Exercise II
Consider this program
public int myStrangeSum (int num) {
int sum = 0;
int numnum = 2 * num;
for (int i = 0; i < numnum; i++) {
for (int j = 0; j < 5; j++) {
for (int k = i; k > 1; k--) {
sum++;
}
}
}
return sum;
}
Exercise Count the number of steps taken by the above program. Write the count as a function $T(N)$ where $N$ is the size of the "input". You need to determine what quantity is a suitable choice for the "size of input" here.
Solution
The run time can be described as a function of $N$, the value of variable num.
public int myStrangeSum (int num) {
int sum = 0; // 1
int numnum = 3 * num; // 2
for (int i = 0; i < numnum; i++) { // 1 + (3N + 1) + 3N
for (int j = 0; j < 5; j++) { // (1 + 6 + 5) * (3N)
for (int k = i; k > 1; k--) { // (tricky! let's call it x) * 15N
sum++; // 1 * x * 15N
}
}
}
return sum; // 1
}
The variable k takes the values of i:
i | # iteration inner loop (not considering middle loop) | values of k |
|---|---|---|
0 | will not run | 0 |
1 | will not run | 1 |
2 | once | 2, 1 |
3 | twice | 3, 2, 1 |
4 | 3 | 4, 3, 2, 1 |
| $\vdots$ | $\vdots$ | $\vdots$ |
N | $N-1$ | N $\dots$ 1 |
| $\vdots$ | $\vdots$ | $\vdots$ |
3N - 1 | $3N-2$ | (3N-1) $\dots$ 1 |
Let's see how many times each statement in the most inner loop will be executed for different values of i:
i | int k = i | k > 1 | k-- | sum++ |
|---|---|---|---|---|
0 | $1$ | $1$ | $0$ | $0$ |
1 | $1$ | $1$ | $0$ | $0$ |
2 | $1$ | $2$ | $1$ | $1$ |
3 | $1$ | $3$ | $2$ | $2$ |
4 | $1$ | $4$ | $3$ | $3$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |
N | $1$ | $N$ | $N-1$ | $N-1$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |
3N - 1 | $1$ | $3N - 1$ | $3N-2$ | $3N-2$ |
Let's add up the number of times each statement is executed throughout the execution of the code snippet.
| Expression | Total count |
|---|---|
int k = i | $5 \times (1 + 1 + 1 + \dots + 1) = 15N$ |
k > 1 | $5 \times (1 + 2 + 3 + \dots + (3N-1)) = 5 \times \left ( \frac{9N^2}{2} + \frac{3N}{2} \right )$ |
k-- | $5 \times (1 + 2 + 3 + \dots + (3N-2)) = 5 \times \left ( \frac{9N^2}{2} - \frac{9N}{2} + 1 \right )$ |
sum++ | $5 \times (1 + 2 + 3 + \dots + (3N-2)) = 5 \times \left ( \frac{9N^2}{2} - \frac{9N}{2} + 1 \right )$ |
The $5$ multiplier is the effect of the second loop.
To calculate the arithmetic sums, I've used the Gauss formula:
$$ \sum_{i=1}^{n} = \frac{n(n + 1)}{2} $$
Whew! Let's add up all the counts (for the entire program):
$$ 42N + 3 + \left (\frac{135n^2 - 75n + 20}{2} \right ) + 1 $$
Simplifying the expression, we get the following:
$$ T(N) = \frac{135N^2+9N+28}{2} $$
I hope that you will appreciate how tedious it can be to work out precisely the number of RAM instructions executed. To be honest, I am uncertain as to whether I've made a mistake in counting!
As it turns out we will not need such precise calculation. It is sufficient for us to know the run time is roughly proportional to the square of the size of input! More on this to come!