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# Hands up!

## Question

Three persons( A,B,and C) are not allowed to speak,but to raise a hand  refering to ( Yes or No ), you don`t know which hand means Yes,and which means No.

How can you know  the persons which is which by asking only three( yes or no) questions.

You may ask one ,two or all of them...as you like.

any  question should be asked to a single person.

Edited by wolfgang

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Wolfgang, have you seen my attempt at this? Is that correct?

I read it.... but it is not correct... because in this case ,for the 3rd question, the 1st person may be A,,, or may not...hmmm...then what?

If the 1st person is not A, it is known within first two questions. I have annotated my solution with how the possibilities change with each question, see below (ABC means the possibility that, "A is sitting in chair 1, B in chair 2, C in chair 3").

Q1. [ABC, ACB, BAC, BCA, CAB, CBA] (initially, all possibilities are open) Ask the person in chair-1: "Is the truth value of 'you are person A' the same as truth value of 'you are right-handed'"?

Q1R.Q2 (if the person raised the right hand for Q1) [ABC, ACB, BAC, BCA, CAB, CBA] Ask the person in chair-2: "Is the truth value of 'you are person B' the same as truth value of 'you are right-handed'"?

Q1R.Q2R. [ABC, ACB, BAC, BCA, CAB, CBA] Arrangement is: A-1, B-2, C-3

Q1R.Q2L. [ABC, ACB, BAC, BCA, CAB, CBA] Arrangement is: A-1, C-2, B-3

Q1L.Q2 [ABC, ACB, BAC, BCA, CAB, CBA] (if the person raised the left hand for Q1) Ask the person in chair-2: "Is the truth value of 'you are person A' the same as truth value of 'you are right-handed'"?

Q1L.Q2R.Q3 [ABC, ACB, BAC, BCA, CAB, CBA] Ask the person in chair-3: "Is the truth value of 'you are person C' the same as truth value of 'you are right-handed'"?

Q1L.Q2R.Q3R [ABC, ACB, BAC, BCA, CAB, CBA] Arrangement is: B-1, A-2, C-3

Q1L.Q2R.Q3L [ABC, ACB, BAC, BCA, CAB, CBA] Arrangement is: C-1, A-2, B-3

Q1L.Q2L.Q3 [ABC, ACB, BAC, BCA, CAB, CBA] Ask the person in chair-2: "Is the truth value of 'you are person C' the same as truth value of 'you are right-handed'"?

Q1L.Q2L.Q3R [ABC, ACB, BAC, BCA, CAB, CBA] Arrangement is: B-1, C-2, A-3

Q1L.Q2L.Q3L [ABC, ACB, BAC, BCA, CAB, CBA] Arrangement is: C-1, B-2, A-3

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What are we trying to learn?

Their identity (which is which in some sense)?

Or the hand, for each person, that means Yes?

Ask each a question for which there can be only one answer.

Is the sky blue? The hand each one raises is their Yes hand.

Or, could you ever truthfully claim to be a liar? The raised hand is the No hand.

that they all tell the truth.
If one or more might be a liar or a random answerer, the questions must be of a different type.
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I want their identity.   and the letters are instead of names.

so I want to know who is A, who is B, and who is C

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OK so I map 123 into ABC where 123 signifies the order that I question them, and I can rely on their being truthful when they answer?

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I guess I need another clarification:

Is the left/right hand meaning of yes/no in some order the SAME for all the robots?

It seems that it must be, but I have to ask.

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I guess I need another clarification:

Is the left/right hand meaning of yes/no in some order the SAME for all the robots?

It seems that it must be, but I have to ask.

No,,,,it is not  the same

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This is a challenging puzzle. If I understand correctly,

1. I ask someone a question then write "1" on his forehead.
I might ask someone else a question then write "2" on his forehead.
I then write "3" on the last person's forehead.

2. If A raises his left hand it might mean No.
If B raises his left hand it might mean Yes.
Similarly for person C.

3. After I ask my third question, I will state, for example,
Number 1 is B; Number 2 is C; and Number 3 is A.

Is that the idea?

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I don't know how to solve this, but I may be able to show that it is not solvable with three questions..

Assume A, B and C are sitting in a room on chairs labeled 1, 2 and 3 in some order. Each of these can either be "left-handed" (raising left hand for 'Yes'), or "right-handed" (raising right hand for 'Yes').

