Exponentiation order: Difference between revisions
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(5**3)**2 ───► 15625 |
(5**3)**2 ───► 15625 |
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5**(3**2) ───► 1953125 |
5**(3**2) ───► 1953125 |
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</pre> |
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=={{header|Ring}}== |
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<lang ring> |
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see "(5^3)^2 =>" + pow(pow(5,3),2) + nl |
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see "5^(3^2) =>" + pow(5,pow(3,2)) + nl |
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</lang> |
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Output: |
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<pre> |
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�(5^3)^2 =>15625 |
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5^(3^2) =>1953125 |
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</pre> |
</pre> |
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Revision as of 17:05, 11 August 2017
You are encouraged to solve this task according to the task description, using any language you may know.
This task will demonstrate the order of exponentiation (xy) when there are multiple exponents.
(Many programming languages, especially those with extended-precision integer arithmetic, usually support one of **
, ^
, ↑
or some such for exponentiation.)
- Task requirements
Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).
If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.
Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):
- 5**3**2
- (5**3)**2
- 5**(3**2)
If there are other methods (or formats) of multiple exponentiations, show them as well.
- See also
- MathWorld entry: exponentiation
- Related task
- Rosetta Code task: Arbitrary-precision integers (included)
ALGOL 68
Algol 68 provides various alternative symbols for the exponentiation operator generally, "**", "^" and "UP" can be used. <lang algol68>print( ( "5**3**2: ", 5**3**2, newline ) ); print( ( "(5**3)**2: ", (5**3)**2, newline ) ); print( ( "5**(3**2): ", 5**(3**2), newline ) )</lang>
- Output:
5**3**2: +15625 (5**3)**2: +15625 5**(3**2): +1953125
AWK
<lang AWK>
- syntax: GAWK -f EXPONENTIATION_ORDER.AWK
BEGIN {
printf("5^3^2 = %d\n",5^3^2) printf("(5^3)^2 = %d\n",(5^3)^2) printf("5^(3^2) = %d\n",5^(3^2)) exit(0)
} </lang>
output:
5^3^2 = 1953125 (5^3)^2 = 15625 5^(3^2) = 1953125
C
C does not have an exponentiation operator. The caret operator '^' performs xor bitwise operation in C. The function pow in the standard C Math library takes 2 arguments.
Expressions in C are evaluated by RPN. The RPNs of 5^3^2 and 5^(3^2) are the same and thus also their pow expressions. <lang C> /*Abhishek Ghosh, 20th March 2014, Rotterdam*/
include<stdio.h>
- include<math.h>
int main() { printf("\n5 ^ 3 ^ 2 = %.0f",pow(5,pow(3,2))); /*.0f suppresses decimal output*/ printf("\n(5 ^ 3) ^ 2 = %.0f",pow(pow(5,3),2)); printf("\n5 ^ (3 ^ 2) = %.0f",pow(5,pow(3,2)));
return 0; } </lang>
- Output:
5 ^ 3 ^ 2 = 1953125 (5 ^ 3) ^ 2 = 15625 5 ^ (3 ^ 2) = 1953125
Clojure
Clojure uses prefix notation and expt only takes 2 arguments for exponentiation, so "5**3**2" isn't represented.
<lang clojure>(use 'clojure.math.numeric-tower)
- (5**3)**2
(expt (expt 5 3) 2) ; => 15625
- 5**(3**2)
(expt 5 (expt 3 2)) ; => 1953125
- (5**3)**2 alternative
- use reduce
(reduce expt [5 3 2]) ; => 15625
- 5**(3**2) alternative
- evaluating right-to-left with reduce requires a small modification
(defn rreduce [f coll] (reduce #(f %2 %) (reverse coll))) (rreduce expt [5 3 2]) ; => 1953125</lang>
Common Lisp
Because Common Lisp uses prefix notation and expt
accepts only two arguments, it doesn't have an expression for 5**3**2
. Just showing expressions for the latter two.
