Knuth's power tree

Knuth's power tree
You are encouraged to solve this task according to the task description, using any language you may know.

(Knuth's power tree is used for computing   xⁿ   efficiently using Knuth's power tree.)

Task

Compute and show the list of Knuth's power tree integers necessary for the computation of:

•   xⁿ   for any real   x   and any non-negative integer   n.

Then, using those integers, calculate and show the exact (not approximate) value of (at least) the integer powers below:

•   2ⁿ     where   n   ranges from   0 ──► 17   (inclusive)
  
•   3¹⁹¹
•   1.1⁸¹

A  zero  power is often handled separately as a special case.

Optionally, support negative integers   (for the power).

Example

An example of a small power tree for some low integers:

                                                                    1
                                                                     \
                                                                      2
                          ___________________________________________/ \
                         /                                              \
                        3                                                4
                       / \____________________________________            \
                      /                                       \            \
                     5                                         6            8
                    / \____________                           / \            \
                   /               \                         /   \            \
                  7                 10                      9     12           16
                 /                 //\\                     │      │           /\
                /            _____//  \\________            │      │          /  \
              14            /     /    \        \           │      │         /    \
             /│ \         11    13      15       20        18     24        17    32
            / │  \         │    /\      /\        │        /\      │        /\     │
           /  │   \        │   /  \    /  \       │       /  \     │       /  \    │
         19  21    28     22 23   26  25   30    40     27   36    48     33 34   64
         │   /\    /│\     │  │   /\   │   /\    /│\     │   /\    /│\     │  │   /\
         │  /  \  / │ \    │  │  /  \  │  /  \  / │ \    │  /  \  / │ \    │  │  /  \
        38 35 42 29 31 56 44 46 39 52 50 45 60 41 43 80 54 37 72 49 51 96 66 68 65 128

Where, for the power   43,   following the tree "downwards" from   1:

•   (for   2)   compute square of   X,   store X²
•   (for   3)   compute   X * X²,   store X³
•   (for   5)   compute   X³ * X²,   store X⁵
•   (for 10)   compute square of   X⁵,   store X¹⁰
•   (for 20)   compute square of   X¹⁰,   store X²⁰
•   (for 40)   compute square of   X²⁰,   store X⁴⁰
•   (for 43)   compute   X⁴⁰ * X³   (result).

Note that for every even integer (in the power tree),   one just squares the previous value.

For an odd integer, multiply the previous value with an appropriate odd power of   X   (which was previously calculated).   For the last multiplication in the above example, it would be   (43-40),   or   3.

According to Dr. Knuth (see below),   computer tests have shown that this power tree gives optimum results for all of the   n   listed above in the graph.

For   n   ≤ 100,000,   the power tree method:

•   bests the factor method   88,803   times,
•   ties   11,191   times,
•   loses   6   times.

References

•   Donald E. Knuth's book:   The Art of Computer Programming, Vol. 2, Second Edition, Seminumerical Algorithms, section 4.6.3: Evaluation of Powers.
•   link   codegolf.stackexchange.com/questions/3177/knuths-power-tree     It shows a   Haskell,   Python,   and a   Ruby   computer program example   (but they are mostly   code golf).
•   link   comeoncodeon.wordpress.com/tag/knuth/     (See the section on Knuth's Power Tree.)     It shows a   C++   computer program example.
•   link to Rosetta Code   addition-chain exponentiation.

EchoLisp

Power tree

We build the tree using tree.lib, adding leaves until the target n is found. <lang scheme> (lib 'tree)

displays a chain hit

(define (power-hit target chain) (vector-push chain target) (printf "L(%d) = %d - chain:%a " target (1- (vector-length chain)) chain) (vector-pop chain))

build the power-tree

add 1 level of leaf nodes

display all chains which lead to target

(define (add-level node chain target nums (new)) (vector-push chain (node-datum node)) (cond [(node-leaf? node) ;; add leaves by summing this node to all nodes in chain ;; do not add leaf if number already known (for [(prev chain)] (set! new (+ prev (node-datum node))) (when (= new target) (power-hit target chain )) #:continue (vector-search* new nums) (node-add-leaf node new) (vector-insert* nums new) )] [else ;; not leaf node -> recurse (for [(son (node-sons node))] (add-level son chain target nums )) ]) (vector-pop chain))

add levels in tree until target found

return (number of nodes . upper-bound for L(target))

(define (power-tree target) (define nums (make-vector 1 1)) ;; known nums = 1 (define T (make-tree 1)) ;; root node has value 1 (printf "Looking for %d in %a." target T) (while #t #:break (vector-search* target nums) => (tree-count T) (add-level T init-chain: (make-vector 0) target nums) )) </lang>

Output:

(for ((n (in-range 2 18))) (power-tree n))
L(2) = 1 - chain:#( 1 2)
L(3) = 2 - chain:#( 1 2 3)
[ ... ]

(power-tree 17)
Looking for 17 in (🌴 1).
L(17) = 5 - chain:#( 1 2 4 8 16 17)

(power-tree 81)
Looking for 81 in (🌴 1).
L(81) = 8 - chain:#( 1 2 3 5 10 20 40 41 81)
L(81) = 8 - chain:#( 1 2 3 5 10 20 40 80 81)
L(81) = 8 - chain:#( 1 2 3 6 9 18 27 54 81)
L(81) = 8 - chain:#( 1 2 3 6 9 18 36 72 81)
L(81) = 8 - chain:#( 1 2 4 8 16 32 64 65 81)

(power-tree 191)
Looking for 191 in (🌴 1).
L(191) = 11 - chain:#( 1 2 3 5 7 14 19 38 57 95 190 191)
L(191) = 11 - chain:#( 1 2 3 5 7 14 21 42 47 94 188 191)
L(191) = 11 - chain:#( 1 2 3 5 7 14 21 42 63 126 189 191)
L(191) = 11 - chain:#( 1 2 3 5 7 14 28 31 59 118 177 191)
L(191) = 11 - chain:#( 1 2 3 5 7 14 28 31 62 93 186 191)
L(191) = 11 - chain:#( 1 2 3 5 10 11 22 44 88 176 181 191)

(power-tree 12509) ;; not optimal
Looking for 12509 in (🌴 1).
L(12509) = 18 - chain:#( 1 2 3 5 10 13 26 39 78 156 312 624 1248 2496 2509 3757 6253 12506 12509)
L(12509) = 18 - chain:#( 1 2 3 5 10 15 25 50 75 125 250 500 1000 2000 2003 4003 8006 12009 12509) 

(power-tree 222222)
Looking for 222222 in (🌴 1).
L(222222) = 22 - chain:#( 1 2 3 5 7 14 21 35 70 105 210 420 840 1680 1687 3367 6734 13468 26936 53872 57239 111111 222222) 

Exponentiation

<lang scheme>

j such as chain[i] = chain[i-1] + chain[j]

(define (adder chain i) (for ((j i)) #:break (= [chain i] (+ [chain(1- i)] [chain j])) => j ))

(define (power-exp x chain) (define lg (vector-length chain)) (define pow (make-vector lg x)) (for ((i (in-range 1 lg))) (vector-set! pow i ( * [pow [1- i]] [pow (adder chain i)]))) [pow (1- lg)]) </lang>

Output:

(power-exp 2 #( 1 2 4 8 16 17) )
   → 131072
(power-exp 1.1 #( 1 2 3 5 10 20 40 41 81) )
    → 2253.2402360440283

(lib 'bigint)
bigint.lib v1.4 ® EchoLisp
Lib: bigint.lib loaded.
(power-exp 3 #( 1 2 3 5 7 14 19 38 57 95 190 191) )
    → 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

