Lateral Thinking VI and VII (very difficult)

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Holmes
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Lateral Thinking VI and VII (very difficult)

Post by Holmes »

haha, here are two Lateral Thinking problems, really hard, not kidding, REALLY HARD.

Let's see...


Lateral Thinking VI

Of three men, one always tells the truth, one always tells lies, and one answers "yes" or "no" randomly. Each man knows which one each of the others are. You may ask three yes/no questions, each of which may only be answered by one of the three men, after which you must be able to identify which man is which. How can you do it?



Lateral Thinking VII

You and your spouse invite four other couples to a party. During the course of the conversation, it is discovered that, prior to the party, each person except you was acquainted with a different number of the people present. Assuming the acquaintance relationship is symmetric (i.e., if you are acquainted with someone, that person is also acquainted with you), then how many people did your spouse know prior to the party? How many people did you know?


Good Luck! However, I'm sure you can get it.
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Re: Lateral Thinking VI and VII (very difficult)

Post by ccppfan »

OMG this is way to difficult....  :o

Oh well.... SAYUMI!!! GET HERE!!! NOOOOW!!!

Just kidding. ;)
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Re: Lateral Thinking VI and VII (very difficult)

Post by sstimson »

Holmes wrote:haha, here are two Lateral Thinking problems, really hard, not kidding, REALLY HARD.

Let's see...


Lateral Thinking VI

Of three men, one always tells the truth, one always tells lies, and one answers "yes" or "no" randomly. Each man knows which one each of the others are. You may ask three yes/no questions, each of which may only be answered by one of the three men, after which you must be able to identify which man is which. How can you do it?



Lateral Thinking VII

You and your spouse invite four other couples to a party. During the course of the conversation, it is discovered that, prior to the party, each person except you was acquainted with a different number of the people present. Assuming the acquaintance relationship is symmetric (i.e., if you are acquainted with someone, that person is also acquainted with you), then how many people did your spouse know prior to the party? How many people did you know?


Good Luck! However, I'm sure you can get it.


Number 7
[spoiler]Start like this. There are 5 couples. Say they are Aa Bb Cc Dd Ee with you as A. Your wife at minimum knows 5 ( you and 4 others ) the maximum any one could know is 9. more on this later [/spoiler]

Later
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Re: Lateral Thinking VI and VII (very difficult)

Post by Holmes »

Good start sstimson, but be careful, numbre 7 is tricky in a way.
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Re: Lateral Thinking VI and VII (very difficult)

Post by ayw »

Holmes wrote:
Lateral Thinking VII

You and your spouse invite four other couples to a party. During the course of the conversation, it is discovered that, prior to the party, each person except you was acquainted with a different number of the people present. Assuming the acquaintance relationship is symmetric (i.e., if you are acquainted with someone, that person is also acquainted with you), then how many people did your spouse know prior to the party? How many people did you know?

Good Luck! However, I'm sure you can get it.


Let me do VII first. I found two approaches.

[spoiler]
The situation: The guests and your spouse make up nine people who are acquainted with a unique number of people, ranging from 1 to 9. I have included partners as acquaintances, although the arguments below still hold if I preclude them, in which case we would be talking about the range 0 to 8 for the number of acquaintances. "You", the host, have an equal number of acquaintances as one of the guests. What is that number?

The table: The following table labels everyone present with Letters A to J and shows all combination of acquaintances. "a" denotes acquaintance, "p" partner (who is also an acquaintance), "h" hosts, and the number on the right diagonal is the number of acquaintances the person in that row/column has. It is the only distribution of acquaintances possible, because the labels are interchangeable and we are initially not assuming anything about who is who's partner.

Determining "you": Filling in the table with "a"s from the top with person A having 9 acquaintances, person B having 8, an so on, we arrive at persons E and F having five acquaintances each. This is unavoidable, hence you, the host, must be either E or F, and be acquainted with exactly 5 people, including your spouse.

