AlgorithmsandTheoryofComputation … · 2020. 9. 1. · AlgorithmsandTheoryofComputation...
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Algorithms and Theory of Computation Lecture 7: Shortest Path Problem, Priority Queue Xiaohui Bei MAS 714 September 1, 2020 Nanyang Technological University MAS 714 September 1, 2020 1 / 24
Transcript of AlgorithmsandTheoryofComputation … · 2020. 9. 1. · AlgorithmsandTheoryofComputation...
Algorithms and Theory of Computation
Lecture 7: Shortest Path Problem, Priority Queue
Xiaohui Bei
MAS 714
September 1, 2020
Nanyang Technological University MAS 714 September 1, 2020 1 / 24
Shortest Paths Problems
Shortest Paths ProblemsGiven a (undirected or directed) weighted graph G = (V, E) with edge lengths (or weights).For edge e = (u, v), `(e) = `(u, v) is its length.
1 Given vertices s, t, find a shortest path from s to t.2 Given vertex s, find shortest paths from s to all other vertices.3 Find shortest paths for all pair of vertices.
The length of a path is the sum of its edge weights.
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Single-Source Shortest Path
Single-Source Shortest PathGiven a directed weighted graph G = (V, E) with non-negative edge lengths (or weights). Foredge e = (u, v), `(e) = `(u, v) is its length.
1 Given vertices s, t, find a shortest path from s to t.2 Given vertex s, find shortest paths from s to all other vertices.
Focus on directed graphs.undirected graph problem can be reduced to directed graph problem
Uniform case: all edges lengths are 1.BFS(s) solves the problem in O(m+ n) time.
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Single-Source Shortest Path
Single-Source Shortest PathGiven a directed weighted graph G = (V, E) with non-negative edge lengths (or weights). Foredge e = (u, v), `(e) = `(u, v) is its length.
1 Given vertices s, t, find a shortest path from s to t.2 Given vertex s, find shortest paths from s to all other vertices.
Focus on directed graphs.undirected graph problem can be reduced to directed graph problem
Uniform case: all edges lengths are 1.BFS(s) solves the problem in O(m+ n) time.
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Ideas
Why does BFS work?
BFS(s) explores nodes in increasing distance from s
Let dist(s, v) denote the shortest path length from s to v.
Algorithm: ShortestPathAlgorithm(s):Initialize dist(s, v) = ∞ for each vertex v ∈ V;Initialize S = ∅;while S 6= V do
Find vertex v ∈ V − S that is the closest to s;Update dist(s, v);S = S ∪ v
How to find the next closest vertex?
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Ideas
Why does BFS work?
BFS(s) explores nodes in increasing distance from s
Let dist(s, v) denote the shortest path length from s to v.
Algorithm: ShortestPathAlgorithm(s):Initialize dist(s, v) = ∞ for each vertex v ∈ V;Initialize S = ∅;while S 6= V do
Find vertex v ∈ V − S that is the closest to s;Update dist(s, v);S = S ∪ v
How to find the next closest vertex?
Nanyang Technological University MAS 714 September 1, 2020 4 / 24
Ideas
Why does BFS work?
BFS(s) explores nodes in increasing distance from s
Let dist(s, v) denote the shortest path length from s to v.
Algorithm: ShortestPathAlgorithm(s):Initialize dist(s, v) = ∞ for each vertex v ∈ V;Initialize S = ∅;while S 6= V do
Find vertex v ∈ V − S that is the closest to s;Update dist(s, v);S = S ∪ v
How to find the next closest vertex?
Nanyang Technological University MAS 714 September 1, 2020 4 / 24
Ideas
Why does BFS work?
BFS(s) explores nodes in increasing distance from s
Let dist(s, v) denote the shortest path length from s to v.
Algorithm: ShortestPathAlgorithm(s):Initialize dist(s, v) = ∞ for each vertex v ∈ V;Initialize S = ∅;while S 6= V do
Find vertex v ∈ V − S that is the closest to s;Update dist(s, v);S = S ∪ v
How to find the next closest vertex?Nanyang Technological University MAS 714 September 1, 2020 4 / 24
Finding the ith Closest Vertex
At the beginning of the ith iteration, S already stores the i− 1 closest vertices to s.
