Mixing
3 person Knights and Knaves Problem
$begingroup$
The problem goes as such: politicians never tell the truth and non politicians always tell the truth. A stranger meets 3 natives and asks the first of them "Are you a politician?" And he answers. The second native then reports that the first native denied being a politician. The third native says that the first native is a politician. How many of these 3 are politicians?
I'm essentially confused on how to solve this problem without knowing what the 1st native said.
logic puzzle
asked Jan 23 at 21:55
Bria HoltheBria Holthe
111
$endgroup$
add a comment |
$begingroup$
The problem goes as such: politicians never tell the truth and non politicians always tell the truth. A stranger meets 3 natives and asks the first of them "Are you a politician?" And he answers. The second native then reports that the first native denied being a politician. The third native says that the first native is a politician. How many of these 3 are politicians?
I'm essentially confused on how to solve this problem without knowing what the 1st native said.
logic puzzle
asked Jan 23 at 21:55
Bria HoltheBria Holthe
111
$endgroup$
$begingroup$
To get oriented in the problem, just go case by case. Can all three be politicians? Can all three be non-politicians? There are only $8$ cases so you can do this exhaustively if you can't think of anything else.
$endgroup$
– lulu
Jan 23 at 21:57
$begingroup$
Side note: you do know what the first one said, or at least you can figure it out. What does he reply if he is politician? What does he reply if he isn't?
$endgroup$
– lulu
Jan 23 at 22:00
$begingroup$
You do know what 1 said. 1 said what everyone says when you ask them "Are you a liar". Does anyone ever say "Yes, I am a liar"? Who'd say that?
$endgroup$
– fleablood
Jan 23 at 22:19
add a comment |
$begingroup$
The problem goes as such: politicians never tell the truth and non politicians always tell the truth. A stranger meets 3 natives and asks the first of them "Are you a politician?" And he answers. The second native then reports that the first native denied being a politician. The third native says that the first native is a politician. How many of these 3 are politicians?
I'm essentially confused on how to solve this problem without knowing what the 1st native said.
logic puzzle
asked Jan 23 at 21:55
Bria HoltheBria Holthe
111
$endgroup$
The problem goes as such: politicians never tell the truth and non politicians always tell the truth. A stranger meets 3 natives and asks the first of them "Are you a politician?" And he answers. The second native then reports that the first native denied being a politician. The third native says that the first native is a politician. How many of these 3 are politicians?
I'm essentially confused on how to solve this problem without knowing what the 1st native said.
logic puzzle
logic puzzle
asked Jan 23 at 21:55
Bria HoltheBria Holthe
111
asked Jan 23 at 21:55
Bria HoltheBria Holthe
111
asked Jan 23 at 21:55
Bria HoltheBria Holthe
111
asked Jan 23 at 21:55
Bria HoltheBria Holthe
111
asked Jan 23 at 21:55
Bria HoltheBria Holthe
111
111
$begingroup$
To get oriented in the problem, just go case by case. Can all three be politicians? Can all three be non-politicians? There are only $8$ cases so you can do this exhaustively if you can't think of anything else.
$endgroup$
– lulu
Jan 23 at 21:57
$begingroup$
Side note: you do know what the first one said, or at least you can figure it out. What does he reply if he is politician? What does he reply if he isn't?
$endgroup$
– lulu
Jan 23 at 22:00
$begingroup$
You do know what 1 said. 1 said what everyone says when you ask them "Are you a liar". Does anyone ever say "Yes, I am a liar"? Who'd say that?
$endgroup$
– fleablood
Jan 23 at 22:19
add a comment |
$begingroup$
To get oriented in the problem, just go case by case. Can all three be politicians? Can all three be non-politicians? There are only $8$ cases so you can do this exhaustively if you can't think of anything else.
$endgroup$
– lulu
Jan 23 at 21:57
$begingroup$
Side note: you do know what the first one said, or at least you can figure it out. What does he reply if he is politician? What does he reply if he isn't?
$endgroup$
– lulu
Jan 23 at 22:00
$begingroup$
You do know what 1 said. 1 said what everyone says when you ask them "Are you a liar". Does anyone ever say "Yes, I am a liar"? Who'd say that?
$endgroup$
– fleablood
Jan 23 at 22:19
$begingroup$
To get oriented in the problem, just go case by case. Can all three be politicians? Can all three be non-politicians? There are only $8$ cases so you can do this exhaustively if you can't think of anything else.
$endgroup$
– lulu
Jan 23 at 21:57
$begingroup$
To get oriented in the problem, just go case by case. Can all three be politicians? Can all three be non-politicians? There are only $8$ cases so you can do this exhaustively if you can't think of anything else.
$endgroup$
– lulu
Jan 23 at 21:57
$begingroup$
Side note: you do know what the first one said, or at least you can figure it out. What does he reply if he is politician? What does he reply if he isn't?
