Why is the Pulverizer machine partially correct?

Question

From Eric Lehman et al.'s Mathematics for Computer Science [PDF]:

Problem 9.13. $\,$ Define the Pulverizer state machine to have: $$ \begin{align*} \text{states} ::=& \mathbb{N}^6&\\ \text{start state} ::=& (a, b, 0, 1, 1, 0)&(\text{where }a \ge b > 0)\\ \text{transitions} ::=& (x, y, s, t, u, v) \longrightarrow\\ &(y, \text{rem}(x, y), u-sq, v-tq, s, t )& (\text{for }q=\text{qcnt}(x, y), y > 0). \end{align*} $$ (a) Show that the following properties are preserved invariants of the Pulverizer machine: $$ \begin{align*} \gcd(x, y) &= \gcd(a, b), &(\text{Inv1})\\ sa + t b &= y, \text{and} &(\text{Inv2})\\ ua + vb &= x. &(\text{Inv3})\\ \end{align*} $$ (b) Conclude that the Pulverizer machine is partially correct.

(c) Explain why the machine terminates after at most the same number of transi- tions as the Euclidean algorithm.

where the Pulverizer state machine is defined as follows

Today, the Pulverizer is more commonly known as the “Extended Euclidean Gcd Algorithm,” because it is so close to Euclid’s algorithm.

some notation definition:

We use the notation $\text{qcnt}(n, d )$ for the quotient and $\text{rem}(n, d )$ for the remainder.

I can prove (a).

$\text{rem}(x,y)\ge 0$ and it is strictly decreasing when the machine runs. Does (b) says that we need to let $\text{rem}(x,y)= 0$ as the termination condition which is same as the Euclidean algorithm? This is different from the implicit condition $y=0$ implied by $y>0$.

But if so, then it will cause these 2 algorithms have the same number in (c). If not, it should take more number but not at most. Then (c) is still not as the problem expected.

Implied by wikipedia, the above process adding 2 sequences based on the standard Euclidean algorithm should have "correct" for (b) and "same number" for (c). Then what does the book mean for the above statements?

Could you offer one answer or some hints for (b)? And I will try to understand (c) based on that since someone told me it is not one good habit to ask multiple questions in one post.

Answer 1

For (b): The machine performs the transition $$(x,y,s,t,u,v) \mapsto (x',y',s',t',u',v') = (y, \mathrm{rem}(x,y), u-sq, v-tq, s, t) $$ if $y > 0$; otherwise, we need to understand that no transition takes place, i.e., the machine halts and 'produces a result', namely (the relevant part of) the end state $(x,y,s,t,u,v)$.

The notion of partial correctness is explained as follows on p.182 in your reference: "Partial correctness means that when there is a result, it is correct, but the process might not always produce a result, perhaps because it gets stuck in a loop."

So question (b) amounts to proving: whenever the initial state $(a,b,0,1,1,0)$ is such that the machine halts, the end result is correct. Here, using Part (a), this can be taken to mean (I leave you the task to justify this) that whenever the machine reaches an end state $(x_f,y_f = 0,s_f,t_f,u_f,v_f)$, this state satisfies $$\mathrm{gcd}(a,b) = x_f = u_f a + v_f b$$ so that the machine correctly computed $\mathrm{gcd}(a,b)$, but also correctly expresses $\mathrm{gcd}(a,b)$ as a linear integer combination of $a$ and $b$ (as claimed in Theorem 9.2.2).

For (c): Now, we need to prove that the machine always halts (for the allowed initial states) and, moreover, that it does so in no more than the number of steps used in the standard Euclidean algorithm. Hint: It may help to compare this machine with the machine defined by Equation (9.18) in Problem 9.14.