1. CPS420 Euler Circuit Construction
© Sophie Quigley

2. Original Graph G V(G) = {A, B, C, D, E, F, G, H, I, J}
E(G)={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

3. Graph G’
Let G’ = G
E(G’)= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
V(G’) = {A, B, C, D, E, F, G, H, I, J}

4. Graph G’ Pick a vertex of G’: v=B
E(G’)= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
V(G’) = {A, B, C, D, E, F, G, H, I, J}

5. Circuit C=C’
Let C = C’ = a circuit in G which starts and ends in v = B2C13I8H12B
E(C) = E(C’) = {2, 13, 8, 12 }
V(C) = V(C’) = {B, C, I, H}
E(G’)= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
V(G’) = {A, B, C, D, E, F, G, H, I, J}
Core of proof: how can we be sure that there is such a circuit?
Circuit starts at v and each time it enters a vertex it can also leave it because all the vertices have even degree
When the circuit started at v, because the degree of v is even, there was still at least one remaining untraversed edge leading back to v. Call the other vertex on that edge w. The circuit has to reach w in order to be able to return to v. However, the graph is connected, so this means that there exists a path from any vertex in G (and therefore in the circuit) to w and therefore the circuit can eventually return to w and then v.

6. Iterate on G’ Let G’ be the new graph s.t.
E(G’)= E(G’) – E(C’) = {1, 3, 4, 5, 6, 7, 9, 10, 11, 14, 15, 16, 17 }
V(G’) = V(G’) – {all isolated vertices once edges in E(C’) have been removed}
= {A, B, C, D, E, F, G, H, I, J}  
C = C’= B2C13I8H12B
E(C) = E(C’) = {2, 13, 8, 12 }
V(C) = V(C’) = {B, C, I, H}

7. Iterate on C’ Pick a vertex w from V(C)V(G’) w = H
E(G’)= {1, 3, 4, 5, 6, 7, 9, 10, 11, 14, 15, 16, 17 }
V(G’) = {A, B, C, D, E, F, G, H, I, J}
C = B2C13I8H12B
E(C) = {2, 13, 8, 12 }
V(C) = {B, C, I, H}

8. Iterate on C’ Let C’ = circuit of G’ starting at w C’ = H15E5F6G7H
Integrate C’ into C:
C = B2C13I8H15E5F6G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7 }
V(C) = {B, C, I, H, E, F, G}

9. Iterate on G’ Let G’ be the new graph s.t.
E(G’)= E(G’) – E(C’) = {1, 3, 4, 9, 10, 11, 14, 16, 17 }
V(G’) = V(G’) – {all isolated vertices once edges in E(C’) have been removed}
= {A, B, C, D, E, G, I, J}  
C = B2C13I8H15E5F6G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7 }
V(C) = {B, C, I, H, E, F, G}

10. Iterate on C’ Pick a vertex w from V(C)V(G’) w = I
E(G’)= {1, 3, 4, 9, 10, 11, 14, 16, 17 }
V(G’) = {A, B, C, D, E, G, I, J}
C = B2C13I8H15E5F6G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7 }
V(C) = {B, C, I, H, E, F, G}
Note that G’ is now disconnected.. However, the algorithm iterates on the connected components of G’. Also, because edges are removed from G’ two at a time, ,the vertices in G’ still have even degrees. Therefore by the same reasoning as earlier, circuits starting from vertices in V(C)V(G’) can still be found.

11. Iterate on C’ Let C’ = circuit of G’ starting at w C’ = I11B1A10J9I
Integrate C’ into C:
C = B2C13I11B1A10J9I
8H15E5F6G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7, 11, 1, 10, 9 }
V(C) = {B, C, I, H, E, F, G, J, A}

12. Iterate on G’ Let G’ be the new graph s.t.
E(G’)= E(G’) – E(C’) = {3, 4, 14, 16, 17 }
V(G’) = V(G’) – {all isolated vertices once edges in E(C’) have been removed}
= {C, D, E, I, J}  
C = B2C13I11B1A10J9I
8H15E5F6G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7, 11, 1, 10, 9 }
V(C) = {B, C, I, H, E, F, G, J, A}

13. Iterate on C’ Pick a vertex w from V(C)V(G’) w = G
E(G’)= {3, 4, 14, 16, 17 }
V(G’) = {C, D, E, I, J}
C = B2C13I11B1A10J9I
8H15E5F6G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7, 11, 1, 10, 9 }
V(C) = {B, C, I, H, E, F, G, J, A}

14. Iterate on C’ Let C’ = circuit of G’ starting at w C’ = G16I17G
Integrate C’ into C:
C = B2C13I11B1A10J9I
8H15E5F6G16I17G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7, 11, 1, 10, 9 , 16, 17}
V(C) = {B, C, I, H, E, F, G, J, A}

15. Iterate on G’ Let G’ be the new graph s.t.
E(G’)= E(G’) – E(C’) = {3, 4, 14}
V(G’) = V(G’) – {all isolated vertices once edges in E(C’) have been removed}
= {C, D, E}  
C = B2C13I11B1A10J9I
8H15E5F6G16I17G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7, 11, 1, 10, 9 , 16, 17}
V(C) = {B, C, I, H, E, F, G, J, A}

16. Iterate on C’ Pick a vertex w from V(C)V(G’) w = C E(G’)= {3, 4, 14}
V(G’) = {C, D, E}
C = B2C13I11B1A10J9I
8H15E5F6G16I17G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7, 11, 1, 10, 9 , 16, 17}
V(C) = {B, C, I, H, E, F, G, J, A}

17. Iterate on C’ Let C’ = circuit of G’ starting at w C’ = C3D4E14C
Integrate C’ into C:
C = B2C3D4E14C13I11B1A10J9I
8H15E5F6G16I17G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7, 11, 1, 10, 9, 16, 17, 3, 4, 14}
V(C) = {B, C, I, H, E, F, G, J, A, D}

18. Iterate on G’ Let G’ be the new graph s.t. E(G’)= E(G’) – E(C’) = 
V(G’) = V(G’) – {all isolated vertices once edges in E(C’) have been removed}
= 
C = B2C3D4E14C13I11B1A10J9I
8H15E5F6G16I17G7H12B
E(C) = {2, 13, 8, 12, 15, 5, 6, 7, 11, 1, 10, 9, 16, 17, 3, 4, 14}
V(C) = {B, C, I, H, E, F, G, J, A, D}