Mixing
Generalisation of a generating function
Clash Royale CLAN TAG#URR8PPP
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I have asked a question here.
I want to reproduce the coefficients of a generating function of the form:
$$(1 + x)^2 (1 + x + x^2 + x^3+cdots+x^n)^n-1$$
It is important that $x_0$ and $x_n$ to be strictly 0 or 1 and $x_1$ to $x_2$ can be any number within 0 and n (nothing higher). Here are two examples:
For $n=3$ we have:
In: (1 + x)^2 (1 + x + x^2 + x^3)^2 // Expand
Out: 1 + 4 x + 8 x^2 + 12 x^3 + 14 x^4 + 12 x^5 + 8 x^6 + 4 x^7 + x^8
We can produce the coefficients as:
n = 3;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= 1];
v /. Solve[eqn, constraints, v, Integers], t, 0, 8];
Table[Length[tabel[[i]]], i, Length[tabel]]
which gives:
1, 4, 8, 12, 14, 12, 8, 4, 1
as desired. For $n=4$ one had to add extra constraint and change the $t$ range.
We have:
In: (1 + x)^2 (1 + x + x^2 + x^3 + x^4)^3 // Expand
Out: 1 + 5 x + 13 x^2 + 25 x^3 + 41 x^4 + 58 x^5 + 70 x^6 + 74 x^7 +
70 x^8 + 58 x^9 + 41 x^10 + 25 x^11 + 13 x^12 + 5 x^13 + x^14
thus $t$ should be from 0 to 14,
so we have:
n = 4;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= n, 0 <= v[[5]] <= 1];
v /. Solve[eqn, constraints, v, Integers], t, 0, 14];
Table[Length[tabel[[i]]], i, Length[tabel]]
I wonder if these modifications can be done automatically so that one doesn't have to add a constraint by hand and change the range.
Note: I want the output table to be in a format so that I can see all of the possibilities of the sums.
For instance, for n=3, table[[2]] should give all of the possibilities such that the total is equal to 1.
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
list-manipulation table recursion
edited 55 mins ago
Carl Woll
57.8k273150
asked 2 hours ago
William
37917
add a comment |Â
up vote
1
down vote
favorite
I have asked a question here.
I want to reproduce the coefficients of a generating function of the form:
$$(1 + x)^2 (1 + x + x^2 + x^3+cdots+x^n)^n-1$$
It is important that $x_0$ and $x_n$ to be strictly 0 or 1 and $x_1$ to $x_2$ can be any number within 0 and n (nothing higher). Here are two examples:
For $n=3$ we have:
In: (1 + x)^2 (1 + x + x^2 + x^3)^2 // Expand
Out: 1 + 4 x + 8 x^2 + 12 x^3 + 14 x^4 + 12 x^5 + 8 x^6 + 4 x^7 + x^8
We can produce the coefficients as:
n = 3;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= 1];
v /. Solve[eqn, constraints, v, Integers], t, 0, 8];
Table[Length[tabel[[i]]], i, Length[tabel]]
which gives:
1, 4, 8, 12, 14, 12, 8, 4, 1
as desired. For $n=4$ one had to add extra constraint and change the $t$ range.
We have:
In: (1 + x)^2 (1 + x + x^2 + x^3 + x^4)^3 // Expand
Out: 1 + 5 x + 13 x^2 + 25 x^3 + 41 x^4 + 58 x^5 + 70 x^6 + 74 x^7 +
70 x^8 + 58 x^9 + 41 x^10 + 25 x^11 + 13 x^12 + 5 x^13 + x^14
thus $t$ should be from 0 to 14,
so we have:
n = 4;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= n, 0 <= v[[5]] <= 1];
v /. Solve[eqn, constraints, v, Integers], t, 0, 14];
Table[Length[tabel[[i]]], i, Length[tabel]]
I wonder if these modifications can be done automatically so that one doesn't have to add a constraint by hand and change the range.
Note: I want the output table to be in a format so that I can see all of the possibilities of the sums.
For instance, for n=3, table[[2]] should give all of the possibilities such that the total is equal to 1.
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
list-manipulation table recursion
edited 55 mins ago
Carl Woll
57.8k273150
asked 2 hours ago
William
37917
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have asked a question here.
