Mixing
Partitioning a multiset into multisets of fixed sizes
Clash Royale CLAN TAG#URR8PPP
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Say we have a multiset $S(mathbfd$) where $mathbfd$ is a list of $l$ numbers and the multiplicity of the $i$th element of $S$ is $d_i$. The cardinality $N$ of $S$ is $sum d_i$.
We want to partition $S$ into $m$ multisets of size $k_i$ respectively, so that $sum k_i = sum d_i = N$. How many ways can we do this?
In my mind this is a generalization of the multinomial coefficient $binomnk_1,k_2,ldots,k_m$ representing the number of ways to partition a set of $n=sum k_i$ objects into $m$ bins of sizes $k_i$, to a sort of number like $binommathbfdk_1,k_2,ldots,k_m$ or $binommathbfdmathbfk$ representing the number of ways to partition a multiset of $n=sum k_i = sum d_i$ into $m$ bins of sizes $k_i$.
There are a few special cases that are simpler to calculate:
- If $m=1$, then clearly $k_1 = N$ and you're choosing the whole multiset. So $binommathbfd(N) = 1$
- If $m=2$, then you only have to handle choosing $k_1$ or $k_2$ elements from a multiset, because the rest will be the other set. So, as mentioned below, you can use a generating function and $binommathbfd(k_1,k_2)$ is equal to the coefficient of $x^k_1$ or $x^k_2$ in $prodlimits_i=1^l 1 + x^2 + cdots + x^d_i = prodlimits_i=1^l frac1-x^d_i - 11 - x$. But then you also need to account for the fact that order doesn't matter, which I'm not sure how to do. Like in the first example below, you would find that there are $3$ ways to choose $2$ elements, but there are only $2$ ways to split the multiset because you have to choose 2 of them that are compatible.
Examples
Let's say that $mathbfd = (2, 2)$, so $S(mathbfd)$ might be $a, a, b, b$. Let $k_1 = k_2 = 2$, so we need to find all the ways of splitting $S$ into two sub-multisets of size $2$. There are exactly $2$ ways of doing this: $a,a,b,b$ and $a,b,a,b$, so $binom(2,2)(2,2) = 2$.
Another example: $mathbfd = (2,2)$, so $S(mathbfd)$ could be $a,a,b,b$. Let $k_1 = 1$, $k_2 = 1$, and $k_3 = 2$. There are $3$ ways of doing this: $a,a,b,b$, $b,b,a,a$, and $a,b,a,b$. So $binom(2,2)(1,1,2)=3$.
My Attempts
I've tried to figure this out two ways. The first was to find a recurrence relation and some base cases, kind of how Stirling numbers of the second kind can be computed using the identity $S(n,k) = kS(n-1,k) + S(n-1,k-1)$. I tried to think about what happens if you already have a partition and want to add an element to the original multiset, but then you have to decide which bin that element will go into or whether or not to add a new bin.
I also tried to derive it in the way that multinomial coefficients are derived, by counting the number of ways to fill the first bin, and then the second, and so on. The number of ways to choose $k_1$ elements from the multiset to put in the first bin can be computed by finding the coefficient of $x^k_1$ in $prodlimits_i=1^l 1+x+x^2+cdots+x^d_i$, which isn't explicit but it's a start. But then, depending on which elements you chose, you don't know how to adjust your multiset to reflect the remaining elements.
combinatorics set-partition multisets
asked Jul 19 at 4:17
JJW5432
157111
add a comment |Â
up vote
3
down vote
favorite
Say we have a multiset $S(mathbfd$) where $mathbfd$ is a list of $l$ numbers and the multiplicity of the $i$th element of $S$ is $d_i$. The cardinality $N$ of $S$ is $sum d_i$.
We want to partition $S$ into $m$ multisets of size $k_i$ respectively, so that $sum k_i = sum d_i = N$. How many ways can we do this?
In my mind this is a generalization of the multinomial coefficient $binomnk_1,k_2,ldots,k_m$ representing the number of ways to partition a set of $n=sum k_i$ objects into $m$ bins of sizes $k_i$, to a sort of number like $binommathbfdk_1,k_2,ldots,k_m$ or $binommathbfdmathbfk$ representing the number of ways to partition a multiset of $n=sum k_i = sum d_i$ into $m$ bins of sizes $k_i$.
There are a few special cases that are simpler to calculate:
- If $m=1$, then clearly $k_1 = N$ and you're choosing the whole multiset. So $binommathbfd(N) = 1$
- If $m=2$, then you only have to handle choosing $k_1$ or $k_2$ elements from a multiset, because the rest will be the other set. So, as mentioned below, you can use a generating function and $binommathbfd(k_1,k_2)$ is equal to the coefficient of $x^k_1$ or $x^k_2$ in $prodlimits_i=1^l 1 + x^2 + cdots + x^d_i = prodlimits_i=1^l frac1-x^d_i - 11 - x$. But then you also need to account for the fact that order doesn't matter, which I'm not sure how to do. Like in the first example below, you would find that there are $3$ ways to choose $2$ elements, but there are only $2$ ways to split the multiset because you have to choose 2 of them that are compatible.
Examples
Let's say that $mathbfd = (2, 2)$, so $S(mathbfd)$ might be $a, a, b, b$. Let $k_1 = k_2 = 2$, so we need to find all the ways of splitting $S$ into two sub-multisets of size $2$. There are exactly $2$ ways of doing this: $a,a,b,b$ and $a,b,a,b$, so $binom(2,2)(2,2) = 2$.
Another example: $mathbfd = (2,2)$, so $S(mathbfd)$ could be $a,a,b,b$. Let $k_1 = 1$, $k_2 = 1$, and $k_3 = 2$. There are $3$ ways of doing this: $a,a,b,b$, $b,b,a,a$, and $a,b,a,b$. So $binom(2,2)(1,1,2)=3$.
My Attempts
I've tried to figure this out two ways. The first was to find a recurrence relation and some base cases, kind of how Stirling numbers of the second kind can be computed using the identity $S(n,k) = kS(n-1,k) + S(n-1,k-1)$. I tried to think about what happens if you already have a partition and want to add an element to the original multiset, but then you have to decide which bin that element will go into or whether or not to add a new bin.
I also tried to derive it in the way that multinomial coefficients are derived, by counting the number of ways to fill the first bin, and then the second, and so on. The number of ways to choose $k_1$ elements from the multiset to put in the first bin can be computed by finding the coefficient of $x^k_1$ in $prodlimits_i=1^l 1+x+x^2+cdots+x^d_i$, which isn't explicit but it's a start. But then, depending on which elements you chose, you don't know how to adjust your multiset to reflect the remaining elements.
combinatorics set-partition multisets
asked Jul 19 at 4:17
JJW5432
157111
Partitioning of a set of $n$ distinguished members is counted by the exponential Bell polynomial (and by Stirling number of the second kind and Bell number). This is the set partition problem. Maybe there is a possibility to change the calculation algorithm of this combinatorial numbers to consider multisets.
