Does there exist a way to simplify or build a table to find the truth in a set of given statements involving...
$begingroup$
This problem has got me going in circles for several hours and I'm not sure what to do, the problem is as follows:
At a semiconductor laboratory in Hsinchu a security guard caught five
technicians accessing a high level security area reserved for the most
trusted scientists working in a new integrated circuit for an upcoming
computer. However the security film is not very clear and the security
team can only be sure that four out of five technicians have the
access key card to enter the chipset room.
During interrogation the security team deduces that two of the
technicians are lying and the other are telling the truth.
The answers given by the laboratory technicians were as follows:
Audrey: Gwendolyn does not have an access credential.
Dorothy: I was entrusted an access key.
Marina: Hannah has an access key.
Gwendolyn: Audrey is lying.
Hannah: Dorothy is telling the truth.
Based on this information. Which of the technicians does not have
access to the chipset room?
Typically I would provide something but in this case I'm stuck at the very beginning. So far the only thing that I found it is that such problem seems to be a Knights and knaves kind which is related to logic. But in this case there are five individuals, hence the number of possible combinations would mean $2^5=32$ which raises a flag to me as $32$ combinations seems too big to make it a practical approach to try one by one. Therefore I need help into finding a solution or a method which would ease and simplify or solve this problem easily.
Can somebody help me with this? I'm not very knowledgeable with this type of problems. It would help me a lot to visualize what's going on if the proposed solution would include some sort of table or grid so I could identify the concluded result.
logic puzzle word-problem
edited Feb 15 at 7:59
jvdhooft
5,65961641
asked Feb 9 at 10:57
Chris Steinbeck BellChris Steinbeck Bell
855315
$endgroup$
|
show 10 more comments
$begingroup$
This problem has got me going in circles for several hours and I'm not sure what to do, the problem is as follows:
At a semiconductor laboratory in Hsinchu a security guard caught five
technicians accessing a high level security area reserved for the most
trusted scientists working in a new integrated circuit for an upcoming
computer. However the security film is not very clear and the security
team can only be sure that four out of five technicians have the
access key card to enter the chipset room.
During interrogation the security team deduces that two of the
technicians are lying and the other are telling the truth.
The answers given by the laboratory technicians were as follows:
Audrey: Gwendolyn does not have an access credential.
Dorothy: I was entrusted an access key.
Marina: Hannah has an access key.
Gwendolyn: Audrey is lying.
Hannah: Dorothy is telling the truth.
Based on this information. Which of the technicians does not have
access to the chipset room?
Typically I would provide something but in this case I'm stuck at the very beginning. So far the only thing that I found it is that such problem seems to be a Knights and knaves kind which is related to logic. But in this case there are five individuals, hence the number of possible combinations would mean $2^5=32$ which raises a flag to me as $32$ combinations seems too big to make it a practical approach to try one by one. Therefore I need help into finding a solution or a method which would ease and simplify or solve this problem easily.
Can somebody help me with this? I'm not very knowledgeable with this type of problems. It would help me a lot to visualize what's going on if the proposed solution would include some sort of table or grid so I could identify the concluded result.
logic puzzle word-problem
edited Feb 15 at 7:59
jvdhooft
5,65961641
asked Feb 9 at 10:57
Chris Steinbeck BellChris Steinbeck Bell
855315
$endgroup$
$begingroup$
It seems like some rules have been left out. If, say, I guess that $A$ is the one who had no access then I see that the truth values are $(F,T,T,T,T)$. If I guess that $G$ is the one without access then they are $(T,T,T,F,T)$. Why should I prefer one of those to the other?
$endgroup$
– lulu
Feb 9 at 11:05
$begingroup$
To your question: there are only $5$ possible truth tables here (unless I misunderstand). We know that exactly one of these people lacks access, so there is one truth table for each of the five.
$endgroup$
– lulu
Feb 9 at 11:07
$begingroup$
A condition of the form “at least or at most these many people lied” has to be there in the question. Please don’t go in circles if the question is incompletely stated
$endgroup$
– астон вілла олоф мэллбэрг
Feb 9 at 11:09
$begingroup$
@lulu I'm so sorry. There was a missing information in the problem. I'm doing an edit to include that part so maybe that can help solving the problem. For the other part you mentioned that there are only $5$ possible truth tables. But I meant that there are $32$ possible $T$ and $F$ combinations. That's what I meant. Or could I be mistaken?. Sorry. I'm new on this.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:10
$begingroup$
@ChrisSteinbeckBell But you have external information which lets you narrow down the number of tables. If I flip a fair coin and $A$ says "It's $H$" and $B$ says "It's $T$" Then the only truth tables are $(T,F),(F,T)$.
$endgroup$
– lulu
Feb 9 at 11:13
|
show 10 more comments
$begingroup$
This problem has got me going in circles for several hours and I'm not sure what to do, the problem is as follows:
At a semiconductor laboratory in Hsinchu a security guard caught five
technicians accessing a high level security area reserved for the most
trusted scientists working in a new integrated circuit for an upcoming
computer. However the security film is not very clear and the security
team can only be sure that four out of five technicians have the
access key card to enter the chipset room.
