Here is a recursive solution, it stops processing when the alternating
invariant becomes false. It is similar in principal to the sequential
machine solution but uses $: to implement the machine.
NB. assuming that the right list contains the first one, there are four cases
NB. 0 0 -> true,
Here's an implementation which uses ;:
smalt=: 12 > (3;1,~&>0 1 2 3,1 3 2 3,2 1 3 3,:3)&;:@:#.@|:
Example use:
smalt 1 0 0 1 0 1,:0 1 0 0 1 0
1
smalt 0 1 0 1 0,:1 0 1 0 1
1
smalt 0 0 1 0 0 0 0 1,:1 0 0 0 1 0 0 0
1
smalt 1 0 0 0 1,:0 1 1 0 0
0
smalt 1 0 0 1 0 1 1,: 0 1 0 0 1 0 0
0
> Van: Peter B. Kessler
> Verzonden: maandag 30 juli 2012 22:12
(...)
>
> This doesn't address the case where both inputs have 1's in the same column
>
> ]C=:2 5 $ 1 0 0 1 0 0 1 0 1 0
> 1 0 0 1 0
> 0 1 0 1 0
> (>./-<./)+/\+/1 _1 *C
> 1
>
> The original stateme
Actually, I said something in my previous post which was wrong:
Function code 3 for sequential machine gives state table index.
(This is not useful for a lot of the text processing I would like to
do (where I want the state table index for each character being
processed) but it's fine for this ap
