Whichever model is a better independent measure, impulse (force is roughly constant and effective throughout crunch) or energy (all energy is transferred into squashing distance and no energy lost to earth), I think all of you are calculating volume wrong.

If log 1 is the base of the tree, I got 13th log and 23rd log respectively.

Personally I find the latter assumption too much.

If you look closely all answers (including 4 which looks almost the same) have a strike through them regardless. What do you claim your correct answer for question 1 is?

Humanoid (vague) analysis based on araver's lead:

All clues being equally chaotic, it seems like A E I O U might be 'different', and that consecutive or in-between vowels sometimes have no effect. The effect of vowels seems more complicated than a fixed combination of toggling offset's isz/dsz status and adding on the spot.

For spontaneous (unexplained) changes in offset it seems like adding odd numbers occurs a lot more than adding even numbers (at least in my notation).

All jumps seem to be positive-ended (from our natural POV).

Other analysis:

Following these rules naively (basically araver's observations)

-3 -> 1 when hitting vowel

-1 -> 2 when hitting vowel

-4 -> 1 when hitting vowel

-2 -> 3 when hitting vowel

I get this output:

I have used a very insufficient algorithm for guessing the starting position: 1 if non-vowel, 2 if vowel. The first position is 0.

If I plugged in araver's starting points I'd have more data on where the current algorithm is wrong.

If the first letter is correct, I've assumed it's the correct starting point. (I think this accounts for the outliers.)

The error output then indicates the first discrepancy between our guess and the correct password in the form:

<Letter landed on> <Position of this letter (starting at 0)> / <Length of command> <Projected offset> <Actual offset>

Projected offset is just this: e.g. 2 1 0 -1 -2; always assuming the offset is decremented.

Correct: ARKANSAS

Correct: ILLINOIS

Correct: KENTUCKY

Correct: NEWHAMPSHIRE

Correct: NEWYORK

Correct: NORTHCAROLINA

Correct: NORTHDAKOTA

Correct: VERMONT

Correct: WASHINGTON

Got 9 out of 45 correct

Wrong guesses for:

