Burne's Riddles - Bug?

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

Remove all ads!
  1. Endarire

    Endarire Ronald Rynnwrathi

    Joined:
    Jan 7, 2004
    Messages:
    405
    Likes Received:
    30
    I have tried every answer in Burne's greeting puzzle about Priscilla and the 12 people yet I get a 'no' every time. Should this happen? Also, I do not see any scroll appear when I correctly answer the banner puzzle yet wrongly guess the greeting.

    -EE
     
  2. lexb123

    lexb123 Member

    Joined:
    May 18, 2005
    Messages:
    40
    Likes Received:
    0
    It works

    I forgot the answer.. I think it was 5? But it works. I did em both last game.
     
  3. Shiningted

    Shiningted I want my goat back Global Moderator

    Joined:
    Oct 23, 2004
    Messages:
    11,789
    Likes Received:
    37
    I calculated the first one but i swear i could not get that second one, didn't seem enough info to me... o well, i have met my match there! I was going to try reloading and doing every answer but it seemed cheap (and time-consuming ;-)
     
  4. taltamir

    taltamir Established Member

    Joined:
    May 16, 2004
    Messages:
    741
    Likes Received:
    0
    the second one uses somewhat... fuzzy logic...

    On my third theory on what it was i figured it out... I think it was 6 or 5.

    They explain it once you get it. Basically, burne mistakingly assumes that everyone should greet everyone. If they really did, then he would NOT have gotten different answers. If everyone greated everyone they did not know then everyone would have greated 11 people. (because they only knew their master). Its rather complex.
     
  5. khalamar

    khalamar Member

    Joined:
    Jun 28, 2004
    Messages:
    34
    Likes Received:
    0
    Burne's Riddles

    the answers are 1 , 5
     
  6. Steyr

    Steyr Member

    Joined:
    May 1, 2005
    Messages:
    39
    Likes Received:
    0
    The answer is 5.
    When you recall the rules that wizards would not greet their own apprentice, and if A greets B then B will not re-greet A, then you go around the circle and work out how many greetings were made by each unknown person. It works out that one person greets 10 people, the next person greets 9, then 8, 7, 6, 5, 5, 4, 3, 2, 1, 0. You'll notice how 5 appears twice in the list. That's an anomoly caused by wizards not greeting their own apprentices and is the crux of the solution. Because everybody that that was questioned greeted a different number of people, then the unknown answer must be the duplicate 5.
     
  7. taltamir

    taltamir Established Member

    Joined:
    May 16, 2004
    Messages:
    741
    Likes Received:
    0
    thats not exactly correct. because whenever a person greets you you greet BACK.

    The numbers you provide are of how many people INITIATED a greeting, ie, greeted first. But they should have all greeted 11 times.

    This is basic math here, but you can also achive it by counting.
    Lets say 3 people meet, person 1 greets person 2 and 3, who in turn greet him back. Person 2 greets person 3, he does not regreet person 1 because he already greeted him BACK. BUT, most people still count that as a greeting, saying that person 2 greeted two people, not only one person!
    Person 3, then initiates a greeting with noone, since he was already greeted by 1 and 2 and greeted them back, however, in the end, he greeted 2 people, not 0!
     
  8. Steyr

    Steyr Member

    Joined:
    May 1, 2005
    Messages:
    39
    Likes Received:
    0
    If that were the way the riddle worked then everybody would have greeted the same number of people. That is not how the riddle worked. It was explicitly stated in the riddle that everybody greeted a different number of people. The way I have described it is how the riddle works and how to reach the solution.
     
  9. taltamir

    taltamir Established Member

    Joined:
    May 16, 2004
    Messages:
    741
    Likes Received:
    0
    its not a matter of "how the riddle works" its a matter of which logic was employed by the author of the riddle, in game that would be burne.
    Him saying that each person greeted a different amount of people clues you in to the fact. I am guessing there might be more to it, but it could just be a flawed riddle.

    I am still looking for a coherent intresically logical explanation.
     
  10. Dragonreborn567

    Dragonreborn567 Member

    Joined:
    Jan 16, 2008
    Messages:
    8
    Likes Received:
    1
    I realise this is incredibley outdated, but I came across it and I felt I could clarify a few things, so here goes.

    "An annual conclave of mages is frequently held at Greyhawk's University of Magical Arts, just before the Desportium of Magic, to organise matters and exchange ideas. It is usually attended by representatives from Berbobonc, Dyvers and the surrounding lands but last year's affair was rather quiet. I brought Pishella along as well and it seems all of the other 5 mages who were present also brought their apprentices for exposure, so there were a dozen of us altogether that day. We greeted each other warmly and quickly settle down to discussion. After we had finished, out of curiosity I asked everyone (including Pishella) how many they had greeted. Of course it is common courtesy that anyone greeted would return the greeting and obviously no mage would greet his/her own apprentice. To my surprise, no one made the same number of greetings. My question is how many people did Pishella greet?" This is the Exact question That I screenied and copied out.