I want to reproduce the coefficients of a generating function of the form:
$$(1 + x)^2 (1 + x + x^2 + x^3+cdots+x^n)^n-1$$
It is important that $x_0$ and $x_n$ to be strictly 0 or 1 and $x_1$ to $x_2$ can be any number within 0 and n (nothing higher). Here are two examples:
For $n=3$ we have:
In: (1 + x)^2 (1 + x + x^2 + x^3)^2 // Expand
Out: 1 + 4 x + 8 x^2 + 12 x^3 + 14 x^4 + 12 x^5 + 8 x^6 + 4 x^7 + x^8
We can produce the coefficients as:
n = 3;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= 1];
v /. Solve[eqn, constraints, v, Integers], t, 0, 8];
Table[Length[tabel[[i]]], i, Length[tabel]]
which gives:
1, 4, 8, 12, 14, 12, 8, 4, 1
as desired. For $n=4$ one had to add extra constraint and change the $t$ range.
We have:
In: (1 + x)^2 (1 + x + x^2 + x^3 + x^4)^3 // Expand
Out: 1 + 5 x + 13 x^2 + 25 x^3 + 41 x^4 + 58 x^5 + 70 x^6 + 74 x^7 +
70 x^8 + 58 x^9 + 41 x^10 + 25 x^11 + 13 x^12 + 5 x^13 + x^14
thus $t$ should be from 0 to 14,
so we have:
n = 4;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= n, 0 <= v[[5]] <= 1];
v /. Solve[eqn, constraints, v, Integers], t, 0, 14];
Table[Length[tabel[[i]]], i, Length[tabel]]
I wonder if these modifications can be done automatically so that one doesn't have to add a constraint by hand and change the range.
Note: I want the output table to be in a format so that I can see all of the possibilities of the sums.
For instance, for n=3, table[[2]] should give all of the possibilities such that the total is equal to 1.
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
list-manipulation table recursion
edited 55 mins ago
Carl Woll
57.8k273150
asked 2 hours ago
William
37917
I have asked a question here.
I want to reproduce the coefficients of a generating function of the form:
$$(1 + x)^2 (1 + x + x^2 + x^3+cdots+x^n)^n-1$$
It is important that $x_0$ and $x_n$ to be strictly 0 or 1 and $x_1$ to $x_2$ can be any number within 0 and n (nothing higher). Here are two examples:
For $n=3$ we have:
In: (1 + x)^2 (1 + x + x^2 + x^3)^2 // Expand
Out: 1 + 4 x + 8 x^2 + 12 x^3 + 14 x^4 + 12 x^5 + 8 x^6 + 4 x^7 + x^8
We can produce the coefficients as:
n = 3;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= 1];
v /. Solve[eqn, constraints, v, Integers], t, 0, 8];
Table[Length[tabel[[i]]], i, Length[tabel]]
which gives:
1, 4, 8, 12, 14, 12, 8, 4, 1
as desired. For $n=4$ one had to add extra constraint and change the $t$ range.
We have:
In: (1 + x)^2 (1 + x + x^2 + x^3 + x^4)^3 // Expand
Out: 1 + 5 x + 13 x^2 + 25 x^3 + 41 x^4 + 58 x^5 + 70 x^6 + 74 x^7 +
70 x^8 + 58 x^9 + 41 x^10 + 25 x^11 + 13 x^12 + 5 x^13 + x^14
thus $t$ should be from 0 to 14,
so we have:
n = 4;
m = n + 1;
tabel = Table[
v = Array[x, m, 0];
eqn = Total[v] == t;
constraints =
And[0 <= v[[1]] <= 1, 0 <= v[[2]] <= n, 0 <= v[[3]] <= n,
0 <= v[[4]] <= n, 0 <= v[[5]] <= 1];
v /. Solve[eqn, constraints, v, Integers], t, 0, 14];
Table[Length[tabel[[i]]], i, Length[tabel]]
I wonder if these modifications can be done automatically so that one doesn't have to add a constraint by hand and change the range.
Note: I want the output table to be in a format so that I can see all of the possibilities of the sums.
For instance, for n=3, table[[2]] should give all of the possibilities such that the total is equal to 1.
