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Fibonacci matrix-exponentiation

From Rosetta Code
Fibonacci matrix-exponentiation is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The Fibonacci sequence defined with matrix-exponentiation:


Task

Write a program using   matrix exponentiation   to generate Fibonacci(n) for   n   equal to:

  •                  10
  •                100
  •             1,000
  •           10,000
  •         100,000
  •      1,000,000
  •    10,000,000


Only display the first 20 decimal digits   and   the last 20 decimal digits of each Fibonacci number.

Extra

Generate Fibonacci(216 ), Fibonacci(232) and Fibonacci(264) using the same method or another one.


Related tasks



C#[edit]

Matrix exponentiation[edit]

using System;
using System.IO;
using System.Numerics;
using System.Threading;
using System.Diagnostics;
using System.Globalization;
 
namespace Fibonacci {
class Program
{
private static readonly BigInteger[,] F = { { BigInteger.One, BigInteger.One }, { BigInteger.One, BigInteger.Zero } };
private static NumberFormatInfo nfi = new NumberFormatInfo { NumberGroupSeparator = "_" };
private static BigInteger[,] Multiply(in BigInteger[,] A, in BigInteger[,] B)
{
if (A.GetLength(1) != B.GetLength(0))
{
throw new ArgumentException("Illegal matrix dimensions for multiplication.");
}
var C = new BigInteger[A.GetLength(0), B.GetLength(1)];
for (int i = 0; i < A.GetLength(0); ++i)
{
for (int j = 0; j < B.GetLength(1); ++j)
{
for (int k = 0; k < A.GetLength(1); ++k)
{
C[i, j] += A[i, k] * B[k, j];
}
}
}
return C;
}
private static BigInteger[,] Power(in BigInteger[,] A, ulong n)
{
if (A.GetLength(1) != A.GetLength(0))
{
throw new ArgumentException("Not a square matrix.");
}
var C = new BigInteger[A.GetLength(0), A.GetLength(1)];
for (int i = 0; i < A.GetLength(0); ++i)
{
C[i, i] = BigInteger.One;
}
if (0 == n) return C;
var S = new BigInteger[A.GetLength(0), A.GetLength(1)];
for (int i = 0; i < A.GetLength(0); ++i)
{
for (int j = 0; j < A.GetLength(1); ++j)
{
S[i, j] = A[i, j];
}
}
while (0 < n)
{
if (1 == n % 2) C = Multiply(C, S);
S = Multiply(S,S);
n /= 2;
}
return C;
}
public static BigInteger Fib(in ulong n)
{
var C = Power(F, n);
return C[0, 1];
}
public static void Task(in ulong p)
{
var ans = Fib(p).ToString();
var sp = p.ToString("N0", nfi);
if (ans.Length <= 40)
{
Console.WriteLine("Fibonacci({0}) = {1}", sp, ans);
}
else
{
Console.WriteLine("Fibonacci({0}) = {1} ... {2}", sp, ans[0..19], ans[^20..]);
}
}
public static void Main()
{
Stopwatch stopWatch = new Stopwatch();
stopWatch.Start();
for (ulong p = 10; p <= 10_000_000; p *= 10) {
Task(p);
}
stopWatch.Stop();
TimeSpan ts = stopWatch.Elapsed;
string elapsedTime = String.Format("{0:00}:{1:00}:{2:00}.{3:00}",
ts.Hours, ts.Minutes, ts.Seconds,
ts.Milliseconds / 10);
Console.WriteLine("Took " + elapsedTime);
}
}
}
 
 
Output:
Fibonacci(10) = 55
Fibonacci(100) = 354224848179261915075
Fibonacci(1_000) = 4346655768693745643 ... 76137795166849228875
Fibonacci(10_000) = 3364476487643178326 ... 66073310059947366875
Fibonacci(100_000) = 2597406934722172416 ... 49895374653428746875
Fibonacci(1_000_000) = 1953282128707757731 ... 68996526838242546875
Fibonacci(10_000_000) = 1129834378225399760 ... 86998673686380546875
Took 00:04:00.92

Go[edit]

Matrix exponentiation[edit]

This uses matrix exponentiation to calculate the (2^16)th and (2^32)nd Fibonacci numbers the last of which has more than 897 million digits! To improve performance, I've used a GMP wrapper rather than Go's native 'big.Int' type.

I have not attempted to calculate the (2^64)th Fibonacci number which appears to be well out of reach using this approach.

package main
 
import (
"fmt"
big "github.com/ncw/gmp"
"time"
)
 
type vector = []*big.Int
type matrix []vector
 
var (
zero = new(big.Int)
one = big.NewInt(1)
)
 
func (m1 matrix) mul(m2 matrix) matrix {
rows1, cols1 := len(m1), len(m1[0])
rows2, cols2 := len(m2), len(m2[0])
if cols1 != rows2 {
panic("Matrices cannot be multiplied.")
}
result := make(matrix, rows1)
temp := new(big.Int)
for i := 0; i < rows1; i++ {
result[i] = make(vector, cols2)
for j := 0; j < cols2; j++ {
result[i][j] = new(big.Int)
for k := 0; k < rows2; k++ {
temp.Mul(m1[i][k], m2[k][j])
result[i][j].Add(result[i][j], temp)
}
}
}
return result
}
 
func identityMatrix(n uint64) matrix {
if n < 1 {
panic("Size of identity matrix can't be less than 1")
}
ident := make(matrix, n)
for i := uint64(0); i < n; i++ {
ident[i] = make(vector, n)
for j := uint64(0); j < n; j++ {
ident[i][j] = new(big.Int)
if i == j {
ident[i][j].Set(one)
}
}
}
return ident
}
 
func (m matrix) pow(n *big.Int) matrix {
le := len(m)
if le != len(m[0]) {
panic("Not a square matrix")
}
switch {
case n.Cmp(zero) == -1:
panic("Negative exponents not supported")
case n.Cmp(zero) == 0:
return identityMatrix(uint64(le))
case n.Cmp(one) == 0:
return m
}
pow := identityMatrix(uint64(le))
base := m
e := new(big.Int).Set(n)
temp := new(big.Int)
for e.Cmp(zero) > 0 {
temp.And(e, one)
if temp.Cmp(one) == 0 {
pow = pow.mul(base)
}
e.Rsh(e, 1)
base = base.mul(base)
}
return pow
}
 
func fibonacci(n *big.Int) *big.Int {
if n.Cmp(zero) == 0 {
return zero
}
m := matrix{{one, one}, {one, zero}}
m = m.pow(n.Sub(n, one))
return m[0][0]
}
 
func commatize(n uint64) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
 
func main() {
start := time.Now()
n := new(big.Int)
for i := uint64(10); i <= 1e7; i *= 10 {
n.SetUint64(i)
s := fibonacci(n).String()
fmt.Printf("The digits of the %sth Fibonacci number (%s) are:\n",
commatize(i), commatize(uint64(len(s))))
if len(s) > 20 {
fmt.Printf(" First 20 : %s\n", s[0:20])
if len(s) < 40 {
fmt.Printf(" Final %-2d : %s\n", len(s)-20, s[20:])
} else {
fmt.Printf(" Final 20 : %s\n", s[len(s)-20:])
}
} else {
fmt.Printf(" All %-2d  : %s\n", len(s), s)
}
fmt.Println()
}
 
sfxs := []string{"nd", "th"}
for i, e := range []uint{16, 32} {
n.Lsh(one, e)
s := fibonacci(n).String()
fmt.Printf("The digits of the 2^%d%s Fibonacci number (%s) are:\n", e, sfxs[i],
commatize(uint64(len(s))))
fmt.Printf(" First 20 : %s\n", s[0:20])
fmt.Printf(" Final 20 : %s\n", s[len(s)-20:])
fmt.Println()
}
 
fmt.Printf("Took %s\n\n", time.Since(start))
}
Output:

Timings are for an Intel Core i7 8565U machine, using Go 1.14 on Ubuntu 18.04:

The digits of the 10th Fibonacci number (2) are:
  All 2    : 55

The digits of the 100th Fibonacci number (21) are:
  First 20 : 35422484817926191507
  Final 1  : 5

