unreality Posted June 10, 2009 Report Share Posted June 10, 2009 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? Quote Link to comment Share on other sites More sharing options...
0 Glycereine Posted June 10, 2009 Report Share Posted June 10, 2009 Y>X>Z>W Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 Z is the strongest Z>W>X>Y Quote Link to comment Share on other sites More sharing options...
0 unreality Posted June 10, 2009 Author Report Share Posted June 10, 2009 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! (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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 This one seems fairly simple. z>w w>x x>y therefore, z>w>x>y, as has been stated. I'd like to see the other option. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 the two possibilitiesso 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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 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! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 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 Quote Link to comment Share on other sites More sharing options...
0 hugemonkey Posted June 10, 2009 Report Share Posted June 10, 2009 (edited) 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? 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 June 10, 2009 by hugemonkey Quote Link to comment Share on other sites More sharing options...
0 unreality Posted June 10, 2009 Author Report Share Posted June 10, 2009 (edited) jmotley and hugemonkey: if you follow the link in the beginning of the topic you'll see the catch in the comparison. If the comparer is weaker than one or both of the wizards he/she is comparing, the result will be opposite Edited June 10, 2009 by unreality Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 I concur that YZWX Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 These were really fun. you should do more. YZWX Quote Link to comment Share on other sites More sharing options...
0 Glycereine Posted June 10, 2009 Report Share Posted June 10, 2009 These were really fun. you should do more. YZWX I too thoroughly enjoyed this! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 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 Quote Link to comment Share on other sites More sharing options...
0 Glycereine Posted June 10, 2009 Report Share Posted June 10, 2009 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 10, 2009 Report Share Posted June 10, 2009 (edited) 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 June 10, 2009 by Hdizzy Quote Link to comment Share on other sites More sharing options...
0 Glycereine Posted June 10, 2009 Report Share Posted June 10, 2009 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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 22, 2009 Report Share Posted June 22, 2009 (edited) Z is the strongest wizard Edited June 22, 2009 by rohit_bd Quote Link to comment Share on other sites More sharing options...
Question
unreality
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?
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