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
list-manipulation table recursion
list-manipulation table recursion
edited 55 mins ago
Carl Woll
57.8k273150
asked 2 hours ago
William
37917
edited 55 mins ago
Carl Woll
57.8k273150
asked 2 hours ago
William
37917
edited 55 mins ago
Carl Woll
57.8k273150
edited 55 mins ago
Carl Woll
57.8k273150
edited 55 mins ago
Carl Woll
57.8k273150
57.8k273150
asked 2 hours ago
William
37917
asked 2 hours ago
William
37917
asked 2 hours ago
William
37917
37917
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Update
An even faster method (I also modified the order of each possibility as requested in the comments):
tups[n_] := Values @ GroupBy[
Tuples[Join[Range[2], ConstantArray[Range[n+1], n-1], Range[2]] - 1],
Total
]
A comparison with the accepted answer:
r1 = tups[3]; //AbsoluteTiming
r2 = gen[3]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.000123, Null
0.001075, Null
True
Almost an order of magnitude faster for $n=3$. For $n=7$:
r1 = tups[7]; //AbsoluteTiming
r2 = gen[7]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.195056, Null
29.7257, Null
True
Original method
Here's a modification of @kglr's answer to your linked question:
sums[n_]:= Last @ Reap[
Array[Sow[##, Plus[##]]&, Join[2, ConstantArray[n+1, n-1], 2], 0],
_,
#2&
]
For $n=3$:
sums[3][[2]]
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
And a couple checks:
Length /@ sums[3]
Length /@ sums[4]
1, 4, 8, 12, 14, 12, 8, 4, 1
1, 5, 13, 25, 41, 58, 70, 74, 70, 58, 41, 25, 13, 5, 1
answered 1 hour ago
Carl Woll
57.8k273150
Thanks Carol, while it produces the correct answer, I need the elements to be in right place, as I will use them later as a list for other computation, If you try sums[3][[3]] you'd see the first element is not right, 0, 0, 0, 2 the first and last element of this must be 0 or 1, so the only way to assign 2 is 0, 2, 0, 0 and 0,0,2,0 to produce the sum of 2 without violating the constraints. This was happening at kglr's answer.
– William
37 mins ago
1
@William I modified the code as requested. Note that my answer is over an order of magnitude faster than the accepted answer.
– Carl Woll
22 mins ago
1
Yes, this is much nicer for the task at hand (and I for one upvoted).
– Daniel Lichtblau
19 mins ago
@CarlWoll & Daniel I have changed my acceptation. due to timing. Thanks again.
– William
7 mins ago
add a comment |Â
up vote
2
down vote
One can automate the generating construction readily enough using Product and Sum. Getting total degrees and then the subsets of balues that give them is a bit more work. I show one method below.
gen[n_] := Module[
vars, x, t, genFunc, coeffs,
vars = Array[x, n + 1, 0];
genFunc = (1 + First[vars])*(1 + Last[vars])*
Product[Sum[vars[[j]]^k, k, 0, n], j, 2, n] /.
Thread[vars -> t*vars];
coeffs = GroebnerBasis`DistributedTermsList[genFunc, t][[1]];
Map[#[[1, 1]],
GroebnerBasis`DistributedTermsList[#[[2]], vars][[1, All, 1]] &,
coeffs]
]
Example:
gen[3]
(* Out[59]= 8, 1, 3, 3, 1, 7, 1, 3, 3, 0, 1, 3, 2, 1, 1, 2,
3, 1, 0, 3, 3, 1, 6, 1, 3, 2, 0, 1, 3, 1, 1, 1, 2, 3,
0, 1, 2, 2, 1, 1, 1, 3, 1, 0, 3, 3, 0, 0, 3, 2, 1, 0,
2, 3, 1, 5, 1, 3, 1, 0, 1, 3, 0, 1, 1, 2, 2, 0, 1, 2,
1, 1, 1, 1, 3, 0, 1, 1, 2, 1, 1, 0, 3, 1, 0, 3, 2, 0, 0,
3, 1, 1, 0, 2, 3, 0, 0, 2, 2, 1, 0, 1, 3, 1, 4, 1, 3,
0, 0, 1, 2, 1, 0, 1, 2, 0, 1, 1, 1, 2, 0, 1, 1, 1,
1, 1, 0, 3, 0, 1, 0, 2, 1, 0, 3, 1, 0, 0, 3, 0, 1, 0, 2,
2, 0, 0, 2, 1, 1, 0, 1, 3, 0, 0, 1, 2, 1, 0, 0, 3,
1, 3, 1, 2, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 2,
0, 1, 0, 1, 1, 0, 3, 0, 0, 0, 2, 1, 0, 0, 2, 0, 1, 0, 1,
2, 0, 0, 1, 1, 1, 0, 0, 3, 0, 0, 0, 2, 1, 2, 1, 1, 0,
0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 1, 0, 0,
1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1,
0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 *)
Looks spiffier if one uses TableForm or similar.
answered 41 mins ago
Daniel Lichtblau
45.3k274155
add a comment |Â
up vote
1
down vote
Perhaps I didn't understand your question, but
Table[ CoefficientList[(1 + x)^2 (Sum[x^i, i, 0, n])^(n - 1),x] , n, 1, 5]
evaluates the list of coefficients you're looking for.
answered 2 hours ago
Ulrich Neumann
4,842413
Yes this is wrong. As I said I need to know the possible sum. The table there gives me all the possible sums but yours does not.