ALABAMA

ALASKA

ARIZONA

CALIFORNIA

COLORADO

CONNECTICUT

DELAWARE

FLORIDA

GEORGIA

HAWAII

IDAHO

IOWA

KANSAS

LOUISIANA

MARYLAND

MASSACHUSETTS

MICHIGAN

MISSISSIPPI

MISSOURI

MONTANA

NEVADA

NEWJERSEY

NEWMEXICO

OHIO

OKLAHOMA

OREGON

RHODEISLAND

SOUTHCAROLINA

SOUTHDAKOTA

TENNESSEE

TEXAS

UTAH

VIRGINIA

WESTVIRGINIA

WISCONSIN

WYOMING

Wrongness analysis ---------------

ALABAMA

Wrong starting point

ALASKA

Wrong starting point

ARIZONA

Wrong starting point

CALIFORNIA

Wrong starting point

COLORADO

Wrong starting point

CONNECTICUT

I 7/11 0 3

DELAWARE

Wrong starting point

FLORIDA

Wrong starting point

GEORGIA

Wrong starting point

HAWAII

Wrong starting point

IDAHO

Wrong starting point

IOWA

Wrong starting point

KANSAS

A 1/ 6 0 3

LOUISIANA

Wrong starting point

MARYLAND

A 5/ 8 0 1

MASSACHUSETTS

U 7/13 0 3

MICHIGAN

A 6/ 8 1 2

MISSISSIPPI

I 7/11 0 3

MISSOURI

U 5/ 8 0 2

MONTANA

A 6/ 7 1 2

NEVADA

Wrong starting point

NEWJERSEY

E 7/ 9 0 3

NEWMEXICO

I 6/ 9 1 2

OHIO

Wrong starting point

OKLAHOMA

A 7/ 8 3 -2

OREGON

E 2/ 6 0 11

RHODEISLAND

Wrong starting point

SOUTHCAROLINA

Wrong starting point

SOUTHDAKOTA

Wrong starting point

TENNESSEE

E 7/ 9 0 3

TEXAS

Wrong starting point

UTAH

Wrong starting point

VIRGINIA

I 6/ 8 1 2

WESTVIRGINIA

I 5/12 0 1

WISCONSIN

I 7/ 9 0 3

WYOMING

Wrong starting point

CONNETICUT I 7/11 0 3

KANSAS A 1/ 6 0 3

MARYLAND A 5/ 8 0 1

MASSACHUSETTS U 7/13 0 3

MICHIGAN A 6/ 8 1 2

MISSISSIPI I 7/11 0 3

MISSOURI U 5/ 8 0 2

MONTANA A 6/ 7 1 2

NEWJERSEY E 7/ 9 0 3

NEWMEXICO I 6/ 9 1 2

OKLAHOMA A 7/ 8 3 -2

OREGAN E 2/ 6 0 11

TENNESSEE E 7/ 9 0 3

VIRGINIA I 6/ 8 1 2

WESTVIRGINIA I 5/12 0 1

WISCONSIN I 7/ 9 0 3

Obviously stopping at the first discrepancy is just a hopeful snapshot of a representative discrepancy in the algorithm. This, the assumption of the offset being decremented by default, and ignoring the rest of the actual offset vector could mean that there are no clues in this data (the cost of assuming that first vowels' effects are mostly independent of other vowels' effects and positions.)

All answers to question 1 are wrong.

x^2 + y^2 = 4

at

(-sqrt(2 + sqrt(2)), -sqrt(2 - sqrt(2))) = (-1.84776,-0.765367)

(-sqrt(2 - sqrt(2)), sqrt(2 + sqrt(2))) = (-0.765367, 1.84776)

(sqrt(2 - sqrt(2)), -sqrt(2 + sqrt(2))) = (0.765367, -1.84776)

(sqrt(2 + sqrt(2)), sqrt(2 - sqrt(2))) = (1.84776, 0.765367)

If log 1 is the base of the tree, then log 13 transfers the greatest impulse.

People who have done high school physics will know that collisions are associated with a finite period of time. It is not inappropriate (for general purposes) to model the fly's acceleration as constant during contact.

The fly may end up sticking on the train, or bouncing off it. There are other complications with making use of the provided speeds.

Either way, the "moment" in question is not an instant, so if you took "moment" as "period of time" then the most informative answer might be:

v(t) = (v_f - v_i)/(Δt) * t + v_i, 0 <= t <= Δt

We know v_i = 10 kph.

[Five encrypted thoughts]

The five words corresponding to THOUGHT:

81:87 SHORTY

110:114 NEAR

250:253 SORE

211:215 REAR

228:232 HUNT

Is there slang in this?

If I had nothing else to go on, I would start looking for "near the", "is sore", "hunt the", "a thought", "thought of" "i thought", etc.

[Five encrypted thoughts]

It is easy to encrypt as specified with the aid of a spreadsheet but the encryption will always be a symmetrical string. Given either first half or last half of solution, it would be easy to determine the other half with a spreadsheet. The problem arises that every letter in the encrypted can be any one of at least 13 different combinations. If you do not know one, you cannot positively identify the other. Bottom line, this cannot be decoded by any algorithm. With sufficient hints and basic word composition rules, we could eventually get the answer but it would be an elaborate maze of guesses, especially since no spaces are provided for word division.

Okay, letter by letter is getting me nowhere fast. I think that a brute force with a dictionary could probably break this pretty quickly though, but I do not have access to that other than manually, which would be pretty slow. I can find a word if it is in there somewhere, so I will ask for a word hint please.

Each character can be decoded into one of 26. Even using a dictionary, the biggest problem is not speed (though it is a major issue anyway), but how to recognise words. The most commonly used dictionary has 250000 words, which is too many. If a search algorithm accepts words of any length, then the search would start off creating many trivial words like BA or CA, words that are virtually never used; there is no easy way to distinguish MEANINGFUL words. This might be faintly plausible if I could find a dictionary of common words only.

