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Posts posted by ThunderCloud


This one isn't too hard, but is a fun exercise.
Two motor boats began speeding toward one another at the same time from opposite shores of a river. Upon reaching the opposing shore, each motor boat immediately reversed course and headed back to his original shore. In so doing, the motor boats passed each other twice. The first time they crossed, they were 700 feet from one shore of the river; the second time, they were 300 feet from the opposite shore. Assuming each boat traveled with constant speed, and neglecting any influence of river current, how wide is the river?

Although the Archimedes principle does apply. The volume of water displaced by the weight of the anchor in the boat is equal to the volume of water that would weigh the same as the anchor. Since the anchor is denser than water, this volume is larger than the volume of the anchor. Therefore, more water is displaced by the anchor's weight in the boat than by its volume in the tank, and the water level falls when the anchor drops.

After [(2/13) * ln(Population  1)] days, half the population will be zombies. Reasoning: I assume that each person (human or zombie) encounters 13 other people every day. On average, the proportion of those people who are not already zombies is [(
P  z) / P] (where P is the total population of humans and zombies, and z is the zombie count), and 50% of them will be transformed upon contact. I also assume that the total population P is constant (nobody dies, they just become zombified). Then we have dz/dt = (13/2)*z*((P  z) / P). That is, the rate of change of zombie population is proportional to both the zombie count and the proportion of nonzombie humans each zombie encounters. With a little manipulation, this becomes integrable, and gives (P  z) / z = e^((13/2)t + C). Assuming there is only one zombie at the outset, C = ln(P  1). Solving for when (P  z) / z = 1 (i.e., there are as many humans as zombies) gives the above result. 
22. The difficult cases are the ones where all the weights are the same, or where only one of them is different. A reference weight of 22 allows one to discriminate these cases well. For example, suppose all of the weights are 6 grams except for one which is less. You can determine that all of the others are 6 by observing that 4 of them weigh more than the reference weight, and that they all weigh the same as one another. To determine the weight of the "extra" one, add it to three of the other stones and compare against the reference weight. If it is greater, the extra stone weighs 5 grams; if equal, it weighs 4; if less, combine it with the reference weight and compare against four of the other stones. If it is greater, it is 3 grams; if equal, it is 2 grams; if less, it is 1 gram. Similar reasoning seems to work with stones that are each 5 grams...
 1

You have a dozen (12) stones weighing a whole number of grams between 1 and 6 each. You can obtain one reference weight of your choosing.
What reference weight can you choose to be able to figure out the individual weights of the 12 stones using a balance device for any possibility that may exist therein?
For an encore: what is the maximum weight range of stones (1 to N) that you could solve using 2 reference weights of your choice? Provided you can have as many stones as you need.
I don't believe, I have solved this one myself. We could make it a community project after the first question is answered.
HISTORICAL NOTE:
This problem originated on Brain Den. I constructed it based on Bonanova's problem Weighty Thoughts: http://brainden.com/forum/index.php/topic/4932/?p=84107 few years ago.
Back then limited number of people participated. The solution found was for specific numbers in that problem (range 1 to 5) – not general. I'd like to give it another try.
Using a reference weight of 5 grams, it would be easy to distinguish stones weighing 6 grams and 5 grams from the rest. Stones of lower weight could be sorted according to like weight, and then combined to see which combinations yield totals of 5 or 6+. The only difficulty is distinguishing a case where every stone weighs 3 grams from a case where every stone weighs 4 grams; either way, each is less than the reference weight, each is identical to every other, and any two are greater than the reference weight. The only method i can conceive to decide between these cases is by adding some other unknown but finely granulated weight to the scale (such as sand) to balance it... using this trick, one could determine which is the greater difference: one stone from the reference weight or the reference weight from two stones. If the weight difference between the reference weight and one stone is larger than between two stones and the reference, then the stones all weigh 3 grams; else they weigh 4.

I thought I had this one... but found a flaw in my reasoning. [solution withdrawn]

Assuming that the maiden's goal is merely to get out of the lake without encountering the ogre, that she has no particular direction she is obliged to go.
Wouldn't it behoove the maiden to perform continuous reevaluations of her course and steer to the spot currently opposite the ogre at every moment? I'll bet she'd have to be terribly slow oarswoman to lose this race. (However, I don't have the numbers...
Drat! You're right...
She could "spiral" her way out of the lake at slower speed. The question is how much slower... this problem is trickier than I thought.