During interrogation the security team deduces that two of the
technicians are lying and the other are telling the truth.
The answers given by the laboratory technicians were as follows:
Audrey: Gwendolyn does not have an access credential.
Dorothy: I was entrusted an access key.
Marina: Hannah has an access key.
Gwendolyn: Audrey is lying.
Hannah: Dorothy is telling the truth.
Based on this information. Which of the technicians does not have
access to the chipset room?
Typically I would provide something but in this case I'm stuck at the very beginning. So far the only thing that I found it is that such problem seems to be a Knights and knaves kind which is related to logic. But in this case there are five individuals, hence the number of possible combinations would mean $2^5=32$ which raises a flag to me as $32$ combinations seems too big to make it a practical approach to try one by one. Therefore I need help into finding a solution or a method which would ease and simplify or solve this problem easily.
Can somebody help me with this? I'm not very knowledgeable with this type of problems. It would help me a lot to visualize what's going on if the proposed solution would include some sort of table or grid so I could identify the concluded result.
logic puzzle word-problem
edited Feb 15 at 7:59
jvdhooft
5,65961641
asked Feb 9 at 10:57
Chris Steinbeck BellChris Steinbeck Bell
855315
$endgroup$
This problem has got me going in circles for several hours and I'm not sure what to do, the problem is as follows:
At a semiconductor laboratory in Hsinchu a security guard caught five
technicians accessing a high level security area reserved for the most
trusted scientists working in a new integrated circuit for an upcoming
computer. However the security film is not very clear and the security
team can only be sure that four out of five technicians have the
access key card to enter the chipset room.
During interrogation the security team deduces that two of the
technicians are lying and the other are telling the truth.
The answers given by the laboratory technicians were as follows:
Audrey: Gwendolyn does not have an access credential.
Dorothy: I was entrusted an access key.
Marina: Hannah has an access key.
Gwendolyn: Audrey is lying.
Hannah: Dorothy is telling the truth.
Based on this information. Which of the technicians does not have
access to the chipset room?
Typically I would provide something but in this case I'm stuck at the very beginning. So far the only thing that I found it is that such problem seems to be a Knights and knaves kind which is related to logic. But in this case there are five individuals, hence the number of possible combinations would mean $2^5=32$ which raises a flag to me as $32$ combinations seems too big to make it a practical approach to try one by one. Therefore I need help into finding a solution or a method which would ease and simplify or solve this problem easily.
Can somebody help me with this? I'm not very knowledgeable with this type of problems. It would help me a lot to visualize what's going on if the proposed solution would include some sort of table or grid so I could identify the concluded result.
logic puzzle word-problem
logic puzzle word-problem
edited Feb 15 at 7:59
jvdhooft
5,65961641
asked Feb 9 at 10:57
Chris Steinbeck BellChris Steinbeck Bell
855315
edited Feb 15 at 7:59
jvdhooft
5,65961641
asked Feb 9 at 10:57
Chris Steinbeck BellChris Steinbeck Bell
855315
edited Feb 15 at 7:59
jvdhooft
5,65961641
edited Feb 15 at 7:59
jvdhooft
5,65961641
edited Feb 15 at 7:59
jvdhooft
5,65961641
5,65961641
asked Feb 9 at 10:57
Chris Steinbeck BellChris Steinbeck Bell
855315
asked Feb 9 at 10:57
Chris Steinbeck BellChris Steinbeck Bell
855315
asked Feb 9 at 10:57
Chris Steinbeck BellChris Steinbeck Bell
855315
855315
$begingroup$
It seems like some rules have been left out. If, say, I guess that $A$ is the one who had no access then I see that the truth values are $(F,T,T,T,T)$. If I guess that $G$ is the one without access then they are $(T,T,T,F,T)$. Why should I prefer one of those to the other?
$endgroup$
– lulu
Feb 9 at 11:05
$begingroup$
To your question: there are only $5$ possible truth tables here (unless I misunderstand). We know that exactly one of these people lacks access, so there is one truth table for each of the five.
$endgroup$
– lulu
Feb 9 at 11:07
$begingroup$
A condition of the form “at least or at most these many people lied” has to be there in the question. Please don’t go in circles if the question is incompletely stated
$endgroup$
– астон вілла олоф мэллбэрг
Feb 9 at 11:09
$begingroup$
@lulu I'm so sorry. There was a missing information in the problem. I'm doing an edit to include that part so maybe that can help solving the problem. For the other part you mentioned that there are only $5$ possible truth tables. But I meant that there are $32$ possible $T$ and $F$ combinations. That's what I meant. Or could I be mistaken?. Sorry. I'm new on this.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:10
$begingroup$
@ChrisSteinbeckBell But you have external information which lets you narrow down the number of tables. If I flip a fair coin and $A$ says "It's $H$" and $B$ says "It's $T$" Then the only truth tables are $(T,F),(F,T)$.
