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How to generate a matrix with certain conditions

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2

$begingroup$

I want to generate a $n times n$ matrix.

1. I want the diagonal entries to be all 0


2. I want a random choice of matrix elements with 0 or 1.


3. The probability of having a 1 as a matrix element is $1/m$ and the probability of having a 0 as a matrix element is $1-1/m$.

I used the following command but it is wrong.

A[n_, m_] :=Table[If[i == j, 0,RandomVariate[BernoulliDistribution[m],{n,n}]]]

And I tried to test this command with n=4, m=0.4 but it didn't work.

Could anyone kindly tell me how to do this please?
Thank you!

matrix probability-or-statistics random sampling

share|improve this question

edited 22 hours ago

J. M. is computer-less♦

97.1k10303463

asked yesterday

tiffanytiffany

111

New contributor

tiffany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For a start, your code seems to set the diagonal elements to 1 rather than zero.
  $endgroup$
  – MarcoB
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Same question posted here.
  $endgroup$
  – Rohit Namjoshi
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  tiffany, please go here to get your accounts merged, so you can easily access your question.
  $endgroup$
  – J. M. is computer-less♦
  16 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I want to generate a $n times n$ matrix.

1. I want the diagonal entries to be all 0


2. I want a random choice of matrix elements with 0 or 1.


3. The probability of having a 1 as a matrix element is $1/m$ and the probability of having a 0 as a matrix element is $1-1/m$.

I used the following command but it is wrong.

A[n_, m_] :=Table[If[i == j, 0,RandomVariate[BernoulliDistribution[m],{n,n}]]]

And I tried to test this command with n=4, m=0.4 but it didn't work.

Could anyone kindly tell me how to do this please?
Thank you!

matrix probability-or-statistics random sampling

share|improve this question

edited 22 hours ago

J. M. is computer-less♦

97.1k10303463

asked yesterday

tiffanytiffany

111

New contributor

tiffany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For a start, your code seems to set the diagonal elements to 1 rather than zero.
  $endgroup$
  – MarcoB
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Same question posted here.
  $endgroup$
  – Rohit Namjoshi
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  tiffany, please go here to get your accounts merged, so you can easily access your question.
  $endgroup$
  – J. M. is computer-less♦
  16 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I want to generate a $n times n$ matrix.

1. I want the diagonal entries to be all 0


2. I want a random choice of matrix elements with 0 or 1.


3. The probability of having a 1 as a matrix element is $1/m$ and the probability of having a 0 as a matrix element is $1-1/m$.

I used the following command but it is wrong.

A[n_, m_] :=Table[If[i == j, 0,RandomVariate[BernoulliDistribution[m],{n,n}]]]

And I tried to test this command with n=4, m=0.4 but it didn't work.

Could anyone kindly tell me how to do this please?
Thank you!

matrix probability-or-statistics random sampling

share|improve this question

edited 22 hours ago

J. M. is computer-less♦

97.1k10303463

asked yesterday

tiffanytiffany

111

New contributor

tiffany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

A[n_, m_] :=Table[If[i == j, 0,RandomVariate[BernoulliDistribution[m],{n,n}]]]

share|improve this question

edited 22 hours ago

J. M. is computer-less♦

97.1k10303463

asked yesterday

tiffanytiffany

111

New contributor

tiffany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 22 hours ago

J. M. is computer-less♦

97.1k10303463

asked yesterday

tiffanytiffany

111

New contributor

tiffany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 22 hours ago

J. M. is computer-less♦

97.1k10303463

edited 22 hours ago

J. M. is computer-less♦

97.1k10303463

asked yesterday

tiffanytiffany

111

New contributor

tiffany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

tiffanytiffany

111

4 Answers
4

active

oldest

votes

5

$begingroup$

Binary random variables are often modeled using the BernoulliDistribution. You can use the function RandomVariate to get a matrix of such variables.

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

% // MatrixForm

$$begin{pmatrix}
0&1&1&1&1\
1&0&1&1&1\
1&1&0&1&1\
1&0&1&0&1\
0&1&1&1&0end{pmatrix}$$

You can change the 0.9 to any value (this is your m). The second line sets all the diagonal elements to zero.

