Problem 55280. Count estrangements

Created by ChrisR

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<div style = "text-align: start; line-height: 20.4333px; min-height: 0px; white-space: normal; color: rgb(0, 0, 0); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: 400; text-decoration: rgb(0, 0, 0); white-space: normal; "><div style="block-size: 219px; display: block; min-width: 0px; padding-block-start: 0px; padding-top: 0px; perspective-origin: 407px 109.5px; transform-origin: 407px 109.5px; vertical-align: baseline; "><div style="block-size: 126px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; perspective-origin: 384px 63px; text-align: left; transform-origin: 384px 63px; white-space: pre-wrap; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; "><span style="block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 384px 7.79167px; transform-origin: 384px 7.79167px; unicode-bidi: normal; "><span style="">Recently I made a puzzle for my wife that included a cryptogram, which involves an arrangement of the letters A through Z. I used MATLAB (of course) to permute the letters, but in the first arrangement, H was coded as H. I tried again until MATLAB gave me a </span></span><span style="block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 41.2417px 7.79167px; transform-origin: 41.2417px 7.79167px; unicode-bidi: normal; "><span style="font-style: italic; ">derangement</span></span><span style="block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 306.875px 7.79167px; transform-origin: 306.875px 7.79167px; unicode-bidi: normal; "><span style="">, a permutation such that none of the letters was in its original position. I later noticed that the letter I was coded as S and S was coded as I. Though I used that permutation, I really wanted one in which no two elements are simply swapped. I called such a permutation an </span></span><span style="block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 42.7917px 7.79167px; transform-origin: 42.7917px 7.79167px; unicode-bidi: normal; "><span style="font-style: italic; ">estrangement</span></span><span style="block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 165.708px 7.79167px; transform-origin: 165.708px 7.79167px; unicode-bidi: normal; "><span style="">, and although I later learned of a more technical and mathematical description, I will keep my name. </span></span></div><div style="block-size: 84px; font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-block-end: 9px; margin-block-start: 2px; margin-bottom: 9px; margin-inline-end: 10px; margin-inline-start: 4px; margin-left: 4px; margin-right: 10px; margin-top: 2px; perspective-origin: 384px 42px; text-align: left; transform-origin: 384px 42px; white-space: pre-wrap; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; "><span style="block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 256.575px 7.79167px; transform-origin: 256.575px 7.79167px; unicode-bidi: normal; "><span style="">Write a function to count estrangements—i.e., the permutations of elements in a 1x</span></span><span style="font-family: &quot;STIXGeneral&quot;, &quot;STIXGeneral-webfont&quot;, serif; font-style: italic; font-weight: 400; color: rgb(0, 0, 0);">n</span><span style="block-size: auto; display: inline; margin-block-end: 0px; margin-block-start: 0px; margin-bottom: 0px; margin-inline-end: 0px; margin-inline-start: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; perspective-origin: 122.175px 7.79167px; transform-origin: 122.175px 7.79167px; unicode-bidi: normal; "><span style=""> vector such that (1) no element is in its original position and (2) no two elements are simply swapped. For example, if the vector is [1 2 3 4], then [3 2 4 1] and [4 1 3 2] would not be allowed under condition 1 and [4 3 2 1] and [2 1 4 3] would not be allowed under condition 2. Return the count as a string. </span></span></div></div></div>

Recently I made a puzzle for my wife that included a cryptogram, which involves an arrangement of the letters A through Z. I used MATLAB (of course) to permute the letters, but in the first arrangement, H was coded as H. I tried again until MATLAB gave me a derangement, a permutation such that none of the letters was in its original position. I later noticed that the letter I was coded as S and S was coded as I. Though I used that permutation, I really wanted one in which no two elements are simply swapped. I called such a permutation an estrangement, and although I later learned of a more technical and mathematical description, I will keep my name. 
Write a function to count estrangements—i.e., the permutations of elements in a 1x vector such that (1) no element is in its original position and (2) no two elements are simply swapped. For example, if the vector is [1 2 3 4], then [3 2 4 1] and [4 1 3 2] would not be allowed under condition 1 and [4 3 2 1] and [2 1 4 3] would not be allowed under condition 2. Return the count as a string.

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Last Solution submitted on Apr 20, 2023

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William on 19 Apr 2023

Sorry, but I submitted a phoney solution that passes all of the tests. Since there are no assert() functions, just about anything will pass!

William on 20 Apr 2023

Okay, I have now submitted a correct (I think!) solution, but I think test #14 has an error.

Dyuman Joshi on 22 Apr 2023

William, I have updated the test suit to include the assert() command now and rescored the solutions.

And yes, the test case #14 was incorrect and I have commented it out for now. I'm leaving it to Chris to update that particular test case.

ChrisR on 24 Apr 2023

I don't know how I missed the asserts, but in my test file on my computer, they are missing. Thanks, Dyuman, for fixing that problem.

I also fixed Test 14.

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Problem 55280. Count estrangements

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Problem 55280. Count estrangements

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