . 11
492limsupgval 11950 . . . . . . . . . . . 12
5049ad2antlr 707 . . . . . . . . . . 11
5146, 48, 503brtr4d 4053 . . . . . . . . . 10
5228, 32, 34, 42, 51xrletrd 10493 . . . . . . . . 9
5352expr 598 . . . . . . . 8
5453rexlimdva 2667 . . . . . . 7
5554ralimdva 2621 . . . . . 6
5626, 55mpd 14 . . . . 5
576, 35ax-mp 8 . . . . . 6
58 breq2 4027 . . . . . . 7
5958ralrn 5668 . . . . . 6
6057, 59ax-mp 8 . . . . 5
6156, 60sylibr 203 . . . 4
6214, 27ax-mp 8 . . . . 5
63 infmxrgelb 10653 . . . . 5
648, 62, 63mp2an 653 . . . 4
6561, 64sylibr 203 . . 3
66 xrletri3 10486 . . . 4
6716, 62, 66mp2an 653 . . 3
6819, 65, 67sylanbrc 645 . 2
694, 68eqtrd 2315 1
Colors of variables: wff set class Syntax hints: wi 4 wb 176 wa 358 wceq 1623 wcel 1684 wral 2543 wrex 2544 cin 3151 wss 3152 class class class wbr 4023 cmpt 4077 ccnv 4688 crn 4690 cima 4692 wfn 5250 wf 5251 cfv 5255 (class class class)co 5858 csup 7193 cr 8736 cpnf 8864 cxr 8866 clt 8867 cle 8868 cico 10658 clsp 11944 This theorem is referenced by: mbflimsup 19021 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1533 ax-5 1544 ax-17 1603 ax-9 1635 ax-8 1643 ax-13 1686 ax-14 1688 ax-6 1703 ax-7 1708 ax-11 1715 ax-12 1866 ax-ext 2264 ax-sep 4141 ax-nul 4149 ax-pow 4188 ax-pr 4214 ax-un 4512 ax-cnex 8793 ax-resscn 8794 ax-1cn 8795 ax-icn 8796 ax-addcl 8797 ax-addrcl 8798 ax-mulcl 8799 ax-mulrcl 8800 ax-mulcom 8801 ax-addass 8802 ax-mulass 8803 ax-distr 8804 ax-i2m1 8805 ax-1ne0 8806 ax-1rid 8807 ax-rnegex 8808 ax-rrecex 8809 ax-cnre 8810 ax-pre-lttri 8811 ax-pre-lttrn 8812 ax-pre-ltadd 8813 ax-pre-mulgt0 8814 ax-pre-sup 8815 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3or 935 df-3an 936 df-tru 1310 df-ex 1529 df-nf 1532 df-sb 1630 df-eu 2147 df-mo 2148 df-clab 2270 df-cleq 2276 df-clel 2279 df-nfc 2408 df-ne 2448 df-nel 2449 df-ral 2548 df-rex 2549 df-reu 2550 df-rmo 2551 df-rab 2552 df-v 2790 df-sbc 2992 df-csb 3082 df-dif 3155 df-un 3157 df-in 3159 df-ss 3166 df-nul 3456 df-if 3566 df-pw 3627 df-sn 3646 df-pr 3647 df-op 3649 df-uni 3828 df-iun 3907 df-br 4024 df-opab 4078 df-mpt 4079 df-id 4309 df-po 4314 df-so 4315 df-xp 4695 df-rel 4696 df-cnv 4697 df-co 4698 df-dm 4699 df-rn 4700 df-res 4701 df-ima 4702 df-iota 5219 df-fun 5257 df-fn 5258 df-f 5259 df-f1 5260 df-fo 5261 df-f1o 5262 df-fv 5263 df-ov 5861 df-oprab 5862 df-mpt2 5863 df-1st 6122 df-2nd 6123 df-riota 6304 df-er 6660 df-en 6864 df-dom 6865 df-sdom 6866 df-sup 7194 df-pnf 8869 df-mnf 8870 df-xr 8871 df-ltxr 8872 df-le 8873 df-sub 9039 df-neg 9040 df-ico 10662 df-limsup 11945
Copyright terms: Public domain W3C validator | 2,690 | 5,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-43 | latest | en | 0.104076 |
https://livecasinoguru.com/strategies/roulette/great-square-system.html | 1,579,990,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251681412.74/warc/CC-MAIN-20200125191854-20200125221854-00001.warc.gz | 540,826,242 | 13,356 | GREAT SQUARE SYSTEM
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• System: for 3 up to 7 numbers.
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INTRODUCTION
As most professional gamblers know, the probability that any one of the 37 numbers will be drawn in a single spin is 2.7, or 1/37. The "law of the third" tells us that in a complete cycle of 37 spins, there will be only 24 numbers drawn one or more times, while the other 13 will not appear.
Negative probability is (36/37)37=0.3628 = 36.28%, and 37*0.3628=13.42. 13.42, as you can see, is more likely 14 than 13 numbers.
To obtain more precision, years ago, we found a cycle of 39 spins instead of a cycle of 37. Negative probability is (36/37)39=0.3435 or 34.35%, and 37*0.3435=12.7.
This "long cycle" gives us 13 numbers that will not come out. The following system is based on the "cycle 39" (the Cycle 39 is applicable also for the OO wheel).
It uses only numbers in frequency (normal or super), and excludes the absent ones (joined probability distribution).
THE RULES
Every Game is a complete game - do not use the same Table for different games.
For every LIVE Roulette Game you need a total of 24 spins before to bet and others 5-12 spins (in the worst cases) before winning. Before using the table below for the first time it is important to make further copies of it, since one Table is used and discarded for each game (this advice will avoid you to make errors while recording spins).
The collecting zone
We need to note a total of 24 spins.
We need only the numbers drawn 2 (and only 2) times.
In fact we will bet them immediately according on these rules:
The attack zone
The number of spins to bet is calculated as follows: 36 /n, where n is number of targets obtained. For example, if there are 4 targets then play should stop if no win occurs by the 9th spin, if there are 5 targets play for 8 spins (36/5=7,2 that means 7, if there are 6 targets play for 6 spins, and so on.
Number of targets: Spins to bet/ action
Less than 3 Do not bet
3 12
4 9
5 7
6 6
7 5
More than 7 Do not bet
Usually you will bet 4 or 5 number (the 80% of the times) but sometimes you can also obtain 6-7 or 3 numbers to bet. Remember that you must never bet less than 3 number or more than 7 numbers if they happen. In this (very rare) case start a new game.
The second attack chance
In the most numbers of games you will win before the end of the attack zone. In this case is possible to continue betting on the remaining numbers.
You must only remember that you don't have to continue betting after the end of the attack zone.
You don't have to continue to play if the numbers to bet are less than 3, in fact if you are playing a game and after 24 spins (collecting zone) you obtain 3 numbers to bet for 12 successive spins (attack zone), if you win after 3 spins you remain with only 2 numbers bet, in this case you can't use the second chance.
Instead, if you are playing another game and after the collecting zone you obtain 6 numbers to bet and you win for the first time at the second spin (spin number 26 from the beginning of the game), you can use the second chance for: 7 – 2 = 5 successive spins (7 because we are only play the remaining 5 numbers).
Number of targets: Spins to bet/ action Second chance
Less than 3 Do not bet -
3 12 Not allowed
4 9 12-X
5 7 9-X
6 6 7-X
7 5 6-X
More than 7 Do not bet - Where X is the spins of the attack zone when you won the first time
EXAMPLES
3,6,14,2,27,20,4,2,17,6,11,31,20,34,1,31,13,25,3,17,31,5,5,7 (end of first 24 spins) 15,20,18,15,17 (theoretical attack zone spins)
In this case after the first 24 spins we have 6 numers to bet (2,3,5,6,17,20) for a maximum of 6 spins. At the second spin, we win with the number 20! We want to continue with the second chance, we bet only 5 number (2,3,5,6,17) for 7-2=5 successive spins and we will win again after 3 spins! (at the 29 spin from the beginning of the game).
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Special note: There is no guarantee to the amount of money you will win or lose. All casino games are entertaining games of pure chance and luck. LiveCasinoGuru.com cannot be held responsible for persons having a lot of bad luck or taking risky chances. | 1,354 | 4,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-05 | longest | en | 0.913736 |
https://puzzling.stackexchange.com/questions/28171/titanicker-tic-tac-toe | 1,721,566,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00177.warc.gz | 411,398,644 | 40,835 | # Titanicker Tic-Tac-Toe
This is something of a follow-on to Hugh Meyer's enjoyable "Titanic Tic-Tac-Toe." Fortunately, our intrepid heroes Xavier and Oliver have wrested away a bit of wreckage that some fellow named Jack was holding on to and, as they sit in the icy North Atlantic, they pass the time by playing their favourite game.
Since they are now subject to the random undulations of the waves, they've had to change up their play again. Their rules are now as follows:
• Each player may, before placing a marker, shift the board in any direction, not just to the right.
• As before, the player may only shift a row or column provided that it either (a) removes one of his tokens or (b) does not remove any of his opponent's.
• As before, a player may not play on the same square in consecutive turns.
After some discussion and gentlemanly concession, Xavier and Oliver have added the following stipulation:
• The player may not make a play so as to return the board to state that it has previously been after his turn. If that is the only such move they can make, they lose.
To provide some mathematical clarity, if $S_i$ is the state of the board after the $i$ turns, then for $i \ne j$, $S_i = S_j$ only if $i \not\equiv j \mod 2$.
Their hope is that this will keep them from getting caught in a vicious cycle.
So, will Oliver win a game before the Carpathia arrives to save them? (Will Xavier?)
• I knew I should have copyrighted! :) I was actually thinking of placing them on the Lucitania with the added rule that the second player (only) has a one-time opportunity to rotate the board 90 degrees at the start of any turn. I hoped to be able to work out the consequences in my head. Should anyone be interested, feel free to give it a shot. Commented Mar 2, 2016 at 8:13
• Also, if I could ask for some clarification. I assume that the final rule means that you cannot recreate a state that has existed at any time during the current game. Also that "state" includes whose turn it is to move. Too long for a comment, but it is possible to produce a situation with x's and o's in the same positions but it is the opposite player's move. Is this legal or not? Commented Mar 2, 2016 at 10:40
• Does the rule from the previous question, "A player may not play two successive moves on the same square", still apply? Commented Mar 2, 2016 at 15:46
• @HughMeyers At the moment, I'm leaving that possibility open.
– Matt
Commented Mar 2, 2016 at 15:50
• If a player places their piece in the last free space, without completing a row, does the game end in a draw, or is the next player forced to start their move with a tilt? Commented Mar 2, 2016 at 20:39
There can't be a draw because if the board is full, either you can move the board and get new places to play, or you can't because all positions you can go where already used, and you lose.
So the game always ends with a win for first or second player. So, as it is a finite two players game with no draw, there is exactly one player that has a winning strategy.
Now, you can remark that having a marker on the board is always better than not having one : it allows you to make a line, it prevents the other to make a line through it and it allows you to shift the board. Hence, the first player has a winning strategy, because if it hasn't, then he could play the strategy of the second player before him (it's the same argument than classical tic tac toe in fact).
Hence the first player to play has a winning strategy.
Now, knowing it exists doesn't give it, and it can be a very complex one...
• I'm not convinced that a strategy-stealing argument is valid here because of the rule against repeating positions.
– f''
Commented Jul 7, 2016 at 14:28
there are 3^9 possible boards, (and not all of those are reachable) after at-most a few thousand moves someone will have won, either conventionally, or by snookering their opponent | 947 | 3,911 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-30 | latest | en | 0.97518 |
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# VB-Question 4
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### VB-Question 4 - Page Text Content
BC: Why is a quadratic function more reasonable for this problem than a linear function would be? A quadratic function is more reasonable for this problem than a linear function because it is a parabola and it is declining down to zero. It hits more points on the line than a linear function would. The water doesn't drain at a constant rate. It drains faster as time goes on. Also, if it were a linear function, the line would continue down into negative and the bathtub can't hold negative liters.
FC: Splish! Splash! I'm Taking A Bath! Question #4 Victoria Binda
1: Question #4: Assume that the number of liters of water remaining in the bathtub varies quadratically with the number of minutes which have elapsed since you pulled the plug. -If the tub 38.4, 21.6, and 9.6 liters remaining at 1,2, and 3 respectively, since you pulled the plug, write an equation expressing liter in terms of time. -How much water was in the tub when you pulled the plug? -When will the tub be empty? -In the real world, the number of liters would never be negative. What is the lowest number of liters the model predicts? Is this number reasonable? -Why is a quadratic function more reasonable for this problem than a linear function would be?
2: If the tub has 38.4, 21.6, and 9.6 liters remaining at 1,2, and 3 respectively, since you pulled the plug, write an equation expressing liters in the terms of time. y=2.4x^2-24x+60 This equation was found by first creating 3 ordered pairs. (1,38.4) (2,21.6) (3,9.6) From here, the data needs to be put into a calculator. On the calculator, press After pressing STAT, a menu will come up. When you press edit, L1 and L2 columns will come up. Enter the minutes, 1,2, and 3, into L1 and enter the liters, 38.4,21.6, and 9.6, into L2. From here, calculate the quadratic regression. It will be in the set up of "ax^2+bx+c" When finding the quadratic regression, it can be viewed as a quadratic function on the graph.
3: How much water was in the tub when you pulled the plug? When the plug was pulled, there was 60 liters in the bathtub. This answer was found by looking at the equation, which expresses liters. The equation is in the setup of ax^2+bx+c. You can find out how much water was in the tub when the plug was pulled by looking at "c." In this equation, c is equal to 60.
4: When will the tub be empty? The tub will be empty when there is 0 liters of water left. There will be 0 liters of water in the tub at 5 minutes. This answer was found by looking at the STAT plot data. L2 is the column for liters in the bathtub.In L2, find where it says 0. In L1, it tells the time. According to data, it takes 5 minutes for there to be 0 liters of water left.
5: In the real world, the number of liters would never be negative. What is the lowest number of liters the model predicts? Is this number reasonable? The lowest number of liters the model predicts is 0. Yes, zero is a reasonable number because the tub is empty at zero. It wouldn't be a negative because once the tub is empty, it is empty. You can't have negative liters of water.
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## Statistics
Find the alternative would be selected according to expected value and utility.
For the payoff table below the decision maker will use P (s1) = .15, P (s2) = .5, and P (s3) = .35.
s1 s2 s3 d1 -5000 1000 10,000 d2 -15,000 -2000 40,000
1) describe what alternative would be chosen according to expected value?
2) For the lottery having a payoff of 40000 with probability p and -15,000 with probability (1-p) the decision maker expressed the following indifference probabilities.
Payoff Probability 10,000 .85 1000 .60 -2000 .53 -5000 .50
Let U(40,000) = 10 and U(-15,000) = 0 also find the utility value for each payoff.
c. describe what alternative would be chosen according to expected utility?
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M922110
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02-01-2015, 09:17 AM (This post was last modified: 02-01-2015 09:19 AM by parisse.)
Post: #11
parisse Senior Member Posts: 1,135 Joined: Dec 2013
RE: Problem calculating an expression with roots
(01-31-2015 09:59 PM)retoa Wrote: Wow, how to complicate easy things!
Actually you only have to sum/subtract the exponents...
I'm comparing HP Prime and TI nspire CAS, I have to choose one to propose for the students in the school where I work, so I try some calculation we do at school with both of them.
The TI transforms
$$\frac{\sqrt{3}*\sqrt[3]{2*x}}{\sqrt[4]{3*x}}$$
to
$$2^{1/3}*3^{1/4}*x^{1/12}$$
in no time.
I will not say that I will choose TI because of that, the prime is better then TI in many other things, but I can not understand how a CAS calculator can not simplify something so easy. If the TI does it, then it should be possible also on the Prime hardware.
Of course it's possible, but it's not a priority. It applies only to very specific expressions that you can do by hand. I did not make my CAS with the objective to solve american schoolbool exercices, because I don't know them. It does not mean I will not adapt, it depends on time constraints and priorities.
Quote:Does it mean that the Prime can not simplify any expressions with roots at the denominator or with negative fractional exponents? That would be really limiting for the use in a school...
It's exactly the reverse. The algorithm for simplifying fractional powers on the Prime can handle expressions with + (or -), it's therefore more powerful. Simplifying the example with this algorithm takes too much time on a calc, but it works on a desktop.
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Messages In This Thread Problem calculating an expression with roots - retoa - 01-30-2015, 09:44 PM RE: Problem calculating an expression with roots - salvomic - 01-30-2015, 09:57 PM RE: Problem calculating an expression with roots - retoa - 01-30-2015, 10:02 PM RE: Problem calculating an expression with roots - Mark Hardman - 01-30-2015, 10:37 PM RE: Problem calculating an expression with roots - retoa - 01-30-2015, 10:49 PM RE: Problem calculating an expression with roots - Mark Hardman - 01-30-2015, 11:03 PM RE: Problem calculating an expression with roots - parisse - 01-31-2015, 08:16 AM RE: Problem calculating an expression with roots - rprosperi - 01-31-2015, 04:00 PM RE: Problem calculating an expression with roots - Tim Wessman - 01-31-2015, 11:08 PM RE: Problem calculating an expression with roots - parisse - 02-01-2015, 02:40 PM RE: Problem calculating an expression with roots - retoa - 01-31-2015, 09:59 PM RE: Problem calculating an expression with roots - parisse - 02-01-2015 09:17 AM RE: Problem calculating an expression with roots - retoa - 02-02-2015, 11:49 AM RE: Problem calculating an expression with roots - parisse - 02-02-2015, 12:53 PM RE: Problem calculating an expression with roots - retoa - 02-03-2015, 10:55 AM RE: Problem calculating an expression with roots - lrdheat - 02-01-2015, 11:56 PM RE: Problem calculating an expression with roots - Gilles - 02-02-2015, 03:48 PM RE: Problem calculating an expression with roots - parisse - 02-02-2015, 04:40 PM RE: Problem calculating an expression with roots - Tim Wessman - 02-02-2015, 06:08 PM RE: Problem calculating an expression with roots - parisse - 02-02-2015, 06:25 PM RE: Problem calculating an expression with roots - salvomic - 02-02-2015, 06:10 PM RE: Problem calculating an expression with roots - Snorre - 02-02-2015, 08:20 PM RE: Problem calculating an expression with roots - Han - 02-02-2015, 08:51 PM RE: Problem calculating an expression with roots - salvomic - 02-02-2015, 08:52 PM RE: Problem calculating an expression with roots - Tim Wessman - 02-02-2015, 09:35 PM RE: Problem calculating an expression with roots - Snorre - 02-02-2015, 09:35 PM RE: Problem calculating an expression with roots - parisse - 02-03-2015, 06:40 AM RE: Problem calculating an expression with roots - Snorre - 02-03-2015, 11:59 AM RE: Problem calculating an expression with roots - retoa - 02-03-2015, 01:28 PM RE: Problem calculating an expression with roots - parisse - 02-03-2015, 02:36 PM RE: Problem calculating an expression with roots - salvomic - 05-15-2015, 01:40 PM RE: Problem calculating an expression with roots - akmon - 05-15-2015, 02:43 PM RE: Problem calculating an expression with roots - salvomic - 05-15-2015, 03:08 PM RE: Problem calculating an expression with roots - akmon - 05-15-2015, 03:22 PM RE: Problem calculating an expression with roots - salvomic - 05-15-2015, 03:35 PM RE: Problem calculating an expression with roots - akmon - 05-15-2015, 03:52 PM RE: Problem calculating an expression with roots - salvomic - 05-15-2015, 03:53 PM
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# Untitled
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1. module test;
2.
3. import std.algorithm;
4. import std.stdio;
5. import std.traits;
6. import std.range;
7.
8. struct Permutations(R) {
9. ElementType!R[] _input, _perm;
10. size_t k, n;
11.
12. this(R r) {
13. _input = array(r);
14. _perm = array(r);
15. n = _perm.length;
16. k = n;
17. }
18.
19. this(R r, size_t elems) {
20. _input = array(r);
21. _perm = array(r);
22. n = min(elems, _perm.length);
23. k = n;
24. }
25.
26. ElementType!R[] front() { return _perm;}
27. bool empty() { return (n == 1 && k == 0 )|| (n>1 && k <=1);}
28. @property Permutations save() { return this;}
29. void popFront() {
30. k = n;
31. if (k==0) { n=1;} // permutation of an empty range or of zero elements
32. else {
33. C3: _perm = _perm[1..k] ~ _perm[0] ~ _perm[k..\$];
34. if (_perm[k-1] == _input[k-1]) {
35. k--;
36. if (k > 1) goto C3;
37. }
38. }
39. }
40. }
41.
42. /// ditto
43. Permutations!R permutations(R)(R r) if (isDynamicArray!R) {
44. return Permutations!R(r);
45. }
46.
47. /// ditto
48. Permutations!R permutations(R)(R r, size_t n) if (isDynamicArray!R) {
49. return Permutations!R(r, n);
50. }
51.
52. /// ditto
53. Permutations!(ElementType!R[]) permutations(R)(R r) if (!isDynamicArray!R && isForwardRange!R && !isInfinite!R) {
54. return Permutations!(ElementType!R[])(array(r));
55. }
56.
57. /// ditto
58. Permutations!(ElementType!R[]) permutations(R)(R r, size_t n) if (!isDynamicArray!R && isForwardRange!R && !isInfinite!R) {
59. return Permutations!(ElementType!R[])(array(r), n);
60. }
61.
62.
63. void main()
64. {
65. string[] words = ["foo", "bar", "doo"];
66. auto p1 = permutations(words);
67. writeln(p1);
68. } | 714 | 2,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2013-20 | latest | en | 0.109217 |
https://www.dataunitconverter.com/nibble-per-minute-to-mebibit-per-minute | 1,718,318,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00076.warc.gz | 678,847,943 | 16,502 | # Nibble/Min to Mibit/Min → CONVERT Nibbles per Minute to Mebibits per Minute
expand_more
info 1 Nibble/Min is equal to 0.000003814697265625 Mibit/Min
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
Nibble/Min
## Nibbles per Minute (Nibble/Min) Versus Mebibits per Minute (Mibit/Min) - Comparison
Nibbles per Minute and Mebibits per Minute are units of digital information used to measure storage capacity and data transfer rate.
Nibbles per Minute is one of the very "basic" digital unit where as Mebibits per Minute is a "binary" unit. One Nibble is equal to 4 bits. One Mebibit is equal to 1024^2 bits. There are 262,144 Nibble in one Mebibit. Find more details on below table.
Nibbles per Minute (Nibble/Min) Mebibits per Minute (Mibit/Min)
Nibbles per Minute (Nibble/Min) is a unit of measurement for data transfer bandwidth. It measures the number of Nibbles that can be transferred in one Minute. Mebibits per Minute (Mibit/Min) is a unit of measurement for data transfer bandwidth. It measures the number of Mebibits that can be transferred in one Minute.
## Nibbles per Minute (Nibble/Min) to Mebibits per Minute (Mibit/Min) Conversion - Formula & Steps
The Nibble/Min to Mibit/Min Calculator Tool provides a convenient solution for effortlessly converting data rates from Nibbles per Minute (Nibble/Min) to Mebibits per Minute (Mibit/Min). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Nibble) and target (Mebibit) data units.
Source Data Unit Target Data Unit
Equal to 4 bits
(Basic Unit)
Equal to 1024^2 bits
(Binary Unit)
The formula for converting the Nibbles per Minute (Nibble/Min) to Mebibits per Minute (Mibit/Min) can be expressed as follows:
diamond CONVERSION FORMULA Mibit/Min = Nibble/Min x 4 ÷ 10242
Now, let's apply the aforementioned formula and explore the manual conversion process from Nibbles per Minute (Nibble/Min) to Mebibits per Minute (Mibit/Min). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Mebibits per Minute = Nibbles per Minute x 4 ÷ 10242
STEP 1
Mebibits per Minute = Nibbles per Minute x 4 ÷ (1024x1024)
STEP 2
Mebibits per Minute = Nibbles per Minute x 4 ÷ 1048576
STEP 3
Mebibits per Minute = Nibbles per Minute x 0.000003814697265625
Example : By applying the previously mentioned formula and steps, the conversion from 1 Nibbles per Minute (Nibble/Min) to Mebibits per Minute (Mibit/Min) can be processed as outlined below.
1. = 1 x 4 ÷ 10242
2. = 1 x 4 ÷ (1024x1024)
3. = 1 x 4 ÷ 1048576
4. = 1 x 0.000003814697265625
5. = 0.000003814697265625
6. i.e. 1 Nibble/Min is equal to 0.000003814697265625 Mibit/Min.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Nibbles per Minute to Mebibits per Minute using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Nibble ?
A Nibble is a unit of digital information that consists of 4 bits. It is half of a byte and can represent a single hexadecimal digit. It is used in computer memory and data storage and sometimes used as a basic unit of data transfer in certain computer architectures.
arrow_downward
#### What is Mebibit ?
A Mebibit (Mib or Mibit) is a binary unit of digital information that is equal to 1,048,576 bits and is defined by the International Electro technical Commission(IEC). The prefix 'mebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'megabit' (Mb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Nibbles per Minute (Nibble/Min) to Mebibits per Minute (Mibit/Min)
Apply the formula as shown below to convert from 1 Nibbles per Minute (Nibble/Min) to Mebibits per Minute (Mibit/Min).
A B C
1 Nibbles per Minute (Nibble/Min) Mebibits per Minute (Mibit/Min)
2 1 =A2 * 0.000003814697265625
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Nibbles per Minute (Nibble/Min) to Mebibits per Minute (Mibit/Min) Conversion
You can use below code to convert any value in Nibbles per Minute (Nibble/Min) to Nibbles per Minute (Nibble/Min) in Python.
nibblesperMinute = int(input("Enter Nibbles per Minute: "))
mebibitsperMinute = nibblesperMinute * 4 / (1024*1024)
print("{} Nibbles per Minute = {} Mebibits per Minute".format(nibblesperMinute,mebibitsperMinute))
The first line of code will prompt the user to enter the Nibbles per Minute (Nibble/Min) as an input. The value of Mebibits per Minute (Mibit/Min) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Mebibits(Mibit) are there in a Nibble?expand_more
There are 0.000003814697265625 Mebibits in a Nibble.
#### What is the formula to convert Nibble to Mebibit(Mibit)?expand_more
Use the formula Mibit = Nibble x 4 / 10242 to convert Nibble to Mebibit.
#### How many Nibbles are there in a Mebibit(Mibit)?expand_more
There are 262144 Nibbles in a Mebibit.
#### What is the formula to convert Mebibit(Mibit) to Nibble?expand_more
Use the formula Nibble = Mibit x 10242 / 4 to convert Mebibit to Nibble.
#### Which is bigger, Mebibit(Mibit) or Nibble?expand_more
Mebibit is bigger than Nibble. One Mebibit contains 262144 Nibbles.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,616 | 5,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-26 | latest | en | 0.797675 |
https://ktane.timwi.de/HTML/Diophantine%20Equations.html | 1,685,609,322,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00163.warc.gz | 386,699,802 | 2,585 | Diophantine Equations — Keep Talking and Nobody Explodes Module
## On the Subject of Diophantine Equations
People in ancient Greece were able to solve this, can you?
The module consists of a display, a number pad and a submit button. A display will show an equation in the form
Ax + By + Cz + Dw = N, where A, B, C, D and N are all whole numbers. To solve the module input a specific set of 4 numbers solving the equation.
### Finding an infinite family of solutions
We have an equation in the form: Ax + By + Cz + Dw = N
Compose a 5 by 4 matrix in the following form, where A, B, C, and D are equal to the digits neighbouring x, y, z, and w respectively.
A B C D 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
For each step of the algorithm:
1). Pick the left-most closest number to zero that isn’t zero from the top row, call it M.
2). Pick the left-most non-zero element from the top row from a different than M column, call it K.
3). Find an integer Q, such that K = Q * M + R, where R is non-negative and less than the absolute value of M.
4). Multiply the values of the column that M belongs to by Q. Subtract this column from the column that K belongs to. Make sure that after this, column M is not altered.
Run the algorithm several times until you have a matrix, the top row of which contains a single non-zero number in one of the columns, call it L.
. . . . . . L . . . . . . a1 b1 c1 d1 a2 b2 c2 d2 a3 b3 c3 d3 a4 b4 c4 d4
[1] To subtract one column from another, subtract the number from one column from the number of another column in each row.
[2] To multiply a column of a matrix by a number, multiply every number of the column by that number.
If N is not divisible by L, then solution doesn’t exist, proceed with Inputting Answer section; else there exists an infinite family of solutions x, y, z, w.
x = T1 * a1 + T2 * b1 + T3 * c1 + T4 * d1
y = T1 * a2 + T2 * b2 + T3 * c2 + T4 * d2
z = T1 * a3 + T2 * b3 + T3 * c3 + T4 * d3
w = T1 * a4 + T2 * b4 + T3 * c4 + T4 * d4
Proceed with Getting Values of T section.
### Getting Values of T
• If L belonged to ith column, then Ti = N / L.
• Else if ith symbol of the serial number is a digit, then Ti = that digit.
• Else if ith symbol of the serial number is a letter, then Ti = alphabetical order of the letter modulo 10. | 672 | 2,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-23 | latest | en | 0.853081 |
http://www.docstoc.com/docs/125609038/CE | 1,432,730,195,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928965.67/warc/CC-MAIN-20150521113208-00068-ip-10-180-206-219.ec2.internal.warc.gz | 405,281,392 | 40,143 | # CE by pengtt
VIEWS: 10 PAGES: 2
• pg 1
``` Department of Civil Engineering
King Saud University
Course Description: Statically Indeterminate structures. Force Method: beams, frames, trusses
and composite structures. Slope –Deflection Method: beams and frames
CE 461: Structural Analysis II
including sway and support settlement effects and frames with inclined
(Required for a BSCE degree)
members. Moment Distribution Method: beams and frames. Introduction to
stiffness Method for trusses, beams and frames. Computer applications
3 (3,1,0)
Prerequisite GE 201 (Statics), CE 302 (Mechanics of Materials), CE 361 (Structural
Analysis I)
Prerequisite by Topics:
1. Understanding of equilibrium equations to analyze engineering problems.
2. Determining the internal forces in statically determinate beams, frames
and trusses.
3. Determining the area properties of various cross sections.
4. Determining deformation for statically determinate structures using virtual
work method.
Course Learning Objectives Students completing this course successfully will be able to
structures
2. Understand different methods used for analysis of indeterminate
structures.
3. Formulate the necessary equations to analyze statically indeterminate
structures
4. Apply the different method on different indeterminate structures.
5. Understand the concept used in the structural analysis software.
6. Use updated structural analysis software in solving indeterminate
structures.
7. Submit accurate analysis in an efficient and professional way.
Topics Covered 1. Introduction: Definition of indeterminate structures, degree of
indeterminacy and degree of freedom, review of virtual work method. (3
hours)
2. Force Method, Principle of superposition, compatibility equations. (6hours)
3. Application of force method to indeterminate beams under gravity loads
and support settlement; indeterminate trusses under gravity and wind
loads, support settlement, temperature, and lack of fit; indeterminate
frames and composite structures. (6hours)
4. Slope Deflection Method (SDM), fixed end moments, continuity and rigidity
conditions, equilibrium equations, modified stiffness. (6 hours)
5. Application of SDM to indeterminate beams, frames with and without side
sway, beams and frames with support settlement. Frames with inclined
members. (9 hours)
6. Moment distribution Method (MDM), distribution coefficients and carry over
factors. (2 hours)
7. Application of MDM to indeterminate beams and frames. (4 hours)
8. Introduction to stiffness Method, stiffness matrix, load vector, displacement
vector, structure modeling, local and global coordinates, displacement and
force transformation matrix, joint and member loads. Applications to
trusses, beams and rigid frames. (6 hours)
9. Computer application on structural analysis of indeterminate structures. (3
hours)
Class/ tutorial Schedule Class is held three times per week in 50-minute lecture sessions. There is also
a 50-minute weekly tutorial associated with this course.
Computer Applications Commercial and educational structural software and MS Excel are greatly
encouraged to be used during the course.
Projects None
Contribution of Course to
1. Students learn and practice the analysis process to be involved in
Meeting the Professional
designing various structural components used in professional structural
Component
engineering.
2. Students recognize the role of professional societies in developing new
structural software and updating current knowledge.
Relationship of Course to 1. Students apply algebra, elementary calculus, and principles of mechanics.
Program Outcomes 2. Students are able to identify and formulate an engineering problem and to
develop a solution.
3. Students recognize the importance of analysis in designing structural
components.
4. Students are encouraged to submit accurate analysis in an efficient and
professional way.
5. Students recognize their role with an engineering team carrying other
aspects for analyzing structures, in terms of choosing the structural
systems and the interaction of decisions made by various architectural and
engineering teams.
6. Students are encouraged to recognize the different structural systems and
their range of applications.
7. Students recognize the ethical and professional responsibility in achieving
accurate structural analysis for safe and economical design, and its impact
on the well-being of the society.
8. Students recognize the need for technical updating on a continuing basis,
since the course emphasizes on the changing nature of software.
9. Students recognize the importance of reading and understanding technical
contents in English in order to achieve life–long learning and be able to
carryout their responsibilities.
10. Students recognize the important role of computers in facilitating analysis
and design of structural members and systems.
Textbook(s) and/or Other Structural Analysis, by R.C. Hibbeler, Prentice-Hall
Required Material
Outcome Assessment Two Midterm Exams 35%
Class Quizzes and Attendance 5%
Tutorial Quizzes 10%
Final Exam. 50%
Midterm Exams 1) Sunday 23 / 10 / 1428H ( 4 / 11 / 2007 ) 7:00 – 8:30 pm
2) Tuesday 24 / 11 / 1428H ( 4 / 12 / 2007 ) 7:00 – 8:30 pm
```
To top | 1,118 | 5,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2015-22 | longest | en | 0.791761 |
https://serverfault.com/questions/1001556/is-the-temperature-differential-in-my-server-room-to-large-under-heavy-load | 1,708,970,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00210.warc.gz | 523,066,639 | 35,983 | # Is the temperature differential in my server room to large under heavy load
I have two temperature readings for my server room one measures the ambient temperature the other measures the exhaust for a server cluster. I have noticed recently that the difference between these temperatures can reach near 30F is this to large?
Our server room houses a large server cluster, about a dozen rack mounted servers and is cooled by this a/c unit, The estimated maximum total output from the servers and lighting is 121110btu. There is no air exhaust in the room, I was thinking that adding an exhaust would help but was looking for a second opinion.
Low usage
High Usage
• Looks like you accidently uploaded the same image two times. Feb 3, 2020 at 14:19
• Hi, I like sensors reporting, but is it actually cold or hot in the room if you walk by there ? You miss that information. Feb 3, 2020 at 14:26
• @yagmoth555, The room does not feel hot but the exhaust from the cluster is quite warm Feb 3, 2020 at 14:34
This would seem "within the bounds of normal" - maybe slightly on the warm side depending on how your ventilation is set up and where its measured. It likely also depends on the servers - more power hungry servers = more heat to dissipate.
https://www.stulz.de/en/newsroom/blog/delta-t-91/ is an interesting links talking about temp differences in the SC, and reflects 10-15 degrees C as normal.
Alternatively think of temp differences in hot/cold isle DC or what its like to walk in front if a heater. Remember that a cluster if servers drawing 1kw continuously is more-or-less equivalent to a 1kw heater - and what does it feel like when you walk past one of them.
Lastly, temperature (while important) is probably less important then changes in temperature from a reliability POV
• Thank you for your answer, by changes in temperature would a safe range also be 10-15 degrees C? Feb 5, 2020 at 13:39
• Yes, but I think you misunderstand what I'm sating. Its not the difference between input and output temperature, its the difference in input temp over time - so (for example) having the room at 15c at night going to 40c during the day will put more stress on components then having it consistently at say 23c. (ie an air exhaust might be a step in the wrong direction - especially as it implies an air intake which means less control of incoming temp, humidity and probably more dust, Depending on your goals YMMV though.) Feb 5, 2020 at 19:06
• Yes I understand that you mean the difference over time in the server room, our issue currently is that hot hair seems to pool behind the server and there is nowhere for it to go. But based on your answer it seems that our current max temp of 95F(35C) is not something that we should be worried about? Feb 6, 2020 at 17:07
• @Bjorn Don't understand? Where are you measuring the output? If you were (mentally) replacing your servers with a blow heater - which is what the servers are acting as - wouldn't you expect the output to feel warm? This all sounds normal to me. Feb 6, 2020 at 19:17
• We are measuring the temperature right behind the server, of course we expect it to be warmer than the air going in, I was questioning if the temperature we were reading was too high. As you mentioned it does not matter so much what the temperature is so much as the change in temperature over time is. I worry that our change over time is too high. As you can see in the graphs above we are experiencing a 20 degree(F) change over about a 3 hour time interval, and this happens nearly every day. Sorry if I was not clear in my original questions. Feb 7, 2020 at 20:20 | 853 | 3,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.960441 |
http://math.stackexchange.com/questions/28254/what-is-wrong-with-this-reasoning | 1,467,348,176,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783400031.51/warc/CC-MAIN-20160624155000-00158-ip-10-164-35-72.ec2.internal.warc.gz | 211,209,515 | 17,529 | # what is wrong with this reasoning?
14% of people are left handed, if we want to find the chances of a brother and a sister both being left handed, we might be tempted to multiply .14 by .14. What is wrong with this reasoning when trying to find out the probability of both of them being left handed?
-
Is the probability of a male being left handed, and a female being handed the same? – JavaMan Mar 21 '11 at 7:49
Why should it be wrong? – Did Mar 21 '11 at 7:54
The main problem is that the events that the brother and sister are left handed are not known to be independent (and I would expect them not to be independent if there are either genetic or environmental factors which influence handedness). It is only when you know that events $A$ and $B$ are independent that $P(A\cap B) = P(A)P(B)$.
Similarly, if $5\%$ of the population is named Smith, and the siblings are too young to have changed last names, then the chance they are both named Smith might be close to $5\%$ rather than $5\% \times 5\%$.
A more subtle issue is whether your handedness is independent of the number of siblings you have.
-
You have to take into account the total number of people you took that 14% number from.
Lets say 50% of people are left handed. If you sampled that number from the brother and sister, the probability that both are left handed is 0%.
If the brother and sister are not part of the sample your reasoning would be correct.
We have 0.14=left-handed/total people=L/T
The correct cacluation would be P=L/T*(L-1)/(T-1). For a large sample P approaches (L/T)^2
- | 394 | 1,576 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2016-26 | longest | en | 0.967882 |
https://in.mathworks.com/matlabcentral/profile/authors/9758819 | 1,600,592,952,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400196999.30/warc/CC-MAIN-20200920062737-20200920092737-00496.warc.gz | 440,594,514 | 18,853 | Community Profile
# Bobby Dow
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Given a circular pizza with radius _z_ and thickness _a_, return the pizza's volume. [ _z_ is first input argument.] Non-scor...
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Given a and b, return the sum a+b in c.
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Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
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Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
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How to delete a row of a matrix if value in a particular column is less than a specified value?
say I have a 3x4 matrix data = [5 1 200 33; 3 0.5 100 33; 4.5 1.5 150 33] I want to delete the entire row if the value...
3 years ago | 1 answer | 0 | 339 | 1,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-40 | latest | en | 0.781587 |
https://www.wishfin.com/bank-of-baroda-home-loan-emi-calculator | 1,725,950,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00897.warc.gz | 1,017,794,465 | 24,190 | # Bank of Baroda Home Loan EMI Calculator
₹ 50k ₹ 5Cr
6.0% 30.0%
1 years 30 years
Principal Amount
Interest Amount
Monthly EMI
Principal Amount
Interest Amount
Tenure (Years)
Total Amount
## 2044
Are you looking for a quick and easy way to calculate your EMIs? Try the Bank of Baroda Home Loan EMI Calculator. This efficient online tool quickly provides accurate figures for your monthly EMI, total interest outgo, and the overall amount you'll pay over the life of your loan. It's simple to use and provides the financial details you need instantly. In this article, we’ll learn how to use the Wishfin home loan EMI calculator to get your monthly repayment schedules easily.
## How to Use Wishfin Bank of Baroda Home Loan EMI Calculator?
The Wishfin Bank of Baroda home loan calculator is a helpful tool for borrowers to figure out their monthly loan payments before they apply for a loan. Since home loans can last up to 30 years, it's important to choose a repayment period that suits your budget. To use the calculator, you just need to enter:
• Interest Rate
• Tenure
• Loan Amount
The calculator will quickly tell you your monthly payment amount, the total interest you'll pay, and the total amount you'll end up paying back. It also shows an amortization table, which helps you see how your loan payments break down over time. This tool is available 24*7, making your home loan journey a hassle-free experience.
## Bank of Baroda Home Loan EMI Calculation Formula
The calculator uses the below formula to calculate an EMI.
EMI = [P x R x (1+R)^N]/[(1+R)^N-1]
where,
EMI = Equated Monthly Installment
P = Principal Loan Amount
R = Monthly Interest Rate
N = Number of Monthly Installments
For instance, if you avail a home loan worth ₹45,00,000 for a tenure of 20 years at an agreed-upon interest rate of 9%. So, as per the formula:
• Principal Amount (P): ₹45,00,000
• Interest Rate (R): 9% per annum (which is 9/12/100 = 0.0075 per month)
• Loan Tenure (n): 20 years, or 240 months
Let’s see how it works:
[45,00,000 x 0.0075 x (1 + 0.0075) ^ 240]/[(1+0.0075) ^240 -1] = 40,488
The EMIs you will be required to pay is ₹40,488
## Bank of Baroda Home Loan Amortisation Schedule
In order to have more clarity on the same, let’s just take a look at the example below.
Suppose, Ruchi Verma, a 30-year-old, Senior Content Writer in a Company is looking for a home loan of INR 30 Lakh to buy a home in New Delhi. She has applied for the same and BOB has agreed to offer the home loan at an interest rate of 8.30% per annum for a tenure of 30 years. But before saying yes to the lender, Ruchi wants to know the exact calculations of her EMIs. She wants to know whether her monthly installments are going to be pocket-friendly or not? Thus, she used a home loan EMI Calculator for the same. So, let’s see the results in the table below.
Table Showing EMI, Interest Outgo, and Total Repayment Amount
Loan Amount Interest Rate Tenure Monthly Instalment Total Interest Amount Total Amount
₹ 30,00,000 8.30% 1 ₹ 2,61,382 ₹ 1,36,579 ₹ 31,36,579
₹ 30,00,000 8.30% 2 ₹ 1,36,093 ₹ 2,66,225 ₹ 32,66,225
₹ 30,00,000 8.30% 3 ₹ 94,425 ₹ 3,99,294 ₹ 33,99,294
₹ 30,00,000 8.30% 4 ₹ 73,662 ₹ 5,35,774 ₹ 35,35,774
₹ 30,00,000 8.30% 5 ₹ 61,261 ₹ 6,75,649 ₹ 36,75,649
₹ 30,00,000 8.30% 6 ₹ 53,040 ₹ 8,18,898 ₹ 38,18,898
₹ 30,00,000 8.30% 7 ₹ 47,208 ₹ 9,65,497 ₹ 39,65,497
₹ 30,00,000 8.30% 8 ₹ 42,869 ₹ 11,15,414 ₹ 41,15,414
₹ 30,00,000 8.30% 9 ₹ 39,524 ₹ 12,68,618 ₹ 42,68,618
₹ 30,00,000 8.30% 10 ₹ 36,876 ₹ 14,25,069 ₹ 44,25,069
₹ 30,00,000 8.30% 11 ₹ 34,733 ₹ 15,84,728 ₹ 45,84,728
₹ 30,00,000 8.30% 12 ₹ 32,969 ₹ 17,47,548 ₹ 47,47,548
₹ 30,00,000 8.30% 13 ₹ 31,497 ₹ 19,13,481 ₹ 49,13,481
₹ 30,00,000 8.30% 14 ₹ 30,253 ₹ 20,82,476 ₹ 50,82,476
₹ 30,00,000 8.30% 15 ₹ 29,192 ₹ 22,54,477 ₹ 52,54,477
₹ 30,00,000 8.30% 16 ₹ 28,278 ₹ 24,29,428 ₹ 54,29,428
₹ 30,00,000 8.30% 17 ₹ 27,487 ₹ 26,07,267 ₹ 56,07,267
₹ 30,00,000 8.30% 18 ₹ 26,796 ₹ 27,87,934 ₹ 57,87,934
₹ 30,00,000 8.30% 19 ₹ 26,190 ₹ 29,71,363 ₹ 59,71,363
₹ 30,00,000 8.30% 20 ₹ 25,656 ₹ 31,57,488 ₹ 61,57,488
₹ 30,00,000 8.30% 21 ₹ 25,183 ₹ 33,46,242 ₹ 63,46,242
₹ 30,00,000 8.30% 22 ₹ 24,763 ₹ 35,37,555 ₹ 65,37,555
₹ 30,00,000 8.30% 23 ₹ 24,389 ₹ 37,31,358 ₹ 67,31,358
₹ 30,00,000 8.30% 24 ₹ 24,054 ₹ 39,27,579 ₹ 69,27,579
₹ 30,00,000 8.30% 25 ₹ 23,754 ₹ 41,26,148 ₹ 71,26,148
₹ 30,00,000 8.30% 26 ₹ 23,484 ₹ 43,26,991 ₹ 73,26,991
₹ 30,00,000 8.30% 27 ₹ 23,241 ₹ 45,30,037 ₹ 75,30,037
₹ 30,00,000 8.30% 28 ₹ 23,021 ₹ 47,35,214 ₹ 77,35,214
₹ 30,00,000 8.30% 29 ₹ 22,823 ₹ 49,42,450 ₹ 79,42,450
₹ 30,00,000 8.30% 30 ₹ 22,644 ₹ 51,51,673 ₹ 81,51,673
EMI, Total Interest Outgo, Total Repayment (Interest + Principal)
Year Principal Interest Balance Amount
1 ₹ 23,607 ₹ 2,48,115 ₹ 29,76,392
2 ₹ 25,642 ₹ 2,46,080 ₹ 29,50,750
3 ₹ 27,853 ₹ 2,43,868 ₹ 29,22,896
4 ₹ 30,255 ₹ 2,41,467 ₹ 28,92,640
5 ₹ 32,864 ₹ 2,38,858 ₹ 28,59,775
6 ₹ 35,698 ₹ 2,36,024 ₹ 28,24,076
7 ₹ 38,777 ₹ 2,32,945 ₹ 27,85,299
8 ₹ 42,121 ₹ 2,29,601 ₹ 27,43,178
9 ₹ 45,753 ₹ 2,25,969 ₹ 26,97,425
10 ₹ 49,698 ₹ 2,22,024 ₹ 26,47,726
11 ₹ 53,984 ₹ 2,17,738 ₹ 25,93,742
12 ₹ 58,639 ₹ 2,13,083 ₹ 25,35,103
13 ₹ 63,695 ₹ 2,08,027 ₹ 24,71,407
14 ₹ 69,188 ₹ 2,02,534 ₹ 24,02,219
15 ₹ 75,154 ₹ 1,96,568 ₹ 23,27,064
16 ₹ 81,635 ₹ 1,90,087 ₹ 22,45,429
17 ₹ 88,674 ₹ 1,83,048 ₹ 21,56,755
18 ₹ 96,321 ₹ 1,75,401 ₹ 20,60,433
19 ₹ 1,04,627 ₹ 1,67,095 ₹ 19,55,806
20 ₹ 1,13,649 ₹ 1,58,073 ₹ 18,42,157
21 ₹ 1,23,449 ₹ 1,48,273 ₹ 17,18,707
22 ₹ 1,34,094 ₹ 1,37,628 ₹ 15,84,613
23 ₹ 1,45,657 ₹ 1,26,065 ₹ 14,38,955
24 ₹ 1,58,218 ₹ 1,13,504 ₹ 12,80,736
25 ₹ 1,71,861 ₹ 99,861 ₹ 11,08,875
26 ₹ 1,86,681 ₹ 85,041 ₹ 9,22,194
27 ₹ 2,02,779 ₹ 68,943 ₹ 7,19,414
28 ₹ 2,20,265 ₹ 51,457 ₹ 4,99,149
29 ₹ 2,39,258 ₹ 32,464 ₹ 2,59,890
30 ₹ 2,60,046 ₹ 11,832 ₹ 0
## The Minimum EMI Offered by Bank of Baroda on Home Loans
Here are the monthly payments for different home loan amounts from Bandhan Bank, calculated at interest rates of 8.40% and 12.10% for a period of 20 years:
Interest Rate of 8.40%
Interest Rate of 12.10%
10 lakh loan amount - ₹8,615 p.m.
10 lakh loan amount - ₹11,081 p.m.
12 lakh loan amount - ₹10,338 p.m.
12 lakh loan amount - ₹13,297 p.m.
15 lakh loan amount - ₹12,828 p.m.
15 lakh loan amount - ₹16,621 p.m.
17 lakh loan amount - ₹14,539p.m.
17 lakh loan amount - ₹18,837 p.m.
20 lakh loan amount - ₹17,104 p.m.
20 lakh loan amount - ₹22,161 p.m.
## What are the Main Benefits of Utilizing the Bank of Baroda Home Loan EMI Calculator?
Here are the main benefits of using the Bank of Baroda Home Loan EMI Calculator to help you plan your home purchase effectively:
• Easy Budget Planning: The Bank of Baroda EMI calculator helps you understand exactly how much you will need to pay towards your home loan. When you know the EMI details, you can plan your budget and other financial commitments accordingly.
• Saves Time: Instead of manually calculating EMIs, which can be prone to errors and time-consuming, the calculator provides quick and accurate results. This means you can save a lot of time as well as effort.
• Financial Preparedness: Knowing your monthly outgo in advance helps you assess your readiness for the loan, ensuring you're financially prepared for the long-term commitment.
• Transparency: The calculator gives a clear breakdown of the payment schedule, which includes the principal amount, interest, and the total payable amount. This promotes transparency in your financial dealings.
## Bank of Baroda Home Loan Eligibility
Both salaried and self-employed individuals can apply for Bank of Baroda home loans. Hindu Undivided Families (HUFs) are not eligible to apply for this loan.
Minimum Age: 21 years for applicants and 18 years for co-applicants.
Maximum Age: 70 years
Resident Type: Resident Indians and Non-Resident Indians (NRIs) holding Indian passports/ Persons of Indian origin (PIOs) holding foreign passports/ Overseas Citizens of India (OCI).
Work Experience:
• For Resident Indians:
- Salaried individuals need to be employed for at least 1 year.
- Non-salaried individuals, such as business owners or freelancers, need at least 2 years of experience.
- Up to 3 months of employment gap is acceptable.
• For Salaried NRIs/PIOs/OCIs:
- Must have a stable job with a reputable Indian or foreign company, or a government department abroad for at least 2 years.
Must hold a valid job contract or work permit.
• For Self-employed NRIs/PIOs/OCIs:
- Must have been operating a business, or be self-employed, and living abroad for at least 2 years.
Income Requirements:
• The yearly income of the applicant(s) or co-applicant(s) must be at least Rs. 5 lakh.
• For NRIs, the combined annual income of the applicant and co-applicant should also be at least Rs. 5 lakh to meet these criteria.
## Factors that Influence the Bank of Baroda Home Loan EMI
Here are the key factors that play an integral role in deciding the home loan EMIs that you will be required to pay:
• Loan Amount: The total amount you borrow has a direct impact on your EMI. Larger loan amounts will result in higher EMIs.
• Loan Tenure: The repayment period also plays an important role. Longer loan tenures spread the cost over more months. This reduces your monthly payments but increases the total interest paid over the life of the loan. Shorter tenures increase monthly payments but decrease interest costs.
• Interest Rate: The rate at which interest is charged on your loan amount also affects your monthly payments. Higher interest rate leads to higher EMIs.
• Repayment Frequency : How often you make your loan payments (monthly, quarterly, etc.) can influence your EMI. More frequent payments can reduce the interest cost.
• Existing Financial Obligations : If you have other loans or debts, it might affect your loan eligibility and the EMI you can comfortably afford.
## Fees and Charges Related to Bank of Baroda Home Loans
Processing Fees Up to 0.50% (Min. Rs.8,500; Max. Rs.25,000) Penal Interest Not available Prepayment/Foreclosure Charges NIL
## Drawbacks of Using the Bank Of Baroda Home Loan EMI Calculator
Here are the main disadvantages of using a home loan EMI calculator of Bank of Baroda:
• Only Gives Estimates: The calculated EMIs might not match the actual payments due to factors like loan amount adjustments and interest rate changes.
• Misses Extra Costs: It doesn't include other charges like processing fees and legal costs that are part of getting a home loan.
• Assumes Fixed Rates: Assumes the interest rate won't change, which isn’t the case with floating rate loans.
• Ignores Credit Score: Doesn’t consider your credit score, which can affect the interest rate you get.
• Doesn't Show Tax Benefits: Doesn’t account for tax benefits that can lower your overall loan costs.
• Over-Simplifies: Might oversimplify your financial situation, missing out on important details. | 3,654 | 10,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.906941 |
https://www.thestudentroom.co.uk/showthread.php?t=4170025 | 1,524,781,250,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948549.21/warc/CC-MAIN-20180426203132-20180426223132-00607.warc.gz | 873,079,411 | 38,475 | x Turn on thread page Beta
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# Direction of Particles M1 watch
1. So I'm confused on a question on June 2014 OCR M1 and the mark scheme doesn't help me. It says
Particles P and Q move towards each other with speeds 4m/s and 2m/s along a smooth surface. P has mass 0.2 and Q 0.3. They collide. Show that Q changes direction.
So I can do conservation of linear momentum but I can't figure out how you can know what direction a velocity is going without knowing its actual value. The mark scheme says the marks are as follows:
Calculation for both “before” Momentum (magnitudes)
Compares both terms without arithmetic error
Shows direction of after total momentum conflicts with the before velocity/momentum of Q
I don't understand how you can infer direction with two unknown 'after' velocities.
I have the equation
0.2 = 0.2V1 + 0.3V2
Can anyone explain it better than the mark scheme?
2. Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.
You can also find the Exam Thread list for A-levels here and GCSE here.
Just quoting in Puddles the Monkey so she can move the thread if needed
Spoiler:
Show
(Original post by Puddles the Monkey)
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https://industrialengineer.online/operations-research/simplex-method/ | 1,679,925,678,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00334.warc.gz | 371,942,934 | 54,257 | Operations research
# Simplex Method
Simplex method is a solving problem analytic method of linear programming, able to resolve complex models than the resolved through graphic method.
Simplex method is an iterative method that improves the solution on each step. The mathematical reason of this improvement is that the method consists in walking through a neighbor vertex in such a way that raises or decreases (according to the context of the objective’s function, whether it is maximize or minimize), given that the number of vertex that a solution polyhedron is finite, there always be a solution.
This popular method was created in 1947 by the American George Bernard Dantzig and the Russian Leonid Vasilyevich Kantorovich, with the purpose of creating an algorithm able to fix problems of “m” restrictions and “n” variables.
## ¿What is an identity matrix?
A matrix can be defined as a rectangular arrangement of elements, (o finite list of elements), which can be real numbers or complex numbers, disposed in rows and columns.
The Identity matrix or a sometimes ambiguously called a is square (it has the same number of rows, as it has the same number of columns), of order n that has all the diagonal elements equal to one (1) and all other components equal to zero (0), it’s called Identical matrix or Order “n” Identity, and it is denoted by:
The importance of this matrix theory in the Simplex Method is fundamental, given that the algorithm is based on said theory for itself problem resolution.
## Important considerations when using the simplex method
### Slack and Surplus Variables
The Simplex Method works based on equations and initial restrictions that are model trough lineal programing are not,, for this you have to convert this inequalities in equations using some type of variables called: Slack and Surplus Variables that are related with the resource on which the restrictions makes reference, and that on the final tabulate represents the “Slack or Surplus”, which many resolution of operation programs make reference, this variables get a great added value in the analysis of sensitivity and play a key role in the creation of the Simplex base Identity Matrix.
These variables are usually represented by the letter “S”, they add if the restriction is «<= », and they subtract if the restriction is «>=».
For example:
### Artificial variable / “M” method:
An artificial variable is a math trick to turn «>=» inequalities, into equations, or when equalities appear in the original problem to resolve, the main characteristic of this variables is that they should not be part of the solution, given that they do not represent any resource. The main objective of this variables is the formation of the identity matrix.
These variables are represented by the letter “A”, constraints are always added, their coefficient is M (this is why it is called M Method, where M means a number way too big, and very less attractive to the objective’s function), and the sign in the objective’s function goes counter-wise of itself, that is to say, in problems of maximization it sign is minus (-), and on minimization problems it sign is plus (+), repeat with the objective that it’s value on the solution is zero (0).