It's easy enough to make this into a function:

makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 

                    mat - DiagonalMatrix[Diagonal[mat]])

Then the above example is makeMat[5, 0.9]

share|improve this answer

edited 16 hours ago

J. M. is computer-less♦

97.1k10303463

answered yesterday

bill sbill s

53.7k376153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you bill s How will the command be if I do not restrict what n and m be? For example, I would like to generate a set of commands which I can replace m and n easily by any number.
  $endgroup$
  – tiffany
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany, just do With[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

You can use weight option in RandomChoice

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$

share|improve this answer

answered yesterday

Okkes DulgerciOkkes Dulgerci

5,1591917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Altho using the BernoulliDistribution[] is best, this is likely to be more easy to read for someone who is not accustomed to discrete probability distributions.
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Since all the simple answers have been given, here is a SparseArray[] solution that may be useful if you want to generate large matrices without storing unneeded zero entries:

tiffany[n_Integer?Positive, m_] :=

    SparseArray[{j_, k_} /; j != k && RandomReal[] < 1/m :> 1, {n, n}]

As an example:

BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]

   {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},

    {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},

    {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}

share|improve this answer

answered 16 hours ago

J. M. is computer-less♦J. M. is computer-less

97.1k10303463

$endgroup$

add a comment | 

0

$begingroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered yesterday

Vsevolod A.Vsevolod A.

478211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
  $endgroup$
  – tiffany
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany you set m and n in the first two lines...
  $endgroup$
  – Vsevolod A.
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Just rnd := If[RandomReal[{0, 1}] < 1/m, 1, 0]; will do, since the function never uses its argument.
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

Binary random variables are often modeled using the BernoulliDistribution. You can use the function RandomVariate to get a matrix of such variables.

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

% // MatrixForm

$$begin{pmatrix}
0&1&1&1&1\
1&0&1&1&1\
1&1&0&1&1\
1&0&1&0&1\
0&1&1&1&0end{pmatrix}$$

You can change the 0.9 to any value (this is your m). The second line sets all the diagonal elements to zero.

It's easy enough to make this into a function:

makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 

                    mat - DiagonalMatrix[Diagonal[mat]])

Then the above example is makeMat[5, 0.9]

share|improve this answer

edited 16 hours ago

J. M. is computer-less♦

97.1k10303463

answered yesterday

bill sbill s

53.7k376153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you bill s How will the command be if I do not restrict what n and m be? For example, I would like to generate a set of commands which I can replace m and n easily by any number.
  $endgroup$
  – tiffany
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany, just do With[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

Binary random variables are often modeled using the BernoulliDistribution. You can use the function RandomVariate to get a matrix of such variables.

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

% // MatrixForm

$$begin{pmatrix}
0&1&1&1&1\
1&0&1&1&1\
1&1&0&1&1\
1&0&1&0&1\
0&1&1&1&0end{pmatrix}$$

You can change the 0.9 to any value (this is your m). The second line sets all the diagonal elements to zero.

It's easy enough to make this into a function:

makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 

                    mat - DiagonalMatrix[Diagonal[mat]])

Then the above example is makeMat[5, 0.9]

share|improve this answer

edited 16 hours ago

J. M. is computer-less♦

97.1k10303463

answered yesterday

bill sbill s

53.7k376153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you bill s How will the command be if I do not restrict what n and m be? For example, I would like to generate a set of commands which I can replace m and n easily by any number.
  $endgroup$
  – tiffany
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany, just do With[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Binary random variables are often modeled using the BernoulliDistribution. You can use the function RandomVariate to get a matrix of such variables.

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

% // MatrixForm

$$begin{pmatrix}
0&1&1&1&1\
1&0&1&1&1\
1&1&0&1&1\
1&0&1&0&1\
0&1&1&1&0end{pmatrix}$$

You can change the 0.9 to any value (this is your m). The second line sets all the diagonal elements to zero.