The digits of the 1,000th Fibonacci number (209) are:
  First 20 : 43466557686937456435
  Final 20 : 76137795166849228875

The digits of the 10,000th Fibonacci number (2,090) are:
  First 20 : 33644764876431783266
  Final 20 : 66073310059947366875

The digits of the 100,000th Fibonacci number (20,899) are:
  First 20 : 25974069347221724166
  Final 20 : 49895374653428746875

The digits of the 1,000,000th Fibonacci number (208,988) are:
  First 20 : 19532821287077577316
  Final 20 : 68996526838242546875

The digits of the 10,000,000th Fibonacci number (2,089,877) are:
  First 20 : 11298343782253997603
  Final 20 : 86998673686380546875

The digits of the 2^16nd Fibonacci number (13,696) are:
  First 20 : 73199214460290552832
  Final 20 : 97270190955307463227

The digits of the 2^32th Fibonacci number (897,595,080) are:
  First 20 : 61999319689381859818
  Final 20 : 39623735538208076347

Took 11m0.255916206s


Lucas method[edit]

Although Go supports big.Float, the precision needed to calculate the (2^32)nd Fibonacci number makes the use of Binet's formula impractical. I have therefore used the same method as the Julia entry for my alternative approach which is more than twice as quick as the matrix exponentiation method.

Translation of: Julia
package main
 
import (
"fmt"
big "github.com/ncw/gmp"
"time"
)
 
var (
zero = new(big.Int)
one = big.NewInt(1)
two = big.NewInt(2)
three = big.NewInt(3)
)
 
func lucas(n *big.Int) *big.Int {
var inner func(n *big.Int) (*big.Int, *big.Int)
inner = func(n *big.Int) (*big.Int, *big.Int) {
if n.Cmp(zero) == 0 {
return new(big.Int), big.NewInt(1)
}
t, q, r := new(big.Int), new(big.Int), new(big.Int)
u, v := inner(t.Rsh(n, 1))
t.And(n, two)
q.Sub(t, one)
u.Mul(u, u)
v.Mul(v, v)
t.And(n, one)
if t.Cmp(one) == 0 {
t.Sub(u, q)
t.Mul(two, t)
r.Mul(three, v)
return u.Add(u, v), r.Sub(r, t)
} else {
t.Mul(three, u)
r.Add(v, q)
r.Mul(two, r)
return r.Sub(r, t), u.Add(u, v)
}
}
t, q, l := new(big.Int), new(big.Int), new(big.Int)
u, v := inner(t.Rsh(n, 1))
l.Mul(two, v)
l.Sub(l, u) // Lucas function
t.And(n, one)
if t.Cmp(one) == 0 {
q.And(n, two)
q.Sub(q, one)
t.Mul(v, l)
return t.Add(t, q)
}
return u.Mul(u, l)
}
 
func commatize(n uint64) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
 
func main() {
start := time.Now()
n := new(big.Int)
for i := uint64(10); i <= 1e7; i *= 10 {
n.SetUint64(i)
s := lucas(n).String()
fmt.Printf("The digits of the %sth Fibonacci number (%s) are:\n",
commatize(i), commatize(uint64(len(s))))
if len(s) > 20 {
fmt.Printf(" First 20 : %s\n", s[0:20])
if len(s) < 40 {
fmt.Printf(" Final %-2d : %s\n", len(s)-20, s[20:])
} else {
fmt.Printf(" Final 20 : %s\n", s[len(s)-20:])
}
} else {
fmt.Printf(" All %-2d  : %s\n", len(s), s)
}
fmt.Println()
}
 
sfxs := []string{"nd", "th"}
for i, e := range []uint{16, 32} {
n.Lsh(one, e)
s := lucas(n).String()
fmt.Printf("The digits of the 2^%d%s Fibonacci number (%s) are:\n", e, sfxs[i],
commatize(uint64(len(s))))
fmt.Printf(" First 20 : %s\n", s[0:20])
fmt.Printf(" Final 20 : %s\n", s[len(s)-20:])
fmt.Println()
}
 
fmt.Printf("Took %s\n\n", time.Since(start))
}
Output:
As first version except time is now 5m4.13427997s.


Fibmod method[edit]

This uses the Sidef entry's 'Fibmod' approach to enable the (2^64)th Fibonacci number to be processed. As Go lacks such a function, I have translated the Julia version. I have also had to pull in a third party library to provide functions (such as Log) which Go's big.Float implementation lacks.

The speed-up compared to the other approaches is astonishing!

Library: bigfloat
Translation of: Sidef
Translation of: Julia
package main
 
import (
"fmt"
"github.com/ALTree/bigfloat"
"github.com/ncw/gmp"
"math/big"
"time"
)
 
const (
nd = 20 // number of digits to be displayed at each end
pr = 128 // precision to be used
)
 
var (
one = gmp.NewInt(1)
two = gmp.NewInt(2)
ten = gmp.NewInt(10)
onef = big.NewFloat(1).SetPrec(pr)
tenf = big.NewFloat(10).SetPrec(pr)
ln10 = bigfloat.Log(tenf)
)
 
func fibmod(n, nmod *gmp.Int) *gmp.Int {
if n.Cmp(two) < 0 {
return n
}
fibmods := make(map[string]*gmp.Int)
var f func(n *gmp.Int) *gmp.Int
f = func(n *gmp.Int) *gmp.Int {
if n.Cmp(two) < 0 {
return one
}
ns := n.String()
if v, ok := fibmods[ns]; ok {
return v
}
k, t, u, v := new(gmp.Int), new(gmp.Int), new(gmp.Int), new(gmp.Int)
k.Quo(n, two)
t.And(n, one)
if t.Cmp(one) != 0 {
t.Set(f(k))
t.Mul(t, t)
v.Sub(k, one)
u.Set(f(v))
u.Mul(u, u)
} else {
t.Set(f(k))
v.Add(k, one)
v.Set(f(v))
u.Sub(k, one)
u.Set(f(u))
u.Mul(u, t)
t.Mul(t, v)
}
t.Add(t, u)
fibmods[ns] = t.Rem(t, nmod)
return fibmods[ns]
}
w := new(gmp.Int)
w.Sub(n, one)
return f(w)
}
 
func binetApprox(n *big.Int) *big.Float {
phi, ihp := big.NewFloat(0.5).SetPrec(pr), big.NewFloat(0.5).SetPrec(pr)
root := big.NewFloat(1.25).SetPrec(pr)
root.Sqrt(root)
phi.Add(root, phi)
ihp.Sub(root, ihp)
ihp.Neg(ihp)
ihp.Sub(phi, ihp)
ihp = bigfloat.Log(ihp)
phi = bigfloat.Log(phi)
nn := new(big.Float).SetPrec(pr).SetInt(n)
phi.Mul(phi, nn)
return phi.Sub(phi, ihp)
}
 
func firstFibDigits(n *big.Int, k int) string {
f := binetApprox(n)
g := new(big.Float).SetPrec(pr)
g.Quo(f, ln10)
g.Add(g, onef)
i, _ := g.Int(nil)
g.SetInt(i)
g.Mul(ln10, g)
f.Sub(f, g)
f = bigfloat.Exp(f)
p := big.NewInt(int64(k))
p.Exp(big.NewInt(10), p, nil)
g.SetInt(p)
f.Mul(f, g)
i, _ = f.Int(nil)
return i.String()[0:k]
}
 
func lastFibDigits(n *gmp.Int, k int) string {
p := gmp.NewInt(int64(k))
p.Exp(ten, p, nil)
return fibmod(n, p).String()[0:k]
}
 
func commatize(n uint64) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
 
func main() {
start := time.Now()
n := new(big.Int)
for i := uint64(10); i <= 1e7; i *= 10 {
n.SetUint64(i)
nn := new(gmp.Int)
nn.SetUint64(i)
fmt.Printf("\nThe digits of the %sth Fibonacci number are:\n", commatize(i))
nd2, nd3 := nd, nd
// These need to be preset for i == 10 & i == 100
// as there is no way of deriving the total length of the string using this method.
if i == 10 {
nd2 = 2
} else if i == 100 {
nd3 = 1
}
s1 := firstFibDigits(n, nd2)
if len(s1) < 20 {
fmt.Printf(" All %-2d  : %s\n", len(s1), s1)
} else {
fmt.Printf(" First 20 : %s\n", s1)
s2 := lastFibDigits(nn, nd3)
if len(s2) < 20 {
fmt.Printf(" Final %-2d : %s\n", len(s2), s2)
} else {
fmt.Printf(" Final 20 : %s\n", s2)
}
}
}
 
o := big.NewInt(1)
ord := []string{"th", "nd", "th"}
for i, p := range []uint{16, 32, 64} {
n.Lsh(o, p)
nn := new(gmp.Int)
nn.Lsh(one, p)
fmt.Printf("\nThe digits of the 2^%d%s Fibonacci number are:\n", p, ord[i])
fmt.Printf(" First %d : %s\n", nd, firstFibDigits(n, nd))
fmt.Printf(" Final %d : %s\n", nd, lastFibDigits(nn, nd))
}
fmt.Printf("\nTook %s\n", time.Since(start))
}
Output:
The digits of the 10th Fibonacci number are:
  All 2    : 55

