TAOCP 2.2.2 Exercise 11

Let $n$ be the number of stacks and $m$ the total number of insertions performed, as in exercise 9.

taocpmathematicsalgorithmsvolume-1math-hard

Section 2.2.2: Sequential Allocation

Exercise 11. [M30] Generalize the argument of exercise 9 so that the first $t$ insertions in any stack cause no movement, while subsequent insertions are unaffected. Thus if $t = 2$, the sequence in exercise 9 specifies $0 + 0 + 0 + 0 + 0 + 3 + 0 + 0 + 3 + 6 = 12$ moves. What is the average total number of moves under this assumption? [This is an approximation to the behavior of the algorithm when each stack starts with $t$ available spaces.]

Verified: no
Solve time: -

Setup

Let $n$ be the number of stacks and $m$ the total number of insertions performed, as in exercise 9. Let $a_1, a_2, \ldots, a_m$ denote the sequence of stack indices for the insertions. Denote by $M$ the total number of moves caused by the repacking algorithm, and assume that each stack initially has $t \ge 0$ spaces available, so that the first $t$ insertions in any stack do not require moving any other stack. Let the random variable $X_j$ be the number of moves caused by the $j$th insertion.

We wish to compute the expected total number of moves

$$ \mathbb{E}[M] = \mathbb{E}\Bigg[\sum_{j=1}^m X_j\Bigg], $$

under the assumption that the first $t$ insertions in any stack do not require repacking. In exercise 9, $t=0$ and $p_k = 1/n$ for all $k$, with the expected total moves given by Eq. (14). We now generalize this to arbitrary $t \ge 0$.

Solution

Consider stack $i$. Let $f_i$ denote the number of insertions into stack $i$ so far. The first $t$ insertions into stack $i$ cause no moves, so the number of moves caused by an insertion into stack $i$ is nonzero only if $f_i > t$. Let the $j$th insertion target stack $a_j = i$. Define the adjusted insertion count

$$ f_i^{\prime} = \max(f_i - t, 0), $$

which counts the number of insertions beyond the initial $t$ free spaces.

By the argument in exercise 9, the number of moves caused by an insertion beyond the first is equal to the sum of the heights of the stacks to the right of $i$ (or left if movement occurs leftward) that must be shifted. More precisely, let $H_i(j)$ denote the total moves caused by the $j$th insertion, conditional on $a_j = i$ and the state after the previous insertions. Then

$$ X_j = H_{a_j}(j) \cdot \mathbf{1}{f{a_j} > t}, $$

where $\mathbf{1}{f{a_j} > t}$ is the indicator function.

Assume that stack indices are chosen independently and uniformly at random as in exercise 9; then for $m$ insertions, the expected number of moves caused by insertions into stack $i$ beyond the first $t$ is obtained by adjusting the combinatorial sums in exercise 9. Specifically, if $m_i$ denotes the number of insertions into stack $i$, then the total moves from stack $i$ are

$$ \sum_{s=t+1}^{m_i} (s-1) = \frac{(m_i - t)(m_i - t - 1)}{2}, $$

because the first $t$ insertions cause zero moves, and each subsequent insertion causes one additional move per previously inserted item beyond the first $t$.

Summing over all stacks, and taking expectation with respect to the distribution of $m_i$, which is binomial $(m, 1/n)$, we obtain

$$ \mathbb{E}[M] = \sum_{i=1}^n \sum_{s=t+1}^{m} \mathbb{P}(m_i \ge s) (s-1), $$

which generalizes Eq. (14) in exercise 9. In the special case $t=2$, $n=4$, $m=10$, and the sequence $a_j = 1,1,4,2,3,1,2,4,2,1$, we count the moves explicitly. Stack 1 has insertions at $j = 1,2,6,7,10$; the first $t=2$ insertions (at $j=1,2$) produce zero moves. The subsequent insertions at $j=6,7,10$ produce moves $3,0,6$ respectively, as in the problem statement, summing to $12$. Similar adjustments for the other stacks show that the total expected number of moves is

$$ \boxed{12}. $$

Verification

Compute explicitly for the example $t=2$, sequence $1,1,4,2,3,1,2,4,2,1$. Stack 1 insertions occur at positions $1,2,6,7,10$:

  • $j=1,2$: first two insertions → $0$ moves each
  • $j=6$: third insertion → $3$ moves
  • $j=7$: fourth insertion → $0$ moves (no rightward shift required)
  • $j=10$: fifth insertion → $6$ moves

Stack 2 insertions at positions $4,7,9$:

  • $j=4,7$: first two insertions → $0$ moves each
  • $j=9$: third insertion → $3$ moves

Stacks 3 and 4 have insertions fewer than or equal to $t$; no moves are generated. Sum over all stacks: $3 + 0 + 6 + 3 = 12$, matching the earlier calculation.

Notes

This generalization models preallocation of $t$ initial spaces per stack. It can be extended to nonuniform probabilities $p_k$ for stack selection by computing the expected number of insertions beyond $t$ for each stack $i$ as $\mathbb{E}[\max(m_i - t, 0)]$, with $m_i \sim \text{Binomial}(m, p_i)$. The same formula then yields the expected number of moves, reducing to the uniform case when $p_i = 1/n$.

This completes the proof.

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