D1 January 2011 Q3
3.

A new spanning tree is required which includes the arcs DI and HG, and which has the lowest possible total weight.
| Scheme | Marks |
|---|---|
| CI CD (not DI) EF FI (not EI not DE) \(\left\{\begin{matrix}\text{BC}\\ \text{HI}\end{matrix}\right\}\) (not BI) GF | M1 A1 |
| (not GI not HG) AB | A1 |
| (3) |
Notes
1M1: Kruskal’s algorithm – first 4 arcs selected chosen correctly.
1A1: All eight non-rejected arcs chosen correctly.(Working seen in (a))
2A1: All rejections correct and in correct order and at correct time.
| Scheme | Marks |
|---|---|
| AB BC CI CD FI EF IH FG | M1 A1 A1 |
| (3) |
Notes
1M1: Prim’s algorithm – first four arcs chosen correctly, in order, or first five nodes chosen correctly, in order.{A, B,C,I, D} (arcs not arc lengths)
1A1: First six arcs chosen correctly; all 9 nodes chosen correctly, in order.{A,B,C,I,D,F,E,H,G}[1 2 3 5 7 6 9 8 4]
2A1: cso
| Scheme | Marks |
|---|---|
![]() | B1 |
| Weight: 270 | B1 |
| (2) |
Notes
1B1: cao (condone lack of numbers)
2B1: 270 cao
| Scheme | Marks |
|---|---|
| Start off the tree with DI and HG and then apply Kruskal’s algorithm | B2,1, 0 |
| (2) | |
| (10 marks) |
Notes
1B1: Kruskal’s algorithm + some argument
2B1: Kruskal’s algorithm + start with the two arcs. (o.e)
