Applicationof datastructures

Application of Data Structures

Overview Priority Queue structures Heaps Application: Dijkstra’s algorithm Cumulative Sum Data Structures on Intervals Augmenting data structures with extra info to solve questions

Priority Queue (PQ) Structures Stores elements in a list by comparing a key field Often has other satellite data For example, when sorting pixels by their R value, we consider the R as the key field and GB as satellite data Priority queues allow us to sort elements by their key field.

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

Optional PQ operations Union (a,b) Combines two PQs a and b Search (k) Returns the position of the element in the heap with key value k

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 which allows us to do a search in O(1) time. The handling of this data structure will be assumed and not covered.

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

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 a node 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.

Array implementation Consider a heap of size nheap in 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 can ensure that the number of levels in this heap is O(log n) Some properties wrt Key values: BHeap[x] >= BHeap[x/2], BHeap[x] <= BHeap[x*2], BHeap[x] <= BHeap[x*2+1], BHeap[x*2] ?? BHeap[x*2+1]

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 the variable n refers to nheap

Operations on a BHeap Create is trivial – O(1) time Find_min: Return BHeap[1] O(1) time Insert (element with key value x) nheap++ BHeap[nheap] = x T = nheap While (T != 1 && Bheap[T] < BHeap[T/2]) Swap (Bheap[T], BHeap[T/2] T = T / 2 O(log n) time as the number of levels is O(log n)

Operations on a BHeap ChangeDown (position x, new key value k) Assume: k < existing BHeap[x] BHeap[x] = k T = x While (T != 1 && BHeap[T] < BHeap[T/2]) Swap (BHeap[T], BHeap[T/2]) T = T/2 Complexity: O(log n) This procedure is known as “bubbling up” the heap

Operations on a BHeap ChangeUp (position x, new key value k) Assume: k > existing BHeap[x] BHeap[x] = k Heapify(x) O(log n) as complexity of Heapify is O(log n)

Operations on a BHeap Delete (position x on the heap) BHeap[x] = BHeap[nheap] nheap— Heapify(x) T = x While (T != 1 && BHeap[T] < BHeap[T/2]) Swap (BHeap[T], BHeap[T/2]) T = T / 2 Complexity is O(log n) Why must I do both Heapify and “bubble up”?

Operations on a BHeap Heapify (position x on the heap) T = min(BHeap[x], BHeap[x*2], BHeap[x*2+1]) If (T == BHeap[x]) return; K = position where BHeap[K] = T Swap(BHeap[x], BHeap[K]) Heapify(K) O(log n) as the maximum number of levels in the heap is O(log n) and Heapify only goes through each level at most once

BHeap Operations: Summary Create, Find_min in O(1) time Change (includes both ChangeUp and ChangeDown), Insert, and Delete are O(log n) time Union operations are how long? Insertion: O(n log n) union Heapify: O(n) union

Corollary: Heapsort We can convert an unsorted array to a heap using Heapify (why does this work?): For (i = n/2; i >= 1; i--) Heapify(i) We can then return a sorted list (list initially empty): For (i = 1; i <= n; i++) Append the value of find_min to the list Delete(1) Complexity is O(n log n)

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 root of the other B(n-1) tree. Within the tree, the heap property holds i.e. that the key field of any node is greater than the key field of all its children.

Properties of Binomial Trees The number of nodes in B(k) is exactly 2^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 to right, they form the roots of a B(k-1), B(k-2), …, B(0) tree in that order

Binomial Heaps Binomial Heaps are a forest of binomial trees with the following properties: All the binomial trees are of different sizes The binomial trees are ordered (from left to right) by increasing 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 n nodes 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).

Binomial Heap Implementation Each node will store the following data: Key field 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 to the 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 head pointer that points to the root of the smallest binomial tree (which is the leftmost binomial tree)

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 tree T = h1->leftchild h1->leftchild = h2 h2->parent = h1 H2->next_sibling= T O(1) time

Operations on binomial heaps Create – Create a new binomial heap with one node ( key field set) Set Parent, Leftchild, Next sibling to NIL O(1) time Find_min X = head, min = INFINITY While (X != nil) If (X->key < min) min = X->key X = X->next_sibling Return min O(log n) time as there are at most log n binomial trees (log n bits)

More Operations Merge (h1, h2, L) Given binomial heaps with head pointers h1 and h2, create a list L of all the binomial trees of h1 U h2 arranged in ascending order of size For any order k, there may be zero, one, or two binomial trees of order k in this list.