– IV_
Jul 20 at 14:46
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Say we have a multiset $S(mathbfd$) where $mathbfd$ is a list of $l$ numbers and the multiplicity of the $i$th element of $S$ is $d_i$. The cardinality $N$ of $S$ is $sum d_i$.
We want to partition $S$ into $m$ multisets of size $k_i$ respectively, so that $sum k_i = sum d_i = N$. How many ways can we do this?
In my mind this is a generalization of the multinomial coefficient $binomnk_1,k_2,ldots,k_m$ representing the number of ways to partition a set of $n=sum k_i$ objects into $m$ bins of sizes $k_i$, to a sort of number like $binommathbfdk_1,k_2,ldots,k_m$ or $binommathbfdmathbfk$ representing the number of ways to partition a multiset of $n=sum k_i = sum d_i$ into $m$ bins of sizes $k_i$.
There are a few special cases that are simpler to calculate:
- If $m=1$, then clearly $k_1 = N$ and you're choosing the whole multiset. So $binommathbfd(N) = 1$
- If $m=2$, then you only have to handle choosing $k_1$ or $k_2$ elements from a multiset, because the rest will be the other set. So, as mentioned below, you can use a generating function and $binommathbfd(k_1,k_2)$ is equal to the coefficient of $x^k_1$ or $x^k_2$ in $prodlimits_i=1^l 1 + x^2 + cdots + x^d_i = prodlimits_i=1^l frac1-x^d_i - 11 - x$. But then you also need to account for the fact that order doesn't matter, which I'm not sure how to do. Like in the first example below, you would find that there are $3$ ways to choose $2$ elements, but there are only $2$ ways to split the multiset because you have to choose 2 of them that are compatible.
Examples
Let's say that $mathbfd = (2, 2)$, so $S(mathbfd)$ might be $a, a, b, b$. Let $k_1 = k_2 = 2$, so we need to find all the ways of splitting $S$ into two sub-multisets of size $2$. There are exactly $2$ ways of doing this: $a,a,b,b$ and $a,b,a,b$, so $binom(2,2)(2,2) = 2$.
Another example: $mathbfd = (2,2)$, so $S(mathbfd)$ could be $a,a,b,b$. Let $k_1 = 1$, $k_2 = 1$, and $k_3 = 2$. There are $3$ ways of doing this: $a,a,b,b$, $b,b,a,a$, and $a,b,a,b$. So $binom(2,2)(1,1,2)=3$.
My Attempts
I've tried to figure this out two ways. The first was to find a recurrence relation and some base cases, kind of how Stirling numbers of the second kind can be computed using the identity $S(n,k) = kS(n-1,k) + S(n-1,k-1)$. I tried to think about what happens if you already have a partition and want to add an element to the original multiset, but then you have to decide which bin that element will go into or whether or not to add a new bin.
I also tried to derive it in the way that multinomial coefficients are derived, by counting the number of ways to fill the first bin, and then the second, and so on. The number of ways to choose $k_1$ elements from the multiset to put in the first bin can be computed by finding the coefficient of $x^k_1$ in $prodlimits_i=1^l 1+x+x^2+cdots+x^d_i$, which isn't explicit but it's a start. But then, depending on which elements you chose, you don't know how to adjust your multiset to reflect the remaining elements.
combinatorics set-partition multisets
asked Jul 19 at 4:17
JJW5432
157111
Say we have a multiset $S(mathbfd$) where $mathbfd$ is a list of $l$ numbers and the multiplicity of the $i$th element of $S$ is $d_i$. The cardinality $N$ of $S$ is $sum d_i$.
We want to partition $S$ into $m$ multisets of size $k_i$ respectively, so that $sum k_i = sum d_i = N$. How many ways can we do this?
In my mind this is a generalization of the multinomial coefficient $binomnk_1,k_2,ldots,k_m$ representing the number of ways to partition a set of $n=sum k_i$ objects into $m$ bins of sizes $k_i$, to a sort of number like $binommathbfdk_1,k_2,ldots,k_m$ or $binommathbfdmathbfk$ representing the number of ways to partition a multiset of $n=sum k_i = sum d_i$ into $m$ bins of sizes $k_i$.
There are a few special cases that are simpler to calculate:
- If $m=1$, then clearly $k_1 = N$ and you're choosing the whole multiset. So $binommathbfd(N) = 1$
- If $m=2$, then you only have to handle choosing $k_1$ or $k_2$ elements from a multiset, because the rest will be the other set. So, as mentioned below, you can use a generating function and $binommathbfd(k_1,k_2)$ is equal to the coefficient of $x^k_1$ or $x^k_2$ in $prodlimits_i=1^l 1 + x^2 + cdots + x^d_i = prodlimits_i=1^l frac1-x^d_i - 11 - x$. But then you also need to account for the fact that order doesn't matter, which I'm not sure how to do. Like in the first example below, you would find that there are $3$ ways to choose $2$ elements, but there are only $2$ ways to split the multiset because you have to choose 2 of them that are compatible.
Examples
Let's say that $mathbfd = (2, 2)$, so $S(mathbfd)$ might be $a, a, b, b$. Let $k_1 = k_2 = 2$, so we need to find all the ways of splitting $S$ into two sub-multisets of size $2$. There are exactly $2$ ways of doing this: $a,a,b,b$ and $a,b,a,b$, so $binom(2,2)(2,2) = 2$.
Another example: $mathbfd = (2,2)$, so $S(mathbfd)$ could be $a,a,b,b$. Let $k_1 = 1$, $k_2 = 1$, and $k_3 = 2$. There are $3$ ways of doing this: $a,a,b,b$, $b,b,a,a$, and $a,b,a,b$. So $binom(2,2)(1,1,2)=3$.
My Attempts
I've tried to figure this out two ways. The first was to find a recurrence relation and some base cases, kind of how Stirling numbers of the second kind can be computed using the identity $S(n,k) = kS(n-1,k) + S(n-1,k-1)$. I tried to think about what happens if you already have a partition and want to add an element to the original multiset, but then you have to decide which bin that element will go into or whether or not to add a new bin.
