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Topic: NEW PROBLEM: The Operators of Boolania (Read 3450 times) |
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Eric Yeh
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NEW PROBLEM: The Operators of Boolania
« on: Aug 19th, 2002, 9:49pm » |
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Ok, now that the Gibberland posts are pretty much dying down, here is a new problem! As promised, this one is a bit of a breather -- another one I pulled out of my '98 collection. This is the kind that Jon's computer can wail away on, so I highly recommend doing it without computer aid, for the fun of it. Another perspective for those who find this one too easy: Consider it a test of elegance.  Don't worry though, I still have some tougher ones for sometime down the line...
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THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA THE OPERATORS OF BOOLANIA
There are three omniscient gods sitting in a chamber: And, Or, and Xor. They all answer the truth, but they apply their namesake operations in one of the following ways:
CASE 1 "Backward". The operator is applied to all of the questions that have been asked thus far.
CASE 2 "Universal". The operator is applied to ALL of the questions (you have to pick your questions beforehand).
CASE 3 "Selfless". The operator is applied to all of the questions NOT asked to them (you have to pick your questions beforehand).
For each of the three cases:
Determine with proof the minimum number of questions which will allow you to identify which god is which.
[Notes:
Standard: (Rules that are generally assumed unless otherwise noted.) The gods only answer yes/no questions. Each god answers in the single word of their language as appropriate to the question; i.e. each god always gives one of only two possible responses, one affirmative and one negative (e.g. they would always answer "Yes" rather than "That would be true"). Each question asked must be addressed to a single specific god; asking one question to all the gods would constitute three questions. Asking a single god multiple questions is permissible. The question you choose to ask and the god you choose to address may be dynamically chosen based on the answers to previous questions.
Specific: Because of possible time conflicts, you must determine your questions ahead of time, rather than based on previous answers. However, you are still allowed to choose who you ask each of your three questions to dynamically. Scoping is also dynamic; e.g. the pronoun "you" in a question will always refer to the person to whom you are currently asking a question, not a predetermined person). No time related questions (e.g., "if the answer to my second question was 'no', then X otherwise Y") are permissible, as this could lead to paradoxes within the space-time continuum).]
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As usual, I request that no strong spoilers be posted! Thanks!!!
Happy Puzzling!
Eric
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| « Last Edit: Aug 19th, 2002, 9:55pm by Eric Yeh » |
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #1 on: Aug 19th, 2002, 9:57pm » |
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HEY!! How come GLOW doesn't work on my titles any more??!!! 
(And with my luck, it'll soon be fixed, and everyone reading this post will have no idea what I'm talking about... <SIGH>)
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| « Last Edit: Aug 19th, 2002, 9:58pm by Eric Yeh » |
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william wu
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #2 on: Aug 19th, 2002, 10:25pm » |
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might be clearer to say "but they *all* apply their namesake operations in one of the following ways", so the reader doesn't think that each god could be using a different operation application style
very cool riddles, as always 
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #3 on: Aug 20th, 2002, 5:18am » |
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Will,
Thanks for the commentary and clarification. Yes, the glow seems to be working fine for me now as well.
Best,
Eric
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Jeremiah Smith
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #4 on: Aug 21st, 2002, 10:08pm » |
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Mr. Yeh, you're quite the puzzle-maker. I only wish I had your creativity.
Now, how do the operations work with more than 2 questions?
For example, if you ask 4 questions, how would Xor respond Universally, and the like?
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #5 on: Aug 22nd, 2002, 6:35am » |
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Jeremiah,
Many thanks for the warm comments!! They are very much appreciated.
If you have four questions which a strictly truth-telling (i.e. non-operating) Xor would answer with Q1, Q2, Q3, and Q4, the "operating" Xor would answer Q1 xor Q2 xor Q3 xor Q4, i.e. "false" iff there are an even number of "true"s in the Qi.
(For "selfless" operation, e.g., if you ask him Q2, he will answer Q1 xor Q3 xor Q4.)
Hope that clarifies the problem!
Best,
Eric
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Jonathan_the_Red
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #6 on: Aug 28th, 2002, 5:38pm » |
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Seeing as how this one is languishing, I'll take a stab...
