1. Chapter 5 Arrays and Matrices

2. 5.1 One-Dimensional Arrays
Suppose, you need to store years of 100 cars. Will you define 100 variables?
int y1, y2,…, y100;
An array is an indexed data structure to represent several variables having the same data type: int y[100];
y[0] y[1] y[2] … y[k-1] y[k] y[k+1] … y[98] y[99]

3. One-Dimensional Arrays (cont’d)
An element of an array is accessed using the array name and an index or subscript, for example: y[5] which can be used like a variable
In C, the subscripts always start with 0 and increment by 1, so y[5] is the sixth element
The name of the array is the address of the first element and the subscript is the offset
y[0] y[1] y[2] … y[k-1] y[k] y[k+1] … y[98] y[99]

4. Definition and Initialization
An array is defined using a declaration statement.
data_type array_name[size];
allocates memory for size elements
subscript of first element is 0
subscript of last element is size-1
size must be a constant

5. Example int list[5]; allocates memory for 5 integer variables
subscript of first element is 0
subscript of last element is 4
C does not check bounds on arrays
list[6] =5; /* may give segmentation fault or overwrite other memory locations*/

6. Initializing Arrays Examples:
Arrays can be initialized at the time they are declared.
Examples:
double taxrate[3] ={0.15, 0.25, 0.3};
char list[5] = {‘h’, ’e’, ’l’, ’l’, ’o’};
double vector[100] = {0.0}; /* assigns
zero to all 100 elements */
int s[] = {5,0,-5}; /*the size of s is 3*/

7. Assigning values to an array
For loops are often used to assign values to an array
list[0]
list[1]
list[2]
list[3]
list[4]
Example:
int list[5], i;
for(i=0; i<5; i++){
list[i] = i; }
list[0]
list[3]
list[4]
list[1]
list[2] 1 2 3 4 OR
for(i=0; i<=4; i++){
list[i] = i; }

8. Assigning values to an array
Give a for loop to assign the below values to list
int list[5], i;
for(i=0; i<5; i++){
list[i] = 4-i; }
list[0]
list[3]
list[4]
list[1]
list[2] 4 3 2 1

9. Input from a data file
Arrays are often used to store information from a data file
int k;
double time[10], motion[10];
FILE *sensor3;
sensor3 = fopen(“sensor3.dat”, “r”);
for(k=0; k<10; k++) {
fscanf(sensor3, “%lf %lf”,
&time[k], &motion[k]); }

10. Exercise
Show the contents of the arrays defined in each of the following sets of statements.
int x[10] = {-5, 4, 3};
char letters[] = {'a', 'b', 'c'};
double z[4];

11. Exercise int data[100], i; Store random numbers [0,99] in data
for (i=0; i<100; i++)
data[i] = rand() % 100;
Store random numbers [10,109] in data
data[i] = (rand() % 100) + 10; OR
data[i] = rand_int(10,109);

12. Computations on one-D arrays

13. Find Maximum Find maximum value in data array int data[100], max, i;
for (i=0; i<100; i++)
data[i] = rand_int(10,109);
max = data[0];
for (i=1; i<100; i++){
if (data[i] > max)
max = data[i]; }
printf("Max = %d\n",max);

14. Find average Find average of values in data array
int data[100], sum, i, avg;
for (i=0; i<100; i++)
data[i] = rand_int(10,109);
sum = 0;
for (i=0; i<100; i++){
sum = sum + data[i]; }
avg = (double)sum/100;
printf(“Avg = %lf\n", avg);

15. Number of elements greater than average
After finding the average as shown in previous slide, use the following code
count = 0;
for (i=0; i<100; i++){
if (data[i] > avg)
count++; }
printf(“%d elements are “
“greater than avg”, count);

16. } } } Find pair sum data[0]=5 data[1]=7 data[2]=15 data[3]=5 …
Find sum of every pair in data and write into pair array }
data[0]=5
data[1]=7
data[2]=15
data[3]=5 …
data[98]=3
data[99]=12
pair[0]=12
pair[1]=20 …
pair[49]=15 } . }

17. solution int data[100], pair[50], i; for (i=0; i<100; i++)
data[i] = rand_int(1,100);
for (i=0; i<50; i++){
pair[i]= data[2*i] + data[2*i+1]; }

18. Randomly re-shuffle numbers 30 times
data[0]=5
data[1]=7
data[2]=15
data[3]=5 …
data[98]=3
data[99]=12
data[0]=12
data[1]=7
data[2]=5
data[3]=15 …
data[98]=3
data[99]=5 .

