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If $C\geq N$ , the graph already satisfies the condition when $t=0$ and has no edges, so the answer is $0$ . For the rest of the discussion, assume $C<N$ . Suppose $t>0$ and $2^M\leq t<2^{M+1}$ . Partition the vertices in numerical order into blocks of size $2^{M+1}$ . Each block is exactly one connected component, even if the last block contains fewer vertices. Thus, the number of connected components is

\left\lceil\frac{N}{2^{M+1}}\right\rceil.

This number remains unchanged as $t$ varies from $2^M$ through $2^{M+1}-1$ . Therefore, the smallest suitable $t$ must be a power of two. Start with and keep doubling until

\left\lceil\frac{N}{2p}\right\rceil\leq C.

The first value of $p$ satisfying this condition is the answer. The ceiling can be computed using $\lceil x/y\rceil=\lfloor(x+y-1)/y\rfloor$ for positive integers $x$ and . The time complexity is , and the auxiliary space complexity is .

Proof

Let $L=2^{M+1}$ , and define block $q$ as the set of vertices $v$ satisfying $q L \leq v < (q + 1)$ . Only existing vertices are included, so the last block may contain fewer than vertices. The numbers of any two vertices in different blocks differ in at least one bit at position or higher. Their XOR is therefore at least and greater than . Hence, no edge joins different blocks. Let be the first vertex of a block. Starting from any vertex in that block, repeatedly clear a set bit at a position from through . This eventually reaches . When bit is cleared, the XOR of the two consecutive vertex numbers is , so an edge joins those vertices. The higher bits do not change, so every intermediate vertex remains in the same block. The vertex number only decreases, so it is always at least and at most the original . All intermediate vertices therefore exist even when the last block is incomplete. Every vertex in the block is connected to , making the block a single connected component. There are blocks, so the number of connected components is . This number is unchanged throughout , so only the smallest value in each such interval needs to be checked. Since , the value does not satisfy the condition. Checking in order examines the start of every possible interval in increasing order. The first value that satisfies the condition is therefore the answer.

Solution written by GPT6

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