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So another day then!
Last day's puzzle was first solved by illu, good job!
+ Show Spoiler [solution] +
So the proof is you construct a great circle with 2 points (on the plane of those 2 points and the centre of the sphere), and then either 3 points on same side of hemi, or 2-1. Either way you can find 4 on the same hemisphere.
Today's puzzle is this, bear in mind that I actually have NOT solved it yet, so I'd be very interested in seeing not just the answer but the way of getting to it. The first one to do so will get honorary mention (xD).
So today's puzzle:
If you remember monty hall, it's a show where you get 3 doors, behind 2 doors contain the goat, and behind 1 door contain a car.
This problem is different in that, in addition to the 3 doors with the goats and car, you also have a watch. This watch is magical in that it has 2 lights, here's how they work.
On any given day, blue/green either means true/false, or false/true. The order is randomized each day. For instance, yesterday it might be blue=true/green=false, tomorrow might be blue=false/green=true. They can't be both true or both false.
The watch will answer ANY yes/no question with these 2 lights(bear in mind you might not know if it is true/false since blue/green can stand for either) with 100% accuracy, and you can ask maximum of 2 questions per day.
Your task:
Without the host opening any door(normal monty-hall), how do you ask the watch 2 questions such that you will get the car?
Extra info:
Please put answer in spoilers, as usual, and if you have not seen the problem before, please attempt it anyways, and just post some of your thoughts down and write in your post "I've not done this problem before" or something like that, so other people who's lost can work with you. If you already seen the problem and are just re-stating the answer you already know, please say so as well in spoiler, so I can see if people are genuinely working on these puzzles and solving them actively.
That would be all!
Good luck!
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So here's my attempt :p
You CAN decide if blue/green means true/false by asking the watch trivial questions such as "Is my name Evan", to which it will answer "true", and you can see which light it is.
Although you waste a question that way...
I am thinking asking the watch questions about the watch... It seems like to be the convention, remember the truth_teller/liar puzzle? You ask one what the other thinks on something, and it works out to be an invariant. So we'll see...
Some basic truth about the goats and the car is:
Only 1 car.
No 2 sets of 2 doors can have the same content, that's saying (Door_1 same as Door2) AND (Door_2 same as Door_3) will always return false.
Aside from that I'm pretty stuck hahaa
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+ Show Spoiler +Woah this one is difficult. Some thoughts:
Perhaps asking about two goats? As in, Do door #1 and #2 have two goats behind them? Then you can use process of elimation to figure out where the car is. But then you still have to know about the lights...
Perhaps the probably of switching doors might play into this. I know in the Monty Hall problem, you increase your odds with a door switch after the host reveals one goat. I don't know how this would work in a problem like this but I'm just throwing it out there.
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+ Show Spoiler +
Lets suppose 2 possible lights Black and white. Go to door 1 ask
- If someone asks you (the clock) if theres a car there, will you turn on white light?
Go to door to 2 ask the same.
- If someone asks you (the clock) if theres a car there, will you turn on white light?
If they are both the same color go to door 3. If they are different go to door wich turns white.
Why does this work?
- If the clock lights the same color he was saying false both times. He cant say yes both times.
- If the white light is yes then the question will light white in the car
- If the white light is no (black yes) then the question will light white in the car too.
EDIT: EZPZ, after 10 minutes i made it!!.
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Since this is presented as a Monty Hall problem:
How about asking;
Would I, if picking this door and using the normal Monty Hall-switching strategy given the option of switching, get the car? It would explain the Monty Hall presentation, and might work, but I'm not able to work something out right now.
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Here we go:
+ Show Spoiler +
Go to door 1 and ask:
Would you light green if the car is behind this door?
Go to door 2 and ask:
Would you light green if the car is behind this door?
There are six possible configurations (doors and watch), but only three different watch responses no matter if green is "yes/true" or "no/false".
