hughdbrown

#!/usr/bin/env python3

# This is a simulation of the results for the first problem given here:
# https://fdpinstitute.org/resources/FDP%203.0/2024-Q2/Limited%20Sample%20MC%20Questions%20Q2-2024.pdf
#
# The expected answer given by the FDP Institute is 0.84. This is incorrect, as this simulation shows.
# The problem is that FDP has used the Bayesian formula incorrectly. Instead of using a weighted average
# of outcomes as the prior probability, FDP uses only the P(potted|~Simon) as the prior.

from collections import Counter

hughdbrown

from itertools import cycle

def spiral(n):
    matrix = [[0] * n for _ in range(n)]
    x, y = 0, 0
    x0, y0, xn, yn = 0, 0, n, n
    directions = cycle([
        (1, 0, (0, 0, 1, 0)),
        (0, 1, (0, -1, 0, 0)),
        (-1, 0, (0, 0, 0, -1)),

hughdbrown

❯ ls -alrt
total 0
drwxr-xr-x  152 hughbrown  staff  4864 Jan 15 12:00 ..
drwxr-xr-x    9 hughbrown  staff   288 Jan 15 12:00 .git
-rw-r--r--    1 hughbrown  staff     0 Jan 15 12:00 x644
-rwxr-xr-x    1 hughbrown  staff     0 Jan 15 12:01 x755
-rw-------    1 hughbrown  staff     0 Jan 15 12:01 x600
drwxr-xr-x    6 hughbrown  staff   192 Jan 15 12:01 .

hughdbrown

import math
import hashlib
from bitarray import bitarray

class BloomFilter:

    def __init__(self, n_items, fp_prob):
        '''
        n_items : int
            Number of items expected to be stored in bloom filter

hughdbrown

"""
You're given a list of elements. Each "element" has a unique id and 3 properties.

Two elements E1 and E2 are "similar" if they share any of their 3 properties.
Please write a function that takes a list of elements as input, and returns all "similar" groups of elements.

Example Input: [E1, E2, E3]
  E1: (id1, p1, p2, p3)
  E2: (id2, p1, p4, p5)
  E3: (id3, p6, p7, p8)

hughdbrown

#!/usr/bin/env python3
import sys
from collections import defaultdict

line =  sys.stdin.readline()
n = int(line.split(' ')[0])
d = defaultdict(list)
for line in sys.stdin.readlines():
    src, dst = [int(x) for x in line.split(' ')]
    d[src].append(dst)

hughdbrown

use itertools::Itertools;

pub fn split_paragraphs(lines: &Vec<&str>) -> Vec<String> {
    let mut paragraphs: Vec<String> = vec![];
    for (key, group) in &lines.into_iter().group_by(|x| !x.is_empty()) {
        if key {
            paragraphs.push(group.map(|s: &&str| *s).collect::<Vec<&str>>().join(" "));
        }
    }
    paragraphs

hughdbrown

------------------------------ 'create_raw_lookup': 0:00:00.051461 seconds
[((0, 9), 'a7874d04-3a7c-4ab2-85f3-52de785c87e1'),
 ((10, 19), 'bbf24bb0-e1fd-41f3-ba6a-a983fff7c675'),
 ((20, 29), 'fab7cd13-d273-41a1-b749-e73d22521a04'),
 ((30, 39), '8d688ab2-4df3-4787-b715-d5acf36f43a2'),
 ((40, 49), '81081dca-abf8-46e4-a45b-f25d4e6b4876')]
------------------------------ 'create_expanded_lookup': 0:00:00.014262 seconds
[(0, 'a7874d04-3a7c-4ab2-85f3-52de785c87e1'),
 (1, 'a7874d04-3a7c-4ab2-85f3-52de785c87e1'),
 (2, 'a7874d04-3a7c-4ab2-85f3-52de785c87e1'),

hughdbrown

#!/usr/bin/env python3

from pprint import pprint
from random import randint
from datetime import datetime
from uuid import uuid4

import pandas as pd

NUMS = 10000

hughdbrown

#!/usr/bin/env python3

from timeit import timeit

def naive(values, k):
    for i, v1 in enumerate(values):
        for v2 in values[i + 1:]:
            if v1 + v2 == k:
                return True
    return False