Sudoku Solver in 140 bytes

For those of you who haven't heard of code golf, you should check out <a href="http://140byt.es/" rel="nofollow">http://140byt.es/</a> The point is more to learn more about the language than to make something practical so come at it in a lighthearted way. In this case, the problem is actually pretty damn hard.<p>140 Bytes is Jed Schmidt's side project where people compete to make useful functions as small as possible. Jed is kind of a Javascript powerhouse and also happens to be a really nice guy.<p><a href="https://github.com/jed" rel="nofollow">https://github.com/jed</a><p><a href="https://github.com/jed/140bytes/wiki/Byte-saving-techniques" rel="nofollow">https://github.com/jed/140bytes/wiki/Byte-saving-techniques</a><p><i>Edited to add..</i> As a curiosity, Jed might have the most-forked gist on github: <a href="https://gist.github.com/962807" rel="nofollow">https://gist.github.com/962807</a>

Here's one half the size in k:<p><pre><code> p,:3/:_(p:9\:!81)%3 s:{*(,x)(,/{@[x;y;:;]'&#38;21=x[&#38;|/p[;y]=p]?!10}')/&#38;~x} </code></pre> <a href="http://thesweeheng.wordpress.com/2008/11/30/more-sudoku-solvers-in-k-and-q/" rel="nofollow">http://thesweeheng.wordpress.com/2008/11/30/more-sudoku-solv...</a>

That's really neat, a sudoku solver that fits in a tweet.<p>The license is even nicer, that's a real gem, and I think it ought to be submitted for formal approval as an open source license.

A fork for 139 byes: <a href="https://gist.github.com/1247640" rel="nofollow">https://gist.github.com/1247640</a><p>And I had never heard of "DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE". (<a href="http://en.wikipedia.org/wiki/WTFPL" rel="nofollow">http://en.wikipedia.org/wiki/WTFPL</a>)

This won't work for anything but the most basic puzzles. Brute force just constitutes basic gameplay. To solve a moderately advanced Sudoku puzzle you need to compare potential candidates in groups. Then it gets even more advanced with a handful of trial and error techniques.

Could someone help with the syntax on line 24:<p>g&#38;&#38;R(a)<p>I simply don't understand it as a statement and I don't get in the context of the for-loop

4clojure has a code golf league for its exercise: <a href="http://4clojure.com/" rel="nofollow">http://4clojure.com/</a>

I want to see a minimal solution in APL.

I have to admit, I have never seen<p>.) for-loops used this way<p>.) an OR-operator in a for loop<p>.) +a as a shortcut for a.toString()<p>.) syntax like g&#38;&#38;R(a) (line 24)<p>.) syntax like a[j^i==j] (line 29)<p>Where can I learn/read about advanced stuff like this, searching for this information on google is very hard, as I lack the correct terms ...

What's the minimal `J` solution?