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Mathematics Wolfram Language Symbolic Computations

Construct sparse matrix more efficiently?

Milan Vicentijevic

Posted 10 years ago

Hi. I would like to construct sparse matrix with 9 diagonals. It is organized as 4 blocks with 3 diagonals each. Something like this: (A B);(C D) Currently, I am using For loop for the job, but I suspect there is a more efficient way. My code is: Agornja[z_] := zagornja; Aglavna[z_] := zaglavna; Adonja[z_] := zadonja; Bgornja[z_] := zbgornja; Bglavna[z_] := zbglavna; Bdonja[z_] := zbdonja; Cgornja[z_] := zcgornja; Cglavna[z_] := zcglavna; Cdonja[z_] := zcdonja; Dgornja[z_] := zdgornja; Dglavna[z_] := zdglavna; Ddonja[z_] := zddonja; Where z is just for me to see the incrementation. For loop is: brojelemenata = 10; korak = 1; pocetnatacka = 1; z = pocetnatacka - korak; elementi[prvi_, drugi_] = 0; HC = Array[elementi, {brojelemenata, brojelemenata}]; For[i = 1, i < brojelemenata/2, i++, HC[[i, i]] = Aglavna[z + ikorak]; HC[[i, i + 1]] = Agornja[z + (i + 1)korak]; HC[[i + 1, i]] = Adonja[z + ikorak]; If[i + brojelemenata/2 + 1 <= brojelemenata, HC[[i + brojelemenata/2, i + brojelemenata/2]] = Dglavna[z + ikorak]; HC[[i + brojelemenata/2, i + brojelemenata/2 + 1]] = Dgornja[z + (i + 1)korak]; HC[[i + brojelemenata/2 + 1, i + brojelemenata/2]] = Ddonja[z + ikorak]; If[i + brojelemenata/2 + 1 <= brojelemenata, HC[[i, i + brojelemenata/2]] = Bglavna[z + ikorak]; HC[[i + 1, i + brojelemenata/2]] = Bdonja[z + ikorak]; HC[[i, i + brojelemenata/2 + 1]] = Bgornja[z + (i + 1)korak]; HC[[i + brojelemenata/2, i]] = Cglavna[z + ikorak]; HC[[i + brojelemenata/2, i + 1]] = Cgornja[z + (i + 1)korak]; HC[[i + brojelemenata/2 + 1, i]] = Cdonja[z + ikorak]; ]; ]; ]; HC[[brojelemenata/2, brojelemenata/2]] = Aglavna[z + korakbrojelemenata/2]; HC[[1, brojelemenata/2]] = 0; HC[[brojelemenata/2, 1]] = 0; HC[[brojelemenata/2, brojelemenata]] = Bglavna[z + korakbrojelemenata/2]; HC[[brojelemenata, brojelemenata/2]] = Cglavna[z + korakbrojelemenata/2]; HC[[brojelemenata, brojelemenata]] = Dglavna[z + korakbrojelemenata/2]; HC // MatrixForm Where first 4 lines are number of elements; step; starting point; and z0. The matrix should look like:

Hi. I would like to construct sparse matrix with 9 diagonals. It is organized as 4 blocks with 3 diagonals each. Something like this: (A B);(C D) Currently, I am using For loop for the job, but I suspect there is a more efficient way. My code is:

Agornja[z_] := z*agornja;
Aglavna[z_] := z*aglavna;
Adonja[z_] := z*adonja;

Bgornja[z_] := z*bgornja;
Bglavna[z_] := z*bglavna;
Bdonja[z_] := z*bdonja;

Cgornja[z_] := z*cgornja;
Cglavna[z_] := z*cglavna;
Cdonja[z_] := z*cdonja;

Dgornja[z_] := z*dgornja;
Dglavna[z_] := z*dglavna;
Ddonja[z_] := z*ddonja;

Where z is just for me to see the incrementation. For loop is:

