Saleem Ansari tuxdna

Git references

Git clone without deep history

Git clone without deep history make cloning faster when we only need the latest code.

With full history

First run

$ python3 pivot-pandas.py
        date stock_code  price
0 2018-08-27        001   10.0
1 2018-08-27        002   11.0
2 2018-08-27        003   12.0
3 2018-08-27        004   13.0
4 2018-08-26        001   14.0
5 2018-08-26 002 15.0

Groovy

//System.out.println("Hello World!")
//println("Hello World!")
//
//println "Hello World!"
//

class MyClass {

Given a 6 sided dice, how many number of throws are required to guarantee a 6 as output?

This solution is basically finding expectation over 1 roll, 2 rolls, 3 rolls ... and so on until infinite rolls of dice. Basically SUM( k * P(rolls=k)) for k = 1 to infinity. Now probability we get a 6 in exactly kth roll is defined as P(rolls=k) = q^(k-1) * p, where p is probability we get a 6 in kth roll, and q^(k-1) is probability we do not get a 6 in all k-1 rolls. I guess it still cannot guarantee that we will always get a 6, rather that if we roll on average of 6 times in a row, we will get definitely get a 6. Quite interesting problem indeed.

References:

Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law. https://en.wikipedia.org/wiki/Hofstadter's_law

f(x) = g(x) + delta + lambda*f(x)

where

Estimating hard to invert functions

Works for

n log(n)
n! ( Factorial function )

$ python function_estimation.py

How to run?

Create config.py from config.py.template and fill in your AUTH token, and the target URL.

Then invoke the grpah sync:

$ python3 perform_graph_sync.py package input-file.json maven

Invoking url: http://.../api/v1/graphsync/sync_all/maven/HTTPClient%3AHTTPClient

Create a disk file with random data

$ truncate -s 50G encrypted_backup.disk

Setup encryption for that device file

$ cryptsetup --verbose --verify-passphrase -s 512 luksFormat encrypted_backup.disk

	# find closest pair of points in 2d space in O(nlogn)

	import sys
	import math


	def euclidean_distance(p1, p2):
	x1, y1 = p1
	x2, y2 = p2
	return math.sqrt((x2-x1)2 + (y2 - y1)2)