## Step by step: Simplex Method
### The problem
Company SAMAN Limited. Dedicated to the manufacture of furniture, it has expanded its production in two more lines. Thus currently manufactures tables, chairs, beds and libraries. Each table requires 2 rectangular pieces of 8 pins each, and two square pieces of 4 pins each. Each chair requires 1 rectangular piece of 8 pins and 2 square pieces of 4 pins each; each bed requires 1 square piece of 8 pins, 1 square piece of 4 pins and 2 trapezoidal bases of 2 pins each; and finally each library requires 2 rectangular pieces of 8 pins each, 2 trapezoidal bases of 2 pins each and 4 rectangular pieces of 2 pins each.
Each table’s manufacture cost is \$10000, and are sold in \$30000; each chair’s manufacture cost is \$8000, and are sold in \$28000; each bed’s manufacture cost is \$20000, and are sold in \$40000; each library’s manufacture cost is \$40000, and are sold in \$60000. The main goal of the Company is to maximize profits.
## Step 1: Modeling trough lineal programing:
Variables:
X1 = Quantity of tables to manufacture(units)
X2Quantity of chairs (units)
X3 = Quantity of bed to manufacture (Units)
X4Quantity of libraries to manufacture (units)
Constraints:
2X1 + 1X2 + 1X3 + 2X4 <= 24
2X1 + 2X2 + 1X3 <= 20
2X3 + 2X4 <= 20
4X4 <= 16
Objective function:
ZMAX = 20000X1 + 20000X2 + 20000X3 + 20000X4
## Step 2: To convert inequalities in equations
The goal in this step is to assign a Slack Variable to each resource, given that all constraints are «<=».
2X1 + 1X2 + 1X3 + 2X4 + 1S1 + 0S2 + 0S3 + 0S4 = 24
2X1 + 2X2 + 1X3 + 0X4 + 0S1 + 1S2 + 0S3 + 0S4 = 20
0X1 + 0X2 + 2X3 + 2X4 + 0S1 + 0S2 + 1S3 + 0S4 = 20
0X1 + 0X2 + 0X3 + 4X4 + 0S1 + 0S2 + 0S3 + 1S4 = 16
This way we can see an identity matrix (n=4), form by the Slack Variables which only have a 1 coefficient on their respective resource; for example, the Slack Variable “S1” only has a 1 coefficient on the constraint corresponding to resource 1.
The Objective Function does not suffer any kind of variations:
ZMAX = 20000X1 + 20000X2 + 20000X3 + 20000X4
## Step 3: Define the initial basic solution
The simplex method from an initial basic solution to make all its iterations, this solution is formed with the variables of coefficient different from zero (0) on the identity matrix.
1S1 = 24
1S2 = 20
1S3 = 20
1S4 = 16
Solution (second term): In this row is set the second term of the solution, that is to say, the correct way to do it would it be to set them organized, just as in the definition of constraints.
Cj: This row makes reference to the coefficient that each of this variables of the row “Solution” in the Objective Function.
Variable Solution: In this column is set the initial basic solution, and from this column on each iteration includes variables that will be part of the final solution.
Zj: In this row is set the total contribution, that is to say the addition of the products from the term and Cb.
Cj-Zj: In this row is made the subtraction between the row Cj and the row Zj, it’s meaning is a “Shadow Price”, that is to say, the unearned earnings for each unit of the corresponding variable that does not make part of the solution.
Solución inicial:
## Step 4 : Make the necessary iterations
This is the definitive step through the Simplex Method, consists in trying while the polyhedron goes from a vertex to the other.
The following is the procedure:
1. To evaluate which variable is going in, a what’s the optimal solution.
2. The fact that a variable different is a part of the solution variables implies a series of changes in the Simplex tabulate. The change is the following:
• First thing is not to forget the value of a, corresponding to the variables going in, in this case is “a = 4”..
• The next thing is to start filling in the rest of the table, row by row.
• Repeat this procedure with the two remaining rows.
Once set the values of the matrix, you can calculate until filling the table corresponding to the first iteration.
On this way ends the first iteration, this step will repeat as any times is necessary, and only will be ended according the following criteria.
Maximize Minimize Optimal Solution When all the Cj – Zj are <= 0 When all the Cj – Zj are >= 0
• Continue with the iterations, so we have to repeat the anterior cases:
In this last iteration we can observe that the slogan Cj-Zj <= 0, is met; for exercise which objective function is Maximize, thus que have reached an optimal response.
X1 = 0
X2 = 7
X3 = 6
X4 = 4
With a profit of: \$ 340000
Nevertheless, one the Simplex Method has ended, we can observe an identity matrix in the rectangle determine by the decision variables, the fact that in this case is not shown an identity matrix tells us that exists another optimal solution.
The way of reaching the other solution consists in altering the order on which each of the variables gets into the basic solution, remember that this process was decided by random due to Cj-Zj equality of the initial tabulate. The following is a way to reach the other solution.
We can observe that there is an alternative optimal solution.
X1 = 3 (Cantidad de mesas a producir = 3)
X2 = 4 (Cantidad de sillas a producir = 4)
X3 = 6 (Cantidad de camas a producir = 6)
X4 = 4 (Cantidad de bibliotecas a producir = 4)
Con una utilidad de: \$ 340000
## Minimization problems with the Simplex Method
To resolve minimization problems there are two types of procedures:
• The first, which I personally recommend, is based on an artifice applicable to the algorithm founded in the following mathematical logic: for any function f (x), any point that minimizes f (x) will also maximize a – f (x). Therefore, the procedure to follow is multiply by the negative factor (-1) the whole objective function.
The algorithm is then solved as a maximization problem.
• The second procedure which intends to preserve the minimization consists in applying the decision criteria that we have discussed, in the cases where the variable goes in, goes out and in the case that the optimal solution is found. Let’s remember:
### Bryan Salazar López
By profession, Industrial Engineer, Master in Logistics, specialized in productivity, with interest and experience in modeling processes under sustainability indicators. Founder of Ingenieriaindustrialonline.com, site where research contributions, articles and references are collected.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2,399 | 9,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-14 | latest | en | 0.930866 |
https://www.convertunits.com/from/kip/to/mace | 1,679,303,069,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00353.warc.gz | 770,717,707 | 12,453 | ## Convert kip to mace [China]
kip mace
How many kip in 1 mace? The answer is 8.3290642653447E-6.
We assume you are converting between kip and mace [China].
You can view more details on each measurement unit:
kip or mace
The SI base unit for mass is the kilogram.
1 kilogram is equal to 0.0022046226218488 kip, or 264.69031233457 mace.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kips and mace [China].
Type in your own numbers in the form to convert the units!
## Quick conversion chart of kip to mace
1 kip to mace = 120061.50609 mace
2 kip to mace = 240123.01218 mace
3 kip to mace = 360184.51826 mace
4 kip to mace = 480246.02435 mace
5 kip to mace = 600307.53044 mace
6 kip to mace = 720369.03653 mace
7 kip to mace = 840430.54262 mace
8 kip to mace = 960492.0487 mace
9 kip to mace = 1080553.55479 mace
10 kip to mace = 1200615.06088 mace
## Want other units?
You can do the reverse unit conversion from mace to kip, or enter any two units below:
## Enter two units to convert
From: To:
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 476 | 1,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-14 | latest | en | 0.776988 |
https://forum.freecodecamp.org/t/build-a-probability-calculator-project-build-a-probability-calculator-project-final-test-timing-out/695573 | 1,721,580,131,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00532.warc.gz | 232,765,875 | 7,057 | # Build a Probability Calculator Project - Build a Probability Calculator Project - final test timing out
### Tell us what’s happening:
My experiment keeps timing out. It times out in seconds. I’m not sure what to do about that.
``````import copy
import random
class Hat:
contents = []
def __init__(self, **kwargs):
for key, value in kwargs.items():
for _ in range(value):
self.contents.append(key)
def __str__(self):
pass
def draw(self, number_of_balls):
used = []
result = []
too_big = []
if len(self.contents) < number_of_balls:
too_big = self.contents.copy()
self.contents = []
for _ in range(number_of_balls):
random_ball = random.randint(0, len(self.contents)-1)
while True:
if random_ball in used:
random_ball = random.randint(0, len(self.contents)-1)
else:
used.append(random_ball)
result.append(self.contents[random_ball])
self.contents.pop(random_ball)
break
return result
def experiment(hat, expected_balls, num_balls_drawn, num_experiments):
#initializing the list of expected balls and the total number of passing experiments
expected = []
total = 0
# creating the expected balls list
for key, value in expected_balls.items():
for _ in range(value):
expected.append(key)
# running the experiment the required number of times
for _ in range(0,num_experiments):
single_success = 0
total_success = False
# copyint the volitile hat.contents to another variable to put back later
drawn_contents = hat.contents.copy()
drawn = hat.draw(num_balls_drawn)
for i in range(len(expected)):
if expected[i] in drawn:
popper = drawn.index(expected[i])
drawn.pop(popper)
single_success += 1
if single_success == len(expected):
total_success = True
if total_success:
total += 1
# returning the original contents of hat.contents
hat.contents = drawn_contents.copy()
``````
User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/125.0.0.0 Safari/537.36`
### Challenge Information:
Build a Probability Calculator Project - Build a Probability Calculator Project
When I use the example in the project(the one with 2000 as the num_experiments), it runs without an issue. I get a different probability returned when it runs multiple times. I’m not sure what is timing out.
the console states:
// running tests The
``````experiment
``````
method should return a different probability. (Test timed out) // tests completed
there is nothing listed in the F12 console for the final test
Could you explain how the `draw` method of `Hat` works?
OK. I rewrote the draw, and now I am getting another error that tells me nothing about what is actually wrong. F12 only tells that .59 != 1.
AssertionError: 0.59 != 1.0 within 0.01 delta (0.41000000000000003 difference) : Expected experiment method to return a different probability.
def draw(self, number_of_balls):
used =
result =
too_big =
`````` # checking if the number of balls is more than what the hat contains
if len(self.contents) < number_of_balls:
# had to add this because, even though it was working, the test was failing because contents wasn't empty
too_big = self.contents.copy()
self.contents = []
# now that we know the number of balls is no more than what is in the hat, start looping
for _ in range(number_of_balls):
# get a random item index
random_ball = random.randint(0, len(self.contents)-1)
result.append(self.contents[random_ball])
self.contents.pop(random_ball)
return result
``````
Hmm, this is strange. I created two hats:
``````hat = Hat(black=6, red=4, green=3)
hat2 = Hat(yellow=5, blue=9)
``````
Then tried to draw all balls from the second one:
``````print(hat2.draw(14))
``````
And result contained also balls from first hat…
[‘blue’, ‘yellow’, ‘black’, ‘yellow’, ‘blue’, ‘blue’, ‘green’, ‘blue’, ‘blue’, ‘red’, ‘blue’, ‘red’, ‘black’, ‘black’]
1 Like
Awesome, thanks so much.
1 Like | 963 | 3,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.646594 |
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1. An Introduction to Missing Data1.1 Introduction1.2 Chapter Overview1.3 Missing Data Patterns1.4 A Conceptual Overview of Missing Data Theory1.5 A More Formal Description of Missing Data Theory1.6 Why Is the Missing Data Mechanism Important?1.7 How Plausible Is the Missing at Random Mechanism?1.8 An Inclusive Analysis Strategy1.9 Testing the Missing Completely at Random Mechanism1.10 Planned Missing Data Designs1.11 The Three-Form Design1.12 Planned Missing Data for Longitudinal Designs1.13 Conducting Power Analyses for Planned Missing Data Designs1.14 Data Analysis Example1.15 Summary1.16 Recommended Readings2. Traditional Methods for Dealing with Missing Data2.1 Chapter Overview2.2 An Overview of Deletion Methods2.3 Listwise Deletion2.4 Pairwise Deletion2.5 An Overview of Single Imputation Techniques2.6 Arithmetic Mean Imputation2.7 Regression Imputation2.8 Stochastic Regression Imputation2.9 Hot-Deck Imputation2.10 Similar Response Pattern Imputation2.11 Averaging the Available Items2.12 Last Observation Carried Forward2.13 An Illustrative Simulation Study2.14 Summary2.15 Recommended Readings3. An Introduction to Maximum Likelihood Estimation3.1 Chapter Overview3.2 The Univariate Normal Distribution3.3 The Sample Likelihood3.4 The Log-Likelihood3.5 Estimating Unknown Parameters3.6 The Role of First Derivatives3.7 Estimating Standard Errors3.8 Maximum Likelihood Estimation with Multivariate Normal Data3.9 A Bivariate Analysis Example3.10 Iterative Optimization Algorithms3.11 Significance Testing Using the Wald Statistic3.12 The Likelihood Ratio Test Statistic3.13 Should I Use the Wald Test or the Likelihood Ratio Statistic?3.14 Data Analysis Example 13.15 Data Analysis Example 23.16 Summary3.17 Recommended Readings4. Maximum Likelihood Missing Data Handling 4.1 Chapter Overview4.2 The Missing Data Log-Likelihood4.3 How Do the Incomplete Data Records Improve Estimation?4.4 An Illustrative Computer Simulation Study4.5 Estimating Standard Errors with Missing Data4.6 Observed Versus Expected Information4.7 A Bivariate Analysis Example4.8 An Illustrative Computer Simulation Study4.9 An Overview of the EM Algorithm4.10 A Detailed Description of the EM Algorithm4.11 A Bivariate Analysis Example4.12 Extending EM to Multivariate Data4.13 Maximum Likelihood Software Options4.14 Data Analysis Example 14.15 Data Analysis Example 24.16 Data Analysis Example 34.17 Data Analysis Example 44.18 Data Analysis Example 54.19 Summary4.20 Recommended Readings5. Improving the Accuracy of Maximum Likelihood Analyses5.1 Chapter Overview5.2 The Rationale for an Inclusive Analysis Strategy5.3 An Illustrative Computer Simulation Study5.4 Identifying a Set of Auxiliary Variables5.5 Incorporating Auxiliary Variables Into a Maximum Likelihood Analysis5.6 The Saturated Correlates Model5.7 The Impact of Non-Normal Data5.8 Robust Standard Errors5.9 Bootstrap Standard Errors5.10 The Rescaled Likelihood Ratio Test5.11 Bootstrapping the Likelihood Ratio Statistic5.12 Data Analysis Example 15.13 Data Analysis Example 25.14 Data Analysis Example 35.15 Summary5.16 Recommended Readings6. An Introduction to Bayesian Estimation6.1 Chapter Overview6.2 What Makes Bayesian Statistics Different?6.3 A Conceptual Overview of Bayesian Estimation6.4 Bayes' Theorem6.5 An Analysis Example6.6 How Does Bayesian Estimation Apply to Multiple Imputation?6.7 The Posterior Distribution of the Mean6.8 The Posterior Distribution of the Variance6.9 The Posterior Distribution of a Covariance Matrix6.10 Summary6.11 Recommended Readings7. The Impu
Craig K. Enders, Department of Psychology, Arizona State University, Tempe, USA
#### Reviews
"This is a well-written book that will be particularly useful for analysts who are not PhD statisticians. Enders provides a much-needed overview and explication of the current technical literature on missing data. The book should become a popular text for applied methodologists." - Bengt Muthen, Professor Emeritus, University of California, Los Angeles, USA "A needed and valuable addition to the literature on missing data. The simulations are excellent and are a clear strength of the book." - Alan C. Acock, Distinguished Professor and Knudson Chair in Family Research, Department of Human Development and Family Sciences, Oregon State University, USA "The book contains very accessible material on missing data. I would recommend it to colleagues and students, especially those who do not have formal training in mathematical statistics." - Ke-Hai Yuan, Department of Psychology, University of Notre Dame, USA "Many applied researchers are not trained in statistics to the level that would make the classic sources on missing data accessible. Enders makes a concerted - and successful - attempt to convey the statistical concepts and models that define missing data methods in a way that does not assume high statistical literacy. He writes in a conceptually clear manner, often using a simple example or simulation to show how an equation or procedure works. This book is a refreshing addition to the literature for applied social researchers and graduate students doing quantitative data analysis. It covers the full range of state-of-the-art methods of handling missing data in a clear and accessible manner, making it an excellent supplement or text for a graduate course on advanced, but widely used, statistical methods." - David R. Johnson, Department of Sociology, The Pennsylvania State University, USA "A useful overview of missing data issues, with practical guidelines for making decisions about real-world data. This book is all about an issue that is usually ignored in work on OLS regression - but that most of us spend significant time dealing with. The writing is clear and accessible, a great success for a challenging topic. Enders provides useful reminders of what we need to know and why. I appreciated the interpretation of formulas, terms, and output. This book provides comprehensive and vital information in an easy-to-consume style. I learned a great deal reading it." - Julia McQuillan, Director, Bureau of Sociological Research, and Department of Sociology, University of Nebraska--Lincoln, USA | 1,382 | 6,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-13 | latest | en | 0.634964 |
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# 12 Days of Holiday Freebies Day 2: Count the Gumdrops Gingerbread Man Activity
Happy holidays! It’s day two of our “12 Days of Holiday Freebies” countdown, and today’s freebie is a Count the Gumdrops on the Gingerbread Man Counting Activity and Worksheet! This is a fun numeracy activity to help build math skills for young children in pre-K and Kindergarten!
Here’s the freebie for the second day of our holiday giveaway! This is a fun one, and I really love the way it turned out! All you need to do to prep it is print out the Gingerbread Men, laminate them, and find some pom poms to decorate it with. Easy, right? You can add velcro if you want, but I’m sure the kids will have just as much fun without it.
This activity is designed to give young children a fun way to practice counting and matching sets in a fun way with a Gingerbread Man theme! The worksheet is provided as a follow up activity, and so that the children can practice moving from the concrete to the pictorial representation of the concept. I hope that you and your little ones have fun with it!
Materials: White card stock Pom poms or other manipulatives that children can count with and use to decorate the Gingerbread Man. The pictures only show pom poms, but you may wish to use bits of rick rack, buttons, pipe cleaners, or any other little things you can find! The only limit is your imagination! Optional: velcro
Preparation: Print out the Gingerbread Men cards with the numbers onto white cardstock and laminate them or place into page protectors. You may wish to attach velcro to the laminated papers if you want children to stick the pom poms in place, etc.
To Use: Simply show the children how to count out your manipulatives and decorate the Gingerbread Men with the correct amount of manipulatives.
Variation 1: For extra fun, give the children some pennies and nickels, and then let them “shop” for their decorations first! Then see how many different Gingerbread Men they can decorate!
Variation 2: Have the children re-arrange the pom poms in different ways to represent different addition equations as shown in the picture below.
And I hope you didn’t miss yesterday’s post with the free Gingerbread Man ten frames! If so, grab it at the link above!
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Permutation group
In mathematics, a permutation group is a group G whose elements are permutations of a given set M, and whose group operation is the composition of permutations in G (which are thought of as bijective functions from the set M to itself); the relationship is often written as (G,M). Note that the group of all permutations of a set is the symmetric group; the term permutation group is usually restricted to mean a subgroup of the symmetric group. The symmetric group of n elements is denoted by Sn; if M is any finite or infinite set, then the group of all permutations of M is often written as Sym(M).
The application of a permutation group to the elements being permuted is called its group action; it has applications in both the study of symmetries, combinatorics and many other branches of mathematics.
Examples
Permutations are often written in cyclic form, so that given the set M = {1,2,3,4}, a permutation g of M with g(1) = 2, g(2) = 4, g(4) = 1 and g(3) = 3 will be written as (1,2,4)(3), or more commonly, (1,2,4) since 3 is left unchanged.
Consider the following set of permutations G of the set M = {1,2,3,4}:
• e = (1)(2)(3)(4)
• This is the identity, the trivial permutation which fixes each element.
• a = (12)(3)(4) = (12)
• This permutation interchanges 1 and 2, and fixes 3 and 4.
• b = (1)(2)(34) = (34)
• Like the previous one, but exchanging 3 and 4, and fixing the others.
• ab = (12)(34)
• This permutation, which is the composition of the previous two, exchanges simultaneously 1 with 2, and 3 with 4.
G forms a group, since aa = bb = e, ba = ab, and baba = e. So (G,M) forms a permutation group.
The Rubik's Cube puzzle is another example of a permutation group. The underlying set being permuted is the colored subcubes of the whole cube. Each of the rotations of the faces of the cube is a permutation of the positions and orientations of the subcubes. Taken together, the rotations form a generating set, which in turn generates a group by composition of these rotations. The axioms of a group are easily seen to be satisfied; to invert any sequence of rotations, simply perform their opposites, in reverse order.
The group of permutations on the Rubik's Cube does not form a complete symmetric group of the 20 corner and face cubelets; there are some final cube positions which cannot be achieved through the legal manipulations of the cube.
More generally, every group G is isomorphic to a permutation group by virtue of its action on G as a set; this is the content of Cayley's Theorem.
Isomorphisms
If G and H are two permutation groups on the same set S, then we say that G and H are isomorphic as permutation groups if there exists a bijective map f : SS such that r |-> f −1 o r o f defines a bijective map between G and H; in other words, if for each element g in G, there is a unique hg in H such that for all s in S, (g o f)(s) = (f o hg)(s). In this case, G and H are also isomorphic as groups.
Notice that different permutation groups may well be isomorphic as abstract groups, but not as permutation groups. For instance, the permutation group on {1,2,3,4} described above is isomorphic as a group (but not as a permutation group) to {(1)(2)(3)(4), (12)(34), (13)(24), (14)(23)}. Both are isomorphic as groups to the Klein group V4.
If (G,M) and (H,M) such that both G and H are isomorphic as groups to Sym(M), then (G,M) and (H,M) are isomorphic as permutation groups; thus it is appropriate to talk about the symmetric group Sym(M) (up to isomorphism). | 990 | 3,948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2013-20 | latest | en | 0.918972 |
https://www.jiskha.com/questions/758830/a-tugboat-goes-160-miles-upstream-in-20-hours-the-return-trip-downstream-takes-5-hours | 1,591,349,290,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348496026.74/warc/CC-MAIN-20200605080742-20200605110742-00121.warc.gz | 754,815,581 | 5,607 | # algebra
a tugboat goes 160 miles upstream in 20 hours. The return trip downstream takes 5 hours. Find the speed of the tugboat without a current and the speed of the current.
1. 👍 0
2. 👎 0
3. 👁 148
1. if tugboats speed is s, and current is c,
since the distances are the same, and
distance = speed * time,
(s-c)*20 = (s+c)*5 = 160
20s - 20c = 160
5s + 5c = 160
or,
20s+20c = 640
40s = 800
s = 20
c = 12
check:
160/32 = 5
160/8 = 20
1. 👍 0
2. 👎 0
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asked by suuny on November 20, 2016 | 1,039 | 3,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-24 | latest | en | 0.948472 |
https://www.proprofs.com/quiz-school/story.php?title=nji2ntuzy4r1 | 1,713,578,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00354.warc.gz | 857,587,877 | 98,088 | # 4.6 Permutations
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| By Jensenmath9
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Quizzes Created: 7 | Total Attempts: 3,831
Questions: 10 | Attempts: 837
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• 1.
### Evaluate 6!
Explanation
The question asks to evaluate 6!. The exclamation mark denotes the factorial function, which means multiplying a number by all the positive integers less than it. Therefore, 6! is calculated as 6 x 5 x 4 x 3 x 2 x 1, which equals 720.
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• 2.
### Evaluate P(7, 3)
Explanation
The expression P(7, 3) represents the permutation of selecting 3 items from a set of 7 items in a specific order. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the number of items in the set and r is the number of items being selected. In this case, P(7, 3) = 7! / (7-3)! = 7! / 4! = (7 * 6 * 5 * 4!) / 4! = 7 * 6 * 5 = 210. Therefore, the correct answer is 210.
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• 3.
### How many ways can four people be arranged in a straight line?
Explanation
There are 4 people to be arranged in a straight line. The first person can be chosen in 4 ways, the second person can be chosen in 3 ways (as one person is already placed), the third person can be chosen in 2 ways (as two people are already placed), and the fourth person can be chosen in 1 way (as three people are already placed). Therefore, the total number of ways to arrange the four people is 4 x 3 x 2 x 1 = 24.
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• 4.
### How many different arrangements are there of the letters in the word ZAMBONI ?
Explanation
The word "ZAMBONI" has 7 letters. To find the number of different arrangements, we use the formula for permutations of a set, which is n!, where n is the number of elements in the set. In this case, n = 7. Therefore, the number of different arrangements of the letters in the word "ZAMBONI" is 7! = 5040.
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• 5.
### How many different arrangements are there of the letters in the word ROLLER ?
Explanation
The word "ROLLER" has 6 letters. To find the number of different arrangements, we can use the formula for permutations. Since all the letters are unique, we can arrange them in 6! (6 factorial) ways, which is equal to 720. However, since the letter "L" appears twice, we need to divide the total number of arrangements by 2! (2 factorial) to account for the repeated arrangement. Therefore, the number of different arrangements of the letters in the word "ROLLER" is 720 / 2 = 360.
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• 6.
### 5 people are lined up for a 1000 meter race. Determine the number of different ways the first and second positions can be assigned.
Explanation
There are 5 people lined up for the race, and we need to determine the number of different ways the first and second positions can be assigned. Since the order matters, we can use the permutation formula. The number of ways to choose the first position is 5, and once the first position is assigned, there are 4 people remaining for the second position. Therefore, the total number of different ways is 5 * 4 = 20.
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• 7.
### You disconnect the thermostat from your furnace but neglect to jot down how the wires connect to it. There are three wires coming from the wall (red, black, and white) but four terminals on the thermostat. How many different ways can the wires be connected?
Explanation
There are four terminals on the thermostat, but only three wires coming from the wall. This means that one of the terminals will not be connected to any wire. The wire can be connected to any of the four terminals, so there are four possible options for the first wire. After the first wire is connected, there are three remaining terminals for the second wire to be connected to. Finally, there are two remaining terminals for the third wire to be connected to. Therefore, the total number of different ways the wires can be connected is 4 x 3 x 2 = 24.
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• 8.
### How many arrangements are there of the letters in FERMAT if you use all the letters but each arrangement must end in a vowel?
Explanation
There are 6 letters in the word FERMAT. Out of these, 3 letters (E, A, and T) are vowels. In order for an arrangement to end in a vowel, the last letter must be one of these 3 vowels. Therefore, there are 3 options for the last letter. The remaining 5 letters can be arranged in any order. Using the formula for permutations, we can calculate the total number of arrangements as 3 options for the last letter multiplied by the number of permutations of the remaining 5 letters (5!). This gives us a total of 3 * 5! = 3 * 120 = 360. However, since the question specifies that each arrangement must end in a vowel, we need to subtract the number of arrangements where the last letter is not a vowel. There are 3 consonants in the word FERMAT, and the remaining 4 letters can be arranged in any order. Therefore, there are 3 * 4! = 3 * 24 = 72 arrangements where the last letter is not a vowel. Subtracting this from the total number of arrangements, we get 360 - 72 = 288. However, this calculation includes arrangements where the first letter is a vowel, which is not allowed according to the question. There are 2 vowels (E and A) that can be in the first position, and the remaining 4 letters can be arranged in any order. Therefore, there are 2 * 4! = 2 * 24 = 48 arrangements where the first letter is a vowel. Subtracting this from the total number of arrangements, we get 288 - 48 = 240. So, there are 240 arrangements of the letters in FERMAT if each arrangement must end in a vowel.
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• 9.
### For security, an e-mail program requires you to have a ten-character password adn change it every three weeks. You like your current password, XXQWERTYZZ, because it is relatively easy to remember and type in. At the very least, you would like to keep the same letters when you change your password, How many different passwords can be created using your letters while keeping QWERTY together.
Explanation
Consider QWERTY as a single item. 5!/[(2!)(2!)]
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• 10.
### A basketball team is lining up for a team photo. There are eight players and two coaches. How many ways can they be arranged in a straight line if the two coaches must be beside each other?
Explanation
9! x 2! (consider the two coaches as a single item and then consider the different arrangements of the two coaches after)
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Related Topics | 1,747 | 7,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-18 | latest | en | 0.917151 |
https://www.mrexcel.com/board/threads/using-sumproduct-with-date.1051698/ | 1,580,093,052,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694176.67/warc/CC-MAIN-20200127020458-20200127050458-00382.warc.gz | 1,013,546,535 | 15,963 | # using SUMPRODUCT with DATE
##### New Member
I am building inventory report to show how inventory is doing
When I do counts for each day I enter manual data into the excel table i have build
like in first column I have Date of cycle count, next part#, Stock location, physical count, variance value \$
I tally all the numbers on a separate table to be more readable
I am having trouble with SUM product function and date range
I want to count how many variances are in a month,
The tables I use are
DATE of cycle count and Variance
in the month of January I have entered 700 lines and each line has a variance of either \$0 or \$ range in plus (extra stock) and minus (missing stock)
I want to count the variances how many I have done in January , february, march
originally I use this function =SUMPRODUCT(--(LEN(G23:G10000)>0)) and it calculates perfectly the variances but now I have data entry for February, March, and April
and this function counts all how can I make this work for each separate month?
all The dates are entered in A23:A10000 and variances are in G23:G10000
#### jtakw
##### Well-known Member
Hi,
If I understand correctly, the following will do what you want, might need some tweaking if I misunderstood.
Also, the formula you're using to Count Variances can be done simply by a COUNTA function, as shown in L23 and M23.
<b>Excel 2010</b><table cellpadding="2.5px" rules="all" style=";background-color: rgb(255,255,255);border: 1px solid;border-collapse: collapse; border-color: rgb(187,187,187)"><colgroup><col width="25px" style="background-color: rgb(218,231,245)" /><col /><col /><col /><col /><col /><col /><col /><col /><col /></colgroup><thead><tr style=" background-color: rgb(218,231,245);text-align: center;color: rgb(22,17,32)"><th></th><th>A</th><th>B</th><th>G</th><th>H</th><th>I</th><th>J</th><th>K</th><th>L</th><th>M</th></tr></thead><tbody><tr ><td style="color: rgb(22,17,32);text-align: center;">22</td><td style=";">Date</td><td style="text-align: right;;"></td><td style=";">Variance</td><td style="text-align: right;;"></td><td style=";">Month</td><td style=";">Variance Count</td><td style="text-align: right;;"></td><td style=";">Your sumproduct formula</td><td style=";">Count formula</td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">23</td><td style="text-align: right;;">2/2/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;">1</td><td style="text-align: right;;"></td><td style="text-align: right;;">Feb-18</td><td style="text-align: right;;">2</td><td style="text-align: right;;"></td><td style="text-align: right;;">8</td><td style="text-align: right;;">8</td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">24</td><td style="text-align: right;;">3/5/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;">2</td><td style="text-align: right;;"></td><td style="text-align: right;;">Mar-18</td><td style="text-align: right;;">1</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">25</td><td style="text-align: right;;">2/5/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;">3</td><td style="text-align: right;;"></td><td style="text-align: right;;">Apr-18</td><td style="text-align: right;;">3</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">26</td><td style="text-align: right;;">4/15/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;">4</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">27</td><td style="text-align: right;;">4/7/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">28</td><td style="text-align: right;;">10/11/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;">6</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">29</td><td style="text-align: right;;">4/18/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;">7</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">30</td><td style="text-align: right;;">7/20/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;">8</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td></tr><tr ><td style="color: rgb(22,17,32);text-align: center;">31</td><td style="text-align: right;;">4/1/2018</td><td style="text-align: right;;"></td><td style="text-align: right;;">9</td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td><td style="text-align: right;;"></td></tr></tbody></table><p style="width:5.6em;font-weight:bold;margin:0;padding:0.2em 0.6em 0.2em 0.5em;border: 1px solid rgb(187,187,187);border-top:none;text-align: center;background-color: rgb(218,231,245);color: rgb(22,17,32)">Sheet18</p><br /><br /><table width="85%" cellpadding="2.5px" rules="all" style=";border: 2px solid black;border-collapse:collapse;padding: 0.4em;background-color: rgb(255,255,255)" ><tr><td style="padding:6px" ><b>Worksheet Formulas</b><table cellpadding="2.5px" width="100%" rules="all" style="border: 1px solid;text-align:center;background-color: rgb(255,255,255);border-collapse: collapse; border-color: rgb(187,187,187)"><thead><tr style=" background-color: rgb(218,231,245);color: rgb(22,17,32)"><th width="10px">Cell</th><th style="text-align:left;padding-left:5px;">Formula</th></tr></thead><tbody><tr><th width="10px" style=" background-color: rgb(218,231,245);color: rgb(22,17,32)">L23</th><td style="text-align:left">=SUMPRODUCT(<font color="Blue">--(<font color="Red">LEN(<font color="Green">G23:G10000</font>)>0</font>)</font>)</td></tr><tr><th width="10px" style=" background-color: rgb(218,231,245);color: rgb(22,17,32)">M23</th><td style="text-align:left">=COUNTA(<font color="Blue">G23:G10000</font>)</td></tr><tr><th width="10px" style=" background-color: rgb(218,231,245);color: rgb(22,17,32)">J23</th><td style="text-align:left">=SUMPRODUCT(<font color="Blue">(<font color="Red">MONTH(<font color="Green">A23:A10000</font>)=MONTH(<font color="Green">I23</font>)</font>)*(<font color="Red">G23:G10000<>""</font>)</font>)</td></tr></tbody></table></td></tr></table><br />
##### MrExcel MVP
Avoid testing just month, unless the year must be explicitly ignored... So, for correctness:
=SUMPRODUCT(--(\$A\$23:\$A\$31-DAY(\$A\$23:\$A\$31)+1=E23),--ISNUMBER(\$C\$23:\$C\$31))
where E23 is a first day date like 2018-02-01 (means the month/year of February 2018).
##### New Member
OK thank you let me try
##### New Member
It worked perfectly thank you
#### jtakw
##### Well-known Member
You're welcome, welcome to the forum.
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https://www.teacherspayteachers.com/Product/OWL-about-Missing-Addends-Known-Unknown-Range-of-2-20-QR-Codes-fun-2048242 | 1,485,184,763,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282932.75/warc/CC-MAIN-20170116095122-00325-ip-10-171-10-70.ec2.internal.warc.gz | 979,592,867 | 57,073 | # OWL about Missing Addends (Known Unknown) Range of 2-20 QR Codes fun
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What student doesn't love QR codes? With this set of 24 task cards on Missing Addends (addition) in a range of 2-20, students scan the code to see if they got the correct answer. I wrote this to use as a Station in first grade, but it's up to you! This could also be used for remediation.
The problems are written with the total on top and a variety of places the space is in.
For instance:
___ + 3 + 3 = 10.
or 3 + ___ +3 = 10
or 3 + 3 + ___ = 10
They scan the QR code beneath the problem and get the number 4. Please take a look at the preview for an example card!
This also prints fine in black and white. Just choose 'print grayscale' on your printer.
As always, if you find a mistake in my work please be sure to message me and I will fix it asap! Thank you in advance for taking time to let me know how it works for you!
Here are some ELA activities if you are interested:
Character Traits
* Get ‘Hip' to Character Traits QR Task Cards
* Character Trait Fever QR Task Cards
* Character Traits Stars
Main Idea
* Spot the Main Idea QR Task Cards - Set One
* Spot the Main Idea QR Task Cards - Set Two
Inferencing
Text Features
* Text Features - task cards for scoot or review with or without QR codes
ELA QR Bundle
* ELA QR Bundle: Character Traits, Inferencing, Text Features, Main Idea
Here are some Math activities if you are interested:
* QR Scavenger Hunt addition and subtraction with and without regrouping
* 4 digit addition task cards with and without regrouping; with or w/o QR codes
Bundle
Compare
* Compare Numbers to 50
Fractions
* Fractions QR word problem task cards or scoot - Tek 2.3 A-D
Measure
* First Grade measure QR codes Tek 1.7 A-D
* Second Grade Measure length QR word problem task cards - Tek 2.9 A-E
Money
* Money Addition Word Problems QR Code Scavenger Hunt
* First Grade money center tek 1.4
Multiplication
* Multiplication QR task cards missing factors - commutative property
* Multiplication QR codes Scavenger Hunt one digit times two digit number
* Single Digit times Double Digit Multiplication QR code Scavenger Hunt
* Third Grade Math Tek 3.4d and 3.4e
One More One Less, and or Ten More Ten Less
* 1 more, 1 less, 10 more, 10 less to 120 - self checking QR code task cards
* 1 more, 1less to 20 Kindergarten Tek 2f
* Turkey Trouble 1 more, 1less to 20
* Turkey Trouble 1 more, 1 less to 120
Personal Financial Literacy
* Personal Financial Literacy QR Task Cards Tek 2.11A-C
* Personal Financial Literacy QR Task Cards Tek 1.9 A-D
* Personal Financial Literacy QR Task Cards Tek 2.11 D-F
* Third Grade Personal Financial Literacy Teks 3.9 a-f
* Personal Financial Literacy Vocabulary QR
Place Value
* Third Grade Place Value Teks 3.2 A-D
* Second Grade Place Value Teks 2.2 A-F
* First Grade Place Value Teks 1.2 A-C
* First Grade Place Value Safari Teks 1.2 B 32 cards
* Place value task cards Base Ten Units to 1,200 with QR and matching cards
* Place value task cards Base Ten Units to 1,200 with optional matching cards
Skip Counting
* First Grade - Skip Count to 120 Task Cards and Posters
Solve Word Problems
* Solve word problems within 20 QR task cards - Tek 1.3B or CC 1.OAB.3
* Multiple Math Skills Word Problems QR Task Cards for Scoot - Math Center (Owls)
* Graphs and T-Charts word problems - TEK 1.8a-c CC MD4 - with or w/o QR codes
* Strategies to Solve within 20 - TEK 1.3e,f CC 0A.6 - with or w/o QR codes
STAAR Review NEW TEKS
* Third Grade NEW TEKS STAAR Review
Time
* Clocks Time to Minute - 32 QR code Task Cards
* Task Cards - Time to hour and half hour - self checking QR codes
Arrays
* Task Cards - Arrays to 10 x 10 with or without QR: optional matching cards included
Here are some fun Digital Papers to dress up your worksheets or powerpoint presentations :)
Digital Papers
* 16 Fun Film and Camera Digital Backgrounds
* 16 fun Digital Backgrounds
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4 ratings | 1,347 | 5,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-04 | longest | en | 0.865582 |
https://investmentongold.com/for-newbies/rental-property-return-on-investment.html | 1,620,498,132,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00238.warc.gz | 340,391,105 | 18,310 | # Rental property return on investment
## What is the 2% rule in real estate?
The 2% Rule states that if the monthly rent for a given property is at least 2% of the purchase price, it will likely cash flow nicely. It looks like this: monthly rent / purchase price = X. If X is less than 0.02 (the decimal form of 2%) then the property is not a 2% property.
## What is a good rate of return on an investment property?
Most real estate experts agree anything above 8% is a good return on investment, but it’s best to aim for over 10% or 12%. Real estate investors can find the best investment properties with high cash on cash return in their city of choice using Mashvisor’s Property Finder!
## How do you calculate return on investment property?
To calculate the property’s ROI:
1. Divide the annual return (\$9,600) by the amount of the total investment or \$110,000.
2. ROI = \$9,600 ÷ \$110,000 = 0.087 or 8.7%.
## Why rental properties are a bad investment?
There are four big reasons for this: it likely won’t generate the income you expect, it’s hard to generate a compelling return, a lack of diversification is likely to hurt you in the long run and real estate is illiquid, so you can’t necessarily sell it when you want.
## What is the 70 percent rule?
When determining the maximum price you should consider paying for a property, the 70% Rule of real estate investing dictates that you should pay no more than 70% of the after repair value (ARV), minus repair costs.
You might be interested: Investment blogs for beginners
## What does 7.5% cap rate mean?
With that caveat, to understand a CAP rate you simply take the building’s annual net operating income divided by purchase price. For example, if an investment property costs \$1 million dollars and it generates \$75,000 of NOI (net operating income) a year, then it’s a 7.5 percent CAP rate.4 мая 2017 г.
## What is a reasonable return on investment?
Generally speaking, if you’re estimating how much your stock-market investment will return over time, we suggest using an average annual return of 6% and understanding that you’ll experience down years as well as up years.
## How much cash flow is good for rental property?
The 1% rule is a formula used in rental real estate to determine whether a property is likely to have positive cash flow. The rule states the property’s rental rate should be, at a minimum, 1% of the purchase price. So if a property is for sale for \$200,000 it should produce a rental income of \$2,000 a month or more.
## Is the 1% rule realistic?
@Bryan Beal yes, the 1% rule is realistic in numerous markets, however, every investor is different and has different goals. There are many here that want immediate cash flow and typically the homes that are lower in price will achieve the 1% to 2% but these SFR ‘s typically don’t appreciate as much.
## How do I calculate percentage return on investment?
ROI is calculated by subtracting the initial value of the investment from the final value of the investment (which equals the net return), then dividing this new number (the net return) by the cost of the investment, and, finally, multiplying it by 100.
You might be interested: How does a typical variable life policy investment account grow?
## What are the advantages of owning a rental property?
Here are a few perks to becoming a landlord:
• Passive income source. Perhaps the biggest benefit to owning rental property is that it’s a passive income source. …
• Greater security. …
• Flexibility to sell at the right time. …
• Option to move back. …
• Property value appreciation. …
• Diversification of investments.
## Is rental real estate a good investment?
Conclusion. Rental properties can generate income, but the return on investment doesn’t typically happen right away. Rental property investments are also risky because of how many variables can affect its performance, like the housing market or your ability to keep it rented. | 887 | 3,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-21 | latest | en | 0.887997 |
https://www.tutorialgateway.org/c-program-to-find-sum-of-first-and-last-digit-of-a-number/ | 1,726,723,620,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00147.warc.gz | 964,966,233 | 16,845 | # C Program to Find Sum of First and Last Digit Of a Number
How to write a C Program to Find Sum of First and Last Digit Of a Number with an example?.
## C Program to Find Sum of First and Last Digit Of a Number
This program allows the user to enter any number. And then, it is going to find the Sum of First and Last Digit of the user-entered value.
```#include <stdio.h>
#include <math.h>
int main()
{
int Number, FirstDigit, Count, LastDigit, Sum = 0;
printf("\n Please Enter any Number that you wish : ");
scanf("%d", & Number);
Count = log10(Number);
FirstDigit = Number / pow(10, Count);
LastDigit = Number % 10;
Sum = FirstDigit + LastDigit;
printf(" \n The Sum of First Digit (%d) and Last Digit (%d) of %d = %d", FirstDigit, LastDigit, Number, Sum);
return 0;
}```
In this above C Programming example, Number = 2345
Count = log10(Number) = 3
FirstDigit = 2345 / pow(10, 3) = 2345 / 1000 = 2.345 = 2
LastDigit = 2345 % 10 = 5
Sum = 2 + 5 = 7
## Program to Find Sum of First and Last Digit Of a Number Example 2
This program will use While Loop to find the First Digit of the user entered value. We already explained the Analysis part in our previous articles. So, Please refer First and Last article to understand the same.
```#include <stdio.h>
int main()
{
int Number, FD, LD, Sm = 0;
printf("\n Please Enter any Number that you wish : ");
scanf("%d", & Number);
FD = Number;
while(FD >= 10)
{
FD = FD / 10;
}
LD = Number % 10;
Sm = FD + LastDigit;
printf(" \n The Sum of First Digit (%d) and Last Digit (%d) of %d = %d", FD, LD, Number, Sm);
return 0;
}```
`````` Please Enter any Number that you wish : 45896
The Sum of First Digit (4) and Last Digit (6) of 45896 = 10``````
## C Program to calculate Sum of First and Last Digit Of a Number using Function
This program to calculate sum of first and last digit is the same as above, but this time we divided the code using the Functions concept.
```#include <stdio.h>
int FirDi(int num);
int LstDi(int num);
int main()
{
int Number, First, Last, Sm = 0;
printf("\n Please Enter any Number that you wish : ");
scanf("%d", & Number);
First = FirDi(Number);
Last = LstDi(Number);
Sm = FirstDigit + LastDigit;
printf(" \n The Sum of First (%d) and Last (%d) of %d = %d", First, Last, Number, Sm);
return 0;
}
int FirDi(int num)
{
while(num >= 10)
{
num = num / 10;
}
return num;
}
int LstDi(int num)
{
return num % 10;
}```
`````` Please Enter any Number that you wish : 45862
The Sum of First (4) and Last (2) of 45862 = 6`````` | 753 | 2,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.731092 |
https://wikiwordy.com/word/antisymmetric | 1,685,343,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644683.18/warc/CC-MAIN-20230529042138-20230529072138-00413.warc.gz | 691,741,968 | 42,196 | # Antisymmetric
Word ANTISYMMETRIC Character 13 Hyphenation N/A Pronunciations N/A
## Definitions and meanings of "Antisymmetric"
What do we mean by antisymmetric?
(of a binary relation R on a set S) Having the property that, for any two distinct elements of S, at least one is not related to the other via R; equivalently, having the property that, for any x, y ∈ S, if both xRy and yRx then x=y.
(of certain mathematical objects) Whose sign changes on the application of a matrix transpose or some generalisation thereof:
A play on words meaning asymmetrical or no symmetry related to antisymmetric Urban Dictionary
## Synonyms and Antonyms for Antisymmetric
• Synonyms for antisymmetric
• Antonyms for antisymmetric
## The word "antisymmetric" in example sentences
(A world containing such wonders as Borges's Aleph, where parthood is not antisymmetric, might by contrast be finite and yet atomless.) ❋ Unknown (2009)
Notice that the relation thus defined is asymmetric (rather than antisymmetric): it doesn't permit any object to be existentially dependent upon itself. ❋ Lowe, E. Jonathan (2009)
As already mentioned, however, most contemporary authors are inclined to construe the relation of material constitution as a sui generis, non-mereological relation, or else to treat constitution itself as identity (hence, given (16), as a limit case of an antisymmetric parthood relation; see e.g. Noonan 1993). ❋ Unknown (2009)
The Fock operator (represented as a matrix) also contains some off diagonal elements, corresponding to the fact that you can swap the electrons in any two states without really changing anything (except the overall sign of the wave function), due to the fact that the full multi-electron state must be antisymmetric. ❋ Sean (2008)
For every equation there is a symmetric or antisymmetric equation that link different phenomena. ❋ Unknown (2008)
However any weighted average of a series of numbers, in which the weights are antisymmetric about the central value, gives an estimate of the trend. ❋ Unknown (2007)
The control system went into oscillation 37 myr BP when Antarctica started moving into its present position, the temperature of the ocean and that of the rest of the environment opposing each other in antisymmetric mode. ❋ Unknown (2007)
Back to Wikipedia, this time on fermions: Fermions . . . are particles which form totally-antisymmetric composite quantum states. ❋ Horace Jeffery Hodges (2005)
Fermions have half-integral spin and are described by wavefunctions that are antisymmetric in the exchange of two particles, i.e. the wavefunctions change sign when two particles change places, and they follow what is called ❋ Unknown (1996)
The two particles that are used in this theoretical exploration are in what Christandl calls an "antisymmetric state." ❋ Unknown (2010)
Thus we have seen that by making a measurement which distinguishes between the symmetric and antisymmetric subspaces of our two systems, if we will observe symmetric states half the time and anti-symmetric states the other half the time. ❋ Unknown (2010)
It is tempting to envisage that in human and other vertebrate or invertabrate ghosts with sexual differences a trace of this is carried over, and might be representable by the antisymmetric wave functions characteristic of the Fermi-Dirac statistics. ❋ Unknown (2009)
Order (relation): An irreflexive antisymmetric transitive binary relation on a set. ❋ Unknown (2009)
● Let St and At be the symmetric and antisymmetric products of the total composite system. ❋ Unknown (2009)
A (strict) partial order is which is irreflexive, antisymmetric and transitive. ❋ Unknown (2008)
Order (group theory), the least positive integer (if one exists) such that raising a group element to that power gives the identity Order (relation), an irreflexive antisymmetric transitive binary ❋ Unknown (2008)
Projection operator that projects onto an antisymmetric subspace of a tensor product space of identical linear spaces; in quantum mechanics used to symmetry-adapt fermion wave functions. ❋ Unknown (2008)
y™, and we shall take relation ¤ to be reflexive, antisymmetric and transitive. ❋ Mulligan, Kevin (2007)
(called an antisymmetric tensor of the second order) that combines electricity and magnetism into a single ❋ BANESH HOFFMANN (1968)
The obvious is this: regardless of how one feels about matters of ontology, if ˜part™ stands for the general relation exemplified by (1) - (8) and (12) - (15) above, then it stands for a partial ordering ” a reflexive, transitive, antisymmetric relation: ❋ Unknown (2009)
that pictograph is antisymmetrical ❋ This Guy (2004) | 1,058 | 4,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-23 | longest | en | 0.88895 |
http://passmathsng.com/jambmaths/60?page=1 | 1,524,219,355,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937440.13/warc/CC-MAIN-20180420100911-20180420120911-00094.warc.gz | 239,372,362 | 7,454 | # Jambmaths
Maths Question
Question 41
No 1 2 3 4 5 6 Frequency 30 43 54 40 41 32
A dice is rolled 240 times and the result depicted in the table above. If a pie chart is constructed to represent the data, the angle corresponding to 4 is
Question 42
Interval(Years) 10–12 13–15 16–18 19– 20 21–23 No of Pupils 6 14 15 10 32
The table above shows the frequency distribution of ages (in years) of pupils in a certain secondary, what percentage of the total number of pupils is over 15 but less than 21 years
Question 43
Interval(Years) 10–12 13–15 16–18 19– 20 21–23 No of Pupils 6 14 15 10 32
The table above shows the frequency distribution of ages (in years) of pupils in a certain secondary
The cumulative frequency curve above represent the ages of students in a school which age group do 70% of the students belong
Question 44
Find the sum of the range and mode of the set of numbers 10,5, 10, 9,8, 7,7,10, 8,10, 8,4, 6,9,10,9,10,9,7,10,6,5.
Question 45
If the mean of number 0, x, x + 2, 3x + 6, and 4x + 8 is 4. Find the mean deviation
Question 46
The variance of x, 2x, 3x, 4x and 5x is
Question 47
In how many can a delegation of 3 be chosen from among 5 men and 3 women. If at least one man and at least one woman must be included.
Question 48
In how many ways can the word “MATHEMATICS” be arranged?
Question 49
X and Y are two events. The probability of X or Y is 0.7 and the probability of X is 0.4. If X and Y are independent. Find the probability of Y
Question 50
If U ={x: x is an integer and 1≤ x ≤ 20}
E1 = {x: x is a multiple of 3}
E2 = {x: x is a multiple of 4}
and an integer is picked at random from U, find the probability that it is not in E2 | 564 | 1,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-17 | latest | en | 0.887844 |
https://cracku.in/cat-profit-loss-interest-questions | 1,721,323,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00461.warc.gz | 169,151,594 | 27,385 | # 70+ CAT Profit, Loss and Interest Questions With Video Solutions
To help students overcome this difficulty, we have compiled a comprehensive collection of CAT profit and loss, interest questions from all the CAT previous years' papers in pdf format with detailed video solutions. Every question has video solution that is explained by CAT experts. Over the past few years, CAT Profit and Loss questions have made a recurrent appearance in the Quant section. You can expect around 1-2 questions in the new format of the CAT Quant section. Practising free CAT mock where you'll get a fair idea of how questions are asked, and type of questions asked in CAT Exam. Here download the PDF for free without signing up. Click the link below to download the CAT profit, loss and interest questions PDF with detailed video solutions.
## CAT Profit, Loss And Interest Questions Weightage Over Past 5 Years
Year Weightage 2023 6 2022 4 2021 5 2020 8 2019 7 2018 2
## CAT Profit, Loss and Interest Equations Formulas PDF
Profit and loss, Simple interest and Compound interest are some of the essential topics in the quantitative aptitude section, and it is vital to have a clear understanding of the formulas related to them. To help the aspirants to ace this topic, we have made a PDF containing a comprehensive list of formulas, tips, and tricks that you can use to solve Profit, loss and interest questions with ease and speed. Click on the below link to download the CAT profit, loss and interest formulas PDF.
1. Compound Interest - Instalments
If an amount 'P' is borrowed for 'n' years at r% per annum compounded annually, and x is the installment that is paid at the end of each year, starting from the first year, then:
$$P\ =\ \dfrac{\ x}{1+\dfrac{r}{100}}+\ \dfrac{\ x}{\left(1+\dfrac{r}{100}\right)^{^2}}+...+\ \dfrac{\ x}{\left(1+\dfrac{r}{100}\right)^{^n}}$$
or
$$P\ \left(1+\frac{r}{100}\right)^{n\ }=\ x\ \left(\left(1+\frac{r}{100}\right)^{n-1}+\left(1+\frac{r}{100}\right)^{n-2\ }...\ +1\right)$$
2. Consecutive Profit %
When there are two successive profits of $$x\%$$ and $$y\%$$ then the net percentage profit $$=\ \dfrac{\ x+y+xy}{100}$$.
When there is a profit of $$x\%$$ and loss of $$y\%$$ then net percentage profit or loss $$=\ \dfrac{\ x-y-xy}{100}$$
3. Simple and Compound Interest
The principal amount is P, rate of interest is R and time of loan is T
Simple Interest = $$\dfrac{P*T*R}{100}$$
Amount = Principal + Simple Interest
Compound Interest = $$P(1+\dfrac{R}{100})^{T}$$ - P
For the same principal, positive rate of interest and time period, the compound interest on the loan is always greater than the simple interest.
4. SP - CP - Discount - Marked Price
The cost price of an article is C.P, the selling price is S.P and the marked price is M.P
Profit (Loss) = S.P – C.P
% Profit (Loss) = Profit (Loss)/C.P *100
Discount = M.P – S.P
% Discount = Discount/M.P * 100
Total increase in price due to two subsequent increases of X% and Y% is (X+Y+XY/100)%
## CAT 2023 Profit, Loss and Interest questions
#### Question 1
A merchant purchases a cloth at a rate of Rs.100 per meter and receives 5 cm length of cloth free for every 100 cm length of cloth purchased by him. He sells the same cloth at a rate of Rs.110 per meter but cheats his customers by giving 95 cm length of cloth for every 100 cm length of cloth purchased by the customers. If the merchant provides a 5% discount, the resulting profit earned by him is
#### Question 2
Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. He repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year. Then, the total interest, in rupees, paid over the three years is nearest to
#### Question 3
Gita sells two objects A and B at the same price such that she makes a profit of 20% on object A and a loss of 10% on object B. If she increases the selling price such that objects A and B are still sold at an equal price and a profit of 10% is made on object B, then the profit made on object A will be nearest to
#### Question 4
Minu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit. Then, Kanu sells it back to Minu at 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs.500, then the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu is
#### Question 5
Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is
#### Question 6
Jayant bought a certain number of white shirts at the rate of Rs 1000 per piece and a certain number of blue shirts at the rate of Rs 1125 per piece. For each shirt, he then set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10% and made a total profit of Rs.51000. If he bought both colors of shirts, then the maximum possible total number of shirts that he could have bought is
## CAT 2022 Profit, Loss and Interest questions
#### Question 1
Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is
#### Question 2
Nitu has an initial capital of ₹20,000. Out of this, she invests ₹8,000 at 5.5% in bank A, ₹5,000 at 5.6% in bank B and the remaining amount at x% in bank C, each rate being simple interest per annum. Her combined annual interest income from these investments is equal to 5% of the initial capital. If she had invested her entire initial capital in bank C alone, then her annual interest income, in rupees, would have been
#### Question 3
Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg and makes an overall profit of 64%. Then, Amal’s cost price for syrup, in rupees per kg, is
#### Question 4
Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is
## CAT 2021 Profit, Loss and Interest questions
#### Question 1
Raj invested ₹ 10000 in a fund. At the end of first year, he incurred a loss but his balance was more than ₹ 5000. This balance, when invested for another year, grew and the percentage of growth in the second year was five times the percentage of loss in the first year. If the gain of Raj from the initial investment over the two year period is 35%, then the percentage of loss in the first year is
#### Question 2
Amal purchases some pens at ₹ 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at ₹ 12 each. If the remaining pens are sold at ₹ 11 each, then he makes a net profit of ₹ 300, while he makes a net loss of ₹ 300 if the remaining pens are sold at ₹ 9 each. The wage of the employee, in INR, is
#### Question 3
Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to
#### Question 4
Bank A offers 6% interest rate per annum compounded half-yearly. Bank B and Bank C offer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests ₹ 10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in INR, to Rupa is
#### Question 5
Anil, Bobby, and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil’s share of investment is 70%. His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%. Chintu’s share of profit increases by ₹ 80 if the overall profit goes up from 15% to 17%. The amount, in INR, invested by Bobby is
## CAT 2020 Profit, Loss and Interest questions
#### Question 1
For the same principal amount, the compound interest for two years at 5% per annum exceeds the simple interest for three years at 3% per annum by Rs 1125. Then the principal amount in rupees is
#### Question 2
In the final examination, Bishnu scored 52% and Asha scored 64%. The marks obtained by Bishnu is 23 less, and that by Asha is 34 more than the marks obtained by Ramesh. The marks obtained by Geeta, who scored 84%, is
#### Question 3
A person spent Rs 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at 20% profit and the laptop at 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is
#### Question 4
A person invested a certain amount of money at 10% annual interest, compounded half-yearly. After one and a half years, the interest and principal together became Rs.18522. The amount, in rupees, that the person had invested is
#### Question 5
A man buys 35 kg of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg at this price, and 15 kg at a 10% discount. Accidentally, 3 kg of sugar is wasted. He sells the remaining sugar by raising the marked price by p percent so as to make an overall profit of 15%. Then p is nearest to | 2,631 | 10,083 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.856801 |
www.trivia-library.com | 1,369,480,090,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705939136/warc/CC-MAIN-20130516120539-00031-ip-10-60-113-184.ec2.internal.warc.gz | 779,492,012 | 3,774 | # History of the Metric System Part 2 Non-Metric System
## About the history of the metric system, the simple system of measurement the United States refuses to accept.
Meet the Meter
The metric system, like the U.S. currency, is a decimal system; all the units are related to each other by a factor of 10. Humans have used their 10 fingers to count on since the dawn of history, so the manipulation of units of 10 is almost 2nd nature. To convert from one metric unit to another, all one has to do is add easily memorized prefixes and shift the decimal point. For example, the prefix "centi" means one hundredth: c==1/100==10-2==0.01. "Milli" means one thousandth: m==1/1000==10-3==0.001. One millimeter is 1/1000 of a meter and 1/10 of a centimeter. "Kilo" means one thousand: k==1000==10 3. One thousand grams is one kilogram, which is approximately 2.2 lbs.; a thousand grams is also a liter in liquid measure, which corresponds roughly to one quart. Lists follow giving all the established prefixes and their equivalent values in nonmetric units; these have been prepared not only for weights and measures but for other physical quantities as well.
