15-451 Algorithms 9/02/09 RECITATION NOTES ======================================================================= Any difficulties that came up on mini? Questions on recurrences? If so, could do an example like T(n) = 3T(n/3) + n or T(n) = 2T(n/3) + n, going through recursion tree. Homework 1, matrix game problem: Perhaps save 5 min at the end of section to say a bit about this problem and what it *means* to talk about the value of a randomized strategy. E.g., for the standard evens-odds game (top row of matrix is (-1, 1) and bottom row is (1, -1)) suppose Odelia's strategy is that with probability 90% she is going to play "one" and with 10% probability she is going to play "two". Here are 3-4 possible problems to do (probably isn't time to do them all) ----------------------------------------------------------------------------- Problem 1: Analysis of insertion sort in terms of number of inversions (related to hwk). Insertion-sort works by sorting the first i-1 elements and then inserting the ith element where it belongs among the first i-1. Here is some C code: (think of an array like [4, 2, 5, 3], or you could get 4 students to stand at front of room and sort by name) void insertionsort(int a[], int n) { int i,j; for (i=1; i=0; j--) { if (a[j+1] >= a[j]) break; swap(a,j,j+1); // swap these elements } } } Let's look at running time in terms of the number of comparisons between elements. (This is the number of times the code "a[j+1] >= a[j]" is executed.) (a) If the initial array consists of n elements in sorted order, how many comparisons are made? Answer: If the array is already sorted, then we just make one comparison for each i (breaking out of the loop). So, the total is n-1 comparisons. (b) What if the array is in descending order? Answer: In this case, the ith element will be compared to everything smaller than it (the inner for-loop never breaks), for a total of i comparisons. So the total is n(n-1)/2. (c) More generally, call a pair of indices (i,j) an inversion if i a_j. Show that the number of comparisons is always between I and I + n-1. Answer: There are a couple of ways to see this. One is: each swap reduces the number of inversions by *exactly* 1 [think about it]. For each i, the number of comparisons is equal to, or one greater than, the number of swaps (depending on whether we break out of the loop or not). So, the number of comparisons is between I and I + n-1. ----------------------------------------------------------------------------- Problem 1.5: Sorting by swaps. Imagine a sorting algorithm that somehow picks two elements that are out of order with respect to each other (not necessarily adjacent, as in insertion-sort) and swaps them. Could run algorithm like this with sorting students by name. Question: can we find an invariant that is reduced with each swap (i.e., to show that the procedure has to terminate)? Answer: can again look at inversions, but need to argue it always goes down. ----------------------------------------------------------------------------- Problem 2: Here is a problem related to lecture: what is the expected length of the leftmost branch in the quicksort recursion tree? In other words, you start with a set S of n distinct numbers and then repeat the following until they are all gone: 1. pick a random element p in S. 2. throw out p and all numbers > p from S. What is the expected number of times we repeat this? A first-cut analysis suggests log_2(n) because each p will throw out half the remaining numbers on average. But technically this is not a legal argument. WHY NOT? Because average(f(n')) may not be the same as f(average(n')). And in fact, the answer is really H_n, or about ln(n). Here's a way to prove the correct bound. We are interested in the expected number of pivots ever chosen in step 1. So, let X_i be an indicator RV for the event that the ith smallest element is ever chosen. E.g., the smallest element will for sure be chosen eventually, so E[X_1]=1. How about the 2nd smallest? Can anyone give an argument why E[X_2] = 1/2? More generally, why is E[X_i] = 1/i? Answer: one way to think about this is like the dart game from lecture. Game ends when dart hits some number in {1,...,i} and the chance that this number is i is 1/i. Another way is to imagine we implement step 1 by randomly permuting the elements and then choosing them in that order as pivots, but passing over any elements that are > some pivot chosen so far. Then, the problem boils down to asking: "in a random permutation of 1...n, what is the chance that the element i comes before all of {1,...,i-1}?" Well, since it's a random permutation, each element of {1,...,i} is just as likely as any other to come first, so the answer is 1/i. ----------------------------------------------------------------------------- Problem 3: BACKWARDS-ANALYSIS OF QUICKSORT. Here's a kind of bizarre way of analyzing quicksort: look at the algorithm backwards. (This is in the lecture notes). Actually, to do this analysis, it is better to think of a version of Quicksort that instead of being recursive, at each step it picks a random element of all the non-pivots so far, and then breaks up whatever bucket that element happens to be in. This is easiest to think of if you imagine doing the sorting "in place" in the array. If you think about it, this way of thinking is just rearranging the order in which the work happens, but it doesn't change the number of comparisons done (I mean, you probably wouldn't want to implement the algorithm this way because you'd have to keep track of which elements are previous pivots etc). Maybe do an example so that people are comfortable with this version of the algorithm and that it really is doing the same work but in a different order. The reason this version is nice is that if you imagine watching the pivots get chosen and where they would be on a sorted array, they are coming in a completely random permutation. Looking at the algorithm run backwards, at a generic point in time, we have $k$ pivots (producing $k+1$ buckets) and we undo'' one of our pivot choices at random, merging the two adjoining buckets. (Remember, the pivots are coming in a completely random order, so, viewed backwards, the most recent pivot is equally likely to be any of the k.) Now, the cost for an undo operation is the sum of the sizes of the two buckets joined (since this was the number of comparisons needed to split them). Notice that for each undo operation, if you sum the costs over all of the $k$ possible pivot choices, you count each bucket twice (or just once if it is the leftmost or rightmost bucket) and get a total of $< 2n$. Since we are picking one of these $k$ possibilities at random, the {\em expected} cost for this undo operation is at most $2n/k$. So, adding up over all undo operations, we get $\sum_k 2n/k = 2n H_n$. This is pretty slick, and not the sort of thing you'd think of trying off the top of your head, but it turns out to be useful in analyzing related algorithms in computational geometry.