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I remember answering this one incorrectly and when the answer was explained to me I was annoyed with myself. Here is your chance, with the original word phrasing:

Jack is looking at Anne, but Anne is looking at George. Jack is married, but George is not. Is a married person looking at an unmarried person?

A) Yes
B) No
C) Cannot be determined

The first with the correct alternative with the correct logical explanation ... (you know the drill).

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    $\begingroup$ So, out of curiosity, how did you answer it, and what was your reasoning? $\endgroup$ – Dan Henderson Aug 19 '16 at 15:05
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    $\begingroup$ Wait, this question wasn't asked before on this site?? It's a famous puzzle... $\endgroup$ – Sid Aug 19 '16 at 16:43
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    $\begingroup$ It's worth noting that this puzzle assumes that a person can only be married or unmarried - that is, that they're complementary states. For instance, if one admits "divorced" as a separate category, then the answer is "C", as Anne may be divorced. $\endgroup$ – Glen O Aug 21 '16 at 5:25
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    $\begingroup$ @GlenO All divorced people are still either "married" or "unmarried", it just depends on whether or not they remarried since the divorced. If Anne is divorced the answer is still "A". She is either divorced and unmarried or divorced and married, but no matter what a married person is looking at an unmarried person. $\endgroup$ – Paul Aug 21 '16 at 18:20
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    $\begingroup$ This is getting silly. Stop tryinng to find loopholes. "What if Anne is neither married nor not married?". I say what if Anne is a bird and the concept of married does not apply? What if Jack is jack daniels, Anne is Queen Anne and George is George T. Stagg and the OP just drank all of them and is drunk when writing this question? Everyone understood the idea in this question. No need to try and dissect it. Just enjoy it and move on. Or simply move on. $\endgroup$ – Marius Aug 21 '16 at 18:28
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Answer is

YES.

because,

If Anne is married, she’s looking at George, who is unmarried. If Anne is unmarried, Jack is looking at her.

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    $\begingroup$ It also works for arbitrarily long chains of people, as long as the first is married and the last is not. For any binary string starting with 1 and ending in 0, there must be at least one 10 somewhere. $\endgroup$ – Clement C. Aug 19 '16 at 15:47
  • $\begingroup$ @ClementC. Of course, this is only definable at such (the fact that you can pick a place to start which is 1) as this system is a cycle. $\endgroup$ – Weckar E. Aug 22 '16 at 12:10
  • $\begingroup$ As a red-herring, but to create a full cycle: you could also have the last person looking at the first. $\endgroup$ – Paul Evans Sep 11 at 16:42
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Answer:

A) Yes.

Reasoning:

Case 1. Anne is married. Then Anne (married) is looking at George (not married).
Case 2. Anne is not married. Then Jack (married) is looking at Anne (not married).

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    $\begingroup$ In scenario 2 I think you mean anne is unmarried $\endgroup$ – Beastly Gerbil Aug 19 '16 at 13:49
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Answer:

Cannot be determined

Because:

There may be two Anne's, the first may be married and the second not.

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    $\begingroup$ Good point haha :) $\endgroup$ – Kevin Aug 19 '16 at 17:07
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    $\begingroup$ @Kevin Good if the question were tagged lateral thinking, not so much for logical deduction. $\endgroup$ – Mike Kellogg Aug 19 '16 at 21:21
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    $\begingroup$ Seems like it ought to be specified that there are only 3 people, no? $\endgroup$ – Joe Aug 19 '16 at 21:38
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    $\begingroup$ the language strongly suggest that it's the same Anne so you can't assume there are 2. $\endgroup$ – adhg Aug 21 '16 at 18:36
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    $\begingroup$ The term unmarried does not mean not married, there could be an ambiguous state in between with the right context. Jack may not be a person, but rather a robot. This question may be written in another language where words mean something completely different. Logic itself lies on unproven axioms, and without a proof of those axioms we cannot know it is consistent, and in an inconsistent system any statement is true. The world has billions of people, and certainly somewhere an unmarried person is looking at a married person. -1, you could do better. $\endgroup$ – Yakk Aug 22 '16 at 13:36
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Answer:

A) Yes

Because:

Say the symbol -> stands for 'is looking at'. So Jack -> Anne -> George. If we replace the names of Jack and George with either married of not married, we get married -> Anne -> not married. Since Anne can be married or not married, the possible cases are married -> married -> not married and married -> not married -> not married. In both possible cases a married person is looking at an unmarried person. So the right answer is A.