I also tried to derive it in the way that multinomial coefficients are derived, by counting the number of ways to fill the first bin, and then the second, and so on. The number of ways to choose $k_1$ elements from the multiset to put in the first bin can be computed by finding the coefficient of $x^k_1$ in $prodlimits_i=1^l 1+x+x^2+cdots+x^d_i$, which isn't explicit but it's a start. But then, depending on which elements you chose, you don't know how to adjust your multiset to reflect the remaining elements.
combinatorics set-partition multisets
asked Jul 19 at 4:17
JJW5432
157111
edited Jul 20 at 14:45
asked Jul 19 at 4:17
JJW5432
157111
asked Jul 19 at 4:17
JJW5432
157111
asked Jul 19 at 4:17
JJW5432
157111
157111
Partitioning of a set of $n$ distinguished members is counted by the exponential Bell polynomial (and by Stirling number of the second kind and Bell number). This is the set partition problem. Maybe there is a possibility to change the calculation algorithm of this combinatorial numbers to consider multisets.
– IV_
Jul 20 at 14:46
add a comment |Â
Partitioning of a set of $n$ distinguished members is counted by the exponential Bell polynomial (and by Stirling number of the second kind and Bell number). This is the set partition problem. Maybe there is a possibility to change the calculation algorithm of this combinatorial numbers to consider multisets.
– IV_
Jul 20 at 14:46
Partitioning of a set of $n$ distinguished members is counted by the exponential Bell polynomial (and by Stirling number of the second kind and Bell number). This is the set partition problem. Maybe there is a possibility to change the calculation algorithm of this combinatorial numbers to consider multisets.
– IV_
Jul 20 at 14:46
Partitioning of a set of $n$ distinguished members is counted by the exponential Bell polynomial (and by Stirling number of the second kind and Bell number). This is the set partition problem. Maybe there is a possibility to change the calculation algorithm of this combinatorial numbers to consider multisets.
– IV_
Jul 20 at 14:46
add a comment |Â
1 Answer
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up vote
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It would appear that these are multisets of multisets which may be
enumerated using the Polya Enumeration Theorem (PET). Let the multiset
that is being drawn have factorization
$$prod_k=1^m B_k^sigma_k$$
where $k$ is the value of a term and $sigma_k$ the number of times it
ocurs and recall that we have $l$ types of items forming the source
multiset
$$prod_k=1^l A_k^tau_k.$$
The answer is then given by
$$left[prod_k=1^l A_k^tau_kright]
prod_k=1^m
Zleft(S_sigma_k;
Zleft(S_k; sum_k'=1^l A_k'right)right).$$
In terms of combinatorial classes we have made use of the unlabeled
class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscMSET_=sigma_k
left(textscMSET_=k
left(sum_k'=1^l mathcalA_k'right)right).$$
As an example for $2,2choose 1,1,2 = 3$ we get
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2))
times textscMSET_=1
(textscMSET_=2(mathcalA_1+mathcalA_2)).$$
As an additional example we find for $2,2,4choose 1,1,3,3 = 16$
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2+mathcalA_3))
times textscMSET_=2
(textscMSET_=3(mathcalA_1+mathcalA_2+mathcalA_3)).$$
Here we have used the cycle index of the symmetric group $Z(S_n)$,
which is computed from the recurrence by Lovasz which says that
$$Z(S_n) = frac1n sum_l=1^n a_l Z(S_n-l)
quadtextwherequad
Z(S_0) = 1.$$
For this to be effective we need to compute the iterated cycle index
when $Z(S_k)$ is substituted into $Z(S_sigma_k).$ This is
accomplished with the substitution rule for substitution of the former
into the latter:
$$a_q = Z(S_k;; a_1=a_q, ; a_2=a_2q, ; a_3=a_3q, ; ldots).$$
We have used the notation $Z(S_k; F)$ for substitution of a generating
function and on the previous line, the notation for substitution into
the variables of the cycle index. This is in fact all we need and we
can start computing some of these multiset coefficients. For example
we find for the example given by OP the cycle index
$$Z(B_1^2 B_2) =
1/4,a_1^4+1/2,a_1^2a_2+1/4,a_2^2.$$
Continuing with the example we get
$$Z(B_1^2 B_2; A_1+A_2) =
1/4, left( A_1+A_2 right) ^4
+1/2, left( A_1+A_2 right) ^2
left( A_1^2+A_2^2 right)
\ +1/4, left( A_1^2+A_2^2
right) ^2 \ = A_1^4+2,A_1^3A_2
+3,A_1^2A_2^2+2,A_1A_2
^3+A_2^4$$
and we confirm the value $3$ obtained by OP. This algorithm will make
it possible to compute cycle indices not obtainable by enumeration. As
an extra example we find the following excerpt from the cycle index
for $[2,2,3,5,5]:$
$$Z(B_2^3 B_3 B_5^2) = ldots
+frac 11,a_1^8a_2a_4a_57200
+frac 49,a_1^7a_2^2a_3a_514400
\ +frac 5,a_1^7 a_2a_3^2a_41152
+frac 1021,a_1^6a_2^3a_3a_469120
+frac 43,a_1^7a_2a_4a_617280+ldots$$
Here are some additional values that may assist the reader who is
investigating this problem to verify the results of their approach:
$$1,3,3choose 3,4 = 7, ;
2,3,3choose 4,4 = 5, ;
2,3,3choose 2,2,4 = 16
quadtextandquad
1,2,3,3choose 2,3,4 = 87.$$
The Maple code for this problem was as follows.
with(combinat);
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
expand(1/n*
add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_cycleind_comp :=
proc(idxtrg, idx)
local varstrg, vars, vt, sbstrg, sbs, len;
varstrg := indets(idxtrg);
vars := indets(idx);
sbstrg := ;
for vt in varstrg do
len := op(1, vt);
sbs :=
[seq(v = a[op(1, v)*len], v in vars)];
sbstrg :=
[op(sbstrg),
a[len] = subs(sbs, idx)];
od;
expand(subs(sbstrg, idxtrg));
end;
pet_cycleind_mset :=
proc(msetlst)
option remember;
local mset, res, ent, idxtrg, idx;
mset := convert(msetlst, `multiset`);
res := 1;
for ent in mset do
idx := pet_cycleind_symm(ent[1]);
idxtrg := pet_cycleind_symm(ent[2]);
res := res *
pet_cycleind_comp(idxtrg, idx);
od;
expand(res);
end;
GENF :=
proc(src, msetlst)
local vars, srcp, res, gf, term;
vars := add(A[q], q=1..nops(src));
srcp := mul(A[q]^src[q], q=1..nops(src));
gf := expand(pet_varinto_cind
(vars, pet_cycleind_mset(msetlst)));
if not type(gf, `+`) then
gf := [gf];
fi;
res := 0;
for term in gf do
if type(srcp/term, `polynom`) then
res := res + term;
fi;
od;
res;
end;
The syntax to compute $mathrmAchoose mathrmB$ is documented
by the following examples:
> GENF([1,2,3,3], [2,3,4]);
2 3 3
87 A[1] A[2] A[3] A[4]
> GENF([1,2,3,3], [2,2,5]);
2 3 3
33 A[1] A[2] A[3] A[4]
> GENF([1,1,1,1], [2,2]);
3 A[1] A[2] A[3] A[4].
The last one is $frac12 4choose 2.$
answered Jul 25 at 15:12
Marko Riedel
36.5k333107
This is fantastic! I'm having some trouble understanding why this works, and specifically why this is a multiset of multisets, which I think is related to why we're itersting the cycle index of the symmetric group. What do you mean by a factorization of a multiset, and source multiset?
