Lecture 3: Asymptotic Analysis + Recurrences

Transcription

1 Lecture 3: Asymptotic Aalysis + Recurreces Data Structures ad Algorithms CSE 373 SU 18 BEN JONES 1

2 Warmup Write a model ad fid Big-O for (it i = 0; i < ; i++) { for (it j = 0; j < i; j++) { System.out.pritl( Hello! ); } } Summatio = Defiitio: Summatio b f(i) i=a T() = i i=1 = f(a) + f(a + 1) + f(a + ) + + f(b-) + f(b-1) + f(b) 1 i 1 c j=0 CSE 373 SP 18 - KASEY CHAMPION

3 Simplifyig Summatios for (it i = 0; i < ; i++) { for (it j = 0; j < i; j++) { System.out.pritl( Hello! ); } } T() = 1 i 1 1 c = ci j=0 Summatio of a costat 1 = c i Factorig out a costat = c 1 Gauss s Idetity = c c O( ) CSE 373 WI 18 MICHAEL LEE 3

4 Fuctio Modelig: Recursio public it factorial(it ) { if ( == 0 == 1) { +3 +c retur 1; +1 } else { retur * factorial( 1); } +???? CSE 373 SP 18 - KASEY CHAMPION 4

5 Fuctio Modelig: Recursio public it factorial(it ) { if ( == 0 == 1) { +c retur 1; 1 } else { retur * factorial( 1); } +c +T(-1) T() = C 1 C + T(-1) whe = 0 or 1 otherwise Defiitio: Recurrece Mathematical equatio that recursively defies a sequece The otatio above is like a if / else statemet CSE 373 SP 18 - KASEY CHAMPION 5

6 Ufoldig Method T() = C 1 C + T(-1) whe = 0 or 1 otherwise T(3) = C + T(3 1) = C + (C + T( 1)) = C + (C + (C 1 )) = C + C 1 T() = C C Summatio of a costat T() = C 1 + (-1)C CSE 373 SP 18 - KASEY CHAMPION 6

7 Aoucemets - Course backgroud survey due by Friday - HW 1 is Due Friday - HW Assiged o Friday Parter selectio forms due by 11:59pm Thursday CSE 373 SU 18 BEN JONES 7

8 A Detour o Style - Checkstyle for project - No packages for HW1 - Braces for blocks - Good style is easy to read - Javadoc o public methods (ot eeded if iterface has Javadoc) - Commet o-obvious code - Self-Documetig code is better tha commeted code - Good variable ad method ames go a log way towards this - No magic umbers (umbers larger tha or 3 should probably be class costats uless there s a really good reaso) - No code duplicatio - Use Idioms! ex. for (it I = 0; I < 10; i++) istead of for (it I = 0; I == 9; i = i + 1) amig: CONSTANTS_USE_CAPS, ClassName, methodname CSE 373 SU 18 BEN JONES 8

9 Tree Method Idea: -Sice we re makig recursive calls, let s just draw out a tree, with oe ode for each recursive call. -Each of those odes will do some work, ad (if they make more recursive calls) have childre. -If we ca just add up all the work, we ca fid a big-θ boud. CSE 373 SU 18 - ROBBIE WEBER 9

10 Solvig Recurreces I: Biary Search T = 1 whe 1 TT + 1 otherwise 0. Draw the tree. 1. What is the iput size at level i?. What is the umber of odes at level i? 3. What is the work doe at recursive level i? 4. What is the last level of the tree? 5. What is the work doe at the base case? 6. Sum over all levels (usig 3,5). 7. Simplify CSE 373 SU 18 - ROBBIE WEBER 10

11 Solvig Recurreces I: Biary Search T = 1 whe 1 TT + 1 otherwise Draw the tree. 1. What is the iput size at level i?. What is the umber of odes at level i? 3. What is the work doe at recursive level i? 4. What is the last level of the tree? 5. What is the work doe at the base case? 6. Sum over all levels (usig 3,5). 7. Simplify CSE 373 SU 18 - ROBBIE WEBER 11

12 Solvig Recurreces I: Biary Search 1 whe 1 T = T + 1 otherwise Level Iput Size Work/call Work/level / 1 1 / 1 1 i / i 1 1 log Draw the tree. 1. What is the iput size at level i?. What is the umber of odes at level i? 3. What is the work doe at recursive level i? 4. What is the last level of the tree? 5. What is the work doe at the base case? 6. Sum over all levels (usig 3,5). 7. Simplify log σ = log CSE 373 SU 18 - ROBBIE WEBER 1

13 Solvig Recurreces II: T = 1 whe 1 T + otherwise CSE 373 SU 18 - ROBBIE WEBER 13

14 Tree Method Formulas T = 1 whe 1 T + otherwise How much work is doe by recursive levels (brach odes)? 1. What is the iput size at level i? - i= 0 is overall root level.. At each level i, how may calls are there? 3. At each level i, how much work is doe?? Recursive work = How much work is doe by the base case level (leaf odes)? 4. What is the last level of the tree? (/ i ) = 1 i = i = log 5. What is the work doe at the last level? 6. Combie ad Simplify lastrecursivelevel brachnum i brachwork(i) NoRecursive work = WorkPerBaseCase umbercalls T = log 1 i i + = log + (/ i ) i log 1 1 log = i (/ i ) = i i CSE 373 SU 18 - ROBBIE WEBER 14

