Transcript Lecture6MRM

Priority Queues and Heaps
15-211
Fundamental Data Structures
and Algorithms
Margaret Reid-Miller
29 January 2004
Plan
• Today
 Priority Queues
 Heaps
 Implementation questions
• Reading:
 For today: Chapters 21.1-21.4, 6.9
• Reminder: Start HW3
A Simulation
We want to simulate a chain of events Ei.
Each event Ei is associated with a time ti.
No problem: sort the times, and run
through them in order.
t1 < t2 < t3 < ... < tn
But what if an event can cause the creation
of some new events in the future?
A Simulation
We can still sort the original events by time,
but not the additional ones: we don't yet
know what/when they are.
We need to be able to dynamically
- insert an event into a data structure, and
- extract the next event.
And, of course, the operations should be
cheap.
An Example: Street Crossing
Suppose a robot wants to cross a street. No
jaywalking, it can only move straight forward or
backward.
Maximum speed of the robot is c.
Cars are coming from either side, but the robot
has complete information about when and where
they cross your line.
Problem:
Find a way for the robot to get across without
getting hit.
More Street Crossing
How do we model the given information?
In space-time a car is just a rectangular box.
Movement of the robot is constrained to a lightcone.
space
time
More Street Crossing
Hugging an obstacle (not that unrealistic, just
inflate the obstacles to account for safety buffer).
A Death Trap
Note how obstacles may be “fused” together.
Street Crossing
The times when obstacles appear/disappear are
clearly events.
But there are more: The times when the robot
touches an obstacle.
Or when the light cones from both ends of a
disappearing obstacle meet.
Note that these events have to be computed as
we go along, there is no way to compute them
all at the outset.
Time Travel
Note how the first event happens before the first
obstacle arrives! We need to go backwards in
time, starting at the end of the last obstacle.
Towards An Algorithm
How does this translate into a data
structure?
What are the crucial operations?
What would be a simple (not necessarily
efficient) implementation?
Where is room for improvement?
The priority queue interface
A Priority Queue maintains a dynamic set of
comparable objects where only have access
to the smallest object:
• Insert
• findMin
• deleteMin
As usual, we consider storing only simple items.
But in practice store pairs (key, object).
Priority Queues: Applications
• Event simulations (key is time)
• Shortest paths in a graph
• Huffman codes
• Sorting (sort of)
Not a universal data structure (like hash
tables) but irreplaceable in some cases.
Possible priority queue implementations
• Linked list
 deleteMin
 insert
O(1)
O(N)
or
O(N)
O(1)
• Search trees
 All operations
• Heaps
 deleteMin
 insert
O(log N)
avg (assume random) worst
O(log N)
2.6
O(log N)
O(log N)
O(N)
O(N)
special case:
 buildheap
i.e., insert*N
Possible priority queue implementations
• Linked list
 deleteMin
 insert
O(1)
O(N)
or
O(N)
O(1)
• Search trees
 All operations
• Heaps
 deleteMin
 insert
O(log N)
avg (assume random) worst
O(log N)
2.6
O(log N)
O(log N)
O(N)
O(N)
special case:
 buildheap
i.e., insert*N
Possible priority queue implementations
• Linked list
 deleteMin O(1)
 insert
O(N)
• Search trees
 All operations
• Heaps
or
O(N)
O(1)
O(log N)
avg (assume random)
 deleteMin O(log N)
 insert
2.6
 buildheap O(N)
N inserts
worst
O(log N)
O(log N)
O(N)
A Digression:
Perfect and Complete
Binary Trees
Perfect binary trees
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Perfect binary trees
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Perfect binary trees
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Perfect binary trees
• How many nodes?
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h=3
N=15
Perfect binary trees
• How many nodes?
 In general: N =
= 2h+1 - 1
 Most of the nodes are leaves
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h=3
N=15
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A Serious Proof
Define PBT by structural induction:
- nil is a PBT and H(nil) = 0
- (a,L,R) is a PBT whenever L and R are
PBTs and H(L) = H(R).
Moreover, H( (a,L,R) ) = H(L) + 1.
size(nil) = 0
size( (a,L,R) ) = size(L) + size(R) + 1.
Then size(T) = 2H(T) -1.
Quiz Break
Red-green quiz
• In a perfect binary tree, what is the
sum of the heights of the nodes?
 Give a mathematical characterization,
the prettier the better.
 Give a tight upper bound (in big-O
terms)
Perfect binary trees
• What is the sum of the heights?
S=
< N = O(N)
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h=3
N=15
Complete binary trees
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Complete binary trees
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Representing complete binary trees
• Linked structures?
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Representing complete binary trees
• Linked structures? No!
• Instead, use arrays!
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Representing complete binary trees
• Arrays
 Parent at position i in the array
 Children at positions 2i and 2i+1
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Representing complete binary trees
• Arrays (1-based)
 Parent at position i
 Children at 2i and 2i+1.
