Quicksort 4-18-2013 Th Thursda sday, April 18 18 th th Barben Ro - - PowerPoint PPT Presentation

Quicksort 4-18-2013

Th Thursda sday, April 18 18th

th

Barben Ro Rooms s A&B &B, Cheel

 4:00 00 pm pm–4:4 4:45 5 pm pm Panel discussion on “Technical Career Opportunities”, Moderated by Bob Lockwood ‘86  5:00 00 pm pm–5:45 5:45 pm pm Roundtables:

 Ho

How to Commun municate icate with your r Clien ent – Differences in Internal vs. External Client Development with Chris Snelling ‘86 and Rich Bogart ‘86

 Techn

chnology

• logy Trends

nds in Indus ustr try y with Ron Ayers ‘02 and Chris Fohlin ‘07

 Trick

icks and Tips for the Job Seeker ker with Dan Dedrick ‘06 and Bob Lockwood ‘86  6:00 00 pm – 6:45 5 pm “Building Your Digital Presence”; Presentation by Chris Fohlin ’07

 Employers and clients do their homework on you before you walk

through the door. Does your online presence paint the right picture? Learn best practices for managing, and maximizing, your professional digital image.

)

 Sorting

 Quicksort

 OOP

 inheritance

Reading: Maciel

 Chapter 15, Sorting

Project#2: Evil Hangman, due Wed. 4/24 see the sample output

 Selection Sort ϴ(n (n2) if ( [start,stop) contains more than one element ) { i_max = index of the largest element in [start,stop) swap a[i_max] and a[stop - 1] sort [start, stop-1) }  Insertion Sort ϴ(n (n2) if ( [start,stop) contains more than one element ) { sort [start, stop-1) insert a[stop-1] into [start, stop-1) }

 Mergesort if ( the array contains more than one element ) { sort the first half of the array sort the second half of the array merge the two sorted halves }  Quicksort

 find a “pivot point”  partition around the pivot  sort halves

 example run: (top level of recursion) [ 60 12 37 42 25 38 16 ] [ 60 12 37 42 ] [ 25 38 16 ]

• 1. divide

vide in half lf [ 12 37 42 60 ] ] [ 16 25 38 ]

• 2. sort

rt halves lves [ 12 16 25 37 38 42 60 ]

• 3. merge

erge

 example run: trace [ 60 12 37 42 25 38 16 ] [ 60 12 37 42 ] [ 25 38 16 ] ] [ 60 12 ] [ 37 42 ] [ 25 38 ] [ 16 ] [ 60 ] [ 12 ] [ [ 37 ] [ 42 ] ] [ 25 ] [ 38 ] [ 16 ] [ 12 60 ] [ 37 42 ] [ 25 38 ] [ 16 ] [ 12 37 42 60 ] [ 16 25 38 ] [ 12 16 25 37 38 42 60 ]

 merging algorithm: array [ 60 12 37 42 25 38 16] first rst secon

• nd

result sult

[ 12 37 42 60 ] [ 16 25 38 ] [ ] [ 37 42 60 ] [ 16 25 38 ] [ [ 12 ] [ 37 42 60 ] [ 25 38 ] [ [ 12 16 ] [ 37 42 60 ] [ 38 ] [ [ 12 16 25 ] [ 42 60 ] [ 38 ] [ [ 12 16 25 37 ] [ 42 60 ] [ ] [ [ 12 16 25 37 38 ] [ ] [ ] [ [ 12 16 25 37 38 42 60 ]

 idea:

 find a “pivot point”  partition around the pivot  sort halves

} if ( the array contains more than one element ) { choose a pivot element partition the array around the pivot sort each subarray }

 Partition must rearrange the array so that the following 3 conditions hold:

• 1. the element a[p] is in its final place in the

array, for some p

• 2. all elements in a[first..p-1] are ≤ a[p]
• 3. all elements in a[p+1..last] are ≥ a[p]

 example: [ 30 30 10 14 37 42 13 51 5 30 ] pivot vot

• 1. select a pivot
• 2. partition around the pivot

[ 10 14 13 5 | 30 30 | 37 42 51 30 ]

• 3. sort the partitions

[ 5 10 13 14 30 30 30 37 42 51 ]

 If the pivot point divides the list exactly in half, then the number of comparisons would be N log N  If the pivot point is chosen so that one partition is always empty, then the running time is N2

 The performance of quicksort heavily depends on the pivot point  best st case se: pivot element is the median value in the array (divides the array in half)  worst rst case se: the first element of the array is chosen to be the pivot point, but the array is already sorted  in practice, large parts of an input array can be already in sorted order  choose a random element to be the pivot (but then need a call to a pseudorandom number generator)  compromise: choose the median of the first, middle and last element of the array

 example: [ 60 12 37 42 25 38 16 ] piv ivot

• t
• 1. select a pivot: median of [16 42 60] => 42
• 2. partition around the pivot

[ 12 37 25 38 16 ] 42 [ 60 ]

• 2. sort the partitions

[ 12 16 25 37 38 42 60 ]

[ 60 12 37 42 25 38 16 ] [ 12 37 25 38 16 ] 42 [ 60 ] ] [ 12 ] ] 16 16 [ 37 25 38 ] 42 42 [ 60 ] [ 12 ] ] 16 16 [ 25 ] 37 37 [ [ 38 ] 42 42 [ 60 ] [ 12 ] ] 16 [ 25 37 37 38 ] 42 42 [ 60 ] ] [12 16 16 25 37 37 38 ] 42 42 [ 60 ] [12 16 16 25 37 37 38 42 42 ] [ 60 ] [ 12 16 25 37 38 42 60 ]

