samsonq

Ranked by discretionary assets managed in hedge funds worldwide, in millions, as of June 30, 2018, unless otherwise noted.

Rank Manager                                 Assets     Change from 2017
1    Bridgewater Associates                  $132,756   7.9%
2    AQR Capital Mgmt.                       $83,700    9.2%
3    Man Group                               $59,100    11.3%
4    Renaissance Technologies                $57,000    17.3%
5    Two Sigma Inv./Two Sigma Advisers       $38,800    9.6%
6    Millennium Mgmt.                        $35,314    2.7%
7    Elliott Management                      $35,000    7.0%

samsonq

import numpy as np

from pypfopt.black_litterman import BlackLittermanModel
from pypfopt.efficient_frontier import EfficientFrontier
from pypfopt import plotting

from openbb_terminal.sdk import openbb
import seaborn as sns
sns.set_theme()

samsonq

from collections import defaultdict
from operator import itemgetter
from time import time

from binance.client import Client

FEE = 0.0005
PRIMARY = ['ETH', 'USDT', 'BTC', 'BNB']

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def second_largest(n):
    first = n[0]
    second = n[1]
    if second > first:
        first, second = second, first
    for i in range(2, len(n)):
        if n[i] > first:
            first, second = n[i], first
            if second > first:
                first, second = second, first

samsonq

import numpy as np


def monty_hall():
    doors = np.array([1, 0, 0])
    np.random.shuffle(doors)
    initial_choice = np.random.randint(len(doors))
    goat_door = 0
    for i in range(len(doors)):
        if i == initial_choice:

samsonq

def fib_dp(n, memo):
    if n in memo:
        return memo[n]
    else:
        fib_n = fib_dp(n-1, memo) + fib_dp(n-2, memo)
        memo[n] = fib_n
    return fib_n
    

def fibonacci_dp(n):

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def find_poisoned_bottle(bottles):
    """
    Find position of the poisonous wine bottle in 1000 bottles, using 10 strips, given that there
    is only 1 poisonous bottle of the 1000 bottles.
    
    The logic is to line up all 10 strips and represent them as binary numbers, with each strip in a position for the 
    binary numbers. Then, each wine bottle is labeled from 1-1000. Since 2^10 is the first power of 2 that is greater than 1000,
    10 strips are sufficient and we only need to test once with all 10 strips, and then wait 1 week for the test results to come back.
    Then, aligning the strips in the same order as before, there will be a sequence of positive and negative strips. Converting that
    sequence from binary (with positive as 1's and negative as 0's) back into decimal form, we will have the index of the poisoned 

samsonq

import numpy as np
import random

samsonq

import numpy
import math
from matplotlib import pyplot as plt


def pi_monte_carlo(num_simulations, radius=1):
  return ...


if __name__ == "__main__":

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import math

def selection_sort(a):
    for i in range(len(a)):
        curr_min = i
        for j in range(i, len(a)):
            if a[j] < a[curr_min]:
                curr_min = j
        if curr_min != i:
            a[i], a[curr_min] = a[curr_min], a[i]