CSE 5095 Topics in Big Data Analytics Spring 2014; Homework 1 Solutions

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1 CSE 5095 Topics in Big Data Analytics Spring 2014; Homework 1 Solutions Note: Solutions to problems 4, 5, and 6 are due to Marius Nicolae. 1. Consider the following algorithm: for i := 1 to α n log e n do Pick a random j [1, n]; If a[j] = a[j + 1] or a[j] = a[j 1] then output: Type II and quit; Output: Type I ; Analysis: Note that if the array is of type I, the above algorithm will never give an incorrect answer. probability of an incorrect answer as follows. Thus assume that the array is of type II. We ll calculate the Probability of coming up with the correct answer in one iteration of the for loop is n n = 1 n. Thus, probability of failure in any iteration is 1 1 n. As a consequence, ( ) q probability of failure in q successive iterations is 1 1 n exp( q/ n) (using the fact that (1 1/x) x 1/e for any x > 0). This probability will be n α when q α n log e n. Thus the output of this algorithm is correct with high probability. 2. The algorithm runs in phases. In each phase we eliminate a constant fraction of the input keys that cannot be the element of interest. When the number of remaining keys is n, one of the processors performs an appropriate selection and outputs the right element. To with all the keys are alive. In any phase of the algorithm let N stand for the number of alive keys at the ning of the phase. At the ning of the first phase, N = n. Consider a phase where the number of alive keys is N at the ning of the phase. Let Y be the collection of alive keys. We employ N processors in this phase. Partition the N keys into N parts with N keys in each part. Each processor is assigned a part. Each processor in parallel finds the median of its keys in O( N) time. Let M 1, M 2,..., M N be these group medians. One of the processors finds the median M of these N group medians. This will take O( N) time. Now partition Y into Y 1 and Y 2, where Y 1 = {q Y q < M} and Y 2 = {q Y q > M}. There are 3 cases to consider: Case 1: If Y 1 = i 1, M is the element of interest. In this case, we output 1

2 M and quit. Case 2: If Y 1 i, Y 1 will constitute the alive keys for the next phase. Case 3: If the above two cases do not hold, Y 2 will constitute the collection of alive keys for the next phase. In this case we set i := i Y 1 1. In cases 2 and 3 we can perform the partitions using a prefix computation that can be done in O( N) time using N processors. It is easy to see that Y 1 N and Y 4 2 N. As a result, it follows that the number of 4 alive keys at the end of this phase is 3N. 4 ( N ) Thus we infer that the run time of the algorithm is O + (3/4)N + (3/4)2 N +... = O( N). 3. If we employ k-way merge where k = cm/b, the height of the merge tree will be log(n/b). However, in the worst case we may have to do c passes through the data at log(cm/b) each level of the tree, since we can only keep B/c keys of each run. Thus the worst case number of I/O passes needed is 1 + c log(n/m). log(cm/b) 4. If a leaf can store more keys, insertion happens in a similar way, we just have to redefine what it means that a node is full. Node u is full if it s an internal node with 2t 1 children or if it s a leaf with 4t 3 keys. Algorithm 1: IsFull(u) Data: u: a B-Tree node; Result: True if node u is full, False otherwise; return (leaf u AND n u == 4t 3) OR (NOT leaf u AND n u == 2t 1); Also, for simplicity, we will always make the root to be non-leaf. The other thing to modify is how to split a full leaf. A full leaf, which has 4t 3 keys, will be split into two leafs with 2t 2 keys each. The middle key from the original leaf moves up and becomes a key in the parent node. Let SPLIT NODE be the algorithm discussed in class for splitting a full node. The following algorithm will split a node, taking into account splitting full leafs: 2

3 Algorithm 2: SplitNode(p, i, u) Data: p, u: two nodes such that p =parent(u) and u is the i-th child of p; Result: Splits the node u into two nodes; if leaf u then Create node u ; Copy last 2t 2 keys of u to u ; Insert key k2t 1 u as the i-th key of p; Insert u as the i + 1-th child of p; Remove last 2t 1 keys from u; else SPLIT NODE(p, i, u); The insertion algorithm is then the following: Algorithm 3: Insert(T, k) Data: T : a B-Tree; k: a key; Result: Inserts key k into T ; r :=root(t ); if isfull(r) then Create a new node s; n s := 0; leaf s :=False; c s 1 := r; SplitNode(s, 1, r); root(t ) := s; r := s; InsertNonFull(r, k); 3