The digits of the 100th Fibonacci number are:
  First 20 : 35422484817926191507
  Final 1  : 5

The digits of the 1,000th Fibonacci number are:
  First 20 : 43466557686937456435
  Final 20 : 76137795166849228875

The digits of the 10,000th Fibonacci number are:
  First 20 : 33644764876431783266
  Final 20 : 66073310059947366875

The digits of the 100,000th Fibonacci number are:
  First 20 : 25974069347221724166
  Final 20 : 49895374653428746875

The digits of the 1,000,000th Fibonacci number are:
  First 20 : 19532821287077577316
  Final 20 : 68996526838242546875

The digits of the 10,000,000th Fibonacci number are:
  First 20 : 11298343782253997603
  Final 20 : 86998673686380546875

The digits of the 2^16th Fibonacci number are:
  First 20 : 73199214460290552832
  Final 20 : 97270190955307463227

The digits of the 2^32nd Fibonacci number are:
  First 20 : 61999319689381859818
  Final 20 : 39623735538208076347

The digits of the 2^64th Fibonacci number are:
  First 20 : 11175807536929528424
  Final 20 : 17529800348089840187

Took 6.391894ms

Haskell[edit]

Matrix exponentiation[edit]

import System.CPUTime (getCPUTime)
import Data.List
 
main = do
startTime <- getCPUTime
mapM_ (putStrLn.formatAns).take 7.iterate (*10) $ 10
mapM_ (putStrLn.seeFib) [16,32]
finishTime <- getCPUTime
putStrLn $ "Took " ++ (took startTime finishTime)
 
took t = fromChrono.chrono t
 
fromChrono :: (Integer,Integer,Integer) -> String
fromChrono (m,s,ms) = show m ++ "m" ++ show s ++ "." ++ show ms ++ "s"
 
chrono :: Integer -> Integer -> (Integer,Integer,Integer)
chrono start end = (m,s,ms)
where
tera = 1000000000000
fdt = fromIntegral (end - start) / tera
dt = floor fdt
(m,s) = quotRem dt 60
ms = round $ fromIntegral (round (fdt - (fromIntegral dt))*1000) / 1000
 
bagOf :: Int -> [a] -> [[a]]
bagOf _ [] = []
bagOf n xs = let (us,vs) = splitAt n xs in us : bagOf n vs
 
formatIntegral :: Show a => String -> a -> String
formatIntegral sep = reverse.intercalate sep.bagOf 3.reverse.show
 
formatAns :: Integer -> String
formatAns p = start ++ go x
where
start = "Fibonacci("++ (formatIntegral "_" p) ++ ") = "
x = fib p
tenPow20 = 10^20
tenPow40 = tenPow20^2
go u | u <= tenPow20 = show u
go u | u <= tenPow40 = let (us,vs) = splitAt 20 $ show u in us ++ " ... " ++ vs
go u = (take 20 $ show u) ++ " ... " ++ (show . rem u $ 10^20)
 
seeFib :: Integer -> String
seeFib n = start ++ xs ++ " ... " ++ (show . rem x $ 10^20)
where
start = "Fibonacci(2^" ++ (show n) ++") = "
x = fib (2^n)
xs = take 20 $ show x
 
fib :: Integer -> Integer
fib 0 = 0 -- this line is necessary because "something ^ 0" returns "fromInteger 1", which unfortunately
-- in our case is not our multiplicative identity (the identity matrix) but just a 1x1 matrix of 1
fib n = (last . head . unMat) (Mat [[1, 1], [1, 0]] ^ n)
 
mult :: Num a => [[a]] -> [[a]] -> [[a]]
mult uss vss = map ((\xs -> if null xs then [] else foldl1 (zipWith (+)) xs) . zipWith (flip (map . (*))) vss) uss
 
newtype Mat a = Mat
{ unMat :: [[a]]
} deriving (Eq,Show)
 
instance Num a => Num (Mat a) where
negate xm = Mat $ map (map negate) $ unMat xm
xm + ym = Mat $ zipWith (zipWith (+)) (unMat xm) (unMat ym)
xm * ym = Mat $ mult (unMat xm) (unMat ym)
fromInteger n = Mat [[fromInteger n]]
abs = undefined
signum = undefined
Output:

Timing is on Intel Core i5-4300U CPU, Windows 10 Professional, using GHCi Version 8.6.5:

Fibonacci(10) = 55
Fibonacci(100) = 35422484817926191507 ... 5
Fibonacci(1_000) = 43466557686937456435 ... 76137795166849228875
Fibonacci(10_000) = 33644764876431783266 ... 66073310059947366875
Fibonacci(100_000) = 25974069347221724166 ... 49895374653428746875
Fibonacci(1_000_000) = 19532821287077577316 ... 68996526838242546875
Fibonacci(10_000_000) = 11298343782253997603 ... 86998673686380546875
Fibonacci(2^16) = 73199214460290552832 ... 97270190955307463227
Fibonacci(2^32) = 61999319689381859818 ... 39623735538208076347
Took 5m20.1s

Matrix exponentiation - printing alternative[edit]

import System.CPUTime (getCPUTime)
import Data.List
 
main = do
startTime <- getCPUTime
mapM_ (putStrLn.formatAns).take 7.iterate (*10) $ 10
mapM_ (putStrLn.seeFib) [16,32]
finishTime <- getCPUTime
putStrLn $ "Took " ++ (took startTime finishTime)
 
took t = fromChrono.chrono t
 
fromChrono :: (Integer,Integer,Integer) -> String
fromChrono (m,s,ms) = show m ++ "m" ++ show s ++ "." ++ show ms ++ "s"
 
chrono :: Integer -> Integer -> (Integer,Integer,Integer)
chrono start end = (m,s,ms)
where
tera = 1000000000000
fdt = fromIntegral (end - start) / tera
dt = floor fdt
(m,s) = quotRem dt 60
ms = round $ fromIntegral (round (fdt - (fromIntegral dt))*1000) / 1000
 
bagOf :: Int -> [a] -> [[a]]
bagOf _ [] = []
bagOf n xs = let (us,vs) = splitAt n xs in us : bagOf n vs
 
formatIntegral :: Show a => String -> a -> String
formatIntegral sep = reverse.intercalate sep.bagOf 3.reverse.show
 
formatAns :: Integer -> String
formatAns p = start ++ (startEnd 20 (sizeFib p) num)
where
start = "Fibonacci("++ (formatIntegral "_" p) ++ ") = "
num = fib p
 
seeFib :: Integer -> String
seeFib n = start ++ (startEnd 20 (sizeFib p) num)
where
start = "Fibonacci(2^" ++ (show n) ++") = "
p = 2^n
num = fib p
 
startEnd :: (Integral a, Show a) => a -> a -> a -> String
startEnd ndigit len num | len <= ndigit = show num
startEnd ndigit len num | len <= 2*ndigit = let (us,vs) = genericSplitAt ndigit (show num) in us ++ " ... " ++ vs
startEnd ndigit len num = start ++ " ... " ++ end
where
end = show.rem num $ 10 ^ ndigit
start = show.quot num $ 10 ^ (len - ndigit + 1)
 
