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Changing iteration variable in Do loop

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Changing iteration variable in Do loop

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2

$begingroup$

I want to generate random positive hermitian matrices, I start with such a code

rho11R = Table[RandomReal[], i, 1, 4];
rho10R = Table[RandomComplex[], i, 1, 4];
Do[If[rho11R[[i]] - rho11R[[i]]^2 - 
 rho10R[[i]]*Conjugate[rho10R[[i]]] >= 0, 
 Nothing, rho10R[[i]] = RandomComplex[], --i], i, 1, 
 4];
rhoR = Table[rho11R[[i]], rho10R[[i]], Conjugate[rho10R[[i]]], 
 1 - rho11R[[i]], i, 1, 4]

This does not work, as the matrices I get are in general not positive.
The problem is my If function, which does not seem to change the iteration variable i to i-1, when the condition for positivity is not fulfilled (so that I can get new random variable for rho10R[[I]] and again check the condition for positivity). How can I reach this?

procedural-programming

share|improve this question

asked 8 hours ago

AgnieszkaAgnieszka

1041010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried a While[] loop instead?
  $endgroup$
  – Somos
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume that by positive you mean positive semi-definite, right?
  $endgroup$
  – Roman
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I want to generate random positive hermitian matrices, I start with such a code

rho11R = Table[RandomReal[], i, 1, 4];
rho10R = Table[RandomComplex[], i, 1, 4];
Do[If[rho11R[[i]] - rho11R[[i]]^2 - 
 rho10R[[i]]*Conjugate[rho10R[[i]]] >= 0, 
 Nothing, rho10R[[i]] = RandomComplex[], --i], i, 1, 
 4];
rhoR = Table[rho11R[[i]], rho10R[[i]], Conjugate[rho10R[[i]]], 
 1 - rho11R[[i]], i, 1, 4]

This does not work, as the matrices I get are in general not positive.
The problem is my If function, which does not seem to change the iteration variable i to i-1, when the condition for positivity is not fulfilled (so that I can get new random variable for rho10R[[I]] and again check the condition for positivity). How can I reach this?

procedural-programming

share|improve this question

asked 8 hours ago

AgnieszkaAgnieszka

1041010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried a While[] loop instead?
  $endgroup$
  – Somos
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume that by positive you mean positive semi-definite, right?
  $endgroup$
  – Roman
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I want to generate random positive hermitian matrices, I start with such a code

rho11R = Table[RandomReal[], i, 1, 4];
rho10R = Table[RandomComplex[], i, 1, 4];
Do[If[rho11R[[i]] - rho11R[[i]]^2 - 
 rho10R[[i]]*Conjugate[rho10R[[i]]] >= 0, 
 Nothing, rho10R[[i]] = RandomComplex[], --i], i, 1, 
 4];
rhoR = Table[rho11R[[i]], rho10R[[i]], Conjugate[rho10R[[i]]], 
 1 - rho11R[[i]], i, 1, 4]

This does not work, as the matrices I get are in general not positive.
The problem is my If function, which does not seem to change the iteration variable i to i-1, when the condition for positivity is not fulfilled (so that I can get new random variable for rho10R[[I]] and again check the condition for positivity). How can I reach this?

procedural-programming

share|improve this question

asked 8 hours ago

AgnieszkaAgnieszka

1041010 bronze badges

$endgroup$

rho11R = Table[RandomReal[], i, 1, 4];
rho10R = Table[RandomComplex[], i, 1, 4];
Do[If[rho11R[[i]] - rho11R[[i]]^2 - 
 rho10R[[i]]*Conjugate[rho10R[[i]]] >= 0, 
 Nothing, rho10R[[i]] = RandomComplex[], --i], i, 1, 
 4];
rhoR = Table[rho11R[[i]], rho10R[[i]], Conjugate[rho10R[[i]]], 
 1 - rho11R[[i]], i, 1, 4]

share|improve this question

asked 8 hours ago

AgnieszkaAgnieszka

1041010 bronze badges

share|improve this question

asked 8 hours ago

AgnieszkaAgnieszka

1041010 bronze badges

asked 8 hours ago

AgnieszkaAgnieszka

1041010 bronze badges

asked 8 hours ago

AgnieszkaAgnieszka

1041010 bronze badges

3 Answers
3

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4

$begingroup$

This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.