Determining your spouse: J is acquainted with only one person, A. Therefore A is J's spouse. I is acquainted with two people, A and B, but since A is already taken, B must be I's spouse. Similarly, we can determine the other pairings, H-C, G-D, and find that the host pair is E-F, and both are acquainted with exactly 5 people including themselves. Changing the respective "a"s to "p"s and "h"s results in the table below.

[tt]    |  J |  I |  H |  G |  F |  E |  D |  C |  B |  A |
----+----+----+----+----+----+----+----+----+----+----+
  A |  p |  a |  a |  a |  a |  a |  a |  a |  a |  9
----+----+----+----+----+----+----+----+----+----+
  B |    |  p |  a |  a |  a |  a |  a |  a |  8
----+----+----+----+----+----+----+----+----+
  C |    |    |  p |  a |  a |  a |  a |  7
----+----+----+----+----+----+----+----+
  D |    |    |    |  p |  a |  a |  6
----+----+----+----+----+----+----+
  E |    |    |    |    |  h |  5
----+----+----+----+----+----+
  F |    |    |    |    |  5
----+----+----+----+----+
  G |    |    |    |  4
----+----+----+----+
  H |    |    |  3
----+----+----+
  I |    |  2
----+----+
  J |  1
----+[/tt]

[/spoiler]

Another way of arriving at how many acquaintances "you" as the host have is....

[spoiler]
...the following iterative sequence of arguments, where N=10 and n=1,2,3,...,9:

(This sentence will seem a bit obscure...)
If there were two people having N-n acquaintances, there would be exactly N-n+1 people having at least n+1 acquaintances. That is not possible, because exactly n people have at most n acquaintances.

(But the following sentences should make more sense...)
n=1: If there were two people having nine acquaintances, there would be exactly ten people having at least two acquaintances. That is not possible, because exactly one person has at most one acquaintance.

n=2: If there were two people having eight acquaintances, there would be exactly nine people having at least three acquaintances. That is not possible, because exactly two people have at most two acquaintances.
[tt]    .
    .
    .[/tt]
Only for n=5 the logic does not apply:
If there were two people having five acquaintances, there would be exactly six people having at least five acquaintances. That is possible because two of the six people that have at most five acquaintances, have exactly 5 acquaintances (the supposition in previous sentence), leaving four people with at most four acquaintances (which is ok - cf. consequence in previous sentence).

For n<5 the sequence continues as before.

Therefore, the only number of acquaintances that can be doubly filled is 5, hence "you", the host, have 5 acquaintances. To find how many acquaintances your spouse has, proceed as in the first solution.
[/spoiler]
Last edited by ayw on February 22nd, 2009, 11:53 am, edited 1 time in total.
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Re: Lateral Thinking VI and VII (very difficult)

Post by kat1214young »

Why not just permute the problem? Use permutation to solve the number of acquaintances? I'm too lazy to do it right now... XDDD kidding ;D
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Re: Lateral Thinking VI and VII (very difficult)

Post by ayw »

>>sstimson
sstimson wrote:Number 7
[spoiler]Start like this. There are 5 couples. Say they are Aa Bb Cc Dd Ee with you as A. Your wife at minimum knows 5 ( you and 4 others ) the maximum any one could know is 9. more on this later [/spoiler]

Later


You cannot assume that your spouse knows at minimum 5, since it may have been you who invited all the other couples. You might impose the condition that the combined number of acquaintances between you and your spouse must be at least 5, but  this condition is not needed to solve this problem.

>>kat1214
kat1214young wrote:Why not just permute the problem? Use permutation to solve the number of acquaintances? I'm too lazy to do it right now... XDDD kidding ;D

If you try all permutations you will see that every permutation leads to the same result. The permutations are degenerate because when we label everyone with e.g. letters, and assign the acquaintance status between two labels, we don't really care which label corresponds to how many acquaintances the person with that label has, as long as, in the end, each person except you has a different number (1... 9) of acquaintances.
 
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Re: Lateral Thinking VI and VII (very difficult)

Post by Holmes »

ayw, you don´t stop amzing me, your answer with the table explains the problem briefly, but concrete, it can be understood very well, good job!  :D


Now, let´s see if Lateral Thinking VI can be resolved ([spoiler]of course[/spoiler]), I like this one, because it´s not the traditional " a man always tells lies and the other the truth".
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Re: Lateral Thinking VI and VII (very difficult)

Post by Sayumi »

I'll work on VI on Wednesday

btw good job ayw- let me guess- your good at maths???
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Re: Lateral Thinking VI and VII (very difficult)

Post by ayw »

@Holmes & Sayumi, thanks  :)  yes, i like maths a lot; in fact it's essential in my work. That's why I found VII a bit easier than VI, though they're both fiendishly hard - as Holmes had warned.

I think I cracked VI now too - It took me the best part of the day, during which I should have been working!  :-X. Below are the questions that should do the trick. But I won't reveal the reasoning yet.

Holmes, these two problems are really great! Thanks so much again! I'll test my fellow astronomers with them  ;D

Here we go - I hope I got this right:
Definitions:
[spoiler]Persons A, B, and C. Let us call the one who always lies a person of type L. The person who sometimes lies and sometimes says the truth is of type M. Finally, the person who is always truthful is of type T. As is often the case in these sorts of problems, it doesn't matter to whom we address the first question. All further questions and deductions follow accordingly, and in the end, the labels are interchangeable anyway. Also, to make it easier to pose the first question, let us define an honesty ranking: Type T is more honest than Type M and Type L, and Type M is more honest than Type L.[/spoiler]
Q1.
[spoiler]Ask A: "Is B more honest than C?"
An equivalent question without referring to the honesty ranking is to ask: "Is either 'B of type T and C of type M', or 'B of type M and C of type L'?"[/spoiler]
Q2.
[spoiler]If you got a "yes" from A then ask the following question to C, otherwise ask it to B: "Did A just answer 'yes'?"[/spoiler]
Q3.
[spoiler]Ask the same person you asked Q2: "Is A of type M?"[/spoiler]

These questions are probably not a unique solution. There may be other ones one could use, though I wonder if they are fundamentally different. Even just having the questions above, it may be fun to figure out how to determine what types A, B, and C are.
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Re: Lateral Thinking VI and VII (very difficult)

Post by ayw »

The reasoning to the above solution
[spoiler]
Why such a first question? This question will enable you to ask the next question to a person who is definitely not of type M. Suppose A answers "yes" and the statement that B is more honest than C is true. Then C must be of type L, because A and B must be of either type M or T. Suppose A answers "yes" and the statement is false, that is, in truth B is less honest than C. Then C must be of type T, because A and B must be of either type M or L. So, if you get the answer "yes", then C is either of type T or L and can be addressed with the next question. Applying the same logic to the cases where A answers "no" leads to the conclusion that it is B who is either of type T or L, and who should be the recipient of the next question.

It's easy to determine whether someone is of type T or L. Just ask them a question like "Is Mt Fuji in Japan?" or "Did A just answer 'yes'"?

Now that you've determined whether that person is of type T or L, ask that same person what type one of the others is, e.g., "Is A of type M?"

So, after three questions you know the types of all three persons.
[/spoiler]
Last edited by ayw on February 25th, 2009, 6:15 am, edited 1 time in total.
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Re: Lateral Thinking VI and VII (very difficult)

Post by S.H. »

I'm sorry... But I cant get how will the questions of ayw know who is who..
Because if I answered it...This is what i got..
1. If yes(A is T,L,M)      No(A is T,L,M)


2    (If 1 is "yes")    If Yes(C is T,M)    If No(C is L,M)
      (If 1 is "no")      If Yes(B is L,M)     If No(B is T,M)


3  If Yes(A is T,L,M)    No(A is T,L,M)


List of Combinations..

