### forum-msg-id-133749-edit

Original date:2019-04-14 08:01:15 Edited by: petelomax Subject: Wrap-around search algorithm

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 low = compare(start,current)=direction or not wrappable
limit = iff(low ? start-(wrappable=0)*direction : limit)
for i=current to limit by direction do
if text[i]='1' then current=i return i end if
end for
if wrappable and compare(start,current)!=direction 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
```