Question about O(log n) algorithm for computing a^n

[s3a]

Homework Statement

Let a be an integer and n be a non-negative integer.

Without changing the values of a and n, produce an algorithm, of O(log n) complexity, which computes a^n.

Homework Equations

For integers a, n and i, where n >= 0 and is even, where i >= 1 and where a has no further constraints:
a^n = (a^[2i])^(n/[2i])

For integers a, n and i, where n >= 0 and is odd, where i >= 1 and where a has no further constraints:
a^n = a^(n-i) * a^i

The Attempt at a Solution

The following is the solution I attempted by myself, as well as the one provided by a book I'm using.:

    public static void myWay(int a, int n) {
       
        int b = a;
        int k = n;
       
        while( k != 0 ) {
           
            if(k % 2 == 0) {
               
                b = b * b;
                k = k / 2;
            }
            else {
               
                b = b * a;
                k = k - 1;
            }
        }
       
        System.out.println(b);
    }
   
    public static void wayOfBook(int a, int n) {
       
        int k = n, b = 1, c = a;
       
        while(k != 0) {
           
            if(k % 2 == 0) {
               
                k = k / 2;
                c = c * c;
            }
            else {
               
                k = k - 1;
                b = b * c;
            }
        }
       
        System.out.println(b);
    }

I understand what the algorithm of the book is doing, but what I don't understand is why what I am doing is wrong.

Could someone please point out the flaw with my logic?

For instance, if I pass a = 3 and n = 7, my algorithm prints 3^11 = 177147 (which is incorrect), whereas the book's algorithm prints out 3^7 = 2187 (which is correct).

Any input would be GREATLY appreciated!

 

[rcgldr]

The method is called exponentiation by repeated squaring. One variable is squared every loop, the result variable is multiplied by the repeatedly squared variable for each 1 bit in the exponent, starting with the least significant bit.

https://en.wikipedia.org/wiki/Exponentiation_by_squaring