Algorithms and Data Structures 2014 Exercises week 5

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1 Algorithms and Data Structures 014 Exercises week 5 October, 014 Exercises marked by ( ) are hard, but they might show up on the exam. Exercises marked by ( ) are even harder, but they will not be on the exam. 1. Consider the following function: } void example(int A[], int N) { /* N is the size of the array */ Node p; for (int i = 1; i <= N; i++) p = insert(a,n,i,p); convert(p,a,1); Assuming T insert = O(lg N) and T convert = O(N ), analyze the execution time of the function example in the worst case. Answer. First, we note that the for loop executes N times, no matter what. Then, we write T (n), the execution time function for example: T (N) c 1 (N + 1) + (c lg N)N + c 3 N Here, the comes from the fact that we know T insert = O(lg N) and T convert = O(N ), which means T insert c lg N and T convert c 3 N for some constants c and c 3. Now we can see that T (N) O(N ), because N is the greatest factor in T (N) above, but let us show that indeed T (N) CN (for almost all N). T (N) c 1 (N + 1) + (c lg N)N + c 3 N = c 1 N + c 1 + (c lg N)N + c 3 N c 1 N + c 1 N + (c N)N + c 3 N N N, 1 N, lg N N, for N > 0. = (c 1 + c + c 3 ) N C 1

2 . For each of the following bounds on T, draw a recursion tree and guess the least O-order. Then prove that T is in that O-order using the substitution method. (a) T (n) T (n 1) + n for n 1 (Bubblesort) (b) T (n) T ( n ) + 37 for n 1 (Binary search) (c) (**) T (n) T ( n ) + 1 for n 1 (d) ( ) T (n) T ( 3n 4 ) + 1n for n 1 (Lucky quicksort) (e) ( ) T (n) n + T ( n 5 ) + T ( 3n 4 ) for n 1 (Linearselect) Answer (a) The tree is just a line with n + 1 levels and each level i costs i (let us assume that level 0 costs c 0 ). The total cost c 0 + n i=1 i = c 0 + (n + n)/, where c 0 is the value we assume for T (0) (in the sum you need arithmetic progression formula). We conjecture that T (n) O(n ). Wenow prove it using the substitution method: T (n) T (n 1) + n = c(n 1) + n Induction Hypothesis = cn cn + c + n cn For the last equation to hold we need n c (1 c). (c) Drawing the tree one observes that the tree has lg(n)+1 levels and each level i costs i c for c = 1. This means that the total cost is n c (here you need geometric progression formula). Let us prove the the algorithm is in O(n). This example shows that the substitution method is sometimes not easy to use. If we try to use it here what happens is: T (n) T ( n ) + 1 = c( n ) + 1 Induction Hypothesis cn For the last equation to hold we need c n + 1 0, which is impossible if n 0 and c 0. You could now think that the conjecture O(n) is wrong, which is in fact not true. We can prove it using a different method.

3 I did it by first considering the case n = m. For m, we have T ( m ) T ( m 1 ) + 1 T () m 1 + 1( m ) = T () m 1 + 1( m 1 1) (T () + 1) m 1 (T () + 1) m.... m 1 times Given any n >, there exist m, r N such that m+1 n, m 1 and n = m + r. And thus T (n) = T ( m + r) T ( m+1 ) (T () + 1) m+1 = (T () + 4) m (T () + 4)n, which shows T O(n). 3. For the following problems provide a naive solution and the corresponding dynamic programming solution (either using memoization or the bottom up approach). For each problem analyze the time complexity of both the naive solutions and the DP solution. Write clearly what are the subproblems and what is the number of subproblems you need to solve in order to find the solution of the original problem. (a) Wouter loves to go to the Casino. Tonight the main game is the following. Greta, the assistant, puts down at each round a sequence of cards that have random real numbers printed in them A[1],... A[n]. Each player chooses a contiguous subsequence A[i], A[i+1],..., A[j]. You win if you choose the subsequence for which the sum of elements in the subsequence is maximized. E.g. {5, 5, 3,, 7} should return subsequence {3,, 7}. Answer The naive solution would be: take all possible contiguous subsequences of the given array and compute their sum, then select the max. What is the complexity? For the DP solution, we look at subsequences ending at j (the subproblems are maximum sum of a subsequence ending at j, we can then re-use sums already computed): 3