phi :: Double
phi = (1 + sqrt 5)/2
log10phi = logBase 10 phi
halflog10five =(logBase 10 5)/2
 
sizeFib :: Integral a => a -> a
sizeFib p = ceiling $ (fromIntegral p)*log10phi - halflog10five
 
fib :: Integer -> Integer
fib 0 = 0 -- this line is necessary because "something ^ 0" returns "fromInteger 1", which unfortunately
-- in our case is not our multiplicative identity (the identity matrix) but just a 1x1 matrix of 1
fib n = (last . head . unMat) (Mat [[1, 1], [1, 0]] ^ n)
 
mult :: Num a => [[a]] -> [[a]] -> [[a]]
mult uss vss = map ((\xs -> if null xs then [] else foldl1 (zipWith (+)) xs) . zipWith (flip (map . (*))) vss) uss
 
newtype Mat a = Mat
{ unMat :: [[a]]
} deriving (Eq,Show)
 
instance Num a => Num (Mat a) where
negate xm = Mat $ map (map negate) $ unMat xm
xm + ym = Mat $ zipWith (zipWith (+)) (unMat xm) (unMat ym)
xm * ym = Mat $ mult (unMat xm) (unMat ym)
fromInteger n = Mat [[fromInteger n]]
abs = undefined
signum = undefined
Output:

Timing is on Intel Core i5-4300U CPU, Windows 10 Professional, using GHCi Version 8.6.5:

Fibonacci(10) = 55
Fibonacci(100) = 35422484817926191507 ... 5
Fibonacci(1_000) = 4346655768693745643 ... 76137795166849228875
Fibonacci(10_000) = 3364476487643178326 ... 66073310059947366875
Fibonacci(100_000) = 2597406934722172416 ... 49895374653428746875
Fibonacci(1_000_000) = 1953282128707757731 ... 68996526838242546875
Fibonacci(10_000_000) = 1129834378225399760 ... 86998673686380546875
Fibonacci(2^16) = 7319921446029055283 ... 97270190955307463227
Fibonacci(2^32) = 6199931968938185981 ... 39623735538208076347
Took 3m58.1s

Matrix exponentiation for a symmetric matrix[edit]

We will use a property of symmetric matrices which commute.

If X,Y are two symmetric matrices of same size and if they commute then X*Y is a symmetric matrix.

It means to compute Z = X*Y, only terms on and below the diagonal need to be computed (above = below).

At each step of the exponentiation of a symmetric matric, we multiply 2 symmetric matrices which commute.

-- https://yutsumura.com/symmetric-matrices-and-the-product-of-two-matrices/
import System.CPUTime (getCPUTime)
import Data.List
 
main = do
startTime <- getCPUTime
mapM_ (putStrLn.formatAns).take 7.iterate (*10) $ 10
mapM_ (putStrLn.seeFib) [16,32]
finishTime <- getCPUTime
putStrLn $ "Took " ++ (took startTime finishTime)
 
took t = fromChrono.chrono t
 
fromChrono :: (Integer,Integer,Integer) -> String
fromChrono (m,s,ms) = show m ++ "m" ++ show s ++ "." ++ show ms ++ "s"
 
chrono :: Integer -> Integer -> (Integer,Integer,Integer)
chrono start end = (m,s,ms)
where
tera = 1000000000000
fdt = fromIntegral (end - start) / tera
dt = floor fdt
(m,s) = quotRem dt 60
ms = round $ fromIntegral (round (fdt - (fromIntegral dt))*1000) / 1000
 
bagOf :: Int -> [a] -> [[a]]
bagOf _ [] = []
bagOf n xs = let (us,vs) = splitAt n xs in us : bagOf n vs
 
formatIntegral :: Show a => String -> a -> String
formatIntegral sep = reverse.intercalate sep.bagOf 3.reverse.show
 
formatAns :: Integer -> String
formatAns p = start ++ (startEnd 20 (sizeFib p) num)
where
start = "Fibonacci("++ (formatIntegral "_" p) ++ ") = "
num = fib p
 
seeFib :: Integer -> String
seeFib n = start ++ (startEnd 20 (sizeFib p) num)
where
start = "Fibonacci(2^" ++ (show n) ++") = "
p = 2^n
num = fib p
 
startEnd :: (Integral a, Show a) => a -> a -> a -> String
startEnd ndigit len num | len <= ndigit = show num
startEnd ndigit len num | len <= 2*ndigit = let (us,vs) = genericSplitAt ndigit (show num) in us ++ " ... " ++ vs
startEnd ndigit len num = start ++ " ... " ++ end
where
end = show.rem num $ 10 ^ ndigit
start = show.quot num $ 10 ^ (len - ndigit + 1)
 
phi :: Double
phi = (1 + sqrt 5)/2
log10phi = logBase 10 phi
halflog10five =(logBase 10 5)/2
 
sizeFib :: Integral a => a -> a
sizeFib p = ceiling $ (fromIntegral p)*log10phi - halflog10five
 
fib :: Integer -> Integer
fib 0 = 0
fib n = zeroOne (power (Mat 1 1 1 0) n)
 
data Mat a = Mat {zeroZero :: a, zeroOne :: a, oneZero :: a, oneOne :: a} deriving (Eq,Show)
 
-- for a symmetric matrix
square :: Num a => (Mat a) -> (Mat a)
square (Mat x00 x01 x10 x11) = Mat y00 y10 y10 y11
where
y00 = y10 + y11 -- F_{n+1} = F_{n} + F_{n-1}
y10 = x10*(x00+x11)
y11 = x11*x11+x10*x10
 
-- for 2 symmetric matrices which commute
mult :: Num a => (Mat a) -> (Mat a) -> (Mat a)
mult (Mat x00 x01 x10 x11) (Mat y00 y01 y10 y11) = Mat xy00 xy01 xy01 xy11
where
xy00 = xy01 + xy11 -- F_{n+1} = F_{n} + F_{n-1}
xy01 = x10*y00 + x11*y10
xy11 = x10*y01 + x11*y11
 
power :: Num a => (Mat a) -> Integer -> (Mat a)
power _ n | n < 0 = error "Exception: Negative exponent"
power _ 0 = Mat 1 0 0 1
power m 1 = m
power m n = if even n then w else mult w m
where w = square.power m.quot n $ 2
Output:

Timing is on Intel Core i5-4300U CPU, Windows 10 Professional, using GHCi Version 8.6.5:

Fibonacci(10) = 55
Fibonacci(100) = 35422484817926191507 ... 5
Fibonacci(1_000) = 4346655768693745643 ... 76137795166849228875
Fibonacci(10_000) = 3364476487643178326 ... 66073310059947366875
Fibonacci(100_000) = 2597406934722172416 ... 49895374653428746875
Fibonacci(1_000_000) = 1953282128707757731 ... 68996526838242546875
Fibonacci(10_000_000) = 1129834378225399760 ... 86998673686380546875
Fibonacci(2^16) = 7319921446029055283 ... 97270190955307463227
Fibonacci(2^32) = 6199931968938185981 ... 39623735538208076347
Took 2m6.1s

Java[edit]

Performed the task to use Matrix multiplication to compute Fibonacci numbers.

