Algorithm for finding constellations in a grid
So I dont know if this is the type of post people don't like to see here but I thought I'd give it a shot.
Imagine a $7$ by $7$ grid like in this picture.
Every square can have 1 of five states or colours to make it easier.
the goal is it to create a row of at least three, horizontally or perpendicular, NOT diagonally, of 1 colour. This shall be achieved by swapping to tiles with each other, again not diagonally. as soon as this is accomplished the tiles that meet this condition (being in a row of at least 3 tiles with the same colour), vanish and get replaced by the ones above them. the tiles above them "fall down" if you will. new random tiles that are replacing the missing ones on "top".
I couldn't think of any other constellation, where you can swap some tile to meet the condition, but the three you can see in the picture, except for them rotated by 90, 180 or 270 degrees of course.
my initial thought was to just take a tile and check if any of the surrounding tiles have the same colour and then look from there if any tile adds up to form one of the three constellations.
If not, the next tile will be checked and so on. I realised though that this would be extremely slow, because inevitably there would be checks that wouldn't be necessary anymore.
My question now is if there are any algorithms already that solve a similar problem and if not if anyone would help me out creating one.
algorithms
edited Nov 28 at 14:38
amWhy
191k28224439
asked Sep 28 '16 at 14:58
Tazua
62
add a comment |
So I dont know if this is the type of post people don't like to see here but I thought I'd give it a shot.
Imagine a $7$ by $7$ grid like in this picture.
Every square can have 1 of five states or colours to make it easier.
the goal is it to create a row of at least three, horizontally or perpendicular, NOT diagonally, of 1 colour. This shall be achieved by swapping to tiles with each other, again not diagonally. as soon as this is accomplished the tiles that meet this condition (being in a row of at least 3 tiles with the same colour), vanish and get replaced by the ones above them. the tiles above them "fall down" if you will. new random tiles that are replacing the missing ones on "top".
I couldn't think of any other constellation, where you can swap some tile to meet the condition, but the three you can see in the picture, except for them rotated by 90, 180 or 270 degrees of course.
my initial thought was to just take a tile and check if any of the surrounding tiles have the same colour and then look from there if any tile adds up to form one of the three constellations.
If not, the next tile will be checked and so on. I realised though that this would be extremely slow, because inevitably there would be checks that wouldn't be necessary anymore.
My question now is if there are any algorithms already that solve a similar problem and if not if anyone would help me out creating one.
algorithms
edited Nov 28 at 14:38
amWhy
191k28224439
asked Sep 28 '16 at 14:58
Tazua
62
add a comment |
So I dont know if this is the type of post people don't like to see here but I thought I'd give it a shot.
Imagine a $7$ by $7$ grid like in this picture.
Every square can have 1 of five states or colours to make it easier.
the goal is it to create a row of at least three, horizontally or perpendicular, NOT diagonally, of 1 colour. This shall be achieved by swapping to tiles with each other, again not diagonally. as soon as this is accomplished the tiles that meet this condition (being in a row of at least 3 tiles with the same colour), vanish and get replaced by the ones above them. the tiles above them "fall down" if you will. new random tiles that are replacing the missing ones on "top".
I couldn't think of any other constellation, where you can swap some tile to meet the condition, but the three you can see in the picture, except for them rotated by 90, 180 or 270 degrees of course.
my initial thought was to just take a tile and check if any of the surrounding tiles have the same colour and then look from there if any tile adds up to form one of the three constellations.
If not, the next tile will be checked and so on. I realised though that this would be extremely slow, because inevitably there would be checks that wouldn't be necessary anymore.
My question now is if there are any algorithms already that solve a similar problem and if not if anyone would help me out creating one.
algorithms
edited Nov 28 at 14:38
amWhy
191k28224439
asked Sep 28 '16 at 14:58
Tazua
62
So I dont know if this is the type of post people don't like to see here but I thought I'd give it a shot.
Imagine a $7$ by $7$ grid like in this picture.
Every square can have 1 of five states or colours to make it easier.
the goal is it to create a row of at least three, horizontally or perpendicular, NOT diagonally, of 1 colour. This shall be achieved by swapping to tiles with each other, again not diagonally. as soon as this is accomplished the tiles that meet this condition (being in a row of at least 3 tiles with the same colour), vanish and get replaced by the ones above them. the tiles above them "fall down" if you will. new random tiles that are replacing the missing ones on "top".
I couldn't think of any other constellation, where you can swap some tile to meet the condition, but the three you can see in the picture, except for them rotated by 90, 180 or 270 degrees of course.
my initial thought was to just take a tile and check if any of the surrounding tiles have the same colour and then look from there if any tile adds up to form one of the three constellations.
If not, the next tile will be checked and so on. I realised though that this would be extremely slow, because inevitably there would be checks that wouldn't be necessary anymore.
My question now is if there are any algorithms already that solve a similar problem and if not if anyone would help me out creating one.
algorithms
algorithms
edited Nov 28 at 14:38
amWhy
191k28224439
asked Sep 28 '16 at 14:58
Tazua
62
edited Nov 28 at 14:38
amWhy
191k28224439
asked Sep 28 '16 at 14:58
Tazua
62
edited Nov 28 at 14:38
amWhy
191k28224439
edited Nov 28 at 14:38
amWhy
191k28224439
edited Nov 28 at 14:38
amWhy
191k28224439
191k28224439
asked Sep 28 '16 at 14:58
Tazua
62
asked Sep 28 '16 at 14:58
Tazua
62
asked Sep 28 '16 at 14:58
Tazua
62
62
add a comment |
add a comment |
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