The nonmetric system and its shortcomings can be traced back to historical developments that took place before uniform reference standards could be established. For example, a medieval British ruler changed the Roman mile of 5,000' to 5,280' to make it conformable with the length of 8 furlongs. Another British king proclaimed that 3 kernels of grain--wheat or barley--laid end to end were the equivalent of one inch which, in turn, was 1/12 the length of a human foot. As a result we have remained saddled with a complicated system of units which have no relation to one another. There are troy ounces and avoirdupois ounces and liquid ounces. A quart of water has 57.75 cubic inches, but a quart of dry measure is equivalent to 67.20 cubic inches. Pricing or cost accounting of such irregular units using our decimal currency system is an unavoidably laborious process.
The only reason for the continued use of the nonmetric system is human inertia and an unwillingness to accept change. But, contrary to popular opinion, the metric system is already so well established in the U.S. that its official adoption does not constitute the introduction of a radically new system but merely the recognition of one which is already in use in many areas. For instance, we are used to 8-, 16-, or 35-millimeter film; doctors prescribe, pharmacists fill, and nurses administer medicine in cubic centimeter units. The consumption of electricity is measured in watts and kilowatts, and the engine displacement of automobiles is now commonly given in cubic centimeters. Spurred on by General Motors, Ford, IBM, Honeywell, and scores of other large companies who have announced their orderly conversion to the metric system, subcontractors, suppliers, and machine-tool manufacturers will increasingly work to metric standards. Presumably they will produce goods in nonmetric dimensions too, for some time to come, to satisfy the replacement market. But since that market will disappear in time, production eventually will be geared exclusively to metric standards.
You Are Here: Trivia » History of the Metric System » History of the Metric System Part 2 Non-Metric System
« History of the Metric System Part 1 History of the Metric System Part 3 Will America Change? » | 758 | 3,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2013-20 | longest | en | 0.931835 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-4-section-4-7-inverse-trigonometric-functions-exercise-set-page-629/129 | 1,679,730,420,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00465.warc.gz | 883,436,368 | 12,337 | ## Precalculus (6th Edition) Blitzer
Amplitude $=10$ and Period $=12$
We are given the function: $y=10 \cos (\dfrac{\pi}{6})$ Since, Amplitude $=|10|=10$ Also, Period is: $P=\dfrac{2 \pi}{(\pi/6)}=12$ Therefore, Amplitude $=10$ and Period $=12$ | 94 | 245 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-14 | latest | en | 0.543517 |
https://docs.trifacta.com/pages/diffpagesbyversion.action?pageId=156093053&selectedPageVersions=3&selectedPageVersions=4 | 1,686,119,335,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653608.76/warc/CC-MAIN-20230607042751-20230607072751-00187.warc.gz | 247,558,910 | 22,465 | ## Key
• This line was removed.
• Formatting was changed.
D toc
Excerpt
Extracts the ranked Datetime value from the values in a column, where `k=1` returns the maximum value, when a specified condition is met. The value for `k` must be between 1 and 1000, inclusive. Inputs must be Datetime.
`KTHLARGESTDATEIF` calculations are filtered by a conditional applied to the group.
For purposes of this calculation, two instances of the same value are treated as the same value of `k`. So, if your dataset contains three rows with column values `2020-02-15`, `2020-02-14`, and `2020-02-14`, then `KTHLARGESTDATEIF` returns `2020-02-14` for `k=2` and `2020-02-14` for `k=3`.
Input column must be of Datetime type. Other values column are ignored. If a row contains a missing or null value, it is not factored into the calculation.
Info
NOTE: When added to a transformation, this function is applied to the current sample. If you change your sample or run the job, the computed values for this function are updated. Transformations that change the number of rows in subsequent recipe steps do not affect the values computed for this step.
To perform a simple kth largest calculation on Datetime values without conditionals, use the `KTHLARGESTDATE` function. See KTHLARGESTDATE Function.
For a version of this function that applies to non-Datetime values, see KTHLARGESTIF Function.
D s lang vs sql
D s
snippet Basic
D lang syntax
RawWrangle true ref true pivot value: kthlargestdateif(transDate, 2, salesPerson == 'jsmith') limit:1
kthlargestdateif(transDate, 2, salesPerson == 'jsmith')
Output: Returns the secondmost recent Date (rank=2) from the `transDate` column when the `salesPerson` value is `jsmith`.
D s
snippet Syntax
D lang syntax
RawWrangle true syntax true pivot value:kthlargestdateif(col_ref, limit, test_expression) [group:group_col_ref] [limit:limit_count]
kthlargestdateif(col_ref, limit, test_expression) [group:group_col_ref] [limit:limit_count]
ArgumentRequired?Data TypeDescription
col_refYstringReference to the column you wish to evaluate.
k_integerYintegerThe ranking of the value to extract from the source column
test_expressionYstringExpression that is evaluated. Must resolve to `true` or `false`
D s lang notes
For more information on the `group` and `limit` parameter, see Pivot Transform.
### col_ref
Name of the column whose values you wish to use in the calculation. Inputs must be Datetime values.
D s
snippet usage
Required?Data TypeExample Value
YesString that corresponds to the name of the column`transactionDate`
### k_integer
Integer representing the ranking of the value to extract from the source column.
Info
NOTE: The value for `k` must be an integer between 1 and 1,000 inclusive.
• `k=1` represents the maximum value in the column.
• If k is greater than or equal to the number of values in the column, the minimum value is returned.
• Missing and null values are not factored into the ranking of `k`.
### test_expression
This parameter contains the expression to evaluate. This expression must resolve to a Boolean (`true` or `false`) value.
D s
snippet usage
Required?Data TypeExample Value
YesString expression that evaluates to `true` or `false``(LastName == 'Mouse' && FirstName == 'Mickey') `
D s
snippet Examples
### Example - KTHLARGESTDATE functions
Include Page
EXAMPLE - KTHLARGESTDATE Functions EXAMPLE - KTHLARGESTDATE Functions
D s also
label aggregate | 898 | 3,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-23 | latest | en | 0.598338 |
https://numbermatics.com/n/745251495121/ | 1,680,177,199,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949181.44/warc/CC-MAIN-20230330101355-20230330131355-00110.warc.gz | 470,181,889 | 6,411 | # 745251495121
## 745,251,495,121 is an odd composite number composed of four prime numbers multiplied together.
What does the number 745251495121 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 16 divisors.
745251495121 is an odd composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors.
## Prime factorization of 745251495121:
### 7 × 331 × 2143 × 150091
See below for interesting mathematical facts about the number 745251495121 from the Numbermatics database.
### Names of 745251495121
• Cardinal: 745251495121 can be written as Seven hundred forty-five billion, two hundred fifty-one million, four hundred ninety-five thousand, one hundred twenty-one.
### Scientific notation
• Scientific notation: 7.45251495121 × 1011
### Factors of 745251495121
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 4
• Sum of prime factors: 152572
### Divisors of 745251495121
• Number of divisors d(n): 16
• Complete list of divisors:
• Sum of all divisors σ(n): 854693490688
• Sum of proper divisors (its aliquot sum) s(n): 109441995567
• 745251495121 is a deficient number, because the sum of its proper divisors (109441995567) is less than itself. Its deficiency is 635809499554
### Bases of 745251495121
• Binary: 10101101100001000111001101011100110100012
• Base-36: 9ID3HXW1
### Squares and roots of 745251495121
• 745251495121 squared (7452514951212) is 555399790980085886804641
• 745251495121 cubed (7452514951213) is 413912524617799897078149870491656561
• The square root of 745251495121 is 863279.5000004343
• The cube root of 745251495121 is 9066.3876740773
### Scales and comparisons
How big is 745251495121?
• 745,251,495,121 seconds is equal to 23,696 years, 36 weeks, 12 minutes, 1 second.
• To count from 1 to 745,251,495,121 would take you about forty-seven thousand, three hundred ninety-three years!
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 745251495121 cubic inches would be around 755.5 feet tall.
### Recreational maths with 745251495121
• 745251495121 backwards is 121594152547
• The number of decimal digits it has is: 12
• The sum of 745251495121's digits is 46
• More coming soon!
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APA style:
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Some bits of this website may not work unless you switch it on. | 1,024 | 3,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-14 | latest | en | 0.785297 |
https://www.ca5.co/convert/length/how-many-feet-in-yard | 1,675,123,181,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00620.warc.gz | 696,325,186 | 3,278 | # How many Feet are in a Yard
1 yard (yd) in feet (ft).
## How many feet are in a yard
1 yard is equal to 3 feet:
1 yd = 3 ft | 45 | 129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | latest | en | 0.984465 |
http://explorersub.com/type-1/type-i-error-and-type-ii-error.php | 1,529,948,293,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00523.warc.gz | 99,526,191 | 6,637 | Home > Type 1 > Type I Error And Type Ii Error
Type I Error And Type Ii Error
Contents
Cambridge University Press. These terms are commonly used when discussing hypothesis testing, and the two types of errors-probably because they are used a lot in medical testing. So please join the conversation. Type I Error (False Positive Error) A type I error occurs when the null hypothesis is true, but is rejected. Let me say this again, a type I error occurs when the have a peek at these guys
You can unsubscribe at any time. The incorrect detection may be due to heuristics or to an incorrect virus signature in a database. So setting a large significance level is appropriate. Type II error is the error made when the null hypothesis is not rejected when in fact the alternative hypothesis is true.
Probability Of Type 1 Error
Such tests usually produce more false-positives, which can subsequently be sorted out by more sophisticated (and expensive) testing. Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... jbstatistics 101,105 views 8:11 Statistics 101: Visualizing Type I and Type II Error - Duration: 37:43. What Level of Alpha Determines Statistical Significance?
Medicine Further information: False positives and false negatives Medical screening In the practice of medicine, there is a significant difference between the applications of screening and testing. Statistical calculations tell us whether or not we should reject the null hypothesis.In an ideal world we would always reject the null hypothesis when it is false, and we would not In some cases a Type I error is preferable to a Type II error. Type 1 Error Psychology Common mistake: Confusing statistical significance and practical significance.
This feature is not available right now. Probability Of Type 2 Error Usually a type I error leads one to conclude that a supposed effect or relationship exists when in fact it doesn't. menuMinitab® 17 SupportWhat are type I and type II errors?Learn more about Minitab 17 When you do a hypothesis test, two types of errors are possible: type I and type II. click site Reply Bill Schmarzo says: April 16, 2014 at 11:19 am Shem, excellent point!
Let’s go back to the example of a drug being used to treat a disease. Power Of The Test If a test with a false negative rate of only 10%, is used to test a population with a true occurrence rate of 70%, many of the negatives detected by the Easy to understand! Please enter a valid email address.
Probability Of Type 2 Error
Etymology In 1928, Jerzy Neyman (1894–1981) and Egon Pearson (1895–1980), both eminent statisticians, discussed the problems associated with "deciding whether or not a particular sample may be judged as likely to http://statweb.stanford.edu/~susan/courses/s60/split/node100.html Please select a newsletter. Probability Of Type 1 Error The null hypothesis is true (i.e., it is true that adding water to toothpaste has no effect on cavities), but this null hypothesis is rejected based on bad experimental data. Type 3 Error This will then be used when we design our statistical experiment.
ProfKelley 26,173 views 5:02 z-score Calculations & Percentiles in a Normal Distribution - Duration: 13:40. More about the author This sort of error is called a type II error, and is also referred to as an error of the second kind.Type II errors are equivalent to false negatives. Various extensions have been suggested as "Type III errors", though none have wide use. Reply DrumDoc says: December 1, 2013 at 11:25 pm Thanks so much! Type 1 Error Calculator
• As a result of this incorrect information, the disease will not be treated.
• The results of such testing determine whether a particular set of results agrees reasonably (or does not agree) with the speculated hypothesis.
• However, if everything else remains the same, then the probability of a type II error will nearly always increase.Many times the real world application of our hypothesis test will determine if
• There are two kinds of errors, which by design cannot be avoided, and we must be aware that these errors exist.
• Type II Error (False Negative) A type II error occurs when the null hypothesis is false, but erroneously fails to be rejected. Let me say this again, a type II error occurs
• Joint Statistical Papers.
• Let’s look at the classic criminal dilemma next. In colloquial usage, a type I error can be thought of as "convicting an innocent person" and type II error "letting a guilty person go
External links Bias and Confounding– presentation by Nigel Paneth, Graduate School of Public Health, University of Pittsburgh v t e Statistics Outline Index Descriptive statistics Continuous data Center Mean arithmetic Contents 1 Definition 2 Statistical test theory 2.1 Type I error 2.2 Type II error 2.3 Table of error types 3 Examples 3.1 Example 1 3.2 Example 2 3.3 Example 3 Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. check my blog This error is potentially life-threatening if the less-effective medication is sold to the public instead of the more effective one.
The rate of the typeII error is denoted by the Greek letter β (beta) and related to the power of a test (which equals 1−β). Types Of Errors In Accounting Two types of error are distinguished: typeI error and typeII error. About Today Living Healthy Statistics You might also enjoy: Health Tip of the Day Recipe of the Day Sign up There was an error.
Therefore, you should determine which error has more severe consequences for your situation before you define their risks.
Thus it is especially important to consider practical significance when sample size is large. He’s presented most recently at STRATA, The Data Science Summit and TDWI, and has written several white papers and articles about the application of big data and advanced analytics to drive Reply Recent CommentsBill Schmarzo on Most Excellent Big Data Strategy DocumentHugh Blanchard on Most Excellent Big Data Strategy DocumentBill Schmarzo on Data Lake and the Cloud: Pros and Cons of Putting Types Of Errors In Measurement Cambridge University Press.
Show Full Article Related Is a Type I Error or a Type II Error More Serious? Khan Academy 708,056 views 6:40 Hypothesis Testing: Type I Error, Type II Error - Duration: 5:02. Please select a newsletter. http://explorersub.com/type-1/type-1and-type-2-error-in-statistics.php The probability of rejecting false null hypothesis.
Summary Type I and type II errors are highly depend upon the language or positioning of the null hypothesis. A Type II error is committed when we fail to believe a truth.[7] In terms of folk tales, an investigator may fail to see the wolf ("failing to raise an alarm"). Every experiment may be said to exist only in order to give the facts a chance of disproving the null hypothesis. — 1935, p.19 Application domains Statistical tests always involve a trade-off This is what is known as a Type II error.Type I and Type II Errors ExplainedIn more colloquial terms we can describe these two kinds of errors as corresponding to certain
A Type I error occurs when you are found guilty of a murder that you did not commit. The null hypothesis is false (i.e., adding fluoride is actually effective against cavities), but the experimental data is such that the null hypothesis cannot be rejected. Complete the fields below to customize your content. Kimball, A.W., "Errors of the Third Kind in Statistical Consulting", Journal of the American Statistical Association, Vol.52, No.278, (June 1957), pp.133–142.
Up next Type I Errors, Type II Errors, and the Power of the Test - Duration: 8:11. p.56. In this situation, the probability of Type II error relative to the specific alternate hypothesis is often called β. Security screening Main articles: explosive detection and metal detector False positives are routinely found every day in airport security screening, which are ultimately visual inspection systems.
While most anti-spam tactics can block or filter a high percentage of unwanted emails, doing so without creating significant false-positive results is a much more demanding task. A typeII error occurs when failing to detect an effect (adding fluoride to toothpaste protects against cavities) that is present. Often, the significance level is set to 0.05 (5%), implying that it is acceptable to have a 5% probability of incorrectly rejecting the null hypothesis.[5] Type I errors are philosophically a Sign in to add this to Watch Later Add to Loading playlists... Topics What's New Fed Meeting, US Jobs Highlight Busy Week Ahead Regeneron, Sanofi Drug Hits FDA Snag
Comment on our posts and share! The probability of making a type I error is α, which is the level of significance you set for your hypothesis test. | 1,900 | 8,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-26 | latest | en | 0.874356 |
https://academickids.com/encyclopedia/index.php/Cluster_analysis | 1,632,662,212,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00459.warc.gz | 126,061,728 | 7,891 | # Cluster analysis
Data clustering is a common technique for statistical data analysis, which is used in many fields, including machine learning, data mining, pattern recognition, image analysis and bioinformatics. Clustering is the classification of similar objects into different groups, or more precisely, the partitioning of a data set into subsets (clusters), so that the data in each subset (ideally) share some common trait - often proximity according to some defined distance measure.
Besides the term data clustering, there are a number of terms with similar meanings, including cluster analysis, automatic classification, numerical taxonomy, botryology and typological analysis.
Data clustering algorithms can be hierarchical or partitional. With hierarchical algorithms, successive clusters are found using previously established clusters, whereas partitional algorithms determine all clusters in one go. Hierarchical algorithms can be agglomerative (bottom-up) or divisive (top-down). Agglomerative algorithms begin with each element as a separate cluster and merge them in successively larger clusters. The divisive algorithms begin with the whole set and proceed to divide it into successively smaller clusters.
Contents
## Hierarchical clustering
### Introduction
Hierarchical clustering builds a hierarchy of clusters. The traditional representation of this hierarchy is a tree, with individual elements at one end and a single cluster with every element at the other. Agglomerative algorithms begin at the top of the tree, whereas divisive algorithms begin at the bottom. (In the figure, the arrows indicate an agglomerative clustering.)
Missing image
Hierarchical_clustering_diagram.png
Cutting the tree at a given height will give a clustering at a selected precision. In the above example, cutting after the second row will yield clusters {a} {b c} {d e} {f}. Cutting after the third row will yield clusters {a} {b c} {d e f}, which is a coarser clustering, but with fewer clusters.
### Agglomerative hierarchical clustering
This method builds the hierarchy from the individual elements by progressively merging clusters. Again, we have six elements {a} {b} {c} {d} {e} and {f}. the first step is to determine which elements to merge in a cluster. Usually, we want to take the two closest elements, therefore we must define a distance [itex]d(\mathrm{element}_1,\mathrm{element}_2)[itex] between elements.
Suppose we have merged the two closest elements b and c, we now have the following clusters {a}, {b, c}, {d}, {e} and {f}, and want to merge them further. But to do that, we need to take the distance between {a} and {b c}, and therefore define the distance between two clusters. Usually the distance between two clusters [itex]\mathcal{A}[itex] and [itex]\mathcal{B}[itex] is one of the following:
• the maximum distance between elements of each cluster (also called complete linkage clustering) [itex] \max_{x \in \mathcal{A},\, y \in \mathcal{B}} d(x,y) [itex]
• the minimum distance between elements of each cluster (also called single linkage clustering) [itex] \min_{x \in \mathcal{A},\, y \in \mathcal{B}} d(x,y) [itex]
• the mean distance between elements of each cluster (also called average linkage clustering) [itex] {1 \over {\mathrm{card}(\mathcal{A})\mathrm{card}(\mathcal{B})}}\sum_{x \in \mathcal{A}}\sum_{ y \in \mathcal{B}} d(x,y) [itex]
• the sum of all intra cluster variance
• the increase in variance for the cluster being merged (Ward's criterion)
Each agglomeration occurs at a greater distance between clusters than the previous agglomeration, and one can decide to stop clustering either when the clusters are too far apart to be merged (distance criterion) or when there is a sufficiently small number of clusters (number criterion).
## Partitional clustering
### k-means and derivatives
#### k-means clustering
The k-means algorithm assigns each point to the cluster whose center (or centroid) is nearest. The centroid is the point generated by computing the arithmetic mean for each dimension separately for all the points in the cluster.
Example: The data set has three dimensions and the cluster has two points: X = (x1, y1, z1) and Y = (x2, y2, z2). Then the centroid Z becomes Z = (x3, y3, z3), where x3 = (x1 + x2)/2 and y3 = (y1 + y2)/2 and z3 = (z1 + z2)/2
This is the basic structure of the algorithm (J. MacQueen, 1967):
• Randomly generate k clusters and determine the cluster centers or directly generate k seed points as cluster centers
• Assign each point to the nearest cluster center.
• Recompute the new cluster centers.
• Repeat until some convergence criterion is met (usually that the assignment hasn't changed).
The main advantages of this algorithm are its simplicity and speed, which allows it to run on large datasets. Yet it does not systematically yield the same result with each run of the algorithm. Rather, the resulting clusters depend on the initial assignments. The k-means algorithm maximizes inter-cluster (or minimizes intra-cluster) variance, but does not ensure that the solution given is not a local minimum of variance.
#### Fuzzy c-means clustering
One of the problems of the k-means algorithm is that it gives a hard partitioning of the data, that is to say that each point is attributed to one and only one cluster. But points on the edge of the cluster, or near another cluster may not be as much in the cluster as point in the center of cluster.
Therefore, in fuzzy clustering, each point does not pertain to a given cluster, but has a degree of belonging to a certain cluster, as in fuzzy logic. For each point x we have a coefficient giving the degree of being in the kth cluster [itex]u_k(x)[itex]. Usually, the sum of those coefficients has to be one, so that [itex]u_k(x)[itex] denotes a probability of belonging to a certain cluster:
[itex] \forall x \sum_{k=1}^{\mathrm{num.}\ \mathrm{clusters}} u_k(x) \ =1.[itex]
With fuzzy c-means, the centroid of a cluster is computed as being the mean of all points, weighted by their degree of belonging to the cluster, that is
[itex]\mathrm{center}_k = {{\sum_x u_k(x) x} \over {\sum_x u_k(x)}}.[itex]
The degree of being in a certain cluster is the inverse of the distance to the cluster
[itex]u_k(x) = {1 \over d(\mathrm{center}_k,x)},[itex]
then the coefficients are normalized and fuzzyfied with a real parameter [itex]m>1[itex] so that their sum is 1. So [itex]u_k(x) = \frac{1}{\sum_j \left(\frac{d(\mathrm{center}_k,x)}{d(\mathrm{center}_j,x)}\right)^{\frac{1}{m-1}}}[itex]
The fuzzy c-means algorithm is greatly similar to the k-means algorithm; when [itex]m\leftarrow 1[itex], algorithms are equivalent:
• Choose a number of clusters
• Assign randomly to each point coefficients for being in the clusters
• Repeat until the algorithm has converged (that is, the coefficients' change between two iterations is no more than [itex]\epsilon[itex], the given sensitivity threshold) :
• Compute the centroid for each cluster, using the formula above
• For each point, compute its coefficients of being in the clusters, using the formula above
The fuzzy c-means algorithm minimizes intra-cluster variance as well, but has the same problems as k-means, the minimum is local minimum, and the results depend on the initial choice of weights.
## Applications
In biology has two main applications in the fields of computational biology and bioinformatics.
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy | 1,797 | 7,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-39 | latest | en | 0.858559 |
https://es.planetcalc.com/397/?thanks=1 | 1,643,272,540,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00495.warc.gz | 292,108,692 | 11,046 | # Kinematics Problem: Ball Thrown Straight Up
This online calculator solves kinematics problem from the Svetozarov's book for MEPI matriculates
### Esta página existe gracias a los esfuerzos de las siguientes personas:
#### cruelity0_0
Creado: 2015-12-23 17:41:02, Última actualización: 2021-03-20 15:24:43
Well, we finally got to the acceleration topic, which, as you know, is the change in speed over time.
Condition:
The second body was thrown vertically upward next to the first one after t with the same speed v. How long after throwing the second body and at which height h the two bodies will collide?
Solution:
Distance traveled equation
$s(t)=v_0t+\frac{at^2}{2}$
For the first body
$y(t)=v(t+t_1)-\frac{g(t+t_1)^2}{2}$
For the second body
$y(t)=vt_1-\frac{gt_1^2}{2}$
Accordingly, when they meet, their coordinates will coincide, i.e.
$v(t+t_1)-\frac{g(t+t_1)^2}{2}=vt_1-\frac{gt_1^2}{2}$,
from which
$t_1=\frac{v}{g}-\frac{t}{2}$
After finding time, substitute it in any formula for the distance and find h
Gravitational acceleration is assumed to equal 9.8 m/s2
#### Kinematics. Throwing body up problem
Digits after the decimal point: 2
Time to collision t1, (s)
Collision height h, (m) | 372 | 1,209 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 5, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-05 | longest | en | 0.726253 |
https://www.cnblogs.com/fzl194/p/9693724.html | 1,718,724,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00340.warc.gz | 627,711,562 | 10,947 | # BZOJ 1066 蜥蜴 最大流
https://www.lydsy.com/JudgeOnline/problem.php?id=1066
1 #include<bits/stdc++.h>
2 #define IOS ios::sync_with_stdio(false);//不可再使用scanf printf
3 #define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数，会超时
4 #define Min(a, b) ((a) < (b) ? (a) : (b))
5 #define Mem(a) memset(a, 0, sizeof(a))
6 #define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1))
7 #define MID(l, r) ((l) + ((r) - (l)) / 2)
8 #define lson ((o)<<1)
9 #define rson ((o)<<1|1)
10 #define Accepted 0
12 using namespace std;
14 {
15 int x=0,f=1;char ch=getchar();
16 while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
17 while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
18 return x*f;
19 }
20 typedef long long ll;
21 const int maxn = 1000 + 10;
22 const int MOD = 1000000007;//const引用更快，宏定义也更快
23 const int INF = 1e9 + 7;
24 const double eps = 1e-6;
25 struct edge
26 {
27 int u, v, c, f;
28 edge(int u, int v, int c, int f):u(u), v(v), c(c), f(f){}
29 };
30 vector<edge>e;
31 vector<int>G[maxn];
32 int level[maxn];//BFS分层，表示每个点的层数
33 int iter[maxn];//当前弧优化
34 int m;
35 void init(int n)
36 {
37 for(int i = 0; i <= n; i++)G[i].clear();
38 e.clear();
39 }
40 void addedge(int u, int v, int c)
41 {
42 //cout<<u<<" "<<v<<" "<<c<<endl;
43 e.push_back(edge(u, v, c, 0));
44 e.push_back(edge(v, u, 0, 0));
45 m = e.size();
46 G[u].push_back(m - 2);
47 G[v].push_back(m - 1);
48 }
49 void BFS(int s)//预处理出level数组
50 //直接BFS到每个点
51 {
52 memset(level, -1, sizeof(level));
53 queue<int>q;
54 level[s] = 0;
55 q.push(s);
56 while(!q.empty())
57 {
58 int u = q.front();
59 q.pop();
60 for(int v = 0; v < G[u].size(); v++)
61 {
62 edge& now = e[G[u][v]];
63 if(now.c > now.f && level[now.v] < 0)
64 {
65 level[now.v] = level[u] + 1;
66 q.push(now.v);
67 }
68 }
69 }
70 }
71 int dfs(int u, int t, int f)//DFS寻找增广路
72 {
73 if(u == t)return f;//已经到达源点，返回流量f
74 for(int &v = iter[u]; v < G[u].size(); v++)
75 //这里用iter数组表示每个点目前的弧，这是为了防止在一次寻找增广路的时候，对一些边多次遍历
76 //在每次找增广路的时候，数组要清空
77 {
78 edge &now = e[G[u][v]];
79 if(now.c - now.f > 0 && level[u] < level[now.v])
80 //now.c - now.f > 0表示这条路还未满
81 //level[u] < level[now.v]表示这条路是最短路，一定到达下一层，这就是Dinic算法的思想
82 {
83 int d = dfs(now.v, t, min(f, now.c - now.f));
84 if(d > 0)
85 {
86 now.f += d;//正向边流量加d
87 e[G[u][v] ^ 1].f -= d;
88 //反向边减d，此处在存储边的时候两条反向边可以通过^操作直接找到
89 return d;
90 }
91 }
92 }
93 return 0;
94 }
95 int Maxflow(int s, int t)
96 {
97 int flow = 0;
98 for(;;)
99 {
100 BFS(s);
101 if(level[t] < 0)return flow;//残余网络中到达不了t，增广路不存在
102 memset(iter, 0, sizeof(iter));//清空当前弧数组
103 int f;//记录增广路的可增加的流量
104 while((f = dfs(s, t, INF)) > 0)
105 {
106 flow += f;
107 }
108 }
109 return flow;
110 }
111 struct node
112 {
113 int x, y;
114 node(){}
115 node(int x, int y):x(x), y(y){}
116 bool operator < (const node& a)const
117 {
118 return x < a.x || x == a.x && y < a.y;
119 }
120 };
121 map<node, int>ID;
122 char Map[30][30];
123 int main()
124 {
125 int n, m, d;
126 scanf("%d%d%d", &n, &m, &d);
127 int tot = 0;
128 for(int i = 0; i < n; i++)
129 {
130 scanf("%s", Map[i]);
131 for(int j = 0; j < m; j++)
132 {
133 if(Map[i][j] != '0')
134 {
135 ID[node(i, j)] = ++tot;
136 addedge(tot, tot + 500, Map[i][j] - '0');
137 if(i < d || j < d || n - i <= d || m - j <= d)
138 addedge(tot + 500, 0, INF);
139 }
140 }
141 }
142 for(int i = 0; i < n; i++)
143 {
144 for(int j = 0; j < m; j++)
145 {
146 if(Map[i][j] != '0')
147 for(int x = -d; x <= d; x++)
148 {
149 for(int y = -d; y <= d; y++)
150 {
151 int xx = x + i;
152 int yy = y + j;
153 if(xx >= 0 && xx < n && yy >= 0 && yy < m && (xx != i || yy != j) && Map[xx][yy] != '0')
154 {
155 if(abs(i - xx) + abs(j - yy) <= d)addedge(ID[node(i, j)] + 500, ID[node(xx, yy)], INF);
156 }
157 }
158 }
159 }
160 }
161 int cnt = 0;
162 for(int i = 0; i < n; i++)
163 {
164 scanf("%s", Map[i]);
165 for(int j = 0; j < m; j++)if(Map[i][j] == 'L')
166 {
173 } | 1,732 | 4,812 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.192959 |
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find the opportunity cost
As the French franc appreciates in value relative to the U.S. dollar, what happens to the price of U.S. goods in France? What happens to the price of French goods in the U.S.?
Suppose the Canadian dollar (C\$) price of one British pound is C\$2.414. A hotel room in London costs 110 pounds, while a similar hotel room in Toronto costs C\$260. In which city is the hotel room cheaper, and by how much?
The nation of Turkovakia produces only turkeys. In 2000, turkey was priced at \$6 per unit. In 2005, turkey was priced at \$12 per unit. If 2005 is the base year, then the price index for 2000 is
50
100
150
200
N. Kamazoon can produce the combo of 18 televisions and 24 widgets with its stock of resources. S. Kamazoon can produce 15 televisions and 15 widgets with its resources. Explain why this information does not allow you to establish limits on the terms of trade between these two countries
Solution Preview
1. As the French Franc appreciates in value relative to the US dollar US goods become cheaper in France and French goods become costlier in the U.S.
This can be shown using a simple hypothetical example. Suppose before appreciation each U.S. dollar is equivalent to 1 French franc. Consider two cars: one made by GM in the U.S., and another made by Renault in France. Suppose the cars cost \$10000 and 10000 Francs respectively. Before appreciation (assuming away tariffs and ...
Solution Summary
The solution assists with finding the opportunity cost.
\$2.19 | 364 | 1,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-39 | latest | en | 0.93433 |
https://la.mathworks.com/matlabcentral/cody/problems/28-counting-money/solutions/925705 | 1,603,854,352,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107896048.53/warc/CC-MAIN-20201028014458-20201028044458-00710.warc.gz | 384,819,367 | 16,991 | Cody
# Problem 28. Counting Money
Solution 925705
Submitted on 20 Jul 2016 by Jonathan Perry
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = {'\$12,001.87','\$0.04','\$103,887.55','\$0.32'}; b = 115889.78; assert(abs(moneySum(a)-b) < 1e-4)
z = 12001.87 0.04 103887.55 0.32 b = 1.1589e+05
2 Pass
a = {'\$0.02'}; b = 0.02; assert(abs(moneySum(a)-b) < 1e-4)
z = 0.02 b = 0.0200
3 Pass
a = {'\$81.47','\$12.69','\$91,337.60'}; b = 91431.76; assert(abs(moneySum(a)-b) < 1e-4)
z = 81.47 12.69 91337.60 b = 9.1432e+04
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Additional info for A Bernstein Property of Affine Maximal Hypersurfaces
Sample text
R be such that 2 i θi = 1. Denote According to Lemma 7, the random variable Y = n i=1 θi X i A Berry-Esseen type inequality for convex bodies with an unconditional basis sup |P (ε + Y ≥ t) − t∈R (t)| ≤ Cε2 , (59) with some universal constant C ≥ 1. The random variable Y has an even, log-concave density by Prékopa–Leindler. We may thus apply Lemma 9, and conclude from (59) that sup |P (α ≤ Y ≤ β) − [ (α) − α≤β (β)]| ≤ 2 sup |P (Y ≥ t) − t∈R (t)| ≤ C ε2 . The theorem is thus proven. Appendix: Proof of Theorem 2 With Cédric Villani’s permission, we reproduce below the proof of Theorem 2 from his book [40, Sect.
I+II. Wiley, New York (1971) 15. : A stability result for mean width of L p -centroid bodies. Adv. Math. 214(2), 865–877 (2007) 123 A Berry-Esseen type inequality for convex bodies with an unconditional basis 16. : Introduction to partial differential equations. Princeton University Press, Princeton (1995) 17. : On the correlation for Kac-like models in the convex case. J. Stat. Phys. 74 (1–2), 349–409 (1994) 18. : L 2 estimates and existence theorems for the ∂¯ operator. Acta Math. 113, 89–152 (1965) 19.
J. 14(89), 386–393 (1964) (in Russian) 22. : A central limit theorem for convex sets. Invent. Math. 168, 91–131 (2007) 23. : Power-law estimates for the central limit theorem for convex sets. J. Funct. Anal. 245, 284–310 (2007) 24. : Inequalities between Dirichlet and Neumann eigenvalues. Arch. Rational Mech. Anal. 94(3), 193–208 (1986) 25. : Géométrie des groupes de transformations. Travaux et Recherches Mathématiques, III. Dunod, Paris (1958) [An English translation was published by Noordhoff International Publishing, Leyden (1977)] 26.
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# How is 71 a prime number?
Updated: 10/25/2022
Wiki User
11y ago
It is prime because only 1 and 71 multiply together to get 71. No other whole numbers multiply to equal 71.
Wiki User
11y ago
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### Is 71 a prime?
71 is a prime number.
### Is 71 prime?
Yes, 71 is a prime number.
### Prime factors of 71?
71 is a prime number. Its only prime factor is 71.
### What is the prime factorization of 71?
71 is a prime number.
### What are the prime factors for 71?
71 is a prime number.
### What is the prime factorization for 71?
71 is a prime number.
### Is 71 prime or composant?
71 is a Prime Number.
### What are the prime numbers 71?
71 is, itself, a prime number.
71
71 is prime.
### What are the prime power factorization of 71?
1 and 71 because 71 IS a prime number.
### Does 71 have any relatively prime numbers?
71 is a prime number | 296 | 1,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-40 | latest | en | 0.912715 |
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User: according to newton's first law of motion for every action there is an equal and opposite reaction? true or false
Weegy: False. Whenever a particle A exerts a force on another particle B, B simultaneously exerts a force on A with the same magnitude in the opposite direction. The strong form of the law further postulates that these two forces act along the same line. [ This law is often simplified into the sentence, "To every action there is an equal and opposite reaction." ]
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The formula for calculating the distance travelled by a moving body that is uniformly accelerated is s = ut+ 1 2at2, Where, s = Distance travelled by the body.
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,382 | 5,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.915766 |
https://earnandexcel.com/how-is-vlookup-used-to-compare-two-columns-in-excel/ | 1,552,926,644,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201455.20/warc/CC-MAIN-20190318152343-20190318174343-00230.warc.gz | 484,785,425 | 51,341 | # How is VLOOKUP Used to Compare Two Columns in Excel?
This short and sweet Excel training session, I will reveal a very quick look at how you can compare two columns in Excel using the VLOOKUP function.
## How Can We Compare Two Columns in Excel?
In my online Excel classes, you’ll likely come to appreciate how important making accurate comparisons is. Here we have two sheets, one in green and one in blue, with a list of references in both. In column B for sheet 1 we would like to find out if any of them appear on sheet 2. We can do this with VLOOKUP, which checks for each reference we specify. Anything that isn’t found will return a result of #n/a, so we know that some are not found on that sheet.
The same principle applies to the sheet 2 below. So, let’s continue our Excel training by trying to compare two columns in Excel using VLOOKUP.
Immediately we can see that Apple, Pear and Passion Fruit appear in sheet 1 but not sheet 2. Equally, Avocado, Grapes and Melon appear in sheet 2 but not sheet 1.
A word of caution when doing this: If an extra space finds itself in the data (before or after the lookup value), it can give you an incorrect result. This is a point that will be reiterated numerous times during any quality course on Excel.
One way to take care of that is to use the trim formula as we have mentioned in our ‘what can go wrong with the VLOOKUP formula’ article. Or you can take some Excel for beginners courses to learn this function in greater detail.
## How Do I Use the Approximate Match Option Instead of Exact Match When I Want to Compare Two Columns in Excel?
When working with VLOOKUP, you should typically use the exact match, 0 or “false” for range_lookup, so why would you need an approximate match? This portion of your Excel training will teach you the more advanced features of the program and how, even when they seem unimportant, they can change your experience with Microsoft Excel. An approximate match is very useful for calculating a result from a series of data with tiered data, such as commission rates, exam grades and the like. You can achieve the same with a nested “if” statement but that can get ridiculously messy if you have several tiered scores. VLOOKUP with approximate match makes quick work of this.
One very important thing about using approximate match is that in order for the function to work effectively, you must store your data (grading system in this case) in ascending order. We mentioned this at the very beginning of the article. What approximate match will effectively do is to find the next largest value that is less than your lookup_value, so in our example it will find that 66 on row 6 is greater than our lookup_value, therefore it will revert to the previous figure above on row 5. In this case it’s 51, and the grade on that row = D.
And there you have it, a simple Excel tutorial on how to compare two columns using VLOOKUP.
## The Importance of Excel Training in Expanding your Knowledge
If you already know the simplest ways of achieving the basic desired goal you have in mind for your spreadsheet, that’s great! But with the help of my Excel training, you can reap the added benefits of learning the more complex solutions to these everyday Excel functions! And when you know more, you can do more. Once you understand the basic functions of VLOOKUP, you could very well be ready for more advanced Excel training courses.
2018-06-06T21:15:44+00:00Excel Training| | 763 | 3,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-13 | longest | en | 0.910132 |
http://www.acmerblog.com/hdu-4585-shaolin-7629.html | 1,503,344,180,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00659.warc.gz | 454,516,504 | 13,012 | 2015
09-17
# Shaolin
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1945 Accepted Submission(s): 836
Problem Description
Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shaolin temple every year, trying to be a monk there. The master of Shaolin evaluates a young man mainly by his talent on understanding the Buddism scripture,
but fighting skill is also taken into account.
When a young man passes all the tests and is declared a new monk of Shaolin, there will be a fight , as a part of the welcome party. Every monk has an unique id and a unique fighting grade, which are all integers. The new monk must fight with a old monk whose
fighting grade is closest to his fighting grade. If there are two old monks satisfying that condition, the new monk will take the one whose fighting grade is less than his.
The master is the first monk in Shaolin, his id is 1，and his fighting grade is 1,000,000,000.He just lost the fighting records. But he still remembers who joined Shaolin earlier, who joined later. Please recover the fighting records for him.
Input
There are several test cases.
In each test case:
The first line is a integer n (0 <n <=100,000),meaning the number of monks who joined Shaolin after the master did.(The master is not included).Then n lines follow. Each line has two integer k and g, meaning a monk’s id and his fighting grade.( 0<= k ,g<=5,000,000)
The monks are listed by ascending order of jointing time.In other words, monks who joined Shaolin earlier come first.
The input ends with n = 0.
Output
A fight can be described as two ids of the monks who make that fight. For each test case, output all fights by the ascending order of happening time. Each fight in a line. For each fight, print the new monk’s id first ,then the old
monk’s id.
Sample Input
3
2 1
3 3
4 2
0
Sample Output
2 1
3 2
4 2
#include<iostream>
#include<cmath>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<cstdio>
using namespace std;
struct Node
{
Node* son[2];
int rank;
int size;
int key;
bool operator<(const Node &a)const
{
return rank<a.rank; //重载小于好，用于维护堆的性质
}
int cmp(int x)const
{
if(x==key) return -1; //比较函数，用于确定寻找的值的位置，0表示左儿子，1表示右儿子，-1代表当前节点
return x<key?0:1;
}
void maintain()//这里仅仅需要维护size
{
size = 1 + son[0]->size + son[1]->size;
}
};
struct Treap
{
Node * root;
Node *null = new Node();//定义空结点，可以有效避免NULL带来的问题
//初始化Treap
void initial()
{
srand(time(NULL));
root = null;
}
//旋转操作，d=0时代表左旋，d=1时代表右旋
void rotate(Node* &o,int d)
{
Node* k = o->son[d^1];
o->son[d^1] = k -> son[d];
k->son[d] = o;
o->maintain();//注意，必须先维护o再维护k，因为o是子节点，不优先维护，会导致父节点出错
k->maintain();
o = k;
}
void ins(Node* &o,int x)
{
if(o==null)
{
o = new Node();
o->son[0] = o->son[1] = null;
o->rank = rand();
o->key = x;
o->size = 1;
}
else
{
int d = o->cmp(x);
ins(o->son[d],x);
o->maintain();
if(o < o->son[d])
rotate(o,d^1);//当位于左儿子时，需要进行的是右旋，当位于右儿子时应当进行左旋，因此通项为rotate(o,d^1)
}
}
void del(Node* &o,int x)
{
int d = o->cmp(x);
if(d==-1)//已经找到待删除节点，将会有两种情况
{
if(o->son[0] == null) o = o->son[1];//左儿子为空，直接连接父节点和右儿子
else if(o->son[1] == null) o= o->son[0];//右儿子为空，同上
else
{
d = o->son[0] < o->son[1] ? 0:1;//在子节点中找到一个rank大的节点，将其旋转到当前节点，那么当前节点应当位于相反的子节点
rotate(o,d);
del(o->son[d],x);
}
}
else
del(o->son[d],x);
if(o!=null) o->maintain();//删除节点后，需要对size进行维护。需要注意的是，如果节点为空时，就不需要维护了
}
int kth(Node* o,int k)//寻找第k大数
{
if(o == null || k<=0 || k>o->size) return -1;//k过大或者过小都不能找到，Treap中没有节点也无需找
int s = (o->son[1] == null?0:o->son[1]->size);//获得右儿子的size
if(k == s+1) return o->key;//当k为 s + 1时，说明当前节点就是答案
else
if(k<=s) return kth(o->son[1],k);//当k<s时，说明第k大数在右子树，k值无需更改
else
return kth(o->son[0],k-s-1);//当k>s+1时，第k大数在左子树，需要减去右子树的值和节点本身
}
int find(Node* o,int k)//返回该数是第几大
{
if(o == null) return -1;
int d = o->cmp(k);
if(d == -1) return 1 + o->son[1]->size;
else if(d == 1) return find(o->son[d],k);
else
{
int tmp = find(o->son[d],k);
if(tmp == -1) return -1;
else
return tmp + 1 + o->son[1]->size;
}
}
} treap;
int id[5000005];
int main()
{
int n;
while(scanf("%d",&n)&&n)
{
treap.initial();
int x,y;
scanf("%d%d",&x,&y);
treap.ins(treap.root,y);
id[y] = x;
printf("%d %d\n",x,1);
for(int i =2;i<=n;i++)
{
scanf("%d%d",&x,&y);
id[y] = x;
treap.ins(treap.root,y);
int t = treap.find(treap.root,y);
int ans1,ans2,ans;
ans1 = treap.kth(treap.root,t-1);
ans2 = treap.kth(treap.root,t+1);
if(ans1 != -1)
{
if(ans2==-1) ans = ans1;
else
if(ans1 - y >= y-ans2) ans = ans2; else ans = ans1;
}
else
ans =ans2;
printf("%d %d\n",x,id[ans]);
}
}
return 0;
} | 1,686 | 4,539 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-34 | longest | en | 0.946118 |
https://cheatsheeting.com/show.html?sheet=petabyte-to-exabit-conversions | 1,675,325,861,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00709.warc.gz | 180,346,591 | 9,144 | Home > Conversions (Computer data storage) > Conversion tables from/to petabyte > PB to Eb Conversion Cheat Sheet (Interactive)
To build or customize your cheat sheet (table below) adjust the values (From, Step, Decimals) in this form and hit the Update button. You could also enter the values to convert and print directly on the table From: Step: Decimals: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
[Formula: Eb = PB x 0.0078125] [Printer friendly] [Exabits to Petabytes]
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PB Eb
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# Petabytes to Exabits Conversion Table
PB = 0.0078125 Eb
# How to convert from Petabytes to Exabits
Since 1 petabyte is equal to 0.0078125 exabits, we could say that n petabytes are equal to 0.0078125 times n exabits. In other words, we could use the following formula:
exabits = petabytes x 0.0078125
For example, let's say that we want to convert 2 petabytes to exabits. Then, we just replace petabytes in the abovementioned formula with 2:
exabits = 2 x 0.0078125
That is, 2 petabytes are equal to 0.015625 exabits. | 637 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-06 | latest | en | 0.259951 |
http://www.physicsforums.com/showthread.php?p=4089258 | 1,369,513,093,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706298270/warc/CC-MAIN-20130516121138-00016-ip-10-60-113-184.ec2.internal.warc.gz | 651,853,529 | 11,518 | ## Calculating the resistance when the switch is off
1. The problem statement, all variables and given/known data
EDIT: The chain is connected to a battery, thus the current in this chain is direct.
When the electric chain's switch (J) is on, resistance Rab (from dots A to B) is equal to 80 Ohms.
What is the electric chain's resistance when the switch (J) is off?
2. Relevant equations
3. The attempt at a solution
I thought that the resistors are connected in parallel , but my teacher said that when the switch if on resistors are connected in parallel and when it's off it's in series.
Is this true, cause I think it's parallel, because the wires split either way, no matter if the switch is on or not...
Don't know for sure what's the connection type in this chain, so...
PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
Recognitions: Homework Help When the switch J is closed, isn't there a direct path from A to B passing right through it? How can the resistance then be 80Ω? Can you confirm the question statement and diagram?
Quote by gneill When the switch J is closed, isn't there a direct path from A to B passing right through it? How can the resistance then be 80Ω? Can you confirm the question statement and diagram?
Well the book says exactly that. In my view this is reasonable because only some electricity is flowing through the switch when it's on, not all of it...
Anyway, does anyone has the answer to my original questions?
Recognitions:
Homework Help
## Calculating the resistance when the switch is off
Quote by kakadas Well the book says exactly that. In my view this is reasonable because only some electricity is flowing through the switch when it's on, not all of it...
No, an ideal switch and wires have zero resistance. The switch will form a short circuit, and ALL of the current will pass through it. So, either the text is wrong, or there is more to this problem than has been stated. Is the switch non-ideal with some resistance of its own? Could it be that the switch should really be oriented horizontally across the circuit rather than vertically?
Recognitions:
Homework Help
Quote by kakadas Don't know for sure what's the connection type in this chain, so...
No one knows, because you have got the diagram wrong, kakadas. As gneill observed.
Anyway, does anyone has the answer to my original questions?
I agree with the other posters. The original question, the diagram and your teachers comment are all inconsistent with each other.
Can you scan that page of the book?
PS Are you sure the switch shouldn't be between the other two corners of the square?
Quote by CWatters I agree with the other posters. The original question, the diagram and your teachers comment are all inconsistent with each other. Can you scan that page of the book? PS Are you sure the switch shouldn't be between the other two corners of the square?
I'm 100% sure of that. I will upload a scan ASAP.
Oh I think I forgot to mention a crucial part of this chain (Or maybe not, I'm just learning..): It's connected to a battery, thus the current is direct.
EDIT:
Well maybe You're a bit confused because English is not my native language, so...
But I really think that I translated the original question correctly...
Here's the Photo of the chain from my book:
I really hope someone will answer my questions.
When the electric chain's switch (J) is on, resistance Rab (from dots A to B) is equal to 80 Ohms. What is the electric chain's resistance when the switch (J) is off? 2. Relevant equations 3. The attempt at a solution I thought that the resistors are connected in parallel , but my teacher said that when the switch if on resistors are connected in parallel and when it's off it's in series. -------------------------- Parallel circuit $R_{total}=\frac{R_1 R_2 R_3}{R_1 + R_2 +R_3}$ Any factor in the numerator with a zero value will result in total zero resistance.
Recognitions: Homework Help According to the diagram, RAB must be zero when the switch (j) is closed (on). Is it possible that there is a language confusion regarding the switch positions? "closed" or "on" means that the contacts are touching and current can flow. "Open" or "off" means that the contacts are not touching and current cannot flow --- open circuit. By the way, out of curiosity, what is the original language of the text?
Quote by gneill According to the diagram, RAB must be zero when the switch (j) is closed (on). Is it possible that there is a language confusion regarding the switch positions? "closed" or "on" means that the contacts are touching and current can flow. "Open" or "off" means that the contacts are not touching and current cannot flow --- open circuit. By the way, out of curiosity, what is the original language of the text?
That was exactly what i meant to say..
But does the current not flow when the switch is off? I think it still flows, because IMO the resistors are connected in parallel, so I think that the current flows either way, no matter in what state the switch is.
If I'm wrong, Could anyone prove to me that the current doesn't flow when the switch is off?
Recognitions: Homework Help Yes the current flows when the switch is off (open) --- It then must flow through the resistor network. But your problem statement contends that the resistance is 80 Ohms when the switch is closed (on). That is not possible with the given information.
I agree. Looks like ON and OFF may have been confused but... If it's 80 Ohms with the switch OFF (open) that would allow R to be calculated but when the switch is ON (conducting) the equivalent resistance is ZERO. The teachers comment about it changing from serial to parallel depending on the switch position makes no sense.
Quote by CWatters I agree. Looks like ON and OFF may have been confused but... If it's 80 Ohms with the switch OFF (open) that would allow R to be calculated but when the switch is ON (conducting) the equivalent resistance is ZERO. The teachers comment about it changing from serial to parallel depending on the switch position makes no sense.
Why is that so, please explain in more detail?
BTW does it make a difference if the current is DC or AC? In my case it's DC.
With the switch OFF you have two resistors in series, in parallel with, two resistors in series. With the switch ON the resistors are irrelevant. All the current flows through the switch. You could replace the resistors with ones that were much bigger, much smaller, or a short circuit and it would make no difference. The equivalent resistance is zero Ohms. Makes no difference if it's AC or DC.
I Just want to clearinfy that By telling that the switch is ON I meant that it's CONDUCTING. OFF means that it's open and NOT CONDUCTING.
' As others said, there are errors in the data and in the drawing. It is waste of time computing on it. I guess the problem meant for Wheatstone Bridge. http://en.wikipedia.org/wiki/Wheatstone_bridge
Quote by kakadas I Just want to clearinfy that By telling that the switch is ON I meant that it's CONDUCTING. OFF means that it's open and NOT CONDUCTING.
That's what I had assumed as well.
Tags parallel, resistors, series, switch | 1,641 | 7,339 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2013-20 | latest | en | 0.951111 |
https://rosettacode.org/wiki/AVL_tree/C | 1,719,336,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866218.13/warc/CC-MAIN-20240625171218-20240625201218-00411.warc.gz | 442,728,601 | 35,923 | # AVL tree/C
### Code
```#include <stdio.h>
#include <stdlib.h>
#include <time.h>
struct node {
int height;
struct node * kid[2];
} dummy = { 0, 0, {&dummy, &dummy} }, *nnil = &dummy;
// internally, nnil is the new nul
struct node *new_node(int value)
{
struct node *n = calloc(1, sizeof *n);
*n = (struct node) { value, 1, {nnil, nnil} };
return n;
}
int max(int a, int b) { return a > b ? a : b; }
inline void set_height(struct node *n) {
n->height = 1 + max(n->kid[0]->height, n->kid[1]->height);
}
inline int ballance(struct node *n) {
return n->kid[0]->height - n->kid[1]->height;
}
// rotate a subtree according to dir; if new root is nil, old root is freed
struct node * rotate(struct node **rootp, int dir)
{
struct node *old_r = *rootp, *new_r = old_r->kid[dir];
if (nnil == (*rootp = new_r))
free(old_r);
else {
old_r->kid[dir] = new_r->kid[!dir];
set_height(old_r);
new_r->kid[!dir] = old_r;
}
return new_r;
}
{
struct node *root = *rootp;
int b = ballance(root)/2;
if (b) {
int dir = (1 - b)/2;
if (ballance(root->kid[dir]) == -b)
rotate(&root->kid[dir], !dir);
root = rotate(rootp, dir);
}
if (root != nnil) set_height(root);
}
// find the node that contains value as payload; or returns 0
struct node *query(struct node *root, int value)
{
return root == nnil
? 0
? root
}
void insert(struct node **rootp, int value)
{
struct node *root = *rootp;
if (root == nnil)
*rootp = new_node(value);
else if (value != root->payload) { // don't allow dup keys
}
}
void delete(struct node **rootp, int value)
{
struct node *root = *rootp;
// if this is the node we want, rotate until off the tree
if (nnil == (root = rotate(rootp, ballance(root) < 0)))
return;
}
// aux display and verification routines, helpful but not essential
struct trunk {
struct trunk *prev;
char * str;
};
void show_trunks(struct trunk *p)
{
if (!p) return;
show_trunks(p->prev);
printf("%s", p->str);
}
// this is very haphazzard
void show_tree(struct node *root, struct trunk *prev, int is_left)
{
if (root == nnil) return;
struct trunk this_disp = { prev, " " };
char *prev_str = this_disp.str;
show_tree(root->kid[0], &this_disp, 1);
if (!prev)
this_disp.str = "---";
else if (is_left) {
this_disp.str = ".--";
prev_str = " |";
} else {
this_disp.str = "`--";
prev->str = prev_str;
}
show_trunks(&this_disp);
if (prev) prev->str = prev_str;
this_disp.str = " |";
show_tree(root->kid[1], &this_disp, 0);
if (!prev) puts("");
}
int verify(struct node *p)
{
if (p == nnil) return 1;
int h0 = p->kid[0]->height, h1 = p->kid[1]->height;
int b = h0 - h1;
if (p->height != 1 + max(h0, h1) || b < -1 || b > 1) {
show_tree(p, 0, 0);
abort();
}
return verify(p->kid[0]) && verify(p->kid[1]);
}
#define MAX_VAL 32
int main(void)
{
int x;
struct node *root = nnil;
srand(time(0));
for (x = 0; x < 10 * MAX_VAL; x++) {
// random insertion and deletion
if (rand()&1) insert(&root, rand()%MAX_VAL);
else delete(&root, rand()%MAX_VAL);
verify(root);
}
puts("Tree is:");
show_tree(root, 0, 0);
puts("\nQuerying values:");
for (x = 0; x < MAX_VAL; x++) {
struct node *p = query(root, x);
if (p) printf("%2d found: %p %d\n", x, p, p->payload);
}
for (x = 0; x < MAX_VAL; x++) {
delete(&root, x);
verify(root);
}
puts("\nAfter deleting all values, tree is:");
show_tree(root, 0, 0);
return 0;
}
```
Output:
```Tree is:
.--0
.--2
| `--4
| `--5
.--6
| | .--10
| `--12
| | .--13
| `--15
---17
| .--18
| .--19
| .--21
| | | .--24
| | `--25
`--26
`--27
`--30
Querying values:
0 found: 0x180a210 0
2 found: 0x180a1b0 2
4 found: 0x180a010 4
5 found: 0x180a0d0 5
6 found: 0x180a0f0 6
10 found: 0x180a190 10
12 found: 0x180a150 12
13 found: 0x180a250 13
15 found: 0x180a050 15
17 found: 0x180a270 17
18 found: 0x180a1d0 18
19 found: 0x180a170 19
21 found: 0x180a070 21
24 found: 0x180a030 24
25 found: 0x180a0b0 25
26 found: 0x180a130 26
27 found: 0x180a1f0 27
30 found: 0x180a110 30
After deleting all values, tree is:
```
### Efficient AVL tree
The following example implements an AVL tree without the need of calculating the height of the nodes (which can be quite time consuming if the tree gets large)! It is based on an example of AVL tree in C# (see [1]).
