1. Analysis of Algorithms CS 477/677 Final Exam Review Instructor: George Bebis

2. 2 The Heap Data Structure Def: A heap is a nearly complete binary tree with the following two properties: –Structural property: all levels are full, except possibly the last one, which is filled from left to right –Order (heap) property: for any node x Parent(x) ≥ x Heap 5 7 8 4 2

3. 3 Array Representation of Heaps A heap can be stored as an array A. –Root of tree is A[1] –Parent of A[i] = A[  i/2  ] –Left child of A[i] = A[2i] –Right child of A[i] = A[2i + 1] –Heapsize[A] ≤ length[A] The elements in the subarray A[(  n/2  +1).. n] are leaves The root is the max/min element of the heap A heap is a binary tree that is filled in order

4. 4 Operations on Heaps (useful for sorting and priority queues) –MAX-HEAPIFY O(lgn) –BUILD-MAX-HEAP O(n) –HEAP-SORT O(nlgn) –MAX-HEAP-INSERT O(lgn) –HEAP-EXTRACT-MAX O(lgn) –HEAP-INCREASE-KEY O(lgn) –HEAP-MAXIMUM O(1) –You should be able to show how these algorithms perform on a given heap, and tell their running time

5. 5 Lower Bound for Comparison Sorts Theorem: Any comparison sort algorithm requires  (nlgn) comparisons in the worst case. Proof: How many leaves does the tree have? –At least n! (each of the n! permutations if the input appears as some leaf)  n! –At most 2 h leaves  n! ≤ 2 h  h ≥ lg(n!) =  (nlgn) h leaves

6. 6 Linear Time Sorting Any comparison sort will take at least nlgn to sort an array of n numbers We can achieve a better running time for sorting if we can make certain assumptions on the input data: –Counting sort: each of the n input elements is an integer in the range [0, r] and r=O(n) –Radix sort: the elements in the input are integers represented as d-digit numbers in some base-k where d=Θ(1) and k =O(n) –Bucket sort: the numbers in the input are uniformly distributed over the interval [0, 1)

7. 7 Analysis of Counting Sort Alg.: COUNTING-SORT(A, B, n, k) 1.for i ← 0 to r 2. do C[ i ] ← 0 3.for j ← 1 to n 4. do C[A[ j ]] ← C[A[ j ]] + 1 5. C[i] contains the number of elements equal to i 6.for i ← 1 to r 7. do C[ i ] ← C[ i ] + C[i -1] 8. C[i] contains the number of elements ≤ i 9.for j ← n downto 1 10. do B[C[A[ j ]]] ← A[ j ] 11. C[A[ j ]] ← C[A[ j ]] - 1  (r)  (n)  (r)  (n) Overall time:  (n + r)

8. 8 RADIX-SORT Alg.: RADIX-SORT (A, d) for i ← 1 to d do use a stable sort to sort array A on digit i 1 is the lowest order digit, d is the highest-order digit  (d(n+k))

9. 9 Analysis of Bucket Sort Alg.: BUCKET-SORT(A, n) for i ← 1 to n do insert A[i] into list B[  nA[i]  ] for i ← 0 to n - 1 do sort list B[i] with quicksort sort concatenate lists B[0], B[1],..., B[n -1] together in order return the concatenated lists O(n)  (n) O(n)  (n)

10. 10 Hash Tables Direct addressing (advantages/disadvantages) Hashing –Use a function h to compute the slot for each key –Store the element (or a pointer to it) in slot h(k) Advantages of hashing –Can reduce storage requirements to ( |K| ) –Can still get O(1) search time in the average case

11. 11 Hashing with Chaining How is the main idea? Practical issues? Analysis of INSERT, DELETE Analysis of SEARCH –Worst case –Average case (both successful and unsuccessful)

12. 12 Designing Hash Functions The division method h(k) = k mod m The multiplication method h(k) =  m (k A mod 1)  Universal hashing –Select a hash function at random, from a carefully designed class of functions Advantage: fast, requires only one operation Disadvantage: certain values of m give are bad (powers of 2) Disadvantage: Slower than division method Advantage: Value of m is not critical: typically 2 p Advantage: provides good results on average, independently of the keys to be stored

13. 13 Open Addressing Main idea Different implementations –Linear probing –Quadratic probing –Double hashing Know how each one of them works and their main advantages/disadvantages –How do you insert/delete? –How do you search? –Analysis of searching

14. 14 Binary Search Tree Tree representation: –A linked data structure in which each node is an object Binary search tree property: –If y is in left subtree of x, then key [y] ≤ key [x] –If y is in right subtree of x, then key [y] ≥ key [x] 2 3 5 5 7 9

15. 15 Operations on Binary Search Trees –SEARCH O(h) –PREDECESSOR O(h) –SUCCESOR O(h) –MINIMUM O(h) –MAXIMUM O(h) –INSERT O(h) –DELETE O(h) –You should be able to show how these algorithms perform on a given binary search tree, and tell their running time

16. 16 Red-Black-Trees Properties Binary search trees with additional properties: 1.Every node is either red or black 2.The root is black 3.Every leaf ( NIL ) is black 4.If a node is red, then both its children are black 5.For each node, all paths from the node to descendant leaves contain the same number of black nodes

17. 17 Properties of Red-Black-Trees Any node with height h has black-height ≥ h/2 The subtree rooted at any node x contains at least 2 bh(x) - 1 internal nodes No path is more than twice as long as any other path  the tree is balanced –Longest path: h <= 2bh(root) –Shortest path: bh(root)