19. solution int data[100], i, j, k, tmp; for (i=0; i<100; i++)
data[i] = rand_int(1,109);
for (n=0; n<30; n++){
i=rand_int(0,99);
j=rand_int(0,99);
tmp = data[i];
data[i] = data[j];
data[j] = tmp; }

20. Copy array1 to array2 in reverse order1 2 3 4 5
array1 6 3 1 9 7 2
array2 2 7 9 1 3 6

21. Print sum of top-bottom pairs…
A[49]=5
A[50]=3
A[98]=4
A[99]=5 + + + ….

22. Random numbers from an irregular range
Suppose you want to generate 50 random numbers, but you want to chose them uniformly from a set of given numbers like
Can you do this using arrays?

23. } } } Group avg Suppose we have a sorted array of hundred grades.….
Grade[9]
Grade[10] …
Grade[19]
Grade[20]
Grade[90]
Grade[99]
Group avg }
Suppose we have a sorted array of hundred grades.
We want to find the average of top ten, second top ten students etc. }

24. Arrays as Function Arguments

25. Function Arguments
Individual elements of an array can be passed as regular arguments.
void donothing(int a, int b) { … }
int main(void)
/* Declare variables and functions */
int array[5] = {1,2,3,4,5};
donothing(array[2], array[4]);
Calls donothing(3, 5);

26. Passing Arrays to Functions
Arrays are always pass by reference
Modifications to the array are reflected to main program
The array name is the address of the first element
The maximum size of the array must be specified at the time the array is declared.
The actual number of array elements that are used will vary, so the actual size of the array is usually passed as another argument to the function

27. Exercise a[0]=3 a[1]=5 c=? 8 b= n=2 i=0 1 2 sum=0 3 8 main() {
int a[2]={3, 5};
int c;
c = sum_arr(a, 2) }
int sum_arr(int b[], int n)
int i, sum=0;
for(i=0; i < n; i++)
sum = sum + b[i];
return(sum);
a[0]=3
a[1]=5
c=? 8 b=
n=2 i=
sum=

28. Exercise a[0]=3 20 a[1]=5 c=? 8 b= n=2 i=0 1 2 sum=0 3 8 main() {
int a[2]={3, 5};
int c;
c = sum_arr(a, 2) }
int sum_arr(int b[], int n)
int i, sum=0;
for(i=0; i < n; i++)
sum = sum + b[i];
b[0] = 20;
return(sum);
a[0]=3 20
a[1]=5
c=? 8 b=
n=2 i=
sum=

29. Exercise Write a function to find maximum value in the array data
int main() {
int data[100],i, max;
for (i=0; i<100; i++)
data[i] = rand() % 100;
max = maximum(data,100);
printf("Max = %d\n",max);
return(0); }
int maximum(int fdata[],
int n) {
int i, fmax;
fmax = fdata[0];
for (i=0; i<n; i++)
if(fdata[i] > fmax)
fmax = fdata[i];
return(fmax); }

30. Exercise What is the output of the following program? int main() {
void print(int pdata[], int n) { int i; for (i=0; i<n; i++) printf("data[%d]=%d\n", i,pdata[i]); return; } void modify(int fdata[], int n) for (i=0; i<n; i++) fdata[i] = 1;
int main() {
int data[10];
for (i=0; i<10; i++)
data[i] = rand() % 100;
print(data,10);
modify(data,10);
return(0); }

