Discussion
Bidirectional search reduces \( b^{d} \) to \( 2b^{d/2} \). Compare the two for b = 10, d = 6.
Answer: A
Evaluating both expressions:
\[ 2 \times 10^{3} = 2000 \text{ against } 10^{6} \]
so two thousand against one million, a saving of five hundred fold. The gain grows with depth, which is why the technique is worth its considerable implementation difficulty on deep problems. Halving an exponent transforms a problem's feasibility in a way that no constant factor ever can.
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