Proof of induction on pseudocode

问题

Given the pseudocode

MUL(a,b) 
   x=a
   y=0
   WHILE x>=b DO
      x=x-b
      y=y+1
   IF x=0 THEN
      RETURN(true)
   ELSE
      RETURN(false)

Let x(n) and y(n) denote the value of x and y after the while loop has run n times.

I have to show by the proof of induction that

x(n) + b*y(n) = a

What I've done:

P(n): x(n) + by(n) = a

Let a and b be arbitrary numbers then the first loop will give x(1) = a - b and y(1) = 0 + 1 = 1

P(1): x(1) + by(1) = a <=> a = a

so P(1) is true.

Assume P(n) is true. We want to show that P(n+1) is also true.

For step n + 1 the loop will give x(n+1) = x(n) - b and y(n+1) = y(n) + 1

P(n+1): x(n+1) + by(n+1) = a <=> x(n) + by(n) = a

Using the assumption that P(n) is true, it follows that P(n+1) is also true, and the proof is complete.

My question: Since this is my first time using the proof of induction on a pseudocode, I'm not sure how to go about it. I just want to know if this is the right way to work around the problem, and if not what should the process be like?

Thanks

回答1:

As you got it (almost) right in your question, answering this feels like overkill, but I'll do it anyway:

Your approach is fine, although you should not ignore the case where no loop iterations are executed at all, which is the case when a < b and corresponds to P₀. If you prove P₁ is true, and P_n+1 is true when P_n is true, you don't say anything about P₀.

So with that slight modification, the derivation of the proof by induction goes as follows:

Define P_n as the value of the expression x + b*y after n iterations of the loop have been executed:

P_n : x_n + b.y_n = a

It is to be proved that P_n is true for all n >= 0

1. Base Case: P₀

Note that this is a possible case: when the condition of the while loop is first evaluated, no iterations have been performed yet, so n is 0, yet we do have the values for the two variables at play: x₀ and y₀.

The pre-condition for the loop is determined by the assignment statements that precede the loop:

x=a
y=0

So we have:

P₀ : x₀ + b.y₀ = a + b.0 = a

2. Inductive Step: P_n+1

Here we assume that P_n is true for a given n:

P_n : x_n + b.y_n = a

This is the pre-condition when starting the next iteration of the loop, in which the following statements are executed:

x=x-b
y=y+1

By substitution we get the post-condition for that particular iteration which by definition is P_n+1:

P_n+1 : (x_n - b) + b.(y_n + 1) = x_n + b.y_n = P_n = a

来源：https://stackoverflow.com/questions/40117582/proof-of-induction-on-pseudocode