– William
2 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Update
An even faster method (I also modified the order of each possibility as requested in the comments):
tups[n_] := Values @ GroupBy[
Tuples[Join[Range[2], ConstantArray[Range[n+1], n-1], Range[2]] - 1],
Total
]
A comparison with the accepted answer:
r1 = tups[3]; //AbsoluteTiming
r2 = gen[3]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.000123, Null
0.001075, Null
True
Almost an order of magnitude faster for $n=3$. For $n=7$:
r1 = tups[7]; //AbsoluteTiming
r2 = gen[7]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.195056, Null
29.7257, Null
True
Original method
Here's a modification of @kglr's answer to your linked question:
sums[n_]:= Last @ Reap[
Array[Sow[##, Plus[##]]&, Join[2, ConstantArray[n+1, n-1], 2], 0],
_,
#2&
]
For $n=3$:
sums[3][[2]]
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
And a couple checks:
Length /@ sums[3]
Length /@ sums[4]
1, 4, 8, 12, 14, 12, 8, 4, 1
1, 5, 13, 25, 41, 58, 70, 74, 70, 58, 41, 25, 13, 5, 1
answered 1 hour ago
Carl Woll
57.8k273150
Thanks Carol, while it produces the correct answer, I need the elements to be in right place, as I will use them later as a list for other computation, If you try sums[3][[3]] you'd see the first element is not right, 0, 0, 0, 2 the first and last element of this must be 0 or 1, so the only way to assign 2 is 0, 2, 0, 0 and 0,0,2,0 to produce the sum of 2 without violating the constraints. This was happening at kglr's answer.
– William
37 mins ago
1
@William I modified the code as requested. Note that my answer is over an order of magnitude faster than the accepted answer.
– Carl Woll
22 mins ago
1
Yes, this is much nicer for the task at hand (and I for one upvoted).
– Daniel Lichtblau
19 mins ago
@CarlWoll & Daniel I have changed my acceptation. due to timing. Thanks again.
– William
7 mins ago
add a comment |Â
up vote
2
down vote
accepted
Update
An even faster method (I also modified the order of each possibility as requested in the comments):
tups[n_] := Values @ GroupBy[
Tuples[Join[Range[2], ConstantArray[Range[n+1], n-1], Range[2]] - 1],
Total
]
A comparison with the accepted answer:
r1 = tups[3]; //AbsoluteTiming
r2 = gen[3]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.000123, Null
0.001075, Null
True
Almost an order of magnitude faster for $n=3$. For $n=7$:
r1 = tups[7]; //AbsoluteTiming
r2 = gen[7]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.195056, Null
29.7257, Null
True
Original method
Here's a modification of @kglr's answer to your linked question:
sums[n_]:= Last @ Reap[
Array[Sow[##, Plus[##]]&, Join[2, ConstantArray[n+1, n-1], 2], 0],
_,
#2&
]
For $n=3$:
sums[3][[2]]
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
And a couple checks:
Length /@ sums[3]
Length /@ sums[4]
1, 4, 8, 12, 14, 12, 8, 4, 1
1, 5, 13, 25, 41, 58, 70, 74, 70, 58, 41, 25, 13, 5, 1
answered 1 hour ago
Carl Woll
57.8k273150
Thanks Carol, while it produces the correct answer, I need the elements to be in right place, as I will use them later as a list for other computation, If you try sums[3][[3]] you'd see the first element is not right, 0, 0, 0, 2 the first and last element of this must be 0 or 1, so the only way to assign 2 is 0, 2, 0, 0 and 0,0,2,0 to produce the sum of 2 without violating the constraints. This was happening at kglr's answer.
– William
37 mins ago
1
@William I modified the code as requested. Note that my answer is over an order of magnitude faster than the accepted answer.
– Carl Woll
22 mins ago
1
Yes, this is much nicer for the task at hand (and I for one upvoted).
– Daniel Lichtblau
19 mins ago
@CarlWoll & Daniel I have changed my acceptation. due to timing. Thanks again.
