the fast you are willing to prototype, the more willing you are to fail, the faster you will get to a successful partial solution you can refine and repeat.

how to prototype faster?

In order of decreasing slowness—-

• build out the whole product…
• building the minimum viable product…
• skeleton prototyping (Figma)…
• Pen and paper…
• Talking about it

The trade-off: each level gives increased fidelity: its closer to what will actually ship, so you can get better+detailed feedback.

Coping with NP Completeness

It’s possible to solve NP complete problems!

• average case/worst case complexity: it’s possible to solve SAT for a lot of problems which solves the average case problems (“Heuristics vs. Algorithms”)
• special cases: 2SAT, subset sum, etc. can be solved in very special cases

A workshop hosted by PSC about Spark.

Contents:

• Big Data

Chapter 3

Problem 3.10

Part a

Notably, the slope field is symmetric across the \(y\) axis, and repeats with every \(m\pi\) interval about the line \(\frac{\pi}{4}\).

Part b

We have a stationary value at \(y = \frac{\pi}{4}\). Beyond that, as initial \(x>0, y<\frac{\pi}{4}\), solutions will all trend towards \(y=\frac{\pi}{4}\) as \(t \to \infty\), because the derivative is positive for that entire region. For \(x>0, \frac{\pi}{2}>y> \frac{\pi}{4}\), the function will also trend towards \(\frac{\pi}{4}\), as the slope is negative for that entire region. This pattern repeats for all \(y_0+m\pi\). That is, for instance, for \(y\) between \(m\pi+\frac{\pi}{4} < y < m\pi + \frac{\pi}{2}\), \(y\) will trend towards \(m\pi + \frac{\pi}{4}\). For initial \(t<0, y < \frac{\pi}{4}\), most solutions will trend towards \(-\infty\) as the region has negative slope. Yet, as \(t_0 \approx 0\), the function will never hit the singularity point of \(y = -\frac{\pi}{2}\) before traveling to the \(t>0\) side, resulting in it trending towards \(+\infty\). Finally, for initial \(y>\frac{\pi}{4}, t<0\), the function will reach \(+\infty\) because it will hit the positive singularity at \(\frac{\pi}{2}\).