## FAQ#

Each assignment will have an FAQ linked at the top. You can also access it by adding “/faq” to the end of the URL. The FAQ for Lab 25 is located here.

## Introduction #

As usual, pull the files from the skeleton and make a new IntelliJ project.

In today’s lab, we’ll be discussing sorting, algorithms for rearranging elements in a collection to be in a specific order. There are many problems you can more easily solve with a sorted collection, including performing binary search in $$O(\log N)$$ time, efficiently identifying adjacent pairs within a list, finding the $$k^{th}$$ largest element, and so forth.

There are several kinds of sorting algorithms, each of which is appropriate for different situations. At the highest level, we will distinguish between two types of sorting algorithms:

• Comparison-based sorts, which rely on making pairwise comparisons between elements.
• Counting-based sorts, which group elements based on their individual digits before sorting and combining each group. Counting sorts do not need to compare individual elements to each other.

In this lab, we will discuss several comparison-based sorts including insertion sort, selection sort, merge sort, and quicksort. Why all the different sorts? Each sort has a different set of advantages and disadvantages: under certain conditions, one sort may be faster than the other, or one sort may take less memory than the other, and so forth. When working with large datasets (or even medium and small datasets), choosing the right sorting algorithm can make a big difference. Along the way, we’ll develop an intuition for how each sort works by exploring examples and writing our own implementations of each sort.

Here is a nice visualizer for all of the sorts we are going to cover today.

## Order and Stability #

To put elements in order implies that we can enforce an ordering between any two elements. Given any two elements in a list, according to total order, we should be able to decide which of the two elements is larger or smaller than the other.

However, it’s also possible that neither element is necessarily larger or smaller than the other. For instance, if we wish to determine the ordering between two strings, ["sorting", "example"], and we want to order by the length of the string, it’s not clear which one should come first because both strings are of the same length 7.

In this case, we can defer to the notion of stability: if a sort is stable, then it will preserve the relative orderings between elements in the list. In the above example then, the resultant array will be ["sorting", "example"] in a stable sort rather than ["example", "sorting"] as is possible in an unstable sort. Remember that, according to our total order by the length of the strings, the second list is still considered correctly sorted even though the relative order of equivalent elements is not preserved.

What is the benefit of stable sorting? It allows us to sort values based off multiple attributes. For example, we could stably sort a library catalog by alphabetical order, then by genre, author, etc. Without stable sorting, we are not guaranteed that the relative ordering of the previous sorts would persist so it is possible that the catalog would only be sorted by our last sort.

Consider the following example where we sort a list of animals by alphabetical order and then length of string.

Original collection:

cow
giraffe
octopus
cheetah
bat
ant


Sort by alphabetical order:

ant
bat
cheetah
cow
giraffe
octopus


Stable sort by length of string:

ant
bat
cow
cheetah
giraffe
octopus


Now the collection is sorted by length and elements with the same length are in alphabetical order with each other. If our sorting algorithm was not stable, then we would potentially lose the alphabetical information we achieved in the previous sort.

## Space Complexity #

Thus far, in this class, we’ve mostly talked about time complexity. Similarly to how we can do asymptotic analysis for runtime, we can also analyze how much space (i.e. memory) a given algorithm uses. For sorting algorithms, one common trait we look for is if the algorithm is in-place.

Any sorting algorithm that takes in an input of size N is going to have to work with some amount of memory proportional to N to store the size of the input itself. An in-place algorithm is one that doesn’t use a significant amount of additional memory. In this class, this means that the algorith must use a constant amount of additional memory (for example, a few variables to keep track of your current index or something). Another way to think about it is if we can do the entire algorithm within the original given list, without creating an additional data structure.

Note that the definition of in-place can vary–though in our class we say only a constant amount of space counts, some measures say using a logarithmic amount is okay. If you see other articles online, just be wary!

