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Find maximum of the output from reduce
How Can I use Solve/Reduce OutputFailure message from ReduceExtract desired solutions from ReduceFinding the least positive integer satisfying a quantified statementhow do I control the output of Reduce function?Using the output of ReduceIncomplete and weird output from ReduceUsing Solve returns unnecessary Root, overcomplicated formula, and erroneous negative valueHow to analyse huge output from Reduce systematically?Make Reduce produce nicer output
$begingroup$
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 Mar 7 at 1:28
gagansogaganso
1307
$endgroup$
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 Mar 7 at 1:28
gagansogaganso
1307
$endgroup$
1
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
Mar 7 at 1:52
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
Mar 7 at 1:58
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 Mar 7 at 1:28
gagansogaganso
1307
$endgroup$
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 Mar 7 at 1:28
gagansogaganso
1307
asked Mar 7 at 1:28
gagansogaganso
1307
edited Mar 7 at 1:57
gaganso
asked Mar 7 at 1:28
gagansogaganso
1307
asked Mar 7 at 1:28
gagansogaganso
1307
asked Mar 7 at 1:28
gagansogaganso
1307
1307
1
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
Mar 7 at 1:52
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
Mar 7 at 1:58
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
Mar 7 at 1:52
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
Mar 7 at 1:58
1
1
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
Mar 7 at 1:52
$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
Mar 7 at 1:52
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
Mar 7 at 1:58
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
Mar 7 at 1:58
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 Mar 7 at 2:13
bbgodfreybbgodfrey
44.9k1059110
$endgroup$
$begingroup$
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
Mar 7 at 2:21
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
Mar 7 at 2:22
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 Mar 7 at 2:29
kglrkglr
189k10206424
$endgroup$
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 Mar 7 at 2:13
bbgodfreybbgodfrey
44.9k1059110
$endgroup$
$begingroup$
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
Mar 7 at 2:21
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
Mar 7 at 2:22
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 Mar 7 at 2:13
bbgodfreybbgodfrey
44.9k1059110
$endgroup$
$begingroup$
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
Mar 7 at 2:21
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
Mar 7 at 2:22
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 Mar 7 at 2:13
bbgodfreybbgodfrey
44.9k1059110
$endgroup$
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 Mar 7 at 2:13
bbgodfreybbgodfrey
44.9k1059110
edited Mar 7 at 2:18
answered Mar 7 at 2:13
bbgodfreybbgodfrey
44.9k1059110
answered Mar 7 at 2:13
bbgodfreybbgodfrey
44.9k1059110
answered Mar 7 at 2:13
bbgodfreybbgodfrey
44.9k1059110
44.9k1059110
$begingroup$
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
Mar 7 at 2:21
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
Mar 7 at 2:22
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
Mar 7 at 2:21
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
Mar 7 at 2:22
$begingroup$
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
Mar 7 at 2:21
$begingroup$
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
Mar 7 at 2:21
1
1
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
Mar 7 at 2:22
$begingroup$
@gaganso Precisely so.
$endgroup$
– bbgodfrey
Mar 7 at 2:22
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 Mar 7 at 2:29
kglrkglr
189k10206424
$endgroup$
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 Mar 7 at 2:29
kglrkglr
189k10206424
$endgroup$
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 Mar 7 at 2:29
kglrkglr
189k10206424
$endgroup$
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 Mar 7 at 2:29
kglrkglr
189k10206424
edited Mar 7 at 3:00
answered Mar 7 at 2:29
kglrkglr
189k10206424
answered Mar 7 at 2:29
kglrkglr
189k10206424
answered Mar 7 at 2:29
kglrkglr
189k10206424
189k10206424
add a comment |
add a comment |
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$begingroup$
Several symbols in your code are undefined. Please provide the definitions to aid the reader in answering your question.
$endgroup$
– bbgodfrey
Mar 7 at 1:52
$begingroup$
@bbgodfrey, sorry about that. I have updated the question now.
$endgroup$
– gaganso
Mar 7 at 1:58