4 Algorithm 4: InsertNonFull(u, k) Data: u: a non full B-Tree node; k: a key; Result: Inserts key k into the subtree rooted at u; if leaf u then Insert k at the right place; else Choose i s.t. ki 1 u k < ki u ; if IsFull(c u i ) then SplitNode(u, i, c u i ); Update i s.t. ki 1 u k < ki u ; InsertNonFull(c u i, k); 5. Dijkstra s algorithm can be described as follows: Algorithm 5: Dijkstra(V, E, s) Data: (V, E): a graph; s: a source node; let w(u, v) be the weight of edge (u, v); Result: array d where d u is the length of the shortest path from s to u; for u in V do d u := ; d s := 0; Create a priority queue Q to store pairs of the form (node, distance); Insert the pair (s, 0) into Q; while Q not empty do (u, r) := ExtractMin(Q); for every child c of u do if d c > d u + w(u, c) then d c := d u + w(u, c); Insert(Q, (c, d c )); // update distance if c present We assume that we can store the priority queue in memory (O( V )). The algorithm will read the neighbors of each node at most once. Therefore, the total number of I/Os 4

5 is u E degu ( B = O E ). + V B 6. We apply the LMM algorithm with l = m = M. We assume known that we can merge M sequences of length M each in 3 passes through the data. The pseudocode of the algorithm is given below: Algorithm 6: Sort(X, N) Data: X: array of elements; N = M 2 : number of elements in X; Result: sorted array X; // First Pass; Split the input into M runs of length M each; Sort each run and unshuffle it into m = M sequences of length M each; // Second Pass; Merge groups of l = M unshuffled sequences (in memory); // Third Pass; Shuffle groups of m = M merged sequences of length M each; At the same time clean up the dirty regions; At this point we have M sorted runs of length M M each; // Third Pass (can be done with the previous pass); Unshuffle each run of length M M into m = M sequences of length M each ; // Fourth, Fifth and Sixth Pass; Merge groups of l = M unshuffled sequences of length M each; // Seventh Pass; Shuffle groups of m = M merged sequences of length M M each; Clean up dirty regions; For an arbitrary N, the general principle is to first merge M sequences of length M each, then merge M sequences of length M M each and so on. Let K stand for M and let T (u, v) be the number of passes required to merge u sorted sequences of length v each. Then we have the familiar formulas: 5

6 T (K, M) = 3 T (K, K i M) = 2 + T (K, K i 1 M) = 2i + 3 T (K c, M) = T (K, M) + T (K, KM) + T (K, K 2 M) T (K, K c 1 ) c 1 = (2i + 3) = c 2 + 2c i=0 However, as we saw in the previous pseudocode, when we compute T (K c, M) we can overlap the unshuffling at the ning of a T (K, K i M) computation with the shuffling done at the end of the previous T (K, K i 1 M) computation. Therefore, the last equation becomes: T (K c, M) = T (K, M) T (K, K c 1 ) (c 1) = c 2 + c + 1 Therefore the number of passes for M 2 and M 3 elements are: T (M 2 ) = T (M, M) = T (K 2, M) = = 7 T (M 3 ) = T (M 2, M) = T (K 4, M) = = 21 In general, for a given N, if K c log N/M = N/M it means that c = 2 and the number of log M passes to sort N elements is: T (N) = T (K c, M) = 4 ( ) 2 log N/M log N/M + 2 log M log M Let the input strings be S 1, S 2,..., S k with k i=1 S i = M. Build a generalized suffix tree for these strings in O(M) time. Let the suffixes be labelled with (i, j) where i refers to S i and j refers to the jth suffix in S i. Perform a depth first traversal in this tree. When we reach a leaf labelled (i, 1) for some i, this leaf corresponds to the entire string S i. This leaf might have more than one labels. Let these labels (in addition to (i, 1)) be (i 1, l 1 ), (i 2, l 2 ),..., (i q, l q ). Clearly, all the strings S i1, S i2,..., S iq have S i as a substring. Output all of these strings as those that contain S i. Check if the edge to this leaf s parent is labeled with $. If not, proceed with the traversal. If yes, let x be the parent 6

7 of this leaf. Also, let c 1, c 2,..., c r be the other children of x. Traverse through all the subtrees rooted at these children. All the leaves in these subtrees also correspond to strings that have S i as a substring. Output these strings as well (as those that contain S i ) and proceed with the traversal. The entire algorithm can be implemented to run in time O(M + k 2 ). 8. Let S 1, S 2,..., S k be the given input strings. Let S i = n i, for 1 i k. For any two strings S i and S j we can compute the longest common substring between them in O(n i + n j ) time, for 1 i, j k. Use this algorithm to compute the longest common substring between every pair of strings. The total run time is O( k i=1 k j=1 (n i+n j )) = O(kM). 9. Note that on a common CRCW PRAM we can compute the minimum or maximum of n integers (in the range [1, n O(1) ]) in O(1) time using n processors. Let T be the text and P be the pattern with T = m and P = n. We can use binary search on the suffix array. In any iteration of binary search, we have to compare the pattern P with a suffix T i of the text. This comparison involves the identification of the smallest integer q such that P [q] T i [q]. This can be done in O(1) time using the above algorithm. Thus the entire binary search takes O(log m) time. 7

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