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Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
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#1
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Math help needed.
Can't seem to figure out an answer. Permutation/combination stuff.
Suppose I have n red balls, x yellow balls, y green balls, and z orange balls. (no blue. G) I need a formula for how many unique ways there are to arrange them. Even better would be a formula for how many unique ways to arrange them on the perimeter of a circle, so that no pattern can be duplicated by rotating the circle. Extra credit for a link to an algorithm to generate the patterns. Racking my brain, and Googling for the last week. |
#2
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Math help needed.
On Sat, 11 Jul 2015 23:29:00 -0400, Steve Walker wrote:
Can't seem to figure out an answer. Permutation/combination stuff. Suppose I have n red balls, x yellow balls, y green balls, and z orange balls. (no blue. G) I need a formula for how many unique ways there are to arrange them. Even better would be a formula for how many unique ways to arrange them on the perimeter of a circle, so that no pattern can be duplicated by rotating the circle. Extra credit for a link to an algorithm to generate the patterns. Racking my brain, and Googling for the last week. If you had N = x + y + z unique balls*, then there would be N! (N factorial) ways to arrange them. Within the yellow balls, there would be x! unique ways to arrange them, but you lose that. Ditto green and orange. So there are ((x+y+z)!) / ((x!)(y!)(z!)) unique ways to arrange the balls in a line. I'm in a different building from my statistics book; if I remember I'll look it up and tell you the correct name for the function. For two numbers x and y, the spoken expression is "x+y choose x" -- but I can't remember the formal name to look up. On finding the number of combinations that go away if you arrange them in a circle -- I'm pretty sure that's hard. For a start, if x+y+z is prime then there are no rotational symmetries to be had at all. Then, if x+y+z and x are coprime, there are no rotational symmetries (or y, or z). But if x+y+z and x do have common factors, the location of the yellow balls in the circle does suddenly matter. It's enough to make me run away screaming, or demand to be paid by the hour to figure it out. * How unique you are if you have N 2 balls is another matter. 0 and 2 cover most of the population, 1 a small but significant fraction, but for N 2 there is a sharply diminishing proportion -- at N 5 you're well past "unusual" and into "astonishing" or perhaps even "frighteningly pitiful". -- www.wescottdesign.com |
#3
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Math help needed.
On 2015-07-12, Tim Wescott wrote:
On Sat, 11 Jul 2015 23:29:00 -0400, Steve Walker wrote: Can't seem to figure out an answer. Permutation/combination stuff. Suppose I have n red balls, x yellow balls, y green balls, and z orange balls. (no blue. G) I need a formula for how many unique ways there are to arrange them. Even better would be a formula for how many unique ways to arrange them on the perimeter of a circle, so that no pattern can be duplicated by rotating the circle. Extra credit for a link to an algorithm to generate the patterns. Racking my brain, and Googling for the last week. If you had N = x + y + z unique balls*, then there would be N! (N factorial) ways to arrange them. Within the yellow balls, there would be x! unique ways to arrange them, but you lose that. Ditto green and orange. So there are ((x+y+z)!) / ((x!)(y!)(z!)) unique ways to arrange the balls in a line. This is correct. i I'm in a different building from my statistics book; if I remember I'll look it up and tell you the correct name for the function. For two numbers x and y, the spoken expression is "x+y choose x" -- but I can't remember the formal name to look up. On finding the number of combinations that go away if you arrange them in a circle -- I'm pretty sure that's hard. For a start, if x+y+z is prime then there are no rotational symmetries to be had at all. Then, if x+y+z and x are coprime, there are no rotational symmetries (or y, or z). But if x+y+z and x do have common factors, the location of the yellow balls in the circle does suddenly matter. It's enough to make me run away screaming, or demand to be paid by the hour to figure it out. * How unique you are if you have N 2 balls is another matter. 0 and 2 cover most of the population, 1 a small but significant fraction, but for N 2 there is a sharply diminishing proportion -- at N 5 you're well past "unusual" and into "astonishing" or perhaps even "frighteningly pitiful". |
#4
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Math help needed.
On Sat, 11 Jul 2015 22:59:36 -0500, Tim Wescott wrote:
On Sat, 11 Jul 2015 23:29:00 -0400, Steve Walker wrote: Can't seem to figure out an answer. Permutation/combination stuff. Suppose I have n red balls, x yellow balls, y green balls, and z orange balls. (no blue. G) I need a formula for how many unique ways there are to arrange them. Even better would be a formula for how many unique ways to arrange them on the perimeter of a circle, so that no pattern can be duplicated by rotating the circle. Extra credit for a link to an algorithm to generate the patterns. Racking my brain, and Googling for the last week. If you had N = x + y + z unique balls*, then there would be N! (N factorial) ways to arrange them. Within the yellow balls, there would be x! unique ways to arrange them, but you lose that. Ditto green and orange. So there are ((x+y+z)!) / ((x!)(y!)(z!)) unique ways to arrange the balls in a line. I'm in a different building from my statistics book; if I remember I'll look it up and tell you the correct name for the function. For two numbers x and y, the spoken expression is "x+y choose x" -- but I can't remember the formal name to look up. On finding the number of combinations that go away if you arrange them in a circle -- I'm pretty sure that's hard. For a start, if x+y+z is prime then there are no rotational symmetries to be had at all. Then, if x+y+z and x are coprime, there are no rotational symmetries (or y, or z). But if x+y+z and x do have common factors, the location of the yellow balls in the circle does suddenly matter. It's enough to make me run away screaming, or demand to be paid by the hour to figure it out. If x, y and z have some lowest common denominator that is greater than one, then there are arrangements of them around a circle that would be symmetrical in rotation, but others that would not. E.g., if x, y and z were all multiples of six, then there would be arrangements of them in a circle that would not be rotationally symmetric, would be symmetric on rotations of 180 degrees only, or 120 degrees, or 60 degrees. But putting numbers on that is still something to make my head hurt. -- www.wescottdesign.com |
#5
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Math help needed.
