this post was submitted on 24 May 2026

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Mildly Infuriating

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Find Cow (lemmy.zip)

submitted 2 days ago by favoredponcho@lemmy.zip to c/mildlyinfuriating@lemmy.world

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[–] Snazz@lemmy.world 6 points 2 days ago (1 children)

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.

[–] MatSeFi@lemmy.liebeleu.de 2 points 1 day ago

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

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 🤔