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Algoirthms > Shortest paths algorithms > Flashcards

Shortest paths algorithms Flashcards

1
Q

What is the generic algorithm for finding delta(s,v) for all v’s?

A

Init:

  • d(s)=0, d(v)=infinity

loop:

  • while (exists an edge (u,v) for which d(v)>d(u)+w(u,v):
    • find such an edge and call Relax(u,v,w)

end:

  • return d(v) for every v
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2
Q

Lower bound claim

All along the run of the algorithm, for all v, the inequality: d(v)>=delta(s,v) is satisfied, and if d(v)

proof that claim.

A

Using induction on the number of Relax operations

  • base: after initilization
    • v is not s, delta(s,v)
    • d(s)=0 = delta(s,s)
  • Induction assumption
    • by the end of the i-1 step, the assumption is satisfied.
  • step
    • Let’s look at Relax on the ith step.
    • case 1: no update
      • array d stays the same
    • case 2: updated
      • d(v)-the only one that has changed.
      • d(v)=d(u)+w(u,v)
      • d(u)>=delta(s,u) (by I.A)
      • delta(s,u)+w(u,v) >= delta(s,v)
        • observation.
      • d(v)>= delta(s,u)+w(u,v)>=delta(s,v)
      • since d(v) is updated:
        • d(v)
      • d(u) - a weight of some Psu.
      • d(v)=d(u)+w(u,v)*-**weight of Psu+(u,v)*
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3
Q

What is the correctness statement for the generic algorithm?

A

If through the run of the algorithm, for every edge (u,v), d(v)<=d(u)+w(u,v) is satisfied(that is, the algorith, stops), then, for every v, d(v)=delta(s,v).

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4
Q

What is the correctness statement proof for the generic algorithm, in case it stops?

A
  • Assume the algorithm stopped
    • for all (u,v) in E, d(v)<=d(u)+w(u,v)
  • Assume in negation: there exists x in V s.t:
    • d(x) != delta(s,x)
  • Let Psx the lightest path from s to x.
  • let v be the first vertex in P s.t. d(v) != delta(s,v)
    • by the lower bound claim: d(v)>delta(s,v)
  • let u be the preceding vertex to v in P
    • d(u) = delta(s,u)
  • By sub-paths of a shortests path prinicipal
    • delta(s,u)+w(u,v)=delta(s,v).
  • d(v)<=d(u)+w(u,v)=delta(s,u)+w(u,v)=delta(s,v)
    • contradiction
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5
Q

What is the “full” correctness statement of the generic algorithm?

A
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6
Q

What is the exact definition of

rooted tree in vertex s

A

Directed graph is called rooted tree in vertex s if every vertex in the graph is accessible from s, and it is accessible only via one path

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7
Q

Which data structure can help us maintain a rooted tree in vertex s?

A

It can be maintaind via array of length |V|

  • The index of a cell represents some vertex*
  • The content of cell v is former(v)*
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8
Q

“lightest paths tree T for graph G and source s”

what is it?

A

It is a rooted tree in vertex s which satisfies:

for every v, the only path from s to v in T is the lightest path from s to v in G.

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9
Q

General algorithm, what claims are used in order to prove the correctness statement?

A
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10
Q

Generic algorithm, how the first claim is proved?

A
  • N.T.F - d(u) = delta(s,u)
  • lower bound claim - d(u)>=delta(s,u)
  • d(v)=d(u)+w(u,v) = delta(s,v)
    • last Relax operation -> d(v)=delta(s,u)
  • delta(s,v)<= delta(s,u)+w(u,v)
  • d(u)+w(u,v) = delta(s,v)<=delta(s,u)+w(u,v)
    • d(u)<=delta(s,u)
  • d(u)>=delta(s,u) & d(u)<=delta(s,u)
    • d(u) = delta(s,u)
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11
Q

General algorithm, proof for the 2nd claim.

What is the induction based on

A

It is based on the size of V’. That is, |V’|.

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12
Q

Genereal algorithm, proof of the 2nd claim.

What is the base case for the induction proof?

