Arrays and Structures Arrays Polynomial representation Polynomial add

Arrays and Structures

Arrays • Polynomial representation • Polynomial add, subtract, multiply, …

Sparse Matrices Matrix table of values 00304 00570 00000 02600 Column 4 Row 2 4 x 5 matrix 4 rows 5 columns 20 elements 6 nonzero elements

Sparse Matrices Sparse matrix #nonzero elements/#elements is small. Examples: • Diagonal § Only elements along diagonal may be nonzero § n x n matrix ratio is n/n 2 = 1/n • Tridiagonal • Only elements on 3 central diagonals may be nonzero • Ratio is (3 n-2)/n 2 = 3/n – 2/n 2

Sparse Matrices • Lower triangular (? ) • Only elements on or below diagonal may be nonzero • Ratio is n(n+1)(2 n 2) ~ 0. 5 These are structured sparse matrices. Nonzero elements are in a well-defined portion of the matrix.

Sparse Matrices An n x n matrix may be stored as an n x n array. This takes O(n 2) space. The example structured sparse matrices may be mapped into a 1 D array so that a mapping function can be used to locate an element quickly; the space required by the 1 D array is less than that required by an n x n array.

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 pointers => 4 million bytes § total number of nonempty flight lists = 20, 000 (say) § need at most 20, 000 list pointers => at most 80, 000 bytes

The sparse matrix • Introduction – In mathematics, a matrix contains m rows and n columns of elements, we write m n to designate a matrix with m rows and n columns. sparse matrix 5*3 15/15 6*6 8/36

The sparse matrix – A minimal set of operations • • Matrix creation Addition Multiplication Transpose

The sparse matrix • The standard representation of a matrix is a two dimensional array defined as a[MAX_ROWS][MAX_COLS] – We can locate quickly any element by writing a[i ][ j ] • Sparse matrix wastes space – We must consider alternate forms of representation. – Our representation of sparse matrices should store only nonzero elements. – Each element is characterized by <row, col, value>.

The sparse matrix • We implement the Create operation as below:

The sparse matrix • 3 -Column representation (coordinated list) • Figure shows how the sparse matrix of prior matrix is represented in the array a. – Represented by a two-dimensional array. – Each element is characterized by <row, col, value>. # of rows (columns) # of nonzero terms

Single linear list Representation of Sparse Matrices Single linear list in row-major order. scan the nonzero elements of the sparse matrix in rowmajor order (i. e. , scan the rows left to right beginning with row 1 and picking up the nonzero elements) each nonzero element is represented by a triple (row, column, value) the list of triples is stored in a 1 D array

Single Linear List Example 00304 00570 00000 02600 list = row 1 1 2 2 4 4 column 3 5 3 4 2 3 value 3 4 5 7 2 6

/ C++ program for Sparse Matrix Representation // using Array #include<stdio. h> int main() { // Assume 4 x 5 sparse matrix int sparse. Matrix[4][5] = { {0 , 3 , 0 , 4 }, {0 , 5 , 7 , 0 }, {0 , 0 , 0 }, {0 , 2 , 6 , 0 } }; // Making of new matrix int compact. Matrix[3][size]; int k = 0; for (int i = 0; i < 4; i++) for (int j = 0; j < 5; j++) if (sparse. Matrix[i][j] != 0) { compact. Matrix[0][k] = i; compact. Matrix[1][k] = j; compact. Matrix[2][k] = sparse. Matrix[i][j]; k++; } for (int i=0; i<3; i++) { for (int j=0; j<size; j++) printf("%d ", compact. Matrix[i][j]); int size = 0; for (int i = 0; i < 4; i++) for (int j = 0; j < 5; j++) if (sparse. Matrix[i][j] != 0) size++; printf("n"); } return 0; } // number of columns in compact. Matrix (size) must be equal to number of non zero elements in sparse. Matrix Output: 001133 242312 345726

One Linear List Per Row 00304 00570 00000 02600 row 1 = [(3, 3), (5, 4)] row 2 = [(3, 5), (4, 7)] row 3 = [] row 4 = [(2, 2), (3, 6)]

Single Linear List • Array representation – 1 D Array of triples of type term • int row, col, value • Size of 1 D array generally not predictable at time of initialization. – Start with some default capacity/size (say 10) – Increase capacity as needed – Use REALLOC

Approximate Memory Requirements 500 x 500 matrix with 1994 nonzero elements, 4 bytes per element 2 D array 500 x 4 = 1 million bytes 1 D array of triples 3 x 1994 x 4 = 23, 928 bytes row, col, value