$endgroup$
– lulu
Feb 9 at 11:13
|
show 10 more comments
$begingroup$
It seems like some rules have been left out. If, say, I guess that $A$ is the one who had no access then I see that the truth values are $(F,T,T,T,T)$. If I guess that $G$ is the one without access then they are $(T,T,T,F,T)$. Why should I prefer one of those to the other?
$endgroup$
– lulu
Feb 9 at 11:05
$begingroup$
To your question: there are only $5$ possible truth tables here (unless I misunderstand). We know that exactly one of these people lacks access, so there is one truth table for each of the five.
$endgroup$
– lulu
Feb 9 at 11:07
$begingroup$
A condition of the form “at least or at most these many people lied” has to be there in the question. Please don’t go in circles if the question is incompletely stated
$endgroup$
– астон вілла олоф мэллбэрг
Feb 9 at 11:09
$begingroup$
@lulu I'm so sorry. There was a missing information in the problem. I'm doing an edit to include that part so maybe that can help solving the problem. For the other part you mentioned that there are only $5$ possible truth tables. But I meant that there are $32$ possible $T$ and $F$ combinations. That's what I meant. Or could I be mistaken?. Sorry. I'm new on this.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:10
$begingroup$
@ChrisSteinbeckBell But you have external information which lets you narrow down the number of tables. If I flip a fair coin and $A$ says "It's $H$" and $B$ says "It's $T$" Then the only truth tables are $(T,F),(F,T)$.
$endgroup$
– lulu
Feb 9 at 11:13
$begingroup$
It seems like some rules have been left out. If, say, I guess that $A$ is the one who had no access then I see that the truth values are $(F,T,T,T,T)$. If I guess that $G$ is the one without access then they are $(T,T,T,F,T)$. Why should I prefer one of those to the other?
$endgroup$
– lulu
Feb 9 at 11:05
$begingroup$
It seems like some rules have been left out. If, say, I guess that $A$ is the one who had no access then I see that the truth values are $(F,T,T,T,T)$. If I guess that $G$ is the one without access then they are $(T,T,T,F,T)$. Why should I prefer one of those to the other?
$endgroup$
– lulu
Feb 9 at 11:05
$begingroup$
To your question: there are only $5$ possible truth tables here (unless I misunderstand). We know that exactly one of these people lacks access, so there is one truth table for each of the five.
$endgroup$
– lulu
Feb 9 at 11:07
$begingroup$
To your question: there are only $5$ possible truth tables here (unless I misunderstand). We know that exactly one of these people lacks access, so there is one truth table for each of the five.
$endgroup$
– lulu
Feb 9 at 11:07
$begingroup$
A condition of the form “at least or at most these many people lied” has to be there in the question. Please don’t go in circles if the question is incompletely stated
$endgroup$
– астон вілла олоф мэллбэрг
Feb 9 at 11:09
$begingroup$
A condition of the form “at least or at most these many people lied” has to be there in the question. Please don’t go in circles if the question is incompletely stated
$endgroup$
– астон вілла олоф мэллбэрг
Feb 9 at 11:09
$begingroup$
@lulu I'm so sorry. There was a missing information in the problem. I'm doing an edit to include that part so maybe that can help solving the problem. For the other part you mentioned that there are only $5$ possible truth tables. But I meant that there are $32$ possible $T$ and $F$ combinations. That's what I meant. Or could I be mistaken?. Sorry. I'm new on this.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:10
$begingroup$
@lulu I'm so sorry. There was a missing information in the problem. I'm doing an edit to include that part so maybe that can help solving the problem. For the other part you mentioned that there are only $5$ possible truth tables. But I meant that there are $32$ possible $T$ and $F$ combinations. That's what I meant. Or could I be mistaken?. Sorry. I'm new on this.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:10
$begingroup$
@ChrisSteinbeckBell But you have external information which lets you narrow down the number of tables. If I flip a fair coin and $A$ says "It's $H$" and $B$ says "It's $T$" Then the only truth tables are $(T,F),(F,T)$.
$endgroup$
– lulu
Feb 9 at 11:13
$begingroup$
@ChrisSteinbeckBell But you have external information which lets you narrow down the number of tables. If I flip a fair coin and $A$ says "It's $H$" and $B$ says "It's $T$" Then the only truth tables are $(T,F),(F,T)$.
$endgroup$
– lulu
Feb 9 at 11:13
|
show 10 more comments
3 Answers
3
active
oldest
votes
$begingroup$
With the new information (that exactly $2$ are lying) we can solve the problem.
There are $5$ states to consider according to whichever lacked access. We list them all:
If $A$ lacked access: $(F,T,T,T,T)$
If $D$ lacked access: $(F,F,T,T,F)$
If $M$ lacked access: $(F,T,T,T,T)$
If $G$ lacked access $(T,T,T,F,T)$
If $H$ lacked access: $(F,T,F, T,T)$
By inspection, Hannah is the guilty party.
