analytics
Wednesday, October 2, 2013
fertilizer from sunlight
So I thought, have a pond that grows algae, the algae use water and sunlight to create carbohydrates, feed the algae and water to a tank of yeast, the yeast eat the carbohydrates and produce alcohol, keep the alcohol and feed the dead yeast to bacteria, the bacteria eat the dead yeast to produce methane, keep the methane, and the dead bacteria can be used as fertilizer for plants. So from sunlight and water you get alcohol and methane some of which is used as fuels to drive the pumps that feed from one tank to another and you end up with fertilizer.
separating noise from information in a picture
Suppose you have a picture like:
It seems to work as an edge detector...
It's easy for a person to tell where the two interesting objects are in this picture but I found a way for the computer to do it as well.
Step 1: Make a new image where you look at each pixel in the original image and add up how far it's value is from it's neighbors. If it is exactly the same color as all it's neighbors that is a 0 if every surrounding pixel is exactly the opposite color that is a 255 (this means white in computer graphics).
This has the effect of making all contiguous areas of color or gradients solid black.
Step 2:Make a new picture where the new pixels are the average of the old pixels and their neighbors. Sort of a blur.
Step 4. Blur again but only keep pixels that are between 0 black and 128 grey, everything else is made white.
Now do the same thing again but only keep pixels that are between 0 black and 64 grey. All else stays white.
Now you should have a nice binary map of where the interesting parts of the original picture are with the noise removed.
So it's kind of interesting to see what this process does to normal pictures...
Tuesday, October 1, 2013
3d keyboard
My idea was that there would be this ball with buttons on it for all the keys on a keyboard inside each triangle and the triangles stick out a bit so you can hold it easily without pressing any on accident, you press inside each triangle for the key and a little harder to do shift+the key. So you could type using only one hand while the other one does the mouse is the idea. I don't know what the learning curve would be like or if people could get pretty good at it or whether it would just drive anyone insane that tried to use it lol.
maybe a better idea than the last thing, this is a ball you hold in your fist so your fingertips are on the five buttons, and it has a gyroscope in it so it can tell which way you are holding it in space like an ipad. And depending on which way you rotate it, it changes what the five buttons do.Like in one position the five buttons could be e,t,a,o,space, and if you rotate it a bit in a certain direction they change to n,s,h,i,backspace.
maybe a better idea than the last thing, this is a ball you hold in your fist so your fingertips are on the five buttons, and it has a gyroscope in it so it can tell which way you are holding it in space like an ipad. And depending on which way you rotate it, it changes what the five buttons do.Like in one position the five buttons could be e,t,a,o,space, and if you rotate it a bit in a certain direction they change to n,s,h,i,backspace.
Sunday, September 29, 2013
Graph Isomorphism matrix
If you have a graph like:
Note that some of the edges are weighted, one vertex connects to itself, and one connection can cross another.
We can associate a connectivity matrix M with this graph like so:
Note that some of the edges are weighted, one vertex connects to itself, and one connection can cross another.
We can associate a connectivity matrix M with this graph like so:
Here, there is a in cell (x,y) if x and y are connected and the weight if any replaces the 1.
Also a permutation matrix of the same size can be defined as any matrix that has one 1 in every row and every column, for example:
A nice theorem about these matrices is that if two graphs are isomorphic we can write:
P*M*P = N
where M is the connectivity matrix of graph M, and N is the connectivity matrix of graph N, and P is any permutation matrix.
Proof:
P*M*P is the matrix you get when you first permute any number of rows of the matrix, and then do the same permutation on the columns. And this is exactly the same as taking the graph associated with the connectivity matrix and permuting the labeling any way. And of course two graphs are isomorphic if there is a way to permute the labeling so that you generate the same connectivity matrix.
This isn't an if and only if because I found two non-isomorphic graphs that are related in this way by a permutation matrix.
Friday, September 27, 2013
maybe spacetime under gravity?
I found this nice relationship between the unit circle and a distribution I name in the picture...
In words it is a distribution that is tangent to the circle at 0*, and has the same area as the circle from -infinity to infinity. So it's pretty interesting on it's own but I found it thinking about how relativity is always explained as how mass warps spacetime like a ball on a rubber sheet. I had the idea that maybe a unit mass of a unit area displaces a unit area of spacetime from minus infinity to infinity. I don't know enough differential geometry to say whether this fits with the very complex relativity formulas or not but it was just a thought.
IMPORTANT NOTE:
The area of the formula for the green region from minus infinity to infinity is pi, but subtracting the bottom half of the circle's area of pi/2 gives pi/2. As I'm considering it as a displacement it makes sense this way.
* Dr. Rose helped me with this part I originally though it stayed tangent to the circle over an interval...
IMPORTANT NOTE:
The area of the formula for the green region from minus infinity to infinity is pi, but subtracting the bottom half of the circle's area of pi/2 gives pi/2. As I'm considering it as a displacement it makes sense this way.
* Dr. Rose helped me with this part I originally though it stayed tangent to the circle over an interval...
Monday, September 23, 2013
Rough modeling of rows of Pascal's triangle as a distribution
I found a rough correspondence between the formula under the Consider: and Pascal's triangle...
Also, one can solve for the middle number approximately:
for r = 11
the right answer is 252 so pretty close.
One practical reason for this approach is for example finding the middle number of row r of pascal's triangle, ordinarily you would use the combinatoric formula r!/((r-1!)*(r!)) but eventually the factorial involves numbers too large for Maple to calculate, but using my integral formula for the middle number only involves calculating 2^r which it can do more easily.
Sunday, September 22, 2013
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