Jump to content
BrainDen.com - Brain Teasers
  • 0


unreality
 Share

Question

20 answers to this question

Recommended Posts

  • 0
Y>X>Z>W
:)

there are actually two correct possibilities, one of which you listed. However, in both possibilities, Y is the strongest. Good job! :D (and there's nothing circular about it ;D)

Z is the strongest

Z>W>X>Y

Remember that Y compared Z and W and got Z>W.

However, in your scenario, Y is less than Z & W and should therefore say that W>Z instead of Z>W.

Link to comment
Share on other sites

  • 0

the two possibilities

so you have three steps A being that the answer is correct B being that it isnt if you assume for step one is correct then the next must be wrong and then you get to choose for the last one so

ABA

or ABB

yields y>z>w>x

and y>x>z>w

if you try B starting you get

BA then it is impossible because at this point you know w is the strongest but the third choices preclude that possibility

good one i had fun

Link to comment
Share on other sites

  • 0

Y is the strongest ... when u take this to be true and proceed, u will get

Y>Z>W and Y>Z>X ... compn between W and X not possible.

Now, for any other one to be the strongest, some or the other assumptions become false

Link to comment
Share on other sites

  • 0
http://brainden.com/forum/index.php?showtopic=8415

Using the same method in that post to compare wizardry strength, four wizards duel it out: W, X, Y and Z.

* Y compares Z,W and gets Z>W

* W compares X,Y and gets X>Y

* Z compares W,X and gets W>X

Who is the strongest wizard?

These requirements can be expressed as:

Y > Z > W or (W > Z and W > Y)

and

W > X > Y or (Y > X and Y > W)

and

Z > W > X or (X > W and X > Z)

- Assume Y > Z > W

- - Assume W > X > Y

- - - Then Y > Z > W > X > Y, inconsistent

- - Therefore Y > X and Y > W

- - - Assume Z > W > X

- - - - Then Y > Z > W > X, consistent

- - - Now assume X > W and X > Z

- - - - Then Y > X > Z > W, consistent

- Assume W > Z and W > Y

- - Assume W > X > Y

- - - Assume Z > W > X

- - - - Then W > Z > W, inconsistent

- - - Therefore X > W and X > Z

- - - - Then X > W > X, inconsistent

- - Therefore Y > X and Y > W

- - - Then W > Y > W, inconsistent

Thus, there are two (and only two) possible answers: Y > Z > W > X or Y > X > Z > W. Either way, Y is the strongest wizard. I probably didn't need to go through that whole tree, but I thought it laid out nicely in case anyone was curious.

Link to comment
Share on other sites

  • 0

There are 2 possible orders:

Y > Z > W > X

or

Y > X > Z > W

Both options matches with the 3 comparison rules.

However the question is who is the strongest, thus answer is:

Y is the Strongest!

Link to comment
Share on other sites

  • 0
http://brainden.com/forum/index.php?showtopic=8415

Using the same method in that post to compare wizardry strength, four wizards duel it out: W, X, Y and Z.

* Y compares Z,W and gets Z>W

* W compares X,Y and gets X>Y

* Z compares W,X and gets W>X

Who is the strongest wizard?

How do you get that Y is the strongest. The way it looks to me is that Z is the strongest and Y is the weakest.

Z>W

X>Y

W>X

So if X>Y and W>X that means that both X&W>Y Now show that Z>W that means that Z&W&X>Y

Link to comment
Share on other sites

  • 0
http://brainden.com/forum/index.php?showtopic=8415

Using the same method in that post to compare wizardry strength, four wizards duel it out: W, X, Y and Z.

* Y compares Z,W and gets Z>W

* W compares X,Y and gets X>Y

* Z compares W,X and gets W>X

Who is the strongest wizard?

Huh? :huh:

if Z is greater than W and W is greater than X and X is greater than Y then Z>W>X>Y.

So it would go:

1st strongest Z

2nd W

3rd X

and 4th Y

I am I not understanding something about this?

*edited: I am an idiot for not following the link that shows that the wizards would lie about the observed strength of others in certain situations.*

Edited by hugemonkey
Link to comment
Share on other sites

  • 0
How do you get that Y is the strongest. The way it looks to me is that Z is the strongest and Y is the weakest.

Z>W

X>Y

W>X

So if X>Y and W>X that means that both X&W>Y Now show that Z>W that means that Z&W&X>Y

From the first puzzle:

Called a "mind duel", the two wizards being compared stand in opposite circles and just stand there, staring at each other. A wizard stronger than both of them stands in between them and compares the strength of their magical fields to determine the stronger of the two.

If the wizard doing the comparison is stronger than both of them, the comparer correctly determines the stronger of the two dueling wizards.

However if the comparer is weaker than one or both of the dueling wizards, then they get the opposite answer; that is, they think that the weaker of the two wizards is actually the stronger one.

Link to comment
Share on other sites

  • 0

I seem to have come rather late to this debate.

It took me a couple of minutes to sort all the comparisons out.

The solution is already out there - but it was necessary to read the earlier problem.

Here I have another question. Is it possible to frame the problem slightly differently and either come up with a paradox that two or more wizards are "strongest" or to have a situation where all possible orders are ruled out?

^_^

Donjar

Link to comment
Share on other sites

  • 0
I seem to have come rather late to this debate.

It took me a couple of minutes to sort all the comparisons out.

The solution is already out there - but it was necessary to read the earlier problem.

Here I have another question. Is it possible to frame the problem slightly differently and either come up with a paradox that two or more wizards are "strongest" or to have a situation where all possible orders are ruled out?

^_^

Donjar

It's most definitely possible, but the idea was for it to be solvable.

Example:

A measures B, C result B

B measures C, D result C

C measures D, A result D

D measures A, B result A

No wizard can be the strongest, they are always contradicted.

Link to comment
Share on other sites

  • 0

I came up with three answers, but I can't figure out why my third one doesn't work (since everyone else seems to think only 2 possibilities work) . . .

(1) Y X Z W

(2) Y Z W X

(3) Y Z X W

Can anyone enlighten me? Thanks.

EDIT: nevermind - I see why the 3rd one doesn't work.

Edited by Hdizzy
Link to comment
Share on other sites

  • 0
I came up with three answers, but I can't figure out why my third one doesn't work (since everyone else seems to think only 2 possibilities work) . . .

Can anyone enlighten me? Thanks.

(1) Y X Z W

(2) Y Z W X

(3) Y Z X W

If Z was greater than X and W, then W has to be above X because Z himself said so :)

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...