To stress: we need some rule to let us know how to evaluate the various True-False configurations. A priori, we have no idea what they might mean. A natural rule would be, say, "the person who lacked access is lying, everyone else is telling the truth". That rule doesn't lead to a unique solution here (though you can narrow down the list of suspects to $A,G$. As it stands, we are given the rule "exactly two people are lying". Happily, that rule does lead to a unique solution.
answered Feb 9 at 11:27
lulululu
43.3k25080
$endgroup$
$begingroup$
For some reason I suspected that Gwendolyn could be the one telling lies. But this was more of intuition or just assigning a psicological method of defense, neither any logical aspect on it. For the rest looking at the tables. I'm still confused. What does the order of trues and falses mean. Each statement?. But from looking at this I can understand the source of your initial confusion. For starters, do I get the impression that there can only be one $F$ value for each row?. If so meant that what to choose $G$ or $A$? as each one satisfies that condition.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:47
$begingroup$
$D$ is the row with more $F$ and it doesn't seem to be logical way as there are only two possible $textrm{F's}$ not three, so the remaning $H$ should be the answer. Again, with no prior experience in these kinds of problems that would be my assumption based on your answer. Still with few words I believe that a more graphical approach or a step-by-step solution would assess better what I requested rather a straightforward answer, which isn't bad as it solves the problem but doesn't aid much to the OP (me) to understand what to think and what to look first.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:51
1
$begingroup$
Just go statement by statement. Don't focus on who is lying, at least at first. Focus on who lacked access. Let's say $A$ lacked access. Then $A$ is lying (since $G$ did have access), $D$ is telling the truth, $M$ is telling the truth, $G$ is telling the truth, and $H$ is telling the truth. We now just list those values as $(F,T,T,T,T)$.
$endgroup$
– lulu
Feb 9 at 11:52
1
$begingroup$
I don't understand. A priori we have no reason to assume that anybody is lying or that anybody is telling the truth. Maybe all five are lying, maybe all five are telling the truth. Who knows? We need some rule to go by. Here the rule is "exactly two people are lying." Fine. We can work with that. But we need some sort of rule. Once I have the rule, then I just look to see which state of the world satisfies that rule. Here, only the state "Hannah lacked access" satisfies the rule.
$endgroup$
– lulu
Feb 9 at 12:05
1
$begingroup$
You could try different rules, but I doubt that a randomly chosen rule would lead to a unique solution. If, say, you said "the person who lacked access is lying, everyone else is telling the truth" then the best we can say is that it's one of $A,G$ who lacked access.
$endgroup$
– lulu
Feb 9 at 12:08
|
show 7 more comments
$begingroup$
If G is telling the truth then A is lying and vice versa. So exactly one of A or G is a liar.
Since there are exactly two liars then exactly one of D, M and H is the second liar.
D and H are either both telling the truth or both lying. But if they are both liars then there are three liars altogether. So they are both telling the truth.
Therefore the second liar is M, so Hannah does not have an access key.
answered Feb 9 at 11:28
gandalf61gandalf61
9,219825
$endgroup$
add a comment |
$begingroup$
You need to write five tables and then see if the person is lying or not based in this table, the exercise ask for a table that has two people lying, is this case would be the table $(H, H, H, H, N)$ h stands for has the key and n for does not have the key. So the last person does not have the credential to access this laboratory.
answered Feb 9 at 11:31
CristianoCristiano
406
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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active
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active
oldest
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$begingroup$
With the new information (that exactly $2$ are lying) we can solve the problem.
There are $5$ states to consider according to whichever lacked access. We list them all:
If $A$ lacked access: $(F,T,T,T,T)$
If $D$ lacked access: $(F,F,T,T,F)$
If $M$ lacked access: $(F,T,T,T,T)$
If $G$ lacked access $(T,T,T,F,T)$
If $H$ lacked access: $(F,T,F, T,T)$
By inspection, Hannah is the guilty party.
To stress: we need some rule to let us know how to evaluate the various True-False configurations. A priori, we have no idea what they might mean. A natural rule would be, say, "the person who lacked access is lying, everyone else is telling the truth". That rule doesn't lead to a unique solution here (though you can narrow down the list of suspects to $A,G$. As it stands, we are given the rule "exactly two people are lying". Happily, that rule does lead to a unique solution.
answered Feb 9 at 11:27
lulululu
43.3k25080
$endgroup$
$begingroup$
For some reason I suspected that Gwendolyn could be the one telling lies. But this was more of intuition or just assigning a psicological method of defense, neither any logical aspect on it. For the rest looking at the tables. I'm still confused. What does the order of trues and falses mean. Each statement?. But from looking at this I can understand the source of your initial confusion. For starters, do I get the impression that there can only be one $F$ value for each row?. If so meant that what to choose $G$ or $A$? as each one satisfies that condition.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:47
$begingroup$
$D$ is the row with more $F$ and it doesn't seem to be logical way as there are only two possible $textrm{F's}$ not three, so the remaning $H$ should be the answer. Again, with no prior experience in these kinds of problems that would be my assumption based on your answer. Still with few words I believe that a more graphical approach or a step-by-step solution would assess better what I requested rather a straightforward answer, which isn't bad as it solves the problem but doesn't aid much to the OP (me) to understand what to think and what to look first.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:51
1
$begingroup$
Just go statement by statement. Don't focus on who is lying, at least at first. Focus on who lacked access. Let's say $A$ lacked access. Then $A$ is lying (since $G$ did have access), $D$ is telling the truth, $M$ is telling the truth, $G$ is telling the truth, and $H$ is telling the truth. We now just list those values as $(F,T,T,T,T)$.
