Fibonacci matrix-exponentiation: Difference between revisions

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The Fibonacci sequence defined with matrix-exponentiation:

{\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{n}={\begin{pmatrix}F_{n+1}&F_{n}\\F_{n}&F_{n-1}\end{pmatrix}}.

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 and 10_000_000.

Display only the 20 first digits and 20 last digits of each Fibonacci number.

Extra

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

Related tasks

C#

Matrix exponentiation

<lang csharp>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(1), B.GetLength(0)];
           for (int i = 0; i < A.GetLength(0); ++i)
           {
               for (int j = 0; j < B.GetLength(1); ++j)
               {
                   for (int k = 0; k < B.GetLength(0); ++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);
       }
   }

}

</lang>

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

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.

Library: GMP(Go wrapper)

<lang go>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))

}</lang>

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

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

<lang go>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))

}</lang>

Output:

As first version except time is now 5m4.13427997s.

Fibmod method

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

<lang go>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))

}</lang>

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

Matrix exponentiation

<lang Haskell>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</lang>

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

<lang Haskell>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</lang>

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

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. <lang Haskell>-- 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</lang>

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

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

Implemented fib and fibMod.

<lang java> 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);
   }

} </lang>

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

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. <lang julia># 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))

</lang>

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

Perl

Translation of: Sidef

<lang perl>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;

}</lang>

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

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. <lang perl6>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**(D×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

}</lang>

Output:

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

Phix

Library: 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.) <lang Phix>include mpfr.e mpfr_set_default_prec(-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</lang>

Output:

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

matrix exponentiation (2^16)

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. <lang Phix>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 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</lang>

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]

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.

Sidef

Computing the n-th Fibonacci number, using matrix-exponentiation (this function is also built-in): <lang ruby>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]</lang>

First and last 20 digits of the n-th Fibonacci number: <lang ruby>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}"
   }

}</lang>

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: <lang ruby>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}"

}</lang>

Output:

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