Sum of list numbers smaller than one goal
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I have five values that I would like to add them so that they can be equal to or less than 3000.
v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Subsets [{v1, v2, v3, v4, v5}]
For example:
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v3
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v4
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v5
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2 + v2
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2 + v2 + v2
-> 2940
All are possibilities, among other possibilities.
Is there a feature that validates these possibilities?
combinatorics
asked Jan 21 at 18:15
LCarvalhoLCarvalho
5,68142886
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add a comment |
$begingroup$
I have five values that I would like to add them so that they can be equal to or less than 3000.
v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Subsets [{v1, v2, v3, v4, v5}]
For example:
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v3
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v4
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v5
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2 + v2
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2 + v2 + v2
-> 2940
All are possibilities, among other possibilities.
Is there a feature that validates these possibilities?
combinatorics
asked Jan 21 at 18:15
LCarvalhoLCarvalho
5,68142886
$endgroup$
add a comment |
$begingroup$
I have five values that I would like to add them so that they can be equal to or less than 3000.
v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Subsets [{v1, v2, v3, v4, v5}]
For example:
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v3
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v4
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v5
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2 + v2
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2 + v2 + v2
-> 2940
All are possibilities, among other possibilities.
Is there a feature that validates these possibilities?
combinatorics
asked Jan 21 at 18:15
LCarvalhoLCarvalho
5,68142886
$endgroup$
I have five values that I would like to add them so that they can be equal to or less than 3000.
v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Subsets [{v1, v2, v3, v4, v5}]
For example:
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v3
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v4
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v5
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2 + v2
-> 2940
v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v1 + v2 + v2 + v2
-> 2940
All are possibilities, among other possibilities.
Is there a feature that validates these possibilities?
combinatorics
combinatorics
asked Jan 21 at 18:15
LCarvalhoLCarvalho
5,68142886
asked Jan 21 at 18:15
LCarvalhoLCarvalho
5,68142886
edited Jan 21 at 19:18
LCarvalho
asked Jan 21 at 18:15
LCarvalhoLCarvalho
5,68142886
asked Jan 21 at 18:15
LCarvalhoLCarvalho
5,68142886
asked Jan 21 at 18:15
LCarvalhoLCarvalho
5,68142886
5,68142886
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Append 1 to the list of vs and use FrobeniusSolve:
w = {v1, v2, v3, v4, v5, 1};
res = FrobeniusSolve[w, 3000][[2;;, ;; 5]];
Length[res]
1345
Short @ res
{{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1336>>,{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
And @@ Thread[Total /@ res <= 3000]
True
Up to ordering, res is the same as Henrik's resulttable:
Sort[res] == Sort[resulttable]
True
An alternative way to use IntegerPartitions using w:
res2 = DeleteCases[Rest@IntegerPartitions[3000, All, w], 1, 2];
Length@res2
1345
restab = Map[Lookup[Counts[#], Most @ w, 0] &] @ res2;
Sort[res] == Sort[restab]
True
To get the totals that can be obtained using vs:
Sort[DeleteDuplicates[res.Most[w]]] (* or *)
Sort[3000 - DeleteDuplicates@Rest[FrobeniusSolve[w, 3000]][[All, -1]]]
{140, 280, 420, 560, 700, 840, 980, 1120, 1260, 1400, 1540, 1680,
1820, 1960, 2100, 2240, 2380, 2520, 2660, 2800, 2940}
Sort @ Counts[res.Most[w]]
<|140 -> 1, 280 -> 2, 420 -> 3, 560 -> 5, 700 -> 7, 840 -> 10,
980 -> 13, 1120 -> 18, 1260 -> 23, 1400 -> 30, 1540 -> 37,
1680 -> 47, 1820 -> 57, 1960 -> 70, 2100 -> 84, 2240 -> 101,
2380 -> 119, 2520 -> 141, 2660 -> 164, 2800 -> 192, 2940 -> 221|>
Update: You can also use Reduce and Solve:
v = {v1, v2, v3, v4, v5};
xv = Array[x, 5];
resReduce = Reduce[{v.xv <= 3000, ##&@@Thread[xv >= 0]}, xv, Integers][[2;;, All, -1]] /.