A
  • |V’| = 1.*

That is:

  • V’ = {s}
  • E’ = empty set.

Trivial.

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13
Q
  • Generic algorithm, proof of the 2nd claim.*
  • what is the step in the induction?*
A
  • we look at the next vertex v to be added into V’. That is, the next vertex for which d(v) = delta(s,v).*
  • N.T.S: T=(V’ U {v}, E’ U {former(v),v})* is a shortest paths tree.
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14
Q

Generic algorithm, proof of the 2nd claim.

What is the proof for the step phase of the induction?

A

According to the 1st claim: d(v)=delta(s,v) means:

  • former(v) = delta(s,former(v))
  • therefore, former(v) is in V’ too.

Proof: T is a tree.

  • (V’,E’) is a tree.
  • former(v) is in V’
  • we added an edge from former(v) to a new vertex v in V’.

proof: T is a shortest paths tree from s

  • I.A holds for all other vertices. N.T.F for v.
  • The path Psv in T is composed of:
    • Psformer(v) + (former(v),v)
  • by the assumption: d(v)=delta(s,v)
  • by claim 1: d(former(v)) = delta(s,former(v))
  • w(Psv)=w(Psformer(v))+w(former(v),v) =
  • delta(s,former(v))+w(fomer(v),v)=d(v) = delta(s,v)
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15
Q

Generic algorithm, what is the proof for the correctness statement, using the claims.

A
  • If the algorithm stops, then we proved that for all v, d(v) = delta(s,v) is satisfied.*
  • Therefore, the Tree V’=V E’={(former(v),v): v!=s} can be created, and by the 2nd claim - G = (V’,E’) is a lightests paths tree from s in G.*.
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16
Q

Dijstra, which graphs can it handle?

A

Directed or undirected graphs with non-negative weights.

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17
Q

Dijstra, which problem can it solve?

A

Finding the shortest paths from source s to all vertices.

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18
Q

_Dijstra, write exactly the Relax(u,v,w(u,v)) procedure_

A
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19
Q

Dijstra, show me the initilization step.

You are given G=(V,E),s,w

A
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20
Q

Dijstra, what is the loop step in the algorithm?

A
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21
Q

How is Dijkstra’s algorithm implemented?

what’s his run-time? divide it to operation on Q and operations on the array d

A
  • Q is maintained by a minimum heap in which d(v) is a key.*
  • we push mininum d(v) from Q using ExtractMin.*
    1. Q*
  • Initilization of Q - O(|V|)
  • There are exactly |V| Extract-Min operations.
  • There are at most |E| Relax operations, which decreases the value of d(v). This implies O(|E|) decrease-key(v) operations (log(|V|)). In total - O(|E|log|V|).
  1. d
  • Initializing d and s - O(|V|)
  • |E| Relax operations on d(v), each O(1). O(|E|)
  • In total: O(|E|+|V|).

In total: O(|E| log|V|)

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22
Q

Dijkstra, what is the lower bound claim?

A

d(v)>=delta(s,v)

That is, d(v) is always greater equal to the value of the lightest path.

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23
Q

Dijkstra, What is the correctness statement?

A

When the algorithm is done, for all v, d(v)=delta(s,v)

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24
Q
  • Dijkstra, which observation is used in the proof?*
  • Explain it.*
A

Once vertex u is being added to S, d(u) does not change.

Explanation:

  • After u is being added to S, we execute Relax operation on vertex v from U, which share an edge with u. The updated value is d(v), and is again, is not chosen from S.
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25
Q

Dijkstra, what is the first claim?

A

Claim 1:

  • After Relax(u,v), d(v)<= d(u)+w(u,v) is satisfied through all the remaining steps of the algorithm.
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26
Q

Diskjstra, what is the proof for the first claim?

A

By the end of Relax(u,v):

  • d(v)=min{d(v),d(u)+w(u),v} is satisfied.
    • That is, d(v)<= d(u)+w(u,v)
  • Relax is performed due to the insertion of u into S.
  • According to the observation: d(u) will not further change.
  • In contrast, d(v) might change, but only decrease, so the inequality holds.
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27
Q

Dijkstra, what is the helping statement?