It's easy enough to make this into a function:

makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 

                    mat - DiagonalMatrix[Diagonal[mat]])

Then the above example is makeMat[5, 0.9]

share|improve this answer

edited 16 hours ago

J. M. is computer-less♦

97.1k10303463

answered yesterday

bill sbill s

53.7k376153

$endgroup$

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

% // MatrixForm

makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 

                    mat - DiagonalMatrix[Diagonal[mat]])

share|improve this answer

edited 16 hours ago

J. M. is computer-less♦

97.1k10303463

answered yesterday

bill sbill s

53.7k376153

edited 16 hours ago

J. M. is computer-less♦

97.1k10303463

edited 16 hours ago

J. M. is computer-less♦

97.1k10303463

answered yesterday

bill sbill s

53.7k376153

answered yesterday

bill sbill s

53.7k376153

4

$begingroup$

You can use weight option in RandomChoice

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$

share|improve this answer

answered yesterday

Okkes DulgerciOkkes Dulgerci

5,1591917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Altho using the BernoulliDistribution[] is best, this is likely to be more easy to read for someone who is not accustomed to discrete probability distributions.
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

You can use weight option in RandomChoice

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$

share|improve this answer

answered yesterday

Okkes DulgerciOkkes Dulgerci

5,1591917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Altho using the BernoulliDistribution[] is best, this is likely to be more easy to read for someone who is not accustomed to discrete probability distributions.
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can use weight option in RandomChoice

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$

share|improve this answer

answered yesterday

Okkes DulgerciOkkes Dulgerci

5,1591917

$endgroup$

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

share|improve this answer

answered yesterday

Okkes DulgerciOkkes Dulgerci

5,1591917

answered yesterday

Okkes DulgerciOkkes Dulgerci

5,1591917

answered yesterday

Okkes DulgerciOkkes Dulgerci

5,1591917

2

$begingroup$

Since all the simple answers have been given, here is a SparseArray[] solution that may be useful if you want to generate large matrices without storing unneeded zero entries:

tiffany[n_Integer?Positive, m_] :=

    SparseArray[{j_, k_} /; j != k && RandomReal[] < 1/m :> 1, {n, n}]

As an example:

BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]

   {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},

    {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},

    {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}

share|improve this answer

answered 16 hours ago

J. M. is computer-less♦J. M. is computer-less

97.1k10303463

$endgroup$

add a comment | 

2

$begingroup$

Since all the simple answers have been given, here is a SparseArray[] solution that may be useful if you want to generate large matrices without storing unneeded zero entries:

tiffany[n_Integer?Positive, m_] :=

    SparseArray[{j_, k_} /; j != k && RandomReal[] < 1/m :> 1, {n, n}]

As an example:

BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]

   {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},

    {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},

    {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}

share|improve this answer

answered 16 hours ago

J. M. is computer-less♦J. M. is computer-less

97.1k10303463

$endgroup$

add a comment | 

$begingroup$

Since all the simple answers have been given, here is a SparseArray[] solution that may be useful if you want to generate large matrices without storing unneeded zero entries:

tiffany[n_Integer?Positive, m_] :=

    SparseArray[{j_, k_} /; j != k && RandomReal[] < 1/m :> 1, {n, n}]

As an example:

BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]

   {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},

    {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},

    {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}

share|improve this answer

answered 16 hours ago

J. M. is computer-less♦J. M. is computer-less

97.1k10303463

$endgroup$

tiffany[n_Integer?Positive, m_] :=

    SparseArray[{j_, k_} /; j != k && RandomReal[] < 1/m :> 1, {n, n}]

BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]

   {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},

    {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},

    {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}

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answered 16 hours ago

J. M. is computer-less♦J. M. is computer-less

97.1k10303463

answered 16 hours ago

J. M. is computer-less♦J. M. is computer-less

97.1k10303463

answered 16 hours ago

J. M. is computer-less♦J. M. is computer-less

97.1k10303463

0

$begingroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

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answered yesterday

Vsevolod A.Vsevolod A.

478211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
  $endgroup$
  – tiffany
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany you set m and n in the first two lines...
  $endgroup$
  – Vsevolod A.
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Just rnd := If[RandomReal[{0, 1}] < 1/m, 1, 0]; will do, since the function never uses its argument.
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered yesterday

Vsevolod A.Vsevolod A.

478211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
  $endgroup$
  – tiffany
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany you set m and n in the first two lines...
  $endgroup$
  – Vsevolod A.
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Just rnd := If[RandomReal[{0, 1}] < 1/m, 1, 0]; will do, since the function never uses its argument.
  $endgroup$
  – J. M. is computer-less♦
  22 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered yesterday

Vsevolod A.Vsevolod A.

478211

$endgroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered yesterday

Vsevolod A.Vsevolod A.

478211

answered yesterday

Vsevolod A.Vsevolod A.

478211

answered yesterday

Vsevolod A.Vsevolod A.

478211