Example:

For[
 i = 1,
 i <= 4,
 i++,
 If[
 i == 2,
 Print[i++],
 Print[i]
 ]
 ]

This prints 1, 2, 4.

Another option is to use While, as Somos wrote in a comment.

share|improve this answer

answered 7 hours ago

C. E.C. E.

53.3k33 gold badges102102 silver badges210210 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:

randommatrix[n_Integer?Positive] := 
 RandomVariate[NormalDistribution[], n, n, 2].1, I

randomHermitian[n_Integer?Positive] := 
 ConjugateTranspose[#].# & @ randommatrix[n]

In this way you don't need to reject anything.

Test:

randomHermitian[10] // Eigenvalues
(* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
 12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)

There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.

share|improve this answer

answered 7 hours ago

RomanRoman

14.3k11 gold badge1919 silver badges5151 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
  $endgroup$
  – mikado
  1 hour ago
  

  
  
  
  

add a comment | 

0

$begingroup$

The following is a more "functional" way of doing acceptance/rejection

mtx := RandomComplex[1+I, 2, 2]

pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]]; 
 If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]

You can then generate a list of positive definite matrices of any length e.g.

Table[pmtx, 20] 

(Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)

share|improve this answer

answered 2 hours ago

mikadomikado

7,29011 gold badge99 silver badges2929 bronze badges

$endgroup$

add a comment | 

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4

$begingroup$

This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.

Example:

For[
 i = 1,
 i <= 4,
 i++,
 If[
 i == 2,
 Print[i++],
 Print[i]
 ]
 ]

This prints 1, 2, 4.

Another option is to use While, as Somos wrote in a comment.

share|improve this answer

answered 7 hours ago

C. E.C. E.

53.3k33 gold badges102102 silver badges210210 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.

Example:

For[
 i = 1,
 i <= 4,
 i++,
 If[
 i == 2,
 Print[i++],
 Print[i]
 ]
 ]

This prints 1, 2, 4.

Another option is to use While, as Somos wrote in a comment.

share|improve this answer

answered 7 hours ago

C. E.C. E.

53.3k33 gold badges102102 silver badges210210 bronze badges

$endgroup$

add a comment | 

$begingroup$

This is one the cases where For can be helpful (generally it's just more complicated.) With For, you can manipulate the iteration variable, which you cannot do with Do.

Example:

For[
 i = 1,
 i <= 4,
 i++,
 If[
 i == 2,
 Print[i++],
 Print[i]
 ]
 ]

This prints 1, 2, 4.

Another option is to use While, as Somos wrote in a comment.

share|improve this answer

answered 7 hours ago

C. E.C. E.

53.3k33 gold badges102102 silver badges210210 bronze badges

$endgroup$

For[
 i = 1,
 i <= 4,
 i++,
 If[
 i == 2,
 Print[i++],
 Print[i]
 ]
 ]

share|improve this answer

answered 7 hours ago

C. E.C. E.

53.3k33 gold badges102102 silver badges210210 bronze badges

answered 7 hours ago

C. E.C. E.

53.3k33 gold badges102102 silver badges210210 bronze badges

answered 7 hours ago

C. E.C. E.

53.3k33 gold badges102102 silver badges210210 bronze badges

4

$begingroup$

A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:

randommatrix[n_Integer?Positive] := 
 RandomVariate[NormalDistribution[], n, n, 2].1, I

randomHermitian[n_Integer?Positive] := 
 ConjugateTranspose[#].# & @ randommatrix[n]

In this way you don't need to reject anything.

Test:

randomHermitian[10] // Eigenvalues
(* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
 12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)

There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.

share|improve this answer

answered 7 hours ago

RomanRoman

14.3k11 gold badge1919 silver badges5151 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
  $endgroup$
  – mikado
  1 hour ago
  

  
  
  
  

add a comment | 

4

$begingroup$

A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:

randommatrix[n_Integer?Positive] := 
 RandomVariate[NormalDistribution[], n, n, 2].1, I

randomHermitian[n_Integer?Positive] := 
 ConjugateTranspose[#].# & @ randommatrix[n]

In this way you don't need to reject anything.