More Operations Merge (h1, h2, L) Assume that NIL is a node of infinitely small order L = empty While (h1 != NIL || h2 != NIL) If (h1->degree < h2->degree) Append the (binomial)tree with root h1 to L h1 = h1->next_sibling Else Apply above steps to h2 instead

More Operations Union (h1, h2) The fundamental operation involving binomial heaps Takes two binomial heaps with head pointers h1 and h2 and creates a new binomial heap of the union of h1 and h2

More Operations Union (h1, h2) Start with empty binomial heap Merge (h1, h2, L) Go by increasing k in the list L until L is empty If there is exactly one or exactly three (how can this happen?) binomial trees of order k in L, append one binomial tree of order k to the binomial heap and remove that tree from L If there are two trees of order k, remove both trees, use Link to form a tree of order (k+1) and pre-pend this tree to L Union is O(log n)

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 only key fields and satellite data! What’s the complexity if you physically change the node?) O (log n) time

More Operations Delete (node at position x) Deleting position x from the heap 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 Form a new binomial heap with the children of the root of this binomial tree the roots in the new binomial heap Remove the original binomial tree from the original binomial heap Union (original heap, new heap) O(log n) complexity

More Operations ChangeUp (node at position X, new value) Delete (X) Insert (new value) O (log n) time

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 more prudent (saves time coding and debugging) to use a binary heap instead Unless there are MANY Union operations

Application of heaps: Dijkstra The following describes how Dijkstra’s algorithm can be coded with a binary heap Initializing phase: Let n be the number of nodes Create a heap of size n, all key fields initialized to INFINITY Change_val (s, 0) where s is the source node

Running of Dijkstra’s algorithm While (heap is not empty) X = node corresponding to find_min value Delete (position of X in heap = 1) For all nodes k that are adjacent to X If (cost[X] + distance[X][k] < cost[k]) ChangeDown (position of k in heap, cost[X] + distance[X][k])

Analysis of running time At most n nodes are deleted O(n log n) Let m be the number of edges. Each edge is relaxed 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 case m is about O(n^2) which leads to a running time of O(n^2 log n)

Cumulative Sum on Intervals Problem: We have a line that runs from x coordinate 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 and the interval [a,b] keeps changing?

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 change C(x), C(x+1), C(x+2), …, C(N) Any change in gold results in an update in O(N) time

Cumulative Sum Tree We can use the binary representation of any number to 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

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 ]

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 x except the least significant bit of x (the rightmost 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]

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 of C2 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 b can be done in O(log N)

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 the following way: While (x <= N) Update C2(x) Add the value of the least significant bit of x to x This runs in O(log N) time Hence updates to g can also be done in O(log n) time, which is a great improvement over the O(N) needed for an array.

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 very simply: 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

Sum of Intervals Tree Another way to solve the question is to use a “Sum of 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

Sum of Intervals Tree How to find C(x) = g(1) + g(2) + … + g(x)? We trace from the root downwards L = 1, R = N, C = 0 While (L != R) M = (L + R) / 2 If (M < x) C += value of (L,R) Set L and R to the left child of the current node Else Set L and R to the right child of the current node C += value at (L,R) [ or (L,L) or (R,R) as L = R ] Time complexity: O(log n)

Sum of Intervals Tree What happens when g(x) is changed? Trace from (x,x) upwards to the root Let L = R = x While (L,R) is not the root Update the value of (L,R) Set (L,R) to the parent of (L,R) 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

Augmenting Data Structures It is often useful to change the data structure in some way, by adding additional data in each 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 sum problems We can use properties of elements in the interval (L,R) that are related to L and R.

Other data structures Balanced (and unbalanced) binary trees Red-Black trees 2-3-4 trees Splay trees Suffix Trees Fibonacci Heaps

Applicationof datastructures

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