1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Permutations .How do u find the number of paths in a 3D object?

  1. May 3, 2005 #1
    How do u find the number of paths in a 3D object?...let say a cube...from A to B....
  2. jcsd
  3. May 4, 2005 #2


    User Avatar

    I don't understand?

    maby you want the distance from A to B in 3D? Its the same formula as in 2D just an extra variable
    squareroot(x^2 + y^2 + z^2)

    assuming B is relative to A and those are B's cooridinates to A.

    Hope this helps!
  4. May 4, 2005 #3
    It doesnt matter whether the object is 3D or 2D. What does matter is how many nodes and how are their interconnections. You can draw a 3D cube in a 2D way and calculate the number of paths from A to B without affecting anything. (All u have to make sure that u are representing every edge of 3D object in 2D diagram)

    -- AI
  5. May 4, 2005 #4
    how do u know how to draw it in 2D?...and from which point to which point?
  6. May 4, 2005 #5
    I'm also not sure where you're going with all of this. There's an infinite amount of paths between any two points in R^3, same thing with R^2. If you're talking about like, a lattice coordinate plane, this is different. A lattice coordinate plane only has "points" at integers, so it's actually feasible to count the number of paths between two points in a cube that is set in the lattice plane. If this is what you want (or if it isn't), please clarify.
  7. May 4, 2005 #6
    ../|.......... /|

    Above i represent a cube in 2D (albeit in a very shabby way), but u would see that i have done 2D representation of the 3D cube and each node of the cube and each edge of the cube can be uniquely mapped to this figure.

    -- AI
    Last edited: May 4, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook