Multidimensional Arrays

Multidimensional array is the array with two or more dimensions.For example: char box [3] [3] defines a two-dimensional array and box[2][1] is an element in row 2 , column 1andchar box[][3] can be used in the function prototype

note that only the first dimension can be omitted

Multidimensional Arrays• For example : double table [NROWS][NCOLS]; Can be used as parameter in the function

prototype as:void process_martrix( int in[ ][4], int out[ ][4], int nrows)

Two Dimensional Array• Char box [3] [3]

0

1

2

0 1 2

Row

Column

box [1] [2]

/* * Checks whether a box is completely filled */intfilled(char box[3][3]) /* input - box to check */{ int r,c, /* row and column subscripts */ ans; /* whether or not box is filled. */ /* Assumes box is filled until blank is found */ ans = 1; /* Resets ans to zero if a blank is found */ for (r = 0; r < 3; ++r) for (c = 0; c < 3; ++c) if (box[r][c] == ' ') ans = 0; return (ans);}

Arrays with Several Dimensions

• int soil_type[4] [7] [MAXDEPTH]

Case Study: Cellular Telephone System

• Problem: Finding the best way to build a cellular network. There is some marketing data that predicts the demand will be at tree time of interest. There are only 10 transmitters and there is a need for a program to help analyzing call demand data.

Case Study: Cellular Telephone System

• Analysis: There should be three matrices shows traffic density for each time of the day:

Input: int commuters[GRID_SIZE][GRID_SIZE] int salesforce[GRID_SIZE][GRID_SIZE] int weekend[GRID_SIZE][GRID_SIZE]Output:int summed_data[GRID_SIZE][GRID_SIZE] int location_i, location_j

Case Study: Cellular Telephone System• Design: initial algorithm:1. Get traffic data for three time period2. Get the weights from user3. Multiply weight by each matrix entry and

store the sum in the summed data4. Find highest valued cells in the summed

data and display them as the pair of location_i and location_j

• Implementation

Filling the multidimensional array/* Fills 3 GRID_SIZE x GRID_SIZE arrays with traffic data from

TRAFFIC_FILE*/voidget_traffic_data(int commuters[GRID_SIZE][GRID_SIZE], /* output */ int salesforce[GRID_SIZE][GRID_SIZE], /* output */ int weekend[GRID_SIZE][GRID_SIZE]) /* output */{ int i, j; /* loop counters */ FILE *fp; /* file pointer */ fp = fopen(TRAFFIC_FILE, "r"); for (i = 0; i < GRID_SIZE; ++i) for (j = 0; j < GRID_SIZE; ++j) fscanf(fp, "%d", &commuters[i][j]); for (i = 0; i < GRID_SIZE; ++i) for (j = 0; j < GRID_SIZE; ++j) fscanf(fp, "%d", &salesforce[i][j]); for (i = 0; i < GRID_SIZE; ++i) for (j = 0; j < GRID_SIZE; ++j) fscanf(fp, "%d", &weekend[i][j]); fclose(fp);}

/* Computes and displays the weighted, summed_data */

for (i = 0; i < GRID_SIZE; ++i) for (j = 0; j < GRID_SIZE; ++j) summed_data[i][j] = commuter_weight* commuters[i][j] + salesforce_weight * salesforce[i][j] + weekend_weight * weekend[i][j];

Modifying the multidimensional array

/* Finds the NUM_TRANSMITTERS highest values in the summed_data matrix.Temporarily stores the coordinates in location_i and location_j, and then displays the resulting locations */

printf("\n\nLocations of the %d transmitters:\n\n", NUM_TRANSMITTERS);

for (tr = 1; tr <= NUM_TRANSMITTERS; ++tr) { current_max = SELECTED; /* Starts off our search with a value that is known to be too low. */ for (i = 0; i < GRID_SIZE; ++i) { for (j = 0; j < GRID_SIZE; ++j) { if (current_max < summed_data[i][j]) { current_max = summed_data[i][j]; location_i = i; location_j = j; } } }

Searching in the multidimensional array

Printing the contents of multidimensional array/* * Displays contents of a GRID_SIZE x GRID_SIZE matrix of

integers */voidprint_matrix(int matrix[GRID_SIZE][GRID_SIZE]){ int i, j; /* loop counters */ for (i = 0; i < GRID_SIZE; ++i) { for (j = 0; j < GRID_SIZE; ++j) printf("%3d ", matrix[i][j]); printf("\n"); }}

Vectors

• Vector: a mathematical object consisting of a sequence of numbers./* a vector <4, 12, 19> */int vect[3] = {4, 12, 19};

• Differences between vector and array: 1- an n_dimensional vector is represented

in C as a one dimensional array of size n. 2- vect3 is vect[2] in C

Vectors

• Calculating scalar product: <1,2,4>. <2,3,1> = 1*2 + 2*3 +4*1=12In C:

sum_prod = 0;for (k=0; k<n; k++)

sum_prod += x[k] * w[k];

