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Find maximum of the output from reduce
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$begingroup$
I am trying to reduce a function in two variables($n_1$ and $n_2$) whose domain is the set of Integers. I get a long list of pairs of values for these two variables(instead of a range). This could be because the range of $n_2$ changes for each $n_1$. I just want the maximum value of $n_1$ and $n_2$. Can you please guide me?
driftParamSet = 1.9 - 0.2 Subscript[n, 2] + Subscript[n, 1] (-0.2 + (2.91434*10^-16 Subscript[n, 1])/(1. Subscript[n, 1] + 1.5 Subscript[n, 2]));
drift[Gamma] = 17;
Reduce[driftParamSet> -drift[Gamma] && Subscript[n, 1]>= 0 && Subscript[n, 2]>= 0,{Subscript[n, 1],Subscript[n, 2]}, Integers];
Current output:
$n_1=0land n_2=1left|n_1=0land n_2=2right|n_1=0land n_2=3|n_1=0land n_2=4
\......\left|n_1=0land n_2=90right|n_1=0land n_2=91|\.....\
left(n_1=92land n_2=2right)lor left(n_1=93land n_2=0right)lor left(n_1=93land n_2=1right)lor left(n_1=94land n_2=0right)$
Expected output:
$n_1$=94 and $n_2=$91
equation-solving functions
asked 2 hours ago
gagansogaganso
1207
$endgroup$
add a comment |
$begingroup$
I am trying to reduce a function in two variables($n_1$ and $n_2$) whose domain is the set of Integers. I get a long list of pairs of values for these two variables(instead of a range). This could be because the range of $n_2$ changes for each $n_1$. I just want the maximum value of $n_1$ and $n_2$. Can you please guide me?
driftParamSet = 1.9 - 0.2 Subscript[n, 2] + Subscript[n, 1] (-0.2 + (2.91434*10^-16 Subscript[n, 1])/(1. Subscript[n, 1] + 1.5 Subscript[n, 2]));
drift[Gamma] = 17;
Reduce[driftParamSet> -drift[Gamma] && Subscript[n, 1]>= 0 && Subscript[n, 2]>= 0,{Subscript[n, 1],Subscript[n, 2]}, Integers];
Current output:
$n_1=0land n_2=1left|n_1=0land n_2=2right|n_1=0land n_2=3|n_1=0land n_2=4
\......\left|n_1=0land n_2=90right|n_1=0land n_2=91|\.....\
left(n_1=92land n_2=2right)lor left(n_1=93land n_2=0right)lor left(n_1=93land n_2=1right)lor left(n_1=94land n_2=0right)$
Expected output:
$n_1$=94 and $n_2=$91
equation-solving functions
asked 2 hours ago
gagansogaganso
1207
$endgroup$
1
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
1 hour ago
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
1 hour ago
add a comment |
$begingroup$
I am trying to reduce a function in two variables($n_1$ and $n_2$) whose domain is the set of Integers. I get a long list of pairs of values for these two variables(instead of a range). This could be because the range of $n_2$ changes for each $n_1$. I just want the maximum value of $n_1$ and $n_2$. Can you please guide me?
driftParamSet = 1.9 - 0.2 Subscript[n, 2] + Subscript[n, 1] (-0.2 + (2.91434*10^-16 Subscript[n, 1])/(1. Subscript[n, 1] + 1.5 Subscript[n, 2]));
drift[Gamma] = 17;
Reduce[driftParamSet> -drift[Gamma] && Subscript[n, 1]>= 0 && Subscript[n, 2]>= 0,{Subscript[n, 1],Subscript[n, 2]}, Integers];
Current output:
$n_1=0land n_2=1left|n_1=0land n_2=2right|n_1=0land n_2=3|n_1=0land n_2=4
\......\left|n_1=0land n_2=90right|n_1=0land n_2=91|\.....\
left(n_1=92land n_2=2right)lor left(n_1=93land n_2=0right)lor left(n_1=93land n_2=1right)lor left(n_1=94land n_2=0right)$
Expected output:
$n_1$=94 and $n_2=$91
equation-solving functions
asked 2 hours ago
gagansogaganso
1207
$endgroup$
I am trying to reduce a function in two variables($n_1$ and $n_2$) whose domain is the set of Integers. I get a long list of pairs of values for these two variables(instead of a range). This could be because the range of $n_2$ changes for each $n_1$. I just want the maximum value of $n_1$ and $n_2$. Can you please guide me?
