Make perfect square when smooth number is >N

Author: WrathfulSpatula

FindAFactor/_find_a_factor.cpp

@@ -827,8 +827,8 @@ struct Factorizer {
         const BigInteger bIndex = dis(gen) % threadRange;
         const BigInteger halfBIndex = threadOffset + (bIndex >> 1U) + 1U;
         const BigInteger bNum = (bIndex & 1U) ? halfBIndex : (batchTotal - halfBIndex);
-        BigInteger n = forwardFn((bNum * wheelEntryCount) + (dis(gen) % wheelEntryCount));
-        const std::vector<size_t> pfv = factorizationVector(&n);
+        const BigInteger n = forwardFn((bNum * wheelEntryCount) + (dis(gen) % wheelEntryCount));
+        const std::vector<size_t> pfv = factorizationVector(n);
         if (!pfv.size()) {
           continue;
         }
@@ -837,6 +837,20 @@ struct Factorizer {
           fv[pi] += pfv[pi];
         }
       }
+      // We actually want not just a smooth number,
+      // but a smooth perfect square.
+      for (size_t pi = 0U; pi < primes.size(); ++pi) {
+        size_t& vp = fv[pi];
+        if (vp & 1U) {
+          // If the prime factor component parity is odd,
+          // multiply by the prime once to make it even.
+          perfectSquare *= primes[pi];
+          ++vp;
+        }
+        // The parity is necessarily even in this factor, by now.
+        // Divide by 2 to get the count of square prime factors.
+        vp >>= 1U;
+    }
 
       const size_t batchLimit = smoothBatchLimit * (1ULL << batchPower);
       batchPower = (batchPower + 1U) % batchSizeVariance;
@@ -957,8 +971,7 @@ struct Factorizer {
   }
 
   // Compute the prime factorization.
-  std::vector<size_t> factorizationVector(BigInteger* numPtr) {
-    BigInteger& num = *numPtr;
+  std::vector<size_t> factorizationVector(BigInteger num) {
     std::vector<size_t> vec(primes.size(), 0);
     while (true) {
       // Proceed in steps of the GCD with the smooth prime wheel radius.
@@ -990,20 +1003,6 @@ struct Factorizer {
       // because it is not smooth.
       return std::vector<size_t>();
     }
-    // We actually want not just a smooth number,
-    // but a smooth perfect square.
-    for (size_t pi = 0U; pi < primes.size(); ++pi) {
-      size_t& vp = vec[pi];
-      if (vp & 1U) {
-        // If the prime factor component parity is odd,
-        // multiply by the prime once to make it even.
-        num *= primes[pi];
-        ++vp;
-      }
-      // The parity is necessarily even in this factor, by now.
-      // Divide by 2 to get the count of square prime factors.
-      vp >>= 1U;
-    }
 
     // This number is necessarily a smooth perfect square.
     return vec;

pyproject.toml

@@ -8,7 +8,7 @@ build-backend = "setuptools.build_meta"
 
 [project]
 name = "FindAFactor"
-version = "4.7.9"
+version = "4.7.10"
 requires-python = ">=3.8"
 description = "Find any nontrivial factor of a number"
 readme = {file = "README.txt", content-type = "text/markdown"}

setup.py

@@ -40,7 +40,7 @@ def build_extension(self, ext):
 
 setup(
     name='FindAFactor',
-    version='4.7.9',
+    version='4.7.10',
     author='Dan Strano',
     author_email='stranoj@gmail.com',
     description='Find any nontrivial factor of a number',