## Friday, May 31, 2013

### solving for the graceful tree of a graph

The goal of finding a graceful graph is finding numbers to fill in the letters of this graph:

such that the differences between them (to label the edges) are 1,2,3,4,5. Or for larger graphs following that same pattern.

First one can observe that one equation can be formed from the information as:

But there are five unknowns and that is only one equation, so the way I found to provide the other equations is simply to raise this basic equation to higher powers. So I use:

This provides many solutions but the first one given is:

This solution does work. There might be other ways to find a set of nonlinearly-related equations using the information given that work even better.

## Sunday, May 26, 2013

### Acid/base battery part 2

\
Above is the battery after running for 24 hours. It held between a quarter and a third of an amp for at least a couple hours, and today 24 hours later it is still reading a few milliamps. These readings were taken by simply placing the electrodes of the voltmeter in each beaker.  The jar in the left held a green liquid aluminum chloride and the one on the right held a whitish aluminum hydroxide. These were prepared by dissolving an excess of aluminum foil in HCl and NaOH and decanting each liquid part into a beaker. Then a tube full of water was carefully inserted between them to form a bridge. You might be able to see in the picture above that eventually an unknown to me orange precipitate forms in the one on the right and a dark precipitate forms in the one on the left.
The trick was finding an acid and a base that conduct electricity well, I found it hard to find information on the internet, but aluminum chloride has about 750 ohms of resistance and aluminum hydroxide has 14,000. So an obvious way to improve the battery would be to find a better base for conducting electricity than aluminum hydroxide.

## Monday, May 13, 2013

### Redistribution network

I was thinking to study a network like so: suppose you have a graph, this one is simple but they don't have to be:

Now consider this matrix, the letters on the left are just labeling the rows:

Now imagine what each cell in the matrix represents is how much of each different variable each node has in its inventory. Here I've started every node off with 1 unit of it's own variable, but they could start in any way.

Then they go through this cycle, every node divides what it has in it's inventory among the nodes it is connected to. For example A in the graph diagram is connected to B,E, and F. and it has in it's inventory A, so it will give (1/3)A to B, (1/3)A to E, and (1/3)A to F. At the same time every node distributes what is in it's inventory to all the other nodes. So after one iteration you have:
So F has 5 connections to A, D, E, G, and H. It gets a share from each of those nodes that depends on the number of nodes those nodes are connected to. Like row F and column E is 1/4 because E is connected to 4 nodes of which F is one, and E had previously just had one E in it's row.

After another iteration (rounded to maximum of 3 decimal places):
[0.261, 0.083, 0.111, 0.113, 0.133, 0.05 , 0.1     , 0.05  , 0.0   ,  0.0    ],
[0.083, 0.305, 0.0    , 0.145, 0.083, 0.116, 0.0    , 0.0    , 0.083, 0.0    ],
[0.111, 0.0    , 0.277, 0.062, 0.145, 0.05  , 0.0    , 0.062, 0.062, 0.125],
[0.15  , 0.194, 0.083, 0.258, 0.05  , 0.05  , 0.1    , 0.113, 0.083, 0.125],
[0.177, 0.11  , 0.194, 0.05  , 0.279, 0.116, 0.1    , 0.05  , 0.063, 0.0    ],
[0.083, 0.194, 0.083, 0.063, 0.145, 0.316, 0.125, 0.125, 0.125, 0.125],
[0.066, 0.0    , 0.0    , 0.05  , 0.05  , 0.05  , 0.225, 0.05  , 0.063, 0.125],
[0.066, 0.0    , 0.083, 0.113, 0.05  , 0.1    , 0.1    , 0.363, 0.125, 0.125],
[0.0    , 0.111, 0.083, 0.083, 0.063, 0.1    , 0.125, 0.125, 0.333, 0.125],
[0.0    , 0.0    , 0.083, 0.063, 0.0    , 0.05  , 0.125, 0.063, 0.063, 0.25  ]

Here row A and column A is .261 because A gets 1/3 of every part of B's inventory but B had 1/3 A so A gets 1/9 from B. And similarly 1/12 and 1/15 from E and F, added together that is .261.

After 100 iterations:

[[0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088],
[0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088],
[0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088],
[0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118],
[0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118],
[0.147, 0.147, 0.147, 0.147, 0.147, 0.147, 0.147, 0.147, 0.147, 0.147]
[0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058],
[0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118],
[0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118],
[0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058],
[0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058],

It ends up all of these numbers are approached as limits! 30 Iterations is correct to about 3 decimal places and they get closer and closer.

**EDIT**
I found a way to find these numbers exactly. Imagine that the process iterated until it reached the limits. That would mean in another iteration, the amount going into each node would match the amount going out. That would mean that the new amount in A for example would be 1/3 B, 1/4 E, 1/5 F. So we can write out the equations:
Here there would be 10 equations one for each node, but you remove one or otherwise the solutions would be 0 for all variables. In the removed equations place you put A+B+C+D+E+F+G+H+J+K=1. That's the other constraint on the system. Then the solutions are:

And it appears that the bottom number of these fractions is twice the number of edges in the graph and the top number is the number of edges coming from the vertex.

## Monday, May 6, 2013

### Concerning isometries

I noticed if you have two graphs labeled any way, like:

You can set up a set of equations like so for each of them, for the graph on the left it is:
a) a+m=b+c+f
b) b+m=a+c+d+e
c) c=m+a+b+d+e+f
d) d=m+b+c+e
e) e=m+b+c+d+f
f)  f=m+a+c+e
Where each lettered equation is a letter for a vertex +m = the vertices that lettered vertex is connected to summed.
The set of equations for the figure on the right in the above image is:
a) a+m=b+e+f
b) b+m=a+c+d+e+f
c) c+m=b+d+e
d) d+m=b+c+f
e) e+m=a+b+c+f
f) f+m=a+b+d+e

Now solving each system respectively gives:

It's not too hard to see that these are the same solutions but with some of the letters having exchanged roles.

The idea is the graph is uniquely identified by the system of equations up to renaming vertices, and the solution for each variable is unique for the system of equations up to renaming variables, so isometrical graphs have the same set of solutions up to renaming variables.

** I did find that it can't always distinguish between a graph and the same graph where a vertex is connected to itself but it seems to work for simple graphs