Using Linked Lists In linked list, each node has four fields. These four fields are defined as: • Row: Index of row, where non-zero element is located • Column: Index of column, where non-zero element is located • Value: Value of the non zero element located at index – (row, column) • Next node: Address of the next node

#include<stdio. h> #include<stdlib. h> // Node to represent sparse matrix struct Node { int value; int row_position; int column_postion; struct Node *next; }; // Function to create new node void create_new_node(struct Node** start, int non_zero_element, int row_index, int column_index ) { struct Node *temp, *r; temp = *start; if (temp == NULL) { // Create new node dynamically temp = (struct Node *) malloc (sizeof(struct Node)); temp->value = non_zero_element; temp->row_position = row_index; temp->column_postion = column_index; temp->next = NULL; *start = temp; } else { while (temp->next != NULL) temp = temp->next; // Create new node dynamically r = (struct Node *) malloc (sizeof(struct Node)); r->value = non_zero_element; r->row_position = row_index; r->column_postion = column_index; r->next = NULL; temp->next = r; } }

// This function prints contents of linked list // starting from start void Print. List(struct Node* start) { struct Node *temp, *r, *s; temp = r = start; printf("row_position: "); while(temp != NULL) { // Driver of the program int main() { // Assume 4 x 5 sparse matrix int sparse. Matric[4][5] = { {0 , 3 , 0 , 4 }, {0 , 5 , 7 , 0 }, {0 , 0 , 0 }, {0 , 2 , 6 , 0 } }; printf("%d ", temp->row_position); temp = temp->next; /* Start with the empty list */ struct Node* start = NULL; } printf("n"); printf("column_postion: "); while(r != NULL) { printf("%d ", r->column_postion); r = r->next; } printf("n"); printf("Value: "); while(s != NULL) { printf("%d ", s->value); s = s->next; } printf("n"); } for (int i = 0; i < 4; i++) for (int j = 0; j < 5; j++) // Pass only those values which are non - zero if (sparse. Matric[i][j] != 0) create_new_node(&start, sparse. Matric[i][j], i, j); Print. List(start); return 0; }

Approximate Memory Requirements 500 x 500 matrix with 1994 nonzero elements, 4 bytes per element

The sparse matrix • Transpose a Matrix – For each row i • take element <i, j, value> and store it in element <j, i, value> of the transpose. • difficulty: where to put <j, i, value> (0, 0, 15) ====> (0, 0, 15) (0, 3, 22) ====> (3, 0, 22) (0, 5, -15) ====> (5, 0, -15) (1, 1, 11) ====> (1, 1, 11) Move elements down very often. – For all elements in column j, • place element <i, j, value> in element <j, i, value>

The sparse matrix – Transpose a Matrix transpose

Matrix Transpose 00304 00570 00000 02600 0002 3506 0700 4000

Matrix Transpose 00304 00570 00000 02600 row 1 1 2 2 4 4 column 3 5 3 4 2 3 value 3 4 5 7 2 6 0000 0002 3506 0700 4000 2 3 3 3 4 5 4 1 2 4 2 1 2 3 5 6 7 4

Fast Transpose

Matrix Transpose 00304 00570 00000 02600 0002 3506 0700 4000 Step 1: #nonzero in each row of transpose. = #nonzero in each column of original matrix = [0, 1, 3, 1, 1] Step 2: Start of each row of transpose = sum of size of preceding rows of transpose = [0, 0, 1, 4, 5] row 1 1 2 2 4 4 column 3 5 3 4 2 3 value 3 4 5 7 2 6 Step 3: Move elements, left to right, from original list to transpose list.

Matrix Transpose Step 1: #nonzero in each row of transpose. = #nonzero in each column of original matrix = [0, 1, 3, 1, 1] Step 2: Start of each row of transpose = sum of size of preceding rows of transpose (accumulative row values) = [0, 0, 1, 4, 5, 6] Complexity m x n original matrix t nonzero elements Step 1: O(n+t) Step 2: O(n) Step 3: O(t) Overall O(n+t) Step 3: Move elements, left to right, from original list to transpose list. 00304 00570 00000 02600 0002 3506 0700 4000

The sparse matrix • Matrix multiplication – Definition: Given A and B where A is m n and B is n p, the product matrix D has dimension m p. Its <i, j> element is for 0 i < m and 0 j < p. – Example:

Runtime Performance Matrix Transpose 500 x 500 matrix with 1994 nonzero elements Run time measured on a 300 MHz Pentium II PC 2 D array Sparse. Matrix 210 ms 6 ms

Performance Matrix Addition. 500 x 500 matrices with 1994 and 999 nonzero elements 2 D array Sparse. Matrix 880 ms 18 ms