$endgroup$
– lulu
Feb 9 at 11:52
1
$begingroup$
I don't understand. A priori we have no reason to assume that anybody is lying or that anybody is telling the truth. Maybe all five are lying, maybe all five are telling the truth. Who knows? We need some rule to go by. Here the rule is "exactly two people are lying." Fine. We can work with that. But we need some sort of rule. Once I have the rule, then I just look to see which state of the world satisfies that rule. Here, only the state "Hannah lacked access" satisfies the rule.
$endgroup$
– lulu
Feb 9 at 12:05
1
$begingroup$
You could try different rules, but I doubt that a randomly chosen rule would lead to a unique solution. If, say, you said "the person who lacked access is lying, everyone else is telling the truth" then the best we can say is that it's one of $A,G$ who lacked access.
$endgroup$
– lulu
Feb 9 at 12:08
|
show 7 more comments
$begingroup$
With the new information (that exactly $2$ are lying) we can solve the problem.
There are $5$ states to consider according to whichever lacked access. We list them all:
If $A$ lacked access: $(F,T,T,T,T)$
If $D$ lacked access: $(F,F,T,T,F)$
If $M$ lacked access: $(F,T,T,T,T)$
If $G$ lacked access $(T,T,T,F,T)$
If $H$ lacked access: $(F,T,F, T,T)$
By inspection, Hannah is the guilty party.
To stress: we need some rule to let us know how to evaluate the various True-False configurations. A priori, we have no idea what they might mean. A natural rule would be, say, "the person who lacked access is lying, everyone else is telling the truth". That rule doesn't lead to a unique solution here (though you can narrow down the list of suspects to $A,G$. As it stands, we are given the rule "exactly two people are lying". Happily, that rule does lead to a unique solution.
answered Feb 9 at 11:27
lulululu
43.3k25080
$endgroup$
$begingroup$
For some reason I suspected that Gwendolyn could be the one telling lies. But this was more of intuition or just assigning a psicological method of defense, neither any logical aspect on it. For the rest looking at the tables. I'm still confused. What does the order of trues and falses mean. Each statement?. But from looking at this I can understand the source of your initial confusion. For starters, do I get the impression that there can only be one $F$ value for each row?. If so meant that what to choose $G$ or $A$? as each one satisfies that condition.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:47
$begingroup$
$D$ is the row with more $F$ and it doesn't seem to be logical way as there are only two possible $textrm{F's}$ not three, so the remaning $H$ should be the answer. Again, with no prior experience in these kinds of problems that would be my assumption based on your answer. Still with few words I believe that a more graphical approach or a step-by-step solution would assess better what I requested rather a straightforward answer, which isn't bad as it solves the problem but doesn't aid much to the OP (me) to understand what to think and what to look first.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:51
1
$begingroup$
Just go statement by statement. Don't focus on who is lying, at least at first. Focus on who lacked access. Let's say $A$ lacked access. Then $A$ is lying (since $G$ did have access), $D$ is telling the truth, $M$ is telling the truth, $G$ is telling the truth, and $H$ is telling the truth. We now just list those values as $(F,T,T,T,T)$.
$endgroup$
– lulu
Feb 9 at 11:52
1
$begingroup$
I don't understand. A priori we have no reason to assume that anybody is lying or that anybody is telling the truth. Maybe all five are lying, maybe all five are telling the truth. Who knows? We need some rule to go by. Here the rule is "exactly two people are lying." Fine. We can work with that. But we need some sort of rule. Once I have the rule, then I just look to see which state of the world satisfies that rule. Here, only the state "Hannah lacked access" satisfies the rule.
$endgroup$
– lulu
Feb 9 at 12:05
1
$begingroup$
You could try different rules, but I doubt that a randomly chosen rule would lead to a unique solution. If, say, you said "the person who lacked access is lying, everyone else is telling the truth" then the best we can say is that it's one of $A,G$ who lacked access.
$endgroup$
– lulu
Feb 9 at 12:08
|
show 7 more comments
$begingroup$
With the new information (that exactly $2$ are lying) we can solve the problem.
There are $5$ states to consider according to whichever lacked access. We list them all:
If $A$ lacked access: $(F,T,T,T,T)$
If $D$ lacked access: $(F,F,T,T,F)$
If $M$ lacked access: $(F,T,T,T,T)$
If $G$ lacked access $(T,T,T,F,T)$
If $H$ lacked access: $(F,T,F, T,T)$
By inspection, Hannah is the guilty party.
