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Design and Analysis of Algorithms
Instructor: Sharma Thankachan
Lecture 16: Single-Source Shortest-Path
Slides modified from Dr. Hon, with permission
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About this lecture • What is the problem about ?
• Dijkstra’s Algorithm [1959]
• ~ Prim’s Algorithm [1957]
• Folklore Algorithm for DAG [???]
• Bellman-Ford Algorithm• Discovered by Bellman [1958], Ford [1962]
• Allowing negative edge weights
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• Let G = (V,E) be a weighted graph• the edges in G have weights• can be directed/undirected• can be connected/disconnected
• Let s be a special vertex, called source
Target: For each vertex v, compute the length of shortest path from s to v
Single-Source Shortest Path
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• E.g.,
Single-Source Shortest Path
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8
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8 79
10
144
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2
67s
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8 79
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67s 0
4 12 19
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119
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Relax• A common operation that is used in the
three algorithms is called Relax : when a vertex v can be reached from the source with a certain distance, we examine an outgoing edge, say (v,w), and check if we can improve w
• E.g., 4
8
11
8
1
2
67s 0
4 ?
?
? ?
vCan we improve this?
Can we improve these?
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Dijkstra’s AlgorithmDijkstra(G, s)
For each vertex v, Mark v as unvisited, and set d(v) = 1 ;
Set d(s) = 0 ;while (there is unvisited vertex) {
v = unvisited vertex with smallest d ;Visit v, and Relax all its outgoing edges;
}return d ;
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Example
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2
67s
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67 1
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Relax
1s
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Example
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Relax
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Example
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Relax
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Example
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Relax
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s
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Example
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Relax
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21s
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Example
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Relax
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Example
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Relax
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Example
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Relax
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Example
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Relax
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CorrectnessTheorem:
The kth vertex closest to the source s is selected at the kth step inside the while loop of Dijkstra’s algorithmAlso, by the time a vertex v is selected, d(v) will store the length of the shortest path from s to v
How to prove ? (By induction)
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Proof• Both statements are true for k = 1 ;• Let vj = jth closest vertex from s• Now, suppose both statements are true
for k = 1, 2, …, r-1• Consider the rth closest vertex vr
• If there is no path from s to vr
è d(vr) = 1 is never changed• Else, there must be a shortest path
from s to vr ; Let vt be the vertex immediately before vr in this path
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• Then, we have t £ r-1 (why??)
è d(vr) is set correctly once vt is selected, and the edge (vt,vr) is relaxed (why??)
è After that, d(vr) is fixed (why??)
è d(vr) is correct when vr is selected ; also, vr must be selected at the rth step, because no unvisited nodes can have a smaller d value at that time
Thus, the proof of inductive case completes
Proof (cont)
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Performance• Dijkstra’s algorithm is similar to Prim’s• By using Fibonacci Heap,
• Relax ó Decrease-Key• Pick vertex ó Extract-Min
• Running Time: • O(V) Insert/Extract-Min• At most O(E) Decrease-Keyè Total Time: O(E + V log V)
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Finding Shortest Path in DAGWe have a faster algorithm for DAG :DAG-Shortest-Path(G, s)
Topological Sort G ;For each v, set d(v) = 1 ; Set d(s) = 0 ;for (k = 1 to |V|) {
v = kth vertex in topological order ;Relax all outgoing edges of v ;
}return d ;
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Example
Topological Sort
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s4 3 2 6
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s
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Example
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s4 3 2 6
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1 0 1 1 1 1
Process this node
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s4 3 2 6
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Relax
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Example
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s4 3 2 6
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s4 3 2 6
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Relax
1 1 10 3 11
Process this node
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Example
Process this node 8
s4 3 2 6
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Relax
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Example
Process this node 8
s4 3 2 6
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Relax
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Example
Process this node
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Example
Process this node
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CorrectnessTheorem:
By the time a vertex v is selected, d(v) will store the length of the shortest path from s to v
How to prove ? (By induction)
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Proof• Let vj = jth vertex in the topological order• We will show that d(vk) is set correctly
when vk is selected, for k = 1,2, …, |V|• When k = 1,
vk = v1 = leftmost vertexIf it is the source, d(vk) = 0If it is not the source, d(vk) = 1è In both cases, d(vk) is correct (why?)