F#

Integer Exponentiation

<lang fsharp> // Integer exponentiation using Knuth power tree. Nigel Galloway: October 29th., 2020 let kT α=let n=Array.zeroCreate<int*int list>((pown 2 (α+1))+1)

        let fN g=let rec fN p=[yield g+p; if p>0 then let g,_=n.[p] in yield! fN g] in (g+g)::(fN (fst n.[g]))|>List.rev
        let fG g=[for α,β in g do for g in β do let p,_=n.[g] in n.[g]<-(p,fN g|>List.filter(fun β->if box n.[β]=null then n.[β]<-(g,[]); true else false)); yield n.[g]]
        let rec kT n g=match g with 0->() |_->let n=fG n in kT n (g-1)
        let fE X g=let α=let rec fE g=[|yield g; if g>1 then yield! fE (fst n.[g])|] in fE g
                   let β=Array.zeroCreate<bigint>α.Length
                   let l=β.Length-1
                   β.[l]<-bigint (X+0)
                   for e in l-1.. -1..0 do β.[e]<-match α.[e]%2 with 0->β.[e+1]*β.[e+1] |_->let l=α.[e+1] in β.[e+1]*β.[α|>Array.findIndex(fun n->l+n=α.[e])]
                   β.[0]
        n.[1]<-(0,[2]); n.[2]<-(1,[]); kT [n.[1]] α; (fun n g->if g=0 then 1I else fE n g) 

let xp=kT 11 [0..17]|>List.iter(fun n->printfn "2**%d=%A\n" n (xp 2 n)) printfn "3**191=%A" (xp 3 191) </lang> {out}

2**0=1
2**1=2
2**2=4
2**3=8
2**4=16
2**5=32
2**6=64
2**7=128
2**8=256
2**9=512
2**10=1024
2**11=2048
2**12=4096
2**13=8192
2**14=16384
2**15=32768
2**16=65536
2**17=131072

3**191=13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Float Exponentiation

<lang fsharp> // Float exponentiation using Knuth power tree. Nigel Galloway: October 29th., 2020 let kTf α=let n=Array.zeroCreate<int*int list>((pown 2 (α+1))+1)

         let fN g=let rec fN p=[yield g+p; if p>0 then let g,_=n.[p] in yield! fN g] in (g+g)::(fN (fst n.[g]))|>List.rev
         let fG g=[for α,β in g do for g in β do let p,_=n.[g] in n.[g]<-(p,fN g|>List.filter(fun β->if box n.[β]=null then n.[β]<-(g,[]); true else false)); yield n.[g]]
         let rec kT n g=match g with 0->() |_->let n=fG n in kT n (g-1)
         let fE X g=let α=let rec fE g=[|yield g; if g>1 then yield! fE (fst n.[g])|] in fE g
                    let β=Array.zeroCreate<float>α.Length
                    let l=β.Length-1
                    β.[l]<-X
                    for e in l-1.. -1..0 do β.[e]<-match α.[e]%2 with 0->β.[e+1]*β.[e+1] |_->let l=α.[e+1] in β.[e+1]*β.[α|>Array.findIndex(fun n->l+n=α.[e])]
                    β.[0]
         n.[1]<-(0,[2]); n.[2]<-(1,[]); kT [n.[1]] α; (fun n g->if g=0 then 1.0 else fE n g) 

let xpf=kTf 11 printfn "1.1**81=%f" (xpf 1.1 81) </lang>

Output:

1.1**81=2253.240236

Go

Translation of: Kotlin

<lang go>package main

import (

   "fmt"
   "math/big"

)

var (

   p   = map[int]int{1: 0}
   lvl = [][]int{[]int{1}}

)

func path(n int) []int {

   if n == 0 {
       return []int{}
   }
   for {
       if _, ok := p[n]; ok {
           break
       }
       var q []int
       for _, x := range lvl[0] {
           for _, y := range path(x) {
               z := x + y
               if _, ok := p[z]; ok {
                   break
               }
               p[z] = x
               q = append(q, z)
           }
       }
       lvl[0] = q
   }
   r := path(p[n])
   r = append(r, n)
   return r

}

func treePow(x float64, n int) *big.Float {

   r := map[int]*big.Float{0: big.NewFloat(1), 1: big.NewFloat(x)}
   p := 0
   for _, i := range path(n) {
       temp := new(big.Float).SetPrec(320)
       temp.Mul(r[i-p], r[p])
       r[i] = temp
       p = i
   }
   return r[n]

}

func showPow(x float64, n int, isIntegral bool) {

   fmt.Printf("%d: %v\n", n, path(n))
   f := "%f"
   if isIntegral {
       f = "%.0f"
   }
   fmt.Printf(f, x)
   fmt.Printf(" ^ %d = ", n)
   fmt.Printf(f+"\n\n", treePow(x, n))

}

func main() {

   for n := 0; n <= 17; n++ {
       showPow(2, n, true)
   }
   showPow(1.1, 81, false)
   showPow(3, 191, true)

}</lang>

Output:

0: []
2 ^ 0 = 1

1: [1]
2 ^ 1 = 2

2: [1 2]
2 ^ 2 = 4

3: [1 2 3]
2 ^ 3 = 8

4: [1 2 4]
2 ^ 4 = 16

5: [1 2 4 5]
2 ^ 5 = 32

6: [1 2 4 6]
2 ^ 6 = 64

7: [1 2 4 6 7]
2 ^ 7 = 128

8: [1 2 4 8]
2 ^ 8 = 256

9: [1 2 4 8 9]
2 ^ 9 = 512

10: [1 2 4 8 10]
2 ^ 10 = 1024

11: [1 2 4 8 10 11]
2 ^ 11 = 2048

12: [1 2 4 8 12]
2 ^ 12 = 4096

13: [1 2 4 8 12 13]
2 ^ 13 = 8192

14: [1 2 4 8 12 14]
2 ^ 14 = 16384

15: [1 2 4 8 12 14 15]
2 ^ 15 = 32768

16: [1 2 4 8 16]
2 ^ 16 = 65536

17: [1 2 4 8 16 17]
2 ^ 17 = 131072

81: [1 2 4 8 16 32 64 80 81]
1.100000 ^ 81 = 2253.240236

191: [1 2 4 8 16 32 64 128 160 176 184 188 190 191]
3 ^ 191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Groovy

Translation of: Java

<lang groovy>class PowerTree {

   private static Map<Integer, Integer> p = new HashMap<>()
   private static List<List<Integer>> lvl = new ArrayList<>()

   static {
       p[1] = 0

       List<Integer> temp = new ArrayList<Integer>()
       temp.add 1
       lvl.add temp
   }

   private static List<Integer> path(int n) {
       if (n == 0) return new ArrayList<Integer>()
       while (!p.containsKey(n)) {
           List<Integer> q = new ArrayList<>()
           for (Integer x in lvl.get(0)) {
               for (Integer y in path(x)) {
                   if (p.containsKey(x + y)) break
                   p[x + y] = x
                   q.add x + y
               }
           }
           lvl[0].clear()
           lvl[0].addAll q
       }
       List<Integer> temp = path p[n]
       temp.add n
       temp
   }

   private static BigDecimal treePow(double x, int n) {
       Map<Integer, BigDecimal> r = new HashMap<>()
       r[0] = BigDecimal.ONE
       r[1] = BigDecimal.valueOf(x)

       int p = 0
       for (Integer i in path(n)) {
           r[i] = r[i - p] * r[p]
           p = i
       }
       r[n]
   }

   private static void showPos(double x, int n, boolean isIntegral) {
       printf("%d: %s\n", n, path(n))
       String f = isIntegral ? "%.0f" : "%f"
       printf(f, x)
       printf(" ^ %d = ", n)
       printf(f, treePow(x, n))
       println()
       println()
   }

   static void main(String[] args) {
       for (int n = 0; n <= 17; ++n) {
           showPos 2.0, n, true
       }
       showPos 1.1, 81, false
       showPos 3.0, 191, true
   }

}</lang>

Output:

