Linked Lists

1
Linked Lists

• CSC 172
• SPRING 2002
• LECTURE 3

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Agenda

• Average case lookup unsorted list array
• Amortized insert for array
• Sorted insert
• Binary search

3
Workshop sign-up

• Still time
• Dave Feil-Seifer
• Ross Carmara

4
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?

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Lookup Array

• public boolean lookup(Object o)
• for (int j 0 j lt lengthj)
• if (datumj.equals(o)) return true
• return false

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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

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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?

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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

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Average case analysis

• Expected work (1/n)1 (1/n)2 . . .
  (1/n)n
• Expected work (1/n)(12n)
• Expected work (1/n)

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In order to calculate expected work

• Prove
• Then,
• Expected work (1/n)(n(n1)/2)) (n1)/2
• Or about n/2
• So, proof is important Thus, chapter 2

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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

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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 dont have to do this every time, right

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Cost of inserting 16 items array2
capacity
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Cost of inserting 16 items array2
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Cost of inserting 16 items array2
So, what is the average (amortized) run time?
We will formalize this proof Later in the course.
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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?

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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?)

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So, suppose we kept the list sorted

• Each insert now costs n
• But what happens to lookup?

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Guessing Game

• Im thinking of a number 1ltxlt64
• Ill tell you higher or lower
• How many guesses do you need?
• 29
• 47
• 3
• You know with higher or lower you can divide
• The remaining search space in half (i.e.
  log2(64)6)

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Binary Search

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

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Lookup on sorted array

• public boolean lookup(Object o)
• return lookupRange(Object o,0,length-1)

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LookupRange Binary Search

• public static void
• lookupRange(Object o, int lower,int higher)
• if (lower gt higher) return false
• else
• int mid (low high)/2
• if (datummid.compareTo(o) lt 0)
• return lookupRange(o,lower,mid)
• else if (dataummid.compareTo(o) gt 0)
• return lookupRange(o,mid1,higher)
• else return true // o datummid

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So,

• For the price of n instead of 1 on insertion
• We can go from n/2 to log2(n) on lookup.
• Which is better?
• Constant insert n lookup
• n insert log2(n) lookup

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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
  someones phone number in the phone book.
• You can save time by using binary search.