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Topological sort more formally
• Is it possible to execute all the tasks in G in an order
that respects all the precedence requirements given
by the graph edges?
• The answer is "yes" if and only if the directed graph
G has no cycle!
(otherwise we have a deadlock)
• Such a G is called a Directed Acyclic Graph, or just a
DAG
Edge classification by DFS
Edge (u,v) of G is classified as a:
(1) Tree edge iff u discovers v during the DFS: P[v] = u
If (u,v) is NOT a tree edge then it is a:
(2) Forward edge iff u is an ancestor of v in the DFS tree
(3) Back edge iff u is a descendant of v in the DFS tree
(4) Cross edge iff u is neither an ancestor nor a
descendant of v
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DAGs and back edges
• Can there be a back edge in a DFS on a DAG?
• NO! Back edges close a cycle!
• A graph G is a DAG <=> there is no back edge
classified by DFS(G)
What is a DAG?
• A directed acyclic graph (DAG) is a directed graph without cycles.
• DAGs arrise in modeling many problems that involve prerequisite constraints (construction
projects, course prerequisites, document version control, compilers, etc.)
• Some properties of DAGs:
– Every DAG must have at least one vertex with in-degree zero and at least one vertex
with out-degree zero
• A vertex with in-degree zero is called a source vertex, a vertex with out-degree zero
is called a sink vertex.
– G is a DAG iff each vertex in G is in its own strongly connected component
– Every edge (v, w) in a DAG G has finishingTime[w] < finishingTime[v] in a DFS traversal of
G
Example:
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Topological sort Algorithm
• TOPOLOGICAL-SORT(G):
1) call DFS(G) to compute finishing times f[v] for
each vertex v
2) as each vertex is finished, insert it onto the front
of a linked list
3) return the linked list of vertices
Topological Sort Algorithms: DFS based algorithm
Topological-Sort(G)
{
1. Call dfs AllVertices on G to compute f[v] for each vertex v
2. If G contains a back edge (v, w) (i.e., if f[w] > f[v]) , report error ;
3. else, as each vertex is finished prepend it to a list; // or push in stack
4. Return the list; // list is a valid topological sort
}
• Running time is O(V+E), which is the running time for DFS.
Topological order: A C D B E H F G
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Time complexity of TS(G)
• Running time of topological sort:
Θ(n + m)
where n=|V| and m=|E|
• Why? Depth first search takes Θ(n + m) time
in the worst case, and inserting into the front
of a linked list takes Θ(1) time
Proof of correctness
• Theorem: TOPOLOGICAL-SORT(G) produces a
topological sort of a DAG G
• The TOPOLOGICAL-SORT(G) algorithm does a DFS
on the DAG G, and it lists the nodes of G in order
of decreasing finish times f[]
• We must show that this list satisfies the
topological sort property, namely, that for every
edge (u,v) of G, u appears before v in the list
• Claim: For every edge (u,v) of G: f[v] < f[u] in DFS
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Proof of correctness
“For every edge (u,v) of G, f[v] < f[u] in this DFS”
• The DFS classifies (u,v) as a tree edge, a
forward edge or a cross-edge (it cannot be a
back-edge since G has no cycles):
i. If (u,v) is a tree or a forward edge ⇒ v is a
descendant of u ⇒ f[v] < f[u]
ii. If (u,v) is a cross-edge
Proof of correctness
“For every edge (u,v) of G: f[v] < f[u] in this DFS”
ii. If (u,v) is a cross-edge:
• as (u,v) is a cross-edge, by definition, neither u
is a descendant of v nor v is a descendant of u:
d[u] < f[u] < d[v] < f[v]
or
d[v] < f[v] < d[u] < f[u]
since (u,v) is an edge, v is
surely discovered before
u's exploration completes
f[v] < f[u]
Q.E.D. of Claim
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Proof of correctness
• TOPOLOGICAL-SORT(G) lists the nodes of G
from highest to lowest finishing times
• By the Claim, for every edge (u,v) of G:
f[v] < f[u]
⇒ u will be before v in the algorithm's list
• Q.E.D of Theorem
END
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Unit ii divide and conquer -3
- 1. 8/31/2020 1 Topological Sort Material prepared by Dr. N. Subhash Chandra from Web resources Topological sort • We have a set of tasks and a set of dependencies (precedence constraints) of form “task A must be done before task B” • Topological sort: An ordering of the tasks that conforms with the given dependencies • Goal: Find a topological sort of the tasks or Decide that there is no such ordering 1 2
- 2. 8/31/2020 2 Example: Topological order: linear ordering of vertices of a graph according prerequisites. Sort 1: C1, C2, C4, C5, C3, C6, C8, C7, C10, C13, C12, C14, C15, C11, C9 Sort 2: C4, C5, C2, C1, C6, C3, C8, C15, C7, C9, C10, C11, C13, C12, C14 Definition of Topological Sort • Given a directed graph G = (V, E) a topological sort of G is an ordering of V such that for any edge (u, v), u comes before v in the ordering. • Example1: • Example2: 3 4
- 3. 8/31/2020 3 Definition of Topological Sort • Example3: The graph in (a) can be topologically sorted as in (b) (a) (b) Topological Sort is not unique • Topological sort is not unique. • The following are all topological sort of the graph below: s1 = {a, b, c, d, e, f, g, h, i} s2 = {a, c, b, f, e, d, h, g, i} s3 = {a, b, d, c, e, g, f, h, i} s4 = {a, c, f, b, e, h, d, g, i} etc. 5 6