Implemented fib and fibMod.

 
import java.math.BigInteger;
import java.util.Arrays;
 
public class FibonacciMatrixExponentiation {
 
public static void main(String[] args) {
BigInteger mod = BigInteger.TEN.pow(20);
for ( int exp : Arrays.asList(32, 64) ) {
System.out.printf("Last 20 digits of fib(2^%d) = %s%n", exp, fibMod(BigInteger.valueOf(2).pow(exp), mod));
}
 
for ( int i = 1 ; i <= 7 ; i++ ) {
BigInteger n = BigInteger.TEN.pow(i);
System.out.printf("fib(%,d) = %s%n", n, displayFib(fib(n)));
}
}
 
private static String displayFib(BigInteger fib) {
String s = fib.toString();
if ( s.length() <= 40 ) {
return s;
}
return s.substring(0, 20) + " ... " + s.subSequence(s.length()-20, s.length());
}
 
// Use Matrix multiplication to compute Fibonacci numbers.
private static BigInteger fib(BigInteger k) {
BigInteger aRes = BigInteger.ZERO;
BigInteger bRes = BigInteger.ONE;
BigInteger cRes = BigInteger.ONE;
BigInteger aBase = BigInteger.ZERO;
BigInteger bBase = BigInteger.ONE;
BigInteger cBase = BigInteger.ONE;
while ( k.compareTo(BigInteger.ZERO) > 0 ) {
if ( k.mod(BigInteger.valueOf(2)).compareTo(BigInteger.ONE) == 0 ) {
BigInteger temp1 = aRes.multiply(aBase).add(bRes.multiply(bBase));
BigInteger temp2 = aBase.multiply(bRes).add(bBase.multiply(cRes));
BigInteger temp3 = bBase.multiply(bRes).add(cBase.multiply(cRes));
aRes = temp1;
bRes = temp2;
cRes = temp3;
}
k = k.shiftRight(1);
BigInteger temp1 = aBase.multiply(aBase).add(bBase.multiply(bBase));
BigInteger temp2 = aBase.multiply(bBase).add(bBase.multiply(cBase));
BigInteger temp3 = bBase.multiply(bBase).add(cBase.multiply(cBase));
aBase = temp1;
bBase = temp2;
cBase = temp3;
}
return aRes;
}
 
// Use Matrix multiplication to compute Fibonacci numbers.
private static BigInteger fibMod(BigInteger k, BigInteger mod) {
BigInteger aRes = BigInteger.ZERO;
BigInteger bRes = BigInteger.ONE;
BigInteger cRes = BigInteger.ONE;
BigInteger aBase = BigInteger.ZERO;
BigInteger bBase = BigInteger.ONE;
BigInteger cBase = BigInteger.ONE;
while ( k.compareTo(BigInteger.ZERO) > 0 ) {
if ( k.mod(BigInteger.valueOf(2)).compareTo(BigInteger.ONE) == 0 ) {
BigInteger temp1 = aRes.multiply(aBase).add(bRes.multiply(bBase)).mod(mod);
BigInteger temp2 = aBase.multiply(bRes).add(bBase.multiply(cRes)).mod(mod);
BigInteger temp3 = bBase.multiply(bRes).add(cBase.multiply(cRes)).mod(mod);
aRes = temp1;
bRes = temp2;
cRes = temp3;
}
k = k.shiftRight(1);
BigInteger temp1 = aBase.multiply(aBase).add(bBase.multiply(bBase)).mod(mod);
BigInteger temp2 = aBase.multiply(bBase).add(bBase.multiply(cBase)).mod(mod);
BigInteger temp3 = bBase.multiply(bBase).add(cBase.multiply(cBase)).mod(mod);
aBase = temp1;
bBase = temp2;
cBase = temp3;
}
return aRes.mod(mod);
}
 
}
 
Output:
Last 20 digits of fib(2^32) = 39623735538208076347
Last 20 digits of fib(2^64) = 17529800348089840187

fib(10) = 55
fib(100) = 354224848179261915075
fib(1,000) = 43466557686937456435 ... 76137795166849228875
fib(10,000) = 33644764876431783266 ... 66073310059947366875
fib(100,000) = 25974069347221724166 ... 49895374653428746875
fib(1,000,000) = 19532821287077577316 ... 68996526838242546875
fib(10,000,000) = 11298343782253997603 ... 86998673686380546875

Julia[edit]

Because Julia uses the GMP library for its BigInt type, a BigInt cannot be larger than about 2^(2^37). This prevents generation of the 2^64-th fibonacchi number, due to BigInt overflow. The Binet method actually overflows even with the 2^32-nd fibonacchi number, so the Lucas method is used as the alternative method.

# Here is the matrix Fibonacci formula as specified to be used for the solution.
const b = [big"1" 1; 1 0]
matrixfibonacci(n) = n == 0 ? 0 : n == 1 ? 1 : (b^(n+1))[2,2]
 
# This exact Binet Fibonacci formula is not used due to BigFloat exponent size limitations.
binetfibonacci(n) = ((1+sqrt(big"5"))^n-(1-sqrt(big"5"))^n)/(sqrt(big"5")*big"2"^n)
 
# Use the exponent size limiting variant of the Binet formula seen in the Sidef example.
function firstbinet(bits, ndig=20)
logφ = big"2"^bits * log(10, (1 + sqrt(BigFloat(5.0))) / 2)
mantissa = logφ - trunc(logφ) + ndig + 1
return string(BigInt(round((10^mantissa - 10^(-mantissa)) / sqrt(BigFloat(5.0)))))[1:ndig]
end
 
# The fibmod function has no builtin in Julia, so here is one.
function fibmod(n::BigInt, nmod::BigInt)
n < 2 && return n
fibmods = Dict{BigInt, BigInt}()
function f(n::BigInt)
n < 2 && return 1
haskey(fibmods, n) && return fibmods[n]
k = div(n, 2)
fibmods[n] = iseven(n) ?
(f(k) * f(k) + f(k - 1) * f(k - 1)) % nmod :
(f(k) * f(k + 1) + f(k - 1) * f(k)) % nmod
end
f(n - 1)
end
lastfibmod(bits, ndig=21) = string(fibmod(big"2"^bits, big"10"^(ndig+1)))
 
# See Wikipedia on Lucas function for the algorithm below.
# inner -> F(n/2), F(n/2 - 1), L(n) = F(n) + 2F(n-1), and L(n/2) * F(n/2) = F(n)
function lucasfibonacci(n)
function inner(n)
if n == 0
return big"0", big"1"
end
u, v = inner(n >> 1)
q = (n & 2) - 1
u *= u
v *= v
return isodd(n) ? (BigInt(u + v), BigInt(3 * v - 2 * (u - q))) :
(BigInt(2 * (v + q) - 3 * u), BigInt(u + v))
end
u, v = inner(n >> 1)
l = 2*v - u # the lucas function
if isodd(n)
q = (n & 2) - 1
return v * l + q
end
return u * l
end
 
m2s(bits) = string(matrixfibonacci(big"2"^bits))
l2s(bits) = string(lucasfibonacci(big"2"^bits))
firstlast(s) = (length(s) < 40 ? s : s[1:20] * "..." * s[end-20+1:end])
 
println("N", " "^23, "Matrix", " "^40, "Lucas", " "^40, "Mod\n", "-"^145)
println("2^16 ", rpad(firstlast(m2s(16)), 45), rpad(firstlast(l2s(16)), 45),
rpad(firstlast(firstbinet(16) * lastfibmod(16)), 45))
println("2^32 ", rpad(firstlast(m2s(32)), 45), rpad(firstlast(l2s(32)), 45),
rpad(firstlast(firstbinet(32) * lastfibmod(32)), 45))
println("2^64 ", " "^90, rpad(firstlast(firstbinet(64) * lastfibmod(64)), 45))
 
Output:
N                       Matrix                                        Lucas                                        Mod
-------------------------------------------------------------------------------------------------------------------------------------------------
2^16  73199214460290552832...97270190955307463227  73199214460290552832...97270190955307463227  73199214460290552832...97270190955307463227
2^32  61999319689381859818...39623735538208076347  61999319689381859818...39623735538208076347  61999319689381859818...39623735538208076347
2^64                                                                                            11175807536929528424...17529800348089840187

Nim[edit]

Translation of: Wren
Translation of: Go

Using the Lucas method.