## Discussion: Sorting by Hand #

With your partner, discuss how you would sort a hand of 13 playing cards if you are dealt the cards one-by-one. Your hand should end up sorted first by suit, and then by rank within each suit.

Then discuss how you would sort a pile of 300 CS 61BL exams by student ID. If it’s different than your card-sorting algorithm of the previous step, explain why.

Afterwards, discuss with your partner and roughly describe an algorithm to formalize your sort. Can you tell if one is faster than the other? How so?

## Insertion Sort #

The first comparison-based sort we’ll learn is called an insertion sort. The basic idea for insertion sort can be formally summed up as follows:

1. Start a loop over all items in your collection.
2. For each item in the collection, insert it into its correct place among all the items you’ve looked at so far. You can do this by swapping it backwards with the item before it, until you reach an item that is larger.

You might have intuitively come up with insertion sort when we asked you how to sort cards. This is like when you sort cards by continually putting the next card in the right spot in a group of sorted cards that you’re holding.

Here is code that applies the insertion sort algorithm (to sort from small to large) to an array named arr.

public static void insertionSort(int[] arr) {
for (int i = 1; i < arr.length; i++) {
for (int j = i; j > 0 && arr[j] < arr[j - 1]; j--) {
swap(arr, j, j - 1);
}
}
}


Note that insertion sort is stable. We never swap elements if they are equal so the relative order of equal elements is preserved.

Now that you’ve read the above explanation, we recommend watching this video to solidify your understanding.

## Check your understanding: Insertion Sort #

Analyze the insertionSort code given above carefully. In particular, consider the comparison step in the for loop condition.

### Practice: #

For the following questions, discuss with your partner and verify your answers by highlighting the line under the question.

Assume we have an array of $$N$$ integers. What would the array have to look like before we ran insertion sort that would make insertion sort run the fastest, i.e. minimizing the number of steps needed?

Sorted List.

What is the runtime of running insertion sort on this array?

Theta(N).

What type of initial ordering of a list would maximize the number of comparisons and result in the slowest runtime?

Reverse Sorted Array.

What is the runtime of running insertion sort on the type of array you identified above?

Theta(N^2).

### Exercise: insertionSort#

Above, we have provided code for insertion sort on an array. Now, let’s try writing insertion sort for linked lists!

The file DLList.java contains an implementation of a doubly-linked list along with method headers for different sorts. Complete the implementation of insertionSort by writing the insertionSortHelper method, which will insert a single element to an already sorted DLList.

This method should be non-destructive, so the original DLList should not be modified.

## Selection Sort #

Selection sort on a collection of $$N$$ elements can be described by the following pseudocode:

for each element in the collection:
find the smallest remaining element, E, in the *unsorted* collection
remove E and add E to the end of the *sorted* collection
repeat unsorted collection's original length number of times (or repeat until unsorted collection has no more elements)


Given below is an implementation of the selection sort algorithm for arrays. You can also see a version for linked lists completed in DLList.java.

// At the beginning of every iteration, elements 0 ... k - 1 are in sorted
// order.
for (int k = 0; k < values.length; k++) {

int min = values[k];
int minIndex = k;

// Find the smallest element among elements k ... values.length - 1.
for (int j = k + 1; j < values.length; j++) {
if (values[j] < min) {
min = values[j];
minIndex = j;
}
}

// Put MIN in its proper place at the end of the sorted collection.
// Elements 0 ... k are now in sorted order.
swap(values, minIndex, k);
}


In our code for selection sort we swap the minimum element in the unsorted collection with the element at the beginning of the sorted collection. This can rearrange the relative ordering of equal elements. Thus, selection sort is unstable.

After reading the above, we recommend watching this video on selection sort!

