[seqfan] Re: a partition sequence

[franktaw at netscape.net]

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.

Franklin T. Adams-Watters

-----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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