import strformat, strutils, times
import bignum
 
let
One = newInt(1)
Two = newInt(2)
Three = newInt(3)
 
proc lucas(n: Int): Int =
 
proc inner(n: Int): (Int, Int) =
if n.isZero: return (newInt(0), newInt(1))
var t = n shr 1
var (u, v) = inner(t)
t = n and Two
let q = t - One
var r = newInt(0)
u *= u
v *= v
t = n and One
if t == One:
t = (u - q) * Two
r = v * Three
result = (u + v, r - t)
else:
t = u * Three
r = v + q
r *= Two
result = (r - t, u + v)
 
var t = n shr 1
let (u, v) = inner(t)
let l = v * Two - u
t = n and One
if t == One:
let q = n and Two - One
return v * l + q
 
return u * l
 
let start = now()
 
var n: Int
var i = 10
while i <= 10_000_000:
n = newInt(i)
let s = $lucas(n)
 
echo &"The digits of the {($i).insertSep}th Fibonacci number ({($s.len).insertSep}) are:"
if s.len > 20:
echo &" First 20 : {s[0..19]}"
if s.len < 40:
echo &" Final {s.len-20:<2} : {s[20..^1]}"
else:
echo &" Final 20 : {s[^20..^1]}"
else:
echo &" All {s.len:<2}  : {s}"
echo()
i *= 10
 
for e in [culong 16, 32]:
n = One shl e
let s = $lucas(n)
echo &"The digits of the 2^{e}th Fibonacci number ({($s.len).insertSep}) are:"
echo &" First 20 : {s[0..19]}"
echo &" Final 20 : {s[^20..^1]}"
echo()
 
echo &"Took {now() - start}"
Output:
The digits of the 10th Fibonacci number (2) are:
  All 2    : 55

The digits of the 100th Fibonacci number (21) are:
  First 20 : 35422484817926191507
  Final 1  : 5

The digits of the 1_000th Fibonacci number (209) are:
  First 20 : 43466557686937456435
  Final 20 : 76137795166849228875

The digits of the 10_000th Fibonacci number (2_090) are:
  First 20 : 33644764876431783266
  Final 20 : 66073310059947366875

The digits of the 100_000th Fibonacci number (20_899) are:
  First 20 : 25974069347221724166
  Final 20 : 49895374653428746875

The digits of the 1_000_000th Fibonacci number (208_988) are:
  First 20 : 19532821287077577316
  Final 20 : 68996526838242546875

The digits of the 10_000_000th Fibonacci number (2_089_877) are:
  First 20 : 11298343782253997603
  Final 20 : 86998673686380546875

The digits of the 2^16th Fibonacci number (13_696) are:
  First 20 : 73199214460290552832
  Final 20 : 97270190955307463227

The digits of the 2^32th Fibonacci number (897_595_080) are:
  First 20 : 61999319689381859818
  Final 20 : 39623735538208076347

Took 6 minutes, 18 seconds, 643 milliseconds, 622 microseconds, and 965 nanoseconds

Perl[edit]

Translation of: Sidef
use strict;
use warnings;
 
use Math::AnyNum qw(:overload fibmod floor);
use Math::MatrixLUP;
 
sub fibonacci {
my $M = Math::MatrixLUP->new([ [1, 1], [1,0] ]);
(@{$M->pow(shift)})[0][1]
}
 
for my $n (16, 32) {
my $f = fibonacci(2**$n);
printf "F(2^$n) = %s ... %s\n", substr($f,0,20), $f % 10**20;
}
 
sub binet_approx {
my($n) = @_;
use constant PHI => sqrt(1.25) + 0.5 ;
use constant IHP => -(sqrt(1.25) - 0.5);
(log(PHI)*$n - log(PHI-IHP))
}
 
sub nth_fib_first_k_digits {
my($n,$k) = @_;
my $f = binet_approx($n);
floor(exp($f - log(10)*(floor($f / log(10) + 1))) * 10**$k)
}
 
sub nth_fib_last_k_digits {
my($n,$k) = @_;
fibmod($n, 10**$k);
}
 
print "\n";
for my $n (16, 32, 64) {
my $first20 = nth_fib_first_k_digits(2**$n, 20);
my $last20 = nth_fib_last_k_digits(2**$n, 20);
printf "F(2^$n) = %s ... %s\n", $first20, $last20;
}
Output:
F(2^16) = 73199214460290552832 ... 97270190955307463227
F(2^32) = 61999319689381859818 ... 39623735538208076347

F(2^16) = 73199214460290552832 ... 97270190955307463227
F(2^32) = 61999319689381859818 ... 39623735538208076347
F(2^64) = 11175807536929528424 ... 17529800348089840187

Phix[edit]

Library: Phix/mpfr
Translation of: Sidef
Since I don't have a builtin fibmod, I had to roll my own, and used {n,m} instead of(/to mean) 2^n+m, thus avoiding some 2^53 native atom limits on 32-bit.

(mpz and mpfr variables are effectively pointers, and therefore simply won't work as expected/needed should you try and use them as keys to a cache.)

without javascript_semantics -- (no mpfr_log(), mpfr_exp() under pwa/p2js)
requires("1.0.0") -- (mpfr_set_default_prec[ision] has been renamed)
include mpfr.e
mpfr_set_default_precision(-40)
 
constant PHI = mpfr_init(1.25),
         IHP = mpfr_init(),
         HLF = mpfr_init(0.5),
         L10 = mpfr_init(10)
mpfr_sqrt(PHI,PHI)
mpfr_sub(IHP,PHI,HLF)
mpfr_add(PHI,PHI,HLF)
mpfr_neg(IHP,IHP)
mpfr_log(L10,L10)
 
procedure binet_approx(mpfr r, integer n)
    mpfr m = mpfr_init()
    mpfr_ui_pow_ui(m,2,n) -- (n as in 2^n here)
    mpfr_log(r,PHI)
    mpfr_mul(r,r,m)
    mpfr_sub(m,PHI,IHP)
    mpfr_log(m,m)
    mpfr_sub(r,r,m)
    m = mpfr_free(m)
end procedure
 
function nth_fib_first_k_digits(integer n, k)
    mpfr {f,g,h} = mpfr_inits(3)
    binet_approx(f,n)
    mpfr_div(g,f,L10)
    mpfr_add_si(g,g,1)
    mpfr_floor(g,g)
    mpfr_mul(g,L10,g)
    mpfr_sub(f,f,g)
    mpfr_exp(f,f)
    mpfr_ui_pow_ui(h,10,k)
    mpfr_mul(f,f,h)
    mpfr_floor(f,f)
    string fmt = sprintf("%%%d.0Rf",k)
    return mpfr_sprintf(fmt,f)
end function
 
integer cache = new_dict()
-- key of {n,m} means 2^n+m (where n and m are integers, n>=0, m always 0 or -1)
setd({0,0},mpz_init(0),cache)  -- aka fib(0):=0
setd({1,-1},mpz_init(1),cache) --     fib(1):=1
setd({1,0},mpz_init(1),cache)  --     fib(2):=1
 
procedure mpz_fibmod(mpz f, p, integer n, m=0)
    sequence key = {n,m}
    if getd_index(key,cache)!=NULL then
        mpz_set(f,getd(key,cache))
    else
        mpz {f1,f2} = mpz_inits(2)
        n -= 1
        mpz_fibmod(f1,p,n)
        mpz_fibmod(f2,p,n,-1)
        if m=-1 then
            -- fib(2n-1) = fib(n)^2 + fib(n-1)^2
            mpz_mul(f1,f1,f1)
            mpz_mul(f2,f2,f2)
            mpz_add(f1,f1,f2)
        else
            -- fib(2n) = (2*fib(n-1)+fib(n))*fib(n)
            mpz_mul_si(f2,f2,2)
            mpz_add(f2,f2,f1)
            mpz_mul(f1,f2,f1)
        end if
        mpz_mod(f1,f1,p) 
        setd(key,f1,cache)
        mpz_set(f,f1)   
    end if
end procedure
 
function nth_fib_last_k_digits(integer n, k)
    mpz {f,p} = mpz_inits(2)
    mpz_ui_pow_ui(p,10,k)
    mpz_fibmod(f,p,n)
    return mpz_get_str(f)
end function
 
constant tests = {16,32,64}
for i=1 to length(tests) do
    integer n = tests[i]
    string first20 = nth_fib_first_k_digits(n, 20),
           last20 = nth_fib_last_k_digits(n, 20)
    printf(1,"F(2^%d) = %s ... %s\n",{n,first20,last20})
end for
Output:
F(2^16) = 73199214460290552832 ... 97270190955307463227
F(2^32) = 61999319689381859818 ... 39623735538208076347
F(2^64) = 11175807536929528424 ... 17529800348089840187

matrix exponentiation (2^16)[edit]