0: []
2 ^ 0 = 1

1: [1]
2 ^ 1 = 2

2: [1, 2]
2 ^ 2 = 4

3: [1, 2, 3]
2 ^ 3 = 8

4: [1, 2, 4]
2 ^ 4 = 16

5: [1, 2, 4, 5]
2 ^ 5 = 32

6: [1, 2, 4, 6]
2 ^ 6 = 64

7: [1, 2, 4, 6, 7]
2 ^ 7 = 128

8: [1, 2, 4, 8]
2 ^ 8 = 256

9: [1, 2, 4, 8, 9]
2 ^ 9 = 512

10: [1, 2, 4, 8, 10]
2 ^ 10 = 1024

11: [1, 2, 4, 8, 10, 11]
2 ^ 11 = 2048

12: [1, 2, 4, 8, 12]
2 ^ 12 = 4096

13: [1, 2, 4, 8, 12, 13]
2 ^ 13 = 8192

14: [1, 2, 4, 8, 12, 14]
2 ^ 14 = 16384

15: [1, 2, 4, 8, 12, 14, 15]
2 ^ 15 = 32768

16: [1, 2, 4, 8, 16]
2 ^ 16 = 65536

17: [1, 2, 4, 8, 16, 17]
2 ^ 17 = 131072

81: [1, 2, 4, 8, 16, 32, 64, 80, 81]
1.100000 ^ 81 = 2253.240236

191: [1, 2, 4, 8, 16, 32, 64, 128, 160, 176, 184, 188, 190, 191]
3 ^ 191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Haskell

Works with: GHC version 8.8.1

Library: containers version 0.6.2.1

<lang haskell>{-# LANGUAGE ScopedTypeVariables #-}

module Rosetta.PowerTree

   ( Natural
   , powerTree
   , power
   )  where

import Data.Foldable (toList) import Data.Map.Strict (Map) import qualified Data.Map.Strict as Map import Data.Maybe (fromMaybe) import Data.List (foldl') import Data.Sequence (Seq (..), (|>)) import qualified Data.Sequence as Seq import Numeric.Natural (Natural)

type M = Map Natural S type S = Seq Natural

levels :: [M] levels = let s = Seq.singleton 1 in fst <$> iterate step (Map.singleton 1 s, s)

step :: (M, S) -> (M, S) step (m, xs) = foldl' f (m, Empty) xs

 where
   f :: (M, S) -> Natural -> (M, S)
   f (m', ys) n = foldl' g (m', ys) ns
     where
       ns :: S
       ns = m' Map.! n

       g :: (M, S) -> Natural -> (M, S)
       g (m, zs) k =
           let l = n + k
           in  case Map.lookup l m of
                   Nothing -> (Map.insert l (ns |>  l) m, zs |> l)
                   Just _  -> (m, zs)

powerTree :: Natural -> [Natural] powerTree n

   | n <= 0    = []
   | otherwise = go levels
 where
   go :: [M] -> [Natural]
   go []       = error "impossible branch"
   go (m : ms) = fromMaybe (go ms) $ toList <$> Map.lookup n m

power :: forall a. Num a => a -> Natural -> a power _ 0 = 1 power a n = go a 1 (Map.singleton 1 a) $ tail $ powerTree n

 where
   go :: a -> Natural -> Map Natural a -> [Natural] -> a
   go b _ _ []       = b
   go b k m (l : ls) =
       let b' = b * m Map.! (l - k)
           m' = Map.insert l b' m
       in  go b' l m' ls</lang>

Output:

Library: numbers version 3000.2.0.2

(The CReal type from package numbers is used to get the exact result for the last example.)

powerTree 0 = [], power 2 0 = 1
powerTree 1 = [1], power 2 1 = 2
powerTree 2 = [1,2], power 2 2 = 4
powerTree 3 = [1,2,3], power 2 3 = 8
powerTree 4 = [1,2,4], power 2 4 = 16
powerTree 5 = [1,2,3,5], power 2 5 = 32
powerTree 6 = [1,2,3,6], power 2 6 = 64
powerTree 7 = [1,2,3,5,7], power 2 7 = 128
powerTree 8 = [1,2,4,8], power 2 8 = 256
powerTree 9 = [1,2,3,6,9], power 2 9 = 512
powerTree 10 = [1,2,3,5,10], power 2 10 = 1024
powerTree 11 = [1,2,3,5,10,11], power 2 11 = 2048
powerTree 12 = [1,2,3,6,12], power 2 12 = 4096
powerTree 13 = [1,2,3,5,10,13], power 2 13 = 8192
powerTree 14 = [1,2,3,5,7,14], power 2 14 = 16384
powerTree 15 = [1,2,3,5,10,15], power 2 15 = 32768
powerTree 16 = [1,2,4,8,16], power 2 16 = 65536
powerTree 17 = [1,2,4,8,16,17], power 2 17 = 131072
powerTree 191 = [1,2,3,5,7,14,19,38,57,95,190,191], power 3 191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347
powerTree 81 = [1,2,3,5,10,20,40,41,81], power 1.1 81 = 2253.240236044012487937308538033349567966729852481170503814810577345406584190098644811

J

Generally, this task should be accomplished in J using x^n . Here we take an approach that's more comparable with the other examples on this page.

This task is a bit verbose...

We can represent the tree as a list of indices. Each entry in the list gives the value of n for the index a+n. (We can find a using subtraction.)

<lang J>knuth_power_tree=:3 :0

 L=: P=: %(1+y){._ 1
 findpath=: ]
 while. _ e.P do.
   for_n.(/: findpath&>)I.L=>./L-._ do.
     for_a. findpath n do.
        j=. n+a
        l=. 1+n{L
        if. j>y do. break. end.
        if. l>:j{ L do. continue. end.
        L=: l j} L
        P=: n j} P
     end.
     findpath=: [: |. {&P^:a:
   end.
 end.
 P

)

usepath=:4 :0

 path=. findpath y
 exp=. 1,({:path)#x
 for_ex.(,.~2 -~/\"1])2 ,\path  do.
   'ea eb ec'=. ex
   exp=.((ea{exp)*eb{exp) ec} exp
 end.
 {:exp

)</lang>

Task examples:

<lang J> knuth_power_tree 191 NB. generate sufficiently large tree 0 1 1 2 2 3 3 5 4 6 5 10 6 10 7 10 8 16 9 14 10 14 11 13 12 15 13 18 14 28 15 28 16 17 17 21 18 36 19 26 20 40 21 40 22 30 23 42 24 48 25 48 26 52 27 44 28 38 29 31 30 56 31 42 32 64 33 66 34 46 35 57 36 37 37 50 38 76 39 76 40 41 41 43 42 80 43 84 44 47 45 70 46 62 47 57 48 49 49 51 50 100 51 100 52 70 53 104 54 104 55 108 56 112 57 112 58 61 59 112 60 120 61 120 62 75 63 126 64 65 65 129 66 67 67 90 68 136 69 138 70 140 71 140 72 144 73 144 74 132 75 138 76 144 77 79 78 152 79 152 80 160 81 160 82 85 83 162 84 168 85 114 86 168 87 105 88 118 89 176 90 176 91 122 92 184 93 176 94 126 95 190

  findpath 0

0

  2 usepath 0

1

  findpath 1

1

  2 usepath 1

2

  findpath 2

1 2

  2 usepath 2

4

  findpath 3

1 2 3

  2 usepath 3

8

  findpath 4

1 2 4

  2 usepath 4

16

  findpath 5

1 2 3 5

  2 usepath 5

32

  findpath 6

1 2 3 6

  2 usepath 6

64

  findpath 7

1 2 3 5 7

  2 usepath 7

128

  findpath 8

1 2 4 8

  2 usepath 8

256

  findpath 9

1 2 3 6 9

  2 usepath 9

512

  findpath 10

1 2 3 5 10

  2 usepath 10

1024

  findpath 11

1 2 3 5 10 11

  2 usepath 11

2048

  findpath 12

1 2 3 6 12

  2 usepath 12

4096

  findpath 13

1 2 3 5 10 13

  2 usepath 13

8192

  findpath 14

1 2 3 5 7 14

  2 usepath 14

16384

  findpath 15

1 2 3 5 10 15

  2 usepath 15

32768

  findpath 16

1 2 4 8 16

  2 usepath 16

65536

  findpath 17

1 2 4 8 16 17

  2 usepath 17

131072

  findpath 191

1 2 3 5 7 14 19 38 57 95 190 191

  3x usepath 191

13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

  findpath 81

1 2 3 5 10 20 40 41 81

  (x:1.1) usepath 81

2253240236044012487937308538033349567966729852481170503814810577345406584190098644811r1000000000000000000000000000000000000000000000000000000000000000000000000000000000 </lang>

Note that an 'r' in a number indicates a rational number with the numerator to the left of the r and the denominator to the right of the r. We could instead use decimal notation by indicating how many characters of result we want to see, as well as how many characters to the right of the decimal point we want to see.

