Evaluation of the Course (Modified)

1
Evaluation of the Course (Modified)

• Course work 30
• Four assignments (25)
• 7.5 5 points for each of the first two three
  assignments
• 10 points for the last assignment
• One term paper (5) (week13 Friday)
• Find an open problem from internet.
• State the problem definition in English.
• Write the definition mathematically.
• Summarize the current status
• No more than 1 page
• A final exam 70

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Single source shortest path with negative cost
edges
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Shortest Paths Dynamic Programming

• Def. OPT(i, v)length of shortest s-v path P
  using at most i edges.
• Case 1 P uses at most i-1 edges.
• OPT(i, v) OPT(i-1, v)
• Case 2 P uses exactly i edges.
• If (w, v) is the last edge, then OPT use the
  best s-w path using at most i-1 edges and edge
  (w, v).
• Remark if no negative cycles, then OPT(n-1,
  v)length of shortest s-v path.
  

?
Cwv
s
w
v
OPT(0, s)0.
4
Shortest Paths implementation

• Shortest-Path(G, t)
• for each node v ? V
• M0, v ?
• M0, s 0
• for i 1 to n-1
• for each node w ? V
• Mi, w Mi-1, w
• for each edge (w, v) ? E
• Mi, v min Mi, v, Mi-1, w
  cwv
• Analysis. O(mn) time, O(n2) space.
• m--no. of edges, nno. of nodes
• Finding the shortest paths. Maintain a
  "successor" for each table entry.

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Shortest Paths Practical implementations

• Practical improvements.
• Maintain only one array Mv shortest v-t path
  that we have found so far.
• No need to check edges of the form (w, v) unless
  Mw changed in previous iteration.
• Theorem. Throughout the algorithm, Mv is the
  length of some s-v path, and after i rounds of
  updates, the value Mv ? the length of shortest
  s-v path using ? i edges.
• Overall impact.
• Memory O(m n).
• Running time O(mn) worst case, but substantially
  faster in practice.

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Bellman-Ford Efficient Implementation

• Push-Based-Shortest-Path(G, s, t)
• for each node v ? V
• Mv ?
• successorv empty
• Ms 0
• for i 1 to n-1
• for each node w ? V
• if (Mw has been updated in
  previous iteration)
• for each node v such
  that (w, v) ? E
• if (Mv gt Mw
  cwv)
• Mv
  Mw cwv
• 
  successorv w
• If no Mw value changed in
  iteration i, stop.

Note Dijkstras Algorithm select a w with the
smallest Mw .
Time O(mn), space O(n).
7
(a)
8
5
u
v
6
8
-2
6
-3
8
0
7
s
-4
2
7
7
8
9
x
y
(b)
9
5
u
v
6
4
-2
6
-3
8
0
7
s
-4
2
7
7
2
9
x
y
(c)
10
5
u
v
2
4
-2
6
-3
8
0
7
s
-4
2
7
2
7
9
x
y
(d)
11
5
u
v
2
4
-2
6
-3
8
0
7
s
-4
2
7
7
-2
y
x
9
(e)
vertex s u v x y
d 0 2 4 7 -2 successor s v x s
u
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Corollary If negative-weight circuit exists in
the given graph, in the n-th iteration, the cost
of a shortest path from s to some node v will be
further reduced.

• Demonstrated by the following example.

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5
1
6
-2
8
7
7
9
2
2
5
-8
An example with negative-weight cycle
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i1
15
i2
16
i3
17
i4
18
5
6
11
1
6
-2
0
8
12
7
7
9
2
6
15
2
5
-8
8
1
i5
19
i6
20
5
6
11
1
6
-2
0
8
12
7
7
9
2
5
15
2
5
-8
8
0
x
i7
21
5
6
11
1
6
-2
0
8
12
7
7
9
2
5
15
2
5
-8
7
0
x
i8
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Dijkstras Algorithm (Recall)

• Dijkstras algorithm assumes that w(e)?0 for each
  e in the graph.
• maintain a set S of vertices such that
• Every vertex v ?S, dv?(s, v), i.e., the
  shortest-path from s to v has been found. (Intial
  values Sempty, ds0 and dv?)
• (a) select the vertex u?V-S such that
• dumin dxx ?V-S. Set
  SS?u
• (b) for each node v adjacent to u do
  RELAX(u, v, w).
• Repeat step (a) and (b) until SV.

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Continue

• DIJKSTRA(G,w,s)
• INITIALIZE-SINGLE-SOURCE(G,s)
• S
• Q VG
• while Q
• do u EXTRACT -MIN(Q)
• S S u
• for each vertex v ? Adju
• do RELAX(u,v,w)

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u
v
s
y
x
(a)
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u
v
1
10/s
8
10
9
0
s
3
4
6
2
7
5
5/s
8
2
y
x
(b)
(s,x) is the shortest path using one edge. It is
also the shortest path from s to x.
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u
v
1
14/x
8/x
10
9
0
s
3
4
6
2
7
5
7/x
5/s
2
y
x
(c)
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u
v
1
13/y
8/x
10
9
0
s
3
4
6
2
7
5
7/x
5/s
2
y
x
(d)
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(e)
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u
v
1
9/u
8/x
10
9
0
s
3
4
6
2
7
5
7/x
5/s
2
y
x
(f)
Backtracking v-u-x-s
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• Theorem Consider the set S at any time in the
  algorithms execution. For each v?S, the path Pv
  is a shortest s-v path.
• Proof We prove it by induction on S.
• If S1, then the theorem holds. (Because ds0
  and Ss.)
• Suppose htat the theorem is true for Sk for
  some kgt0.
• Now, we grow S to size k1 by adding the node v.

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• Proof (continue)
• Now, we grow S to size k1 by adding the node v.
• Let (u, v) be the last edge on our s-v path Pv.
• Consider any other path from P s,,x,y, , v.
  (red in the Fig.)
• y is the first node that is not in S and x?S.
• Since we always select the node with the
  smallest value d in the algorithm, we have
  dv?dy.
• Moreover, the length of each edge is ?0.
• Thus, the length of P?dy?dv.
• That is, the length of any path ?dv.
• Therefore, our path Pv is the shortest.

y
x
s
If y does not exist, dv is the smallest length
for paths from s to v using red nodes only since
we did relax.from every red node to v.
u
v
Set S
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The algorithm does not work if there are negative
weight edges in the graph

• .

u
-10
2
v
s
1
S-gtv is shorter than s-gtu, but it is longer than
s-gtu-gtv.
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0-1 version
v/w 1, 3, 3.6, 3.66, 4.
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