I... This is frustrating. You keep posting puzzles with multiple solutions, and I keep telling you about it, but it feels like you haven't gotten the message. Here's how I tried to solve this puzzle. The upper- and lower-left corners fell quickly, and there I got stuck. I made several counts and eventually determined that the absolutely necessary polyominoes (given the no-rectangle rule) totaled 94, and later I proved there must be an additional hexomino, giving a total area of 100. Then I managed to solve the puzzle this far:
As I hinted earlier, this doesn't have a unique solution. One group of alternate solutions is provided by putting a 5 at R2C3; another group has an 8 at R3C10. Neither of those is the solution I think you had in mind. Now, I do see the D4 symmetry already in the grid, so I know adding two clues to cut the number of solutions to 1 would have to reduce the symmetry. Adding four clues in symmetric positions would maintain the symmetry, even though I don't know how you'd react to adding that many. Well, I should just tell you my fix. Put a 6 in R9C7, probably with the corresponding clues in your choice of symmetric positions. (The other possible value in that spot, 4, would still lead to multiple solutions, even if you fixed R7C9 as another symmetric clue.)
You are right,neither of the two resulted from my logic.My solution had a 6 in R9C7,which fixes the entire grid(damn i didnt notice the multiple branches possible with a 4 in the same cell,despite having solved this 4 times after it was constructed). This also shows why you could conclude that R2C3=5 or R3C10=8 was not my intended solution.
I... This is frustrating. You keep posting puzzles with multiple solutions, and I keep telling you about it, but it feels like you haven't gotten the message.
ReplyDeleteHere's how I tried to solve this puzzle. The upper- and lower-left corners fell quickly, and there I got stuck. I made several counts and eventually determined that the absolutely necessary polyominoes (given the no-rectangle rule) totaled 94, and later I proved there must be an additional hexomino, giving a total area of 100. Then I managed to solve the puzzle this far:
4555566666
44____44_6
343_4_343_
3344_8_3__
6666686__8
65558864_8
55848_6__8
33844464_8
34888___5_
44488_4455
As I hinted earlier, this doesn't have a unique solution. One group of alternate solutions is provided by putting a 5 at R2C3; another group has an 8 at R3C10. Neither of those is the solution I think you had in mind.
Now, I do see the D4 symmetry already in the grid, so I know adding two clues to cut the number of solutions to 1 would have to reduce the symmetry. Adding four clues in symmetric positions would maintain the symmetry, even though I don't know how you'd react to adding that many.
Well, I should just tell you my fix. Put a 6 in R9C7, probably with the corresponding clues in your choice of symmetric positions. (The other possible value in that spot, 4, would still lead to multiple solutions, even if you fixed R7C9 as another symmetric clue.)
You are right,neither of the two resulted from my logic.My solution had a 6 in R9C7,which fixes the entire grid(damn i didnt notice the multiple branches possible with a 4 in the same cell,despite having solved this 4 times after it was constructed). This also shows why you could conclude that R2C3=5 or R3C10=8 was not my intended solution.
ReplyDelete