Distribution of 0 digits in factorial series: Difference between revisions
Distribution of 0 digits in factorial series (view source)
Revision as of 19:46, 10 June 2021
, 2 years agoAdded Go
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permanently falls below 0.16. This task took many hours in the Python example, though I wonder if there is a faster
algorithm out there.
=={{header|Go}}==
{{libheader|GMP(Go wrapper)}}
Brute force as I'll be surprised if there is a faster 'exact' algorithm for this task.
However, the combination of a native code compiler and GMP really cuts down the times (2.8 seconds for the basic task and 182.5 seconds for the stretch goal). Expect these times to be reduced further by the fastest languages.
<lang go>package main
import (
"fmt"
big "github.com/ncw/gmp"
"rcu"
)
func main() {
fact := big.NewInt(1)
sum := 0.0
first := int64(0)
firstRatio := 0.0
fmt.Println("The mean proportion of zero digits in factorials up to the following are:")
for n := int64(1); n <= 50000; n++ {
fact.Mul(fact, big.NewInt(n))
bytes := []byte(fact.String())
digits := len(bytes)
zeros := 0
for _, b := range bytes {
if b == '0' {
zeros++
}
}
sum += float64(zeros)/float64(digits)
ratio := sum / float64(n)
if n == 100 || n == 1000 || n == 10000 {
fmt.Printf("%6s = %12.10f\n", rcu.Commatize(int(n)), ratio)
}
if first > 0 && ratio >= 0.16 {
first = 0
firstRatio = 0.0
} else if first == 0 && ratio < 0.16 {
first = n
firstRatio = ratio
}
}
fmt.Printf("%6s = %12.10f", rcu.Commatize(int(first)), firstRatio)
fmt.Println(" (stays below 0.16 after this)")
fmt.Println(sum/ 50000)
}</lang>
{{out}}
<pre>
The mean proportion of zero digits in factorials up to the following are:
100 = 0.2467531862
1,000 = 0.2035445511
10,000 = 0.1730038482
47,332 = 0.1599999958 (stays below 0.16 after this)
</pre>
=={{header|Python}}==
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