Part A: Determining fair rank between teams(no ties)
Let win=3, draw=1 and lose=0. The reason for this weighting method is because it is natural for the loser not to get a point. Also, there must be a visible gap between winners and draw players. This can be proved by contradiction proof. Blue | B | Crimson | C | Green | G | Orange | O | Red | R | Yellow | Y |
Assume that win=2 draw=1 and lose=0. If the supremacy matrix is calculated, tie always occurs, which proves that if there is a small gap between win and draws, that supremacy matrix isn’t valid.
This is the teams and their initial letters, arranged by alphabetical order.
Then, the diagraph below is converted into the following matrix. M= …show more content…
MPS Task 2
B=Blues | HL=Highlanders | BR=Brumbies | L=Lions | BU=Bulls | R=Reds | C=Chiefs | SH=Sharks | CH=Cheetahs | ST=Stormers | CR=Crusaders | WA=Wawratahs | H=Hurricanes | WF=Western Force |
Let the teams be called with following table.
Below is the result of 6 rounds. Round 1 | | Round 4 | | 12/02/2010 | WF vs BR | 5/3/10 | C vs R | | CH vs BU | | BR vs L | | B vs H | 6/3/10 | ST vs HL | 13/02/2010 | L vs ST | | CH vs H | | R vs WA | | CR vs B | | SH vs C | | WA vs SH | | CR vs HL | Round 2 | | Round 5 | | 19/02/2010 | SH vs CH | 12/3/10 | C vs CR | | HL vs B | | WA vs L | | R vs CR | 13/3/10 | BU vs HL | | L vs C | | ST vs H | 20/02/2010 | BU vs BR | | BR vs SH | | ST vs WA | `14/3/10 | R vs WF | Round 3 | | Round 6 | | 26/2/10 | ST vs BR | 19/3/10 | BU vs H | | CR vs SH | | B vs BR | 27/2/10 | CH vs HL | 20/3/10 | ST vs CH | | R vs B | | CR vs L | | WF vs C | | WF vs WA | | BU vs WA | | HL vs SH | | H vs L | | CH vs R |
The result was converted into the following table.
| B | BR | BU | C | CH | CR | H | HL | L | R | SH | ST |