<lang lisp>(expt (expt 5 3) 2)
(expt 5 (expt 3 2))</lang>
- Output:
15625 1953125
D
<lang d>void main() {
import std.stdio, std.math, std.algorithm;
writefln("5 ^^ 3 ^^ 2 = %7d", 5 ^^ 3 ^^ 2); writefln("(5 ^^ 3) ^^ 2 = %7d", (5 ^^ 3) ^^ 2); writefln("5 ^^ (3 ^^ 2) = %7d", 5 ^^ (3 ^^ 2)); writefln("[5, 3, 2].reduce!pow = %7d", [5, 3, 2].reduce!pow);
}</lang>
- Output:
5 ^^ 3 ^^ 2 = 1953125 (5 ^^ 3) ^^ 2 = 15625 5 ^^ (3 ^^ 2) = 1953125 [5, 3, 2].reduce!pow = 15625
EchoLisp
<lang scheme>
- the standard and secure way is to use the (expt a b) function
(expt 5 (expt 3 2)) ;; 5 ** ( 3 ** 2)
→ 1953125
(expt (expt 5 3) 2) ;; (5 ** 3) ** 2
→ 15625
- infix EchoLisp may use the ** operator, which right associates
(lib 'match) (load 'infix.glisp)
(5 ** 3 ** 2)
→ 1953125
((5 ** 3) ** 2)
→ 15625
(5 ** (3 ** 2))
→ 1953125
</lang>
Fortran
<lang Fortran>write(*, "(a, i0)") "5**3**2 = ", 5**3**2 write(*, "(a, i0)") "(5**3)**2 = ", (5**3)**2 write(*, "(a, i0)") "5**(3**2) = ", 5**(3**2)</lang>
- Output:
5**3**2 = 1953125 (5**3)**2 = 15625 5**(3**2) = 1953125
FreeBASIC
<lang freebasic>' FB 1.05.0
' The exponentation operator in FB is ^ rather than **. ' In the absence of parenthesis this operator is ' left-associative. So the first example ' will have the same value as the second example.
Print "5^3^2 =>"; 5^3^2 Print "(5^3)^2 =>"; (5^3)^2 Print "5^(3^2) =>"; 5^(3^2) Sleep</lang>
- Output:
5^3^2 => 15625 (5^3)^2 => 15625 5^(3^2) => 1953125
Haskell
Haskell has three infix exponentiation operators dealing with different domains:
λ> :i (^) (^) :: (Num a, Integral b) => a -> b -> a -- Defined in ‘GHC.Real’ infixr 8 ^ λ> :i (**) class Fractional a => Floating a where ... (**) :: a -> a -> a ... -- Defined in ‘GHC.Float’ infixr 8 ** λ> :i (^^) (^^) :: (Fractional a, Integral b) => a -> b -> a -- Defined in ‘GHC.Real’ infixr 8 ^^
All of them are right-associative.
λ> 5^3^2 1953125 λ> (5^3)^2 15625 λ> 5^(3^2) 1953125 λ> 5**3**2 == 5**(3**2) True
However natural chaining of (^^) operator is impossible:
5^^3^^2 = 5^^(3^^2)
but (3^^2) is not Integral any longer, so evaluation leads to the type error. Left-assiciative chain is Ok:
λ> (5^^3)^^2 15625.0 λ> ((5^^3)^^2)^^4 5.9604644775390624e16
Io
Io> 5**3**2 ==> 15625 Io> (5**3)**2 ==> 15625 Io> 5**(3**2) ==> 1953125 Io> 5 pow(3) pow(2) ==> 15625 Io> 5 **(3) **(2) ==> 15625 Io> Number getSlot("**") == Number getSlot("pow") ==> true Io>
Operators in Io are implemented as methods. Here the **
method is the same as the pow
method. Syntax sugar converts "normal" mathematical expressions to messages.
J
J uses the same evaluation order for exponentiation as it does for assignment. That is to say: the bottom up view is right-to-left and the top-down view is left-to-right.
<lang J> 5^3^2 1.95312e6
(5^3)^2
15625
5^(3^2)
1.95312e6</lang>
Java
Java has no exponentiation operator, but uses the static method java.lang.Math.pow(double a, double b). There are no associativity issues.