What do we know about the ith closest vertex?
Claim 1Let P be a shortest path from s to v where v is the ith closest vertex. Then all intermediatevertices in P belong to S.
Proof: Let v ′ be an intermediate vertex in path P, then dist(s, v ′) < dist(s, v). Thus v ′ ∈ S.
CorollaryAt each step, the next closest vertex is adjacent to S.
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Finding the ith Closest Vertex
At the beginning of the ith iteration, S already stores the i− 1 closest vertices to s.
What do we know about the ith closest vertex?
Claim 1Let P be a shortest path from s to v where v is the ith closest vertex. Then all intermediatevertices in P belong to S.
Proof: Let v ′ be an intermediate vertex in path P, then dist(s, v ′) < dist(s, v). Thus v ′ ∈ S.
CorollaryAt each step, the next closest vertex is adjacent to S.
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Finding the ith Closest Vertex
At the beginning of the ith iteration, S already stores the i− 1 closest vertices to s.
What do we know about the ith closest vertex?
Claim 1Let P be a shortest path from s to v where v is the ith closest vertex. Then all intermediatevertices in P belong to S.
Proof: Let v ′ be an intermediate vertex in path P, then dist(s, v ′) < dist(s, v). Thus v ′ ∈ S.
CorollaryAt each step, the next closest vertex is adjacent to S.
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Finding the ith Closest Vertex
At the beginning of the ith iteration, S already stores the i− 1 closest vertices to s.
What do we know about the ith closest vertex?
Claim 1Let P be a shortest path from s to v where v is the ith closest vertex. Then all intermediatevertices in P belong to S.
Proof: Let v ′ be an intermediate vertex in path P, then dist(s, v ′) < dist(s, v). Thus v ′ ∈ S.
CorollaryAt each step, the next closest vertex is adjacent to S.
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Finding the ith Closest Vertex
Claim 2If s = v0 → v1 → · · · → vk = v is a shortest path from s to v,then for any 1 ≤ i < k:
s = v0 → v1 → · · · → vi is a shortest path from s to vi
Assume that S contains the i− 1 closest vertices to s. For each u ∈ V − S, let
π(u) = mina∈S
(dist(s, a) + `(a, u))
LemmaIf u is the ith closest vertex to s, then π(u) = dist(s, u).
CorollaryThe ith closest vertex to s is the vertex u ∈ V − S such that π(u) = minv∈V−S π(v).
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Finding the ith Closest Vertex
Claim 2If s = v0 → v1 → · · · → vk = v is a shortest path from s to v,then for any 1 ≤ i < k:
s = v0 → v1 → · · · → vi is a shortest path from s to vi
Assume that S contains the i− 1 closest vertices to s. For each u ∈ V − S, let
π(u) = mina∈S
(dist(s, a) + `(a, u))
LemmaIf u is the ith closest vertex to s, then π(u) = dist(s, u).
CorollaryThe ith closest vertex to s is the vertex u ∈ V − S such that π(u) = minv∈V−S π(v).
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Finding the ith Closest Vertex
Claim 2If s = v0 → v1 → · · · → vk = v is a shortest path from s to v,then for any 1 ≤ i < k:
s = v0 → v1 → · · · → vi is a shortest path from s to vi
Assume that S contains the i− 1 closest vertices to s. For each u ∈ V − S, let
π(u) = mina∈S
(dist(s, a) + `(a, u))
LemmaIf u is the ith closest vertex to s, then π(u) = dist(s, u).
CorollaryThe ith closest vertex to s is the vertex u ∈ V − S such that π(u) = minv∈V−S π(v).
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Finding the ith Closest Vertex
Claim 2If s = v0 → v1 → · · · → vk = v is a shortest path from s to v,then for any 1 ≤ i < k:
s = v0 → v1 → · · · → vi is a shortest path from s to vi
Assume that S contains the i− 1 closest vertices to s. For each u ∈ V − S, let
π(u) = mina∈S
(dist(s, a) + `(a, u))
LemmaIf u is the ith closest vertex to s, then π(u) = dist(s, u).
