

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 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
This is actually genius, I’m not gonna lie.
Lmao! Reminds me of findtheinvisiblecow.com. Similar
i almost lost focus before finding it because of all the owo’s
That rectangle with the exclamation mark is for spoilers.
Spoiler
like this
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 endThe 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 🤔
- fill the grid completely at random
Top of the page says find 1 cow, so yes
It’s right under COCOOW
I found coooooooc :)
Many, many cocs.
Wow!


Put it in spoiler tags, please!

Put it in spoiler tags, please!
I gave up on the 6. line.
Keep going! It’s on the 15th line, almost on the far right hand side of the page.
Please use the spoiler tag!