{And | Or -> List};
resSolve = xv /. Rest@Solve[{v.xv <= 3000, ## & @@ Thread[xv >= 0]}, xv, Integers];
res == resReduce == resSolve
True
answered Jan 21 at 19:34
kglrkglr
183k10201416
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add a comment |
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a = {140, 280, 420, 560, 700};
results = Join @@ DeleteCases[
IntegerPartitions[#, {1, ∞}, a] & /@ Range[3000],
{}
];
And @@ Thread[Total /@ results <= 3000]
resulttable = Map[Lookup[Counts[#], a, 0] &, results];
resulttable // Short
True
{{1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, <<1341>> , {19, 1, 0, 0, 0}, {21, 0, 0, 0, 0}}
answered Jan 21 at 18:41
Henrik SchumacherHenrik Schumacher
52.9k471148
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v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Select[Subsets[{v1, v2, v3, v4, v5}], Total[#] <= goal &]
Are you sure your "goal" is 3000? Every subset totals less than that.
And what does 21 * v1 in your question refer to? What is $21$, and why multiplication?
answered Jan 21 at 18:38
David G. StorkDavid G. Stork
24.4k22153
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a = {140, 280, 420, 560, 700};
coef = Tuples[Range[0, 21], 5];
vec = coef.a;
Extract[coef, Position[Ramp[vec - 3000], 0]] // Short
{{0,0,0,0,0},{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1335>>,{18,1,0,0,0},{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
Pick[vec, Ramp[vec - 3000], 0] // Short
{0,700,1400,2100,2800,560,1260,1960,2660,1120,1820,2520,1680,2380,<<1318>>,2940,2800,2380,2940,2800,2660,2940,2520,2940,2800,2660,2940,2800,2940}
answered Jan 21 at 19:45
Okkes DulgerciOkkes Dulgerci
4,9521817
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Could use integer linear programming via Maximize, for example. The constraint I use is overkill, all we really need to enforce is that values be nonnegative.
vals = {140, 280, 420, 560, 700};
goal = 3000;
vars = Array[v, Length[vals]];
Maximize[{vars.vals,
Flatten[{vars.vals <= goal,
MapIndexed[0 <= # <= Floor[goal/vals[[#2[[1]]]]] &,
vars]}]}, vars, Integers]
(* Out[1287]= {2940, {v[1] -> 10, v[2] -> 4, v[3] -> 1,
v[4] -> 0, v[5] -> 0}} *)
answered Jan 22 at 15:42
Daniel LichtblauDaniel Lichtblau
46.9k276163
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Append 1 to the list of vs and use FrobeniusSolve:
w = {v1, v2, v3, v4, v5, 1};
res = FrobeniusSolve[w, 3000][[2;;, ;; 5]];
Length[res]
1345
Short @ res
{{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1336>>,{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
And @@ Thread[Total /@ res <= 3000]
True
Up to ordering, res is the same as Henrik's resulttable:
Sort[res] == Sort[resulttable]
True
An alternative way to use IntegerPartitions using w:
res2 = DeleteCases[Rest@IntegerPartitions[3000, All, w], 1, 2];
Length@res2
1345
restab = Map[Lookup[Counts[#], Most @ w, 0] &] @ res2;
Sort[res] == Sort[restab]
True
To get the totals that can be obtained using vs:
Sort[DeleteDuplicates[res.Most[w]]] (* or *)
Sort[3000 - DeleteDuplicates@Rest[FrobeniusSolve[w, 3000]][[All, -1]]]
{140, 280, 420, 560, 700, 840, 980, 1120, 1260, 1400, 1540, 1680,
1820, 1960, 2100, 2240, 2380, 2520, 2660, 2800, 2940}
Sort @ Counts[res.Most[w]]
<|140 -> 1, 280 -> 2, 420 -> 3, 560 -> 5, 700 -> 7, 840 -> 10,
980 -> 13, 1120 -> 18, 1260 -> 23, 1400 -> 30, 1540 -> 37,
1680 -> 47, 1820 -> 57, 1960 -> 70, 2100 -> 84, 2240 -> 101,
2380 -> 119, 2520 -> 141, 2660 -> 164, 2800 -> 192, 2940 -> 221|>
Update: You can also use Reduce and Solve:
v = {v1, v2, v3, v4, v5};
xv = Array[x, 5];
resReduce = Reduce[{v.xv <= 3000, ##&@@Thread[xv >= 0]}, xv, Integers][[2;;, All, -1]] /.
