MrRubix's Riddle Thread

[One Winged Angel]

Back to the drawing board :/

[s0ulst0n3]

Originally Posted by heyitsmee

so Shash Soulstone and laggy at night around the campfire working their wood
while laggy pays to enjoy others wood because he cant get his own wood
ok

MrRubix_MK5 i hope you had a camcorder this will be hotter then paris

Hawt.

[Fractal_Monkey]

Reach, I think it's 51.

If there are six points all connected to each other by lines, then there will be 15 lines in the diagram. The maximum number of intersections of those 15 lines would be if each line intersected the other 14, which gives sum(1:14) = 105.

However, at each of the six given points, there are 5 intersecting lines. So we have to subtract the maximum possible number of intersections of 5 lines (10) at each point, and replace it with the one forced mass intersection. This gives a total maximum of 105 - (6x9) = 51 intersections.

Don't ask for a diagram, my solution doesn't give an arrangement of points that satisfies these conditions.

[Reach]

Originally Posted by Fractal_Monkey

Reach, I think it's 51.

If there are six points all connected to each other by lines, then there will be 15 lines in the diagram. The maximum number of intersections of those 15 lines would be if each line intersected the other 14, which gives sum(1:14) = 105.

However, at each of the six given points, there are 5 intersecting lines. So we have to subtract the maximum possible number of intersections of 5 lines (10) at each point, and replace it with the one forced mass intersection. This gives a total maximum of 105 - (6x9) = 51 intersections.

Don't ask for a diagram, my solution doesn't give an arrangement of points that satisfies these conditions.

I don't need the diagram. This is correct and the best way to solve the problem. Usually I would give the problem with more dots so people can't draw diagrams, but allowing them to do so is a good trick to keep people from solving it, since most people jump right to diagrams and can only end up getting 49 or 50 intersections...not the optimal 51.

Exceptionally good thinking.

Since Rubix hasn't posted a new puzzle, I'll post one:

Five identical cubes, each having a volume of 1 cubic inch, are fused together to form a solid object. Adjoining cubes are fused together in such a way that their contiguous faces cover one another entirely without any overlap. No cube is connected to the rest of the object at only a finite number of points or lines. How many different shapes could these 5-cubic-inch objects have?

[MrRubix_MK5]

I had 51 actually, miscounted one of my points. I used a diagram. Not very elegant, but it worked.

[MahouMinachan]

it's weird, because I used to play this game Block Out, and it used blocks exactly what Reach is talking about. I can only remember 17 of them though...

[MrRubix_MK5]

Are we assuming that a shape is all that matters? (ie. an arrangement isomorphic to another is considered the same arrangement? Do the cube positions themselves matter? As in, if we had two cubes, would we say there's only 1 position?)

[Reach]

Originally Posted by MrRubix_MK5

Are we assuming that a shape is all that matters? (ie. an arrangement isomorphic to another is considered the same arrangement? Do the cube positions themselves matter? As in, if we had two cubes, would we say there's only 1 position?)

If there were two cubes, there would be only one distinct shape. The rule is essentially this: If a piece can be rotated through three dimensional space to fit exactly another piece, they are identical shapes and this counts as only one shape.

Otherwise the answer would become ambiguous, since there are a lot of configurations that appear different but rotate to become the same.

[MrRubix_MK5]

New riddle up

[MrRubix_MK5]

Reach: I'm getting 29 shapes here

Will explain logic if correct

[Reach]

Three men begin atop a mountain slope during a snowy night. Each man sits on top of their own flatbed cart on wheels (think Bugs Bunny cartoon, here). One of the men, Man #3, however, is stationed within a mountain tunnel, where no snow can pass through. Each man/cart is identical in every way, and their distances down to the bottom of the slope is the same. Each cart, at time 0, has no snow. They all start to go down the slope. Man #1 decides to sleep and does not do anything about the snow that falls onto his cart. Man #2 is constantly sweeping snow off his cart perpendicular to the direction of motion. Man #3, in the tunnel, obviously does nothing as he has no snow. Who makes it down to the bottom of the slope first, second, and third, and why?

Well, this requires a few assumptions on the physics end. I'll assume we can neglect friction in the tunnel. I'll assume assume we can neglect air resistance in the tunnel. Thus, Man 3 will beat Man 2, as Man 3 should be approximately the same weight, should experience less friction and MUCH less air resistance.

Now we have to consider whether man 1 will beat man 2. This could be complicated, depending on how we look at the physics, but I will assume that man 1 will beat man 2 for the following reason: If friction is treated as coulombic, man 1 should start going faster due to his gain in mass. However, this depends on how fast you're going. =/

Finally, man 3 should beat man 1 because his terminal velocity will be significantly higher (Man 3 does not face wind resistance under my assumption).


Thus Man 3 > Man 1 > Man 2.

This could be wrong, depending on whether or not my assumptions are correct.



Rubix, 29 is correct. Dispatched that one quickly D:

[MrRubix_MK5]

I had made a few edits already to the post, Reach, but no, friction won't matter here.

Again I can only say one is correct if the answer and logic is correct, so I will say no, that answer's not quite right

[Reach]

Originally Posted by MrRubix_MK5

I had made a few edits already to the post, Reach, but no, friction won't matter here.

Again I can only say one is correct if the answer and logic is correct, so I will say no, that answer's not quite right

Well assuming no air resistance changes things, since now the man in the tunnel loses his primary advantage for realistically creaming the other two.


Man 1 should win in this case. Realistically, it would depend on how long the hill was, but given his mass continues to increase and he faces no air resistance he should cruise past the other two. I could probably work this out mathematically but I won't waste time.

Now who wins between man 2 and 3 is a bit trickier in this scenario. I would say pushing mass off perpendicular to the direction of motion like that is going to slow you down. Man 3 should beat man 2.

So man 1 > man 3 > man 2


If not, then I'll let someone else tackle this, as I have trouble neglecting variables and then making sense of the problem.

[MrRubix_MK5]

Hint: Since this is a problem about mass and speed, all other variables negligible, think about a conservation law that includes these two variables.

[Reach]

Originally Posted by MrRubix_MK5

Hint: Since this is a problem about mass and speed, all other variables negligible, think about a conservation law that includes these two variables.

I considered momentum immediately. I don't like this problem. You can probably tell based on my puzzle style that I prefer very concretely defined variables XD

[MrRubix_MK5]

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]

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]

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]

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

[MahouMinachan]

I kinda did #2 mathematically...