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Is the sequence [1,0] a subsequence of [A,B,C]? Where 1=married and 0=not.

[A,B,C] is either [1,1,0] or [1,0,0]. Both contain [1,0]. Yes.

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Proved the answer in coq:

Inductive person : Type :=
  | jack : person
  | anne : person
  | george : person.
Parameter married : person -> bool.
Parameter looking_at : person -> person -> bool.

Goal
  looking_at jack anne = true ->
  looking_at anne george = true ->
  married jack = true ->
  married george = false ->
  exists p q, married p = true /\ married q = false /\ looking_at p q = true.
Proof with auto.
  intros.
  destruct (married anne) eqn:H3.
    (* Anne is married *)
    exists anne, george...
    (* Anne is unmarried *)
    exists jack, anne...
Qed.

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    $\begingroup$ Please check it's all still aligned as you expect. $\endgroup$ – Jonathan Allan Aug 22 '16 at 3:53
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    $\begingroup$ @JonathanAllan Ah, nice. Had to escape some *s, but otherwise good. Thanks $\endgroup$ – Michael Mrozek Aug 22 '16 at 8:14
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The answer marked correct is correct to the spirit of the question. But since this is a puzzle, its fair to point out that there are cases where it might not be answerable.

This is a legal question and it requires that consistent laws apply to all participants to be unambiguously answerable. We don't know that's the case here.

Consider this scenario: Jack is in South Africa which recognizes his marriage and is looking at unmarried Anne who is standing a few feet away in Botswana where Jack's marriage is not recognized. In South Africa we have a married Jack looking at an unmarried Anne. As his gaze crosses the border the law changes and it becomes the gaze of an unmarried man on an unmarried woman.

You can keep the distracting gender/fidelity subtext and resolve the legal ambiguity by replacing "is married" with "is wearing a wedding ring".

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    $\begingroup$ I get your point here... but do look at the tags. it says 'logical deduction' not "lateral thinking" $\endgroup$ – Sid Aug 20 '16 at 13:16
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    $\begingroup$ But... the definition of what a wedding ring is may vary between South Africa and Botswana, where the two are standing :-p Point is, if you want to nitpick, you're pretty much making it difficult to answer period. Define looking, define person... define is (temporal distortion anyone!?!) Once in a while a little straightforward logic is a beautiful thing (even if I did manage to get it wrong, humbling isn't so bad!) $\endgroup$ – JeopardyTempest Aug 22 '16 at 7:40
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Jack(male, married) -> Anne(female, ?) -> George(male, unmarried). Where "->" means "is looking at"; "?" is one marital status among ("married" or "unmarried" or "civil partnership" or "married and then survived the spouse").

Anne is not determined, so it cannot be determined. If it were a Boolean problem, the answer would be "yes". However, real world problems are rarely binary.

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    $\begingroup$ Full marks for political correctness, but none for logic. Whatever other status they might have, everyone is either married or unmarried. Male or female makes no difference. $\endgroup$ – DJClayworth Aug 20 '16 at 4:06
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    $\begingroup$ Agree with DJClayworth, except I'm not sure it's full marks for political correctness, either. The sexes of Jack, Anne, and George are not given, only their names. $\endgroup$ – 6005 Aug 21 '16 at 9:12
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Answer: Yes. Because: George is the child of Jack and Anne. Married mother is looking at unmarried child.

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    $\begingroup$ This is not a lateral-thinking puzzle. What makes you think George is the child of the other two? $\endgroup$ – elias Aug 20 '16 at 1:27

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