The requirement is to find out who is sitting in which chair (there are 3! = 6 possible permutations here), and the handedness of each person (there are 2x2x2 = 8 possibilities here).

So we have a total of 3!.2.2.2 = 48 possible "states" for the room. We need at-least 6 binary bits for encoding a number that can take values upto 48.

By asking three yes/no questions (whatever they maybe) to a person in a specific chair, we get only 3 bits of information out of the room (did they raise left hand or right hand?). From these, we can NOT fully determine which "state" the room is in.

If only A and B are there (giving 2!.2.2 = 8 states, requiring 3 bits), then it is possible to solve with three questions.

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And of course, if there is no "C",

a. Ask person in chair 1: Is 1+1=2? The hand he/she raises signifies 'Yes'

b. Repeat (a) for person in chair 2.

c. Finally ask the person in chair 1, "Are you A?".

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The problem is still solvable. Remember that you only need to tell which person of #1, #2, and #3 corresponds to A, B, and C -- it doesn't ask you to also solve whether the left hand or the right hand of each person signifies truth. The three yes/no questions would have to distinguish between six different states, which is (at least theoretically) possible. But you are correct: if you also need to figure out whether the left or the right hand signifies yes or no for even one of the three people, then you would have to distinguish between 6 x 2 = 12 possible states, which can't be done with three yes/no questions.

That gets to the fundamental "trick" of solving the problem.

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In other words, you can all the time ask only one person - in terms of this exercise it makes no difference.

Edited by koren
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The problem is still solvable. Remember that you only need to tell which person of #1, #2, and #3 corresponds to A, B, and C -- it doesn't ask you to also solve whether the left hand or the right hand of each person signifies truth. The three yes/no questions would have to distinguish between six different states, which is (at least theoretically) possible. But you are correct: if you also need to figure out whether the left or the right hand signifies yes or no for even one of the three people, then you would have to distinguish between 6 x 2 = 12 possible states, which can't be done with three yes/no questions.

That gets to the fundamental "trick" of solving the problem.

Ah, ok.. I misunderstood. As you have stated, it can certainly be solved with 3 questions.

If we need to ask a question 'Q' to a person, we'll ask a different question Q' that will cause the person will raise the right hand (say) if Q is true; left hand if otherwise

We can make a truth table to do this..

a. If Q is true and X is right-handed, he/she should raise right hand ([1 1] => 1)

b. If Q is true and X is left-handed, he/she should raise right hand ([1 0] => 0)

c. If Q is false and X is right-handed, he/she should raise left hand ([0 1] => 0)

d. If Q is false and X is left-handed, he/she should raise left hand ([0 0] => 1)

This is nothing but an xor operation, can also be stated as is Q' = "the truth value of Q the same as the truth value of saying 'you are right-handed'"

Q1. Ask the person in chair-1: "Is the truth value of 'you are person A' the same as truth value of 'you are right-handed'"?

Q1R.Q2 (if the person raised the right hand for Q1) Ask the person in chair-2: "Is the truth value of 'you are person B' the same as truth value of 'you are right-handed'"?

Q1R.Q2R. Arrangement is: A-1, B-2, C-3

Q1R.Q2L. Arrangement is: A-1, C-2, B-3

Q1L.Q2 (if the person raised the left hand for Q1) Ask the person in chair-2: "Is the truth value of 'you are person A' the same as truth value of 'you are right-handed'"?

Q1L.Q2R.Q3 Ask the person in chair-3:  "Is the truth value of 'you are person C' the same as truth value of 'you are right-handed'"?
Q1L.Q2R.Q3R Arrangement is: B-1, A-2, C-3
Q1L.Q2R.Q3L Arrangement is: C-1, A-2, B-3
Q1L.Q2L.Q3 Ask the person in chair-2:  "Is the truth value of 'you are person C' the same as truth value of 'you are right-handed'"?
Q1L.Q2L.Q3R Arrangement is: B-1, C-2, A-3
Q1L.Q2L.Q3L Arrangement is: C-1, B-2, A-3

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The problem is still solvable. Remember that you only need to tell which person of #1, #2, and #3 corresponds to A, B, and C -- it doesn't ask you to also solve whether the left hand or the right hand of each person signifies truth. The three yes/no questions would have to distinguish between six different states, which is (at least theoretically) possible. But you are correct: if you also need to figure out whether the left or the right hand signifies yes or no for even one of the three people, then you would have to distinguish between 6 x 2 = 12 possible states, which can't be done with three yes/no questions.

