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Re: Re: generating submultisets with repeated elements
monochrome wrote:
> 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}}
>
A very small modification of your method will work for general sets.
multiSetsK[set_,k_] := With[{diffset=Range[k]-1},
Map[set[[#-diffset]]&, Subsets[Range[Length[set]+k-1], {k}]]]
multiSetsToK[set_,k_] := Join @@ Table[multiSetsK[set,j],{j,k}]
In[27]:= set = {1,0,-7,"dog",913,"cat"};
InputForm[multiSetsToK[set,3]]
In[28]:=
Out[28]//InputForm=
{{1}, {0}, {-7}, {"dog"}, {913}, {"cat"}, {1, 1}, {1, 0}, {1, -7},
{1, "dog"}, {1, 913}, {1, "cat"}, {0, 0}, {0, -7}, {0, "dog"}, {0, 913},
{0, "cat"}, {-7, -7}, {-7, "dog"}, {-7, 913}, {-7, "cat"}, {"dog",
"dog"},
{"dog", 913}, {"dog", "cat"}, {913, 913}, {913, "cat"}, {"cat", "cat"},
{1, 1, 1}, {1, 1, 0}, {1, 1, -7}, {1, 1, "dog"}, {1, 1, 913}, {1, 1,
"cat"},
{1, 0, 0}, {1, 0, -7}, {1, 0, "dog"}, {1, 0, 913}, {1, 0, "cat"},
{1, -7, -7}, {1, -7, "dog"}, {1, -7, 913}, {1, -7, "cat"},
{1, "dog", "dog"}, {1, "dog", 913}, {1, "dog", "cat"}, {1, 913, 913},
{1, 913, "cat"}, {1, "cat", "cat"}, {0, 0, 0}, {0, 0, -7}, {0, 0, "dog"},
{0, 0, 913}, {0, 0, "cat"}, {0, -7, -7}, {0, -7, "dog"}, {0, -7, 913},
{0, -7, "cat"}, {0, "dog", "dog"}, {0, "dog", 913}, {0, "dog", "cat"},
{0, 913, 913}, {0, 913, "cat"}, {0, "cat", "cat"}, {-7, -7, -7},
{-7, -7, "dog"}, {-7, -7, 913}, {-7, -7, "cat"}, {-7, "dog", "dog"},
{-7, "dog", 913}, {-7, "dog", "cat"}, {-7, 913, 913}, {-7, 913, "cat"},
{-7, "cat", "cat"}, {"dog", "dog", "dog"}, {"dog", "dog", 913},
{"dog", "dog", "cat"}, {"dog", 913, 913}, {"dog", 913, "cat"},
{"dog", "cat", "cat"}, {913, 913, 913}, {913, 913, "cat"},
{913, "cat", "cat"}, {"cat", "cat", "cat"}}
To my surprise, it seems quite efficient (after seeing some of the other
entrants, I had pretty much given up in the speed category).
In[5]:= Timing[Length[s7 = multiSetsToK[Range[20],7]]]
Out[5]= {1.49377, 888029}
In[6]:= Timing[Length[s8 = multiSetsToK[Range[20],8]]]
Out[6]= {5.42518, 3108104}
In[7]:= Timing[Length[s10 = multiSetsToK[Range[15],10]]]
Out[7]= {5.97709, 3268759}
Caveat: It seems to be memory hungry. I ran the largest examples on a 64
bit machine. Even packing the subsets via Developer`ToPackedArray does
not salvage it; something (maybe Part?) is later unpacking.
Daniel Lichtblau
Wolfram Research
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