The example distinguish between the tree implementation itself (see below) and the data to be stored in the tree (see example below).
AvlTree.h
```#ifndef AVLTREE_INCLUDED
#define AVLTREE_INCLUDED
typedef struct Tree *Tree;
typedef struct Node *Node;
Tree Tree_New (int (*comp)(void *, void *), void (*print)(void *));
void Tree_Insert (Tree t, void *data);
void Tree_DeleteNode (Tree t, Node node);
Node Tree_SearchNode (Tree t, void *data);
Node Tree_FirstNode (Tree t);
Node Tree_LastNode (Tree t);
Node Tree_PrevNode (Tree t, Node n);
Node Tree_NextNode (Tree t, Node n);
void Tree_Print (Tree t);
void *Node_GetData (Node n);
#endif
```
AvlTree.c
```#include <stdio.h>
#include <stdlib.h>
#include "AvlTree.h"
//
// Private datatypes
//
struct Tree {
Node root;
int (*comp) (void *, void *);
void (*print) (void *);
};
struct Node {
Node parent;
Node left;
Node right;
void *data;
int balance;
};
struct trunk {
struct trunk *prev;
char *str;
};
//
// Declaration of private functions.
//
void Tree_InsertBalance (Tree t, Node node, int balance);
void Tree_DeleteBalance (Tree t, Node node, int balance);
Node Tree_RotateLeft (Tree t, Node node);
Node Tree_RotateRight (Tree t, Node node);
Node Tree_RotateLeftRight (Tree t, Node node);
Node Tree_RotateRightLeft (Tree t, Node node);
void Tree_Replace (Node target, Node source);
Node Node_New (void *data, Node parent);
void print_tree (Tree t, Node n, struct trunk *prev, int is_left);
//----------------------------------------------------------------------------
// Tree_New --
//
// Creates a new tree using the parameters 'comp' and 'print' for
// comparing and printing the data in the nodes.
//
Tree Tree_New (int (*comp)(void *, void *), void (*print)(void *)) {
Tree t;
t = malloc (sizeof (*t));
t->root = NULL;
t->comp = comp;
t->print = print;
return t;
}
// Tree_Insert --
//
// Insert new data in the tree. If the data is already in the tree,
// nothing will be done.
//
void Tree_Insert (Tree t, void *data) {
if (t->root == NULL) {
t->root = Node_New (data, NULL);
} else {
Node node = t->root;
while (node != NULL) {
if ((t->comp) (data, node->data) < 0) {
Node left = node->left;
if (left == NULL) {
node->left = Node_New (data, node);
Tree_InsertBalance (t, node, -1);
return;
} else {
node = left;
}
} else if ((t->comp) (data, node->data) > 0) {
Node right = node->right;
if (right == NULL) {
node->right = Node_New (data, node);
Tree_InsertBalance (t, node, 1);
return;
} else {
node = right;
}
} else {
node->data = data;
return;
}
}
}
}
// Tree_DeleteNode --
//
// Removes a given node from the tree.
//
void Tree_DeleteNode (Tree t, Node node) {
Node left = node->left;
Node right = node->right;
Node toDelete = node;
if (left == NULL) {
if (right == NULL) {
if (node == t->root) {
t->root = NULL;
} else {
Node parent = node->parent;
if (parent->left == node) {
parent->left = NULL;
Tree_DeleteBalance (t, parent, 1);
} else {
parent->right = NULL;
Tree_DeleteBalance (t, parent, -1);
}
}
} else {
Tree_Replace (node, right);
Tree_DeleteBalance (t, node, 0);
toDelete = right;
}
} else if (right == NULL) {
Tree_Replace (node, left);
Tree_DeleteBalance (t, node, 0);
toDelete = left;
} else {
Node successor = right;
if (successor->left == NULL) {
Node parent = node->parent;
successor->parent = parent;
successor->left = left;
successor->balance = node->balance;
if (left != NULL) {
left->parent = successor;
}
if (node == t->root) {
t->root = successor;
} else {
if (parent->left == node) {
parent->left = successor;
} else {
parent->right = successor;
}
}
Tree_DeleteBalance (t, successor, -1);
} else {
while (successor->left != NULL) {
successor = successor->left;
}
Node parent = node->parent;
Node successorParent = successor->parent;
Node successorRight = successor->right;
if (successorParent->left == successor) {
successorParent->left = successorRight;
} else {
successorParent->right = successorRight;
}
if (successorRight != NULL) {
successorRight->parent = successorParent;
}
successor->parent = parent;
successor->left = left;
successor->balance = node->balance;
successor->right = right;
right->parent = successor;
if (left != NULL) {
left->parent = successor;
}
if (node == t->root) {
t->root = successor;
} else {
if (parent->left == node) {
parent->left = successor;
} else {
parent->right = successor;
}
}
Tree_DeleteBalance (t, successorParent, 1);
}
}
free (toDelete);
}
// Tree_SearchNode --
//
// Searches the tree for a node containing the given data.
//
Node Tree_SearchNode (Tree t, void *data) {
Node node = t->root;
while (node != NULL) {
if ((t->comp) (data, node->data) < 0) {
node = node->left;
} else if ((t->comp) (data, node->data) > 0) {
node = node->right;
} else {
return node;
}
}
return NULL;
}
// Tree_Print --
//
// Prints an ASCII representation of the tree on screen.
//
void Tree_Print (Tree t) {
print_tree (t, t->root, 0, 0);
fflush (stdout);
}
// Tree_FirstNode --
//
// Returns the node containing the smallest key.
//
Node Tree_FirstNode (Tree t) {
Node node = t->root;
while ((node != NULL) && (node->left != NULL)) {
node = node->left;
}
return node;
}
// Tree_LastNode --
//
// Returns the node containing the biggest key.
//
Node Tree_LastNode (Tree t) {
Node node = t->root;
while ((node != NULL) && (node->right != NULL)) {
node = node->right;
}
return node;
}
// Tree_PrevNode --
//
// Returns the predecessor of the given node.
//
Node Tree_PrevNode (Tree t, Node n) {
Node nTemp;
if (n->left != NULL) {
n = n->left;
while (n->right != NULL) {
n = n->right;
}
} else {
nTemp = n;
n = n->parent;
while ((n != NULL) && (n->left == nTemp)) {
nTemp = n;
n = n->parent;
}
}
return n;
}
// Tree_NextNode --
//
// Returns the follower of the given node.
//
Node Tree_NextNode (Tree t, Node n) {
Node nTemp;
if (n->right != NULL) {
n = n->right;
while (n->left != NULL) {
n = n->left;
}
} else {
nTemp = n;
n = n->parent;
while ((n != NULL) && (n->right == nTemp)) {
nTemp = n;
n = n->parent;
}
}
return n;
}
// Node_GetData --
//
// Returns the data in a node.
//
void *Node_GetData (Node n) {
return n->data;
}
//----------------------------------------------------------------------------
//
// Internal functions.
//
void Tree_InsertBalance (Tree t, Node node, int balance) {
while (node != NULL) {
balance = (node->balance += balance);
if (balance == 0) {
return;
} else if (balance == -2) {
if (node->left->balance == -1) {
Tree_RotateRight (t, node);
} else {
Tree_RotateLeftRight (t, node);
}
return;
} else if (balance == 2) {
if (node->right->balance == 1) {
Tree_RotateLeft (t, node);
} else {
Tree_RotateRightLeft (t, node);
}
return;
}
Node parent = node->parent;
if (parent != NULL) {
balance = (parent->left == node) ? -1 : 1;
}
node = parent;
}
}
void Tree_DeleteBalance (Tree t, Node node, int balance) {
while (node != NULL) {
balance = (node->balance += balance);
if (balance == -2) {
if (node->left->balance <= 0) {
node = Tree_RotateRight (t, node);
if (node->balance == 1) {
return;
}
} else {
node = Tree_RotateLeftRight (t, node);
}
} else if (balance == 2) {
if (node->right->balance >= 0) {
node = Tree_RotateLeft (t, node);
if (node->balance == -1) {
return;
}
} else {
node = Tree_RotateRightLeft (t, node);
}
} else if (balance != 0) {
return;
}
Node parent = node->parent;
if (parent != NULL) {
balance = (parent->left == node) ? 1 : -1;
}
node = parent;
}
}
void Tree_Replace (Node target, Node source) {
Node left = source->left;
Node right = source->right;
target->balance = source->balance;
target->data = source->data;
target->left = left;
target->right = right;
if (left != NULL) {
left->parent = target;
}
if (right != NULL) {
right->parent = target;
}
}
Node Tree_RotateLeft (Tree t, Node node) {
Node right = node->right;
Node rightLeft = right->left;
Node parent = node->parent;
right->parent = parent;
right->left = node;
node->right = rightLeft;
node->parent = right;
if (rightLeft != NULL) {
rightLeft->parent = node;
}
if (node == t->root) {
t->root = right;
} else if (parent->right == node) {
parent->right = right;
} else {
parent->left = right;
}
right->balance--;
node->balance = -right->balance;
return right;
}
Node Tree_RotateRight (Tree t, Node node) {
Node left = node->left;
Node leftRight = left->right;
Node parent = node->parent;
left->parent = parent;
left->right = node;
node->left = leftRight;
node->parent = left;
if (leftRight != NULL) {
leftRight->parent = node;
}
if (node == t->root) {
t->root = left;
} else if (parent->left == node) {
parent->left = left;
} else {
parent->right = left;
}
left->balance++;
node->balance = -left->balance;
return left;
}
Node Tree_RotateLeftRight (Tree t, Node node) {
Node left = node->left;
Node leftRight = left->right;
Node parent = node->parent;
Node leftRightRight = leftRight->right;
Node leftRightLeft = leftRight->left;
leftRight->parent = parent;
node->left = leftRightRight;
left->right = leftRightLeft;
leftRight->left = left;
leftRight->right = node;
left->parent = leftRight;
node->parent = leftRight;
if (leftRightRight != NULL) {
leftRightRight->parent = node;
}
if (leftRightLeft != NULL) {
leftRightLeft->parent = left;
}
if (node == t->root) {
t->root = leftRight;
} else if (parent->left == node) {
parent->left = leftRight;
} else {
parent->right = leftRight;
}
if (leftRight->balance == 1) {
node->balance = 0;
left->balance = -1;
} else if (leftRight->balance == 0) {
node->balance = 0;
left->balance = 0;
} else {
node->balance = 1;
left->balance = 0;
}
leftRight->balance = 0;
return leftRight;
}
Node Tree_RotateRightLeft (Tree t, Node node) {
Node right = node->right;
Node rightLeft = right->left;
Node parent = node->parent;
Node rightLeftLeft = rightLeft->left;
Node rightLeftRight = rightLeft->right;
rightLeft->parent = parent;
node->right = rightLeftLeft;
right->left = rightLeftRight;
rightLeft->right = right;
rightLeft->left = node;
right->parent = rightLeft;
node->parent = rightLeft;
if (rightLeftLeft != NULL) {
rightLeftLeft->parent = node;
}
if (rightLeftRight != NULL) {
rightLeftRight->parent = right;
}
if (node == t->root) {
t->root = rightLeft;
} else if (parent->right == node) {
parent->right = rightLeft;
} else {
parent->left = rightLeft;
}
if (rightLeft->balance == -1) {
node->balance = 0;
right->balance = 1;
} else if (rightLeft->balance == 0) {
node->balance = 0;
right->balance = 0;
} else {
node->balance = -1;
right->balance = 0;
}
rightLeft->balance = 0;
return rightLeft;
}
void print_trunks (struct trunk *p) {
if (!p) {
return;
}
print_trunks (p->prev);
printf ("%s", p->str);
}
void print_tree (Tree t, Node n, struct trunk *prev, int is_left) {
if (n == NULL) {
return;
}
struct trunk this_disp = { prev, " " };
char *prev_str = this_disp.str;
print_tree (t, n->right, &this_disp, 1);
if (!prev) {
this_disp.str = "---";
} else if (is_left) {
this_disp.str = ".--";
prev_str = " |";
} else {
this_disp.str = "`--";
prev->str = prev_str;
}
print_trunks (&this_disp);
(t->print) (n->data);
printf (" (%+d)\n", n->balance);
if (prev) {
prev->str = prev_str;
}
this_disp.str = " |";
print_tree (t, n->left, &this_disp, 0);
if (!prev) {
puts ("");
}
}
Node Node_New (void *data, Node parent) {
Node n;
n = malloc (sizeof (*n));
n->parent = parent;
n->left = NULL;
n->right = NULL;
n->data = data;
n->balance = 0;
return n;
}
```
And here's the example which shows how to use the package. It creates in an endless loop random numbers between 0..999 and stores the number (the key) together with its square root (the value) in the tree. If an element with the given key is already in tree, it will be deleted.
```#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <math.h>
#include "AvlTree.h"
#define MAX_KEY_VALUE 1000
typedef struct NodeData {
int key;
double value;
} *NodeData;
int comp (void *a1, void *a2) {
NodeData nd1 = (NodeData) a1;
NodeData nd2 = (NodeData) a2;
if (nd1->key < nd2->key) {
return -1;
} else if (nd1->key > nd2->key) {
return +1;
} else {
return 0;
}
}
void print (void *a) {
NodeData nd = (NodeData) a;
printf ("%3d \"%6.3f\"", nd->key, nd->value);
}
int main (int argc, char **argv) {
Tree tree;
Node node;
NodeData nd1 = NULL, nd2;
tree = Tree_New (comp, print);
while (1) {
if (nd1 == NULL) {
nd1 = malloc (sizeof (*nd1));
}
nd1->key = rand () % MAX_KEY_VALUE;
nd1->value = sqrt (nd1->key);
if ((node = Tree_SearchNode (tree, nd1)) != NULL) {
printf (">>> delete key %d\n\n", nd1->key);
nd2 = Node_GetData (node);
Tree_DeleteNode (tree, node);
free (nd2);
} else {
printf (">>> insert key %d\n\n", nd1->key);
Tree_Insert (tree, nd1);
nd1 = NULL;
}
Tree_Print (tree);
sleep (2);
}
return 0;
}
```
After a number of iterations, the tree will look similar to this:
Output:
``` .--980 "31.305" (+0)
.--972 "31.177" (+0)
| `--944 "30.725" (+0)
.--933 "30.545" (+0)
| | .--932 "30.529" (+0)
| `--929 "30.480" (+0)
| `--920 "30.332" (+0)
.--917 "30.282" (+0)
| | .--876 "29.597" (+0)
| | .--868 "29.462" (+1)
| `--862 "29.360" (+1)
| `--847 "29.103" (+0)
.--843 "29.034" (-1)
| | .--752 "27.423" (+0)
| | .--743 "27.258" (-1)
| | | | .--727 "26.963" (+0)
| | | `--711 "26.665" (+0)
| | | `--710 "26.646" (+0)
| `--705 "26.552" (-1)
| | .--700 "26.458" (-1)
| | | `--675 "25.981" (+0)
| | .--635 "25.199" (+0)
| | | | .--629 "25.080" (+0)
| | | `--624 "24.980" (+1)
| `--592 "24.331" (+1)
| | .--551 "23.473" (+0)
| `--549 "23.431" (+1)
---529 "23.000" (+0)
| .--508 "22.539" (-1)
| | `--491 "22.159" (+0)
| .--484 "22.000" (+1)
| | `--431 "20.761" (+0)
| .--427 "20.664" (+0)
| | | .--425 "20.616" (+0)
| | | .--397 "19.925" (+1)
| | `--393 "19.824" (+0)
| | `--366 "19.131" (-1)
| | `--363 "19.053" (+0)
`--351 "18.735" (-1)
| .--344 "18.547" (+0)
| .--330 "18.166" (+1)
| .--310 "17.607" (+1)
| | `--301 "17.349" (+0)
`--262 "16.186" (-1)
| .--236 "15.362" (+0)
| .--223 "14.933" (+1)
`--164 "12.806" (-1)
| .--134 "11.576" (-1)
| | `--126 "11.225" (+0)
`--119 "10.909" (+0)
`--108 "10.392" (-1)
`-- 30 " 5.477" (+0)
``` | 5,991 | 18,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.306005 |
https://socratic.org/questions/expansion-of-x-1-4 | 1,576,481,652,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541318556.99/warc/CC-MAIN-20191216065654-20191216093654-00542.warc.gz | 543,867,015 | 6,138 | # Expansion of (x-1)^4?
Apr 8, 2018
${\left(x - 1\right)}^{4} \equiv {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 1$
#### Explanation:
We can expand the expression using the binomial theorem:
${\left(x - 1\right)}^{4} \equiv {\sum}_{r = 0}^{4} \left(\begin{matrix}n \\ r\end{matrix}\right) {\left(x\right)}^{r} {\left(- 1\right)}^{n - r}$
$\text{ } = \left(\begin{matrix}4 \\ 0\end{matrix}\right) {\left(x\right)}^{4} {\left(- 1\right)}^{0} + \left(\begin{matrix}4 \\ 1\end{matrix}\right) {\left(x\right)}^{3} {\left(- 1\right)}^{1} +$
$\text{ } \left(\begin{matrix}4 \\ 2\end{matrix}\right) {\left(x\right)}^{2} {\left(- 1\right)}^{2} + \left(\begin{matrix}4 \\ 3\end{matrix}\right) {\left(x\right)}^{1} {\left(- 1\right)}^{3} +$
$\text{ } \left(\begin{matrix}4 \\ 4\end{matrix}\right) {\left(x\right)}^{0} {\left(- 1\right)}^{4}$
$\text{ } = \left(1\right) \left({x}^{4}\right) \left(1\right) + \left(4\right) \left({x}^{3}\right) \left(- 1\right) + \left(6\right) \left({x}^{2}\right) \left(1\right) +$
$\text{ } \left(4\right) \left(x\right) \left(- 1\right) + \left(1\right) \left(1\right) \left(1\right)$
$\text{ } = {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 1$
We could also you the appropriate row from Pascal's Triangle to gain the coefficients. | 539 | 1,256 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2019-51 | latest | en | 0.326456 |
https://www.eleccircuit.com/low-dropout-voltage-regulators-circuits/ | 1,717,064,805,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00147.warc.gz | 637,942,718 | 31,602 | # 5V Low Dropout Regulator Circuit using transistor and LED
This is a fixed DC regulator, a 5V low dropout regulator type. It uses a transistor and LED as main parts. It requires an input voltage of at least 6V and there will be 1V lost inside the circuit.
Therefore, the output voltage will always be 5V at 0.5A.
### Why should we use a 5V Low Dropout Regulator?
Assuming that we need to use a digital circuit which is required a 5V regulated power supply. But we have a 6V battery that is incompatible with our intended uses.
I have many ideas but which is better?
• First, I will use a resistor to lower down voltage. But it also lower the current and become not constant.
• Second, I will connect a 1N5402 diode in series to reduce the voltage. In principle, it will reduce a voltage down about 0.7V, bringing the output voltage down to 5.3V, close to what we needed. But it does not work, because the output current got lower.
• Third, Use 7805 in a 5V regulator circuit. It can output a high current. But we cannot use 7805 because It requires an input voltage of more than 7.3V.
• Fourth, We need to use a high current circuit. Therefore, the 5V low dropout regulator is better in this case, or “LDO Regulator” in short.
The LDO Regulator can be created in many ways. Nowadays popular way is to use IC because of its convenience and high efficiency.
However, This circuit we are making consists of the transistor and LED. Because we want to make great use of old and common components in our inventory.
Read first for beginners: How do transistor circuits work
### How does transistor LDO Regulator works
In a general, the regulator circuit has an input voltage that must be 3V higher than the output.
But this is a special circuit, 6V battery input is only 1V higher than the 5V output.
As the circuit above, the load connects the collector of the Q4 output transistor. This makes the transistor works in saturation, it causes a voltage drop across the collector and the emitter to be very low. Therefore, it can let more current passing through.
The current passing through depends on the type of transistors, in this case, is about 0.5A. And this transistor has a saturation voltage of 0.2V. Also, the voltage across R6 (current limiting resistor) is to be accounted for.
When the voltage across R6 reaches 0.7V, the Q3 transistor will get activated. At the same time, the output current is limited by R6.
There will also be some current flows through R1 to LED1, causing It to glow up. The LED1 has two functions, first is to indicate if the circuit working or not. The second is to keep a reference voltage that has been set at 1.9V, which is across the emitter of Q1 and ground.
The base current of Q1 is received from the output through the voltage divider circuit consisting of R4, VR1, and R5. The Q1 transistor conducts more or less depending on the difference between the reference voltage and the output voltage.
The result of the C-E of Q1 current makes the Q2 provide the current to control the base of the Q4 power transistor.
The capacitor C1 filters the output current to smooth up.
If we cannot find a transistor BD140. We may replace it with BD136, or BD138. However, these transistor has saturation voltage that is slightly higher than BD438.
For LED1 can be green or red to determine the reference voltage. Different color LEDs have different reference voltages. But we can always change an output voltage by adjusting VR1’s resistance.
The transistor Q4 is quite hot, therefore, it needs to mount a heatsink. We may use a TIP42 or MJE2955 instead (We can buy it later down the line) to increase the output current up to 2A.
### Components list
Semiconductors:
• Q1, Q3: 2N3906, 45V 0.1A, PNP TO-92 Transistor
• Q2: 2N3904, 45V 0.1A, NPN TO-92 Transistor
• Q4: BD140, 80V 1.5A PNP Transistor,TO-225
• LED1: Green 3mm LEDs. Quantity: 1
0.25W metal/carbon film Resistors, tolerance: 5%, (Unless stated otherwise)
• R1: 1K
• R2: 10K
• R3: 470Ω
• R4: 1.2K
• R5: 560Ω
• R6: 1.2Ω 1W
Electrolytic Capacitors
• C1: 10µF 25V
Miscellaneous:
• Perforated board, Wires, Heatsink for Q2
### Building and Testing the circuit
This circuit has a few components. Thus, we may assemble them on the perforated PCB. It may save some time to build.
We measure the voltage at various points as the circuit diagram below.
My daughter has tested it on a 10Ω resistor as a load. It can power a current of about 0.5A, and the output voltage reduces to 4.9V. It is close to when no load (5V).
This is the first LDO Regulator circuit, it uses a transistor as a key component. Next time, we will try making by using IC.
What is more? See More Power supply circuits
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### 1 thought on “5V Low Dropout Regulator Circuit using transistor and LED”
1. Hi.
I need circuit low volt cut off. 12v battery when 10v than cut off. | 1,312 | 5,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.911845 |
http://etutorials.org/Programming/Software+engineering+and+computer+games/Part+I+Software+Engineering+and+Computer+Games/Chapter+17.+Selection+games/17.2+The+PickNPop+implementation/ | 1,495,885,968,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608953.88/warc/CC-MAIN-20170527113807-20170527133807-00471.warc.gz | 148,193,610 | 15,970 | # 17.2 The PickNPop implementation
A lot of the work of designing software goes into improving the way that the program looks onscreen. Software engineering is a little like theater, or like stage-magic. Your goal is to give the user the illusion that your program is a very solid, tangible kind of thing. Getting everything in place requires solid design and a lot of tweaking.
One thing differentiating PickNPop from Spacewar and Airhockey is that we chose to make the _border of the world have a non-zero z size so that the shapes can pass above and below each other when we show the game in the OpenGL 3D mode.
#### Making the score come out even
Though not all games must have a numerical score, if you have one, then it should be easy to understand. On the one hand you might require that your game events have simple, round-number score values assigned to them. On the other hand you might require that your maximum possible game score total be a round easy number like 100, 1000, or even 1,000,000. If you are able to control the number of things that can happen in your game, then you can satisfy both conditions. If not, then you have to settle for one of the conditions: round-number values or round-number maximum score.
In PickNPop, we allow for varying sizes of worlds, and, since the game might still be developed further, we allow for recompiling the program with different values of JEWEL_PERCENT. So it's not possible both to have round-number values and to have a round-number max score.
Our decision here was to go for the round-number maximum score. In the CPopDoc::seedBubbles(int gametype, int count) method we figure out how many jewels and peanuts to make, and then we figure out how much they should be worth, and finally we calculate a _scorecorrection value that we add in at the game's end to make it possible for the user's score to exactly equal the nice round number MAX_SCORE.
In cGamePickNPop::seedCritters() we compute the peanutstoadd peanuts and jewelstoadd jewels needed, and then bury the jewels 'under' the peanuts by adding them in second. The default behavior of cBiota is to draw the earlier array members after the later array members. When using the two-dimensional cGraphicsMFC, this causes a 'painter's algorithm' effect of having the later-listed critters appear behind the earlier-listed ones. When using the three-dimensional cGraphicsOpenGL, the critters are actually sorted according to the z-value of their _position values. The cheap and dirty cGame::zStackCritters() call gives the critters different z-values, again arranging them so the earlier-listed critters have larger z-values than the later-listed critters' z-values and end up appearing on top in the default view from up on the positive side of the z-axis.
```void cGamePickNPop::seedCritters()
{
/* First we'll set the _bubble array to have room for count
bubbles. Then we'll add jewels and peanuts, randomizing their
radii, positions, and colors as we go along. In the case of
PGT_3D, we go back and change the radii at the end. */
int i;
Real jewelprobability = cGamePickNPop::JEWEL_WEIGHT;
int jewelvalue(0), peanutvalue(0);
cCritter *pcritternew;
/* I use the jewelprobability to decide how many jewels and how
many peanuts to have. These are the jewelstoadd and
peanutstoadd numbers. We think of randomly drawing from this
supply and adding them into the game. I want my standard game
score to be MAX_SCORE, with JEWEL_GAME_WEIGHT portion of the
score coming from the jewels and the rest and from the
peanuts. The scores have to be integers, so it may be that the
total isn't quite MAX_SCORE, so I will give the rest to the
user as game-end bonus. */
//----------Get the counts and the scorevalues ready----------
jewelvalue =
int(_maxscore*cGamePickNPop::JEWEL_GAME_SCORE_WEIGHT)/
peanutvalue = (_maxscore ?
_scorecorrection = _maxscore ? (jewelstoadd*jewelvalue +
/* We'll add this in at the end, so that user's maximum
score is the same as the targeted _maxscore). */
//--------------------Renew the _bubble contents ----------
_pbiota->purgeNonPlayerNonWallCritters();
// Need to delete any from last round
/* Regarding the stacking, it's worth mentioning that
cBiota::draw draws the critters in reverse order, last
index to first, so the first-added members appear on top
in 2D. We want the peanuts "on top", so we add them first.
Of course in 3D, the zStackCritters is going to take care
of this irregardless of what order the critters are drawn. */
{
pcritternew = new cCritterPeanut(this); /* White bubble that
we call a "Peanut", can't move out of _packingbox */
pcritternew->setValue(peanutvalue);
}
{ /* Make a pcritternew and then add it into _bubble at the
bottom of loop. */
pcritternew = new cCritterJewel(this); /* Colored bubble
that we call a "Jewel", can move all over within
_border.*/
pcritternew->setValue(jewelvalue);
}
zStackCritters();
} ```
#### The world rectangles
In PickNPop we want to try and fit our game as nicely as possible into our window. We give the CDocument a cGraphicRealBox _packingbox and _targetbox field. These are to be rectangles that fit nicely inside the _border. Rather than setting their values with brute numbers, we set their values as proportions of the _border. The cRealBox::innerBox function returns a cRealBox slightly inside the caller box. And we give them some nice colors and edges.
#### Converting a critter
One of the parts of the code the author initially had trouble with was in the cCritterJewel method where we react to moving the critter inside the _targetbox. Here we have to replace one class of object by a different class of object, while still having the object be in some ways the 'same.' It turns out that you can't do this with something so simple as a type-cast of the sort you'd use to turn an int into a float. Class instances carry too much baggage for that. What we do instead is to create a brand-new object which copies the desired properties of the object that you wanted to 'cast.' We do this by means of a cCritterUnpackedJewel copy constructor.
```void cCritterJewel::update(CPopView *pactiveview)
{
cGamePickNPop *pgamepnp = NULL;
//(1) Apply force if turned on.
cCritter::update(pactiveview); //Always call this.
cVector safevelocity(_velocity); /* To be safe, don't let any z
get into velocity. */
safevelocity.setZ(0.0);
setVelocity(safevelocity);
//(2) Check if in targetbox, and if so, replace yourself with a good
// jewel.
if (pgame()->IsKindOf(RUNTIME_CLASS(cGamePickNPop)))
/* We need to do the cast to access the targetbox field, and to
be safe we check that the cast will work. */
pgamepnp = (cGamePickNPop*)(pgame());
else
return;
cRealBox effectivebox = pgamepnp-
if (!effectivebox.inside(_position))
return;
//Reaction to being inside _targetbox.
playSound("Ding");
cCritterUnpackedJewel *pcritternew =
new cCritterUnpackedJewel(this); //Copy constructor
pcritternew->setMoveBox(pgamepnp->targetbox());
pcritternew->setDragBox(pgamepnp->targetbox());
delete_me(); /* Just tell cBiota to just remove the old critter.
Don't use the overridden cCritterJewel::die to make a noise
and subtract _value from score.*/
critter.
} ```
The delete_me makes a service request to the _pownerbiota cBiota object. The add_me makes a service request as well, but since pcritternew isn't yet a member of _pownerbiota, we need to pass this pointer into the add_me method.
Abbreviations
Part II: Software Engineering and Computer Games Reference
Appendix A. The Windows keycodes
Appendix B. The Pop help file
Appendix C. Summary of the controls for Visual Studio | 1,882 | 7,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-22 | longest | en | 0.909356 |
http://www.gap-system.org/ForumArchive/Waki.1/Katsushi.1/About_Tw.1/1.html | 1,534,309,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209884.38/warc/CC-MAIN-20180815043905-20180815063905-00264.warc.gz | 500,093,056 | 1,514 | > < ^ Date: Tue, 11 Apr 2000 14:23:52 +0100
> < ^ From: Katsushi Waki <slwaki@cc.hirosaki-u.ac.jp >
Dear Forum,
Could you tell me why I can not get Cohomology from the following.
```(This is a log of GAP4r1.)
gap> g:=GL(2,2);;
gap> pgl:=Action(g,Orbit(g,One(GF(2))*[1,0],OnLines),OnLines);;
gap> iso_g:=IsomorphismPcGroup(pgl);;
gap> G:=Image(iso_g);;
gap> mats:=MutableCopyMat(GeneratorsOfGroup(g));;
gap> M:=GModuleByMats(mats,GF(2));;
gap> TwoCoboundaries( G, M );
[ [ 0*Z(2), Z(2)^0, 0*Z(2), 0*Z(2), 0*Z(2), 0*Z(2) ],
[ 0*Z(2), 0*Z(2), Z(2)^0, 0*Z(2), 0*Z(2), 0*Z(2) ],
[ 0*Z(2), 0*Z(2), 0*Z(2), 0*Z(2), Z(2)^0, 0*Z(2) ],
[ 0*Z(2), 0*Z(2), 0*Z(2), 0*Z(2), 0*Z(2), Z(2)^0 ] ]
gap> TwoCocycles( G, M );
[ [ 0*Z(2), Z(2)^0, 0*Z(2), 0*Z(2), 0*Z(2), 0*Z(2) ],
[ 0*Z(2), 0*Z(2), Z(2)^0, 0*Z(2), Z(2)^0, Z(2)^0 ] ]
gap> cc := TwoCohomology( G, M );;
AppendList: <list2> must be a list (not a object (positional)) at
Append( cat, lst );
Concatenation( List( Wvectors, function ( v )
return zero;
end ), BasisVectors( canbas ) ) called from
NaturalHomomorphismBySubspaceOntoFullRowSpace( co, cb ) called from
<function>( <arguments> ) called from read-eval-loop
Entering break read-eval-print loop, you can 'quit;' to quit to outer loop,
or you can return a list for <list2> to continue
```
Katsushi Waki + Hirosaki University +
> < [top] | 609 | 1,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-34 | latest | en | 0.382762 |
https://www.halfbakery.com/idea/Infinite_20depth_20of_20field_20contact_20lens?op=aye | 1,638,219,990,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00345.warc.gz | 910,833,833 | 12,084 | h a l f b a k e r y
Romantic, but doomed to fail.
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# Infinite depth of field contact lens
Facetious.
(+2, -1) [vote for, against]
So, I was trying to think of how to make a contact lens (or any lens) which would give you an infinite depth of field.
The topic of "pinhole lenses" has already been discussed. In theory, a pinhole lens is great, but a greater depth of field is paid for by a much lower total light intensity. Pinhole glasses have multiple pinholes, and they work, but not very well.
So.
First, we're going to make a single pinhole lens. Imagine a small cube of transparent material (acrylic or whatever), maybe 1mm on a side. Coat the face of it with black paint, and then make a pinhole in the middle of the paint. You now have a pinhole camera (of sorts). If the back of the cube were a screen, you would see projected on to it a perfect image of what the camera was looking at, albeit very dim (also upside down, but hey, details).
However, there is no actual "screen" on the back of this camera. Instead, the back of the camera has a slight curvature on it, so that it acts as a micro-lens, projecting the perfect image (which has been "focussed" by the pinhole) onto the retina. This "micro lens" doesn't have to do much focussing (except for compensating for anything that the eye's own lens does).
So far, so pointless. We are still looking at the world through a single small pinhole, and the image (although in focus throughout) will be very dim.
Now, though, we make a second cube, similar to the first one, and we glue the two of them side by side. We also make a micro-lens on the back of this second cube. This micro lens is similar to the first one, except that it has a tilt on it so that it projects the second image superimposed on the image from the first one.
So we now have two pinhole images, both in perfect full- depth focus, and projected superimpositionally on the retina.
And so we continue. By the time we've finished, we have a contact lens which consists of perhaps a hundred individual pinhole camera facets, with all 100 images projected in alignment on the retina.
(Note: in case there's any doubt, each pinhole image covers the full field of view, just as a "normal" pinhole camera images the whole scene in front of it. It's not like pinhole glasses, where you see only a small part of the field of view through each pinhole.)
Such a contact lens would give you an acceptably bright image (perhaps 20% of the normal light intensity, which would not be noticeable - a well-lit room is tens of times dimmer than daylight) covering the whole field of view, and with everything being simultaneously in focus.
— MaxwellBuchanan, Jul 06 2011
MB's slightly less famous eye experimenter http://gorm.wordpre...ith-a-blunt-needle/
[Ling, Jul 10 2011]
MB's slightly less than successful experiment http://i923.photobu...%20MkI/IMG_5419.jpg
[MaxwellBuchanan, Jul 10 2011]
My initial response to the final paragraph was a grumpy "Oh yeah? Prove it." But since I have no data to the contrary, I'll restrain myself and try a nap instead.
It did remind me of a pair of glasses a friend of mine had in the early seventies, that did something similar but without the projections in alignment on a common spot. The image was of having compound eyes like a fly.
— normzone, Jul 06 2011
Light refracted through a glass of water or a particularly curved windscreen on a car might play merry hell with these, as it does with similar optics such as certain (older) telephoto lenses and roof-prism binocs. I don't know this for certain, I'm just postulating. Knowing my wife's troubles with contacts before her lasik, I might also say these should be daytime-use only devices.
— Alterother, Jul 06 2011
hmmmm, clever…
— hippo, Jul 06 2011
//It did remind me of a pair of glasses...// Yes, I have just bought a pair out of curiosity, and they are hopeless. They do give an uncanny depth of field, but the pinholes are too far in front of the eyes, and any part of the scene is viewed through only one pinhole per eye, so it is like looking at the world through a collander.
However, by converging all the full-field images from multiple pinholes close to the cornea, I think this effect could be eliminated.
— MaxwellBuchanan, Jul 06 2011
Incidentally, I am not sure of the exact lens geometry which would be needed to combine all the pinhole images as I described, but I suspect it might turn out to be a regular convex lens.
In this case, then, what we would have is basically a regular "positive" contact lens, but with the front face being opaque apart from multiple pinholes.
— MaxwellBuchanan, Jul 06 2011
IMHO, light (falling on retina) from one big lens will always be more than 100 pinholes combined. Hence resulting image will always be dimmer.
Focal lenght of a pinhole is determined by hole diameter. Smaller the hole, smaller the focal length. Hence there will be restriction on how much smaller holes can be made. I am not sure, if it will be possible to fit 100 holes.
Superimposing 100 images can be trickier than using a mere convex lense. I think a simple lense may not work.
— VJW, Jul 06 2011
[Max] I think your logic is flawed here. For instance, if you replace a pupil 2 mm in diameter with a pair of pinholes 2 mm apart, the effective depth of field will be about the same, except that out-of-focus objects will now appear to be doubled, rather than blurred. As the number of pinholes increases, the result will approach a dimmer version of an image with the same depth of field as a pupil whose size is the same as the total extent of the pinhole field, rather than one the size of an individual pinhole.
To put it another way, your 100 images are taken from 100 vantage points, so they can only be made to coincide for a single focal plane; objects at other distances will be blurred due to parallax. Or rather, the 100 images of objects at other distances will not coincide, due to parallax.
— spidermother, Jul 06 2011
Yes, the image will definitely still be dimmer than normal, but very large changes in light intensity are not that noticeable (hence my point that light intensity outdoors is much much higher than in a well-lit room). Of course some of this is compensated for by the eye's iris, but not much; most of the compensation is perceptual.
Focal length: smaller holes give greater depth of field, until you hit a diffraction limit. I'm not sure what size hole would be optimal.
Superimposing 100 images: well, a regular lens (including that in the healthy eye) does exactly this and more; it is effectively superimposing an infinite number of images (ie, from light rays hitting all parts of the lens). Hence my suggestion that the geometry to superimpose 100 individual images may indeed just be a regular lens.
— MaxwellBuchanan, Jul 06 2011
I think you misunderstand me. Not only will the image be dimmer, but the depth of field will be no better than that of a single opening whose diameter is the same as the total size of the field of pinholes.
— spidermother, Jul 06 2011
[spidermother] I was replying to [VJW]'s anno (you were writing yours while I was replying).
Your point about depth of field is interesting and probably right... but there's got to be some edge. Basically, a pinhole trades brightness for depth of field, and there ought to be a way to capitalize on that. We can handle some loss of brightness without really missing it. But, I take your point that I may not have the right geometry here.
At a cruder level, those of us with lousy eyes (near- and far-sighted) already use the pinhole effect: focussing errors are much less severe in very bright light, but only because the iris contracts. The effect is a sort of pseudo- pinhole, where only the central part of the lens (which does less refraction, and where refractive errors are lower) is used.
But in the end, this argument converges with the "pinhole contacts" idea that was discussed elsewhere.
I'm going to go and play with contact lenses and acrylic paint.
(More generally, I've been thinking about ways to use photonic materials to create an infinite depth of field, and this idea seemed the simplest and crudest.)
— MaxwellBuchanan, Jul 06 2011
— spidermother, Jul 06 2011
I took optics was several decades ago, but isn't the whole depth of field issue a result of spherical lens grinding? (A compromise as parabolic grinding is so hard.) So can't you solve the issue by using parabolic mirrors, parabolic lenses or spherical lenses doped to alter radial refractive index to mimic parabolic lenses? The later should be easy as radial doped glass is what is used to create single mode fiber-optic cable.
— MisterQED, Jul 06 2011
No, depth of field is simply a consequence of aperture size. You are thinking of spherical aberration.
— spidermother, Jul 06 2011
What [spidermother] said. You can have a perfect lens (at least in theory) which will focus parallel incoming rays of light (as from a distant source) at a certain point. However, if the incoming rays of light are diverging (as from a close source), the will be focussed further back (ie, the near object will be out of focus in the original plane).
Conversely, if you make a stronger lens to bring the diverging rays into focus, it will over-refract the parallel rays, so the distant object is out of focus.
— MaxwellBuchanan, Jul 06 2011
Yes, to be perfectly focussed at all distances, all of the light has to pass through a single point. A pinhole is a good approximation of that point. Two or more pinholes is not.
Perhaps if there were some way of increasing the amount of light through (head-mounted floodlights?), or increasing the sensitivity of the receptors (drugs? Fitting babies with these pinhole devices at birth so they develop with extra sensitivity?), then monopinholeous contact lenses may be plausible.
— pocmloc, Jul 06 2011
Hang on. I'm not sure I agree with that.
I make a good, old-fashioned pinhole camera with a very small pinhole. I take a photo.
I make another identical camera, and take an identical photo. Both photos will be fully in focus (at least, both will be very sharp) at all depths.
I then superimpose those two images. Is the result less sharp than the original? Technically, this is no different from taking a single exposure for twice as long, if the image superimposition is accurate.
In practice, of course, there is the matter of superimposing all the pinhole images. However, as I mentioned earlier, this is no different from what a regular lens does (superimposing light rays which originate from a single point but enter the lens at different points).
What counts, though, is that a lens which will superimpose these images will work for all focal lengths.
But I am not convinced by my own argument, and it does seem elegant that two pinholes would give a resolution similar to a single larger pinhole.
So, help me out here. Why does superimposing two identical images result in a blurrier image and, if it does, why is this different from a single exposure of twice the length?
[EDIT - don't forget, we're not talking about two pinholes forming a single image; we have two pinholes forming independent images which are then aligned. I agree that a single pinhole camera with two pinholes will be a mess.]
— MaxwellBuchanan, Jul 06 2011
[Note: I am just waiting for the paint to dry on a pair of contacts. So, if I do not make any further annotations, please contact me by braille.]
— MaxwellBuchanan, Jul 06 2011
Two identical images can be superimposed, but two images taken from slightly different positions can't. If you line up the foreground, the background will be slightly out, and vice versa.
— spidermother, Jul 07 2011
//In theory, a pinhole lens is great//
Not so much. They always give blurrier picture compared to simple lense. Photos have a dreamy look; Borders are always darker compared to center, unlike simple lens.
Superimposing : If you draw the ray diagram, I think it will be clearer. For a lense, these will 100 point sources, very close to the lense. I think it will be a mess. Also, output from the lense will be near parellel rays since, these 100 pinholes will be quite close to focal point of the lense.
— VJW, Jul 07 2011
Another approach would be to muck about with the chemistry of your retina to make it much more sensitive to light. Then, a pinhole opening would give you the infinite depth of field you're after and be bright enough. A mechanical approach could be tried too - cats have a reflective layer behind their retinas, so that light passes through the light- sensitive cells twice. How hard would it be to insert a reflective layer behind the human retina?
— hippo, Jul 07 2011
Now, a 3D retina might allow independance from the lens...
— Ling, Jul 10 2011
Hmmm. Well, after some extensive research at the MaxCo. Optical Laboratory (involving some old contacts, acrylic spray paint, and sophisticated pinhole-making technology), I have to report poor results on the pinhole front.
Aside from the problem that contacts slide around on the cornea, it seems that diffraction becomes a major problem when the (single) pinhole is small enough to give a good image.
This doesn't entirely explain why the normal pupil, when constricted by bright light, doesn't give diffractive problems, but perhaps this is because (a) it's still larger than my pinholes and (b) it's closer to the lens (?).
The quest continues. Meanwhile, [-].
— MaxwellBuchanan, Jul 10 2011
a
— VJW, Jul 10 2011
[MB]... any relations to Isaac Newton? Link.
— Ling, Jul 10 2011
Oddly, he's great^n cousin. We're not in touch, however.
(Painting contacts is easy; the acrylic paint is only on the outer surface and, once thoroughly dried, is stable and non-irritating. Photo of MkI in link.)
— MaxwellBuchanan, Jul 10 2011
Good God...looks like something out of Terminator. Or perhaps more fittingly: Half a hangover.
— Ling, Jul 10 2011
I tried wearing two such contacts. However, when I disovered that I was trying to take photos using the TV remote, I reverted to one.
— MaxwellBuchanan, Jul 10 2011
what [spidermother]'s been saying. Yes, each pinhole image is crystal clear *but* they're not all the same image, so they wont superimpose. Blurryness is just a lot of crystal-clear superimposed images that aren't fully keyed.
e.g. An artificial eye has a pupil of aperture 10mm. Incoming light-rays can enter anywhere between point Point A is at 0mm (on the left) and point B is 10mm (on the right) this pupil forms the "waist" of a cone of light. Point the eye at an equilateral triangle 10m away. Draw a line from one corner of the equilateral triangle through Point A until the ray hits the cornea. Now do the same, but through Point B. The two points hit the cornea at some distance apart. This distance is a measure of how blurry the resultant image is going to be.
Now repeat the process with a pinhole camera projecting one image over Point A, and another one Projecting another (crystal clear) image over Point B. Both images are clearly resolved, but by the time they are projected back onto the cornea - they don't overlap anymore.
Creating a lens to bend the two pinhole projections back into a single resolvable image is doing exactly what a pair of spectacles does - only in discrete, compound fashion, rather than over a continuum. The question is, does that help?
— zen_tom, Jul 12 2011
// Creating a lens to bend the two pinhole projections back into a single resolvable image is doing exactly what a pair of spectacles does - only in discrete, compound fashion, rather than over a continuum. The question is, does that help?//
Yes, it should. It's relatively easy to create a normal contact lens which does this for a single focal length, not for all focal lengths simultaneously. I was hoping that, because each pinhole image has infinite (or at least good) depth of field, a single lens could converge them all to give a unified full-depth image.
However, my experiments to date suggest that diffraction is the real killer (I think someone mentioned this up there).
I'm now seriously thinking about how to make a contact lens with autofocus, or at least manually- settable focus. I may post this as another idea.
— MaxwellBuchanan, Jul 12 2011
[annotate]
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Helper I
## Equivalent of =SUMPRODUCT?
Looking for an PBI equivilant of sum product to work out weighted mean. Been stuck trouble shooting solutions and now I just think I'm overthinking this...
Identifier Scores Weight 1 0.67 4% 2 0.67 4% 3 0.63 19% 4 0.6 24% 5 0.59 20% 6 0.33 4% 7 0.14 8% 8 0.11 11% 9 0 5% 10 0 2%
Weighted Average (value im looking for) = 0.46
Average = 0.37
Bonus if it can be dynamic (e.g. user filters for identifier 1,2,3 and caculates weight and then weighted mean)
1 ACCEPTED SOLUTION
Super User I
Hi @Rewind
A measure like this should do the trick, assuming the weighting is to be done row-by-row over your table:
``````Weighted Average =
VAR WeightedSum =
SUMX (
YourTable,
YourTable[Scores] * YourTable[Weight]
)
VAR TotalWeight =
SUM ( YourTable[Weight] )
RETURN
DIVIDE ( WeightedSum, TotalWeight )``````
The measure iterates over your table, summing Scores * Weight, then divides by the total Weight.
This will also be dynamic in that it will respond to any filters you apply.
Regards,
Owen
Owen Auger
My Blog
2 REPLIES 2
Super User I
Hi @Rewind
A measure like this should do the trick, assuming the weighting is to be done row-by-row over your table:
``````Weighted Average =
VAR WeightedSum =
SUMX (
YourTable,
YourTable[Scores] * YourTable[Weight]
)
VAR TotalWeight =
SUM ( YourTable[Weight] )
RETURN
DIVIDE ( WeightedSum, TotalWeight )``````
The measure iterates over your table, summing Scores * Weight, then divides by the total Weight.
This will also be dynamic in that it will respond to any filters you apply.
Regards,
Owen
Owen Auger
My Blog
Helper I
Thanks Owen, this solved my problem. Should have come here a few hours ago 😅
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https://www.jiskha.com/display.cgi?id=1403056480 | 1,503,510,649,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123312.44/warc/CC-MAIN-20170823171414-20170823191414-00024.warc.gz | 913,985,861 | 3,945 | # Math
posted by .
– 11 is the __________ term of the arithmetic sequence – 847, – 836, – 825, – 814, – 803, …
• Math -
use
term(n) = a + (n-1)d
here a = -847 , d = 11 , n = ?? , and term(n) = -11
solve :
-11 = -847 + (n-1)(11)
which should be pretty straightforward to do.
## Similar Questions
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More Similar Questions | 725 | 2,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-34 | latest | en | 0.938215 |
https://mathoverflow.net/questions/198232/does-independence-of-the-sequence-fa-i-b-imply-the-sequence-is-independent | 1,606,595,560,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195745.90/warc/CC-MAIN-20201128184858-20201128214858-00351.warc.gz | 399,433,850 | 29,363 | # Does independence of the sequence $f(A_i, B)$ imply the sequence is independent of $B$?
Suppose $B, \{A_i: i \in \omega\}$ are i.i.d. random variables with uniform distributions on $[0,1]$. If $f$ is a map such that $\{f(A_i, B): i \in \omega\}$ are independent, must $\{f(A_i, B): i \in \omega\}$ also be independent of $B$?
• I can prove it for discrete uniform distributions. I wonder if that helps for continuous uniform distributions. – Brendan McKay Feb 23 '15 at 14:16
• Do you think there is a counterexample for general (i.e. non-uniform) distributions? – usul Feb 23 '15 at 17:22
• No, I think any distributions at all are ok, even a different one for each of the arguments. See the stuff I added to my answer. – Brendan McKay Feb 24 '15 at 1:19
It is true.
Let $\chi_Z(x,y)$ be the indicator function of the set $\lbrace (x,y) \,|\, f(x,y)\in Z\rbrace$ for some set $Z$.
The condition that the events $f(A_1,B)\in Z$ and $f(A_2,B)\in Z$ are independent is that $$\int_{[0,1]^3} \chi_Z(x,y)\chi_Z(x'y)\,dxdx'dy = \int_{[0,1]^2} \chi_Z(x,y) \,dxdy \ \ \int_{[0,1]^2} \chi_Z(x',y') \,dx'dy'.$$ Define $g(y) = \int_0^1 \chi_Z(x,y)\,dx$. Then the above condition can be rearranged into $$\int_{[0,1]^2} g(y) (g(y)-g(y'))\, dy dy' = 0.$$ That's just a change of variables from $$\int_{[0,1]^2} g(y') (g(y)-g(y'))\, dy dy' = 0,$$ so we can subtract the two to get $$\int_{[0,1]^2} (g(y)-g(y'))^2 \,dydy' = 0$$ which by positivity means $g(y)$ is constant a.e.. Harking back to the definition of $\chi_Z$ and noting that the set $Z$ was arbitrary, we see that $f(A_1,B)$ is independent of $B$.
I have only considered pairwise independence, but once you find that $g(y)$ is independent of $y$ it makes the whole lot of them independent. At least, it seems so at 3am...
ADDED after waking up: At the moment I can't see how uniform distributions are needed for this argument. Just consider any vertical measure $\mu$ and horizontal measure $\nu$ and replace $dx$ by $d\mu$, $dy$ by $d\nu$, etc, in the above argument. Still works, doesn't it?
MORE ADDED: Nate's question leads me to note the following, which surely must be well-known in the theory of exchangeable random variables (but I didn't know it).
Theorem. Let $X,Y$ be random variables such that $(X,Y)$ and $(Y,X)$ have the same distribution. Then $X$ and $Y$ are independent iff for all measurable sets $S$, $P(X\in S \wedge Y\in S)=P(X\in S)^2$.
Proof. Let $S,T$ be measurable sets. First assume $S$ and $T$ are disjoint. Since by the exchangeability assumption $P(X\in T\wedge Y\in S)=P(X\in S\wedge Y\in T)$, we have \begin{align} 2 P(X\in S\wedge Y\in T)&=P(X,Y\in S\cup T) - P(X,Y\in S) - P(X,Y\in T) \\ &= P(X\in S\cup T)^2 - P(X\in S)^2 - P(X\in T)^2 \\ &= (P(X\in S) + P(X\in T))^2 - P(X\in S)^2 - P(X\in T)^2 \\ &= 2 P(X\in S) P(X\in T) \\ &= 2 P(X\in S) P(Y\in T). \end{align} If $S$ and $T$ are not disjoint, break $P(X\in S\wedge Y\in T)$ into four disjoint cases like $P(X\in S\wedge Y\in T\setminus S)$ and apply the above to each. It works.
• In general, to show two random variables are independent, you have to show $P(X \in U, Y \in V) = P(X \in U) P(Y \in V)$ for $U,V$ ranging over an appropriately rich collection of sets. Why does it suffice to prove the special case where $U=V=Z$? Sets of the form $Z \times Z$ don't generate the product $\sigma$-algebra, do they? – Nate Eldredge Feb 24 '15 at 1:52
• @Nate : What you say is correct, but I don't think I'm making that mistake. Independence of $X$ and $Y$ implies that the events $X\in Z$ and $Y\in Z$ are independent; the converse is false but I don't use it. Once we know $g(y)$ is a.e. constant, the distributions of $f(A,B)$ conditional on any two non-null $B$-events are the same. But let me know if you still think there's a problem. – Brendan McKay Feb 24 '15 at 2:27
• @Nate : To my surprise, the converse is true in this case, see my addition. – Brendan McKay Feb 24 '15 at 4:08 | 1,325 | 3,935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-50 | latest | en | 0.859857 |
https://www.jiskha.com/similar?question=hey+guys%2C+i+follow+you+up+to+the+point+of+intergrating+the+final+line.+How+would+I+intergrate+%5Bx%5E-2.e%5E-x%5D+dx+thx&page=293 | 1,563,549,768,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526254.26/warc/CC-MAIN-20190719140355-20190719162355-00444.warc.gz | 751,943,924 | 23,141 | # hey guys, i follow you up to the point of intergrating the final line. How would I intergrate [x^-2.e^-x] dx thx
49,088 questions, page 293
1. ## Physics
Please help me!! I need this as soon as possible. 2.A billiard ball at rest is struck by the cue ball of the same mass whose speed is 5 m/s .After an elastic collision the cue ball goes off at 50 degree with respect to its original direction of motion and
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3. ## Chemistry
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4. ## English
I added a few more sentences. I hope you can help me express them correctly. Thank you. 1) Do you like soccer (Italian football)? Is soccer popular in America? 2) Can you watch the Italian football season in America? (Do you follow the Italian football
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5. ## college
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8. ## Chemistry HELP!
A student placed 10.5g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100.- mL mark on the neck of the flask was reached. The flask was then shaken until
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11. ## Chemistry
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Coasting due west on your bicycle at 7.9 , you encounter a sandy patch of road 7.0 across. When you leave the sandy patch your speed has been reduced by 2.0 to 5.9 . 1. Assuming the sand causes a constant acceleration, what was the bicycle's acceleration
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Find the equation of the tangent line 5x^2+y^2=14 at (1,3). P.S: It's calculus. So far I know this much 10x+2y(dy/dx)=o -10x -10x ___________________ 2y(dy/dx)= -10x (dy/dx)= -10x/2y= -5x/y m= -5(1)/3= -5/3 y=mx+b 3= -5/3(1)+b then what?
