Addition-chain exponentiation

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Addition-chain exponentiation is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
This page uses content from Wikipedia. The original article was at Addition-chain exponentiation. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)

In cases of special objects (such as with matrices) the operation of multiplication can be excessively expensive. In these cases the operation of multiplication should be avoided or reduced to a minimum.

In mathematics and computer science, optimal addition-chain exponentiation is a method of exponentiation by positive integer powers that requires a minimal number of multiplications. It works by creating a shortest addition chain that generates the desired exponent. Each exponentiation in the chain can be evaluated by multiplying two of the earlier exponentiation results. More generally, addition-chain exponentiation may also refer to exponentiation by non-minimal addition chains constructed by a variety of algorithms (since a shortest addition chain is very difficult to find).

The shortest addition-chain algorithm requires no more multiplications than binary exponentiation and usually less. The first example of where it does better is for a15, where the binary method needs six multiplies but a shortest addition chain requires only five:

a^{15} = a \times (a \times [a \times a^2]^2)^2  \! (binary, 6 multiplications)
a^{15} = a^3 \times ([a^3]^2)^2  \! (shortest addition chain, 5 multiplications)

On the other hand, the addition-chain method is much more complicated, since the determination of a shortest addition chain seems quite difficult: no efficient optimal methods are currently known for arbitrary exponents, and the related problem of finding a shortest addition chain for a given set of exponents has been proven NP-complete.

Table demonstrating how to do Exponentiation using Addition Chains
Number of
Multiplications
Actual
Exponentiation
Specific implementation of
Addition Chains to do Exponentiation
0 a1 a
1 a2 a × a
2 a3 a × a × a
2 a4 (a × a→b) × b
3 a5 (a × a→b) × b × a
3 a6 (a × a→b) × b × b
4 a7 (a × a→b) × b × b × a
3 a8 ((a × a→b) × b→d) × d
4 a9 (a × a × a→c) × c × c
4 a10 ((a × a→b) × b→d) × d × b
5 a11 ((a × a→b) × b→d) × d × b × a
4 a12 ((a × a→b) × b→d) × d × d
5 a13 ((a × a→b) × b→d) × d × d × a
5 a14 ((a × a→b) × b→d) × d × d × b
5 a15 ((a × a→b) × b × a→e) × e × e
4 a16 (((a × a→b) × b→d) × d→h) × h

The number of multiplications required follows this sequence: 0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 5, 5, 6, 5, 6, 6, 6, 5, 6, 6, 6, 6, 7, 6, 7, 5, 6, 6, 7, 6, 7, 7, 7, 6, 7, 7, 7, 7, 7, 7, 8, 6, 7, 7, 7, 7, 8, 7, 8, 7, 8, 8, 8, 7, 8, 8, 8, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 8, 8, 8, 9, 7, 8, 8, 8, 8, 8, 8, 9, 8, 9, 8, 9, 8, 9, 9, 9, 7, 8, 8, 8, 8...

This sequence can be found at: http://oeis.org/A003313

Task requirements:

Using the following values: A=\begin{bmatrix}
  \sqrt{\frac{1}{2}} & 0          &\sqrt{\frac{1}{2}} & 0          & 0 & 0 &\\
   0          &\sqrt{\frac{1}{2}} & 0          &\sqrt{\frac{1}{2}} & 0 & 0 &\\
   0          &\sqrt{\frac{1}{2}} & 0          &-\sqrt{\frac{1}{2}} & 0 & 0 &\\
  -\sqrt{\frac{1}{2}} & 0          &\sqrt{\frac{1}{2}} & 0          & 0 & 0 &\\
   0          & 0          & 0          & 0          & 0 & 1 &\\
   0          & 0          & 0          & 0          & 1 & 0 &\\
\end{bmatrix} m = 31415 and n = 27182

Repeat task Matrix-exponentiation operator, except use addition-chain exponentiation to better calculate:

Am, An and A ^ {n \times m}.

As an easier alternative to doing the matrix manipulation above, generate the addition-chains for 31415 and 27182 and use addition-chain exponentiation to calculate these two equations:

  • 1.0000220644541631415
  • 1.0000255005525127182

Also: Display a count of how many multiplications were done in each case.

