Burne's Riddles - Bug?

Discussion in 'The Temple of Elemental Evil' started by Endarire, May 26, 2005.

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  1. The Royal Canadian

    The Royal Canadian Established Member

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    Hi rt142
    You are correct Burne didn't say anything about who apprentices did or did not greet. However, if you assume that each apprentice traveled to the conclave with his/her master (a logical assumption to make) and you know that no master greeted his/her apprentice, then it is also logical to assume that no apprentice greeted his/her master.
     
  2. Allyx

    Allyx Master Crafter Global Moderator Supporter

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    I think you'll find, that Burne's riddles had nothing to do with Troika, it was an early Co8 mod which gave players the opportunity to destroy the golden skull without the need of a wizard or sorcerer who has chosen both fireball and gust of wind spells and scribe scroll feat.
     
  3. Necroticpus

    Necroticpus Cthulhu Ftaghn!

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    Did you guys change the answer in the latest version of your mod to 10, 11, 12.67, 3.142536 or anything else? I haven't played it yet.
     
  4. Allyx

    Allyx Master Crafter Global Moderator Supporter

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    I've not played it myself yet, but I strongly suspect it remains unchanged.
     
  5. Gaear

    Gaear Bastard Maestro Administrator

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    Nope, it's now 3.142536.

    zhuge should have called the quest 'Burne's Giggles' for the amount of consternation it seems to cause everybody.
     
  6. oudeis

    oudeis Member

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    So these riddles are part of the endgame? I've never seen them.
     
  7. Shiningted

    Shiningted I want my goat back Global Moderator

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    Making the answer a randomly determined number WOULD put the challenge back into it... :eyebrow:
     
  8. GuardianAngel82

    GuardianAngel82 Senior Member

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    Choosing a random number isn't a challenge. (Except for the Mersenne Twister) :p
     
  9. Dragonreborn567

    Dragonreborn567 Member

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    Apprentice 6 greets apprentice 1,2,3,4,5 and mage 1,2,3,4,5. (10)
    Apprentice 5 greets apprentice 1,2,3,4,6 and mage 1,2,3,4. (9)
    Apprentice 4 greets apprentice 1,2,3,5,6 and mage 1,2,3. (8)
    Apprentice 3 greets apprentice 1,2,4,5,6 and mage 1,2. (7)
    Apprentice 2 greets apprentice 1,3,4,5,6 and mage 1. (6)
    Apprentice 1 (Pishella) greets apprentice 2,3,4,5,6. (5)
    Mage 6 greets NO ONE. (0)
    Mage 5 greets apprentice 6. (1)
    Mage 4 greets apprentice 5,6. (2)
    Mage 3 greets apprentice 4,5,6. (3)
    Mage 2 greets apprentice 3,4,5,6. (4)
    Mage 1 (Burne) greets apprentice 2,3,4,5,6. (5)
    The question is not how many greetings were initiated, but rather how many were given. You return a greeting, once you are greeted. The quote "Of course it is common courtesy that anyone greeted would return the greeting", thus, you greet when you are greeted. "My question is how many people did Pishella greet?" NOT how many did she initiate the greeting with. Thus, the only possible layout of greetings is roughly what mine is. Pishella DOES however have to be the 5, because there are 11 answers, and 12 people. Thus, Burne has to be the other 5, so that the 11 people answering his question having unique answers works out.
    Also, Burne never did explicitly say that an apprentice couldn't greet their master, that is true. But, he did say a master does not greet their apprentice, and you return a greeting, therefore, an apprentice and master do not greet EACH OTHER. At all.
    Yes, it is a complex question. Yes, it is confusing. But YES, it does work out.
     
    Last edited: Dec 24, 2010
  10. Forgalz

    Forgalz Established Member

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    If this is a mod, might it be possible to alter the question text? I think only a very minor change would be required to make this crystal clear:
    Rationale for #1: There is obviously some confusion on whether mages would greet back their apprentices, or whether the second part of the sentence is supposed to explain that they are dispensed from exercising "common courtesy" towards their apprentices.

    Rationale for #2: Since Burne knows what he himself did, the "no one" could be read to apply to all 12 persons present, creating a very similar puzzle with no unique solution.
     
    Last edited: Dec 28, 2010
  11. Zombra

    Zombra Member

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    Yep, the problem is insoluble as presented, because of a bit of sloppy language. Forgalz is correct.

    "To my surprise, no one made the same number of greetings"

    should be changed to

    "To my surprise, no two people I asked made the same number of greetings".

    Whoops, I didn't even read the post fully before making the same correction in bold. :p

    Well anyway, the correction still hasn't been made since 2010. Hope this helps. :)
     
  12. Flop

    Flop Member

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    Sorry for resurrecting this thread, but I still don't understand the logic. To return to rtl42's argument, why does the total number of initiated greetings of each wizard-apprentice pair need to equal 10? It's obvious that Burne must have initiated 5 greetings, but I don't see why his apprentice must therefore also have initiated 5.
     