driftParamSet = 1.9 - 0.2 Subscript[n, 2] + Subscript[n, 1] (-0.2 + (2.91434*10^-16 Subscript[n, 1])/(1. Subscript[n, 1] + 1.5 Subscript[n, 2]));
drift[Gamma] = 17;
Reduce[driftParamSet> -drift[Gamma] && Subscript[n, 1]>= 0 && Subscript[n, 2]>= 0,{Subscript[n, 1],Subscript[n, 2]}, Integers];
Current output:
$n_1=0land n_2=1left|n_1=0land n_2=2right|n_1=0land n_2=3|n_1=0land n_2=4
\......\left|n_1=0land n_2=90right|n_1=0land n_2=91|\.....\
left(n_1=92land n_2=2right)lor left(n_1=93land n_2=0right)lor left(n_1=93land n_2=1right)lor left(n_1=94land n_2=0right)$
Expected output:
$n_1$=94 and $n_2=$91
equation-solving functions
equation-solving functions
asked 2 hours ago
gagansogaganso
1207
asked 2 hours ago
gagansogaganso
1207
edited 1 hour ago
gaganso
asked 2 hours ago
gagansogaganso
1207
asked 2 hours ago
gagansogaganso
1207
asked 2 hours ago
gagansogaganso
1207
1207
1
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
1 hour ago
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
1 hour ago
add a comment |
1
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
1 hour ago
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
1 hour ago
1
1
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
1 hour ago
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
1 hour ago
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
1 hour ago
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let the large result of Reduce be rs. Then the maximum of each quantity is determined by
Max@Cases[rs, Equal[Subscript[n, 1], z_] -> z, Infinity]
(* 94 *)
Max@Cases[rs, Equal[Subscript[n, 2], z_] -> z, Infinity]
(* 94 *)
not 91 as speculated in the question. The corresponding terms in rs can be obtained by
Position[rs, 94, Infinity]
(* {{94, 2, 2}, {4559, 1, 2}} *)
rs[[94]]
(* Subscript[n, 1] == 0 && Subscript[n, 2] == 94 *)
rs[[4559]]
(* Subscript[n, 1] == 94 && Subscript[n, 2] == 0 *)
answered 1 hour ago
bbgodfreybbgodfrey
44.8k958110
$endgroup$
$begingroup$
thank you! To understand this better, the Cases[] function with the specified parameter creates a list of values of n1/n2 and the Max[] function operates on this list to give the maximum?
$endgroup$
– gaganso
1 hour ago
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
1 hour ago
add a comment |
$begingroup$
An alternative is to use Solve after Rationalizeing input expressions:
driftParamSet = Rationalize[1.9 - 0.2 n2 +
n1 (-0.2 + (2.91434*10^-16 n1)/(1. n1 + 1.5 n2)), 10^-16]
driftγ = 17;
solutions = Solve[driftParamSet > -driftγ && n1 >= 0 && n2 >= 0, {n1, n2}, Integers];
Max /@ Transpose[{n1, n2} /. solutions]
{94, 94}
Yet another approach is using ArgMax:
Extract[ArgMax[{#, driftParamSet > -driftγ && n1 >= 0 && n2 >= 0}, {n1, n2}, Integers]& /@
{n1, n2}, {{1, 1}, {-1, -1}}]
{94, 94}
answered 59 mins ago
kglrkglr
187k10203421
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let the large result of Reduce be rs. Then the maximum of each quantity is determined by
Max@Cases[rs, Equal[Subscript[n, 1], z_] -> z, Infinity]
(* 94 *)
Max@Cases[rs, Equal[Subscript[n, 2], z_] -> z, Infinity]
(* 94 *)
not 91 as speculated in the question. The corresponding terms in rs can be obtained by
Position[rs, 94, Infinity]
(* {{94, 2, 2}, {4559, 1, 2}} *)
rs[[94]]
(* Subscript[n, 1] == 0 && Subscript[n, 2] == 94 *)
rs[[4559]]
(* Subscript[n, 1] == 94 && Subscript[n, 2] == 0 *)
answered 1 hour ago
bbgodfreybbgodfrey
44.8k958110
$endgroup$
$begingroup$
thank you! To understand this better, the Cases[] function with the specified parameter creates a list of values of n1/n2 and the Max[] function operates on this list to give the maximum?
