How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Tcl 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 Tcl 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 Tcl programming language
Source code in the tcl programming language
proc ProperDivisors {n} {
if {$n == 1} {return 0}
set divs 1
set sum 1
for {set i 2} {$i*$i <= $n} {incr i} {
if {! ($n % $i)} {
lappend divs $i
incr sum $i
if {$i*$i<$n} {
lappend divs [set d [expr {$n / $i}]]
incr sum $d
}
}
}
list $sum $divs
}
proc cmp {i j} { ;# analogous to [string compare], but for numbers
if {$i == $j} {return 0}
if {$i > $j} {return 1}
return -1
}
proc classify {k} {
lassign [ProperDivisors $k] p ;# we only care about the first part of the result
dict get {
1 abundant
0 perfect
-1 deficient
} [cmp $k $p]
}
puts "Classifying the integers in \[1, 20_000\]:"
set classes {} ;# this will be a dict
for {set i 1} {$i <= 20000} {incr i} {
set class [classify $i]
dict incr classes $class
}
# using [lsort] to order the dictionary by value:
foreach {kind count} [lsort -stride 2 -index 1 -integer $classes] {
puts "$kind: $count"
}
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