You obviously can't do better than three questions. Therefore a solution that does it in three needs no proof of minimality.
Here's a solution for case 1: Backward. Hidden by color.
And = A
Or = O
Xor = X
Question 1: [to God A]: Are you Xor?
Question 2: [to God B]: Is 2+2 equal to 4?
Question 3: [to God B if answer to Q2 is Y, God C otherwise] Is And to the left of Or?
Answer YYY: Gods are X,O,A
Answer YNY: Gods are X,A,O
YYN and YNN are not possible (Q3 will be asked of Or in either case)
The remaining cases depend on the truth table of the ternary XOR operator. Going under the assumption that XOR(A, B, C) is true iff an odd number of (A, B, C) are true, they are:
Answer NYY: Gods are A, X, O
Answer NYN: Gods are O, X, A
Answer NNY: Gods are A, O, X
Answer NNN: Gods are O, A, X
If instead A XOR B XOR C is true iff ( (A AND B AND C) OR (~A AND ~B AND ~C) ), just reverse A and O in the above four answers.
I haven't really tried the other two cases yet, but I'll get to 'em.
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Jonathan_the_Red
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #7 on: Aug 28th, 2002, 5:40pm » |
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Er... I just noticed that the puzzle talks about the Gods answering "in their own language". I don't think the solution above works with the iff trick... Eric, do the Gods not speak English? I'll have to play around a bit if they don't.
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #8 on: Aug 28th, 2002, 9:26pm » |
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No, this one's in English. The "standard note" is just that -- a generic one to handle all cases.
Good luck with the next two, the difficulty escalates.
Best,
Eric
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #9 on: Aug 29th, 2002, 6:02am » |
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Jonathan,
I just had a chance to look at this a little more carefully.
on Aug 28th, 2002, 5:38pm, Jonathan_the_Red wrote:| Seeing as how this one is languishing, I'll take a stab.... |
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Yes, I was starting to get worried! Thanks for playing!
on Aug 28th, 2002, 5:38pm, Jonathan_the_Red wrote:| You obviously can't do better than three questions. Therefore a solution that does it in three needs no proof of minimality. |
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True.
on Aug 28th, 2002, 5:38pm, Jonathan_the_Red wrote:Question 1: [to God A]: Are you Xor?
Question 2: [to God B]: Is 2+2 equal to 4?
Question 3: [to God B if answer to Q2 is Y, God C otherwise] Is And to the left of Or?
Going under the assumption that XOR(A, B, C) is true iff an odd number of (A, B, C) are true, they are:
Answer NYY: Gods are A, X, O
Answer NYN: Gods are O, X, A
Answer NNY: Gods are A, O, X
Answer NNN: Gods are O, A, X
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I'm a little confused here. In the case of AOX, for example, you will get NYY, not NNY as you listed, no? Because Or always responds Y after the 2+2=4. Similarly, AXO seems to answer NNY rather than NYY (the second response being Y xor Y, and the third being N or Y or Y). Am I misunderstanding your question set?
on Aug 28th, 2002, 5:38pm, Jonathan_the_Red wrote:The remaining cases depend on the truth table of the ternary XOR operator.
If instead A XOR B XOR C is true iff ( (A AND B AND C) OR (~A AND ~B AND ~C) ), just reverse A and O in the above four answers. |
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Interesting, I have never seen it defined that way. Is that something that is sometimes done?? It doesn't even generalize to the binary operator case!
on Aug 28th, 2002, 5:38pm, Jonathan_the_Red wrote:| I haven't really tried the other two cases yet, but I'll get to 'em. |
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Good luck!
Eric
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Jonathan_the_Red
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #10 on: Aug 29th, 2002, 1:58pm » |
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on Aug 29th, 2002, 6:02am, Eric Yeh wrote:
I'm a little confused here. In the case of AOX, for example, you will get NYY, not NNY as you listed, no? Because Or always responds Y after the 2+2=4. Similarly, AXO seems to answer NNY rather than NYY (the second response being Y xor Y, and the third being N or Y or Y). Am I misunderstanding your question set?
