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Lecture Overview

  • Recipes for time complexity analysis
  • Testing checklist

Time Complexity Analysis Recipes

Simple Statements

  • Most are O(1)
  • Some can be more expensive: method and function calls, object constructors.
    1. If the method is a standard one (get on LinkedList), lookup up the time complexity.
    2. If you have access to the code for the method, analyze the code.
    3. If all else fails, do empirical tests that measure the time for different input sizes.

For many calls on the same line, add their time complexity. (See the next item.)

Composing Statements Sequentially

Recipe:

time_of_sequence = time_of_first + time_of_second + ...

Example:

public static void flood(WaterColor color,
                        LinkedList<Coord> flooded_list,
                        Tile[][] tiles, Integer board_size) {
    HashSet<Coord> flooded = new HashSet<>(flooded_list);   // O(n)
    for (int i = 0; i != flooded_list.size(); ++i) { ... }  // O(n^2)
}

The construction of the flooded HashSet is O(n). The for loop is O(n^2). Adding those together yields O(n^2).

If-Then-Else Statements

Recipe:

time_of_if_then_else = time_of_condition + time_of_then + time_of_else

Example:

static boolean is_rotated(int[] A_orig, int[] A_new) {
    if (A_orig.length < 2) {
        return Arrays.equals(A_orig, A_new);
    } else {
        boolean result = A_new[0] == A_orig[A_orig.length - 1];
        for (int i = 0; i != A_orig.length - 1; ++i) {
            result = result && (A_orig[i] == A_new[i + 1]);
        }
        return result;
    }
}

The time of the condition A_orig.length < 2 is O(1). The time of the then-branch Arrays.equals(A_orig, A_new) is O(n). The time of the else-branch is O(n) (we shall discuss loops below).

So the time of the if-then-else is O(1) + O(n) + O(n) = O(n)

If-Then Statements

Recipe:

time_of_if_then = time_of_condition + time_of_then

Loops

Recipe:

time_of_loop = number_iterations * time_of_body

Example:

for (int i = 0; i != A_orig.length - 1; ++i) {
    result = result && (A_orig[i] == A_new[i + 1]);
}

number of iterations is O(n) time of body is O(1) time of for loop is O(n) * O(1) = O(n)

Example (nested loops):

int i = n;
while (i > 0) {
    for (int j = 0; j < n; j++)
        System.out.println("*");
    i = i / 2;
}

number of iterations of while loop is O(log(n)) time of body of while is the sum of time of for loop: ? the statement i = i / 2;: O(1)

number of iterations of for loop is O(n) time of body of for is O(1) time of for loop is O(n) * O(1) = O(n)

time of while loop is O(log(n)) * O(n) = O(n log(n))

Recursive Functions

Recipe:

time_of_function = recursion_depth * time_per_level

To determine recursion_depth, analyze how much the input size changes in the recursive calls.

  • If size gets smaller by 1 (or any constant amount), recursion depth is O(n).
  • If size get divided in half (or more), recursion depth is O(log n).

To determine the time_per_level:

time_per_level = number_calls_per_level (not big-O) * time_per_call

To determine number_calls occuring within each level, one needs to analyze the code to see how many recursive calls can be spawned by one call to the recursive function.

To determine time_per_call, analyze the time complexity of one call to the recursive function, ignoring the recursive calls.

One complication is that the number of calls in each level often changes (e.g. doubling at each level: 1, 2, 4, 8, ..., 2^d) and the time_per_call also changes because the input size gets smaller (e.g. cut in half at each level: n, n/2, n/4, ..., 1), but in many examples, these two affects cancel out and the time per level stays the same. (If not, one needs to use a more advanced analysis called the Master Theorem.)

Example:

static Node append(Node N1, Node N2) {
    if (N1 == null)
        return N2;
    else {
        return new Node(N1.data, append(N1.next, N2));
    }
}

analysis:

  • recursion_depth = O(n), input size reduced by one: append(N1.next, N2)

  • number_calls_per_level = 1 (only one call to append)

  • time_per_call = O(1) (allocate one node)

  • time_per_level = number_calls_per_level * time_per_call = O(1)

  • time_of_function = recursion_depth * time_per_level = O(n) * O(1) = O(n)

Example:

Assume the tree is AVL, so it is balanced.

