Why does square based stacking game always reach the same number of maximum stackings?
$begingroup$
The square stacking games works like this: theres a number NP which is the number of pegs on a board. On those pegs balls with numbers can be placed. But they can only be placed on a peg if they are either empty or have a ball on it whose number added to the one of the ball to be placed results in a number whose square root is a whole number (meaning if newball+oldball: 0,1,4,9,16,25,36,49,64...).
Each new balls value is one more than the previous. So we place ball 1 then ball 2 then ball 3 then ball 4 etc.
Now I have been tasked to develop an algorithm that given NP pegs determines how many balls can be placed.
So what I did is write a program the just takes a ball, determines all possible places it can be placed (except on empty pegs, there it always takes the first available) and then splits of into different paths through recursion determining all possible combinations of placement and then out of those finds the highest it got. Basicly brute force, since I have no idea how to optimise this problem, since for example placing a 3 on the 1 may prevent 8 from finding a partner, etc.
But what I found out is that now matter how I stack the blocks, it always seams to find a way to the the highest possible amount of blocks no matter what, even if I just always choose the first available block.
So I changed my algoerithm to always only choose the first available path and the result is always correct.
Of course, this doesn't make any sense. Balls should always block positions takeable by other balls and the combination of the the top layer of numbers is always completely different after a few balls.
So how can the end result always be the same?
Is there some mathematical principle I'm overlooking?
logic algorithms square-numbers
asked Jan 4 at 9:54
user1510024user1510024
32
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add a comment |
$begingroup$
The square stacking games works like this: theres a number NP which is the number of pegs on a board. On those pegs balls with numbers can be placed. But they can only be placed on a peg if they are either empty or have a ball on it whose number added to the one of the ball to be placed results in a number whose square root is a whole number (meaning if newball+oldball: 0,1,4,9,16,25,36,49,64...).
Each new balls value is one more than the previous. So we place ball 1 then ball 2 then ball 3 then ball 4 etc.
Now I have been tasked to develop an algorithm that given NP pegs determines how many balls can be placed.
So what I did is write a program the just takes a ball, determines all possible places it can be placed (except on empty pegs, there it always takes the first available) and then splits of into different paths through recursion determining all possible combinations of placement and then out of those finds the highest it got. Basicly brute force, since I have no idea how to optimise this problem, since for example placing a 3 on the 1 may prevent 8 from finding a partner, etc.
But what I found out is that now matter how I stack the blocks, it always seams to find a way to the the highest possible amount of blocks no matter what, even if I just always choose the first available block.
So I changed my algoerithm to always only choose the first available path and the result is always correct.
Of course, this doesn't make any sense. Balls should always block positions takeable by other balls and the combination of the the top layer of numbers is always completely different after a few balls.
So how can the end result always be the same?
Is there some mathematical principle I'm overlooking?
logic algorithms square-numbers
asked Jan 4 at 9:54
user1510024user1510024
32
$endgroup$
add a comment |
$begingroup$
The square stacking games works like this: theres a number NP which is the number of pegs on a board. On those pegs balls with numbers can be placed. But they can only be placed on a peg if they are either empty or have a ball on it whose number added to the one of the ball to be placed results in a number whose square root is a whole number (meaning if newball+oldball: 0,1,4,9,16,25,36,49,64...).
Each new balls value is one more than the previous. So we place ball 1 then ball 2 then ball 3 then ball 4 etc.
Now I have been tasked to develop an algorithm that given NP pegs determines how many balls can be placed.
So what I did is write a program the just takes a ball, determines all possible places it can be placed (except on empty pegs, there it always takes the first available) and then splits of into different paths through recursion determining all possible combinations of placement and then out of those finds the highest it got. Basicly brute force, since I have no idea how to optimise this problem, since for example placing a 3 on the 1 may prevent 8 from finding a partner, etc.
But what I found out is that now matter how I stack the blocks, it always seams to find a way to the the highest possible amount of blocks no matter what, even if I just always choose the first available block.
So I changed my algoerithm to always only choose the first available path and the result is always correct.
Of course, this doesn't make any sense. Balls should always block positions takeable by other balls and the combination of the the top layer of numbers is always completely different after a few balls.
So how can the end result always be the same?
Is there some mathematical principle I'm overlooking?
logic algorithms square-numbers
asked Jan 4 at 9:54
user1510024user1510024
32
$endgroup$
The square stacking games works like this: theres a number NP which is the number of pegs on a board. On those pegs balls with numbers can be placed. But they can only be placed on a peg if they are either empty or have a ball on it whose number added to the one of the ball to be placed results in a number whose square root is a whole number (meaning if newball+oldball: 0,1,4,9,16,25,36,49,64...).
Each new balls value is one more than the previous. So we place ball 1 then ball 2 then ball 3 then ball 4 etc.
Now I have been tasked to develop an algorithm that given NP pegs determines how many balls can be placed.
So what I did is write a program the just takes a ball, determines all possible places it can be placed (except on empty pegs, there it always takes the first available) and then splits of into different paths through recursion determining all possible combinations of placement and then out of those finds the highest it got. Basicly brute force, since I have no idea how to optimise this problem, since for example placing a 3 on the 1 may prevent 8 from finding a partner, etc.
But what I found out is that now matter how I stack the blocks, it always seams to find a way to the the highest possible amount of blocks no matter what, even if I just always choose the first available block.
So I changed my algoerithm to always only choose the first available path and the result is always correct.
Of course, this doesn't make any sense. Balls should always block positions takeable by other balls and the combination of the the top layer of numbers is always completely different after a few balls.
So how can the end result always be the same?
Is there some mathematical principle I'm overlooking?
logic algorithms square-numbers
logic algorithms square-numbers
asked Jan 4 at 9:54
user1510024user1510024
32
asked Jan 4 at 9:54
user1510024user1510024
32
asked Jan 4 at 9:54
user1510024user1510024
32
asked Jan 4 at 9:54
user1510024user1510024
32
asked Jan 4 at 9:54
user1510024user1510024
32
32
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