greatest number of balls in a box
$begingroup$
What is the greatest number of balls of radius $1/2$ that can be placed within a rectangular box of size $10*10*1$ ?
I start considering that
(1) I can reduce the problem in 2D
(2) The centres of the balls must lie in a square $9*9$
Now I think I have to find a tassellation for this square in order to find the greatest number of centres I can place in it but I don't know how to proceed...
geometry
asked Dec 30 '18 at 15:39
LanceLance
10112
$endgroup$
add a comment |
$begingroup$
What is the greatest number of balls of radius $1/2$ that can be placed within a rectangular box of size $10*10*1$ ?
I start considering that
(1) I can reduce the problem in 2D
(2) The centres of the balls must lie in a square $9*9$
Now I think I have to find a tassellation for this square in order to find the greatest number of centres I can place in it but I don't know how to proceed...
geometry
asked Dec 30 '18 at 15:39
LanceLance
10112
$endgroup$
1
$begingroup$
Of interest may be en.wikipedia.org/wiki/Circle_packing_in_a_square
$endgroup$
– coffeemath
Dec 30 '18 at 15:47
add a comment |
$begingroup$
What is the greatest number of balls of radius $1/2$ that can be placed within a rectangular box of size $10*10*1$ ?
I start considering that
(1) I can reduce the problem in 2D
(2) The centres of the balls must lie in a square $9*9$
Now I think I have to find a tassellation for this square in order to find the greatest number of centres I can place in it but I don't know how to proceed...
geometry
asked Dec 30 '18 at 15:39
LanceLance
10112
$endgroup$
What is the greatest number of balls of radius $1/2$ that can be placed within a rectangular box of size $10*10*1$ ?
I start considering that
(1) I can reduce the problem in 2D
(2) The centres of the balls must lie in a square $9*9$
Now I think I have to find a tassellation for this square in order to find the greatest number of centres I can place in it but I don't know how to proceed...
geometry
geometry
asked Dec 30 '18 at 15:39
LanceLance
10112
asked Dec 30 '18 at 15:39
LanceLance
10112
asked Dec 30 '18 at 15:39
LanceLance
10112
asked Dec 30 '18 at 15:39
LanceLance
10112
asked Dec 30 '18 at 15:39
LanceLance
10112
10112
1
$begingroup$
Of interest may be en.wikipedia.org/wiki/Circle_packing_in_a_square
$endgroup$
– coffeemath
Dec 30 '18 at 15:47
add a comment |
1
$begingroup$
Of interest may be en.wikipedia.org/wiki/Circle_packing_in_a_square
$endgroup$
– coffeemath
Dec 30 '18 at 15:47
1
1
$begingroup$
Of interest may be en.wikipedia.org/wiki/Circle_packing_in_a_square
$endgroup$
– coffeemath
Dec 30 '18 at 15:47
$begingroup$
Of interest may be en.wikipedia.org/wiki/Circle_packing_in_a_square
$endgroup$
– coffeemath
Dec 30 '18 at 15:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A rectangular pack would allow $10 times 10=100$ balls. Packing problems are hard. Packomania shows how to pack $106$ balls of radius greater than $0.05$ in a unit square.
answered Dec 30 '18 at 15:47
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
How can I prove 106 is the best? It can't be so hard, it's an old IMO longlist problem
$endgroup$
– Lance
Dec 30 '18 at 15:54
2
$begingroup$
Many of the configurations on packomania are not proven to be the best. $106$ is not listed in boldface, so the site does not consider it proven. Many of them are just the result of a lot of trying different starting patterns, then "shaking" the circles trying to improve the radius. That is why the problems are hard. There might be a better configuration out there that we have not found yet.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:58
$begingroup$
105 is not difficult to find - alternate rows with 10 balls and rows with 9 balls. This occupies $1+5sqrt 3$ from the height. The remaining place of width $0.34$ is encouraging ... the packing from Packomania is better :)
$endgroup$
– user376343
Dec 31 '18 at 0:38
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A rectangular pack would allow $10 times 10=100$ balls. Packing problems are hard. Packomania shows how to pack $106$ balls of radius greater than $0.05$ in a unit square.
answered Dec 30 '18 at 15:47
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
How can I prove 106 is the best? It can't be so hard, it's an old IMO longlist problem
$endgroup$
– Lance
Dec 30 '18 at 15:54
2
$begingroup$
Many of the configurations on packomania are not proven to be the best. $106$ is not listed in boldface, so the site does not consider it proven. Many of them are just the result of a lot of trying different starting patterns, then "shaking" the circles trying to improve the radius. That is why the problems are hard. There might be a better configuration out there that we have not found yet.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:58
$begingroup$
105 is not difficult to find - alternate rows with 10 balls and rows with 9 balls. This occupies $1+5sqrt 3$ from the height. The remaining place of width $0.34$ is encouraging ... the packing from Packomania is better :)
$endgroup$
– user376343
Dec 31 '18 at 0:38
add a comment |
$begingroup$
A rectangular pack would allow $10 times 10=100$ balls. Packing problems are hard. Packomania shows how to pack $106$ balls of radius greater than $0.05$ in a unit square.
