Thanks for all the suggestions. That gave me the general idea.
My take using boxing (slower)
*sep=.10 #.^:_1 ]*
*#a=.n#~+./"1[1=>#&.>~.&.>{3,\"1 sep n=.1e5 to 999999*
*33219*
So there are 33,219 six-digit integers that have 3 or more identical
consecutive digits.
* 30 timex '#a=.n#~+./"1[1=>#&.>~.&.>{3,\"1 sep n=.1e5 to 999999'*
*2.76304*
Not very fast.
Skip Cave
Cave Consulting LLC
On Sat, Jul 3, 2021 at 6:13 PM Skip Cave <[email protected]> wrote:
> Can I use cut to find all the 6-digit integers that have three consecutive
> identical digits? Or is there a more efficient way?
>
> sep=:10#.^:_1] NB. Separate digits.
>
> to=:[+i.@:>:@-~ NB. Generate integer ranges.
>
>
> $ sn=.sep n=.1e5 to 999999 NB. Generate all the 6 digit integers and
> separate their digits.
>
> 900000 6
>
> Now what? Can I apply cut to sn to find all 6-digit integers with three
> consecutive identical digits? Or is there a better way?
>
>
> 123334 is correct
>
> 122344 is not correct
>
> 121212 is not correct
>
> 111222 is correct
>
> 112122 is not correct
>
> 555432 is correct
>
>
> NB. Test vector: tv=. 123334 122344 121212 111222 112122 555432
>
> Skip
>
>
> Skip Cave
> Cave Consulting LLC
>
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