A

When vertex u is being added to S d(u)=delta(s,u) is satisfied.

That is, d(u) holds the value for the lightest path

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28
Q

Dijkstra, what is the proof for the correctness statement?

A

Proof of the correctness statement

  • In each step a vertex shifts from Q to S
    • After |V| iterations, S=V and the algorithm stops.
  • By the helping statement, when vertex u enters S:
    • d(u) = delta(s,u)
    • By the observation, this value does not change.
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29
Q

Dijkstra, what is the proof for the helping statement?

A

We prove using induction on the algorithm steps - i:

  • basis i = 1. when u=s.
    • d(u) = 0
    • delta(s,s) = 0
  • Induction assumption:
    • after i-1 iterations, for all u in S, d(u)=delta(s,u)
  • step:
    • let u be that ith vertex chosed to enter S
    • Assume u is accessible from S.
    • Let Psu be some lightest path from s to u.
    • Let (x,y) be an edge in Psu, x in S, y in U.
      • Psx is Reisha of lightest path
        • w(Psx) = delta(s,x).
      • for x in S, the induction assumption holds
        • d(x) = delta(s,x)
      • w(Pyu)=>0
        • because lights are non-negative
      • by claim 1 on (x,y):
        • d(y)<=d(x)+w(x,y)
      • The algoirthm choose u in this step thus:
        • d(y)>=d(u)
      • In total:

delta(s,u) = w(Psu) = w(Psx) + w(x,y) + w(Pyu)

  • w(Psx) + w(x,y) + w(Pyu) = delta(s,x)+w(x,y)+w(Pyu)*
  • delta(s,x)+w(x,y)+w(Pyu) = d(x) + w(x,y) + w(Pyu)*
  • d(x) + w(x,y) + w(Pyu) >= d(x)+w(x,y)>= d(y) >= d(u)*
30
Q

What is the bottle neck problem?

A

Definition:

  • For source vertex s and vertex u:
    • bottle neck in path P is:
      • w(minimal edge in P)
    • B(s,u):
      • B(s,s) = infinity
      • Largest bottle neck within all paths from s to u
      • If no such path exists - minus infinity

The problem:

  • Input: weighted graph G=(V,E,w) and source vertex s
  • Find B(S,u) for every u in V.
31
Q
  • bottle neck problem:*
  • explain BRelax(u,v,w)
A

BRelxa(u,v,w):

  • There’s an edge between u and v
  • b[v] - current largest bottle neck from s to u.
  • Check if there’s a path s to u with a larger bottle neck:
    • if b[v]
    • b[v] = min{b[u],w(u,v)}
32
Q
  • BottleNeck(G, w, s)*
  • Explain the initialization step.*
A
33
Q
  • BottleNeck(G, w, s)*
  • Explain the loop step.*
A
34
Q

BottleNack

  • What is the main statement?
  • what is claim 1?
  • what is claim 2?
  • what is the observation?
A
35
Q

In the proof of the correntness statement there are three cases we need to check.

What are they?

A
  • We need to prove that for any vertex u, the moment it enters S, b[u]=B(s,u) is satisfied.*
  • Note: by the observation this equality hold until the end of the run.*

Cases to check:

  • u=s
  • No path from s to u
  • There’s a path from s to u, and u is not s.
36
Q
  • BottleNeck corretness statement proof*
  • How is the first case proved?*
A

s = u.

  • We set b[s] = infinity at the initialization step.
  • s is the first to enter S
  • B(s,s) = infinity by definition.
37
Q

BottleNeck corretness statement proof

How is the second case proved?

A

There is no path from s to u

  • B(s,u) = minus infinity
  • The 2nd claim tells us that b[u]<=B(s,u), always.
  • b[u] = minus infinity
38
Q

BottleNeck corretness statement proof

  • Third case:*
  • it is proved using induction. induction on what?
A

On the order of removal of vertices from Q

AKA, u = EXTRACT-MAX(Q)

39
Q

BottleNeck corretness statement proof

  • third case:*
  • what is the induction proof?
A
40
Q

BottleNeck, what is the proof for claim 1?

