### Wrap-around search algorithm

- Posted by petelomax in April
- 1088 views

This proved far harder than I ever imagined possible, any improvements welcome.

The following is a simplified testbed.

First simplification: text is "0000".."1111" and "find" is whether or not text[1..4] is '1'.

Second simplification: we set "start" explicitly in the testing loop, maybe unlike real use.

(Technically we shouldn't need to reset wrappable in this testbed, but abs. rqd. in real use)

Starting from 1..4, with N=sum of 1s, we want f3find() to get N/eof/N/eof, iyswim.

The challenge is(was) to get all 64 test cases to work, then repeat searching backwards.

Anyway, a second set of eyes, before I try applying this to Edix/Edita, anything that simplifies it or makes it any easier to understand, or OE-compatible, thankx.

sequence text integer start, current, wrappable = 1 function f3find(integer direction) -- -- direction should be +/-1. [existing code uses 0 to mean "from line 1"] -- return "next/prev" '1' in text, wrapping around relative to start. -- return -1 at "eof", aka "start", allowing restart. -- integer limit = iff(direction<0?1:length(text)), begin = iff(direction>0?1:length(text)) current += direction bool high = wrappable and compare(start,current)!=direction limit = iff(high ? limit : start-(wrappable=0)*direction) for i=current to limit by direction do if text[i]='1' then current=i return i end if end for if high then wrappable = 0 limit = start-direction for i=begin to limit by direction do if text[i]='1' then current=i return i end if end for end if current = start-direction wrappable = 1 return -1 end function constant TRIES=4 integer fails = 0, total = 0 for i=0 to 15 do text = sprintf("%04b",i) integer N = sum(sq_eq(text,'1')) ?{text,N} for direction=+1 to -1 by -2 do for j=1 to 4 do start = j current = j-direction wrappable = 1 sequence s = {} for t=1 to (N+1)*TRIES do s &= f3find(direction) end for total += 1 if s[$]!=-1 or sum(sq_eq(s,-1))!=TRIES then s &= {"9/0"} fails += 1 ?{"s=",s} end if -- ?s end for end for end for printf(1,"fails: %d, pass: %d/%d\n",{fails,total-fails,total})

output:

{"0000",0} {"0001",1} {"0010",1} {"0011",2} {"0100",1} {"0101",2} {"0110",2} {"0111",3} {"1000",1} {"1001",2} {"1010",2} {"1011",3} {"1100",2} {"1101",3} {"1110",3} {"1111",4} fails: 0, pass: 128/128