### Disussion: Runtime #

Now, let’s determine the asymptotic runtime of selection sort. One may observe that, in the first iteration of the loop, we will look through all $$N$$ elements of the array to find the minimum element. On the next iteration, we will look through $$N - 1$$ elements to find the minimum. On the next, we’ll look through $$N - 2$$ elements, and so on. Thus, the total amount of work will be the $$N + (N - 1) + ... + 1$$, no matter what the ordering of elements in the array or linked list prior to sorting.

Hence, we have an $$\Theta(N^2)$$ algorithm, equivalent to insertion sort’s normal case. But notice that selection sort doesn’t have a better case, while insertion sort does.

Can you come up with any reason we would want to use selection sort over insertion sort? Discuss with your partner.

## HeapSort #

For an introduction, watch this.

Recall the basic structure for selection sort

for each element in the collection:
find the smallest remaining element, E, in the unsorted collection
remove E and add E to the end of the sorted collection


Adding something to the end of a sorted array or linked list can be done in constant time. What hurt our runtime was finding the smallest element in the collection, which always took linear time in an array.

Is there a data structure we can use that allows us to find and remove the smallest element quickly? A heap will! Removal of the largest element from a heap of $$N$$ elements can be done in time proportional to $$\log N$$, allowing us to sort our elements in $$O(N \log N)$$ time. Recall that we can also build a heap in linear time using the bottom-up heapify algorithm. This step is only done once, so it doesn’t make our overall runtime worse than $$O(N \log N)$$ that we previously established. So, once the heap is created, sorting can be done in $$O(N \log N)$$.

### Out of scope Note #

Note, the runtime is not exactly as simple as $$N \log N$$ because later removals only need to sift through a smaller and smaller heap. (For more information, see this Stack Overflow post.)

HeapSort is not stable because the heap operations (recall bubbleUp and bubbleDown) can change the relative order of equal elements.

## New Idea: “Divide and Conquer” #

The first few sorting algorithms we’ve introduced work by iterating through each item in the collection one-by-one. With insertion sort and selection sort, both maintain a “sorted section” and an “unsorted section” and gradually sort the entire collection by moving elements over from the unsorted section into the sorted section. Another approach to sorting is by way of divide and conquer. Divide and conquer takes advantage of the fact that empty collections or one-element collections are already sorted. This essentially forms the base case for a recursive procedure that breaks the collection down to smaller pieces before merging adjacent pieces together to form the completely sorted collection.

The idea behind divide and conquer can be broken down into the following 3-step procedure.

1. Split the elements to be sorted into two collections.
2. Sort each collection recursively.
3. Combine the sorted collections.

Compared to selection sort, which involves comparing every element with every other element, divide and conquer can reduce the number of unnecessary comparisons between elements by sorting or enforcing order on sub-ranges of the full collection. The runtime advantage of divide and conquer comes largely from the fact that merging already-sorted sequences is very fast.

Two algorithms that apply this approach are merge sort and quicksort.

## Merge Sort #

Merge sort works by executing the following procedure until the base case of an empty or one-element collection is reached.

1. Split the collection to be sorted in half.
2. Recursively call merge sort on each half.
3. Merge the sorted half-lists.

The reason merge sort is fast is because merging two lists that are already sorted takes linear time proportional to the sum of the lengths of the two lists. In addition, splitting the collection in half requires a single pass through the elements. The processing pattern is depicted in the diagram below. Each level in the diagram is a collection of processes that all together run in linear time. Since there are $$2 \log N$$ levels with each level doing work proportional to $$N$$, the total time is proportional to $$N \log N$$.

Merge sort is stable as long as we make sure when merging two halves together that we favor equal elements in the left half.

Now, watch this video on mergeSort before attempting the exercise below!

## Exercise: mergeSort#

To test your understanding of merge sort, fill out the mergeSort method in DLList.java. Be sure to take advantage of the provided merge method!

This method should be non-destructive, so the original DLList should not be modified.