– JJW5432
Jul 26 at 0:14
Thank you. The cycle index of the symmetric group implements the multiset operator in the symbolic method. Now your subsets are multisets and since we may choose them more than once we get multisets of multisets. The support of a multiset is a partition and in cycle indices these are usually written in factorized form, hence I chose to use this notation here as well.
– Marko Riedel
Jul 26 at 9:22
For more information consult Wikipedia on the symbolic method.
– Marko Riedel
Jul 26 at 9:31
I've read the first chapter of Flajolet and Sedgewick's "Analytic Combinatorics" on the symbolic method, so now I understand more about combinatorial classes and generator functions. Can you explain why the construction $$deftextsc#1dosc#1csod defdosc#1#2csodrm #1small #2 textscMSET_=sigma_k left(textscMSET_=k left(sum_k'=1^l mathcalA_k'right)right)$$ is the correct one?
– JJW5432
Jul 28 at 3:39
This part follows by inspection for the user of the symbolic method. I have added two examples to clarify the notation. The book by Harary and Palmer Graphical Enumeration is a useful reference for the nested cycle indices. I also checked my work using several enumeration routines for cycle indices and multisets.
– Marko Riedel
Jul 28 at 12:50
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It would appear that these are multisets of multisets which may be
enumerated using the Polya Enumeration Theorem (PET). Let the multiset
that is being drawn have factorization
$$prod_k=1^m B_k^sigma_k$$
where $k$ is the value of a term and $sigma_k$ the number of times it
ocurs and recall that we have $l$ types of items forming the source
multiset
$$prod_k=1^l A_k^tau_k.$$
The answer is then given by
$$left[prod_k=1^l A_k^tau_kright]
prod_k=1^m
Zleft(S_sigma_k;
Zleft(S_k; sum_k'=1^l A_k'right)right).$$
In terms of combinatorial classes we have made use of the unlabeled
class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscMSET_=sigma_k
left(textscMSET_=k
left(sum_k'=1^l mathcalA_k'right)right).$$
As an example for $2,2choose 1,1,2 = 3$ we get
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2))
times textscMSET_=1
(textscMSET_=2(mathcalA_1+mathcalA_2)).$$
As an additional example we find for $2,2,4choose 1,1,3,3 = 16$
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2+mathcalA_3))
times textscMSET_=2
(textscMSET_=3(mathcalA_1+mathcalA_2+mathcalA_3)).$$
Here we have used the cycle index of the symmetric group $Z(S_n)$,
which is computed from the recurrence by Lovasz which says that
$$Z(S_n) = frac1n sum_l=1^n a_l Z(S_n-l)
quadtextwherequad
Z(S_0) = 1.$$
For this to be effective we need to compute the iterated cycle index
when $Z(S_k)$ is substituted into $Z(S_sigma_k).$ This is
accomplished with the substitution rule for substitution of the former
into the latter:
$$a_q = Z(S_k;; a_1=a_q, ; a_2=a_2q, ; a_3=a_3q, ; ldots).$$
We have used the notation $Z(S_k; F)$ for substitution of a generating
function and on the previous line, the notation for substitution into
the variables of the cycle index. This is in fact all we need and we
can start computing some of these multiset coefficients. For example
we find for the example given by OP the cycle index
$$Z(B_1^2 B_2) =
1/4,a_1^4+1/2,a_1^2a_2+1/4,a_2^2.$$
Continuing with the example we get
$$Z(B_1^2 B_2; A_1+A_2) =
1/4, left( A_1+A_2 right) ^4
+1/2, left( A_1+A_2 right) ^2
left( A_1^2+A_2^2 right)
\ +1/4, left( A_1^2+A_2^2
right) ^2 \ = A_1^4+2,A_1^3A_2
+3,A_1^2A_2^2+2,A_1A_2
^3+A_2^4$$
and we confirm the value $3$ obtained by OP. This algorithm will make
it possible to compute cycle indices not obtainable by enumeration. As
an extra example we find the following excerpt from the cycle index
for $[2,2,3,5,5]:$
$$Z(B_2^3 B_3 B_5^2) = ldots
+frac 11,a_1^8a_2a_4a_57200
+frac 49,a_1^7a_2^2a_3a_514400
\ +frac 5,a_1^7 a_2a_3^2a_41152
+frac 1021,a_1^6a_2^3a_3a_469120
+frac 43,a_1^7a_2a_4a_617280+ldots$$
Here are some additional values that may assist the reader who is
investigating this problem to verify the results of their approach:
$$1,3,3choose 3,4 = 7, ;
2,3,3choose 4,4 = 5, ;
2,3,3choose 2,2,4 = 16
quadtextandquad
1,2,3,3choose 2,3,4 = 87.$$
The Maple code for this problem was as follows.
with(combinat);
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
expand(1/n*
add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_cycleind_comp :=
proc(idxtrg, idx)
local varstrg, vars, vt, sbstrg, sbs, len;
varstrg := indets(idxtrg);
vars := indets(idx);
sbstrg := ;
for vt in varstrg do
len := op(1, vt);
sbs :=
[seq(v = a[op(1, v)*len], v in vars)];
sbstrg :=
[op(sbstrg),
a[len] = subs(sbs, idx)];
od;
expand(subs(sbstrg, idxtrg));
end;
pet_cycleind_mset :=
proc(msetlst)
option remember;
local mset, res, ent, idxtrg, idx;
mset := convert(msetlst, `multiset`);
res := 1;
for ent in mset do
idx := pet_cycleind_symm(ent[1]);
idxtrg := pet_cycleind_symm(ent[2]);
res := res *
pet_cycleind_comp(idxtrg, idx);
od;
expand(res);
end;
GENF :=
proc(src, msetlst)
local vars, srcp, res, gf, term;
vars := add(A[q], q=1..nops(src));
srcp := mul(A[q]^src[q], q=1..nops(src));
gf := expand(pet_varinto_cind
(vars, pet_cycleind_mset(msetlst)));
if not type(gf, `+`) then
gf := [gf];
fi;
res := 0;
for term in gf do
if type(srcp/term, `polynom`) then
res := res + term;
fi;
od;
res;
end;
The syntax to compute $mathrmAchoose mathrmB$ is documented
by the following examples:
> GENF([1,2,3,3], [2,3,4]);
2 3 3
87 A[1] A[2] A[3] A[4]
> GENF([1,2,3,3], [2,2,5]);
2 3 3
33 A[1] A[2] A[3] A[4]
> GENF([1,1,1,1], [2,2]);
