Hill sorting correctness (Correctness of ShellSort)

Science Hill sort, felt very nature orderly keeping of magic, but you can never find mathematical proof. The last of Computer Programming to find the (apparently I have not read this book) at the Donald E. Knuth's The Art, now extract and organize it.

 

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.

Theorem　　 If one sequence is k ordered in h After sorting, holding k ordered.

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。