Thus, for example:

<lang J> 90j83 ": (x:1.1) usepath 81

 2253.24023604401248793730853803334956796672985248117050381481057734540658419009864481100</lang>

Java

Translation of: Kotlin

<lang Java>import java.math.BigDecimal; import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map;

public class PowerTree {

   private static Map<Integer, Integer> p = new HashMap<>();
   private static List<List<Integer>> lvl = new ArrayList<>();

   static {
       p.put(1, 0);

       ArrayList<Integer> temp = new ArrayList<>();
       temp.add(1);
       lvl.add(temp);
   }

   private static List<Integer> path(int n) {
       if (n == 0) return new ArrayList<>();
       while (!p.containsKey(n)) {
           List<Integer> q = new ArrayList<>();
           for (Integer x : lvl.get(0)) {
               for (Integer y : path(x)) {
                   if (p.containsKey(x + y)) break;
                   p.put(x + y, x);
                   q.add(x + y);
               }
           }
           lvl.get(0).clear();
           lvl.get(0).addAll(q);
       }
       List<Integer> temp = path(p.get(n));
       temp.add(n);
       return temp;
   }

   private static BigDecimal treePow(double x, int n) {
       Map<Integer, BigDecimal> r = new HashMap<>();
       r.put(0, BigDecimal.ONE);
       r.put(1, BigDecimal.valueOf(x));

       int p = 0;
       for (Integer i : path(n)) {
           r.put(i, r.get(i - p).multiply(r.get(p)));
           p = i;
       }
       return r.get(n);
   }

   private static void showPow(double x, int n, boolean isIntegral) {
       System.out.printf("%d: %s\n", n, path(n));
       String f = isIntegral ? "%.0f" : "%f";
       System.out.printf(f, x);
       System.out.printf(" ^ %d = ", n);
       System.out.printf(f, treePow(x, n));
       System.out.println("\n");
   }

   public static void main(String[] args) {
       for (int n = 0; n <= 17; ++n) {
           showPow(2.0, n, true);
       }
       showPow(1.1, 81, false);
       showPow(3.0, 191, true);
   }

}</lang>

Output:

0: []
2 ^ 0 = 1

1: [1]
2 ^ 1 = 2

2: [1, 2]
2 ^ 2 = 4

3: [1, 2, 3]
2 ^ 3 = 8

4: [1, 2, 4]
2 ^ 4 = 16

5: [1, 2, 4, 5]
2 ^ 5 = 32

6: [1, 2, 4, 6]
2 ^ 6 = 64

7: [1, 2, 4, 6, 7]
2 ^ 7 = 128

8: [1, 2, 4, 8]
2 ^ 8 = 256

9: [1, 2, 4, 8, 9]
2 ^ 9 = 512

10: [1, 2, 4, 8, 10]
2 ^ 10 = 1024

11: [1, 2, 4, 8, 10, 11]
2 ^ 11 = 2048

12: [1, 2, 4, 8, 12]
2 ^ 12 = 4096

13: [1, 2, 4, 8, 12, 13]
2 ^ 13 = 8192

14: [1, 2, 4, 8, 12, 14]
2 ^ 14 = 16384

15: [1, 2, 4, 8, 12, 14, 15]
2 ^ 15 = 32768

16: [1, 2, 4, 8, 16]
2 ^ 16 = 65536

17: [1, 2, 4, 8, 16, 17]
2 ^ 17 = 131072

81: [1, 2, 4, 8, 16, 32, 64, 80, 81]
1.100000 ^ 81 = 2253.240236

191: [1, 2, 4, 8, 16, 32, 64, 128, 160, 176, 184, 188, 190, 191]
3 ^ 191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Julia

Translation of: Java

Module: <lang julia>module KnuthPowerTree

const p = Dict(1 => 0) const lvl = 1

function path(n)

   global p, lvl
   iszero(n) && return Int[]
   while n ∉ keys(p)
       q = Int[]
       for x in lvl[1], y in path(x)
           if (x + y) ∉ keys(p)
               p[x + y] = x
               push!(q, x + y)
           end
       end
       lvl[1] = q
   end
   return push!(path(p[n]), n)

end

function pow(x::Number, n::Integer)

   r = Dict{typeof(n), typeof(x)}(0 => 1, 1 => x)
   p = 0
   for i in path(n)
       r[i] = r[i - p] * r[p]
       p = i
   end
   return r[n]

end

end # module KnuthPowerTree</lang>

Main: <lang julia>using .KnuthPowerTree: path, pow

for n in 0:17

   println("2 ^ $n:\n - path: ", join(path(n), ", "), "\n - result: ", pow(2, n))

end

for (x, n) in ((big(3), 191), (1.1, 81))

   println("$x ^ $n:\n - path: ", join(path(n), ", "), "\n - result: ", pow(x, n))

end</lang>

Output:

2 ^ 0:
 - path:
 - result: 1
2 ^ 1:
 - path: 1
 - result: 2
2 ^ 2:
 - path: 1, 2
 - result: 4
2 ^ 3:
 - path: 1, 2, 3
 - result: 8
2 ^ 4:
 - path: 1, 2, 4
 - result: 16
2 ^ 5:
 - path: 1, 2, 3, 5
 - result: 32
2 ^ 6:
 - path: 1, 2, 3, 6
 - result: 64
2 ^ 7:
 - path: 1, 2, 3, 5, 7
 - result: 128
2 ^ 8:
 - path: 1, 2, 4, 8
 - result: 256
2 ^ 9:
 - path: 1, 2, 3, 6, 9
 - result: 512
2 ^ 10:
 - path: 1, 2, 3, 5, 10
 - result: 1024
2 ^ 11:
 - path: 1, 2, 3, 5, 10, 11
 - result: 2048
2 ^ 12:
 - path: 1, 2, 3, 6, 12
 - result: 4096
2 ^ 13:
 - path: 1, 2, 3, 5, 10, 13
 - result: 8192
2 ^ 14:
 - path: 1, 2, 3, 5, 7, 14
 - result: 16384
2 ^ 15:
 - path: 1, 2, 3, 5, 10, 15
 - result: 32768
2 ^ 16:
 - path: 1, 2, 4, 8, 16
 - result: 65536
2 ^ 17:
 - path: 1, 2, 4, 8, 16, 17
 - result: 131072
3 ^ 191:
 - path: 1, 2, 3, 5, 7, 14, 19, 38, 57, 95, 190, 191
 - result: 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347
1.1 ^ 81:
 - path: 1, 2, 3, 5, 10, 20, 40, 41, 81
 - result: 2253.2402360440283

Kotlin

Translation of: Python

<lang scala>// version 1.1.3

import java.math.BigDecimal

var p = mutableMapOf(1 to 0) var lvl = mutableListOf(listOf(1))

fun path(n: Int): List<Int> {

   if (n == 0) return emptyList<Int>()
   while (n !in p) {
       val q = mutableListOf<Int>()
       for (x in lvl[0]) {
           for (y in path(x)) { 
               if ((x + y) in p) break
               p[x + y] = x
               q.add(x + y)
           } 
       }
       lvl[0] = q
   }
   return path(p[n]!!) + n

}

fun treePow(x: Double, n: Int): BigDecimal {

   val r = mutableMapOf(0 to BigDecimal.ONE, 1 to BigDecimal(x.toString()))
   var p = 0
   for (i in path(n)) {
       r[i] = r[i - p]!! * r[p]!!
       p = i
   }
   return r[n]!!

