lecture 24: merge sort –or– lessons from roman empire

LECTURE 24:MERGE SORT –OR–LESSONS FROM ROMAN EMPIRE

CSC 213 – Large Scale Programming

Today’s Goals

Review past discussion of data sorting algorithms Weaknesses of past approaches & when we

use them Can we find limit to how long sorting

needs? What does this mean for sorting & those

past sorts Get good idea of how merge sort is

executed What is algorithm and what will this require? What are execution trees & how they show

runtime?

Ghosts of Sorts Past

We have already seen & discussed 4 sorts Bubble-sort -- O(n2) time sort; slowest

sort Selection-sort -- O(n2) time sort; PQ

concept Insertion-sort -- O(n2) time sort; PQ

concept Heap-sort -- O(n log n) time sort; requires

PQ All of these sorts of limited usefulness

Counting Comparisons

Consider sort as a path in a decision tree Nodes are single decision needed for

sorting

yes no

Is xi > xj ?

Counting Comparisons

Consider sort as a path in a decision tree Nodes are single decision needed for

sorting Traveling from root to leaf sorts data Tree’s height is lower-bound on sorting

complexity

xi < xj ?

xa < xb ?

xc < xd ? xc < xd ?xc < xd ? xc < xd ?

xa < xb ?

Decision Tree Height

Unique leaf for each ordering of data initially Needed to ensure we sort different inputs

differently Consider 4, 5 as data to be sorted using a

tree Could be entered in 2 possible orders: 4, 5

or 5, 4 Need two leaves for this sort unless (4 < 5)

== (5 < 4)


Unique leaf for each ordering of data initially Needed to ensure we sort different inputs

differently Consider 4, 5 as data to be sorted using a

tree Could be entered in 2 possible orders: 4, 5

or 5, 4 Need two leaves for this sort unless (4 < 5)

== (5 < 4)

For sequence of n numbers, can arrange n! ways Tree with n! leaves needed to sort n

numbers Given this many leaves, what is height of

the tree?


With n! external nodes, binary tree’s height is:minimum height (time)

log (n!)

n!

xi < xj ?

xa < xb ?

xc < xd ? xc < xd ?xc < xd ? xc < xd ?

xa < xb ?

The Lower Bound

But what does O(log(n!)) equal?

n! = n * n-1 * n -2 * n-3 * n/2 * … * 2 * 1 n! ≤ (½*n)½*n (½ of series is larger than ½*n)

log(n!) ≤ log((½*n)½*n)

log(n!) ≤ ½*n * log(½*n)

O(log(n!)) ≤ O(½*n * log(½*n))

The Lower Bound

But what does O(log(n!)) equal?

n! = n * n-1 * n -2 * n-3 * n/2 * … * 2 * 1 n! ≤ (½*n)½*n (½ of series is larger than ½*n)

log(n!) ≤ log((½*n)½*n)

log(n!) ≤ ½*n * log(½*n)

O(log(n!)) ≤ O(n log n)

Lower Bound on Sorting

Smallest number of comparisons is tree’s height Decision tree sorting n elements has n!

leaves At least log(n!) height needed for this many

leaves As we saw, this simplifies to at most O(n log

n) height O(n log n) time needed to compare data!

Means that heap-sort is fastest possible (in big-Oh)

Pain-in-the- to code & requires external heap

Lower Bound on Sorting

Smallest number of comparisons is tree’s height Decision tree sorting n elements has n!

leaves At least log(n!) height needed for this many

leaves As we saw, this simplifies to at most O(n log

n) height O(n log n) time needed to compare data!

Means that heap-sort is fastest possible (in big-Oh)

Pain-in-the- to code & requires external heap

Is there a simple sort using only Sequence?

Julius, Seize Her!

Formula to Roman success Divide peoples before an attack Then conquer weakened armies

Common programming paradigm Divide: split into 2 partitions Recur: solve for partitions Conquer: combine solutions

Divide-and-Conquer

Like all recursive algorithms, need base case Has immediate solution to a simple problem Work is not easy and sorting 2+ items takes

work Already sorted 1 item since it cannot be out

of order Sorting a list with 0 items even easer

Recursive step simplifies problem & combines it Begins by splitting data into two equal Sequences

Merges subSequences after they have been sorted

Merge-Sort

Algorithm mergeSort(Sequence<E> S, Comparator<E> C)if S.size() < 2 then // Base case

return Selse // Recursive case

// Split S into two equal-sized partitions S1 and S2

mergeSort(S1, C)mergeSort(S2, C)S merge(S1, S2, C)return S

Merging Sorted Sequences

Algorithm merge(S1, S2, C)Sequence<E> retVal = // Code instantiating a Sequencewhile !S1.isEmpty() && ! S2.isEmpty()if C.compare(S1.get(0), S2.get(0)) < 0 retVal.insertLast(S1.removeFirst())else retVal.insertLast(S2.removeFirst())endifendwhile !S1.isEmpty() retVal.insertLast(S1.removeFirst())endwhile !S2.isEmpty() retVal.insertLast(S2.removeFirst()) endreturn retVal

Execution Tree

Depicts divide-and-conquer execution Recursive call represented by each oval

node Original Sequence shown at start At the end of the oval, sorted Sequence

shown Initial call at root of the (binary) tree

Bottom of the tree has leaves for base cases

Execution Example

Not in a base case7 2 9 4 3 8 6 1 1 2 3 4 6 7 8

9

Execution Example

Not in a base case, so split into S1 & S2

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

Execution Example

Recursively call merge-sort on S1

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

Execution Example


7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

Execution Example

Not in a base case, so split into S1 & S2

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

Execution Example


7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

Execution Example


7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

Execution Example

Still no base case, so split again & recurse on S1

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

7 7

Execution Example

Enjoy the base case – literally no work to do!

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

7 7

Execution Example

Recurse on S2 and solve for this base case

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

7 7 2 2

Execution Example

Merge the two solutions to complete this call

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

7 7 2 2

Execution Example

Recurse on S2 and sort this subSequence

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

9 4 4 9

7 7 2 2

Execution Example

Split into S1 & S2 and solve the base cases

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

9 4 4 9

9 9

4 47 7 2

2

Execution Example

Merge the 2 solutions to sort this Sequence

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

9 4 4 9

9 9

4 47 7 2

2

Execution Example

I feel an urge, an urge to merge

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

9 4 4 9

9 9

4 47 7 2

2

Execution Example

Let's do the merge sort again! (with S2)

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

9 4 4 9

3 8 6 1 1 3 6 8

9 9

4 47 7 2

2

Execution Example


7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

9 4 4 9

3 8 6 1 1 3 6 8

3 8 3 8

8 8

3 3

9 9

4 47 7 2

2

Execution Example


7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

9 4 4 9

3 8 6 1 1 3 6 8

3 8 3 8

6 1 1 6

6 6

1 1

8 8

3 3

9 9

4 47 7 2

2

Execution Example

Merge the last call to get the final result

7 2 9 4 2 4 7 9

7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9

7 2 2 7

9 4 4 9

3 8 6 1 1 3 6 8

3 8 3 8

6 1 1 6

6 6

1 1

8 8

3 3

9 9

4 47 7 2

2

For Next Lecture

New weekly assignment for week was posted Discussing sorts which have few concepts

to code Will return soon enough; do not worry

about it Keep reviewing requirements for

program #2 Preliminary deadlines arrive before final

version Time spent on requirements & design saves

coding Reading on quick sort for this Friday

Guess what? It can be really, really, quick Severe drawbacks also exist; read to

understand this

lecture 24: merge sort –or– lessons from roman empire

Documents