Amortized Analysis
1 Flexible Arrays
2 Expanding Flexible Arrays
3 Shrinking Flexible Arrays
4 Cascading Arrays
5 Handing In

Amortized Analysis

For this assignment, write all time complexities in the form \(O([k \rightarrow k])\), define what each of your variables represent, and do not use \(n\) as a variable. Some examples of incorrect time complexities are: \([k \rightarrow k]\), \(O[k \rightarrow k]\), or \(O([k])\). Be sure to explain your reasoning behind each answer.

1 Flexible Arrays

Suppose we want an ordered, variable-size data structure that gives quick access to all elements. Specifically, we need to be able to:
  • insert elements at one end.

  • get the kth element.

In Pyret, lists support insertion (and deletion) at one end in constant time, but getting the \(k^{th}\) element takes \(O([k \rightarrow k])\) time. This means lists alone won’t be sufficient.

Arrays (known as vectors in some languages) support accessing elements in constant time, but are fixed in size. However, the fixed size of arrays is a major limitation. Once an array is full, we can no longer insert elements.

To solve this problem, we start out with a vector of a certain size (assume a starting size of 1) and replace it with a bigger array when it is full. Here is a pseudocode description of this “flexible array”:


    currentArray = new array[1]

    nextUnusedIndex = size of currentArray



    if nextUnusedIndex > size of currentArray:

        newArray = new array[size of currentArray + 1]

        copy each element of currentArray to newArray

        nextUnusedIndex = size of currentArray + 1

        currentArray = newArray



        currentArray[nextUnusedIndex] = elem

        nextUnusedIndex = nextUnusedIndex + 1



    get kth element of currentArray

Answer the following questions:
  1. What is the worst case time complexity of makeFlexibleArray? insertAtEnd? get?

  2. Is the amortized complexity of a sequence of \(k\) insertAtEnd operations any better? Assume that you start with the array of size 1.

2 Expanding Flexible Arrays

Fortunately, there is a way to improve the amortized cost per operation of the flexible array. We change insertAtEnd so that the array doubles in size when it is full.

makeFlexibleArray():  -- unchanged

get(k): -- unchanged


    ONLY change is

        newArray = new array[size of currentArray + 1]

    is replaced by

        newArray = new array[size of currentArray * 2]

Answer the following question:
  1. Assume that you start with an empty array. What is the amortized cost per operation of \(k\) insertAtEnd operations?

3 Shrinking Flexible Arrays

Suppose we want to extend our expanding flexible array to also support removing elements at the end. Specifically, we add the following operation:


    delete last element of currentArray

To conserve memory, we may also want to shrink the array if it gets too empty. Here are two choices:
  • When the array is half-full, shrink its size to half of what it was.

  • When the array is one-fourth full, shrink its size to half of what it was.

Shrinking the array takes constant time. But you should assume that it notifies the operating system that it is taking less space, so the system is free to allocate the space beyond the end of the shrunk region. That is, future allocation must again copy all elements.

Answer the following question:

  1. Which option is more efficient, and why? Give an example of a sequence of insertions and deletions with the corresponding amortized analysis to support your claim.

4 Cascading Arrays

Suppose we need a slightly different functionality from our data structure. Instead of looking up the \(k^{th}\) element, we need to be able to search for elements. Specifically, we need to be able to:
  • Insert an element.

  • Determine whether a given element is in the structure.

One way to do this is to keep an array of elements in sorted order. Constant time access allows us to quickly search for elements using a binary search in \(O([k \rightarrow log k])\) time, but inserting elements is slow, \(O([k \rightarrow k])\), due to the requirement that the array remain sorted: when we add an element, all the elements after it in order must be shifted one array index to the right.

Here is a better design for a data structure that provides this functionality: Keep a list of arrays, numbered \(0\) through \(n\), where array \(k\) is of size \(2^k\). We will maintain the invariant that all arrays are either empty or full, and all full arrays are in sorted order. For example:

Array 0: [3]

Array 1: [1, 6]

Array 2: [_, _, _, _]

Array 3: [2, 5, 7, 8, 10, 12, 20, 42]

To insert an element, we first create an array of size \(1\) containing only that element. Then we iterate through the list of arrays, starting with the smallest. If array \(k\) is empty, we replace it with the full new array and we are done. Otherwise, we merge array \(k\) into the new array, doubling its size. Then we set array \(k\) to be empty and move on to array \(k + 1\).

To insert the element 4 into the example above, we would create the array [4], then merge [4] with [3] to get [3, 4], then merge that with [1, 6] to get [1, 3, 4, 6], and finally place [1, 3, 4, 6] in position 2, resulting in the following configuration:

Array 0: [_]

Array 1: [_, _]

Array 2: [1, 3, 4, 6]

Array 3: [2, 5, 7, 8, 10, 12, 20, 42]

The pseudocode looks like this:


    listOfArrays = empty


    foreach array in listOfArrays:

        binary search for elem in array


    newArray = [elem]

    foreach array in listOfArrays:

        if array is empty:

            replace array with newArray in listOfArrays



            newArray = merge array with newArray

            array = empty

    add newArray to end of listOfArrays

Answer the following questions:

  1. What is the time complexity of search?

  2. Assume that you start with an empty array. What is the amortized cost per operation of \(s\) insert operations?

  3. Can you think of a data structure you already know that provides this functionality with a better time complexity? Naming the structure is fine; we don’t expect a detailed response.

5 Handing In

Please turn in a hard copy of your answers for each of the questions, cleanly broken up and labelled by section. You can turn in your homework into the cs019 handin box on the second floor of the CIT, just outside the Fishbowl. Please do not hand-write your analyses unless your writing is extraordinarily legible; the burden of legibility is on you, not our responsibility. Please submit on letter-sized paper with all mathematical content formatted appropriately. Put your name and login at the top of each page.