15 Solvig Recurreces III c c T = 5 whe 4 3T 4 + c otherwise c c cc 4 c c c cc c c c 4 c c Aswer the followig questios: 1. What is iput size o level i?. Number of odes at level i? 3. Work doe at recursive level i? 4. Last level of tree? 5. Work doe at base case? 6. What is sum over all levels? CSE 373 SU 18 - ROBBIE WEBER 15

16 Solvig Recurreces III T = 5 whe 4 3T 4 + c otherwise 1. Iput size o level i? 4 i Level (i) Number of Nodes Work per Node Work per Level 0 1 c c. How may calls o level i? 3 i 1 3 c 4 3 c 3. How much work o level i? 3 i c 4 i = 3 i c 3 c 4 3 i 3 i c 4 i 3 i c c 4. What is the last level? Whe 4 i = 4 log 4 1 Base = log log log A. How much work for each leaf ode? B. How may base case calls? 3 log 4 1 = 3log 4 3 power of a log x log b y = y log b x 5 6. Combiig it all together = log T = log 4 3 i c log 4 3 CSE 373 SU 18 - ROBBIE WEBER

17 Solvig Recurreces III 7. Simplify log 4 T = 3 i c log 4 3 factorig out a costat b b cf(i) = c f(i) log 4 T = c 3 i log 4 3 i=a i=a fiite geometric series 1 x i = x 1 x 1 Closed form: log T = c log 4 3 If we re tryig to prove upper boud T c 3 i log 4 3 ifiite geometric series x i = 1 1 x whe -1 < x < 1 T c 1 T O( ) log 4 3 CSE 373 SU 18 - ROBBIE WEBER 17

18 Aother Example 1 if = 1 T = ቐ if = T + 4 otherwise CSE 373 SU 18 - ROBBIE WEBER 18

19 Is there a easier way? We do all that effort to get a exact formula for the umber of operatios, But we usually oly care about the Θ boud. There must be a easier way Sometimes, there is! CSE 373 SU 18 - ROBBIE WEBER 19

20 Master Theorem Give a recurrece of the followig form: d T = ቐ whe some costat at b + c otherwise Where a, b, c, ad d are all costats. The big-theta solutio always follows this patter: If log b a < c the T is Θ c If log b a = c the T is Θ c log If log b a > c the T is Θ log b a CSE 373 SU 18 - ROBBIE WEBER 0

21 Apply Master Theorem Give a recurrece of the form: d whe some costat T = at b + c otherwise If log T is Θ c b a < c the If log b a = c the T is Θ c log If log b a > c the T is Θ log b a T = 1 whe 1 T + otherwise log b a = c log = 1 a = b = c = 1 d = 1 T is Θ c log Θ 1 log CSE 373 SU 18 - ROBBIE WEBER 1

22 Reflectig o Master Theorem Give a recurrece of the form: d whe some costat T = at b + c otherwise If log T is Θ c b a < c the If log b a = c the T is Θ c log If log b a > c height log b a the brachwork c log b a leafwork d log b a T is Θ log b a The case - Recursive case coquers work more quickly tha it divides work - Most work happes ear top of tree - No recursive work i recursive case domiates growth, c term The case - Work is equally distributed across levels of the tree - Overall work is approximately work at ay level x height The log b a < c log b a = c log b a > c case - Recursive case divides work faster tha it coquers work - Most work happes ear bottom of tree - Work at base case domiates. CSE 373 SU 18 - ROBBIE WEBER

23 Beefits of Solvig By Had If we had the Master Theorem why did we do all that math??? Not all recurreces fit the Master Theorem. -Recurreces show up everywhere i computer sciece. -Ad they re ot always ice ad eat. It helps to uderstad exactly where you re spedig time. -Master Theorem gives you a very rough estimate. The Tree Method ca give you a much more precise uderstadig. CSE 373 SU 18 - ROBBIE WEBER 3

24 Amortizatio What s the worst case for isertig ito a ArrayList? -O(). If the array is full. Is O() a good descriptio of the worst case behavior? -If you re worried about a sigle isertio, maybe. -If you re worried about doig, say, isertios i a row. NO! Amortized bouds let us study the behavior of a buch of cosecutive calls. CSE 373 SU 18 - ROBBIE WEBER 4

25 Amortizatio The most commo applicatio of amortized bouds is for isertios/deletios ad data structure resizig. Let s see why we always do that doublig strategy. How log i total does it take to do isertios? We might eed to double a buch, but the total resizig work is at most O() Ad the regular isertios are at most O 1 = O() So isertios take O() work total Or amortized O(1) time. CSE 373 SU 18 - ROBBIE WEBER 5

26 Amortizatio Why do we double? Why ot icrease the size by 10,000 each time we fill up? How much work is doe o resizig to get the size up to? Will eed to do work o order of curret size every 10,000 iserts σ i 10,000 10,000 = O( ) The other iserts do O work total. The amortized cost to isert is O = O(). Much worse tha the O(1) from doublig! CSE 373 SU 18 - ROBBIE WEBER 6