{And | Or -> List};
resSolve = xv /. Rest@Solve[{v.xv <= 3000, ## & @@ Thread[xv >= 0]}, xv, Integers];
res == resReduce == resSolve
True
answered Jan 21 at 19:34
kglrkglr
183k10201416
$endgroup$
add a comment |
$begingroup$
Append 1 to the list of vs and use FrobeniusSolve:
w = {v1, v2, v3, v4, v5, 1};
res = FrobeniusSolve[w, 3000][[2;;, ;; 5]];
Length[res]
1345
Short @ res
{{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1336>>,{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
And @@ Thread[Total /@ res <= 3000]
True
Up to ordering, res is the same as Henrik's resulttable:
Sort[res] == Sort[resulttable]
True
An alternative way to use IntegerPartitions using w:
res2 = DeleteCases[Rest@IntegerPartitions[3000, All, w], 1, 2];
Length@res2
1345
restab = Map[Lookup[Counts[#], Most @ w, 0] &] @ res2;
Sort[res] == Sort[restab]
True
To get the totals that can be obtained using vs:
Sort[DeleteDuplicates[res.Most[w]]] (* or *)
Sort[3000 - DeleteDuplicates@Rest[FrobeniusSolve[w, 3000]][[All, -1]]]
{140, 280, 420, 560, 700, 840, 980, 1120, 1260, 1400, 1540, 1680,
1820, 1960, 2100, 2240, 2380, 2520, 2660, 2800, 2940}
Sort @ Counts[res.Most[w]]
<|140 -> 1, 280 -> 2, 420 -> 3, 560 -> 5, 700 -> 7, 840 -> 10,
980 -> 13, 1120 -> 18, 1260 -> 23, 1400 -> 30, 1540 -> 37,
1680 -> 47, 1820 -> 57, 1960 -> 70, 2100 -> 84, 2240 -> 101,
2380 -> 119, 2520 -> 141, 2660 -> 164, 2800 -> 192, 2940 -> 221|>
Update: You can also use Reduce and Solve:
v = {v1, v2, v3, v4, v5};
xv = Array[x, 5];
resReduce = Reduce[{v.xv <= 3000, ##&@@Thread[xv >= 0]}, xv, Integers][[2;;, All, -1]] /.