    Fact #1) common courtesy means everyone who was greeted returned the greeting.
    Fact #2) No mage would greet his/her own apprentice (not explicitly said, but hopefully obvious, no one greets themselves either).
    Fact #3) Each person greeted a different number of people.
    Fact #4) 12 people all together, 11 people answered Burne's question.
    From this, one can deduce everything needed to answer the question. Keep in mind that one person can only greet 10 people because they don't greet themselves, or their companions. (E.g. Burne doesn't greet himself or Pishella, but could still greet the other 5 wizards and the 5 other apprentices, although

    "There are 12 conclave members comprising of 6 mage-apprentice pairs. You asked 11 people and all gave different answers. This could happen only if someone did not greet anyone at all, someone made 1 greeting, another made 3 and so on till the person who made the maximum of 10 greetings. The person who gave 10 greetings obviously greeted everyone except his own companion, so the person who did not greet anyone must be the companion of the person who gave 10 greetings, since everyone else exchanged greetings with him. Similarly, the person who gave 9 greetings would have a companion who only exchanged 1 greeting and so on. Therefore we would have pairs with 8 and 2 greetings, 7 and 3 greetings, 6 and 4 greetings and 5 and 5 greetings respectively. However since nobody made the same number of greetings, the pair who made 5 and 5 greetings must have been you and Pishella (the only one you didn't ask the question was yourself). Therefore, Pishella exchanged 5 greetings with the others." The exact answer, again screenied and copied out.

    If you take the time to read the question, you should have been able to get the answer after thinking it through, or for the lazy people who just kept guess-checking or looking up the answer, the answer your character gives should have made it clear. the only thing they didn't specifically say was that a person doesn't greet themself, but that's rather obvious. Taltamir is wrong in that Burne never says everyone should greet everyone. He says everyone returns a greeting. Also, they would only greet 10 times, not 11, because they only greet people who are not in their pair (I.E. themself and their companion), and 12-2=10. Also, your analogy doesn't work because of what Steyr told you. The "wizards not greeting their own apprentices is the crux of the solution" means that saying "three people greet each other, so everyone greets 2 people" kind of doesn't really mean anything. And Endarire, you might not be able to get the answer because maybe your wizard doesn't have high enough intelligence. I believe that's how you get the answer option, not just correctly guessing the number. Again, sorry to rehash old posts, but This is when I came across it, so... Anyways, hope I cleared this up for everone.
     
  11. rtl42

    rtl42 Member

    Joined:
    Sep 12, 2009
    Messages:
    14
    Likes Received:
    0
    Just as a warning, this post is quite long, so if you're not interested in an in-depth explanation, just skip to the very bottom.

    I think this is the part of the riddle that is flawed, and I would like to post the following as a counterexample.

    Let's go over the assumptions in the riddle given or implied by Burne:
    (I) No master-apprentice pair greets each other.

    (II) Everyone exchanged greetings, regardless of whether they initiated the greeting or not, and did so only once; i.e. in any table/matrix representing the greetings of those gathered, all entries are defined, and can only be 0 ("a greeting was not initiated") or 1 ("a greeting was initiated").

    (III) When Burne asked everyone else at the gathering how many greetings they had each initiated, no two people initiated the same number of greetings; i.e. if we list the total number of greetings initiated by each person at the gathering, and then exclude Burne's "total" value, then in the remaining set, there are no repetitions.

    In his own wording, Burne messes up (I) by only saying "obviously no master greets his own apprentice". If we were to follow what Burne said, then strictly speaking, the maximum possible number of total initiated greetings is 11, not 10 (as is described in the in-game solution). However, by using (I), we can then derive that the minimum and maximum possible number of initiated greetings are 0 and 10, respectively.
    Therefore, if we consider everyone except Burne, then by the pigeonhole principle, there is only one person who initiated 0 greetings, only one person who initiated 1 greeting, ..., and only one person who initiated 10 greetings. And this is the idea which the in-game solution (mostly) alludes to.

    Now, here's my proposed counterexample: suppose you are given the following table/matrix that indicates which person initiated a greeting with someone else:

    Code:
         M_1  A_1  M_2  A_2  M_3  A_3  M_4  A_4  M_5  A_5  M_6  A_6    Total
    M_1   0    0    0    1    0    1    0    1    0    0    0    0       3
    A_1   0    0    0    1    1    1    1    1    1    1    1    1       9
    M_2   1    1    0    0    1    1    1    1    1    1    1    1       10
    A_2   0    0    0    0    0    1    0    1    0    0    0    0       2
    M_3   1    0    0    1    0    0    1    1    1    1    1    1       8
    A_3   0    0    0    0    0    0    0    0    0    0    0    0       0
    M_4   1    0    0    1    0    1    0    0    1    1    1    1       7
    A_4   0    0    0    0    0    1    0    0    0    0    0    0       1
    M_5   1    0    0    1    0    1    0    1    0    0    0    0       4
    A_5   1    0    0    1    0    1    0    1    0    0    1    0       5
    M_6   1    0    0    1    0    1    0    1    1    0    0    0       5
    A_6   1    0    0    1    0    1    0    1    1    1    0    0       6
    The matrix is to be read as follows:
    • M_i, A_i represent the i-th master, apprentice (respectively)
    • Entries with a 1 represent the statement: "M_i/A_i (row) initiated a greeting with M_j/A_j (column)"
    • Entries with a 0 represent the statement: "M_i/A_i (row) was greeted by M_j/A_j (column)" (strictly speaking, it should be "...did not initiate a greeting with...", but by assumption (II), this is equivalent to "...was greeted by...")
    • The Total column shows the total number of greetings initiated by M_i/A_i (row)