On Sat, 11 Jul 2015 23:29:00 -0400, Steve Walker wrote:
Can't seem to figure out an answer. Permutation/combination stuff. Suppose I have n red balls, x yellow balls, y green balls, and z orange balls. (no blue. G) I need a formula for how many unique ways there are to arrange them. Even better would be a formula for how many unique ways to arrange them on the perimeter of a circle, so that no pattern can be duplicated by rotating the circle. Extra credit for a link to an algorithm to generate the patterns. Racking my brain, and Googling for the last week. Tim Wescott showed a correct technique for the first part of your question (the number of unique ways to arrange the balls in a line) for three colors of balls, with answer (x+y+z)! / (x! y! z!). For four colors of balls, with n, x, y, and z members respectively, the count is (n+x+y+z)! / (n! x! y! z!). For the number of unique ways to arrange them on a circle, see wikipedia at https://en.wikipedia.org/wiki/Necklace_%28combinatorics%29 -- jiw |
#6
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Math help needed.
Ignoramus10431 wrote in
news On 2015-07-12, Tim Wescott wrote: On Sat, 11 Jul 2015 23:29:00 -0400, Steve Walker wrote: Can't seem to figure out an answer. Permutation/combination stuff. Suppose I have n red balls, x yellow balls, y green balls, and z orange balls. (no blue. G) I need a formula for how many unique ways there are to arrange them. Even better would be a formula for how many unique ways to arrange them on the perimeter of a circle, so that no pattern can be duplicated by rotating the circle. Extra credit for a link to an algorithm to generate the patterns. Racking my brain, and Googling for the last week. If you had N = x + y + z unique balls*, then there would be N! (N factorial) ways to arrange them. Within the yellow balls, there would be x! unique ways to arrange them, but you lose that. Ditto green and orange. So there are ((x+y+z)!) / ((x!)(y!)(z!)) unique ways to arrange the balls in a line. This is correct. No, it's not. The OP has a total of K = n + x + y + z balls in *four* different colors. |
#7
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Math help needed.
On Sun, 12 Jul 2015 13:31:37 +0000, Doug Miller wrote:
Ignoramus10431 wrote in news On 2015-07-12, Tim Wescott wrote: On Sat, 11 Jul 2015 23:29:00 -0400, Steve Walker wrote: Can't seem to figure out an answer. Permutation/combination stuff. Suppose I have n red balls, x yellow balls, y green balls, and z orange balls. (no blue. G) I need a formula for how many unique ways there are to arrange them. Even better would be a formula for how many unique ways to arrange them on the perimeter of a circle, so that no pattern can be duplicated by rotating the circle. Extra credit for a link to an algorithm to generate the patterns. Racking my brain, and Googling for the last week. If you had N = x + y + z unique balls*, then there would be N! (N factorial) ways to arrange them. Within the yellow balls, there would be x! unique ways to arrange them, but you lose that. Ditto green and orange. So there are ((x+y+z)!) / ((x!)(y!)(z!)) unique ways to arrange the balls in a line. This is correct. No, it's not. The OP has a total of K = n + x + y + z balls in *four* different colors. Whoops -- I missed that. Please don't ask me how. choices = K! / ((n!)(x!)(y!)(z!)) Eliminating the circularly symmetric possibilities is still a bitch. -- www.wescottdesign.com |
#8
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Math help needed.
On 7/11/2015 11:59 PM, Tim Wescott wrote:
On Sat, 11 Jul 2015 23:29:00 -0400, Steve Walker wrote: Can't seem to figure out an answer. Permutation/combination stuff. Suppose I have n red balls, x yellow balls, y green balls, and z orange balls. (no blue. G) I need a formula for how many unique ways there are to arrange them. Even better would be a formula for how many unique ways to arrange them on the perimeter of a circle, so that no pattern can be duplicated by rotating the circle. Extra credit for a link to an algorithm to generate the patterns. Racking my brain, and Googling for the last week. If you had N = x + y + z unique balls*, then there would be N! (N factorial) ways to arrange them. Within the yellow balls, there would be x! unique ways to arrange them, but you lose that. Ditto green and orange. So there are ((x+y+z)!) / ((x!)(y!)(z!)) unique ways to arrange the balls in a line. That's the formula I was trying to suss out. Thanks. I'm in a different building from my statistics book; if I remember I'll look it up and tell you the correct name for the function. For two numbers x and y, the spoken expression is "x+y choose x" -- but I can't remember the formal name to look up. On finding the number of combinations that go away if you arrange them in a circle -- I'm pretty sure that's hard. For a start, if x+y+z is prime then there are no rotational symmetries to be had at all. Then, if x+y+z and x are coprime, there are no rotational symmetries (or y, or z). But if x+y+z and x do have common factors, the location of the yellow balls in the circle does suddenly matter. It's enough to make me run away screaming, or demand to be paid by the hour to figure it out. I've done the screaming. G The circular part is just secondary curiosity. Still working on the algorithm for the combinations. The application is for a slot car track program I'm working on, just because I can. Not going to market it, just personal curiosity. * How unique you are if you have N 2 balls is another matter. 0 and 2 cover most of the population, 1 a small but significant fraction, but for N 2 there is a sharply diminishing proportion -- at N 5 you're well past "unusual" and into "astonishing" or perhaps even "frighteningly pitiful". |
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