Re: [math-fun] one-to-one relation (odd partitions) <-> (non-repeatpartitions)

22 Aug 2005

      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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