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eater than n. Based on thất test, return the correct resuit. r example, a call to simpleSqrt(8) would recursively call simpleSqrt(7) and get back 2 as the answer. Then we would square (2+1) 3 to get 9. Since 9 is bigger than 8, we know that 3 is too big, so return 2 in this case.

eater than n. Based on thất test, return the correct resuit. r example, a call to simpleSqrt(8) would recursively call simpleSqrt(7) and get back 2 as the answer. Then we would square (2+1) 3 to get 9. Since 9 is bigger than 8, we know that 3 is too big, so return 2 in this case.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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I cant use the sqrt() library function

**Write the definition of a recursive function** `int simpleSqrt(int n)` The function returns the integer square root of n, meaning the biggest integer whose square is less than or equal to n. You may assume that the function is always called with a nonnegative value for n. Use the following algorithm: - If n is 0 then return 0. - Otherwise, call the function recursively with n-1 as the argument to get a number t. Check whether or not t+1 squared is strictly greater than n. Based on that test, return the correct result. For example, a call to simpleSqrt(8) would recursively call simpleSqrt(7) and get back 2 as the answer. Then we would square (2+1) = 3 to get 9. Since 9 is bigger than 8, we know that 3 is too big, so return 2 in this case. On the other hand, a call to simpleSqrt(9) would recursively call simpleSqrt(8) and get back 2 as the answer. Again we would square (2+1) = 3 to get back 9. So 3 is the correct return value in this case.
Transcribed Image Text:**Write the definition of a recursive function** `int simpleSqrt(int n)` The function returns the integer square root of n, meaning the biggest integer whose square is less than or equal to n. You may assume that the function is always called with a nonnegative value for n. Use the following algorithm: - If n is 0 then return 0. - Otherwise, call the function recursively with n-1 as the argument to get a number t. Check whether or not t+1 squared is strictly greater than n. Based on that test, return the correct result. For example, a call to simpleSqrt(8) would recursively call simpleSqrt(7) and get back 2 as the answer. Then we would square (2+1) = 3 to get 9. Since 9 is bigger than 8, we know that 3 is too big, so return 2 in this case. On the other hand, a call to simpleSqrt(9) would recursively call simpleSqrt(8) and get back 2 as the answer. Again we would square (2+1) = 3 to get back 9. So 3 is the correct return value in this case.
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