"You want to prove that [n-1] breaks result in [n] pieces. In another formulation: [PiecesAfterBreaks(n-1)=n] Base case: For [n=1] it works, because if you don't break the bar at all, you're left with what you originally had - one bar. [PiecesAfterBreaks(0)=1] For the step case [P(n+1)], it works because after one break of a bar you get a bar which resembles the bar you just broke, plus an additional bar. [PiecesAfterBreaks(n+1)=PiecesAfterBreaks(n)+1]. This equation holds down until [n=0]"
- Rami Botros
"The only reason you were so highly up-modded is that people liked your joke."
- Rami Botros
MIT course notes on the fundamentals of electrical engineering (circuits, signals, etc.) using the Python language (PDF link) - http://www.reddit.com/r...
Itzhak Perlman plays J.S Bach's Partita no.2 for solo violin - ciaconna. It's hard to fathom that something this virtuosic was written around 1720. - http://www.reddit.com/r...