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105 Comments
- sp0rk, on 10/12/2007, -5/+42Keep trying, you'll figure it out one day.
- beelz, on 10/12/2007, -21/+46One answer for them all. 42.
- ChzPlz, on 10/12/2007, -3/+27This headline gets truncated to "Very Difficult Anal..." on my newsfeed. Is there any other kind?
- smellinator, on 10/12/2007, -0/+10Wrong. The problem never said that the counterfeit coin is *lighter*! They just said that the weight is different.
- spectre_25gt, on 10/12/2007, -0/+9I can't believe this reference still gets modded up.
- smellinator, on 10/12/2007, -1/+10***** SPOILER ALERT *****
Another attempt at an explanation to number 1:
Number the bottles from 1 to 1000, in Binary. (i.e. 0000000001 to 1111101000). There are 10 digits in each binary bottle number. Take your 10 prisoners, and have them sample any wine that has a bottle number with a "1" in their respective digit (500 sips of wine). If he's alive after 23 hours, then you know that EVERY bottle that he sampled is fine - even though he's drunk. If he's dead, then you know that every bottle that he didn't sample is fine. Either way, with each prisoner you eliminate half the bottles from contention.
By looking at who died, you can construct which bottle number is bad. If, say, the prisoner who sampled all the odd numbered bottles died, and the prisoner who sampled all the bottles numbered 512 and above died (and everyone else lived), then you know that the bottle number is 1000000001, or bottle number 513. The more prisoners who die, the more ones there are in the binary representation of the poison bottle number.
Does that explain it better? - kankerfist, on 10/12/2007, -1/+8He doesn't die for 23 hours, so 1 prisoner can only test 1 bottle before the celebration.
- ardellin, on 10/12/2007, -0/+6Never go against a Sicilian when death is on the line!
- mvnicosia, on 10/12/2007, -0/+6What about the time it would take to consume all the wine? I think that might put them behind schedule for the party. If I had thousands of prisoners, 999 would get one suitably sized drink (to ensure the poison works) and then I'd wait it out. If no one died, I'd throw out the last bottle and hope that one of them hadn't spent the last few years building up an immunity to iocaine powder.
- kirakun, on 10/12/2007, -0/+5but you still need 10 binary digits to cover all numbers from 1 thru 999.
- chris9902, on 10/12/2007, -2/+7agghh, my Sunday brain hurts.
- smellinator, on 10/12/2007, -0/+3shoot, all my plus signs disappeared! ... stupid digg comments! And EDIT went away!
This may be more readable. sorry for the duplication. Mod me Thumbs down on the parent!
I liked the card trick (#5).
My son and I did this. (He humored me for Father's Day, the good kid!) Here are the specifics.
***** SPOILER ALERT *****
We decided on the following "rules":
- Ace is high (which is relevant for calculating highest and lowest cards).
- Suits are in this order: Spades (highest), Hearts, Clubs, Diamonds (lowest). (can be any convention as long as you both remember it).
- The setup man is given 5 cards, selected by the audience participant. 4 cards are passed to the magician. The 5th (the mystery card) is given back to the audience participant.
- When the magician receives the cards, the 1st card flipped determines the suit of the mystery card.
- The next three cards are the "point" cards (explained below).
- The numerical lowest of the 3 point cards is considered first. Depending on its position, it's either worth 1, 3, or 5 points (depending on if it's flipped 2nd, 3rd, or 4th). If there's a tie for lowest numerical value (like a pair of 6's, for instance), then the suit is used to determine which is lower.
- The numerical highest card is then considered. It can be in one of the 2 remaining positions, if you don't consider the 1st flipped card and the lowest flipped point card. The value is either 0 or 1.
- Total the point values. This total can range from 1 (=1 plus 0) to 6 (=5 plus 1).
- Add the point value to the first flipped card, wrapping around at the Ace. So Queen of Clubs plus 6 points equals the 5 of clubs. Don't change the suit.
Using this technique, the setup man can communicate to the magician the exact card chosen every time. All the setup man needs to do is to view the 5 cards, and choose two that are the same suit (there must be at least two cards among the five of one suit), and choose the card which is 1 to 6 points less than another card of the same suit (which must always occur, if you consider a circular number line. example, 2 and 9 are seven apart (9 minus 2), or six apart if you wrap around ([13 plus] 2 minus 9)). Then position the remaining cards correctly.
With a smart kid (16 yr old), it takes about 2 minutes of explanation and 1 minute of practice. Then we were able to perform it flawlessly about 95% of the time, with most of the mistakes being mine! - smellinator, on 10/12/2007, -0/+3@uzig: Read the answer on the webpage. They prove you wrong in the answer. (actually, technically you are correct (no "more" than 4), but they can do it in less.)
