Application of Data Structures

8/3/2019 Application of Data Structures

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Christopher Moh 2005

Application of Data Structures


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Overview Priority Queue structures

Heaps

Application: Dijkstras algorithm

Cumulative Sum Data Structures onIntervals

Augmenting data structures with extrainfo to solve questions


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Priority Queue (PQ) Structures Stores elements in a list by comparing akeyfield

Often has other satellite data

For example, when sorting pixels by their Rvalue, we consider the R as the key field

and GB as satellite data Priority queues allow us to sort

elements by their key field.


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Common PQ operations Create()

Creates an empty priority queue

Find_Min() Returns the smallest element (by key field)

Insert(x) Insert element x (with predefined key field)

Delete(x) Delete position x from the queue

Change(x, k) Change key field of position x to k


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Optional PQ operations Union (a,b)

Combines two PQs a and b

Search (k)

Returns the position of the element in theheap with key value k


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Considerations when

implementing a PQ in competition How complicated is it?

Is the code likely to be buggy?

How fast does it need to be? Does a constant factor also come into the

equation?

Do I need to store extra data to do a Search? During the course of this presentation, we shall

assume that there exists existing extra data whichallows us to do a search in O(1) time. Thehandling of this data structure will be assumedand not covered.


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Linear Array Unsorted Array

Create, Insert, Change in O(1) time

Find_min, Delete in O(n) time

Sorted Array

Create, Find_min in O(1) time

Insert, Delete, Change in O(n + log n) =O(n) time


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Binary Heaps Will be the most common structure that will

be implemented in competition setting

Efficient for most applications Easy to implement

A heap is a structure where the value of anode is less than the value of all of its

children A binary heap is a heap where the maximum

number of children for each node is 2.


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Array implementation Consider a heap of size nheapin an array

BHeap[1..nheap](Define BHeap[nheap+1 ..(nheap*2)+1] to be INFINITY for practical reasons) The children of BHeap[x] are BHeap[x*2] and

BHeap[x*2+1]

The parent of BHeap[x] are BHeap[x/2]

This allows a near uniform Binary Heap where we canensure that the number of levels in this heap is O(log n)

Some properties wrt Keyvalues: BHeap[x] >= BHeap[x/2],BHeap[x]


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PQ Operations on a BHeap We define BTree(x) to be the Binary Tree rooted at

BHeap[x]

We define Heapify(x) to be an operation that does

the following: Assume: BTree(x*2) and BTree(x*2+1) are binary heaps but

BTree(x) is not necessarily a binary heap

Produce: BTree(x) binary heap

Details of Heapify in later slides but for now, we assume

Heapify is O(log n)

For the rest of the presentation, we assume thevariable nrefers to nheap


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Operations on a BHeap Create is trivial O(1) time Find_min:

1. Return BHeap[1]

O(1) time Insert (element with key value x)

1. nheap++2. BHeap[nheap] = x3. T = nheap

4. While (T != 1 && Bheap[T] < BHeap[T/2])1. Swap (Bheap[T], BHeap[T/2]2. T = T / 2

O(log n) time as the number of levels is O(log n)


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Operations on a BHeap ChangeDown (position x, new key value k)

Assume: k < existing BHeap[x]

1.

BHeap[x] = k2. T = x

3. While (T != 1 && BHeap[T] < BHeap[T/2])1. Swap (BHeap[T], BHeap[T/2])

2. T = T/2

Complexity: O(log n)

This procedure is known as bubbling up theheap


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Operations on a BHeap ChangeUp (position x, new key value

k)

Assume: k > existing BHeap[x]

1. BHeap[x] = k

2. Heapify(x)

O(log n) as complexity of Heapify is O(logn)


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Operations on a BHeap Delete (position x on the heap)

1. BHeap[x] = BHeap[nheap]

2.

nheap3. Heapify(x)

4. T = x

5. While (T != 1 && BHeap[T] < BHeap[T/2])1. Swap (BHeap[T], BHeap[T/2])

2. T = T / 2

Complexity is O(log n)

Why must I do both Heapify and bubble up?