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No, I was just being a gimboid, yet again. I think I mistranscribed my solution, and I've unfortunately discarded my notes. I'll reconstruct it and repost it later.
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Interesting, I have never seen it defined that way. Is that something that is sometimes done?? It doesn't even generalize to the binary operator case!
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Er, what I meant was:
If instead A XOR B XOR C is truefalse iff ( (A AND B AND C) OR (~A AND ~B AND ~C) ), just reverse A and O in the above four answers.
This does generalize to the binary case. It's basically saying that XOR(p1, p2, p3, p4, ... pn) is true when at least one of the propositions is true, but not all of them. So which version of XOR did you have in mind?
My track record for posting working solutions isn't the greatest, but here's what I believe to be a valid solution for the Universal case, hidden by color:
Since each God will answer all of the questions, there's absolutely no benefit in asking any single God more than one question (unless you need more than three, which we don't.) So ask these questions to three different Gods:
Q1: Is two plus two equal to four?
Q2: Is two plus two equal to five?
Q3: Is God B Xor if and only if And is to the left of Or?
This produces the following unique answers:
A O X N Y Y
O A X Y N N
A X O N N Y
O X A Y Y N
X A O Y N Y
X O A N Y N
By the way, my solutions so far have been done without the aid of a computer. I'm going to try to keep it kosher this time
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #11 on: Aug 29th, 2002, 2:05pm » |
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Fair enough -- and ye, I figured most likely you just meant "false iff..." instead of "true iff...". Still, I've never heard of the operator being defined in that way. Can anyone else corrobrate?
In any case, I did indeed intend the former -- see my Reply #5.
I don't have time to check your soln right now, but will get around to it some time.
Best,
Eric
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Jonathan_the_Red
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #12 on: Aug 29th, 2002, 3:50pm » |
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Clarification re. Selfless:
The Gods are A, B, and C. I ask three questions, directing them to A, A, and B respectively.
B will answer Q1 op Q2. Will A twice answer Q3, or will he answer Q2 op Q3 followed by Q1 op Q3?
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #13 on: Aug 29th, 2002, 3:56pm » |
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The latter. Thanks for the clarification.
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Eric Yeh
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Is Boolania too easy??
« Reply #14 on: Sep 9th, 2002, 6:59am » |
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Hey guys,
Hope I didn't scare everyone off with my request not to post solutions! I'd still love to know how people are doing with this one!! Let me know if it was too easy!!! Or if everyone who was interested is already done with it. In either case, I would be happy to post a new puzzle... (But I'm also happy to wait if people are still mopping up on this one! )
Happy puzzling,
Eric
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| « Last Edit: Sep 9th, 2002, 7:00am by Eric Yeh » |
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TimMann
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Backward case
« Reply #15 on: Sep 11th, 2002, 12:19am » |
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To avoid spoilers I could just post some inane comment like "that was easy", but instead I'll post a solution hidden by color.
Here's one solution, which was pretty
much the first thing I tried.
1) Left: Is AND on the left?
2a) If answer 1 was yes, AND is on the left,
and the "yes" anded with his next answer
will leave that answer unchanged, so next ask:
"Left: Is OR in the middle?"
This resolves the other two positions; done.
2b) If answer 1 was no, OR or XOR is on the left,
and the "no" ored or xored with his next
answer will leave that answer unchanged, so
next ask "Left: Is XOR on the left?"
3a) If answer 2b was no, OR is on the left,
and the two nos ored with his next answer
will leave that answer unchanged, so
next ask "Left: Is AND in the middle?"
This resolves the other two positions; done.
3b) If answer 2b was yes, XOR is on the left,
and the no + yes xored with his next answer
will complement that answer, so next ask
"Left: Is AND in the middle?" Complement
the answer you get to resolve the other
two positions; done.
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| « Last Edit: Sep 11th, 2002, 12:24am by TimMann » |
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TimMann
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Universal case
« Reply #16 on: Sep 11th, 2002, 1:17am » |
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Jonathan's solution to the Universal case checks out, but here's a different one that I think is interesting because it points toward an approach to the Selfless case as well.
Consider this question.
