Using good method to produce a regular matrix
1
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
matrix
asked 6 hours ago
KarryMaKarryMa
112
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KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
|
show 4 more comments
1
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
matrix
asked 6 hours ago
KarryMaKarryMa
112
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
$endgroup$
– KarryMa
6 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
$endgroup$
– Henrik Schumacher
6 hours ago
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
6 hours ago
|
show 4 more comments
1
1
1
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
matrix
asked 6 hours ago
KarryMaKarryMa
112
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The matrixform is as follow, and how can I use good method to produce it?
matrix
matrix
asked 6 hours ago
KarryMaKarryMa
112
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
KarryMaKarryMa
112
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
KarryMaKarryMa
112
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
KarryMaKarryMa
112
asked 6 hours ago
KarryMaKarryMa
112
112
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
$endgroup$
– KarryMa
6 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
$endgroup$
– Henrik Schumacher
6 hours ago
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
6 hours ago
|
show 4 more comments
$begingroup$
If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
$endgroup$
– KarryMa
6 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
$endgroup$
– Henrik Schumacher
6 hours ago
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
6 hours ago
$begingroup$
If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
$endgroup$
– KarryMa
6 hours ago
$begingroup$
If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
$endgroup$
– KarryMa
6 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].$endgroup$
– Henrik Schumacher
6 hours ago
3
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]$endgroup$
– LouisB
6 hours ago
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]$endgroup$
– LouisB
6 hours ago
|
show 4 more comments
2 Answers
2
active
oldest
votes
2
$begingroup$
IntegerDigits[{12,24,4,8,10,20},3,3]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
also..
s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];
Flatten[{{s[[1]]},Reverse@Rest@s},2]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
answered 6 hours ago
J42161217J42161217
4,193324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
1 hour ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
1 hour ago
add a comment |
0
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 2 hours ago
MikeYMikeY
3,758916
$endgroup$
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
1 hour ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
37 mins ago
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
2
$begingroup$
IntegerDigits[{12,24,4,8,10,20},3,3]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
also..
s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];
Flatten[{{s[[1]]},Reverse@Rest@s},2]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
answered 6 hours ago
J42161217J42161217
4,193324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
1 hour ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
1 hour ago
add a comment |
2
$begingroup$
IntegerDigits[{12,24,4,8,10,20},3,3]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
also..
s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];
Flatten[{{s[[1]]},Reverse@Rest@s},2]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
answered 6 hours ago
J42161217J42161217
4,193324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
1 hour ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
1 hour ago
add a comment |
2
2
2
$begingroup$
IntegerDigits[{12,24,4,8,10,20},3,3]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
also..
s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];
Flatten[{{s[[1]]},Reverse@Rest@s},2]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
answered 6 hours ago
J42161217J42161217
4,193324
$endgroup$
IntegerDigits[{12,24,4,8,10,20},3,3]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
also..
s = Transpose[Permutations /@ {{1, 1, 0}, {2, 2, 0}}];
Flatten[{{s[[1]]},Reverse@Rest@s},2]
{{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
answered 6 hours ago
J42161217J42161217
4,193324
edited 1 hour ago
answered 6 hours ago
J42161217J42161217
4,193324
answered 6 hours ago
J42161217J42161217
4,193324
answered 6 hours ago
J42161217J42161217
4,193324
4,193324
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
1 hour ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
1 hour ago
add a comment |
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
1 hour ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
1 hour ago
2
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
@KarryMa Alternatively to
PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].$endgroup$
– Coolwater
1 hour ago
$begingroup$
@KarryMa Alternatively to
PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].$endgroup$
– Coolwater
1 hour ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
1 hour ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
1 hour ago
add a comment |
0
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 2 hours ago
MikeYMikeY
3,758916
$endgroup$
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
1 hour ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
37 mins ago
add a comment |
0
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 2 hours ago
MikeYMikeY
3,758916
$endgroup$
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
1 hour ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
37 mins ago
add a comment |
0
0
0
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 2 hours ago
MikeYMikeY
3,758916
$endgroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
{{a,a,0},{0,a,a},{a,0,a}}// ArrayFlatten
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 2 hours ago
MikeYMikeY
3,758916
edited 38 mins ago
answered 2 hours ago
MikeYMikeY
3,758916
answered 2 hours ago
MikeYMikeY
3,758916
answered 2 hours ago
MikeYMikeY
3,758916
3,758916
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
1 hour ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
37 mins ago
add a comment |
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
1 hour ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
37 mins ago
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
1 hour ago
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
1 hour ago
$begingroup$
My purpose was to introduce
ArrayFlatten. Not enough info to algorithmically determine the order of perms.$endgroup$
– MikeY
37 mins ago
$begingroup$
My purpose was to introduce
ArrayFlatten. Not enough info to algorithmically determine the order of perms.$endgroup$
– MikeY
37 mins ago
add a comment |
KarryMa is a new contributor. Be nice, and check out our Code of Conduct.
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KarryMa is a new contributor. Be nice, and check out our Code of Conduct.
KarryMa is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
If the picture can't show,then here is the matrixform:H={{1,1,0},{2,2,0},{0,1,1},{0,2,2},{1,0,1},{2,0,2}}
$endgroup$
– KarryMa
6 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
6 hours ago
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].$endgroup$
– Henrik Schumacher
6 hours ago
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]$endgroup$
– LouisB
6 hours ago