Mixed Doubles

Three married couples were the only members to turn up at the spoofball club one rainy night. Despairing at length of any others arriving, they decided to hold an impromptu tournament, in which each married couple in turn played two mixed pairs each representing both the other families. As the players were of unequal strength, they resorted to handicapping, marking one man and one woman scratch, and giving one married couple the highest handicap each. As soon as either side had scored 6 the game was over, and the side with the larger joint handicap then added to its score the difference between its own and the other side's joint handicap.

The final result was most curious. The unfortunate Bakers were always on the side that scored less in play. All the scores-by-play were different, and the handicapping was so good that the final weighted scored (i.e. after the joint-handicap difference had been applied to the scores-by-play) were with one exception six-all, the exception being that when the Carvers played against Baker and Mrs. Archer the handicap-difference of 2 left them still leading. The Archers' joint handicap was smaller than the Carvers'.

Question: What was each individual handicap, and what points were scored by play in each game?

Solution

A solution can readily be found by inspection and/or brute force. What is less obvious is if it is the only solution. Here I offer a slightly unusual approach which not only produces the answer, but in passing demonstrates that it is unique.

The approach uses linear algebra (simultaneous equations to you and me) and the answer kind of drops out 'by magic'. It only needs O-level maths (if that).

Problem restatement

We don't know anyone's handicap, so we use variables to represent them.

We let A represent Mr. Archer's handicap, and a represent Mrs. Archer's handicap; likewise for Baker's (B) and Mrs. Baker's (b) and Carver's (C) and Mrs. Carver's (c).

We also don't know the scores for the various matches; so we assign variables to these too:

Team 1		Score 1	Score 2	Team 2
Archer	Mrs. Archer	6	s	Baker	Mrs. Carver
Archer	Mrs. Archer	6	t	Carver	Mrs. Baker
Baker	Mrs. Baker	u	6	Archer	Mrs. Carver
Baker	Mrs. Baker	v	6	Carver	Mrs. Archer
Carver	Mrs. Carver	6	w	Archer	Mrs. Baker
Carver	Mrs. Carver	6	x	Baker	Mrs. Archer

The idea is to write down equations which relate the variables we have and to work out constraints on them.

Constraints

The constraints we have are as follows:

each of A, a, B, b, C, c are greater than or equal to zero.
One of {A, B, C} is zero;
similarly one of {a, b, c} is zero.
One of the pairs {(A, a), (B, b), (C, c)} has both coordinates greater than the other two pairs.
The set {s, t, u, v, w, x} is a permutation of {0, 1, 2, 3, 4, 5}.
In addition, we have that x must be less than four, since adding two to it (the handicap difference in the Cc-Ba game) leaves the resulting score less than six.

We will be coming back to these constraints to rule out candidate solutions to the set of linear equations that we will derive next.

Equations

These form the meat of the solution. For the first five games listed above, we know that the handicap adjustments left the scores at 6-all. For the last game, we know that the handicap difference was exactly 2.

We can therefore write down the following six equations:

Eq. No.	Equation
1	6 = s + B + c - (A + a)
2	6 = t + b + C - (A + a)
3	6 = u + B + b - (A + c)
4	6 = v + B + b - (C + a)
5	6 = w + A + b - (C + c)
6	2 = B + a - (C + c)

The first step is to re-write these equations to put s, ..., w on the RHS alone.

Eq. No.	Equation
7	A + a - B - c + 6 = s
8	A + a - b - C + 6 = t
9	A - B - b + c + 6 = u
10	a - B - b + C + 6 = v
11	- A - b + C + c + 6 = w
12	- a - B + C + c + 2 = 0

Now, we combine these equations to remove occurrences of A from all but number 7.

Eq. No.	From eq.s	Equation
13	7	A + a - B - c + 6 = s
14	7 - 8	- B + b + C - c = s - t
15	7 - 9	a + b -2c = s - u
16	10	a - B - b + C + 6 = v
17	7 + 11	a - B - b + C +12 = s + w
18	12	- a - B + C + c + 2 = 0

This process is called diagonalisation; we'll repeat it a number of times. This time, we look to remove a from all equations apart from 15.