– William
7 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Update
An even faster method (I also modified the order of each possibility as requested in the comments):
tups[n_] := Values @ GroupBy[
Tuples[Join[Range[2], ConstantArray[Range[n+1], n-1], Range[2]] - 1],
Total
]
A comparison with the accepted answer:
r1 = tups[3]; //AbsoluteTiming
r2 = gen[3]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.000123, Null
0.001075, Null
True
Almost an order of magnitude faster for $n=3$. For $n=7$:
r1 = tups[7]; //AbsoluteTiming
r2 = gen[7]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.195056, Null
29.7257, Null
True
Original method
Here's a modification of @kglr's answer to your linked question:
sums[n_]:= Last @ Reap[
Array[Sow[##, Plus[##]]&, Join[2, ConstantArray[n+1, n-1], 2], 0],
_,
#2&
]
For $n=3$:
sums[3][[2]]
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
And a couple checks:
Length /@ sums[3]
Length /@ sums[4]
1, 4, 8, 12, 14, 12, 8, 4, 1
1, 5, 13, 25, 41, 58, 70, 74, 70, 58, 41, 25, 13, 5, 1
answered 1 hour ago
Carl Woll
57.8k273150
Update
An even faster method (I also modified the order of each possibility as requested in the comments):
tups[n_] := Values @ GroupBy[
Tuples[Join[Range[2], ConstantArray[Range[n+1], n-1], Range[2]] - 1],
Total
]
A comparison with the accepted answer:
r1 = tups[3]; //AbsoluteTiming
r2 = gen[3]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.000123, Null
0.001075, Null
True
Almost an order of magnitude faster for $n=3$. For $n=7$:
r1 = tups[7]; //AbsoluteTiming
r2 = gen[7]; //AbsoluteTiming
Sort /@ r1 === Sort /@ Reverse @ r2[[All, 2]]
0.195056, Null
29.7257, Null
True
Original method
Here's a modification of @kglr's answer to your linked question:
sums[n_]:= Last @ Reap[
Array[Sow[##, Plus[##]]&, Join[2, ConstantArray[n+1, n-1], 2], 0],
_,
#2&
]
For $n=3$:
sums[3][[2]]
0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0
And a couple checks:
Length /@ sums[3]
Length /@ sums[4]
1, 4, 8, 12, 14, 12, 8, 4, 1
1, 5, 13, 25, 41, 58, 70, 74, 70, 58, 41, 25, 13, 5, 1
answered 1 hour ago
Carl Woll
57.8k273150
edited 16 mins ago
answered 1 hour ago
Carl Woll
57.8k273150
answered 1 hour ago
Carl Woll
57.8k273150
answered 1 hour ago
Carl Woll
57.8k273150
57.8k273150
Thanks Carol, while it produces the correct answer, I need the elements to be in right place, as I will use them later as a list for other computation, If you try sums[3][[3]] you'd see the first element is not right, 0, 0, 0, 2 the first and last element of this must be 0 or 1, so the only way to assign 2 is 0, 2, 0, 0 and 0,0,2,0 to produce the sum of 2 without violating the constraints. This was happening at kglr's answer.
– William
37 mins ago
1
@William I modified the code as requested. Note that my answer is over an order of magnitude faster than the accepted answer.
– Carl Woll
22 mins ago
1
Yes, this is much nicer for the task at hand (and I for one upvoted).
– Daniel Lichtblau
19 mins ago
@CarlWoll & Daniel I have changed my acceptation. due to timing. Thanks again.
– William
7 mins ago
add a comment |Â
Thanks Carol, while it produces the correct answer, I need the elements to be in right place, as I will use them later as a list for other computation, If you try sums[3][[3]] you'd see the first element is not right, 0, 0, 0, 2 the first and last element of this must be 0 or 1, so the only way to assign 2 is 0, 2, 0, 0 and 0,0,2,0 to produce the sum of 2 without violating the constraints. This was happening at kglr's answer.
– William
37 mins ago
1
@William I modified the code as requested. Note that my answer is over an order of magnitude faster than the accepted answer.
– Carl Woll
22 mins ago
1
Yes, this is much nicer for the task at hand (and I for one upvoted).
– Daniel Lichtblau
19 mins ago
@CarlWoll & Daniel I have changed my acceptation. due to timing. Thanks again.
– William
7 mins ago
Thanks Carol, while it produces the correct answer, I need the elements to be in right place, as I will use them later as a list for other computation, If you try sums[3][[3]] you'd see the first element is not right, 0, 0, 0, 2 the first and last element of this must be 0 or 1, so the only way to assign 2 is 0, 2, 0, 0 and 0,0,2,0 to produce the sum of 2 without violating the constraints. This was happening at kglr's answer.