$endgroup$
– gaganso
1 hour ago
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
1 hour ago
add a comment |
$begingroup$
Let the large result of Reduce be rs. Then the maximum of each quantity is determined by
Max@Cases[rs, Equal[Subscript[n, 1], z_] -> z, Infinity]
(* 94 *)
Max@Cases[rs, Equal[Subscript[n, 2], z_] -> z, Infinity]
(* 94 *)
not 91 as speculated in the question. The corresponding terms in rs can be obtained by
Position[rs, 94, Infinity]
(* {{94, 2, 2}, {4559, 1, 2}} *)
rs[[94]]
(* Subscript[n, 1] == 0 && Subscript[n, 2] == 94 *)
rs[[4559]]
(* Subscript[n, 1] == 94 && Subscript[n, 2] == 0 *)
answered 1 hour ago
bbgodfreybbgodfrey
44.8k958110
$endgroup$
$begingroup$
thank you! To understand this better, the Cases[] function with the specified parameter creates a list of values of n1/n2 and the Max[] function operates on this list to give the maximum?
$endgroup$
– gaganso
1 hour ago
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
1 hour ago
add a comment |
$begingroup$
Let the large result of Reduce be rs. Then the maximum of each quantity is determined by
Max@Cases[rs, Equal[Subscript[n, 1], z_] -> z, Infinity]
(* 94 *)
Max@Cases[rs, Equal[Subscript[n, 2], z_] -> z, Infinity]
(* 94 *)
not 91 as speculated in the question. The corresponding terms in rs can be obtained by
Position[rs, 94, Infinity]
(* {{94, 2, 2}, {4559, 1, 2}} *)
rs[[94]]
(* Subscript[n, 1] == 0 && Subscript[n, 2] == 94 *)
rs[[4559]]
(* Subscript[n, 1] == 94 && Subscript[n, 2] == 0 *)
answered 1 hour ago
bbgodfreybbgodfrey
44.8k958110
$endgroup$
Let the large result of Reduce be rs. Then the maximum of each quantity is determined by
Max@Cases[rs, Equal[Subscript[n, 1], z_] -> z, Infinity]
(* 94 *)
Max@Cases[rs, Equal[Subscript[n, 2], z_] -> z, Infinity]
(* 94 *)
not 91 as speculated in the question. The corresponding terms in rs can be obtained by
Position[rs, 94, Infinity]
(* {{94, 2, 2}, {4559, 1, 2}} *)
rs[[94]]
(* Subscript[n, 1] == 0 && Subscript[n, 2] == 94 *)
rs[[4559]]
(* Subscript[n, 1] == 94 && Subscript[n, 2] == 0 *)
answered 1 hour ago
bbgodfreybbgodfrey
44.8k958110
edited 1 hour ago
answered 1 hour ago
bbgodfreybbgodfrey
44.8k958110
answered 1 hour ago
bbgodfreybbgodfrey
44.8k958110
answered 1 hour ago
bbgodfreybbgodfrey
44.8k958110
44.8k958110
$begingroup$
thank you! To understand this better, the Cases[] function with the specified parameter creates a list of values of n1/n2 and the Max[] function operates on this list to give the maximum?
$endgroup$
– gaganso
1 hour ago
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
1 hour ago
add a comment |
$begingroup$
thank you! To understand this better, the Cases[] function with the specified parameter creates a list of values of n1/n2 and the Max[] function operates on this list to give the maximum?
$endgroup$
– gaganso
1 hour ago
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
1 hour ago
$begingroup$
thank you! To understand this better, the Cases[] function with the specified parameter creates a list of values of n1/n2 and the Max[] function operates on this list to give the maximum?