Note: There are two ways to approach this task:

  • Brute force - try every permutation possible and pick one with the least number of multiplications. If the brute force is a simpler algorithm, then present it as a subtask under the subtitle "Brute force", eg ===Brute Force===.
  • Some clever algorithm - the wikipedia page has some hints, subtitle the code with the name of algorithm.

Note: Binary exponentiation does not usually produce the best solution. Provide only optimal solutions.

Kudos (κῦδος) for providing a routine that generate sequence A003313 in the output.

Contents

[edit] C

Using complex instead of matrix. Requires Achain.c. It takes a long while to compute the shortest addition chains, such that if you don't have the chain lengths precomputed and stored somewhere, you are probably better off with a binary chain (normally not shortest but very simple to calculate) whatever you intend to use the chains for.

#include <stdio.h>
 
#include "achain.c" /* not common practice */
 
/* don't have a C99 compiler atm */
typedef struct {double u, v;} cplx;
 
inline cplx c_mul(cplx a, cplx b)
{
cplx c;
c.u = a.u * b.u - a.v * b.v;
c.v = a.u * b.v + a.v * b.u;
return c;
}
 
cplx chain_expo(cplx x, int n)
{
int i, j, k, l, e[32];
cplx v[32];
 
l = seq(n, 0, e);
 
puts("Exponents:");
for (i = 0; i <= l; i++)
printf("%d%c", e[i], i == l ? '\n' : ' ');
 
v[0] = x; v[1] = c_mul(x, x);
for (i = 2; i <= l; i++) {
for (j = i - 1; j; j--) {
for (k = j; k >= 0; k--) {
if (e[k] + e[j] < e[i]) break;
if (e[k] + e[j] > e[i]) continue;
v[i] = c_mul(v[j], v[k]);
j = 1;
break;
}
}
}
printf("(%f + i%f)^%d = %f + i%f\n",
x.u, x.v, n, v[l].u, v[l].v);
 
return x;
}
 
int bin_len(int n)
{
int r, o;
for (r = o = -1; n; n >>= 1, r++)
if (n & 1) o++;
return r + o;
}
 
int main()
{
cplx r1 = {1.0000254989, 0.0000577896},
r2 = {1.0000220632, 0.0000500026};
int n1 = 27182, n2 = 31415, i;
 
init();
puts("Precompute chain lengths");
seq_len(n2);
 
chain_expo(r1, n1);
chain_expo(r2, n2);
puts("\nchain lengths: shortest binary");
printf("%14d %7d %7d\n", n1, seq_len(n1), bin_len(n1));
printf("%14d %7d %7d\n", n2, seq_len(n2), bin_len(n2));
for (i = 1; i < 100; i++)
printf("%14d %7d %7d\n", i, seq_len(i), bin_len(i));
return 0;
}

output

...
Exponents:
1 2 4 8 10 18 28 46 92 184 212 424 848 1696 3392 6784 13568 27136 27182
(1.000025 + i0.000058)^27182 = -0.000001 + i2.000001
Exponents:
1 2 4 8 16 17 33 49 98 196 392 784 1568 3136 6272 6289 12561 25122 31411 31415
(1.000022 + i0.000050)^31415 = -0.000001 + i2.000000

chain lengths: shortest binary
         27182      18      21
         31415      19      24
             1       0       0
             2       1       1
             3       2       2
             4       2       2
      ...
            89       9       9
            90       8       9
            91       9      10
            92       8       9
            93       9      10
      ...

[edit] Go

A non-optimal solution.

/*
Continued fraction addition chains, as described in "Efficient computation
of addition chains" by F. Bergeron, J. Berstel, and S. Brlek, published in
Journal de théorie des nombres de Bordeaux, 6 no. 1 (1994), p. 21-38,
accessed at http://www.numdam.org/item?id=JTNB_1994__6_1_21_0.
*/

package main
 
import (
"fmt"
"math"
)
 
// Representation of addition chains.
// Notes:
// 1. While an []int might represent addition chains in general, the
// techniques here work only with "star" chains, as described in the paper.
// Knowledge that the chains are star chains allows certain optimizations.
// 2. The paper descibes a linked list representation which encodes both
// addends of numbers in the chain. This allows additional optimizations, but
// for the purposes of the RC task, this simpler representation is adequate.
type starChain []int
 
