The New Sudoku Players' Forum

[Smythe Dakota]

For any finite number of coins 3 or larger, it is always possible to determine which is the false coin, and whether it is lighter or heavier.

That's assuming the puzzle follows the rule -- all the true coins weigh the same, there is one and only one false coin, and the false coin has a different weight from the true coins.

An upper bound for X is N-1. Just pick one coin and weigh each of the remaining N-1 against it.

Bill Smythe

[7b53]

there is no need to bother with "lighter or heavier".
as this was never the question.

since the false coin could be either lighter or heavier to begin with.
how is it possible to say at the end that it must be heavier or it must be lighter.

a valid solution should automatically shows at the end,
whether it's lighter or heavier base on the outcome from the scale.

[Smythe Dakota]

.... a valid solution should automatically shows at the end, whether it's lighter or heavier base on the outcome from the scale.

Not true.

Assuming the premises are met -- i.e. there is only one false coin, the true coins are all the same weight, and the false coin is not the same weight -- there are versions of the problem where the false coin can be identified in W weighings, but a (W+1)st weighing would be needed to determine whether the false coin is lighter or heavier.

The simplest example is 4 coins, where you are limited to 2 weighings.

The solution is to weigh first A against B, then A against C.

• If A and B are not the same, the false coin must be either A or B. Then the second weighing tells you which it is:
  • If A and C are not the same, the false coin must be A. In this case, you also know whether A is lighter or heavier.
  • If A and C are the same, the false coin must be B. In this case, you also know whether B is lighter or heavier.
• If A and B are the same, the false coin must be either C or D. Then the second weighing tells you which it is:
  • If A and C are not the same, the false coin must be C. In this case, you also know whether C is lighter or heavier.
  • If A and C are the same, the false coin must be D. In this case, you still don't know whether D is lighter or heavier.

Thus, in one of the four possible outcomes, it is not known after two weighings whether the false coin is lighter or heavier, even though it is known which coin is false.

Bill Smythe

[7b53]

7b53 wrote: .... a valid solution should automatically shows at the end, whether it's lighter or heavier base on the outcome from the scale.


Not true.

can this be true for the 12 coins puzzle ??

if a coin never touches the scale, does it consider to be an "outcome from the scale"?

from your 4 coins example, if i would change the limit to 3 weighings,
do i have to include "please say at the end whether it is lighter or heavier"?

7b53

[Smythe Dakota]

.... can this be true for the 12 coins puzzle ?? ....

I'm sure it is. This problem barely has a solution at all. If you look at all the internet stuff, you'll see that a mighty struggle is required just to identify the false coin in three weighings. To also find whether it is ligher or heavier, in all scenarios, would surely require a fourth weighing.

.... if a coin never touches the scale, does it consider to be an "outcome from the scale"? ....

Oh, I don't know. The original problem never used the phrase "outcome from the scale". As long as you identify the false coin, you have satisfied the conditions stated in the puzzle, even if you still don't know whether it is lighter or heavier.

.... from your 4 coins example, if i would change the limit to 3 weighings, do i have to include "please say at the end whether it is lighter or heavier"? ....

In my opinion. yes. If you don't include that phrase, you don't have a good puzzle, because a solution exists with just 2 weighings.

When you invent a puzzle, you should be careful to say exactly what you mean.

When you try to solve a puzzle, you should be careful to figure out exactly what is being asked for.

That's what makes a good puzzle. The statement in the problem can be technically correct, but easily misinterpreted by a solver who doesn't read the problem carefully. That's what the 12 coins puzzle is all about.

Bill Smythe

[Pat]

The 12 coins puzzle barely has a solution at all;
a mighty struggle is required
just to identify the false coin in three weighings.

To also find whether it is ligher or heavier, in all scenarios,
would surely require a fourth weighing.

no,
3 weighings solve it.

see "The twelve-coin problem"
in http://en.wikipedia.org/wiki/Balance_puzzle

[7b53]

nice to know there actually can be more than one solutions for this 12 coins puzzle.
thanks to champagne and other web's info.

although the first scale should only be 4 coins at both side.
it's the second weighing where you can place 3 or 4 coins at each side of the scale.
both(3 or 4 coins) can produce a valid solution.

as for 81 coins problem..............still working on it.

[7b53]

believe it or not, the 81 coins is much simpler than the 12 coins one.
no need to exchanged or getting rid of coins. pretty much straight forward.

answer: 5 weighing. and yes, it will shows whether is heavier or lighter...how weird !