CorollaryThe ith closest vertex to s is the vertex u ∈ V − S such that π(u) = minv∈V−S π(v).
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Algorithm
Algorithm: ShortestPathAlgorithm(s):Initialize π(v) = ∞ for each vertex v ∈ V;Initialize S = ∅, π(s) = 0;while S 6= V do
Find vertex u ∈ V − S such that π(u) = minv∈V−S π(v);dist(s, u) = π(u);S = S ∪ u;foreach vertex v ∈ V − S do
π(v) = mina∈S(dist(s, a) + `(a, v));
Correctness: by induction using previous lemmas.Running Time: O(n(n+m)) time.
n outer iterations. O(m+ n) time in each iteration.
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Algorithm
Algorithm: ShortestPathAlgorithm(s):Initialize π(v) = ∞ for each vertex v ∈ V;Initialize S = ∅, π(s) = 0;while S 6= V do
Find vertex u ∈ V − S such that π(u) = minv∈V−S π(v);dist(s, u) = π(u);S = S ∪ u;foreach vertex v ∈ V − S do
π(v) = mina∈S(dist(s, a) + `(a, v));
Correctness: by induction using previous lemmas.
Running Time: O(n(n+m)) time.n outer iterations. O(m+ n) time in each iteration.
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Algorithm
Algorithm: ShortestPathAlgorithm(s):Initialize π(v) = ∞ for each vertex v ∈ V;Initialize S = ∅, π(s) = 0;while S 6= V do
Find vertex u ∈ V − S such that π(u) = minv∈V−S π(v);dist(s, u) = π(u);S = S ∪ u;foreach vertex v ∈ V − S do
π(v) = mina∈S(dist(s, a) + `(a, v));
Correctness: by induction using previous lemmas.Running Time: O(n(n+m)) time.
n outer iterations. O(m+ n) time in each iteration.Nanyang Technological University MAS 714 September 1, 2020 7 / 24
Example
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More Efficient Implementation
Critical Optimization For each unexplored vertex v, explicitly maintain π(v) instead ofcomputing directly from formula:
π(v) = minu∈S
(dist(s, u) + `(u, v)).
For each v /∈ S, π(v) can only decrease (because S can only increase).More specifically, suppose u is added to S and there is an edge (u, v) leaving u. Then, itsuffices to update
π(v) = minπ(v), dist(s, u) + `(u, v)
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More Efficient Implementation
Critical Optimization For each unexplored vertex v, explicitly maintain π(v) instead ofcomputing directly from formula:
π(v) = minu∈S
(dist(s, u) + `(u, v)).
For each v /∈ S, π(v) can only decrease (because S can only increase).More specifically, suppose u is added to S and there is an edge (u, v) leaving u. Then, itsuffices to update
π(v) = minπ(v), dist(s, u) + `(u, v)
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Dijkstra’s Algorithm
1 eliminate π(v) and let dist(s, v) maintain it2 update dist values after adding v by scanning edges out of v
Algorithm: Dijkstra(s):Initialize dist(s, v) = ∞ for each vertex v ∈ V;Initialize S = ∅, dist(s, s) = 0;while S 6= V do
Find vertex u ∈ V − S such that dist(s, u) = minv∈V−S dist(s, v);S = S ∪ u;foreach vertex v ∈ Adj(u) do
dist(s, v) = min dist(s, v), dist(s, u) + `(u, v);
How to maintain dist values efficiently? Priority Queues
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Dijkstra’s Algorithm
1 eliminate π(v) and let dist(s, v) maintain it2 update dist values after adding v by scanning edges out of v
Algorithm: Dijkstra(s):Initialize dist(s, v) = ∞ for each vertex v ∈ V;Initialize S = ∅, dist(s, s) = 0;while S 6= V do
Find vertex u ∈ V − S such that dist(s, u) = minv∈V−S dist(s, v);S = S ∪ u;foreach vertex v ∈ Adj(u) do
dist(s, v) = min dist(s, v), dist(s, u) + `(u, v);
How to maintain dist values efficiently?