// ⊗= operator. modifies receiver.
func (s1 *starChain) cMul(s2 starChain) {
p := *s1
i := len(p)
n := p[i-1]
p = append(p, s2[1:]...)
for ; i < len(p); i++ {
p[i] *= n
}
*s1 = p
}
 
// ⊕= operator. modifies receiver.
func (p *starChain) cAdd(j int) {
c := *p
*p = append(c, c[len(c)-1]+j)
}
 
// The γ function described in the paper returns a set of numbers in general,
// but certain γ functions return only singletons. The dichotomic strategy
// is one of these and gives good results so it is the one used for the
// RC task. Defining the package variable γ to be a singleton allows some
// simplifications in the code.
var γ singleton
 
type singleton func(int) int
 
func dichotomic(n int) int {
return n / (1 << uint((λ(n)+1)/2))
}
 
// integer log base 2
func λ(n int) (a int) {
for n != 1 {
a++
n >>= 1
}
return
}
 
// minChain as described in the paper.
func minChain(n int) starChain {
switch a := λ(n); {
case n == 1<<uint(a):
r := make(starChain, a+1)
for i := range r {
r[i] = 1 << uint(i)
}
return r
case n == 3:
return starChain{1, 2, 3}
}
return chain(n, γ(n))
}
 
// chain as described in the paper.
func chain(n1, n2 int) starChain {
q, r := n1/n2, n1%n2
if r == 0 {
c := minChain(n2)
c.cMul(minChain(q))
return c
}
c := chain(n2, r)
c.cMul(minChain(q))
c.cAdd(r)
return c
}
 
func main() {
m := 31415
n := 27182
show(m)
show(n)
show(m * n)
showEasier(m, 1.00002206445416)
showEasier(n, 1.00002550055251)
}
 
func show(e int) {
fmt.Println("exponent:", e)
s := math.Sqrt(.5)
a := matrixFromRows([][]float64{
{s, 0, s, 0, 0, 0},
{0, s, 0, s, 0, 0},
{0, s, 0, -s, 0, 0},
{-s, 0, s, 0, 0, 0},
{0, 0, 0, 0, 0, 1},
{0, 0, 0, 0, 1, 0},
})
γ = dichotomic
sc := minChain(e)
fmt.Println("addition chain:", sc)
a.scExp(sc).print("a^e")
fmt.Println("count of multiplies:", mCount)
fmt.Println()
}
 
var mCount int
 
func showEasier(e int, a float64) {
fmt.Println("exponent:", e)
γ = dichotomic
sc := minChain(e)
fmt.Printf("%.14f^%d: %.14f\n", a, sc[len(sc)-1], scExp64(a, sc))
fmt.Println("count of multiplies:", mCount)
fmt.Println()
}
 
func scExp64(a float64, sc starChain) float64 {
mCount = 0
p := make([]float64, len(sc))
p[0] = a
for i := 1; i < len(p); i++ {
d := sc[i] - sc[i-1]
j := i - 1
for sc[j] != d {
j--
}
p[i] = p[i-1] * p[j]
mCount++
}
return p[len(p)-1]
}
 
func (m *matrix) scExp(sc starChain) *matrix {
mCount = 0
p := make([]*matrix, len(sc))
p[0] = m.copy()
for i := 1; i < len(p); i++ {
d := sc[i] - sc[i-1]
j := i - 1
for sc[j] != d {
j--
}
p[i] = p[i-1].multiply(p[j])
mCount++
}
return p[len(p)-1]
}
 
func (m *matrix) copy() *matrix {
return &matrix{append([]float64{}, m.ele...), m.stride}
}
 
