How to resolve the algorithm Count in factors step by step in the jq programming language

Published on 12 May 2024 09:40 PM
#Jq

How to resolve the algorithm Count in factors step by step in the jq 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 jq programming language

Source code in the jq programming language

# To take advantage of gojq's arbitrary-precision integer arithmetic:
def power($b): . as $in | reduce range(0;$b) as $i (1; . * $in);

# Input: a non-negative integer determining when to stop
def count_in_factors:
  "1: 1",
  (range(2;.) | "\(.): \([factors] | join("x"))");

def count_in_factors($m;$n):
  if  . == 1 then  "1: 1" else empty end,
  (range($m;$n) | "\(.): \([factors] | join("x"))");

10 | count_in_factors,
"",
count_in_factors(2144; 2145),
"",
(2|power(100) | count_in_factors(.; .+ 2))

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