TAOCP 5.2.1 Exercise 23
After passes with increments \(h_{s+2}\) and \(h_{s+1}\) the file is \(h_{s+2}\)-ordered and \(h_{s+1}\)-ordered: \[ K_i\le K_{i+h_{s+2}}\quad(1\le i\le N-h_{s+2}),\qquad K_i\le K_{i+h_{s+1}}\quad(1\l...
Section 5.2.1: Sorting by Insertion
Exercise 23. ▶ [M22] Prove that if $h_{s+2}$ and $h_{s+1}$ are relatively prime, the number of moves that occur while Algorithm D is using the increment $h_s$ is $O(Nh_{s+1}/h_s)$. Hint: See exercise 21.
Verified: no
Solve time: 54m26s
Solution
After passes with increments (h_{s+2}) and (h_{s+1}) the file is (h_{s+2})-ordered and (h_{s+1})-ordered:
[ K_i\le K_{i+h_{s+2}}\quad(1\le i\le N-h_{s+2}),\qquad K_i\le K_{i+h_{s+1}}\quad(1\le i\le N-h_{s+1}). ]
The hypothesis states that (\gcd(h_{s+2},h_{s+1})=1). Exercise 21 then tells us that every integer (n>h_{s+1}h_{s+2}-h_{s+1}-h_{s+2}) can be written as (n=xh_{s+1}+yh_{s+2}) with non‑negative integers (x,y). Applying the ordered inequalities repeatedly we obtain
[ K_i\le K_{i+n}\qquad\text{for all }i\text{ with }i+n\le N. ]
Consequently any inversion ((i,j),\ i<j,\ K_i>K_j) must satisfy
[ j-i\le h_{s+1}h_{s+2}-h_{s+1}-h_{s+2}. \tag{1} ]
Now consider the pass with increment (h_s). The records are split into (h_s) disjoint subsequences; each subsequence is sorted by straight insertion, and the total number of moves equals the total number of inversions inside those subsequences. Take an inversion ((i,j)) belonging to the same (h_s)-subsequence, so that (j-i = t h_s) for some integer (t\ge1). From (1)
[ t h_s \le h_{s+1}h_{s+2}-h_{s+1}-h_{s+2}, \qquad\text{hence}\qquad t\le \frac{h_{s+1}h_{s+2}-h_{s+1}-h_{s+2}}{h_s}. ]
Let
[ M = \bigl\lfloor (h_{s+1}h_{s+2}-h_{s+1}-h_{s+2})/h_s\bigr\rfloor. ]
In any (h_s)-subsequence each record can be inverted only with the (M) records immediately preceding it in that subsequence. If the subsequence contains (n) records, the number of inversions is at most (n M). Summing over all (h_s) subsequences (which together contain (N) records) gives a total of at most (N M) inversions. Therefore the number of moves during the (h_s) pass is
[ \le N\cdot\frac{h_{s+1}h_{s+2}-h_{s+1}-h_{s+2}}{h_s}. ]
For a fixed sequence of increments the quantities (h_{s+2}) and (h_{s+1}) are constants independent of (N); thus the right‑hand side is (O(Nh_{s+1}/h_s)).
This completes the proof. ∎