– William
37 mins ago
Thanks Carol, while it produces the correct answer, I need the elements to be in right place, as I will use them later as a list for other computation, If you try sums[3][[3]] you'd see the first element is not right, 0, 0, 0, 2 the first and last element of this must be 0 or 1, so the only way to assign 2 is 0, 2, 0, 0 and 0,0,2,0 to produce the sum of 2 without violating the constraints. This was happening at kglr's answer.
– William
37 mins ago
1
1
@William I modified the code as requested. Note that my answer is over an order of magnitude faster than the accepted answer.
– Carl Woll
22 mins ago
@William I modified the code as requested. Note that my answer is over an order of magnitude faster than the accepted answer.
– Carl Woll
22 mins ago
1
1
Yes, this is much nicer for the task at hand (and I for one upvoted).
– Daniel Lichtblau
19 mins ago
Yes, this is much nicer for the task at hand (and I for one upvoted).
– Daniel Lichtblau
19 mins ago
@CarlWoll & Daniel I have changed my acceptation. due to timing. Thanks again.
– William
7 mins ago
@CarlWoll & Daniel I have changed my acceptation. due to timing. Thanks again.
– William
7 mins ago
add a comment |Â
up vote
2
down vote
One can automate the generating construction readily enough using Product and Sum. Getting total degrees and then the subsets of balues that give them is a bit more work. I show one method below.
gen[n_] := Module[
vars, x, t, genFunc, coeffs,
vars = Array[x, n + 1, 0];
genFunc = (1 + First[vars])*(1 + Last[vars])*
Product[Sum[vars[[j]]^k, k, 0, n], j, 2, n] /.
Thread[vars -> t*vars];
coeffs = GroebnerBasis`DistributedTermsList[genFunc, t][[1]];
Map[#[[1, 1]],
GroebnerBasis`DistributedTermsList[#[[2]], vars][[1, All, 1]] &,
coeffs]
]
Example:
gen[3]
(* Out[59]= 8, 1, 3, 3, 1, 7, 1, 3, 3, 0, 1, 3, 2, 1, 1, 2,
3, 1, 0, 3, 3, 1, 6, 1, 3, 2, 0, 1, 3, 1, 1, 1, 2, 3,
0, 1, 2, 2, 1, 1, 1, 3, 1, 0, 3, 3, 0, 0, 3, 2, 1, 0,
2, 3, 1, 5, 1, 3, 1, 0, 1, 3, 0, 1, 1, 2, 2, 0, 1, 2,
1, 1, 1, 1, 3, 0, 1, 1, 2, 1, 1, 0, 3, 1, 0, 3, 2, 0, 0,
3, 1, 1, 0, 2, 3, 0, 0, 2, 2, 1, 0, 1, 3, 1, 4, 1, 3,
0, 0, 1, 2, 1, 0, 1, 2, 0, 1, 1, 1, 2, 0, 1, 1, 1,
1, 1, 0, 3, 0, 1, 0, 2, 1, 0, 3, 1, 0, 0, 3, 0, 1, 0, 2,
2, 0, 0, 2, 1, 1, 0, 1, 3, 0, 0, 1, 2, 1, 0, 0, 3,
1, 3, 1, 2, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 2,
0, 1, 0, 1, 1, 0, 3, 0, 0, 0, 2, 1, 0, 0, 2, 0, 1, 0, 1,
2, 0, 0, 1, 1, 1, 0, 0, 3, 0, 0, 0, 2, 1, 2, 1, 1, 0,
0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 1, 0, 0,
1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1,
0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 *)
Looks spiffier if one uses TableForm or similar.
answered 41 mins ago
Daniel Lichtblau
45.3k274155
add a comment |Â
up vote
2
down vote
One can automate the generating construction readily enough using Product and Sum. Getting total degrees and then the subsets of balues that give them is a bit more work. I show one method below.
gen[n_] := Module[
vars, x, t, genFunc, coeffs,
vars = Array[x, n + 1, 0];
genFunc = (1 + First[vars])*(1 + Last[vars])*
Product[Sum[vars[[j]]^k, k, 0, n], j, 2, n] /.