- 4. 8/31/2020 4 Applications of topological sorting A common application of topological sorting is in scheduling a sequence of jobs. •The jobs are represented by vertices, and there is an edge from x to y if job x must be completed before job y can be started •For example, in constructing a building, the basement must be completed before the first floor, which must be completed before the second floor and so on. •A topological sort gives an order in which we should perform the jobs. Topological sort more formally • Suppose that in a directed graph G = (V, E) vertices V represent tasks, and each edge (u, v)∊E means that task u must be done before task v • What is an ordering of vertices 1, ..., |V| such that for every edge (u, v), u appears before v in the ordering? • Such an ordering is called a topological sort of G Note: there can be multiple topological sorts of G 7 8
- 5. 8/31/2020 5 Topological sort more formally • Is it possible to execute all the tasks in G in an order that respects all the precedence requirements given by the graph edges? • The answer is "yes" if and only if the directed graph G has no cycle! (otherwise we have a deadlock) • Such a G is called a Directed Acyclic Graph, or just a DAG Edge classification by DFS Edge (u,v) of G is classified as a: (1) Tree edge iff u discovers v during the DFS: P[v] = u If (u,v) is NOT a tree edge then it is a: (2) Forward edge iff u is an ancestor of v in the DFS tree (3) Back edge iff u is a descendant of v in the DFS tree (4) Cross edge iff u is neither an ancestor nor a descendant of v 9 10
- 6. 8/31/2020 6 Edge classification by DFS b a Tree edges Forward edges Back edges Cross edges c c The edge classification depends on the particular DFS tree! Edge classification by DFS b a Tree edges Forward edges Back edges Cross edges c b a c Both are valid The edge classification depends on the particular DFS tree! 11 12
- 7. 8/31/2020 7 DAGs and back edges • Can there be a back edge in a DFS on a DAG? • NO! Back edges close a cycle! • A graph G is a DAG <=> there is no back edge classified by DFS(G) What is a DAG? • A directed acyclic graph (DAG) is a directed graph without cycles. • DAGs arrise in modeling many problems that involve prerequisite constraints (construction projects, course prerequisites, document version control, compilers, etc.) • Some properties of DAGs: – Every DAG must have at least one vertex with in-degree zero and at least one vertex with out-degree zero • A vertex with in-degree zero is called a source vertex, a vertex with out-degree zero is called a sink vertex. – G is a DAG iff each vertex in G is in its own strongly connected component – Every edge (v, w) in a DAG G has finishingTime[w] < finishingTime[v] in a DFS traversal of G Example: 13 14
- 8. 8/31/2020 8 Topological sort: Source removal Method Example 15 16
- 9. 8/31/2020 9 Example: Find Topological sort A B C D E A 0 1 1 1 0 B 0 0 1 1 0 C 0 0 0 0 1 D 0 0 0 0 1 E 0 0 0 0 0 Topological order: A B C D E A B D C E Exercise problems 17 18
- 10. 8/31/2020 10 Topological sort: DFS Method Example Top of Stack Adjacen t Stack Pop A B A B C AB C E ABC E F ABCE F --- ABCEF F E C B D B A G ---- --- D ---- __ --- -- ABCE ABC ABD AB AB A G E C --- D B A G Topological sorted : G ABDCEF 19 20
- 12. 8/31/2020 12 Topological sort Algorithm • TOPOLOGICAL-SORT(G): 1) call DFS(G) to compute finishing times f[v] for each vertex v 2) as each vertex is finished, insert it onto the front of a linked list 3) return the linked list of vertices Topological Sort Algorithms: DFS based algorithm Topological-Sort(G) { 1. Call dfs AllVertices on G to compute f[v] for each vertex v 2. If G contains a back edge (v, w) (i.e., if f[w] > f[v]) , report error ; 3. else, as each vertex is finished prepend it to a list; // or push in stack 4. Return the list; // list is a valid topological sort } • Running time is O(V+E), which is the running time for DFS. Topological order: A C D B E H F G 23 24
- 13. 8/31/2020 13 Time complexity of TS(G) • Running time of topological sort: Θ(n + m) where n=|V| and m=|E| • Why? Depth first search takes Θ(n + m) time in the worst case, and inserting into the front of a linked list takes Θ(1) time Proof of correctness • Theorem: TOPOLOGICAL-SORT(G) produces a topological sort of a DAG G • The TOPOLOGICAL-SORT(G) algorithm does a DFS on the DAG G, and it lists the nodes of G in order of decreasing finish times f[] • We must show that this list satisfies the topological sort property, namely, that for every edge (u,v) of G, u appears before v in the list • Claim: For every edge (u,v) of G: f[v] < f[u] in DFS 25 26
- 14. 8/31/2020 14 Proof of correctness “For every edge (u,v) of G, f[v] < f[u] in this DFS” • The DFS classifies (u,v) as a tree edge, a forward edge or a cross-edge (it cannot be a back-edge since G has no cycles): i. If (u,v) is a tree or a forward edge ⇒ v is a descendant of u ⇒ f[v] < f[u] ii. If (u,v) is a cross-edge Proof of correctness “For every edge (u,v) of G: f[v] < f[u] in this DFS” ii. If (u,v) is a cross-edge: • as (u,v) is a cross-edge, by definition, neither u is a descendant of v nor v is a descendant of u: d[u] < f[u] < d[v] < f[v] or d[v] < f[v] < d[u] < f[u] since (u,v) is an edge, v is surely discovered before u's exploration completes f[v] < f[u] Q.E.D. of Claim 27 28
- 15. 8/31/2020 15 Proof of correctness • TOPOLOGICAL-SORT(G) lists the nodes of G from highest to lowest finishing times • By the Claim, for every edge (u,v) of G: f[v] < f[u] ⇒ u will be before v in the algorithm's list • Q.E.D of Theorem END 29