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Sparse Matrices

This article explores different examples of sparse matrices and their representations, including diagonal, tridiagonal, lower triangular, airline flight matrix, and web page matrix. It also discusses various representations such as single linear list, array linear list, chain representation, one linear list per row, orthogonal list representation, and circular list variation.

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Sparse Matrices

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  1. Sparse Matrices sparse … many elements are zero dense … few elements are zero

  2. Example Of Sparse Matrices diagonal tridiagonal lower triangular (?) These are structured sparse matrices. May be mapped into a 1D array so that a mapping function can be used to locate an element.

  3. Unstructured Sparse Matrices Airline flight matrix. airports are numbered 1 through n flight(i,j)= list of nonstop flights from airport i to airport j n = 1000 (say) n x n array of list references => 4 million bytes total number of flights = 20,000 (say) need at most 20,000 list references => at most 80,000 bytes

  4. Unstructured Sparse Matrices Web page matrix. web pages are numbered 1 through n web(i,j)= number of links from page i to page j Web analysis. authority page … page that has many links to it hub page … links to many authority pages

  5. Web Page Matrix n = 2 billion (and growing by 1 million a day) n x n array of ints => 16 * 1018 bytes (16 * 109 GB) each page links to 10 (say) other pages on average on average there are 10 nonzero entries per row space needed for nonzero elements is approximately 20 billion x 4 bytes = 80 billion bytes (80 GB)

  6. Representation Of Unstructured Sparse Matrices Single linear list in row-major order. scan the nonzero elements of the sparse matrix in row-major order each nonzero element is represented by a triple (row, column, value) the list of triples may be an array list or a linked list (chain)

  7. list = row 1 1 2 2 4 4 column 3 5 3 4 2 3 value 3 4 5 7 2 6 Single Linear List Example 0 0 3 0 4 0 0 5 7 0 0 0 0 0 0 0 2 6 0 0

  8. element 0 1 2 3 4 5 row 1 1 2 2 4 4 column 3 5 3 4 2 3 value 3 4 5 7 2 6 Array Linear List Representation row 1 1 2 2 4 4 list = column 3 5 3 4 2 3 value 3 4 5 7 2 6

  9. row col value next Chain Representation Node structure.

  10. 1 3 1 5 2 3 2 4 4 2 4 3 3 4 5 7 2 6 null firstNode Single Chain row 1 1 2 2 4 4 list = column 3 5 3 4 2 3 value 3 4 5 7 2 6

  11. One Linear List Per Row row1 = [(3, 3), (5,4)] row2 = [(3,5), (4,7)] row3 = [] row4 = [(2,2), (3,6)] 0 0 3 0 4 0 0 5 7 0 0 0 0 0 0 0 2 6 0 0

  12. next col value Array Of Row Chains Node structure.

  13. null 3 3 5 4 null 3 5 4 7 null 2 2 3 6 row[] Array Of Row Chains 0 0 3 0 4 0 0 5 7 0 0 0 0 0 0 0 2 6 0 0 null

  14. row col value down next Orthogonal List Representation Both row and column lists. Node structure. struct NodeS { int r; // row number int c; // column number struct NodeS *rptr; // row pointer struct NodeS *cptr; // column header int data; // actual data };typedef struct NodeS NodeS;

  15. 1 3 1 5 3 4 n 2 3 2 4 5 7 n null 4 2 4 3 2 6 n Row Lists 0 0 3 0 4 0 0 5 7 0 0 0 0 0 0 0 2 6 0 0

  16. 1 3 1 5 3 4 n 2 3 2 4 5 7 4 2 4 3 2 6 n n Column Lists 0 0 3 0 4 0 0 5 7 0 0 0 0 0 0 0 2 6 0 0

  17. 1 3 1 5 3 4 n n 2 3 2 4 5 7 n null 4 2 4 3 2 6 n n n row[] Orthogonal Lists 0 0 3 0 4 0 0 5 7 0 0 0 0 0 0 0 2 6 0 0

  18. Variations May use circular lists instead of chains.

  19. Sparse matrices • Represent each column of a sparse matrix as a circularly linked list with a head node. • A similar representation for each row of a sparse matrix. • Each node has a tag field that is used to distinguish between head nodes and entry nodes. • Each head node has three additional fields: down, right, and next. • down field: links into a column list • right field: links into a row list • next field: links the head nodes together • The head node for row i is also the head node for column i, and the total number of head nodes is max {number of rows, number of columns}.

