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Re: Partitioning a list into unequal partitions


x=Range[11];

k=0;
Rest[NestList[
    (k++;Take[x,{k(k-1)/2+1,k(k+1)/2}])&, 
    x,Floor[(Sqrt[1+8Length[x]]-1)/2]]]

{{1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10}}


Bob Hanlon

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)

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


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