MrRubix's Riddle Thread

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

For the cubes problem I built off an initial chain N and considered all cases 2<=x<=N

Assume you have a maximum N cubes, with an initial chain X cubes long, and leftover cubes L=N-X. The number of unique available areas to attach the L cubes to the initial chain would be represented by A=round(X/2), so if your initial chain is even, you can attach unique sub-structures to half the available faces, and if it's odd, you don't neglect the middle of the initial chain.

A given available side (denoted as A) here has to be taken care of recursively for all attached substructures, where each recursion also needs accounting for all unique non-isomorphic permutations of additional cubes attached to other faces. Also, substructures forming sub-X contiguous chains outside of the initial chain also need accounting for to avoid overlapping with earlier cases (since it would have been considered as another initial chain case prior). This isn't so much a problem when N=5, but it would be absolutely disgusting later on and I'd probably need to write a program to crunch to crunch the recursions for me.

For N=5, if X=5, L=0, and there is only one arrangement for this chain.
For N=5, if X=4, L=1, then A=round(4/2)=2, and there are two arrangements here.
For N=5, if X=3, L=2, then A=round(3/2)=2, and there are seventeen arrangements here.
For N=5, if X=2, L=3, then A=round(2/2)=1, and there are nine arrangements here.

1+2+17+9=29

[MahouMinachan]

Re: MrRubix's Riddle Thread

ugh... the only thing bothering me about this new one is Man 2... Man 1 and Man 3 are easy to figure out.

Here's what I have so far:

Man 1: F=ma (m and a are dependent to t)
so I = (dm/dt) * a + mv

Man 3: F=ma (m is constant, a is dependent to t)
so I = mv

So Man 1 is faster than Man 3.

Man 2... I'm still figuring out.

The only thing I could think of for Man 2 is that his mass deviates but slowly gains speed faster than Man 3 because of snow, and that the perpendicular force doesn't affect his vertical force in anyway, but not as fast as Man 1 since his cart's total weight gains faster than Man 2's cart.

So:
1st: Man 1
2nd: Man 2
3rd: Man 3

[ieatyourlvllol]

Re: MrRubix's Riddle Thread

I'll take a crack at this one, too...

Man #2, Man #1, Man #3

Man #3 has no extraneous factors affecting his velocity. Man #1 inherits the downward momentum of the snow as well as the consequent additional mass. Man #2 does the same while converting otherwise static mass into a means of propulsion via displacement. According to the conventions of conservation, the momentum of the snow tossed corresponds to the gain in cart velocity (under the assumption of ideal transfer). In essence, Man #2 is contributing his own energy to the equation, meaning he should get to the bottom first. The results should hold true regardless of slope grade, since the momentum vector for snow mass and that for snow tossed will always remain parallel.

I have this strange feeling that I'm going to be foiled by acceleration. =/

Seeing as how nobody bothered to post the straightforward mathematical process for solving #2, I'll go ahead and paste my own work directly from notepad for anyone who is interested.


Label designations:

walker1 + walker2 = Fractal and Emerald (interchangeable)
pair1 = Tass + walker2
pair2 = Tass + walker1


Action sequence:

Phase 1
-walker1 travels at 3mph until receiving bike at transaction
-pair1 travels at 5mph until separating

*pair1 separates*

Phase 2
-walker2 continues along course at 3mph
-Tass travels on bike in opposite direction at 15mph
-walker1 continues along course at 3mph

*Tass and walker1 rendezvous for transaction*

Phase 3
-walker2 continues along course at 3mph
-pair 2 travels 5mph towards destination

*walker2 and pair2 arrive at 30-mile mark at the same time*


Equations for deriving solution:

3/5 = (walker2 remaining distance at transaction) / (pair2 remaining distance at transaction)

3(pair2 remaining distance at transaction) = 5(walker2 remaining distance at transaction)

3(30-pair2 distance at transaction) = 5(30-walker2 distance at transaction)


Mathematical Operations:

d = walker1 distance at separation

d/3 = time of separation

5d/3 = pair1 distance at separation

2d/3 = distance between walker1 and pair1

d/9 = walker1 distance until given bike

5d/9 = Tass backwards displacement until giving bike

10d/9 = pair2 distance at transaction

10d/27 = time of transaction

d/27 = time elapsed between separation and transaction

d/9 = walker2 additional distance until transaction

16d/9 = walker2 distance at transaction

3(30-10d/9) = 5(30-16d/9)

90 - 30d/9 = 150 - 80d/9

60 = 50d/9

50d = 540

d = 10.8 mi

time of separation = d/3 = 10.8mi/3mph = 3.6 hr

walker2 distance at transaction = 16d/9 = 172.8mi/9 = 19.2mi

walker2 remaining distance at transaction = 30mi - 19.2mi = 10.8mi

walker2 remaining time at transaction = 10.8mi/3mph = 3.6 hr

walker2 remaining time at separation = 12mi/3mph = 4hr

walker2 time between separation and transaction = 4hr - 3.6hr = 0.4hr

total time elapsed = 3.6hr + 0.4hr + 3.6hr = 7.6hr

[s0ulst0n3]

Re: MrRubix's Riddle Thread

Isn't this more like physics instead of a riddle?