- lokoluis15, on 10/12/2007, -0/+3@deesine
The're just testing the wine, i.e. taking sips. If they drank the whole bottle there would only be one left for the celebration! Pretty much defying the point of thesting them in the first place. I think it's plausible for 10 guys to take 999 sips of wine in an hour - Rabid_Llama, on 10/12/2007, -0/+3Nope, I thought that for a second too. So, the last guy in line counts up the number of, say, blue hats in front of him. Let's say that there's 2 blue hats and 97 red hats in front of him. So he says "Blue", meaning "even" (which is what they figured out beforehand). So, the guy in front of him, if he sees 96 red hats and 2 blue hats, sees that the number is still even, and can then say that he has a red hat.
It's a parity check, basically. - jasqwerty, on 10/12/2007, -0/+3Ha, 3 is so unfair. You get to magically solder together any wires arbitrarily. What the hell?
- scooter9900, on 10/12/2007, -0/+3Read the question again...it says *around* the 23rd hour, so NO, the answer is not wrong and your method would not work.
- Zanneth, on 10/12/2007, -0/+2I thought most of these questions were oddly worded.
- Rabid_Llama, on 10/12/2007, -0/+2You want to know which start corresponds to which end. You know the cables WORK, that's given, but you have to figure out which one is which.
And if you start adding extra stuff to logic puzzles, they're just dumb. What you would REALLY do is hook up one end entirely, and send a different signal (sine wave, or whatever) over each. Then, on the other end, they just test each signal and order them up. - rezophonic, on 10/12/2007, -0/+2And why the hell does it say:
"camels (being somewhat marsupial in nature) can jump over each other..."
What does being a marsupial have to do with jumping? You can't take the kangaroo into consideration and ignore things like the koala and the opossum. All being a marsupial means is that they have a pouch--which camels indeed don't. - smellinator, on 10/12/2007, -2/+4The answer to number 4 is on the right track, but worded wrong!
**** SPOILER ALERT *****
'The first peasant counts all the hats in front and then answers "Red" if the number is odd or "Blue" if the number is even.' .... if the first peasant (meaning the one in the back of the line - the *last* peasant, or the first one to be challenged) counts the number of hats in front of him, he's always going to arrive at an odd number... 99! I think the answer is supposed to say that he counts all the ones of a particular, pre-defined color. - deesine, on 10/12/2007, -5/+7So all 10 guys are going to drink 999 bottles in 1 hour?
Ya, ok.
Great puzzle. - rezophonic, on 10/12/2007, -0/+2@ analgesia:
They have to flip the switch twice to account for the position the first switch would have started in. For instance, if the leader were to just count flipping the switch down 22 times, then if the switch started in the up position, he would have counted that time, also, incorrectly assuming that all had flipped a switch when really all but one had. However, if he were to count 23 times allowing for the flip to start in the up position, but it started in the down position, once they reached 22 there would be no new time and they would be stuck forever.
So, by making everyone flip the switch twice, if the leader counted 44 flips but included the initial up state, the only thing that would happen is that one prisoner would have only flipped the switch once. - smellinator, on 10/12/2007, -0/+2@lukas:
> "smellinator, no, i still dont understand, but now I want to have your babies."
Sorry, my babies are safe and sound at home. You can't have them. Plus, the one's got a little colic - you sure you want him? - nitsuj, on 10/12/2007, -1/+3pigphuker,
Thanks for showing us a window into your desperate sex-free virgin life. May you be forever locked in the vinegar strokes of your solo-gratification. - cibby, on 10/12/2007, -0/+2Man, I love puzzles like this...
- uziq, on 10/12/2007, -0/+2fine, you win. :P
3 is the answer. - smellinator, on 10/12/2007, -1/+3@Intrepion:
"if you had only 9 people, the most you could test is 513 ( 2^9 combinations 1 no-drink state )"
Count again. With 9 people, you can only test 512 bottles. 1-511, plus 1 no-drink state (either number zero, or number 512, your choice). You were close, but off by one.
With 1 tester, you can test 2 bottles (including 1 no-drink). With 2, 4 bottles (including 1 no-drink). With 3, 8 bottles. With 4, 16.... with 9, 512. - lukas88, on 10/12/2007, -0/+2@alkali, yes it does, it says the celebration is "tomarrow". Hence, 24 hour deadline and only 1 hour to do the drinking.
At one point I was thinking along the same lines as you, using time to determine the bottle. But since it doesnt say the poison takes exactly 23 hours (it says around 23 hours), that is out of the question. - DoctorIan, on 10/12/2007, -0/+2If two people were wearing black, the third would know for a fact he is wearing a white hat as there are only two black hats. From the perspective of the two people wearing black hats in that situation, they would see 1 black and 1 white, with the white-hat person eventually figuring out that he must have a white hat. This would happen really quick because there's only 2 black hats so it's pretty simple for the white-hatter to see.