{And | Or -> List};
resSolve = xv /. Rest@Solve[{v.xv <= 3000, ## & @@ Thread[xv >= 0]}, xv, Integers];
res == resReduce == resSolve
True
answered Jan 21 at 19:34
kglrkglr
183k10201416
$endgroup$
add a comment |
$begingroup$
Append 1 to the list of vs and use FrobeniusSolve:
w = {v1, v2, v3, v4, v5, 1};
res = FrobeniusSolve[w, 3000][[2;;, ;; 5]];
Length[res]
1345
Short @ res
{{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1336>>,{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
And @@ Thread[Total /@ res <= 3000]
True
Up to ordering, res is the same as Henrik's resulttable:
Sort[res] == Sort[resulttable]
True
An alternative way to use IntegerPartitions using w:
res2 = DeleteCases[Rest@IntegerPartitions[3000, All, w], 1, 2];
Length@res2
1345
restab = Map[Lookup[Counts[#], Most @ w, 0] &] @ res2;
Sort[res] == Sort[restab]
True
To get the totals that can be obtained using vs:
Sort[DeleteDuplicates[res.Most[w]]] (* or *)
Sort[3000 - DeleteDuplicates@Rest[FrobeniusSolve[w, 3000]][[All, -1]]]
{140, 280, 420, 560, 700, 840, 980, 1120, 1260, 1400, 1540, 1680,
1820, 1960, 2100, 2240, 2380, 2520, 2660, 2800, 2940}
Sort @ Counts[res.Most[w]]
<|140 -> 1, 280 -> 2, 420 -> 3, 560 -> 5, 700 -> 7, 840 -> 10,
980 -> 13, 1120 -> 18, 1260 -> 23, 1400 -> 30, 1540 -> 37,
1680 -> 47, 1820 -> 57, 1960 -> 70, 2100 -> 84, 2240 -> 101,
2380 -> 119, 2520 -> 141, 2660 -> 164, 2800 -> 192, 2940 -> 221|>
Update: You can also use Reduce and Solve:
v = {v1, v2, v3, v4, v5};
xv = Array[x, 5];
resReduce = Reduce[{v.xv <= 3000, ##&@@Thread[xv >= 0]}, xv, Integers][[2;;, All, -1]] /.
{And | Or -> List};
resSolve = xv /. Rest@Solve[{v.xv <= 3000, ## & @@ Thread[xv >= 0]}, xv, Integers];
res == resReduce == resSolve
True
answered Jan 21 at 19:34
kglrkglr
183k10201416
$endgroup$
Append 1 to the list of vs and use FrobeniusSolve:
w = {v1, v2, v3, v4, v5, 1};
res = FrobeniusSolve[w, 3000][[2;;, ;; 5]];
Length[res]
1345
Short @ res
{{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1336>>,{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
And @@ Thread[Total /@ res <= 3000]
True
Up to ordering, res is the same as Henrik's resulttable:
Sort[res] == Sort[resulttable]
True
An alternative way to use IntegerPartitions using w:
res2 = DeleteCases[Rest@IntegerPartitions[3000, All, w], 1, 2];
Length@res2
1345
restab = Map[Lookup[Counts[#], Most @ w, 0] &] @ res2;
Sort[res] == Sort[restab]
True
To get the totals that can be obtained using vs:
Sort[DeleteDuplicates[res.Most[w]]] (* or *)
Sort[3000 - DeleteDuplicates@Rest[FrobeniusSolve[w, 3000]][[All, -1]]]
{140, 280, 420, 560, 700, 840, 980, 1120, 1260, 1400, 1540, 1680,
1820, 1960, 2100, 2240, 2380, 2520, 2660, 2800, 2940}
Sort @ Counts[res.Most[w]]
<|140 -> 1, 280 -> 2, 420 -> 3, 560 -> 5, 700 -> 7, 840 -> 10,
980 -> 13, 1120 -> 18, 1260 -> 23, 1400 -> 30, 1540 -> 37,
1680 -> 47, 1820 -> 57, 1960 -> 70, 2100 -> 84, 2240 -> 101,
2380 -> 119, 2520 -> 141, 2660 -> 164, 2800 -> 192, 2940 -> 221|>
Update: You can also use Reduce and Solve:
v = {v1, v2, v3, v4, v5};
xv = Array[x, 5];
resReduce = Reduce[{v.xv <= 3000, ##&@@Thread[xv >= 0]}, xv, Integers][[2;;, All, -1]] /.
