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July 5th, 2018, 03:44 PM   #1
jks
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Math Focus: Electrical Engineering Applications
Fun with World Cup Group Play

Attached are the point totals of this year's World Cup group (round robin) play.

Just for fun, and looking only at the number of points for each team, in which groups may:

1) The outcomes of all 6 matches be determined?
2) The outcomes of 2 matches be determined?
3) The outcome of 1 match be determined?
4) The outcome of no matches be determined?

To clarify: Suppose that 3 different results of the 6 matches produces the given point total for a group. If the outcome of only 1 of the matches is the same for all 3, then we can say that only 1 outcome may be determined.

Some may find this very easy, but I have to admit that I bungled one of them. To check, I wrote a script.

Edit: I forgot to mention (for those that do not know), the winner of a match gets 3 points, the loser 0 points, and 1 point to each team for a tie.
Attached Images
File Type: jpg wc1.JPG (52.3 KB, 3 views)
File Type: jpg wc2.JPG (55.5 KB, 2 views)

Last edited by jks; July 5th, 2018 at 04:26 PM.
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July 15th, 2018, 04:43 PM   #2
jks
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Well the World Cup is over and congratulations to France.

I thought I would go ahead and post my results to the question that I asked above (these results my need further checking, but I did check them against the actual results for at least one level of a sanity check).

Note that I use T for a tie (draw) instead of D because of notation in my program (not shown).

First, for groups A and G where the points are 9,6,3,0. Let's just take group A:

Code:
U -   9     3W
R -   6     2W
SA - 3     1W (or 3T, which is shown to be wrong - remember we are looking only at points, not the W, D (T), or L
E -   0
SA must have 1W since there cannot be an odd number of ties. Obviously, there is only one possible outcome as U won all 3 matches, R lost to U and beat SA and E. SA lost to U and R and beat E. And E lost all of its matches.

So we can determine the outcomes for groups A and G.

----------
Next, let's take groups C and E where the points are 7,5,3,1. Taking group C:

Code:
F - 7    2W 1T
D - 5   1W 2T
P - 3    1W (or 3T, which is shown to be wrong)
A - 1    1T
P must have 1W since 3T would give an odd number of ties.

The matches are:
FD - this must be a tie since D never loses. So F must beat P and A.
FP - F wins
FA - F wins
DP - D must win since D never loses and P never ties.
DA - tie since that is all that is left for D
PA - P wins since this is all that is left for both.

So we can determine all of the outcomes for groups C and E.

----------
For group D we have:

Code:
C - 9   3W
A - 4   1W 1T
N - 3   1W (or 3T, which is shown to be wrong)
I - 1    1T
N must have 1W or there would be an odd number of ties. The matches are:
CA - C wins since C wins all 3 matches.
CN - C wins
CI - C wins
AN - A wins, due to the outcomes of CA and AI
AI - tie, since these are the only 2 teams with a tie.
NI - N wins, which is all that is left for N.

So we can determine all of the outcomes for group D.

----------
For group B we have:

Code:
S - 5    1W 2T
P - 5    1W 2T
I - 4    1W 1T
M - 1   1T
The matches are:
SP - must be a tie since neither loses.
SI
SM
PI
PM
IM - I wins as I cannot beat S or P since they do not lose.

Only 2 match outcomes may be determined as both of the following results end up in the points being correct:

Code:
SP_T  SI_T  SM_S  PI_P  PM_T  IM_I
SP_T  SI_S  SM_T  PI_T  PM_P  IM_I
(a T after the underscore denotes a tie, otherwise the winning side is after the underscore).

----------
For group H we have:

Code:
C - 6   2W
J - 4   1W 1T
S - 4  1W 1T
P - 3  1W (or 3T, which is shown to be wrong)
P must have 1W since 3T would give an odd number of ties.

Here we can only say that 1 match may be determined, which is that the JS match must have been a tie since these are the only teams with a tie.

There are 3 outcomes that produce the points:
Code:
CJ_C  CS_C  CP_P  JS_T  JP_J  SP_S
CJ_C  CS_S  CP_C  JS_T  JP_J  SP_P
CJ_J  CS_C  CP_C  JS_T  JP_P  SP_S
----------
For group F we have:
Code:
S - 6    2W
M - 6   2W
K - 3    1W (or 3T, which is shown to be wrong)
G - 3    1W (or 3T, which is shown to be wrong)
With 18 total points for the 6 matches, all must have had winners so there can be no ties.

There are 4 outcomes that produce the points:
Code:
SM_S  SK_S  SG_G  MK_M  MG_M  KG_K
SM_S  SK_K  SG_S  MK_M  MG_M  KG_G
SM_M  SK_S  SG_S  MK_M  MG_G  KG_K
SM_M  SK_S  SG_S  MK_K  MG_M  KG_G
Therefore, no matches can be determined.
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