Month: June 2020

Preposterous Primes

This scrupulous challenge is an algebraic one, and quite interesting.

Let a “prime factor connection” be a connection from two whole numbers A and B, such that:

  1. A and B share a common prime factor C
  2. A plus C equals B
  3. A divided by C is not prime

 For any whole number N, let the “prime connected set” of N be the set of all whole numbers that satisfy the following:

  1. For any number X in this set, N is part of its prime connected set
  2. All numbers prime connected to N are in this set
  3. This set is the minimum possible size given the first two requirements

 Let S(N) be the size of the prime connected set of N.

Does S(N) reach infinity eventually, or is it a fast growing but strictly finite function? If it reaches infinity, can you find the smallest value at which it is infinite? Try to prove it is so. If S(N) never reaches infinity, try to prove this. Good luck!

SOLUTION

Solution to Fascinating Frequencies BONUS

\text{let }x,\,y,\,z\in\R,\;x\leq y\leq z\\[8pt] \text{if }\;\exists!\;\{x,\,y,\,z\}: \operatorname fa=\operatorname gb=\operatorname hc\;\; \forall\;\;\{a,\,b,\,c\}\equiv\{x,\,y,\,z\},\\[4pt] \text{then }\;f\bigcirc g\bigcirc h\\[8pt] \text{if }\ f\bigcirc g\bigcirc h\;\ \text{and }\ f’\circledcirc g’\circledcirc h’,\ \;\text{then}\;f\circledcirc g\circledcirc h

We’re trying to find a set of three functions that satisfied a certain property. Look closely at it, and you see that what it really means is that, firstly, all three lines are concurrent at exactly three points. Then, their derivatives have the same property. This is recursive. Consider that the derivative of a sine wave is also a sine wave. We then have a relatively simple solution to the problem, but which is tricky to figure out.

\left. \begin{array}{l} \text{if $0\le x\lt3\pi$:} & \sin x \\ \text{if $x\ge3\pi$:} & f(-x) \\ \text{if $x\lt0$:} & e^x \\ \end{array} \right\} =f(x) \\[8pt] g(x) = -f(x)\\[8pt] h(x) = 0\\[16pt] f\circledcirc g\circledcirc h

Fascinating Frequencies BONUS

This challenge has calculus and algebra.

We begin this problem by defining a few variables and functions, see below.
Let a ‘functional triangle’ be a relationship between three functions shown, denoted by the two large circles.
The properties of such are shown.
Then, a ‘meta-functional triangle’ is denoted in the same manner be a functional triangle with the given additional property.
The challenge is to find a meta-functional triangle. Good luck!

\text{let }x,\,y,\,z\in\R,\;x\leq y\leq z\\[8pt] \text{if }\;\exists!\;\{x,\,y,\,z\}: \operatorname fa=\operatorname gb=\operatorname hc\;\; \forall\;\;\{a,\,b,\,c\}\equiv\{x,\,y,\,z\},\\[4pt] \text{then }\;f\bigcirc g\bigcirc h\\[8pt] \text{if }\ f\bigcirc g\bigcirc h\;\ \text{and }\ f’\circledcirc g’\circledcirc h’,\ \;\text{then}\;f\circledcirc g\circledcirc h
SOLUTION

Solution to Galvanizing Growth

We’re trying to solve a rather insane system of equations. Due to the size of this problem, It’s publishing was delayed. Also, the solution is not typed, that would take a very long time, but shown is a scan of my rough work.

The initial equation system problem was:

a_{n1}=125\ e^\frac{-c_n}{100}\\[8pt] a_{n2}=5\ e^\frac{-c}{5}\\[16pt] g_n=\left\lfloor a_{n1}s_n{\left({3-2\sqrt2\operatorname{erfc^{-1}}{(r)}\over200}\right)}\left(1-\frac{s_n}{p_n}\right)\right\rfloor\\[16pt] g_{a_b}=\left\lfloor a_{b2}s_b{\left({3-2\sqrt2\operatorname{erfc^{-1}}{(r)}\over200}\right)}\left(1-\frac{s_a}{p_a}\right)\right\rfloor\\[16pt] i_n=s_n+10\left(s_n-\frac{p_n}5+|s_n-\frac{p_n}5|\right)\\[16pt] x=5-3\sqrt2\operatorname{erfc^{-1}}{(2r)}\\[8pt] y=11-5\sqrt2\operatorname{erfc^{-1}}{(2r)}

The solution is:

solution_to_galvanizing_growthDownload

Galvanizing Growth Python Simulation

import pandas as pd
from random import random
from math import floor, log
import matplotlib
from matplotlib import pyplot as plt
from numpy.random import normal
group = {}
growth = {}
countries = [('China', 137e7, 274e6), ('India', 127e7, 254e6), ('United States', 324e6, 648e5), ('Indonesia', 258e6, 516e5), ('Brazil', 206e6, 412e5), ('Pakistan', 202e6, 404e5), ('Nigeria', 186e6, 372e5), ('Bangladesh', 156e6, 312e5), ('Russia', 142e6, 284e5), ('Japan', 127e6, 254e5), ('Mexico', 123e6, 246e5), ('Philippines', 103e6, 206e5), ('Ethiopia', 102e6, 204e5), ('Vietnam', 953e5, 1906e4), ('Egypt', 947e5, 1894e4), ('Iran', 828e5, 1656e4), ('DR Congo', 813e5, 1626e4), ('Germany', 807e5, 1614e4), ('Turkey', 803e5, 1606e4), ('Thailand', 682e5, 1364e4), ('France', 668e5, 1336e4), ('United Kingdom', 644e5, 1288e4), ('Italy', 62e6, 124e5), ('Burma', 569e5, 1138e4), ('South Africa', 543e5, 1086e4), ('Tanzania', 525e5, 105e5), ('South Korea', 509e5, 1018e4), ('Spain', 486e5, 972e4), ('Colombia', 472e5, 944e4), ('Kenya', 468e5, 936e4), ('Ukraine', 442e5, 884e4), ('Argentina', 439e5, 878e4), ('Algeria', 403e5, 806e4), ('Poland', 385e5, 77e5), ('Uganda', 383e5, 766e4), ('Iraq', 381e5, 762e4), ('Sudan', 367e5, 734e4), ('Canada', 354e5, 708e4), ('Morocco', 337e5, 674e4), ('Afghanistan', 333e5, 666e4), ('Malaysia', 31e6, 62e5), ('Venezuela', 309e5, 618e4), ('Peru', 307e5, 614e4), ('Uzbekistan', 295e5, 59e5), ('Nepal', 29e6, 58e5), ('Saudi Arabia', 282e5, 564e4), ('Yemen', 274e5, 548e4), ('Ghana', 269e5, 538e4), ('Mozambique', 259e5, 518e4), ('North Korea', 251e5, 502e4)]
groupKeys = []
scale = 50
for i, j, k in countries:
  group[i] = [[0, j, 0, k]]
  growth[i] = [0]
  groupKeys.append(i)
group[groupKeys[0]] = [[13, countries[0][1]-13, 0, countries[0][2]]]
growth[groupKeys[0]] = [13]
e = 2.7183
def eDistribute(bottom, top, original):
  if(top == bottom):
    return top
  return top*e**((bottom-original)/(top-bottom))
def grow(sets, keys, past, index, spread=lambda a,b:1, expose=lambda a:1, expand=0.05, cut=5, spreadFactor=0.5, growthFactor=10):
  spreadSum = 0
  for i in keys:
    if(sets[i][index][0]+sets[i][index][1] == 0):
      adjExpand = 0
    else:
      adjExpand = expand * sets[i][index][1]/(sets[i][index][0]+sets[i][index][1])
    if(len(sets[i]) == index+1):
      sets[i].append([sets[i][index][0], sets[i][index][1], sets[i][index][2], sets[i][index][3]])
      past[i].append(0)
    for j in keys:
      if(i != j):
        currentSpread = spread(past[i], past[j])
        spreadSum -= currentSpread
        todaySpread = round(normal(1.5, 1)*eDistribute(0, spreadFactor, currentSpread)*sets[i][index][0]*expand)
        if(todaySpread > 0):
          if(len(sets[j]) == index+1):
            sets[j].append([sets[j][index][0]+todaySpread, sets[j][index][1]-todaySpread, sets[j][index][2], sets[j][index][3]])
            past[j].append(todaySpread)
          else:
            sets[j][index+1][0] += todaySpread
            sets[j][index+1][1] -= todaySpread
            if(sets[j][index+1][1] < 0):
              sets[j][index+1][0] += sets[j][index+1][1]
              sets[j][index+1][1] = 0
            past[j][index+1] += todaySpread
    currentSpread = expose(past[i])
    spreadSum -= currentSpread
    newGrowth = round(normal(1.5, 1)*eDistribute(0, growthFactor, currentSpread)*sets[i][index][0]*adjExpand)
    sets[i][index+1][0] += newGrowth
    sets[i][index+1][1] -= newGrowth
    past[i][index+1] += newGrowth
    if(sets[i][index+1][1] < 0):
      sets[i][index+1][0] += sets[i][index+1][1]
      past[i][index+1] -= sets[i][index+1][1]
      sets[i][index+1][1] = 0
    if(index >= cut):
      if(past[i][index-cut] > sets[i][index+1][0]):
        past[i][index-cut] = sets[i][index+1][0]
      sets[i][index+1][0] -= past[i][index-cut]
      sets[i][index+1][2] += past[i][index-cut]
  for one in sets:
    spreadSum += sets[one][index+1][0]
    if(sets[one][index+1][0] > sets[one][index+1][3]):
      spreadSum += 10*(sets[one][index+1][0]-sets[one][index+1][3])
  return spreadSum

impact = 0
impacts = {}
spreadFactor = 0.5
growthFactor = 1.25
meantime = 11
incperiod = 8
strictness = 100, 70
def adj(maxVal):
  def decorator(f):
    def wrapper(*args, **kwargs):
      raw = f(*args, **kwargs)
      if(raw < maxVal):
        return maxVal - raw
      return 0
    return wrapper
  return decorator
compensate = growthFactor*scale + spreadFactor*scale**2
def testAlgs(ban, close):
  i = 0
  impact = 0
  while i < 10:
    j = 0
    while i < 150:
      impacts[str(i)] = impact
      impact = floor(impact + compensate + grow(group, groupKeys, growth, i, spread=lambda a,b:ban(a, b, i-incperiod), expose = lambda a:close(a, i-incperiod), expand = 0.029, cut=meantime+incperiod, spreadFactor=spreadFactor, growthFactor=growthFactor))
      j += 1
  return impact / 10