}

fun showPow(x: Double, n: Int, isIntegral: Boolean = true) {

   println("$n: ${path(n)}")
   val f = if (isIntegral) "%.0f" else "%f"
   println("${f.format(x)} ^ $n = ${f.format(treePow(x, n))}\n")

}

fun main(args: Array<String>) {

   for (n in 0..17) showPow(2.0, n)
   showPow(1.1, 81, false)
   showPow(3.0, 191)

}</lang>

Output:

0: []
2 ^ 0 = 1

1: [1]
2 ^ 1 = 2

2: [1, 2]
2 ^ 2 = 4

3: [1, 2, 3]
2 ^ 3 = 8

4: [1, 2, 4]
2 ^ 4 = 16

5: [1, 2, 4, 5]
2 ^ 5 = 32

6: [1, 2, 4, 6]
2 ^ 6 = 64

7: [1, 2, 4, 6, 7]
2 ^ 7 = 128

8: [1, 2, 4, 8]
2 ^ 8 = 256

9: [1, 2, 4, 8, 9]
2 ^ 9 = 512

10: [1, 2, 4, 8, 10]
2 ^ 10 = 1024

11: [1, 2, 4, 8, 10, 11]
2 ^ 11 = 2048

12: [1, 2, 4, 8, 12]
2 ^ 12 = 4096

13: [1, 2, 4, 8, 12, 13]
2 ^ 13 = 8192

14: [1, 2, 4, 8, 12, 14]
2 ^ 14 = 16384

15: [1, 2, 4, 8, 12, 14, 15]
2 ^ 15 = 32768

16: [1, 2, 4, 8, 16]
2 ^ 16 = 65536

17: [1, 2, 4, 8, 16, 17]
2 ^ 17 = 131072

81: [1, 2, 4, 8, 16, 32, 64, 80, 81]
1.100000 ^ 81 = 2253.240236

191: [1, 2, 4, 8, 16, 32, 64, 128, 160, 176, 184, 188, 190, 191]
3 ^ 191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Nim

Translation of: Kotlin

Library: bignum

This is a translation of the Kotlin/Python program. The algorithm is the same but there are some changes: replacement of global variables by a tree object, use of longer names, replacement of the "lvl" list of lists by a simple list, use of generics to handle the case of big integers...

<lang Nim>import tables import bignum

type

 Id = Natural    # Node identifier (= exponent value).

 Tree = object
   parents: Table[Id, Id]  # Mapping node id -> parent node id.
   lastLevel: seq[Id]      # List of node ids in current last level.

func initTree(): Tree =

 ## Return an initialized tree.
 const Root = Id(1)
 Tree(parents: {Root: Id(0)}.toTable, lastLevel: @[Root])

func path(tree: var Tree; id: Id): seq[Id] =

 ## Return the path to node with given id.

 if id == 0: return

 while id notin tree.parents:
   # Node "id" not yet present in the tree: build a new level.
   var newLevel: seq[Id]
   for x in tree.lastLevel:
     for y in tree.path(x):
       let newId = x + y
       if newId in tree.parents: break   # Node already created.
       # Create a new node.
       tree.parents[newId] = x
       newLevel.add newId
   tree.lastLevel = move(newLevel)

 # Node path is the concatenation of parent node path and node id.
 result = tree.path(tree.parents[id]) & id

func treePow[T: SomeNumber | Int](tree: var Tree; x: T; n: Natural): T =

 ## Compute x^n using the power tree.
 let one = when T is Int: newInt(1) else: T(1)
 var results = {0: one, 1: x}.toTable  # Intermediate and last results.
 var k = 0
 for i in tree.path(n):
   results[i] = results[i - k] * results[k]
   k = i
 return results[n]

proc showPow[T: SomeNumber | Int](tree: var Tree; x: T; n: Natural) =

 echo n, " → ", ($tree.path(n))[1..^1]
 let result = tree.treePow(x, n)
 echo x, "^", n, " = ", result

when isMainModule:

 var tree = initTree()
 for n in 0..17: tree.showPow(2, n)
 echo ""
 tree.showPow(1.1, 81)
 echo ""
 tree.showPow(newInt(3), 191)</lang>

Output:

0 → []
2^0 = 1
1 → [1]
2^1 = 2
2 → [1, 2]
2^2 = 4
3 → [1, 2, 3]
2^3 = 8
4 → [1, 2, 4]
2^4 = 16
5 → [1, 2, 4, 5]
2^5 = 32
6 → [1, 2, 4, 6]
2^6 = 64
7 → [1, 2, 4, 6, 7]
2^7 = 128
8 → [1, 2, 4, 8]
2^8 = 256
9 → [1, 2, 4, 8, 9]
2^9 = 512
10 → [1, 2, 4, 8, 10]
2^10 = 1024
11 → [1, 2, 4, 8, 10, 11]
2^11 = 2048
12 → [1, 2, 4, 8, 12]
2^12 = 4096
13 → [1, 2, 4, 8, 12, 13]
2^13 = 8192
14 → [1, 2, 4, 8, 12, 14]
2^14 = 16384
15 → [1, 2, 4, 8, 12, 14, 15]
2^15 = 32768
16 → [1, 2, 4, 8, 16]
2^16 = 65536
17 → [1, 2, 4, 8, 16, 17]
2^17 = 131072

81 → [1, 2, 4, 8, 16, 32, 64, 80, 81]
1.1^81 = 2253.240236044025

191 → [1, 2, 4, 8, 16, 32, 64, 128, 160, 176, 184, 188, 190, 191]
3^191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Perl

<lang perl>my @lvl = [1]; my %p = (1 => 0);

sub path {

   my ($n) = @_;
   return () if ($n == 0);
   until (exists $p{$n}) {
       my @q;
       foreach my $x (@{$lvl[0]}) {
           foreach my $y (path($x)) {
               my $z = $x + $y;
               last if exists($p{$z});
               $p{$z} = $x;
               push @q, $z;
           }
       }
       $lvl[0] = \@q;
   }
   (path($p{$n}), $n);

}

sub tree_pow {

   my ($x, $n) = @_;
   my %r = (0 => 1, 1 => $x);
   my $p = 0;
   foreach my $i (path($n)) {
       $r{$i} = $r{$i - $p} * $r{$p};
       $p = $i;
   }
   $r{$n};

}

sub show_pow {

   my ($x, $n) = @_;
   my $fmt = "%d: %s\n" . ("%g^%s = %f", "%s^%s = %s")[$x == int($x)] . "\n";
   printf($fmt, $n, "(" . join(" ", path($n)) . ")", $x, $n, tree_pow($x, $n));

}

show_pow(2, $_) for 0 .. 17; show_pow(1.1, 81); {

   use bigint (try => 'GMP');
   show_pow(3, 191);

}</lang>

Output:

0: ()
2^0 = 1
1: (1)
2^1 = 2
2: (1 2)
2^2 = 4
3: (1 2 3)
2^3 = 8
4: (1 2 4)
2^4 = 16
5: (1 2 4 5)
2^5 = 32
6: (1 2 4 6)
2^6 = 64
7: (1 2 4 6 7)
2^7 = 128
8: (1 2 4 8)
2^8 = 256
9: (1 2 4 8 9)
2^9 = 512
10: (1 2 4 8 10)
2^10 = 1024
11: (1 2 4 8 10 11)
2^11 = 2048
12: (1 2 4 8 12)
2^12 = 4096
13: (1 2 4 8 12 13)
2^13 = 8192
14: (1 2 4 8 12 14)
2^14 = 16384
15: (1 2 4 8 12 14 15)
2^15 = 32768
16: (1 2 4 8 16)
2^16 = 65536
17: (1 2 4 8 16 17)
2^17 = 131072
81: (1 2 4 8 16 32 64 80 81)
1.1^81 = 2253.240236
191: (1 2 4 8 16 32 64 128 160 176 184 188 190 191)
3^191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Phix

