Ramsey's theorem: Difference between revisions
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(Updated and hopefully fixed the D entry) |
(→{{header|J}}: test if correct) |
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=={{header|J}}== |
=={{header|J}}== |
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{{incorrect|J|The task has been changed to also require demonstrating that the graph is a solution.}} |
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Interpreting this task as "reproduce the output of all the other examples", then here's a stroll to the goal through the J interpreter: <lang j> i.@<.&.(2&^.) N =: 17 NB. Count to N by powers of 2 |
Interpreting this task as "reproduce the output of all the other examples", then here's a stroll to the goal through the J interpreter: <lang j> i.@<.&.(2&^.) N =: 17 NB. Count to N by powers of 2 |
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1 2 4 8 |
1 2 4 8 |
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1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 _ 1 |
1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 _ 1 |
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1 1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 _</lang> |
1 1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 _</lang> |
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Yes, this is really how you solve problems in J. |
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To test if all combinations of 4 rows and columns contain both a 0 and a 1 |
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<lang j> |
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comb=: 4 : 0 M. NB. All size x combinations of i.y |
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if. (x>:y)+.0=x do. i.(x<:y),x else. (0,.x comb&.<: y),1+x comb y-1 end. |
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) |
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NB. returns 1 iff the subbmatrix of y consisting of the columns and rows labelled x contains both 1 and 0 |
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checkRow =. 4 : 0 "1 _ |
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*./ 0 1 e. ,x{"1 x{y |
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) |
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*./ (4 comb 17) checkRow a |
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1 |
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</lang> |
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=={{header|Mathematica}}== |
=={{header|Mathematica}}== |
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