1. Unit 3 Chinese Postman

4. Non - Eulerian Semi - Eulerian Eulerian

5. In a Chinese Postman problem the optimum route is only possible if the graph is EULERIAN, starting and finishing at the same point.

6. a) As there are 4 odd vertices at A, B, C and I the graph is neither eulerian or semi – eulerian.

8. The COMBINATIONS are AB + CI = 100 + 440 = 540 AC + BI = 150 + 450 = 600 AI + BC = 380 + 120 = 500 REPEAT the SHORTEST DISTANCE AI + BC = 380 + 120 = 500 TOTAL DISTANCE 2090 + 500 = 2590 The ODD VERTICES are noted down A, B, C, I

9. The Examiner is looking for the following points The ODD VERTICES are noted down A, B, C, I The COMBINATIONS with DISTANCES AB + CI = 100 + 440 = 540 AC + BI = 150 + 450 = 600 AI + BC = 380 + 120 = 500 REPEAT the SHORTEST DISTANCE AI + BC = 380 + 120 = 500 TOTAL DISTANCE 2090 + 500 = 2590

11. TOTAL NUMBER OF STATUES SEEN IS = B C D E F G H I J 2 + 2 + 3 + 2 + 2 + 3 + 1 + 2 + 1 = 18 Statues As the number of statues seen is found by dividing the edges (INCLUDING THE REPEATED EDGES) at each vertex by 2

14. ODD vertices at A, C, D, F AC + DF AF + CD AD + CF AC (14) + 32 AF (10) +36 AD (26) + 50 Repeat AC + DF = 32 DF (18) = CD (26) = CF (24) = Total route is 150 + (AC + DF) 32 = 182 km

15. As he is starting at an ODD vertex and completing his journey at an ODD vertex (A and C) he only needs to repeat FD Total = 150 + 18 = 168 km

16. ci) She needs to start at an ODD vertex and finish at an ODD vertex and REPEAT the SMALLEST value of the pairs of ODD VERTICES Smallest distance is AF (10) Distance therefore is 150 + 10 = = 160 km cii) If she repeats AF she must start / finish at C and D

17. What the examiner is looking for You have stated the ODD VERTICES You have stated the different combinations and their distances You have added the new edges to your diagram You have stated which combination is repeated You have stated the total distance, showing the sum