To stress: we need some rule to let us know how to evaluate the various True-False configurations. A priori, we have no idea what they might mean. A natural rule would be, say, "the person who lacked access is lying, everyone else is telling the truth". That rule doesn't lead to a unique solution here (though you can narrow down the list of suspects to $A,G$. As it stands, we are given the rule "exactly two people are lying". Happily, that rule does lead to a unique solution.
answered Feb 9 at 11:27
lulululu
43.3k25080
$endgroup$
With the new information (that exactly $2$ are lying) we can solve the problem.
There are $5$ states to consider according to whichever lacked access. We list them all:
If $A$ lacked access: $(F,T,T,T,T)$
If $D$ lacked access: $(F,F,T,T,F)$
If $M$ lacked access: $(F,T,T,T,T)$
If $G$ lacked access $(T,T,T,F,T)$
If $H$ lacked access: $(F,T,F, T,T)$
By inspection, Hannah is the guilty party.
To stress: we need some rule to let us know how to evaluate the various True-False configurations. A priori, we have no idea what they might mean. A natural rule would be, say, "the person who lacked access is lying, everyone else is telling the truth". That rule doesn't lead to a unique solution here (though you can narrow down the list of suspects to $A,G$. As it stands, we are given the rule "exactly two people are lying". Happily, that rule does lead to a unique solution.
answered Feb 9 at 11:27
lulululu
43.3k25080
edited Feb 9 at 12:25
answered Feb 9 at 11:27
lulululu
43.3k25080
answered Feb 9 at 11:27
lulululu
43.3k25080
answered Feb 9 at 11:27
lulululu
43.3k25080
43.3k25080
$begingroup$
For some reason I suspected that Gwendolyn could be the one telling lies. But this was more of intuition or just assigning a psicological method of defense, neither any logical aspect on it. For the rest looking at the tables. I'm still confused. What does the order of trues and falses mean. Each statement?. But from looking at this I can understand the source of your initial confusion. For starters, do I get the impression that there can only be one $F$ value for each row?. If so meant that what to choose $G$ or $A$? as each one satisfies that condition.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:47
$begingroup$
$D$ is the row with more $F$ and it doesn't seem to be logical way as there are only two possible $textrm{F's}$ not three, so the remaning $H$ should be the answer. Again, with no prior experience in these kinds of problems that would be my assumption based on your answer. Still with few words I believe that a more graphical approach or a step-by-step solution would assess better what I requested rather a straightforward answer, which isn't bad as it solves the problem but doesn't aid much to the OP (me) to understand what to think and what to look first.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:51
1
$begingroup$
Just go statement by statement. Don't focus on who is lying, at least at first. Focus on who lacked access. Let's say $A$ lacked access. Then $A$ is lying (since $G$ did have access), $D$ is telling the truth, $M$ is telling the truth, $G$ is telling the truth, and $H$ is telling the truth. We now just list those values as $(F,T,T,T,T)$.
$endgroup$
– lulu
Feb 9 at 11:52
1
$begingroup$
I don't understand. A priori we have no reason to assume that anybody is lying or that anybody is telling the truth. Maybe all five are lying, maybe all five are telling the truth. Who knows? We need some rule to go by. Here the rule is "exactly two people are lying." Fine. We can work with that. But we need some sort of rule. Once I have the rule, then I just look to see which state of the world satisfies that rule. Here, only the state "Hannah lacked access" satisfies the rule.
$endgroup$
– lulu
Feb 9 at 12:05
1
$begingroup$
You could try different rules, but I doubt that a randomly chosen rule would lead to a unique solution. If, say, you said "the person who lacked access is lying, everyone else is telling the truth" then the best we can say is that it's one of $A,G$ who lacked access.
$endgroup$
– lulu
Feb 9 at 12:08
|
show 7 more comments
$begingroup$
For some reason I suspected that Gwendolyn could be the one telling lies. But this was more of intuition or just assigning a psicological method of defense, neither any logical aspect on it. For the rest looking at the tables. I'm still confused. What does the order of trues and falses mean. Each statement?. But from looking at this I can understand the source of your initial confusion. For starters, do I get the impression that there can only be one $F$ value for each row?. If so meant that what to choose $G$ or $A$? as each one satisfies that condition.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:47
$begingroup$
$D$ is the row with more $F$ and it doesn't seem to be logical way as there are only two possible $textrm{F's}$ not three, so the remaning $H$ should be the answer. Again, with no prior experience in these kinds of problems that would be my assumption based on your answer. Still with few words I believe that a more graphical approach or a step-by-step solution would assess better what I requested rather a straightforward answer, which isn't bad as it solves the problem but doesn't aid much to the OP (me) to understand what to think and what to look first.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:51
1
$begingroup$
Just go statement by statement. Don't focus on who is lying, at least at first. Focus on who lacked access. Let's say $A$ lacked access. Then $A$ is lying (since $G$ did have access), $D$ is telling the truth, $M$ is telling the truth, $G$ is telling the truth, and $H$ is telling the truth. We now just list those values as $(F,T,T,T,T)$.