  13. Forgalz

    Forgalz Established Member

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    That is not quite how the logic flows here. After the analysis, it should be "obvious" (a bold word seeing the rest of this thread) that there is exactly one apprentice-master pair that exchanged 5-5 greetings AND nobody else exchanged 5 greetings (leave the "initiated" out of it, it does not matter who initated the greeting). Since Burne states that everybody ELSE (in the intended interpretation of his words) exchanged a DIFFERENT NUMBER of greetings, it follows that this pair must be HIM and Phishella (otherwise there would be two people among the others who exchanged the same number of greetings). Therefore, Phishella exchanged 5 greetings.
     
  14. Pygmy

    Pygmy Established Member Supporter

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    I am not at all sure that mage-apprentice pairs adding up 10 is absolutely necessary

    Consider the maximum no. of greetings initiated: 10
    Minimum no. of greetings initiated:0

    10,9,8,7,6,5,4,3,2,1,0 gives you slots for 10 respondents + Burne himself
    Pischella must have same number of greetings as one of the eleven

    Each incremental decrease corresponds to "9" not being able to greet "10" etc....

    The presence of mages and apprentices screws up the incremental decrease unless Pischella is also "5" such that if a mage is in position 10,9,8,7,6 or 5 their apprentice is in 5,4,3,2,1 or 0

    The point being that there are 6 places for the mage/apprentice above the correction and 6 places for their "partner" below

    All pairs must be "across" the repetiition such that correction is made to ensure that 10 - No. of greetings received = No. of greetings given

    The repetition at "5" adds 1 onto "greetings given" to compensate for the fact that "greetings received" has been reduced by 1.

    Obviously Burne is the only other person who can be "5" because there would be no decrement of "hand shakes given" between them
     
  15. Corwyn

    Corwyn Gnoll Pincushion

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    This is not a beginner logic puzzle, for sure. But neither is it as complicated and unclear as some may insist. (Sudoku, anyone?) RTL42's matrix and proposed counter example is wrong (in effect, he inserted an error into the puzzle and then proceeded to show there's an error in the puzzle). The puzzle is not flawed as he suggests, but like all logic puzzles, the author only gives the bare minimum info to solve it.

    Your question: Why does the total number of "initiated" (rtl42's inserted error) greetings of each wizard-apprentice pair need to equal 10?
    Answer: (First, remove "initiated") Because it's the only solution that satisfies all stated requirements of 1) no greeting own apprentice/master, and 2) every one reported to Burne a unique number of greetings. 3) a "greeting" is defined, by Burne, as a mutual exchange, given and reciprocated.

    Supposing Person X (whether master or apprentice, it doesn't matter) is the one with 10 greetings, which breaks down into 5 greetings with Masters plus 5 greetings with Apprentices other than his own master/apprentice --> this being the only way to total 10 greetings in a room of 12 people (not counting Self and own Apprentice/Master). So Person X is, and will remain, the only person in the room with the max possible 10 greetings.

    If Person X greeted all the Masters and Apprentices except his own, then his own is --and will remain-- the only person with zero greetings. One, and only one, person must have zero greetings. If the person with zero greetings were someone else's companion, then there wouldn't be 10 people for X to greet (this is the key to the puzzle). Thus Master X and Apprentice X have 10/0 greetings, in some order.

    [Wait for it .... If that makes sense to you, then you don't need to read the rest of this post, as the mental light bulb should now be bright and steady.]

    Continue the pattern and it apply it to the 9/1 pairing, and the 8/2, and so on.

    [Edit, I'll spell out the next step, just in case the light bulb is struggling.] The person (euphemistically labeled Person Y) who greets 9, greets everyone else in the room except for the aforementioned person with 0 (zero) greetings and their own traveling companion, who has 1 greeting (from Person X). Again, if the person with 1 greeting were someone else's companion, then there wouldn't be 9 people remaining for Y to greet.

    So, Person Y is --and will remain-- the only person with 9 greetings and his traveling companion is --and will remain-- the only person with only 1 greeting, as Everyone else now has at least 2 greetings (one each from Person X and Person Y).
    So, traveling companions Master Y and Apprentice Y have 9/1 greetings, in some order.

    And so on ....

    Conclusion: The only way to comply with the given clues of "no greeting own apprentice" and "everyone greets a different number" (other than Burne, of course), and "Greeting being a reciprocal exchange" results in each pair of traveling companions (Master/Apprentice) totaling 10 greetings.

    So, when we get to the 5/5 pairing, we know it must be Burne (since he is the only anomaly in the pattern of 'everyone greets a different number'), and we know he must be paired with Pischella because every other Master is already paired with their own Apprentice.
     
    Last edited: Dec 31, 2013
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