TAOCP 6.3 Exercise 45

Section 6.3: Digital Searching

Exercise 45. [M25] [M25] If the seven keys of Fig. 33 are inserted in random order by the algorithm of exercise 15, what is the probability of obtaining the tree shown?

Verified: yes
Solve time: 1m24s

Solution

Let $T$ be the binary search tree shown in Fig. 33, and let $v$ range over its nodes. For each node $v$, let $S(v)$ denote the set of keys in the subtree rooted at $v$, and write

$$ n(v)=|S(v)|,\quad n_L(v)=|S(\text{left}(v))|,\quad n_R(v)=|S(\text{right}(v))|. $$

A random insertion order corresponds to a uniformly random permutation of the $7$ keys. The algorithm of exercise 15 builds a binary search tree by inserting keys sequentially into an initially empty tree, so each permutation produces a unique resulting shape.

Fix a node $v$. In any insertion sequence that yields the given tree $T$, the key stored at $v$ must be inserted before all keys in $S(v)$ that lie in its subtrees do not affect its position once $v$ is placed. The relative order of insertions inside $S(v)$ determines how the left and right subtrees of $v$ are formed.

Among the $n(v)!$ possible insertion orders of the keys in $S(v)$, exactly those sequences that interleave the $n_L(v)!$ orders of the left subtree and the $n_R(v)!$ orders of the right subtree while preserving internal orders of each subtree produce the fixed split at $v$. The number of such interleavings equals the number of ways to choose positions for the $n_L(v)$ left-subtree insertions among the $n_L(v)+n_R(v)$ non-root insertions, hence

$$ \binom{n_L(v)+n_R(v)}{n_L(v)}. $$

This argument applies recursively at every node, and the choices at distinct nodes are independent once subtree sizes are fixed, since each subtree is generated entirely from the relative order of its own keys. Multiplying contributions over all nodes gives the total number of permutations that produce $T$:

$$ N(T)=\prod_{v}\binom{n_L(v)+n_R(v)}{n_L(v)}. $$

Since all $7!$ permutations are equally likely, the required probability is

$$ \frac{N(T)}{7!}=\frac{1}{7!}\prod_{v}\binom{n_L(v)+n_R(v)}{n_L(v)}. $$

Thus the probability of obtaining the tree in Fig. 33 is

$$ \boxed{\frac{1}{7!}\prod_{v}\binom{n_L(v)+n_R(v)}{n_L(v)}}. $$

This completes the proof. ∎