31. Skip Study 5.2, 5.3, and 5.5 from the textbook.
We will go over section 5.4 Statistical measurements in class

32. 5.4 Statistical measurements
In engineering, analyzing the statistical characteristics of data is important
Suppose we have an array of measurements (double data)
Let us develop functions to compute max, min, mean (average), median, variance, std-dev of these measurements

33. max() and min() double max(double x[], int n) {
/* Declare variables. */
int k;
double max_x;
/* Determine maximum
value in the array. */
max_x = x[0];
for (k=1; k <= n-1; k++)
if (x[k] > max_x)
max_x = x[k];
/* Return maximum value.*/
return max_x; }
double min(double x[], int n) {
/* Declare variables. */
int k;
double min_x;
/* Determine minimum
value in the array. */
min_x = x[0];
for (k=1; k < n; k++)
if (x[k] < min_x)
min_x = x[k];
/* Return minimum value.*/
return min_x; }

34. mean() double mean(double x[], int n) { /* Declare variables. */
int k;
double sum;
/* Determine sum of
values in the array. */
sum = x[0];
for (k=1; k<=n-1; k++)
sum = sum + x[k];
/* Return avg value. */
return sum/n; }

35. median() /* values in x must be sorted ! */
double median(double x[], int n) {
/* Declare variables. */
int k;
double median_x;
/* Determine median of
values in the array. */
k = floor(n/2);
if( n % 2 != 0)
median_x = x[k];
else
median_x = (x[k-1]+x[k])/2;
/* Return median value. */
return median_x; }
Example:
x[]={7, 9, 15, 27, 29};
n=5;
median_x=15
x[]={3, 6, 7, 9};
n=4;
median_x=(6+7)/2
median_x=6.5

36. variance() double variance(double x[], int n) {
/* Declare variables. */
int k;
double mu, sum=0;
mu = mean(x, n);
for (k=0; k<=n-1; k++)
sum = sum + pow(x[k]-mu, 2);
/* Return variance value. */
return sum/(n-1); }

37. std_dev() double std_dev(double x[], int n) {
/* Return standard deviation */
return sqrt(variance(x, n)); }

38. Name
Addr
Content
Lecture 23
Lecture++;

39. 5.6 Sorting an array 1 2 3 4 5 6 3 1 9 7 2 1 3 6 9 7 2 1 2 6 9 7 3 1 2 3 9 7 6 1 2 3 6 7 9 1 2 3 6 7 9

40. Selection Sort (solution 1)
void selection_sort(double x[], int n) { int k,j,m; double temp; for(k=0; k<=n-2; k++) { m = k; for(j=m+1; j<=n-1; j++) if(x[j] < x[m]) m = j; temp = x[k]; x[k] = x[m]; x[m] = temp }
m = find_min_pos(x, n, k);
swap(x, k, m);

41. Selection Sort (solution 2)
void selection_sort(double x[], int n) { int k, m; for(k=0; k<=n-2; k++){ m = find_min_pos(x, n, k); swap(x, k, m); }

42. Selection Sort cont’d
int find_min_pos(double fx[], int fn, int fk) { int j; int m=fk; for (j=m+1; i<=fn-1; j++) if (fx[j] < fx[m]) m = j; return(m); }

43. Selection Sort cont’d void swap(double sx[], int sk, int sm) {
double temp;
temp = sx[sk];
sx[sk] = sx[sm];
sx[sm] = temp;
return; }

44. Reverse an array 1 2 3 4 5 6 3 1 9 7 2 2 7 9 1 3 6

45. Reverse an Array void reverse(double x[], int n) { int i=0, j=n-1;
while (i<j) {
swap(x,i,j);
i = i + 1;
j = j - 1; }
return;

46. Merge two sorted array
Assume we have A and B arrays containing sorted numbers
For example
A = { 3, 5, 7, 9, 12}
B = {4, 5, 10}
Merge these two arrays as a single sorted array C, for example
C = {3, 4, 5, 5, 7, 9, 10, 12}

