CSE 332. Data Structures and Parallelism

Transcription

1 Aam Blak Lecture 6a Witer 2017 CSE 332 Data Structures a Parallelism

2 CSE 332: Data Structures a Parallelism More Recurreces T () T (/2) T (/2) T (/4) T (/4) T (/4) T (/4)

3 P1 De-Brief 1 You i somethig substatial! You worke with real worl software You hoe your ebuggig skills You trasitioe from 143 to 332 You ejoye it?? (okay, ot the ebuggig, but... ) Oh, some presets... tokes++ EX06 Now Due Moay; EX07 is ea :( While we re here...

4 Solvig the reverse Recurrece 2 0 if = 0 T () = c 0 +c 1 +T( 1) otherwise T () = (c 0 +c 1 ) + T ( 1) = (c 0 +c 1 ) + (c 0 +c 1 ( 1)) + T ( 2) = (c 0 +c 1 ) + (c 0 +c 1 ( 1)) + (c 0 +c 1 ( 2))++(c 0 +c 1 (1))+ 0 1 = i=0 1 = i=0 (c 0 +c 1 ( i)) c 0 + i=0 c 1 ( i) + 0 = c 0 +c 1 i + 0 i=1 = c 0 +c 1 ( (+1) ) = O( 2 )

5 Solvig Liear Recurreces 3 A recurrece where we solve some costat piece of the problem (e.g. -1, -2, etc.) is calle a Liear Recurrece. We solve these like we i above by Urollig the Recurrece. This is a facy way of sayig plug the efiitio ito itself util a patter emerges. Now, back to mergesort.

6 Aalyzig Merge Sort 4 Merge Sort 1 sort(l) { 2 if (L.size() < 2) { 3 retur L; 4 } 5 else { 6 it mi = L.size() / 2; 7 retur merge( 8 sort(l.sublist(0, mi)), 9 sort(l.sublist(mi, L.size())) 10 ); 11 } 12 } First, we ee to fi the recurrece: 0 if = 0 T () = 1 if = 1 c 0 +c 1 +2T (/2) otherwise This recurrece is t liear! This is a ivie a coquer recurrece.

7 Aalyzig Merge Sort 5 0 if = 0 T () = 1 if = 1 c 0 +c 1 +2T (/2) otherwise This time, there are multiple possible approaches: Urollig the Recurrece T () = (c 2 +c 1 ) + 2(c 2 +c 1 +2T (/4)) = (c 2 +c 1 ) + 2(c 2 +c 1 +2(c 2 +c 1 +2T (/8))) = c 2 +2c 2 + 4c 2 ++argh+ This works, but I rarely recomme it. Isight: We re brachig i this recurrece. So, represet it as a tree!

8 Merge Sort: Solvig the Recurrece 6 0 if = 0 T () = 1 if = 1 c 0 +c 1 +2T (/2) otherwise T ()

9 Merge Sort: Solvig the Recurrece 7 0 if = 0 T () = 1 if = 1 c 0 +c 1 +2T (/2) otherwise T (/2) T (/2)

10 Merge Sort: Solvig the Recurrece 8 0 if = 0 T () = 1 if = 1 c 0 +c 1 +2T (/2) otherwise /2 /2 T (/4) T (/4) T (/4) T (/4)

11 Merge Sort: Solvig the Recurrece 9 0 if = 0 T () = 1 if = 1 c 0 +c 1 +2T (/2) otherwise /2 /2 /4 /4 /4 /4

12 Merge Sort: Solvig the Recurrece 10 0 if = 0 T () = 1 if = 1 c 0 +c 1 +2T (/2) otherwise /2 /2 lg() /4 /4 /4 /4 Sice the recursio tree has height lg() a each row oes work, it follows that T () O(lg()).

13 sum Examples #1 11 Fi A Big-Oh Bou For The Worst Case Rutime 1 sum() { 2 if ( < 2) { 3 retur ; 4 } 5 retur 2 + sum( 2); 6 } 0 if = 0 T () = 0 if = 1 c 0 +T( 2) otherwise T () = c 0 +c 0 + +c = c 0 ( 2 )+ 0 = O()

14 sum Examples #2 12 Fi A Big-Oh Bou For The Worst Case Rutime 1 biarysearch(l, value) { 2 if (L.size() == 0) { 3 retur false; 4 } 5 else if (L.size() == 1) { 6 retur L[0] == value; 7 } 8 else { 9 it mi = L.size() / 2; 10 if (L[mi] < value) { 11 retur biarysearch(l.sublist(mi + 1, L.size()), value); 12 } 13 else { 14 retur biarysearch(l.sublist(0, mi), value); 15 } 16 } 17 } 0 if = 0 T () = 1 if = 1 c 0 +T(/2) otherwise

15 sum Examples # if = 0 T () = 1 if = 1 c 0 +T(/2) otherwise c 0 c 0 lg() c 0 1 So, T () = c 0 (lg() 1)+ 1 = O(lg).

