The Sum-XOR Dialogue

Represent all possible pairs as a bipartite multigraph.

Create a left vertex $S_s$ for every possible sum $s$, and a right vertex $Y_y$ for every possible XOR value $y$. For every possible pair $(a,b)$, add an edge between $S_{a+b}$ and $Y_{a\mathbin{\mathrm{xor}}b}$.

At the beginning of a day, the remaining edges are exactly the pairs consistent with every earlier statement. Dadas knows the pair exactly when the vertex for the observed sum has degree $1$. Mog knows the pair exactly when the vertex for the observed XOR has degree $1$.

Because the statements are simultaneous, all edges incident to a degree-$1$ vertex at the beginning of a day must be removed simultaneously. The answer is therefore the peeling round in which the actual edge is removed. An edge that remains in the graph's $2$-core forever has answer $-1$.

Assume $X\ge 10$. Let $P$ be the largest power of two not exceeding $X$.

The only sum vertices of initial degree $1$ correspond to the following four pairs:

• $(1,2)$ with sum $3$.
• $(1,3)$ with sum $4$.
• $(X-2,X)$ with sum .

The initial degree-$1$ XOR vertices are classified as follows.

• If $X=P$, every pair $(i,P)$ with $1\le i<P$ is unique.
• If $X=P$, only and are unique.

For the proof, fix an XOR value $y$. If $y<P$, the highest bit is preserved. Since $P\ge 8$, every occurring XOR value has at least two representations inside the lower block.

If $y\ge P$, write $y=P+z$. Write the larger number as $P+t$. The smaller number must then be $$, where . Only is invalid because it produces .

• If $X-P=0$, there is only one possible $t$, so $(z,P)$ is unique.
• If $X-P=1$, exactly leave one valid pair, namely and .

Checking the vertices that can lose degree in each round gives the following table.

When $X=P$, the first round removes every $(i,P)$. Consequently, only $(P-3,P-1)$ remains at sum , and only remains at sum . These two edges are removed on day .