47. 5.7 Search Algorithms Unordered list Ordered list Linear search
In a loop compare each element in array with the value you are looking for ()
Ordered list
A better solution is known as Binary search (but we will skip it this time, it is like looking for a word in a dictionary)

48. Unordered list – linear search
int search1(int x[], int n, int value) {
int i;
for(i=0; i < n; i++) {
if (x[i]== value)
return i; }
return(-1);

49. Ordered list – linear search
int search2(int x[], int n, int value) {
int i;
for(i=0; i < n; i++) {
if (x[i]== value)
return i;
else if (x[i] > value)
break; }
return(-1);

50. Strings (strings are discussed in Section 6.6 of the textbook )
A string is an array of characters
char data[10] = “Hello”;
char data2[] = {‘H’, ‘e’, ‘l’, ‘l’, ‘o’, ‘\0’}
Use printf to print strings
printf(“%s”,data);
Can be accessed char by char
data[0] is first character
End of String Symbol H e l l o \0
data

51. Strings Each character has an integer representation a b c d e … … … …z
………………………112 A B C D E … … … … Z
……………………… 90 1 2 3 4 5 6 7 8 9 \0 \n 10

52. Strings Characters can be interpreted as integers char c = ‘A’;
printf(“%c \n”,c);
prints A
printf(“%d \n”,c);
prints 65
Printf(“%c \n”,65);

53. Exercise
Write a function to count the number of characters in a string.
Idea: count the number of characters before \0 H e l l o \0

54. Solution
int count_letters(char cdata[]) { int i=0; while (cdata[i] != '\0') i = i + 1; return(i); }

55. Exercise Write a function that prints a string in reverse
Idea: find the end of the string and print the characters backwards. H e l l o \0
Output: olleH

56. Solution
void print_reverse(char pdata[]) { int size,position; size = count_letters(pdata); position = size - 1; while (position >= 0) { printf("%c",pdata[position]); position = position -1; } printf("\n"); return;

57. Exercise
Write a function that compares 2 strings S1 and S2 using lexicographic order.
Idea: compare character by character
Return
a neg value if S1 < S2,
0 if S1 == S2
a pos value if S1 > S2 H e l l o \0 H e l o o \0
l < o in lexicographic order

58. Solution (incomplete)
int compare(char cdata1[], char cdata2[]) { int i= 0; while (cdata1[i] == cdata2[i]) i = i + 1; return (cdata1[i] - cdata2[i]); }

59. Solution (complete)
int compare(char cdata1[], char cdata2[]) { int i= 0; while (cdata1[i] != ‘\0’ && cdata2[i] != ‘\0’ && cdata1[i] == cdata2[i]) i = i + 1; return (cdata1[i] - cdata2[i]); }

60. Exercise: Spell out a number in text using arrays of strings
Write a program that reads a number between 1 and 999 from user and spells out it in English.
For example:
453  Four hundred fifty three
37  Thirty seven
204  Two hundred four

61. More Examples of one dimensional arrays

62. Trace a program
Trace the following program and show how variables change in memory.
int main() {
int x[5]={3, 5, 3, 4, 5};
int i, j, count;
for(i = 0; i < 5; i++){
count = 1;
for(j = i+1; j < 5; j++){
if (x[i] == x[j]) count++; }
printf(“%d %d \n”, x[i], count);
x[0] 3
x[1] 5
x[2]
x[3] 4
x[4] i ? j
count

63. Count how many times a value appears in the array
int find_count(int cdata[], int n, int x) { int count = 0; int i; for (i = 0; i<n; i++){ if (cdata[i] == x) count = count + 1; } return (count); 6 3 1 9 7 2 4 5
find_count(A,6,2) returns 1
find_count(A,4,2) returns 0

64. Insert unique RNs int A[6]; for (i=0; i<6; i++) A[i] = rand() % 10;
Insert random numbers into an array so that each number appears in the array at most once.
int A[6];
for (i=0; i<6; i++)
A[i] = rand() % 10; 6 3 1 9 2 A
Produces Duplicates

