Hi,
I'm having a little problem understanding this (from Viktor Cerovski's
solution):
>
> diffdig =: *./@~:@(#.^:_1)
Why is a scalar left argument sufficient in this case
(10&#.^:_1) 123
1 2 3
but when I omit the power adverb and use the inverse function directly
I need a vector to get the same result?
10#:123
3
10 10 10#:123
1 2 3
Isn't #:^:_1 the same as #. since they are inverse functions?
Rudi
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