## Quicksort #

Another example of dividing and conquering is the quicksort algorithm, which proceeds as follows:

1. Split the collection to be sorted into three collections by partitioning around a pivot (or “divider”). One collection consists of elements smaller than the pivot, the second collection consists of elements equal to the pivot, and the third consists of elements greater than or equal to the pivot.
2. Recursively call quicksort on the each collection.
3. Merge the sorted collections by concatenation.

Specifically, this version of quicksort is called “three-way partitioning quicksort” due to the three partitions that the algorithm makes on every call.

Here’s an example of how this might work, sorting an array containing 3, 1, 4, 5, 9, 2, 8, 6. 1. Choose 3 as the pivot. (We’ll explore how to choose the pivot shortly.)
2. Put 4, 5, 9, 8, and 6 into the “large” collection and 1 and 2 into the “small” collection. No elements go in the “equal” collection.
3. Sort the large collection into 4, 5, 6, 8, 9; sort the small collection into 1, 2; combine the two collections with the pivot to get 1, 2, 3, 4, 5, 6, 8, 9.

Depending on the implementation, quicksort is not stable because when we move elements to the left and right of our pivot the relative ordering of equal elements can change.

Before moving on to the next part of the lab, check out this video to solidify your understanding of quicksort. Note this was taken from last years lecture, so you can stop after the section on quicksort. That is, you can stop at 1:41:00.

## Exercise: quicksort#

Some of the code is missing from the quicksort method in DLList.java. Fill in the function to complete the quicksort implementation.

Be sure to use the supplied helper methods, namely append and addLast! This method should be non-destructive, so the original DLList should not be modified.

## Discussion: Quicksort #

### Discussion 1: Runtime #

First, let’s consider the best-case scenario where each partition divides a range optimally in half. Using some of the strategies picked up from the merge sort analysis, we can determine that quicksort’s best case asymptotic runtime behavior is $$O(N \log N)$$. Discuss with your partner why this is the case, and any differences between quicksort’s best case runtime and merge sort’s runtime.

However, quicksort is faster in practice and tends to have better constant factors (which aren’t included in the big-Oh analysis). To see this, let’s examine exactly how quicksort works.

We know concatenation in a linked list can be done in constant time or linear time if it’s an array. Partitioning can be done in time proportional to the number of elements $$N$$. If the partitioning is optimal and splits each range more or less in half, we have a similar logarithmic division of levels downward like in merge sort. On each division, we still do the same linear amount of work as we need to decide whether each element is greater or less than the pivot.

However, once we’ve reached the base case, we don’t need as many steps to reassemble the sorted collection. Remember that with merge sort, while each list of one element is sorted with respect to itself, the entire set of one-element lists is not necessarily in order which is why there are $$\log N$$ steps to merge upwards in merge sort. This isn’t the case with quicksort as each element is in order. Thus, merging in quicksort is simply one level of linear-time concatenation.

Unlike merge sort, quicksort has a worst-case runtime different from its best-case runtime. Suppose we always choose the first element in a range as our pivot. Then, which of the following conditions would cause the worst-case runtime for quicksort? Discuss with your partner, and verify your understanding by highlighting the line below for the answer.

Sorted or Reverse Sorted Array. This is because the pivot will always be an extreme value (the largest or smallest unsorted value) and we will thus have N recursive calls, rather than log(n)..

What is the runtime of running quicksort on this array?

Theta(N^2).

Under these conditions, does this special case of quicksort remind you of any other sorting algorithm we’ve discussed in this lab? Discuss with your partner.

We see that quicksort’s worst case scenario is pretty bad… You might be wondering why we’d even bother with it then! However, though it’s outside the scope of this class for you to prove why, we can show that on average, quicksort has $$O(N \log(N))$$ runtime! In practice, quicksort ends up being very fast.

### Discussion 2: Choosing a Pivot #

Given a random collection of integers, what’s the best possible choice of pivot for quicksort that will break the problem down into $$\log N$$ levels? Discuss with your partner and describe an algorithm to find this pivot element. What is its runtime? It’s okay if you think your solution isn’t the most efficient.