Priority Queues
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Dijkstra’s Algorithm
1 eliminate π(v) and let dist(s, v) maintain it2 update dist values after adding v by scanning edges out of v
Algorithm: Dijkstra(s):Initialize dist(s, v) = ∞ for each vertex v ∈ V;Initialize S = ∅, dist(s, s) = 0;while S 6= V do
Find vertex u ∈ V − S such that dist(s, u) = minv∈V−S dist(s, v);S = S ∪ u;foreach vertex v ∈ Adj(u) do
dist(s, v) = min dist(s, v), dist(s, u) + `(u, v);
How to maintain dist values efficiently? Priority QueuesNanyang Technological University MAS 714 September 1, 2020 10 / 24
Priority Queues
Priority QueuesStore a set S of n elements, where each element v ∈ S has an associated real/integer keyk(v), with the following operations:
1 Make-Queue: create an empty queue2 Find-Min: find the minimum key in S
3 Extract-Min: remove v ∈ S with the smallest key and return it4 Insert(v, k(v)): insert new element v with key k(v) to S
5 Delete(v): remove element v from S
6 Decrease-Key(v, k ′(v)): decrease key of v from k(v) to k ′(v)
Decrease-Key is implemented via delete and insert.
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Priority Queues
Applications:Prim’s MST algorithmDijkstra’s shortest-path algorithmheapsortonline medianetc.
Implementations:binary heapsd-ary heapsbinomial heapsFibonacci heaps
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Priority Queues
Applications:Prim’s MST algorithmDijkstra’s shortest-path algorithmheapsortonline medianetc.
Implementations:binary heapsd-ary heapsbinomial heapsFibonacci heaps
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Complete Binary Tree
A binary tree is a tree in which each vertex has at most two children.
A complete tree is a binary tree such that every level, except possibly the last level, iscompletely filled, and all vertices are as far left as possible.
PropertyThe depth of a complete binary tree with n vertices is blog2 nc.
Proof. The depth increases (by 1) only when n is a power of 2.
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Complete Binary Tree
A binary tree is a tree in which each vertex has at most two children.
A complete tree is a binary tree such that every level, except possibly the last level, iscompletely filled, and all vertices are as far left as possible.
PropertyThe depth of a complete binary tree with n vertices is blog2 nc.
Proof. The depth increases (by 1) only when n is a power of 2.Nanyang Technological University MAS 714 September 1, 2020 13 / 24
Binary Tree in Nature
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Binary Heap
Binary HeapA Binary Heap is a complete binary tree such that for every element v at some vertex, theelement w at this vertex’s parent satisfies k(w) ≤ k(v).
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parent
child child
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Explicit ImplementationPointer representation. Vertex has a pointer to parent and two children.
maintain number of elements n
maintain pointer to root vertex
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Implicit ImplementationArray representation. Indices start at 1.
take vertices in level orderparent of vertex at k is at bk/2cchildren of vertex at k are at 2k and 2k+ 1
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Binary Heap: Insert
Insert. Add element in new vertex at end; repeatedly exchange new element with element inits parent until heap order is restored.
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swim up
Time = O(logn)
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Binary Heap: Insert
Insert. Add element in new vertex at end; repeatedly exchange new element with element inits parent until heap order is restored.
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swim up
Time = O(logn)
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Binary Heap: Insert
Insert. Add element in new vertex at end; repeatedly exchange new element with element inits parent until heap order is restored.
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Time = O(logn)
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Binary Heap: Insert
Insert. Add element in new vertex at end; repeatedly exchange new element with element inits parent until heap order is restored.
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swim up
Time = O(logn)
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Binary Heap: Extract-Min
Extract-Min. Exchange element in root vertex with last vertex; repeatedly exchange element inroot with its smaller child until heap order is restored.
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sink down
Time = O(logn)
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Binary Heap: Extract-Min
Extract-Min. Exchange element in root vertex with last vertex; repeatedly exchange element inroot with its smaller child until heap order is restored.
