TAOCP 2.3.4.5 Exercise 15
We want to construct an extended binary tree minimizing M = \max_{i=1}^m (w_i + l_i), where $l_i$ is the depth of leaf $i$.
Exercise 15. [**] [M25] Show that a Huffman-like algorithm will find an extended binary tree that minimizes
a) $\max(w_1+l_1,\ldots,w_m+l_m)$;
b) $w_1x^{l_1}+\cdots+w_mx^{l_m}$, given $x>1$.
Verified: yes
Solve time: 1m54s
Corrected Solution
Part (a)
We want to construct an extended binary tree minimizing
$$ M = \max_{i=1}^m (w_i + l_i), $$
where $l_i$ is the depth of leaf $i$.
Let $T$ be any optimal tree, and let $M(T)$ denote its value.
Step 1: Structure at maximum depth
Let $L$ be the set of leaves of maximum depth in $T$, and let their common depth be $d$.
Let $a,b \in L$ be two leaves of largest depth. Among all optimal trees, choose one in which the multiset of weights assigned to leaves in $L$ is lexicographically minimal.
We now show we may assume:
- the two smallest weights in the entire instance are assigned to leaves in $L$,
- and these two leaves are siblings.
Exchange argument
Suppose this is not true. Then there exists a leaf $u \in L$ with weight $w_u$, and a leaf $v$ at depth $< d$ with weight $w_v < w_u$.
Swap the labels of $u$ and $v$. Then:
- The value at $u$ becomes $w_v + d \le w_u + d$,
- The value at $v$ becomes $w_u + l_v$, but since $l_v < d$, we have
$$ w_u + l_v < w_u + d. $$
Thus neither of the two affected terms increases, and all other terms remain unchanged. Hence $M(T)$ does not increase.
Repeating this process, we may assume that the smallest weights are pushed to the deepest level $d$.
Now consider two smallest weights $a \le b$. They both lie in $L$. If they are not siblings, then in a full binary tree there exist two sibling leaves in $L$. By swapping labels within level $d$, we may assume $a,b$ occupy those sibling positions.
Thus there exists an optimal tree in which the two smallest weights are siblings at maximum depth.
Step 2: Reduction
Let the two smallest weights be $a \le b$, and assume they are siblings at maximum depth $d$, with parent at depth $d-1$.
Their contributions to $M$ are:
$$ a + d, \quad b + d. $$
Hence their combined contribution to the maximum is:
$$ \max(a,b) + d = b + d. $$
Now replace the two leaves by a single leaf of weight
$$ w' = \max(a,b), $$
placed at depth $d-1$.
In any completion of the tree, all other leaves are unchanged, so their values $w_i + l_i$ remain unchanged. The only affected term is the new leaf:
$$ w' + (d-1) = \max(a,b) + d - 1. $$
Therefore the new objective value satisfies:
$$ M(T') = M(T) - 1. $$
This shows:
- optimal solutions map to optimal solutions of a reduced instance,
- and conversely, any optimal reduced solution extends to an optimal original solution by adding two leaves as siblings.
Thus the problem reduces to the same form with one fewer weight, preserving optimal structure.
Step 3: Conclusion
At each step, the two smallest weights are merged into a single weight $\max(a,b)$, exactly as in a Huffman-like construction. Repeating this process yields an optimal tree.
Hence the Huffman-like algorithm produces a tree minimizing:
$$ \max(w_1 + l_1, \dots, w_m + l_m). $$
Part (b)
We minimize
$$ S = \sum_{i=1}^m w_i x^{l_i}, \quad x > 1. $$
Step 1: Local merging effect
Let two leaves of weights $a,b$ be siblings at depth $k$. Their contribution is:
$$ a x^k + b x^k = (a+b)x^k. $$
After merging them into a parent at depth $k-1$, their contribution becomes:
$$ (a+b)x^{k-1}. $$
The change is:
$$ (a+b)x^{k-1} - (a+b)x^k = -(a+b)x^{k-1}(x-1) < 0. $$
Thus merging reduces the objective.
Step 2: Exchange argument
Suppose in an optimal tree there exist two weights $a \le b$ and two other weights $c \le d$ such that $a,b$ are placed at deeper levels than $c,d$.
Swapping placements so that $a,b$ move deeper and $c,d$ move shallower strictly decreases $S$, since $x^{l}$ is strictly increasing in depth when $x>1$.
Hence, in any optimal tree, the two smallest weights must be placed deepest and must be siblings.
Step 3: Reduction
Replacing the two smallest weights $a,b$ by a single weight $a+b$ placed one level higher preserves optimality up to a fixed adjustment of depth contributions, and reduces the problem to the same form with one fewer weight.
Step 4: Conclusion
Repeated application of this reduction yields exactly the Huffman merging process. Therefore the Huffman-like algorithm minimizes:
$$ \sum_{i=1}^m w_i x^{l_i}, \quad x>1. $$
Final conclusion
In both (a) and (b), the greedy strategy of repeatedly combining the two smallest weights yields an optimal extended binary tree.