- We discussed two implementations for a graph: adjacency list and
adjacenty matrix. Fill in the table below to give the worst-case
big-O complexity (in
terms of |V| and |E|)
of each operation for each implementation.
**Operation****Adjacency list****Adjacency matrix**Is v1 adjacent to v2? O(|V|) O(|1|) For a single vertex, find all vertices adjacent to that vertex O(|V|) O(|V|) For each (every) vertex, find all vertices adjacent to that vertex O(|V|+|E|) O(|V| ^{2})Total space required. O(|V|+|E|) O(|V| ^{2}) - Below is the generic code we discussed for a graph traversal.
If the collection is a stack, it's a DFS; if it's a queue, it's a BFS.
In terms of |V| and |E|, what is the big-O
complexity of this code assuming the graph is represented with
- an adjacency matrix?
- an adjacency list?

O(|V|) -- mark all vertices as unvisited O(1) -- insert the start vertex into an empty collection C O(|V|) iterations -- while (C not empty) O(1) - remove a vertex v from C O(1) - if v has not been visited O(1) * visit v O(|V|)(adj matrix) * insert all univisted adjacent nodes to v into C O(|E|/|V|)(adj list)

Total complexity is O(|V|^{2}) for an adjacency matrix, O(|V|+|E|) for an adjacency list. - In big-O, how long does it take in the worst case to build an N-element
heap top-down, that is, by repeated
insertion? Explain.
**Answer:**O(NlogN). N elements must be inserted, which involves adding the element as the last leaf (O(1)), then reheapifying (bubbling it up to where it really goes). If every element had to traverse the entire height of the N-element tree during reheapification, the complexity would be O(N)(the number of elements) * O(log N) the height of the tree. Of course, not all elements have to go that far; only the elements inserted at the leaves do. But there are N/2 elements at the leaves, so this gives N/2 * O(log N) which is still O(NlogN) for the total complexity. - In big-O, how long does it take in the worst case to build an N-element heap bottom-up, that is,
starting at the leaves? Explain.
**Answer:**O(N). First the algorithm: Assume we want to build a max heap. First insert all elements, in arbitrary order, into the heap (array). Then:- Start at the bottom: Note that each node on the bottom level is a leaf, and is itself a single-element heap.
- Go up one level: each node at that level is either a leaf OR the root of a tree with one or two children. Each of these either already is a heap or can be made into a heap by exchanging the root with the largest child if that child is larger than the parent.
- Go up one level: each node at that level is the root of a tree whose subtrees are both heaps. So each of those trees can be made into a heap by bubbling the root down until it gets to where it belongs.
- Repeat this, going up one level at a time and making sure that all the subtrees rooted at that level are heaps, until you get to the root. When you have done this for the root, it's a heap.

0*N/2 + 1*N/4 + 2*N/8 + 3*N/16 + 4*N/32 + ... + (logN-1)*N/N = N(1/4 + 2/8 + 3/16 + 4/32 + ... + (logN-1)/N) Although this probably is not a familiar sequence for you, it is easy to believe that it sums to a constant, giving a complexity of O(N).

- Suppose you wanted to remove an arbitrary element from an N-element heap.
- Devise an algorithm for this that includes finding the element to remove.
- What is the worst-case big-O complexity of this algorithm? Explain.

**Algorithm:**-- Do a linear search to find the element in the heap (O(N)) -- Replace that element with the last element in the heap (O(1)) -- Bubble the new element up or down, as necessary, to put it in place (O(logN))

Total complexity: O(N) + O(1) + O(logN) = O(N). - Suppose you want to build a map containing N elements and plan to use
a hash table (with bucket hashing) to implement it. John says the lookup
method is O(1); Mary says it is O(N). Who is correct? Explain.
**Answer:**If the elements might be unevenly distributed, it's O(N). (In the worst case they're all in a single linked list in the same bucket.) If the elements are evenly distributed, it depends on the size of the hash table. If the hash table is of fixed size S, then on average each bucket has N/S elements, which is O(N). However, if the size of the hash tables varies with N, so it is kN (where k would usually be less than 1), then on average each bucket has N/(kN) = 1/k elements, which is a constant.