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My first instinct was to go forward from A, which is a mess because multiplying by 2 can blow up fast and you end up with a huge search space.
Recognize this as a shortest path problem on integers and solve it using BFS from A to B, or equivalently work backwards from B to A using greedy division when even. For the given cases, compute the minimum operations by hand or with the algorithm, and state the time complexity as O(log B) for the greedy approach or O(B) for BFS.
Pro tip: When B > A, working backwards from B to A is more efficient: if B is even, divide by 2; otherwise subtract 1. This greedy strategy is optimal because dividing by 2 reduces the value faster than subtracting 1, and it avoids exploring many states.
Confirm that operations are subtract 1 and multiply by 2, and that A and B are positive integers. Note that if A >= B, the only option is repeated subtraction.
For small B, BFS from A to B works. For large B, use the reverse greedy approach: from B, if even divide by 2, else subtract 1, until reaching A. This is optimal and runs in O(log B) time.
Compute the minimum operations for the four given pairs using the chosen method, showing the sequence of operations and counting steps.
State that BFS takes O(B) time and space, while the greedy reverse approach takes O(log B) time and O(1) space. Mention that the greedy is optimal for this problem.
Double-check each case by forward simulation to ensure the number of operations matches the reverse count, and conclude with the final answers.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.