$endgroup$
– lulu
Jan 23 at 22:00
$begingroup$
Side note: you do know what the first one said, or at least you can figure it out. What does he reply if he is politician? What does he reply if he isn't?
$endgroup$
– lulu
Jan 23 at 22:00
$begingroup$
You do know what 1 said. 1 said what everyone says when you ask them "Are you a liar". Does anyone ever say "Yes, I am a liar"? Who'd say that?
$endgroup$
– fleablood
Jan 23 at 22:19
$begingroup$
You do know what 1 said. 1 said what everyone says when you ask them "Are you a liar". Does anyone ever say "Yes, I am a liar"? Who'd say that?
$endgroup$
– fleablood
Jan 23 at 22:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Case 1: No. 1 is a liar.
"Are you a liar?"
No. 1 think. "Hmm, I am but I better lie" and says "No, I'm not". That's a lie.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's true.
You have one liar.
Case 2: No. 1 is not a liar.
"Are you a liar?"
No.1 thinks "No. I'm not and I better tell the truth" and says "No, I'm not". That's true.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's a lie.
You have 1 liar.
So... You have one liar. Either 1 is a liar and 3 is a truther. Or 1 is a truther and 3 is a liar. 2 is definitely a truther.
answered Jan 23 at 22:18
fleabloodfleablood
72.3k22687
$endgroup$
$begingroup$
Could perhaps add some insight into how you came up with this argument/proof? For example, why did you do a case split, and why did you choose do this on whether or not no. 1 is a liar? Perhaps that would help the OP... Thanks!
$endgroup$
– Marnix Klooster
Feb 2 at 18:09
$begingroup$
Insight? I didn't have ANY insight. I just imagined myself in the situation and tried to consider what would happen. I ask a person "Are you a liar" and imagine what s/he would respond... well, that depends on if s/he is a liar so I considered each case.
$endgroup$
– fleablood
Feb 2 at 18:31
add a comment |
$begingroup$
Let's write $;P(x);$ for "$;x;$ is a politician". We are told that, when $;x;$ says $;phi;$, then $;phi;$ is true if and only if $;x;$ is not a politician. That is,
$$
lnot P(x) equiv underline{phi}
$$
(where I've underlined $;x;$'s statement for clarity, this line does not have any formal meaning).
Naming the politicians $;A,B,C;$ in order, the second's statement translates to
$$
tag{1} lnot P(B) equiv underline{(lnot P(A) equiv underline{lnot P(A)})}
$$
Now simplify the right hand side, and the conclusion about $;B;$ is...?
For the third, we get
$$
lnot P(C) equiv underline{P(A)}
$$
What does this tell us about the combination of $;A;$ and $;C;$?
Add up those sub-answers and you get your answer.
The nice thing about this style of proof is that there is no need for any case analysis, which (in my opinion) makes the argument clearer.
answered Jan 23 at 22:33
Marnix KloosterMarnix Klooster
4,22122149
$endgroup$
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Case 1: No. 1 is a liar.
"Are you a liar?"
No. 1 think. "Hmm, I am but I better lie" and says "No, I'm not". That's a lie.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's true.
You have one liar.
Case 2: No. 1 is not a liar.
"Are you a liar?"
No.1 thinks "No. I'm not and I better tell the truth" and says "No, I'm not". That's true.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's a lie.
You have 1 liar.
So... You have one liar. Either 1 is a liar and 3 is a truther. Or 1 is a truther and 3 is a liar. 2 is definitely a truther.
answered Jan 23 at 22:18
fleabloodfleablood
72.3k22687
$endgroup$
$begingroup$
Could perhaps add some insight into how you came up with this argument/proof? For example, why did you do a case split, and why did you choose do this on whether or not no. 1 is a liar? Perhaps that would help the OP... Thanks!
$endgroup$
– Marnix Klooster
Feb 2 at 18:09
$begingroup$
Insight? I didn't have ANY insight. I just imagined myself in the situation and tried to consider what would happen. I ask a person "Are you a liar" and imagine what s/he would respond... well, that depends on if s/he is a liar so I considered each case.
$endgroup$
– fleablood
Feb 2 at 18:31
add a comment |
$begingroup$
Case 1: No. 1 is a liar.
"Are you a liar?"
No. 1 think. "Hmm, I am but I better lie" and says "No, I'm not". That's a lie.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's true.
You have one liar.
Case 2: No. 1 is not a liar.
"Are you a liar?"
No.1 thinks "No. I'm not and I better tell the truth" and says "No, I'm not". That's true.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's a lie.
You have 1 liar.