That gets to the fundamental "trick" of solving the problem.

Ah, ok.. I misunderstood. As you have stated, it can certainly be solved with 3 questions.

If we need to ask a question 'Q' to a person, we'll ask a different question Q' that will cause the person will raise the right hand (say) if Q is true; left hand if otherwise

We can make a truth table to do this..

a. If Q is true and X is right-handed, he/she should raise right hand ([1 1] => 1)

b. If Q is true and X is left-handed, he/she should raise right hand ([1 0] => 0)

c. If Q is false and X is right-handed, he/she should raise left hand ([0 1] => 0)

d. If Q is false and X is left-handed, he/she should raise left hand ([0 0] => 1)

This is nothing but an xor operation, can also be stated as is Q' = "the truth value of Q the same as the truth value of saying 'you are right-handed'"

Q1. Ask the person in chair-1: "Is the truth value of 'you are person A' the same as truth value of 'you are right-handed'"?

Q1R.Q2 (if the person raised the right hand for Q1) Ask the person in chair-2: "Is the truth value of 'you are person B' the same as truth value of 'you are right-handed'"?

Q1R.Q2R. Arrangement is: A-1, B-2, C-3

Q1R.Q2L. Arrangement is: A-1, C-2, B-3 = or  A-1, B-2, C-3 and B is left-handed =

Q1L.Q2 (if the person raised the left hand for Q1) Ask the person in chair-2: "Is the truth value of 'you are person A' the same as truth value of 'you are right-handed'"?

Q1L.Q2R.Q3 Ask the person in chair-3:  "Is the truth value of 'you are person C' the same as truth value of 'you are right-handed'"?
Q1L.Q2R.Q3R Arrangement is: B-1, A-2, C-3
Q1L.Q2R.Q3L Arrangement is: C-1, A-2, B-3
Q1L.Q2L.Q3 Ask the person in chair-2:  "Is the truth value of 'you are person C' the same as truth value of 'you are right-handed'"?
Q1L.Q2L.Q3R Arrangement is: B-1, C-2, A-3
Q1L.Q2L.Q3L Arrangement is: C-1, B-2, A-3

Further on  in the same manner...

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The problem is still solvable. Remember that you only need to tell which person of #1, #2, and #3 corresponds to A, B, and C -- it doesn't ask you to also solve whether the left hand or the right hand of each person signifies truth. The three yes/no questions would have to distinguish between six different states, which is (at least theoretically) possible. But you are correct: if you also need to figure out whether the left or the right hand signifies yes or no for even one of the three people, then you would have to distinguish between 6 x 2 = 12 possible states, which can't be done with three yes/no questions.

That gets to the fundamental "trick" of solving the problem.

Ah, ok.. I misunderstood. As you have stated, it can certainly be solved with 3 questions.

If we need to ask a question 'Q' to a person, we'll ask a different question Q' that will cause the person will raise the right hand (say) if Q is true; left hand if otherwise

We can make a truth table to do this..

a. If Q is true and X is right-handed, he/she should raise right hand ([1 1] => 1)

b. If Q is true and X is left-handed, he/she should raise right hand ([1 0] => 0)

c. If Q is false and X is right-handed, he/she should raise left hand ([0 1] => 0)

d. If Q is false and X is left-handed, he/she should raise left hand ([0 0] => 1)

This is nothing but an xor operation, can also be stated as is Q' = "the truth value of Q the same as the truth value of saying 'you are right-handed'"

Q1. Ask the person in chair-1: "Is the truth value of 'you are person A' the same as truth value of 'you are right-handed'"?

Q1R.Q2 (if the person raised the right hand for Q1) Ask the person in chair-2: "Is the truth value of 'you are person B' the same as truth value of 'you are right-handed'"?

Q1R.Q2R. Arrangement is: A-1, B-2, C-3

Q1R.Q2L. Arrangement is: A-1, C-2, B-3 = or  A-1, B-2, C-3 and B is left-handed =

Q1L.Q2 (if the person raised the left hand for Q1) Ask the person in chair-2: "Is the truth value of 'you are person A' the same as truth value of 'you are right-handed'"?

