Math with K'nex

[Chasney913]

OK, so I'll try to explain this the best I can.

Construct a 3x3x3 cube out of blue rods and use two blue connectors snapped together as vertices. There should also be central vertices, so 27x2 blue connectors will be used.
If you don't have K'nex, visualize (and Google).

The blue connectors that form the vertices can have 3 possible configurations, because the diagonal connections can only occupy two planes as there are only two connectors.

Thusly, towards a square (see below), the diagonal connection can face or not face, depending on what configuration it is in.

There are 36 squares formed that are 1x1, assuming a blue rod is 1 unit long. A square is considered "filled" if the connections are in one of the following: (8 denoting a possible connection, based on configuration, 0 denoting a impossible connection)

8----0 or 8---0 (or any rotations thereof)
|***|***|***|
|***|***|***| (The asterisks are not important, they are there for placeholders.)
|***|***|***|
0----0***0---8

Considering that any given vertex can create a connection on two planes at once, but
cannot on the third, is there a set number of filled squares? I believe it to be 8. Is there anyway to prove it?

If you have no idea what I'm talking about, I'll try to explain further, however, I need to find a way to create and share virtual K'nex models.

I've attached a picture to show square configurations. The dark blue crosses are dark blue connectors not oriented to the plane the square is on, so it is a 0. The others are dark blue connectors oriented to the plane the square is on, so it is a 8. Like above.

[attachment=1916:attachment]

Converted to a GIF (96.65% compression). –Goose

[BlackOpSource]

I looked into what you're talking about once. I don't think there is a way to do virtual K'nex.

Also, you're correct to assume I have no idea what you're talking about.

However, I did once build a binary adder out of K'nex. It added one-digit binary numbers (input via golf balls), and output a two-digit binary number. I felt cool for a week about that. That was what really got me into computers.

[1337-time]
~814c|<0p50urc3
[/that was fun]

[DarkerLine]

If your numbers for the number of squares and the number of correctors are correct, I'd describe your construction as a 2x2x2 cube, rather than a 3x3x3, because the length of an edge is 2 blue rods.

In this case, I believe I've shown there isn't a set number of filled squares. Suppose all the connectors are oriented the same way. Then any square is either in a plane the connectors are oriented to (in which case all vertices are "8" and it's not filled) or it's in the plane the connectors aren't oriented to (in which case all vertices are "0" and it's not filled). So no square is filled in this configuration.

But it's clearly possible to create a configuration in which filled squares exist, and this means the number of filled squares can vary.

Disclaimer: the above proof is built on absolutely no K'nex experience.

[Chasney913]

Thanks, that seems correct from my two minutes of thinking about it. And yes, I was talking about a 2x2x2 cube, but I was thinking connectors not rods, and got all messed up. But you knew what I meant.

@BlackOpSource Do you still have the binary adder? I'd like to see it if you do, or know how it worked.

[DarkerLine]

So what prompted you to describe those two types of squares and not others as 'filled'?

[BlackOpSource]

No. My mom gave all my K'nex away last year (Sniff.... we had a good run: 1996-2006.). But here's how it worked.

I was too lazy to build it so you just dropped one ball on the rails. It consisted of two logic gates, XOR and AND, and to indicate a "1" as input, you just dropped a golf ball onto a track leading into each. For a "0", you did nothing. Each of these gates was built into one housing, and, because of their mechanical nature, they rotated during operation, outputting to a needle on the front that pointed one way or another- for zero or one (The default state for both was zero; the machine needed to be reset each time.). The XOR gate consisted of an upside-down "T" that pivoted about the "intersection" each time a ball was dropped into it. The AND gate consisted of a basket into which balls fell, which was on a lever, counterweighted by the large wheel pieces (You know, with the tires and whatnot.). The AND gate would only rotate if two balls were in the basket, since one was not heavy enough. So, replacing the golf balls with ones and zeros, we get....

XOR
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0

AND
0 + 0 = 0
0 + 1 = 0
1 + 0 = 0
1 + 1 = 1

Because of the way the gates were arranged in the housing, the indicator for the output of AND was to the left of XOR, so for the whole machine....

0 + 0 = 00 = decimal 0
0 + 1 = 01 = decimal 1
1 + 0 = 01 = decimal 1
1 + 1 = 10 = decimal 2

Just let me know if you want more detail on the mechanical operations; especially with the XOR, I don't know that I was all that clear, really.

~BlackOpSource

[Chasney913]

I decided filled vs not filled while making the cube. Basically, if I could connect a rod, either at both ends or at one end, the cube was filled. However, if two rods could be in the square, they would conflict, and so I removed them both. The two arrangements are the only two that have only one rod in them. (The rods are connected diagonally, if you haven't guessed.)

@BlackOpSource That's perfectly clear, although from my interpretation, you need to drop a ball representing each digit twice. It would be more awesome (in my opinion) to have something like such: http://woodgears.ca/marbleadd/index.html
However, I think it would be difficult to do with K'nex, because according to the site, it was challenging enough with easily-sculpted wood, let alone fixed length and angle plastic.

In a unrelated matter, does anyone have any good chemistry tutorials? Grade 11 & 12, I guess. My friends are taking it this semester, and I'm usually good at that sort of thing, so they ask me to help. Unfortunately, I've yet to actually take chemistry, so I need a good, in-depth tutorial. Almost a lesson, I guess. It would be greatly appreciated, both by me, and my friends.

[BlackOpSource]

I'm not sure I understand....

To actually operate the machine, you dropped a ball into each of the two parts of the machine for every "one". So to add one, you would expend two balls from some hypothetical cache of golf balls you might have.