The watch may show
green/blue so the car is behind door 1
blue/green so the car is behind door 2
blue/blue so the car is behind door 3
Just test it with excel
config d1 d2 d3 g b a#1 a#2
#1 1 0 0 1 0 g b
#2 1 0 0 0 1 g b
#3 0 1 0 1 0 b g
#4 0 1 0 0 1 b g
#5 0 0 1 1 0 b b
#6 0 0 1 0 1 b b
d=door (1=car), g=green b=blue (1=yes), a=answer
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+ Show Spoiler +
In this case, the answer to the question "is there a car behind this door?" is already an XOR between the value of the door and the value of the watch: the answer changes whenever one input is constant and the other input is inverted. So to ask for the plain value of the door, we can factor out the watch by XOR'ing the value of the watch to the answer of "is there a car behind this door?".
For example, one can ask "Do the statements 'the watch answers blue to: is there a car behind door #1?' and 'blue represents true on the watch today' have the same truth value?" Here blue will indicate a car and green will indicate a goat.
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+ Show Spoiler +
1.) "do door 1 and 2 have the same prize?"
2.) "if the last answer was no, does door 1 have the car?"
if watch doesn't light up after question 2, then the car is behind door 3. if it does light up, compare the color to the "no answer" of question 1, and you will know if it is behind door 1 or 2
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travis:
+ Show Spoiler +I don't expect this to work out. The watch can't decide to light up or not. It just lights green or blue, i guess.
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+ Show Spoiler +My attempt I will name the doors A, B and C Question 1 Ask some trivial question such as is today sunday to get which colour is true/false. Now choose door A for consideration Question 2 Ask the watch if it is your interest to switch to door B. If the car is behind A it will display false, if the car is behind door B it will show true. If the car is behind C then the watch will not be able to answer your question and I assume nothing will light up. Pick the door with the car 
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On May 04 2009 04:28 no_re wrote:+ Show Spoiler +My attempt I will name the doors A, B and C Question 1 Ask some trivial question such as is today sunday to get which colour is true/false. Now choose door A for consideration Question 2 Ask the watch if it is your interest to switch to door B. If the car is behind A it will display false, if the car is behind door B it will show true. If the car is behind C then the watch will not be able to answer your question and I assume nothing will light up. Pick the door with the car
+ Show Spoiler + no, it'd be false if it was behind c, because it'd be in your interest to switch to c, so that doesn't work
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Better method:
Ask the watch if your name is your name. Determine which colour is true.
Ask the watch if a highly volatile stock on the market will end the day higher than it is now.
Bet the farm if yes.
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On May 04 2009 05:14 SpiritoftheTunA wrote:Show nested quote +On May 04 2009 04:28 no_re wrote:+ Show Spoiler +My attempt I will name the doors A, B and C Question 1 Ask some trivial question such as is today sunday to get which colour is true/false. Now choose door A for consideration Question 2 Ask the watch if it is your interest to switch to door B. If the car is behind A it will display false, if the car is behind door B it will show true. If the car is behind C then the watch will not be able to answer your question and I assume nothing will light up. Pick the door with the car + Show Spoiler + no, it'd be false if it was behind c, because it'd be in your interest to switch to c, so that doesn't work
+ Show Spoiler + question 2 was asking whether to switch to door b specifically, not to switch from a to one of the other 2
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On May 04 2009 05:35 no_re wrote:Show nested quote +On May 04 2009 05:14 SpiritoftheTunA wrote:On May 04 2009 04:28 no_re wrote:+ Show Spoiler +My attempt I will name the doors A, B and C Question 1 Ask some trivial question such as is today sunday to get which colour is true/false. Now choose door A for consideration Question 2 Ask the watch if it is your interest to switch to door B. If the car is behind A it will display false, if the car is behind door B it will show true. If the car is behind C then the watch will not be able to answer your question and I assume nothing will light up. Pick the door with the car + Show Spoiler + no, it'd be false if it was behind c, because it'd be in your interest to switch to c, so that doesn't work + Show Spoiler + question 2 was asking whether to switch to door b specifically, not to switch from a to one of the other 2
+ Show Spoiler + the question asks if it's in your interest. no gain = not in your interest = false will light up
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+ Show Spoiler +I basically agree with kernipu and datscilly's solutions, Malongo seems very slightly off (I guess not really, but if "they both turn the same color" said color would have ot be black). datscilly I think has the most mathematically correct answer as to why this is the solution, though it could be worded a bit better. As someone who also hadn't done this before, the way I arrived at the idea is because it's the only way to get enough data. If you simply ask the watch questions about the car, because you don't know which is true and which is false, there are only 2 possible results you can get from your questions: both lights the same, or both lights different. Similarly, if you waste a question to determine the truth values first, you get only 2 possible results: true or false to one question. A single binary answer is not enough data to determine which of three items is correct. Therefore, you need to ask the watch questions about the color it would answer to hypothetical questions in order to gain context of the colors without wasting questions, thus giving you 4 possible results: BB, BG, GB, GG, which is enough to choose between 3 doors.