18. 18 Upper bound on the height of Red-Black-Trees Lemma: A red-black tree with n internal nodes has height at most 2lg(n + 1). Proof: n Add 1 to both sides and then take logs: n + 1 ≥ 2 b ≥ 2 h/2 lg(n + 1) ≥ h/2  h ≤ 2 lg(n + 1) root l r height(root) = h bh(root) = b number n of internal nodes ≥ 2 b - 1 ≥ 2 h/2 - 1 since b  h/2

19. 19 Operations on Red-Black Trees –SEARCH O(h) –PREDECESSOR O(h) –SUCCESOR O(h) –MINIMUM O(h) –MAXIMUM O(h) –INSERT O(h) –DELETE O(h) Red-black-trees guarantee that the height of the tree will be O(lgn) You should be able to show how these algorithms perform on a given red-black tree (except for delete), and tell their running time

20. 20 Adj. List - Adj. Matrix Comparison ComparisonBetter Faster to test if (x, y) exists? Faster to find vertex degree? Less memory on sparse graphs? Faster to traverse the graph? matrices lists lists (m+n) vs. n 2 Adjacency list representation is better for most applications Graph representation: adjacency list, adjacency matrix

21. 21 Minimum Spanning Trees Given: A connected, undirected, weighted graph G = (V, E) A minimum spanning tree: 1.T connects all vertices 2.w(T) = Σ (u,v)  T w(u, v) is minimized a bcd e ggf i 4 8 7 8 11 1 2 7 2 4 14 9 10 6

22. 22 Correctness of MST Algorithms (Prim’s and Kruskal’s) Let A be a subset of some MST (i.e., T), (S, V - S) be a cut that respects A, and (u, v) be a light edge crossing (S, V-S). Then (u, v) is safe for A. Proof: Let T be an MST that includes A –edges in A are shaded Case1: If T includes (u,v), then it would be safe for A Case2: Suppose T does not include the edge (u, v) Idea: construct another MST T’ that includes A  { (u, v) } u v S V - S

23. 23 PRIM( V, E, w, r ) 1. Q ←  2. for each u  V 3. do key[u] ← ∞ 4. π[u] ← NIL 5. INSERT( Q, u ) 6. DECREASE-KEY( Q, r, 0 ) ► key[r] ← 0 7. while Q   8. do u ← EXTRACT-MIN( Q ) 9. for each v  Adj[u] 10. do if v  Q and w(u, v) < key[v] 11. then π[v] ← u 12. DECREASE-KEY( Q, v, w(u, v) ) O(V) if Q is implemented as a min-heap Executed |V| times Takes O(lgV) Min-heap operations: O(VlgV) Executed O(E) times Constant Takes O(lgV) O(ElgV) Total time: O(VlgV + ElgV) = O(ElgV) O(lgV)

24. 24 1.A ←  2.for each vertex v  V 3. do MAKE-SET( v ) 4.sort E into non-decreasing order by w 5.for each (u, v) taken from the sorted list 6. do if FIND-SET( u )  FIND-SET( v ) 7. then A ← A  { (u, v) } 8. UNION( u, v ) 9.return A Running time: O(V+ ElgE+ElgV )=O( ElgE ) – dependent on the implementation of the disjoint-set data structure KRUSKAL( V, E, w ) O(V) O(ElgE) O(E) O(lgV)

25. 25 Shortest Paths Problem Variants of shortest paths problem Effect of negative weights/cycles Notation –d[v]: estimate –δ(s, v): shortest-path weight Properties –Optimal substructure theorem –Triangle inequality –Upper-bound property –Convergence property –Path relaxation property

26. 26 Relaxation Relaxing an edge (u, v) = testing whether we can improve the shortest path to v found so far by going through u If d[v] > d[u] + w(u, v) we can improve the shortest path to v  update d[v] and  [v] 59 2 uv 57 2 uv RELAX(u, v, w) 56 2 uv 56 2 uv After relaxation: d[v]  d[u] + w(u, v)

27. 27 Single Source Shortest Paths Bellman-Ford Algorithm –Allows negative edge weights –TRUE if no negative-weight cycles are reachable from the source s and FALSE otherwise –Traverse all the edges |V – 1| times, every time performing a relaxation step of each edge Dijkstra’s Algorithm –No negative-weight edges –Repeatedly select a vertex with the minimum shortest-path estimate d[v] – uses a queue, in which keys are d[v]

28. 28 BELLMAN-FORD( V, E, w, s ) 1. INITIALIZE-SINGLE-SOURCE(V, s) 2. for i ← 1 to |V| - 1 3. do for each edge (u, v)  E 4. do RELAX(u, v, w) 5. for each edge (u, v)  E 6. do if d[v] > d[u] + w(u, v) 7. then return FALSE 8. return TRUE Running time: O(V+VE+E)=O(VE)  (V) O(V) O(E) O(VE)

29. 29 Dijkstra (G, w, s) 1. INITIALIZE-SINGLE-SOURCE( V, s ) 2. S ←  3. Q ← V[G] 4. while Q   5. do u ← EXTRACT-MIN(Q) 6. S ← S  { u } 7. for each vertex v  Adj[u] 8. do RELAX( u, v, w ) 9. Update Q (DECREASE_KEY) Running time: O(VlgV + ElgV) = O(ElgV)  (V) O(V) build min-heap Executed O(V) times O(lgV) O(E) times (total) O(lgV) O(VlgV) O(ElgV)

30. 30 Correctness Bellman-Ford’s Algorithm: Show that d[v]= δ (s, v), for every v, after |V-1| passes. Dijkstra’s Algorithm: For each vertex u  V, we have d[u] = δ(s, u) at the time when u is added to S.

31. 31 NP-completeness Algorithmic vs Problem Complexity Class of “P” problems Tractable/Intractable/Unsolvable problems NP algorithms and NP problems P=NP ? Reductions and their implication NP-completeness and examples of problems How do we prove a problem NP-complete? Satisfiability problem and its variations