Talk:Anonymous recursion: Difference between revisions

Revision as of 22:30, 27 April 2015

Revision as of 18:39, 11 May 2020

Line 71:

I came up with a simpler lambda calculus solution for factorial. All I did was take the anonymous recursion version, and replace "lambda a, b" with "lambda a: lambda b:", and "(a, b)" with "(a)(b)".

<sourcesyntaxhighlight lang=python>

(lambda f: lambda x: f(f)(x)) \

(lambda f: lambda x: 1 if x == 0 else x * f(f)(x-1))

</syntaxhighlight>

</source>

But, this is original research I guess, so I won't put it on the page. Also the existing explanation might help people trying to understand the Y combinator. <small class="autosigned">—&nbsp;Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/15.203.233.79|15.203.233.79]] ([[User talk:15.203.233.79|talk]]) 22:19, 27 April 2015 (UTC)</small><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->