3 A[1] A[2] A[3] A[4].
The last one is $frac12 4choose 2.$
answered Jul 25 at 15:12
Marko Riedel
36.5k333107
This is fantastic! I'm having some trouble understanding why this works, and specifically why this is a multiset of multisets, which I think is related to why we're itersting the cycle index of the symmetric group. What do you mean by a factorization of a multiset, and source multiset?
– JJW5432
Jul 26 at 0:14
Thank you. The cycle index of the symmetric group implements the multiset operator in the symbolic method. Now your subsets are multisets and since we may choose them more than once we get multisets of multisets. The support of a multiset is a partition and in cycle indices these are usually written in factorized form, hence I chose to use this notation here as well.
– Marko Riedel
Jul 26 at 9:22
For more information consult Wikipedia on the symbolic method.
– Marko Riedel
Jul 26 at 9:31
I've read the first chapter of Flajolet and Sedgewick's "Analytic Combinatorics" on the symbolic method, so now I understand more about combinatorial classes and generator functions. Can you explain why the construction $$deftextsc#1dosc#1csod defdosc#1#2csodrm #1small #2 textscMSET_=sigma_k left(textscMSET_=k left(sum_k'=1^l mathcalA_k'right)right)$$ is the correct one?
– JJW5432
Jul 28 at 3:39
This part follows by inspection for the user of the symbolic method. I have added two examples to clarify the notation. The book by Harary and Palmer Graphical Enumeration is a useful reference for the nested cycle indices. I also checked my work using several enumeration routines for cycle indices and multisets.
– Marko Riedel
Jul 28 at 12:50
 |Â
show 8 more comments
up vote
1
down vote
accepted
It would appear that these are multisets of multisets which may be
enumerated using the Polya Enumeration Theorem (PET). Let the multiset
that is being drawn have factorization
$$prod_k=1^m B_k^sigma_k$$
where $k$ is the value of a term and $sigma_k$ the number of times it
ocurs and recall that we have $l$ types of items forming the source
multiset
$$prod_k=1^l A_k^tau_k.$$
The answer is then given by
$$left[prod_k=1^l A_k^tau_kright]
prod_k=1^m
Zleft(S_sigma_k;
Zleft(S_k; sum_k'=1^l A_k'right)right).$$
In terms of combinatorial classes we have made use of the unlabeled
class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscMSET_=sigma_k
left(textscMSET_=k
left(sum_k'=1^l mathcalA_k'right)right).$$
As an example for $2,2choose 1,1,2 = 3$ we get
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2))
times textscMSET_=1
(textscMSET_=2(mathcalA_1+mathcalA_2)).$$
As an additional example we find for $2,2,4choose 1,1,3,3 = 16$
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2+mathcalA_3))
times textscMSET_=2
(textscMSET_=3(mathcalA_1+mathcalA_2+mathcalA_3)).$$
Here we have used the cycle index of the symmetric group $Z(S_n)$,
which is computed from the recurrence by Lovasz which says that
$$Z(S_n) = frac1n sum_l=1^n a_l Z(S_n-l)
quadtextwherequad
Z(S_0) = 1.$$
For this to be effective we need to compute the iterated cycle index
when $Z(S_k)$ is substituted into $Z(S_sigma_k).$ This is
accomplished with the substitution rule for substitution of the former
into the latter:
$$a_q = Z(S_k;; a_1=a_q, ; a_2=a_2q, ; a_3=a_3q, ; ldots).$$
We have used the notation $Z(S_k; F)$ for substitution of a generating
function and on the previous line, the notation for substitution into
the variables of the cycle index. This is in fact all we need and we
can start computing some of these multiset coefficients. For example
we find for the example given by OP the cycle index
$$Z(B_1^2 B_2) =
1/4,a_1^4+1/2,a_1^2a_2+1/4,a_2^2.$$
Continuing with the example we get
$$Z(B_1^2 B_2; A_1+A_2) =
1/4, left( A_1+A_2 right) ^4
+1/2, left( A_1+A_2 right) ^2
left( A_1^2+A_2^2 right)
\ +1/4, left( A_1^2+A_2^2
right) ^2 \ = A_1^4+2,A_1^3A_2
+3,A_1^2A_2^2+2,A_1A_2
^3+A_2^4$$
and we confirm the value $3$ obtained by OP. This algorithm will make
it possible to compute cycle indices not obtainable by enumeration. As
an extra example we find the following excerpt from the cycle index
for $[2,2,3,5,5]:$
$$Z(B_2^3 B_3 B_5^2) = ldots
+frac 11,a_1^8a_2a_4a_57200
+frac 49,a_1^7a_2^2a_3a_514400
\ +frac 5,a_1^7 a_2a_3^2a_41152
+frac 1021,a_1^6a_2^3a_3a_469120
+frac 43,a_1^7a_2a_4a_617280+ldots$$
Here are some additional values that may assist the reader who is
investigating this problem to verify the results of their approach:
$$1,3,3choose 3,4 = 7, ;
2,3,3choose 4,4 = 5, ;
2,3,3choose 2,2,4 = 16
quadtextandquad
1,2,3,3choose 2,3,4 = 87.$$
The Maple code for this problem was as follows.