{And | Or -> List};
resSolve = xv /. Rest@Solve[{v.xv <= 3000, ## & @@ Thread[xv >= 0]}, xv, Integers];
res == resReduce == resSolve
True
answered Jan 21 at 19:34
kglrkglr
183k10201416
edited Jan 22 at 14:35
answered Jan 21 at 19:34
kglrkglr
183k10201416
answered Jan 21 at 19:34
kglrkglr
183k10201416
answered Jan 21 at 19:34
kglrkglr
183k10201416
183k10201416
add a comment |
add a comment |
$begingroup$
a = {140, 280, 420, 560, 700};
results = Join @@ DeleteCases[
IntegerPartitions[#, {1, ∞}, a] & /@ Range[3000],
{}
];
And @@ Thread[Total /@ results <= 3000]
resulttable = Map[Lookup[Counts[#], a, 0] &, results];
resulttable // Short
True
{{1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, <<1341>> , {19, 1, 0, 0, 0}, {21, 0, 0, 0, 0}}
answered Jan 21 at 18:41
Henrik SchumacherHenrik Schumacher
52.9k471148
$endgroup$
add a comment |
$begingroup$
a = {140, 280, 420, 560, 700};
results = Join @@ DeleteCases[
IntegerPartitions[#, {1, ∞}, a] & /@ Range[3000],
{}
];
And @@ Thread[Total /@ results <= 3000]
resulttable = Map[Lookup[Counts[#], a, 0] &, results];
resulttable // Short
True
{{1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, <<1341>> , {19, 1, 0, 0, 0}, {21, 0, 0, 0, 0}}
answered Jan 21 at 18:41
Henrik SchumacherHenrik Schumacher
52.9k471148
$endgroup$
add a comment |
$begingroup$
a = {140, 280, 420, 560, 700};
results = Join @@ DeleteCases[
IntegerPartitions[#, {1, ∞}, a] & /@ Range[3000],
{}
];
And @@ Thread[Total /@ results <= 3000]
resulttable = Map[Lookup[Counts[#], a, 0] &, results];
resulttable // Short
True
{{1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, <<1341>> , {19, 1, 0, 0, 0}, {21, 0, 0, 0, 0}}
answered Jan 21 at 18:41
Henrik SchumacherHenrik Schumacher
52.9k471148
$endgroup$
a = {140, 280, 420, 560, 700};
results = Join @@ DeleteCases[
IntegerPartitions[#, {1, ∞}, a] & /@ Range[3000],
{}
];
And @@ Thread[Total /@ results <= 3000]
resulttable = Map[Lookup[Counts[#], a, 0] &, results];
resulttable // Short
True
{{1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, <<1341>> , {19, 1, 0, 0, 0}, {21, 0, 0, 0, 0}}
answered Jan 21 at 18:41
Henrik SchumacherHenrik Schumacher
52.9k471148
edited Jan 21 at 19:09
answered Jan 21 at 18:41
Henrik SchumacherHenrik Schumacher
52.9k471148
answered Jan 21 at 18:41
Henrik SchumacherHenrik Schumacher
52.9k471148
answered Jan 21 at 18:41
Henrik SchumacherHenrik Schumacher
52.9k471148
52.9k471148
add a comment |
add a comment |
$begingroup$
v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Select[Subsets[{v1, v2, v3, v4, v5}], Total[#] <= goal &]
Are you sure your "goal" is 3000? Every subset totals less than that.
And what does 21 * v1 in your question refer to? What is $21$, and why multiplication?
answered Jan 21 at 18:38
David G. StorkDavid G. Stork
24.4k22153
$endgroup$
add a comment |
$begingroup$
v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Select[Subsets[{v1, v2, v3, v4, v5}], Total[#] <= goal &]
Are you sure your "goal" is 3000? Every subset totals less than that.
And what does 21 * v1 in your question refer to? What is $21$, and why multiplication?
answered Jan 21 at 18:38
David G. StorkDavid G. Stork
24.4k22153
$endgroup$
add a comment |
$begingroup$
v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Select[Subsets[{v1, v2, v3, v4, v5}], Total[#] <= goal &]
Are you sure your "goal" is 3000? Every subset totals less than that.
