[Double Liar Paradox (Jourdain's paradox)]

What underlies paradoxes of this type is the syntactical rule

that a declarative sentence is by its nature an assertion of

some particular truth. To use a presumed assertion of

truth to deny that same truth is paradoxical: One cannot

convey usable knowledge by asserting a denial. Nor can one

meaningfully deny a truth: the coin has two paradoxical

sides:

[1] "I am asserting a falsehood." or "I am lying."

[2] "I am not asserting something that is true." or "I am not telling the truth."

Putting it another way, it's physically possible to speak the

words, "I am lying." But when one undertakes a linear

analysis of what has happened when the words are spoken,

one is drawn into the syntactical analogy of a Moebius Strip:

a piece of paper having a physical connection of its two sides.

The circular reasoning forced on the mind by a linear

analysis of such statements creates a pleasantly frustrating

tease, and the desire for consistency and meaning leaves

one in a disturbingly uncomfortable state.

Long live paradoxes...

[Pouring water VI.]

Alternatively you can use the following solution:

1. Fill C with B leaving 4 litres in B

B starts with 3 liters.

Filling C [takes 1 liter] leaves 2 [not 4] liters in B.

[Weighing IV.]

Each weighing has three outcomes:

Left side is [lighter than] [equal to] [heavier than] Right side.

Three weighings can thus discern among [3]x[3]x[3]= 27 cases.

We have only 24 cases:

one of 12 balls is heavier or lighter than the rest.

So we can solve the problem, so long as ...

[1] The first weighing reduces the cases to no more than 9.

[2] The second weighing reduces the cases to no more than 3.

[3] The third weighing then distinguishes among 3 or fewer cases.

First weighing:

Set aside four balls. Why?

Because, if the first weighing balances, we have 8 [fewer than 9] cases:

One of the 4 excluded balls is heavier or lighter.

[1] Weigh 1 2 3 4 5 6 7 8

Outcome 1[a] 1 2 3 4 balances 5 6 7 8.

Since only one ball is odd, 1 2 3 4 5 6 7 8 must all be normal.

We have 8 cases: 9 10 11 or 12 is H or L

[shorthand: H=heavier; L=lighter]

[2] Weigh 1 2 3 9 10 11 [we know 1 2 3 are normal]

If this balances, 9 10 11 are also normal, and 12 is H or L.

[3] Weigh 12 [any other ball]

If 12 rises, then 12L; if 12 falls, then 12H.

If [2] 9 10 11 rises, we have 3 cases: 9L, 10L or 11L

[3] Weigh 9 10.

If [3] balances, then 11L;

If 10 rises, then 10L; If 9 rises, then 9L

If [2] 9 10 11 falls, we have 3 cases: 9H, 10H or 11H

[3] Weigh 9 10.

If [3] balances, then 11H;

If 10 falls, then 10H; If 9 falls, then 9H

end of Outcome 1[a]: balance.

Outcome 1: 1 2 3 4 falls, and 5 6 7 8 rises.

This gives 8 cases: 1H, 2H, 3H, 4H, 5L, 6L, 7L, or 8L

The second weighing in this case becomes tricky.

Remember each of its three outcomes can lead to no

more than 3 cases for the third weighing to resolve.

Again, we exclude a number of balls and involve the others.

We exclude any three of the balls. Why? Because if the

other [included] balls balance, we have exactly 3 cases.

Without loss of generality we exclude balls 1 2 3.

Since that leaves an odd number of balls, 4 5 6 7 8,

we need to use one of the normal balls.

Finally we choose which three to weigh against the others.

And here's the only hard part of this problem.

We must mix some of the possibly light balls with some

of the possibly heavy balls. Otherwise, one of the

outcomes of the second weighing will leave us with more

than 3 cases, and the third weighing will not resolve this.

[2] weigh 4[H] 5[L] 6[L] 1[normal] 7[L] 8[L]

in parentheses I've indicated the POSSIBLE cases

that we have determined:

4[H] means 4 is heavier if it's not normal.

Outcome 2[a]: 4 5 6 balances 1 7 8

These balls are all normal.

We have 3 cases: 1H 2H or 3H.

[3] weigh 1 2

If [3] balances, then 1 and 2 are normal, and 3H

if 2 falls, then 2H

if 1 falls, then 1H

Outcome 2b: 4 5 6 falls, and 1 7 8 rises.

We have 3 cases: 4H 7L or 8L

[3] weigh 7 8

if [3] balances, then 7 and 8 are normal, and 4H

if 7 rises, then 7L

if 8 rises, then 8L

Outcome 2[c]: 4 5 6 rises, and 1 7 8 falls.

We have 2 cases: 5L or 6L.

[3] weigh 1[normal] 5[L]

If 5 rises, then 5L

If balance, then 5 is normal, and 6L

Now we can go back and take the remaining case

Outcome 1[c]: 1 2 3 4 rises, and 5 6 7 8 falls.

to distinguish among the remaining 8 cases:

1L 2L 3L 4L 5H 6H 7H and 8H

exactly as we analyzed Outcome 1b.

Simply substitute H for L and v.v.

Problem solved.

[Barbera]

The puzzle asks: Who am I to Barbara - NOT who is Barbara to me. Hence the correct answer is ... Daughter. Unless of course I am a male. Then the correct answer is not in the choices listed, it's Son-in-Law.