16 Master Theorem 14 Cosier a recurrece of the form: if = 1 T () = at ( b )+c otherwise The, If log b (a) < c, the T () = Θ( c ). If log b (a) = c, the T () = Θ( c lg()). If log b (a) > c, the T () = Θ( log b (a) ). Saity Check: For Merge Sort, we have a = 2,b = 2,c = 1. The, log 2 (2) = 1 = 1. So, T () = lg.

17 Provig the First Case of Master Theorem 15 if = 1 T () = at ( b )+c otherwise We assume that log b (a) < c. The, urollig the recurrece, we get: T () = c +at (/b) = c +a((/b) c +at (/b 2 )) = c +a(/b) c +a 2 (/b 2 ) c + +a log b () (/b log b ) c log b () = i=0 a i ( c b ic ) log b () = c ( a i i=0 b c ) = c ( a b c ) log b ()+1 1 ( a b c ) 1 c (( a b c ) log b () ) c

18 Aam Blak Lecture 6b Witer 2017 CSE 332 Data Structures a Parallelism

19 CSE 332: Data Structures a Parallelism Amortize Aalysis

20 Stack ADT & ArrayStack Aalysis 1 Stack ADT push(val) pop() peek() isempty() As val to the stack. Returs the most-recet item ot alreay reture by a pop. (Errors if empty.) Returs the most-recet item ot alreay reture by a pop. (Errors if empty.) Returs true if all iserte elemets have bee reture by a pop. Let s aalyze the time complexity for these various methos. (You kow how they work, because you just implemete them!) Metho Time Complexity isempty() Θ(1) peek() Θ(1) pop() Θ(1) push(val)?? push is actually slightly more iterestig.

21 Aalyzig push for a ArrayStack 2 Best Case Worst Case Isight: Our aalysis seems wrog. Sayig liear time feels wrog.

22 Aalyzig push for a ArrayStack 2 Best Case There s more space i the uerlyig array! The, it s Ω(1). Worst Case Isight: Our aalysis seems wrog. Sayig liear time feels wrog.

23 Aalyzig push for a ArrayStack 2 Best Case There s more space i the uerlyig array! The, it s Ω(1). Worst Case If there s o more space, we ouble the size of the array, a copy all the elemets. So, it s O() Isight: Our aalysis seems wrog. Sayig liear time feels wrog.

24 Aalyzig push for a ArrayStack 3 This is where amortize aalysis comes i. Sometimes, we have a very rare expesive operatio that we ca charge to other operatios. Ituitio: Ret, Tuitio You pay oe big sum for a log perio of time, but you ca affor it because it happes very rarely. Back to ArrayStack Say we have a full Stack of size. The, cosier the ext pushes: The ext push will take O() (to resize the array to size 2) The 1 operatios after that will all be O(1), because we kow we have eough space Cosierig these operatios i aggregate, we have operatios that take (c 0 +c 1 )+( 1) c 2 time. So, how log oes each operatio take: (c 0 +c 1 )+( 1) c 2 max(c 0,c 2 )+c 1 = max(c 0,c 2 )+c 1 = O(1)

25 Aalyzig push for a ArrayStack 4 What happes if we chage our resize rule to each of the followig: Which is better 2, 3 2, or 5? Java uses 3 2 to miimize waste space.

26 Aalyzig push for a ArrayStack 4 What happes if we chage our resize rule to each of the followig: +1 This is really ba! We ca oly amortize over the sigle operatio which gives us: 1 = O() Which is better 2, 3 2, or 5? Java uses 3 2 to miimize waste space.

27 Aalyzig push for a ArrayStack 4 What happes if we chage our resize rule to each of the followig: +1 This is really ba! We ca oly amortize over the sigle operatio which gives us: 1 = O() 3 2 This still works. Now, we go over the ext 3 2 operatios: 5 +(/2 1) 1 2 = O(1) Which is better 2, 3 2, or 5? Java uses 3 2 to miimize waste space.

28 Aalyzig push for a ArrayStack 4 What happes if we chage our resize rule to each of the followig: +1 This is really ba! We ca oly amortize over the sigle operatio which gives us: 1 = O() 3 2 This still works. Now, we go over the ext 3 2 operatios: 5 This is goo too: +(/2 1) 1 2 = O(1) +(4 1) 1 = O(1) 4 Which is better 2, 3 2, or 5? Java uses 3 2 to miimize waste space.