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8 9 10
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Representing complete binary trees
• Arrays (1-based)
 Parent at position i
 Children at 2i and 2i+1.
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4 5 6 7
8 9 10
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Representing complete binary trees
• Arrays (1-based)
 Parent at position i
 Children at 2i (and 2i+1).
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8 9 10
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Representing complete binary trees
• Arrays (1-based)
 Parent at position i
 Children at 2i and 2i+1.
public class BinaryHeap {
private Comparable[] heap;
private int size;
public BinaryHeap(int capacity) {
size=0;
heap = new Comparable[capacity+1];
}
. . .
Representing complete binary trees
• Arrays
 Parent at position i
 Children at 2i and 2i+1.
• Example: find the leftmost child
int left=1;
for(; left<size; left*=2);
return heap[left/2];
• Example: find the rightmost child
int right=1;
for(; right<size; right=right*2+1);
return heap[(right-1)/2];
Quiz Break
• What index is the parent of node i?
• What would the array representation
be for a complete tertiary tree?
I.e., In a 3-way complete tree which
indices are the children of node i?
What is its parent index?
Implementing Priority
Queues with Binary Heaps
Binary heaps: the invariant
• Representation invariants
1. Structure property
• Complete binary tree (i.e. the elements of
the heap are stored in positions 1…size of
the array)
2. Heap order property
• Parent keys less than children keys
Heaps
• Representation invariant
1. Structure property
• Complete binary tree
• Hence: efficient compact representation
2. Heap order property
• Parent keys less than children keys
• Hence: rapid insert, findMin, and deleteMin
• O(log(N)) for insert and deleteMin
• O(1) for findMin
The heap order property
• Each parent is less than each of its
children.
• Hence: Root is less than every other
node.
 (obvious proof by induction)
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Operating with heaps
Representation invariant:
• All methods must:
1. Produce complete binary trees
2. Guarantee the heap order property
• All methods may assume
1. The tree is initially complete binary
2. The heap order property holds
Constructor method
• All methods must:
1. Produce complete binary trees
• Trivially true
2. Guarantee the heap order property
• Also trivially true
• This is the base case
findMin ()
• The code
public boolean isEmpty() {
return size == 0;
}
public Comparable findMin() {
if(isEmpty()) return null;
return heap[1];
}
• Does not change the tree
 Trivially preserves the invariant
insert (Comparable x)
• Process
1. Create a “hole” for x at the next tree leaf
heap[size+1]
This preserves the completeness of the tree.
2. Percolate the hole up the tree until the
heap order property is satisfied.
This assures the heap order property is satisfied.
insert (Comparable x)
• Process
1. Create a “hole” at the next tree cell for x.
heap[size+1]
This preserves the completeness of the tree
assuming it was complete to begin with.
2. Percolate the hole up the tree until the
heap order property is satisfied.
This assures the heap order property is
satisfied assuming it held at the outset.
Complete the binary tree
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Not really there...
Percolation up
• Bubble the hole up the tree until the heap
order property is satisfied.
hole = 11
HOP false
hole = 5
HOP false
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Percolation up
• Bubble the hole up the tree until the heap
order property is satisfied.
hole = 5
HOP false
hole = 2
HOP true
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done
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Percolation up
public void insert(Comparable x)
throws Overflow
{
if(isFull()) throw new Overflow();
int hole = ++size;
for(;
hole>1 && x.compareTo(heap[hole/2])<0;
hole/=2)
heap[hole] = heap[hole/2];
heap[hole] = x;
}
Percolation up
public void insert(Comparable x)
throws Overflow
{
if(isFull()) throw new Overflow();
int hole = ++size;
for(;
hole>1 && x.compareTo(heap[hole/2])<0;
hole/=2)
Integer division
heap[hole] = heap[hole/2];
heap[hole] = x;
}
deleteMin()
/**
* Remove the smallest item from the priority queue.
* @return the smallest item, or null, if empty.
*/
public Comparable deleteMin( )
{
if(isEmpty()) return null;
Comparable min = heap[1];
heap[1] = heap[size--];
percolateDown(1);
return min;
}
Grab min
element
Temporarily
place last
element at top
!!!
Complete binary tree
• Establish the compete binary tree
property by moving leaf to root.
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Percolation down
• Bubble the transplanted leaf value down
the tree until the heap order property is
satisfied.
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Percolation down
• Bubble the transplanted leaf value down
the tree until the heap order property is
satisfied.
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done
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percolateDown(int hole)
private void percolateDown( int hole )
{
int child = hole;
Initially 1
Comparable tmp = heap[hole];
for( ; hole*2 <= size; hole=child )
Start at left child
{
child = hole * 2;
Is there a right child?
if(child!=size &&
heap[child+1].compareTo(heap[child]) < 0)
child++;
Select smaller child
if(array[child].compareTo(tmp) < 0)
heap[hole] = heap[child];
else break;
}
}
heap[hole] = tmp;
Bubble up if smaller than tmp
Exit loop if not bubbling
Finally, place the orphan
deleteMin ()
• Observe that both components of
the representation invariant are
preserved by deleteMin.