$endgroup$
– gaganso
1 hour ago
$begingroup$
thank you! To understand this better, the Cases[] function with the specified parameter creates a list of values of n1/n2 and the Max[] function operates on this list to give the maximum?
$endgroup$
– gaganso
1 hour ago
1
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
1 hour ago
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
1 hour ago
add a comment |
$begingroup$
An alternative is to use Solve after Rationalizeing input expressions:
driftParamSet = Rationalize[1.9 - 0.2 n2 +
n1 (-0.2 + (2.91434*10^-16 n1)/(1. n1 + 1.5 n2)), 10^-16]
driftγ = 17;
solutions = Solve[driftParamSet > -driftγ && n1 >= 0 && n2 >= 0, {n1, n2}, Integers];
Max /@ Transpose[{n1, n2} /. solutions]
{94, 94}
Yet another approach is using ArgMax:
Extract[ArgMax[{#, driftParamSet > -driftγ && n1 >= 0 && n2 >= 0}, {n1, n2}, Integers]& /@
{n1, n2}, {{1, 1}, {-1, -1}}]
{94, 94}
answered 59 mins ago
kglrkglr
187k10203421
$endgroup$
add a comment |
$begingroup$
An alternative is to use Solve after Rationalizeing input expressions:
driftParamSet = Rationalize[1.9 - 0.2 n2 +
n1 (-0.2 + (2.91434*10^-16 n1)/(1. n1 + 1.5 n2)), 10^-16]
driftγ = 17;
solutions = Solve[driftParamSet > -driftγ && n1 >= 0 && n2 >= 0, {n1, n2}, Integers];
Max /@ Transpose[{n1, n2} /. solutions]
{94, 94}
Yet another approach is using ArgMax:
Extract[ArgMax[{#, driftParamSet > -driftγ && n1 >= 0 && n2 >= 0}, {n1, n2}, Integers]& /@
{n1, n2}, {{1, 1}, {-1, -1}}]
{94, 94}
answered 59 mins ago
kglrkglr
187k10203421
$endgroup$
add a comment |
$begingroup$
An alternative is to use Solve after Rationalizeing input expressions:
driftParamSet = Rationalize[1.9 - 0.2 n2 +
n1 (-0.2 + (2.91434*10^-16 n1)/(1. n1 + 1.5 n2)), 10^-16]
driftγ = 17;
solutions = Solve[driftParamSet > -driftγ && n1 >= 0 && n2 >= 0, {n1, n2}, Integers];
Max /@ Transpose[{n1, n2} /. solutions]
{94, 94}
Yet another approach is using ArgMax:
Extract[ArgMax[{#, driftParamSet > -driftγ && n1 >= 0 && n2 >= 0}, {n1, n2}, Integers]& /@
{n1, n2}, {{1, 1}, {-1, -1}}]
{94, 94}
answered 59 mins ago
kglrkglr
187k10203421
$endgroup$
An alternative is to use Solve after Rationalizeing input expressions:
driftParamSet = Rationalize[1.9 - 0.2 n2 +
n1 (-0.2 + (2.91434*10^-16 n1)/(1. n1 + 1.5 n2)), 10^-16]
driftγ = 17;
solutions = Solve[driftParamSet > -driftγ && n1 >= 0 && n2 >= 0, {n1, n2}, Integers];
Max /@ Transpose[{n1, n2} /. solutions]
{94, 94}
Yet another approach is using ArgMax:
Extract[ArgMax[{#, driftParamSet > -driftγ && n1 >= 0 && n2 >= 0}, {n1, n2}, Integers]& /@
{n1, n2}, {{1, 1}, {-1, -1}}]
{94, 94}
answered 59 mins ago
kglrkglr
187k10203421
edited 28 mins ago
answered 59 mins ago
kglrkglr
187k10203421
answered 59 mins ago
kglrkglr
187k10203421
answered 59 mins ago
kglrkglr
187k10203421
187k10203421
add a comment |
add a comment |
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1
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
1 hour ago
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
1 hour ago