If I were the girl in the boat, I think I'd start by centering the boat in the lake. Then, I would row directly away from where the ogre is standing. The girl will have to row a distance equal to the radius of the lake (
r), in less time than than the ogre can run a distance equal to half the circumference of the lake (Pi*r). Since it will take the ogre (Pi*r / 4) hours to complete the distance (assuming the radius is expressed in miles, of course), the girl must row with minimum speed S so that r / S = Pi * r / 4. With a little algebra, S = 4 / Pi mph. 
Although I believe the logic to answer the original problem was present, it was distributed among several postings... It did not seem right to mark any single post as the best answer. Therefore, here is the bonus round.
If approached correctly, this version is not much harder than the original.
The puzzle:
Three perfect logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e.1, 2, 3, ...), and that the sum of the integers on all three stickers is either 1002 or 1003. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' (or say "I don't know"). The question was repeated, again in turn, until EACH of the three logicians had either named their number or given up. All three stickers actually had the same number written on them. Who among the three logicians was able to deduce his number, and who among them gave up? (Furthermore, how did each answer?)

Yeah deciphering is not necessary, but it makes it mentally easier for me, at least, lol, to keep track of, and somehow I feel more accomplished for getting an extra piece of information ;P.
Check out the puzzle I linked...the setup there (where True and Lie can also answer any yesno question...they are omniscient apparently about Random's answers) disallows you to know which word stands for which, since there were only two possible answers. In these puzzles 8 possible arrangements of True/Lie/Random, so 2^3=8 will resolve which arrangement with no information to spare. This puzzle there are three potential answers: a word, a word that is different than the previous word, and lack of an answer, so there is more possible information gained.
Nice puzzle tho, it was fun, thanks .
I will take a look at it, thanks. Glad you enjoyed the problem. Btw, I only count 6 possible arrangements for True/Lie/Random (3!) ...?

Lol, well as an engineer, I do try to be as
lazy*cough* efficient as possible...Denoting the gods as A,B, and C for simplicity...
1st question: Ask A, "Would B answer yes (or 'whatever in his language means yes' if you need to be that specific) to the question 'Is C the random god?'"
If he can't answer, then B is the random god. In this case, there are a couple ways you could go. The simplest IMO is probably to ask A, like, "Would you answer yes if I asked you 'Is the sky blue'?" or something like that that tells you which of his answers means yes (or no), and then for the last question ask "Is the sky blue" to determine which of A and C is the liar and truthteller.
If A did answer the first question, then either A or C is the random god, and B is definitely not the random god. In this case, take note of what he just said, and move on to B. Ask B, "Would C answer yes (or 'whatever in his language means yes') to the question 'Did A just say [insert whatever A just said]'?" (or 'Is the sky blue' or any question that is true I suppose... )
If B can't answer question 2, then C is the random god, and whatever A answered previously means 'no' in A's language. So move back to A and ask him 'Is the sky blue' or some such to determine which of A and B is the liar and the truthteller.
If B did answer question 2, A is the random god and whatever B just answered means 'no'. So then ask him 'Is the sky blue' to determine which of B and C is the liar and the truthteller.
(Of course, if you want to be technical you can always replace 'Is the sky blue' with a more undisputable mathematical question, like "does 1+1=2?" or logical question, like "Does yes mean yes?", but I just like to use it in these problems )
Well done. My answer was very similar to this.
The only real difference was my third question in the case where B is not random. After determining which of (A, C) is random using question #2, I then ask the nonrandom god whose answer I have heard (either A or B) "If I were to ask you whether you always told the truth, would you say (insert word that this god has previously uttered)?" If he replies with the same word again, then he is the god of Truth; else he is the god of Lies.
Although it is possible to decipher at least one god's language during the course of the three questions (as you demonstrated), it is actually not necessary... which leaves me with the feeling that the problem can be made still harder, somehow....

