# Calibans Will Solution

## preamble cop out

I believe this solution may be non-optimal, but I think it is correct. It is based on one from brainyplanet. However, I am not happy with the brainyplanet solution. The weakest part is its (huge?) jump when it states the possible combinations for U and T. I have shown how these are derived, but I get a larger number of possibilities than them. Also, I have defined 'T' in a different way to them, as I believe their definition is wrong. (Their T is my G, and yet they claim that T cannot contain l, hmmm).

You can read another solution on Richard Harter's site. His solution is shorter, more wordy, and less formal (not that mine is formal!). If you like the wordy approach, I recommend Richard's.

## The Puzzle

No person who has seen me in a green tie is to choose before Low.

If Y.Y. was not in Oxford in 1920 the first chooser never lent me an umbrella.

If Y.Y. or 'Critic' has second choice, 'Critic' comes before the one who first fell in love."

## Definitions

• l = Low
• c = Critic
• y = Y.Y.
• G = the set of people who saw the tie
• T = G intersection { c, y } i.e. those who saw the tie excluding Low - i.e. T = G - { l }
• U = the set of people who lent Caliban an umbrella
• {} = the empty set.

### Formalised version of the three statements (1), (2) and (3)

```(1) for all x in T, x > l
(2) for all x in U, x > 1
(3) L = 2 or C < 3```

For clarification of (3), the third sentance can be transformed to :

`if l != 2, then c < 3`

which can then be chaned to (as stated above)

`l = 2 or c < 3`

## Someone saw the tie.

If neither Y.Y nor Critic saw the green tie, then (1) is superflous, therefore at least one must have.

`(a) T != {}`

## Someone lent Caliban an umbrella.

If nobody lent Caliban an umbrella (or if YY was in oxford), then sentance 2 is redundant (and so the reverse must be true). i.e. yy was not in oxford, and somebody lent Caliban an umbrella.

`(b) U != {}`

## Someone lent an umbrella, that did not see the tie

```if, for all x in T, x exists in U
then (1) implies (2).

If (1) implies (2), then (2) is superfluous, and we aren't allowed superfluous statements, therefore :

(c) there exists an x in T, such that x is not in U.```

## Low did not lend Caliban an umbrella

```if l is in U, then from (2), l > 1

and this implies (3) (i.e. 3 is superfluous), which isn't allowed. Nice!

(d) l is not in U```

### Possible values for a set

With our three people, there are the following 8 combinations for a set of people :

```{}
{ l }
{ y }
{ y, l }
{ c }
{ c, l }
{ c, y }
{ c, y, l }```

### Possible values for set U

Here's a table of all 8 possible values for set U, and on thlets use the above to limit the number of possible values. Remember this is a set, it is NOT ordered.

```{} (b)
{ l } (d)
{ y }
{ y, l } (d)
{ c }
{ c, l } (d)
{ c, y }
{ c, y, l } (d)```

So we have three possibilities U = { y }, U = { c } or U = { c, y }

### Possible values for set T

T is defined as not containing l, so all possible combos are :

```{}
{ y }
{ c }
{ y, c }```

This is not so easy to break down, as T is dependant on U, so lets take the three possibilies of U, and find all of the final combinations.

## If U = { y }

```{} (a)
{ y } (c)
{ c }
{ y, c } (c)```
`A valid comination is : U = { y } and T = { c }`

## If U = { c }

```{} (a)
{ y }
{ c } (c)
{ y, c } (c)```
`A valid comination is : U = { c } and T = { y }`

## If U = { y, c }

```{} (a)
{ y }
{ c }
{ y, c } (c)```
```A valid comination is : U = { y, c } and T = { y }
A valid comination is : U = { y, c } and T = { c }```

### So all valid combinations of U and T

```U = { y } and T = { c }
U = { c } and T = { y }
U = { y, c } and T = { y }
U = { y, c } and T = { c }```

### Initial Limiting of the Possible Orders

The complete set of possible orders are :

```c l y
c y l
l c y
l y c
y c l
y l c```

Given (b) and (1), then l < 3

so we can remove c y l and y c l,

Given (3) we can remove l y c, leaving :

```c l y
l c y
y l c```

## The final hurdle

Now lets taking each valid combination of U and T in turn, and see which order the heirs can choose their books. As if by magic, the order is always the same :-)

For each one, list all of the possible orderings, and next to each, say which rule prevent it from being valid. If we end up ruling out all orderings bar one, then we have succeded.

### U = { y } and T = { c }

```c l y (1)
l c y
y l c (2)```

### U = { c } and T = { y }

```c l y (2)
l c y
y l c (1)```

### U = { y, c } and T = { y }

```c l y (2)
l c y
y l c (2)```

### U = { y, c } and T = { c }

```c l y (2)
l c y
y l c (2)```

So, regardless of which combination of U and T we choose, the ordering is always ; l, c, y

## Summary

The order is Low, Critic, Y.Y.

Y.Y. was the first to fall in love.

Y.Y. was not in oxford in 1920.

Its impossible to know who lent the umbrella, and who saw the green tie (I think!).