Q1: Is the following true?
a) You are on the left and AND is sitting leftward of OR, or
b) You are in the middle and OR is sitting leftward of XOR, or
c) You are on the right and XOR is sitting leftward of AND.
Note: "sitting leftward" here is strictly linear: AOX, AXO, and XAO are the cases where AND is sitting leftward of OR.
One solution is to ask Q1 once and ask "Q2: Are you AND?" twice. You must speak to a different god each time, but it doesn't matter in what order you speak to them or in what order you use the questions.
Why does this work? AND's answer to Q2 is yes, so when you speak to him, what he says is the answer to (Q1 AND Q2 AND Q2) = (Q1 AND yes AND yes) = Q1. On the other hand, for OR, Q2's answer is no, so what he says is (Q1 OR no OR no) = Q1. For XOR, Q2's answer is no, so what he says is (Q1 XOR no XOR no) = Q1 as well.
Thus you've gotten each god to answer Q1 truthfully. Due to the guards ("you are on the left and..."), the question gives you different information from each god, just enough to identify them. The inner questions could have been any set of three questions that distinguish the 6 seating orders, but I liked the symmetry of the three given above.
Next to think about: what else could you do with question Q1?
More on the above:
A second way to use Q1 would be to ask it three times. Since (p AND p AND p) = p, (p OR p OR p) = p, and (p XOR p XOR p) = p for any truth value p, again this gets each god to simply answer Q1 itself.
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| « Last Edit: Sep 11th, 2002, 1:59am by TimMann » |
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TimMann
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Selfless case
« Reply #17 on: Sep 11th, 2002, 1:26am » |
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This follows on from my previous post. I don't yet have a clue whether or not it is optimal, but extending the previous solution to Universal gives a solution to Selfless.
Note that Eric's clarification to Selfless is very important. To state it another way, in Selfless, each god's answer is his operator applied to all questions on your list other than the one you are addressing to him right now, including questions you may address to him in the past or future. The statement on the main puzzle page (and the name "Selfless") makes it sound like questions you ask(ed) of this god in the past or future are also excluded. But that would create temporal paradoxes, so it wouldn't make sense as part of the rules.
Now a spoiler -- a possibly non-optimal solution to Selfless:
Question Q1 from my previous post immediately provides a 4-question solution to Selfless. Simply ask Q1 four times, at least once of each god. Then each answer is again the xor of three copies of Q1, which is just Q1 again.
The second time you ask Q1 of the same god, you already know what he's going to say (same as last time you asked!), but you have to ask it anyway. If you had the question on your list but you walk away without asking it, the gods will strike you down in wrath. Well, or at least they'll bill you for the 4 questions that were on your list anyway.
I don't know yet if there is a 3-question solution to Selfless. Must sleep now...
Eric, thanks for the cool puzzles.
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TimMann
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Selfless case improved
« Reply #18 on: Sep 11th, 2002, 2:20am » |
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Guess I didn't need to sleep on that one. Here is a better solution to Selfless.
Ask this question once of each god:
"Q3: Is OR not sitting in the next position cyclically after yours?" By "cyclically after" I mean that the next position after the last position is the first position again.
XOR's reply is (Q3 XOR Q3) = No.
Because OR is sitting in his own position, not in the position cyclically after his own, Q3 is true for him, so he replies (Q3 OR Q3) = (Yes OR Yes) = Yes.
AND's answer to Q3 distinguishes the cases AOX, XAO, OXA (where he says No) from the cases OAX, XOA, AXO (where he says Yes). His reply is (Q3 AND Q3) = Q3.
Making a table, this works:
A O X -> N Y N
A X O -> Y N Y
O A X -> Y Y N
O X A -> Y N N
X A O -> N N Y
X O A -> N Y Y
Comment on my thought process: this was inspired by Jonathan's solution to Universal -- not by the particular questions, but by the structure of the mapping between yes/no replies and positions -- combined with my previous idea (spoiler: i.e., asking the same question all three times).
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| « Last Edit: Sep 11th, 2002, 2:31am by TimMann » |
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #19 on: Sep 11th, 2002, 6:22am » |
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Tim,
Nice job with all the answers! Very elegant.