Translation of: Go

Library: Phix/mpfr

with javascript_semantics
constant p = new_dict({{1,0}})
sequence lvl = {1}
 
function path(integer n)
    if n=0 then return {} end if
    while getd_index(n,p)=NULL do
        sequence q = {}
        for i=1 to length(lvl) do
            integer x = lvl[i]
            sequence px = path(x)
            for j=1 to length(px) do
                integer y = x+px[j]
                if getd_index(y,p)!=NULL then exit end if
                setd(y,x,p)
                q &= y
            end for
        end for
        lvl = q
    end while
    return path(getd(n,p))&n
end function
 
include mpfr.e
mpfr_set_default_precision(500)
 
function treepow(object x, integer n, sequence pn = {})
-- x can be atom or string (but not mpfr)
-- (asides: sequence r uses out-by-1 indexing, ie r[1] is for 0.
--          sequence c is used to double-check we are not trying
--          to use something which has not yet been calculated.)
    if pn={} then pn=path(n) end if
    sequence r = {mpfr_init(1),mpfr_init(x)},
             c = {1,1}&repeat(0,max(0,n-1))
    for i=1 to max(0,n-1) do r &= mpfr_init() end for
    integer p = 0
    for i=1 to length(pn) do
        integer pi = pn[i]
        if c[pi-p+1]=0 then ?9/0 end if
        if c[p+1]=0 then ?9/0 end if
        mpfr_mul(r[pi+1],r[pi-p+1],r[p+1])
        c[pi+1] = 1
        p = pi
    end for
--  string res = shorten(trim_tail(mpfr_sprintf("%.83Rf",r[n+1]),".0"))
    string res = shorten(trim_tail(mpfr_get_fixed(r[n+1],83),".0"))
    r = mpfr_free(r)
    return res
end function
 
procedure showpow(object x, integer n)
    sequence pn = path(n)
    string xs = iff(string(x)?x:sprintf("%3g",x))
    printf(1,"%48V : %3s ^ %d = %s\n", {pn,xs,n,treepow(x,n,pn)})
end procedure
 
for n=0 to 17 do
    showpow(2,n)
end for
showpow("1.1",81)
showpow(3,191)

Output:

                                              {} :   2 ^ 0 = 1
                                             {1} :   2 ^ 1 = 2
                                           {1,2} :   2 ^ 2 = 4
                                         {1,2,3} :   2 ^ 3 = 8
                                         {1,2,4} :   2 ^ 4 = 16
                                       {1,2,4,5} :   2 ^ 5 = 32
                                       {1,2,4,6} :   2 ^ 6 = 64
                                     {1,2,4,6,7} :   2 ^ 7 = 128
                                       {1,2,4,8} :   2 ^ 8 = 256
                                     {1,2,4,8,9} :   2 ^ 9 = 512
                                    {1,2,4,8,10} :   2 ^ 10 = 1024
                                 {1,2,4,8,10,11} :   2 ^ 11 = 2048
                                    {1,2,4,8,12} :   2 ^ 12 = 4096
                                 {1,2,4,8,12,13} :   2 ^ 13 = 8192
                                 {1,2,4,8,12,14} :   2 ^ 14 = 16384
                              {1,2,4,8,12,14,15} :   2 ^ 15 = 32768
                                    {1,2,4,8,16} :   2 ^ 16 = 65536
                                 {1,2,4,8,16,17} :   2 ^ 17 = 131072
                        {1,2,4,8,16,32,64,80,81} : 1.1 ^ 81 = 2253.240236044012487937308538033349567966729852481170503814810577345406584190098644811
  {1,2,4,8,16,32,64,128,160,176,184,188,190,191} :   3 ^ 191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Python

<lang python>from __future__ import print_function

1. remember the tree generation state and expand on demand

def path(n, p = {1:0}, lvl=1): if not n: return [] while n not in p: q = [] for x,y in ((x, x+y) for x in lvl[0] for y in path(x) if not x+y in p): p[y] = x q.append(y) lvl[0] = q

return path(p[n]) + [n]

def tree_pow(x, n):

   r, p = {0:1, 1:x}, 0
   for i in path(n):
       r[i] = r[i-p] * r[p]
       p = i
   return r[n]

def show_pow(x, n):

   fmt = "%d: %s\n" + ["%g^%d = %f", "%d^%d = %d"][x==int(x)] + "\n"
   print(fmt % (n, repr(path(n)), x, n, tree_pow(x, n)))

for x in range(18): show_pow(2, x) show_pow(3, 191) show_pow(1.1, 81)</lang>

Output:

0: []
2^0 = 1

1: [1]
2^1 = 2

2: [1, 2]
2^2 = 4

<... snipped ...>

17: [1, 2, 4, 8, 16, 17]
2^17 = 131072

191: [1, 2, 3, 5, 7, 14, 19, 38, 57, 95, 190, 191]
3^191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

81: [1, 2, 3, 5, 10, 20, 40, 41, 81]
1.1^81 = 2253.240236

Racket

Translation of: Python

<lang Racket>#lang racket

(define pow-path-cache (make-hash '((0 . (0)) (1 . (0 1)))))

(define pow-path-level '(1))

(define (pow-path-extend!)

 (define next-level
   (for*/fold ([next-level '()])
              ([x (in-list pow-path-level)]
               [y (in-list (pow-path x))]
               [s (in-value (+ x y))]
               #:when (not (hash-has-key? pow-path-cache s)))
     (hash-set! pow-path-cache s (append (hash-ref pow-path-cache x) (list s)))
     (cons s next-level)))
 (set! pow-path-level (reverse next-level)))

(define (pow-path n)

 (let loop ()
   (unless (hash-has-key? pow-path-cache n)
     (pow-path-extend!)
     (loop)))
(hash-ref pow-path-cache n))

(define (pow-tree x n)

 (define pows (make-hash `((0 . 1) (1 . ,x))))
 (for/fold ([prev 0])
           ([i (in-list (pow-path n))])
   (hash-set! pows i (* (hash-ref pows (- i prev)) (hash-ref pows prev)))
   i)
 (hash-ref pows n))

(define (show-pow x n)

 (printf "~a: ~a\n" n (cdr (pow-path n)))
 (printf "~a^~a = ~a\n" x n (pow-tree x n)))

(for ([x (in-range 18)])

 (show-pow 2 x))

(show-pow 3 191) (show-pow 1.1 81)</lang>

Output:

0: ()
2^0 = 1
1: (1)
2^1 = 2
2: (1 2)
2^2 = 4
3: (1 2 3)
2^3 = 8
4: (1 2 4)
2^4 = 16
5: (1 2 3 5)
2^5 = 32
6: (1 2 3 6)
2^6 = 64
7: (1 2 3 5 7)
2^7 = 128
8: (1 2 4 8)
2^8 = 256
9: (1 2 3 6 9)
2^9 = 512
10: (1 2 3 5 10)
2^10 = 1024
11: (1 2 3 5 10 11)
2^11 = 2048
12: (1 2 3 6 12)
2^12 = 4096
13: (1 2 3 5 10 13)
2^13 = 8192
14: (1 2 3 5 7 14)
2^14 = 16384
15: (1 2 3 5 10 15)
2^15 = 32768
16: (1 2 4 8 16)
2^16 = 65536
17: (1 2 4 8 16 17)
2^17 = 131072
191: (1 2 3 5 7 14 19 38 57 95 190 191)
3^191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347
81: (1 2 3 5 10 20 40 41 81)
1.1^81 = 2253.2402360440283

