Talk:Anonymous recursion: Difference between revisions
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:The J solution is incorrect because it does not check for a negative sign, it simply implements the fibonacci function. Using $: is problematic because it is a reference to the largest verb which means that it will re-execute the entire function including the check if such a check is included. I have fixed the Mathematica example so that the check is performed only once, #0 is local to the anonymous function it is referenced in, this is the fibonacci function and not the function with the check. --[[User:Zeotrope|Zeotrope]] 16:18, 8 January 2011 (UTC) |
:The J solution is incorrect because it does not check for a negative sign, it simply implements the fibonacci function. Using $: is problematic because it is a reference to the largest verb which means that it will re-execute the entire function including the check if such a check is included. I have fixed the Mathematica example so that the check is performed only once, #0 is local to the anonymous function it is referenced in, this is the fibonacci function and not the function with the check. --[[User:Zeotrope|Zeotrope]] 16:18, 8 January 2011 (UTC) |
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::Since the J solution acts like an identity when called with negative arguments the check for negative numbers can be performed after the fibonacci function is called, signalling a domain error if the result is negative: <lang J>fibN =: 3 : '[:^:(0&>) (-&2 +&$: -&1)^:(1&<) M. y'"0</lang> --[[User:Zeotrope|Zeotrope]] |
::Since the J solution acts like an identity when called with negative arguments the check for negative numbers can be performed after the fibonacci function is called, signalling a domain error if the result is negative: <lang J>fibN =: 3 : '[:^:(0&>) (-&2 +&$: -&1)^:(1&<) M. y'"0</lang> Alternatively using a control structure: <lang J>(3 : 'if. y<0 do. <''negative argument'' else. <(-&2 +&$: -&1)^:(1&<) M. y end.'"0)</lang> --[[User:Zeotrope|Zeotrope]] 19:48, 8 January 2011 (UTC) |
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