// code below copied from matrix multiplication task
type matrix struct {
ele []float64
stride int
}
 
func matrixFromRows(rows [][]float64) *matrix {
if len(rows) == 0 {
return &matrix{nil, 0}
}
m := &matrix{make([]float64, len(rows)*len(rows[0])), len(rows[0])}
for rx, row := range rows {
copy(m.ele[rx*m.stride:(rx+1)*m.stride], row)
}
return m
}
 
func (m *matrix) print(heading string) {
if heading > "" {
fmt.Print(heading, "\n")
}
for e := 0; e < len(m.ele); e += m.stride {
fmt.Printf("%6.3f ", m.ele[e:e+m.stride])
fmt.Println()
}
}
 
func (m1 *matrix) multiply(m2 *matrix) (m3 *matrix) {
m3 = &matrix{make([]float64, (len(m1.ele)/m1.stride)*m2.stride), m2.stride}
for m1c0, m3x := 0, 0; m1c0 < len(m1.ele); m1c0 += m1.stride {
for m2r0 := 0; m2r0 < m2.stride; m2r0++ {
for m1x, m2x := m1c0, m2r0; m2x < len(m2.ele); m2x += m2.stride {
m3.ele[m3x] += m1.ele[m1x] * m2.ele[m2x]
m1x++
}
m3x++
}
}
return m3
}

Output (manually wrapped at 80 columns.)

exponent: 31415
addition chain: [1 2 4 5 10 20 25 50 55 110 220 245 490 980 1960 3920 7840
15680 31360 31415]
a^e
[ 0.707  0.000  0.000 -0.707  0.000  0.000] 
[ 0.000  0.707  0.707  0.000  0.000  0.000] 
[ 0.707  0.000  0.000  0.707  0.000  0.000] 
[ 0.000  0.707 -0.707  0.000  0.000  0.000] 
[ 0.000  0.000  0.000  0.000  0.000  1.000] 
[ 0.000  0.000  0.000  0.000  1.000  0.000] 
count of multiplies: 19

exponent: 27182
addition chain: [1 2 4 8 10 18 28 46 92 184 212 424 848 1696 3392 6784 13568
27136 27182]
a^e
[-0.500 -0.500 -0.500  0.500  0.000  0.000] 
[ 0.500 -0.500 -0.500 -0.500  0.000  0.000] 
[-0.500 -0.500  0.500 -0.500  0.000  0.000] 
[ 0.500 -0.500  0.500  0.500  0.000  0.000] 
[ 0.000  0.000  0.000  0.000  1.000  0.000] 
[ 0.000  0.000  0.000  0.000  0.000  1.000] 
count of multiplies: 18

exponent: 853922530
addition chain: [1 2 4 5 7 12 24 48 96 103 206 309 412 721 1133 1854 3708 4841
9682 19364 21218 26059 52118 104236 208472 416944 833888 1667776 3335552 6671104
13342208 26684416 53368832 106737664 213475328 426950656 853901312 853922530]
a^e
[-0.500  0.500 -0.500  0.500  0.000  0.000] 
[-0.500 -0.500 -0.500 -0.500  0.000  0.000] 
[-0.500 -0.500  0.500  0.500  0.000  0.000] 
[ 0.500 -0.500 -0.500  0.500  0.000  0.000] 
[ 0.000  0.000  0.000  0.000  1.000  0.000] 
[ 0.000  0.000  0.000  0.000  0.000  1.000] 
count of multiplies: 37

exponent: 31415
1.00002206445416^31415: 1.99999999989447
count of multiplies: 19

exponent: 27182
1.00002550055251^27182: 1.99999999997876
count of multiplies: 18

[edit] MATLAB / Octave

I assume that Matlab and Octave have about such optimization already included. On a single core of an "Pentium(R) Dual-Core CPU E5200 @ 2.50GHz", the computation of 100000 repetitions of the matrix exponentiation A27182 and A31415 take about 2 and 2.2 seconds, resp.

x = sqrt(.5);
A = [x,0,x,0,0,0;0,x,0,x,0,0; 0,x,0,-x,0,0;-x,0,x,0,0,0;0,0,0,0,0,1; 0,0,0,0,1,0];A
t = cputime(); for k=1:100000,x1=A^27182;end; cputime()-t,x1,
t = cputime(); for k=1:100000,x2=A^31415;end; cputime()-t,x2,

Output:

Octave3.4.3>  t = cputime(); for k=1:100000,x1=A^27182;end; cputime()-t,x1,
ans =  1.9900
x1 =