Raku

(formerly Perl 6) Paths are random. It is possible replace .pick(*) with .reverse if you want paths as in Perl, or remove it for Python paths. <lang perl6>use v6;

sub power-path ($n ) {

   state @unused_nodes = (2,);
   state @power-tree = (False,0,1);
   
   until @power-tree[$n].defined {
       my $node = @unused_nodes.shift;

       for  $node X+ power-path($node).pick(*) {
           next if @power-tree[$_].defined;
           @unused_nodes.push($_);
           @power-tree[$_]= $node;
       }        
   }

   ( $n, { @power-tree[$_] } ...^ 0 ).reverse;

}

multi power ( $, 0 ) { 1 }; multi power ( $n, $exponent ) {

   state  %p;
   my     %r =  %p{$n}  // ( 0 => 1, 1 => $n ) ;  

   for power-path( $exponent ).rotor( 2 => -1 ) -> ( $p, $c ) {
       %r{ $c } = %r{ $p } * %r{ $c - $p }
   }

   %p{$n} := %r ;
   %r{ $exponent }

}

say 'Power paths: ', pairs map *.&power-path, ^18; say '2 ** key = value: ', pairs map { 2.&power($_) }, ^18;

say 'Path for 191: ', power-path 191; say '3 ** 191 = ', power 3, 191; say 'Path for 81: ', power-path 81; say '1.1 ** 81 = ', power 1.1, 81; </lang>

Output:

Power paths: (0 => () 1 => (1) 2 => (1 2) 3 => (1 2 3) 4 => (1 2 4) 5 => (1 2 3 5) 6 => (1 2 3 6) 7 => (1 2 3 6 7) 8 => (1 2 4 8) 9 => (1 2 3 6 9) 10 => (1 2 3 5 10) 11 => (1 2 3 6 9 11) 12 => (1 2 3 6 12) 13 => (1 2 3 6 12 13) 14 => (1 2 3 6 12 14) 15 => (1 2 3 6 9 15) 16 => (1 2 4 8 16) 17 => (1 2 4 8 16 17))
2 ** key = value: (0 => 1 1 => 2 2 => 4 3 => 8 4 => 16 5 => 32 6 => 64 7 => 128 8 => 256 9 => 512 10 => 1024 11 => 2048 12 => 4096 13 => 8192 14 => 16384 15 => 32768 16 => 65536 17 => 131072)
Path for 191: (1 2 3 6 9 18 27 54 108 162 189 191)
3 ** 191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347
Path for 81: (1 2 3 6 9 18 27 54 81)
1.1 ** 81 = 2253.24023604401

REXX

This REXX version supports results up to 1,000 decimal digits   (which can be expanded with the   numeric digits nnn   REXX statement.

Also, negative powers are supported. <lang rexx>/*REXX program produces & displays a power tree for P, and calculates & displays X^P.*/ numeric digits 1000 /*be able to handle some large numbers.*/ parse arg XP /*get sets: X, low power, high power.*/ if XP= then XP='2 -4 17 3 191 191 1.1 81' /*Not specified? Then use the default.*/

         /*────── X LP HP   X  LP  HP    X  LP  ◄── X, low power, high power ··· repeat*/
    do  until XP=
    parse var XP    x pL pH   XP;    x= x / 1   /*get X, lowP, highP; and normalize X. */
    if pH=  then pH= pL                       /*No highPower?  Then assume lowPower. */

      do e=pL  to pH;           p= abs(e) / 1   /*use a range of powers;   use  │E│    */
      $= powerTree(p);          w= length(pH)   /*construct the power tree, (pow list).*/
                                                /* [↑]  W≡length for an aligned display*/
         do i=1  for words($);  @.i= word($, i) /*build a fast Knuth's power tree array*/
         end   /*i*/

      if p==0  then do;  z= 1;  call show;  iterate;  end  /*handle case of zero power.*/
      !.= .;   z= x;     !.1= z;     prv= z     /*define/construct the first power of X*/

          do k=2  to words($);       n= @.k     /*obtain the power (number) to be used.*/
          prev= k - 1;     diff= n - @.prev     /*these are used for the odd powers.   */
          if n//2==0  then z= prv ** 2          /*Even power?   Then square the number.*/
                      else z= z * !.diff        /* Odd   "        "  mult. by pow diff.*/
          !.n= z                                /*remember for other multiplications.  */
          prv= z                                /*remember for squaring the numbers.   */
          end   /*k*/
      call show                                 /*display the expression and its value.*/
      end       /*e*/
    end         /*until XP ···*/

exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ powerTree: arg y 1 oy; $= /*Z is the result; $ is the power tree.*/

          if y=0 | y=1  then return y           /*handle special cases for zero & unity*/
          #.= 0;   @.= 0;    #.0= 1             /*define default & initial array values*/
                                                /* [↓]  add blank "flag" thingy──►list.*/
                  do  while \(y//2);  $= $ ' '  /*reduce "front" even power #s to odd #*/
                  if y\==oy  then $= y $        /*(only)  ignore the first power number*/
                  y= y % 2                      /*integer divide the power (it's even).*/
                  end   /*while*/

          if $\==  then $= y $                /*re─introduce the last power number.  */
          $= $ oy                               /*insert last power number 1st in list.*/
          if y>1  then do      while  @.y==0;            n= #.0;        m= 0
                         do    while  n\==0;             q= 0;          s= n
                           do  while  s\==0;             _= n + s
                           if @._==0  then do;  if q==0  then m_= _;
                                                #._= q;  @._= n;        q= _
                                           end
                           s= @.s
                           end   /*while s¬==0*/
                         if q\==0  then do;   #.m= q;   m= m_;   end
                         n= #.n
                         end     /*while n¬==0*/
                       #.m= 0
                       end       /*while @.y==0*/
          z= @.y
                           do  while z\==0;   $= z $;   z= @.z;  end /*build power list*/
          return space($)                                            /*del extra blanks*/

/*──────────────────────────────────────────────────────────────────────────────────────*/ show: if e<0 then z=format(1/z, , 40)/1; _=right(e, w) /*use reciprocal? */

     say left('power tree for '  _  " is: "  $,60)  '═══'  x"^"_  ' is: '  z;      return</lang>

output   when using the default inputs:

power tree for  -4  is:  1 2 4                               ═══ 2^-4  is:  0.0625
power tree for  -3  is:  1 2 3                               ═══ 2^-3  is:  0.125
power tree for  -2  is:  1 2                                 ═══ 2^-2  is:  0.25
power tree for  -1  is:  1                                   ═══ 2^-1  is:  0.5
power tree for   0  is:  0                                   ═══ 2^ 0  is:  1
power tree for   1  is:  1                                   ═══ 2^ 1  is:  2
power tree for   2  is:  1 2                                 ═══ 2^ 2  is:  4
power tree for   3  is:  1 2 3                               ═══ 2^ 3  is:  8
power tree for   4  is:  1 2 4                               ═══ 2^ 4  is:  16
power tree for   5  is:  1 2 3 5                             ═══ 2^ 5  is:  32
power tree for   6  is:  1 2 3 6                             ═══ 2^ 6  is:  64
power tree for   7  is:  1 2 3 5 7                           ═══ 2^ 7  is:  128
power tree for   8  is:  1 2 4 8                             ═══ 2^ 8  is:  256
power tree for   9  is:  1 2 3 6 9                           ═══ 2^ 9  is:  512
power tree for  10  is:  1 2 3 5 10                          ═══ 2^10  is:  1024
power tree for  11  is:  1 2 3 5 10 11                       ═══ 2^11  is:  2048
power tree for  12  is:  1 2 3 6 12                          ═══ 2^12  is:  4096
power tree for  13  is:  1 2 3 5 10 13                       ═══ 2^13  is:  8192
power tree for  14  is:  1 2 3 5 7 14                        ═══ 2^14  is:  16384
power tree for  15  is:  1 2 3 5 10 15                       ═══ 2^15  is:  32768
power tree for  16  is:  1 2 4 8 16                          ═══ 2^16  is:  65536
power tree for  17  is:  1 2 4 8 16 17                       ═══ 2^17  is:  131072
power tree for  191  is:  1 2 3 5 7 14 19 38 57 95 190 191   ═══ 3^191  is:  13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347
power tree for  81  is:  1 2 3 5 10 20 40 41 81              ═══ 1.1^81  is:  2253.240236044012487937308538033349567966729852481170503814810577345406584190098644811