Thread[vars -> t*vars];
coeffs = GroebnerBasis`DistributedTermsList[genFunc, t][[1]];
Map[#[[1, 1]],
GroebnerBasis`DistributedTermsList[#[[2]], vars][[1, All, 1]] &,
coeffs]
]
Example:
gen[3]
(* Out[59]= 8, 1, 3, 3, 1, 7, 1, 3, 3, 0, 1, 3, 2, 1, 1, 2,
3, 1, 0, 3, 3, 1, 6, 1, 3, 2, 0, 1, 3, 1, 1, 1, 2, 3,
0, 1, 2, 2, 1, 1, 1, 3, 1, 0, 3, 3, 0, 0, 3, 2, 1, 0,
2, 3, 1, 5, 1, 3, 1, 0, 1, 3, 0, 1, 1, 2, 2, 0, 1, 2,
1, 1, 1, 1, 3, 0, 1, 1, 2, 1, 1, 0, 3, 1, 0, 3, 2, 0, 0,
3, 1, 1, 0, 2, 3, 0, 0, 2, 2, 1, 0, 1, 3, 1, 4, 1, 3,
0, 0, 1, 2, 1, 0, 1, 2, 0, 1, 1, 1, 2, 0, 1, 1, 1,
1, 1, 0, 3, 0, 1, 0, 2, 1, 0, 3, 1, 0, 0, 3, 0, 1, 0, 2,
2, 0, 0, 2, 1, 1, 0, 1, 3, 0, 0, 1, 2, 1, 0, 0, 3,
1, 3, 1, 2, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 2,
0, 1, 0, 1, 1, 0, 3, 0, 0, 0, 2, 1, 0, 0, 2, 0, 1, 0, 1,
2, 0, 0, 1, 1, 1, 0, 0, 3, 0, 0, 0, 2, 1, 2, 1, 1, 0,
0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 1, 0, 0,
1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1,
0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 *)
Looks spiffier if one uses TableForm or similar.
answered 41 mins ago
Daniel Lichtblau
45.3k274155
add a comment |Â
up vote
2
down vote
up vote
2
down vote
One can automate the generating construction readily enough using Product and Sum. Getting total degrees and then the subsets of balues that give them is a bit more work. I show one method below.
gen[n_] := Module[
vars, x, t, genFunc, coeffs,
vars = Array[x, n + 1, 0];
genFunc = (1 + First[vars])*(1 + Last[vars])*
Product[Sum[vars[[j]]^k, k, 0, n], j, 2, n] /.
Thread[vars -> t*vars];
coeffs = GroebnerBasis`DistributedTermsList[genFunc, t][[1]];
Map[#[[1, 1]],
GroebnerBasis`DistributedTermsList[#[[2]], vars][[1, All, 1]] &,
coeffs]
]
Example:
gen[3]
(* Out[59]= 8, 1, 3, 3, 1, 7, 1, 3, 3, 0, 1, 3, 2, 1, 1, 2,
3, 1, 0, 3, 3, 1, 6, 1, 3, 2, 0, 1, 3, 1, 1, 1, 2, 3,
0, 1, 2, 2, 1, 1, 1, 3, 1, 0, 3, 3, 0, 0, 3, 2, 1, 0,
2, 3, 1, 5, 1, 3, 1, 0, 1, 3, 0, 1, 1, 2, 2, 0, 1, 2,
1, 1, 1, 1, 3, 0, 1, 1, 2, 1, 1, 0, 3, 1, 0, 3, 2, 0, 0,
3, 1, 1, 0, 2, 3, 0, 0, 2, 2, 1, 0, 1, 3, 1, 4, 1, 3,
0, 0, 1, 2, 1, 0, 1, 2, 0, 1, 1, 1, 2, 0, 1, 1, 1,
1, 1, 0, 3, 0, 1, 0, 2, 1, 0, 3, 1, 0, 0, 3, 0, 1, 0, 2,
2, 0, 0, 2, 1, 1, 0, 1, 3, 0, 0, 1, 2, 1, 0, 0, 3,
1, 3, 1, 2, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 2,
0, 1, 0, 1, 1, 0, 3, 0, 0, 0, 2, 1, 0, 0, 2, 0, 1, 0, 1,
2, 0, 0, 1, 1, 1, 0, 0, 3, 0, 0, 0, 2, 1, 2, 1, 1, 0,
0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 1, 0, 0,
1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1,
0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 *)
Looks spiffier if one uses TableForm or similar.
answered 41 mins ago
Daniel Lichtblau
45.3k274155
One can automate the generating construction readily enough using Product and Sum. Getting total degrees and then the subsets of balues that give them is a bit more work. I show one method below.
gen[n_] := Module[
vars, x, t, genFunc, coeffs,
vars = Array[x, n + 1, 0];
genFunc = (1 + First[vars])*(1 + Last[vars])*
Product[Sum[vars[[j]]^k, k, 0, n], j, 2, n] /.
