Answer: True
Explanation:
Subset sum problem and Knapsack problem can be solved using dynamic programming.
In case of Knapsack problem there is a set of weights associative with objects and a set of profits associated with each object and a total capacity of knapsack let say C. With the help of dynamic programming we try to include object's weight such that total profit is maximized without fragmenting any weight of objects and without exceeding the capacity of knapsack, it is also called as 0/1 knapsack problem.
Similar to knapsack problem, in subset sum problem there is set of items and a set of weights associated with the items and a capacity let say C, task is to choose the subset of items such that total sum of weights associated with items of subset is maximized without exceeding the total capacity.
On the basis of above statements we can say that subset sum problem is generalization of knapsack problem.
The answer is C, ROM often stores the basic instructions a computer needs when powering on, part if the BIOS.
Answer: Program for bit stuffing in C
#include<stdio.h>
int main()
{
int i=0,count=0;
char data[50];
printf("Enter the Bits: ");
scanf("%s",data); //entering the bits ie. 0,1
printf("Data Bits Before Bit Stuffing:%s",databits);
printf("\nData Bits After Bit stuffing :");
for(i=0; i<strlen(data); i++)
{
if(data[i]=='1')
count++;
else
count=0;
printf("%c",data[i]);
if(count==4)
{
printf("0");
count=0;
}
}
return 0;
}
Explanation:
bit stuffing is the insertion of non-information bits during transmission of frames between sender and receiver. In the above program we are stuffing 0 bit after 4 consecutive 1's. So to count the number of 1's we have used a count variable. We have used a char array to store the data bits . We use a for loop to iterate through the data bits to stuff a 0 after 4 consecutive 1's.
