help with prime pairs code

[Salih Dincer]

On Friday, 31 January 2025 at 20:05:54 UTC, Jabari Zakiya wrote:
> I'm converting Ruby code to D and am running into issues.
> Would appreciate what needs to be done to get it to compile/run 
> correctly.

First of all, I see that you don't use reserve() at all in 
arrays. Since I work on a mobile platform (by Dcoder from India), 
I have limited opportunities. So would you try the code below; 
does it work quickly, safely and with accurate results.

```d
import std.stdio, std.conv, std.range, std.algorithm;
import std.datetime.stopwatch : StopWatch;

void main(string[] args)
{
   alias T = uint;
   auto inputs = ["5_005_446", "65_537"];
   auto nums   = inputs.map!(i => i.filter!(n => n != '_'));
   auto n      = nums.map!(f => f.to!T())[0];

   auto timer = StopWatch();
        timer.start();

        n.prime_pairs_lohi();
        timer.peek.writeln();
}

void prime_pairs_lohi(T)(T n)
{
   if (n & 1 || n < 4) { return writeln("Input not even n > 2"); }
   if (n <= 6) { writeln([n, 1]); writeln([n/2, n/2]); 
writeln([n/2, n/2]); return; }

   T ndiv2 = n / 2;
   T rhi   = n - 2;
   T[] lhr;// = iota(3, ndiv2, 2).filter!(e => gcd(e, n) == 
1).array;/*
       lhr.reserve(n / 4); // SDB at 79 added!

   auto factors = n.getFactors!100;
   for (T i = 3; i < ndiv2; i += 2)
   {
     bool coprime = true;
     foreach (p; factors)
     {
       if (i % p == 0)
       {
         coprime = false;
         break;
       }
     }
     if (coprime) lhr ~= i;
   }//*/

   // identify and store all powers and cross-products of the lhr 
members < n-2
   T[] lhr_del;                      // lhr multiples, not part of 
a pcp
       lhr_del.reserve(n / 2); // SDB at 79 added!

   foreach(i, r; lhr) {              // iterate thru lhr to find 
prime multiples
     auto rmax = rhi / r;            // ri can't multiply r with 
values > this
     if (r < rmax) lhr_del ~= (r*r < ndiv2) ? r*r : n - r*r; // 
for r^2 multiples
     if (lhr[i+1] > rmax) break;     // exit if product of 
consecutive r’s > n-2
     foreach(ri; lhr[i+1..$]) {      // for each residue in 
reduced list
       if (ri > rmax) break;         // exit for r if 
cross-product with ri > n-2
       lhr_del ~= (r*ri < ndiv2) ? r*ri : n - r*ri;         // 
store value if < n-2
   } }

   // remove from lhr its lhr multiples, the pcp remain
   lhr_del.sort!"a < b";
   lhr = setDifference(lhr, lhr_del).array;

   writeln([n,     lhr.length]);     // show n and pcp prime pairs 
count
   writeln([lhr[0],  n-lhr[0]]);     // show first pcp prime pair 
of n
   writeln([lhr[$-1],n-lhr[$-1]]);   // show last  pcp prime pair 
of n
}

auto getFactors(int limit, T)(T n)
{
   T[] factors;
   T f = 3;
   T x = n;/*
   T x = n;
   while (x % 2 == 0) {
     factors ~= 2;
           x /= 2;
   }//*/
   while (f < limit && f * f <= x)
   {
     if (x % f == 0)
     {
       if (!factors.canFind(f))
       {
         factors ~= f;
       }
       x /= f;
     } else {
       f += 2;
     }
   }
   //if (x > 1) factors ~= x;

   return factors;
}

auto gcd(T)(T a, T b)
{
   while (b != 0)
   {
     auto temp = b;
     //while (a >= b) a -= b;/*
     a %= b; //*/
     b = a;
     a = temp;
   }
   return cast(T)a;
}
```

SDB at 79