And what does 21 * v1 in your question refer to? What is $21$, and why multiplication?
answered Jan 21 at 18:38
David G. StorkDavid G. Stork
24.4k22153
$endgroup$
v1 = 140; v2 = 280; v3 = 420; v4 = 560; v5 = 700; goal = 3000;
Select[Subsets[{v1, v2, v3, v4, v5}], Total[#] <= goal &]
Are you sure your "goal" is 3000? Every subset totals less than that.
And what does 21 * v1 in your question refer to? What is $21$, and why multiplication?
answered Jan 21 at 18:38
David G. StorkDavid G. Stork
24.4k22153
answered Jan 21 at 18:38
David G. StorkDavid G. Stork
24.4k22153
answered Jan 21 at 18:38
David G. StorkDavid G. Stork
24.4k22153
answered Jan 21 at 18:38
David G. StorkDavid G. Stork
24.4k22153
24.4k22153
add a comment |
add a comment |
$begingroup$
a = {140, 280, 420, 560, 700};
coef = Tuples[Range[0, 21], 5];
vec = coef.a;
Extract[coef, Position[Ramp[vec - 3000], 0]] // Short
{{0,0,0,0,0},{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1335>>,{18,1,0,0,0},{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
Pick[vec, Ramp[vec - 3000], 0] // Short
{0,700,1400,2100,2800,560,1260,1960,2660,1120,1820,2520,1680,2380,<<1318>>,2940,2800,2380,2940,2800,2660,2940,2520,2940,2800,2660,2940,2800,2940}
answered Jan 21 at 19:45
Okkes DulgerciOkkes Dulgerci
4,9521817
$endgroup$
add a comment |
$begingroup$
a = {140, 280, 420, 560, 700};
coef = Tuples[Range[0, 21], 5];
vec = coef.a;
Extract[coef, Position[Ramp[vec - 3000], 0]] // Short
{{0,0,0,0,0},{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1335>>,{18,1,0,0,0},{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
Pick[vec, Ramp[vec - 3000], 0] // Short
{0,700,1400,2100,2800,560,1260,1960,2660,1120,1820,2520,1680,2380,<<1318>>,2940,2800,2380,2940,2800,2660,2940,2520,2940,2800,2660,2940,2800,2940}
answered Jan 21 at 19:45
Okkes DulgerciOkkes Dulgerci
4,9521817
$endgroup$
add a comment |
$begingroup$
a = {140, 280, 420, 560, 700};
coef = Tuples[Range[0, 21], 5];
vec = coef.a;
Extract[coef, Position[Ramp[vec - 3000], 0]] // Short
{{0,0,0,0,0},{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1335>>,{18,1,0,0,0},{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
Pick[vec, Ramp[vec - 3000], 0] // Short
{0,700,1400,2100,2800,560,1260,1960,2660,1120,1820,2520,1680,2380,<<1318>>,2940,2800,2380,2940,2800,2660,2940,2520,2940,2800,2660,2940,2800,2940}
answered Jan 21 at 19:45
Okkes DulgerciOkkes Dulgerci
4,9521817
$endgroup$
a = {140, 280, 420, 560, 700};
coef = Tuples[Range[0, 21], 5];
vec = coef.a;
Extract[coef, Position[Ramp[vec - 3000], 0]] // Short
{{0,0,0,0,0},{0,0,0,0,1},{0,0,0,0,2},{0,0,0,0,3},{0,0,0,0,4},{0,0,0,1,0},<<1335>>,{18,1,0,0,0},{19,0,0,0,0},{19,1,0,0,0},{20,0,0,0,0},{21,0,0,0,0}}
Pick[vec, Ramp[vec - 3000], 0] // Short
{0,700,1400,2100,2800,560,1260,1960,2660,1120,1820,2520,1680,2380,<<1318>>,2940,2800,2380,2940,2800,2660,2940,2520,2940,2800,2660,2940,2800,2940}
answered Jan 21 at 19:45
Okkes DulgerciOkkes Dulgerci
4,9521817
answered Jan 21 at 19:45
Okkes DulgerciOkkes Dulgerci
4,9521817
answered Jan 21 at 19:45
Okkes DulgerciOkkes Dulgerci
4,9521817
answered Jan 21 at 19:45
Okkes DulgerciOkkes Dulgerci
4,9521817
4,9521817
add a comment |
add a comment |
$begingroup$
Could use integer linear programming via Maximize, for example. The constraint I use is overkill, all we really need to enforce is that values be nonnegative.
