Handout: How to calculate time complexity? CSE 101 Winter 2014

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2 Hadout: How to calculate time complexity? CSE 101 Witer 014 Evaluatig sums Arithmetic progressio Geometric progressio T = T 1 + c T = T 1 c We have: +c +c +c +c T 0 T 1 T 1 T T = rst + c (# of times c has bee added) rst + last T 0 + T T = (# of terms) Example: = 1 + c c c c T 0 T 1 T 1 T We have (c 6= 1): (# of times c has bee multiplied) T = rst c T 0 + T T = rst c(# of terms) 1 c 1 Example (c 6= 1): 1 + c + c + + c = c+1 1 c 1 Examples A) Fast multiplicatio Algorithm The idea is to square a to get a (oe multiplicatio) ad whe is odd, evaluate a 1 the same way ad the multiply it by a to get a (oe multiplicatio) Pseudo-code procedure mult(a, ): if == 1: retur a else if is eve: b = mult(a, / ) retur b * b else retur a * mult(a, - 1) Time complexity equatio Let T () be the umber of multiplicatios We assume without loss of geerality that is power of This will ot iuece the al boud i ay importat way, after all, is at most a multiplicative factor away from some power of At each step is divided by T () = T + 1

3 Hadout: How to calculate time complexity? CSE 101 Witer 014 Evaluatig T () Total work for step Total work for step Work at step (1 multiplicatio) The Master Theorem gives us a = 1 b = d = 0 so log b a = log 1 = 0 = d so T () = O(log ) B) Tower of Haoi: Lookig at the recursio tree Algorithm The algorithm requires that to move disks, we move rst 1 disks, the 1 ad ally 1 agai Time complexity equatio Let T () be the umber of moves T () = T ( 1) T ( 1) Total work for disks Total work for 1 disks Move 1 disk Total work for 1 disks We ca't apply Master Theorem o the complexity equatio T () = T ( evaluate T () to be able to give a big-o boud 1) + 1 so we have to The recursio tree Let's cosider the recurrece sub-problem tree: Level 0 : 1 problem Level 1 : problems 1 1 Level : 4 problems Level k: k Level 1 problems k k k k At the k-th level, there are k sub-problems that all require 1 move to get to level k 1 So, there are w(k) = k moves at level k (for level k oly) Google "tower of haoi recursive solutio" if you do't remember the solutio 3

4 Hadout: How to calculate time complexity? CSE 101 Witer 014 Summig to evaluate T () As T () is the sum of the moves of each level, we get: T () = We recogise here the sum of the rst terms of a geometric progressio whose ratio is ad rst term is 1 so we get: Therefore, T () = O( ) T () = 1 1 = 1 C) Tower of Haoi: Guessig with the first terms the prove Whe you have the recursio equatio T () = T ( by lookig at the rst terms: 1) + 1, you ca try to guess the value of T () T (0) = 0; T (1) = 1; T () = 3; T (3) = 7; T (4) = 15; T (5) = 31; T (6) = 63; : : : We ca guess it may be T () = 1, but we have to prove it! Let's do it by recursio o : Base case T (0) = 0 OK! Iductive hypothesis Let's suppose T () = 1 for a 0 Iductive step We have: T ( + 1) = T () + 1 = ( 1) + 1 (id hyp) = +1 1 Coclusio We have proved that T () = 1 for all 0 Therefore, T () = O( ) D) Tower of Haoi: Guessig by urollig the prove Aother way we ca guess the solutio is by urollig the recurrece, by substitutig it ito itself: T () = T ( 1) + 1 = ( T ( ) + 1) + 1 = 4 T ( ) + 3 = 4( T ( 3) + 1) + 3 = T ( 3) + 7 : : : Here we ca guess a ew recurrece: T () = k T ( k) + ( k 1) But we have also to prove it! Let's do it by recursio o k: 4

5 Hadout: How to calculate time complexity? CSE 101 Witer 014 Base case For k = 0, we have 0 T ( 0) + ( 0 1) = T () Ok! Iductive hypothesis Let's suppose T () = k T ( k) + ( k 1) for a k 0 Iductive step We have: T () = k T ( k) + ( k 1) = k ( T ( k 1) + 1) + ( k 1) itial recursio = k+1 T ( (k + 1)) + ( k+1 1) Coclusio We have proved that T () = k T ( k) + ( k 1) for k 0 Therefore, for k = we get T () = 1 so T () = O( ) E) Closest Pair of Poits i a Plae Let's evaluate the time complexity of the closest pair of poits i a plae solutio usig Divide ad Coquer is a power of We assume without loss of geerality that is power of This will ot iuece the al boud i ay importat way, after all, is at most a multiplicative factor away from some power of Sortig the poits before solvig First, before startig to solve the problem, we sort twice: Oce by the x-coordiate ad oce by the y-coordiate No further sortig is required at subsequet recursive steps Usig a stadard sortig algorithm requires O( log ), for example via mergesort Solvig For a set of poits i the plae (let T () be the time complexity): We partitio the curret set ito sub-sets deed by media x-coordiate As the poits have bee sorted by x-coordiate, dig the media requires oly a costat time operatio Recursively compute the closest pair distaces for the sub-sets T ( ) + T ( ) This requires a complexity of Deal with the middle ribbo Worst case sceario: all poits from the left side sub-problem ( poits) are i the ribbo ad for each of them, we have to check a maximum of 6 poits i the right side As the poits are sorted by y-coordiates ad we step through at most, checkig the middle ribbo takes at most a liear amout of time Fially take the miimum distace amog: the smallest of the right side, the smallest of the left side, ad the smallest of the middle ribbo That takes a costat time: O(1) 5

6 Hadout: How to calculate time complexity? CSE 101 Witer 014 We get: T () = O() + T + T + O() + O(1) Work for poits Splittig the plae (x-coord media) Work for the left side Work for the right side Work for the ribbo Combiig results (mi of 3 umbers) It ca be simplied i: Master Theorem gives us a = b = d = 1 T () = T + O() so log b a = log 1 = 1 = d so T () = O( log ) Coclusio We did steps: before solvig, we sorted the poits with a complexity of O( log ), ad we solved the problem with a complexity of T () = O( log ) So the overall time complexity is O( log ) F) Recursio tree agai We are give: T () = T + log We assume without loss of geerality that is power of This will ot iuece the al boud i ay importat way, after all, is at most a multiplicative factor away from some power of We ca't use the Master Theorem so we have to evaluate T () to be able to give the complexity i big-o terms Let's cosider the recurrece sub-problem tree: Level 0 : 1 problem Level 1 : problems = = Level : 4 problems =4 =4 =4 =4 Level k: k problems = k = k = k = k Level log

7 Hadout: How to calculate time complexity? CSE 101 Witer 014 As we kow the size of the sub-problems at level k is k accordig to the complexity equatio: The work for oe sub-problem of level k is, k log k = 1 k log() k As there is k sub-problems at level k, the total work made at level k is: k 1 k log() k = log() k # of problems at level k work for each problem of level k total work made at level k Therefore: T () = (work at level 0) + (work at level 1) + + (work at level log() - 1) = log() + log() = log() 1 + 1! log() We saw i the homework that we had: m = O(log m) I our previous sum, m = log so we get T () = O( log(log )) 7

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