Omega notation. Transitivity etc.

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1 Omega notation Big-Omega: Lecture 2, Sept. 25, 2014 f () n (()) g n const cn, s.t. n n : cg() n f () n Small-omega: f () n (()) g n const c, n s.t. n n : cg() n f () n Intuition (works most of the time): O: o: : : 21 Transitivity etc. Most rules apply: Example: transitivity Not all rules apply! a b, b c a c f Og ( ), g Oh ( ) f Oh ( ) Proof: f Og ( ) const cn 1, 1 s.t. n n1: 0 f( n) cgn 1 ( ) g Oh ( ) const c, n s.t. n n: 0 gn ( ) chn ( ) Take n3 max( n1, n2), c3 cc 1 2 Then: n n3: 0 f( n) cg 1 ( n) cch 1 2 ( n) ch 3 ( n) f( n) O( h( n)) QED f, g s.t. f O( g) and g O( f ) example: f n, g n1 sinn 22 1

2 Theta notation Theta: f ( n) ( g( n)) const c, c, n s.t. n n : cg( n) f ( n) c g( n) Often confused with Big-Oh notation! Example: n2/2 2 n ( n2) Proof: take n 8, then for n n : n /2 2 n n /4 n /4 2 n n /4 8 n/4 2 n n /4 n n n On the other hand, we have: 2/ 2 2 2/ 2 n n n n c c Thus: 2/4 2/2 2 2/2 i.e. 1 1/4, 2 1/2. Claim: Low order terms do not matter. Needs a proof! (HW?) 23 Simple Theorem Claim: f(n) = O(g(n)) and g(n) = O(f(n)) f(n) = (g(n)) Proof: n1, c1 s.t. n n 1: 0 f( n) c1g( n) n, c s.t. n n : 0 g( n) c f( n) n max( n, n ):0 1 g() n f() n c g() n QED c

3 Summary Remember the definitions. Formally prove from definitions. Use intuition from the properties of,, etc. Consider behavior of f(n)/g(n) as n 25 Example of an algorithm Stable Marriage n men and n women Each woman ranks all men and each man ranks all women Find a way to match (marry) all men and women such that there are no two pairs (m,w) and (m,w ) that are married and such that» m prefers w to w» w prefers m to m In other words, m will steal w from m and w will agree m m w w Red line shows instability 26 3

4 Discussion Bipartite graph Matching = legal set of marriages (no polygamy) Perfect matching = all men/women married Looking for perfect matching with stability constraint (similarity to shortest path, minimum spanning tree) Is there a perfect matching? Is there a stable perfect matching? Always or only for some input data? Brute force approach? Gale-Shapley algorithm As long as there is a man that is not engaged» Pick free man m» m tries to propose to the next woman w on his list (going down in terms of preferences) that he did not propose to yet» if w is free, then m and w become engaged else (w is engaged to m ) if w prefers m to m,» then m remains free» else m&w become engaged, m becomes free (m steals w from m ) 28 4

5 Some useful claims Claim 1: Once woman is proposed to for the first time (and becomes engaged), she never becomes free. Sequence of her partners improves (in terms of her preference list) Claim 2: The sequence of women a man m proposes to gets worse and worse (in terms of his preference list) Claim 3: If at some point m is free, than he has not yet proposed to all the women on his list Proof:» Assume m has proposed to all women already» Since he is still free, all women are engaged (see Claim 1)» But the number of men and women is the same - contradiction 29 Analysis Termination: follows from Claim 3 (At this point we have a perfect matching) Correctness: Assume that the algorithm is incorrect, m m i.e. there exist 2 engaged couples (m,w), (m,w ) such that» m prefers w to w» w prefers m to m w w» By construction, m last proposal was to w» Did m propose to w beforehand? If not, then m prefers w to w contradiction using claim 2 If yes, then w rejected him contradiction using claim 1 (woman improves her partners) 30 5

6 Running time analysis Need to find measure of progress Consider number of possible proposal (m,w) pairs» Proposal pairs never repeat» Total n 2 possible proposal pairs» n 2 iterations» Each iteration O(1) time» Total running time O(n 2 ) 31 Analyzing Insertion Sort as a Recursive Algorithm Basic idea: divide and conquer» Divide into 2 (or more) subproblems.» Solve each subproblem recursively.» Combine the results. Insertion sort is just a bad divide & conquer!» Subproblems: (a) last element (b) all the rest» Combine: find where to put the last element 32 6

7 Recursion for Insertion Sort We get a recursion for the running time T(n): Tn ( 1) n forn 1 Tn ( ) 1 forn 1 Tn ( ) Tn ( 1) n Tn ( 2) ( n 1) n Tn ( 3) ( n 2) ( n 1) n... n i i 1 2 ( n ) Formal proof: by induction. Another way of looking: split into n subproblems, merge one by one. 33 Improving the insertion sort Simple insertion sort is good only for small n. Balance sorting vs. merging: Merge equal size chunks. How to merge: (details of what happens when i or j reach end of the arrays are omitted) i=1, j=1 for k=1 to 2n if A(i)<B(j) then C(k)=A(i) i++ else C(k)=B(j) j++ end O(n) time to merge 34 7

8 Analysis Iterative approach:» Merge size-1 chunks into size-2 chunks» Merge size-2 chunks into size-4 chunks» etc. n ( ) (1) n (2) n Tn merge merge merge(4) Overall: ( nlog n) Intuitively right, but needs proof! 35 Analyzing Recursive Merge-Sort Another approach: recursive.» Divide into 2 equal size parts.» Sort each part recursively.» Merge. Recursion is a way of thinking. Easy to design recursive algorithms. We directly get the following recurrence: ½ 2T (n/2) + Θ(n) n>1 T (n) = 1 n =1 How to formally solve recurrence?» For example, does it matter that we have (n) instead of an exact expression??» Does it matter that we sometimes have n not divisible by 2?? 36 8

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