# MAW Chapter 7: Sorting writing questions

## Solutions

including: MAW 7.1, 7.2, 7.3, 7.4, 7.5.a, 7.9, 7.10, 7.11, 7.12, 7.13

### MAW 7.1

Sort the sequence 3,1,4,1,5,9,2,6,5 using insertion sort.

| index    | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| original | 3 | 1 | 4 | 1 | 5 | 9 | 2 | 6 | 5 |
|----------|---|---|---|---|---|---|---|---|---|
| pass 1   | 1 | 3 | 4 | 1 | 5 | 9 | 2 | 6 | 5 |
| pass 2   | 1 | 3 | 4 | 1 | 5 | 9 | 2 | 6 | 5 |
| pass 3   | 1 | 1 | 3 | 4 | 5 | 9 | 2 | 6 | 5 |
| pass 4   | 1 | 1 | 3 | 4 | 5 | 9 | 2 | 6 | 5 |
| pass 5   | 1 | 1 | 3 | 4 | 5 | 9 | 2 | 6 | 5 |
| pass 6   | 1 | 1 | 2 | 3 | 4 | 5 | 9 | 6 | 5 |
| pass 7   | 1 | 1 | 2 | 3 | 4 | 5 | 6 | 9 | 5 |
| pass 8   | 1 | 1 | 2 | 3 | 4 | 5 | 5 | 6 | 9 |


### MAW 7.2

What is the running time of insertion sort if all keys are equal?

If you take a look at the code on p. 220, you can see that inner for loop checks A[j-1] > tmp and it will fail immediately. Thus, the running time is $$O(N)$$.

### MAW 7.3

Suppose we exchange elements $$A[i]$$ and $$A[i+k]$$, which were originally out of order. Prove that at least 1 and at most $$2k-1$$ inversions are removed.

The inversion that existed between $$A[i]$$ and $$A[i+k]$$ is removed. This shows at least one inversion is removed. Now let's consider $$A[i], A[i+1], \dots, A[i+k-1], A[i+k]$$, Suppose $$A[i]$$ is greater than $$A[i+1], \dots, A[i+k]$$ and $$A[i+k]$$ is smaller than $$A[i], \dots, A[i+k-1]$$. In this case, by swapping $$A[i]$$ and $$A[i+k]$$, we fix $$2k-1$$ inversions ($$-1$$ is that $$A[i]$$ greater than $$A[i+k]$$ and $$A[i+k]$$ smaller than $$A[i]$$ points to the same inversion).

Another way to think about $$2k-1$$ is that for each of the $$k-1$$ elements $$A[i+1], A[i+2], \dots, A[i+k-1]$$, at most two inversions can be removed by exchange. For instance, for $$A[i+1]$$, two inversions are $$A[i]$$ and $$A[i+1]$$, and $$A[i+1]$$ and $$A[i+k]$$ (i.e. for sequence 10,4,3, by swapping 10 and 3, we remove inversion {10,4} and {4,3}). Thus, a maximum of $$2(k-1)+1 = 2k-1$$.

### MAW 7.4

Show the result of running Shellsort on the input 9,8,7,6,5,4,3,2,1 using the increments {1,3,7}

| index        | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| original     | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
|--------------|---|---|---|---|---|---|---|---|---|
| after 7-sort | 2 | 1 | 7 | 6 | 5 | 4 | 3 | 9 | 8 |
| after 3-sort | 2 | 1 | 4 | 3 | 5 | 7 | 6 | 9 | 8 |
| after 1-sort | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |


### MAW 7.5.a

What is the running time of Shellsort using the two-increment sequence {1,2}?

The answer is $$\Theta(N^2)$$. Let's first show the lower bound. By the conclusion of 7.3, we know that The 2-sort removes at most only three (i.e. $$k=2$$) inversions at a time. In addition, a pass with increment $$h_k$$ consists of $$h_k$$ insertion sorts of about $$N/h_k$$ elements. Then, by theorem 7.2, we know that the algorithm is $$\Omega(N^2)$$. By the same argument, the 2-sort is two insertion sorts of size $$N/2$$, so the cost of that pass is $$O(N^2)$$. The 1-sort is also $$O(N^2)$$, so the upper bound for the algorithm is $$O(N^2)$$.

### MAW 7.9

Determine the running time (i.e. number of swaps) of Shellsort for

a. sorted input

$$O(N \log N)$$. No exachanges acutally done in each each pass but we will still need to go through the second for loop, which indicates that each pass takes $$O(N)$$. There are total $$O(\log N)$$ passes and the answer follows.

b. reverse-ordered input

$$O(N \log N)$$. It is easy to show that after an $$h_k$$ sort, no element is farther than $$h_k$$ from its rightful position. Thus, if the increments satisfy $$h_{k+1} \le ch_k$$ for a constant $$c$$, which implies $$O(\log N)$$ increments, then the bound follows.