Also, I like L's idea. 
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ok, well if my last answer doesn't work - here goes
(this took me forever lol, didn't really know what i was doing at all)
+ Show Spoiler +
1.) ask "(Does Blue = yes and the car is behind door 1 or 2) OR (Green = yes and car is behind door 3)"
[If the color you get is blue then ask:
2.) "(Does green = yes and the car is behind door 1) OR (blue = yes and the car is behind door 2)"
if the color for this is Blue - the car is behind door 2
if the color for this is Green - the car is behind door 1]
[If the color you get is green then the door is behind 3]
hopefully this is right. im not gonna check it again lol, so here goes
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keep in mind my "OR" statements are part of 1 large question
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+ Show Spoiler +
Statement 1: If I say that the car is behind door A, then the watch will light blue
Statement 2: If I say that the car is behind door B, then the watch will light blue
How this works:
First statement
If the car is behind door A and blue = true: the watch will turn blue
If the car is behind door A and blue = false: the watch will still turn blue
If the car is behind door B or C and blue = true: the watch will turn green
If the car is behind door B or C and blue = false: the watch will still turn green
Second statement
If the car is behind door B and blue = true: the watch will turn blue
If the car is behind door A or C and blue = false: the watch will still turn blue
If the car is behind door B and blue = true: the watch will turn green
If the car is behind door A or C and blue = false: the watch will still turn green
With this, we can know for sure which door the car is behind because
If statement 1 made it turn blue, then the car is behind door A
If statement 2 made it turn blue, then the car is behind door B
If both statement made it turn green, then the car is behind door C.
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Kau
Canada3502 Posts
+ Show Spoiler +Case 1: Blue = True, Green = False, Car = Door 1
Question 1: If I asked if the car is behind Door 1, would you turn green?
Result: Green
Question 2: If I asked if the car is behind Door 2, would you turn green?
Result: Blue
Case 2: Blue = True, Green = False, Car = Door 2
Question 1: If I asked if the car is behind Door 1, would you turn green?
Result: Blue
Question 2: If I asked if the car is behind Door 2, would you turn green?
Result: Green
Case 3: Blue = True, Green = False, Car = Door 3
Question 1: If I asked if the car is behind Door 1, would you turn green?
Result: Blue
Question 2: If I asked if the car is behind Door 2, would you turn green?
Result: Blue
Case 4: Blue = False, Green = True, Car = Door 1
Question 1: If I asked if the car is behind Door 1, would you turn green?
Result: Green
Question 2: If I asked if the car is behind Door 2, would you turn green?
Result: Blue
Case 5: Blue = False, Green = True, Car = Door 2
Question 1: If I asked if the car is behind Door 1, would you turn green?
Result: Blue
Question 2: If I asked if the car is behind Door 2, would you turn green?
Result: Green
Case 6: Blue = False, Green = True, Car = Door 3
Question 1: If I asked if the car is behind Door 1, would you turn green?
Result: Blue
Question 2: If I asked if the car is behind Door 2, would you turn green?
Result: Blue
If car behind Door X is true, would you show Color(true)? True
If car behind Door X is false, would you show Color(true)? False
If car behind Door X is true, would you show Color(false)? False
If car behind Door X is false, would you show Color(false)? True
So when you ask if the car is behind Door X, you know that if the color matches the color you specified, then the car is behind Door X.
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EPT![[image loading]](http://www.ubergizmo.com/photos/2006/6/binary-watch.jpg)
