3 ms·

As others have mentioned, it does seem fairly easy once you get a feel for the game mechanics. I might suggest that you have each level include a max number th

As others have mentioned, it does seem fairly easy once you get a feel for the game mechanics. I might suggest that you have each level include a max number that each tile has to have for a correct solution. I found myself trying to get the lowest number, but I almost get everything to say 5, but messed up and missed one tile that had to get bumped to 6, it was then easy to just make one more pass and set everything to 6. If I had to restart to get the 5 I think it might be a slight increase to the difficulty.

hoopism 12y ago
I agree with this... seemed tough till I realized how forgiving it was. I gave up after few minutes... maybe later levels introduce this complexity?
- tehaugmenter 12y ago
  Levels 51+ introduce a black and white mechanic that decrements or increment the tile instead of them all just increment. This does add a level of complexity.
Chinjut 12y ago
Hm... You're saying it can be possible to solve two different ways with two different numbers? Suppose the corners and the center are black and the other squares are white. Then every two moves in a row adds one to black and one to white. So an even length sequence keeps the difference between the black sum and the white sum the same. In the end, with all values set to X, this difference equals 5X - 4X = X. Thus, for an even-length game, we know from the start what X must be: the difference between the black and white sums. This means there's no room for a wrong move. Any solution has a prescribed number of times to hit each tile. (FWIW, the total number of moves will be 9X - the starting sum = 9(B - W) - (B + W) = 8B - 10W) So far, we've assumed we'll be playing an even-length game. Might we actually be playing an odd-length game? Sure. In that case, we also know from the start what X should be (since it amounts to making one move, adding 1 to the color not started on, and then playing an even-length game). Is there an easy way to see from the start whether we play an even- or odd-length game? Is it possible that both are allowed for a given board? It's not yet clear to me. I'll reply with updates. But at any rate, there are at most two different values you can solve with, one for even-length solutions and another for odd-length solutions (and these two values will be next to each other). [This is all for the version of the game in the early levels. I see from other comments that the game changes at level 51, which I haven't played to yet. I also see that "black" and "white" start actually being used in the game at that point, making this explanation potential confusing.]
- Chinjut 12y ago
  I suppose, if the game doesn't end immediately upon solution, of course it could be possible to have both an even- and an odd-length solution: if it's possible to solve ending on a white square, then it's also possible to give a longer solution consisting of the first solution followed by winding around hitting every square on the board once more.

hoopism 12y ago

I agree with this... seemed tough till I realized how forgiving it was. I gave up after few minutes... maybe later levels introduce this complexity?

tehaugmenter 12y ago
Levels 51+ introduce a black and white mechanic that decrements or increment the tile instead of them all just increment. This does add a level of complexity.

Chinjut 12y ago

Hm... You're saying it can be possible to solve two different ways with two different numbers? Suppose the corners and the center are black and the other squares are white. Then every two moves in a row adds one to black and one to white. So an even length sequence keeps the difference between the black sum and the white sum the same. In the end, with all values set to X, this difference equals 5X - 4X = X. Thus, for an even-length game, we know from the start what X must be: the difference between the black and white sums. This means there's no room for a wrong move. Any solution has a prescribed number of times to hit each tile. (FWIW, the total number of moves will be 9X - the starting sum = 9(B - W) - (B + W) = 8B - 10W) So far, we've assumed we'll be playing an even-length game. Might we actually be playing an odd-length game? Sure. In that case, we also know from the start what X should be (since it amounts to making one move, adding 1 to the color not started on, and then playing an even-length game). Is there an easy way to see from the start whether we play an even- or odd-length game? Is it possible that both are allowed for a given board? It's not yet clear to me. I'll reply with updates. But at any rate, there are at most two different values you can solve with, one for even-length solutions and another for odd-length solutions (and these two values will be next to each other). [This is all for the version of the game in the early levels. I see from other comments that the game changes at level 51, which I haven't played to yet. I also see that "black" and "white" start actually being used in the game at that point, making this explanation potential confusing.]

Chinjut 12y ago
I suppose, if the game doesn't end immediately upon solution, of course it could be possible to have both an even- and an odd-length solution: if it's possible to solve ending on a white square, then it's also possible to give a longer solution consisting of the first solution followed by winding around hitting every square on the board once more.