Somewhat closer to the original specification, but certainly not recommended for 2^32, let alone 2^64...
Contains copies of routines from Matrix-exponentiation_operator#Phix, but modified to use gmp.

with javascript_semantics
include mpfr.e
 
constant ZERO = mpz_init(0),
         ONE = mpz_init(1)
 
function identity(integer n)
    sequence res = repeat(repeat(ZERO,n),n)
    for i=1 to n do
        res[i][i] = ONE
    end for
    return res
end function
 
function matrix_mul(sequence a, sequence b)
    if length(a[1])!=length(b) then
        crash("matrices cannot be multiplied")
    end if
    sequence c = repeat(repeat(ZERO,length(b[1])),length(a))
    for i=1 to length(a) do
        for j=1 to length(b[1]) do
            for k=1 to length(a[1]) do
                mpz cij = mpz_init()
                mpz_mul(cij,a[i][k],b[k][j])
                mpz_add(cij,cij,c[i][j])
                c[i][j] = cij
            end for
        end for
    end for
    return c
end function
 
function matrix_exponent(sequence m, integer n)
    integer l = length(m)
    if l!=length(m[1]) then crash("not a square matrix") end if
    if n<0 then crash("negative exponents not supported") end if
    sequence res = identity(l)
    while n do
        if and_bits(n,1) then
            res = matrix_mul(res,m)
        end if
        n = floor(n/2)
        if n=0 then exit end if -- (avoid unnecessary matrix_mul)
        m = matrix_mul(m,m)
    end while
    return res
end function
 
function fibonacci(integer n)
    sequence m = {{ONE, ONE},
                  {ONE, ZERO}}
    m = matrix_exponent(m,n-1)
    mpz res = m[1][1]
    return res
end function
 
atom t0 = time()
mpz fn = fibonacci(power(2,16))
string s = mpz_get_str(fn)
string e = elapsed(time()-t0)
printf(1,"fibonnaci(2^16) = %s [%s]\n",{shorten(s),e})
 
for i=1 to iff(platform()=JS?6:7) do
    t0 = time()
    integer n = power(10,i)
    fn = fibonacci(n)
    s = mpz_get_str(fn)
    t0 = time()-t0
    e = iff(t0>0.1?" ["&elapsed(t0)&"]":"")
    printf(1,"fibonnaci(%,d) = %s%s\n",{n,shorten(s),e})
end for
Output:
fibonnaci(2^16) = 73199214460290552832...97270190955307463227 (13,696 digits) [0s]
fibonnaci(10) = 55
fibonnaci(100) = 354224848179261915075
fibonnaci(1,000) = 43466557686937456435...76137795166849228875 (209 digits)
fibonnaci(10,000) = 33644764876431783266...66073310059947366875 (2,090 digits)
fibonnaci(100,000) = 25974069347221724166...49895374653428746875 (20,899 digits)
fibonnaci(1,000,000) = 19532821287077577316...68996526838242546875 (208,988 digits) [0.2s]
fibonnaci(10,000,000) = 11298343782253997603...86998673686380546875 (2,089,877 digits) [9.0s]

Note that under pwa/p2js 10^6 takes 11.4s so we'll stop there...
Change that loop to 8 and a 9 year old 3.3GHz i3 also eventually gets:

fibonnaci(100,000,000) = 47371034734563369625...06082642167760546875 (20,898,764 digits) [14 minutes and 24s]

Clearly 2^32 (897 million digits, apparently) is a tad out of bounds, let alone 2^64.

Raku[edit]

(formerly Perl 6) Following the general approach of Sidef, and relying on Perl for fibmod function. Borrowed/adapted routines from Ramanujan's_constant task to allow FatRat calculations throughout. Does not quite meet task spec, as n^64 results not working yet.

use Math::Matrix;
use Inline::Perl5;
my $p5 = Inline::Perl5.new();
$p5.use( 'Math::AnyNum' );
 
constant D = 53; # set the size of FatRat calcluations
 
# matrix exponentiation
sub fibonacci ($n) {
my $M = Math::Matrix.new( [[1,1],[1,0]] );
($M ** $n)[0][1]
}
 
# calculation of 𝑒
sub postfix:<!> (Int $n) { (constant f = 1, |[] 1..*)[$n] }
sub 𝑒 (--> FatRat) { sum map { FatRat.new(1,.!) }, ^D }
 
# calculation of π
sub π (--> FatRat) {
my ($a, $n, $g, $z, $pi) = 1, 1, sqrt(1/2.FatRat), 0.25;
 
for ^5 {
given [ ($a + $g)/2, sqrt $a × $g ] {
$z -= (.[0] - $a)**2 × $n;
$n += $n;
($a, $g) = @$_;
$pi = ($a ** 2 / $z).substr: 0, 2 + D
}
}
$pi.FatRat
}
 
# square-root: accepts/return FatRat
multi sqrt(FatRat $r --> FatRat) {
FatRat.new: sqrt($r.nude[0] × 10**(2) div $r.nude[1]), 10**D
}
 
# square-root: accepts/return Int
multi sqrt(Int $n --> Int) {
my $guess = 10**($n.chars div 2);
my $iterator = { ( $^x + $n div ($^x) ) div 2 };
my $endpoint = { $^x == $^y|$^z };
min ($guess, $iterator$endpoint)[*-1, *-2]
}
 
# arithmetic-geometric mean: accepts/returns FatRat
sub AG-mean(FatRat $a is copy, FatRat $g is copy --> FatRat) {
($a, $g) = ($a + $g)/2, sqrt $a × $g until $a - $g < 10**-D;
$a
}
 
# override built-in definitions with 'FatRat' versions
constant 𝑒 = &𝑒();
constant π = &π();
 
# approximation of natural log, accepts any numeric, returns FatRat
sub log-approx ($x --> FatRat) {
constant ln2 = 2 * [+] (((1/3).FatRat**(2*$_+1))/(2*$_+1) for 0..D); # 1/3 = (2-1)/(2+1)
π / (2 × AG-mean(1.FatRat, 2.FatRat**(2-D)/$x)) - D × ln2
}
 
# power function, with exponent less than zero: accepts/returns FatRat
multi infix:<**> (FatRat $base, FatRat $exp is copy where * < 0 --> FatRat) {
constant ε = 10**-D;
my ($low, $high) = 0.FatRat, 1.FatRat;
my $mid = $high / 2;
my $acc = my $sqr = sqrt($base);
$exp = -$exp;
 
while (abs($mid - $exp) > ε) {
$sqr = sqrt($sqr);
if ($mid <= $exp) { $low = $mid; $acc ×= $sqr }
else { $high = $mid; $acc ×= 1/$sqr }
$mid = ($low + $high) / 2
}
 
(1/$acc).substr(0, D).FatRat
}
 
sub binet_approx (Int $n --> FatRat) {
constant PHI = sqrt(1.25.FatRat) + 0.5 ;
constant IHP = -(sqrt(1.25.FatRat) - 0.5);
$n × log-approx(PHI) - log-approx(PHI - IHP)
}
 
sub nth_fib_first_k_digits ($n,$k) {
my $f = binet_approx($n);
my $log10 = log-approx(10);
floor( 𝑒**($f - $log10×(floor($f/$log10 + 1))) × 10**$k)
}
 
my &nth_fib_last_k_digits =
$p5.run('sub { my($n,$k) = @_; Math::AnyNum::fibmod($n, 10**$k) }');
 
# matrix exponentiation is very inefficient, n^64 not feasible
for (16, 32) -> $n {
my $f = fibonacci(2**$n);
printf "F(2^$n) = %s ... %s\n", substr($f,0,20), $f % 10**20
}
 
# this way is much faster, but not yet able to handle 2^64 case
for 16, 32 -> $n {
my $first20 = nth_fib_first_k_digits(2**$n, 20);
my $last20 = nth_fib_last_k_digits(2**$n, 20);
printf "F(2^$n) = %s ... %s\n", $first20, $last20
}
Output:
F(2^16) = 73199214460290552832 ... 97270190955307463227
F(2^32) = 61999319689381859818 ... 39623735538208076347
F(2^16) = 73199214460290552832 ... 97270190955307463227
F(2^32) = 61999319689381859818 ... 39623735538208076347

Sidef[edit]