In the given situation you see two whites. If you had a black hat on, then think from one of the white-hat person's perspective. He sees 1 white and 1 black. BUT, the white-hat person he sees doesn't say he has a white hat which he would if there were two black hats.
So in this situation nobody says that they know their hat is white. Therefore because they can't figure it out, the situation must be that everyone is seeing 2 white hats.
It's a bit presumptuous because it relies on people working it out at a given rate, and being clever enough for the first half but too stupid for the last step of logic. - stesun, on 10/12/2007, -1/+3No information theory is from our old friend shannon. But take the first one, calculating the information that the 1000 bottles contains in binary you take the 2 logarithm of 1000, log2(1000) = 9.96. So you need atleast 10 ppl to get that information. The coin task is the same log2(12) = 3.58, meaning you need to use the scale atleast 4 times to always get the solution.
- chesterjosiah, on 10/12/2007, -0/+2Awesome puzzles, but man, this guy's explanations are just terrible! Too many commas concatenating sentences that shouldn't be together. Can someone besides the author of the page explain the solution to The Warden (#6 on the very difficult page)?
- Ordinant, on 10/12/2007, -0/+2mvnicosia, the goal is to preserve as much wine as possible for the party. You aren't going to make the prisoners drink fistfuls each. It would be one pre-poured sip from that prisoner's designated number of bottles, then move on.
Churches manage to get communion wafers down 1000 parishioners in about 15 minutes. A despot with 1000 expendable prisoners can certainly do better than that. - smellinator, on 10/12/2007, -0/+2@ sirnicholai
> Maybe I'm stupid, but isn't there an easier way to do this?
> Each guy could just speak the color of the hat directly in front of him.
Without commenting on stupidity...
That wouldn't work. Yes, each person would know their color hat. But each one would be declaring their guess of their own colored hat to be the color of the one that they see in front of them. So if the hats alternated colors, then all the peasants would be killed.
The goal is to maximize how many peasants survive - not just communicate which color hat someone has. - rezophonic, on 10/12/2007, -0/+1Is it just me, or do an unusual amount of the logic problems result in death? Is the author trying to hint at killing off the unintelligent through riddles? Genocide based on puzzle solving ability instead of race?
Or should I just make sure I always know the color of my own hat? - Lifestory, on 10/12/2007, -0/+1by the time i solve one of these problems, i probably could have finish all those chores the long way..
- jasqwerty, on 10/12/2007, -0/+1And this is accomplished how exactly? We have to assume you somehow have enough bend in the wire, and enough 'not bend', as well as length in general, to somehow connect 15 wires together.
- triforcer, on 10/12/2007, -0/+1the analytical puzzles were like doing word problems in math, i prefer the other puzzles on the site. much more fun.
- WrecksTXP, on 10/12/2007, -0/+1Oops. My bad.
- y2048, on 10/12/2007, -0/+1if you do the math, you'll find the child is currently -0.75 years old, which is -9 months.
in other words, the father is currently getting busy with mom. - Ordinant, on 10/12/2007, -0/+1He would have been funnier if his answer to "Where's the father?" was "in flagrante delicto."
- bradx3, on 10/12/2007, -0/+1I have discovered a truly marvelous proof of these, which this comment is too narrow to contain.
- Rabid_Llama, on 10/12/2007, -0/+1Hah, there's a critical problem with the "bundled wire" one. You have to make a circuit, so saying you hook a battery up to one wire and it can light the bulb is completely wrong.
- lukas88, on 10/12/2007, -0/+1smellinator, no, i still dont understand, but now I want to have your babies.
- lukas88, on 10/12/2007, -0/+1I have answered a few of the very easy/easy ones wrong too! However, somehow I am proud of myself just because I answered a few of them right.
- ztoner, on 10/12/2007, -0/+1@ dorkafork
Maybe I am doing the math wrong, but.... (math spelled out as plus signs seem to disappear.)
neg three fourths plus 6 is five and a quarter
five and a quarter times five is 26 years, 3 months. Not 27, which is how old the mother would be 6 years later.
That said, I like the problem. I may have to re-write it for my bio students... 8 ) - recursive, on 10/12/2007, -0/+1The scale has 3 possible outcomes, not 2: Balanced. Left side heavier. Right side heavier.
- recursive, on 10/12/2007, -0/+1You don't have time to wait a day for each bottle. You have to find the bottle before the celebration tomorrow.
- recursive, on 10/12/2007, -0/+1You don't have to solder them. Just connect them by touching them together.
- Barnstormer, on 10/12/2007, -0/+1It's already been observed that it is impractical to have 10 prisoners each drink wine from 500 bottles in 4 hours. If I were advising the king I would say get 1000 prisoners and give them each one sip. You'll have your answer quicker and with less chance of cross-contaminating the bottles.
Also, don't get into a land war in Asia. -
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