answered Dec 30 '18 at 15:47
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
How can I prove 106 is the best? It can't be so hard, it's an old IMO longlist problem
$endgroup$
– Lance
Dec 30 '18 at 15:54
2
$begingroup$
Many of the configurations on packomania are not proven to be the best. $106$ is not listed in boldface, so the site does not consider it proven. Many of them are just the result of a lot of trying different starting patterns, then "shaking" the circles trying to improve the radius. That is why the problems are hard. There might be a better configuration out there that we have not found yet.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:58
$begingroup$
105 is not difficult to find - alternate rows with 10 balls and rows with 9 balls. This occupies $1+5sqrt 3$ from the height. The remaining place of width $0.34$ is encouraging ... the packing from Packomania is better :)
$endgroup$
– user376343
Dec 31 '18 at 0:38
add a comment |
$begingroup$
A rectangular pack would allow $10 times 10=100$ balls. Packing problems are hard. Packomania shows how to pack $106$ balls of radius greater than $0.05$ in a unit square.
answered Dec 30 '18 at 15:47
Ross MillikanRoss Millikan
297k23198371
$endgroup$
A rectangular pack would allow $10 times 10=100$ balls. Packing problems are hard. Packomania shows how to pack $106$ balls of radius greater than $0.05$ in a unit square.
answered Dec 30 '18 at 15:47
Ross MillikanRoss Millikan
297k23198371
answered Dec 30 '18 at 15:47
Ross MillikanRoss Millikan
297k23198371
answered Dec 30 '18 at 15:47
Ross MillikanRoss Millikan
297k23198371
answered Dec 30 '18 at 15:47
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
How can I prove 106 is the best? It can't be so hard, it's an old IMO longlist problem
$endgroup$
– Lance
Dec 30 '18 at 15:54
2
$begingroup$
Many of the configurations on packomania are not proven to be the best. $106$ is not listed in boldface, so the site does not consider it proven. Many of them are just the result of a lot of trying different starting patterns, then "shaking" the circles trying to improve the radius. That is why the problems are hard. There might be a better configuration out there that we have not found yet.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:58
$begingroup$
105 is not difficult to find - alternate rows with 10 balls and rows with 9 balls. This occupies $1+5sqrt 3$ from the height. The remaining place of width $0.34$ is encouraging ... the packing from Packomania is better :)
$endgroup$
– user376343
Dec 31 '18 at 0:38
add a comment |
$begingroup$
How can I prove 106 is the best? It can't be so hard, it's an old IMO longlist problem
$endgroup$
– Lance
Dec 30 '18 at 15:54
2
$begingroup$
Many of the configurations on packomania are not proven to be the best. $106$ is not listed in boldface, so the site does not consider it proven. Many of them are just the result of a lot of trying different starting patterns, then "shaking" the circles trying to improve the radius. That is why the problems are hard. There might be a better configuration out there that we have not found yet.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:58
$begingroup$
105 is not difficult to find - alternate rows with 10 balls and rows with 9 balls. This occupies $1+5sqrt 3$ from the height. The remaining place of width $0.34$ is encouraging ... the packing from Packomania is better :)
$endgroup$
– user376343
Dec 31 '18 at 0:38
$begingroup$
How can I prove 106 is the best? It can't be so hard, it's an old IMO longlist problem
$endgroup$
– Lance
Dec 30 '18 at 15:54
$begingroup$
How can I prove 106 is the best? It can't be so hard, it's an old IMO longlist problem
$endgroup$
– Lance
Dec 30 '18 at 15:54
2
2
$begingroup$
Many of the configurations on packomania are not proven to be the best. $106$ is not listed in boldface, so the site does not consider it proven. Many of them are just the result of a lot of trying different starting patterns, then "shaking" the circles trying to improve the radius. That is why the problems are hard. There might be a better configuration out there that we have not found yet.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:58
$begingroup$
Many of the configurations on packomania are not proven to be the best. $106$ is not listed in boldface, so the site does not consider it proven. Many of them are just the result of a lot of trying different starting patterns, then "shaking" the circles trying to improve the radius. That is why the problems are hard. There might be a better configuration out there that we have not found yet.
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:58
$begingroup$
105 is not difficult to find - alternate rows with 10 balls and rows with 9 balls. This occupies $1+5sqrt 3$ from the height. The remaining place of width $0.34$ is encouraging ... the packing from Packomania is better :)
$endgroup$
– user376343
Dec 31 '18 at 0:38
$begingroup$
105 is not difficult to find - alternate rows with 10 balls and rows with 9 balls. This occupies $1+5sqrt 3$ from the height. The remaining place of width $0.34$ is encouraging ... the packing from Packomania is better :)
$endgroup$
– user376343
Dec 31 '18 at 0:38
add a comment |
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1
$begingroup$
Of interest may be en.wikipedia.org/wiki/Circle_packing_in_a_square
$endgroup$
– coffeemath
Dec 30 '18 at 15:47