factors.Score: Avoid overflow by using math/big

When using really large numbers, factors.Score could overflow, which
would cause an incorrect result.  Using arbitrary-precision arithmetic
fixes this.  I only do so above 2^28, since below then factor sums are
guaranteed to not overflow, and normal arithmetic is faster.

1 files changed, 26 insertions, 1 deletions

diff --git a/factors/score.go b/factors/score.go
index 0759a74..c777d11 100644
--- a/factors/score.go
+++ b/factors/score.go
@@ -1,6 +1,9 @@
 package factors
 
-import "math"
+import (
+	"math"
+	"math/big"
+)
 
 // Score returns a "factor score" equal to the sum of the reciprocoals
 // of the number n's factors.
@@ -20,6 +23,8 @@ import "math"
 func Score(n uint) float64 {
 	if n == 0 {
 		return math.NaN()
+	} else if n > maxSmall {
+		return bigScore(n)
 	}
 
 	factorSum := uint(0)
@@ -29,6 +34,26 @@ func Score(n uint) float64 {
 	return float64(factorSum) / float64(n)
 }
 
+const maxSmall = 1 << 28 - 1
+
+func bigScore(n uint) float64 {
+	factorSum := new(big.Int)
+
+	// initialize bigFactor like this to avoid reallocating memory
+	bigFactor := new(big.Int).SetUint64(uint64(n))
+
+	for _, f := range Factors(n) {
+		bigFactor.SetUint64(uint64(f))
+		factorSum.Add(factorSum, bigFactor)
+	}
+
+	bigN := new(big.Int).SetUint64(uint64(n))
+	score := new(big.Rat).SetFrac(factorSum, bigN)
+
+	score64, _ := score.Float64()
+	return score64
+}
+
 // BasicRank returns a rank describing how well a radix handles the simplest
 // fractions (1/2, 1/3, 1/4 and 1/5).  Zero and one are not true radices,
 // but because this rank otherwise only depends on a radix's remainder