Re: Re: generating submultisets with repeated elements

*To*: mathgroup at smc.vnet.net*Subject*: [mg103885] Re: [mg103849] Re: generating submultisets with repeated elements*From*: David Bevan <david.bevan at pb.com>*Date*: Sat, 10 Oct 2009 07:10:51 -0400 (EDT)*References*: <ha4r9k$d0h$1@smc.vnet.net> <200910071101.HAA00387@smc.vnet.net>

Bayard, To handle general sets you do something like: multiSets[s_List, k] := s[[#]]& /@ multiSets[Length[s], k] multiSets[n_, k_] := f /@ Subsets[Range[n + k - 1], {k}]] David %^> > -----Original Message----- > From: monochrome [mailto:bayard.webb at gmail.com] > Sent: 9 October 2009 12:16 > To: mathgroup at smc.vnet.net > Subject: [mg103849] Re: generating submultisets with repeated elements > > There is one thing that I glossed over here. This solution works only > on the indices of the sets. This will go horribly wrong if you use it > with general sets like {"cat","dog",...,"hamster"}. Worse yet, if you > used this with sets of numbers, the method would calculate, but return > garbage. Here's why... > > There is an algorithm to generate all subsets of a set that goes as > follows: > 1. List the elements > 2. Over the first k elements, draw an arrow pointing right > 3a. Capture the elements with arrows as a subset > 3b. Move the rightmost free arrow to the right (an arrow is free if > points to an empty space) > 3c. If, after moving the arrow, it isn't free, reverse its direction > 3d. Working to the right, move all free arrows that lie to the right > of the arrow you just moved until they are no longer free and reverse > them. (Note that they will all be pointing left when you start this > step, and all point right at the end.) > Repeat 3 until there are no free arrows. > > The difference between sets and multisets lies in steps 3b and 3d. To > get multisets you allow arrows to lie over the same element, either at > the end of the list or at the element that moves in step 3b. > > The method I proposed uses the index positions of the k arrows and the > function accounts for the differences noted above. So while KSubsets > will work on general sets, my solution requires that you work with > indices. You could modify the solution to use set elements and use the > Position[] function to recover the index, or use the solution as is > and do a substitution at the end. > > This may have been obvious to everyone, but I felt I should have > stated it more clearly. Also, the algorithm is what I can remember > from an excerpt from a book by Brualdi. I haven't seen it for fifteen > years and had to recreate it from what I could remember. If there is a > mistake in my explanation, it's mine not Brualdi's. I had actually > stumbled on the function first, then had to recreate this algorithm to > figure out why it worked. > > Bayard > > > On Oct 8, 4:57 am, David Bevan <david.be... at pb.com> wrote: > > That's an interesting bijection I wasn't aware of. Thanks. > > > > David %^> > > > > > > > > > -----Original Message----- > > > From: monochrome [mailto:bayard.w... at gmail.com] > > > Sent: 7 October 2009 12:02 > > > To: mathgr... at smc.vnet.net > > > Subject: Re: generating submultisets with repeated elements > > > > > I did a little research and found out that there are Choose(n+k-1, k) > > > multisets of size k from a set of size n. This made me think that > > > there should be a mapping from the k-subsets of n+k-1 to the k- > > > multisets of n. A few quick examples led me to the following function= : > > > > > f[set_] := Table[set[[i]] - (i - 1), {i, Length[set]}] > > > > > This allows the following construction using the KSubsets function > > > from Combinatorica: > > > > > << "Combinatorica`"; > > > n = 6; > > > k = 3; > > > set = Range[n + k - 1]; > > > Map[f, KSubsets[set, k]] > > > > > ===OUTPUT=== > > > {{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}, {1, 1, 5}, {1, 1, 6}, {1= , > > > 2, 2}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 2, 6}, {1, 3, 3}, {1= = > , > > > 3, 4}, {1, 3, 5}, {1, 3, 6}, {1, 4, 4}, {1, 4, 5}, {1, 4, 6}, {1, 5= = > , > > > 5}, {1, 5, 6}, {1, 6, 6}, {2, 2, 2}, {2, 2, 3}, {2, 2, 4}, {2, 2= = > , > > > 5}, {2, 2, 6}, {2, 3, 3}, {2, 3, 4}, {2, 3, 5}, {2, 3, 6}, {2, 4, > > > 4}, {2, 4, 5}, {2, 4, 6}, {2, 5, 5}, {2, 5, 6}, {2, 6, 6}, {3, 3, > > > 3}, {3, 3, 4}, {3, 3, 5}, {3, 3, 6}, {3, 4, 4}, {3, 4, 5}, {3, 4, > > > 6}, {3, 5, 5}, {3, 5, 6}, {3, 6, 6}, {4, 4, 4}, {4, 4, 5}, {4, 4, > > > 6}, {4, 5, 5}, {4, 5, 6}, {4, 6, 6}, {5, 5, 5}, {5, 5, 6}, {5, 6, > > > 6}, {6, 6, 6}} > >

**References**:**Re: generating submultisets with repeated elements***From:*monochrome <bayard.webb@gmail.com>

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