2
$\begingroup$

I played the arcade game Klaxxon a lot (comparable mechanics to Candy Crush, long before the latter even existed) and always wondered if I could get a monster super ultra bonus by making a 3 x 3 unicolor block. (Probably this is impossible; while Klaxxon also kills diagonals, the board is only 5 x 5 - surely too small, also tiles only come from above.)

Let's assume that diagonals don't count, the board has arbitrarily large size, you have as many colors as you like, any 3 or more adjacent (row or column) same colors vanish (you must specify the order of concurrent vanishes; best would be "mark all vanishable, nothing else vanishes before all marked vanished"), blocks fall down into the holes generated, after all come to rest the next vanishing check starts, the starting configuration is stable, and now you switch two colors and all hell breaks loose. Observe:

    c c
    ded
  a eae a
  b dbd b
  ccdcdcc 
  bxxbxxb
aabaabaabaa

Swap a with the e over it. Boom! Boom! Boom! Row of 11 a. Obviously, the construction can be extended ad infinitum to get an arbitrarily long single row of a's.

Can you make a 2 x infinity block at some time? (I'm fairly sure it suffices to set x=a and fill in the rest of a from above.) 3 x infinity? Maybe even infinity x infinity?

$\endgroup$
3
  • $\begingroup$ "any >3 adjacent same colors vanish" - does this mean Puyo Puyo style (where the only thing that matters is the group size) or Bejweled-style (where the three must be in a row)? In other words, does a 2×2 square vanish or not? $\endgroup$
    – Deusovi
    Commented Mar 7, 2023 at 13:52
  • $\begingroup$ Is ">3 adjacent" supposed to be "≥3 adjacent"? $\endgroup$
    – Laurel
    Commented Mar 7, 2023 at 16:31
  • $\begingroup$ Sorry, someone ate a =. Of course >=3. Edited. @Deusovi - Bejeweled, 2x2 doesn't vanish. $\endgroup$ Commented Mar 7, 2023 at 17:26

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.