The simplest method of computing b? requires n?1 multiplication operations, but it can be computed more efficiently than that, as illustrated by the following example. To compute 21??, note that 100 = 64 + 32 + 4. Compute the following in order:22 = 4

(22)2 = 2? = 16

(2?)2 = 2? = 256

(2?)2 = 21? = 65,536

(21?)2 = 232 = 4,294,967,296

(232)2 = 2?? = 18,446,744,073,709,551,616

2?? × 232 × 2? = 21?? = 1,267,650,600,228,229,401,496,703,205,376This series of steps only requires 8 multiplication operations instead of 99 (since the last product above takes 2 multiplications).

In general, the number of multiplication operations required to compute b? can be reduced to Θ(log n) by using exponentiation by squaring or (more generally) addition-chain exponentiation. Finding the minimal sequence of multiplications (the minimal-length addition chain for the exponent) for b? is a difficult problem for which no efficient algorithms are currently known (see Subset sum problem), but many reasonably efficient heuristic algorithms are available.[29]

平方求冪這一頁很難總結，但它基本上是2?==(2?)2==(22)2，所以不需要計算2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256，而是可以計算2 × 2 = 4; 4 × 4 = 16; 16 × 16 = 256。在

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