Matrices• Matrix: a mathematical object consisting of a rectangular

arrangement of numbers called the element of matrix../* a matrix 3 6

4 5int x[2][2] = {{3, 6}, {4, 5}};

3 6

4 5

x

Matrices• Multiplying a matrix by a vector A * X = V 1 1 1 5 2 3 1 1 10 multiplication 1 -1 -1 * 2 = -3 on 0 1 2 2 6 the right• In C for each member of V:

v[i] = 0;for (k=0; k<n; k++)

v[k] += a[i][k] * x[k];

/* Computes the product of M-by-N matrix a and the N-dimensional vector x. The result is stored in the output parameter v, an M-dimensional vector.*/

void mat_vec_prod(double v[], /* M-dimensional vector */ double a[M][N], /* M-by-N matrix */ double x[]) /* N-dimensional vector */ { int i, k; for (i = 0; i < M; ++i) { v[i] = 0; for (k = 0; k < N; ++k) { v[i] += a[i][k] * x[k]; } } }

Matrix Multiplication 1 1 1 2 0 1 6 0 0 2 3 1 * 1 -1 0 = 10 -2 1 1 -1 -1 3 1 -1 -2 0 2

for ( i=0; i< m , ++i) { for (j=0; j<p; ++j) { …….. compute c[i][j]…. }}

/* Multiplies matrices A and B yielding product matrix C */ void mat_prod(double c[M][P], /* output - M by P matrix */ double a[M][N], /* input - M by N matrix */ double b[N][P]) /* input - N by P matrix */ { int i, j, k; for (i = 0; i < M; ++i) { for (j = 0; j < P; ++j) { c[i][j] = 0; for (k = 0; k < N; ++k) c[i][j] += a[i][k] * b[k][j]; } } }

Solving System of Linear Equations

• To solve many problems such as force equation in three-dimensional system we need to solve a three linear equations:

A X = Y 1 1 1 x1 4 2 3 1 * x2 = 9 1 -1 -1 x3 -2It is multiplication of a matrix by a vector on the

right

Gaussian Elimination

• Gaussian elimination can be used to solve a linear equation.

• The algorithm for Gussian elimination is:1. Transform the original system into scaled

triangular form.2. Solve for xi by back substitution

Gaussian Elimination

triangular form 1 1 1 x1 4 0 1 -1 * x2 = 1 0 0 1 x3 1 back substitution x1 + x2 + x3 = 4 x2 - x3 = 1 x3 = 1

Gaussian Elimination

• For doing that we need to triangularizing the augmented matrix by following operations:

1. Multiply any row of aug by nonzero number2. Add to any row of aug a multiple of other rows3. Swap any two rows• If system has a unique solution, we can get

the system into desired form by this three operations.

/* * Performs pivoting with respect to the pth row and the pth column * If no nonzero pivot can be found, FALSE is sent back through piv_foundp */void pivot(double aug[N][N+1], /* input/output - augmented matrix */ int p, /* input - current row */ int *piv_foundp) /* output - whether or not nonzero pivot found*/{ double xmax, xtemp; int j, k, max_row; /* Finds maximum pivot*/ xmax = fabs(aug[p][p]); max_row = p; for (j = p+1; j < N; ++j) { if (fabs(aug[j][p]) > xmax) { xmax = fabs(aug[j][p]); max_row = j; } }

/* Swaps rows if nonzero pivot was found */ if (xmax == 0) { *piv_foundp = FALSE; } else { *piv_foundp = TRUE; if (max_row != p) { /* swap rows */ for (k = p; k < N+1; ++k) { xtemp = aug[p][k]; aug[p][k] = aug[max_row][k]; aug[max_row][k] = xtemp; } } }}

/* * Performs back substitution to compute a solution vector to a system of * linear equations represented by the augmented matrix aug. Assumes that * the coefficient portion of the augmented matrix has been triangularized, * and its diagonal values are all 1. */voidback_sub(double aug[N][N+1], /* input - scaled, triangularized augmented matrix */ double x[N]) /* output - solution vector */{ double sum; int i, j; x[N - 1] = aug[N - 1][N]; for (i = N - 2; i >= 0; --i) { sum = 0; for (j = i + 1; j < N; ++j) sum += aug[i][j] * x[j]; x[i] = aug[i][N] - sum; }}

Common Programming Errors• Use constants for each dimension’s size when

declaring multidimensional array• When declaring the array as a parameter of a

function if you omit the first dimension all other dimensions must be supplied

• Since access to the elements of a multidimensional array requires nested counting loops it is easy to make out-of-range error.

• Since using multidimensional arrays as local variables requires large memory space, you may need to tell to operating system to increase stack size when the program is running