Linked Lists CSC 172 SPRING 2002 LECTURE 3

Agenda  Average case lookup – unsorted list & array  Amortized insert for array  Sorted insert  Binary search

Workshop sign-up Still time : Dave Feil-Seifer Ross Carmara

Average Case Analysis  Consider “lookup” on either a linked list or array  We said “n” run time  Under what conditions do we really get “n”  What do we get when the element is the first in list?  Second.. ?  Third.. ?  Middle.. ?  Last?

Lookup: Array public boolean lookup(Object o){ for (int j = 0 ; j < length;j++) if (datum[j].equals(o)) return true; return false; }

Lookup: Linked List public boolean lookup(Object o){ Object temp = head; while (temp != null){ if ((temp.data).equals(o)) return true; temp = temp.next; } return false; }

Average case  On a list of length n in how many different places can the item be?  If the list is unorganized (unsorted/random) what is the chance that the item is at any one location?  What is the probability of getting “heads” on a coin toss?  What is the probability of getting “3” rolling a die?

Average case analysis  For all the possible locations, multiply the chance of dealing with that location, times the amount of work we have to do at it. First location work = (1/n) * 1 Second location work = (1/n) * 2 Third location work = (1/n) * 3 …. But we have to add them all up

Average case analysis Expected work = (1/n)*1 + (1/n)* (1/n)*n Expected work = (1/n)*(1+2+…+n) Expected work = (1/n)*

 Prove  Then, Expected work == (1/n)*(n(n+1)/2)) == (n+1)/2 Or about n/2 So, proof is important: Thus, chapter 2 In order to calculate expected work

Amortized Analysis Amortize? to deaden? Latin : ad-, to + mortus dead Modern usage: payment by installment - sort of like a student loan, where you... never mind The idea is to spread payments out

Amortized analysis  Spread the “cost” out over all the “events”  Consider insert() on an array  We know the worst case is when we have to expand()  Cost of “n”  But, we don’t have to do this every time, right

Cost of inserting 16 items – array[2] capacity

Cost of inserting 16 items – array[2]

Cost of inserting 16 items – array[2] So, what is the average (amortized) run time? We will formalize this proof Later in the course.

Recap  Insert on a linked list?  Constant - 1  Insert on an array?  Worst case n  Average case constant – 2  Lookup on a list?  n/2  Lookup on an array  n/2  Can we do better?

What if we kept the array sorted?  How do we do it?  How much does it cost?  Imagine inserting “5” into  We have to “shift” (could you write this?)

So, suppose we kept the list sorted  Each insert now costs “n”  But what happens to lookup?

Guessing Game  I’m thinking of a number 1<=x<=64  I’ll tell you “higher” or “lower”  How many guesses do you need?  29  47 33 You know with “higher” or “lower” you can divide The remaining search space in half (i.e. log 2 (64)==6)

Binary Search  Recursively divide the array half until you find the item (if it is sorted).

Lookup on sorted array public boolean lookup(Object o) { return lookupRange(Object o,0,length-1); }

LookupRange: Binary Search public static void lookupRange(Object o, int lower,int higher){ if (lower > higher) return false; else { int mid = (low + high)/2; if (datum[mid].compareTo(o) < 0) return lookupRange(o,lower,mid); else if (dataum[mid].compareTo(o) > 0) return lookupRange(o,mid+1,higher); else return true; // o == datum[mid] }

So,  For the price of “n” instead of “1” on insertion We can go from “n/2” to log 2 (n) on lookup.  Which is “better”?  Constant insert & n lookup  n insert & log 2 (n) lookup

All very well and good  But can it help me get a date?  YES!  In order to get a date you have to look up someone’s phone number in the phone book.  You can save time by using binary search.