Eq. No.	From eq.s	Equation
19	13	A + a - B - c + 6 = s
20	14	- B + b + C - c = s - t
21	15	a + b -2c = s - u
22	15 - 16	B +2b - C -2c - 6 = s - u - v
23	17 - 16	6 = s + w - v
24	15 + 18	- B + b + C - c + 2 = s - u

Finally we subtract equation 20 from equation 24, and add equations 20 and 22 to get:

Eq. No.	From eq.s	Equation
25	24 - 20	2 = t - u
26	20 + 22	3b -3c - 6 = 2s - t - u - v

Combining the reduced equations with the constraints to produce potential solutions by inspection

Now, equations 23 and 25 give us some simple relationships between the final scores. We can tabulate possible values for s, w and v and similarly for t and u:

t - u = 2
t	u
2	0
3	1
4	2
5	3

s + w - v = 6
s	w	v
5	1	0
4	2	0
2	4	0
1	5	0
5	2	1
4	3	1
3	4	1
2	5	1
5	3	2
3	5	2
5	4	3
4	5	3

Now, we look to combine these two tables. For example, we can reject the row (s=5, w=2, v=1) because none of the possible pairs of values for t and u fit. This combination produces fewer possibilities, which are tabulated below. We include a sixth column for completeness, in which the value for x is tabulated.

s	w	v	t	u	x
5	1	0	4	2	3
4	2	0	3	1	5
4	2	0	5	3	1
2	4	0	3	1	5
2	4	0	5	3	1
1	5	0	4	2	3
4	3	1	2	0	5
3	4	1	2	0	5
5	4	3	2	0	1
4	5	3	2	0	1

We can reject the rows above which do not have x less than 4. The resulting possibilities are tabulated below, along with a new column, the RHS (Right Hand Side) of equation 26. The reason for this will be explained in a moment.

s	w	v	t	u	x	2s - t - u - v
5	1	0	4	2	3	4
4	2	0	5	3	1	0
2	4	0	5	3	1	-4
1	5	0	4	2	3	-4
5	4	3	2	0	1	5
4	5	3	2	0	1	3

Now, the LHS of equation 26 is 3b - 3c - 6. We know that b and c are whole numbers, so this must work out to a multiple of three! That is why we tabulated values for the RHS of this equation above; it allows us to see that we can reject all but two of the possible scoring arrangements for the six matches. Given only two alternatives, we can proceed to work backwards through the equations we've derived, working out values for the various handicaps. We do this for each alternative.

Alternative 1	Alternative 2
s=4, t=5, u=3, v=0, w=2, x=1	s=4, t=2, u=0, v=3, w=5, x=1
Equation 26 gives us:
3b - 3c - 6 = 0 b = c + 2	3b - 3c - 6 = 3 b = c + 3
Substituting into equation 21 gives us:
a + b - 2c = 1 a + c + 2 - 2c = 1 a = c - 1	a + b - 2c = 4 a + c + 3 - 2c = 4 a = c + 1
Since one of {a, b, c} is zero and the others are positive, this gives us:
a=0, b=3, c=1	a=1, b=3, c=0
We put these values into equation 24:
-B + C + 4 = 1 B = C + 3	-B + C + 5 = 4 B = C + 1
Equation 19:
A - B + 5 = 4 A - C - 3 + 5 = 4 A = C + 2	A - B + 7 = 4 A - C - 1 + 7 = 4 A = C - 2
Since one of {A, B, C} is zero and the others are positive, this gives us:
A=2, B=3, C=0	A=0, B=3, C=2

We can now use the constraint that the Archers had a smaller joint handicap than the Carvers to reject alternative 1. This gives us the final solution.

Summary of solution

Family	Husband's handicap	Wife's handicap
Archer	0	1
Baker	3	3
Carver	2	0

Team 1		Score 1	Score 2	Team 2
Archer	Mrs. Archer	6	4	Baker	Mrs. Carver
Archer	Mrs. Archer	6	2	Carver	Mrs. Baker
Baker	Mrs. Baker	0	6	Archer	Mrs. Carver
Baker	Mrs. Baker	3	6	Carver	Mrs. Archer
Carver	Mrs. Carver	6	5	Archer	Mrs. Baker
Carver	Mrs. Carver	6	1	Baker	Mrs. Archer

s	w	v	t	u	x
5	1	0	4	2	3
4	2	0	3	1	5
4	2	0	5	3	1
2	4	0	3	1	5
2	4	0	5	3	1
1	5	0	4	2	3
4	3	1	2	0	5
3	4	1	2	0	5
5	4	3	2	0	1
4	5	3	2	0	1

s	w	v	t	u	x
5	1	0	4	2	3
4	2	0	3	1	5
4	2	0	5	3	1
2	4	0	3	1	5
2	4	0	5	3	1
1	5	0	4	2	3
4	3	1	2	0	5
3	4	1	2	0	5
5	4	3	2	0	1
4	5	3	2	0	1

s	w	v	t	u	x
5	1	0	4	2	3
4	2	0	3	1	5
4	2	0	5	3	1
2	4	0	3	1	5
2	4	0	5	3	1
1	5	0	4	2	3
4	3	1	2	0	5
3	4	1	2	0	5
5	4	3	2	0	1
4	5	3	2	0	1