A.  A is T,L,M; C is T,M; B is T,L,M
B.  A is T,L,M; C is L,M; B is T,L,M
C.  A is T,L,M; B is L,M; C is T,L,M
D.  A is T,L,M; B is T,M; C is T,L,M


Sorry...But I still cant get it even with the reasoning of ayw

Can someone explain how will it work? thx ^^ 
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Re: Lateral Thinking VI and VII (very difficult)

Post by Sayumi »

Okay, as promised I did LT VI today (it actually got me through all of politics and half of maths on my way to the bus stop I finally got an idea for the last step and tried it out while I was sitting in the bus-I think the guy sittung next to me thought I was weird-being so enthusiastic about maths :D) Anyway, I really liked this one, it's definitely my favorite LT out of the ones I've done.
It's a bit different from ayw's one but the general idea is similar
[spoiler]
1.Same as ayw: Ask A: Is B more honest than C?
2. Go to the person who is less honest!!!-You will always end up going to either L or T (T will send you L, not M; L will send you to T, not M and M can only send you to T or L) and ask a random question (like is 1+1=2 ?) to find out if he is T or L.
3. Ask the same person: Is A more honest that B/C (whoever you are not talking to at the moment). Revers he answer if you are talking to L. Now you've got everything you need. You know either T or L from step 2 and who is more honest out of the other two.

It took me quite a long time because I always processed to the more honest one after 1. and couldn't get M out of the equation... (I think I must have tried 4 or 5 different questions for 2 and actually wrote down all of the possible answers....)
[/spoiler]
I haven't actually read ayw's full answer yet- I'll do that now. I just really wanted to post this  ;D

EDIT: Okay I just read ayw's answer and we've actually got the same one (seems to happen a lot  8) ) except for having slightly different questions for 2 and 3. The long explantion (okay it wasn't really long) with lots of colorful ABCLTMs confused me for a minute into thinking it was anotherway of solving this.
Last edited by Sayumi on February 25th, 2009, 8:23 am, edited 1 time in total.
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Re: Lateral Thinking VI and VII (very difficult)

Post by Holmes »

I read all the answers and I think they are very good reasonings and are ok.

Here is the "correct" answer of the problem:

[spoiler]There are six possible scenarios. Let's call the first man A, the second man B, and the third man C. The six scenarios, then, are:

Scenario A B C
I  Truthteller -  Liar -  Random Man
II  Truthteller-  Random Man -  Liar
III  Liar - Truthteller -  Random Man
IV  Liar - Random Man -  Truthteller
V  Random Man -  Truthteller -  Liar
VI  Random Man -  Liar -  Truthteller

Follow these steps to determine which possibility listed above is correct:

1) Ask A, "Is B more likely to tell the truth than C?"
If yes, go to step 2.
If no, go to step 5.
2) Ask C, "Are you the random man?"
If yes, go to step 3.
If no, go to step 4.
3) Ask C, "Is A the truthteller?"
If yes, then scenario V is the case.
If no, then scenario II is the case.
4) Ask C, "Is A the liar?"
If yes, then scenario IV is the case.
If no, then scenario VI is the case.
5) Ask B, "Are you the random man?"
If yes, go to step 6.
If no, go to step 7.
6) Ask B, "Is A the truthteller?"
If yes, then scenario VI is the case.
If no, then scenario I is the case.
7) Ask B, "Is A the liar?"
If yes, then scenario III is the case.
If no, then scenario V is the case.
[/spoiler]
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Re: Lateral Thinking VI and VII (very difficult)

Post by ayw »

Sayumi's answer, mine, and Holmes' "correct" one are essentially the same, though Sayumi's explanation is clearest. Yeah, the colours in my post don't really help  :-\

@S. H. It is not important what A might be after the first question, but that the next question goes to someone who is either T or L.

A will either say the truth or lie. For the moment it's not important what type A is. Ask the first question and if the answer is...

    yes and A said the truth then (C is L)
    yes and A lied then (C is T)

    no and A said the truth then (B is L)
    no and A lied then (B is T)

So, if the answer is yes, then ask the next question to C. If it is no then ask the next question to B. In either case, you will ask the next question to someone who is either T or L, and you can easily determine with one question which (e.g. "Is 1+1=2?")
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