"Random"=R=Random answerer
T=Truth teller
L=Liar
1. To #1: If I ask 2 if he is random, what will he say?
Answers 2 is not random go to question 4
No answer 2 is random go to question 2
2. To #1: If I ask 3 if 2+2=4, what will he say?
Answer will be “no” go to question 3
3. To #1: Does 2+2=4?
Answers same as question 2 1L, 2R, 3T
Answers different from question 2 1T, 2R, 3L
4. To #2: If I ask 3 if he is random, what will he say?
Answers 1 is random go to question 5
No answer 3 is random go to question 6
5. To #2: Does 2+2=4?
Answers same as question 5 1R, 2T, 3L
Answers different from question 5 1R, 2L, 3T
To #2: Does 2+2=4?
Answers same as question 1 1L, 2T, 3R
Answers different from question 1 1T, 2L, 3R
The algorithm is selfreferential in question 5 (probably a typo)...

ThunderCloud, you don't specifically limit our questions to be of the Yes/No variety, but it seems implied?
If I understand your question correctly: the gods of Truth and Lies will not respond to questions for which an appropriate answer is neither "Yes" nor "No." The "god of Randomness" will answer any question with a "Yes" or "No" response.

It does say in the OP that "the gods respond only to yesno questions"...(paragraph 2, 4th sentence)
I meant to ask, what happens if you ask the lying god, "Is your answer to this question a lie?"
True would answer "no" definitely, but it's a paradox for Lie.
Ask two of them, "will the random god answer whatever means 'yes' in his language to the next question?", only Random can answer, so this tells you who Random is.
Use last question as above if acceptable.
This is a different solution than the one I had in mind, and I like it.
I had not considered exploiting that aspect of the lying god...Having said that, I know at least one solution that does not rely on this technique. So, to make life difficult: suppose that the lying god, when thusly cornered, answers at random and considers either response to be a "lie". Can you still identify the gods?

Hope I didn't mix up the questions after rearranging them a billion times. . .If non yes/no questions are allowed. . .1. To #1: What is 2+2?
Answer #1 is random answerer See question 4.
No answer See question 2.
2. To #1: If I ask #2 if 2+2=4, what will he say?
No answer#2 is random answerer See question 3.
Answer The word given in answer means no. See question 3.
3. To #1: Does 2+2=4?
Answers same as before #1 is liar if no answer to question 2, 2 is random answerer, #3 is truth teller. If answer to #2, switch #2 and #3.
Answers with different word #1 is truth teller if no answer to question 2, #2 is random answerer, #3 is liar. If answer to #2, switch #2 and #3.
4. To #2: Are you the truth teller? Given answer means yes. See question 5.
5. To #2: Does 2+2=4
Answers same as before #1 is random answerer, #2 is truth teller, #3 is liar.
Answers with different word #1 is random answerer, #2 is liar, #3 is truth teller.
There is a path that does not quite complete: If in (1) there is no answer, and in (2) there is no answer, then in (3) there can be no comparison to "the same answer as before."

4. If I were to ask the lying god about a question the truth telling god would have not been able to answer is it possible for him to lie by answering
Asking the lying god
If the god to you left (who just happen to be the truth telling God) was asked if his answer would mean no what would he reply.
Now the truth telling god cannot reply to the question "will your answer mean no". Can the lying god whom you are addressing then say his equivalent of "Yes" or "No" as he chooses (and thus lying) or would he also be constrained not to answer?
In this case he would not answer. Neither the "god of Lies" nor the "god of Truth" can respond to a question to which there is no definite answer.

I see...
Actually the puzzle I linked had the same setup for 2. @bonchan.
A couple questions to TC:
1. Do the other two gods know what Random is going to answer before hand?
2. Are you allowed to ask questions that potentially do not have a definite answer? Can you take the totem's lack of an answer as an information source?
3. Are you allowed to ask questions that potentially create a paradox?
If you ask "Do you always lie?" to the lying god, does he not answer or his head explode ;P?
You could potentially use a strategy of using nonanswers to ferret out the random god with the first two questions and the lying god with the last.
In order:
(1) No.
(2) Yes.
(3) I am not sure what you mean...
If you ask the "god of Lies" the question "Do you always lie?", the question has a definite answer (i..e it is Yes), therefore the god will answer with his word for "No."
Such a strategy is what I am seeking.