4 MaxSum(j) A[j] if j=0 = max( MaxSum(j 1) + A[j] Extend window ending at position j-1, A[j] start a new window at position j ) otherwise With each element of MaxSum, you also keep the starting element of the sum (the same as for MaxSum(j 1) or j if you restart). At the end, scan MaxSum for the maximum value and return it together with the starting and ending indexes. Alternatively, you could keep track of the maximum value as you create MaxSum. The array MaxSum has size n and evaluating each element takes constant. Scanning MaxSum also takes O(n) so in total we have a O(n) algorithm. (b) At some point Greta decided to change the rules and asked the players to, given a sequence of n cards A[1],... A[n] (still with real numbers), pick a subsequence of cards (not necessarily contiguous) of maximum length, so that the values in the cards the player picked form a strictly increasing sequence. (c) At the end of his visit to the Casino Wouter changes the casino chips he has into real money. Imagine that the Casino s cashier has n types of coins of values v 1 < v <... < v n (all natural numbers). Assume v 1 = 1, so you can always make change for any amount of chips. Wouter hates to carry coins so he asks the cashier to give him as few coins as possible. Give an algorithm which makes change for any amount with as few coins as possible. (d) (*) Greta s aunt wants to build a new fence to protect her chickens. She was given by her neighbour who just finish renovating his house a truck with n types of rectangular 3-D bricks. She measured all the bricks and registered for each type i, the height h i, the width w i and the depth d i (all real numbers). She then wondered what would be the highest stack of bricks that she could build, having as rule that one can only stack a brick on top of another one if the dimensions of the base of the lower brick are each strictly larger than those of the base of the higher brick. Of course, you can rotate a brick so that any of the 6 sides functions as its base. It is also allowed to use multiple times the the same type of brick. 4. The (always used as example) Fibonacci recurrence f(n + 1) = f(n) + f(n 1) can be represented in the form a matrix as follows: ( ) ( ) ( ) 1 1 f(n) f(n + 1) = 1 0 f(n 1) f(n) where ( ) ( f(0) 1 =. f(1) 1) 4

5 ( ) 1 1 Let A =. One can easily observe that in order to compute f(n+1) 1 0 we can use the equality ( ) ( ) ( ) f(n) f(n + 1) A = = A n f(1) f(n 1) f(n) f(0) So all that we have to do is compute the n th power of the matrix A. Provide a O(lg n) time algorithm that does this. Answer Hint: use recursive doubling; to find A n, one can do A n/ A n/ (take care if n is odd). 5. (**) Hacker team (Kleinberg & Tardos): Suppose you are managing a consulting team of expert computer hackers, and each week you have to choose a job for them to undertake. The set of possible jobs is divided into those that are low-stress (e.g. setting up a Web site for a class at the local elementary school) and those that are high-stress (e.g., protecting the dean s most valuable secrets.) The basic question each week is whether to take on a low-stress job or a high-stress job. If you select a low-stress job for your team in week i, then you get a revenue of l i > 0 dollars; if you select a high-stress job, you get a revenue of h i > 0 dollars. High-stress jobs typically pay more. The catch, however, is that in order for the team to take on a high-stress job in week i, it is required that they do no job (of either type) in week i 1; they need a full week of prep time to get ready for the crushing stress level. On the other hand, it is okay for them to take a low-stress job in week i even if they have done a job (of either type) in week i 1. So, given a sequence of n weeks, a plan is specified by a choice of low-stress, high-stress or none for each of the n weeks, with the property that if high-stress is chosen for week i > 1, then none has to be chosen for week i 1. (It is okay to choose a high-stress job in week 1.) The value of the plan is determined in the natural way; for each i, you add l i to the value if you choose low-stress in week i, and you add h i to the value if you choose high-stress in week i. (You add 0 if you choose none in week i.) Given sets of values l 1, l,..., l n and h 1, h,..., h n, find a plan of maximum value (such a plan is called optimal.) Give an efficient algorithm to take the input values and return the value of an optimal plan. 5