Test:

randomHermitian[10] // Eigenvalues
(* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
 12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)

There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.

share|improve this answer

answered 7 hours ago

RomanRoman

14.3k11 gold badge1919 silver badges5151 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While the matrices you generate are Hermitian by construction, they may not be exactly, because of rounding errors. It is good practice to force to Hermitian, by averaging with the conjugate transpose.
  $endgroup$
  – mikado
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

A much easier way to construct random positive semi-definite Hermitian matrices is to start with Gaussian random matrices and Hermitian-square them:

randommatrix[n_Integer?Positive] := 
 RandomVariate[NormalDistribution[], n, n, 2].1, I

randomHermitian[n_Integer?Positive] := 
 ConjugateTranspose[#].# & @ randommatrix[n]

In this way you don't need to reject anything.

Test:

randomHermitian[10] // Eigenvalues
(* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
 12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)

There is of course the question of the measure (distribution) from which you pull the random matrices; you'd have to be more specific in your question to address this point.

share|improve this answer

answered 7 hours ago

RomanRoman

14.3k11 gold badge1919 silver badges5151 bronze badges

$endgroup$

randommatrix[n_Integer?Positive] := 
 RandomVariate[NormalDistribution[], n, n, 2].1, I

randomHermitian[n_Integer?Positive] := 
 ConjugateTranspose[#].# & @ randommatrix[n]

randomHermitian[10] // Eigenvalues
(* 71.4553, 53.6575, 46.3275, 31.8263, 21.4754,
 12.9687, 7.36107, 4.40568, 1.23665, 0.199904 *)

share|improve this answer

answered 7 hours ago

RomanRoman

14.3k11 gold badge1919 silver badges5151 bronze badges

answered 7 hours ago

RomanRoman

14.3k11 gold badge1919 silver badges5151 bronze badges

answered 7 hours ago

RomanRoman

14.3k11 gold badge1919 silver badges5151 bronze badges

0

$begingroup$

The following is a more "functional" way of doing acceptance/rejection

mtx := RandomComplex[1+I, 2, 2]

pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]]; 
 If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]

You can then generate a list of positive definite matrices of any length e.g.

Table[pmtx, 20] 

(Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)

share|improve this answer

answered 2 hours ago

mikadomikado

7,29011 gold badge99 silver badges2929 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

The following is a more "functional" way of doing acceptance/rejection

mtx := RandomComplex[1+I, 2, 2]

pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]]; 
 If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]

You can then generate a list of positive definite matrices of any length e.g.

Table[pmtx, 20] 

(Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)

share|improve this answer

answered 2 hours ago

mikadomikado

7,29011 gold badge99 silver badges2929 bronze badges

$endgroup$

add a comment | 

$begingroup$

The following is a more "functional" way of doing acceptance/rejection

mtx := RandomComplex[1+I, 2, 2]

pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]]; 
 If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]

You can then generate a list of positive definite matrices of any length e.g.

Table[pmtx, 20] 

(Note: I'm not addressing the question of whether this gives an appropriate distribution of random matrices)

share|improve this answer

answered 2 hours ago

mikadomikado

7,29011 gold badge99 silver badges2929 bronze badges

$endgroup$

mtx := RandomComplex[1+I, 2, 2]

pmtx := Module[m = mtx, m1, m1 = Chop[m + ConjugateTranspose[m]]; 
 If[PositiveDefiniteMatrixQ[m1], m1, pmtx]]

Table[pmtx, 20] 

share|improve this answer

answered 2 hours ago

mikadomikado

7,29011 gold badge99 silver badges2929 bronze badges

answered 2 hours ago

mikadomikado

7,29011 gold badge99 silver badges2929 bronze badges

answered 2 hours ago

mikadomikado

7,29011 gold badge99 silver badges2929 bronze badges