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?enter image description here






matrix







share|improve this question







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
    16 hours ago












  • $begingroup$
    KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
    $endgroup$
    – Henrik Schumacher
    15 hours ago












  • $begingroup$
    Great,thank you very much!
    $endgroup$
    – KarryMa
    15 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
    15 hours ago






  • 3




    $begingroup$
    Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
    $endgroup$
    – LouisB
    15 hours ago
















1












$begingroup$


The matrixform is as follow, and how can I use good method to produce it?enter image description here






matrix







share|improve this question







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
    16 hours ago












  • $begingroup$
    KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
    $endgroup$
    – Henrik Schumacher
    15 hours ago












  • $begingroup$
    Great,thank you very much!
    $endgroup$
    – KarryMa
    15 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
    15 hours ago






  • 3




    $begingroup$
    Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
    $endgroup$
    – LouisB
    15 hours ago














1












1








1





$begingroup$


The matrixform is as follow, and how can I use good method to produce it?enter image description here






matrix







share|improve this question







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?enter image description here






matrix




matrix






share|improve this question







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.











share|improve this question







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.









share|improve this question




share|improve this question






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 16 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
    16 hours ago












  • $begingroup$
    KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
    $endgroup$
    – Henrik Schumacher
    15 hours ago












  • $begingroup$
    Great,thank you very much!
    $endgroup$
    – KarryMa
    15 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
    15 hours ago






  • 3




    $begingroup$
    Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
    $endgroup$
    – LouisB
    15 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
    16 hours ago












  • $begingroup$
    KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
    $endgroup$
    – Henrik Schumacher
    15 hours ago












  • $begingroup$
    Great,thank you very much!
    $endgroup$
    – KarryMa
    15 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
    15 hours ago






  • 3




    $begingroup$
    Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
    $endgroup$
    – LouisB
    15 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
16 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
16 hours ago














$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
15 hours ago






$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
15 hours ago














$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
15 hours ago




$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
15 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
15 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
15 hours ago




3




3




$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
15 hours ago




$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
15 hours ago










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}}







share|improve this answer











$endgroup$









  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    15 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
    10 hours ago










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    10 hours ago



















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


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$






share|improve this answer











$endgroup$













  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    9 hours ago












Your Answer





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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}}







share|improve this answer











$endgroup$









  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    15 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
    10 hours ago










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    10 hours ago
















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}}







share|improve this answer











$endgroup$









  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    15 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
    10 hours ago










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    10 hours ago














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}}







share|improve this answer











$endgroup$



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}}








share|improve this answer














share|improve this answer



share|improve this answer








edited 10 hours ago

























answered 15 hours ago









J42161217J42161217

4,248324




4,248324








  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    15 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
    10 hours ago










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    10 hours ago














  • 2




    $begingroup$
    I like the answer,thanks!
    $endgroup$
    – KarryMa
    15 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
    10 hours ago










  • $begingroup$
    yes, you are so right!
    $endgroup$
    – J42161217
    10 hours ago








2




2




$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
15 hours ago




$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
15 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
10 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
10 hours ago












$begingroup$
yes, you are so right!
$endgroup$
– J42161217
10 hours ago




$begingroup$
yes, you are so right!
$endgroup$
– J42161217
10 hours ago











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


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$






share|improve this answer











$endgroup$













  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    9 hours ago
















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


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$






share|improve this answer











$endgroup$













  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    9 hours ago














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


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$






share|improve this answer











$endgroup$



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


$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$







share|improve this answer














share|improve this answer



share|improve this answer








edited 9 hours ago

























answered 12 hours ago









MikeYMikeY

3,768916




3,768916












  • $begingroup$
    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
    10 hours ago










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    9 hours 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
    10 hours ago










  • $begingroup$
    My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
    $endgroup$
    – MikeY
    9 hours 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
10 hours 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
10 hours ago












$begingroup$
My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
9 hours ago




$begingroup$
My purpose was to introduce ArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
9 hours ago










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.












KarryMa is a new contributor. Be nice, and check out our Code of Conduct.
















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Vallis Paradisi

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Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
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Tabula Rosettana

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Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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