希尔排序的正确性 (Correctness of ShellSort)

学希尔排序的时候，觉得有序性保持的性质十分神奇，但哪里都找不到数学证明。最后在Donald E. Knuth的The Art of Computer Programming中找到了（显然我没有读过这套书），现摘录并整理之。

 

Theorem K. If a k-ordered file is h-sorted, it remains k-ordered.

Thus a file that is first 7-sorted, then 5-sorted, becomes both 7-ordered and 5-ordered. And if we 3-sort it, the result is ordered by 7s, 5s, and 3s. Examples of this remarkable property can be seen in Table 4 on page 85.

Proof. Exercise 20 shows that Theorem K is a consequence of the following fact:

Lemma L. Let m, n, r be nonnegative integers, and let (x₁ , ... , x_m+r) and (y₁ , ... , y_n+r) be any sequences of numbers such that

　　　　　　　　　　　　　　　　y₁ ≤ x_m+1 ,　　y₂ ≤ x_m+2 ,　　... ,　　y_r ≤ x_m+r　　　　　　　　　　　　　　　　(7)

If the x's and y's are sorted independently, so that x₁ ≤ ... ≤ x_m+r and y₁ ≤ ... ≤ y_n+r , the relations (7) will still be valid.

Proof. All but m of the x’s are known to be dominate (that is, to be greater than or equal to) some y, where distinct x’s dominate distinct y’s. Let 1 ≤ j ≤ r. Since x_m+j after sorting dominates m + j of the x’s, it dominates at least j of the y’s; therefore it dominates the smallest j of the y’s; hence x_m+j ≥ y_j after sorting.

Exercise 20. Show that Theorem K follows from Lemma L.

Solution. (This is much harder to write down than to understand.) Assume that a k-ordered file R₁ , ... , R_N has been h-sorted, and let 1 ≤ i ≤ N - k; we want to show that K_i ≤ K_i+k . Find u, v such that i ≡ u and i + k ≡ v (modulo h), 1 ≤ u, v ≤ h; and apply Lemma L with x_j = K_v+(j-1)h , y_j = K_u+(j-1)h . Then the first r elements K_u , K_u+h , ... , K_u+(r-1)h of the y’s are respectively ≤ the last r elememts K_u+k , K_u+k+h , ... , K_u+k+(r-1)h of the x’s, where r is the greatest integer such that u + k + (r - 1)h ≤ N.

定理　　如果一个序列是k有序的，在h排序后它保持k有序。

证明　　引理：设m, n, r为非负整数，(x₁ , ... , x_m+r)与(y₁ , ... , y_n+r)为满足以下性质的序列：

y₁ ≤ x_m+1 ,　　y₂ ≤ x_m+2 ,　　... ,　　y_r ≤ x_m+r

那么如果x和y被分别排序，使x₁ ≤ ... ≤ x_m+r且y₁ ≤ ... ≤ y_n+r ，则以上关系仍成立。

除m个以外所有x都大于等于某个y，其中不同的x大于等于不同的y。设1 ≤ j ≤ r。由于排序后x_m+j大于等于x中的m + j个，它至少大于y中的j个。因此它大于等于y中最小的j个，故排序后x_m+j ≥ y_j，引理得证。

假设k有序的序列R₁ , ... , R_N被h排序，设1 ≤ i ≤ N - k，即证K_i ≤ K_i+k。存在u、v使i ≡ u，i + k ≡ v (mod h)，其中1 ≤ u，v ≤ h。将x_j = K_v+(j-1)h ，y_j = K_u+(j-1)h代入引理。则y中的前r个元素K_u , K_u+h , ... , K_u+(r-1)h分别小于等于x中的后r个元素K_u+k , K_u+k+h , ... , K_u+k+(r-1)h ，其中r是满足u + k + (r - 1)h ≤ N的最大整数。证毕。

因此一开始7有序的序列在5排序后同时成为7有序与5有序的。如果再3排序，结果是7、5与3有序的。