    So for example, the entry in row 1 (M_1), column 8 (A_4) means "Master 1 initiated a greeting with Apprentice 4"; likewise, an entry like row 12 (A_6), column 2 (A_1) means "Apprentice 6 was greeted by Apprentice 1".
    If you check the corresonding diagonally "opposite" entry, you'll see that Apprentice 4 was greeted by Master 1, and Apprentice 1 did indeed greet Apprentice 6.
    And as an example of the Total column, we find that the Total of A_6 = 6 because A_6 initiated 6 greetings (with M_1, A_2, A_3, A_4, M_5, and A_5).

    With that said, let's additionally suppose that Burne is labelled M_6. Then:
    • Assumptions (I) and (II) are clearly satisfied (no companions interact with each other; all entries are defined, and are either 0 or 1).
    • If we look at the Total values for everyone except Burne (M_6), then each number from 0 to 10 is listed only once, so assumption (III) is satisfied!
    • Prishella is A_6.
    • Prishella initiated 6 greetings!

    Alternately, re-name M_5 -> A_5, re-name A_5 -> M_5, and suppose Burne is (this "new") M_5; then, again, there are no contradictions with assumptions (I)-(III), and this time, Prishella initiated 4 greetings!

    In other words, the statement:
    is a falsely-made deduction. What it's trying to say is: "if P is the person who made 10 greetings, and Q is his companion, then if Q made at least one greeting, there is a contradiction to one of our assumptions"; however, in my counterexample, we can see that although M_2 is the one who made 10 greetings, the fact that his companion (A_2) made 2 greetings leads to no contradictions. Thus, one cannot logically deduce the above-quoted statement: it must be assumed, if it is to be considered any further.

    In fact, what the in-game solution does is assume the following particular interaction/greeting matrix:
    If person P has a total # of greetings X, then his companion Q has a total of Y = 10 - X
    Given this, it is trivial to see that Pishella would have initiated 5 greetings.

    =======================================
    tl;dr
    In conclusion, there is no unique answer, and imho, it's highly likely that Prishella's total value could be constructed to be any number between 0 and 10.

    Unfortunately, I'm not sure what there is to be done about this. I haven't thought about how to fix the riddle so that "5" is still the correct answer, although I'm not sure it's possible without making a "trivializing" assumption.
     
    Last edited: Sep 14, 2009
  12. The Royal Canadian

    The Royal Canadian Established Member

    Joined:
    Oct 29, 2005
    Messages:
    741
    Likes Received:
    2
    Hi RT142
    I hate to say this, but I think you are totally wrong here is why
    1.) There is a total of 12 people in the room ( 6 mages plus 6 apprentices )
    2.) As Burne says, no Mage would greet his/her apprentice, and no apprentice would greet his/her master.
    3.) You can also assume that nobody (unless insane) would greet himself
    Therefore the maximum number of people a mage or mage's apprentice could greet is 10. For sake of simplicity the equation for the maximum number of people a mage could greet can be written as follows: 12 - 1 (mage) - 1 (mage's apprentice) = 10
    TRC
     
  13. rtl42

    rtl42 Member

    Joined:
    Sep 12, 2009
    Messages:
    14
    Likes Received:
    0
    ???

    Burne never said "no apprentice greets his master", he only said "no master greets his own apprentice": that's why i made the comment about Burne not giving correct info.

    the exact wording of the riddle is quoted in the post above my previous one, in case you're curious.

    i suppose i didn't mention self-greeting, but i guess you could it include as, like, "assumption 0" since it's kind of implicit in the definition of the word "greeting".

    anyways, i didn't just say "the maximum number is not 10", i said "the maximum number is 10 assuming masters and apprentices do not greet each other". so i don't see how i'm "totally wrong", could you explain what you mean?
     
    Last edited: Sep 15, 2009
  14. Necroticpus

    Necroticpus Cthulhu Ftaghn!

    Joined:
    Jul 2, 2009
    Messages:
    1,379
    Likes Received:
    0
    Anyone who is arguing for no significant reason is about as wrong as one could be without them being twins. The answer IS 5. These are the facts and they are undisputed. The game accepts the answer 5 and not 10 or 11 or 12.975 or 3.142536... This should not be that difficult to understand. No matter what you think the answer is, you are wrong. It is 5 and will never be anything else as Troika is defunct and Atari is not going to put out anymore ToEE expansions or anything. Use your youthful exuberence and love of life to actually go out-of-ddors and experience life. You'll be a much happier person. ;)
     
  15. rtl42

    rtl42 Member

    Joined:
    Sep 12, 2009
    Messages:
    14
    Likes Received:
    0
    i can't tell if you're serious or not, the mix-up is too strong.
     
Our Host!