65. Insert unique RNs (cont’s)
Insert random numbers into an array so that each number appears in the array at most once.
void initialize(int idata[], int n) {
int i=0, j, elem;
while (i<=n) {
elem = rand() % 10;
if (find_count(idata,i,elem) == 0) {
idata[i] = elem;
i = i + 1; } 6 3 1 9 A
Try a new number

66. Intersection Set
Suppose we have two sets (groups) represented by A and B
E.g., A is the set of students taking Math,
B is the set of students taking Science.
Find set C, the intersection of A and B, i.e., students taking both Math and Science
For each element ID in A
Search that ID in B
if found, put ID into C 4 5 8

67. Use arrays to represent A and B Hand example6 3 1 9 7 2 4 2 5 6 1 8 B
i=0
i=1
j=0
j=3
i=2
i=3
i=4
i=5
i=6
j=1
j=2
j=6
j=1
j=2
j=3
j=4
j=5
j=0
j=1
j=2
j=3
j=4
j=0
k=0
k=1
k=2
k=3 6 1 2 C

68. Solution int intersection(int A[],int B[], int C[], int n) {
int i=0, j=0, k=0;
for(i=0; i < n; i++) {
for(j=0; j < n; j++) {
if (A[i]==B[j]){
C[k]=A[i];
k++;
break; }
return(k); 6 3 1 9 7 2 4 5 8

69. Another Solution int intersection(int A[], int B[], int C[], int n) {
int i=0, k=0, elem;
while (i < n) {
elem = A[i];
if(find_count(B,n,elem) == 1) {
C[k] = elem;
k = k + 1; }
i = i + 1;
return(k); 6 3 1 9 7 2 4 5 8

70. What if A and B were sorted?1 2 3 6 7 9 1 2 4 5 6 8
Will the previous solution work?
Yes, but we could find intersection set faster!
How?
See next slide 1 2 6

71. int sorted_intersection(int A[],int B[],
int C[], int n) {
int i=0, j=0, k=0;
while( i < n && j < n ) {
if (A[i]==B[j]){
C[k]=A[i];
k++; i++; j++;
} else if (A[i] < B[j]) {
i++;
} else { /* A[i] > B[j] */
j++; }
return(k);

72. Exercise: union or difference
As in previous example suppose two sets are given as arrays.
Find union and difference
For example
A={3,4,5} and B={2,3,5,7}
A U B = {2,3,4,5,7}
A – B = {4}
B – A = {2,7}

73. Exercise: Histogram
Suppose somehow we read npts integer numbers from a file into an array declared by int x[100]. We know that all the values are integers between 0 and 10. Now suppose we want to find how many 0’s ,1’s, …, 10’s exist in the array x.
Write a function that will take x and npts as the parameters and prints out the number of 0’s ,1’s, …, 10’s that exist in the array x

74. solution void my_function(int x[], int npt) { int i, hist[11]={0};
for(i=0; i < npt; i++)
hist[x[i]]++;
for(i=0; i < 11; i++)
printf(“%d appears %d times\n”,
i, hist[i]);
return; }

75. Number of Even values in a given range
Suppose somehow we read 100 values from a file into an array declared by int x[100]. We are now interested in finding the number of even values in x that are happen to be in a given range, say int Low=22, High=53.
Write a function that will take x, Low, High as parameters and return the number of even values in x that are happen to be between Low and High.

