How to resolve the algorithm Count in factors step by step in the CoffeeScript programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Count in factors step by step in the CoffeeScript programming language
Table of Contents
Problem Statement
Write a program which counts up from 1, displaying each number as the multiplication of its prime factors. For the purpose of this task, 1 (unity) may be shown as itself.
2 is prime, so it would be shown as itself. 6 is not prime; it would be shown as
2 × 3
{\displaystyle 2\times 3}
. 2144 is not prime; it would be shown as
2 × 2 × 2 × 2 × 2 × 67
{\displaystyle 2\times 2\times 2\times 2\times 2\times 67}
.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Count in factors step by step in the CoffeeScript programming language
Source code in the coffeescript programming language
count_primes = (max) ->
# Count through the natural numbers and give their prime
# factorization. This algorithm uses no division.
# Instead, each prime number starts a rolling odometer
# to help subsequent factorizations. The algorithm works similar
# to the Sieve of Eratosthenes, as we note when each prime number's
# odometer rolls a digit. (As it turns out, as long as your computer
# is not horribly slow at division, you're better off just doing simple
# prime factorizations on each new n vs. using this algorithm.)
console.log "1 = 1"
primes = []
n = 2
while n <= max
factors = []
for prime_odometer in primes
# digits are an array w/least significant digit in
# position 0; for example, [3, [0]] will roll as
# follows:
# [0] -> [1] -> [2] -> [0, 1]
[base, digits] = prime_odometer
i = 0
while true
digits[i] += 1
break if digits[i] < base
digits[i] = 0
factors.push base
i += 1
if i >= digits.length
digits.push 0
if factors.length == 0
primes.push [n, [0, 1]]
factors.push n
console.log "#{n} = #{factors.join('*')}"
n += 1
primes.length
num_primes = count_primes 10000
console.log num_primes
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