Download Multidimensional Arrays

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 1
and
char 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
Row
1
Column
2
0
1
2
box [1] [2]
/*
* Checks whether a box is completely filled
*/
int
filled(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
•
1.
2.
3.
Design: initial algorithm:
Get traffic data for three time period
Get the weights from user
Multiply weight by each matrix entry and
store the sum in the summed data
4. 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*/
void
get_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);
}
Modifying the multidimensional array
/* 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];
Searching in 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;
}
}
}
Printing the contents of multidimensional array
/*
* Displays contents of a GRID_SIZE x GRID_SIZE matrix of
integers
*/
void
print_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=12
In 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 5
int x[2][2] = {{3, 6}, {4, 5}};
x
3
6
4
5
Matrices
• Multiplying a matrix by a vector
A
*
X=
V
1 1 1
5
2 3 1
1
10
1 -1 -1 *
2 =
-3
0 1 2
2
6
• In C for each member of V:
v[i] = 0;
for (k=0; k<n; k++)
v[k] += a[i][k] * x[k];
multiplication
on
the right
/* Computes the product of M-by-N matrix a and the Ndimensional 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
2 3 1 * 1 -1 0 = 10 -2
1 -1 -1
3 1 -1
-2 0
for ( i=0; i< m , ++i) {
for (j=0; j<p; ++j) {
…….. compute c[i][j]….
}
}
0
1
2
/* 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
-2
It 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
0 1 -1 *
x2
0 0 1
x3
back substitution
x 1 + x2 + x3 = 4
x2 - x3 = 1
x3 = 1
=
4
1
1
Gaussian Elimination
•
For doing that we need to triangularizing the
augmented matrix by following operations:
1. Multiply any row of aug by nonzero number
2. Add to any row of aug a multiple of other rows
3. 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.
*/
void
back_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