solve this riddle about a CUBE!?

Question

paths are drawn from one vertex of a cube to an opposite vertex in such a way that the path never goes tghrough the same point twice. how many such paths are there?

Answer 1

your question is soooo unclear what you're talking about?
i guess it's either one or infinity

Answer 2

it depends what you mean by "oposite vertex". if you're talking about THE oposite vertex, then there's only 1 possible path. And that's assuming your paths are straight lines (something you didn't mention in your riddle)

now, if you relax that "oposite vertex" condition, then since you have 8 vertexes in a cube, you'll have 7 possible paths.

Well, as yuo can see, the REAL riddle is to understand what the f*** you mean in your riddle : you're not making any sense at all.

other possible interpretation: you absolutely want your paths to go from one vertex to another (a neighbour) and then FINALLY get to THE OPPOSITE vertex (in which case, SAY IT!!!). And then, there is 3 choices for the 1st move, then 2 choices, then 1, then 1, then 1, then 1, then 1. That makes a total of 6 possibilitie paths. But that's still assuming your question have this particular meaning.

Answer 3

Six (provided the starting point was already chosen - since there are eight points, the total number of paths is 8*6=48, or 24 if you consider start -> end identical to end -> start).

The second point can be chosen in three ways (all the three neighbors are equal because of symmetry). The third point can be chosen in two ways: three neighbors but the starting point is not allowed. Again, for reasons of symmetry both are equal.

But now, the rest of the path is determined. First, you must go to the fourth point in the square formed by the first three points since otherwise that point will be left as a dead-end. For the fifth point you have only one choice. For the sixth point you have two choices but one of them is the end point. And for the seventh point you have only one choice.

Answer 4

Are you talking about only one pair of opposite vertices, or the 4 pairs of opposite vertices. Also, when you say a path cannot pass through the same point twice, does that include the opposite vertices themselves or not?

The answers could be 1, 4 or infinite because you did not say the paths had to be straight lines.

Answer 5

I count eighteen paths. Here's how I figure it.

Draw a cube.
Let the bottom face be a square with corners A, B, C, D.
Let the top face be a square with corners E, F, G, H - with E above A, F above B, G above C, and H above D.

Suppose you want to start at corner A, and end at corner G.

There are three possible paths that start with ADC:
ADCG
ADCBFG
ADCBFEHG

By symmetry, there must also be three paths that start with ADH.

By further symmetry, since there are six paths that start with AD, there must also be six paths each that start with AB and AE.

That adds up to eighteen possible paths from A to G. Assuming, of course, that you must travel along the edges of the cube. :-)

Answer 6

Unless you have some other limiting condition, the number is infinite. If you have trouble visualizing this with a cube, try imagining a sphere around the cube with a diameter equal to the diagonal of the cube. There are an infinite number of great circles of "longitude" that will connect one vertex to it's opposite vertex.

Answer 7

3

Answer 8

4

Answer 9

Assuming that you are travelling along the edges of the cube...
There are 6 paths of length 3
0 of length 4
6 of length 5
0 of length 6
6 of length 7
and no longer ones

Answer 10

only 1 path

Answer 11

The path must necessarily pass through the centre of the cube. Hence the answer is Only one path.