Math 275 – Graph Theory – Fall 2014
Solutions to practice problems
X3. For the graph shown below, find a minimum weight spanning tree by using (i) Kruskal’s algorithm,
and (ii) the Jarnı́k-Prim algorithm, starting at a. In each case show the edges of the tree on the graph, and
provide a list of the edges of the tree in the order in which you added them to the tree.
m 1
ℓ
13
4
21 5
n
j 3
i
10
25
h
18
13
g
k
15
10
f
21
19
d
7
e
6
18
9
c
17
b
15
11
a
Note: The class web site contains a file that you can use to print out copies of the graph.
Solution. [4+4=8] (i) See at left below; the order of edges chosen is indicated by w(i) where i is the time
chosen. Edges were considered in order of increasing weight; crossed out edges would have formed a cycle:
ℓm : 1, jn : 3, ℓn : 4, jm : 5, be : 6, de : 7, bd : 9, f k : 10, ih : 10, ab : 11, gh : 13, jℓ : 13, ac : 15, f g : 15,
bc : 17, ec : 18, f j : 18, df : 19 and stop because we now have a spanning tree; no need to consider
more edges.
So edges in order chosen are: ℓm, jn, ℓn, be, de, f k, ih, ab, gh, ac, f g, f j, df .
m 1(1) ℓ
4(3)
13
21 5
n
j 3(2)
i
10(7)
i
10(9)
25
h
g
15(11)
19(13)
d
g
10(6)
21
7(5)
9
d
18
k
f
b
10(6)
21
7(3)
e
6(2)
18
9
c
17
15(7)
19(5)
e
6(4)
18(10)
13(8)
k
f
25
h
18(12)
13(9)
m1(13) ℓ
4(12)
13
21 5
n
j 3(11)
c
17
b
15(10)
11(8)
15(4)
11(1)
a
a
1
(ii) See at right above; the order of edges chosen is indicated by w/i where i is the time chosen. Here are
the edges considered at each time step; the chosen one is underlined.
1: ab, ac
5: df , ef
9: f j, hi, kj
13: jm, ℓm
2: ac, bc, bd, be
6: f g, f j, f k
10: f j, ij, kj
3: ac, bc, bd, ce, de, ef 7: f g, f j, kj
11: jℓ, jm, jn
4: ac, bc, ce, df, ef
8: f j, gh, kj
12: jℓ, jm, ℓn
So edges in order chosen are: ab, be, de, ac, df , f k, f g, gh, hi, f j, jn, ℓn, ℓm.
X6. Apply Dijkstra’s Algorithm to the weighted digraph shown below to find shortest paths from a to
every other vertex. At each step draw a separate copy of the graph, showing the current outbranching, with
permanent arcs solid and tentative arcs dashed, and with the value of ℓ(v) shown next to each vertex v.
Note: the class web site contains a PDF file with a page containing multiple copies of this
digraph, which you may use for your answer.
2
b
d
27
a
17
5
14
8
f
3
7
13
12
c
11
e
4
Solution. [8] We show vertices as v/ℓ where v is the vertex name and ℓ is the (tentative) distance to v from
the root (a). If we do not show ℓ then ℓ(v) = ∞.
At the first iteration we choose the root a as the unmarked vertex with smallest ℓ and update ℓ(b), ℓ(c).
At the second iteration we choose c as the unmarked vertex with smallest ℓ and update ℓ(b), ℓ(d) and ℓ(e).
2
2
b/27
b/26
d/20
d
27
17
27
17
5
5
a/0
a/0
f
f
14
3
14
3
8
7
8
7
13
13
12
11
12
11
e
c/12
c/12
e/25
4
4
At the third iteration we choose d and update ℓ(b), ℓ(e) and ℓ(f ), and at the fourth iteration we choose
e and update ℓ(f ).
2
2
b/25
d/20
b/25
d/20
27
17
5
a/0
14
8
7
c/12
4
14
11
8
f /34
7
3
13
12
e/23
17
5
a/0
3
13
12
27
f /37
c/12
4
11
e/23
At the fifth iteration we choose b and do not change ℓ, and at the sixth and final iteration we choose f
and do not change ℓ. Now all vertices are marked with their distance from a, and shortest paths from a are
as shown.
2
2
b/25
27
d/20
17
5
a/0
14
8
c/12
4
14
11
c/12
3
17
8
f /34
7
3
13
12
e/23
d/20
5
a/0
3
13
12
27
f /34
7
2
b/25
4
11
e/23