Lecture 7: Solving Recurrences

Transcription

1 Lecture 7: Solvig Recurreces CSE 7: Data Structures ad Algorithms CSE 7 19 WI KASEY CHAMPION 1

2 Warm Up Writig Recurreces CSE 7 19 WI KASEY CHAMPION 2

3 Admiistriva HW 2 Part 1 due Friday git ruers will get overloaded o Friday, pla accordigly No Kasey office hours Friday CSE 7 19 WI KASEY CHAMPION

4 Solvig Recurreces CSE 7 19 WI KASEY CHAMPION 4

5 Review: Modelig Recursio Write a mathematical model of the followig code public it factorial(it ) { } } if ( == 0 == 1) { retur 1; } else { +1 retur * factorial(1); +2 4 &h(" " = 0,1! " = $ 2 +! " 1 /0h(1&2( + What is the Big O? CSE 7 19 WI KASEY CHAMPION 5

6 Solvig Recurreces How do we go from code model to Big O? 1. Explore the recursive patter by tracig through the a few levels of recursio 2. Write a ew model of the rutime or work doe for the patter i terms of the level of recursio i. Use algebra (ad likely a summatio) to simplify the T recursive call out of your ew model 4. Use algebra to simplify dow to the closed form so you ca easily idetify the Big O CSE 7 19 WI KASEY CHAMPION 6

7 Urollig Method Walk through fuctio defiitio util you see a patter 4 &h(" " = 0,1! " = $ 2 +!("" 1) /0h(1&2( 4 &h(" " = 0,1! " =! " = $ 2 +! " 1 /0h(1&2(! " = !(" 1 1) i = 1 4 &h(" " = 0,1 1 +! " 1 /0h(1&2(! " = !(" 2)! " = $! " = !(" 2 1) i = 2! " = !(" )! " = $ 4 &h(" " = 0,1 1 +! " 1 /0h(1&2(! " = !(" 1) i =! " = ! 1 = i =? T(i) = T(1) whe i= 1 1 recursive cases 1 base case >?=! " = 4 + : 2 ;<= Summatio of a costat > : " = 4 + 2(" 1)! 4 = 2 +! 4 1 = ! 1 = ! 2 1 = = CSE 7 19 WI KASEY CHAMPION 7

8 Urollig Method 5 Miutes Walk through fuctio defiitio util you see a patter! " = 1 %h'" " = 0 2! " 1 + 1,h'.%/0' i = 1 = 2! " i = 2 = 2 2! " = 2 1! " i = = 2 1 2! " = 2! " i = 4 = 2 2! " = " i = i = 2 7! " / Fiite Geometric Series <85 : 7;6 = 7 = =< 1 = 1 = 2 < + 2< = 2< + 2 < 1 = 2 <> = 2 7! " / + :?;6 2? <85 <85 = 2 <! " " + : 2? = 2 < 1 + :?;6?;6 CSE 7 19 WI KASEY CHAMPION 2? 8

9 Solvig Recurreces How do we go from code model to Big O? 1. Explore the recursive patter 2. Write a ew model i terms of i. Use algebra simplify the T away 4. Use algebra to fid the closed form Usig urollig method 1. Plug defiitio ito itself to write out first few levels of recursio 2. Simplify away parethesis but leave separate terms to help idetify patter i terms of i. Plug i a value of i to solve for base case, write summatio represetig recursive work 4. Usig summatio idetities as appropriate reduced to closed form CSE 7 19 WI KASEY CHAMPION 9

10 Tree Method Draw out call stack, how much work does each call do? 1 )h+" " 1! " = ' 2! " 2 + ".h+/)01+! " " 2 + "! " 2 + " 1. Draw a overall root represetig the start of your family of recursive calls 2. How much work is doe by the top recursive level?. How much of that work is delegated to dowstream recursive calls? 4. How much work is doe by each of those child recursive calls? 5. How much of that work is delegated to dowstream recursive calls? What does the last row of the tree look like? 8. Sum up all the work!! " 4 +! " " 24 + " 2 2 "! "! " 4 4 4! "! 4 + "! " 4 + " 2 2 "! " "! " CSE 7 SP 18 KASEY CHAMPION 10

11 Tree Method & ' = 1 )h+' ' 1 2& ' 2 + './h+0)12+ How much work How may pieces of work at each level? doe by each piece? How much work across each level?

12 Tree Method Formulas O = = 1 )h"= = 1 2O = 2 + = *Hh"%)'&" How much work is doe by recursive levels (brach odes)? 1. How may recursive calls are o the ith level of the tree? umbernodesperlevel(i) = 2 i i = 0 is overall root level 2. At each level i, how may iputs does a sigle ode process? iputsperrecursivecall(i) = (/ 2 i ). How may recursive levels are there? brachcout = log 2 1 Based o the patter of how we get dow to base case :!"#$%&'(" )*%+ =./0 ;%<=#h?$@ ' ;%<=#ha*%+(') How much work is doe by the base case level (leaf odes)? 1. How much work is doe by a sigle leaf ode? 2. How may leaf odes are there? leafwork = 1 UVW X 4YZ O(= > 1) =./0 2. = 2. leafcout = 2 log 2 =?*=!"#$%&'(" )*%+ = D"<EA*%+ D"<EG*$=H = D"<EA*%+ ;%<=#h?$@ 49IJKLKMN O = 1 = 1 2 M8[ \4 = = H*H<D )*%+ = %"#$%&'(" )*%+ + =*=%"#$%&'(" )*%+ = UVW X 4YZ O = =./0 2. = 2. + = = = log \ = + = CSE 7 SP 18 KASEY CHAMPION 12

