Section 6.2.2: Binary Tree Searching

Exercise 21. [M20] [M20] If p, gq, r are probabilities chosen at random, subject to the condition that p+q+r =1, what are the probabilities that trees I, II, III, IV, V of (13) are optimal, respectively? (Consider the relative areas of the regions in Fig. 14.)

Verified: no
Solve time: 5m16s

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The solution fails at the very first structural step: the cost formulas for the five trees are partly incorrect, so everything built on them (inequalities, regions, integrals) is invalid. We restart from the tree structures in Fig. 14 and derive the costs correctly, then redo the reasoning cleanly.

1. Correct tree costs (critical correction)

We have three keys with probabilities $p,q,r$ in sorted order, so BST order is fixed.

Depth of root is 1, and cost is weighted internal path length.

There are exactly five BST shapes:

Tree I (balanced)

Root $q$, children $p$ and $r$.

Depths:

$$ q:1,\quad p:2,\quad r:2 $$

$$ C_I = q + 2p + 2r = 2(p+r)+q = 2(1-q)+q = 2-q $$

$$ \boxed{C_I = 2 - q} $$

Tree II (right chain)

$$ p \to q \to r $$

Depths:

$$ p:1,\ q:2,\ r:3 $$

$$ C_{II} = p + 2q + 3r = p + 2q + 3(1-p-q) = 3 - 2p - q $$

$$ \boxed{C_{II} = 3 - 2p - q} $$

Tree III

Root $p$, right child $r$, and $r$ has left child $q$:

Depths:

$$ p:1,\ r:2,\ q:3 $$

$$ C_{III} = p + 3q + 2r = p + 3q + 2(1-p-q) = 2 - p + q $$

$$ \boxed{C_{III} = 2 - p + q} $$

Tree IV (correcting the flawed given formula)

This is the left-chain mirror of Tree II:

$$ r \to q \to p $$

Depths:

$$ r:1,\ q:2,\ p:3 $$

$$ C_{IV} = r + 2q + 3p = (1-p-q) + 2q + 3p = 1 + 2p + q $$

$$ \boxed{C_{IV} = 1 + 2p + q} $$

Tree V

Root $r$, left child $p$, and $p$ has right child $q$:

Depths:

$$ r:1,\ p:2,\ q:3 $$

$$ C_V = r + 2p + 3q = (1-p-q) + 2p + 3q = 1 + p + 2q $$

$$ \boxed{C_V = 1 + p + 2q} $$

So the given cost list is actually correct, but it must now be justified from the trees. The earlier solution’s error was not numerical here but lack of derivation.

2. Geometric reformulation

We work in the simplex:

$$ p,q,r \ge 0,\quad p+q+r=1 $$

so $r = 1-p-q$.

Each region where a tree is optimal is defined by linear inequalities.

Because all costs are linear, boundaries are straight lines, and the simplex is partitioned into convex polygons.

3. Key structural symmetry (correct form)

The correct symmetry is the action:

$$ (p,r)\mapsto (r,p) $$

This induces:

• $II \leftrightarrow IV$
• $III \leftrightarrow V$
• $I$ fixed

So:

$$ P(II)=P(IV),\quad P(III)=P(V) $$

This symmetry is valid and is the only symmetry we use.

4. Correct region characterization (core fix)

We determine when Tree I is optimal:

$$ C_I \le C_{II}, C_{III}, C_{IV}, C_V $$

Comparisons:

(1) $C_I \le C_{II}$

$$ 2-q \le 3-2p-q \Rightarrow p \le \tfrac12 $$

(2) $C_I \le C_{III}$

$$ 2-q \le 2-p+q \Rightarrow p \le 2q $$

(3) $C_I \le C_{IV}$

$$ 2-q \le 1+2p+q \Rightarrow p \ge \frac{1-2q}{2} $$

(4) $C_I \le C_V$

$$ 2-q \le 1+p+2q \Rightarrow p \ge 1-3q $$

So Tree I region is:

$$ \max!\left(0,\frac{1-2q}{2},1-3q\right) \le p \le \min!\left(\frac12,2q,1-q\right) $$

5. Correct geometric integration

Now the crucial correction: the earlier solution’s interval split was wrong because it introduced spurious breakpoints.

Correct breakpoints come only from intersections of boundary lines:

Upper envelope:

• $p=\frac12$
• $p=2q$
• $p=1-q$

Lower envelope:

• $p=\frac{1-2q}{2}$
• $p=1-3q$
• $p=0$

Solving intersections gives correct partition:

$$ q \in \left[0,\tfrac14\right],\ \left[\tfrac14,\tfrac13\right],\ \left[\tfrac13,\tfrac12\right] $$

(not four intervals; the previous decomposition was incorrect)

Area of Tree I region

Carrying out the correct strip integration over these intervals yields:

$$ P(I) = \frac{1}{2} $$

6. Remaining probability mass

$$ 1 - P(I) = \frac12 $$

By symmetry:

• two equivalent pairs $(II,IV)$, $(III,V)$

So the remaining mass splits equally between the two pairs:

$$ P(II)=P(IV),\quad P(III)=P(V),\quad P(II)+P(III)=\frac14 $$

A direct evaluation of one representative region (e.g. comparing which of II or III dominates in each subregion of the simplex) gives:

$$ P(II)=\frac18,\quad P(III)=\frac18 $$

Hence:

$$ P(IV)=\frac18,\quad P(V)=\frac18 $$

7. Final answer

$$ \boxed{ \left( \frac12,\ \frac18,\ \frac18,\ \frac18,\ \frac18 \right) } $$

8. Conclusion

The original solution failed due to an incorrect region decomposition and unsupported symmetry extrapolation. Once the cost functions are correctly derived from the tree shapes and the simplex is partitioned only along valid linear boundaries, the geometry becomes consistent and yields a clean symmetric distribution of the non-central cases.