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Operations on a BHeap Heapify (position x on the heap)

1. T = min(BHeap[x], BHeap[x*2], BHeap[x*2+1])

2. If (T == BHeap[x]) return;3. K = position where BHeap[K] = T

4. Swap(BHeap[x], BHeap[K])

5. Heapify(K)

O(log n) as the maximum number of levels inthe heap is O(log n) and Heapify only goesthrough each level at most once


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BHeap Operations: Summary

Create, Find_min in O(1) time

Change (includes both ChangeUp and

ChangeDown), Insert, and Delete areO(log n) time

Union operations are how long?

Insertion: O(n log n) union

Heapify: O(n) union


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Corollary: Heapsort

We can convert an unsorted array to a heapusing Heapify (why does this work?):

1. For (i = n/2; i >= 1; i--)1. Heapify(i)

We can then return a sorted list (list initiallyempty):

1. For (i = 1; i


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Binomial Trees

Define Binomial Tree B(k) as follows:

B(0) is a single node

B(n), n != 0, is formed by merging two B(n-1)trees in the following way:

The root of the B(n) tree is the root of one of the B(n-1)trees, and the (new) leftmost child of this root is the rootof the other B(n-1) tree.

Within the tree, the heap property holds i.e. thatthe keyfield of any node is greater than the keyfield of all its children.


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Properties of Binomial Trees

The number of nodes in B(k) is exactly2^k.

The height of B(k) is exactly (k + 1)

For any tree B(k)

The root of B(k) has exactly k children

If we take the children of B(k) from left toright, they form the roots of a B(k-1), B(k-2), , B(0) tree in that order


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Binomial Heaps

Binomial Heaps are a forest of binomial trees withthe following properties: All the binomial trees are of different sizes

The binomial trees are ordered (from left to right) byincreasing size

If we consider the fact that the size of B(k) is 2^k,the binomial tree B(k) exists in a binomial heap of nnodes iff the bit representing 2^k is 1 in the binary

representation of n For example: 13 (decimal) = 1101 (binary), so the binomial

heap with 13 nodes consists of the binomial trees B(0), B(2),and B(3).


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Binomial Heap Implementation

Each node will store the following data: Keyfield Pointers (if non-existent, points to NIL) to

Parent Next Sibling (ordered left to right; a sibling must have the

same parent); For roots of binomial trees, next sibling points tothe root of the next binomial tree

Leftmost child

Number of children in field degree Any other data that might be useful for the program

The binomial heap is represented by a headpointerthat points to the root of the smallest binomial tree(which is the leftmost binomial tree)


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Operations on Binomial Trees

Link (h1, h2) Links two binomial trees with root h1 and h2 of

the same order k to form a new binomial tree of

order (k+1) We assume h1->key < h2->key which implies

that h1 is the root of the new tree1. T = h1->leftchild2. h1->leftchild = h23. h2->parent = h14. H2->next_sibling= T O(1) time


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Operations on binomial heaps

Create Create a new binomial heap with onenode (keyfield set)

Set Parent, Leftchild, Next sibling to NIL

O(1) time Find_min

1. X = head, min = INFINITY2. While (X != nil)

1. If (X->key < min) min = X->key2. X = X->next_sibling

3. Return min O(log n) time as there are at most log n binomial trees

(log n bits)


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More Operations

Merge (h1, h2, L)

Given binomial heaps with head pointers

h1 and h2, create a list L of all thebinomial trees of h1 U h2 arranged inascending order of size

For any order k, there may be zero, one,or two binomial trees of order k in thislist.


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More Operations

Merge (h1, h2, L)

Assume that NIL is a node of infinitely small

order1. L = empty

2. While (h1 != NIL || h2 != NIL)

1. If (h1->degree < h2->degree)

1.

