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Contest 10 - Three Logic Questions with a Twist (Ended)
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Weicong Sng RJ: Park Min Young CO: Psychotic Nutcase Post Rating: 0 + / - Total Posts: 59 Karma: 28 Joined: Apr 10, 2012 |
Posted on Apr 20, 2012 1. Switch on A and B, wait for a period of time just so that the bulb will be sufficiently hot, then switch off B and switch on C and immediately move on to room B. If its switched on, hot = A. lukewarm (you just switched it on so I would expect a temperature rise) = C. If its switched off, hot = B. room temperature = D.2. No one will ever leave. Why? Because 1. they do not know the total numbers of blue eyes and brown eyes on the island. Ratan Joyce arriving on the island telling everyone that someone has blue eyes doesn't trigger anything. A blue eyed person would've seen 49 blue eyed and 50 brown eyed guys on the island. A brown eyed person would've seen 50 blue and 49 brown eyed person on the island. Ratan Joyce's statement would've been refering to just any one of the 49/50 blue eyed guys on the island. There is no actual trigger. (crap why is my answer so different from the rest) 3.Just ask if there's an odd number of liars. If they said yes, it doesn't matter if they are actual liars or truth-ers. This is because, if the one saying yes is a liar, the island you are in is even, and therefore the statue is on this island. If the one saying yes is a truth-er, it means that since there is an odd number of liars, there are an even number of truth-ers, which is the island you're in. and conversely if you get no for an answer, it the answerer was a truth-er, it also mean that there is an odd number of truth-ers in the island you're in, and the statue is on the other island. if the answerer is a liar, it means that the island you're in actually has an odd number of liars, and the statue is on the other island. |
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Innocent Bystander RJ: Matthew Matician Post Rating: 0 + / - Total Posts: 76 Karma: 54 Joined: Apr 2, 2012 |
Posted on Apr 20, 2012 Sorry!If it makes you feel any better, 1) I'm a professional mathematician and 2) the third one really stumped me for an hour. I liked that one a lot--aside from my first answer being wrong, something was really nagging at me that everyone's 2-question answers weren't making use of parity. If you'll let me award a couple million dollar prizes (through, say, purchasing overpriced B2B items with your blessing), I'd be happy to post another couple of questions to keep the contest rolling. |
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Weicong Sng RJ: Park Min Young CO: Psychotic Nutcase Post Rating: 0 + / - Total Posts: 59 Karma: 28 Joined: Apr 10, 2012 |
Posted on Apr 20, 2012 Can someone enlighten me about Q2?
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Innocent Bystander RJ: Matthew Matician Post Rating: 0 + / - Total Posts: 76 Karma: 54 Joined: Apr 2, 2012 |
Posted on Apr 20, 2012 Weicong Sng: the answer to #2 works like this:It's a proof by induction on the number of blue-eyed people, where the premise is: if there are exactly n blue-eyed people, they all leave on the n-th night. One important note: every person on the island knows that the number blue-eyed people can be one of two consecutive numbers. For instance, with 70 brown and 30 blue, each brown would see 30 blue and know that there are either 30 or 31 blue-eyes. A blue-eyed person would see 29 blue and know that there are either 29 or 30 blue-eyes. With 99 brown and 1 blue, from the perspective of the blue-eyed guy/gal, there are either 0 or 1 blue-eyed people. That is the base case (for n = 1): if only one person has blue eyes, then from that person's perspective, the moment Ratan announces that # blue eyes > 0, he (his name is George) knows it must be him. Then, we assume that the statement is true for n and prove that it implies n+1 is true as well. That is, we assume: "if there are n blue-eyed people, they will leave on the nth night" Now for the n+1 case: we have n+1 people with blue eyes. Each blue-eyed person on our island knows there are either n or (n+1) blue-eyed people (they themselves being the (n+1) person). Once no one leaves on the nth night, which means that, by our assumption, there are not exactly n blue-eyed people. This means that they then realize the next morning, since no one left, that there must be n+1 blue eyes. That means they know they are blue-eyed. That means they leave on the n+1 night. That finishes our proof by induction. If there are x blue-eyed people, each brown-eyed person knows there are x or x+1 blue eyed people, so when night 50 rolls around, the 50 brown-eyed people are thinking: there are either 50 or 51 blue-eyed people. That is, they don't know their own color. The next day, once all the blue-eyed people depart, they realize their own eyes must be brown. |
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Weicong Sng RJ: Park Min Young CO: Psychotic Nutcase Post Rating: 0 + / - Total Posts: 59 Karma: 28 Joined: Apr 10, 2012 |
Posted on Apr 20, 2012 omg i finally get it. thanks.
