Problem 45193. Fun with permutations

Created by William

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{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?><w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>There are factorial(N) permutations of the numbers from 1 to N. For each of these permutations, we can find the set of indexes j that will return the numbers to their original order, 1:N. For example, if we generate a random permutation p with p = randperm(1:N) then the set of indices j that will return p to the original order p(j) = 1:N may be found by [~,j] = sort(p).</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>It sometimes happens that this set of indices j and the corresponding permutation p are the same. For example, if p = 1:N, then j = 1:N as well; and if p = N:-1:1, then j is also N:-1:1. However, for N = 3, there are 6 permutations in all, and it happens that only 4 of them have this property.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>Your task is to determine, given the value of N, the number of permutations p of the array 1:N for which the permutation p and the indexes j are the same.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>input: N, the range of the integers, output: The number of permutations of 1:N for which isequal(p,j)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>HINT: For N &lt; 12, you can probably solve this by checking each of the individual permutations, but for larger N this may prove to be too time-consuming.</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

There are factorial(N) permutations of the numbers from 1 to N. For each of these permutations, we can find the set of indexes j that will return the numbers to their original order, 1:N. For example, if we generate a random permutation p with p = randperm(1:N) then the set of indices j that will return p to the original order
p(j) = 1:N may be found by [~,j] = sort(p).It sometimes happens that this set of indices j and the corresponding permutation p are the same. For example, if p = 1:N, then j = 1:N as well; and if p = N:-1:1, then
j is also N:-1:1. However, for N = 3, there are 6 permutations in all, and it happens that only 4 of them have this property.Your task is to determine, given the value of N, the number of permutations p of the array 1:N for which the permutation p and the indexes j are the same.input: N, the range of the integers,
output: The number of permutations of 1:N for which isequal(p,j)HINT: For N &lt; 12, you can probably solve this by checking each of the individual permutations, but for larger N this may prove to be too time-consuming.

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Last Solution submitted on Jan 06, 2021

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ChrisR on 6 Jan 2021 at 3:58

I couldn't figure out why my numbers were larger than the answers in the test suite and finally realized that most of those answers were products of the unique factors of the numbers of permutations.

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Problem 45193. Fun with permutations

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