LECTURE 40:SELECTIONCSC 212 – Data Structures

Sequence of Comparable elements available Only care implementation has O(1) access

time Elements unordered within the Sequence

Easy finding smallest & largest elements What about if we want the kth largest

element? Real function used surprisingly often

Statistical analyses Database lookups Ranking teams in league

Selection Problem

Could sort Collection as a first step Then just return kth element

Sorting is slow Sorting takes at least O(n*log n) time

Ordered Collection faster for selection time Selection becomes simple O(1) process O(n) insertion time would also result Usually this is not a winning tradeoff

Selection Problem

7 4 9 6 2 2 4 6 7 9k=3 7 4 9 6 27 4 9 6 2 2 4 6 7 9

Works like binary search finds specific value Come from same family of algorithms

Divide-and-conquer algorithms work similarly Divide into easier sub-problems to be

solved Recursively solve sub-problems Conquer by combining solutions to the

sub-problems

Quick-Select

Also known as pruneprune-and--and-searchsearch Read this as divide-and-conquer

Prune: Pick a pivot & split into 2 Collections L will get all elements less than pivot Equal and larger elements go into G Keep pivot element off to the side

Search: Solve only for Collection with solution When k < L.size(), continue search in L pivot is solution when k = L.size()+1 Else, element in G solves this problem

Quick-Select

Quick-Select In Pictures

x

x

L G

if k < L.size() then return quickSelect(k, L)

if k > L.size() then k = k – (L.size() + 1) return quickSelect(k, G)

if k == L.size() then return x

Quick-Select Visualization

Draw Collection and k for each recursive call

k=5, C=(7 4 9 3 2 6 5 1 8)

5

k=2, C=(7 4 9 6 5 8)

k=2, C=(7 4 6 5)

k=1, C=(7 6 5)

Quick-Select Running Time Each call of the algorithm would take

time:

So the worst-case running time:

We would expect the running time to be:

Expected Running Time

Recursive call for Collection of size s Good call: L & G each have less than ¾ * s

elements Bad call: More than ¾ * s elements for L or

G(Note: they cannot both be larger than ¾*s)

7 9 7

7 2 9 4 3 7 6 1

2 4 3 1 7 9 4 3 7 61

7 2 9 4 3 7 6 1

Good call Bad call

How often can we expect to make a good call? Ultimately it will all depend on selection of

pivot ½ of possible pivots would create good split So with random guess, get good call 50% of

the time

Expected Running Time

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Good pivotsBad pivots Bad pivots

Probabilities

Probability Fact #1: After 2 coin tosses, expect to see heads at

least once Probability Fact #2:

Expectations are additive

E(X + Y) = E(X) + E(Y)

E(c * X) = c * E(X)

Let T(n) = expectedexpected execution time

T(n) < b * n * (# calls before good call) + T(¾ * n)T(n) < b * n * 2 + T(¾ * n)T(n) < b * n * 2 + b * ¾ * n * 2 + T((¾)2 * n)T(n) < b * n * 2 + b * ¾ * n * 2 + b * (¾)2 * n * 2 +…T(n) < O(n) + O(n) + O(n) +…

… then a mathematical miracle occurs…T(n) = O(n)

More Big-Oh

Finish week #15 assignment Due on Friday at 5PM

Before Next Lecture…