nilsso

import typing


T = typing.TypeVar("T")


def not_none(v: T | None) -> T:
    assert v is not None
    return v

nilsso

#!/usr/bin/env python
import functools
import argparse
import json

from pathlib import Path
import typing

import csv

nilsso

#!/usr/bin/env python
"""Text file cleaning utility (primarily for secured roll preparation)."""
from pathlib import Path
import typing


def clean_file(
    in_path: Path,
    out_path: Path,
    in_encoding: str,

nilsso

n,atomic,1,2,3,4,7,8,11,13,14
1131,true,0,0,1,0,0,0,0,1,1
951171,true,0,0,1,0,0,0,0,1,3
799934811,true,0,0,1,0,0,0,0,1,5
672745176051,true,0,0,1,0,0,0,0,1,7
6591,true,0,0,1,0,0,0,0,3,0
5543031,true,0,0,1,0,0,0,0,3,2
4661689071,true,0,0,1,0,0,0,0,3,4
3920480508711,true,0,0,1,0,0,0,0,3,6
3297124107825951,true,0,0,1,0,0,0,0,3,8

nilsso

n,atomic,1,2,3,4,7,8,11,13,14
1131,true,0,0,1,0,0,0,0,1,1
951171,true,0,0,1,0,0,0,0,1,3
799934811,true,0,0,1,0,0,0,0,1,5
672745176051,true,0,0,1,0,0,0,0,1,7
6591,true,0,0,1,0,0,0,0,3,0
5543031,true,0,0,1,0,0,0,0,3,2
4661689071,true,0,0,1,0,0,0,0,3,4
3920480508711,true,0,0,1,0,0,0,0,3,6
3297124107825951,true,0,0,1,0,0,0,0,3,8

nilsso

nilsso

macro_rules! ichain {
    ($a:expr, $b:expr) => {
        $a.chain($b)
    };
    ($a:expr, $b:expr, $($others:expr),+) => {
        $a.chain(ichain!($b, $($others),+))
    };
}

nilsso

import numpy as np

import matplotlib as mpl
from matplotlib import pyplot as plt

import sympy
from sympy import lambdify, latex
from sympy.parsing.sympy_parser import parse_expr

import argparse

nilsso

Crates

Dependencies

Intellij IDEA with Intellij Rust sometimes can't pickup on re-exported libraries for auto-completion. To get around this, adding a dependency and renaming it can work. For example Amethyst re-exports specs to ecs, so adding a dependency ecs = { package = "specs", version = "0.14" } gets auto-completion working on ecs::.

nilsso

Proposition. Prove $(a,b)=(a,b+a)$ for all $a,b\in\mathbb{Z}$

Proof. For some $x,y\in\mathbb{Z}$. $$ d_1=(a,b)=ax+by $$

Thus $d_1$ divides both $a$ and $b$.

\begin{align*} d_2=(a,b+a)&=ax+(b+a)y\