$endgroup$
– lulu
Feb 9 at 11:52
1
$begingroup$
I don't understand. A priori we have no reason to assume that anybody is lying or that anybody is telling the truth. Maybe all five are lying, maybe all five are telling the truth. Who knows? We need some rule to go by. Here the rule is "exactly two people are lying." Fine. We can work with that. But we need some sort of rule. Once I have the rule, then I just look to see which state of the world satisfies that rule. Here, only the state "Hannah lacked access" satisfies the rule.
$endgroup$
– lulu
Feb 9 at 12:05
1
$begingroup$
You could try different rules, but I doubt that a randomly chosen rule would lead to a unique solution. If, say, you said "the person who lacked access is lying, everyone else is telling the truth" then the best we can say is that it's one of $A,G$ who lacked access.
$endgroup$
– lulu
Feb 9 at 12:08
$begingroup$
For some reason I suspected that Gwendolyn could be the one telling lies. But this was more of intuition or just assigning a psicological method of defense, neither any logical aspect on it. For the rest looking at the tables. I'm still confused. What does the order of trues and falses mean. Each statement?. But from looking at this I can understand the source of your initial confusion. For starters, do I get the impression that there can only be one $F$ value for each row?. If so meant that what to choose $G$ or $A$? as each one satisfies that condition.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:47
$begingroup$
For some reason I suspected that Gwendolyn could be the one telling lies. But this was more of intuition or just assigning a psicological method of defense, neither any logical aspect on it. For the rest looking at the tables. I'm still confused. What does the order of trues and falses mean. Each statement?. But from looking at this I can understand the source of your initial confusion. For starters, do I get the impression that there can only be one $F$ value for each row?. If so meant that what to choose $G$ or $A$? as each one satisfies that condition.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:47
$begingroup$
$D$ is the row with more $F$ and it doesn't seem to be logical way as there are only two possible $textrm{F's}$ not three, so the remaning $H$ should be the answer. Again, with no prior experience in these kinds of problems that would be my assumption based on your answer. Still with few words I believe that a more graphical approach or a step-by-step solution would assess better what I requested rather a straightforward answer, which isn't bad as it solves the problem but doesn't aid much to the OP (me) to understand what to think and what to look first.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:51
$begingroup$
$D$ is the row with more $F$ and it doesn't seem to be logical way as there are only two possible $textrm{F's}$ not three, so the remaning $H$ should be the answer. Again, with no prior experience in these kinds of problems that would be my assumption based on your answer. Still with few words I believe that a more graphical approach or a step-by-step solution would assess better what I requested rather a straightforward answer, which isn't bad as it solves the problem but doesn't aid much to the OP (me) to understand what to think and what to look first.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:51
1
1
$begingroup$
Just go statement by statement. Don't focus on who is lying, at least at first. Focus on who lacked access. Let's say $A$ lacked access. Then $A$ is lying (since $G$ did have access), $D$ is telling the truth, $M$ is telling the truth, $G$ is telling the truth, and $H$ is telling the truth. We now just list those values as $(F,T,T,T,T)$.
$endgroup$
– lulu
Feb 9 at 11:52
$begingroup$
Just go statement by statement. Don't focus on who is lying, at least at first. Focus on who lacked access. Let's say $A$ lacked access. Then $A$ is lying (since $G$ did have access), $D$ is telling the truth, $M$ is telling the truth, $G$ is telling the truth, and $H$ is telling the truth. We now just list those values as $(F,T,T,T,T)$.
$endgroup$
– lulu
Feb 9 at 11:52
1
1
$begingroup$
I don't understand. A priori we have no reason to assume that anybody is lying or that anybody is telling the truth. Maybe all five are lying, maybe all five are telling the truth. Who knows? We need some rule to go by. Here the rule is "exactly two people are lying." Fine. We can work with that. But we need some sort of rule. Once I have the rule, then I just look to see which state of the world satisfies that rule. Here, only the state "Hannah lacked access" satisfies the rule.
$endgroup$
– lulu
Feb 9 at 12:05
$begingroup$
I don't understand. A priori we have no reason to assume that anybody is lying or that anybody is telling the truth. Maybe all five are lying, maybe all five are telling the truth. Who knows? We need some rule to go by. Here the rule is "exactly two people are lying." Fine. We can work with that. But we need some sort of rule. Once I have the rule, then I just look to see which state of the world satisfies that rule. Here, only the state "Hannah lacked access" satisfies the rule.
$endgroup$
– lulu
Feb 9 at 12:05
1
1
$begingroup$
You could try different rules, but I doubt that a randomly chosen rule would lead to a unique solution. If, say, you said "the person who lacked access is lying, everyone else is telling the truth" then the best we can say is that it's one of $A,G$ who lacked access.
$endgroup$
– lulu
Feb 9 at 12:08
$begingroup$
You could try different rules, but I doubt that a randomly chosen rule would lead to a unique solution. If, say, you said "the person who lacked access is lying, everyone else is telling the truth" then the best we can say is that it's one of $A,G$ who lacked access.
$endgroup$
– lulu
Feb 9 at 12:08
|
show 7 more comments
$begingroup$
If G is telling the truth then A is lying and vice versa. So exactly one of A or G is a liar.
