List comprehensions: Difference between revisions
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=={{header|Picat}}== |
=={{header|Picat}}== |
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===List comprehensions=== |
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<lang Picat>pyth(N) = [[A,B,C] : A in 1..N, B in A..N, C in B..N, A**2 + B**2 == C**2].</lang> |
<lang Picat>pyth(N) = [[A,B,C] : A in 1..N, B in A..N, C in B..N, A**2 + B**2 == C**2].</lang> |
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===Array comprehensions=== |
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Picat also has array comprehensions. Arrays are generally used for faster access (using <code>{}</code> instead of <code>[]</code>). |
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<lang Picat>pyth(N) = {{A,B,C} : A in 1..N, B in A..N, C in B..N, A**2 + B**2 == C**2}.</lang> |
<lang Picat>pyth(N) = {{A,B,C} : A in 1..N, B in A..N, C in B..N, A**2 + B**2 == C**2}.</lang> |
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===findall/2=== |
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A related construct is findall/2 to get all solutions for the specific goal at the second parameter. Here this is shown with member/2 for generating the numbers to test (which for this task is fairly inefficient). |
A related construct is <code>findall/2</code> to get all solutions for the specific goal at the second parameter. Here this is shown with <code>member/2</code> for generating the numbers to test (which for this task is fairly inefficient). |
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<lang>pyth(N) = findall([A,B,C], (member(A,1..N), member(B,1..N), member(C,1..N), A < B, A**2 + B**2 == C**2)).</lang> |
<lang>pyth(N) = findall([A,B,C], (member(A,1..N), member(B,1..N), member(C,1..N), A < B, A**2 + B**2 == C**2)).</lang> |
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=={{header|PicoLisp}}== |
=={{header|PicoLisp}}== |
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