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33. ## physics
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36. ## Physics
Two blocks move along a linear path on a nearly frictionless air track. One block, of mass 0.107 kg, initially moves to the right at a speed of 4.70 m/s, while the second block, of mass 0.214 kg, is initially to the left of the first block and moving to
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37. ## Math, mathematics of finance
I am stuck on these two math questions. If nyone could help me solve them it would be greatly appreciated! Here is the information: Engineering estimates indicate that the variable cost of manufacturing a new product will be \$35 per unit.Based on market
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38. ## chemistry
The vapor pressure of water at 80 degrees c is 355mmHg. A 100cm3 vessel contained wateraturated oxygen at 80 degrees c., the total gas pressure being 760 mmHg. The contents of the vessel were transferred by pumping to a 50cm3 vessel at the same
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39. ## physics
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40. ## PHYSICS 1301 (COLLEGE)
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43. ## Math
How do you factor a polynomial? Using the FOIL method works with some simple polynomials. FOIL stands for First, Outside, Inside, Last. For example: x^2-4x+4 factors to (x-2)(x-2) Check your work by multiplying them all. x times x is first, x times -2 is
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44. ## Language arts
Characters: (in order of appearance) 1. JASON: a boy of about 14. He is a student in Ms. Smith’s English class. He regularly misbehaves in order to get attention, and he doesn’t apply himself to his schoolwork. 2. THE CLASS: 25 seventh-graders, male
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45. ## English
Tomorrow we have several events. I'd like you to know the schedule of our school. From June 2 through June 4, second year students have a school excursion (trip), which is a kind of study tour. They have a day off on June 5th. So you don't need to go to
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46. ## Food Chemistry
A heat exchanger is used to warm apple cider using steam as the heat source. The cider is heated from an initial temperature of 4°C to a final temperature of 65°C. The steam enters the heat exchanger as 50% quality steam and exits as water condensate at
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47. ## Math
A heat exchanger is used to warm apple cider using steam as the heat source. The cider is heated from an initial temperature of 4°C to a final temperature of 65°C. The steam enters the heat exchanger as 50% quality steam and exits as water condensate at
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48. ## CHEMISTRY
Observations Mass of Fe metal:100 MG Volume of water in Calorimeter: 100 ML Mass of water in Calorimeter (use 1g per ml): 100G Initial Temperature of metal: 20 Initial Temperature of water: 20 Final Temperature (both metal and water): 100 Determine the
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49. ## AP math how to word my answer no equations
The problem states that I have to come up with best possible predictor of a 100yr flood plan. Well I found that best projection and I have to state why. Could I say: The projection from the power trend line is a better indicator because with the
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50. ## Physics
A football is kicked off with an initial speed of 63 ft/s at a projection angle of 45 degrees. A receiver on the goal line 67 yd away in the direction of the kick starts running to meet the ball at that instant. What must be his minimum speed (in
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51. ## Production and Operations management
A ball bearing assembly firm wants to set up an assembly line which must have an output of 60 units per hour. The work elements are A to J and the task times are 40, 30, 50, 40, 6, 25, 15, 20, 18 and 30 seconds respectively. a. Determine the cycle time
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52. ## math
A vector can be used to represent the path of a drill tip used to bore a deep mine shaft in Sudbury one quarter of the way to the centre of the Earth. Represent the vector using a directed line segment and Cartesian co-ordinates and describe which
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53. ## geography
Will you please check my answers? 1.What states border Lake Michigan? Wisconsin, Illinois, Indiana, and Michigan 2. How many states are on the 60 degrees N line of latitude? Just one state ( Alaska ) 3. What do the following countries have in common: Chad,
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54. ## physics
A football is kicked off with an initial speed of 65 ft/s at a projection angle of 45 degrees. A receiver on the goal line 63 yd away in the direction of the kick starts running to meet the ball at that instant. What must be his minimum speed (in
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55. ## GLOBE/GEOGRAPHY HELP
If the sun is directly overhead at 12 noon at your location and GMT (Greenwich Mean Time) is 10:00pm, what is your approximate longitude? Check this site to see what place is 10 hours behind London and the GMT. (Broken Link Removed) Then, you can use a map
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56. ## chemistry
You are conducting experiments with your x-ray diffractometer. (a) A specimen of molybdenum is exposed to a beam of monochromatic x-rays of wavelength set by the Kα line of silver. Calculate the value of the smallest Bragg angle, θhkl, at which you can
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57. ## math
The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 4 sin ¦Ðt + 3 cos ¦Ðt, where t is measured in seconds. (a)Find the average velocity during each time period. (i) [1, 2]
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58. ## math
which of the following types of information is not suited for display on a double line graph A: the change in rainfall based on the change in temperature B: the number of and different colors of car in parking lot C: the increase in value of jewelry for
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Which of the following is an alternative to the bureaucratic paradigm? A. An organization is defined by the resources it controls B> An organization sticks to routine and focuses on its own needs C. An organization empowers front line employees D. An
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60. ## Cash Budgeting Question
Cash Budgeting – Helen Bowers, owner of Helen’s Fashion Designs, is planning to request a line of credit from her bank. She has estimated the following sales forecasts for the firm for parts of 2006 and 2007: May 2006 \$180,000 June
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61. ## physics
A regulation 145 g baseball can be hit at speeds of 100 mph. If a line drive is hit essentially horizontally at this speed and is caught by a 65 kg player who has leapt directly upward into the air, what horizontal speed (in cm/s) does he acquire by
asked by Student1 on October 28, 2009
62. ## Calculus and vectors
A vector can be used to represent the path of a drill tip used to bore a deep mine shaft in Sudbury one quarter of the way to the centre of the Earth. Represent the vector using a directed line segment and Cartesian co-ordinates and describe which
asked by Court on September 27, 2013
63. ## math
A designer enlarged both the length and the width of a rectangular carpet by 60 percent. The new carpet was too large so the designer was asked to reduce its length and width by 25 percent. By what percent was the area of the final design greater than the
asked by jenny on June 3, 2016
64. ## Physics
A lighted candle is placed 38 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens of focal length 10 cm that is 5 cm from the diverging lens. The final image is real, inverted, and 19 cm beyond the
asked by Ali on August 3, 2015
65. ## Need help Statistic
please check answer Unit 4 Test Multiple Choice 1. Which choices listed below indicate that a linear model is not the best fit for a dataset? Choose all that apply. (3 points) • Scatterplot shows a strong linear pattern. • Scatterplot shows a curve
asked by Shantel on January 6, 2014
66. ## chemistry
what is the mass (by difference) of a sample if the initial mass of the sample in the bottle was 29.89-g and the final mass of the bottle and sample was 22.01-g ?
asked by Omar on December 4, 2014
67. ## Religions of the World
For my final project I have to conduct an interview with someone with a different religion other than my own and visit their place of worship. I have to pick a religion..Any Suggestions? and also.. I need to conduct an interview asking 10 questions. Three
asked by Jen on June 19, 2009
68. ## English
Journal Entry 3: Prewriting and Thesis Statement Brainstorm: Review the description of brainstorming in your textbook on page 111. Then write a list of all the social media and social networking websites and apps you might use to connect with friends and
asked by anonymous on October 27, 2016
69. ## accouting
- Hanna Manufacturing – Due Hanna Manufacturing manufactures components for the farming industry and is considering replacing its existing equipment with new and more high technology machinery, despite the fact that its existing eq uipment is only a few
70. ## English
I still have some doubts about a summary I made of Macbeth's plot. I hope you can have a look at it, too. I included various alternatives. Please, tell me if they are all possible. 1)The play opens with the news that Macbeth has succeeded in defending
asked by Mike on February 13, 2011
71. ## algebra
3. Moving Costs Moving company A charges \$80 plus \$55 an hour to move households across town. Moving Company B charges \$75 an hour for cross-town moves. For what lengths of time is Moving Company B more expensive than Moving Company A? Write an inequality
asked by Tammy on August 21, 2011
72. ## English...check my work
parts of speech 1.Mason is fantastic at gossamer billowing. my answer: Mason=noun, is=verb fantastic=adjective at=preposition gossamer=adjective billowing=noun We get to go home early on Wednesday. my answer: We=pronoun get to go=verb home=noun
asked by Mark on March 2, 2017
73. ## english grammar check please
parts of speech 1.Mason is fantastic at gossamer billowing. my answer: Mason=noun, is=verb fantastic=adjective at=preposition gossamer=adjective billowing=noun We get to go home early on Wednesday. my answer: We=pronoun get to go=verb home=noun
asked by Mark on March 1, 2017
74. ## Geometry
The midpoint of line AB has coordinates of (5,-1). If the coordinates of A are (2,-3), what are the coordinates of B? Any help is appreciated!
asked by Gabs on November 8, 2010
Find the slope of the line through the given points. ( 8, -2) and ( 8, -1) Since 8-8 is 0 would the slope be 0 or would there be no slope?
asked by Lyndse on December 20, 2009
76. ## Geometry
M is the midpoint of line AN, A has coordinates (-6,-6) and M has coordinates (1,2). Find the coordinates of N.
asked by W Haley on September 27, 2017
The line y = 3x + 5 crosses the y axis at ______? When it crosses the y-axis x has to be 0, so what is y when x=0 in the equation y = 3x + 5 Isn't it 5? Hmm, you don't sound too confident with your answer, but yes, it's 5. If you have an equation y=mx+b, m
asked by Linda on October 11, 2006
78. ## limerick poem
There once was a man who had ripped pants He kept his legs close whenever he danced He never got new ones cause he had to pay I don't have no money is what he would say i got that so far i know i need one more line but i don't understand what i should say
asked by 2phoneeeeee on November 10, 2014
79. ## Algebra II
Find the equation of the line that passes through the points (1,4) and (2,-8). Question options: y = -3x + 9 y = 6x + 5 y = -12x + 16 y = 12x - 8 < this one is wrong.. I thought I had it right so how do I find the correct answer? i'm not getting any of the
asked by Kate on February 1, 2017
80. ## physics
city b is north of city a and the 400 mile trip by care takes 8 the cities are 320 miles apart in a straight line what is the velocity of a trip from city a to city b
asked by kanisha on February 24, 2016
81. ## physics science
3. An unknown element produces a spectral line of 425nm. A. Calculate the energy of the emitted photon in joules. B. What is the energy of the emitted photon in electron-volts?
asked by suszan on January 23, 2012
82. ## Electrical Circuits
A length of copper telephone line has a resistance of 24Ω at 20 degree Celsius. What it its resistance on a hot summer day when its temperature rises to 36 degree Celsius.
asked by Anonymous on December 8, 2016
83. ## math
a worker on the production line is paid a base salary of \$210 per week plus \$.55 for each unit produced. One week the worker earnedh \$329.35. How many units were produced.
asked by Anonymous on September 18, 2010
Find the points where the tangent line is horizontal or vertical. Find the area under one period of the curve. x= r(deta - cos deta) Y= r(1- sin deta) deta= pie/3
asked by stan on August 28, 2010
how to fid g(4), g'(4) and g''(4) when graph of the function f shown above consist of six line segments. Let g be the function given by g(x))=___ f(t) dt. The ___ is form like f with an x variable at the right top corner and a 0 in the right low corner.
asked by Hugo on April 23, 2009
For #1-4, write an equation of the line that passes through each pair of points. Show all of your work to receive full credits. 1.) (9, -2) and (4, 3) Answer: 9 = -1x + -2 2.) (3, 5) and (2, -2) Answer: 3 = 7x + 5 3.) (4, 3) and (2, 0) Answer: 4 = 2/3x + 3
asked by Sharpshooter on September 25, 2017
87. ## Math
The teacher gave this pattern to the class: the first term is 5 and the rule is add 4, subtract 1. Each student says one number. The first student says 5. Victor is tenth in line. What number should Victor say?
asked by Jayden on April 21, 2015
88. ## congress and law
Why hasn’t the president simply vetoed such laws since Congress was most often unable to override a veto? Will a line-item veto now give the president too much of an advantage over Congress?
asked by elley on November 22, 2008
89. ## Math
and my last problem i need help with. i don't know how to even begin settin the problem up to figure out the answer. E coli, a typical bacterius, is about 2 micrometers long. how many E coli could you line up, end-to-end, in one mm?
asked by amy on June 20, 2010
90. ## Math Rational Functions-Please Clarify =)
I have a question about the rational function I recently posted. Would the range still be (-∞,2)U(2,∞) if the rational function is what Reiny posted "(2x^2-18)/(x^2+3x-10)." The y-intercept confuses me because its (0,1.8) and when I look at the line it
asked by Enlia on October 15, 2012
Write the standard line notation for each cell below. IO3-(aq) + Fe2+(aq)--> Fe3+(aq) +I2(aq) ________________________________________ H2O2 + 2 H+ + 2 e- --> 2 H2O E = 1.78 V O2 + 2 H+ + 2 e- --> H2O2 E= 0.68 V ________________________________________ Mn2+
asked by Janvi on March 6, 2012
92. ## Math
The teacher gave this pattern to the class: the first term is 5 and the rule is add 4, subtract 1. Each student says one number. The first student says 5. Victor is tenth in line. What number should Victor say?
asked by Jayden on April 21, 2015
93. ## math
Which graph is more appropriate for displaying the number of individuals of differing ages at a picnic? A histogram, bar graph, or line graph? Why would it not be appropriate to use the ones not picked? I am confused... I thought it would be bar graph as
asked by m on November 17, 2010
94. ## Trigonometry
Angle of Depression Find the angle of depression from the top of the lighthouse 250 feet above water level to the water line of a ship 2.5 miles offshore.
asked by AwesomeGuy on February 4, 2013
95. ## math
a line segment in a coordinate plane with a length of 76 units was rotated 60 deg about the origin and then translate 10 units to the left. what would be a resulting length of the segment after these transformation
asked by johnsa on March 14, 2017
96. ## Math
The teacher gave this pattern to the class: the first term is 5 and the rule is add 4, subtract 1. Each student says one number. The first student says 5. Victor is tenth in line. What number should Victor say?
asked by Chris on May 7, 2014
97. ## Math
If the following integers were plotted on a number line with twenty-five, which integer would appear on the same side of zero and have a greater distance from zero? 1. opposite of eighteen 2. negative one 3. opposite of negative thirty-one 4. twenty-four
asked by Tom on September 9, 2013
98. ## economics
What are the components of aggregate expenditure? In the model developed in this chapter, which components vary with changes in the level of real GDP? What determines the slope of the aggregate expenditure line?
asked by kwame kinker on March 26, 2011
99. ## Calculus 1
The acceleration function (in m/s^2) and the initial velocity are given for a particle moving along a line. a(t)=t+4, v(0)=4, 0 ≤ t ≤ 11 (a) Find the velocity at time t. (b) Find the distance traveled during the give time interval.
asked by TayB on December 1, 2015
100. ## Dynamics
A particle moving along a straight line is subjected to a deceleration a=-2v^3 m/s2. If it has a velocity v=8 m/s and a position x=10 m when t=0, determine its velocity and position when t=4s. I tried to integrate the deceleration equation, but it didn't
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http://theskepticalzone.com/wp/?p=2592 | 1,397,780,830,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00508-ip-10-147-4-33.ec2.internal.warc.gz | 209,402,655 | 32,038 | # The eleP(T|H)ant in the room
The pattern that signifies Intelligence?
Winston Ewert has a post at Evolution News & Views that directly responds to my post here, A CSI Challenge which is nice. Dialogue is good. Dialogue in a forum where we can both post would be even better. He is extremely welcome to join us here
In my Challenge, I presented a grey-scale photograph of an unknown item, and invited people to calculate its CSI. My intent, contrary to Ewert’s assumption, was not:
…to force an admission that such a calculation is impossible or to produce a false positive, detecting design where none was present.
but to reveal the problems inherent in such a calculation, and, in particular, the problem of computing the probability distribution of the data under the null hypothesis: The eleP(T|H)ant in the room
In his 2005 paper, Specification: the Pattern that Signifies Intelligence, Dembski makes a bold claim: that there is an identifiable property of certain patterns that “Signifies Intelligence”. Dembski spends the major part of his paper on making three points:
• He takes us on a rather painful walk-through Fisherian null hypothesis testing, which generates the probability (the “p value”) that we would observe our data were the null to be true, and allows us to “reject the null” if our data fall in the tails of the probability distribution where the p value falls below our “alpha” criterion: the “rejection region”.
• He argues that if we set the “alpha” criterion at which we reject a Fisherian null as 1/[the number of possible events in the history of the universe], no way jose will we ever see the observed pattern under that null. tbh I’d be perfectly happy to reject a non-Design null at a much more lenient alpha that that.
• He defines a pattern as being Specified if it is both
• One of a very large number of patterns that could be made from the same elements (Shannon Complexity)
• One of a very small subset of those patterns that can be defined as, or more simply than the pattern in question (Kolmogorov compressibility)
He then argues that if a pattern is one of a very small specifiable subset of patterns that could be produced under some non-Design null hypothesis, and that subset is less than 1/[the number of possible events in the history of the universe] of the whole set, it has CSI and we must conclude Design.
The problem, however, as I pointed out in a previous post, Belling the Cat, is that the problem with CSI is not computing the Specification (well, it’s a bit of a problem, but not insuperable) nor with deciding on an alpha criterion (and, as I said, I’d be perfectly happy with something much more lenient – after all, we frequently accept an alpha of .05 in my field (making appropriate corrections for multiple comparisons) and even physicists only require 5 sigma. The problem is computing the probability of observing your data under the null hypothesis of non-Design.
Ewert points out to me that Dembski has always said that the first step in the three-step process of design detection is:
1. Identify the relevant chance hypotheses.
2. Reject all the chance hypotheses.
3. Infer design.
and indeed he has. Back on the old EF days, the first steps were to rule out “Necessity” which can often produce patterns that are both complex and compressible (indeed, I’d claim my Glacier is one) as well as “Chance”, and to conclude, if these explanations were to rejected, Design. And I fully understand why, for the sake of algebraic elegance, Dembski has decided to roll Chance and Necessity up together in a single null.
But the first task is not merely to identify the “relevant [null] chance hypothesis” but to compute the expected probability distribution of our data under that null, which we need in order to compute the the probability of observing our data under that null, neatly written as P(T|H), and which I have referred to as the eleP(T|H)ant in the room (and, being rather proud of my pun, have repeated it in this post title). P(T|H) is the Probability that we would observe the Target (i.e. a member of the Specified subset of patterns) given the null Hypothesis.
And not only does Dembski not tell us how to compute that probability distribution, describing H in a throwaway line as “the relevant chance hypothesis that takes into account Darwinian and other material mechanisms”, but by characterising it as a “chance” hypothesis, he implicitly suggests that the probability distribution under a null hypothesis that posits “Darwinian and other material mechanisms” is not much harder to compute than that in his toy example, i.e. the probability distribution under the null that a coin will land heads and tails with equal probability, in which the null can be readily computed using the binomial theorem.
Which of course it is not. And what is worse is that using the Fisherian hypothesis testing system that Dembski commends to us, our conclusion, if we reject the null,is, merely that we have, well, rejected the null. If our null is “this coin is fair”, then the conclusion we can draw from rejecting this null is easy: “this coin is not fair”. It doesn’t tell us why it is not fair – whether by Design, Skulduggery, or indeed Chance (perhaps the coin was inadvertently stamped with a head on both sides). We might have derived our hypothesis from a theory (“this coin tosser is a shyster, I bet he has weighted his coin”), in which case rejecting the null (usually written H0), and accepting our “study hypothesis” (H1) allows us to conclude that our theory is supported. But it does not allow us to reject any hypothesis that was not modelled as the null.
Ewert accepts this; indeed he takes me to task for misunderstanding Dembski on the matter:
We have seen that Liddle has confused the concept of specified complexity with the entire design inference. Specified complexity as a quantity gives us reason to reject individual chance hypotheses. It requires careful investigation to identify the relevant chance hypotheses. This has been the consistent approach presented in Dembski’s work, despite attempts to claim otherwise, or criticisms that Dembski has contradicted himself.
Well, no, I haven’t. I’m not as green as I’m cabbage-looking. I have not “confused the concept of Specified Complexity with the entire design inference”. Nor, even am I confused as to whether Dembski is confused. I think he is very much aware of the eleP(T|H)ant in the room, although I’m not so sure that all his followers are similarly unconfused – I’ve seen many attempts to assert that CSI is possessed by some biological phenomenon or other, with calculations to back up the assertion, and yet in those calculations no attempt has been made to computed P(T|H) under any hypothesis other than random draw. In fact, I think CSI, or FCSI, or FCO are a perfectly useful quantities when computed under the null of random draw, as both Durston et al (2007) and Hazen et al 2007 do. They just don’t allow us to reject any null other than random draw. And this is very rarely a “relevant” null.
It doesn’t matter how “consistent” Dembski has been in his assertion that Design detection requires “careful investigation to identify the relevant chance hypothesis”. Unless Dembski can actually compute the probability distribution under the null that some relevant chance hypothesis is true, he has no way to reject it.
However, let’s suppose that he does manage to compute the probability distribution under some fairly comprehensive null that includes “Darwinian and other material mechanisms”. Under Fisherian hypothesis testing, still, all he is entitled to do is to reject that null, not reject all non-Design hypotheses, including those not included in the rejected “relevant null hypothesis”.
Ewert defends Dembski on this:
But what if the actual cause of an event, proceeding from chance or necessity, is not among the identified hypotheses? What if some natural process exists that renders the event much more probable than would be expected? This will lead to a false positive. We will infer design where none was actually present. In the essay “Specification,” Dembski discusses this issue:
Thus, it is always a possibility that [the set of relevant hypotheses] omits some crucial chance hypothesis that might be operating in the world and account for the event E in question.
The method depends on our being confident that we have identified and eliminated all relevant candidate chance hypotheses. Dembski writes at length in defense of this approach.
But how does Dembski defend this approach? He writes
At this point, critics of specified complexity raise two objections. First, they contend that because we can never know all the chance hypotheses responsible for a given outcome, to infer design because specified complexity eliminates a limited set of chance hypotheses constitutes an argument from ignorance.
Yes, indeed, this critic does. But Dembski counters:
In eliminating chance and inferring design, specified complexity is not party to an argument from ignorance. Rather, it is underwriting an eliminative induction. Eliminative inductions argue for the truth of a proposition by actively refuting its competitors (and not, as in arguments from ignorance, by noting that the proposition has yet to be refuted). Provided that the proposition along with its competitors form a mutually exclusive and exhaustive class, eliminating all the competitors entails that the proposition is true.
OK, but…
But eliminative inductions can be convincing without knocking down every conceivable alternative,a point John Earman has argued effectively. Earman has shown that eliminative inductions are not just widely employed in the sciences but also indispensable to science.
Hold it right there. When Earman makes his plea for eliminative induction, he says:
Even if we can never get down to a single hypothesis, progress occurs if we succeed in eliminating finite or infinite chunks of the possibility space. This presupposes of course that we have some kind of measure, or at least topology, on the space of possibilities.
Earman gives as an example a kind of “hypothesis filter” whereby hypotheses are rejected at each of a series of stages, none of which non-specific “Design” would even pass, as each requires candidate theories to make specific predictions. Not only that, but Earman’s approach is in part a Bayesian one, an approach Dembski specifically rejects for design detection. Just because Fisherian hypothesis testing is essentially eliminative (serial rejection of null hypotheses) does not mean that you can use it for eliminative induction when the competing hypotheses do not form an exhaustive class, and Dembski offers no way of doing so.
In other words, not only does Dembski offer no way of computing the probability distribution under P(T|H) unless H is extremely limited, thereby precluding any Design inference anyway, he also offers no way of computing the topology of the space of non-Design Hypotheses, and thus no way of systematically eliminating them other than one-by-one, never knowing what proportion of viable hypotheses have been eliminated at any stage. In other words, his is, indeed, an argument from ignorance. Earman’s essay simply does not help him.
Suffice it to say, by refusing the eliminative inductions by which specified complexity eliminates chance, one artificially props up chance explanations and (let the irony not be missed) eliminates design explanations whose designing intelligences don’t match up conveniently with a materialistic worldview.
The irony misser here, of course, is Dembski. Nobody qua scientist has “eliminated” a “design explanation”. The problem for Dembski is not that those with a “materialistic worldview” have eliminated Design, but that the only eliminative inductionist approach he cites (Earman’s) would eliminate his Design Hypothesis out of the gate. That’s not because there aren’t perfectly good ways of inferring Design (there are), but because by refusing to make any specific Design-based predictions, Dembski’s hypothesis remains (let the irony not be missed) unfalsifable.
But until he deals with the eleP(T|H)ant, that’s a secondary problem.
Edited for typos and clarity
## 62 thoughts on “The eleP(T|H)ant in the room”
1. Well, we know Dembski is arguing from a foregone conclusion, and constructing a system of rationalizations. And we know that Dembski knows that his foregone conclusion is unable to actually explain anything, rendering it incapable of making any predictions.
I suppose it’s fun to poke holes in it, and find the errors in the rationalization. But underneath all this verbiage, both the motivations and the errors are fairly transparent. We know 2+2 does not equal four becauise God Said So. And therefore we can conclude that 2+2=22, which we knew all along anyway, because we’ve eliminated all other competing sums. Trust me.
2. Lizzie,
However, let’s suppose that he does manage to compute the probability distribution under some fairly comprehensive null that includes “Darwinian and other material mechanisms”.
It’s ironic that ID proponents are always demanding mutation-by-mutation accounts of how this or that biological feature evolved, because that is the level of detail they must provide in order to justify the values they assign to P(T|H). It’s even worse for them, in fact, because P(T|H) must encompass all possible evolutionary pathways to a given endpoint.
P.S. Winston’s last name is “Ewert”, with two E’s.
3. I haven’t read it yet, but I would rather appreciate it if you’d correct the spelling of my last name. Thanks!
4. A very good and pointed summary.
When we discussed this before, I summarized the situation thus:
a) We are trying to see whether natural selection and random mutation (and similar evolutionary forces) can explain why we have an adaptation that is as good as it is.
b) If not, then we conclude for Design.
So according to Dembski’s protocol we
1. Look at the adapation and what might have brought it about.
2. See if we can rule out RM+NS.
3. If we can, then we conclude for Design.
and, oh yes, in that case, and in that case only, we declare that points 1, 2, 3 show that it has CSI.
Now note that the declaration that it has CSI is simply an afterthought, and does not even occur until we have already reached stage 3. So the concept of CSI is not at all central to the design detection.
There are many of Dembski’s friends at UD who have declared, loudly and proudly, that CSI is a property that can be detected without knowing how an object came about. And that having CSI shows that the object was designed.
Apparently they have all been wrong all this time, which speaks for a charge of lack of clarity against Dembski’s works.
5. Joe,
There are many of Dembski’s friends at UD who have declared, loudly and proudly, that CSI is a property that can be detected without knowing how an object came about.
They are technically correct. You don’t have to know how an object actually originated to decide that it has CSI, but you do have to know the probability that it could have been produced via “Darwinian and other material mechanisms” — P(T|H), in other words.
The problem, as you say, is that CSI is an afterthought. You already have to know that something could not have evolved before you attribute CSI to it. Thus CSI is useless for demonstrating that something could not have evolved.
I pointed this out to Dembski in 2006 and to many other ID proponents since then. I’ve never seen him, or them, acknowledge the circularity.
6. But they already DO know something (life) could not have evolved. All that CSI life has is just a way of expressing and reflecting this knowledge.
I hope we sometimes step back from the scientific approach where conclusions come last, to recognize that this simply isn’t the case for Dembski at all. For Dembski and all the UD folks, their conclusions drive their reasoning and not the other way around.
7. This is a very nice summary of the ID/creationist position.
As Flint points out, they already know the “answer” they want. The game, therefore, is to bury those preconceptions under a pile of “math” while at the same time not learning enough science to see any “distracting” possibilities in the natural world.
In fact getting fundamental scientific concepts wrong and foisting those misconceptions on the unsuspecting public simply makes ID/creationists arguments seem more plausible to those with little or no scientific education. The science – as ID/creationists tell it – can’t possibly get the job done; as anyone off the street can see (this was actually a recent argument by Sewell).
The real story is pretty much the opposite of the Dembski narrative; knowing the relevant science puts more possibilities in front of us than we can check in a lifetime. Finding the “recipe” of life is a daunting task because there are so many possible lines to explore. Thus, part of the research involves checking other places that might harbor life. This is done in order to try to bracket the problem.
I think Dembski’s filter might better be turned around. Eliminate all the deities that may or may not have been involved in the origins and evolution of life. There are not only thousands of them; there are thousands of interpretations of the characteristics of each of these deities.
Since sectarians cannot agree among themselves – even to the point of centuries of bloodshed – what even one of these deities is like; we place deities at the extreme low probability end of the spectrum, leave the ID/creationist misconceptions and mischaracterizations of science behind, and carry on doing science with a clear conscience. Life is short; and there are many scientific avenues to explore.
Just because ID/creationists are unable to come up with scientific research programs doesn’t mean that more knowledgeable and talented people can’t.
8. Thanks for noting the spelling – fixed, and apologies to Ewert.
9. Dembski is notorious for scoffing that
ID is not a mechanistic theory, and it’s not ID’s task to match your pathetic level of detail in telling mechanistic stories.
His statement was mocked for obvious reasons, but it was also unintentionally prophetic. He’s right that ID’s job isn’t to match evolution’s “pathetic level of detail” — ID has to exceed that level of detail in order to establish the value of P(T|H). Without a value for P(T|H), or at least a defensible upper bound on its value, the presence of CSI can never be demonstrated — by Dembski’s own rules.
Think of what that would involve in the case of biology. You’d not only have to identify all possible mutational sequences leading to the feature in question — you’d also have to know the applicable fitness landscapes at each stage, which would mean knowing things like the local climatic patterns and the precise evolutionary histories of the other organisms in the shared ecosystem.
If he didn’t realize it then, Dembski must certainly see by now that it’s a quixotic and hopeless task. That may be why he’s moved on to “the search for a search”.
10. I can’t remember whether you specifically addressed Ewert’s earlier EnV piece, Information, Past and Present (linked to in his response to mine), Joe?
11. I don’t know if anyone else has read Earman’s essay/book (there’s a pdf of it, linked to in the OP), but it would be interesting to apply his eliminative induction to the essays in the “Cornell” collection
For instance, we could eliminate all the YEC theories on the first pass. Probably most of the Separate Creation for Humans theories on a second, including Adam and Noah.
We’d probably be left with Behe, by which time we would have eliminated a human-health-prioritising God.
And we still wouldn’t have eliminated any non-materialist theory except on grounds of being incomplete, which all scientific theories are anyway.
12. No, I haven’t yet replied to “Information, Past and Present”. I have been a little busy, but I will get to it. The issues there are (1) whether Dembski was clear about CSI having the P(T|H) step in writings such as the book No Free Lunch, and (2) whether he has any place in his Search For a Search where he shows that natural selection cannot have produced the adaptations that we see. I have been reading my way through NFL. Little matters like research, summer teaching, and grant-writing keep getting in the way.
The remaining big issue is the one that you are so nicely dealing with here, whether the version of Dembski’s argument that he gives in his 2006 paper has any way of dealing with the EleP(T|H)ant in the room. As you argue, it doesn’t. And that relegates the CSI step to total irrelevance. Not one of these folks has demonstrated that we need the concept of CSI to do their design inference.
13. Welcome to TSZ, Mr Ewert. Lizzie has already corrected the mis-spelling. Apologies that your comment lay pending for a while. Any further comment should appear immediately. Nothing personal, it’s the norm for any new registration.
14. Winston Ewert has commented upthread but was held in moderation. So, no headsup needed.
15. Richardthughes:
Exactly so, and more. They need all *possible* accounts and paths.
True, but even with that “pathetic level of detail” the lottery winner fallacy is still present. Considering one isolated biological artifact ignores the fact that evolutionary mechanisms are capable of generating a phenomenal range of outcomes. The implicit assumption that humans or bacterial flagella or beetles are a target or intended outcome is not valid.
16. The important concept for both biologists and IDists is whether the history of a genome contains any implausible steps.
Not knowing the history in pathetic detail makes this determination impossible.
I haven’t read it yet, but I would rather appreciate it if you’d correct the spelling of my last name. Thanks!
Yes, welcome!
And apologies again for mis-spelling your name. I’m usually good at getting names right, but I should be lest trusting of my own accuracy!
18. I don’t see why the assumption that humans are an intended outcome is not valid. I don’t think validity is relevant when intention cannot be either established or discarded.
But I don’t think ID people are concerned with a history of “plausible steps”. In their model, there was only a single step – poof – and whether or not you find this plausible is up to you.
Evidence really doesn’t much matter to a model not built on evidence.
19. True, but even with that “pathetic level of detail” the lottery winner fallacy is still present. Considering one isolated biological artifact ignores the fact that evolutionary mechanisms are capable of generating a phenomenal range of outcomes. The implicit assumption that humans or bacterial flagella or beetles are a target or intended outcome is not valid.
Their lottery winner fallacy has in recent years (especially since Edwards v. Aguillard in 1986) been focused on things like molecular assemblies; such as proteins and DNA. While this singles out the molecules of life – and replicating molecules in particular – it now places the ID issue among all molecular assemblies.
And this is where ID/creationist arguments really get weird; they have to assert that certain molecular assemblies are due to “chance and necessity,” but others above a certain threshold of complexity require assistance from some intelligent input.
Where along this chain of increasing level of complexity do the laws of physics and chemistry stop and intelligence has to take over in order to do the job that physics and chemistry “cannot do?”
The sample space in all these CSI calculations is always an “ideal gas” of inert atoms that are supposed to just come together into some precisely specified configuration. Strings of letters and numbers are now used as stand-ins for atoms and molecules in these calculations.
But if one is going to play this game, why not calculate the CSI of a rock (we did that here and on Panda’s Thumb)? Why can’t a specified rock be the target of an atomic/molecular assembly? We are at the atomic/molecular level now. Rocks and crystals self assemble; and crystals in particular can replicate under suitable conditions. In fact, DNA is a quasi-crystal; so why not include rocks?
Where is the cutoff between assemblies that can occur due to “chance and necessity” – I really dislike that mischaracterization – and those that require intelligent guidance?
Think about just how weird this ID picture is. There are all these atomic and molecular assemblies out there in the universe that can come together by “chance and necessity;” but there is this island of assemblies that suddenly are the result of ideal gases of inert atoms and molecules being jockeyed into specified positions.
It’s as though atoms and molecules that are about to assemble into certain specified arrangements suddenly lose all their properties and have to seek help from some intelligent being. How do they do that?
20. heh. I’m a good googler.
21. Intelligence (on observation) needs fuel – molecular fuel. We can certainly apply a certain amount of intent to generate assemblies that did not ‘just happen’. But we had lunch, and it wasn’t free.
Even weirder is the insistence that the very existence of atoms themselves required intelligence to bring it about. I really don’t know how you’d distinguish that kind of ID!
22. Isn’t the essence of the Probability Bound calculation that they are arguing that a result this good (this far out into the tail of the distribution) cannot occur even once in the whole history of the universe? So that is the border between “chance and necessity” and design?
Or do I misunderstand your argument.
23. Flint: “I don’t see why the assumption that humans are an intended outcome is not valid. I don’t think validity is relevant when intention cannot be either established or discarded.”
Fair enough. Let me rephrase as: The implicit assumption that humans or bacterial flagella or beetles are a target or intended outcome must be made explicit and supported. Otherwise the lottery winner fallacy is still present.
24. Hi. I gave up looking at ID some years ago. But I came across your interesting blog, and couldn’t resist commenting.
First, on “eliminative inductions”. I think we might reasonably consider various non-design hypotheses, reject them, and then infer design. But I suggest that if we’re justified in doing so it’s because we are (perhaps intuitively) weighing the merits of the design hypothesis against the plausibility of there being some further unconsidered explanation. Dembski wants us to accept the design hypothesis without considering its merits. No thanks. I like to consider the merits of a hypothesis before I accept it.
Dembski is extremely vague about how to apply his method to real cases. The bacterial flagellum was supposed to be his flag-ship case. But in that case he hardly applied his own method. Instead he relied primarily on an argument from irreducible complexity. The one “chance hypothesis” he considered was purely random combination of parts, which no one seriously proposes, so it was effectively irrelevant. And even in considering that hypothesis he omitted most of the apparatus of his Fisher-based statistical method.
I won’t address here the validity of that method for eliminating individual hypotheses. But it seems that Ewert has failed to apply it correctly.
Ewert: In the case of the model used in “Specification: The Pattern that Signifies Intelligence,” we need to determine the specification resources. This is defined as being the number of patterns at least as simple as the one under consideration. We can measure the simplicity of the image by how compressible it is using PNG compression. A PNG file representing the image requires 3,122,824 bits. Thus we conclude that there are 2 to the 3,122,824th power simpler or equally simple images.
In fact, Dembski defines the specificational resources as “the number of patterns for which S’s semiotic description of them is at least as simple as S’s semiotic description of T.” Surely Ewert should be calculating the complexity of the description of his chosen rejection region, not the complexity of the image. Unfortunately he doesn’t identify what rejection region (T) he is using, nor provide a probability calculation or a calculation of replicational resources. We have no way of checking his calculations
For two of his hypotheses Ewert calculates negative figures for so-called “specified complexity”, which correspond to probabilities greater than 1! If the untransformed numbers (before applying the -log2 transformation) are really probabilities, they shouldn’t exceed 1. If they’re not really probabilities, then there’s even less justification for applying this transformation than there was when Dembski claimed to be transforming probabilities into information measures. Either way, the transformation by -log2 is entirely superfluous, and this isn’t a genuine measure of complexity.
25. Joe Felsenstein: Mike Elzinga, Isn’t the essence of the Probability Bound calculation that they are arguing that a result this good (this far out into the tail of the distribution) cannot occur even once in the whole history of the universe? So that is the border between “chance and necessity” and design?Or do I misunderstand your argument.
On page 23 of his “Specification” paper Dembski uses a legitimate physics calculation by Seth Lloyd in Physical Review Letters that estimates that it takes 10^120 logical operations to specify the entire universe; which includes everything in it. Here is the sentence from page 23 of Dembski’s “Specification” paper.
Theoretical computer scientist Seth Lloyd has shown that 10^120 constitutes the maximal number of bit operations that the known, observable universe could have performed throughout its entire multi-billion year history.
So all Dembski is doing with his CSI is taking the logarithm to base 2 of Np, where N = 10^120, and then asserting that p has to be greater than or equal to 1/10^120 in order for there to be at least one occurrence of a specified event such as the origin of a living molecule.
Taking log to base 2 and calling it CSI simply obscures this simple calculation as well as the assertion Dembski is making in his calculations.
Ironically, Lloyd’s calculation already includes life forms; which obviously means that p is much greater than 1/N because life forms are already an existing subset of the universe; but Dembski didn’t appear to notice this.
26. Some posts moved. As usual, no ethical judgment implied, just tidying up.
27. I honestly find the Seth Lloyd thing (even if correct, which I understand is in doubt) completely irrelevant. As I’ve said, I’d be perfectly happy with a much more lenient rejection alpha. 5 sigma would do me fine.
It’s how you actually compute the probability distribution that concerns me, not where you make the cut-off having computed it.
28. I would hazard a speculation about that. I suspect that Dembski picked that number because it came from a paper in PRL.
Lloyd’s estimate is interesting because it is a legitimate calculation that tries to get a handle on how much computing power is needed to do a complete simulation of the universe and its history.
It’s a rough estimate at best; but it shows how information is used properly in physics and computing. For example, “information” is connected to entropy, and thereby properly to energy states by the amount of energy required to flip bits in a computer.
If I am recalling correctly, in earlier attempts by ID/creationists, they didn’t want to leave any doubt about a bound on a probability (CSI), so they had some of their own estimates of how much “information” is contained in the universe. None of these estimates had anything to do with energy; they were just enumerations of configurations.
By picking a paper from PRL that happened along at a propitious moment in the evolution of his calculations, Dembski borrows “legitimacy” from a paper in a prestigious physics journal.
I don’t believe Dembski read or understood Lloyd’s paper; I think he just copied what he wanted from the abstract.
That’s my speculation. Could be wrong, but it would be consistent with ID/creationist history and tactics.
29. P.S. My point above about the bacterial flagellum was based on Dembski’s treatment of the subject in NFL. I’d forgotten that he returned to that subject in the “Specification” paper. There he attempts to apply the latest version of his statistical method to “an evolutionary chance hypothesis”. I think his attempt to measure specificational resources, based on the number of words in the description “bidirectional rotary motor-driven propeller”, is pretty weak. But more important, he is unable to calculate P(T|H). So he hasn’t managed to infer design in biology, even with his own method.
I think this inability to calculate P(T|H) in the case that matters most to him is the reason why he has been moving away from this empirical method towards an attempt to make an in-principle argument based on search algorithms. He thinks that can free him of the pesky need to provide a probability calculation or any other empirical evidence. Of course, it’s pie in the sky.
30. It’s how you actually compute the probability distribution that concerns me, not where you make the cut-off having computed it.
I think your analysis shows very nicely the equivalent way of looking at the probabilities given the hypothesis of “chance” and necessity.
At, say your five sigma alpha, there would have to be a large number of trials in order to produce an instance that far out in the tail of the distribution.
If there hasn’t been enough time in the history of the universe to run enough trials that would produce a specified event, then the probability of that event is certainly far out on the tail of some distribution.
But as you have pointed out, Dembski can’t tell us what that distribution looks like. He simply declares that chance configurations of certain atoms and molecules can’t occur in the history of the universe given some upper limit on how fast trials can take place. If I recall correctly, that frequency was something like the inverse of the Planck time.
31. Welcome!
32. TSZ is back? Grats.
Lizzie: In my Challenge, I presented a grey-scale photograph of an unknown item, and invited people to calculate its CSI.
Why do you expect people to be able to calculate the CSI of some of some unknown item when you can’t even accurately calculate the CSI for your own CSI demonstration?
33. Mung,
“People”, including ID proponents, can’t determine the CSI of known biological items, much less unknown ones, because they can’t determine P(T|H).
The reason that ID proponents can’t determine CSI in actual biological cases is that they can’t calculate P(T|H) when H is “the relevant chance hypothesis that takes into account Darwinian and other material mechanisms.”
Instead, they typically compute P(T|H) using a much simpler H: the hypothesis that something came about through pure random luck with no selection. It’s much easier to use this H, but the answers are bogus because the simpler H is not the real H, which involves not only randomness but also non-random selection.
Lizzie’s earlier post was written when she still believed that the H in P(T|H) was pure randomness. Most likely she believed it because so many ID supporters have (mistakenly) interpreted it that way.
However, Dembski specifies that the H in P(T|H) is “the relevant chance hypothesis that takes into account Darwinian and other material mechanisms.”
Can you calculate it for an actual biological feature, Mung?
I remember that you had a lot of trouble grasping the concept of P(T|H) when we last discussed it. (At one point you even thought that T|H was a fraction, with T in the numberator and H in the denominator!)
P(T|H) is what’s known as a conditional probability. You might want to grab a beginner’s book on probability and read about it before commenting further.
34. Mung:
TSZ is back? Grats.
Lizzie: In my Challenge, I presented a grey-scale photograph of an unknown item, and invited people to calculate its CSI.
Why do you expect people to be able to calculate the CSI of some of some unknown item when you can’t even accurately calculate the CSI for your own CSI demonstration?
Well, Mung, the “people” in question—the people to whom Lizzie directed her query—are ID-pushers who are stridently insistent that yes, this CSI thingie damn well is a real thing, and this CSI thingie damn well is measurable, and this CSI thingie damn well can be used to distinguish between Design and Not-Design.
This being the case, it makes perfect sense to ask CSI-lovin’ ID-pushers to, like, determine the CSI of an arbitrary target (a photographic image, in this case). And if someone who isn’t a CSI-lovin’ ID-pusher can’t use CSI the way that CSI-lovin’ ID-pushers claim CSI can be used… so fucking what? The question is whether or not CSI-lovin’ ID-pushers can use CSI the way that CSI-lovin’ ID-pushers claim CSI can be used!
And in this case, it seems that CSI-lovin’ ID-pushers… couldn’t use CSI the way that CSI-lovin’ ID-pushers claim CSI can be used.
Tell me, Mung: If the proponents of a putative ‘scientific theory’ can’t substantiate the claims they make in support of their putative ‘theory’, exactly why should anybody else give two wet farts in a hurricane about said putative ‘theory’?
35. Could someone clarify this for me?
Upthread it seems that several people are saying that every possible mutational sequence (sequence in time) occurring in a set of nucleotides represents an H in the sense of “P(T|H)”. That is to say, any investigator wanting to eliminate “chance and necessity” or any other non-design cause, would need to work through all possible mutational sequences to prove they couldn’t have done it?
36. timothya,
That is to say, any investigator wanting to eliminate “chance and necessity” or any other non-design cause, would need to work through all possible mutational sequences to prove they couldn’t have done it?
It depends on what you mean by “work through”.
Dembski’s approach depends on being able to eliminate all non-design explanations, so every possible non-design cause must at least be considered. However, it may be possible to reject some of them without doing a detailed analysis.
For example, the probability of the vertebrate eye evolving in a single generation is vanishingly small. It’s not impossible, but the associated probability is so small as to be negligible. It will have almost no effect on the overall P(T|H) and can therefore be neglected.
The problem for Dembski et al is that even without considering these vastly improbable outliers, the difficulty in calculating P(T|H) for a complicated biological structure is overwhelming. The required information is simply not available. That’s why IDers haven’t done it, and that’s why no one expects them to.
37. CSI as a practical calculation in biology would appear to suffer fatally from combinatorial explosion, since results come from series of events. Going backwards, each genetic change A depends upon the prior situation B. P=(A|B) – the probability of A given B. Since B was itself the result of a prior (x|y), we have a nesting of probabilities P=(A|(B|(C|(D| … )))), and could rapidly exceed the UPB for any series if ‘this’ is the target. But one would need to know the density of outcomes with this level of ‘surprise’ or greater in the (vast) overall space of event series to know how unlikely this one was.
One could try applying the notion to one’s own genome, for example. One’s genetic constitution is the result of that of one’s parents, which in turn derives from theirs … each is a probabilistic sampling of a prior situation. You don’t have to go back very far before the number of ‘possible individuals’ who aren’t me exceeds the UPB – and keeps growing. Yet here I am, improbably.
38. Very nicely articulated and more clear than my somewhat terse comment on the Lottery Winner fallacy. Evolutionary mechanisms operating in the environment we observe have a vast number of potential results. To be useful, Dembski’s calculations must include something like your “density of outcomes with this level of ‘surprise’ or greater.”
That strikes me as at least as difficult to calculate as P(T|H) for a single artifact.
39. Allan Miller: But one would need to know the density of outcomes with this level of ‘surprise’ or greater in the (vast) overall space of event series to know how unlikely this one was
Well, essentially, this is what Dembski is getting at with his concept of “Specification”.
But giving it a name doesn’t make it calculable.
40. Considered more generally, essentially everything that happens in life is the result of a sequence of contingent prior events, every one of which is vanishingly unlikely, and the particular sequence even moreso. It takes only a very short time for everything that happens to exceed the UPB, from any given prior set of conditions. Trying to use math to eliminate all of reality from instant to instant as “too unlikely” is prima facie folly.
Faced with all this, most of us here shrug because that’s how reality MUST operate. Some regard every event from the quantum level on up as being Divinely guided in real time (and who can say they’re wrong?) The UD people try to eliminate all this by (1) asserting POOF as the historical mechanism; and (2) constraining the scope of their vision so as to eliminate all the “noise” which is the substance of reality.
At some point, “goddidit” morphed from a one-size-fits-all non-explanation of what’s not understood, into an actual reified “invisible superman” intelligent agent. Working backwards to demonstrate the reality of the imaginary is guaranteed to be both nonsense, and incurable.
41. Mung:
TSZ is back? Grats.
Lizzie: In my Challenge, I presented a grey-scale photograph of an unknown item, and invited people to calculate its CSI.
Why do you expect people to be able to calculate the CSI of some of some unknown item when you can’t even accurately calculate the CSI for your own CSI demonstration?
Hi, Mung. As others have said, I calculated the CSI for the demonstration you linked to on the assumption that H was “random draw”. This is what Kairosfocus, for example, does here, as you can see from the fact that he takes (at the bottom of his OP) Durston et al’s fits for random-draw, without alteration, and interprets them as estimates of p(T|H).
However, at the time when I wrote that OP, I had indeed overlooked the fact that Dembski adds the rider “that takes into account Darwinian and other material mechanisms” to his description of the “relevant chance hypothesis”.
My challenge (which you are welcome to take up) is how you compute an appropriate P(T|H) for the “relevant Chance hypothesis” where this is not restricted to random-draw.
For example, for a biological phenomenon.
42. Well, essentially, this is what Dembski is getting at with his concept of “Specification”.
“Specification” is Dembski’s attempt at dealing with the fact that vastly improbable things happen all the time. Problem is, specifications are usually too specific.
For example, Dembski knows that he would be committing the lottery winner fallacy if he claimed that the bacterial flagellum, exactly as it appears today, was evolution’s “target”. Instead, he broadens the specification to include any “bidirectional rotary motor-driven propeller.”
But this is still far too specific. Even “propulsion system” is too specific, because evolution didn’t set out to produce a propulsion system. Evolution’s only “target” is differential reproductive advantage, and even then the word “target” is too strong.
Allan’s formulation is closer to what Dembski should have been shooting for:
But one would need to know the density of outcomes with this level of ‘surprise’ or greater in the (vast) overall space of event series to know how unlikely this one was.
43. It had occurred to me that this probability calculation of a specified target at the end of a long chain of contingencies was at the heart of what ID/creationists were actually trying to assert happens; but then I thought that was just too …, well, stupid. I figured they simply meant something like tornados-in-a-junkyard or things falling out of an ideal gas of inert stuff.
But it appears you may be right; they really do calculate the probability of a specified event at the end of a chain of contingencies.
My impression of how they “take into account Darwinian and other material mechanisms” is that they always misrepresent them with some type of emotionally loaded caricature that appeals to their sectarian base. These mechanisms can’t possibly work because they are the result of “materialistic thinking,” which rules out design from the beginning.
It may be too much to expect that ID/creationist “arguments” would be consistent among themselves. Historically they appear to be the result of their latest emotional outbursts at “Darwinists,” “materialists,” and atheists.
If there is anything consistent in ID/creationist thinking, it would be circularity. They already have the answer they want; they just have to court-proof it by making it look like science in order to get it into public education.
Their latest push – if we are to take UD as a barometer of their thinking – appears to be making “materialism” to be a competing “philosophy” in the most pejorative sense. “Materialism” clouds the mind and makes it impossible for “materialists” to understand simple CSI calculations. According to UD, we haven’t deconstructed CSI here because we don’t understand CSI.
The culture war continues.
44. Me:
But one would need to know the density of outcomes with this level of ‘surprise’ or greater in the (vast) overall space of event series to know how unlikely this one was.
Taken in isolation, I now struggle to parse my own statement!
The difficulty is in factoring in the role of selection. When there is a beneficial phenotype, the ‘surprising’ result is its extinction. But that still can happen with nonzero probability, because s values are typically small. Yet, because there are many different genetic paths to an equivalent phenotype, and many potential phenotypes that could be beneficial, the population has multiple bites at the cherry, and favourable accessible phenotypes are likely to arise.
With a process that favours serial adaptation, the vast ‘random’ space is substantially reduced and channeled. Nonetheless, the challenge of calculating likelihood is in no way diminished, not least because historic benefit depends on lost information.
45. I would describe the calculation as the probability that a fitness as great as this, or greater could arise as a result of the processes of evolution — including natural selection — which are encompassed in the “chance” hypothesis H. The T is the set of all outcomes that have as high a fitness or higher than the observed fitness.
It has nothing much to do with “surprise” except insofar as a very high fitness is surprising. Dembski has often identified the scale as one of compressibility, but this is not fundamental to his argument, as witness the identification of his quantities with the quantities computed by Hazen and Durston, which are not expressed in terms of compressibility. There is also the issue that if we take compressibility to be fundamental, we find that an organism whose structure is so simple that it could not survive has higher compressibility than one which actually can survive.
And of course, even if we can compute the probability P(T|H), and it is so small that we can conclude in favor of design, there is still at that point nothing to be gained by further expressing this probability as showing that CSI is present. The fundamental argument is made by the lowness of this tail probability, and the CSI part is redundant.
46. In short, Dembski is hosed.
1) His concept of specification is too narrow, but even if it weren’t,
2) P(T|H) can’t be computed for realistic biological cases, but even if it could,
3) you have to answer the relevant question — “Could this have evolved?” — without the use of CSI, when you calculate P(T|H),
4) so the concept of CSI adds nothing, and if you invoke it the entire argument becomes circular: X couldn’t have evolved, so it must have CSI; X has CSI, therefore it couldn’t have evolved.
Dembski’s “solution” to these problems:
A. Retreat. Renounce the explanatory filter but affirm the value of CSI, as if these were separable concepts.
B. Give up arguing that evolution cannot produce adaptive complexity, without actually admitting that it can.
C. Argue instead that any adaptive complexity produced by evolution was already implicit in the environment, and that a Designer must have placed it there.
Not much of an improvement, but at least he’s fighting a different set of battles.
47. Not much of an improvement, but at least he’s fighting a different set of battles.
One of the themes of ID/creationism over the years has been to get everyone else – especially the general public – to use ID/creationist concepts and vocabulary. ID/creationist debaters always pulled the debate onto their territory and flooded the discussion with their words and their concepts.
Their entire political push is to get that same set of concepts and vocabulary into the public schools; and get that stuff into the discussions before students have had a chance to learn the real science concepts and vocabulary.
One cannot help noticing just how much time is spent discussing Dembski’s and other ID/creationist papers even though there is nothing of any value in them. In the ID/creationist world, that is considered a success because it allows them to claim that there is an intense scientific discussion going on in science that is being kept from beginning students.
Back in the 1970s, 80s, and 90s, when I was giving talks to the public, I made it a point to never use ID/creationist vocabulary. I discussed the relevant science and the ID/creationist distortions explicitly. My church audiences recognized the political ploys ID/creationists were using; and they did not approve.
If any of this ever came up with my students or my coworkers, I used the misconceptions as a foil for teaching the real concepts. I never took any ID/creationist concepts seriously; and I made it a point early on that I would never debate one of them.
ID/creationists have wanted their concepts and vocabulary to have gravitas and be treated with respect and deference. Exposing them for what they really are highlights the political nature of ID/creationism.
I suspect it is pretty obvious that I have little respect for ID/creationists; but they started the fight.
48. That’s been a bit exasperating for decades. Scientists started out regarding creation “science” as so obviously silly that they entered debates confident that they’d have no difficulty crushing the creationist with facts, observations, evidence, theories, hypotheses, and the scientific method.
And that arsenal was hopelessly outgunned by such devastating weapons as defining the concepts, framing the discussion, establishing the language and vocabulary, and from that pad launching into a rapid fire assault of misquotes, lies, half-truths, misdirection, equivocation, redirection, and more lies. Even the “debate” format was carefully tilted toward such techniques, rendering careful explanation of even a single point boringly incomprehensible to an audience selected for ignorance of the necessary background.
I think most scientists decided the fight was someone else’s, and chose to ignore it. But even here at TSZ, I notice that what people are saying, over and over ad nauseum, is that ID conclusions cannot be drawn from their evidence. As if that were ever their intent. We see nearly every penny creationists can raise going into PR events, political campaigns, and propaganda. And we keep saying “hey, wait, that’s bad science!” — as though that matters, while the elected McLeroys strive to control the contents of national high school science texts.
And while half the US population, convinced they must choose between evolution and God, have not “converted” to “evolutionism.” Of course, they don’t know what evolution IS, since in much of the country it’s not covered in school. Why anger parents and make administration and teaching harder?
49. There are a number of problems with that analysis in the case of the Higgs boson. This Higgs experiment is taking place at two different detectors, the ATLAS and the CMS; each detector using a completely different decay mode for the Higgs. This has been done deliberately in order to account for systematic issues.
Both detectors are reporting results at the same energy and at nearly the same 5 or 6 sigma significance.
Secondly, these results are accumulating as they are being watched. The decision to hold off reporting them until 5 or 6 sigma was because the experimentalists wanted both the time and the data in order to make many crosschecks of their methods of analysis and to be sure that everything is working at the level of sensitivity they need. They are extremely aware of the potential for being wrong, so those other crosschecks are important even though they have not been mentioned explicitly in press reports.
Thirdly, the particle is only tentatively being called the Higgs; physicists are quite aware of the possibility it might be something else. However, the result is consistent with the probabilities of the various decay modes for the Higgs; and two decay modes are being monitored by those two detectors.
Fourthly, the particle needs further study in order to determine its properties such as its spin. That will be done when the LHC is brought back on line for the next phase of operating at higher energy. Many checks of the machine have to take place before ramping up its energy. Further checks of the detectors will also take place in this down time. And there will be many crosschecks of the data in upcoming runs because of the need to determine the particle’s properties.
Fifthly, there are other possible Higgs for which a search will be made; this isn’t the end of the search. Other theories are also being tested at the LHC.
This is not only a difficult experiment, it is a crucial experiment; and it is being done with money paid out by many governments. These experimentalists have learned from the politics of big science that it is deadly to be wrong when that much money and importance ride on the results.
Physicists are light years beyond the sloppy use of statistics Ziliak sees in some areas of testing. They are not as naive as this Stephen Ziliak seems to imply.
50. The reason why it is important to explain clearly why Dembski’s argument doesn’t work is that there are lots of pro-ID folks who repeatedly, loudly, proudly declare that see, there is this thing called CSI. That we can decide whether CSI is present in a life form without having to know how the life form originated. That CSI can only arise by Design.
They are wrong about that, but the argument is phrased as a technical scientific argument, and so needs to be answered clearly, with good science. And this has to be done repeatedly, since the pro-ID people do not seem often to notice counterarguments.
Dembski may have effectively retreated from much of his argument, but the pro-ID crowd has not noticed that, either.
51. Joe Felsenstein:
The reason why it is important to explain clearly why Dembski’s argument doesn’t work is that there are lots of pro-ID folks who repeatedly, loudly, proudly declare that see, there is this thing called CSI.That we can decide whether CSI is present in a life form without having to know how the life form originated.That CSI can only arise by Design.
They are wrong about that, but the argument is phrased as a technical scientific argument, and so needs to be answered clearly, with good science. And this has to be done repeatedly, since the pro-ID people do not seem often to notice counterarguments.
Dembski may have effectively retreated from much of his argument, but the pro-ID crowd has not noticed that, either.
The other often-repeated claim is that “evolution can’t add information to the genome”.
One thing my CSI demonstration showed is that it can. Unless information produced by evolution doesn’t count
52. Dembski, Marks and co. argue that the information is not being created even if it is being put into the genome by natural selection. They argue that the differences in fitness between genotypes contain the information to start with.
However you used the formulation “add information to the genome” which allows for their interpretation. The important issue is not whether information was already lying around out there, but that natural selection can cause it to be in the genome when it wasn’t before.
Your demonstration, and ones I did too (here, in the section on Generating Specified Information, and here) show the information coming to be in the genome. All the objections to those demonstrations were simple attempts to divert the discussion to something else.
53. Joe Felsenstein:
Dembski, Marks and co. argue that the information is not being created even if it is being put into the genome by natural selection. They argue that the differences in fitness between genotypes contain the information to start with.
