Question

1.Take this recursive Fibonacci implementation and convert it into the caching based version discussed in class. Implement your caching to store a maximum of 5 values. Create 2 variations: one that stores all values and one that only stores even values. Make sure that you don't leave any gaps in your cache — if you have 5 cache entries you must cache 5 unique and valid values. Compare the caching implementations to the recursive implementation using the time utility. How do the caching implementations affect the results? Why do the different caching strategies show different results? What happens if you increase the maximum number of values to 10? Note that you can define different constants at compile time using something like -DCACHE_SIZE=10. Test the implementations with a variety of values between 32 and 52. The program is designed to take the desired Fibonacci number via the command line. To get the 42nd Fibonacci number you would use ./fib 42. To run the time utility you would enter the following on the command line: time ./fib 42

fib.c:

#include <stdio.h>

#include <stdlib.h>

long fib(int n)

{

long result;

  

if (n==0)

result = 0;

else if (n==1 || n==2)

result = 1;

else

result = fib(n-1) + fib(n-2);

  

return result;

}

int main( int argc, char *argv[] )

{

// we really should check the input...

int fibNum = atoi(argv[1]);

  

printf("The %d Fibonacci number is %ld\n", fibNum, fib(fibNum));

  

return EXIT_SUCCESS;

}