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By Meli & Amy & Meggie & Bex. What is route inspection hmmmm??? Objective: Is to go along every single edge and end up back to where you started from.

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Presentation on theme: "By Meli & Amy & Meggie & Bex. What is route inspection hmmmm??? Objective: Is to go along every single edge and end up back to where you started from."— Presentation transcript:

1 By Meli & Amy & Meggie & Bex

2 What is route inspection hmmmm??? Objective: Is to go along every single edge and end up back to where you started from. Aim: Is to find the quickest route from a given starting point. REMEMBER route inspection only works if the graph is “Eulerian”

3 How To Do A Route Inspection

4 Step 1 Note Down All The Odd Vertices Odd Vertices => B, D, E, F 3 3 5 17 4 4 2 6 71 3 1 2 3 1 4 2 3 1 34 2 3 1 5 34 2 3 1 6 5 34 2 3 17 6 5 34 2 3 1 4 7 6 5 34 2 3 1

5 Step 2 & 3 Pair Up Odd Vertices BE – 5 (Find the shortest routes) Don’t forget there might be more than one different way of getting from 1 letter to another. Make sure to check all the different ways 4 7 6 5 34 2 3 1 BD – 6 BF – 2 DE – 3 DF – 7 EF – 4 BD + EF = 10 BE + DF = 12 BF + DE = 5

6 Step 4 Add up all the weights on the graph Total Weights => 1 + 7 + 4 + 3 + 4 + 3 + 2 + 5 + 6 = 35 4 7 6 5 34 2 3 1

7 Step 5 Add on repeated edges 35 + 5 = 40 4 7 6 5 34 2 3 1 3 22

8 Step 6 State A Route 4 7 6 5 34 2 3 1 2 3 Example: FABEFBCDEBD

9 Practise Questions 1. Below is the shape of a university campus. Each building is given a letter. As you can see there is the distance between each building written on the diagram. Using route inspection find out the shortest route, starting at A. 67 25 89 37 12 42 7 96 30 55 40 10 90 A B D C E F G I H 22

10 1.Odd vertices are BCDFGH, BC=37 CD=42 DF=70 FG=25 GH=32 BD=79 CF=52 DG=52 FH=50 BF=95 CG= 34 DH=40 BG=77 CH=22 BH=65 BC+DF+GH= 37+70+32=139 BD+CH+FG= 79+22+25=126 BF+CG+DH= 95+34+40=169 BG+CD+FH= 77+42+50=169 BH+CF+DG=65+52+52= 169 Practise question answers 67 25 89 37 12 42 7 96 30 55 40 10 90 A B D C E F G I H 22 Total of all weights= 622 Repeated edges = 126 748 and a possible route = ABICBDCHIDEFIGFGHA

11 Practise Questions continued… 2. Use route inspection to solve this problem: 11 B C D A EF 12 14 18 13 12 18 15 19 17 10

12 Practise Question answer 2. Odd vertices are ACEF AC=24CE=13EF=10 AE=11CF=23 AF=19 AC+EF=24+10=34 AE+CF=11+23=34 AF+CE=19+13=32 Total of all edges = 159 Repeated edges = 32 11 B C D A EF 12 14 18 13 12 18 15 19 17 10 191

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