Q1L.Q2R.Q3 Ask the person in chair-3:  "Is the truth value of 'you are person C' the same as truth value of 'you are right-handed'"?
Q1L.Q2R.Q3R Arrangement is: B-1, A-2, C-3
Q1L.Q2R.Q3L Arrangement is: C-1, A-2, B-3
Q1L.Q2L.Q3 Ask the person in chair-2:  "Is the truth value of 'you are person C' the same as truth value of 'you are right-handed'"?
Q1L.Q2L.Q3R Arrangement is: B-1, C-2, A-3
Q1L.Q2L.Q3L Arrangement is: C-1, B-2, A-3

Further on  in the same manner...

No, that would be an incorrect observation.

Suppose B was left handed. And suppose B is in chair - 2.

The question we are asking the person in chair - 2 is: "Is the truth value of 'you are person B' the same as truth value of 'you are right-handed'"?. For the person in chair - 2, the first part evaluates to "true" ('you are person B'). But the second part evaluates to "false" ('you are right-handed'). So the answer to the compound question is 'false'. B uses right-hand to show 'false'. Hence B raises his/her right hand.

So if the person in chair - 2 raises his/her right hand in response to the question, the person is B.

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This is a challenging puzzle. If I understand correctly,

1. I ask someone a question then write "1" on his forehead.

I might ask someone else a question then write "2" on his forehead.

I then write "3" on the last person's forehead.

2. If A raises his left hand it might mean No.

If B raises his left hand it might mean Yes.

Similarly for person C.

3. After I ask my third question, I will state, for example,

Number 1 is B; Number 2 is C; and Number 3 is A.

Is that the idea?

yes,,,,thats the idea

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Three persons( A,B,and C) are not allowed to speak,but to raise a hand  refering to ( Yes or No ), you don`t know which hand means Yes,and which means No.

How can you know  the persons which is which by asking only three( yes or no) questions.

You may ask one ,two or all of them...as you like.

any  question should be asked to a single person.

The solution I propose is based on a certain question  configuration:

"Will you raise your right hand on the question: (yes/no question)?"

A positive response in this case - the right hand raising , the negative - left and no matter to whom you addressed the question.

All the rest is not difficult.

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Three persons( A,B,and C) are not allowed to speak,but to raise a hand  refering to ( Yes or No ), you don`t know which hand means Yes,and which means No.

How can you know  the persons which is which by asking only three( yes or no) questions.

You may ask one ,two or all of them...as you like.

any  question should be asked to a single person.

The solution I propose is based on a certain question  configuration:

"Will you raise your right hand on the question: (yes/no question)?"

A positive response in this case - the right hand raising , the negative - left and no matter to whom you addressed the question.

All the rest is not difficult.

now you know one of them,as right handed... what about the two others? and how to confirm their identity?

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I do not know who is right-handed, etc. To my question everyone raises the right hand when they say YES. For example, a question for the left-hander:  "Will you raise your right hand on the question: "2>1?"?". In normal mode, for "2>1?" he will raise his left hand, in the construction of my question he, as well as the right-hander, raises his right.

Let's see how it works:

1Q to #1: "Will you raise your right hand on the question: "are you A?"?"

Right hand to 1Q - the person is A

2Q tu #2: "Will you raise your right hand on the question: "are you B?"?"

Right hand - the person is B and #3 is C

Left hand -   the person is C and #3 is B

Left hand to 1Q -  the person is B or C

2Q tu #1: "Will you raise your right hand on the question: "are you B?"?"

Right hand to 2Q - the person is B

3Q tu #2: "Will you raise your right hand on the question: "are you A?"?"

Right hand - the person is A and #3 is C

Left hand -   the person is C and #3 is A

Left hand to 2Q-   the person is C

3Q tu #2: "Will you raise your right hand on the question: "are you A?"?"

Right hand - the person is A and #3 is B

Left hand -   the person is B and #3 is A

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I do not know who is right-handed, etc. To my question everyone raises the right hand when they say YES. For example, a question for the left-hander:  "Will you raise your right hand on the question: "2>1?"?". In normal mode, for "2>1?" he will raise his left hand, in the construction of my question he, as well as the right-hander, raises his right.

Let's see how it works:

1Q to #1: "Will you raise your right hand on the question: "are you A?"?"

Right hand to 1Q - the person is A

2Q tu #2: "Will you raise your right hand on the question: "are you B?"?"

Right hand - the person is B and #3 is C

Left hand -   the person is C and #3 is B

Left hand to 1Q -  the person is B or C

2Q tu #1: "Will you raise your right hand on the question: "are you B?"?"

Right hand to 2Q - the person is B

3Q tu #2: "Will you raise your right hand on the question: "are you A?"?"

