For this editorial, denote ~f(x, y)~ as the interactor's response after making a query with ~l = x~ and ~r = y~. Also, ~a_1, a_2, \dots, a_N~ is the sequence which we are trying to find. Finally, let ~\oplus~ denote the bitwise XOR operator.

~M = 1~

If we do not restrict ourselves to maximising the value of ~M~, then there are many possible solutions, one of which will be described here.

First, we can query ~f(1, 2)~, ~f(1, 3)~ and ~f(2, 3)~, which allows us to find ~a_1~, ~a_2~ and ~a_3~.

Then, for each ~x~ from ~4~ up to ~N~, we can find ~a_x~ by calculating ~a_{x-1} \oplus f(x-1, x)~.

~M = \lfloor \frac{N-1}{2} \rfloor~

Again, there are many constructions which produce a value of ~M = \lfloor \frac{N-1}{2} \rfloor~. It turns out that this is the maximum possible value of ~M~. A proof of this will follow after an example of a possible construction.

First, find the value of ~f(i, N)~ for ~1 \le N \le \lceil \frac{N+1}{2} \rceil~. This lets us work out the value of ~a_x~ for ~1 \le x \le \lceil \frac{N-1}{2} \rceil~, since ~a_x = f(x, N) \oplus f(x+1, N)~.

Then, we can work out the value of ~a_x~ for ~\lceil \frac{N+1}{2} \rceil \le x < N~ with one additional query for each ~x~, since ~a_x = f(1, x) \oplus a_1 \oplus a_2 \oplus \dots \oplus a_{x-1}~. Cumulative prefix XOR sums can be used to speed up this computation.

Finally, we can work out ~a_N~ as equal to ~f(1, N) \oplus a_1 \oplus a_2 \oplus \dots \oplus a_{N-1}~. Note that ~f(1, N)~ has already been found from the first step, so this does not require an extra query.

This solves the problem with ~N~ queries and ~M = \lfloor \frac{N-1}{2} \rfloor~.

Now, let's consider why this value of ~M~ is maximal. Let ~p_x~ represent the prefix XOR sequence, where ~p_x = a_1 \oplus a_2 \oplus \dots \oplus a_x~. For convenience, let ~p_0 = 0~. Note that being able to work out sequence ~p~ is a necessary and sufficient condition for being able to work out sequence ~a~, so we will turn our attention to finding sequence ~p~.

We can see that ~f(l, r) = p_{l-1} \oplus p_r~. Therefore, if we know ~f(l, r)~ and one of ~p_{l-1}~ or ~p_r~, then we can work out the other value out of ~p_{l-1}~ or ~p_r~.

This motivates the following idea: create a graph containing ~N+1~ nodes labelled ~0, 1, 2, \dots, N~. Whenever we make a query asking for ~f(l, r)~, then add a bidirectional edge between ~l-1~ and ~r~. Then, two nodes ~x~ and ~y~ are connected if and only if knowing the value of ~p_x~ allows us to figure out ~p_y~, and vice versa.

After performing ~N~ queries, we require the graph to be connected; otherwise, there is no unique solution. Since there are ~N+1~ nodes and ~N~ edges, the edges must form a spanning tree of this graph. This also explains why the minimal number of queries to find sequence ~a~ is ~N~.

Recall that the value of ~M~ is the minimum value of ~|x-y|-1~, for all edges ~(x, y)~ in the graph. Now, consider node ~\lfloor \frac{N}{2} \rfloor~. Since the edges form a spanning tree, this node must be connected to another node. However, the minimum value of ~|\lfloor \frac{N}{2} \rfloor - x|-1~ for ~0 \le x \le N~ is ~\lfloor \frac{N-1}{2} \rfloor~. This proves that ~M~ cannot exceed ~\lfloor \frac{N-1}{2} \rfloor~. On the other hand, we have already shown via construction that this value of ~M~ is attainable, so ~\lfloor \frac{N-1}{2} \rfloor~ is the maximum possible value of ~M~.

This graph idea also lends itself to a different construction. First, generate any spanning tree of the ~N+1~ nodes such that ~|x-y|-1 \ge \lfloor \frac{N-1}{2} \rfloor~ for all edges ~(x, y)~. For each edge ~(x, y)~ in this graph with ~x < y~, let it have a weight of ~f(x+1, y)~. We know that ~p_0 = 0~. Performing a DFS/BFS on this graph, we can work out the entire sequence ~p~, as ~p_x~ is simply the XOR sum of all edges on the path from node ~0~ to node ~x~. Once we know ~p~, we can easily find sequence ~a~, since ~a_x = p_{x-1} \oplus p_x~ for all ~1 \le x \le N~.