76. Exercise: strings (char array)
Write a function to check if v=“abcd” appears in a given string A?

77. 5.8 Matrices (2D-array) 4 1 2 -1 3 in memory
A matrix is a set of numbers arranged in a grid with rows and columns.
A matrix is defined using a type declaration statement.
datatype array_name[row_size][column_size];
int matrix[3][4]; 4 1 2 -1 3 4 1 2 -1 3
Row 0
Row 1
Row 2
Column 3
Column 0
Column 1
Column 2
in memory

78. Accessing Array Elements
int matrix[3][4];
matrix has 12 integer elements
matrix[0][0] element in first row, first column
matrix[2][3] element in last row, last column
matrix is the address of the first element
matrix[1] is the address of the Row 1
matrix[1] is a one dimensional array (Row 1)

79. Initialization
int x[4][4] = { {2, 3, 7, 2}, {7, 4, 5, 9}, {5, 1, 6, -3}, {2, 5, -1, 3}}; int x[][4] = { {2, 3, 7, 2},

80. Initialization
int i, j, matrix[3][4]; for (i=0; i<3; i++) for (j=0; j<4; j++) matrix[i][j] = i;
matrix[i][j] = j; j 1 2 i j 1 2 i 1 2 1 2 3

81. Exercise int i, j, x[4][3]; for(i=0; i<4; i++)
Write the nested loop to initialize a 2D array as follow
int i, j, x[4][3];
for(i=0; i<4; i++)
for(j=0; j<3; j++)
x[i][j] = i+j; 1 2 3 4 5

82. 2-Dim Arrays as Arguments to Functions
void print_m(int m[3][4],
int r, int c)
void print_m(int m[][4],
int r, int c) {
int i,j;
for (i=0; i < r; i++) {
for (j=0; j < c; j++)
printf("%.5d ",m[i][j]);
printf("\n"); }
return;
int i, j, matrix[3][4];
for (i=0; i<3; i++)
for (j=0; j<4; j++)
matrix[i][j] = i;
print_m(matrix, 3, 4);

83. Computations on 2D arrays

84. Max in 2D Find the maximum of int matrix[3][4] int max = matrix[0][0];
int max = matrix[0][0];
for (i=0; i<3; i++)
for (j=0; j<4; j++)
if (matrix[i][j] > max)
max = matrix[i][j]; 1 2 -1 2 4 3 1 -1 3 1 2

85. Find a value in 2D int count = 0;
Find the number of times x appears in int matrix[3][4]
int count = 0;
for (i=0; i<3; i++)
for (j=0; j<4; j++)
if (matrix[i][j] == x)
count = count + 1; 1 2 -1 2 4 3 1 -1 3 1 2

86. Matrix sum Compute the addition of two matrices + = 0 1 2 3 0 1 2 3 1 2 3 -1 3 1 3 3 3 + -1 2 4 3 1 4 2 = 6 6 3 1 1 1 -1 3 1 2 1 1 3 2 4 4 2 2 2

87. solution int matrix1[3][4], matrix2[3][4], sum[3][4];
// initialize matrix1 and matrix2
for (i=0; i<3; i++)
for (j=0; j<4; j++)
sum[i][j]= matrix1[i][j]+matrix2[i][j];

88. Exchange Two Rows 4 6 2 5 3 8 1 4 6 2 1 8 5 3

89. Transpose
void transpose(int a[NROWS][NCOLS], int b[NCOLS][NROWS]) { /* Declare Variables. */ int i, j; /* Transfer values to the transpose matrix. */ for(i=0; i<NROWS; i++) { for(j=0; j<NCOLS; j++) { b[j][i] = a[i][j]; } return; a 1 5 3 4 2 6 b 1 4 5 2 3 6

90. mine sweeper If m[i][j] is 0, there is no mine in cell m[i][j]
If m[i][j] is 1, there is a mine in cell m[i][j]
Print the number of mines around cell m[i][j]
[i][j]

91. Solution (1) - incomplete
count=0;
if( m[i-1][j-1] ) count++;
if( m[i-1][j] ) count++;
if( m[i-1][j+1] ) count++;
if( m[i][j-1] ) count++;
if( m[i][j+1] ) count++;
if( m[i+1][j-1] ) count++;
if( m[i+1][j] ) count++;
if( m[i+1][j+1] ) count++;