No, there will only be 2 instances of me: the one that origanaly put in the money, and the one who puts in the interest, because he said "I will like to take out all money but £100", ensuring that ALL interest is taken out, not just the 1st set of interest.

No, assuming the future isn't changed by time travel, you can't only have 2 instances of yourself going to the bank. If you only went twice, then you only had £100 + £3, yielding £106.03, which is clearly not infinite. If you claim to get infinite interest, then you must have entered the bank on infinite occasions.

Your argument for 2 instances of you fails because then you can't explain why your interest would be infinite.

[Think about these]

7. First of all, according to general relativity, nothing can move at the speed of light, and nothing can exceed the speed of light. There's no such thing as "above the speed of light". With that concluded, number 7 is deemed impossible.

Your objection to "above the speed of light" is debatable (http://en.wikipedia.org/wiki/Faster_than_light). Besides, there are times in modern developments when we consider "was Einstein wrong", unless you suggest physicists have learned nothing since him.

[Five encrypted thoughts]

Interesting, but I am sceptical about whether word hints will help solve subsequent parts of the message unless there are a lot of words given. I'm going for bruteforce.

Look. Imagine I put in £100. Let's asume the intrest was 3% each year. In 1 year, I would have £103 (from the interest), and I say "I would ike to take out all money but £100" I obviously have £3 with me now. If I take it back in time to when I origanaly put £100 in, and put my £3 in there too, I would have £103 in. But then, when the interest is put on, 3% more will mean I would of had £106.03, which means instead of taking out £3, I took out £6.03, and put that in the bank instead of the £3, so I have £106.09, so I end up with the interst of £109.27, so I put in £9.27 and have £109.27 in the bank...

My money is constantly growing, however, it is all instantly hapening at the same time. Basicly, I wave an infinite amount of money!!!

Due to the natural law of conservation of time, you will end up forming an infinite queue at the bank consisting of you and yourself, thus taking forever to generate the infiniteth dollar.

[If You Can]

chromium breathalyzer

juice box

pipette

By object, do you mean physical?

What's a WAI?

Yeah I meant even cases where number of troublemakers is equal or less than the number of honest persons.

It is not solvable with 2 people, and 4 may be impossible too (perhaps an informal proof may arise, or simply a brute-force program demonstrates this one and we give up on the rest). If it is solvable, I wonder if there is a general pattern with even cases, particularly one with polynomial cost.

My bad, I only read the start and end of plainglazed's post, but k-man's proof that you can add the unpaired one puts away any doubts.

Just wondering, can even cases be solved with limited questions?

Thanks but Plainglazed finished it off. And glad to see you understand the answer.

You all seem too happy to ditch the puzzle before solving it; Plainglazed notes (99 - 1) / 2 = 49, which is odd (PUN!)!

No one has demonstrated a complete solution yet: a complete solution needs induction or a general algorithm, along with stating the money spent in the worst case scenario.

While collectively, 3M, 5M and 7M cases have been solved fully, independent arguments were used to solve the worst case; there has been no induction. "3M, 5M, 7M have a pattern" is an illegitimate observation since the related series is actually 1M, 3M, 7M, 15M, ...

[5M is not a base case of 7M's worst case; nor is 9M related to 7M SINCE (9 - 1) IS A MULTIPLE OF 4!

Number of people: 1 3 7 15 31 63

Cost to shift/reduce: 0 1 3 7 15 31

Total cost to solve: 0 1 4 11 26 57]

New interest:

2M (1 Honest 1 Troubler)?, 4M (2H 2T)?, hence 9M?

These even-numbered groups seem less favourable.

It should be clear by now that with our popular algorithm the worst case is when, in an odd-sized group arranged for pairing off, every person in a pair that you question happens to be a Troubler and says Yes their 'partner' is honest. This minimises your information gain.

This is a nice challenging puzzle.

There must be a way to establish FACTS.

There must be an interrogation system that can be used repeatedly to establish facts.

