6 ms·

Thank you for putting my 16 LEDs Puzzle on Hacker News! Yes, the principle is similar to Lights Out, but in my puzzle pressing a button toggles its LED and all

Thank you for putting my 16 LEDs Puzzle on Hacker News! Yes, the principle is similar to Lights Out, but in my puzzle pressing a button toggles its LED and all the other LEDs on the same diagonal(s). I found that this rule is harder to solve, even on a 4x4 grid.

tromp 5y ago
Using diagonals decomposes the puzzle into 2 disjoint puzzles according to the black and white squares in a checkerboard pattern. One puzzle can be marked as _ A _ B C _ D _ _ E _ F B _ G _ If we denote the corresponding buttons in lowercase, we have the equation A 1 0 1 1 0 1 0 a B 0 1 0 1 1 0 0 b C 1 0 1 0 1 0 1 c D = 1 1 0 1 1 1 0 d E 0 1 1 1 1 0 1 e F 1 0 0 1 0 1 1 f G 0 0 1 0 1 1 1 g Where A=1 if and only if light A needs to change. This matrix over GF(2) is not invertible; its rank is only 5, so the lights cannot be changed arbitrarily. By Gaussian elimination, one finds that button f does the same as buttons a b e together, and button g the same as b c d together. This reduces the equation to A 0 1 1 1 1 a B 0 1 0 1 1 b C = 1 1 0 1 1 c D 0 0 0 1 0 d E 0 0 0 0 1 e and the solution comes out as e = E d = D c = A B b = B D E a = B C Solving for the remaining buttons yields e = E d = D c = A B b = B D E a = B C
- tromp 5y ago
  Oops; I can't edit my post anymore. It went south starting from > This reduces the equation to We instead just ignore buttons f and g, and invert the equation A 1 0 1 1 0 a B 0 1 0 1 1 b C = 1 0 1 0 1 c D 1 1 0 1 1 d E 0 1 1 1 1 e to a 0 1 0 1 0 A b 1 1 1 0 0 B c = 0 1 0 0 1 C d 1 0 0 1 1 D e 0 0 1 1 1 E Showing which buttons to press in terms of exclusive-or of wrong lights, e.g. press button a if exactly one of lights B and D is wrong.

tromp 5y ago

Using diagonals decomposes the puzzle into 2 disjoint puzzles according to the black and white squares in a checkerboard pattern. One puzzle can be marked as _ A _ B C _ D _ _ E _ F B _ G _ If we denote the corresponding buttons in lowercase, we have the equation A 1 0 1 1 0 1 0 a B 0 1 0 1 1 0 0 b C 1 0 1 0 1 0 1 c D = 1 1 0 1 1 1 0 d E 0 1 1 1 1 0 1 e F 1 0 0 1 0 1 1 f G 0 0 1 0 1 1 1 g Where A=1 if and only if light A needs to change. This matrix over GF(2) is not invertible; its rank is only 5, so the lights cannot be changed arbitrarily. By Gaussian elimination, one finds that button f does the same as buttons a b e together, and button g the same as b c d together. This reduces the equation to A 0 1 1 1 1 a B 0 1 0 1 1 b C = 1 1 0 1 1 c D 0 0 0 1 0 d E 0 0 0 0 1 e and the solution comes out as e = E d = D c = A B b = B D E a = B C Solving for the remaining buttons yields e = E d = D c = A B b = B D E a = B C

tromp 5y ago
Oops; I can't edit my post anymore. It went south starting from > This reduces the equation to We instead just ignore buttons f and g, and invert the equation A 1 0 1 1 0 a B 0 1 0 1 1 b C = 1 0 1 0 1 c D 1 1 0 1 1 d E 0 1 1 1 1 e to a 0 1 0 1 0 A b 1 1 1 0 0 B c = 0 1 0 0 1 C d 1 0 0 1 1 D e 0 0 1 1 1 E Showing which buttons to press in terms of exclusive-or of wrong lights, e.g. press button a if exactly one of lights B and D is wrong.