  -0.50000  -0.50000  -0.50000   0.50000   0.00000   0.00000
   0.50000  -0.50000  -0.50000  -0.50000   0.00000   0.00000
  -0.50000  -0.50000   0.50000  -0.50000   0.00000   0.00000
   0.50000  -0.50000   0.50000   0.50000   0.00000   0.00000
   0.00000   0.00000   0.00000   0.00000   1.00000   0.00000
   0.00000   0.00000   0.00000   0.00000   0.00000   1.00000

Octave3.4.3> t = cputime(); for k=1:100000,x2=A^31415;end; cputime()-t,x2,
ans =  2.2000
x2 =

   0.70711   0.00000   0.00000  -0.70711   0.00000   0.00000
   0.00000   0.70711   0.70711   0.00000   0.00000   0.00000
   0.70711   0.00000   0.00000   0.70711   0.00000   0.00000
   0.00000   0.70711  -0.70711   0.00000   0.00000   0.00000
   0.00000   0.00000   0.00000   0.00000   0.00000   1.00000
   0.00000   0.00000   0.00000   0.00000   1.00000   0.00000

[edit] Racket

The addition chains correspond to binary exponentiation.

 
#lang racket
(define (chain n)
 ; computes a simple addition chain for n
(cond [(= n 1) '()]
[(even? n) (define n/2 (/ n 2))
(cons (list n n/2 n/2) (chain n/2))]
[(odd? n) (define n-1 (- n 1))
(cons (list n n-1 1) (chain (- n 1)))]))
 
(define mult
(let ([n 0])
(λ xs
(cond [(equal? xs (list 'count)) n]
[(equal? xs (list 'reset)) (set! n 0)]
[else (set! n (+ n 1))
(apply * xs)]))))
 
(define (expt/chain x n chain)
 ; computes x^n using the addition chain
(define ht (make-hash))
(hash-set! ht 1 x)
(define (expt1 n)
(or (hash-ref ht n #f)
(let ()
(define x^n
(match (assoc n chain)
[(list _ s t) (mult (expt1 s) (expt1 t))]))
(hash-set! ht n x^n)
x^n)))
(expt1 n))
 
(define (test x n)
(displayln (~a "Chain for " n "\n" (chain n)))
(mult 'reset)
(displayln (~a x " ^ " n " = " (expt/chain x n (chain n))))
(displayln (~a "Multiplications: " (mult 'count)))
(newline))
 
(test 1.00002206445416 31415)
(test 1.00002550055251 27182)
 

Output:

 
Chain for 31415
((31415 31414 1) (31414 15707 15707) (15707 15706 1) (15706 7853 7853) (7853 7852 1) (7852 3926 3926) (3926 1963 1963) (1963 1962 1) (1962 981 981) (981 980 1) (980 490 490) (490 245 245) (245 244 1) (244 122 122) (122 61 61) (61 60 1) (60 30 30) (30 15 15) (15 14 1) (14 7 7) (7 6 1) (6 3 3) (3 2 1) (2 1 1))
1.00002206445416 ^ 31415 = 1.9999999998913485
Multiplications: 24
 
Chain for 27182
((27182 13591 13591) (13591 13590 1) (13590 6795 6795) (6795 6794 1) (6794 3397 3397) (3397 3396 1) (3396 1698 1698) (1698 849 849) (849 848 1) (848 424 424) (424 212 212) (212 106 106) (106 53 53) (53 52 1) (52 26 26) (26 13 13) (13 12 1) (12 6 6) (6 3 3) (3 2 1) (2 1 1))
1.00002550055251 ^ 27182 = 1.9999999999774538
Multiplications: 21
 

[edit] Tcl

Using code at Matrix multiplication#Tcl and Matrix Transpose#Tcl (not shown here).