Right hand - the person is A and #3 is C

Left hand -   the person is C and #3 is A

Left hand to 2Q-   the person is C

3Q tu #2: "Will you raise your right hand on the question: "are you A?"?"

Right hand - the person is A and #3 is B

Left hand -   the person is B and #3 is A

What if the 1st person is left handed and he is B?

for the 1st question, he can not raise his right hand because he is not A, neither to raise his left hand which means (yes),and that is not true.

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Wolfgang, have you seen my attempt at this? Is that correct?

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I do not know who is right-handed, etc. To my question everyone raises the right hand when they say YES. For example, a question for the left-hander:  "Will you raise your right hand on the question: "2>1?"?". In normal mode, for "2>1?" he will raise his left hand, in the construction of my question he, as well as the right-hander, raises his right.

Let's see how it works:

1Q to #1: "Will you raise your right hand on the question: "are you A?"?"

Right hand to 1Q - the person is A

2Q tu #2: "Will you raise your right hand on the question: "are you B?"?"

Right hand - the person is B and #3 is C

Left hand -   the person is C and #3 is B

Left hand to 1Q -  the person is B or C

2Q tu #1: "Will you raise your right hand on the question: "are you B?"?"

Right hand to 2Q - the person is B

3Q tu #2: "Will you raise your right hand on the question: "are you A?"?"

Right hand - the person is A and #3 is C

Left hand -   the person is C and #3 is A

Left hand to 2Q-   the person is C

3Q tu #2: "Will you raise your right hand on the question: "are you A?"?"

Right hand - the person is A and #3 is B

Left hand -   the person is B and #3 is A

What if the 1st person is left handed and he is B?

for the 1st question, he can not raise his right hand because he is not A, neither to raise his left hand which means (yes),and that is not true.

There is no contradiction here. If first person is left handed and he is B, he has to answer the question 'will I raise my right hand for the question "are you A?"?'. Answer is 'Yes, right hand will be raised if the question was "Are you A"'.. Since B is left handed, he will finally raise his left hand to signify "Yes".

I think my approach is essentially the same.

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Wolfgang, have you seen my attempt at this? Is that correct?

I read it.... but it is not correct... because in this case ,for the 3rd question, the 1st person may be A,,, or may not...hmmm...then what?

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I do not know who is right-handed, etc. To my question everyone raises the right hand when they say YES. For example, a question for the left-hander:  "Will you raise your right hand on the question: "2>1?"?". In normal mode, for "2>1?" he will raise his left hand, in the construction of my question he, as well as the right-hander, raises his right.

Let's see how it works:

1Q to #1: "Will you raise your right hand on the question: "are you A?"?"

Right hand to 1Q - the person is A

2Q tu #2: "Will you raise your right hand on the question: "are you B?"?"

Right hand - the person is B and #3 is C

Left hand -   the person is C and #3 is B

Left hand to 1Q -  the person is B or C

2Q tu #1: "Will you raise your right hand on the question: "are you B?"?"

Right hand to 2Q - the person is B

3Q tu #2: "Will you raise your right hand on the question: "are you A?"?"

Right hand - the person is A and #3 is C

Left hand -   the person is C and #3 is A

Left hand to 2Q-   the person is C

3Q tu #2: "Will you raise your right hand on the question: "are you A?"?"

Right hand - the person is A and #3 is B

Left hand -   the person is B and #3 is A

What if the 1st person is left handed and he is B?

for the 1st question, he can not raise his right hand because he is not A, neither to raise his left hand which means (yes),and that is not true.

There is no contradiction here. If first person is left handed and he is B, he has to answer the question 'will I raise my right hand for the question "are you A?"?'. Answer is 'Yes, right hand will be raised if the question was "Are you A"'.. Since B is left handed, he will finally raise his left hand to signify "Yes".

I think my approach is essentially the same.

nice trial... but take it in an easier way..

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I think this is pretty simple. Isn't Koren correct?

Wolfgang, if the first person uses his left hand to signify a positive response, and he is B, then to the question, "Would you raise your right hand if I asked you if you were A?", he would raise his left hand, which Koren would take to mean that he is not A, which is correct.

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This problem is pretty similar to a host of puzzles that are variants of a classic puzzle known as "The Hardest Logic Puzzle Ever". I think one presented on BrainDen's official site.

What Koren is doing is employing a device called a counterfactual, which is the general solution to the aforementioned puzzle.

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