Nested loops into mathematical model to count number of operations

Question

I\'m reading the book \'Algorithms - Fourth edition\' by Sedgewick and Wayne and I must admit that some parts in the \"Analysis of Algorithms\" chapter are confusing me! Thi

Answer 1

You can use Sigma notation and discover how to come up with the formula mentioned in your book:

Answer 2

The 6 is derived from 3! (three factorial derived from three loops).

Consider the outer-most loop on, say, an array of 5 elements. You're counting N=5 for that part of the equation, but you'll only reach your inner loop for values i=0, i=1 or i=2. Similarly, you're representing the next loop by (N-1), but you'll only reach your inner loop for values j=1, j=2, or j=3 and not the four values implied by (N-1).

Dividing by 6 (for three loops) compensates for the values that would run out of values in the array before reaching the inner-most loop.

Answer 3

As we know...

1+2+3+4...+N => N(N-1)/2

Similarly, the innermost loop works something like

1.n+2(N-1)+3(N-2)+...N.1 => N(N-1)(N-2)/6

Here is a proof for this.

Answer 4

If you can understand the N(N-1)(N-2) part, here's a thought:

Take a combination of 3 numbers, i, j, k, whatever 3 that fall into the range 0 <= i,j,k < N and different one from another (this is also taken care in the code and that's why the formula is N(N-1)(N-2) and not N^3.

Now, lets say the numbers are 13, 17, 42. It doesn't really matters whoch numbers they are. In how many ways can you put them in line?

13-17-42
13-42-17
17-13-42
17-42-13
42-13-17
42-17-13

Six!

How many of these ways can appear in the code? Only one! (that's taken care in the initializaton of j and k).

So, the total number of N(N-1)(N-2) should be divided by 6.