Appending a counter is the easiest. Since we never reverse more elements
than there are in the list, we don't even have to bother with removing
the counter. Here's my code.
topswap =: [: {: [: (>:@{: _1} ((|.@{. , }.)~ {.))^:(1~:{.)^:_ ,&0
topswap 2 4 1 3
4
The part inside the while loop is (>:@{: _1} ((|.@{. , }.)~ {.)). You'll
recognize the right half of the fork as topswap from your message, and
the rest simply replaces the last element with its successor.
Then we only have to maximize this over all permutations. To generate
all permutations of (>:i.y), we have
allperms =: i.@! A. >:@i.
So one solution to the problem is
topswaps =: [: >./ topswap"1@:allperms
Of course this is rather inefficient. For an efficient solution I advise
copying the C algorithm.
Marshall
On Mon, Nov 26, 2012 at 09:19:37PM -0500, David Ward Lambert wrote:
> Reference: http://rosettacode.org/wiki/Topswops
> What are good ways to count iterations?
>
> While=: 2 : 'u^:v^:_'
>
> topswop=: (|.@:{. , }.)~ {.
>
> topswop While (1 ~: {.) 2 4 1 3
> 1 2 3 4
>
> These ideas seem like structural nightmares:
> * separate data and counter with boxes;
> * append a counter to the data;
> * explicit code with a global noun;
> * +/@:(0 = (0,~topswop)While(1 ~: {.)) NB. <laughs>.
>
> The boxes or explicit code seem most general.
>
> Thank you, Dave.
>
> Bonus: find a definition topswop using C.
>
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