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Topic: Solder Cube (Read 5678 times) |
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jkemp
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"Say you have some bendable wires (any number, any length). What is the minimum number of solder connections needed to make a cube? Prove it. Also, what is the minimum number of wires necessary? Prove it."
I think I need more info, because why not make such a cube out of a single wire, with no solder connections? Assuming the wire will keep its shape after bending, and that the wire can bend back on itself.
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otter
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Re: Solder Cube
« Reply #1 on: Aug 8th, 2002, 7:30am » |
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> Say you have some bendable wires (any number, any length). What is the minimum number
> of solder connections needed to make a cube? Prove it. Also, what is the minimum number
> of wires necessary? Prove it.
Two wires, 4 solder joints.
I'm assuming here that a solder joint is only required to connect separate wires, not to connect a single wire to itself. (If this is not the case then 8 solder joints are required).
Since each of the six corners has an odd number of vertices, Euler's theorem dictates that the cube cannot be constructed with a single wire. However, two sets of two faces each can be constructed, each from a single wire (thus 2 wires). The faces are at right angles to each other. This gives you 4 faces of the cube (unassembled). When soldered at the 4 corners to form a cube, the remaining two faces of the cube are formed automatically.
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otter
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Re: Solder Cube
« Reply #2 on: Aug 8th, 2002, 7:55am » |
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Oops! My bad. This construct gives two vertices that contain 2 wires. Looks like 4 wires and 3 solder joints is the best bet.
on Aug 8th, 2002, 7:30am, otter wrote:> Say you have some bendable wires (any number, any length). What is the minimum number
> of solder connections needed to make a cube? Prove it. Also, what is the minimum number
> of wires necessary? Prove it.
Two wires, 4 solder joints.
I'm assuming here that a solder joint is only required to connect separate wires, not to connect a single wire to itself. (If this is not the case then 8 solder joints are required).
Since each of the six corners has an odd number of vertices, Euler's theorem dictates that the cube cannot be constructed with a single wire. However, two sets of two faces each can be constructed, each from a single wire (thus 2 wires). The faces are at right angles to each other. This gives you 4 faces of the cube (unassembled). When soldered at the 4 corners to form a cube, the remaining two faces of the cube are formed automatically. |
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We shall not cease from exploration. And the end of all our exploring will be to arrive where we started
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Dan A.
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I think one wire should be sufficient to build your cube, with no overlapping edges. Of course, this requires that diagonals are allowed.
Let's define the verticies of a cube to be the following (please, oh please let them be using a fixed-width font):
5------6
/| /|
1------2 |
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| 7----|-8
|/ |/
3------4
If you connect the verticies together in the following order:
1->2->4->3->1->5->6->8->7->5->2->6->4->8->3->7
you end up with a cube. Yes, you'll have diagonals on faces (1,2,6,5), (2,4,8,6), and (3,4,8,7) but that doesn't change the fact that it is a cube. Is it possible to do this in less than 15 line segments? As for the proof, I challenge anyone out there to construct a cube out of zero wires!
This cube, and all others, will require 8 solder joints. You'll notice that three segments connect at each of the 8 verticies. By bending a wire, you can make two segments without a solder joint, but three is impossible. Thus, at a minimum you will need one solder joint per vertex.
(I don't know how others are approaching the problem, but I was visualizing it as attempting to draw a cube without picking up your pencil. Each time you pick up your pencil, that's another wire that's needed.)
Standard disclaimers concerning question interpretation apply.
DSA
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Icarus
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Re: Solder Cube
« Reply #4 on: Oct 17th, 2002, 8:06pm » |
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My interpretation of the problem was that each edge of the cube can be formed by a single wire (no back tracking or allowing a second wire to take the same path), and wires can only follow the edges of the cube. Also, whenever a wire ends at a vertex, you will need a solder joint there to hold it solid.
Given those conditions: since each vertex is the meeting of 3 edges, one wire has to end there, so all eight vertices require a solder joint. Also this gives you a minimum of 8 wire ends, so at least 4 wires are required.
Running one wire around the top, down a side edge, then around the bottom, and using the other 3 wires to make the remaining three edges, shows that the minimum is obtainable.
So, given this interpretation of the problem, the optimum is 4 wires and 8 joints.
If you allow diagonals, but require rigid joints, then you get Dan A.'s solution of 1 wire and 8 joints. (15 segments is the absolute minimum.)
If you allow loose joints (soldering necessary only to hold separate wires together), but no diagonals or doubling of edges, then 4 wires are required, as in the strict case. But with 4 wires, there are only 8 wire ends, so each vertex must consist of 1 wire "passing through" and one wire end. Thus any joint can only join two wires, and a minimum of 3 solder joints are needed to unite all 4 wires, as Otter states.
If you allow both diagonals (or doubled edges) and loose joints, then obviously you need only 1 wire and no joints.
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