However you used the formulation “add information to the genome” which allows for their interpretation. The important issue is not whether information was already lying around out there, but that natural selection can cause it to be in the genome when it wasn’t before.
Your demonstration, and ones I did too (here, in the section on Generating Specified Information, and here) show the information coming to be in the genome. All the objections to those demonstrations were simple attempts to divert the discussion to something else.
OK, point taken.
I still haven’t really got my head around the Search for a Search argument.
The WEASEL example is obviously irrelevant as the optimal solution is encoded in the fitness function. It’s a special case, but Dembski seems to generalise from it to all evolutionary processes.
Part of the problem I think is that the Search + Target metaphor is a really misleading one.
54. Joe Felsenstein
I would describe the calculation as the probability that a fitness as great as this, or greater could arise as a result of the processes of evolution — including natural selection — which are encompassed in the “chance” hypothesis H. The T is the set of all outcomes that have as high a fitness or higher than the observed fitness.
What I was groping to articulate is the fact that any retrospective calculation would need to include further history – the traits that impress are more than just single runs of mutation-to-fixation. Which renders the calculation nigh impossible. From a given start point, a given set of genotypes is accessible in a given time, and T of those will have equal or greater fitness than a given result. But the serial operation of the ‘chance’ mechanism already eliminated a large part of the total space, by going somewhere else first – most likely somewhere with increased fitness.
Certainly, if you look at it purely in terms of the fitnesses of the various possible genotypes in a single trait of interest, it is very hard even in that narrow view to exclude evolution as a cause for a present high value, given that higher fitness is the very thing that is rewarded with increased representation in the gene pool – NS is a fitness-maximising engine, and probably better at it than any Designer. It often strikes me as a peculiarity of ID that an agent must optimise for the very thing that the environment will do for free.
The refuge must lie in islands – the processes of evolution are in fact so effective that high fitness becomes a trap.
55. Yes indeed! Thanks! | 12,678 | 60,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-15 | longest | en | 0.923583 |
https://www.daniweb.com/programming/software-development/threads/409082/make-a-print-statement-when-popping-from-stack | 1,656,230,825,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037649.11/warc/CC-MAIN-20220626071255-20220626101255-00472.warc.gz | 776,588,631 | 16,897 | I need to create a print statement as "n" is being popped from the stack, just like I have one for when it is being pushed on to the stack. I understand what is happening on that final return when in the base case, but I don't know how to show the items being popped, how would I do this? Thanks in advance.
``````/**
*
* @author harryluthi
* @assignment 3
* @precoditions none
*
* SumOfSquares will calculate and return the sum of squares
* while also telling the user know when an item is pushed
* and popped from the stack.
*
*/
public class SumOfSquares {
private static int square;
public static int ss(int n) {
if (n == 0) {
System.out.println("Base case: n = 0");
System.out.print("Sum of squares is "+square);
return square;
}
else {
while (n > 0) {
System.out.println("Pushing call onto stack. n = " + n);
square = n*n + square;
return square + ss(n-1);
}
}
return -1;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
ss(4);
}
}``````
Here is the console display after code runs, followed by what I need it to ultimately display.
Pushing call onto stack. n = 4
Pushing call onto stack. n = 3
Pushing call onto stack. n = 2
Pushing call onto stack. n = 1
Base case: n = 0
Sum of squares is 30
Need:
Pushing call onto stack. n = 4
Pushing call onto stack. n = 3
Pushing call onto stack. n = 2
Pushing call onto stack. n = 1
Base case: n = 0
Popping previous call from stack. n = 1
Popping previous call from stack. n = 2
Popping previous call from stack. n = 3
Popping previous call from stack. n = 4
Sum of squares is 30
## All 6 Replies
you don't have a stack.
first thing to do is add one.
a stack is like an array, you can add things to them, take them out of it again, step by step.
for instance: if you add four elements to a stack,
and you take three elements away, without knowing exactly what you take out of it, the last element will always be equal to what you stored in there as first element.
in your code above, you may be able to 'simulate' it a bit with fixed values, but it still won't be a stack. a stack would be able to do the same if you add random values, that is something your code is not able to do.
Oh of course I should do the obvious and create a stack. Thank for the quick help
if you *need* a stack for your assignement then go for it, but as far as i can see the recursive solution that you are using is fine, the thing is you are printing out "(Sum of squares is "+square)" inside your ss() method while really you would want to print this when the function is over, since it returns a int. The problem with doing this with your current code is that your function operates on a global variable, so the recursive work isn't really used corectly.
first, drop the while, its useless, as you use a "return" statement in it, it never loops, and you do not want to loop in a recursive solution.
if you want the output to look like the one you ased for, drop the while, drop the "square" variable, simply work with n within the method, and in your main use "System.out.println("Sum of squares is " + ss(4));"
lastly, in the "else" part of your method, after printing the "push" call the function again with n-1 , store that call's return value , print "poping : (that value you jsut stored)" then return it.
here is a correct recursive version of your code, without a stack.
``````public class SumOfSquares {
public static int ss(int n) {
System.out.println("Beginning ss(" + n + ").");
if( n < 0 ) return -1;
if (n == 0) {
System.out.println("Base case : n = 0");
System.out.println("ss(" + n +") returning : " + n);
return n;
}
else {
int x;
System.out.println("Calling ss(" + (n-1) + ").");
x = n*n + ss(n-1);
System.out.println("ss(" + n +") returning : " + x);
return x;
}
}
public static void main(String[] args) {
System.out.println("Sum of squares is : " + ss(4));
}
}``````
output will be :
Beginning ss(4).
Calling ss(3).
Beginning ss(3).
Calling ss(2).
Beginning ss(2).
Calling ss(1).
Beginning ss(1).
Calling ss(0).
Beginning ss(0).
Base case: n = 0
ss(0) returning : 0
ss(1) returning : 1
ss(2) returning : 5
ss(3) returning : 14
ss(4) returning : 30
Sum of squares is : 30
Don't need a stack, just figured it out.
``````import java.util.Scanner;
/**
*
* @author harryluthi
* @assignment 3
* @precoditions none
*
* SumOfSquares will calculate and return the sum of squares
* while also telling the user know when an item is pushed
* and popped from the stack.
*
*/
class SumOfSquares {
private static int square;
public static void ss(int n) {
if (n == 0) {
System.out.println("Base case: n = 0");
System.out.println("Sum of squares is " + square);
}
else {
System.out.println("Pushing item on to stack. n = "+n);
square = n*n + square;
ss(n-1);
System.out.println("Popping previous item from stack. n = "+n);
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println("Running test driver...");
System.out.println("Enter number to compute the sum of squares on: ");
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
ss(n);
}
}``````
Philipe, thanks for your help though!! That is basically what I was looking for.
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We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, learning, and sharing knowledge. | 1,447 | 5,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.905726 |
http://slideplayer.com/slide/7638665/ | 1,571,835,871,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833766.94/warc/CC-MAIN-20191023122219-20191023145719-00535.warc.gz | 171,893,277 | 27,057 | # BCT 2083 DISCRETE STRUCTURE AND APPLICATIONS
## Presentation on theme: "BCT 2083 DISCRETE STRUCTURE AND APPLICATIONS"— Presentation transcript:
BCT 2083 DISCRETE STRUCTURE AND APPLICATIONS
BCT2083 DISCRETE STRUCTURE & APPLICATIONS CHAPTER 4 TREES BCT 2083 DISCRETE STRUCTURE AND APPLICATIONS SITI ZANARIAH SATARI FIST/FSKKP UMP I0910 CHAPTER 4 1
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
CONTENT CHAPTER 4 TREES 4.1 Introduction to Trees 4.2 Application of Trees 4.3 Tree Traversal 4.4 Trees and Sorting (Data structure) 4.5 Spanning Trees 4.6 Minimum Spanning Trees CHAPTER 4 2
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
CHAPTER 4 TREES 4.1 INTRODUCTION TO TREES Define and recognize a tree Define and recognize a rooted tree Define and recognize a m-ary tree CHAPTER 4 3
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
A Tree 4.1 INTRODUCTION TO TREES A tree is a connected undirected graph with no simple circuit. An undirected graph is a tree if and only if there is a unique simple path between any two of its vertices. THEOREM: A tree with n vertices has n – 1 edges. EXAMPLES a b c a b c a b c d e f d e f d e f NOT A TREE A TREE NOT A TREE CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.1 4.1 INTRODUCTION TO TREES Which of the following graphs are trees? a b c a b c d e d e FIGURE 1 FIGURE 2 a v2 v1 v3 c d e v5 b v4 CHAPTER 4 FIGURE 3 FIGURE 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
A Rooted Tree 4.1 INTRODUCTION TO TREES A rooted tree is a tree in which one vertex has been designated as the root and every edge is directed away from the root. Different choice of root produce different rooted tree EXAMPLES c f g a a b d d c b e b d a e f g e f g c A TREE with root a A TREE with root c A TREE CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Properties of Rooted Trees 4.1 INTRODUCTION TO TREES Parent – A vertex other than root is a parent if it has one or more children The parent of c is b Children – If A is a vertex with successors B and C, then B and C are the children of A. The children of a is b, f and g Siblings – Children with the same parent vertex. h, i and j are siblings Level – the length of the unique path from the root to a vertex Vertex a is at level 0 Vertices d and e is at level 3 EXAMPLES a b g f j c h i d e k l m Height – The maximum level of all the vertices The height of this tree is 3. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Properties of Rooted Trees 4.1 INTRODUCTION TO TREES Ancestor of a vertex (v) – the vertices in the path from the root to this vertex excluding this vertex. The ancestors of e are c, b and a Descendent of a vertex (v) – vertices that have v as ancestor. The descendants of b are c, d and e Leaf – A vertex with no children The leaves are d, e, f, i, k, l and m Internal Vertices – vertices that have children The internal vertices are a, b, c, g, h and j EXAMPLES a b g f j c h i d e k l m A subtree rooted at g Subtree – A subgraph of the tree consisting of a root and its descendent and all edges incident to these descendent. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.1 4.1 INTRODUCTION TO TREES Answer these questions about the rooted tree illustrated. Which vertex is the root? Which vertices are internal? Which vertices are leaves? Which vertices are children of g? Which vertex is the parent of o? Which vertices are siblings of e? Which vertices are ancestors of m? Which vertices are descendants of d? a b d Find a subgraph. Find the level of each vertex. What is the height of this tree? Is this tree a balanced tree? c e g f h i o j k l m n p r q CHAPTER 4 s
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
m-ary Tree 4.1 INTRODUCTION TO TREES A rooted tree is called an m-ary tree if every vertex has no more than m children. The tree is called a full m-ary tree if every internal vertex has exactly m children. A rooted m-ary tree is balanced if all leaves are at levels h or h-1. EXAMPLES c a a a b e c d e b d c b d f g k h i j l m f g e f g A full binary TREE A 3-ary TREE A full 4-ary TREE CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
m-ary Tree 4.1 INTRODUCTION TO TREES THEOREM 1: A full m-ary tree with i internal vertices contains n = mi + 1 vertices THEOREM 2: A full m-ary tree with n vertices has internal vertices and leaves. i internal vertices has vertices and leaves. l leaves has vertices and internal vertices. THEOREM 3: There are at most leaves in an m-ary tree of height h. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXAMPLE: m-ary tree 4.1 INTRODUCTION TO TREES Suppose that someone starts a chain letter. Each person who receives the letter is asked to send it on to four other people. Some people do this, but others do not send any letters. How many people have seen the letter, including the first person, if no one receives more than one letter and if the chain letter ends after there have been 100 people, who read it but did not send it out? Solution: Send letter to four other people ⇛ 4-ary tree ⇛ m = 4 People send out the letter ⇛ internal vertices ⇛ i People do not send out the letter ⇛ leaves ⇛ l = 100 By theorem 2 (iii): Number of people have seen this letter ⇛ n = (4·100 – 1) / (4 – 1) = 133 Number of people send out the letter ⇛ i = 133 – 100 = 33 CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.1 4.1 INTRODUCTION TO TREES A chain letter starts with a person sending a letter out to 10 others. Each person is asked to send the letter out to 10 others, and each letter contains a list of previous six people in the chain. Unless there are fewer than six names in the list, each person sends one dollar to the first person from the list, moves up each of other five names one position, and insert his names at the end of this list. If no person breaks the chain and no one receives more than one letter, how much money will a person in the chain ultimately receive? CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.1 : EXTRA 4.1 INTRODUCTION TO TREES PAGE : 693, 694 and 695 Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
CHAPTER 4 TREES 4.2 APPLICATION OF TREES Introduce Binary Search Trees Introduce Decision Trees Introduce Game Trees CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Binary Search Trees 4.2 APPLICATION OF TREES EXAMPLE A binary tree in which each child of a vertex is designated as a right or left child No vertex has more than one right child or left child Each vertex is labeled with a key Vertices are assigned keys so that the key of a vertex is both larger than the keys of all vertices in its left subtree and smaller than the keys of all vertices in its right subtree. Binary search tree for the words mathematics, physics, geography, zoology, meteorology, geology, psychology, and chemistry using alphabetical order mathematics geography physics chemistry geology zoology metereology psycology CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Decision Trees 4.2 APPLICATION OF TREES EXAMPLE A rooted tree in which each internal vertex corresponds to a decision, with a subtree at these vertices for each possible outcome of decision. The possible solutions of the problem correspond to the paths to the leaves of this rooted tree. A Decision tree that orders the elements of the list a, b, c a : b a > b a < b a : c b : c a > c a < c b > c b < c b : c c > a > b a : c c > b > a b > c b < c a < c a > c a > b > c a > c > b b > a > c b > c > a CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Game Trees 4.2 APPLICATION OF TREES Use trees to analyze certain types of games Vertices represent the positions can be in as it progresses Edges represent legal moves between this position Games tree is infinite if the games they represent never end EXAMPLE A Game tree that represents the first two level of the tic-tac-toe game. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.2 : EXTRA 4.2 APPLICATION OF TREES PAGE : 708, 709 and 710 Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
CHAPTER 4 TREES 4.3 TREE TRAVERSAL Determine preorder traversal, inorder traversal and postorder traversal of an ordered rooted tree. Determine the infix, prefix, and postfix form of an expression. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Tree Traversal 4.3 TREE TRAVERSAL Ordered trees are often used to restore data/info. Tree traversal is a procedure for systematically visiting each vertex of an ordered rooted tree to access data. If the tree is label by Universal Address System we can totally order the vertices using lexicographic ordering Example: 0 < 1 < 1.1 < 1.2 < < 1.3 < 2 < 3 < 3.1 < 3.1.1 < < < < 4 < 4.1 Tree traversal algorithm Preorder, inorder and postorder traversal CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Preorder Traversal 4.3 TREE TRAVERSAL Let T be an ordered rooted tree with root r. If T consists only of r, then r is the preorder traversal of T. If T1, T2, …, Tn are subtrees at r from left to right in T, then the preorder traversal begins by visiting r, continues by traversing T1 in preorder, then T2 in preorder, and so on until Tn is traversed in preorder. STEP 1 Visit r r TIPS Preorder Traversal: Visit root, visit subtrees left to right T1 T2 Tn STEP Visit T1 in preorder STEP Visit T2 in preorder STEP n Visit Tn in preorder CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXAMPLE: Preorder Traversal 4.3 TREE TRAVERSAL a a• b• e f• c• d• g h• i• T d b c e f k m g h l i j j k m l n ITERATION 2 p o n p o a• b• e• j• k f• c• d• g• l• m• h• i• a• b c• d n ITERATION 3 p o e f g h i a• b• e• j• k• n• o• p• f• c• d• g• l• m• h• i• k j m l ITERATION 4 n p o ITERATION 1 The preorder traversal of T CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Inorder Traversal 4.3 TREE TRAVERSAL Let T be an ordered rooted tree with root r. If T consists only of r, then r is the inorder traversal of T. If T1, T2, …, Tn are subtrees at r from left to right in T, then the inorder traversal begins by traversing T1 in inorder, then visiting r, continues by traversing T2 in inorder, and so on until Tn is traversed in inorder. STEP 2 Visit r r TIPS Inorder Traversal: Visit leftmost subtree, Visit root, Visit other subtrees left to right. T1 T2 Tn STEP Visit T1 in inorder STEP Visit T2 in inorder STEP n Visit Tn in inorder CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXAMPLE: Inorder Traversal 4.3 TREE TRAVERSAL a e b• f• a• c• g d• h• i• T d b c e f k m g h l i j j k m l n ITERATION 2 p o n p o j• e• b• f• a• c• k l• g• m• d• h• i• b a• c• d n ITERATION 3 p o e f g h i j• e• n• k• o• p• b• f• a• c• l• g• m• d• h• i• k j m l ITERATION 4 n p o ITERATION 1 The inorder traversal of T CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Postorder Traversal 4.3 TREE TRAVERSAL Let T be an ordered rooted tree with root r. If T consists only of r, then r is the postorder traversal of T. If T1, T2, …, Tn are subtrees at r from left to right in T, then the preorder traversal begins by traversing T1 in postorder, then T2 in postorder, and so on until Tn is traversed in postorder and ends by visiting r. STEP n+1 Visit r r TIPS Postorder Traversal: Visit subtrees left to right, Visit root. T1 T2 Tn STEP Visit T1 in postorder STEP Visit T2 in postorder STEP n Visit Tn in postorder CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXAMPLE: Postorder Traversal 4.3 TREE TRAVERSAL a e f• b• c• g h• i• d• a• T d b c e f k m g h l i j j k m l n ITERATION 2 o p n p o j• e• f• b• c• l• m• g• h• i• d• a• k b c• d a• n ITERATION 3 o p e f g h i j• n• o• p• k• e• f• b• c• l• m• g• h• i• d• a• k j m l ITERATION 4 n p o ITERATION 1 The preorder traversal of T CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.3 4.3 TREE TRAVERSAL Determine the order in which a preorder, inorder, and postorder traversal visits the vertices of the following rooted tree. a a b d b g f c e g f j c h h i i j k l o m d e p k l n m r q s FIGURE 1 FIGURE 2 CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Represent Expression by Rooted Tree 4.3 TREE TRAVERSAL We can represent complicated expression (propositions, sets, arithmetic) using ordered rooted trees. EXAMPLE: A binary tree representing ((x+y)↑2)+((x-4)/3) + / _ + 2 3 x y x 4 CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Infix, Prefix & Postfix Notation 4.3 TREE TRAVERSAL We obtain the Infix form of an expression when we traverse its rooted tree in Inorder. The infix form for expression ((x+y)↑2)+((x-4)/3) is x + y ↑2 + x – 4 / 3 or ((x+y)↑2)+((x-4)/3) We obtain the Prefix form of an expression when we traverse its rooted tree in Preorder. The prefix form for expression ((x+y)↑2)+((x-4)/3) is + ↑ + x y 2 / - x 4 3 We obtain the Postfix form of an expression when we traverse its rooted tree in Postorder. The postfix form for expression ((x+y)↑2)+((x-4)/3) is x y + 2 ↑ x 4 – 3 / + CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Evaluating Prefix Expression 4.3 TREE TRAVERSAL Working right to left and performing operations using the operands on the right. EXAMPLE: The value of the prefix expression + - * / ↑ is 3 + - * / ↑ 2 3 4 + - * 2 ↑ 3 = 8 2 * 3 = 6 + - * / 8 4 8/4 =2 6 – 5 =1 + - * + 1 2 2 * 3 = 6 1 + 2 = 3 CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Evaluating Postfix Expression 4.3 TREE TRAVERSAL Working left to right and performing operations using the operands on the left. EXAMPLE: The value of the postfix expression * - 4 ↑ 9 3 / + is 4 7 2 3 * - 4 ↑ 9 3 / + 1 4 ↑ 9 3 / + 2 * 3 = 6 1 ↑ 4 = 1 ↑ 9 3 / + 1 9 3 / + 7 - 6 = 1 9/3 = 3 1 4 ↑ 9 3 / + 1 3 + 1 ↑ 4 = 1 1 + 3 = 4 CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.3 4.3 TREE TRAVERSAL Represent the following expression using binary trees. Then write these expression in infix, prefix and postfix notations. ((x+2)↑3)*(y – (3+x)) – 5 (A∩B) – (A∪(B – A)) What is the value of these expression in prefix expression? + – ↑3 2 ↑ 2 3 / 6 – 4 2 * ↑ What is the value of these expression in postfix expression? 3 2 * 2 ↑ 5 3 – 8 4 / * – 9 3 / – * CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.3 : EXTRA 4.3 TREE TRAVERSAL PAGE : 722, 723 and 724 Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
CHAPTER 4 TREES 4.5 SPANNING TREES Find spanning trees of a simple graph CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
Spanning Trees 4.5 SPANNING TREES Let G be a simple graph. A spanning tree of G is a subgraph of G that is a tree containing every vertex of G. A simple graph is connected if and only if it has a spanning tree. Applied in IP multitasking. a b c d a b c d Not Spanning tree e f g a b c d e f g Spanning tree A simple graph e f g CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.5 4.5 SPANNING TREES Find a spanning tree for the following graphs. b a b f e a c g h d e d c FIGURE 1 FIGURE 2 b a b a e c c d d FIGURE 3 FIGURE 4 CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
EXERCISE 4.5 : EXTRA 4.5 SPANNING TREES PAGE : 734, 735, 736 and 737 Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
CHAPTER 4 TREES 4.6 MINIMUM SPANNING TREES Find minimum spanning tree using Prim’s algorithm Find minimum spanning tree using Kruskal’s algorithm CHAPTER 4
4.6 MINIMUM SPANNING TREES
BCT2083 DISCRETE STRUCTURE & APPLICATIONS Minimum Spanning Trees 4.6 MINIMUM SPANNING TREES A minimum spanning tree in a connected weighted graph is a spanning tree that has the smallest possible sum of weights of it edges. Two algorithms can be used: Prim’s Algorithm Robert Prim, 1957 Kruskal’s Algorithm Joseph Kruskal, 1956 CHAPTER 4
4.6 MINIMUM SPANNING TREES
BCT2083 DISCRETE STRUCTURE & APPLICATIONS Prim’s Algorithm 4.6 MINIMUM SPANNING TREES Chose an edge with the least weight. Include it in spanning tree, T. Select an edge of least weight that is incident with a vertex of an edge in T. If it does not create a cycle (simple circuit) with the edges in T, then include it in T; otherwise discard it. Repeat STEPS 3 and 4 until T contains n-1 edges. There may be more than one minimum spanning tree for a given connected weighted simple graph. If there are two edges with similar smallest weight, chose either one. CHAPTER 4
4.6 MINIMUM SPANNING TREES
BCT2083 DISCRETE STRUCTURE & APPLICATIONS EXAMPLE: Prim’s Algorithm 4.6 MINIMUM SPANNING TREES c CHOICE EDGE WEIGHT SPANNING TREE 1 {a, b} 7 2 {a, d} 10 3 {b, e} 11 4 {e, c} 12 13 13 b 10 b d 16 7 12 15 7 a 10 11 a b d e 12 7 10 FIGURE 1 a The minimum spanning tree is given by: b d 7 10 c 11 a e d 12 b c d 12 b 7 10 11 e a 7 10 11 e Total weight = 40 a CHAPTER 4
4.6 MINIMUM SPANNING TREES
BCT2083 DISCRETE STRUCTURE & APPLICATIONS Kruskal’s Algorithm 4.6 MINIMUM SPANNING TREES Arrange the edges in G in increasing order. Chose an edge with the minimum weight. Include it in spanning tree, T. Add an edge of least weight to T. If it does not create a cycle (simple circuit) with the edges in T, then include it in T; otherwise discard it. Repeat STEPS 4 and 5 until T contains n-1 edges. There may be more than one minimum spanning tree for a given connected weighted simple graph. If there are two edges with similar smallest weight, chose either one. CHAPTER 4
4.6 MINIMUM SPANNING TREES
BCT2083 DISCRETE STRUCTURE & APPLICATIONS EXAMPLE: Kruskal’s Algorithm 4.6 MINIMUM SPANNING TREES c CHOICE EDGE WEIGHT SPANNING TREE 1 {a, b} 7 2 {b, d} 10 3 {b, e} 11 4 {e, c} 12 13 13 b 10 b d 16 7 12 15 7 a 10 11 a b d e 12 10 7 FIGURE 1 a The minimum spanning tree is given by: b d 10 7 c 11 a e b d c 10 12 b 7 d 10 11 e 12 a 7 11 e Total weight = 40 a CHAPTER 4
4.6 MINIMUM SPANNING TREES
BCT2083 DISCRETE STRUCTURE & APPLICATIONS EXERCISE 4.6 4.6 MINIMUM SPANNING TREES Construct a minimum spanning tree for each of the following connected weighted graphs using Prim’s and Kruskal’s algorithm. a b c 8 3 11 f 1 6 1 2 1 e 2 8 a b e 5 3 7 6 f 8 1 2 1 3 h g 1 5 d 7 c d FIGURE 1 FIGURE 2 CHAPTER 4
4.6 MINIMUM SPANNING TREES
BCT2083 DISCRETE STRUCTURE & APPLICATIONS EXERCISE 4.6 : EXTRA 4.6 MINIMUM SPANNING TREES PAGE : 742 and 743 Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007. CHAPTER 4
BCT2083 DISCRETE STRUCTURE & APPLICATIONS
CHAPTER 4 TREES Tree is a connected undirected graph with no simple circuits Trees have been employed to solve problems in a wide variety of disciplines. SUMMARY CHAPTER 4
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https://forum.allaboutcircuits.com/threads/sending-a-single-pulse-when-circuit-is-powered-and-inverse-single-pulse-when-power-is-cut.179004/ | 1,713,095,440,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00064.warc.gz | 243,273,521 | 25,073 | # sending a single pulse when circuit is powered, and inverse single pulse when power is cut
#### Dongiulio
Joined May 12, 2021
4
Hello,
I'm trying to work my way around an automated irrigation system I'd like to build, I bought a bunch of VHM-014 (12 v) circuits so theoretically I should have everything I need for it.
Except that I'm struggling to find the right valves to actually open/shut the water flow.
Initially I bought some 12v solenoid valves, but quickly realised that with water pressure they won't open just on battery power (8 AA), and I tried with a 2A 12v transformer, it wouldn't open either (it does open when no/little pressure is applied).
I disassembled an old Gardena T14 which works with a 9v battery and figured that this device uses a latching 9v valve, that opens with an impulse in one direction and closes with an inverted impulse (i.e. switching + a -).
My VHM-014 are simpler than that and simply open a relay when they figure we need watering and close it when they figure they watered enough.
So I thought I need to explore converting this VHM-014 output so that when the relay closes and my circuit is energised (12v) it needs to send out a single pulse to the latching valve with a given + and - orientation.
Then when the circuit loses power it needs to send another pulse with + and - inverted, maybe using a capacitor?
technically I could use the same 12v source to power the circuit so to have power to send the closing pulse. Ideally I'd like this system to use the least possible current, so that I won't have to replace batteries too often.
something like this:
I need to fill the gap between D1 and D2 (through the solenoid) and equally through D3 and D4 (also through the solenoid)
U4 would be activated by the VHM-014 when watering is needed
can anyone help me out here?
thanks,
UPDATE: I've just learned of the H bridge, with relays, that can invert polarity from a power source to a consumer (ex. a latching valve) building one of these I would just need some kind of system that when current is applied only a pulse is sent, then stop.
with two relays and two of these systems I could complete my setup.
any ideas about how to build such a device to just send a pulse?
#### Tonyr1084
Joined Sep 24, 2015
7,886
Welcome to AAC.
I disassembled an old Gardena T14 which works with a 9v battery and figured that this device uses a latching 9v valve, that opens with an impulse in one direction and closes with an inverted impulse (i.e. switching + a -).
The reason for this is to preserve battery life. If the battery was used to "HOLD" the valve open then the battery would drain and die quickly. So a short "ON" pulse to turn the water on and a short - reversed polarity - "OFF" pulse to turn it off means the battery can last the whole season. Possibly longer. Exactly as you stated.
However, you mentioned water pressure. What is that pressure? Most electric water valves use a small spring to hold the valve shut. When pressure is equalized on both sides of the diaphragm the spring is all that is needed to shut the flow off. However, if you have a very low pressure then even when commanded to be "ON" the water pressure alone might not be enough to open the valve. So what is the pressure we're talking about?
when the circuit loses power it needs to send another pulse with + and - inverted, maybe using a capacitor?
If the battery loses power the capacitor will likely also lose power. Unless you use a diode to prevent back feeding the battery from the capacitor. But even then it's likely the capacitor might not have sufficient power by the time the battery has lost its power.
As for using the H bridge - yes, that can be done. But if you want low current draw you should consider using MOSFET's. They don't use current they use voltage to switch on and off. Relays are going to drain your battery faster than just using the valve and a momentary pulse.
There are plenty of battery powered sprinkler timers and valves on the market. You should be able to find one and copy its design. However, I don't understand why you're trying to re-invent the sprinkler control system when so many others have done it for you. But if it's your wish to do so - I can't say "Don't". However, if you're thinking about commercial applications (making a profit) you could be violating copyrights laws.
#### Dongiulio
Joined May 12, 2021
4
However, you mentioned water pressure. What is that pressure? Most electric water valves use a small spring to hold the valve shut. When pressure is equalized on both sides of the diaphragm the spring is all that is needed to shut the flow off. However, if you have a very low pressure then even when commanded to be "ON" the water pressure alone might not be enough to open the valve. So what is the pressure we're talking about?
The water pressure is about 3/4 bar, which should be well within the range of my valve, still for some reasons it won't open. I tried it using a long garden hose so I was able with a trick to gradually reduce the pressure applied to the valve, and found that it will open on battery power when very little (non zero) pressure is present. I didn't replicate the experiment with the transformer.
If the battery loses power the capacitor will likely also lose power. Unless you use a diode to prevent back feeding the battery from the capacitor. But even then it's likely the capacitor might not have sufficient power by the time the battery has lost its power.
it's not the battery losing power, it's the VHM-014 opening the relay, so it should be full power, then after a moment no power.
I'll study MOSFETs I'm not familiar with them.
There are plenty of battery powered sprinkler timers and valves on the market. You should be able to find one and copy its design. However, I don't understand why you're trying to re-invent the sprinkler control system when so many others have done it for you. But if it's your wish to do so - I can't say "Don't". However, if you're thinking about commercial applications (making a profit) you could be violating copyrights laws.
I know of many battery powered sprinkler systems, all work with a timer though.
Because water is a precious resource with a time only dependent approach, I'll be likely to waste it.
This is because it will be really hard to get the perfect timing, and the amount of watering will be dependent on the weather, on rainy days (wetter) or windy days (dryer) with a timer based approach I will be inevitably overwatering or under-watering respectively, wasting water or ruining my garden which is a precious resource too (to me at least).
What I'm trying to achieve is to water my garden solely depending on soil moisture, and that's what the VHM-014 is designed for. It wasn't probably designed for being operated on batteries (that's why it works with 12v, and not 9v where batteries are easier to find).
I don't know of any watering system that works with a soil moisture sensor.
I'm very curious but very new too to electronics, and I'm looking for a convenient way to set this up. It is turning out to be way harder than I thought initially.
#### Tonyr1084
Joined Sep 24, 2015
7,886
I'm certainly not the expert here but I would imagine that monitoring soil moisture will also use battery power. My first inkling to approach soil monitoring would be a comparator circuit where one leg monitors the voltage passed through the soil and compares it to a reference voltage you set. When the reference voltage goes higher then the comparator output can go high (or low if designed that way) and trigger the water valve. Again, I'm not the expert on such things. But if battery power source is unlimited (solar for instance) then I might approach it that way. MIGHT!
3/4 bar is just below 11 PSI. Typical water pressure is 45 to 90 PSI. Water valves are designed for such circumstances.
#### Dongiulio
Joined May 12, 2021
4
I'm certainly not the expert here but I would imagine that monitoring soil moisture will also use battery power. My first inkling to approach soil monitoring would be a comparator circuit where one leg monitors the voltage passed through the soil and compares it to a reference voltage you set. When the reference voltage goes higher then the comparator output can go high (or low if designed that way) and trigger the water valve. Again, I'm not the expert on such things. But if battery power source is unlimited (solar for instance) then I might approach it that way. MIGHT!
The moisture measure is done via a forked sensor that I believe measures resistance between the two sides, infinite resistance = super dry, low resistance = super wet. so yeah uses battery power, but I don't think it uses a lot of it.
I already have the circuit that triggers the valve when watering is needed, but this just works by powering a relay (i.e. closing the circuit to energise the solenoid valve) to start watering, keeping it powered until the soil is moist enough, then just reopening the circuit in the relay. I can't change this cirquit, but because this configuration is made for normal solenoid valves it uses a lot of battery power, I'd like to modify it so that it can use latching valves, to do so I'm thinking about adding an extra circuit which I need to produce, that:
• when the relay closes the circuit (i.e. start watering signal) I just send a quick impulse to the latching valve (to open the water flow) and then stop energy flow almost right away, no need to waste battery power, the latching valve is already open and doesn't need additional current)
• when the relay opens the circuit (i.e. watered enough signal), my circuit sees a cut in energy and sends another quick impulse (with polarity switched + becomes - and - becomes +) that will shit the latching valve (stopping the water flow), then stop passing current to the latching valve, it has already switched, doesn't require any additional power
Another idea I could use the relay as a switch, when closed send a pulse, when open send another pulse.
I could even unsolder the relay from the VHM-014 and hack my circuit in, so that when there's tension between the two ends of where the solenoid used to be I could send a pulse and another when that tension goes off.
3/4 bar is just below 11 PSI. Typical water pressure is 45 to 90 PSI. Water valves are designed for such circumstances.
My calculator says that 3/4 bars is between 43 and 59 PSI
#### Ya’akov
Joined Jan 27, 2019
9,117
Traditionally, this sort of automation would be done with time delay relays of various sorts, possibly mounted on a DIN rail. All sensors and signals you need are available in this form, and with the cheap Chinese options, they won't cost much.
Take a look at DIN mount time delay relays, in particular single shot types that will generate a pulse on getting power. DIN mount power supplies are also cheap and readily available. | 2,463 | 10,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.960854 |
https://www.instructables.com/community/How-to-figure-out-schematic-of-8x8-LED-matrix/ | 1,618,696,464,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00542.warc.gz | 921,747,571 | 14,852 | 15884Views10Replies
# How to figure out schematic of 8x8 LED matrix? Answered
I don't have a datasheet for the bicolor LED matrix that I got on eBay.
It has 24 (12x12) pins, but no way for me to tell which one is pin 1.
Any clues? :)
Tags:
The forums are retiring in 2021 and are now closed for new topics and comments.
Normally bicolour led matricies have
http://oomlout.com/DATASHEETS/LED-8X8M-03.pdf
both leds in series, 8 rows and 16 columns = 24 pins.
As for pin 1, look around for a square dip trace on the back of the board, there MUST be something to indicate pin 1.
Thank you! Still unsure tho, how to trace it :( There's absolutely no indication of Pin1. I took some pictures, maybe it will help? :) I attached marked piece of foam to right most pin under the lettering.
I've tried to trace pins by applying power via resistor, and mapped some of the dots, but for all I know I might have this thing upside down and not know it :(
Any tips for finding out Pin1?
One thing I figured out for sure is that LED rows are parallel to the pin rows, and that helps! :)
So far I have this idea:
If Pin1 is one of the pins on ends I have about 1 in 8 chances to figure it out (1 in 4 but it might be either cathode or anode so...). So my plan is:
1. I will choose a dot I want to light up (i.e. Red, Row 5, Column 4)
2. Choose one of the 5 corner pins and connect it to the ground.
3. Try connecting all other pins to +5V via a resistor until LED from step 1 is lit.
4. If none get lit, connect that corner pin to +5V via resistor and probe all other pins to Ground.
5. Continue 1-5 for 3 remaining corner pins.
Does my logic make sense?
So, top picture showing the JX serial number
With that facing you, the row along that side SHOULD be pins 1-12 going left to right closest to the black text. The other row runs right to left counting 13-24.
Try jabbing around based on the datasheet like thatand see if that helps.
Well I finally figured it out. That is I had to create my own "data sheet", because one from Sure LED didn't fit at all...
I marked pin under "HG" letters as Pin1 and went from there.
4 leftmost pins (12,11,19,9 and 13, 14, 15,16) are common anodes.
1 thru 8 are Green LED cathode, and 17 thru 24 are Red LED cathodes. Going from there I made a table and eventually changed pins on the image from Datasheet. And finally came up with this. Row closes to the black letters is row 8, and left most column is column 1.
Thanks man! Unfortunately it didn't work :( That pin on the left of the lettering is actually anode, so it can't be pin1.
Either Pin1 is not in the corner or Datasheet that I'm using is incorrect for this model. According to datasheet PIN1 controls all Green lights in 5th Column. I tried all 4 corner pins and none of them lights anything close to 5th column... Crap, I guess I'll have to write diagram for it from scratch, oh well :)
I wish there was EDIT option... I wanted to say that in Step1 I will choose LED that connected to PIN1 on the oomlout DataSheet (provided that Datasheet is correct)...
quote from a sparkfun comment:
"If you look at the bottom, there are pin numberes in the epoxy. rule of them, if you hold the display so the black model number on the side is facing down, then pin 1 will be bottom left, but hold it upside down and look for yourself (I’m going off the Sure Electronics LE-MM103 displays)"
Would that be?
dude i am new to programming with microcontrollers can you help me how to program with picaxe 18 m
Forget PIC if are new! Get Arduino, it's very user friendly. | 930 | 3,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-17 | latest | en | 0.94353 |
https://homeworkspool.com/jefferson-labs-a-nonprofit-organization-estimates-that-it-can-save-33-090-a-year-in-cash-operating-costs-for-the-next-years-if-it-buys/ | 1,701,659,304,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100523.4/warc/CC-MAIN-20231204020432-20231204050432-00475.warc.gz | 363,199,495 | 13,450 | # Jefferson Labs , a nonprofit organization , estimates that it can save \$33 090 a year in cash operating costs for the next & years if it buys
Jefferson Labs , a nonprofit organization , estimates that it can save \$33 090 a year in cash operating costs
for the next & years if it buys a special-purpose eye – testing mach
testing machine at a cost
cost of \$140 000 . No terminal disposal value is
expected . Jefferson Labs required rate of return is 12% . Assume all cash flows occur
at year – end except for initial investment amounts . Jefferson Labs uses straight line depreciation
resent Value of \$1 table Present Value of Annuity of \$1 table Future Value of ST table Future Value of Annuity of ST table
Requirement 1 Calculate the following for the special – purpose eye – testing machine :
a . Net present value ( NPV ) ( Use factors to three decimal places X X X X and use a minus sign or parentheses for a negative net present value Enter the net present
I value of the investment rounded to the nearest whole dollar . )
The net present value is s
Calculate the following for the
for the special – purpose eye – testing machine
a . Net present value
b . Payback period
C . Internal rate of return
Accrual accounting rate of return based on net initial investment
Accrual accounting rate of return based on average investme
2 . What other factors should Jefferson
Labs consider in deciding whether to
purchase the special-purpose eye- testing machine ? | 321 | 1,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-50 | latest | en | 0.807586 |
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A130139 Let f denote the map that replaces k with the concatenation of its proper divisors, written in increasing order, each divisor being written in base 10 in the normal way. Then a(n) = prime reached when starting at 2n+1 and iterating f. 5
1, 3, 5, 7, 3, 11, 13, 1129, 17, 19, 37, 23, 5, 313, 29, 31, 311, 1129, 37, 313, 41, 43 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS If 2n+1 is 1 or a prime, set a(n) = 2n+1. If no prime is ever reached, set a(n) = -1. LINKS EXAMPLE n = 7: 2n+1 = 15 = 3*5 -> 35 = 5*7 -> 57 = 3*19 -> 319 = 11*29 -> 1129, prime, so a(7) = 1129. CROSSREFS Cf. A130140, A130141, A130142. A bisection of A120716. Sequence in context: A099984 A130141 A130142 * A204938 A101088 A134487 Adjacent sequences: A130136 A130137 A130138 * A130140 A130141 A130142 KEYWORD base,more,nonn AUTHOR N. J. A. Sloane, Jul 30 2007 EXTENSIONS The value of a(22) is currently unknown. STATUS approved
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Last modified December 11 10:48 EST 2018. Contains 318049 sequences. (Running on oeis4.) | 537 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-51 | latest | en | 0.815801 |
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Get Representation Theory essential facts below. View Videos or join the Representation Theory discussion. Add Representation Theory to your PopFlock.com topic list for future reference or share this resource on social media.
Representation Theory
Representation theory studies how algebraic structures "act" on objects. A simple example is how the symmetries of regular polygons, consisting of reflections and rotations, transform the polygon.
Representation theory is a branch of mathematics that studies abstract algebraic structures by representing their elements as linear transformations of vector spaces,[1] and studies modules over these abstract algebraic structures.[2][3] In essence, a representation makes an abstract algebraic object more concrete by describing its elements by matrices and their algebraic operations (for example, matrix addition, matrix multiplication). The theory of matrices and linear operators is well-understood, so representations of more abstract objects in terms of familiar linear algebra objects helps glean properties and sometimes simplify calculations on more abstract theories.
The algebraic objects amenable to such a description include groups, associative algebras and Lie algebras. The most prominent of these (and historically the first) is the representation theory of groups, in which elements of a group are represented by invertible matrices in such a way that the group operation is matrix multiplication.[4][5]
Representation theory is a useful method because it reduces problems in abstract algebra to problems in linear algebra, a subject that is well understood.[6] Furthermore, the vector space on which a group (for example) is represented can be infinite-dimensional, and by allowing it to be, for instance, a Hilbert space, methods of analysis can be applied to the theory of groups.[7][8] Representation theory is also important in physics because, for example, it describes how the symmetry group of a physical system affects the solutions of equations describing that system.[9]
Representation theory is pervasive across fields of mathematics for two reasons. First, the applications of representation theory are diverse:[10] in addition to its impact on algebra, representation theory:
Second, there are diverse approaches to representation theory. The same objects can be studied using methods from algebraic geometry, module theory, analytic number theory, differential geometry, operator theory, algebraic combinatorics and topology.[14]
The success of representation theory has led to numerous generalizations. One of the most general is in category theory.[15] The algebraic objects to which representation theory applies can be viewed as particular kinds of categories, and the representations as functors from the object category to the category of vector spaces.[5] This description points to two obvious generalizations: first, the algebraic objects can be replaced by more general categories; second, the target category of vector spaces can be replaced by other well-understood categories.
## Definitions and concepts
Let V be a vector space over a field F.[6] For instance, suppose V is Rn or Cn, the standard n-dimensional space of column vectors over the real or complex numbers, respectively. In this case, the idea of representation theory is to do abstract algebra concretely by using n × n matrices of real or complex numbers.
There are three main sorts of algebraic objects for which this can be done: groups, associative algebras and Lie algebras.[16][5]
This generalizes to any field F and any vector space V over F, with linear maps replacing matrices and composition replacing matrix multiplication: there is a group GL(V,F) of automorphisms of V, an associative algebra EndF(V) of all endomorphisms of V, and a corresponding Lie algebra gl(V,F).
### Definition
There are two ways to say what a representation is.[17] The first uses the idea of an action, generalizing the way that matrices act on column vectors by matrix multiplication. A representation of a group G or (associative or Lie) algebra A on a vector space V is a map
${\displaystyle \Phi \colon G\times V\to V\quad {\text{or}}\quad \Phi \colon A\times V\to V}$
with two properties. First, for any g in G (or a in A), the map
{\displaystyle {\begin{aligned}\Phi (g)\colon V&\to V\\v&\mapsto \Phi (g,v)\end{aligned}}}
is linear (over F). Second, if we introduce the notation g · v for ${\displaystyle \Phi }$ (g, v), then for any g1, g2 in G and v in V:
${\displaystyle (1)\quad e\cdot v=v}$
${\displaystyle (2)\quad g_{1}\cdot (g_{2}\cdot v)=(g_{1}g_{2})\cdot v}$
where e is the identity element of G and g1g2 is the product in G. The requirement for associative algebras is analogous, except that associative algebras do not always have an identity element, in which case equation (1) is ignored. Equation (2) is an abstract expression of the associativity of matrix multiplication. This doesn't hold for the matrix commutator and also there is no identity element for the commutator. Hence for Lie algebras, the only requirement is that for any x1, x2 in A and v in V:
${\displaystyle (2')\quad x_{1}\cdot (x_{2}\cdot v)-x_{2}\cdot (x_{1}\cdot v)=[x_{1},x_{2}]\cdot v}$
where [x1, x2] is the Lie bracket, which generalizes the matrix commutator MNNM.
The second way to define a representation focuses on the map ? sending g in G to a linear map ?(g): V -> V, which satisfies
${\displaystyle \varphi (g_{1}g_{2})=\varphi (g_{1})\circ \varphi (g_{2})\quad {\text{for all }}g_{1},g_{2}\in G}$
and similarly in the other cases. This approach is both more concise and more abstract. From this point of view:
### Terminology
The vector space V is called the representation space of ? and its dimension (if finite) is called the dimension of the representation (sometimes degree, as in [18]). It is also common practice to refer to V itself as the representation when the homomorphism ? is clear from the context; otherwise the notation (V,?) can be used to denote a representation.
When V is of finite dimension n, one can choose a basis for V to identify V with Fn, and hence recover a matrix representation with entries in the field F.
An effective or faithful representation is a representation (V,?), for which the homomorphism ? is injective.
### Equivariant maps and isomorphisms
If V and W are vector spaces over F, equipped with representations ? and ? of a group G, then an equivariant map from V to W is a linear map ?: V -> W such that
${\displaystyle \alpha (g\cdot v)=g\cdot \alpha (v)}$
for all g in G and v in V. In terms of ?: G -> GL(V) and ?: G -> GL(W), this means
${\displaystyle \alpha \circ \varphi (g)=\psi (g)\circ \alpha }$
for all g in G, that is, the following diagram commutes:
Equivariant maps for representations of an associative or Lie algebra are defined similarly. If ? is invertible, then it is said to be an isomorphism, in which case V and W (or, more precisely, ? and ?) are isomorphic representations, also phrased as equivalent representations. An equivariant map is often called an intertwining map of representations. Also, in the case of a group G, it is on occasion called a G-map.
Isomorphic representations are, for practical purposes, "the same"; they provide the same information about the group or algebra being represented. Representation theory therefore seeks to classify representations up to isomorphism.
### Subrepresentations, quotients, and irreducible representations
If ${\displaystyle (V,\psi )}$ is a representation of (say) a group ${\displaystyle G}$, and ${\displaystyle W}$ is a linear subspace of ${\displaystyle V}$ that is preserved by the action of ${\displaystyle G}$ in the sense that for all ${\displaystyle w\in W}$ and ${\displaystyle g\in G}$, ${\displaystyle g\cdot W\in W}$ (Serre calls these ${\displaystyle W}$ stable under ${\displaystyle G}$[18]), then ${\displaystyle W}$ is called a subrepresentation: by defining
${\displaystyle \phi :G\to {\text{Aut}}(W)}$
where ${\displaystyle \phi (g)}$ is the restriction of ${\displaystyle \psi (g)}$ to ${\displaystyle W}$, ${\displaystyle (W,\phi )}$ is a representation of ${\displaystyle G}$ and the inclusion of ${\displaystyle W\hookrightarrow V}$ is an equivariant map. The quotient space ${\displaystyle V/W}$ can also be made into a representation of ${\displaystyle G}$. If ${\displaystyle V}$ has exactly two subrepresentations, namely the trivial subspace {0} and ${\displaystyle V}$ itself, then the representation is said to be irreducible; if ${\displaystyle V}$ has a proper nontrivial subrepresentation, the representation is said to be reducible.[19] The definition of an irreducible representation implies Schur's lemma: an equivariant map
${\displaystyle \alpha :(V,\psi )\to (V',\psi ')}$
between irreducible representations is either the zero map or an isomorphism, since its kernel and image are subrepresentations. In particular, when ${\displaystyle V=V'}$, this shows that the equivariant endomorphisms of ${\displaystyle V}$ form an associative division algebra over the underlying field F. If F is algebraically closed, the only equivariant endomorphisms of an irreducible representation are the scalar multiples of the identity. Irreducible representations are the building blocks of representation theory for many groups: if a representation ${\displaystyle V}$ is not irreducible then it is built from a subrepresentation and a quotient that are both "simpler" in some sense; for instance, if ${\displaystyle V}$ is finite-dimensional, then both the subrepresentation and the quotient have smaller dimension. There are counterexamples where a representation has a subrepresentation, but only has one non-trivial irreducible component. For example, the additive group ${\displaystyle (\mathbb {R} ,+)}$ as a two dimensional representation
${\displaystyle \phi (a)={\begin{bmatrix}1&a\\0&1\end{bmatrix}}}$
This group has the vector ${\displaystyle {\begin{bmatrix}1&0\end{bmatrix}}^{T}}$ fixed by this homomorphism, but the complement subspace maps to
${\displaystyle {\begin{bmatrix}0\\1\end{bmatrix}}\mapsto {\begin{bmatrix}a\\1\end{bmatrix}}}$
giving only one irreducible subreprentation. This is true for all unipotent groups[20]pg 112.
### Direct sums and indecomposable representations
If (V,?) and (W,?) are representations of (say) a group G, then the direct sum of V and W is a representation, in a canonical way, via the equation
${\displaystyle g\cdot (v,w)=(g\cdot v,g\cdot w).}$
The direct sum of two representations carries no more information about the group G than the two representations do individually. If a representation is the direct sum of two proper nontrivial subrepresentations, it is said to be decomposable. Otherwise, it is said to be indecomposable.
### Complete reducibility
In favorable circumstances, every finite-dimensional representation is a direct sum of irreducible representations: such representations are said to be semisimple. In this case, it suffices to understand only the irreducible representations. Examples where this "complete reducibility" phenomenon occur include finite and compact groups, and semisimple Lie algebras.
In cases where complete reducibility does not hold, one must understand how indecomposable representations can be built from irreducible representations as extensions of a quotient by a subrepresentation.
### Tensor products of representations
Suppose ${\displaystyle \phi _{1}:G\rightarrow \mathrm {GL} (V_{1})}$ and ${\displaystyle \phi _{2}:G\rightarrow \mathrm {GL} (V_{2})}$ are representations of a group ${\displaystyle G}$. Then we can form a representation ${\displaystyle \phi _{1}\otimes \phi _{2}}$ of G acting on the tensor product vector space ${\displaystyle V_{1}\otimes V_{2}}$ as follows:[21]
${\displaystyle (\phi _{1}\otimes \phi _{2})(g)=\phi _{1}(g)\otimes \phi _{2}(g)}$.
If ${\displaystyle \phi _{1}}$ and ${\displaystyle \phi _{2}}$ are representations of a Lie algebra, then the correct formula to use is[22]
${\displaystyle (\phi _{1}\otimes \phi _{2})(X)=\phi _{1}(X)\otimes I+I\otimes \phi _{2}(X)}$.
In general, the tensor product of irreducible representations is not irreducible; the process of decomposing a tensor product as a direct sum of irreducible representations is known as Clebsch-Gordan theory.
In the case of the representation theory of the group SU(2) (or equivalently, of its complexified Lie algebra ${\displaystyle \mathrm {sl} (2;\mathbb {C} )}$), the decomposition is easy to work out.[23] The irreducible representations are labeled by a parameter ${\displaystyle l}$ that is a non-negative integer or half integer; the representation then has dimension ${\displaystyle 2l+1}$. Suppose we take the tensor product of the representation of two representations, with labels ${\displaystyle l_{1}}$ and ${\displaystyle l_{2},}$ where we assume ${\displaystyle l_{1}\geq l_{2}}$. Then the tensor product decomposes as a direct sum of one copy of each representation with label ${\displaystyle l}$, where ${\displaystyle l}$ ranges from ${\displaystyle l_{1}-l_{2}}$ to ${\displaystyle l_{1}+l_{2}}$ in increments of 1. If, for example, ${\displaystyle l_{1}=l_{2}=1}$, then the values of ${\displaystyle l}$ that occur are 0, 1, and 2. Thus, the tensor product representation of dimension ${\displaystyle 3\times 3=9}$ decomposes as a direct sum of a 1-dimensional representation ${\displaystyle (l=0),}$ a 3-dimensional representation ${\displaystyle (l=1),}$ and a 5-dimensional representation ${\displaystyle (l=2)}$.
## Branches and topics
Representation theory is notable for the number of branches it has, and the diversity of the approaches to studying representations of groups and algebras. Although, all the theories have in common the basic concepts discussed already, they differ considerably in detail. The differences are at least 3-fold:
1. Representation theory depends upon the type of algebraic object being represented. There are several different classes of groups, associative algebras and Lie algebras, and their representation theories all have an individual flavour.
2. Representation theory depends upon the nature of the vector space on which the algebraic object is represented. The most important distinction is between finite-dimensional representations and infinite-dimensional ones. In the infinite-dimensional case, additional structures are important (for example, whether or not the space is a Hilbert space, Banach space, etc.). Additional algebraic structures can also be imposed in the finite-dimensional case.
3. Representation theory depends upon the type of field over which the vector space is defined. The most important cases are the field of complex numbers, the field of real numbers, finite fields, and fields of p-adic numbers. Additional difficulties arise for fields of positive characteristic and for fields that are not algebraically closed.
### Finite groups
Group representations are a very important tool in the study of finite groups.[24] They also arise in the applications of finite group theory to geometry and crystallography.[25] Representations of finite groups exhibit many of the features of the general theory and point the way to other branches and topics in representation theory.
Over a field of characteristic zero, the representation of a finite group G has a number of convenient properties. First, the representations of G are semisimple (completely reducible). This is a consequence of Maschke's theorem, which states that any subrepresentation V of a G-representation W has a G-invariant complement. One proof is to choose any projection ? from W to V and replace it by its average ?G defined by
${\displaystyle \pi _{G}(x)={\frac {1}{|G|}}\sum _{g\in G}g\cdot \pi (g^{-1}\cdot x).}$
?G is equivariant, and its kernel is the required complement.
The finite-dimensional G-representations can be understood using character theory: the character of a representation ?: G -> GL(V) is the class function ??: G -> F defined by
${\displaystyle \chi _{\varphi }(g)=\mathrm {Tr} (\varphi (g))}$
where ${\displaystyle \mathrm {Tr} }$ is the trace. An irreducible representation of G is completely determined by its character.
Maschke's theorem holds more generally for fields of positive characteristic p, such as the finite fields, as long as the prime p is coprime to the order of G. When p and |G| have a common factor, there are G-representations that are not semisimple, which are studied in a subbranch called modular representation theory.
Averaging techniques also show that if F is the real or complex numbers, then any G-representation preserves an inner product ${\displaystyle \langle \cdot ,\cdot \rangle }$ on V in the sense that
${\displaystyle \langle g\cdot v,g\cdot w\rangle =\langle v,w\rangle }$
for all g in G and v, w in W. Hence any G-representation is unitary.
Unitary representations are automatically semisimple, since Maschke's result can be proven by taking the orthogonal complement of a subrepresentation. When studying representations of groups that are not finite, the unitary representations provide a good generalization of the real and complex representations of a finite group.
Results such as Maschke's theorem and the unitary property that rely on averaging can be generalized to more general groups by replacing the average with an integral, provided that a suitable notion of integral can be defined. This can be done for compact topological groups (including compact Lie groups), using Haar measure, and the resulting theory is known as abstract harmonic analysis.
Over arbitrary fields, another class of finite groups that have a good representation theory are the finite groups of Lie type. Important examples are linear algebraic groups over finite fields. The representation theory of linear algebraic groups and Lie groups extends these examples to infinite-dimensional groups, the latter being intimately related to Lie algebra representations. The importance of character theory for finite groups has an analogue in the theory of weights for representations of Lie groups and Lie algebras.
Representations of a finite group G are also linked directly to algebra representations via the group algebra F[G], which is a vector space over F with the elements of G as a basis, equipped with the multiplication operation defined by the group operation, linearity, and the requirement that the group operation and scalar multiplication commute.
### Modular representations
Modular representations of a finite group G are representations over a field whose characteristic is not coprime to |G|, so that Maschke's theorem no longer holds (because |G| is not invertible in F and so one cannot divide by it).[26] Nevertheless, Richard Brauer extended much of character theory to modular representations, and this theory played an important role in early progress towards the classification of finite simple groups, especially for simple groups whose characterization was not amenable to purely group-theoretic methods because their Sylow 2-subgroups were "too small".[27]
As well as having applications to group theory, modular representations arise naturally in other branches of mathematics, such as algebraic geometry, coding theory, combinatorics and number theory.