protected Node<K> find(K key, Node<K> curr, Node<K> parent) {
    if (curr == null)
        return parent;
    else if (lessThan.test(key, curr.data))
        return find(key, curr.left, curr);
    else if (lessThan.test(curr.data, key))
        return find(key, curr.right, curr);
    else
        return curr;
}

analysis:

  • recursion_depth = O(log(n)), input size cut in half: find(key, curr.left, curr)

  • number_calls_per_level = 1 (only one call to find)

  • time_per_call = O(1) (assuming call to test is O(1))

  • time_per_level = number_calls_per_level * time_per_call = O(1)

  • time_of_function = recursion_depth * time_per_level = O(log(n)) * O(1) = O(log(n))

Example:

static Node merge_sort(Node N) {
    if (N == null || N.next == null) {
        return N;
    } else {
        int n = Utils.length(N);
        Node left = merge_sort(Utils.take(N, n / 2));
        Node right = merge_sort(Utils.drop(N, n / 2));
        return merge(left, right);
    }
}

analysis:

  • recursion_depth = O(log(n)), input size cut in half: merge_sort(Utils.take(N, n / 2)

  • number_calls_per_level = 1, 2, 4, 8, ... doubling because there are 2 recursive calls to merge_sort

  • time_per_call = n, n/2, n/4, n/8, ... (Utils.length + Utils.take + Utils.drop + merge)

  • time_per_level = number_calls_per_level * time_per_call = O(n)

  • time_of_function = recursion_depth * time_per_level = O(log(n)) * O(n) = O(n log(n))

Testing Checklist

  1. common cases (find in lecture, textbook, internet)
  2. corner cases
  3. code coverage -> more cases
  4. big, randomized input

Let's go through some of the autograder's tests for the MergeSortList and NextPrevBinaryTree labs.

Common Case Test for merge

@Test
public void merge_small() {
    int[] A = {1,3,4,6,8,8,12,13};
    int[] B = {2,4,5,7,9,11,14};
    Node N1 = Utils.array_to_list(A);
    Node N2 = Utils.array_to_list(B);
    Node M = MergeSort.merge(N1, N2);
    assertArrayEquals(A, Utils.list_to_array(N1));
    assertArrayEquals(B, Utils.list_to_array(N2));
    assertTrue(Utils.is_sorted(M));
    assertTrue(Utils.is_permutation(M, Utils.append(N1, N2)));
}

Common Case Test for merge_in_place

@Test
public void merge_in_place() {
    int[] A = {1,3,4,6,8,8,12,13};
    int[] B = {2,4,5,7,9,11,14};
    Node N1 = Utils.array_to_list(A);
    Node N2 = Utils.array_to_list(B);
    Node M = MergeSort.merge_in_place(N1, N2);
    assertTrue(Utils.is_sorted(M));
    assertTrue(Utils.is_permutation(M, Utils.append(Utils.array_to_list(A),
                                             Utils.array_to_list(B))));
    // Check that the merge was done in place.
    assertTrue(Utils.equals(M, N1));
}

Big, Randomized Test for merge sort

@Test
public void sort_big() {
    Random r = new Random(0);
    for (int n = 0; n != 100; ++n) {
        int[] A = new int[n];
        for (int i = 0; i != n; ++i)
            A[i] = r.nextInt(50);
        Node N = Utils.array_to_list(A);
        Arrays.sort(A);
        N = MergeSort.sort(N);
        int[] B = Utils.list_to_array(N);
        assertArrayEquals(A, B);
    }
}

Common and Big Tests for BinaryTree Iterator

private void test_next_helper(ArrayList<Integer> expected,
                              BinaryTree<Integer> T) {
    int j = 0;
    for (Iterator<Integer> i = T.begin(); !i.equals(T.end()); i.advance()) {
        assertTrue(j != expected.size()); // to catch error in Iter.equals
        assertEquals(expected.get(j).intValue(), i.get().intValue());
        ++j;
    }
}

public void test_advance() throws Exception {
    Integer expected[] = {2, 5, 5, 6, 7, 8};
    test_next_helper(new ArrayList<Integer>(Arrays.asList(expected)), T);
}

@Test
public void test_advance_big() throws Exception {
    Random r = new Random(0);
    for (int n = 0; n != 100; ++n) {
        ArrayList<Integer> expected = new ArrayList<>();
        for (int i = 0; i != n; ++i)
            expected.add(r.nextInt(100));
        BinaryTree<Integer> bigT = new BinaryTree<>(expected);
        test_next_helper(expected, bigT);
    }
}