Kotlin
Kotlin does not have a dedicated exponentiation operator and we would normally use Java's Math.pow function instead. However, it's possible to define an infix function which would look like an operator and here we do so for integer base and exponent. For simplicity we disallow negative exponents altogether and consider 0 ** 0 == 1. Associativity would, of course, be the same as for a normal function call. <lang scala>// version 1.0.5-2
infix fun Int.ipow(exp: Int): Int = when {
exp < 0 -> throw IllegalArgumentException("negative exponents not allowed") exp == 0 -> 1 else -> { var ans = 1 var base = this var e = exp while(e != 0) { if (e and 1 == 1) ans *= base e = e shr 1 base *= base } ans }
}
fun main(args: Array<String>) {
println("5**3**2 = ${5 ipow 3 ipow 2}") println("(5**3)**2 = ${(5 ipow 3) ipow 2}") println("5**(3**2) = ${5 ipow (3 ipow 2)}")
}</lang>
- Output:
5**3**2 = 15625 (5**3)**2 = 15625 5**(3**2) = 1953125
Lua
<lang Lua>print("5^3^2 = " .. 5^3^2) print("(5^3)^2 = " .. (5^3)^2) print("5^(3^2) = " .. 5^(3^2))</lang>
- Output:
5^3^2 = 1953125 (5^3)^2 = 15625 5^(3^2) = 1953125
Lua also has math.pow(a, b), which is identical to pow(a, b) in C. Since function arguments are contained in brackets anyway, the associativity of nested uses of math.pow will be obvious.
Mathematica / Wolfram Language
<lang Mathematica>a = "5^3^2"; Print[a <> " = " <> ToString[ToExpression[a]]] b = "(5^3)^2"; Print[b <> " = " <> ToString[ToExpression[b]]] c = "5^(3^2)"; Print[c <> " = " <> ToString[ToExpression[c]]]</lang>
- Output:
5^3^2 = 1953125 (5^3)^2 = 15625 5^(3^2) = 1953125
OCaml
OCaml language has '**' as an exponentiation symbol for floating point integers <OCaml code>
- 5. ** 3. ** 2. ;;
- 5. **( 3. ** 2.) ;;
- (5. ** 3. ) **2. ;;
- Output:
- : float = 1953125. - : float = 1953125. - : float = 15625.
PARI/GP
Exponentiation is right-associative in GP. <lang parigp>f(s)=print(s" = "eval(s)); apply(f, ["5^3^2", "(5^3)^2", "5^(3^2)"]);</lang>
- Output:
5^3^2 = 1953125 (5^3)^2 = 15625 5^(3^2) = 1953125
Perl
<lang perl>say "$_ = " . eval($_) for qw/5**3**2 (5**3)**2 5**(3**2)/;</lang>
- Output:
5**3**2 = 1953125 (5**3)**2 = 15625 5**(3**2) = 1953125
Perl 6
<lang perl6>use MONKEY-SEE-NO-EVAL; sub demo($x) { say " $x\t───► ", EVAL $x }
demo '5**3**2'; # show ** is right associative demo '(5**3)**2'; demo '5**(3**2)';
demo '[**] 5,3,2'; # reduction form, show only final result demo '[\**] 5,3,2'; # triangle reduction, show growing results</lang>
- Output:
5**3**2 ───► 1953125 (5**3)**2 ───► 15625 5**(3**2) ───► 1953125 [**] 5,3,2 ───► 1953125 [\**] 5,3,2 ───► 2 9 1953125
Note that the reduction forms automatically go right-to-left because the base operator is right-associative. Most other operators are left-associative and would automatically reduce left-to-right instead.
Unicode postfix exponents are supported as well: <lang perl6>demo '(5³)²'; demo '5³²';</lang>
- Output:
(5³)² ───► 15625 5³² ───► 23283064365386962890625
The form without parentheses ends up raising to the 32nd power. Nor are you even allowed to parenthesize it the other way: 5(³²) would be a syntax error. Despite all that, for programs that do a lot of squaring or cubing, the postfix forms can enhance both readability and concision.
Phix
Phix has a power function rather than an infix power operator, hence there is no possible confusion. <lang Phix>?power(power(5,3),2) ?power(5,power(3,2))</lang>
- Output:
15625 1953125
PicoLisp
The PicoLisp '**' exponentiation function takes 2 arguments <lang PicoLisp>: (** (** 5 3) 2) -> 15625
- (** 5 (** 3 2))
-> 1953125</lang>
Python
<lang python>>>> 5**3**2 1953125 >>> (5**3)**2 15625 >>> 5**(3**2) 1953125 >>> # The following is not normally done >>> try: from functools import reduce # Py3K except: pass
>>> reduce(pow, (5, 3, 2)) 15625 >>> </lang>
Racket
<lang racket>#lang racket
- 5**3**2 depends on associativity of **
- Racket's (scheme's) prefix function
- calling syntax only allows for pairs of arguments for expt.
- So no can do for 5**3**2
- (5**3)**2
(displayln "prefix") (expt (expt 5 3) 2)
- (5**3)**2
(expt 5 (expt 3 2))
- There is also a less-used infix operation (for all functions, not just expt)... which I suppose
- might do with an airing. But fundamentally nothing changes.