### Unitary representations
A unitary representation of a group G is a linear representation ? of G on a real or (usually) complex Hilbert space V such that ?(g) is a unitary operator for every g ? G. Such representations have been widely applied in quantum mechanics since the 1920s, thanks in particular to the influence of Hermann Weyl,[28] and this has inspired the development of the theory, most notably through the analysis of representations of the Poincaré group by Eugene Wigner.[29] One of the pioneers in constructing a general theory of unitary representations (for any group G rather than just for particular groups useful in applications) was George Mackey, and an extensive theory was developed by Harish-Chandra and others in the 1950s and 1960s.[30]
A major goal is to describe the "unitary dual", the space of irreducible unitary representations of G.[31] The theory is most well-developed in the case that G is a locally compact (Hausdorff) topological group and the representations are strongly continuous.[11] For G abelian, the unitary dual is just the space of characters, while for G compact, the Peter-Weyl theorem shows that the irreducible unitary representations are finite-dimensional and the unitary dual is discrete.[32] For example, if G is the circle group S1, then the characters are given by integers, and the unitary dual is Z.
For non-compact G, the question of which representations are unitary is a subtle one. Although irreducible unitary representations must be "admissible" (as Harish-Chandra modules) and it is easy to detect which admissible representations have a nondegenerate invariant sesquilinear form, it is hard to determine when this form is positive definite. An effective description of the unitary dual, even for relatively well-behaved groups such as real reductive Lie groups (discussed below), remains an important open problem in representation theory. It has been solved for many particular groups, such as SL(2,R) and the Lorentz group.[33]
### Harmonic analysis
The duality between the circle group S1 and the integers Z, or more generally, between a torus Tn and Zn is well known in analysis as the theory of Fourier series, and the Fourier transform similarly expresses the fact that the space of characters on a real vector space is the dual vector space. Thus unitary representation theory and harmonic analysis are intimately related, and abstract harmonic analysis exploits this relationship, by developing the analysis of functions on locally compact topological groups and related spaces.[11]
A major goal is to provide a general form of the Fourier transform and the Plancherel theorem. This is done by constructing a measure on the unitary dual and an isomorphism between the regular representation of G on the space L2(G) of square integrable functions on G and its representation on the space of L2 functions on the unitary dual. Pontrjagin duality and the Peter-Weyl theorem achieve this for abelian and compact G respectively.[32][34]
Another approach involves considering all unitary representations, not just the irreducible ones. These form a category, and Tannaka-Krein duality provides a way to recover a compact group from its category of unitary representations.
If the group is neither abelian nor compact, no general theory is known with an analogue of the Plancherel theorem or Fourier inversion, although Alexander Grothendieck extended Tannaka-Krein duality to a relationship between linear algebraic groups and tannakian categories.
Harmonic analysis has also been extended from the analysis of functions on a group G to functions on homogeneous spaces for G. The theory is particularly well developed for symmetric spaces and provides a theory of automorphic forms (discussed below).
### Lie groups
A Lie group is a group that is also a smooth manifold. Many classical groups of matrices over the real or complex numbers are Lie groups.[35] Many of the groups important in physics and chemistry are Lie groups, and their representation theory is crucial to the application of group theory in those fields.[9]
The representation theory of Lie groups can be developed first by considering the compact groups, to which results of compact representation theory apply.[31] This theory can be extended to finite-dimensional representations of semisimple Lie groups using Weyl's unitary trick: each semisimple real Lie group G has a complexification, which is a complex Lie group Gc, and this complex Lie group has a maximal compact subgroup K. The finite-dimensional representations of G closely correspond to those of K.
A general Lie group is a semidirect product of a solvable Lie group and a semisimple Lie group (the Levi decomposition).[36] The classification of representations of solvable Lie groups is intractable in general, but often easy in practical cases. Representations of semidirect products can then be analysed by means of general results called Mackey theory, which is a generalization of the methods used in Wigner's classification of representations of the Poincaré group.
### Lie algebras
A Lie algebra over a field F is a vector space over F equipped with a skew-symmetric bilinear operation called the Lie bracket, which satisfies the Jacobi identity. Lie algebras arise in particular as tangent spaces to Lie groups at the identity element, leading to their interpretation as "infinitesimal symmetries".[36] An important approach to the representation theory of Lie groups is to study the corresponding representation theory of Lie algebras, but representations of Lie algebras also have an intrinsic interest.[37]
Lie algebras, like Lie groups, have a Levi decomposition into semisimple and solvable parts, with the representation theory of solvable Lie algebras being intractable in general. In contrast, the finite-dimensional representations of semisimple Lie algebras are completely understood, after work of Élie Cartan. A representation of a semisimple Lie algebra g is analysed by choosing a Cartan subalgebra, which is essentially a generic maximal subalgebra h of g on which the Lie bracket is zero ("abelian"). The representation of g can be decomposed into weight spaces that are eigenspaces for the action of h and the infinitesimal analogue of characters. The structure of semisimple Lie algebras then reduces the analysis of representations to easily understood combinatorics of the possible weights that can occur.[36]
#### Infinite-dimensional Lie algebras
There are many classes of infinite-dimensional Lie algebras whose representations have been studied. Among these, an important class are the Kac-Moody algebras.[38] They are named after Victor Kac and Robert Moody, who independently discovered them. These algebras form a generalization of finite-dimensional semisimple Lie algebras, and share many of their combinatorial properties. This means that they have a class of representations that can be understood in the same way as representations of semisimple Lie algebras.
Affine Lie algebras are a special case of Kac-Moody algebras, which have particular importance in mathematics and theoretical physics, especially conformal field theory and the theory of exactly solvable models. Kac discovered an elegant proof of certain combinatorial identities, Macdonald identities, which is based on the representation theory of affine Kac-Moody algebras.
#### Lie superalgebras
Lie superalgebras are generalizations of Lie algebras in which the underlying vector space has a Z2-grading, and skew-symmetry and Jacobi identity properties of the Lie bracket are modified by signs. Their representation theory is similar to the representation theory of Lie algebras.[39]
### Linear algebraic groups
Linear algebraic groups (or more generally, affine group schemes) are analogues in algebraic geometry of Lie groups, but over more general fields than just R or C. In particular, over finite fields, they give rise to finite groups of Lie type. Although linear algebraic groups have a classification that is very similar to that of Lie groups, their representation theory is rather different (and much less well understood) and requires different techniques, since the Zariski topology is relatively weak, and techniques from analysis are no longer available.[40]
### Invariant theory
Invariant theory studies actions on algebraic varieties from the point of view of their effect on functions, which form representations of the group. Classically, the theory dealt with the question of explicit description of polynomial functions that do not change, or are invariant, under the transformations from a given linear group. The modern approach analyses the decomposition of these representations into irreducibles.[41]
Invariant theory of infinite groups is inextricably linked with the development of linear algebra, especially, the theories of quadratic forms and determinants. Another subject with strong mutual influence is projective geometry, where invariant theory can be used to organize the subject, and during the 1960s, new life was breathed into the subject by David Mumford in the form of his geometric invariant theory.[42]
The representation theory of semisimple Lie groups has its roots in invariant theory[35] and the strong links between representation theory and algebraic geometry have many parallels in differential geometry, beginning with Felix Klein's Erlangen program and Élie Cartan's connections, which place groups and symmetry at the heart of geometry.[43] Modern developments link representation theory and invariant theory to areas as diverse as holonomy, differential operators and the theory of several complex variables.
### Automorphic forms and number theory
Automorphic forms are a generalization of modular forms to more general analytic functions, perhaps of several complex variables, with similar transformation properties.[44] The generalization involves replacing the modular group PSL2 (R) and a chosen congruence subgroup by a semisimple Lie group G and a discrete subgroup ?. Just as modular forms can be viewed as differential forms on a quotient of the upper half space H = PSL2 (R)/SO(2), automorphic forms can be viewed as differential forms (or similar objects) on ?\G/K, where K is (typically) a maximal compact subgroup of G. Some care is required, however, as the quotient typically has singularities. The quotient of a semisimple Lie group by a compact subgroup is a symmetric space and so the theory of automorphic forms is intimately related to harmonic analysis on symmetric spaces.
Before the development of the general theory, many important special cases were worked out in detail, including the Hilbert modular forms and Siegel modular forms. Important results in the theory include the Selberg trace formula and the realization by Robert Langlands that the Riemann-Roch theorem could be applied to calculate the dimension of the space of automorphic forms. The subsequent notion of "automorphic representation" has proved of great technical value for dealing with the case that G is an algebraic group, treated as an adelic algebraic group. As a result, an entire philosophy, the Langlands program has developed around the relation between representation and number theoretic properties of automorphic forms.[45]
### Associative algebras
In one sense, associative algebra representations generalize both representations of groups and Lie algebras. A representation of a group induces a representation of a corresponding group ring or group algebra, while representations of a Lie algebra correspond bijectively to representations of its universal enveloping algebra. However, the representation theory of general associative algebras does not have all of the nice properties of the representation theory of groups and Lie algebras.
#### Module theory
When considering representations of an associative algebra, one can forget the underlying field, and simply regard the associative algebra as a ring, and its representations as modules. This approach is surprisingly fruitful: many results in representation theory can be interpreted as special cases of results about modules over a ring.
#### Hopf algebras and quantum groups
Hopf algebras provide a way to improve the representation theory of associative algebras, while retaining the representation theory of groups and Lie algebras as special cases. In particular, the tensor product of two representations is a representation, as is the dual vector space.
The Hopf algebras associated to groups have a commutative algebra structure, and so general Hopf algebras are known as quantum groups, although this term is often restricted to certain Hopf algebras arising as deformations of groups or their universal enveloping algebras. The representation theory of quantum groups has added surprising insights to the representation theory of Lie groups and Lie algebras, for instance through the crystal basis of Kashiwara.
## Generalizations
### Set-theoretic representations
A set-theoretic representation (also known as a group action or permutation representation) of a group G on a set X is given by a function ? from G to XX, the set of functions from X to X, such that for all g1, g2 in G and all x in X:
${\displaystyle \rho (1)[x]=x}$
${\displaystyle \rho (g_{1}g_{2})[x]=\rho (g_{1})[\rho (g_{2})[x]].}$
This condition and the axioms for a group imply that ?(g) is a bijection (or permutation) for all g in G. Thus we may equivalently define a permutation representation to be a group homomorphism from G to the symmetric group SX of X.
### Representations in other categories
Every group G can be viewed as a category with a single object; morphisms in this category are just the elements of G. Given an arbitrary category C, a representation of G in C is a functor from G to C. Such a functor selects an object X in C and a group homomorphism from G to Aut(X), the automorphism group of X.
In the case where C is VectF, the category of vector spaces over a field F, this definition is equivalent to a linear representation. Likewise, a set-theoretic representation is just a representation of G in the category of sets.
For another example consider the category of topological spaces, Top. Representations in Top are homomorphisms from G to the homeomorphism group of a topological space X.
Two types of representations closely related to linear representations are:
### Representations of categories
Since groups are categories, one can also consider representation of other categories. The simplest generalization is to monoids, which are categories with one object. Groups are monoids for which every morphism is invertible. General monoids have representations in any category. In the category of sets, these are monoid actions, but monoid representations on vector spaces and other objects can be studied.
More generally, one can relax the assumption that the category being represented has only one object. In full generality, this is simply the theory of functors between categories, and little can be said.
One special case has had a significant impact on representation theory, namely the representation theory of quivers.[15] A quiver is simply a directed graph (with loops and multiple arrows allowed), but it can be made into a category (and also an algebra) by considering paths in the graph. Representations of such categories/algebras have illuminated several aspects of representation theory, for instance by allowing non-semisimple representation theory questions about a group to be reduced in some cases to semisimple representation theory questions about a quiver.
## Notes
1. ^ "The Definitive Glossary of Higher Mathematical Jargon -- Mathematical Representation". Math Vault. 2019-08-01. Retrieved .
2. ^ Classic texts on representation theory include Curtis & Reiner (1962) and Serre (1977). Other excellent sources are Fulton & Harris (1991) and Goodman & Wallach (1998).
3. ^ "representation theory in nLab". ncatlab.org. Retrieved .
4. ^ For the history of the representation theory of finite groups, see Lam (1998). For algebraic and Lie groups, see Borel (2001).
5. ^ a b c Etingof, Pavel; Golberg, Oleg; Hensel, Sebastian; Liu, Tiankai; Schwendner, Alex; Vaintrob, Dmitry; Yudovina, Elena (January 10, 2011). "Introduction to representation theory" (PDF). www-math.mit.edu. Retrieved .
6. ^ a b There are many textbooks on vector spaces and linear algebra. For an advanced treatment, see Kostrikin & Manin (1997).
7. ^
8. ^ a b c Teleman, Constantin (2005). "Representation Theory" (PDF). math.berkeley.edu. Retrieved .
9. ^ a b
10. ^ Lam 1998, p. 372.
11. ^ a b c
12. ^
13. ^
14. ^ See the previous footnotes and also Borel (2001).
15. ^ a b
16. ^ Fulton & Harris 1991, Simson, Skowronski & Assem 2007, Humphreys 1972.
17. ^ This material can be found in standard textbooks, such as Curtis & Reiner (1962), Fulton & Harris (1991), Goodman & Wallach (1998), Gordon & Liebeck (1993), Humphreys (1972), Jantzen (2003), Knapp (2001) and Serre (1977).
18. ^ a b
19. ^ The representation {0} of dimension zero is considered to be neither reducible nor irreducible, just like the number 1 is considered to be neither composite nor prime.
20. ^ Humphreys, James E. (1975). Linear Algebraic Groups. New York, NY: Springer New York. ISBN 978-1-4684-9443-3. OCLC 853255426.
21. ^ Hall 2015 Section 4.3.2
22. ^ Hall 2015 Proposition 4.18 and Definition 4.19
23. ^ Hall 2015 Appendix C
24. ^
25. ^
26. ^ Serre 1977, Part III.
27. ^
28. ^ See Weyl 1928.
29. ^
30. ^
31. ^ a b
32. ^ a b
33. ^
34. ^
35. ^ a b
36. ^ a b c
37. ^
38. ^
39. ^
40. ^
41. ^
42. ^
43. ^
44. ^
45. ^ | 8,305 | 37,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 74, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-45 | latest | en | 0.895973 |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_2 | 1,708,493,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00548.warc.gz | 118,830,591 | 11,855 | # 2007 AIME II Problems/Problem 2
## Problem
Find the number of ordered triples $(a,b,c)$ where $a$, $b$, and $c$ are positive integers, $a$ is a factor of $b$, $a$ is a factor of $c$, and $a+b+c=100$.
## Solution
Denote $x = \frac{b}{a}$ and $y = \frac{c}{a}$. The last condition reduces to $a(1 + x + y) = 100$. Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$.
Subtracting the one, we see that $x + y = \{0,1,3,4,9,19,24,49,99\}$. There are exactly $n - 1$ ways to find pairs of $(x,y)$ if $x + y = n$. Thus, there are $0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}$ solutions of $(a,b,c)$.
Alternatively, note that the sum of the divisors of $100$ is $(1 + 2 + 2^2)(1 + 5 + 5^2)$ (notice that after distributing, every divisor is accounted for). This evaluates to $7 \cdot 31 = 217$. Subtract $9 \cdot 2$ for reasons noted above to get $199$. Finally, this changes $1 \Rightarrow -1$, so we have to add one to account for that. We get $\boxed{200}$.
~ pi_is_3.14
## See also
2007 AIME II (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | 473 | 1,329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 27, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-10 | latest | en | 0.727621 |
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# CHAPTER 1
## PHYSICAL QUANTITIES AND UNITS
1
2
Physical Quantities & SI Units
Physics is the study of how the universe/world behaves and
how the laws of nature operate
Physics is a mathematical science. The underlying concepts and
principles have a mathematical basis.
3
Physical Quantities
Physics involves the study of physical quantities and its measurement
Accurate measurement is very important in science particularly physics,
known as the scientific method
Scientific method: observe, measure, collect data & analyse to discover a
pattern to make it a theory, and then law otherwise repeat or reject
A physical quantity is a quantity that can be measured e.g. length, mass or
time of fall
Physical quantities have a numerical value and unit but not always
Some quantities have no units e.g. pi, ratios, radian, strain
A physical quantity can be divided into base quantities and derived
quantities
4
Base Quantities
Base quantities are the quantities that are conventionally accepted as
functionally independent of one another.
It is a quantity that cannot be defined in terms of other physical quantities
nor is it derived from other units, i.e. it is independent of other units.
5
Common language of measurement units
Same as spoken languages, different systems of measurement evolved
throughout the world
Examples: foot, furlongs, cubit, gantang, pounds, carats, grains, kati etc
Foot is the length of King Henry VIIs foot
Although units of measurement can be converted between systems it is
cumbersome and far better to have just one system
Hence the System International (SI) system was born in 1960
6
7 base quantities
The Systme International (SI) is based on 7 fundamental or base quantities
and its units are given below:
Quantity Name of unit Unit symbol
Length metre m
Mass kilogram kg
Time second s
Electric current Ampere A
Temperature Kelvin K
Amount of substance mole mol
Luminous intensity candela cd
7
Beware !
A distance of thirty metres should be written as 30 m and not 30 ms or 30
m s
The letter s is never included in a unit for the plural
If a space is left between 2 letters, the letters denote different units
So, 30 m s would mean thirty metre seconds and 30 ms would mean 30
milliseconds
8
Derived quantities and derived units
A derived quantity is a physics quantity that consists of some combination of base
units
It is a quantity which is derived from the base quantities and is a combination of base
units through multiplying and/or dividing them, but never added or subtracted
All derived units are expressible as products or quotients of the base units
e.g N kg m s
-2
and J kg m
2
s
-2
.
SI derived units are units of measurement defined in the International System of
Units (SI).
They are derived from the seven base units and can be expressed in base-unit
equivalents
Most derived units have a special name
The names of all SI units are written in lowercase. The unit symbols of units named
after persons are always spelled with an initial capital letter (e.g., hertz, Hz; but
meter, m).
The exception is degrees Celsius, which refers to degrees on the Celsius temperature
scale.
9
Derived quantity & equations
A derived quantity has a defining equation which defines the quantity in
terms of other quantities.
It enables us to express a derived unit in terms of base-unit equivalent.
Example: F = ma ; Newton = kg m s
-2
P = F/A ; Pascal = kg m s
-2
/m
2
= kg m
-1
s
-2
10
Some derived units
Derived quantity Base equivalent units _______ Symbol
area square meter m
volume cubic meter m
speed, velocity meter per second m/s or m s
-1
acceleration meter per second squared m/s/s or m s
-2
density kilogram per cubic meter kg m
-3
amount concentration mole per cubic meter mol m
-3
force kg m s
-2
Newton
work/energy kg m
2
s
-2
Joule
power kg m
2
s
-3
Watt
pressure kg m
-1
s
-2
Pascal
frequency s
-1
Hertz
11
Magnitude/size
Magnitudes of physical quantities range from very very large to very very small.
E.g. mass of sun is 10
30
kg and mass of electron is 10
-31
kg.
Hence, prefixes are used to describe these magnitudes.
Common prefixes
12
13
Order of magnitude in metres
Earth to universe 1.4 x 10
26
Earth to Sun 1.5 x 10
11
Length of car 4
Diameter of hair 5 x 10
-4
Diameter of an atom 3 x 10
-10
Diameter of a nucleus 6 x 10
-15
14
Scientific notation
Large and small values are usually expressed in scientific notation i.e. as a
simple number multiplied by a power of ten
A value expressed in the A x 10
n
form where 1 s A < 10 is called the
standard form scientific notation.
There is far less chance of making a mistake with the number of zeroes
E.g 154 000 000 would be written as 1.54 x 10
8
0.00034 would be written as 3.4 x 10
-4
15
Conversions
Since there are so many base units and derived units, and orders of
magnitudes, conversions from one unit to another is inevitable
Let us try some conversions;
a) 30 mm
2
= ? m
2
b) 865 km h
-1
= ? m s
-1
c) 300 g cm
-3
= ? kg m
-3
16
a) 30 mm
2
= ? m
2
( ) ( )
2
3
2
m 10 mm 1
=
2 6 2
m 10 mm 1
=
2 5 2 6 2
m 10 3.0 or m 10 30 mm 30
=
17
b) 865 km h
-1
= ? m s
-1
1 1
s m 240 h km 865
=
18
c) 300 g cm
-3
= ? kg m
-3
-3 5 3
m kg 10 3.0 cm g 300 =
## 1.2 Dimensions of Pyhsical Quantities
19
20
Dimension Of Physical Quantities
The dimensions of a physical quantity shows the relation between that
quantity to the base quantities
e.g the dimensions of area are Length x Length hence area is the
product of Length.
The dimensions of a physical quantity are a combination of the physical
quantities, raised to the appropriate powers, which are used to define the
physical quantity.
Dimensions are not units.
e.g. dimension of length is L, mass is M, time is T etc
Dimensional analysis is used
to determine the unit of the physical quantity.
to determine whether a physical equation is correct or not
dimensionally by using the principle of homogeneity.
but not if a formula is valid or not
21
Dimension Symbol
Mass M
Length L
Time T
Electric current I
Temperature
Amount of substance N
22
23
Equations
For any equation to make sense, each term involved in the equation must have the
same base units
Look at this equation: 3 kg + 6 kg = 9 m
The numbers are correct but the units make it nonsense
A term in an equation is a group of numbers and symbols, and each of the terms is
added to, or subtracted from other terms
In any equation where each term has the same base units, the equation is said to be
homogeneous or balanced
An equation is homogeneous if quantities on BOTH sides of the equation have the
same unit.
e.g. s = ut + at
2
LHS : unit of s = m
RHS : unit of ut = m s
-1
x s = m
unit of at
2
= m s
-2
x s
2
= m
Unit on LHS = unit on RHS, hence equation is homogeneous
A homogeneous equation may not be physically correct but a physically correct
equation is definitely homogeneous
e.g. a) s = 2ut + at
2
is homogeneous but not physically correct
(correct equation is s = ut + at),
b) F = ma is homogeneous and physically correct (try it!)
24
a) To confirm if equation is homogeneous
E.g: The period of oscillation of a simple pendulum is dependent on the
length l and acceleration of free fall, g.
Is the equation;
T = 2t (l/g) or T = 2t (g/l) ?
Take the first equation T = 2t (l/g)
LHS : unit of T = s
RHS : unit of 2t (l/g) = [ m/(m s
-2
) ]
= s
Equation is homogeneous since unit on LHS = RHS
25
b) To find units of constant
E.g: Newtons Universal Law of gravitation, says that the gravitational
force between two objects is given by the formula
F = GMm/r
2
where F - force,
G - Universal Gravitational constant,
M, m - masses of objects,
r - distance apart.
Find the units of G.
Solution
To find units of G : Rearrange the equation.
G = Fr
2
/Mm
therefore unit of G = N m
2
kg
-2
26
Determining the dimension and unit
Determine the dimension and the S.I. unit for the following quantities:
a. Velocity
b. Acceleration
c. Momentum
d. Pressure
e. Force
27
Example solution for velocity
..... dimension
Hence the S.I. unit of velocity is m s
-1
.
| |
| |
| | interval time
nt displaceme in change
Velocity =
| |
| |
| | t
s
v
A
A
= | |
1
LT
T
L
= = v
28
Exercises
1. The moment of inersia, I, of a uniform rod of mass m and length l about an
axis perpendicular to one end of the rod is given by I=1/3 ml. Find
(a) the units of I and
(b) the dimensions of I
2. Under uniform acceleration, motion of an object with velocity, v, is
represented by v = a + bx where a and b are constants and x is a variable
for displacement. If both a and b have dimensions, find the dimensions of
(a) a
(b) bx
(c) b
3. In a simple pendulum experiment, a student makes an assumption that the
period of oscillation of the pendulum, T, is related to the mass, m, of the
pendulum bob, the length, l, of the string and also the acceleration due to
gravity, g. Derive an equation for T by dimensional analysis.
1.3 Scalars and Vectors
29
A scalar quantity is a quantity that has magnitude only and has
no direction in space
Scalars
Examples of Scalar Quantities:
Length
Area
Volume
Time
Mass
A vector quantity is a quantity that has both magnitude and a
direction in space
Vectors
Examples of Vector Quantities:
Displacement
Velocity
Acceleration
Force
Vector diagrams are shown
using an arrow
The length of the arrow
represents its magnitude
The direction of the arrow
shows its direction
Vector Diagrams
VECTOR APPLICATION
ADDITION: When two (2) vectors point in the SAME direction,
EXAMPLE: A man walks 46.5 m east, then another 20 m east.
Calculate his displacement relative to where he started.
66.5 m, E
MAGNITUDE relates to the
size of the arrow and
DIRECTION relates to the
way the arrow is drawn
46.5 m, E
+
20 m, E
VECTOR APPLICATION
SUBTRACTION: When two (2) vectors point in the OPPOSITE
direction, simply subtract them.
EXAMPLE: A man walks 46.5 m east, then another 20 m west.
Calculate his displacement relative to where he started.
26.5 m, E
46.5 m, E
-
20 m, W
Vectors in opposite directions:
6 m s
-1
10 m s
-1
= 4 m s
-1
6 N 10 N = 4 N
Resultant of Two Vectors
Vectors in the same direction:
6 N 4 N = 10 N
6 m
= 10 m
4 m
The resultant is the sum or the combined effect of
two vector quantities
The Parallelogram Law
When two vectors are joined
tail to tail
Complete the parallelogram
The resultant is found by
drawing the diagonal
When two vectors are joined
Draw the resultant vector by
completing the triangle
When resolving a vector into components we
are doing the opposite to finding the resultant
We usually resolve a vector into components
that are perpendicular to each other
Resolving a Vector Into Perpendicular
Components
y
x
Here a vector v is resolved into
an x component and a y
component
If a vector of magnitude v and makes an angle with the
horizontal then the magnitude of the components are:
x = v Cos
y = v Sin
Calculating the Magnitude of the Perpendicular
Components
y=v Sin
x=v Cos
y
Proof:
v
x
Cos = u
u vCos x =
v
y
Sin = u
u vSin y =
x
NON-COLLINEAR VECTORS
When two (2) vectors are PERPENDICULAR to each other,
you must use the PYTHAGOREAN THEOREM
Example: A man travels 120 km east
then 160 km north. Calculate his
resultant displacement.
120 km, E
160 km, N
the hypotenuse is
called the RESULTANT
HORIZONTAL COMPONENT
VERTICAL
COMPONENT
S
T
A
R
T
FINISH
c
2
= a
2
+ b
2
c = a
2
+ b
2
c = resul tant = 120
( )
2
+ 160
( )
2
| |
c = 200km
In the example, DISPLACEMENT asked for and since it is a
VECTOR quantity, we need to report its direction.
N
S
E
W
N of E
E of N
S of W
W of S
N of W
W of N
S of E
E of S
NOTE: When drawing a right
triangle that conveys some type
of motion, you MUST draw your
N of E
NEED A VALUE ANGLE!
Just putting N of E is not good enough (how far north of
east ?). We need to find a numeric value for the direction.
N of E
160 km, N
120 km, E
To find the value of
the angle we use a
Trig function called
TANGENT.
Tanu =
opposite side
=
160
120
=1.333
u = Tan
1
(1.333) = 53.1
o
u
200 km
So the COMPLETE final answer is :
200 km, 53.1 degrees North of East
components?
Suppose a person walked 65 m, 25 degrees East of North. What were
his horizontal and vertical components?
65 m
25
H.C. = ?
V.C = ?
The goal: ALWAYS MAKE A
RIGHT TRIANGLE!
To solve for components, we
often use the trig functions
sine and cosine.
E m C H opp
hypotenuse
side opposite
hypotenuse
, 47 . 27 25 sin 65 . .
, 91 . 58 25 cos 65 . .
sin cos
sine cosine
= = =
= = =
= =
= =
u u
u u
Example
A boat moves with a velocity of 15 m/s, N in a river which flows
with a velocity of 8.0 m/s, west. Calculate the boat's resultant
velocity with respect to due north.
15 m/s, N
8.0 m/s, W
R
v
u
1 . 28 ) 5333 . 0 (
5333 . 0
15
8
/ 17 15 8
1
2 2
= =
= =
= + =
Tan
Tan
s m R
v
u
u
17 m/s, @ 28.1 degrees West of North
Scalar product of two coplanar vectors
Scalar product or dot
product of two coplanar
vector a and b is written as
A.B (read as A dot B)
The product is a scalar given
by
where is the angle
between the two vectors.
A.B=B.A
44
B
A
The work done by a force F
when the displacement of its
point of application s is
given by the scalar product
of F and s.
45
Cross Product of Two Coplanar Vectors
46
Resolving Vector
47
Example
48
49 | 3,806 | 13,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-43 | latest | en | 0.908323 |
https://books.google.ie/books?id=UhgPAAAAIAAJ&dq=editions:UOM39015065618988&output=html_text&lr= | 1,653,186,660,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543264.49/warc/CC-MAIN-20220522001016-20220522031016-00623.warc.gz | 192,009,768 | 10,863 | # Introduction and books 1,2
The University Press, 1908 - Mathematics, Greek
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### Contents
INTRODUCTION 1 29 7 99 17
BOOK 413 Book II 420
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##### 1 Answer
Jul 22, 2016
$y = - 2$
#### Explanation:
For this problem it's good to know some trigonometric identities. For example.
Remember that in polar coordinates, we write
$x = r \cos \left(\theta\right)$, $y = r \sin \left(\theta\right)$, and ${r}^{2} = {x}^{2} + {y}^{2}$
Since we know that $\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$, we can try to rewrite our polar equation.
$r = - 2 \csc \left(\theta\right) \to r = - 2 \cdot \frac{1}{\sin} \left(\theta\right)$
If we multiply both sides by $\sin \left(\theta\right)$, we can simply replace it with $y$:
$r = - 2 \cdot \frac{1}{\sin} \left(\theta\right)$
$r \sin \left(\theta\right) = - 2$
Thus, in cartesian form, we get
$y = - 2$ | 277 | 783 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-04 | latest | en | 0.697897 |
http://codeforces.com/blog/Destopia | 1,660,075,380,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00786.warc.gz | 11,448,187 | 16,025 | ### Destopia's blog
By Destopia, history, 2 months ago,
Given an array of $n \leq 10^5$ integers. $a_1, a_2, \dots a_n$. $f(i)$ is defined as follows:
• Take the first $i$ elements $a_1, a_2, a_3, \dots a_i$, sort them in nondecreasing order. Let's call resulting prefix after sort $s$
• Let $f(i) = 1 \times s_1 + 2 \times s_2 + 3 \times s_3 + \dots + i \times s_i$.
We're asked to calculate $f(1) + f(2) + \dots + f(n)$ $\bmod 10^9 + 7$.
Example $n = 4$ and array $[4, 3, 2, 1]$
$f(1) = 1 \times 4 = 4$
$f(2) = 1 \times 3 + 2 \times 4 = 11$,
$f(3) = 1 \times 2 + 2 \times 3 + 3 \times 4 = 20$
$f(4) = 1 \times 1 + 2 \times 2 + 3 \times 3 + 4 \times 4 = 30$
$f(1) + f(2) + f(3) + f(4) = 4 + 11 + 20 + 30 = 65$.
Read more »
• +9
By Destopia, history, 3 months ago,
Consider following code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, k;
cin >> n >> k;
vector<int> a(n);
int sum = 0;
for (auto &it : a) {
cin >> it;
sum += it;
}
cout << sum << "\n";
for (int i = 0; i < n; i++) {
cout << a[i] << " ";
}
cout << endl;
}
Input like (or anything greater than INT_MAX into k)
5 1234567891564
1 2 3 4 5
makes the program print
0
0 0 0 0 0
What actually happens? We don't use the value of $k$ at all.
Read more »
• +3
By Destopia, history, 3 months ago,
I've solved a lot of two pointers problems, I found that every problem I solved with two pointers is also solvable with binary search, is that true?
Read more »
• +3
By Destopia, history, 4 months ago,
Let's say I want to check whether an integer is prime or not.
Implementation $1$:
bool is_prime1(long long n) {
if (n < 2)
return false;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
Implementation $2$:
bool is_prime2(long long n) {
if (n < 2)
return false;
for (long long i = 2; i <= n / i; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
Which implementation is better?
Read more »
• 0
By Destopia, history, 6 months ago,
Is there any way to submit past year problems? It seems there's only a practice problem which is duplicated every year and every one get almost a perfect score in it.
Read more »
• +6
By Destopia, history, 15 months ago,
Given N strings N <= 1000. The task is to choose a subsequence such that the concatenation of the chosen elements gives the maximum lexicographical string. For example N = 3, ["ab", "ac", "b"] answer = "b", N = 2 ["z", "za"] answer is "zza". How to solve this problem?
Read more »
• -17
By Destopia, history, 2 years ago,
int n = 1000;
int cnt = 0;
for (int i = 0; i < n; i++)
cnt++;
Is the above code O(n) or O(1)? Could anyone verify this?
Read more »
• -14
By Destopia, history, 4 years ago,
I am a manager of a group, and want to add some contest as training. Whenever I type the contest name and click add, I get a message "contest not found". What should I do?
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# Note for MATH 263A with Professor Holston at Ohio-section 1.6 notes
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Limits invoivmg infinity Chris Hoiston Limits Involving Infinity Chris Holston lelts lnvolvmg lnfinm Vertical Asymptotes cm Holston Preliminaries We write lim fx 00 when as X approaches the XHa number a from the left the function becomes increasingly large lelts lnvolvmg lnfinm Vertical Asymptotes Chris Holston WWW We write lim fx 00 when as X approaches the Xaa number a from the left the function becomes increasingly large lim fx 700 means almost the same thing The XHE function becomes increasingly large in magnitude except the function values are all negative lelts lnvolvmg lnfinm Vertical Asymptotes Chris Holston WWW We write lim fx 00 when as X approaches the Xaa number a from the left the function becomes increasingly large lim fx 700 means almost the same thing The XHE function becomes increasingly large in magnitude except the function values are all negative limJr fx ioo are Xgta defined similarly lelts lnvolvmg lnfinm Vertical Asymptotes Chris Holston WWW We write lim fx 00 when as X approaches the Xaa number a from the left the function becomes increasingly large lim fx 700 means almost the same thing The XHE function becomes increasingly large in magnitude except the function values are all negative limJr fx ioo are Xgta defined similarly Definition We say the line X a is a vertical asymptote for fx if either lim fx ioo or lingr fx ioo XH XHa lelts lnvolvmg lnfinm Vertical Asymptotes Chris Holston WWW We write lim fx 00 when as X approaches the Xaa number a from the left the function becomes increasingly large lim fx 700 means almost the same thing The XHE function becomes increasingly large in magnitude except the function values are all negative limJr fx ioo are Xgta defined similarly Definition We say the line X a is a vertical asymptote for fx if either lim fx ioo or lingr fx ioo XH XHa Remarks Nothing in this definition requires that a is not in the domain of fx lelts lnvolvmg lnfinm Vertical Asymptotes Chris Holston WWW We write lim fx 00 when as X approaches the Xaa number a from the left the function becomes increasingly large lim fx 700 means almost the same thing The XHE function becomes increasingly large in magnitude except the function values are all negative limJr fx ioo are Xgta defined similarly Definition We say the line X a is a vertical asymptote for fx if either lim fx ioo or lingr fx ioo XH XHa Remarks Nothing in this definition requires that a is not in the domain of fx The function need not approach the vertical asymptote from both sides lelts lnvolvmg lnfinm Vertical Asymptotes Chris Holston WWW We write lim fx 00 when as X approaches the Xaa number a from the left the function becomes increasingly large lim fx 700 means almost the same thing The XHE function becomes increasingly large in magnitude except the function values are all negative limJr fx ioo are Xgta defined similarly Definition We say the line X a is a vertical asymptote for fx if either lim fx ioo or lingr fx ioo XH XHa Remarks Nothing in this definition requires that a is not in the domain of fx The function need not approach the vertical asymptote from both sides A function can have infinitely many vertical asymptotes EX fx tanX Limits invoivmg infinity Two sided Limits Chris Hoiston Preiiminzries Iim fx ioo are defined in the same way as finite limits xgta Limits invoivmg infinity Two sided Limits Chris Hoiston Preiiminzries Iim fx ioo are defined in the same way as finite limits xgta 1 Examples IIm 7 oo xaox Limits invoivmg infinity Two sided Limits Chris Hoiston Preiiminzries Iim fx ioo are defined in the same way as finite limits xgta 1 Examples IIm 7 oo xaox l DNE Iim xaox umus nvohmg n nm Horizontal Asymptotes cm Ho ston Prehmmznes im fx L means as we take increasingly large values Xgtoo of X to the quotfar right of the number line the function approaches the real number L lelts lnvolvmg lnfinm Cm Hm Horizontal Asymptotes Preliminaries lim fx L means as we take increasingly large values of X to the quotfar right of the number line the function approaches the real number L lim fx L is defined similarly lelts lnvolvmg lnfinm Horizontal Asymptotes Chris Holston Preliminaries lim fx L means as we take increasingly large values Xgtoo of X to the quotfar right of the number line the function approaches the real number L lim fx L is defined Xaioo similarly Definition We say the line y L is a horizontal asymptote for fx is either lim fx L or lim fx L Xgtoo lelts lnvolvmg lnfinm Horizontal Asymptotes cm Holston Preliminaries lim fx L means as we take increasingly large values of X to the quotfar right of the number line the function approaches the real number L lim fx L is defined similarly Definition We say the line y L is a horizontal asymptote for fx is either lim fx L or lim fx L Xgtoo Remarks Nothing in this definition requires that the function can never attain the value L lelts lnvolvmg lnfinm Horizontal Asymptotes cm Holston Preliminaries lim fx L means as we take increasingly large values Xgtoo of X to the quotfar right of the number line the function approaches the real number L lim fx L is defined similarly Definition We say the line y L is a horizontal asymptote for fx is either lim fx L or lim fx L Xgtoo Remarks Nothing in this definition requires that the function can never attain the value L The function need not approach the horizontal asymptote from both sides lelts lnvolvmg lnfinm Horizontal Asymptotes cm Holston Preliminaries lim fx L means as we take increasingly large values of X to the quotfar right of the number line the function approaches the real number L lim fx L is defined similarly Definition We say the line y L is a horizontal asymptote for fx is either lim fx L or lim fx L Xgtoo Remarks Nothing in this definition requires that the function can never attain the value L The function need not approach the horizontal asymptote from both sides A function can have no more than 2 horizontal asymptotes Limits invoivmg infinity Limits at Infinity Chris Hoiston Preiiminzries Not all functions have horizontal asymptotes Examples Iim x2 00 Xgtoo Limits invoivmg infinity Limits at Infinity Chris Hoiston Preiiminzries Not all functions have horizontal asymptotes Examples Iim x2 00 Xgtoo Iim x2 oo Xaioo Limits invoivmg infinity Limits at Infinity Chris Hoiston Preiiminzries Not all functions have horizontal asymptotes Examples Iim x2 00 Xgtoo Iim x2 oo Xaioo Iim 7X3 foo Xgtoo Limits invoivmg infinity Limits at Infinity Chris Hoiston Preiiminzries Not all functions have horizontal asymptotes Examples Iim x2 00 Xgtoo Iim x2 oo Xaioo Iim 7X3 foo Xgtoo Iim 7X3 oo xaioo Limits invoivmg infinity Limits at Infinity Chris Hoiston Preiiminzries Not all functions have horizontal asymptotes Examples Iim x2 00 Xgtoo Iim x2 oo Xaioo Iim 7X3 foo Xgtoo Iim 7X3 oo Xaioo Iim sinx DNE Xgtoo errts lmolvmg ln nm Calculating LImIts lnvolvmg lnflnlty cm Holston Prellmmzrles The following properties illustrates that O and 00 are like inverses Limits lnvolvrng infinity Calculating Limits Involving Infinity Chris Holston Pieiiiiiiiiziies The following properties illustrates that O and 00 are like inverses 1 If 11 fXioo then Iim O Xaioo m Limits lnvolvrng infinity WWW Calculating Limits Involving Infinity Pieiiiiiiiiziies The following properties illustrates that O and 00 are like inverses 1 If 11 fxioo then Iim O Xaioo m If Iim fx 0 then Iim L ioo and Xgta XHa f X IimJr fx ioo although Iim fx might still not exist Limits invoivmg infinity Chris Hoiston 8 Exerci ses Sketch the graph of an example of a function 2 that satisfies all of the given conditions Iim fx foo Iim fx 2f0 07 f is even Xgt3 14 Fi d the limit Jim 55 x w 5 Aghl b 14 Fi d the limit Hm 55 x w 5 4amp9 Aghl b1 b 16 4gu452 E 16 4gu452 E 18 Fii d he limit 5 Aghl b1 b 18 4gu452 E 4gu452 E Limits invoivmg infinity Chris Hoiston Exerci 595 Find the limit Ii t2 2 m taioot3t271 0 Limits invoivmg infinity Chris Hoiston Exerci 595 Find the limit Iim X2 3X7 X2 bx Limits invoivmg infinity Chris Hoiston Exerci 595 Find the limit Iim X2 3X7 X2 bx aib 2
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Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com | 2,702 | 10,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2016-44 | latest | en | 0.90571 |
http://betterlesson.com/lesson/reflection/20092/making-the-most-out-of-the-time-we-have | 1,487,835,409,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171162.4/warc/CC-MAIN-20170219104611-00166-ip-10-171-10-108.ec2.internal.warc.gz | 25,246,695 | 20,500 | ## Reflection: Intervention and Extension Review: Arcs, Angles, Chords, Tangents, and Proof - Section 1: Arcs and Angles Menu
In the menu activity, students move freely from station to station, working on various problems of their own choosing. A change I made during this lesson was to use this time to check in with students who I know have been struggling with some of the current circles topics. I felt that this was an extremely valuable use of time because I now gained an opportunity to work individually with students who need the most help but who often shy away from asking questions in class. Additionally, I felt this was an ideal way to provide individualized help without calling attention to the student in front of his/her peers.
Making the Most Out of the Time We Have
Intervention and Extension: Making the Most Out of the Time We Have
# Review: Arcs, Angles, Chords, Tangents, and Proof
Unit 9: Discovering and Proving Circles Properties
Lesson 6 of 8
## Big Idea: In a kinesthetic menu activity, students will be able to choose from a variety of problems to work on to review for their upcoming Circles Assessment.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, Geometry, circles, properties of circles, circle constructions
55 minutes
### Jessica Uy
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Nothing to upload details close | 480 | 2,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-09 | latest | en | 0.900792 |
https://www.physicsforums.com/threads/please-if-anyone-can-help-me-to-solve-this-differential-equation.590034/ | 1,519,532,037,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816094.78/warc/CC-MAIN-20180225031153-20180225051153-00751.warc.gz | 928,437,269 | 16,200 | Please if anyone can help me to solve this differential equation.
1. Mar 24, 2012
artan
Please if anyone can help me to solve this differential equation.
2. Mar 25, 2012
JJacquelin
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
Hello,
Two obvious solution of the homogeneous part of the equation are y=x and y=x². The third is a special function.
Nevertheless, particular solutions for the whole ODE can be derived :
Let y=x*f(x) and solve the ODE which unknown is f(x).
3. Mar 25, 2012
artan
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
I can find that first and second obvious solution of the homogeneous part x and x^2,but how can I find the third,it is something of x^x.
Can anyone explain how to find the particular solution showing me some steps of the solution.Thank you
4. Mar 25, 2012
artan
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
Can you write the special function and show me some steps how to find particular solution.Thank you
5. Mar 25, 2012
kosovtsov
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
The general solution to your ODE is as follows
$y(x) = -3x(x-3)\ln(x)+[(-\frac{1}{2}x^2+x)\int_{-x}^∞ \frac{\exp(-t)}{t}dt -\frac{1}{2}\exp(x)(x-1)]C_1-\frac{x^3+9}{2}+x^2C_2+xC_3$
where $C_i$ are arbitrary constants.
6. Mar 25, 2012
JJacquelin
7. Mar 25, 2012
artan
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
I started solving this DE this way:
y=x^m
y'=m*x^(m-1)
y''=m(m-1)x^(m-2)
y'''=m(m-1)(m-2)x^(m-3)
and replace them in DE we get:
(m-1)(m-2)(m*x^(m-1)-x^m)=0
so we get :
y1=x
y2=x^2
m*x^(m-1)-x^m=0 this is the third solution but i dont know how to find it
thank you.
8. Mar 25, 2012
JJacquelin
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
You suppose that the third solution is on the patern y=x^m which is not the case. As a consequence this method cannot lead to the third solution.
By chance, the first and the second solution are on the patern y=x^m, so leading to m=1 and m=2. But obviously not the third.
9. Mar 25, 2012
artan
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
MR.Kosovtsov can you write the whole method how did you get the solution,not just the final solution.
I will be grateful.
Thank you.
Last edited: Mar 25, 2012
10. Mar 25, 2012
artan
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
which method should i use to get the solution.
Can you write for me the beginig of the method you use?
I will be grateful.
Thank you for the help.
Last edited: Mar 25, 2012
11. Mar 25, 2012
JJacquelin
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
Hi !
In attachment, the solution for the homogeneous ODE.
Same method for the complete ODE.
Attached Files:
• Homogeneous ODE.JPG
File size:
17.5 KB
Views:
126
12. Mar 25, 2012
artan
Re: need:x^2y'''-x^2y''+2xy'-2y=x^3+3x
Thank you for the help
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 1,050 | 2,818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-09 | longest | en | 0.855498 |
https://nl.mathworks.com/matlabcentral/cody/problems/838-check-if-number-exists-in-vector/solutions/805284 | 1,606,762,453,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00716.warc.gz | 417,257,796 | 16,644 | Cody
# Problem 838. Check if number exists in vector
Solution 805284
Submitted on 12 Jan 2016 by William
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% a = 1; b = [1,2]; y_correct = 1; assert(isequal(existsInVector(a,b),y_correct))
2 Pass
%% a = 12; b = [1,3,4,5,6,7,8,1,2]; y_correct = 0; assert(isequal(existsInVector(a,b),y_correct))
3 Pass
%% a = -1; b = [1,2]; y_correct = 0; assert(isequal(existsInVector(a,b),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 214 | 688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-50 | latest | en | 0.67209 |
http://archives.cpajournal.com/2000/0800/dept/d86200b.htm | 1,586,050,350,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370526982.53/warc/CC-MAIN-20200404231315-20200405021315-00494.warc.gz | 12,329,311 | 6,257 | ### FEDERAL TAXATION
August 2000
PENALTY-FREE EARLY RETIREMENT
By Joseph W. Bencivenga, CPA, O’Connor Davies & Company LLP
With the stock market posting impressive returns in the 1990s, many individuals find their retirement plans flush with high-flying investments and contemplate early retirement. However, there is one formidable obstacle: the 10% penalty for early withdrawals from retirement plans.
Is there a way to circumvent this penalty? Yes. A little-known exception for withdrawals that constitute “substantially equal periodic payments” may be just what the accountant ordered for early retirees.
Avoiding the Early Withdrawal Penalty
IRC section 72(t)(2)(A)(iv) provides an exception for distributions that are part of a series of substantially equal periodic payments (not less frequent than annually) made for the life (or life expectancy) of the employee or the joint lives (or joint life expectancies) of employee and beneficiary. This interesting exception can be the key to early retirement.
Neither the IRC nor the Treasury Regulations define “substantially equal periodic payments.” To assist those grappling with this term, the IRS has issued Notice 89-25. Question 12 of the notice provides three methods for determining substantially equal periodic payments:
• Minimum distribution method,
• Amortization of account balance method, and
• Annuity factor method.
Minimum Distribution Method
The minimum distribution method permits the calculation of payments to be made using a method that would be acceptable for the purposes of calculating the minimum distribution required under IRC section 401(a)(9). The payment may be based on the life expectancy of the employee or the joint life and last survivor expectancy of the employee and the named beneficiary.
An example of the minimum distribution method calculation follows: Assume a 50-year-old taxpayer with an IRA account balance of \$2 million in 2000. Using the single life expectancy table (Table I) from Publication 590, this taxpayer is expected to live for 33.1 years; \$2 million divided by 33.1 years equals \$60,423. Therefore, the taxpayer must distribute \$60,423 from her IRA account for 2000. Because the single life expectancy factor changes as a taxpayer gets older, the minimum distribution amount must be recalculated each year.
Amortization of Account Balance
The amortization of account balance method amortizes the taxpayer’s account balance over a number of years equal to the life expectancy of the account owner or the joint life and last survivor expectancy of the account owner and beneficiary at an interest rate that does not exceed a reasonable interest rate on the date payments commence.
Assume a 50-year-old individual with a life expectancy of 33.1 years, an account balance of \$2 million, and an assumed interest rate of 8%. Amortizing the \$2 million account balance over 33.1 years at 8% comes to \$173,580 annually. Distributing \$173,580 annually would therefore satisfy the substantially equal periodic payment provision.
Annuity Factor
The third method allows payments to be treated as substantially equal periodic payments if the amount to be distributed annually is determined by dividing the taxpayer’s account balance by an annuity factor (the present value of an annuity of \$1 per year beginning at the taxpayer’s age attained in the first distribution year and continuing for the life of the taxpayer). The annuity factor is derived using a standard mortality table and a reasonable interest rate.
Given an annuity factor of 11.109 for a \$1 per year annuity for a 50-year-old individual (calculated using an interest rate of 8% and the UP-1984 Mortality Table), a \$2 million account balance would yield annual distribution of \$180,035 (\$2 million divided by 11.109).
Section 72(t)(4)(A) Pitfalls
IRC section 72(t)(4)(A) must be followed carefully to avoid drastic penalties that would nullify the advantages to this early retirement strategy. Section 72(t)(4)(A) states that if the series of payments is subsequently modified (other than by reason of death or disability) before the end of a five-year period beginning with the date of the first payment and after the employee attains age 59 1/2, then the taxpayer is subject to the 10% penalty. The penalty will also apply if the five-year period has elapsed and the series of payments is modified before the employee attains age 59 1/2.
Section 72(t)(4)(A) describes the penalty resulting from a subsequent modification as follows: The taxpayer’s tax for the first taxable year in which such modification occurs is increased by an amount equal to the tax which would have been imposed, plus interest for the deferral period. Section 72(t)(4)(B) defines the deferral period to be the period beginning with the taxable year in which the distribution would have been includable in gross income and ending with the taxable year in which the modification occurs.
Example 1. Assume a taxpayer, upon turning 54 on January 1, 1996, commenced annual withdrawals of \$50,000 from an IRA that constituted a series of substantially equal periodic payments. He withdrew \$50,000 on January 1, 1996, 1997, and 1998. On January 1, 1999, he withdrew \$60,000, impermissibly modifying the stream of payments.
Because he subsequently modified his payments before the close of the five-year period beginning with the date of the first payment, he would be subject to a penalty totaling \$21,000, plus interest, calculated as follows:
1996 \$5,000 (10% of \$50,000) 1997 \$5,000 1998 \$5,000 1999 \$6,000 (10% of \$60,000)
Example 2. Assume the same facts in the first example, except the taxpayer turns 58 on January 1, 1996. Her penalty would be \$10,000, plus interest, calculated as follows:
1996 (at age 58) \$5,000 (10% \$50,000) 1997 (at age 59) \$5,000
There would be no penalty thereafter, because by January 1, 1998, the date of her next withdrawal, the penalty applies only to those withdrawals made before the taxpayer reaches the age of 59 1/2. The penalty still applies for the withdrawals in 1996 and 1997 because the series of payments was modified before the end of the five-year period beginning with the date of the first payment.
Example 3. Assume a taxpayer, upon turning 51 on January 1, 1993, commenced annual withdrawals of \$50,000 from an IRA that constituted a series of substantially equal periodic payments. She withdrew \$50,000 each January 1 until January 1, 1999, when she impermissibly modified the stream of payments by withdrawing \$60,000.
The taxpayer would be subject to the penalty because the series of payments was modified after the close of the five-year period but before she attained age 59 1/2. The penalty would be \$36,000, plus interest, calculated as follows:
1993–1998 \$5,000 each year (10% of \$50,000) 1999 \$6,000 (10% of \$60,000)
Further Clarification Needed
The statute and regulations do not define the term “subsequent modification.” As a result, taxpayers have no authoritative source to evaluate a proposed modification and must turn to precedent or private letter rulings.
In private letter rulings, the Tax Court and the IRS appear to allow modifications that do not depart from the spirit of Notice 89-25. Modifications unforeseen at the time payments commenced, such as those resulting from a reasonable cost-of-living modification or divorce, were deemed in private letter rulings to be reasonable modifications not violating the intent of Notice 89-25. Conversely, taxpayers that tried to circumvent the intent of the notice were subject to the harsh consequences of section 72(t)(4).
For potential early retirees, the substantially equal periodic payment exception may provide funding. To be on the safe side, it is imperative to not modify the substantially equal periodic payments. Considering the risks of a misstep, a private letter ruling on the specific facts of a case would most likely be worth the cost.
Editors:
Edwin B. Morris, CPA
Rosenberg, Neuwirth & Kuchner
Contributing Editor:
Ira Inemer, CPA
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27: Two graphs with the same degree sequence edges in F are added to F such that the four edges involved are incident with the same four vertices. The Havel-Hakimi Theorem Let H be a graph containing four distinct vertices u, v, w and x such that uv, wx ∈ E(H) and uw, vx ∈ / E(G). 28, where a dashed line means no edge). This produces a new graph G having the same degree sequence as H. u u v ... ...... . . . . ..... .... ...... .... . . . ...... w x v ...... ..... . . . . .. ...........
An ) or a1 a2 · · · an such that ai is 0 or 1 for 1 ≤ i ≤ n (commonly called n-bit strings), such that two vertices are adjacent if and only if the corresponding ordered n-tuples differ at precisely one coordinate. The graph Qn is an nregular graph of order 2n . 23, where their vertices are labeled by n-bit strings. The graphs Qn are often called hypercubes. 110 10 11 ............ ... ............. ............ .............. 1 00 01 0 . ... Q1 : .............. ............ .... . ...
32: The graphs in Exercise 2 3. Let s : 2, 2, 2, 2, 2, 2, 2, 2, 2 and let Gs be the set of all graphs with degree sequence s. Let G be a graph with V (G) = Gs where two vertices F and H in G are adjacent if F can be transformed into H by a single 2-switch. Which familiar graph is G isomorphic to? 4. 2. DEGREE SEQUENCES (2) two vertices F and H of G are adjacent if F can be transformed into H by a single 2-switch and (3) G contains a triangle. | 777 | 3,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-04 | longest | en | 0.926603 |
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posted by on .
solve.
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multiply the whole equation by 6x.
you would then get rid of all the denominators, having 18 - 2x = x.
and then just solve for x | 63 | 188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-22 | latest | en | 0.948048 |
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###### Question:
On June 30, 2020, Sarasota Company issued $3,340,000 face value of 14%, 20-year bonds at$3,842,540, a yield of 12%. Sarasota uses the effective- interest method to amortize bond premium or discount. The bonds pay semiannual interest on June 30 and December 31. (a) Prepare the journal entries to record the following transactions. (Round answer to 0 decimal places, e.g. 38,548. If no entry is required, select "No Entry" for the account titles and enter O for the amounts. Credit account titles are automatically indented when amount is entered. Do not indent manually.) (1) The issuance of the bonds on June 30, 2020. (2) The payment of interest and the amortization of the premium on December 31, 2020. (3) The payment of interest and the amortization of the premium on June 30, 2021. (4) The payment of interest and the amortization of the premium on December 31, 2021. No. Account Titles and Explanation Debit Credit Date (1) June 30, 2020 (2) December 31, 2020 (3) June 30, 2021
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Spatial Principal Component Analysis of Head-Related Transfer Functions and Its Domain Dependency
Written By
Shouichi Takane
Reviewed: March 10th, 2022 Published: April 21st, 2022
DOI: 10.5772/intechopen.104449
From the Edited Volume
Principal Component Analysis [Working Title]
Prof. Fausto Pedro García Márquez
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Abstract
In this chapter, the Principal Component Analysis (PCA) was adopted to spatial variation of Head-Related Transfer Function (HRTF) or its corresponding inverse Fourier Transform, called Head-Related Impulse Response (HRIR), in order to compactly represent their spatial variation. This is called the Spatial PCA (SPCA). The SPCA was carried out for a database of HRTFs in all directions by selecting the domain as one of the HRIRs, the complex HRTFs, the frequency amplitudes of HRTFs, log-amplitudes of HRTFs, and complex logarithm of HRTFs. The minimum phase approximation was incorporated for the frequency amplitudes and log-amplitudes of HRTFs. Comparison of the accuracies in both time and frequency domains taking into account their influence on subjective evaluation showed that the log-amplitudes and complex logarithm of HRTFs are suitable for the SPCA of HRTFs.
Keywords
• spatial principal component analysis
• head-related transfer function
• head-related impulse response
• domain
• compact representation
1. Introduction
1.1 Head-related transfer function (HRTF)
Head-Related Transfer Function (HRTF) is defined as an acoustic transfer function from sound acquired at a center point when a listener is absent to that acquired at the listener’s ear [1] in a free field (a field without any reflection). A sample of its illustration is depicted in Figure 1. As in Figure 1(a), a microphone is located at the center of a subject’s head with the subject absent. The output YAzis obtained as the response to the input Xzby using the z-transform as follows:
YAz=MzHAzSzXz,E1
where Mzand Szare system functions corresponding to the microphone and loudspeaker, respectively. As in Figure 1(b), the same microphone is located at the subject’s ear. The output YEzis also obtained as the response to the same input Xzfed to the same loudspeaker as follows:
YEz=MzHEzSzXz.E2
the z-transform of HRTF, Hz, is acquired from YAzand YEzas follows:
Hz=YEzYAz=HEzHAz.E3
Computation of Eq. (3) eliminates the system functions of Mzand Szwhen the same microphone and loudspeaker are used for the acquisition of the HRTF, except the case that either of these system functions has zeros. The HRTF is obtained as Hzz=expwhere jis imaginary unit, ω=2πfis the angular frequency and fis the frequency. Time domain representation (impulse response) corresponding to Hzis called as the Head-Related Impulse Response (HRIR).
