COMP 1672 Sections 1 & 2
Homework #4 Solutions

1.  Do problem 2 on page 496
    a.  The base case is if u is equal to 0 or 1
        One of two output statements executes
    b.  The general case is if u is neither 0 nor 1
        u is decremented and the function is called recursively
    c.  The letter 'B' is output

2.  a.  Func is called exactly once, returning in the first base case.
        The number 2 is output.
    b.  Func is called exactly once, returning in the second base case.
        The number 3 is output.
    c.  Func is called with the value of 2, the final return is
        executed, which returns Func(1) + Func (0).  By parts a and b,
        this is 2+3 or 5
    d.                          Func(5)  calls:
                Func(4)           +              Func(3)
              Func(4) calls:                   Func(3) calls:
        Func(3)   +     Func(2)         Func(2)    +     Func(1)
     Func(3) calls:i    returns 5       returns 5       returns 3 
     Func(2)+Func(1)     by part c      by part c       by part b
     =5+3=8 (by c and b) 
             Func(4) gets back               Func(3) gets back
             8    +      5=13                5      +      3 = 8
             and returns it                   and returns it
                               Func(5) gets back
                       13         +          8 = 21, which gets displayed
                          
3.  Here is just one solution.  There are many!
#include <iostream>
#include <string>
using namespace std;

bool is_palindrome(string s)
{
    if (s.length()==0 || s.length() == 1)
        return true;
    if (s[0] != s[s.length()-1]) return false;
    return is_palindrome(s.substr(1,s.length()-2));
}

int main()
{
    string str;
    cout << "Input the string you want to check to see if it's a palindrome: ";
    cin >> str;
    cout << str  << (is_palindrome(str)?" is":" isn't") << " a palindrome.\n";
    return 0;
}