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Time = O(logn)
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Binary Heap: Extract-Min
Extract-Min. Exchange element in root vertex with last vertex; repeatedly exchange element inroot with its smaller child until heap order is restored.
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sink down
Time = O(logn)
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Binary Heap: Extract-Min
Extract-Min. Exchange element in root vertex with last vertex; repeatedly exchange element inroot with its smaller child until heap order is restored.
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Time = O(logn)
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Binary Heap: Extract-Min
Extract-Min. Exchange element in root vertex with last vertex; repeatedly exchange element inroot with its smaller child until heap order is restored.
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Time = O(logn)
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Binary Heap: Extract-Min
Extract-Min. Exchange element in root vertex with last vertex; repeatedly exchange element inroot with its smaller child until heap order is restored.
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Time = O(logn)
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Binary Heap: Delete and Find-Min
Delete. Exchange the element in this vertex with the last vertex; either swim up or sink downthe vertex until heap order is restored.
Time = O(logn)
Decrease-Key. Repeatedly swim up the vertex until heap order is restored.Time = O(logn)
Find-Min. Return the element in the root vertex.Time = O(1)
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Binary Heap: Analysis
Dijkstra’a Algorithm with a priority queueO(n) Insert operationsO(n) Extract-Min operationsO(m) Decrease-Key operations
Dijkstra’s AlgorithmDijkstra’s algorithm can be implemented in O((n+m) logn) time.
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Heapify
Heapify. Given n elements, construct a binary heap containing them.
Can be done in O(n logn) time by inserting each element.Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
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Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
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Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
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Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
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Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
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Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
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Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
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Nanyang Technological University MAS 714 September 1, 2020 22 / 24
Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
8
12
7
14 11
21
15 22
9
3 2615
21
3
9
12
Nanyang Technological University MAS 714 September 1, 2020 22 / 24
Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
8
12
7
14 11
21
15 22
9
3 2615
21
3
9
7
12
Nanyang Technological University MAS 714 September 1, 2020 22 / 24
Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
8
12
7
14 11
21
15 22
9
3 2615
21
3
9
7
11
12
Nanyang Technological University MAS 714 September 1, 2020 22 / 24
Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
8
12
7
14 11
21
15 22
9
3 2615
21
3
9
7
11
12
8
Nanyang Technological University MAS 714 September 1, 2020 22 / 24
Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
8
12
7
14 11
21
15 22
9
3 2615
21
3
9
7
11
12
3
8
Nanyang Technological University MAS 714 September 1, 2020 22 / 24
Heapify
Heapify. Given n elements, construct a binary heap containing them.Can be done in O(n logn) time by inserting each element.
Bottom-up Method. For i = n to 1, repeatedly exchange the element in vertex i with itssmaller child until subtree rooted at i is heap-ordered.
8
12
7
14 11
21
15 22
9
3 2615
21
3
9
7
11
12
3
8
Nanyang Technological University MAS 714 September 1, 2020 22 / 24
Heapify
TheoremGiven n elements, a binary heap containing those n elements can be constructed in O(n) time.
Intuition:There are at most dn/2h+1e vertices of height h.The amount of work to sink a vertex is proportional to its height h.
Nanyang Technological University MAS 714 September 1, 2020 23 / 24
Heapify
TheoremGiven n elements, a binary heap containing those n elements can be constructed in O(n) time.
Intuition:There are at most dn/2h+1e vertices of height h.The amount of work to sink a vertex is proportional to its height h.
Nanyang Technological University MAS 714 September 1, 2020 23 / 24
More on Dijkstra and Priority Queues
1 Dijkstra’s algorithm gives shortest paths from s to all vertices in V.
2 Dijkstra’s algorithm can be implemented in O(n logn+m) time using Fibonacci Heaps.If m = Ω(nlogn), running time is linear in input size.
3 Data structures are complicated to analyze/implement. Recent work has obtained datastructures that are easier to analyze and implement, and perform well in practice.Rank-Pairing Heaps (European Symposium on Algorithms, September 2009)
Nanyang Technological University MAS 714 September 1, 2020 24 / 24