To find an honest one you must sift out the others, using the 'honest majority' to your advantage. To sift out 49 troublemakers with 100 questions you have to aim to learn 1 fact per 2 questions. (How can you learn 1 fact with 2 questions as in the above spoiler?).

Thanks, it seems that I was a little too quick on the trigger. I was curious as to why I had $2 left over! I expect there is a solution, but ya gotta spend every last dollar.

I'll get back to thinking about it. Thanks again!

I haven't thought about the worst case scenario yet, but in general I think the amount of money left is arbitrary.

There is no way to be certain given those conditions. The best that can be hoped for is to have a fair chance of being pretty sure, since we are told the majority are honest, so we know over half are honest. So I would choose one villager and ask ten others if that one is honest. If most of them say no, I would say that is a bad start and find another villager to ask another ten about, until I find one that most of the ten say is honest. After finding such a villager, proceed to ask another ten about that same one, and so on but stopping when my money is at 100, and hopefully the majority will agree and I will have found an honest one. That method has no guarantees and works better the more the honest people outnumber the troublemakers.

I can see no other methods that would work. If we knew the exact numbers of troublemakers, or if they always lied, then there might be another way. Or if we could continue asking questions without having to pay another dollar for extra questions, there could be a way then by pairing them off and eliminating pairs when one says the other is not honest and then remixing the pairs up. Any two who say each other are honest must either be both honest or both troublemakers, so by mixing pairs up after each round of questions, the troublemakers would eventually be eliminated and all that is left are the honest ones who outnumber them. But that would cost a lot of money to do.

All of your conclusions are wrong or off-track, except the bolded thought.

[Murder in the Desert]

I think the story given by the OP (haven't read the book) is a bit vague; not all logicians can come to a conclusion based on that only (but some can).

For example, how come C dies of thirst but A and B don't? What were the chances of them all finding water?

My opinion anyway:

Loosely speaking, C was already slowly dying of thirst, and drinkable water would save him. Neither A nor B directly/intentionally caused C's condition of thirst (at least not that we know of), which will kill him.

Supposing more water was impossible to find, A's actions guaranteed that C would die. We cannot know whether A intended C to drink the poison or knew that even without drinking the poison, C would die. We could also imagine that B saved C from the poisoned water, but guaranteed he die of thirst instead.

My guess at an answer: Although A's actions guaranteed that C would die (this does not equate to "C is a dead murdered man"), C did not die as a result of A's actions but from B's. Furthermore, C died the way B intended.

Conclusion:

B intended for C to die of thirst (and he is not directly responsible for the thirst) by getting rid of water, the only thing that could save C. I don't know much about law, but I would consider this to be murder committed by B, since they intelligently planned and ensured for C to die of thirst.

lol got me good. I somehow doubt it would be convincing standing in front of it though.

The Shape Shifter

[Think about these]

4) When she is standing in front of her dead grandmother, I don't think she would be so willing to accept the "she never killed her grandmother" theory.

In modern science fiction time travel terms, "traveling back in time" is transferring matter (~teleportation/creation/destruction...) to a different time in an equivalent parallel universe (it was identical and equivalent until the time travel occurred; that then changes the whole future after the destination time.)

Universe #1 is the "original" universe.

Universe #2 is a parallel universe.

-Girl travels back in time. She is now in universe #2. She cannot get back to #1 even if she tried to return immediately (she would find the "back to the future" is different to her original "present").

-Girl kills grandmother.

There is nothing questionable about it other than in the physics of time travel; her existence is not paradoxical as she simply appeared from nothing in #2.

I opened the spoilers in your signature by Tab + Enter, but naturally I did not stop in time and gave you a reputation mark - actually I don't know whether it was +1 or -1 due to a bug.

The two lines are parallel. If he walks along the second line, it is implied that the first line would still be visible without having to walk between them.

Just for fun:

1) A straight line is an arc of a circle with infinite radius.

2) Starting at any point on the circumference of a circle, you can take either direction along it and eventually reach your starting point.

3) Therefore, if the man travels to the left (away from the second line) for an infinite period of time, he may eventually see the second line and even return to the starting point.