with(combinat);
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
expand(1/n*
add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_cycleind_comp :=
proc(idxtrg, idx)
local varstrg, vars, vt, sbstrg, sbs, len;
varstrg := indets(idxtrg);
vars := indets(idx);
sbstrg := ;
for vt in varstrg do
len := op(1, vt);
sbs :=
[seq(v = a[op(1, v)*len], v in vars)];
sbstrg :=
[op(sbstrg),
a[len] = subs(sbs, idx)];
od;
expand(subs(sbstrg, idxtrg));
end;
pet_cycleind_mset :=
proc(msetlst)
option remember;
local mset, res, ent, idxtrg, idx;
mset := convert(msetlst, `multiset`);
res := 1;
for ent in mset do
idx := pet_cycleind_symm(ent[1]);
idxtrg := pet_cycleind_symm(ent[2]);
res := res *
pet_cycleind_comp(idxtrg, idx);
od;
expand(res);
end;
GENF :=
proc(src, msetlst)
local vars, srcp, res, gf, term;
vars := add(A[q], q=1..nops(src));
srcp := mul(A[q]^src[q], q=1..nops(src));
gf := expand(pet_varinto_cind
(vars, pet_cycleind_mset(msetlst)));
if not type(gf, `+`) then
gf := [gf];
fi;
res := 0;
for term in gf do
if type(srcp/term, `polynom`) then
res := res + term;
fi;
od;
res;
end;
The syntax to compute $mathrmAchoose mathrmB$ is documented
by the following examples:
> GENF([1,2,3,3], [2,3,4]);
2 3 3
87 A[1] A[2] A[3] A[4]
> GENF([1,2,3,3], [2,2,5]);
2 3 3
33 A[1] A[2] A[3] A[4]
> GENF([1,1,1,1], [2,2]);
3 A[1] A[2] A[3] A[4].
The last one is $frac12 4choose 2.$
answered Jul 25 at 15:12
Marko Riedel
36.5k333107
This is fantastic! I'm having some trouble understanding why this works, and specifically why this is a multiset of multisets, which I think is related to why we're itersting the cycle index of the symmetric group. What do you mean by a factorization of a multiset, and source multiset?
– JJW5432
Jul 26 at 0:14
Thank you. The cycle index of the symmetric group implements the multiset operator in the symbolic method. Now your subsets are multisets and since we may choose them more than once we get multisets of multisets. The support of a multiset is a partition and in cycle indices these are usually written in factorized form, hence I chose to use this notation here as well.
– Marko Riedel
Jul 26 at 9:22
For more information consult Wikipedia on the symbolic method.
– Marko Riedel
Jul 26 at 9:31
I've read the first chapter of Flajolet and Sedgewick's "Analytic Combinatorics" on the symbolic method, so now I understand more about combinatorial classes and generator functions. Can you explain why the construction $$deftextsc#1dosc#1csod defdosc#1#2csodrm #1small #2 textscMSET_=sigma_k left(textscMSET_=k left(sum_k'=1^l mathcalA_k'right)right)$$ is the correct one?
– JJW5432
Jul 28 at 3:39
This part follows by inspection for the user of the symbolic method. I have added two examples to clarify the notation. The book by Harary and Palmer Graphical Enumeration is a useful reference for the nested cycle indices. I also checked my work using several enumeration routines for cycle indices and multisets.
– Marko Riedel
Jul 28 at 12:50
 |Â
show 8 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It would appear that these are multisets of multisets which may be
enumerated using the Polya Enumeration Theorem (PET). Let the multiset
that is being drawn have factorization
$$prod_k=1^m B_k^sigma_k$$
where $k$ is the value of a term and $sigma_k$ the number of times it
ocurs and recall that we have $l$ types of items forming the source
multiset
$$prod_k=1^l A_k^tau_k.$$
The answer is then given by
$$left[prod_k=1^l A_k^tau_kright]
prod_k=1^m
Zleft(S_sigma_k;
Zleft(S_k; sum_k'=1^l A_k'right)right).$$
In terms of combinatorial classes we have made use of the unlabeled
class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscMSET_=sigma_k
left(textscMSET_=k
left(sum_k'=1^l mathcalA_k'right)right).$$
As an example for $2,2choose 1,1,2 = 3$ we get
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2))
times textscMSET_=1
(textscMSET_=2(mathcalA_1+mathcalA_2)).$$
As an additional example we find for $2,2,4choose 1,1,3,3 = 16$
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2+mathcalA_3))
times textscMSET_=2
(textscMSET_=3(mathcalA_1+mathcalA_2+mathcalA_3)).$$
Here we have used the cycle index of the symmetric group $Z(S_n)$,
which is computed from the recurrence by Lovasz which says that
$$Z(S_n) = frac1n sum_l=1^n a_l Z(S_n-l)
quadtextwherequad
Z(S_0) = 1.$$
For this to be effective we need to compute the iterated cycle index
when $Z(S_k)$ is substituted into $Z(S_sigma_k).$ This is
accomplished with the substitution rule for substitution of the former
into the latter:
$$a_q = Z(S_k;; a_1=a_q, ; a_2=a_2q, ; a_3=a_3q, ; ldots).$$
We have used the notation $Z(S_k; F)$ for substitution of a generating
function and on the previous line, the notation for substitution into
the variables of the cycle index. This is in fact all we need and we
can start computing some of these multiset coefficients. For example
we find for the example given by OP the cycle index
$$Z(B_1^2 B_2) =
1/4,a_1^4+1/2,a_1^2a_2+1/4,a_2^2.$$
Continuing with the example we get
$$Z(B_1^2 B_2; A_1+A_2) =
1/4, left( A_1+A_2 right) ^4
+1/2, left( A_1+A_2 right) ^2
left( A_1^2+A_2^2 right)
\ +1/4, left( A_1^2+A_2^2
right) ^2 \ = A_1^4+2,A_1^3A_2
+3,A_1^2A_2^2+2,A_1A_2
^3+A_2^4$$
and we confirm the value $3$ obtained by OP. This algorithm will make
it possible to compute cycle indices not obtainable by enumeration. As
an extra example we find the following excerpt from the cycle index
for $[2,2,3,5,5]:$
$$Z(B_2^3 B_3 B_5^2) = ldots
+frac 11,a_1^8a_2a_4a_57200
+frac 49,a_1^7a_2^2a_3a_514400
\ +frac 5,a_1^7 a_2a_3^2a_41152
+frac 1021,a_1^6a_2^3a_3a_469120
+frac 43,a_1^7a_2a_4a_617280+ldots$$
Here are some additional values that may assist the reader who is
investigating this problem to verify the results of their approach:
$$1,3,3choose 3,4 = 7, ;
2,3,3choose 4,4 = 5, ;
2,3,3choose 2,2,4 = 16
quadtextandquad
1,2,3,3choose 2,3,4 = 87.$$
The Maple code for this problem was as follows.
