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QuestionFind an Euler circuit using Fleury's Algorithm for edges: FG, FE, FI, FJ, ED, EH, EI, DF, DH, DI, JG, IH, GJ. List the vertices visited.

Studdy Solution

STEP 1

Assumptions1. The graph has the following edges FG, FE, FI, FJ, ED, EH, EI, DF, DH, DI, JG, IH, GJ. . We need to find an Euler circuit, which means we need to visit every edge exactly once and return to the starting vertex.
3. We will use Fleury's Algorithm to find the Euler circuit.

STEP 2

First, we need to check if an Euler circuit exists in the graph. According to the graph theory, an Euler circuit exists if and only if all vertices in the graph have an even degree.

STEP 3

Let's calculate the degree of each vertex in the graph. The degree of a vertex is the number of edges connected to it.
Degree(F)=5Degree(F) =5Degree(E)=Degree(E) =Degree(I)=Degree(I) =Degree(J)=3Degree(J) =3Degree(D)=Degree(D) =Degree(H)=3Degree(H) =3Degree(G)=3Degree(G) =3

STEP 4

Since vertices J, H, and G have odd degrees, an Euler circuit does not exist in the graph. However, an Euler path, which visits every edge exactly once without necessarily returning to the starting vertex, does exist because there are exactly two vertices with odd degrees. We will adjust our goal to find an Euler path instead.

STEP 5

Now, let's use Fleury's Algorithm to find an Euler path. We start from one of the vertices with an odd degree, let's say vertex J.

STEP 6

From J, we can go to either F or G. We choose F because removing the edge J does not disconnect the graph.
Path=J−FPath = J - F

STEP 7

From F, we can go to, I, or G. We choose because removing the edge FE does not disconnect the graph.
Path=J−F−Path = J - F -

STEP 8

From, we can go to D, H, or I. We choose D because removing the edge ED does not disconnect the graph.
Path=J−F−−DPath = J - F - - D

STEP 9

From D, we can go to F, H, or I. We choose F because removing the edge DF does not disconnect the graph.
Path=J−F−−D−FPath = J - F - - D - F

STEP 10

From F, we can go to I or G. We choose I because removing the edge FI does not disconnect the graph.
Path=J−F−−D−F−IPath = J - F - - D - F - I

STEP 11

From I, we can go to, H, or D. We choose because removing the edge IE does not disconnect the graph.
Path=J−F−−D−F−I−Path = J - F - - D - F - I -

STEP 12

From, we can go to H. So, we go to H.
Path=J−F−−D−F−I−−HPath = J - F - - D - F - I - - H

STEP 13

From H, we can go to I or D. We choose I because removing the edge HI does not disconnect the graph.
Path=J−F−−D−F−I−−H−IPath = J - F - - D - F - I - - H - I

STEP 14

From I, we can go to D. So, we go to D.
Path=J−F−−D−F−I−−H−I−DPath = J - F - - D - F - I - - H - I - D

STEP 15

From D, we can go to H. So, we go to H.
Path=J−F−−D−F−I−−H−I−D−HPath = J - F - - D - F - I - - H - I - D - H

STEP 16

From H, we can only go to G. So, we go to G.
Path=J−F−−D−F−I−−H−I−D−H−GPath = J - F - - D - F - I - - H - I - D - H - G

STEP 17

From G, we can go to J or F. We choose J because removing the edge GJ does not disconnect the graph.
Path=J−F−−D−F−I−−H−I−D−H−G−JPath = J - F - - D - F - I - - H - I - D - H - G - J

STEP 18

From J, we can only go to G. So, we go to G.
Path=J−F−−D−F−I−−H−I−D−H−G−J−GPath = J - F - - D - F - I - - H - I - D - H - G - J - G

STEP 19

From G, we can only go to F. So, we go to F.
Path=J−F−−D−F−I−−H−I−D−H−G−J−G−FPath = J - F - - D - F - I - - H - I - D - H - G - J - G - FThe Euler path through the graph is J - F - - D - F - I - - H - I - D - H - G - J - G - F.

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