è Base case is correct
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Proof (cont)• Now, suppose the statement is true for
k = 1, 2, …, r-1• Consider the vertex vr
• If there is no path from s to vr
è d(vr) = 1 is never changed• Else, we shall use similar arguments as
proving the correctness of Dijkstra’s algorithm …
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• First, let vt be the vertex immediately before vr in the shortest path from s to vr
è t £ r-1è d(vr) is set correctly once vt is selected,
and the edge (vt,vr) is relaxedè After that, d(vr) is fixedè d(vr) is correct when vr is selected
Thus, the proof of inductive case completes
Proof (cont)
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Performance• DAG-Shortest-Path selects vertex
sequentially according to topological order• no need to perform Extract-Min
• We can store the d values of the vertices in a single array è Relax takes O(1) time
• Running Time: • Topological sort : O(V + E) time• O(V) select, O(E) Relax : O(V + E) timeè Total Time: O(V + E)
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Handling Negative Weight Edges• When a graph has negative weight edges,
shortest path may not be well-defined
v4
8
11-7
s
-7
What is the shortest path from s to v?
E.g.,
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Handling Negative Weight Edges• The problem is due to the presence of a
cycle C, reachable by the source, whose total weight is negativeè C is called a negative-weight cycle
• How to handle negative-weight edges ??è if input graph is known to be a DAG,
DAG-Shortest-Path is still correctè For the general case, we can use
Bellman-Ford algorithm
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Bellman-Ford AlgorithmBellman-Ford(G, s) // runs in O(VE) time
For each v, set d(v) = 1 ; Set d(s) = 0 ;for (k = 1 to |V|-1)
Relax all edges in G in any order ;/* check if s reaches a neg-weight cycle */
for each edge (u,v), if (d(v) > d(u) + weight(u,v))
return “something wrong !!” ;return d ;
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Example 14
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Relax all
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Relax all10
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Example 1After the 4th Relax all
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After checking, we found that there is nothing wrong è distances are correct
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Example 24
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Example 2After the 4th Relax all
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After checking, we found that something must be wrong è distances are incorrect
This edge shows something must
be wrong …
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Correctness (Part 1)Theorem:
If the graph has no negative-weight cycle, then for any vertex v with shortest path from s consists of k edges, Bellman-Fordsets d(v) to the correct value after the kth
Relax all (for any ordering of edges in each Relax all )
How to prove ? (By induction)
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CorollaryCorollary: If there is no negative-weight
cycle, then when Bellman-Ford terminates, d(v) £ d(u) + weight(u,v)
for all edge (u,v)
Proof: By previous theorem, d(u) and d(v) are the length of shortest path from s to u and v, respectively. Thus, we must have
d(v) £ length of any path from s to vè d(v) £ d(u) + weight(u,v)
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“Something Wrong” LemmaLemma: If there is a negative-weight cycle,
then when Bellman-Ford terminates, d(v) > d(u) + weight(u,v)
for some edge (u,v)
How to prove ? (By contradiction)
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• Firstly, we know that there is a cycle C = (v1, v2, … , vk, v1)
whose total weight is negative
• That is, Si = 1 to k weight(vi, vi+1) < 0
• Now, suppose on the contrary that d(v) £ d(u) + weight(u,v)
for all edge (u,v) at termination
Proof
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• Can we obtain another bound for Si = 1 to k weight(vi, vi+1) ?
• By rearranging, for all edge (u,v)weight(u,v) ³ d(v) - d(u)
è Si = 1 to k weight(vi, vi+1)
³ Si = 1 to k (d(vi+1) - d(vi)) = 0 (why?)
è Contradiction occurs !!
Proof (cont)
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Correctness (Part 2)
Theorem: There is a negative-weight cycle in the input graph if and only if when Bellman-Ford terminates,
d(v) > d(u) + weight(u,v) for some edge (u,v)
• Combining the previous corollary and lemma, we have:
 as pivot•Idea 2: Process](https://img.dokumen.tips/doc/110x75/5fa5a0c444ecf844063c21a0/instructor-sharma-thankachan-lecture-5-sharmacop3503lectureslecture5pdf-13.jpg)
![Design and Analysis of Algorithms - UCF Computer Sciencesharma/COP3503lectures/lecture6.pdfBucket Sort •Input: Array A[1..n] of nelements, each is drawn uniformly at random from](https://img.dokumen.tips/doc/110x75/610f07b4273b224a753c7394/design-and-analysis-of-algorithms-ucf-computer-sharmacop3503lectureslecture6pdf.jpg)
