# 15-122 Principles of Imperative Computation Recitation 2 - Fri Jan 14

## Course Website

Look at the course website for assignments, schedule of topics, notes from previous classes and recitations, and sample code completed in class. The website also has references for the C0 language and for the C0 libraries, along with additional resources you may need.

## Bulletin Board

Remember that there is a bulletin board for this course so you can ask general questions to the course staff and others in the class. Please remember to use the bulletin board appropriately and do not post homework answers or inappropriate messages.

To access the bulletin board, access your andrew mail using SquirrelMail by going to webmail.andrew.cmu.edu. Once you're there, click the Folders link and then go to the Subscribe section. Highlight academic.cs.15-122 and click Subscribe. You should then see the bboard listed in your folders on the left side of the webpage. You can read messages and post messages just like you would for email.

If you want to post a new message, send email to post+academic.cs.15-122@andrew.cmu.edu or if you want to reply to a message on the board, use Reply All to send to the board and the person who posted that message.

## Reasoning with Invariants

In class, we discussed how to implement a exponentiation function as follows:

```int fastpow (int x, int y)
//@requires y >= 0;
//@ensures \result == pow(x, y);
{
int r = 1;
int b = x;
int e = y;
while (e > 0)
//@loop_invariant r * pow(b,e) == pow(x,y);
//@loop_invariant e >= 0;
{
if (e % 2 == 1) r = r * b;
b = b * b;
e = e / 2;
}
//@assert e == 0;
return r;
}
```

Recall that a loop invariant is a condition that is true immediately before a loop condition is tested, both before you enter the loop and at the end of each iteration (if the loop runs).

Let's show that this function does return the correct result given its precondition (requires) and its loop invariants.

1. Show that the loop invariants hold before the loop condition is tested for the first time.

Just before the loop condition is tested for the first time, we know that r = 1, b = x, e = y, and y >= 0.

Looking at the first invariant, we need to show r * pow(b,e) == pow(x,y). We prove this by transforming the left side into the right side using what we know to be true about the values of variables and mathematical identities.

```r * pow(b,e) == 1 * pow(x,y)    (by substitution)
1 * pow(x,y) == pow(x,y)        (1 * x = x (multiplicative identity))
```

The second invariant we need to show is e >= 0. For this proof we use the information we know about y from the precondition

```y >= 0  (@REQUIRES)
e >= 0  (since e = y)
```

2. Show that the loop invariants hold just before the loop condition is tested after each iteration. Equivalently: if the loop invariants hold at the start of an iteration, show that they hold at the end of that iteration. (Consider only one iteration or pass through the loop, not the entire execution of the loop.)

Looking at the first invariant, assume at the start of an iteration that:

```r * pow(b,e) == pow(x,y)
```

There are two cases to consider here.

1. Consider the case where e is even. If e is even, we can say that e = 2f for some integer f. (That is, e is two times some other integer for it to be even.)

Show that r' * pow(b',e') == pow(x',y') where variables with primes (') indicate the new values of the variables at the end of that iteration.

Since e is even, the if statement does not change r so at the end of the iteration, we know that:

```r' = r
b' = b * b
e' = e / 2 = 2f / 2 = f
x' = x
y' = y
```

Now we show that the invariant after the loop runs is true after one iteration if we assume it is true before that iteration runs:

```r' * pow(b',e') == r * pow(b*b,f)    (by substitution)
r * pow(b*b,f)  == r * pow(b,2f)     (by definition of a^b)
r * pow(b,2f) == r * pow(b,e)        (by substitution)
r * pow(b,e) == pow(x,y)             (from loop invariant)
pow(x,y) = pow(x',y')                (by substitution)
```

What we're showing is that the invariant after the iteration is true as long as the invariant holds before that iteration.

Looking at the other invariant (when e is even), we need to show that e' >= 0 assuming e >= 0.

```e >= 0    (from loop invariant)
2f >= 0   (by substitution)
f >= 0    (by definition of integer division)
e' >= 0   (by substitution)
```

So we've shown that the loop invariants hold at the end of an iteration if they hold at the start of an iteration and e is even.

2. Consider the case where e is odd. If e is odd, we can say that e = 2f + 1 for some integer f. (That is, e is two times some other integer plus 1 for it to be odd.)

Show that r' * pow(b',e') == pow(x',y'). Since e is odd, the if statement does change r this time so at the end of the iteration, we know that:

```r' = r * b
b' = b * b
e' = e / 2 = (2f + 1)/ 2 = f     (WHY?)
x' = x
y' = y

```

Now we show that the invariant after the loop runs is true after one iteration if we assume it is true before that iteration runs:

```r' * pow(b',e') == r * b * pow(b*b,f)    (by substitution)
r * b * pow(b*b,f)  == r * b * pow(b,2f) (by definition of a^b)
r * b * pow(b,2f) == r * pow(b,2f+1)     (by definition of a^b)
r * pow(b,2f+1) == r * pow(b,e)          (by substitution)
r * pow(b,e) == pow(x,y)                 (from loop invariant)
pow(x,y) = pow(x',y')                    (by substitution)
```

We also have to show that e' >= 0 if e >= 0 when e is odd.

```e >= 0      (from loop invariant)
2f+1 >= 0   (by substitution)
f >= 0      (by definition of integer division)
e' >= 0     (by substitution)
```

So we've shown that the loop invariants hold at the end of an iteration if they hold at the start of an iteration and e is odd.

Before we move on, what have we shown so far? The invariants hold just before you enter the loop. They also hold at the end of each iteration if they hold at the start of each iteration. This implies that the invariants hold at the end of the first iteration (since the loop condition doesn't change anything in the invariant). But since the invariants hold at the end of the first iteration, they must hold at the start of the second iteration (again since the loop condition doesn't change any variable values). Thus, they must also hold at the end of the second iteration, and so on. What we've shown is that the loop invariant holds at the start and end of EVERY iteration.

3. Show that if the loop terminates, we achieve our desired postcondition (given in the ensures annotation).

If the loop terminates, the loop invariants

```r * pow(b,e) == pow(x,y)  &&  e >= 0
```

should still hold (why?). What else holds? The negation of the loop condition must also be true. That is, !(e > 0) == e <= 0. What can we infer from this?

```r * pow(b,e) == pow(x,y) && e >= 0 && e <= 0
r * pow(b,e) == pow(x,y) && e == 0
r * pow(b,0) == pow(x,y)
r * 1 == pow(x,y)
r == pow(x,y)

```

This says that immediately after the loop terminates, r is equal to pow(x,y). But this is exactly what the function returns as specified by our ensures annotation, and the function performs no other operations after the loop terminates (other than returning the answer), so we've shown that the function does return the correct answer... if the loop terminates.

4. Argue that the loop terminates.

To prove that the function terminates, we need to convince ourselves that e <= 0 after a finite number of loop iterations. One way to show this is to prove that e is strictly decreasing on each execution of the loop. In this example, e is divided by 2 each time through the loop, so we need to prove e/2 < e. Since integer division rounds down, we know this is true. Formally, we can prove this by analyzing the even and odd cases:

In either case, we know that we entered the loop, so e > 0

1. e is even, so e = 2f for f < e (strictly greater because e > 0). Then e/2 = 2f/2 = f < e.
2. e is odd, so e = 2f+1 for f < e (strictly greater because e > 0). Then e/2 = (2f+1)/2 = f < e.

So we've shown that given the precondition, we can use the loop invariants to show that the postcondition follows and that the function is correct.
written by Tom Cortina, 1/14/11