Um...I'm getting deja vu...is it something like
Something like, but not exactly like. It is a variation. Each god speaks his own individual language (so there are three unknown languages), and you do not know the words each will use in advance.

This puzzle is a variation on the "Hardest Logic Puzzle Ever", as can be seen on Wikipedia. I've found it to be a fun problem to play with.
There are three gods, each of whom speaks through his respective totem. One god always tells the truth, one always lies, and one answers entirely at random. The three totems are unlabeled, so you do not know which god is which. The gods respond only to yesno questions, and may only be addressed individually via the querant's choice of totem. Furthermore, each god answers in his own personal language, and you know nothing in advance about any of the three gods' languages, save that each includes distinct words for "yes" and "no". Your task is to correctly ascribe each totem to its respective god with only three questions. What are your questions, and how are they directed?
Note: Because the "god of Truth" must always tell the truth, and the "god of Lies" must always lie, neither god is able to respond to a question which lacks a definite answer. The "god of Randomness", however, will respond to any question  his response is unrelated to the content of the question, but is instead prompted by the fact that he was asked one.

I think you are assuming too muchIf the answer is 25%, then because 2 out of 4 answer choices are "25%", the answer must actually be 50%. This is a contradiction. So the answer cannot be 25%.
If the answer is 50% (or 100%), then because 50% (or 100%) is 1 out of 4 answer choices, the answer must be 25%. This is again a contradiction. So the answer cannot be 50% (nor 100%).
So none of the provided answer choices are correct. Therefore the probability of choosing the correct answer is 0%.
I did assume that if (a) is correct then (d) is also correct, and vice versa. If this is not the case, then exactly one of them must be correct, and the answer is 25%.

If the answer is 25%, then because 2 out of 4 answer choices are "25%", the answer must actually be 50%. This is a contradiction. So the answer cannot be 25%.
If the answer is 50% (or 100%), then because 50% (or 100%) is 1 out of 4 answer choices, the answer must be 25%. This is again a contradiction. So the answer cannot be 50% (nor 100%).
So none of the provided answer choices are correct. Therefore the probability of choosing the correct answer is 0%.

Let it inside a plastic bag, tie a string on the opening (let no air),
fish in the bag of water is heavier than the bag of water without the fish.
Tie the end of string to a weigh hook with the fish bag fully submerged.
I've been pondering this one... and I am not sure if it would work without lifting the bag out of the aquarium (at least if I am picturing it correctly). If the bag (with or without the fish) is submerged in the aquarium, then a weight hook would be useful for measuring its
buoyancy, but not its weight. A bag full of water, submerged in water, would have virtually zero apparent weight (the plastic of the bag and the material of the string would contribute a little, technically); if the fish had his air bladder balanced right, the bag with the fish inside would also have virtually zero apparent weight, at least while underwater. (In fact, if the fish had an overinflated air bladder, the bag with fish inside might appear to have negative weight while submerged, as the fish would tend to float and push the bag to the surface.) 
You would also need to account for the water displaced by the fish to get an accurate calculation. So, a measure of the fish's volume would be required. I believe this is also the case for the "fish in a plastic bag" solution  it is important to know how much water is displaced by the fish in order to determine the weight.
To modify my approach: capture fish in open, buoyant pot. Measure interior and exterior water levels. Release fish, fill pot with water until exterior water level is the same as before (pot sits as deeply in the water of the aquarium as before). Measure the new interior water level, and calculate the amount of "extra" water added to the pot to reach the same weight as when the pot held the fish (including the volume of water the fish originally displaced). Then, using the known (or easily measurable) density of the water, you can determine the weight of the "extra" water, and this is the weight of the fish.

Capture the fish inside an open, buoyant container (something like a large pot with some inherent buoyancy), so that the fish is contained within the pot which is now floating in the aquarium. Measure the water level both inside and outside the pot. Release the fish, and fill the pot with aquarium water until it sits as deeply in the water as before (the "outside" water level is the same). Note the water level within the pot again, and how this differs from the interior water level as measured with the fish. You can determine the weight of the additional water by calculating its volume (current interior water depth minus interior water depth with fish included, multiplied by the crosssectional area of the pot...) and multiplying by the known density of water. This is the weight of the fish.
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