Re: Binary iteration question

[Laura]

Rians excellent logic can be refactored to use the Over iterator. (You might not be familiar with Over with a ternary or quaternary function.)

 

q)g:{"b"$0^fills@/[;(x,y+1){where x=x msum y}'1 notz;:;1 0]count[z]#0N} 
q)g[3;2] 
00001111000111001101000b 00000011110001111111110b 
q)g[3;2] 
110010111001111b 000000001111111b

 

The key concept here is that the first argument of Amend At Over @/ is the initial state: count[z]#0N. The other (right) arguments are same-length lists, or atoms.  Over works through the argument lists in succession.

Once again we see the Zen monks as a point-free alternative to writing (z;not z).

The refactoring here doesnt save much time, but spotting opportunities like this improves your ability to find iterator solutions, some of which will save you significant CPU.