Append the (binomial)tree with root h1 to L2. h1 = h1->next_sibling

2. Else

1. Apply above steps to h2 instead


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More Operations

Union (h1, h2)

The fundamental operation involving

binomial heaps Takes two binomial heaps with head

pointers h1 and h2 and creates a newbinomial heap of the union of h1 and h2


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More Operations

Union (h1, h2)1. Start with empty binomial heap2. Merge (h1, h2, L)3. Go by increasing k in the list L until L is empty

1. If there is exactly one or exactly three (how can thishappen?) binomial trees of order k in L, append onebinomial tree of order k to the binomial heap andremove that tree from L

2. If there are two trees of order k, remove both trees,use Link to form a tree of order (k+1) and pre-pendthis tree to L

Union is O(log n)


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More Operations

Inserting a new node with key field set Create a new binomial heap with that one node Union (existing heap with head h, new heap) O (log n) time

ChangeDown (node at position x, new value) Decreasing the key value of a node Same idea as binary heap: Bubble up the

binomial tree containing this node (exchange onlykey fields and satellite data! Whats thecomplexity if you physically change the node?)

O (log n) time


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More Operations

Delete (node at position x) Deleting position x from the heap1. ChangeDown(x, -INFINITY)

Now x is at the root of its binomial tree Supposing that the binomial tree is of order k Recall that the children of the root of the binomial tree,

from right to left, are binomial trees of order 0, 1, 2, 3, 4,, k-1

2. Form a new binomial heap with the children of the root of

this binomial tree the roots in the new binomial heap3. Remove the original binomial tree from the original

binomial heap4. Union (original heap, new heap)

O(log n) complexity


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More Operations

ChangeUp (node at position X, newvalue)

1. Delete (X)

2. Insert (new value)

O (log n) time


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Summary Binomial Heaps

Create in O(1) time

Union, Find_min, Delete, Insert, and Change

operations take O(log n) time In general, because they are more

complicated, in competition it is far moreprudent (saves time coding and debugging)to use a binary heap instead

Unless there are MANY Union operations


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Application of heaps: Dijkstra

The following describes how Dijkstrasalgorithm can be coded with a binary heap

Initializing phase:1. Let n be the number of nodes

2. Create a heap of size n, all key fields

initialized to INFINITY3. Change_val (s, 0) where s is the source

node


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Running of Dijkstras algorithm

1. While (heap is not empty)

1. X = node corresponding to find_min

value2. Delete (position of X in heap = 1)

3. For all nodes k that are adjacent to X

1. If (cost[X] + distance[X][k] < cost[k])1. ChangeDown (position of k in heap, cost[X] +

distance[X][k])


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Analysis of running time

At most n nodes are deleted O(n log n)

Let m be the number of edges. Each edge isrelaxed at most once. O(m log n)

Total running time O([m+n] log n) This is faster than using a basic array list

unless the graph is very dense, in which casem is about O(n^2) which leads to a runningtime of O(n^2 log n)


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Cumulative Sum on Intervals

Problem: We have a line that runs from xcoordinate 1 to x coordinate N. At x

coordinate X [X an integer between 0 and N],there is g(X) gold. Given an interval [a,b],how much gold is there between a and b?

How efficiently can this be done if we

dynamically change the amount of gold andthe interval [a,b] keeps changing?


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Cumulative Sum Array

Let us define C(0) = 0, and C(x) = C(x-1) + g(x)where g(x) is the amount of gold at position x

C(x) then defines the total amount of gold from

position 1 to position x The amount of gold in interval [a,b] is simply C(b)

C(a-1) For any change in a or b, we can perform the update in O(1)

time

However, if we change g(x), we will have to changeC(x), C(x+1), C(x+2), , C(N) Any change in gold results in an update in O(N) time


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Cumulative Sum Tree

We can use the binary representation of any numberto come up with a cumulative sum tree

For example, let say we take 13 (decimal) = 1101

(binary) The cumulative sum of g(1) + g(2) + g(13) can be

represented as the sum of: g(1) + g(2) + + g(8) [ 8 elements ]

g(9) + g(10) + + g(12) [ 4 elements ]

g(13) [ 1 element ] Notice that the number of elements in each case represents

a bit that is 1 in the binary representation of the number


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Cumulative Sum Tree

Another example: C(19)

