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Unformatted text preview: Advanced Algorithms Lecture Outline January 10, 2011 Example. Prove that parenleftbigg n parenrightbigg parenleftbigg n 1 parenrightbigg + parenleftbigg n 2 parenrightbigg ... + ( 1) n parenleftbigg n n parenrightbigg = 0 Solution. One way to solve this problem is by substituting x = 1 in the previous example. When x = 1 the above equation becomes n = 0 = n summationdisplay k =0 parenleftbigg n k parenrightbigg ( 1) k . A combinatorial argument however will give us more insight into what the expression means. We now present a combinatorial proof of the claim. We want to prove that parenleftbigg n parenrightbigg + parenleftbigg n 2 parenrightbigg + parenleftbigg n 4 parenrightbigg + ... = parenleftbigg n 1 parenrightbigg + parenleftbigg n 3 parenrightbigg + parenleftbigg n 5 parenrightbigg + ... Consider a set X = { x 1 ,x 2 ,x 3 ,... ,x n } . We want to show that the total number of subsets of X that have even size equals the total number of subsets of X that have odd size. We will now show that both these quantities equal 2 n 1 from which the claim follows. An evensized subset of X can be constructed as follows. Step 1. Decide whether x 1 belongs to the subset of not. Step 2. Decide whether x 2 belongs to the subset of not. . . . Step n . Decide whether x n belongs to the subset of not. Note that there are 2 choices for each of the first n 1 steps but exactly one choice for performing step n . This is because if we have choose an even number of elements from X \ { x n } then we must leave out x n otherwise we must include x n in the subset. Hence using the multiplication rule the total number of evensized subsets of X equals 2 n 1 . Since we know that the total number of subsets of X is 2 n , the total number of oddsized subsets of X is 2 n 2 n 1 = 2 n 1 . Before we move on to the next topic we will prove a couple of identities using counting techniques. Specifically, we will use the following technique. To prove an identity we will pose a counting question. We will then answer the question in two ways, one answer will correspond to LHS and the other would correspond to the RHS. Example. Prove that n k =0 ( n k ) = 2 n . 2 Lecture Outline January 10, 2011 Solution. We pose the following counting question. Given a set S of n distinct elements how many subsets are there of the set S ? From an earlier lecture we know that the answer is 2 n . This gives us the RHS. Another way to compute the answer to the question is as follows. The set P ( S ) of all possible subsets can be partitioned into S ,S 1 ,... ,S n , where S i , 0 i n , is the set of all subsets of S that have cardinality i . Thus  P ( S )  =  S  +  S 1  + ... +  S n  = parenleftbigg n parenrightbigg + parenleftbigg n 1 parenrightbigg + ... + parenleftbigg n n parenrightbigg = n summationdisplay k =0 parenleftbigg n k parenrightbigg = LHS This proves the claim....
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 Winter '10
 RajivSir
 Algorithms, Prime number, Rational number, Generalized Pigeonhole Principle

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