TAOCP 6.2.1 Exercise 9

Let Algorithm B denote the standard binary search of Section 6.

taocpmathematicsalgorithmsvolume-3medium

Section 6.2.1: Searching an Ordered Table

Exercise 9. [20] [20] Is there any value of N > 1 for which Algorithm B and C are exactly equivalent, in the sense that they will both perform the same sequence of comparisons for all search arguments?

Verified: yes
Solve time: 1m21s

Solution

Let Algorithm B denote the standard binary search of Section 6.2.1, and Algorithm C the uniform binary search defined by the table $\mathrm{DELTA}[j]$ in equation (6).

Equivalence for all search arguments means that for every key $K$, both algorithms generate exactly the same sequence of indices $i$ in stepwise comparisons, hence they induce identical comparison trees.

Root comparison

In Algorithm B, the first comparison is made at the middle position of the table $1,\dots,N$, hence at

$$ i_B(1)=\left\lfloor \frac{N+1}{2}\right\rfloor. $$

In Algorithm C, step C1 sets

$$ i_C(1)=\mathrm{DELTA}[1]=\frac{N+2^{1-1}}{2^1}=\frac{N+1}{2}. $$

Thus the first comparison index is the same in both algorithms if and only if $\frac{N+1}{2}$ is an integer, hence $N$ is odd. In that case,

$$ i_B(1)=i_C(1)=\frac{N+1}{2}. $$

Let

$$ N=2M-1 $$

so that the root is at $i=M$.

Necessary structure for full equivalence

After the first comparison, Algorithm B restricts the search to either the left interval of size $M-1$ or the right interval of size $M-1$. The subsequent comparisons in both cases are identical up to relabeling, so the comparison tree of Algorithm B is recursively perfectly symmetric. Hence for all subproblems to match between Algorithms B and C, the left and right subtrees after the root must again be identical in structure and in probe positions.

This forces the same condition at the next level: each subtree of size $M-1$ must itself split into two equal subtrees under the same rule. Thus $M-1$ must again be of the form $2M_1-1$ for some integer $M_1$. Hence

$$ M=2M_1, \qquad N=2M-1=4M_1-1. $$

Repeating this argument at every level yields a chain of equalities

$$ N+1 = 2M = 2^2 M_1 = \cdots = 2^k, $$

for some integer $k$, terminating when the remaining subtree size becomes $1$. Therefore

$$ N+1=2^k, \qquad N=2^k-1. $$

Sufficiency when $N=2^k-1$

Assume $N=2^k-1$. Then every recursive subdivision in Algorithm B produces subintervals of sizes

$$ 2^{k-1}-1,\ 2^{k-2}-1,\ \dots,\ 1, $$

with perfect symmetry at each level.

For Algorithm C, the values of $\mathrm{DELTA}[j]$ become

$$ \mathrm{DELTA}[j]=\frac{(2^k-1)+2^{j-1}}{2^j}=2^{k-j}-\frac{1}{2}+\frac{1}{2^j}, $$

and since each stage of the algorithm operates on intervals whose lengths are exactly powers of two minus one, each computed shift matches exactly the midpoint chosen by Algorithm B at the corresponding level. Thus every comparison index selected by Algorithm C coincides with the midpoint selection of Algorithm B throughout all recursive calls. Hence both algorithms generate identical comparison sequences for every search argument.

Necessity when $N\neq 2^k-1$

If $N+1$ is not a power of two, then at some level the recursive subdivision in Algorithm B produces subintervals whose sizes differ from perfect symmetry. In particular, there exists a level at which a subinterval of size $t$ splits into two parts whose sizes are not equal.

Algorithm C, however, uses a fixed predetermined sequence $\mathrm{DELTA}[j]$ independent of the actual remaining interval length, so its probes depend only on level, not on the exact residual size. At the first level where the subtree sizes differ from a power-of-two-minus-one structure, the midpoint chosen by Algorithm B depends on the actual endpoints, while the index chosen by Algorithm C is fixed by the global table and cannot adjust to the imbalance. This forces a divergence in at least one of the two branches (left or right), producing a different next comparison index for some search argument $K$.

Hence identical comparison sequences for all $K$ cannot occur unless every recursive interval remains perfectly balanced, which is equivalent to $N=2^k-1$.

Conclusion

There exists a value of $N>1$ for which Algorithms B and C are exactly equivalent if and only if

$$ N=2^k-1 \quad \text{for some integer } k\ge 2. $$

In this case both algorithms induce the same perfect binary comparison tree and therefore perform identical sequences of comparisons for every search argument.

$$ \boxed{N=2^k-1} $$