brojelemenata = 10;
korak = 1;
pocetnatacka = 1;
z = pocetnatacka - korak;
elementi[prvi_, drugi_] = 0;
HC = Array[elementi, {brojelemenata, brojelemenata}];
For[i = 1, i < brojelemenata/2, i++,
  HC[[i, i]] = Aglavna[z + i*korak];
  HC[[i, i + 1]] = Agornja[z + (i + 1)*korak];
  HC[[i + 1, i]] = Adonja[z + i*korak];
  If[i + brojelemenata/2 + 1 <= brojelemenata,
   HC[[i + brojelemenata/2, i + brojelemenata/2]] = 
    Dglavna[z + i*korak];
   HC[[i + brojelemenata/2, i + brojelemenata/2 + 1]] = 
    Dgornja[z + (i + 1)*korak];
   HC[[i + brojelemenata/2 + 1, i + brojelemenata/2]] = 
    Ddonja[z + i*korak];
   If[i + brojelemenata/2 + 1 <= brojelemenata,
    HC[[i, i + brojelemenata/2]] = Bglavna[z + i*korak];
    HC[[i + 1, i + brojelemenata/2]] = Bdonja[z + i*korak];
    HC[[i, i + brojelemenata/2 + 1]] = Bgornja[z + (i + 1)*korak];
    HC[[i + brojelemenata/2, i]] = Cglavna[z + i*korak];
    HC[[i + brojelemenata/2, i + 1]] = Cgornja[z + (i + 1)*korak];
    HC[[i + brojelemenata/2 + 1, i]] = Cdonja[z + i*korak];
    ];
   ];
  ];
HC[[brojelemenata/2, brojelemenata/2]] = 
  Aglavna[z + korak*brojelemenata/2];
HC[[1, brojelemenata/2]] = 0;
HC[[brojelemenata/2, 1]] = 0;
HC[[brojelemenata/2, brojelemenata]] = 
  Bglavna[z + korak*brojelemenata/2];
HC[[brojelemenata, brojelemenata/2]] = 
  Cglavna[z + korak*brojelemenata/2];
HC[[brojelemenata, brojelemenata]] = 
  Dglavna[z + korak*brojelemenata/2];
HC // MatrixForm

Where first 4 lines are number of elements; step; starting point; and z0. The matrix should look like: enter image description here

POSTED BY: Milan Vicentijevic

2 Replies

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Sander Huisman

Sander Huisman, University of Twente

Posted 10 years ago

There is an 'easier' way perhaps: ClearAll[SubMatrix] SubMatrix[funcs:{f1_,f2_,f3_},n_:5]:=Total[MapThread[DiagonalMatrix[#1/@Range[#3,#3+n-1],#2,n]&,{funcs,{-1,0,1},{1,1,2}}]] Agornja[z_]:=zagornja; Aglavna[z_]:=zaglavna; Adonja[z_]:=zadonja; Bgornja[z_]:=zbgornja; Bglavna[z_]:=zbglavna; Bdonja[z_]:=zbdonja; Cgornja[z_]:=zcgornja; Cglavna[z_]:=zcglavna; Cdonja[z_]:=zcdonja; Dgornja[z_]:=zdgornja; Dglavna[z_]:=zdglavna; Ddonja[z_]:=zddonja; funcs={{{Adonja,Aglavna,Agornja},{Bdonja,Bglavna,Bgornja}},{{Cdonja,Cglavna,Cgornja},{Ddonja,Dglavna,Dgornja}}}; matrix=Flatten[Map[SubMatrix,funcs,{2}],{{1,3},{2,4}}]; matrix//Grid This solution is tailor-made for this case, it might be slightly different for other cases. But it does work for the above case. Each submatrix is created first, and then using flatten these 'blocks' are merged to a big matrix.

There is an 'easier' way perhaps:

ClearAll[SubMatrix]
SubMatrix[funcs:{f1_,f2_,f3_},n_:5]:=Total[MapThread[DiagonalMatrix[#1/@Range[#3,#3+n-1],#2,n]&,{funcs,{-1,0,1},{1,1,2}}]]

Agornja[z_]:=z*agornja;
Aglavna[z_]:=z*aglavna;
Adonja[z_]:=z*adonja;

Bgornja[z_]:=z*bgornja;
Bglavna[z_]:=z*bglavna;
Bdonja[z_]:=z*bdonja;

Cgornja[z_]:=z*cgornja;
Cglavna[z_]:=z*cglavna;
Cdonja[z_]:=z*cdonja;

Dgornja[z_]:=z*dgornja;
Dglavna[z_]:=z*dglavna;
Ddonja[z_]:=z*ddonja;

funcs={{{Adonja,Aglavna,Agornja},{Bdonja,Bglavna,Bgornja}},{{Cdonja,Cglavna,Cgornja},{Ddonja,Dglavna,Dgornja}}};

matrix=Flatten[Map[SubMatrix,funcs,{2}],{{1,3},{2,4}}];
matrix//Grid

This solution is tailor-made for this case, it might be slightly different for other cases. But it does work for the above case. Each submatrix is created first, and then using flatten these 'blocks' are merged to a big matrix.

POSTED BY: Sander Huisman

Milan Vicentijevic

Posted 10 years ago

Thank you. This is very helpful.

POSTED BY: Milan Vicentijevic

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