92. What if [i][j] is not in the middle?

93. Solution (1) – complete NR: is number of rows NC: number of columns
count=0;
if( i>0 && j>0 && m[i-1][j-1] ) count++;
if( i>0 && m[i-1][j] ) count++;
if( i>0 && j<NC-1 && m[i-1][j+1] ) count++;
if( j>0 && m[i][j-1] ) count++;
if( j<NC-1 && m[i][j+1] ) count++;
if( i<NR-1 && j>0 && m[i+1][j-1] ) count++;
if( i<NR-1 && m[i+1][j] ) count++;
if( i<NR-1 && j<NC-1 && m[i+1][j+1] ) count++;

94. Solution (2) NR: is number of rows, NC: number of columns
int r, c, count=0;
for(r=i-1; r <= i+1; r++) {
if (r < 0 || r >= NR) continue;
for(c=j-1; c <= j+1; c++) {
if (c < 0 || c >= NR) continue;
if (c == c) continue;
if( m[r][c]) count++; }

95. Example: Resize a picture
A b&w picture is usually represented using a two-dimensional array, where each element (called pixel) of the array is an integer number denoting the intensity of the light at a given pixel. Suppose we have a b&w picture with the size of 100x200 pixels and we want to reduce its size to 50x100 pixels.
For this, we may consider 4 pixels from the original picture and take their average to create one pixel in the new picture. For example:
4x6 original picture can be reduced to 2x3 resized picture
Write a function that takes orig[100][200] and resized[50][100] as parameters and determines the values in resized picture as described above. 4 5 3 2 1 6 10 4 3 1 5

96. Matrix multiplication
double a[3][2], b[2][4], c[3][4];
Find c = a * b; 3 4 5 2 1 6 22 29 45 35 18 40 47 21 26 33 43 49 2 3 7 1 4 5 6 8 = x
3*2 + 4*4=22
3*3 + 4*5=29
3*7 + 4*6=45
3*1 + 4*8=35
5*2 + 2*4=18
5*3 + 2*5=40
5*7 + 2*6=47
5*1 + 2*8=21
1*2 + 6*4=26
1*3 + 6*5=33
1*7 + 6*6=43
1*1 + 6*8=49

97. Matrix Multiplication cont’dj j 22 29 45 35 18 40 47 21 26 33 43 49 3 4 5 2 1 6 1 2 1 2 2 3 7 1 4 5 6 8 = x i i
j=0
c[i][j] =
a[i][k=0]*b[k=0][j] +
a[i][k=1]*b[k=1][j] 2 4 3 4 k =
i=0 x k

98. Matrix Multiplication cont’d
#define N 3
#define M 2
#define L 4
void matrix_mul(a[N][M], int b[M][L], int c[N][L]) {
int i, j, k;
for(i=0; i < N; i++) {
for(j=0; j < L; j++) {
c[i][j] = 0;
for(k=0; k < M; k++) {
c[i][j] = c[i][j] + a[i][k] * b[k][j]; }
return;

99. Exercise: Find possible values for cell s[i][j] in Sudoku4 1 7 6 8 3 5 9
[i][j] 2

100. Exercise: Dynamic programming1 2 3 4 5
max +
A[i][j] = max{A[i-1][j-1], A[i-1][j]+A[i][j-1]}

101. Exercise: Walks on 2d Arrays write a code to print the values in the given order

102. Example: Closest Points1 2 3
Suppose somewhat we read 100 points in a 2-dimentional space (each point is represented by (x,y)) and we store these points in an array declared by double p[100][2] (you can think that p[i][0] is the x value and p[i][1] is the y value of ith point).
We are now interested in finding the closest two points. For example, if p[3][2] = {{1,1},
{2,1},
{1,3} };
Then we will say that points (1,1) and (2,1) are the closest two points.
Write a function that takes p[100][2] as a parameter and prints out the closest two points. Suppose you have the following function:
double distance (double p1[ ], double p2[ ]) {
return sqrt( (p1[0]-p2[0])*(p1[0]-p2[0]) + (p1[1]-p2[1])*(p1[1]-p2[1])); }