[seqfan] Re: a partition sequence

franktaw at netscape.net franktaw at netscape.net
Mon Oct 3 03:30:12 CEST 2011

```After further consideration, I believe that David's algorithm does
always work. Let me describe essentially the same algorithm
constructively instead of as a series of modifications:

Process the parts in decreasing order of size. At each step, place all
the parts of the size being processed in the left-most column that does
not already have parts with that frequency.

Sketch of proof that this always works: after placement, the column (if
not the first) will still be no longer than the preceding column, since
the set of frequencies in the column is a subset of those for the
previous column; and since the parts are being processed in decreasing
order, they will be less than any preceding part size in the same row.

---
This would produce the same results as David's algorithm if his step
(3) was replaced by 3') if past the end of partition, stop.

-----Original Message-----
From: David Scambler <dscambler at bmm.com>
To: seqfan <seqfan at list.seqfan.eu>
Sent: Sun, Oct 2, 2011 6:43 pm
Subject: [seqfan] Re: a partition sequence

Try this algorithm:

1) arrange the parts in columns of like magnitude
2) k <= 1
3) if well-formed stop
4) k <= k+1
5) move column k to underneath left-most column without the frequency
of column
k
6) go to 3)

e.g.

4^2,3,2^2,1^3

Start:
4321
4.21
...1

Move col 2 under col 1
4.21
4.21
3..1

Move col 3 under col 1

4.2.
4.2.
3...
1...
1...
1...

dave

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```