Knights and Knaves question Announcing the arrival of Valued Associate #679: Cesar Manara ...
"Destructive force" carried by a B-52?
Unix AIX passing variable and arguments to expect and spawn
How to create a command for the "strange m" symbol in latex?
Is "ein Herz wie das meine" an antiquated or colloquial use of the possesive pronoun?
Putting Ant-Man on house arrest
When does Bran Stark remember Jamie pushing him?
Can this water damage be explained by lack of gutters and grading issues?
Coin Game with infinite paradox
Married in secret, can marital status in passport be changed at a later date?
How was Lagrange appointed professor of mathematics so early?
What came first? Venom as the movie or as the song?
Knights and Knaves question
lm and glm function in R
Will I be more secure with my own router behind my ISP's router?
Like totally amazing interchangeable sister outfit accessory swapping or whatever
How to mute a string and play another at the same time
Can a Wizard take the Magic Initiate feat and select spells from the Wizard list?
What is the evidence that custom checks in Northern Ireland are going to result in violence?
/bin/ls sorts differently than just ls
Can 'non' with gerundive mean both lack of obligation and negative obligation?
Should man-made satellites feature an intelligent inverted "cow catcher"?
Who's this lady in the war room?
What documents does someone with a long-term visa need to travel to another Schengen country?
Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?
Knights and Knaves question
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)About the island of Knights and KnavesAbout Knights and Knaves and their consistencyThe way to Acarien, with Knights and KnavesKnights , Knaves and Spies - Part 1Knights and knaves at a partyMeta Knights and Knaves Puzzle with HatsKnights, Knaves and Normals - the tough oneKnights Knaves and SpiesSolve the following knights and knaves problem
$begingroup$
You are on a land inhabited by Knights and Knaves. Knights will always tell the truth and knaves always lie.
You meet three inhabitants(A, B, and C), and ask how many of them are knaves. A answers so quietly, so you ask B what A had said. B says that A had said that exactly two of them were knaves. C says B is lying.
Is it possible to know what A is?
Further, what are B and C.
logical-deduction liars
New contributor
$endgroup$
add a comment |
$begingroup$
You are on a land inhabited by Knights and Knaves. Knights will always tell the truth and knaves always lie.
You meet three inhabitants(A, B, and C), and ask how many of them are knaves. A answers so quietly, so you ask B what A had said. B says that A had said that exactly two of them were knaves. C says B is lying.
Is it possible to know what A is?
Further, what are B and C.
logical-deduction liars
New contributor
$endgroup$
$begingroup$
Please don't change your question after someone has answered it. Now the answer references D, E, and F, and makes no sense. I'm editing the answer to match the revised question, but in the future it's best not to make changes that make the answers obsolete ... especially when the changes are superficial like this.
$endgroup$
– Rubio♦
31 mins ago
add a comment |
$begingroup$
You are on a land inhabited by Knights and Knaves. Knights will always tell the truth and knaves always lie.
You meet three inhabitants(A, B, and C), and ask how many of them are knaves. A answers so quietly, so you ask B what A had said. B says that A had said that exactly two of them were knaves. C says B is lying.
Is it possible to know what A is?
Further, what are B and C.
logical-deduction liars
New contributor
$endgroup$
You are on a land inhabited by Knights and Knaves. Knights will always tell the truth and knaves always lie.
You meet three inhabitants(A, B, and C), and ask how many of them are knaves. A answers so quietly, so you ask B what A had said. B says that A had said that exactly two of them were knaves. C says B is lying.
Is it possible to know what A is?
Further, what are B and C.
logical-deduction liars
logical-deduction liars
New contributor
New contributor
edited 1 hour ago
user58804
New contributor
asked 2 hours ago
user58804user58804
212
212
New contributor
New contributor
$begingroup$
Please don't change your question after someone has answered it. Now the answer references D, E, and F, and makes no sense. I'm editing the answer to match the revised question, but in the future it's best not to make changes that make the answers obsolete ... especially when the changes are superficial like this.
