1) In network figure below, use Floyd's algorithm to determine the shortest routes between each of the following pairs of nodes: 10 5 15 1.1.) From node 5 to node 1. a) 5-1,5-4-1, 5-4-2-1, distance 12 b) 5-1,5-4-2,5-4-2-1, distance 12 1.2. From node 3 to node 5. a) 3-5, 3-5, distance 15 b) 3-5, 3-4-5, distance = 10 c) 3-5, 4-3-5, distance = 21 c) 5-1,5-4-1,5-4-3-1, distance 20 5-1,5-4-3,5-4-2-1, distance 20 d) d) 3-5, 3-4, 3-4-5, distance = 10

Database System Concepts
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Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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1) In network figure below, use Floyd's algorithm to determine the shortest
routes between each of the following pairs of nodes:
2
Love
10
15
1.1.) From node 5 to node 1.
a) 5-1, 5-4-1, 5-4-2-1,
distance 12
b) 5-1, 5-4-2,5-4-2-1,
distance 12
1.2. From node 3 to node 5.
a) 3-5, 3-5, distance 15
b) 3-5, 3-4-5, distance 10
c) 3-5, 4-3-5, distance = 21
1.3. From node 5 to node 3.
a) 5-3, distance 15
b) 5-3,5-4-2-1-3, distance
= 22
1.4) From node 5 to node 2
a) 5-2,5-3-4-2, distance = 26
b) 5-2,5-4-2, distance = 9
5
c) 5-1,5-4-1, 5-4-3-1,
distance 20
d) 5-1, 5-4-3,5-4-2-1,
distance 20
d) 3-5, 3-4, 3-4-5, distance =
10
c) 5-3,5-4-3, distance = 10
d) 5-3, 3-4-5, distance 10
c) 5-2,5-4-3-1-2, distance
d) 5-2,5-3-1-2, distance = 28
= 23
Transcribed Image Text:1) In network figure below, use Floyd's algorithm to determine the shortest routes between each of the following pairs of nodes: 2 Love 10 15 1.1.) From node 5 to node 1. a) 5-1, 5-4-1, 5-4-2-1, distance 12 b) 5-1, 5-4-2,5-4-2-1, distance 12 1.2. From node 3 to node 5. a) 3-5, 3-5, distance 15 b) 3-5, 3-4-5, distance 10 c) 3-5, 4-3-5, distance = 21 1.3. From node 5 to node 3. a) 5-3, distance 15 b) 5-3,5-4-2-1-3, distance = 22 1.4) From node 5 to node 2 a) 5-2,5-3-4-2, distance = 26 b) 5-2,5-4-2, distance = 9 5 c) 5-1,5-4-1, 5-4-3-1, distance 20 d) 5-1, 5-4-3,5-4-2-1, distance 20 d) 3-5, 3-4, 3-4-5, distance = 10 c) 5-3,5-4-3, distance = 10 d) 5-3, 3-4-5, distance 10 c) 5-2,5-4-3-1-2, distance d) 5-2,5-3-1-2, distance = 28 = 23
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