 Quicksort

 worst case: ϴ(n

(n2)

 best & average case: ϴ(n

(n log n)

 easy to write recursively

 Mergesort

 all cases: ϴ(n

(n log n)

 disadvantage: needs extra space proportional to n  good when sequential access is required (e.g. sort a

linked list)

 easy to write recursively

 use a better partitioning element (pivot) to avoid the worst case  median-of-3 partitioning works well in practice  use a simple sort for small partitions (can reduce running time by 20%)

 e.g. if size of array ≤ M then use insertion sort

 The C++ standard library (in <algorithm>) provides the following functions:

template<typename RI> void sort(RI first, RI last); template<typename RI, typename Compare> void sort(RI first, RI last, Compare comp);

 The template parameter RI is a random-access

iterator.

 The data to be sorted is in the range first .. last,

where last is one past the last data value.

 In practice you would tend to use these

 Assume that array items contains 16 integers.

 sort(

t(items, items, items s + 16 16);

• will sort the whole array.

 sort(i

rt(item tems, , items ems + + 8);

• will sort the first half.

 sort(i

rt(item tems, , items ems + + 16, gr greater< ater<int int>()) >());

• will sort in descending order.

 Assume that v is a vector. sort( rt(v. v.beg begin in(), (), v. v.end nd() ()); ); // sorts rts ve vector

• r v

Objects, classes, and inheritance, plus An object is a software entity that combines state and behavior. A class describes the state (member data) and services (member functions) provided by objects that are instances of that class. Classes can be related by inheritance. polymorphism and dynamic binding. Goals of OO: abstraction, encapsulation, comprehensibility, changeability, and reusability.

We’ve now looked at classes and objects. What about inheritance, polymorphism and dynamic binding?

 Object-oriented programming (OOP) is popular because:

 It enables reuse of previous code saved as classes

 Inheritance and hierarchical organization capture idea:

 One thing is a refinement or extension of another

UML Class Diagram class Arrow : public Line {...} An Arrow “is a” Line Arrow

IS-A (inheritance)

Line important to make this public class Line {...}

FilledCircle Line Point

HAS-A

(composition)

2

UML Class Diagram class Line { private: Point endpt1, endpt2; A Line “has a” Point

Circle FilledCircle Point

IS-A

(inheritance)

HAS-A

(composition)

1

base class or superclass derived class or subclass

 Confusing has-a and is-a leads to misusing inheritance  Model a has-a relationship with an attribute (data member/instance variable)

class C { ... private: B part; ...}

 Model an is-a relationship with inheritance

 If every C is-a B then model C as a derived class

(also called subclass) of B

 Show this: in C include : public B:

class C : public B { ... }

/** the Point class represents a 2D point */ class Point { private: int xcoord, ycoord; public: Point() : xcoord(0), ycoord(0) {} Point(int newx, int newy) : xcoord(newx), ycoord(newy) {} int getX() { return xcoord; } int getY() { return ycoord; } void display(); };

class Circle { private: Point center; float radius: public: Circle(): center(Point()), radius(1.0f) {} Circle(int x, int y, float newradius) : center(Point(x,y)), radius(newradius) {} float getRadius() { return radius; } double computeArea(); void display(); };

class FilledCircle public Circle { private: string color; public: FilledCircle(): Circle(), color(“black”) {} FilledCircle(int x, int y, float newradius, string newcolor) : Circle(x, y, newradius), color(newcolor) {} };

/** test program for shapes */ int main() { /* Test the Point class, starting with constructors & accessors: */ Point p1; Point p2(100, 200); Point p3(50, 50); cout << "Point 1: " << p1.getX() << ":" << p1.getY() << endl; cout << "Point 2: " << p2.getX() << ":" << p2.getY() << endl; cout << "Point 3: " << p3.getX() << ":" << p3.getY() << endl; /* test Point's display method */ p2.display(); /* Next test the Circle class, starting with constructors: */ /* (continued on next slide) */

/* test program for shapes, continued */ Circle c1(100, 40, 2.5f); Circle c2; /* test Circle's accessors and method computeArea */ cout << "Circle c1 radius: " << c1.getRadius() << endl; cout << “Area of circle c1 is: " << c1.computeArea() << endl; /* test Circle's display method */ c1.display(); /* Test the FilledCircle class, starting with the constructors */ FilledCircle fc3; FilledCircle fc4(20, 30, 2.5f, "red"); /* continued on next slide */

/* test program for shapes, continued */ /* Test the FilledCircle class, starting with the constructors */ FilledCircle fc3; FilledCircle fc4(20, 30, 2.5f, "red"); /* Test the accessors and inherited method computeArea */ cout << "Filled Circle fc3 color is " << fc3.getColor() << endl; cout << "fc3 area: " << fc3.computeArea() << endl; cout << "Filled Circle fc4 color is " << fc4.getColor() << endl; /* Test the FilledCircle's display method */ fc4.display(); /* Testing completed */ return 0; }

 Sorting

 Maciel: Chapter 15

 Software Life Cycle

 Maciel: Chapter 11

 Error Checking

 Maciel, Chapter 4