  20. Each entry node has six fields: tag, row, col, down, right, value. • down field: links to the next nonzero term in the same column • right field: links to the next nonzero term in the same row • A num_rows × num_cols matrix with num_terms nonzero terms needs max{num_rows, num_cols} + num_terms + 1 nodes. • Total storage will be less than num_rows × num_cols when num_terms is sufficiently small.

  21. 4 4 0 2 11 1 1 1 0 12 5 1 2 -4 3 3 -15

  22. struct NodeS { int r; // row number int c; // column number struct NodeS *rptr; // row pointer struct NodeS *cptr; // column header int data; // actual data }; typedef struct NodeS NodeS;

  23. NodeS* create(NodeS *h) { NodeS *ptr, *ptr1, *ptr2, *node; NodeS *row_ptr, *column_ptr, *row_header, *column_header; int i,j,d; int r, c; printf("\nEnter the no. of rows ::"); scanf("%d", &r); printf("Enter the no. of columns ::"); scanf("%d", &c); // create the head node h=(NodeS *)malloc(sizeof(NodeS)); h->r=r; h->c=c; h->rptr=h; h->cptr=h; h->data=0;

  24. // create column headers ptr=h; for(i=1;i<=c;i++) { NodeS *node; node=(NodeS *)malloc(sizeof(NodeS)); node->r=0; node->c=i; node->data=0; node->rptr=h; node->cptr=node; ptr->rptr=node; ptr=node; }

  25. // create row headers ptr=h; for(i=1;i<=r;i++) { NodeS *node; node=(NodeS *)malloc(sizeof(NodeS)); node->r=i; node->c=0; node->data=0; node->rptr=node; node->cptr=h; ptr->cptr=node; ptr=node; }

  26. printf("\nNow enter the non zero elements one by one\n"); printf("\nEnter row number,column number, data\n"); printf("Enter (0 0 0) to stop ::"); scanf("%d%d%d",&i, &j, &d); if(i>r || j>c ||i<1 ||j<1) { printf("Error in input"); exit(1); }

  27. while(i&&j&&d) { row_header=h->cptr; column_header=h->rptr; // find the correct row header and column header while(row_header->r<i) row_header=row_header->cptr; while(column_header->c<j) column_header=column_header->rptr;

  28. // find the correct position to insert row_ptr=row_header; while((row_ptr->c)<j) { ptr1=row_ptr; row_ptr=row_ptr->rptr; if(row_ptr==row_header) break; }

  29. column_ptr=column_header; while(column_ptr->r<i) { ptr2=column_ptr; column_ptr=column_ptr->cptr; if(column_ptr==column_header) break; }

  30. node=(NodeS *)malloc(sizeof(NodeS)); node->r=i; node->c=j; node->data=d; ptr1->rptr=node; ptr2->cptr=node; node->rptr=row_ptr; node->cptr=column_ptr;

  31. printf("\nEnter row number,column number,data\n"); printf("Enter (0 0 0) to stop ::"); scanf("%d%d%d", &i, &j, &d); if(i>r || j>c ) { printf(" error input"); exit(1); } } return h; }

  32. void display(NodeS*h) { NodeS *right; right=h->cptr; while(right!=h) { NodeS *r=right; right=right->rptr; while(right!=r) { printf("%d\t%d\t%d\n", right->r, right->c, right->data); right=right->rptr; } right=right->cptr; } }

  33. void add(NodeS*h1,NodeS*h2) { NodeS *r1, *r2, *p1, *p2; if(h1->r==h2->r && h1->c==h2->c) printf("The addition of the two given sparse matrices is ::\n"); else { printf("addition is not possible"); exit(1); } r1=h1->cptr; r2=h2->cptr;

  34. while(r1!=h1) { p1=r1; r1=r1->rptr; p2=r2; r2=r2->rptr; while(r1!= p1 && r2!=p2) { if(r1->c==r2->c) { printf("%d\t%d\t%d\n", r1->r, r1->c, (r1->data+r2->data)); r1=r1->rptr; r2=r2->rptr; } else if(r1->c>r2->c) { printf("%d\t%d\t%d\n", r2->r, r2->c, r2->data); r2=r2->rptr; } else { printf("%d\t%d\t%d\n", r1->r, r1->c, r1->data); r1=r1->rptr; } }

  35. while(r1!=p1) { printf("%d\t%d\t%d\n", r1->r, r1->c, r1->data); r1=r1->rptr; } while(r2!=p2) { printf("%d\t%d\t%d\n", r2->r, r2->c, r2->data); r2=r2->rptr; } r1=r1->cptr; r2=r2->cptr; } }

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