So... You have one liar. Either 1 is a liar and 3 is a truther. Or 1 is a truther and 3 is a liar. 2 is definitely a truther.
answered Jan 23 at 22:18
fleabloodfleablood
72.3k22687
$endgroup$
$begingroup$
Could perhaps add some insight into how you came up with this argument/proof? For example, why did you do a case split, and why did you choose do this on whether or not no. 1 is a liar? Perhaps that would help the OP... Thanks!
$endgroup$
– Marnix Klooster
Feb 2 at 18:09
$begingroup$
Insight? I didn't have ANY insight. I just imagined myself in the situation and tried to consider what would happen. I ask a person "Are you a liar" and imagine what s/he would respond... well, that depends on if s/he is a liar so I considered each case.
$endgroup$
– fleablood
Feb 2 at 18:31
add a comment |
$begingroup$
Case 1: No. 1 is a liar.
"Are you a liar?"
No. 1 think. "Hmm, I am but I better lie" and says "No, I'm not". That's a lie.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's true.
You have one liar.
Case 2: No. 1 is not a liar.
"Are you a liar?"
No.1 thinks "No. I'm not and I better tell the truth" and says "No, I'm not". That's true.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's a lie.
You have 1 liar.
So... You have one liar. Either 1 is a liar and 3 is a truther. Or 1 is a truther and 3 is a liar. 2 is definitely a truther.
answered Jan 23 at 22:18
fleabloodfleablood
72.3k22687
$endgroup$
Case 1: No. 1 is a liar.
"Are you a liar?"
No. 1 think. "Hmm, I am but I better lie" and says "No, I'm not". That's a lie.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's true.
You have one liar.
Case 2: No. 1 is not a liar.
"Are you a liar?"
No.1 thinks "No. I'm not and I better tell the truth" and says "No, I'm not". That's true.
No. 2 says "No. 1 denied it". That's true.
No. 3 says "No. 1 is a liar". That's a lie.
You have 1 liar.
So... You have one liar. Either 1 is a liar and 3 is a truther. Or 1 is a truther and 3 is a liar. 2 is definitely a truther.
answered Jan 23 at 22:18
fleabloodfleablood
72.3k22687
answered Jan 23 at 22:18
fleabloodfleablood
72.3k22687
answered Jan 23 at 22:18
fleabloodfleablood
72.3k22687
answered Jan 23 at 22:18
fleabloodfleablood
72.3k22687
72.3k22687
$begingroup$
Could perhaps add some insight into how you came up with this argument/proof? For example, why did you do a case split, and why did you choose do this on whether or not no. 1 is a liar? Perhaps that would help the OP... Thanks!
$endgroup$
– Marnix Klooster
Feb 2 at 18:09
$begingroup$
Insight? I didn't have ANY insight. I just imagined myself in the situation and tried to consider what would happen. I ask a person "Are you a liar" and imagine what s/he would respond... well, that depends on if s/he is a liar so I considered each case.
$endgroup$
– fleablood
Feb 2 at 18:31
add a comment |
$begingroup$
Could perhaps add some insight into how you came up with this argument/proof? For example, why did you do a case split, and why did you choose do this on whether or not no. 1 is a liar? Perhaps that would help the OP... Thanks!
$endgroup$
– Marnix Klooster
Feb 2 at 18:09
$begingroup$
Insight? I didn't have ANY insight. I just imagined myself in the situation and tried to consider what would happen. I ask a person "Are you a liar" and imagine what s/he would respond... well, that depends on if s/he is a liar so I considered each case.
$endgroup$
– fleablood
Feb 2 at 18:31
$begingroup$
Could perhaps add some insight into how you came up with this argument/proof? For example, why did you do a case split, and why did you choose do this on whether or not no. 1 is a liar? Perhaps that would help the OP... Thanks!
$endgroup$
– Marnix Klooster
Feb 2 at 18:09
$begingroup$
Could perhaps add some insight into how you came up with this argument/proof? For example, why did you do a case split, and why did you choose do this on whether or not no. 1 is a liar? Perhaps that would help the OP... Thanks!
$endgroup$
– Marnix Klooster
Feb 2 at 18:09
$begingroup$
Insight? I didn't have ANY insight. I just imagined myself in the situation and tried to consider what would happen. I ask a person "Are you a liar" and imagine what s/he would respond... well, that depends on if s/he is a liar so I considered each case.
$endgroup$
– fleablood
Feb 2 at 18:31
$begingroup$
Insight? I didn't have ANY insight. I just imagined myself in the situation and tried to consider what would happen. I ask a person "Are you a liar" and imagine what s/he would respond... well, that depends on if s/he is a liar so I considered each case.
$endgroup$
– fleablood
Feb 2 at 18:31
add a comment |
$begingroup$
Let's write $;P(x);$ for "$;x;$ is a politician". We are told that, when $;x;$ says $;phi;$, then $;phi;$ is true if and only if $;x;$ is not a politician. That is,
$$
lnot P(x) equiv underline{phi}
$$
(where I've underlined $;x;$'s statement for clarity, this line does not have any formal meaning).