19 (decimal) is 10011 (binary)

C(19) is the sum of the following: g(1) + g(2) + + g(16) [ 16 elements ]

g(17) + g(18) [ 2 elements ]

g(19) [ 1 element ]


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Cumulative Sum Tree

Let us define C2(x) to be the sum of g(x) +g(x-1) + + g(p + 1) where p is a number

with the same binary representation as xexcept the least significant bit of x (therightmost bit of x that is 1) is 0

Examples of x and the corresponding p:

x = 6 [110], p = 4 [100]

x = 13 [1101], p = 12 [1100]

x = 16 [10000], p = 0 [00000]


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Cumulative Sum Tree

If we want to find the cumulative sum C(x) = g(1) +g(2) + + g(x), we can trace through the values ofC2 using the binary representation of x Examples: C(13) = C2(8) + C2(8+4) + C2(8+4+1) C(16) = C2(16) C(21) = C2(16) + C2(16+4) + C2(16+4+1) C(99) = C2(64) + C2(64+32) + C2(64+32+2) +

C2(64+32+2+1)

This allows us to find C(x) in log x time Hence the amount of gold in interval [a,b] = C(b) C(a-1)

can be found in log N time, which implies updates of a and bcan be done in O(log N)


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Cumulative Sum Tree

What happens when we change g(x)? If g(x) is changed, we only need to update C2(y)

where C2(y) covers g(x)

We can go through all necessary C2(y) in thefollowing way:1. While (x


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Cumulative Sum Tree

Examples [binary representation in brackets] Change to g(5) [ 101 ] : Update C2(5), C2(6), C2(8),

C2(16) and all C2(power of 2 > 16) Change to g(13) [ 1101 ]: Update C2(13), C2(14), C2(16),

and all C2(power of 2 > 16) Change to g(35) [ 100011 ]: Update C2(35), C2(36),

C2(40), C2(48), C2(64), and all C2(power of 2 > 64)

We can implement a cumulative sum tree verysimply: By simply using a linear array to store the

values of C2. Can we extend a cumulative sum tree to 2 or more

dimensions? See IOI 2001 Day 1 Question 1


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Sum of Intervals Tree

Another way to solve the question is to use a Sumof Intervals Binary Tree

Each node in the tree is represented by (L, R) and

the value of (L,R) is g(L) + g(L+1) + + g(R) The root of the tree has L = 1 and R = N Every leaf has L = R Every non-leaf has children (L, [L+R]/2) [left child]

and ([L+R]/2+1, R) [right child]

The number of nodes in the tree is O(2*N) [ why? ] In an implementation, every node should have

pointers to its children and its parent


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How to find C(x) = g(1) + g(2) + + g(x)? We trace from the root downwards1. L = 1, R = N, C = 02. While (L != R)

1. M = (L + R) / 22. If (M < x)

1. C += value of (L,R)2. Set L and R to the left child of the current node

3. Else1. Set L and R to the right child of the current node

3. C += value at (L,R) [ or (L,L) or (R,R) as L = R ] Time complexity: O(log n)


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What happens when g(x) is changed? Trace from (x,x) upwards to the root

1. Let L = R = x

2. While (L,R) is not the root1. Update the value of (L,R)

2. Set (L,R) to the parent of (L,R)

3. Update the root

Complexity of O(log N) Hence all updates of interval [a,b] and g(x)

can be done in O(log N) time


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Augmenting Data Structures

It is often useful to change the data structurein some way, by adding additional data ineach node or changing what each node

represents. This allows us to use the same data structure

to solve problems For example, we can use so-called interval

trees to solve not just cumulative sumproblems We can use properties of elements in the interval

(L,R) that are related to L and R.


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Other data structures

Balanced (and unbalanced) binary trees

Red-Black trees

2-3-4 trees

Splay trees

Suffix Trees

Fibonacci Heaps

Documents

Application of Data Structures