1. Completeness
• The last cell ( heap[size] ) is vacated,
providing the value to percolate down.
• This assures that the tree remains
complete.
2. Heap order property
deleteMin ()
• Observe that both components of
the representation invariant are
preserved by deleteMin.
1. Completeness
• The last cell ( heap[size] ) is vacated,
providing the value to percolate down.
• This assures that the tree remains
complete.
2. Heap order property
deleteMin ()
• Observe that both components of
the representation invariant are
preserved by deleteMin.
1. Completeness
• The last cell ( heap[size] ) is vacated,
providing the value to percolate down.
• This assures that the tree remains
complete.
2. Heap order property
• The percolation algorithm assures that the
orphaned value is relocated to a suitable
position.
buildHeap()
• Start with complete (unordered) tree
• Starting from bottom, repeatedly call
percolateDown()
Void buildHeap () {
for (int i = size/2; i>0; i--)
percolateDown(i);
}
buildHeap() performance
• At each iteration, have to do work
proportional to the height of the
current node
• Therefore, total running time is
bounded by the sum of the heights
of all of the nodes
Another Digression:
Invariants
Representation Invariants
• Necessary for a clear understand an
algorithm or piece of code.
• The most useful kind of comment
you can make in a piece of code!
(Invariants)
Plus ça change, plus c’est la même chose.
• The role of an induction hypothesis
(Invariants and induction)
• Induction hypothesis
 What you are allowed to assume
• At the start
• About the result values of a recursive call
• About the object state when a method is
called
 What you must deliver
• At the end
• Of the result values when the recursive call
returns
• About the object state when the method
returns
(Invariants and induction)
• Induction hypothesis
 What you are allowed to assume
• At the start of a loop iteration
• About the result values of a recursive call
• About the object state when a method is
called
 What you must deliver
• At the end of the loop iteration
• Of the result values when the recursive call
returns
• About the object state when the method
returns
(Invariants and induction)
• Induction hypothesis
 What you are allowed to assume
•
• About the result values of a recursive call
• About the object state when a method is
called
 What you must deliver
•
• Of the result values when the recursive call
returns
• About the object state when the method
returns
(Invariants and induction)
• Induction hypothesis
 What you are allowed to assume
• At the start of a loop iteration
• About the result values of a recursive call
• About the object state when a method is
called
 What you must deliver
• At the end of the loop iteration
• Of the result values when the recursive call
returns
• About the object state when the method
returns
(Invariants and induction)
• Induction hypothesis
 What you are allowed to assume
• At the start of a loop iteration
• About the result values of a recursive call
• About the object state when a method is
called
 What you must deliver
• At the end of the loop iteration
• Of the result values when the recursive call
returns
• About the object state when the method
returns
(Invariants)
Plus ça change, plus c’est la même chose
• The role of an induction hypothesis
• Invariants in programs
 Loop invariants
 Recursion invariants
 Representation invariants
(Representation invariant)
• What must always be true of the data
structure when an operation completes.
• Theorem:
 Suppose each constructor assures the representation
invariant is initially correct.
 Suppose each method preserves the representation
invariant, assuming it is true initially.
 Then the representation invariant will always be true at
the completion of each method.
(Representation invariant)
• What must always be true of the data
structure when an operation completes.
• Theorem:
 Suppose each constructor assures the representation
invariant is initially correct.
 Suppose each method preserves the representation
invariant, assuming it is true initially.
 Then the representation invariant will always be true at
the completion of each method.
(Representation invariants)
• What must always be true of the data
structure when an operation completes.
• Theorem:
 Suppose each constructor assures the representation
invariant is initially correct.
 Suppose each method preserves the representation
invariant, assuming it is true initially.
 Then the representation invariant will always be true at
the completion of each method.
 Assuming the code is not concurrent!
Heapsort
• Obviously we can use a priority
queue to sort . . . just insert all
the keys then do repeated
deleteMins
• However we can take advantage
of the fact that the heap always
uses the first part of the array to
do this with no extra space.
• This is called heapsort.
Heapsort
• Reverse the sense of the heap order (largest at the
root)
• Start from position 0 in the array (children are
2i+1 and 2i+2)
• Call percDown(a, i, len)
Public static void heapsort(Comparable[] a) {
for(int i = a.length/2; i>=0; i--)
percDown(a, i, a.length);
for (int j = a.length-1; j>0; j--) {
swapReferences(a, 0, j);
percDown(a, 0, j);
}
}
Heapsort invariant
• At the start of the loop over j:
 a[j+1]…a[a.length-1] are sorted
 a[0]…a[j] are a heap
Homework 3
On Blackboard.
You’re all working on it already.
Right?