http://highscalability.com/blog/2010/5/5/how-will-memristors-change-everything.html

IMDisk: http://www.ltr-data.se/opencode.html#ImDisk

Process Hacker: http://processhacker.sourceforge.net/

Xplorer2: http://www.zabkat.com/

DebugView: http://technet.microsoft.com/en-us/sysinternals/bb896647.aspx

DLL Dependency Walker: http://www.dependencywalker.com/

SDelete: http://technet.microsoft.com/en-us/sysinternals/bb897443.aspx

Wink: http://www.debugmode.com/wink/

GLSL/PyOpenGL: http://pyopengl.sourceforge.net/context/tutorials/index.xhtml

OCR: http://code.google.com/p/ocropus/

OpenCL+Python: http://clyther.sourceforge.net/

First Attempt:
There is a master key. All licence keys issued are derived from this master key.
Have a counter stored in a file. This counter is incremented by 1 for every new licence key issued.
Use  the master key to encrypt the counter (Use a good symmetric key algorithm). The encrypted counter is issued as a licence key.
To verify that a given licence key was issued by the software maker, the software maker needs to decrypt the licence key using the master key. If the decrypted key is a number less than or equal to the number of licences issued, it is valid.

Disclaimer: This method is probably wrong/does not work/has flaws. Please do not use it for your application. If you do use it and get into trouble, I am not responsible.

Solution:

The first step is to unwrap the tetrahedron. I realized that visualizing in 3D takes some effort and practice. So I cheat by “unwrapping” the 3D model to 2D.

<insert image here>

Assume each side of the tetrahedron is a. Thus we have 4 equilateral triangles of side a.

The height of each triangle will be a sin 60.
Half the base of each triangle will be a/2

…..to be continued.

Thus I intend to try and do the opposite of searching for solution sometimes. I intend to apply “Original Thought” to solve a problem. This blog is to list some of the ideas and solutions that I come up with.