How to resolve the algorithm Iterated digits squaring step by step in the 11l programming language

Published on 12 May 2024 09:40 PM
#11l

How to resolve the algorithm Iterated digits squaring step by step in the 11l programming language

Table of Contents

Problem Statement

If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89: An example in Python:

Or, for much less credit - (showing that your algorithm and/or language is slow): This problem derives from the Project Euler problem 92. For a quick algorithm for this task see the talk page

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Iterated digits squaring step by step in the 11l programming language

Source code in the 11l programming language

F next_step(=x)
   V result = 0
   L x > 0
      result += (x % 10) ^ 2
      x I/= 10
   R result

F check(number)
   V candidate = 0
   L(n) number
      candidate = candidate * 10 + n

   L candidate != 89 & candidate != 1
      candidate = next_step(candidate)

   I candidate == 89
      V digits_count = [0] * 10
      L(d) number
         digits_count[d]++

      V result = factorial(number.len)
      L(c) digits_count
         result I/= factorial(c)
      R result

   R 0

V limit = 100000000
V cache_size = Int(ceil(log10(limit)))
assert(10 ^ cache_size == limit)

V number = [0] * cache_size
V result = 0
V i = cache_size - 1

L
   I i == 0 & number[i] == 9
      L.break
   I i == cache_size - 1 & number[i] < 9
      number[i]++
      result += check(number)
   E I number[i] == 9
      i--
   E
      number[i]++
      L(j) i + 1 .< cache_size
         number[j] = number[i]
      i = cache_size - 1
      result += check(number)

print(result)

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