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MathGroup Archive 2004

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Re: Partitioning a list into unequal partitions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47501] Re: Partitioning a list into unequal partitions
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Wed, 14 Apr 2004 07:17:08 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <c5gfdp$an0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <c5gfdp$an0$1 at smc.vnet.net>, "DIAMOND Mark R." <dot at dot.dot> 
wrote:

> Could someone please show me a simple (non-procedural) way of partitioning a
> list into 1,2,3 ... n disjoint sublists, where the length of the list is
> guaranteed to be correct (i.e. n*(n+1)/2)

If I understand your question (a definite example would have helped), 
for a list, lst, how about

  Sort[ReplaceList[lst, {___, a__, ___} -> {a}]]

This gives a list of length n*(n+1)/2 if lst is of length n.

> I can't see a way, and yet I'm sure there *must* be a one-liner using Fold.

I don't see how Fold will do what you want. Distribute might work. For 
example, 

 PowerSet[t_List]:=Sort[Distribute[{{},{#}} & /@ t,List,List,List,Join]]

gives a sorted list of all subsets of a list. 

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 9380 2734
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AUSTRALIA                            http://physics.uwa.edu.au/~paul


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