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Mister Death RJ: McFlono McFloninoo Post Rating: 0 + / - Total Posts: 266 Karma: 300 Joined: Feb 6, 2012 |
Posted on Apr 21, 2012 Wow, I stop reading the forums for a couple days and totally miss this? Sigh.
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Scott (Admin) RJ: Ratan Joyce CO: Ratan Joyce Post Rating: 0 + / - Total Posts: 1175 Karma: 5083 Joined: Jan 13, 2012 |
Posted on Apr 21, 2012 Wow, I stop reading the forums for a couple days and totally miss this? Sigh. The mathematician said he can write up questions, so hopefully the fun can continue. |
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zxektok megatron RJ: zxektok Post Rating: 0 + / - Total Posts: 140 Karma: 170 Joined: Mar 6, 2012 |
Posted on Apr 23, 2012 i completely disagree with the answer to q1if someone did that in my house to see if a lightswitch worked - i would slap them in the face for taking an hour! We would be still be in the stone age if everything took this long - there are tonnes of ways to solve this problem in a more effective and time efficient manner. Want examples or a list? |
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Innocent Bystander RJ: Matthew Matician Post Rating: 0 + / - Total Posts: 76 Karma: 54 Joined: Apr 2, 2012 |
Posted on Apr 23, 2012 Who said anything about needing to solve the problem in a time-efficient manner?
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zxektok megatron RJ: zxektok Post Rating: 0 + / - Total Posts: 140 Karma: 170 Joined: Mar 6, 2012 |
Posted on Apr 23, 2012 ok then - close the curtains/blinds and the rest of the doors from room A and in room Bor simply if time isnt involved at all - do the problem at night and simply press the switches individually and see what illuminates the room. Takes about 10 seconds. ask your friend to stay in the room put your phone on record viewing the light bulb, and every 10 seconds change switches (can see with recording timer)- if your phone has a mic also - shout "switch 1".....and press switch 1.... ad so on go back after a few seconds and pick up your phone and watch your recording.... still much faster than an hour. Atleast this way if the bulb was actually broken - the person in question wasnt being a pleb for an hour missed these as i am in work pretty much all day and night on weekends anyway - going to cook dinner. |
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Innocent Bystander RJ: Matthew Matician Post Rating: 0 + / - Total Posts: 76 Karma: 54 Joined: Apr 2, 2012 |
Posted on Apr 23, 2012 I think you misunderstand the nature of the question. It's a logic puzzle that's designed to be solved by creative thinking, not by finding loopholes in the rules: "You can’t see whether the light is on or off from Room A, and once you leave Room A, you are not allowed to come back."The fundamental realization is that light bulbs can have more than two states: they can be on/off (a state independent of time), and hot/cool (a state dependent on time). |
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zxektok megatron RJ: zxektok Post Rating: 0 + / - Total Posts: 140 Karma: 170 Joined: Mar 6, 2012 |
Posted on Apr 23, 2012 you can "see" if not in the room if it is dark enough...heat of bulb depends on many things including room temperature the bulb could even be broken and still warmer than room temp (dependant on room) is there no windows? you said room not cell/dungeon you never mentioned light intensity of bulb nor heat output. although i can somewhat appreciate your mathmetical mind - its too inside your own box of parameters. The solution to a problem isnt always one answer and the example of windows is perfectly viable as it was not stated in the question. If the state is indeed "once you leave Room A, you are not allowed to come back" then you could not go back into the room to check the temperature of the bulb indicating more of a schrodinger's cat principle. Unless one would a assume the word "room" was a standard and had windows - not a closed prison cell or basement. You also dont have return to the room to see illumination in a darkened room. Dependant on light intensity of the bulb of course but quite often i can see if a room is lit before i walk in.....even from other rooms. both satisfying the question and NOT RETURNING TO THE ROOM So returning to the question at hand IF "You can’t see whether the light is on or off from Room A, and once you leave Room A, you are not allowed to come back." then you cant see the light from room A you have to leave room A you cannot come back then i firmly believe that everybody is wrong as "you cannot go back to room A" IT does say in the question that "you cannot see the light from *room A*" well simply dont view the light from room A then.... if it is sealed room with no windows and you cannot enter then all switches both do and do not light the bulb until it is viewed and the state becomes one or the other. |
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Scott (Admin) RJ: Ratan Joyce CO: Ratan Joyce Post Rating: 0 + / - Total Posts: 1175 Karma: 5083 Joined: Jan 13, 2012 |
Posted on Apr 23, 2012 I remember putting in incandescent light bulbs for a reason. They are known as "small heating devices" somewhere.http://www.mnn.com/earth-matters/energy/blogs/skirting-eu-law-the-rebranding-of-incandescent-bulbs-as-heat-balls You don't need to go back to room A after doing the correct procedures. |
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