Category: Maths

SOPA


English: gold record (for future use in riaa g...

Image via Wikipedia

For a long time, sitting on the side watching the internet play out. The latest news so far is that SOPA and PIPA have been withdrawn, will probably be re-written in some form and come back in another democratic cycle or so.

If I was US based, then I would take more personal steps in the war-on-copyright/fight-against-piracy. Alas, I’m not so I’m going to put up banners, say the right things and hold double standards on all matters.

I think I should finally weigh in here.

I think I have thought up a method of enforcing copyright that would make the RIAA, MPAA and other such organisations happy.

It just takes a bit of technical knowledge and some thought.

Continue reading

Solution


Let us suppose we are given a set of linear equations \mathbf{A}\mathbf{x}=\mathbf{b} to solve. Here \mathbf{A} represents a square matrix of nth order and \mathbf{x} and \mathbf{b} vectors of nth order. We may either treat this problem as it stands and attempt to find \mathbf{x}, or we may solve the more general problem of finding the inverse of the matrix \mathbf{A}, and then allow it to operate on \mathbf{b} giving the required solution or the equation as \mathbf{x}=\mathbf{A^{-1}}\mathbf{b}. If we are quite certain that we only require the solution to be the one set of equations, the former approach has the advantage of involving less work (about one-third the number of multiplications by almost all methods). If, however, we wish to solve a number of sets of equations with the same matrix \mathbf{A} it is more convenient to work out the inverse and apply it to each of the vectors \mathbf{b}. This involves, in addition, n^2 multiplications and n recordings for each vector, compared with a total of about \frac{1}{3}n^3 multiplications in an independent solution.

— Alan Turing (1948)