A
41
Q

BottleNeck, 2nd claim proof.

Recall: 2nd claim

All along the algorithm, for each vertex v, b[v]<=B(s,v) is satisfied.

A
42
Q
  • What is the algorithm for the attached problem?*
  • What is its run-time?*
A
43
Q
  • What is the main statement?*
  • what are the two claims we use?*
A
44
Q

How is the 1st claim proved?

A
45
Q

How is the 2nd claim proved?

A

_Induction on the length of the path i_

46
Q

How is the main statement proved?

A
47
Q

Which algorithm is designed to find the lightests paths when negative weights are allowed?

A

Bellman-Ford

48
Q

Describe Bellman-Ford’s algorithm

A
49
Q

Analyze Bellman-Ford run-time

A
50
Q
  • Bellman-Ford*
  • Assume all vertices are accessible from s*
  • We can run first BFS to identify the unaccessible vertices and avoid them.*
  • Theorem:
  • There exists a lightest path from s to v, for all v in V, if and only if there is no path from s to v which contains a circle of negative weight.*

Prove the theorem

A
51
Q
  • Bellman-Ford*
  • What are the two claims which help us prove the correctness of BF?*
A
52
Q
  • Bellman-Ford*
  • The 1st claim is proved using a theorem and an observation**. What are they?*
A
53
Q
  • Bellman-Ford*
  • How the theorem of the 1st claim is proved?*
A
54
Q
  • Bellman-Ford,*
  • How the 1st claim is solved*
A
55
Q
  • Bellman-Ford*
  • How is the 2nd claim proved?*
A
56
Q

What is the algoirthm for the attached problem?

A
57
Q
  • Find-Negative-Cycles.*
  • What are the claims we use for the proof?*
A
58
Q
  • Find-Neg-Circle*
  • What is the proof for the 2nd claim?*
A
59
Q
  • Find-Neg-Circle*
  • What are the lemmas we use for the 1st claim?*
A
60
Q
  • Find-Neg-Circle: 1st claim*
  • What is the proof for the 2nd lemma?*
A
61
Q
  • Find-Neg-Circle:*
  • what is the proof for claim 1, based on the two lemmas?*
A
62
Q

How is this problem solved?

A
  • solution:*
  • we’ll use reduction to “shortest path between two vertices” problem.*
63
Q
  • TrafficLightProblem*
  • Explain the input translation phase*
A
64
Q
  • TrafficLightProblem*
  • we run Dijkstra after the input translation step. Then, what is the output translation function?*
A
65
Q

TrafficLightProblem

  • what is the main theorem?
  • what is the helping theorem?
A
66
Q
  • TrafficLightProblem*
  • what is the proof for the helping theorem?*
A
67
Q
  • TrafficLightProblem*
  • what is the proof for the main theorem?*
A
68
Q
  • TrafficLightProblem*
  • what is the run-time of this algorithm?*
A
69
Q

What are some basic assumptions we proved, before jumping into Dijkstra and BF?

A

A path from s to u which passes by a negative weight circle can never produce a “shortest path”, thus noted as negative infinity.

A shortest path enforces all its subpaths to be also shortest paths.

A shortest path is definitley a simple path.

70
Q
  • Relaxation is the only means by which shortest path estimates and predecessors change.*
  • The algorithm differ in how many times they relax each edge and the order in which they relax edges.*
  • What is the difference between Dijkstra and BF regarding the number of edges they relax?*
A
  • Dijkstra relax each edge at most once*
  • BF relax each edge |V|-1 times.*
71
Q
  • Shortest paths:*
  • List all main properties.*
A
  • Triganlge inequality
  • Upper-bound inequlity
  • No-path property
  • Convergence property
  • Path relaxation property
  • Predecessor-subgraph property.
72
Q

Assuming there are no negative cycles, prove BF sets v.d = delta(s,v) for every v in V

A

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