
Mildly Infuriating
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I looked for a few minutes, nearly gave up, and then I found it nearly instantly once I started looking for the correct distance between C - W. Kind of interesting!
This is actually genius, I'm not gonna lie.
Found it

Easy
Nailed it!
I did wonder if that might prove to be the answer, glad it wasn't! 😁

Put it in spoiler tags, please!
Tap for spoiler
Line 15, end of the line
Gawd, it would have been so fucking funny if there actually wasn't a single instance of cow in it, though. Lol
You know about spoiler tags?
owo
wow
woo
COO. Cow in Scots.
Good description. I read this and perfectly estimated the location you described within 5 seconds.

Found it. I wonder what the designer did to make sure to have exactly one cow in there.
What I would have done is started filling in letters randomly and every time a C or W ends up next to an O, choose the same letter or an O to put on the opposite side of the O.
Its hard to prove, but I’m pretty sure there isn't a situation where a space can’t be filled in with this algorithm.
Yes, this would work — but it comes with a subtle statistical bias: the character 'W' ends up underrepresented. With a naïve “avoid COW” approach, only about 25% of the grid will typically be 'W'.
A more elegant solution would be:
- fill the grid completely at random
- search for every "COW" cluster
- whenever one is found, copy a random character from one cell in the cluster into another cell of the same cluster
- Iterate until no "COW" remains
- search for every "COW" cluster
That keeps the distribution much closer to uniform while still guaranteeing a valid puzzle. Then just insert the single "COW" manually wherever you want the hidden solution to be.
Julia code example
s= (320,180) #size
m=rand(['C','O','W'],s) #random init
c=1
while c>0 #iterate till solved
c=0
for i in 1:first(s)
for j in 1:last(s)
#check for 'COW' in each cluster of 3 and copy a character
#from a rendom cell to an other random cell of the cluster if found
if i>2 && m[i-2:i,j] ==['C','O','W'] #vertical
c +=1
r =shuffle([1,2])
m[i-r[1],j] = m[i-r[2],j]
end
if j>2 && m[i,j-2:j] ==['C','O','W'] #horizontal
c +=1
r =shuffle([0,1,2])
m[i,j-r[1]] = m[i,j-r[2]]
end
end
end
end
The neat part is that this preserves an almost perfectly balanced character frequency.
For comparison, the puzzle in the example image seems to contain roughly:
C: ~260 (~25%) O: ~520 (~50%) W: ~244 (~25%)
So the original author clearly used a different generation strategy.
Possibly on purpose: visually, 'C' and 'O' are much easier to confuse than 'W'. Reducing the number of 'W's therefore increases the search difficulty. In that sense, the approach suggested by @Snazz@lemmy.world is probably preferable: keep the distribution mostly balanced, but intentionally bias it just enough to make the puzzle psychologically annoying.
I wonder if there is a non iterative way to generate this puzzle with a 'uniform' character distribution 🤔
This is one of those things a computer could do instantaneously, but a human can not.
Cowever, ask a computer to find a how IRL, and it just sits there on your desk and blinks at you.
I, too, cannot find a how IRL

Had to find it

Use spoiler tags, please!
It’s missing the 1, try again

Put it in spoiler tags, please!
Sorry….
i almost lost focus before finding it because of all the owo's
Wow!

Many, many cocs.
This is impossible, it is not circled in red like it's supposed to.
I found coooooooc :)
It's 15 lines down on the right side
Spoiler tags, please!
I found one!
But first, I found "wowcoc"
Didn't find no cow, found a lot of coc tho
