Problem 803. Twist 'n' Match

Created by内德·古利（Ned Gulley）

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{“关系”：[{“ RelativeType”：“ http：//schemas.mathworks.com/matlab/matlab/code/2013/relationships/document'“：” /matlab/document.xml“}，{“ RelatessType”：“ http：//schemas.mathworks.com/matlab/matlab/code/2013/relationships/outputs/output”，“ targetMode”，“ targetMode”：“ targetMode”：“：”“ RID2”，“ TARGET”：“/MATLAB/output.xml”}]，“ parts”：[{“ parturi”：“/matlab/document.xml”，“关系”：[]：[]，“ contentType”：“：”：“”application/vnd.mathworks.matlab.code.document+xml“，” content“：” <？xml版本= \“ 1.0 \” encoding = \“ utf-8 \”？> \ n <w：document xmlns：w documn xmlns：w= \“ http：//schemas.openxmlformats.org/wordprocessingml/2006/main \”> <w：body> <w：p> <w：p> <w：ppr> <w：pstyle w：pstyle w：pstyle w：val = \ \ v \ \'> </w：ppr> <w：r> <w：t>给定的n和m，构造一个n-n-n矩阵a，使得a当旋转90度并与自身进行比较时，在恰好的m ploce中匹配。<w：r> <w：t>匹配数m的数量如下：</w：t> </w：r> </w：p> <w：p> <w：p> <w：ppr> <w：pstyle w：val = \“代码\”/> </w：ppr> <w：r> <w：t> <！[cdata [m = nnz（rot90（a）== a）]]]> </w：t> </w：r> </w：p><w：p> <w：ppr> <w：pstyle w：val = \“ text \”/> </w：ppr> <w：r> <w：t>您的答案A显然不是唯一的。它必须只符合上面所述的标准。</w：t> </w：r> </w：p> <w：p> <w：ppr> <w：pstyle w：val = \ v \ \“ text \”/> </w：ppr> <w：r> <w：t>示例：</w：t> </w：r> </w：p> <w：p> <w：ppr> <w：ppr> <w：pstyle w：val = \“代码\”/> </w：pPr><w:r><w:t><![CDATA[ Input n = 2, m = 1\n One possible output: a = [ 1 2 \n 1 3 ]\n\n Input n = 3, m = 7\n One possible output: a = [ 0 1 1\n 1 1 1\n 1 1 1 ]]]></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>"}]}

Solve

Solution Stats

216 Solutions
63 Solvers

LastSolutionsubmitted on Apr 12, 2022

Last 200 Solutions

Problem Comments

6 Comments

6 Comments

Show 3个较旧的评论

让·玛丽·塞恩蒂尔（Jean-Marie Sainthillier）2017年2月21日

许多解决方案不适用万博尤文图斯于n = 5和m = 0

让·玛丽·塞恩蒂尔（Jean-Marie Sainthillier）on 22 Feb 2017

一点也不...
对于n-n-n矩阵，如果n为奇数，则为m = 1。

Asif Newazon 1 Feb 2020

ur problems are really interesting

内德·古利（Ned Gulley）2020年2月3日

Thanks Asif!

Rafael S.T. Vieiraon 1 Aug 2020

This problem has some issues because some pairs (n,m) do not have a solution. For instance, it is impossible to find a matrix such that m = n^2 -1 since a rotation pivot does not move. Moreover, there are floor((n^2)/4) cycles of numbers, and when we match the cycle beginning with its end, it creates two matches. It is possible to do some manipulations to have an m greater than n^2 - floor((n^2)/4), but not always.

A C2020年10月26日

can you give an example of a matrix before and after?

Solution Comments

Solution 2765800

1 Comment

1 Comment

Rafael S.T. Vieiraon 1 Aug 2020

我提到的一些情况下，我的代码失败了。当前的领先解决方案做得更好（找到我的代码无法找到的一些解决方案），但有时它会进入无限循环。万博尤文图斯

解决方案2114985

1 Comment

1 Comment

Asif Newazon 2 Feb 2020

i had no idea this was gonna work :p

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