$endgroup$
– Rubio♦
31 mins ago
add a comment |
$begingroup$
Please don't change your question after someone has answered it. Now the answer references D, E, and F, and makes no sense. I'm editing the answer to match the revised question, but in the future it's best not to make changes that make the answers obsolete ... especially when the changes are superficial like this.
$endgroup$
– Rubio♦
31 mins ago
$begingroup$
Please don't change your question after someone has answered it. Now the answer references D, E, and F, and makes no sense. I'm editing the answer to match the revised question, but in the future it's best not to make changes that make the answers obsolete ... especially when the changes are superficial like this.
$endgroup$
– Rubio♦
31 mins ago
$begingroup$
Please don't change your question after someone has answered it. Now the answer references D, E, and F, and makes no sense. I'm editing the answer to match the revised question, but in the future it's best not to make changes that make the answers obsolete ... especially when the changes are superficial like this.
$endgroup$
– Rubio♦
31 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's not possible to know what A is.
C says B is lying. So either C is a knave and B is a knight or C is a knight and B is a knave. Therefore there is one knave among (B,C).
Assume A is a knave. Then there are 2 knaves and A would lie about it. Therefore B is lying about what A said, so A, B are knaves and C is a knight.
Assume A is a knight. Then there is only one knave and B is lying about what A said, so A, C are knights and B is a knave.
In both scenarios B is a knave and C is a knight. A could be either a knight or a knave.
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "559"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
user58804 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f82082%2fknights-and-knaves-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's not possible to know what A is.
C says B is lying. So either C is a knave and B is a knight or C is a knight and B is a knave. Therefore there is one knave among (B,C).
Assume A is a knave. Then there are 2 knaves and A would lie about it. Therefore B is lying about what A said, so A, B are knaves and C is a knight.
Assume A is a knight. Then there is only one knave and B is lying about what A said, so A, C are knights and B is a knave.
In both scenarios B is a knave and C is a knight. A could be either a knight or a knave.
$endgroup$
add a comment |
$begingroup$
It's not possible to know what A is.
C says B is lying. So either C is a knave and B is a knight or C is a knight and B is a knave. Therefore there is one knave among (B,C).
Assume A is a knave. Then there are 2 knaves and A would lie about it. Therefore B is lying about what A said, so A, B are knaves and C is a knight.
Assume A is a knight. Then there is only one knave and B is lying about what A said, so A, C are knights and B is a knave.
In both scenarios B is a knave and C is a knight. A could be either a knight or a knave.
$endgroup$
add a comment |
$begingroup$
It's not possible to know what A is.
C says B is lying. So either C is a knave and B is a knight or C is a knight and B is a knave. Therefore there is one knave among (B,C).
Assume A is a knave. Then there are 2 knaves and A would lie about it. Therefore B is lying about what A said, so A, B are knaves and C is a knight.
Assume A is a knight. Then there is only one knave and B is lying about what A said, so A, C are knights and B is a knave.
In both scenarios B is a knave and C is a knight. A could be either a knight or a knave.
$endgroup$
It's not possible to know what A is.
C says B is lying. So either C is a knave and B is a knight or C is a knight and B is a knave. Therefore there is one knave among (B,C).
Assume A is a knave. Then there are 2 knaves and A would lie about it. Therefore B is lying about what A said, so A, B are knaves and C is a knight.
Assume A is a knight. Then there is only one knave and B is lying about what A said, so A, C are knights and B is a knave.
In both scenarios B is a knave and C is a knight. A could be either a knight or a knave.
edited 29 mins ago
Rubio♦
30.8k567189
30.8k567189
answered 2 hours ago
JayJay
2,8742922
2,8742922
add a comment |
add a comment |
user58804 is a new contributor. Be nice, and check out our Code of Conduct.
user58804 is a new contributor. Be nice, and check out our Code of Conduct.
user58804 is a new contributor. Be nice, and check out our Code of Conduct.
user58804 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Puzzling Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f82082%2fknights-and-knaves-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Please don't change your question after someone has answered it. Now the answer references D, E, and F, and makes no sense. I'm editing the answer to match the revised question, but in the future it's best not to make changes that make the answers obsolete ... especially when the changes are superficial like this.
$endgroup$
– Rubio♦
31 mins ago