[MahouMinachan]

Re: MrRubix's Riddle Thread

I kinda did #2 mathematically...

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

Yeah it is a physics riddle. But this thread is home to riddles of any form

[Reach]

Re: MrRubix's Riddle Thread

Ah, physics riddles. I remember no one was able to solve this puzzle last time I posted it, so maybe I will post it again.

Assume that gravity on or near the airless surface of a certain planet called FFR is such that any free-falling object will reach a velocity of 10 meters per second after one second, 20 meters per second after two seconds, 30 meters per second after three seconds, and so forth. Suppose a frictionless pulley (i.e., a wheel with a grooved rim that, once in motion, does not slow down due to friction), having negligible mass, is attached to a sturdy beam at the top of a vertical, airless mine shaft, and an inelastic rope 1000 meters in length but of negligible weight is threaded through the pulley with identical buckets of negligible weight dangling at each end. In one bucket are fifty stones weighing 1 kilogram apiece and an alien named Rubix weighing 50 kilograms, while in the other bucket is a single stone weighing 50 kilograms and an alien named Reach weighing 50 kilograms. Suppose Rubix commences dropping stones out of its bucket at a rate of one per second. The instant the first stone is dropped, Reach commences climbing the rope directly above its own bucket at a rate of one meter of rope per second. If Rubix starts at rest at a position 20 meters below Reach, how many stones will it have dropped the instant the two buckets are precisely side by side?

[ieatyourlvllol]

Re: MrRubix's Riddle Thread

I got an answer of eleven stones to the above riddle. I'm confident in that value for the steps taken, though I wouldn't be surprised if it is incorrect (just look at the last time I gave "eleven" as a riddle answer). 8D

[Reach]

Re: MrRubix's Riddle Thread

I got an answer of eleven stones to the above riddle. I'm confident in that value for the steps taken, though I wouldn't be surprised if it is incorrect (just look at the last time I gave "eleven" as a riddle answer). 8D

Incorrect.

I think I know how you got that answer though, and you're not entirely far off.

[ieatyourlvllol]

Re: MrRubix's Riddle Thread

After incorporating the 1 m/s climbing rate factor into my calculations, I arrived at a new (and hopefully closer/correct) answer of seven stones. If that figure is wrong, it means I'm missing a critical component somewhere in my operations. =/

[tuv]

Re: MrRubix's Riddle Thread

16?

[Reach]

Re: MrRubix's Riddle Thread

Both are wrong.

Given I am right in assuming what you did initially to get your first answer, you were closer then. However, you could have came up with that answer a different way than what I'm thinking.

If it's any condolence, I gave the problem to someone with their masters in physics and they couldn't get it. It's more about the logic of your approach than anything else.

[andy-o24]

Re: MrRubix's Riddle Thread

Just before dropping the thirteenth stone they should be almost exact assuming that Rubix is at the bottom end of the rope. I've got almost no logic whatsoever but it isn't a random guess.

-o24

[QuadDamage]

Re: MrRubix's Riddle Thread

EDIT: Just answered my own question

[QuadDamage]

Re: MrRubix's Riddle Thread

Hm initial calculations give me around 27 stones, but that seems way off to me.

Will take another look at it later -- not confident in my answer. It's been a while since I've messed around with pulleys/tensions/ropes and that sort of thing. I'm making the assumption that initial acceleration is 0 and that acceleration is uniform for both sides once the masses are no longer equal (as it wouldn't make sense to me otherwise), and I'm making the assumption that distance traveled per second is half the acceleration (for a second's timeframe, that is. Also, the acceleration changes every second). I'm also assuming that the fact Reach is climbing is irrelevant since the total weight is always going to be 100 kg. I'm therefore assuming that each side needs to travel 10m since accelerations are equal, and so I'm basically figuring out the summation of distances traveled until both sides travel 10m as a function of their accelerations.

Changing a few of these assumptions also gets me around 22 stones, but again, not confident. Gonna look into this problem more when I get a chance.