vals = {140, 280, 420, 560, 700};
goal = 3000;
vars = Array[v, Length[vals]];
Maximize[{vars.vals,
Flatten[{vars.vals <= goal,
MapIndexed[0 <= # <= Floor[goal/vals[[#2[[1]]]]] &,
vars]}]}, vars, Integers]
(* Out[1287]= {2940, {v[1] -> 10, v[2] -> 4, v[3] -> 1,
v[4] -> 0, v[5] -> 0}} *)
answered Jan 22 at 15:42
Daniel LichtblauDaniel Lichtblau
46.9k276163
$endgroup$
add a comment |
$begingroup$
Could use integer linear programming via Maximize, for example. The constraint I use is overkill, all we really need to enforce is that values be nonnegative.
vals = {140, 280, 420, 560, 700};
goal = 3000;
vars = Array[v, Length[vals]];
Maximize[{vars.vals,
Flatten[{vars.vals <= goal,
MapIndexed[0 <= # <= Floor[goal/vals[[#2[[1]]]]] &,
vars]}]}, vars, Integers]
(* Out[1287]= {2940, {v[1] -> 10, v[2] -> 4, v[3] -> 1,
v[4] -> 0, v[5] -> 0}} *)
answered Jan 22 at 15:42
Daniel LichtblauDaniel Lichtblau
46.9k276163
$endgroup$
add a comment |
$begingroup$
Could use integer linear programming via Maximize, for example. The constraint I use is overkill, all we really need to enforce is that values be nonnegative.
vals = {140, 280, 420, 560, 700};
goal = 3000;
vars = Array[v, Length[vals]];
Maximize[{vars.vals,
Flatten[{vars.vals <= goal,
MapIndexed[0 <= # <= Floor[goal/vals[[#2[[1]]]]] &,
vars]}]}, vars, Integers]
(* Out[1287]= {2940, {v[1] -> 10, v[2] -> 4, v[3] -> 1,
v[4] -> 0, v[5] -> 0}} *)
answered Jan 22 at 15:42
Daniel LichtblauDaniel Lichtblau
46.9k276163
$endgroup$
Could use integer linear programming via Maximize, for example. The constraint I use is overkill, all we really need to enforce is that values be nonnegative.
vals = {140, 280, 420, 560, 700};
goal = 3000;
vars = Array[v, Length[vals]];
Maximize[{vars.vals,
Flatten[{vars.vals <= goal,
MapIndexed[0 <= # <= Floor[goal/vals[[#2[[1]]]]] &,
vars]}]}, vars, Integers]
(* Out[1287]= {2940, {v[1] -> 10, v[2] -> 4, v[3] -> 1,
v[4] -> 0, v[5] -> 0}} *)
answered Jan 22 at 15:42
Daniel LichtblauDaniel Lichtblau
46.9k276163
answered Jan 22 at 15:42
Daniel LichtblauDaniel Lichtblau
46.9k276163
answered Jan 22 at 15:42
Daniel LichtblauDaniel Lichtblau
46.9k276163
answered Jan 22 at 15:42
Daniel LichtblauDaniel Lichtblau
46.9k276163
46.9k276163
add a comment |
add a comment |
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