Thread[vars -> t*vars];
coeffs = GroebnerBasis`DistributedTermsList[genFunc, t][[1]];
Map[#[[1, 1]],
GroebnerBasis`DistributedTermsList[#[[2]], vars][[1, All, 1]] &,
coeffs]
]
Example:
gen[3]
(* Out[59]= 8, 1, 3, 3, 1, 7, 1, 3, 3, 0, 1, 3, 2, 1, 1, 2,
3, 1, 0, 3, 3, 1, 6, 1, 3, 2, 0, 1, 3, 1, 1, 1, 2, 3,
0, 1, 2, 2, 1, 1, 1, 3, 1, 0, 3, 3, 0, 0, 3, 2, 1, 0,
2, 3, 1, 5, 1, 3, 1, 0, 1, 3, 0, 1, 1, 2, 2, 0, 1, 2,
1, 1, 1, 1, 3, 0, 1, 1, 2, 1, 1, 0, 3, 1, 0, 3, 2, 0, 0,
3, 1, 1, 0, 2, 3, 0, 0, 2, 2, 1, 0, 1, 3, 1, 4, 1, 3,
0, 0, 1, 2, 1, 0, 1, 2, 0, 1, 1, 1, 2, 0, 1, 1, 1,
1, 1, 0, 3, 0, 1, 0, 2, 1, 0, 3, 1, 0, 0, 3, 0, 1, 0, 2,
2, 0, 0, 2, 1, 1, 0, 1, 3, 0, 0, 1, 2, 1, 0, 0, 3,
1, 3, 1, 2, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 2,
0, 1, 0, 1, 1, 0, 3, 0, 0, 0, 2, 1, 0, 0, 2, 0, 1, 0, 1,
2, 0, 0, 1, 1, 1, 0, 0, 3, 0, 0, 0, 2, 1, 2, 1, 1, 0,
0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 1, 0, 0,
1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1,
0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 *)
Looks spiffier if one uses TableForm or similar.
answered 41 mins ago
Daniel Lichtblau
45.3k274155
answered 41 mins ago
Daniel Lichtblau
45.3k274155
answered 41 mins ago
Daniel Lichtblau
45.3k274155
answered 41 mins ago
Daniel Lichtblau
45.3k274155
45.3k274155
add a comment |Â
add a comment |Â
up vote
1
down vote
Perhaps I didn't understand your question, but
Table[ CoefficientList[(1 + x)^2 (Sum[x^i, i, 0, n])^(n - 1),x] , n, 1, 5]
evaluates the list of coefficients you're looking for.
answered 2 hours ago
Ulrich Neumann
4,842413
Yes this is wrong. As I said I need to know the possible sum. The table there gives me all the possible sums but yours does not.
– William
2 hours ago
add a comment |Â
up vote
1
down vote
Perhaps I didn't understand your question, but
Table[ CoefficientList[(1 + x)^2 (Sum[x^i, i, 0, n])^(n - 1),x] , n, 1, 5]
evaluates the list of coefficients you're looking for.
answered 2 hours ago
Ulrich Neumann
4,842413
Yes this is wrong. As I said I need to know the possible sum. The table there gives me all the possible sums but yours does not.
– William
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Perhaps I didn't understand your question, but
Table[ CoefficientList[(1 + x)^2 (Sum[x^i, i, 0, n])^(n - 1),x] , n, 1, 5]
evaluates the list of coefficients you're looking for.
answered 2 hours ago
Ulrich Neumann
4,842413
Perhaps I didn't understand your question, but
Table[ CoefficientList[(1 + x)^2 (Sum[x^i, i, 0, n])^(n - 1),x] , n, 1, 5]
evaluates the list of coefficients you're looking for.
answered 2 hours ago
Ulrich Neumann
4,842413
answered 2 hours ago
Ulrich Neumann
4,842413
answered 2 hours ago
Ulrich Neumann
4,842413
answered 2 hours ago
Ulrich Neumann
4,842413
4,842413
Yes this is wrong. As I said I need to know the possible sum. The table there gives me all the possible sums but yours does not.
– William
2 hours ago
add a comment |Â
Yes this is wrong. As I said I need to know the possible sum. The table there gives me all the possible sums but yours does not.
– William
2 hours ago
Yes this is wrong. As I said I need to know the possible sum. The table there gives me all the possible sums but yours does not.
– William
2 hours ago
Yes this is wrong. As I said I need to know the possible sum. The table there gives me all the possible sums but yours does not.
– William
2 hours ago
add a comment |Â
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