Computing the n-th Fibonacci number, using matrix-exponentiation (this function is also built-in):

func fibonacci(n) {
([[1,1],[1,0]]**n)[0][1]
}
 
say 15.of(fibonacci) #=> [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]

First and last 20 digits of the n-th Fibonacci number:

for n in (16, 32) {
 
var f = fibonacci(2**n)
 
with (20) {|k|
var a = (f // 10**(f.ilog10 - k + 1))
var b = (f % 10**k)
say "F(2^#{n}) = #{a} ... #{b}"
}
}
Output:
F(2^16) = 73199214460290552832 ... 97270190955307463227
F(2^32) = 61999319689381859818 ... 39623735538208076347

More efficient approach, using Binet's formula for computing the first k digits, combined with the built-in method fibmod(n,m) for computing the last k digits:

func binet_approx(n) {
const PHI = (1.25.sqrt + 0.5)
const IHP = -(1.25.sqrt - 0.5)
(log(PHI)*n - log(PHI-IHP))
}
 
func nth_fib_first_k_digits(n,k) {
var f = binet_approx(n)
floor(exp(f - log(10)*(floor(f / log(10) + 1))) * 10**k)
}
 
func nth_fib_last_k_digits(n,k) {
fibmod(n, 10**k)
}
 
for n in (16, 32, 64) {
var first20 = nth_fib_first_k_digits(2**n, 20)
var last20 = nth_fib_last_k_digits(2**n, 20)
say "F(2^#{n}) = #{first20} ... #{last20}"
}
Output:
F(2^16) = 73199214460290552832 ... 97270190955307463227
F(2^32) = 61999319689381859818 ... 39623735538208076347
F(2^64) = 11175807536929528424 ... 17529800348089840187

Wren[edit]

Matrix exponentiation[edit]

Translation of: Go
Library: Wren-big
Library: Wren-fmt

A tough task for Wren which takes just under 3 minutes to process even the 1 millionth Fibonacci number. Judging by the times for the compiled, statically typed languages using GMP, the 10 millionth would likely take north of 4 hours so I haven't attempted it.

import "/big" for BigInt
import "/fmt" for Fmt
 
var mul = Fn.new { |m1, m2|
var rows1 = m1.count
var cols1 = m1[0].count
var rows2 = m2.count
var cols2 = m2[0].count
if (cols1 != rows2) Fiber.abort("Matrices cannot be multiplied.")
var result = List.filled(rows1, null)
for (i in 0...rows1) {
result[i] = List.filled(cols2, null)
for (j in 0...cols2) {
result[i][j] = BigInt.zero
for (k in 0...rows2) {
var temp = m1[i][k] * m2[k][j]
result[i][j] = result[i][j] + temp
}
}
}
return result
}
 
var identityMatrix = Fn.new { |n|
if (n < 1) Fiber.abort("Size of identity matrix can't be less than 1")
var ident = List.filled(n, 0)
for (i in 0...n) {
ident[i] = List.filled(n, null)
for(j in 0...n) ident[i][j] = (i != j) ? BigInt.zero : BigInt.one
}
return ident
}
 
var pow = Fn.new { |m, n|
var le = m.count
if (le != m[0].count) Fiber.abort("Not a square matrix")
if (n < 0) Fiber.abort("Negative exponents not supported")
if (n == 0) return identityMatrix.call(le)
if (n == 1) return m
var pow = identityMatrix.call(le)
var base = m
var e = n
while (e > 0) {
var temp = e & BigInt.one
if (temp == BigInt.one) pow = mul.call(pow, base)
e = e >> 1
base = mul.call(base, base)
}
return pow
}
 
var fibonacci = Fn.new { |n|
if (n == 0) return BigInt.zero
var m = [[BigInt.one, BigInt.one], [BigInt.one, BigInt.zero]]
m = pow.call(m, n - 1)
return m[0][0]
}
 
var n = BigInt.zero
var i = 10
while (i <= 1e6) {
n = BigInt.new(i)
var s = fibonacci.call(n).toString
Fmt.print("The digits of the $,sth Fibonacci number ($,s) are:", i, s.count)
if (s.count > 20) {
Fmt.print(" First 20 : $s", s[0...20])
if (s.count < 40) {
Fmt.print(" Final $-2d : $s", s.count-20, s[20..-1])
} else {
Fmt.print(" Final 20 : $s", s[s.count-20..-1])
}
} else {
Fmt.print(" All $-2d  : $s", s.count, s)
}
System.print()
i = i * 10
}
n = BigInt.one << 16
var s = fibonacci.call(n).toString
Fmt.print("The digits of the 2^16th Fibonacci number ($,s) are:", s.count)
Fmt.print(" First 20 : $s", s[0...20])
Fmt.print(" Final 20 : $s", s[s.count-20..-1])
Output:
The digits of the 10th Fibonacci number (2) are:
  All 2    : 55

The digits of the 100th Fibonacci number (21) are:
  First 20 : 35422484817926191507
  Final 1  : 5

The digits of the 1,000th Fibonacci number (209) are:
  First 20 : 43466557686937456435
  Final 20 : 76137795166849228875

The digits of the 10,000th Fibonacci number (2,090) are:
  First 20 : 33644764876431783266
  Final 20 : 66073310059947366875

The digits of the 100,000th Fibonacci number (20,899) are:
  First 20 : 25974069347221724166
  Final 20 : 49895374653428746875

The digits of the 1,000,000th Fibonacci number (208,988) are:
  First 20 : 19532821287077577316
  Final 20 : 68996526838242546875

The digits of the 2^16th Fibonacci number (13,696) are:
  First 20 : 73199214460290552832
  Final 20 : 97270190955307463227


Lucas method[edit]

Translation of: Go

Much quicker than the matrix exponentiation method taking about 23 seconds to process the 1 millionth Fibonacci number and around 32 minutes to reach the 10 millionth.

Apart from the 2^16th number, the extra credit is still out of reach using this approach and unfortunately the Fibmod method is not available as Wren doesn't currently have a suitable library for arbitrary precision floating point calculations.

import "/big" for BigInt
import "/fmt" for Fmt
 
var lucas = Fn.new { |n|
var inner // recursive function
inner = Fn.new { |n|
if (n == BigInt.zero) return [BigInt.zero, BigInt.one]
var t = n >> 1
var res = inner.call(t)
var u = res[0]
var v = res[1]
t = n & BigInt.two
var q = t - BigInt.one
var r = BigInt.zero
u = u.square
v = v.square
t = n & BigInt.one
if (t == BigInt.one) {
t = u - q
t = BigInt.two * t
r = BigInt.three * v
return [u + v, r - t]
} else {
t = BigInt.three * u
r = v + q
r = BigInt.two * r
return [r - t, u + v]
}
}
var t = n >> 1
var res = inner.call(t)
var u = res[0]
var v = res[1]
var l = BigInt.two * v
l = l - u // Lucas function
t = n & BigInt.one
if (t == BigInt.one) {
var q = n & BigInt.two
q = q - BigInt.one
t = v * l
return t + q
}
return u * l
}
 
var n = BigInt.zero
var i = 10
while (i <= 1e7) {
n = BigInt.new(i)
var s = lucas.call(n).toString
Fmt.print("The digits of the $,sth Fibonacci number ($,s) are:", i, s.count)
if (s.count > 20) {
Fmt.print(" First 20 : $s", s[0...20])
if (s.count < 40) {
Fmt.print(" Final $-2d : $s", s.count-20, s[20..-1])
} else {
Fmt.print(" Final 20 : $s", s[s.count-20..-1])
}
} else {
Fmt.print(" All $-2d  : $s", s.count, s)
}
System.print()
i = i * 10
}
n = BigInt.one << 16
var s = lucas.call(n).toString
Fmt.print("The digits of the 2^16th Fibonacci number ($,s) are:", s.count)
Fmt.print(" First 20 : $s", s[0...20])
Fmt.print(" Final 20 : $s", s[s.count-20..-1])
Output:

As matrix exponentiation method, plus:

The digits of the 10,000,000th Fibonacci number (2,089,877) are:
  First 20 : 11298343782253997603
  Final 20 : 86998673686380546875