How to resolve the algorithm Zumkeller numbers step by step in the zkl programming language

Published on 12 May 2024 09:40 PM
#Zkl

How to resolve the algorithm Zumkeller numbers step by step in the zkl programming language

Table of Contents

Problem Statement

Zumkeller numbers are the set of numbers whose divisors can be partitioned into two disjoint sets that sum to the same value. Each sum must contain divisor values that are not in the other sum, and all of the divisors must be in one or the other. There are no restrictions on how the divisors are partitioned, only that the two partition sums are equal.

Even Zumkeller numbers are common; odd Zumkeller numbers are much less so. For values below 10^6, there is at least one Zumkeller number in every 12 consecutive integers, and the vast majority of them are even. The odd Zumkeller numbers are very similar to the list from the task Abundant odd numbers; they are nearly the same except for the further restriction that the abundance (A(n) = sigma(n) - 2n), must be even: A(n) mod 2 == 0

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Zumkeller numbers step by step in the zkl programming language

Source code in the zkl programming language

fcn properDivs(n){ // does not include n
//   if(n==1) return(T);	// we con't care about this case
   ( pd:=[1..(n).toFloat().sqrt()].filter('wrap(x){ n%x==0 }) )
   .pump(pd,'wrap(pd){ if(pd!=1 and (y:=n/pd)!=pd ) y else Void.Skip })
}
fcn canSum(goal,divs){
   if(goal==0 or divs[0]==goal) return(True);
   if(divs.len()>1){
      if(divs[0]>goal) return(canSum(goal,divs[1,*]));  // tail recursion
      else return(canSum(goal - divs[0], divs[1,*]) or canSum(goal, divs[1,*]));
   }
   False
}
fcn isZumkellerW(n){	// a filter for a iterator
   ds,sum := properDivs(n), ds.sum(0) + n;
   // if sum is odd, it can't be split into two partitions with equal sums
   if(sum.isOdd) return(Void.Skip);
   // if n is odd use 'abundant odd number' optimization
   if(n.isOdd){
      abundance:=sum - 2*n;
      return( if(abundance>0 and abundance.isEven) n else Void.Skip);
   }
   canSum(sum/2,ds) and n or Void.Skip	// sum is even
}

println("First 220 Zumkeller numbers:");
zw:=[2..].tweak(isZumkellerW);
do(11){ zw.walk(20).pump(String,"%4d ".fmt).println() }

println("\nFirst 40 odd Zumkeller numbers:");
zw:=[3..*, 2].tweak(isZumkellerW);
do(4){ zw.walk(10).pump(String,"%5d ".fmt).println() }

println("\nThe first 40 odd Zumkeller numbers which don't end in 5 are:");
zw:=[3..*, 2].tweak(fcn(n){ if(n%5) isZumkellerW(n) else Void.Skip });
do(5){ zw.walk(8).pump(String,"%7d ".fmt).println() }

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