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Joined il y a 5 mois
Cake day: 18 février 2026
  • MatSeFi@lemmy.liebeleu.detoTechnology@lemmy.world•You paid me, a long-time Linux user, to use Windows 11 exclusively for a month: here’s how it wentEnglish
    253·
    il y a 22 jours

    He’s completely overlooked the thing that annoys me the most: the unbelievable number of clicks you need to make in Windows/Microsoft to get anything done. – Saving a file to a folder of your choice:

    • Windows: Click ‘Save’ -> Click ‘Choose a different location’ -> Scroll down to skip all the favourites and default locations -> Click the drive where you want to save the file -> Find the folder -> Click ‘Save’
    • Linux: Click ‘Save’ -> Go to the folder -> press ‘Save’

    Not ot mention my recent attempts to rename a Bluetooth device (two devices of the same type were displayed under the same name, making it impossible to tell them apart) 🤮

  • MatSeFi@lemmy.liebeleu.detoMildly Infuriating@lemmy.world•*Permanently Deleted*English
    2·
    il y a 2 mois

    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 🤔