with(combinat);
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
expand(1/n*
add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_cycleind_comp :=
proc(idxtrg, idx)
local varstrg, vars, vt, sbstrg, sbs, len;
varstrg := indets(idxtrg);
vars := indets(idx);
sbstrg := ;
for vt in varstrg do
len := op(1, vt);
sbs :=
[seq(v = a[op(1, v)*len], v in vars)];
sbstrg :=
[op(sbstrg),
a[len] = subs(sbs, idx)];
od;
expand(subs(sbstrg, idxtrg));
end;
pet_cycleind_mset :=
proc(msetlst)
option remember;
local mset, res, ent, idxtrg, idx;
mset := convert(msetlst, `multiset`);
res := 1;
for ent in mset do
idx := pet_cycleind_symm(ent[1]);
idxtrg := pet_cycleind_symm(ent[2]);
res := res *
pet_cycleind_comp(idxtrg, idx);
od;
expand(res);
end;
GENF :=
proc(src, msetlst)
local vars, srcp, res, gf, term;
vars := add(A[q], q=1..nops(src));
srcp := mul(A[q]^src[q], q=1..nops(src));
gf := expand(pet_varinto_cind
(vars, pet_cycleind_mset(msetlst)));
if not type(gf, `+`) then
gf := [gf];
fi;
res := 0;
for term in gf do
if type(srcp/term, `polynom`) then
res := res + term;
fi;
od;
res;
end;
The syntax to compute $mathrmAchoose mathrmB$ is documented
by the following examples:
> GENF([1,2,3,3], [2,3,4]);
2 3 3
87 A[1] A[2] A[3] A[4]
> GENF([1,2,3,3], [2,2,5]);
2 3 3
33 A[1] A[2] A[3] A[4]
> GENF([1,1,1,1], [2,2]);
3 A[1] A[2] A[3] A[4].
The last one is $frac12 4choose 2.$
answered Jul 25 at 15:12
Marko Riedel
36.5k333107
It would appear that these are multisets of multisets which may be
enumerated using the Polya Enumeration Theorem (PET). Let the multiset
that is being drawn have factorization
$$prod_k=1^m B_k^sigma_k$$
where $k$ is the value of a term and $sigma_k$ the number of times it
ocurs and recall that we have $l$ types of items forming the source
multiset
$$prod_k=1^l A_k^tau_k.$$
The answer is then given by
$$left[prod_k=1^l A_k^tau_kright]
prod_k=1^m
Zleft(S_sigma_k;
Zleft(S_k; sum_k'=1^l A_k'right)right).$$
In terms of combinatorial classes we have made use of the unlabeled
class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscMSET_=sigma_k
left(textscMSET_=k
left(sum_k'=1^l mathcalA_k'right)right).$$
As an example for $2,2choose 1,1,2 = 3$ we get
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2))
times textscMSET_=1
(textscMSET_=2(mathcalA_1+mathcalA_2)).$$
As an additional example we find for $2,2,4choose 1,1,3,3 = 16$
$$textscMSET_=2
(textscMSET_=1(mathcalA_1+mathcalA_2+mathcalA_3))
times textscMSET_=2
(textscMSET_=3(mathcalA_1+mathcalA_2+mathcalA_3)).$$
Here we have used the cycle index of the symmetric group $Z(S_n)$,
which is computed from the recurrence by Lovasz which says that
$$Z(S_n) = frac1n sum_l=1^n a_l Z(S_n-l)
quadtextwherequad
Z(S_0) = 1.$$
For this to be effective we need to compute the iterated cycle index
when $Z(S_k)$ is substituted into $Z(S_sigma_k).$ This is
accomplished with the substitution rule for substitution of the former
into the latter:
$$a_q = Z(S_k;; a_1=a_q, ; a_2=a_2q, ; a_3=a_3q, ; ldots).$$
We have used the notation $Z(S_k; F)$ for substitution of a generating
function and on the previous line, the notation for substitution into
the variables of the cycle index. This is in fact all we need and we
can start computing some of these multiset coefficients. For example
we find for the example given by OP the cycle index
$$Z(B_1^2 B_2) =
1/4,a_1^4+1/2,a_1^2a_2+1/4,a_2^2.$$
Continuing with the example we get
$$Z(B_1^2 B_2; A_1+A_2) =
1/4, left( A_1+A_2 right) ^4
+1/2, left( A_1+A_2 right) ^2
left( A_1^2+A_2^2 right)
\ +1/4, left( A_1^2+A_2^2
right) ^2 \ = A_1^4+2,A_1^3A_2
+3,A_1^2A_2^2+2,A_1A_2
^3+A_2^4$$
and we confirm the value $3$ obtained by OP. This algorithm will make
it possible to compute cycle indices not obtainable by enumeration. As
an extra example we find the following excerpt from the cycle index
for $[2,2,3,5,5]:$
$$Z(B_2^3 B_3 B_5^2) = ldots
+frac 11,a_1^8a_2a_4a_57200
+frac 49,a_1^7a_2^2a_3a_514400
\ +frac 5,a_1^7 a_2a_3^2a_41152
+frac 1021,a_1^6a_2^3a_3a_469120
+frac 43,a_1^7a_2a_4a_617280+ldots$$
Here are some additional values that may assist the reader who is
investigating this problem to verify the results of their approach:
$$1,3,3choose 3,4 = 7, ;
2,3,3choose 4,4 = 5, ;
2,3,3choose 2,2,4 = 16
quadtextandquad
1,2,3,3choose 2,3,4 = 87.$$
The Maple code for this problem was as follows.
with(combinat);
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
expand(1/n*
add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_cycleind_comp :=
proc(idxtrg, idx)
local varstrg, vars, vt, sbstrg, sbs, len;
varstrg := indets(idxtrg);
vars := indets(idx);
sbstrg := ;
for vt in varstrg do
len := op(1, vt);
sbs :=
[seq(v = a[op(1, v)*len], v in vars)];
sbstrg :=
[op(sbstrg),
a[len] = subs(sbs, idx)];
od;
expand(subs(sbstrg, idxtrg));
end;
pet_cycleind_mset :=
proc(msetlst)
option remember;
local mset, res, ent, idxtrg, idx;
mset := convert(msetlst, `multiset`);
res := 1;
for ent in mset do
idx := pet_cycleind_symm(ent[1]);
idxtrg := pet_cycleind_symm(ent[2]);
res := res *
pet_cycleind_comp(idxtrg, idx);
od;
expand(res);
end;
GENF :=
proc(src, msetlst)
local vars, srcp, res, gf, term;
vars := add(A[q], q=1..nops(src));
srcp := mul(A[q]^src[q], q=1..nops(src));
gf := expand(pet_varinto_cind
(vars, pet_cycleind_mset(msetlst)));
if not type(gf, `+`) then
gf := [gf];
fi;
res := 0;
for term in gf do
if type(srcp/term, `polynom`) then
res := res + term;
fi;
od;
res;
end;
The syntax to compute $mathrmAchoose mathrmB$ is documented
by the following examples:
> GENF([1,2,3,3], [2,3,4]);
2 3 3
87 A[1] A[2] A[3] A[4]
> GENF([1,2,3,3], [2,2,5]);
2 3 3
33 A[1] A[2] A[3] A[4]
> GENF([1,1,1,1], [2,2]);
3 A[1] A[2] A[3] A[4].
The last one is $frac12 4choose 2.$
answered Jul 25 at 15:12
Marko Riedel
36.5k333107
edited Jul 28 at 14:06
answered Jul 25 at 15:12
Marko Riedel
36.5k333107
answered Jul 25 at 15:12
Marko Riedel
36.5k333107
answered Jul 25 at 15:12
Marko Riedel
36.5k333107
36.5k333107
This is fantastic! I'm having some trouble understanding why this works, and specifically why this is a multiset of multisets, which I think is related to why we're itersting the cycle index of the symmetric group. What do you mean by a factorization of a multiset, and source multiset?