Sidef

Translation of: zkl

<lang ruby>var lvl = 1 var p = Hash(1 => 0)

func path(n) is cached {

   n || return []
   while (n !~ p) {
       var q = []
       for x in lvl[0] {
           for y in path(x) {
               break if (x+y ~~ p)
               y = x+y
               p{y} = x
               q << y
           }
       }
       lvl[0] = q
   }
   path(p{n}) + [n]

}

func tree_pow(x, n) {

   var r = Hash(0 => 1, 1 => x)
   var p = 0
   for i in path(n) {
       r{i} = (r{i-p} * r{p})
       p = i
   }
   r{n}

}

func show_pow(x, n) {

   var fmt = ("%d: %s\n" + ["%g^%s = %f", "%s^%s = %s"][x.is_int] + "\n")
   print(fmt % (n, path(n), x, n, tree_pow(x, n)))

}

for x in ^18 { show_pow(2, x) } show_pow(1.1, 81) show_pow(3, 191)</lang>

Output:

0: []
2^0 = 1
1: [1]
2^1 = 2
2: [1, 2]
2^2 = 4
3: [1, 2, 3]
2^3 = 8
4: [1, 2, 4]
2^4 = 16
5: [1, 2, 4, 5]
2^5 = 32
6: [1, 2, 4, 6]
2^6 = 64
7: [1, 2, 4, 6, 7]
2^7 = 128
8: [1, 2, 4, 8]
2^8 = 256
9: [1, 2, 4, 8, 9]
2^9 = 512
10: [1, 2, 4, 8, 10]
2^10 = 1024
11: [1, 2, 4, 8, 10, 11]
2^11 = 2048
12: [1, 2, 4, 8, 12]
2^12 = 4096
13: [1, 2, 4, 8, 12, 13]
2^13 = 8192
14: [1, 2, 4, 8, 12, 14]
2^14 = 16384
15: [1, 2, 4, 8, 12, 14, 15]
2^15 = 32768
16: [1, 2, 4, 8, 16]
2^16 = 65536
17: [1, 2, 4, 8, 16, 17]
2^17 = 131072
81: [1, 2, 4, 8, 16, 32, 64, 80, 81]
1.1^81 = 2253.240236
191: [1, 2, 4, 8, 16, 32, 64, 128, 160, 176, 184, 188, 190, 191]
3^191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

Wren

Translation of: Kotlin

Library: Wren-big

Library: Wren-fmt

<lang ecmascript>import "/big" for BigRat import "/fmt" for Fmt

var p = { 1: 0 } var lvl = 1

var path // recursive path = Fn.new { |n|

   if (n == 0) return []
   while (!p.containsKey(n)) {
       var q = []
       for (x in lvl[0]) {
           System.write("") // guard against VM recursion bug
           for (y in path.call(x)) {
               if (p.containsKey(x + y)) break
               p[x + y] = x
               q.add(x + y)
           }
       }
       lvl[0] = q
   }
   System.write("") // guard against VM recursion bug
   var l = path.call(p[n])
   l.add(n)
   return l

}

var treePow = Fn.new { |x, n|

   var r = { 0: BigRat.one, 1: BigRat.fromDecimal(x) }
   var p = 0
   for (i in path.call(n)) {
       r[i] = r[i-p] * r[p]
       p = i
   }
   return r[n]

}

var showPow = Fn.new { |x, n|

   System.print("%(n): %(path.call(n))")
   Fmt.print("$s ^ $d = $s\n", x, n, treePow.call(x, n).toDecimal(6))

}

for (n in 0..17) showPow.call(2, n) showPow.call(1.1, 81) showPow.call(3, 191)</lang>

Output:

0: []
2 ^ 0 = 1

1: [1]
2 ^ 1 = 2

2: [1, 2]
2 ^ 2 = 4

3: [1, 2, 3]
2 ^ 3 = 8

4: [1, 2, 4]
2 ^ 4 = 16

5: [1, 2, 4, 5]
2 ^ 5 = 32

6: [1, 2, 4, 6]
2 ^ 6 = 64

7: [1, 2, 4, 6, 7]
2 ^ 7 = 128

8: [1, 2, 4, 8]
2 ^ 8 = 256

9: [1, 2, 4, 8, 9]
2 ^ 9 = 512

10: [1, 2, 4, 8, 10]
2 ^ 10 = 1024

11: [1, 2, 4, 8, 10, 11]
2 ^ 11 = 2048

12: [1, 2, 4, 8, 12]
2 ^ 12 = 4096

13: [1, 2, 4, 8, 12, 13]
2 ^ 13 = 8192

14: [1, 2, 4, 8, 12, 14]
2 ^ 14 = 16384

15: [1, 2, 4, 8, 12, 14, 15]
2 ^ 15 = 32768

16: [1, 2, 4, 8, 16]
2 ^ 16 = 65536

17: [1, 2, 4, 8, 16, 17]
2 ^ 17 = 131072

81: [1, 2, 4, 8, 16, 32, 64, 80, 81]
1.1 ^ 81 = 2253.240236

191: [1, 2, 4, 8, 16, 32, 64, 128, 160, 176, 184, 188, 190, 191]
3 ^ 191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347

zkl

Translation of: Python

<lang zkl># remember the tree generation state and expand on demand fcn path(n,p=Dictionary(1,0),lvl=List(List(1))){

  if(n==0) return(T);
  while(not p.holds(n)){
     q:=List();
     foreach x,y in (lvl[0],path(x,p,lvl)){
        if(p.holds(x+y)) break;  // not this y

y=x+y; p[y]=x; q.append(y);

     }
     lvl[0]=q
  }
  path(p[n],p,lvl) + n

}

fcn tree_pow(x,n,path){

  r,p:=Dictionary(0,1, 1,x), 0;
  foreach i in (path){ r[i]=r[i-p]*r[p]; p=i; }
  r[n]

}

fcn show_pow(x,n){

  fmt:="%d: %s\n" + T("%g^%d = %f", "%d^%d = %d")[x==Int(x)] + "\n";
  println(fmt.fmt(n,p:=path(n),x,n,tree_pow(x,n,p)))

}</lang> <lang zkl>foreach x in (18){ show_pow(2,x) } show_pow(1.1,81);

var [const] BN=Import("zklBigNum"); // GNU GMP big ints show_pow(BN(3),191);</lang>

Output:

0: L()
2^0 = 1

1: L(1)
2^1 = 2

2: L(1,2)
2^2 = 4

3: L(1,2,3)
2^3 = 8

4: L(1,2,4)
2^4 = 16

5: L(1,2,4,5)
2^5 = 32

6: L(1,2,4,6)
2^6 = 64

7: L(1,2,4,6,7)
2^7 = 128

8: L(1,2,4,8)
2^8 = 256

9: L(1,2,4,8,9)
2^9 = 512

10: L(1,2,4,8,10)
2^10 = 1024

11: L(1,2,4,8,10,11)
2^11 = 2048

12: L(1,2,4,8,12)
2^12 = 4096

13: L(1,2,4,8,12,13)
2^13 = 8192

14: L(1,2,4,8,12,14)
2^14 = 16384

15: L(1,2,4,8,12,14,15)
2^15 = 32768

16: L(1,2,4,8,16)
2^16 = 65536

17: L(1,2,4,8,16,17)
2^17 = 131072

81: L(1,2,4,8,16,32,64,80,81)
1.1^81 = 2253.240236

191: L(1,2,4,8,16,32,64,128,160,176,184,188,190,191)
3^191 = 13494588674281093803728157396523884917402502294030101914066705367021922008906273586058258347