I really like this puzzle. Here is another implementation, with the core idea
stolen from John Randall, who wrote:
>Here's another suggestion, using the binomial coefficients mod 2.
What's nice about this implementation is the core logic is so "loopless". Not
even the loops implicit in ^: " &.> /\ the powering functions of & &: etc.
Ok, you've got !/~ but that doesn't "look loopy" to me.
tri =. [: |: 2 | !/~@:i.
tri_txt =. i.&.- |."_1 ":@:tri
clean =. ([: ~:/\ [: </\ '1' = ])`(' '&,:) }
triangle =: clean@:tri_txt f.
triangle 10
1
1 1
1 0 1
1 1 1 1
1 0 0 0 1
1 1 0 0 1 1
1 0 1 0 1 0 1
1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0 1
1 1 0 0 0 0 0 0 1 1
I'm interested in other ways to transform tri N into triangle N . If, in
tri_txt , we substitute rotate (|.) for shift (|.!.) we can get another
formulation:
tri_txt =: i.&.- |.!.' '"_1 ":@:tri
clean =. ([: +./\."_1 '1' = ])`(' '&,:) }
NB.clean =. ([: +./\. '1' = ])`(' '&,:) }&.|: NB. Equivalent
to preceding line.
But that's not as symmetric, and the speedup is negligible. What are some
other, faster, leaner, cooler or more elegant ways to transform the output of
tri into the output of triangle ?
-Dan
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