The HRTF varies due to the sound source position and has strong individuality in both objective and subjective senses. Therefore a set of HRTFs is ideally acquired individually in all sound source directions. While a study considering the efficient sampling scheme of the HRTF measurement exists [2], data size of such set of HRTFs may become numerous. There also exist many datasets involving the HRTFs (HRIRs) of multiple subjects in multiple sound source directions [3, 4, 5, 6], but the individualization using these datasets seems difficult.
1.2 Virtual auditory display (VAD) utilizing head-related transfer functions
Virtual Auditory Displays (VADs), which is a device or an equipment for presentation of an audition in certain sound field to a listener, have been developed since 1990s [7]. On the other hand, the primitive form of the VADs was proposed in 1960s [1] and Morimoto et al.applied the theory into practice in 1980 [8]. Some of the VADs are known to be based on the synthesis of transfer functions involving the Head-Related Transfer Functions (HRTFs). They require the real-time processing on their variation due to the movement of the listener and/or the sound sources. Takane et al.proposed a theory of VAD named ADVISE (Auditory Display based on VIrtual SpherE model) [9], and reported an elemental implementation of the VAD based on ADVISE [10]. The listener’s own HRTFs in all directions are ideally essential in order to carry out the synthesis. Moreover, various implementations of VADs exist based on the synthesis of binaural sound signals using the HRTFs, [7, 11, 12, 13]. Taking into account a set of HRTFs acquired for an individual in all directions, its data size must be as compact as possible with their synthesis accuracy achieved to some extent.
A possible approach to the compact representation for spatial variation of an individual HRTF is modeling. Haneda et al.proposed the Common Acoustical-Pole and Zero (CAPZ) model [14]. In the CAPZ model, it is assumed that the poles in HRTFs are independent of sound source positions while their zeros are dependent on them. They indicated that the spatial variation of the HRTFs of a dummy head was modeled in acceptable accuracy. Based on this model, Watanabe et al.proposed the interpolation method and this method showed good interpolation accuracy [15]. The CAPZ model is useful for the compact representation of the HRTFs since the source-position-independent poles makes the total number of coefficients for the representation of the HRTFs with their spatial variation. The data amount decreased by using the CAPZ model, however, is up to 50% relative to the case that all HRTFs in all directions are represented by the FIR filters with fixed length.
1.3 Head-related transfer functions and principal component analysis
Another promising method for the compact representation of HRTFs is the Principal Component Analysis (PCA) [16, 17]. In some studies, the PCA has an alternative name, the spatial feature extraction method [18, 19, 20]. Both have their theoretical basis on the PCA or the Singular Value Decomposition (SVD). In these researches, the spatial variation of HRTFs is modeled by using small number of principal components or eigenvectors. Xie called the PCA adopted to the dataset(s) of HRTFs the Spatial PCA (SPCA) of HRTFs [19]. The author uses this name after Xie in this chapter. As a result of the SPCA, a HRTF in a certain direction is represented as the linear combination of relatively small number of fixed Principal Components (PCs), meaning that these components do not change according to the sound source positions against the listener. The coefficients for the PCs represent such variation. This property has a potential for effective real-time processing concerning their spatial variation due to dynamic factors. The VAD that can synthesize the HRTFs from multiple sound sources in real-time is currently available, for example by using the computational power of the Graphics Processing Units (GPUs) [21].
Many researches have been carried out on the SPCA of HRTFs [16, 17, 18, 19, 20], but there are some differences among these studies. One of the obvious differences is the domain to execute the SPCA. Kistler et al.applied the log-amplitude of the HRTF to the SPCA [17], Chen et al.applied the complex-valued frequency spectrum [18], and Xie applied the amplitude of the HRTF with the assumption of the minimum phase approximation [19]. On the other hand, Wu et al.applied the HRIRs [20]. Xie surveyed and summarized those results in his book [22]. These studies indicate that the SPCA can be successively and commonly adopted by using each domain. In contrast, the use of different domains may bring about the different properties in the results of the SPCA. If the HRTF/HRIR can be reconstructed by using the smallest number of PCs in a certain domain, the SPCA in that domain may bring about the most compact representations. There exists a study with the similar purpose. Liang et al.compared between the SPCA of the linear and logarithmic magnitudes of the HRTFs [23]. The conclusion of this research was that the SPCA on the linear magnitudes of the HRTFs was better than that on their logarithmic magnitudes in the reconstruction accuracy of their monaural loudness spectra. However, their used HRTFs were limited only in horizontal plane, and they only dealt with two domains with the assumption of the minimum phase approximation. Furthermore, Takane proposed the new domain for the SPCA, the complex logarithm of the HRTFs [24].
In this chapter, all domains dealt with the previous researches are picked up together and the compactness brought by the SPCA using each domain is compared.
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2. SPCA of HRTFs/HRIRs
2.1 Outline
The SPCA of HRTFs/HRIRs is outlined in this section. It is a matter of course that the SPCA of HRTFs/HRIRs is based on the PCA.
1. Spatial average of a certain set of Mvectors gmm=1Mis calculated as follows:
gav=1Mm=1Mgm.E4
2. Covariance matrix, denoted as R, is obtained by calculating the following equation:
R=1Mm=1MgmgavgmgavH.E5
It is noted that Hindicates the Hermitian transpose. The size of the matrix Ris N×N, where Nindicates the size of the vector gm.
3. The computed matrix Ris decomposed into Npairs of PCs (eigenvectors) and eigenvalues by solving the following eigenvalue problem:
Rqk=λkqk.E6
As a result, a set of the eigenvalues and principal components (PCs), λkand qkk=1N, is obtained. Note that λkare sorted from their largest to smallest, i. e., λ1λ2λ3λN, and the PCs are also arranged to the corresponding eigenvalues.
4. By using the matrix Qwith qkin its column vector, the weighting vector, wm, corresponding to the m-th vector gmis calculated as follows:
wm=QHgmgav.E7
As a result of the SPCA, the weight wmis approximated by using q1qK1KNas follows:
wmK=QKHgmgav,E8
where QKis a matrix with its column vectors q1qK. Length of the vector wmKbecomes K.
In the above-mentioned procedure, the vectors and matrices are assumed to have complex values. The Hermitian transpose His changed to the simple transpose, T, if the values of them are real. The value mreflects the sound source position, and also the individuals if the HRTFs/HRIRs of multiple individuals are used for assembling the covariance matrix.
The m-th vector, gm, is reconstructed by using the PCs as follows:
gmK=QKwmK+gav.E9
The computed vector, gmKin Eq. (9), becomes acceptable approximation when K<N, but this may have acceptable accuracy in principle when the Cumulative Proportion of Variance (CPV) R2Kis close to 1.0. The CPV is defined by using the eigenvalues of the covariance matrix, λkk=1N, as follows:
R2K=k=1Kλkk=1Nλk,E10
where Nis the total number of components, equals to the length of the vector gmm=1M.
2.2 Domains used for the SPCA of the HRTFs/HRIRs
Five domains were applied to the assembly of the covariance matrix, based on the previous researches. Kistler et al.applied the log-amplitudes of the HRTF [17], Xie dealt the amplitudes of the HRTF [19]. The minimum-phase approximation was assumed in these studies. Chen et al.dealt the complex HRTF spectrum [18], and Takane propsed the usage of the complex logarithm of HRTF [24]. At last, Wu et al.applied the time domain representation of the HRTFs, i. e., HRIRs [20]. The domains used in these studies are treated as the modeling “domains” in this chapter. The domain “I” corresponds to the application of the HRIR to the SPCA, the domain “C” corresponds to the application of the complex HRTF, The domain “F” corresponds to that of the amplitude of the HRTF, the domain “L” corresponds to that of the logarithm of the HRTF amplitude, and the domain “CL” corresponds to that of the complex logarithm of the HRTF. This is summarized in Table 1.
DomainName
HRIRI
HRTFC
Amplitude of HRTFF
Log-amplitude of HRTFL
Complex logarithm of HRTFCL
Table 1.
Names of five domains expressing modeling conditions for the SPCA.
The m-th HRIR and HRTF are respectively expressed as hmand Hm, and Hmis further decomposed into its amplitude and phase components as follows:
Hm=AmexpjΘm,E11
where jis the imaginary unit. The complex logarithm of Hmcan be written as follows:
logHm=logAm+junwrapΘm=Lm+junwrapΘm.E12
Here the imaginary part of logHm, equal to the phase of the HRTF, is assumed to be unwrapped [24]. The logarithm of the HRTF amplitude vector is defined as L, i.e.,LmlogAm. It is obvious in the domains C, F, L and CL that the frequency spectrum has the following symmetric relations:
Hmk=HmNk,logHmk=logHmNk,E13
Amk=AmNk,Lmk=LmNkk=1N,E14
where Hmk, Amk, Lmkare the k-th component of the vectors Hm, Amand Lm, respectively, and * denotes the conjugate. The relations in Eqs. (13) and (14) indicate that the vector lengths can be almost halved in these domains. When the covariance matrices assembled in the domains I, C, F, L and CL are respectively denoted as RI, RC, RF, RLand RCL, the size of RIis N×N, while those of RC, RF, RLand RCLare N/2+1×N/2+1. In this point, the domains C, F, L and CL have the advantage in the compactness compared with the domain I. On the other hand, components of RCand RCLare complex while those of the covariance matrices in the other domains are real.
The domains I, C, F, L and CL mean that hm, Hm, Am, Lmand logHmare respectively the used domains for the SPCA. When their approximations are obtained by using the first KPCs, they are respectively denoted as hmIK, HmCK, AmFK, LmLKand logHmCLK. The vectors concerning the HRIR or the HRTF are calculated by using the ones estimated via the SPCA in each domain:
Domain I:From hmIK, HmIKis obtained by using Fast Fourier Transform (FFT), and AmIKis the amplitude corresponding to HmIK.
Domain C:From HmCK, AmCKis obtained by computing the corresponding amplitude. After the length of the vector HmCKis increased by applying the relation of Eq. (13), hmCKis obtained by using inverse FFT (IFFT).
Domain F:From AmFK, HmFKis estimated by computing the following equation with its minimum phase components ΘmFKcalculated using Hilbert transform [25]:
HmFK=AmFKexpjΘmFK.E15
After the length of the vector HmFKis increased by applying the relation of Eq. (14), the IFFT of HmFKreveals the estimates of HRIR, denoted as hmFK.
Domain L:From LmLK, AmLKis obtained by calculating the following equation:
AmLK=expLmLK.E16
Then as in the domain F, HmLKis estimated by computing its minimum phase components ΘmKKusing Hilbert transform as follows:
HmLK=AmLKexpjΘmLK.E17
After the length of the vector HmLKis increased by applying the relation of Eq. (14), the IFFT of HmLKreveals the estimates of HRIR, denoted as hmLK.
Domain CL:From logHmCLK, HmCLKis obtained by calculating the following equation:
HmCLK=explogHmCLK.E18
The following procedure is the same as the domain C. AmCLKis obtained by computing the corresponding amplitude. After the length of the vector HmCLKis increased by applying the relation of Eq. (13), hmCLKis obtained by using inverse FFT (IFFT).
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3. Relation between number of PCs and accuracy
3.1 Conditions of analysis
A database of HRIRs of KEMAR HATS (Head And Torso Simulator) provided by Media lab. of MIT [26] was used. Liang et al.used the same data in their study with the similar purpose to the one in this chapter. While they used the HRTFs only in horizontal plane [23], all data in this database involving 710 pairs of HRIRs (total: 1420) with sampling frequency of 44.1 kHz were used for the investigation in this chapter. Number of HRIRs is 1420, corresponding to Min Eqs. (4) and (5).
The initial delay in each response was extracted, then 256 sample points were taken as the data for the analysis, windowing with latter half of 512-points Blackman-Harris window function adjusting its peak at that of the HRIR. The SPCA was executed by constructing the covariance matrices from the HRIRs (called as domain I), the HRTFs (domain C), the amplitude of HRTFs (domain F), the log-amplitude of HRTFs (domain L), and the complex logarithm of HRTFs (domain CL). The HRTFs/HRIRs in all directions (710 directions×2 ears) were used, and the average vector (Eq. (4)) and the covariance matrix (Eq. (5)) were calculated in each case.
3.2 Cumulative proportion of variance (CPV)
When the PCA is generally utilized for some data, the Cumulative Proportion of Variance (CPV), R2K, defined as Eq. (10), is used for the reference indicating how much variance is covered by using the first KPCs. The change of the CPV with PC(s) in each domain was plotted in Figure 2. It is found out from this figure that the CPV is monotonically increased and converges to 1.0 as the number of component(s) is increased in all domain. Among five domains, the domain CL has the largest CPV value for the first PC. The domain C has the fastest increase of the CPV against the number of PC(s), and its CPV value for the domain C is almost the same as that for the domain CL when the number of components is more than 7. In contrast, the domain L has the slowest increase, especially when the number of components is more than 15. This means that relatively large number of PCs is required to cover a certain proportion of variance in data.
Reference values in the CPV, more than which all the corresponding PCs are discarded, varied in the previous studies. Kistler et al.set this value to 0.90 [17], Chen et al.and Wu et al.set to 0.999 [18, 20], and Xie set this to about 0.98 [19]. Direct comparison of these values is impossible since the amount of data and analyzing purposes were different in those studies, but all of these values are more than 0.9. Therefore the least numbers of components to cover four values of the CPV, 0.90, 0.95, 0.99 and 0.999 for five domains are indicated in Table 2. Seeing Table 2, the domain CL has the smallest values among five domains in all of the CPV values, and the domain C has almost the same property. This means that the variance in the spatial variation of HRTFs can be covered by using relatively small number of PCs in these domains. The domains F and L also have smaller number of PCs when the set CPV value is small. In these domains the major PCs having large corresponding eigenvalues cover the major part of variance in data. The required number of PCs increases in the domain L when the set CPV value is large. Varying the CPV values from 0.90 to 0.999, the required number of PCs becomes five to six times in the domains I, C, F and CL, while more than ten times are required in the domain L.
DomainCPV
0.900.950.990.999
I8102039
C461120
F571431
L6113278
CL241239
Table 2.
The least number of PCs to cover the CPV in each case.
3.3 Reconstruction accuracy in time and frequency domains
The CPV is known to be an effective criterion for the coverage of variance with a certain number of PCs. However, comparison of the CPVs among five domains is impossible since the covariance matrices as the target for the PCA are different from each other. Therefore, the reconstruction accuracy, defined as the accuracy between the original HRTFs/HRIRs and the ones reconstructed with a certain number of PCs in five domains. In this chapter, the following two measures were computed in order to evaluate the reconstruction accuracy for the SPCA in five domains in both time and frequency domains:
Signal-to-Deviation Ratio (SDR):Signal-to-Deviation Ratio (SDR) is defined as the level difference between the energy (Euclid norm) of the original impulse response and that of the deviation:
SDRhĥ=10log10hhĥdB,E19
where hand ĥrespectively indicate the original and the reconstructed HRIRs, indicates the Euclid norm of the vector. The larger SDR corresponds to the closer ĥto h.
Spectral Distortion (SD):Spectral Distortion (SD) is defined as standard deviation in log-amplitudes of two frequency spectra, as follows:
SDAÂ=1Nfk=0Nf120log10AkÂkdB,E20
where Aand Âare the frequency amplitude spectrum of the original and the reconstructed responses, respectively, and Akand Âkare the k-th components of the vectors Aand Â, respectively. The value of Nfis the number of frequency bin closest to 20 kHz. The smaller SD corresponds to the closer Âto A.
Calculating the SDRs for the domains F and L, the corresponding original impulse responses are ones constructed with its minimum phase approximation, which are different from the ones in the domains I, C and CL. It is noted that the SDRs in each domain were computed as how much the reconstructed impulse response differs from the desired one. Such a treatment was not related to the calculation of SDs since the SD is defined by using only magnitude of the original and the reconstructed HRTFs.
3.3.1 Changes of SDR and SD with source direction
Examples of the changes of the SDR with the source direction are plotted in Figures 36. Figures 36 are figures for the SDR and the SD, respectively. For the elevation, elevation angles of 0°, 90° and − 90° respectively correspond to the horizontal plane, above and below the subject. The lines were plotted in these figures with 20° interval from −40° to 80°, namely the 6 lines are in each figure. For the azimuth angles, their arrangements are the same as the original data [26], i. e., 0°, 90°, 180° and 270° respectively correspond to the front, right, back, and left of the subject. In each figure, number of components in each domain was set to the least value satisfying two of the CPVs in Table 2. The first value is 0.95 in Figures 3 and 5, and the second is 0.999 in Figures 4 and 6.
Seeing these figures, macroscopic tendency in the change of the SDR and SD with the number of PCs is similar: the larger CPV value brings about the larger SDR and smaller SD, and these values are roughly the same when the CPV values is set equal among five domains. In contrast, it should be emphasized that the values of SD for the domains L and CL are smaller than those in the other domains, as shown in Figure 5(d),(e) and 6(d),(e). The domains using the real and complex logarithm may give relatively smaller distortions in frequency domain.
Seeing the properties in time domain according to Figures 3 and 4, the values of SDR are gradually higher when the CPV value is larger. When the azimuth corresponds to the contralateral side (around 250° ∼ 300°) and especially at the lower elevation angles (less than 0°), the relatively smaller SDR values are found out also commonly in all domains In these azimuths and elevation angles, the relatively larger SD values are also observed, as shown in Figures 5 and 6. Xie stated the same points in his articles [19, 22]. In those range of directions, the HRIRs are very small in their energy because of the subject’s head making a “shadow” making the sound from the sound source hard to reach especially in the high frequency range [1]. As a result, the HRIRs in those directions are relatively difficult to be reconstructed with a small number of PCs.
3.3.2 Spatial average of SDR and SD
In order to show the macroscopic tendencies of the relation between reconstruction accuracies and domains, accuracies in time and frequency domains with number of PCs set to Kare computed in a certain domain X (X = I,C,F,L,CL), the overall average of SDR and SD were calculated by using the following equations:
AvSDRXK=10log101Mm=1M10SDRhmhmXK/10,E21
AvSDXK=1Mm=1MSDAmAmXK2.E22
Changes of the average SDR and SD with number of component(s) Kin each case are plotted in Figure 7 respectively. It is clearly found from these figures that the reconstruction accuracy improves (the larger SDR and the smaller SD) commonly in all domains as the number of PC(s) increases. This means that the CPV corresponds to the tendency of the average accuracies in both time and frequency domains. However, these values are different among five domains. Seeing Figure 7(a), the largest average SDR is achieved with the domain C in most of number of PCs. The domains I and F have the similar tendency, and the domains L and CL are the lowest SDR values when the number of PC(s) is more than 5. On the other hand, as shown in Figure 7(b), the domains L and CL have exceptionally the lowest SD in almost all number of PCs. The domains L and CL has the best accuracy in frequency domain. The third lowest average SD is obtained in the domain C, and the value is almost the same as in the domains L and CL when the number of PCs are relatively large (≥35).
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4. Discussion
In the previous section, the SDR and SD computed in five domains were compared. It was shown that the larger number of PCs brings about the accurate reconstruction of HRTFs and HRIRs in all domains. As differences in domains in the SPCA of HRTFs and HRIRs, the average SDR has the largest value in the domain C, and the average SD has the smallest value in the domains L and CL for almost any given number of components. Since the HRTFs and HRIRs are used for the sound signals at the listener’s both ears, it is essential to take their subjective evaluation into account. Most of the previous researches dealt with both the objective and subjective evaluation [17, 19, 27, 28]. However, three investigations can be found, in which the relation of the subjective evaluation to the SDR and SD values. Hanazawa et al.reported the results of a hearing experiment in which the relation between the accuracy of the interpolated HRIRs with their proposed method and the sound localization performance when the sound stimuli convoluted with them were presented [29]. They showed that around 6 dB in SDR had insignificant difference from the performance when the sound stimuli convoluted with the original HRIRs. Takane et al.also carried out an experiment to evaluated his proposed estimation method of HRIRs from the impulse responses obtained in ordinary room with reflection [30]. The results showed that the subjects did not detect significant difference between the stimuli synthesized from the estimated and the original HRIRs when the SDR between them was more than about 20 dB. Nishino et al.investigated the interpolation accuracy of HRTFs in the median plane, although they did not carry out the subjective evaluation. The least (best) accuracy of their interpolation method is 2 dB in average SD [31]. Considering these researches, the least numbers of PCs to achieve the average SDR more than 20 dB and the average SD less than 2 dB were checked from the results of the SDR and SD computation. The results are shown in Table 3. It is found out from Table 3 that number of PCs satisfying each condition has contrastive feature. The domains I, C and F have relatively small numbers satisfying the condition of average SDR > 20 dB, meaning that these domain can reconstruct the HRIRs in relatively high accuracy with small numbers of PCs. On the contrary, The domains L and CL have relatively small numbers satisfying the condition of average SD < 2 dB, meaning that these domains can reconstruct the amplitude of HRTFs relatively high accuracy with small numbers of PCs. The domain C has the balanced property in both time and frequency domain accuracies.
DomainAverage SDR > 20 dBAverage SD < 2 dB
I1944
C722
F1939
L2514
CL2514
Table 3.
The least number of components to achieve average SDR > 20 dB and average SD < 2 dB in each domain.
It is known that the frequency-domain spectral features in the HRTFs are important for the sound localization especially in the median plane [1, 32]. Iida et al.proposed a parametric model of the HRTF focused on the peaks and notches in the frequency domain, and showed that the first and second lowest notches (called N1 and N2, respectively) in their frequency spectra contribute to the subjects’ perceived elevation [33]. These results may state that the local frequency domain features may cause the difference in the listener’s perceived direction, and the reconstruction accuracy in frequency domain must be well taken into account to avoid such potential difference. These researches support the accuracy in frequency domain is more important than that in time domain. Based on such subjective properties together with the objective properties shown in this chapter, the domain L and CL are suitable for the SPCA of the HRTFs more than the other domains. On the other hand, the domains L and CL require relatively large number of PCs to achieve the CPV values when the CPV is closer to 1. The previous researches indicate that the reconstruction with relatively small number of PCs could make the HRTFs/HRIRs without audible difference [17, 19, 27, 28], therefore it is expected that the domains L and CL can bring about the acceptable subjective evaluation with small number of PCs. The difference in the domains for the SPCA must be investigated more in detail especially based on the subjective evaluation, which is one of the future studies concerning the contents of this chapter.
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5. Summary
In this chapter, the SPCA of the HRTFs was introduced, and its dependency on the domains, in which the covariance matrices are calculated, was investigated. The following points are the summary of the findings in this chapter:
• The SPCA can be carried out commonly for all domains, i. e., the HRIRs (domain I), the (complex) HRTFs (domain C), the amplitude spectrum of HRTFs (domain F), logarithm of the amplitude spectrum of HRTFs (domain L), and the complex logarithm of HRTFs (domain CL).
• For the domains except the domain I, the covariance matrices can be sized down to about 1/4 of the covariance matrix assembled for the domain I, according to the symmetric property of the frequency spectrum.
• The domains I, C and F have relatively small numbers of PCs in order to achieve high time domain accuracy.
• The domains L and CL have relatively small numbers of PCs in order to achieve high frequency domain accuracy.
• Considering the influence on the subjective evaluation of the reconstructed HRTFs/HRIRs with their SPCA, the domains L and CL, bringing about relatively high accuracy in frequency domain, are more suitable for the SPCA of the HRTFs.
References
1. 1. Blauert J. Spatial Hearing. Cambridge, Massachusetts, USA: MIT Press. 1983
2. 2. Zhang W, Zhang M, Kennedy RA, Abhayapala TD. On high-resolution head-related transfer function measurements: An efficient sampling scheme. IEEE Transductions on Audio, Speech & Language Processing. 2012;20(2):575-584
3. 3. Algazi VR, Duda RO, Thompson DM. The CIPIC HRTF database. In: Proceedings of the 2001 IEEE Workshop of the Applications of Signal Processing to Audio and Acoustics; New Platz. 2001 Cat No. 01TH8575 (5 pages)
4. 4. Watanabe K, Iwaya Y, Suzuki Y, Takane S, Sato S. Dataset of head-related transfer functions measured with a circular loudspeaker array. Acoustical Science & Technology. 2014;35(3):159-165
5. 5. Bomhardt R, Klein MF, Fels J. A high-resolution head-related transfer function and three-dimensional ear model database. Proceedings of Meetings on Acoustics. 2016;29:050002
6. 6. Brinkmann F, Dinakaran M, Pelzer R, Grosche P, Voss D, Weinzierl S. A cross-evaluated database of measured and simulated HRTFs including 3D head meshes, anthropometric features, and headphone impulse responses. Journal of the Audio Engineering Society. 2019;67(9):705-718
7. 7. Begault DR. 3-D Sound for Virtual Reality and Multimedia. Cambridge, Massachusetts, USA: AP Professional. 1994
8. 8. Morimoto M, Ando Y. On the simulation of sound localization. Journal of the Acoustical Society of Japan (E). 1980;1(3):167-174
9. 9. Takane S, Suzuki Y, Miyajima T, Sone T. A new theory for high definition virtual acoustic display named ADVISE. Acoustical Science & Technology. 2003;24(5):276-283
10. 10. Takane S, Takahashi S, Suzuki Y, Miyajima T. Elementary real-time implementation of a virtual acoustic display based on ADVISE. Acoustical Science & Technology. 2003;24(5):304-310
11. 11. Otani M, Hirahara T. A dynamic auditory display: Its design, performance, and problems in HRTF switching. In: Proceedings of Japan-China Joint Conference on Acoustics; 4–6 June 2007; Sendai: Acoustical Society of Japan/Acoustical Society of China. 2007. SS-1-3
12. 12. Yairi S, Iwaya Y, Suzuki Y. Estimation of detection threshold of system latency of virtual auditory display. Applied Acoustics. 2007;68(8):851-863
13. 13. Miller JD. Slab: A software-based real-time virtual acoustic environment rendering system. In: Proceedings of the 7th International Conference on Auditory Display; 29 July-1 August; Espoo: International Conference on Auditory Display. 2001. pp. 279-280.https://smartech.gatech.edu/handle/1853/50648
14. 14. Haneda Y, Makino S, Kaneda Y, Kitawaki N. Common acoustical-pole and zero modeling of head-related transfer functions. IEEE Transductions on Speech and Audio Processing. 1999;7:188-196
15. 15. Watanabe K, Takane S, Suzuki Y. A novel interpolation method of HRTFs based on the common-acoustical-pole and zero model. Acta Acustica united with Acustica. 2005;91:958-966
16. 16. Martens WL. Principal components analysis and resynthesis of spectral cues to perceived direction. In: Proceedings of the International Computer Music Conference; Champaign/Urbana: International Computer Music Conference. 1987. pp. 274-281.https://hdl.handle.net/2027/spo.bbp2372.1987.040
17. 17. Kistler DJ, Wightmann FL. A model of head-related transfer functions based on principal components analysis and minimum-phase reconstruction. Journal of the Acoustical Society of America. 1992;91(3):1637-1647
18. 18. Chen J, Van Veen BD, Hecox KE. A spatial feature extraction and regularization model for the head-related transfer functions. Journal of the Acoustical Society of America. 1995;97(1):439-452
19. 19. Xie B. Recovery of individual head-related transfer functions from a small set of measurements. Journal of the Acoustical Society of America. 2012;132(1):282-294
20. 20. Wu Z, Chan FHY, Lam FK, Chan JCK. A time domain binaural model based on spatial feature extraction for the head-related transfer functions. Journal of the Acoustical Society of America. 1998;102(4):2211-2218
21. 21. Watanabe K, Oikawa Y, Sato S, Takane S, Abe K. Development and performance evaluation of virtual auditory display system to synthesize sound from multiple sound sources using graphics processing unit. In: Proceedings of the 21th International Congress on Acoustics; Montreal: International Commission for Acoustics. 2013 2pEAba12 (7 pages in CD-ROM)
22. 22. Xie B. Head-Related Transfer Function and Virtual Auditory Display. Second ed. Plantation, Florida, USA: J. Ross Pub. 2013
23. 23. Liang Z, Xie B, Zhong X. Comparison of principal components analysis of linear and logarithmic magnitude of head-related transfer functions. In: Proceedings of the 2nd IEEE International Congress on Image and Signal Processing; Tianjin: IEEE. 2009. pp. 1-5.https://doi.org/10.1109/CISP.2009.530427
24. 24. Takane S. Spatial principal component analysis of head-related transfer functions using their complex logarithm with unwrapping of phase. In: Proceedings of the 23rd International Congress on Acoustics; 9–13 September; Aachen. 2019. pp. 3048-3055
25. 25. Oppenheim AV, Schafer RW. Discrete-time signal processing. third ed. Lebanon, Indiana, USA: Prentice Hall. 2010
26. 26. Gardner WG, Martin KD. HRTF measurements of a KEMAR. Journal of the Acoustical Society of America. 1995;97:3907-3908
27. 27. Matsui K, Ando A. Estimation of individualized head-related transfer function based on principal component analysis. Acoustical Science & Technology. 2009;30(5):338-347
28. 28. Fink KJ, Ray L. Individualization of head-related transfer functions using principal component analysis. Applied Acoustics. 2015;87:162-173
29. 29. Hanazawa K, Yanagawa H, Matsumoto M. Subjective evaluations of interpolated binaural impulse responses and their interpolation accuracies. In: Proceedings of the Spring Research Meeting of the Acoustical Society of Japan; Tokyo: Acoustical Society of Japan. 2006. pp. 677-678 (in Japanese)
30. 30. Takane S, Nabatame S, Abe K, Watanabe K, Sato S. Subjective evaluation of HRIRs linearly predicted from impulse responses measured in ordinary sound field. In: Proceedings of the 39th Audio Engineering Society Japan Regional Conference; Osaka: Audio Engineering Society. 2008. pp. 1-8 (in CD-ROM)
31. 31. Nishino T, Kajita S, Takeda K, Itakura F. Interpolating head related transfer functions in median plane. In: Proceedings of the 1999 IEEE Workshop on Applications of Signal Processing to Audio and Acoustics; New Paltz: IEEE. 1999. pp. 17-20.https://doi.org/10.1109/ASPAA.1999.810876
32. 32. Asano F, Suzuki Y, Sone T. Role of spectral cues in median plane localization. Journal of the Acoustical Society of America. 1990;88(1):159-168
33. 33. Iida K, Itoh M, Itagaki A, Morimoto M. Median plane localization using a parametric model of the head-related transfer function based on spectral cues. Applied Acoustics. 2007;68:835-850
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Shouichi Takane
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Return on Assets
0.00% (As of . 20)
Return on assets is calculated as net income divided by its total assets. 's annualized net income for the quarter that ended in . 20 was \$ Mil. 's total assets for the quarter that ended in . 20 was \$ Mil. Therefore, 's annualized return on assests (ROA) for the quarter that ended in . 20 was 0.00%.
CCRT' s 10-Year Return on Assets Range
Min: 0 Max: 0
Current: 0
During the past 0 years, 's highest Return on Assets (ROA) was %. The lowest was %. And the median was %.
CCRT's Return on Assetsis ranked lower than
100% of the Companies
in the Global industry.
( Industry Median: vs. CCRT: )
Definition
's annualized Return on Assets (ROA) for the fiscal year that ended in . 20 is calculated as:
Return on Assets (ROA) (A: . 20 ) = Net Income (A: . 20 ) / Total Assets (A: . 20 ) = / = %
's annualized Return on Assets (ROA) for the quarter that ended in . 20 is calculated as:
Return on Assets (ROA) (Q: . 20 ) = Net Income (Q: . 20 ) / Total Assets (Q: . 20 ) = / = 0.00 %
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
In the calculation of annual return on assets, the net income of the last fiscal year is used. The total assets are from the end of year data. Strictly speaking, average total assets over the fiscal year should be used. In calculating the quarterly data, the Net Income data used here is two times the semi-annual (. 20) net income data. Return on assets is displayed in the 10-year financial page.
Explanation
Return on assets (ROA) measures the rate of return on the total assets (shareholder equity plus liabilities). It measures a firm's efficiency at generating profits from shareholders' equity plus its liabilities. ROA shows how well a company uses what it has to generate earnings. ROAs can vary drastically across industries. Therefore, return on assets should not be used to compare companies in different industries. For retailers, a ROA of higher than 5% is expected. For example, Wal-Mart (WMT) has a ROA of about 8% as of 2012. For banks, ROA is close to their interest spread. A banks ROA is typically well under 2%.
Similar to ROE, ROA is affected by profit margins and asset turnover. This can be seen from the Du Pont Formula:
Return on Assets (ROA) (Q: . 20 ) = Net Income / Total Assets = / = (Net Income / Revenue) * (Revenue / Total Assets) = ( / ) * ( / ) = Net Profit Margin * Asset Turnover = * = 0.00 %
Note: The Net Income data used here is two times the semi-annual (. 20) net income data.
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Be Aware
Like ROE, ROA is calculated with only 12 months data. Fluctuations in the companys earnings or business cycles can affect the ratio drastically. It is important to look at the ratio from a long term perspective. ROA can be affected by events such as stock buyback or issuance, and by goodwill, a companys tax rate and its interest payment. ROA may not reflect the true earning power of the assets. A more accurate measurement is Return on Capital (ROC).
Many analysts argue the higher return the better. Buffett states that really high ROA may indicate vulnerability in the durability of the competitive advantage.
E.g. Raising \$43b to take on KO is impossible, but \$1.7b to take on Moodys is. Although Moodys ROA and underlying economics is far superior to Coca Cola, the durability is far weaker because of lower entry cost.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Annual Data
ROA 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
Semi-Annual Data
ROA 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
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https://datapott.com/validity-and-reliability-in-confirmatory-factor-analysis-cfa/ | 1,720,895,300,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00503.warc.gz | 180,587,460 | 19,626 | # Validity and Reliability in Confirmatory Factor Analysis (CFA)
## Validity and Reliability in Confirmatory Factor Analysis (CFA)
It is absolutely necessary to establish convergent and discriminant validity, as well as reliability, when doing a CFA. If your factors do not demonstrate adequate validity and reliability, moving on to test a causal model will be useless – garbage in, garbage out! There are a few measures that are useful for establishing validity and reliability: Composite Reliability (CR), Average Variance Extracted (AVE), Maximum Shared Variance (MSV), and Average Shared Variance (ASV). The video tutorial will show you how to calculate these values. The thresholds for these values are as follows:
Reliability
• CR > 0.7
Convergent Validity
• AVE > 0.5
Discriminant Validity
• MSV < AVE
• Square root of AVE greater than inter-construct correlations
If you have convergent validity issues, then your variables do not correlate well with each other within their parent factor; i.e, the latent factor is not well explained by its observed variables. If you have discriminant validity issues, then your variables correlate more highly with variables outside their parent factor than with the variables within their parent factor; i.e., the latent factor is better explained by some other variables (from a different factor), than by its own observed variables.
If you need to cite these suggested thresholds, please use the following:
• Hair, J., Black, W., Babin, B., and Anderson, R. (2010). Multivariate data analysis (7th ed.): Prentice-Hall, Inc. Upper Saddle River, NJ, USA.
AVE is a strict measure of convergent validity. Malhotra and Dash (2011) note that “AVE is a more conservative measure than CR. On the basis of CR alone, the researcher may conclude that the convergent validity of the construct is adequate, even though more than 50% of the variance is due to error.” (Malhotra and Dash, 2011, p.702).
• Malhotra N. K., Dash S. (2011). Marketing Research an Applied Orientation. London: Pearson Publishing.
Here is an updated video that uses the most recent Stats Tools Package, which includes a more accurate measure of AVE and
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# What does Newcomb’s paradox teach us?
## 1 - NASA Ames Research Center, MS 269-1, Moffett Field, CA 94035-1000
(650) 604-3362 (V), (650) 604-3594 (F), david.h.wolpert.nasa.gov
2 - Physics and Astronomy Department, University of California,
Irvine, CA 92692
April 7, 2009
Abstract
Newcomb’s paradox highlights an apparent conflict involving the ax-
ioms of game theory. It concerns a game in which you choose to take either
one or both of two closed boxes. However before you choose, a prediction
algorithm deduces your choice, and fills the two boxes based on that deduc-
tion. The paradox is that game theory appears to provide two conflicting
recommendations for what choice you should make in this situation. Here
we analyze Newcomb’s paradox using a recently introduced extension of
game theory in which the players set conditional probability distributions in
a Bayes net. Using this extended game theory, we show that the two game
theory recommendations in Newcomb’s scenario implicitly assume different
Bayes nets relating the random variables of your choice and the algorithm’s
prediction. We resolve the paradox by proving that these two assumed Bayes
nets are incompatible, i.e., the associated assumptions conflict. In doing this
we show that the accuracy of the algorithm’s prediction, which was the focus
of much previous work on Newcomb’s paradox, is irrelevant. We also show
that Newcomb’s paradox is time-reversal invariant; both the paradox and its
resolution are unchanged if the algorithm makes its “prediction” after you
make your choice rather than before.
## Electronic copy available at: http://ssrn.com/abstract=1381295
1 Introduction
Suppose you meet a Wise being (W) who tells you it has put \$1,000 in box A,
and either \$1 million or nothing in box B. This being tells you to either take
the contents of box B only, or to take the contents of both A and B. Suppose
further that the being had put the \$1 million in box B only if a prediction algorithm
designed by the being had said that you would take only B. If the algorithm had
predicted you would take both boxes, then the being put nothing in box B.
Presume that due to determinism, there exists a perfectly accurate prediction
algorithm. Assuming W uses that algorithm, what choice should you make? In
Table 1 we present this question as a game theory matrix involving W’s predic-
tion and your choice. Two seemingly logical answers contradict each other. The
Realist answer is that you should take both boxes, because you have free will, and
your choice occurs after W has already made its prediction. More precisely, if
W predicted you would take A along with B, then taking both gives you \$1,000
rather than nothing. If instead W predicted you would take only B, then taking
both boxes yields \$1,001,000, which again is \$1000 better than taking only B.
The Fearful answer, though, is that W designed a prediction algorithm whose an-
swer will match what you do. So you can get \$1,000 by taking both boxes or get
\$1 million by taking only box B. Therefore you should take only B.
This is Newcomb’s Paradox, a famous logical riddle stated by William New-
comb in 1960 [Nozick(1969), Gardner(1974), Bar-Hillel and Margalit(1972), Campbell and Lanning(1985)
Levi(1982), Collins(2001)]. Newcomb never published the paradox, but had long
conversations about it with with philosophers and physicists such as Robert Noz-
ick and Martin Kruskal, along with Scientific American’s Martin Gardner. Gard-
ner said after his second Scientific American column on Newcomb’s paradox ap-
peared that it generated more mail than any other column.
One of us (Benford) worked with Newcomb, publishing several papers to-
gether, and was a friend until Newcomb died in 1999. We often discussed the
paradox, which Newcomb thought would be his best remembered scientific ac-
complishment. Newcomb invented his paradox to test his own ideas, as a lapsed
Catholic: How much faith do we place in the wise being’s predictive power?
Newcomb’s said that he would just take B; why fight a God-like being? How-
ever Nozick said, “To almost everyone, it is perfectly clear and obvious what
should be done. The difficulty is that these people seem to divide almost evenly
on the problem, with large numbers thinking that the opposing half is just being
silly” [Nozick(1969)].
Nozick also pointed out that two accepted principles of game theory conflict in
Newcomb’s problem. The expected-utility principle, proceeding from the prob-
ability of each outcome, says you should take box B only. But the dominance
principle argues that if one strategy is always better, no matter what the circum-
## Electronic copy available at: http://ssrn.com/abstract=1381295
stances, then you should pick it. No matter what box B contains, you are \$1000
richer if you take both boxes than if you take B only.
Is there really a contradiction? Some philosophers argue that a perfect predic-
tor implies a time machine, since with such a machine causality is reversed, i.e.,
the future causes past events, allowing predictions to be perfect.1 Faced with New-
comb’s seemingly logical paradox, the conclusion must be that perfect prediction
is impossible.
But Nozick stated the problem specifically to exclude backward causation
(and so time travel), because his formulation demands only that the predictions
be of high accuracy, not certain. So this line of reasoning cannot resolve the is-
sue. Worse still, Nozick’s reformulation seems to imply that the (in)fallibility of
W’s prediction provides yet another conundrum, in addition to the one underlying
## 2 Game theory over Bayes nets
Central to Newcomb’s scenario is a prediction process, and its (in)fallibility. Re-
cent work has revealed deep formal connections between prediction and observa-
tion. Amongst other things, this work proves that any given prediction algorithm
must fail on at least one prediction task [Binder(2008), Wolpert(2008)]. Unfor-
tunately, that result doesn’t directly resolve Newcomb’s paradox. However its
proof requires an extension of game theory. And as we demonstrate below, that
extension can be used to resolve Newcomb’s paradox.
In game theory there are several “players”, each with their own preferences
over the values of an underlying set of game variables, {X j }. Every player has their
own “move set”, where each move is a probability distribution relating some of the
variables {X j }. To play the game, the players all independently choose a move (i.e.,
choose a distribution) from their respective move sets. The moves sets are care-
fully designed so that every such joint move by the players uniquely specifies a le-
gal joint probability distribution relating the game’s variables [Fudenberg and Tirole(1991),
Myerson(1991), Osborne and Rubenstein(1994), Koller and Milch(2003)].
A richer mathematics arises if we expand the move sets of the players, so
that some joint moves would violate the laws of probability, and therefore are
impossible. It is this mathematics that is used to prove the fallibility of prediction
in [Binder(2008), Wolpert(2008)].
1
Interestingly, near when Newcomb devised the paradox, he also coauthored a paper prov-
ing that a tachyonic time machine could not be reinterpreted in a way that precludes such para-
doxes [Benford et al.(1970)Benford, Book, and Newcomb]. The issues of time travel and para-
doxes are intertwined.
3
What happens if we apply this mathematics to Newcomb’s paradox? There
are two game variables that are central to Newcomb’s paradox: the God-like being
W’s prediction, g, and the choice you actually make, y. So the player moves will
involve the distribution relating those variables. Since there are only two variables,
there are two ways to decompose that joint probability. These two decompositions
turn out to correspond to the two recommendations for how to answer Newcomb’s
question, one matching the reasoning of Realist and one matching Fearful.
The first way to decompose the joint probability is
## P(y, g) = P(g | y)P(y) (1)
(where we define the right-hand side to equal 0 for any y such that P(y) = 0). Such
a decomposition is known as a “Bayes net” having two “nodes” [Pearl(2000)].
The unconditioned distribution, P(y) is identified with the first, “parent” node,
and the conditional distribution, P(g | y), is identified with the second, “child”
node.
This Bayes net can be used to express Fearful’s reasoning. Fearful interprets
the statement that “W designed a perfectly accurate prediction algorithm” to imply
that W has the power to set the conditional distribution in the child node of the
Bayes net, P(g | y), to anything it wants (for all y such that P(y) , 0). More
precisely, since the algorithm is “perfectly accurate”, Fearful presumes that W
chooses to set P(g | y) = δg,y , the distribution that equals 1 if g = y, zero otherwise.
So Fearful presumes that there is nothing you can do that can affect the values of
P(g | y) (for all y such that P(y) , 0). Instead, you get to choose the unconditioned
distribution in the parent node of the Bayes net, P(y). Intuitively, this choice
constitutes your “free will”.
Fearful’s interpretation of Newcomb’s paradox specifies what aspect of P(y, g)
you can choose, and what aspect is instead chosen by W. Those choices — P(y)
and P(g | y), respectively — are the “moves” that you and W make. It is im-
portant to note that these moves by you and W do not directly specify the two
variables y and g. Rather the moves you and W make specify two different
distributions which, taken together, specify the full joint distribution over y and
g [Koller and Milch(2003)]. This kind of move contrasts with the kind considered
in decision theory [Berger(1985)] or causal nets [Pearl(2000)], where the moves
are direct specifications of the variables (which here are g and y).
In game theory, your task is to make the move that maximizes your expected
payoff under the associated joint distribution. For Fearful, this means choosing
the P(y) that maximizes your expected payoff under the P(y, g) associated with
that choice. Given Fearful’s presumption that the Bayes net of Eq. 1 underlies the
game and that you get to set the distribution at the first nod3e, for you to maximize
expected payoff you should choose P(y) = δy,B , i.e., you should make choice
4
B with probability 1. Your doing so results in the joint distribution P(y, g) =
δg,y δy,B = δg,B δy,B , with payoff 1, 000, 000. This is the formal justification of
Fearful’s recommendation.
The second way to decompose the joint probability is
## P(y, g) = P(y | g)P(g) (2)
(where we define the right-hand side to equal 0 for any g such that P(g) = 0). In
the Bayes net of Eq. 2, the unconditioned distribution identified with the parent
node is P(g), and the conditioned distribution identified with the child node is
P(y | g). This Bayes net can be used to express Realist’s reasoning. Realist
interprets the statement that “your choice occurs after W has already made its
prediction” to mean that you can choose any distribution h(y) and then set P(y | g)
to equal h(y) (for all g such that P(g) , 0). This is how Realist interprets your
having “free will”. (Note that this is a different interpretation of “free will”’ from
the one made by Fearful.) Under this interpretation, W has no power to affect
P(y | g). Rather W gets to set the parent node in the Bayes net, P(g). For Realist,
this is the distribution that you cannot affect. (In contrast, in Fearful’s reasoning,
you set a non-conditional distribution, and it is the conditional distribution that
you cannot affect.)
Realist’s interpretation of Newcomb’s paradox specifies what it is you can fix
concerning P(y, g), and what is fixed by W. Just like under Fearful’s reasoning,
under Realist’s reasoning the “moves” you and W make do not directly specify
the variables g and y. Rather the moves by you and W specify two distributions
which, taken together, specify the full joint distribution. As before, your task is
to choose your move — which now is h(y) — to maximize your expected payoff
under the associated P(y, g). Given Realist’s presumption that the Bayes net of
Eq. 2 underlies the game and that you get to set h, you should choose h(y) = P(y |
g) = δy,AB , i.e., you should make choice AB with probability 1. Doing this results
in the expected payoff 1, 000 P(g = AB) + 1, 001, 000 P(g = B), which is your
maximum expected payoff no matter what the values of P(g = AB) and P(g = B)
are. This is the formal justification of Realist’s recommendation.2
What happens if we try to merge the Bayes net that Fearful presumes to under-
lie the game with the Bayes net that Realist presumes to underlie the game? More
2
In Realist’s Bayes net, given the associated restricted possible form of P(y | g), g and y are
“causally independent”, to use the language of causal nets [Pearl(2000)]. This is consistent with
interpreting Newcomb’s scenario as the game in Table 1. In contrast, in Fearful’s Bayes net, y
“causally influences” g. To cast this kind of causal influence in terms of conventional game theory,
we would have to replace the game in Table 1 with an extensive form game in which you first set
y, and then W moves, having observed y. This alternative game is incompatible with Newcomb’s
stipulation that W moves before you do, not after. This is one of the reasons why it is necessary to
use extended game theory rather than conventional game theory to formalize Fearful’s reasoning.
5
formally, what game arises if we combine your move set under Fearful’s presump-
tion of the underlying Bayes net with your move set under Realist’s presumption,
and do the same for W? As we now how, combining move sets this way gives an
“extended game” of the sort considered in [Wolpert(2008)], with the same kind of
impossibility result as the extended game in [Wolpert(2008)].
First, if W’s move sets P(g | y), as under Fearful’s presumption, then some
of your moves under Realist’s presumption become impossible. (This is true for
almost any P(g | y) that W might choose, and in particular even if W does not
predict perfectly.) More precisely, if P(g | y) is set by W, then the only way that
P(y | g) can be g-independent is if it is one of the two delta functions, δy,AB or δy,B .
(See the appendix for a formal proof.) This contradicts Realist’s presumption that
you can set P(y | g) to any h(y) you desire.3
Similarly, if P(g | y) is fixed by W, as under Fearful’s presumption, then your
(Realist) choice of h affects P(g). In fact, your choice of h fully specifies P(g).4
This contradicts Realist’s presumption that it is W’s move that sets P(g), indepen-
dent of you.
Conversely, if you can set P(y | g) to be an arbitrary g-independent distribution
(as Realist presumes), then what you set it to may affect P(g | y) (in violation of
Fearful’s presumption that P(g | y) is set exclusively by W). In other words, if
your having “free will” means what it does to Realist, then you have the power to
change the prediction accuracy of W (!). As an example, if you set P(y = AB |
g) = 3/4 for all g’s such that P(g) , 0, then P(g | y) cannot equal δg,y .
The resolution of Newcomb’s paradox is now immediate: You can be free to
set P(y) however you want, with P(g | y) set by W, as Fearful presumes, or, as
Realist presumes, you can be free to set P(y | g) to whatever distribution h(y) you
want, with P(g) set by W. It is not possible to play both games simultaneously.5
We emphasize that this impossibility arises for almost any P(g | y) choice
by W, i.e., no matter how accurately W predicts. This means that the stipulation
in Newcomb’s paradox that W predicts perfectly is a red herring. (Interestingly,
Newcomb himself did not insist on such perfect prediction in his formulation of
3
Note that of the two δ functions you can choose in this variant of Newcomb’s scenario, it is
better for you to choose h(y) = δy,B , resulting in a payoff of 1, 000, 000. So your optimal response
to Newcomb’s question for this variant is the same as if you were Fearful.
4
For example, if you set h(y) = δy,AB , then P(g) = δg,AB , and if you set h(y) = δy,B , then
P(g) = δg,B .
5
In a variant of Newcomb’s question, you first choose one of these two presumption, and then
set the associated distribution. If the pre-fixed distribution P(g | y) arising in the first presumption
is δg,y , then your optimal responses depend on the pre-fixed distribution P(g) arising in the second
presumption— a distribution that is not specified in Newcomb’s question. If P(g) obeys P(g =
B) > .999, then your optimal pair of choices are first to choose to set the distribution P(y | g) to
some h(y), and then to set h(y) = δy,AB . If this condition is not met, you should first choose to set
P(y), and then set it to δy,AB .
6
the paradox, perhaps to avoid the time paradox problems.) The crucial impossi-
bility implicit in Newcomb’s question is the idea that at the same time you can
arbitrarily specify “your” distribution P(y | g) and W can arbitrarily specify “his”’
distribution P(g | y). In fact, neither of you two can set your distribution without
possibly affecting the other’s distribution; you and W are inextricably coupled.
Note also that no time variable occurs in our analysis of Newcomb’s paradox.
So that analysis is time-reversal invariant. This means that both the paradox and
its resolution are unchanged if the prediction occurs after your choice rather than
before it. This is even the case if the “prediction” algorithm directly observes
your choice. See [Wolpert(2008)] for more on the equivalence of observation and
prediction and the time-reversal invariance of both.
Newcomb’s paradox has been so vexing that it has led some to resort to non-
Bayesian probability theory in their attempt to understand it [Gibbard and Harper(1978),
Hunter and Richter(1978)], some to presume that payoff must somehow depend
on your beliefs as well as what’s under the boxes [Geanakoplos(1997)], and has
even even led some to claim that quantum mechanics is crucial to understanding
the paradox [Piotrowski and Sladkowski(2002)]. This is all in addition to work on
the paradox based on now-discredited formulations of causality [Jacobi(1993)].
Our analysis shows that the resolution of Newcomb’s paradox is in fact quite
simple. Newcomb’s paradox takes two incompatible interpretations of a question,
with two different answers, and makes it seem as though they are the same inter-
pretation. The lesson of Newcomb’s paradox is just the ancient verity that one
must carefully define all one’s terms.
## ACKNOWLEDGEMENTS: We would like to thank Mark Wilber for helpful
APPENDIX:
In the text, it is claimed that if P(g | y) is pre-fixed, then the only way that P(y | g)
can be g-independent is if it is one of the two delta functions, δy,AB or δy,B . To see
why this is true, combine Eq.’s 1 and 2 of the text to get
## P(y | g)P(g) = P(g | y)P(y).
If for all g such that P(g) , 0, P(y | g) = h(y) for some distribution h, then we can
sum both sides over the two values of g, getting P(y) = h(y). Plugging this back in
shows that for any y such that h(y) , 0, P(g | y) must equal P(g). If there were two
such y’s in the support of h, then P(g | y) would have to be the same distribution
over g for both of those y’s. This is not the case for a perfectly accurate P(g | y)
7
though (for which P(g | y) = δg,y ), nor is it the case for almost all other P(g | y)’s.
The only way to avoid this contradiction is for you to set h(y) so that it equals 0
for one of the two y’s. QED.
References
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dox revisited. British Journal of Philosophy of Science, 23:295–304, 1972.
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W. Newcomb. Physical Review D, 2:263, 1970.
## [Berger(1985)] J. M. Berger. Statistical Decision theory and Bayesian Analysis.
Springer-Verlag, 1985.
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## [Campbell and Lanning(1985)] R. Campbell and S. Lanning. Paradoxes of Ra-
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kinds of expected utility. In Foundations and applications of decision theory.
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## [Hunter and Richter(1978)] D. Hunter and R. Richter. Counterfactuals and new-
comb’s paradox. Synthese, 39:249–261, 1978.
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Decision, 35:1–17, 1993.
8
[Koller and Milch(2003)] D. Koller and B. Milch. Multi-agent influence dia-
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## [Myerson(1991)] Roger B. Myerson. Game theory: Analysis of Conflict. Harvard
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Netherland, 1969.
## [Osborne and Rubenstein(1994)] M. Osborne and A. Rubenstein. A Course in
Game Theory. MIT Press, Cambridge, MA, 1994.
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University Press, 2000.
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kowski. Quantum solution to the newcomb’s paradox.
http://ideas.repec.org/p/sla/eakjkl/10.html, 2002.
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1257–1281, 2008. More recent version at http://arxiv.org/abs/0708.1362.
9
Choose AB Choose B
## Predict B: 1, 001, 000 1, 000, 000
Table 1: The payoff to you for the four combinations of your choice and W’s
prediction.
10
Short title: Newcomb’s paradox resolved
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based on an earlier version of the Wolfram Language.
# DatePlus
DatePlus gives the date n days after date. DatePlusgives the date n units after date. DatePlusgives a date offset by units of each specified size. DatePlus[n]gives the date n days after the current date. DatePlus[offset]gives the date with the specified offset from the current date.
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{y,m,d} year, month, day {y,m} the first day of the specified month {y} January 1 of the year y "string" date as a string () {y,m,d,h,m,s} precise time time absolute time specification
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• DatePlus gives results in the same general format as date.
• When date is a list, the result has the same length as date, possibly extended to include the smallest unit in offset. »
Add 35 days to January 1, 2009:
Use a date string as input:
Subtract from a date:
Add 14 weeks to a date:
Add 34 days to the current time:
Add 35 days to January 1, 2009:
Out[1]=
Use a date string as input:
Out[1]=
Subtract from a date:
Out[1]=
Add 14 weeks to a date:
Out[1]=
Add 34 days to the current time:
Out[1]=
Scope (12)
DatePlus can take dates in the standard format of DateList:
DatePlus can take dates in any format supported by DateString:
If a date is given as , DatePlus returns in the same form:
is interpreted as the first day of the specified month:
{y} is interpreted as January 1 of the specified year:
Add 10 months to a date:
Add 40 weeks to a date:
Add 7 weeks and 2 days:
Lists are extended to include smaller offset units:
Add 1 year to a date list:
Add 1 month and 15 days to the specified date list: | 473 | 1,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-39 | longest | en | 0.814192 |
https://kr.mathworks.com/matlabcentral/cody/problems/1087-magic-is-simple-for-beginners/solutions/520014 | 1,579,765,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00049.warc.gz | 528,520,666 | 15,684 | Cody
# Problem 1087. Magic is simple (for beginners)
Solution 520014
Submitted on 3 Nov 2014 by Guillaume
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 3; y_correct = 15; assert(isequal(magic_sum(n),y_correct))
2 Pass
%% n = 5; y_correct = 65; assert(isequal(magic_sum(n),y_correct))
3 Pass
%% n = 7; y_correct = 175; assert(isequal(magic_sum(n),y_correct))
4 Pass
%% n = 8; y_correct = 260; assert(isequal(magic_sum(n),y_correct))
5 Pass
%% n = 20; y_correct = 4010; assert(isequal(magic_sum(n),y_correct))
6 Pass
%% n = 100; y_correct = 500050; assert(isequal(magic_sum(n),y_correct))
7 Pass
%% n = 200; y_correct = 4000100; assert(isequal(magic_sum(n),y_correct))
8 Pass
%% n = 1000; y_correct = 500000500; assert(isequal(magic_sum(n),y_correct)) | 299 | 902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-05 | latest | en | 0.506932 |
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