However, one cannot talk about shellsort without specifying the increment sequence. If we assume the shell sequence (i.e. $$N/2, N/4, \dots, 2, 1$$), then the running time is $$O(N^2)$$ as suggested by this answer, which I'll copy below for future reference.

Shellsort is just a bunch of insertion sorts. For a given increment $$I$$, there will be $$I$$ subarrays to sort by insertion, each of length $$N/I$$. We know that insertion sort requires time $$O(m^2)$$ to sort a reverse-sorted array of length $$m$$. Here, $$m$$ will be ($$N/I$$) for each subarray. Thus one subarray will cost $$(N/I)^2$$ to sort. There are $$I$$ subarrays, so the total cost will be $$I * (N/I)^2 = N^2/I$$. But that is the cost just for a single increment. The total time for all of the iterations must be $$N^2/(N/2) + N^2/(N/4) + N^2/(N/8) + \dots + N^2/2 + N^2/1 = 2N + 4N + \dots + N^2/2 + N^2/1$$ . If we factor out an $$N$$, we get $$N(2+4+\dots+N/2+N)$$ . In parenthesis is the sum of powers of 2 from 2 to $$N$$, which is approximately equalt to $$2N$$. Therefore, the total cost is $$N(2N) = 2N^2 = O(N^2)$$.

### MAW 7.10

Do either of the following modifications to the Shellsort routine coded in Fig. 7.4 affect the worst case running time?

a. Before line 2, subtract one from Increment if it is even.

The key improvement in terms of the worst case running time lies in the increment sequence. As suggested on p.224,225, we improve the worst time running time from $$O(N^2)$$ to $$O(N^{3/2})$$ by changing the increment sequence into the sequence that consecutive increments have no common factors.

If we follow the modification indicated by this question, it is still possible to have a case that we will have consecutive increments to share a common factor. For instance, if we sort an array with size $$N = 45$$, then with the modification, the increment sequence will be $$45, 21 (22-1), 9, 3, 1$$.

b. Before line 2, add one to Increment if it is even.

In this case, conseuctive increments are relatively prime and by the argument in the proof of theorem 7.4, we can have the worst case running time $$O(N^{3/2})$$.

### MAW 7.11

Show how heapsort processes the input 142, 543, 123, 65, 453, 879, 572, 434, 111, 242, 811, 102.

The input is read in as it appears in the question. Then, we first build the heap with the result looks like

879, 811, 572, 434, 543, 123, 142, 65, 111, 242, 453, 102


$$879$$ is removed from the heap and placed at the end. We'll put | to separate the elements that are sorted and not part of the heap. $$102$$ is placed in the hole and bubbled down, obtaining

811, 543, 572, 434, 453, 123, 142, 65, 111, 242, 102, | 879


continuing the process, we obtain

572, 543, 142, 434, 453, 123, 102, 65, 111, 242, | 811, 879
543, 453, 142, 434, 242, 123, 102, 65, 111, | 572, 811, 879
453, 434, 142, 111, 242, 123, 102, 65, | 543, 572, 811, 879
434, 242, 142, 111, 65, 123, 102, | 453, 543, 572, 811, 879
242, 111, 142, 102, 65, 123, | 434, 453, 543, 572, 811, 879
142, 111, 123, 102, 65, | 242, 434, 453, 543, 572, 811, 879
123, 111, 65, 102, | 142, 242, 434, 453, 543, 572, 811, 879
111, 102, 65, | 123, 142, 242, 434, 453, 543, 572, 811, 879
102, 65, | 111, 123, 142, 242, 434, 453, 543, 572, 811, 879
65, | 102, 111, 123, 142, 242, 434, 453, 543, 572, 811, 879
| 65, 102, 111, 123, 142, 242, 434, 453, 543, 572, 811, 879


### MAW 7.12

a. What is the running time of heapsort for presorted input?

Still $$O(N\log N)$$. Heapsort uses at least (roughly) $$N\log N$$ comparisons on any input, so there are no particularly good inputs.

### MAW 7.13

Sort 3,1,4,1,5,9,2,6 using mergesort

First the sequence {3,1,4,1} is sorted. To do this, the sequence {3,1} is sorted. This involves sorting {3} and {1}, which are base cases, and merging the result to obtain {1,3}. The sequence {4,1} is likewise sorted into {1,4}. Then these two sequences are merged to obtain {1,1,3,4}. The second half is sorted similarly, eventually obtaining {2,5,6,9}. The merge result is then easily computed as {1,1,2,3,4,5,6,9}.

### MAW 7.15

Determine the running time of mergesort for a. sorted input b. reverse-ordered input c. random input

The running time for mergesort is $$O(N \log N)$$ regardless of the input pattern.

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