Some follow-up questions:
1) Can you solve the Backward scenario without dynamic questions?
2) Find with proof the minimum number of Gods you need to speak to in each scenario.
BTW, "is not cyclically after" == "is cyclically before".
Have you solved my other three posted puzzles?
Best,
Eric
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Jonathan_the_Red
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #20 on: Sep 11th, 2002, 11:11am » |
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on Sep 11th, 2002, 6:22am, Eric Yeh wrote:
1) Can you solve the Backward scenario without dynamic questions?
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I can.
1) Is A AND? (asked of A)
2) Is A XOR? (asked of A)
3) Is B XOR? (asked of B)
OAX N N N
OXA N N Y
XOA N Y Y
XAO N Y N
AOX Y N Y
AXO Y N N
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BTW, "is not cyclically after" == "is cyclically before".
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Not exactly.
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Jonathan_the_Red
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #21 on: Sep 11th, 2002, 11:19am » |
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Here's a slightly slicker Backwards solution:
Ask all of these of A.
1) Are you AND?
2) Are you not OR?
3) Is XOR next cyclically left of AND?
OAX N N N
OXA N N Y
XOA N Y Y
XAO N Y N
AOX Y Y N
AXO Y Y Y
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| « Last Edit: Sep 11th, 2002, 11:19am by Jonathan_the_Red » |
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #22 on: Sep 11th, 2002, 11:27am » |
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Yep, now you're talking!! The cool thing is that, within some reasonable definition of equivalence, I believe this solution is unique!! (So I wouldn't say just "slightly" )
Ye ye, I guess what I said about those cyclicals isn't quite true, my mistake.  I just thought it sounded funny grammatically.
Best,
Eric
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| « Last Edit: Sep 11th, 2002, 11:31am by Eric Yeh » |
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TimMann
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #23 on: Sep 11th, 2002, 11:23pm » |
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on Sep 11th, 2002, 6:22am, Eric Yeh wrote:
1) Can you solve the Backward scenario without dynamic questions?
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I'll resist looking at the spoiler and give that one a try later on.
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2) Find with proof the minimum number of Gods you need to speak to in each scenario.
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Backward follows trivially from my dynamic-question solution:
My solution speaks to only one god. Speaking to no gods is obviously out, so 1 is minimal.
Universal is easy:
a) Since you need more than 2 bits of information, and each answer supplies at most one bit of information, you need at least three answers.
b) Every time you speak to a god in the Universal scenario, he must say the same thing, because his answer is always his operator applied to all your questions. So speaking to the same god more than once provides no additional information.
Therefore you have to speak to all three gods at least once each. We have solutions that do that, so 3 is minimal.
For Selfless, I need to think about it some more. Solutions like mine clearly require talking to all three gods. (Since all three questions are the same, talking to the same god more than once gives you the same answer each time, as with Universal.) But I don't have a proof that a solution to Selfless has to work that way.
Quote:
Have you solved my other three posted puzzles?
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I solved Past/Present/Future and Gibberland. PPF was a nice hard puzzle, though it's too bad the solution wasn't pretty or unique. Gibberland was excellent, a real mind-wrecker. Was there another problem of yours too? I don't think I found it.
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| « Last Edit: Sep 11th, 2002, 11:24pm by TimMann » |
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Eric Yeh
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Re: NEW PROBLEM: The Operators of Boolania
« Reply #24 on: Sep 12th, 2002, 6:00am » |
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Tim,
Yup, nice job so far; enjoy those two last follow-ups.
(For the min q's q, I actually meant non-dynamically when possible, but it's too much of a bother to ask you to keep trying these minor variations so I'll refrain! But my general "hierarchy" of responses is:
1) Min questions
2) Non-dynamic questions
3) Non-dynamic answerers
4) Min answerers
Regarding my other puzzle: see the "Puzzle Forum" new problem in the hard threads. Will hasn't promoted it to being shown on the main page yet, perhaps because of the tastelessness.
Best,
Eric
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| « Last Edit: Sep 12th, 2002, 6:00am by Eric Yeh » |
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