13 Tree Method Practice! ", 4 +! " 4 +! " 4 +,"! " = 4 %h'" " 1! " 4 +,"./h'0%12',,,! " 4,! " 4,,,,,,! " ,, Aswer the followig questios: 1. How may odes o each brach level? 2. How much work for each brach ode?. How much work per brach level? 4. How may brach levels? 5. How much work for each leaf ode? 6. How may leaf odes? EXAMPLE PROVIDED BY CS 1 JESSICA SU 1

14 Tree Method Practice 5 " = 4 7h9" " 1 5 " 4 +!"% <=h9>7?@9 5 Miutes 1. How may odes o each brach level? 0 Level (i) Number of Nodes Work per Node Work per Level 2. How much work for each brach ode?! " 4 0 % 0 1!" 2!" 2 1! " 4 %!"%. How much work per brach level? 0! " 4 0 % = 0!" % 2 9! " % 9 256!"% base +,. / ,. / 4. How may brach levels? log. " 1 Combiig it all together 5. How much work for each leaf ode? 4 +, D / EF 5 " = A 0BC 0!" % + 4" +,.G 6. How may leaf odes? +, D / power of a log H +, I J = K +, I L " +, D G CSE 7 SP 18 KASEY CHAMPION 14

15 Tree Method Practice ()* +,.! " = $ %&' % 2" + 4" ()* 67 factorig out a costat > > $ 2?(@) = 2 $?(@) ()* +,.! " = 2" $ %&' % + 4" ()* 67 %&= %&= fiite geometric series,. $ %&' A % = A, 1 A 1 Closed form: ()* +, 1! " = 2" + 4" ()*67 1 If we re tryig to prove upper boud! " = 2" $ B %&' % + 4" ()* 67 ifiite geometric series B $ %&' A % = 1 1 A whe 1 < x < 1! " = 2" 1! " :(" ) 1 + 4" ()* 67 CSE 7 SP 18 KASEY CHAMPION 15

16 Solvig Recurreces How do we go from code model to Big O? 1. Explore the recursive patter 2. Write a ew model i terms of i. Use algebra simplify the T away 4. Use algebra to fid the closed form Usig urollig method 1. Plug defiitio ito itself to write out first few levels of recursio 2. Simplify away parethesis but leave separate terms to help idetify patter i terms of i. Plug i a value of i to solve for base case, write summatio represetig recursive work 4. Usig summatio idetities as appropriate reduced to closed form Usig tree method 1. Plug defiitio ito itself to draw out first few levels of tree 2. Aswer questios about ature of tree to idetify work doe by recursive levels ad base case i terms of i. Combie aswers to questios to complete model i terms of i 4. Usig summatio idetities as appropriate reduced to closed form CSE 7 SP 18 KASEY CHAMPION

17 Is there a easier way? What if you do wat a exact closed form? Sorry, o If we wat to fid a big Θ Sometimes, yes! CSE 7 SP 18 KASEY CHAMPION 17

18 Master Theorem Give a recurrece of the followig form:! " = $ %h'" " = 1 )! " * + ",.h'/%01' The thaks to magical math brilliace we ca kow the followig: If log 7 ) < 9 the! " Θ ", If log 7 ) = 9 the! " Θ ", log : " If log 7 ) > 9 the! " Θ " CSE 7 SP 18 KASEY CHAMPION 18

19 Apply Master Theorem Give a recurrece of the form: 0 %h'" " = 1! " = 1! " 2 + " +,h'%./' If If log! " Θ " 7 1 < 8 the log 7 1 = 8! " Θ " log ; " the! " = 1 %h'" " 1 2! " 2 + " +,h'%./' a = 2 b = 2 c = 1 d = 1 If log 7 1 > 8 the! " Θ " A log 7 1 = 8 log ; 2 = 1! " Θ " log ; " Θ " D log ; " CSE 7 SP 18 KASEY CHAMPION 19

20 Reflectig o Master Theorem Give a recurrece of the form: The log 5 ) < 6! " = $ %h'" " = 1 )! " * + ",.h'/%01' 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 If If If log! " Θ ", 5 ) < 6 the log 5 ) = 6! " Θ ", log 9 " the log 5 ) > 6 the! " Θ " ;<= >? The case log 5 ) = 6 Work is equally distributed across call stack (throughout the tree ) Overall work is approximately work at top level x height h'0fh. log 5 ) */)"6hC/D ", log 5 ) A')BC/D $ " ;<= >? The log 5 ) > 6 case Recursive case divides work faster tha it coquers work Most work happes ear bottom of tree Leaf work domiates brach work CSE 7 SP 18 KASEY CHAMPION 20