Naming the politicians $;A,B,C;$ in order, the second's statement translates to
$$
tag{1} lnot P(B) equiv underline{(lnot P(A) equiv underline{lnot P(A)})}
$$
Now simplify the right hand side, and the conclusion about $;B;$ is...?
For the third, we get
$$
lnot P(C) equiv underline{P(A)}
$$
What does this tell us about the combination of $;A;$ and $;C;$?
Add up those sub-answers and you get your answer.
The nice thing about this style of proof is that there is no need for any case analysis, which (in my opinion) makes the argument clearer.
answered Jan 23 at 22:33
Marnix KloosterMarnix Klooster
4,22122149
$endgroup$
add a comment |
$begingroup$
Let's write $;P(x);$ for "$;x;$ is a politician". We are told that, when $;x;$ says $;phi;$, then $;phi;$ is true if and only if $;x;$ is not a politician. That is,
$$
lnot P(x) equiv underline{phi}
$$
(where I've underlined $;x;$'s statement for clarity, this line does not have any formal meaning).
Naming the politicians $;A,B,C;$ in order, the second's statement translates to
$$
tag{1} lnot P(B) equiv underline{(lnot P(A) equiv underline{lnot P(A)})}
$$
Now simplify the right hand side, and the conclusion about $;B;$ is...?
For the third, we get
$$
lnot P(C) equiv underline{P(A)}
$$
What does this tell us about the combination of $;A;$ and $;C;$?
Add up those sub-answers and you get your answer.
The nice thing about this style of proof is that there is no need for any case analysis, which (in my opinion) makes the argument clearer.
answered Jan 23 at 22:33
Marnix KloosterMarnix Klooster
4,22122149
$endgroup$
add a comment |
$begingroup$
Let's write $;P(x);$ for "$;x;$ is a politician". We are told that, when $;x;$ says $;phi;$, then $;phi;$ is true if and only if $;x;$ is not a politician. That is,
$$
lnot P(x) equiv underline{phi}
$$
(where I've underlined $;x;$'s statement for clarity, this line does not have any formal meaning).
Naming the politicians $;A,B,C;$ in order, the second's statement translates to
$$
tag{1} lnot P(B) equiv underline{(lnot P(A) equiv underline{lnot P(A)})}
$$
Now simplify the right hand side, and the conclusion about $;B;$ is...?
For the third, we get
$$
lnot P(C) equiv underline{P(A)}
$$
What does this tell us about the combination of $;A;$ and $;C;$?
Add up those sub-answers and you get your answer.
The nice thing about this style of proof is that there is no need for any case analysis, which (in my opinion) makes the argument clearer.
answered Jan 23 at 22:33
Marnix KloosterMarnix Klooster
4,22122149
$endgroup$
Let's write $;P(x);$ for "$;x;$ is a politician". We are told that, when $;x;$ says $;phi;$, then $;phi;$ is true if and only if $;x;$ is not a politician. That is,
$$
lnot P(x) equiv underline{phi}
$$
(where I've underlined $;x;$'s statement for clarity, this line does not have any formal meaning).
Naming the politicians $;A,B,C;$ in order, the second's statement translates to
$$
tag{1} lnot P(B) equiv underline{(lnot P(A) equiv underline{lnot P(A)})}
$$
Now simplify the right hand side, and the conclusion about $;B;$ is...?
For the third, we get
$$
lnot P(C) equiv underline{P(A)}
$$
What does this tell us about the combination of $;A;$ and $;C;$?
Add up those sub-answers and you get your answer.
The nice thing about this style of proof is that there is no need for any case analysis, which (in my opinion) makes the argument clearer.
answered Jan 23 at 22:33
Marnix KloosterMarnix Klooster
4,22122149
answered Jan 23 at 22:33
Marnix KloosterMarnix Klooster
4,22122149
answered Jan 23 at 22:33
Marnix KloosterMarnix Klooster
4,22122149
answered Jan 23 at 22:33
Marnix KloosterMarnix Klooster
4,22122149
4,22122149
add a comment |
add a comment |
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$begingroup$
To get oriented in the problem, just go case by case. Can all three be politicians? Can all three be non-politicians? There are only $8$ cases so you can do this exhaustively if you can't think of anything else.
$endgroup$
– lulu
Jan 23 at 21:57
$begingroup$
Side note: you do know what the first one said, or at least you can figure it out. What does he reply if he is politician? What does he reply if he isn't?
$endgroup$
– lulu
Jan 23 at 22:00
$begingroup$
You do know what 1 said. 1 said what everyone says when you ask them "Are you a liar". Does anyone ever say "Yes, I am a liar"? Who'd say that?
$endgroup$
– fleablood
Jan 23 at 22:19