Since there are exactly two liars then exactly one of D, M and H is the second liar.
D and H are either both telling the truth or both lying. But if they are both liars then there are three liars altogether. So they are both telling the truth.
Therefore the second liar is M, so Hannah does not have an access key.
answered Feb 9 at 11:28
gandalf61gandalf61
9,219825
$endgroup$
add a comment |
$begingroup$
If G is telling the truth then A is lying and vice versa. So exactly one of A or G is a liar.
Since there are exactly two liars then exactly one of D, M and H is the second liar.
D and H are either both telling the truth or both lying. But if they are both liars then there are three liars altogether. So they are both telling the truth.
Therefore the second liar is M, so Hannah does not have an access key.
answered Feb 9 at 11:28
gandalf61gandalf61
9,219825
$endgroup$
add a comment |
$begingroup$
If G is telling the truth then A is lying and vice versa. So exactly one of A or G is a liar.
Since there are exactly two liars then exactly one of D, M and H is the second liar.
D and H are either both telling the truth or both lying. But if they are both liars then there are three liars altogether. So they are both telling the truth.
Therefore the second liar is M, so Hannah does not have an access key.
answered Feb 9 at 11:28
gandalf61gandalf61
9,219825
$endgroup$
If G is telling the truth then A is lying and vice versa. So exactly one of A or G is a liar.
Since there are exactly two liars then exactly one of D, M and H is the second liar.
D and H are either both telling the truth or both lying. But if they are both liars then there are three liars altogether. So they are both telling the truth.
Therefore the second liar is M, so Hannah does not have an access key.
answered Feb 9 at 11:28
gandalf61gandalf61
9,219825
answered Feb 9 at 11:28
gandalf61gandalf61
9,219825
answered Feb 9 at 11:28
gandalf61gandalf61
9,219825
answered Feb 9 at 11:28
gandalf61gandalf61
9,219825
9,219825
add a comment |
add a comment |
$begingroup$
You need to write five tables and then see if the person is lying or not based in this table, the exercise ask for a table that has two people lying, is this case would be the table $(H, H, H, H, N)$ h stands for has the key and n for does not have the key. So the last person does not have the credential to access this laboratory.
answered Feb 9 at 11:31
CristianoCristiano
406
$endgroup$
add a comment |
$begingroup$
You need to write five tables and then see if the person is lying or not based in this table, the exercise ask for a table that has two people lying, is this case would be the table $(H, H, H, H, N)$ h stands for has the key and n for does not have the key. So the last person does not have the credential to access this laboratory.
answered Feb 9 at 11:31
CristianoCristiano
406
$endgroup$
add a comment |
$begingroup$
You need to write five tables and then see if the person is lying or not based in this table, the exercise ask for a table that has two people lying, is this case would be the table $(H, H, H, H, N)$ h stands for has the key and n for does not have the key. So the last person does not have the credential to access this laboratory.
answered Feb 9 at 11:31
CristianoCristiano
406
$endgroup$
You need to write five tables and then see if the person is lying or not based in this table, the exercise ask for a table that has two people lying, is this case would be the table $(H, H, H, H, N)$ h stands for has the key and n for does not have the key. So the last person does not have the credential to access this laboratory.
answered Feb 9 at 11:31
CristianoCristiano
406
edited Feb 9 at 11:37
answered Feb 9 at 11:31
CristianoCristiano
406
answered Feb 9 at 11:31
CristianoCristiano
406
answered Feb 9 at 11:31
CristianoCristiano
406
406
add a comment |
add a comment |
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$begingroup$
It seems like some rules have been left out. If, say, I guess that $A$ is the one who had no access then I see that the truth values are $(F,T,T,T,T)$. If I guess that $G$ is the one without access then they are $(T,T,T,F,T)$. Why should I prefer one of those to the other?
$endgroup$
– lulu
Feb 9 at 11:05
$begingroup$
To your question: there are only $5$ possible truth tables here (unless I misunderstand). We know that exactly one of these people lacks access, so there is one truth table for each of the five.
$endgroup$
– lulu
Feb 9 at 11:07
$begingroup$
A condition of the form “at least or at most these many people lied” has to be there in the question. Please don’t go in circles if the question is incompletely stated
$endgroup$
– астон вілла олоф мэллбэрг
Feb 9 at 11:09
$begingroup$
@lulu I'm so sorry. There was a missing information in the problem. I'm doing an edit to include that part so maybe that can help solving the problem. For the other part you mentioned that there are only $5$ possible truth tables. But I meant that there are $32$ possible $T$ and $F$ combinations. That's what I meant. Or could I be mistaken?. Sorry. I'm new on this.
$endgroup$
– Chris Steinbeck Bell
Feb 9 at 11:10
$begingroup$
@ChrisSteinbeckBell But you have external information which lets you narrow down the number of tables. If I flip a fair coin and $A$ says "It's $H$" and $B$ says "It's $T$" Then the only truth tables are $(T,F),(F,T)$.
$endgroup$
– lulu
Feb 9 at 11:13