priority queue ADT heaps 1

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1 COMP 250 Lctur 23 priority quu ADT haps 1 Nov. 1/2,

2 Priority Quu Li a quu, but now w hav a mor gnral dinition o which lmnt to rmov nxt, namly th on with highst priority..g. hospital mrgncy room Assum a st o comparabl lmnts or ys. 2

3 Priority Quu ADT add(lmnt) rmovmin() highst priority = numbr 1 priority p() contains(lmnt) rmov(lmnt) 3

4 How to implmnt a Priority Quu? sortd list? binary sarch tr (last lctur)? balancd binary sarch tr (COMP 251)? hap (nxt 3 lcturs) Not th sam hap you har about in COMP

5 Complt Binary Tr (dinition) c m a g j d j d Binary tr o hight h such that vry lvl lss than h is ull, and all nods at lvl h ar as ar to th lt as possibl 5

6 min Hap (dinition) a b l u m Complt binary tr with uniqu comparabl lmnts, such that ach nod s lmnt is lss than its childrn s lmnt. 6

7 Hap.add(lmnt).g. add( c ) a b l u m 7

8 Hap.add(lmnt).g. add( c ) a b l u m? 8

9 Hap.add(lmnt).g. add( c ) a b l u m c Problm : adding at th nxt availabl slot typically will dstroy th hap proprty. 9

10 W swap c with its parnt. Q: Can this crat a problm with c s ormr sibling, who is now c s child? a b c l u m 10

11 W swap c with its parnt. Q: Can this crat a problm with c s ormr sibling, who is now c s child? A: No. Bcaus c < and < m. Thus, c < m. a b c l u m 11

12 Q: Ar w don? A: Not ncssarily. What about c s parnt? a b c l u m 12

13 W swap c with its (nw) parnt. Now w ar don bcaus c is gratr than its parnt a a c b l u m 13

14 Hap.add(lmnt) add( lmnt ){ cur = nw nod at nxt availabl la position cur.lmnt = lmnt whil (cur!= root) and (cur.lmnt < cur.parnt.lmnt){ swapelmnt(cur, parnt) cur = cur.parnt } } 14

15 Hap.add(lmnt) add( lmnt ){ cur = nw nod at nxt availabl la position cur.lmnt = lmnt whil (cur!= root) and (cur.lmnt < cur.parnt.lmnt){ swapelmnt(cur, parnt) cur = cur.parnt } } 15

16 Hap.add(lmnt) add( lmnt ){ cur = nw nod at nxt availabl la position cur.lmnt = lmnt whil (cur!= root) and (cur.lmnt < cur.parnt.lmnt){ swapelmnt(cur, parnt) // argumnts ar nods cur = cur.parnt } } 16

17 How to build a hap? add( ) add( ) add( ) add( a ) add( g ) 17

18 How to build a hap? add( ) add( ) add( ) add( a ) 18

19 How to build a hap? add( ) add( ) add( ) add( a ) add( g ) 19

20 How to build a hap? add( ) add( ) add( ) add( a ) a add( g ) 20

21 How to build a hap? add( ) add( ) add( ) add( a ) add( g ) g a 21

22 This mthod o building a hap is slow. I will show you a astr mthod two lcturs rom now. 22

23 Hap.rmovMin() rturns root lmnt a c b l u m 23

24 rmovmin() a c b l u m 24

25 rmovmin() Claim: i th root has two childrn, thn th nw root will b gratr than at last on o its childrn. Why? c b a How to solv this problm? m l u 25

26 Swap lmnts with smallr child. rmovmin() b a c l u Kp swapping with smallr child, i ncssary. m 26

27 Lt s do it again. rmovmin() b a c l u m 27

28 Lt s do it again. rmovmin() b b c l u m 28

29 rmovmin() Now swap with smallr child, i ncssary, to prsrv hap proprty. m b c l u 29

30 rmovmin() c b Kp swapping with smallr child, i ncssary. m l u 30

31 rmovmin() c b m l u 31

32 rmovmin(){ tmp = root.lmnt rmov last la nod and put its lmnt into th root cur = root whil ((cur has at last on child) and ( (cur.lmnt > cur.lt.lmnt) or (cur has right child and cur.lmnt > cur.right.lmnt)) ) { minchild = child with th smallr lmnt swapelmnt(cur, minchild) cur = minchild } rturn tmp } 32

33 rmovmin(){ tmp = root.lmnt rmov last la nod and put its lmnt into th root cur = root whil ( (cur has a lt child) and ( (cur.lmnt > cur.lt.lmnt) or (cur has right child and cur.lmnt > cur.right.lmnt)) ) { minchild = child with th smallr lmnt swapelmnt(cur, minchild) cur = minchild } rturn tmp } 33

34 rmovmin(){ tmp = root.lmnt rmov last la nod and put its lmnt into th root cur = root whil ( (cur has a lt child) and ( (cur.lmnt > cur.lt.lmnt) or (cur has right child and cur.lmnt > cur.right.lmnt)) ) { minchild = child with th smallr lmnt swapelmnt(cur, minchild) cur = minchild } rturn tmp } 34

35 add(lmnt) rmovmin() uphap downhap 35

36 Q: What about rmov(lmnt)? 36

37 Q: What about rmov(lmnt)? A: Worst cas Θ(n) Bst cas (not discussd) 37

38 Hap (array implmntation) 1 c 2 3 m a g j Not usd d j d d d d

39 1 a 2 3 b l u Not usd m g n q w z a b l u m g n q w z

40 Nxt two lcturs writ add(lmnt) and rmovmin() using array indics bst and worst cas astr algorithm or building a hap 40

Searching Linked Lists. Perfect Skip List. Building a Skip List. Skip List Analysis (1) Assume the list is sorted, but is stored in a linked list.

Searching Linked Lists. Perfect Skip List. Building a Skip List. Skip List Analysis (1) Assume the list is sorted, but is stored in a linked list. 3 3 4 8 6 3 3 4 8 6 3 3 4 8 6 () (d) 3 Sarching Linkd Lists Sarching Linkd Lists Sarching Linkd Lists ssum th list is sortd, but is stord in a linkd list. an w us binary sarch? omparisons? Work? What if