[math-fun] one-to-one relation (odd partitions) <-> (non-repeat partitions)
It is well known that both are equal in count, but can anyone point me to a one-to-one transformation rule between both sets? example: {{7}, {6,1}, {5,2}, {4,3}, {4,2,1}} to & from {{7}, {5,1,1}, {3,3,1}, {3,1,1,1,1}, {1,1,1,1,1,1,1}} or is this a well known *dumb* question? W.
There's a beautiful correspondence, due to Franklin, I believe. I thought that I first saw it in Hardy & Wright, but I can't find it now. Bend odd parts into symmetrical right-angled gnomons, and then read them off as alternate 45-degree gnomons. Seven isn't really big enough to see the picture, but here it is, followed by a bigger single example: o --- o --- o --- o o --- o --- o --- o | | o o | | | 7 | 43 o o | | | | o o o --- o --- o o --- o --- o | \ | \ o o 511 o o 52 | | \ | | \ o o o o o --- o o --- o | \ | 331 \ 421 o o --- o o o o | \ \ | \ \ o o o o o --- o o --- o | \ | \ o o o o 31111 \ 61 \ o o \ \ o o \ \ o o o o \ \ o o 1111111 \ 7 \ o o \ \ o o \ \ o o \ \ o o \ \ o o 24 = 955311 = 10.7421 amongst other things o - o - o - o - o 10 - o - o - o - o | \ o o - o - o 7 o 4 - o | | | \ \ \ o o o - o - o o o o o 1 | | | | \ \ \ o o o o - o o 2 o o o | | | | \ \ \ o o o o o o o o \ 955311 o 10.7421 o To see that this is a bijection is a bit more tricky. Take the unequal parts and bend them into 45-degree gnomons. But where to bend them? Hint: the corners are at a couple of series of knight's moves apart, and the finished picture must be symmetrical. Here there are 5 parts. the smallest goes in the 5th position, marked `1' in the diagram. [if the part were bigger than 1, then the rest would be a diagonal of dots going southeastwards] Next put the corner of the 2 part at `2', stretch enough of it SEwards to be symmetrical with 1. If there's more (as there will be if the difference of the parts is more than 1) then it hangs down vertically. Then put the corner of the 4 part at `4' and draw enough of it horizonally to make the fig symmetrical with the hanging down part (zero in this case) of the previous one. The rest goes SEwards again. Put the corner of the 7 part at `7' and draw enough of it SEwards to make it symmetrical with the diag part of 4. The rest goes downwards. Finally the angle of the 10 part goes at `10' and enough of it horizontally to be symmetrical with the vert part of 7, and the rest SEwards. In this example, this last leg had 6 dots, and these are the right angles of the 6 odd parts of the partition 955311 to which 10.7421 corresponds. R. On Sat, 20 Aug 2005, wouter meeussen wrote:
It is well known that both are equal in count, but can anyone point me to a one-to-one transformation rule between both sets? example: {{7}, {6,1}, {5,2}, {4,3}, {4,2,1}} to & from {{7}, {5,1,1}, {3,3,1}, {3,1,1,1,1}, {1,1,1,1,1,1,1}}
or is this a well known *dumb* question?
W.
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My version was a bit garbled in the final diagram, so I've tried correcting it: On Mon, 22 Aug 2005, Richard Guy wrote:
There's a beautiful correspondence, due to Franklin, I believe. I thought that I first saw it in Hardy & Wright, but I can't find it now.
Bend odd parts into symmetrical right-angled gnomons, and then read them off as alternate 45-degree gnomons. Seven isn't really big enough to see the picture, but here it is, followed by a bigger single example:
o --- o --- o --- o o --- o --- o --- o | | o o | | | 7 | 43 o o | | | | o o
o --- o --- o o --- o --- o | \ | \ o o 511 o o 52 | | \ | | \ o o o o
o --- o o --- o | \ | 331 \ 421 o o --- o o o o | \ \ | \ \ o o o o
o --- o o --- o | \ | \ o o o o 31111 \ 61 \ o o \ \ o o \ \ o o
o o \ \ o o 1111111 \ 7 \ o o \ \ o o \ \ o o \ \ o o \ \ o o
24 = 955311 = 10.7421 amongst other things
o - o - o - o - o 10 - o - o - o - o | \ o o - o - o 7 o 4 - o | | | \ \ \ o o o - o - o o o o o 1 | | | | \ \ \ o o o o - o o 2 o o o | | | | \ \ \ o o o o o o o o \ 955311 o 10.7421 o
To see that this is a bijection is a bit more tricky. Take the unequal parts and bend them into 45-degree gnomons. But where to bend them? Hint: the corners are at a couple of series of knight's moves apart, and the finished picture must be symmetrical. Here there are 5 parts. the smallest goes in the 5th position, marked `1' in the diagram. [if the part were bigger than 1, then the rest would be a diagonal of dots going southeastwards] Next put the corner of the 2 part at `2', stretch enough of it SEwards to be symmetrical with 1. If there's more (as there will be if the difference of the parts is more than 1) then it hangs down vertically. Then put the corner of the 4 part at `4' and draw enough of it horizonally to make the fig symmetrical with the hanging down part (zero in this case) of the previous one. The rest goes SEwards again. Put the corner of the 7 part at `7' and draw enough of it SEwards to make it symmetrical with the diag part of 4. The rest goes downwards. Finally the angle of the 10 part goes at `10' and enough of it horizontally to be symmetrical with the vert part of 7, and the rest SEwards. In this example, this last leg had 6 dots, and these are the right angles of the 6 odd parts of the partition 955311 to which 10.7421 corresponds.