Translation of: Go
# Continued fraction addition chains, as described in "Efficient computation
# of addition chains" by F. Bergeron, J. Berstel, and S. Brlek, published in
# Journal de théorie des nombres de Bordeaux, 6 no. 1 (1994), p. 21-38,
# accessed at http://www.numdam.org/item?id=JTNB_1994__6_1_21_0.
#
# Uses the dichotomic strategy, which produces good results with simpler
# coding than for a pluggable non-deterministic strategy.
 
package require Tcl 8.5
namespace path {::tcl::mathop ::tcl::mathfunc}
 
proc minchain {n} {
if {!($n & ($n-1))} {
for {set i 1} {$i <= $n} {incr i $i} {lappend c $i}
return $c
} elseif {$n == 3} {
return {1 2 3}
}
return [chain $n [expr {$n >> int(ceil(floor(log($n)/log(2))/2))}]]
}
proc chain {n1 n2} {
set q [expr {$n1 / $n2}]
set r [expr {$n1 % $n2}]
if {$r == 0} {
return [chain.* [minchain $n2] [minchain $q]]
} else {
return [chain.+ [chain.* [chain $n2 $r] [minchain $q]] $r]
}
}
proc chain.+ {ns k} {
return [lappend ns [expr {[lindex $ns end] + $k}]]
}
proc chain.* {ns ms} {
set n_k [lindex $ns end]
foreach m_i $ms {
if {$m_i==1} continue
lappend ns [expr {$n_k * $m_i}]
}
return $ns
}
 
# Generate a lambda term to do exponentiation with a given multiplier command.
# Works by extracting information from the addition chain; the lambda term
# generated is minimal
proc makeExponentiationLambda {n mulfunc} {
set chain [minchain $n]
set cmd {set a0}
set idxes 0
foreach c0 [lrange $chain 0 end-1] c1 [lrange $chain 1 end] {
lappend idxes [lsearch $chain [expr {$c1 - $c0}]]
}
for {set i 1} {$i<[llength $chain]} {incr i} {
set cmd "$mulfunc \[$cmd\] \$a[lindex $idxes $i]"
if {$i in $idxes} {
set cmd "set a$i \[$cmd\]"
}
}
list a0 $cmd
}
 
# Demonstrating application of problem to matrix exponentiation
proc count_mult {a b} {incr ::countMult;matrix_multiply $a $b}
set m 31415
set n 27182
set mn [expr {$m*$n}]
set pow_m [makeExponentiationLambda $m count_mult]
set pow_n [makeExponentiationLambda $n count_mult]
set pow_mn [makeExponentiationLambda $mn count_mult]
 
set rh [expr {sqrt(0.5)}]
set mrh [expr {-$rh}]
set A [subst {
{$rh 0 $rh 0 0 0}
{0 $rh 0 $rh 0 0}
{0 $rh 0 $mrh 0 0}
{$mrh 0 $rh 0 0 0}
{0 0 0 0 0 1}
{0 0 0 0 1 0}
}]
puts "A**$m"; set countMult 0
print_matrix [apply $pow_m $A] %6.3f
puts "$countMult matrix multiplies"
puts "A**$n"; set countMult 0
print_matrix [apply $pow_n $A] %6.3f
puts "$countMult matrix multiplies"
puts "A**$mn"; set countMult 0
print_matrix [apply $pow_mn $A] %6.3f
puts "$countMult matrix multiplies"
Output:
A**31415
 0.707  0.000  0.000 -0.707  0.000  0.000 
 0.000  0.707  0.707  0.000  0.000  0.000 
 0.707  0.000  0.000  0.707  0.000  0.000 
 0.000  0.707 -0.707  0.000  0.000  0.000 
 0.000  0.000  0.000  0.000  0.000  1.000 
 0.000  0.000  0.000  0.000  1.000  0.000 
19 matrix multiplies
A**27182
-0.500 -0.500 -0.500  0.500  0.000  0.000 
 0.500 -0.500 -0.500 -0.500  0.000  0.000 
-0.500 -0.500  0.500 -0.500  0.000  0.000 
 0.500 -0.500  0.500  0.500  0.000  0.000 
 0.000  0.000  0.000  0.000  1.000  0.000 
 0.000  0.000  0.000  0.000  0.000  1.000 
18 matrix multiplies
A**853922530
-0.500  0.500 -0.500  0.500  0.000  0.000 
-0.500 -0.500 -0.500 -0.500  0.000  0.000 
-0.500 -0.500  0.500  0.500  0.000  0.000 
 0.500 -0.500 -0.500  0.500  0.000  0.000 
 0.000  0.000  0.000  0.000  1.000  0.000 
 0.000  0.000  0.000  0.000  0.000  1.000 
37 matrix multiplies
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