– JJW5432
Jul 26 at 0:14
Thank you. The cycle index of the symmetric group implements the multiset operator in the symbolic method. Now your subsets are multisets and since we may choose them more than once we get multisets of multisets. The support of a multiset is a partition and in cycle indices these are usually written in factorized form, hence I chose to use this notation here as well.
– Marko Riedel
Jul 26 at 9:22
For more information consult Wikipedia on the symbolic method.
– Marko Riedel
Jul 26 at 9:31
I've read the first chapter of Flajolet and Sedgewick's "Analytic Combinatorics" on the symbolic method, so now I understand more about combinatorial classes and generator functions. Can you explain why the construction $$deftextsc#1dosc#1csod defdosc#1#2csodrm #1small #2 textscMSET_=sigma_k left(textscMSET_=k left(sum_k'=1^l mathcalA_k'right)right)$$ is the correct one?
– JJW5432
Jul 28 at 3:39
This part follows by inspection for the user of the symbolic method. I have added two examples to clarify the notation. The book by Harary and Palmer Graphical Enumeration is a useful reference for the nested cycle indices. I also checked my work using several enumeration routines for cycle indices and multisets.
– Marko Riedel
Jul 28 at 12:50
 |Â
show 8 more comments
This is fantastic! I'm having some trouble understanding why this works, and specifically why this is a multiset of multisets, which I think is related to why we're itersting the cycle index of the symmetric group. What do you mean by a factorization of a multiset, and source multiset?
– JJW5432
Jul 26 at 0:14
Thank you. The cycle index of the symmetric group implements the multiset operator in the symbolic method. Now your subsets are multisets and since we may choose them more than once we get multisets of multisets. The support of a multiset is a partition and in cycle indices these are usually written in factorized form, hence I chose to use this notation here as well.
– Marko Riedel
Jul 26 at 9:22
For more information consult Wikipedia on the symbolic method.
– Marko Riedel
Jul 26 at 9:31
I've read the first chapter of Flajolet and Sedgewick's "Analytic Combinatorics" on the symbolic method, so now I understand more about combinatorial classes and generator functions. Can you explain why the construction $$deftextsc#1dosc#1csod defdosc#1#2csodrm #1small #2 textscMSET_=sigma_k left(textscMSET_=k left(sum_k'=1^l mathcalA_k'right)right)$$ is the correct one?
– JJW5432
Jul 28 at 3:39
This part follows by inspection for the user of the symbolic method. I have added two examples to clarify the notation. The book by Harary and Palmer Graphical Enumeration is a useful reference for the nested cycle indices. I also checked my work using several enumeration routines for cycle indices and multisets.
– Marko Riedel
Jul 28 at 12:50
This is fantastic! I'm having some trouble understanding why this works, and specifically why this is a multiset of multisets, which I think is related to why we're itersting the cycle index of the symmetric group. What do you mean by a factorization of a multiset, and source multiset?
– JJW5432
Jul 26 at 0:14
This is fantastic! I'm having some trouble understanding why this works, and specifically why this is a multiset of multisets, which I think is related to why we're itersting the cycle index of the symmetric group. What do you mean by a factorization of a multiset, and source multiset?
– JJW5432
Jul 26 at 0:14
Thank you. The cycle index of the symmetric group implements the multiset operator in the symbolic method. Now your subsets are multisets and since we may choose them more than once we get multisets of multisets. The support of a multiset is a partition and in cycle indices these are usually written in factorized form, hence I chose to use this notation here as well.
– Marko Riedel
Jul 26 at 9:22
Thank you. The cycle index of the symmetric group implements the multiset operator in the symbolic method. Now your subsets are multisets and since we may choose them more than once we get multisets of multisets. The support of a multiset is a partition and in cycle indices these are usually written in factorized form, hence I chose to use this notation here as well.
– Marko Riedel
Jul 26 at 9:22
For more information consult Wikipedia on the symbolic method.
– Marko Riedel
Jul 26 at 9:31
For more information consult Wikipedia on the symbolic method.
– Marko Riedel
Jul 26 at 9:31
I've read the first chapter of Flajolet and Sedgewick's "Analytic Combinatorics" on the symbolic method, so now I understand more about combinatorial classes and generator functions. Can you explain why the construction $$deftextsc#1dosc#1csod defdosc#1#2csodrm #1small #2 textscMSET_=sigma_k left(textscMSET_=k left(sum_k'=1^l mathcalA_k'right)right)$$ is the correct one?
– JJW5432
Jul 28 at 3:39
I've read the first chapter of Flajolet and Sedgewick's "Analytic Combinatorics" on the symbolic method, so now I understand more about combinatorial classes and generator functions. Can you explain why the construction $$deftextsc#1dosc#1csod defdosc#1#2csodrm #1small #2 textscMSET_=sigma_k left(textscMSET_=k left(sum_k'=1^l mathcalA_k'right)right)$$ is the correct one?
– JJW5432
Jul 28 at 3:39
This part follows by inspection for the user of the symbolic method. I have added two examples to clarify the notation. The book by Harary and Palmer Graphical Enumeration is a useful reference for the nested cycle indices. I also checked my work using several enumeration routines for cycle indices and multisets.
– Marko Riedel
Jul 28 at 12:50
This part follows by inspection for the user of the symbolic method. I have added two examples to clarify the notation. The book by Harary and Palmer Graphical Enumeration is a useful reference for the nested cycle indices. I also checked my work using several enumeration routines for cycle indices and multisets.
– Marko Riedel
Jul 28 at 12:50
 |Â
show 8 more comments
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Partitioning of a set of $n$ distinguished members is counted by the exponential Bell polynomial (and by Stirling number of the second kind and Bell number). This is the set partition problem. Maybe there is a possibility to change the calculation algorithm of this combinatorial numbers to consider multisets.
– IV_
Jul 20 at 14:46