(displayln "\"in\"fix") ((5 . expt . 3) . expt . 2) (5 . expt . (3 . expt . 2))
- everyone's doing a reduction, it seems
(displayln "reduction") (require (only-in srfi/1 reduce reduce-right)) (reduce expt 1 '(5 3 2)) (reduce-right expt 1 '(5 3 2))</lang>
- Output:
prefix 15625 1953125 "in"fix 15625 1953125 reduction 14134776518227074636666380005943348126619871175004951664972849610340958208 1953125
REXX
<lang rexx>/*REXX program demonstrates various ways of multiple exponentiations. */ /*┌────────────────────────────────────────────────────────────────────┐
│ The REXX language uses ** for exponention. │ │ Also, * * can be used. │ └────────────────────────────────────────────────────────────────────┘*/
say ' 5**3**2 ───► ' 5**3**2 say ' (5**3)**2 ───► ' (5**3)**2 say ' 5**(3**2) ───► ' 5**(3**2)
/*stick a fork in it, we're done.*/</lang>
output
5**3**2 ───► 15625 (5**3)**2 ───► 15625 5**(3**2) ───► 1953125
Ring
<lang ring> see "(5^3)^2 =>" + pow(pow(5,3),2) + nl see "5^(3^2) =>" + pow(5,pow(3,2)) + nl </lang> Output:
�(5^3)^2 =>15625 5^(3^2) =>1953125
Ruby
<lang ruby>ar = ["5**3**2", "(5**3)**2", "5**(3**2)", "[5,3,2].inject(:**)"] ar.each{|exp| puts "#{exp}:\t#{eval exp}"} </lang>
- Output:
5**3**2: 1953125 (5**3)**2: 15625 5**(3**2): 1953125 [5,3,2].inject(:**): 15625
Sidef
In Sidef, the whitespace between the operands and the operator controls the precedence of the operation. <lang ruby>var a = [
'5**3**2', '(5**3)**2', '5**(3**2)', '5 ** 3 ** 2', '5 ** 3**2', '5**3 ** 2', '[5,3,2]«**»',
]
a.each {|e|
"%-12s == %s\n".printf(e, eval(e))
}</lang>
- Output:
5**3**2 == 1953125 (5**3)**2 == 15625 5**(3**2) == 1953125 5 ** 3 ** 2 == 15625 5 ** 3**2 == 1953125 5**3 ** 2 == 15625 [5,3,2]«**» == 15625
Tcl
<lang tcl>foreach expression {5**3**2 (5**3)**2 5**(3**2)} {
puts "${expression}:\t[expr $expression]"
}</lang>
- Output:
5**3**2: 1953125 (5**3)**2: 15625 5**(3**2): 1953125
There's also a binary pow()
expression function that always converts its arguments to floating point numbers and then applies the exponentiation operation; it's now largely obsolete because of the **
operator, but is retained for backward compatibility with older programs.
VBScript
<lang vb> WScript.StdOut.WriteLine "5^3^2 => " & 5^3^2 WScript.StdOut.WriteLine "(5^3)^2 => " & (5^3)^2 WScript.StdOut.WriteLine "5^(3^2) => " & 5^(3^2) </lang>
- Output:
5^3^2 => 15625 (5^3)^2 => 15625 5^(3^2) => 1953125
zkl
zkl does not have an exponentiation operator but floats have a pow method. <lang zkl>println("5 ^ 3 ^ 2 = %,d".fmt((5.0).pow((3.0).pow(2)))); println("(5 ^ 3) ^ 2 = %,d".fmt((5.0).pow(3).pow(2))); println("5 ^ (3 ^ 2) = %,d".fmt((5.0).pow((3.0).pow(2))));</lang>
- Output:
5 ^ 3 ^ 2 = 1,953,125 (5 ^ 3) ^ 2 = 15,625 5 ^ (3 ^ 2) = 1,953,125
Go
<lang Go> package main
import "fmt" import "math" func main() {
var a,b,c float64 a=math.Pow(5,math.Pow(3,2)) b=math.Pow(math.Pow(5,3),2) c=math.Pow(5,math.Pow(3,2)) fmt.Printf("5^3^2 = %.0f\n", a) fmt.Printf("(5^3)^2 = %.0f\n", b) fmt.Printf("5^(3^2) = %.0f\n", c)
} </lang>
- Output:
5^3^2 = 1953125 (5^3)^2= 15625 5^(3^2)= 1953125