How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Forth programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Forth programming language
Table of Contents
Problem Statement
These define three classifications of positive integers based on their proper divisors. Let P(n) be the sum of the proper divisors of n where the proper divisors are all positive divisors of n other than n itself.
6 has proper divisors of 1, 2, and 3. 1 + 2 + 3 = 6, so 6 is classed as a perfect number.
Calculate how many of the integers 1 to 20,000 (inclusive) are in each of the three classes. Show the results here.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Forth programming language
Source code in the forth programming language
CREATE A 0 ,
: SLOT ( x y -- 0|1|2) OVER OVER < -ROT > - 1+ ;
: CLASSIFY ( n -- n') \ 0 == deficient, 1 == perfect, 2 == abundant
DUP A ! \ we'll be accessing this often, so save somewhere convenient
2 / >R \ upper bound
1 \ starting sum, 1 is always a divisor
2 \ current check
BEGIN DUP R@ < WHILE
A @ OVER /MOD SWAP ( s c d m)
IF DROP ELSE
R> DROP DUP >R ( R: d n)
OVER TUCK OVER <> * - ( s c c+?d)
ROT + SWAP ( s' c)
THEN 1+
REPEAT DROP R> DROP A @ ( sum n) SLOT ;
CREATE COUNTS 0 , 0 , 0 ,
: INIT COUNTS 3 CELLS ERASE 1 COUNTS ! ;
: CLASSIFY-NUMBERS ( n --) INIT
BEGIN DUP WHILE
1 OVER CLASSIFY CELLS COUNTS + +! 1-
REPEAT DROP ;
: .COUNTS
." Deficient : " [ COUNTS ]L @ . CR
." Perfect : " [ COUNTS 1 CELLS + ]L @ . CR
." Abundant : " [ COUNTS 2 CELLS + ]L @ . CR ;
20000 CLASSIFY-NUMBERS .COUNTS BYE
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