R.
On Sat, 20 Aug 2005, wouter meeussen wrote:
It is well known that both are equal in count, but can anyone point me to a one-to-one transformation rule between both sets? example: {{7}, {6,1}, {5,2}, {4,3}, {4,2,1}} to & from {{7}, {5,1,1}, {3,3,1}, {3,1,1,1,1}, {1,1,1,1,1,1,1}}
or is this a well known *dumb* question?
W.
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_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Thanks for that. So, there are 2 or more bijections with different results: 10.7421 <--> 7551111111 10.7421 <--> 955311 and possibly www.combinatorics.net/lascoux/articles/SylvesterBij.ps Any others? These form different permutations with (maybe) different cycle structures. Unexplored territory? Too bad most are so hard to program. Wouter. ----- Original Message ----- From: "Richard Guy" <rkg@cpsc.ucalgary.ca> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Monday, August 22, 2005 7:59 PM Subject: Re: [math-fun] one-to-one relation (odd partitions) <-> (non-repeatpartitions) My version was a bit garbled in the final diagram, so I've tried correcting it: On Mon, 22 Aug 2005, Richard Guy wrote:
There's a beautiful correspondence, due to Franklin, I believe. I thought that I first saw it in Hardy & Wright, but I can't find it now.
Bend odd parts into symmetrical right-angled gnomons, and then read them off as alternate 45-degree gnomons. Seven isn't really big enough to see the picture, but here it is, followed by a bigger single example:
o --- o --- o --- o o --- o --- o --- o | | o o | | | 7 | 43 o o | | | | o o
o --- o --- o o --- o --- o | \ | \ o o 511 o o 52 | | \ | | \ o o o o
o --- o o --- o | \ | 331 \ 421 o o --- o o o o | \ \ | \ \ o o o o
o --- o o --- o | \ | \ o o o o 31111 \ 61 \ o o \ \ o o \ \ o o
o o \ \ o o 1111111 \ 7 \ o o \ \ o o \ \ o o \ \ o o \ \ o o
24 = 955311 = 10.7421 amongst other things
o - o - o - o - o 10 - o - o - o - o | \ o o - o - o 7 o 4 - o | | | \ \ \ o o o - o - o o o o o 1 | | | | \ \ \ o o o o - o o 2 o o o | | | | \ \ \ o o o o o o o o \ 955311 o 10.7421 o
To see that this is a bijection is a bit more tricky. Take the unequal parts and bend them into 45-degree gnomons. But where to bend them? Hint: the corners are at a couple of series of knight's moves apart, and the finished picture must be symmetrical. Here there are 5 parts. the smallest goes in the 5th position, marked `1' in the diagram. [if the part were bigger than 1, then the rest would be a diagonal of dots going southeastwards] Next put the corner of the 2 part at `2', stretch enough of it SEwards to be symmetrical with 1. If there's more (as there will be if the difference of the parts is more than 1) then it hangs down vertically. Then put the corner of the 4 part at `4' and draw enough of it horizonally to make the fig symmetrical with the hanging down part (zero in this case) of the previous one. The rest goes SEwards again. Put the corner of the 7 part at `7' and draw enough of it SEwards to make it symmetrical with the diag part of 4. The rest goes downwards. Finally the angle of the 10 part goes at `10' and enough of it horizontally to be symmetrical with the vert part of 7, and the rest SEwards. In this example, this last leg had 6 dots, and these are the right angles of the 6 odd parts of the partition 955311 to which 10.7421 corresponds.
R.
On Sat, 20 Aug 2005, wouter meeussen wrote:
It is well known that both are equal in count, but can anyone point me to a one-to-one transformation rule between both sets? example: {{7}, {6,1}, {5,2}, {4,3}, {4,2,1}} to & from {{7}, {5,1,1}, {3,3,1}, {3,1,1,1,1}, {1,1,1,1,1,1,1}}
or is this a well known *dumb* question?
W.
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_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (2)
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Richard Guy -
wouter meeussen