## Quicksort in Practice #

How fast was the pivot-finding algorithm that you came up with? Finding the exact median of our elements may take so much time that it may not help the overall runtime of quicksort at all. It may be worth it to choose an approximate median, if we can do so really quickly. Options include picking a random element, or picking the median of the first, middle, and last elements. These will at least avoid the worst case we discussed above.

In practice, quicksort turns out to be the fastest of the general-purpose sorting algorithms we have covered so far. For example, it tends to have better constant factors than that of merge sort. For this reason, Java uses this algorithm for sorting arrays of primitive types, such as ints or floats. With some tuning, the most likely worst-case scenarios are avoided, and the average case performance is excellent.

Here are some improvements to the quicksort algorithm as implemented in the Java standard library:

• When there are only a few items in a sub-collection (near the base case of the recursion), insertion sort is used instead.
• For larger arrays, more effort is expended on finding a good pivot.
• Various machine-dependent methods are used to optimize the partitioning algorithm and the swap operation.
• Dual pivots

For object types, however, Java uses a hybrid of merge sort and insertion sort called “Timsort” instead of quicksort. Can you come up with an explanation as to why? Hint: Think about stability!

## Conclusion #

To put together the pieces we saw earlier, watch this video Quicksort versus Mergesort

### Summary #

In this lab, we learned about different comparison-based algorithms for sorting collections. Within comparison-based algorithms, we examined two different paradigms for sorting:

1. Simple sorts like insertion sort and selection sort which demonstrated algorithms that maintained a sorted section and moved unsorted elements into this sorted section one-by-one. With optimization like heapsort or the right conditions (relatively sorted list in the case of insertion sort), these simple sorts can be fast!
2. Divide and conquer sorts like merge sort and quicksort. These algorithms take a different approach to sorting: we instead take advantage of the fact that collections of one element are sorted with respect to themselves. Using recursive procedures, we can break larger sorting problems into smaller subsequences that can be sorted individually and quickly recombined to produce a sorting of the original collection.

Here are several online resources for visualizing sorting algorithms. If you’re having trouble understanding these sorts, use these resources as tools to help build intuition about how each sort works.

To summarize the sorts that we’ve learned, take a look at the following table:

Best Case Runtime Worst Case Runtime Stable In Place Notes
Insertion Sort $$\Theta(N)$$ $$\Theta(N^2)$$ Yes Yes
Selection Sort $$\Theta(N^2)$$ $$\Theta(N^2)$$ No Yes Can be made stable under certain conditions.
Heap Sort $$\Theta(N \log N)$$ $$\Theta(N \log N)$$ No Yes If all elements are equal then runtime is $$\Theta(N)$$. Hard to make stable.
Merge Sort $$\Theta(N \log N)$$ $$\Theta(N \log N)$$ Yes Not usually. Typical implementations are not, and making it in-place is terribly complicated. An optimized sort called “Timsort” is used by Java for arrays of reference types.
Quicksort $$\Theta(N \log N)$$ $$\Theta(N^2)$$ Depends Most implementations use log(N) additional space for the recursive stack frames Stability and runtime depends on partitioning strategy; three-way partition quicksort is stable. If all elements are equal, then the runtime using three-way partition quicksort is $$\Theta(N)$$. Used by Java for arrays of primitive types. Fastest in practice.

You may have noticed that there seems to be a lower bound on how fast our sorting algorithms can go. For comparison based sorts, we can prove the best we can do is $$O(N\log(N))$$. You can watch a very brief video explanation here at timestamp 11:42. You can also read a more in depth proof, if you’re into that kind of thing. Tomorrow, we’ll learn about counting sorts, which can do even better when we’re able to use them.

### Deliverables #

To get credit for this lab:

• Complete the following methods in DLList.java:
• insertionSortHelper
• mergeSort
• quicksort