Posted 19 July 2007 - 06:24 PM

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 [b]8H

exactly as we analyzed Outcome 1b.

Simply substitute H for L and v.v.

Problem solved.