Part a. Make a diagram of memory for point one when n == 2. int expo(int x, int n) { int r, t; if (n == r = 1; else if (n == 1) r = x; else { 0) t = expo (x, n / 2); r = t * t; if (n % 2 == 1) r *= t; } //----- ---point one return r; } int main(void) { int i; i = expo(4, 5); return 0; } Part b. Using divide-and-conquer recursion to solve a problem similar to Prob- lem 3 is somewhat tricky but a good test of programming skill. Write a function definition that provides a divide-and-conquer implementation of the following interface: double max_adj_sum(const double *a, int lo, int hi); // REQUIRES: hi - lo >= 2. Elements a[lo] ... a[hi -1] exist. // PROMISES: Return value is the largest sum that can be made by adding // two elements among a[lo] ... a[hi-1] with adjacent indezes. Hint: It makes sense to have two base cases, not just one.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Part a.
Make a diagram of memory for point one when n == 2.
int expo(int x, int n) {
int r, t;
if (n == 0)
r = 1;
else if (n == 1)
r = x;
else {
t = expo (x, n / 2);
r = t * t;
if (n % 2 == 1)
r *= t;
}
/-
-point one
return r;
}
int main(void) {
int i;
i = expo (4, 5);
return 0;
}
Part b.
Using divide-and-conquer recursion to solve a problem similar to Prob-
lem 3 is somewhat tricky but a good test of programming skill. Write a function definition
that provides a divide-and-conquer implementation of the following interface:
double max_adj_sum(const double *a, int lo, int hi);
// REQUIRES: hi - lo >= 2. Elements a[lo] ... a[hi -1] exist.
// PROMISES: Return value is the largest sum that can be made by adding
two elements among a[lo] ... a[hi-1] with adjacent indexes.
Hint: It makes sense to have two base cases, not just one.
Transcribed Image Text:Part a. Make a diagram of memory for point one when n == 2. int expo(int x, int n) { int r, t; if (n == 0) r = 1; else if (n == 1) r = x; else { t = expo (x, n / 2); r = t * t; if (n % 2 == 1) r *= t; } /- -point one return r; } int main(void) { int i; i = expo (4, 5); return 0; } Part b. Using divide-and-conquer recursion to solve a problem similar to Prob- lem 3 is somewhat tricky but a good test of programming skill. Write a function definition that provides a divide-and-conquer implementation of the following interface: double max_adj_sum(const double *a, int lo, int hi); // REQUIRES: hi - lo >= 2. Elements a[lo] ... a[hi -1] exist. // PROMISES: Return value is the largest sum that can be made by adding two elements among a[lo] ... a[hi-1] with adjacent indexes. Hint: It makes sense to have two base cases, not just one.
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