Distinct power numbers: Difference between revisions
Content added Content deleted
Thundergnat (talk | contribs) m (syntax highlighting fixup automation) |
(Added Quackery.) |
||
| Line 656: | Line 656: | ||
{{out}} |
{{out}} |
||
<pre> [1] 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125</pre> |
<pre> [1] 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125</pre> |
||
=={{header|Quackery}}== |
|||
<syntaxhighlight lang="Quackery"> [ [] swap |
|||
behead swap |
|||
witheach |
|||
[ 2dup = iff |
|||
drop done |
|||
dip join ] |
|||
join ] is unique ( [ --> [ ) |
|||
[] |
|||
4 times |
|||
[ i 2 + |
|||
4 times |
|||
[ dup i 2 + ** |
|||
rot join swap ] |
|||
drop ] |
|||
sort unique echo</syntaxhighlight> |
|||
{{out}} |
|||
<pre>[ 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125 ]</pre> |
|||
=={{header|Raku}}== |
=={{header|Raku}}== |
||
Revision as of 01:07, 6 December 2022
Distinct power numbers 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.
- Task
Compute all combinations of where a and b are integers between 2 and 5 inclusive.
Place them in numerical order, with any repeats removed.
You should get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
11l
print(sorted(Array(Set(cart_product(2..5, 2..5).map((a, b) -> a ^ b)))))- Output:
[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]
Action!
INCLUDE "D2:SORT.ACT" ;from the Action! Tool Kit
INT FUNC Power(INT a,b)
INT res,i
res=1
FOR i=1 TO b
DO
res==*a
OD
RETURN (res)
BYTE FUNC Contains(INT ARRAY a INT count,x)
INT i
FOR i=0 TO count-1
DO
IF a(i)=x THEN
RETURN (1)
FI
OD
RETURN (0)
PROC Main()
INT ARRAY a(100)
INT i,j,x,count
Put(125) PutE() ;clear the screen
count=0
FOR i=2 TO 5
DO
FOR j=2 TO 5
DO
x=Power(i,j)
IF Contains(a,count,x)=0 THEN
a(count)=x
count==+1
FI
OD
OD
SortI(a,count,0)
FOR i=0 TO count-1
DO
PrintI(a(i)) Put(32)
OD
RETURN- Output:
Screenshot from Atari 8-bit computer
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
Ada
with Ada.Text_Io;
with Ada.Containers.Doubly_Linked_Lists;
procedure Power_Numbers is
package Number_Lists is new Ada.Containers.Doubly_Linked_Lists (Integer);
package Number_Sorting is new Number_Lists.Generic_Sorting;
use Number_Lists, Ada.Text_Io;
List : Number_Lists.List;
begin
for A in 2 .. 5 loop
for B in 2 .. 5 loop
declare
R : constant Integer := A**B;
begin
if not List.Contains (R) then
List.Append (R);
end if;
end;
end loop;
end loop;
Number_Sorting.Sort (List);
for E of List loop
Put (Integer'Image (E));
Put (" ");
end loop;
New_Line;
end Power_Numbers;
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
ALGOL 68
BEGIN # show in order, distinct values of a^b where 2 <= a <= b <= 5 #
INT max number = 5;
INT min number = 2;
# construct a table of a ^ b #
INT length = ( max number + 1 ) - min number;
[ 1 : length * length ]INT a to b;
INT pos := 0;
FOR i FROM min number TO max number DO
a to b[ pos +:= 1 ] := i * i;
FOR j FROM min number + 1 TO max number DO
INT prev = pos;
a to b[ pos +:= 1 ] := a to b[ prev ] * i
OD
OD;
# sort the table #
# it is small and nearly sorted so a bubble sort should suffice #
FOR u FROM UPB a to b - 1 BY -1 TO LWB a to b
WHILE BOOL sorted := TRUE;
FOR p FROM LWB a to b BY 1 TO u DO
IF a to b[ p ] > a to b[ p + 1 ] THEN
INT t = a to b[ p ];
a to b[ p ] := a to b[ p + 1 ];
a to b[ p + 1 ] := t;
sorted := FALSE
FI
OD;
NOT sorted
DO SKIP OD;
# print the table, excluding duplicates #
INT last := -1;
FOR i TO UPB a to b DO
INT next = a to b[ i ];
IF next /= last THEN print( ( " ", whole( next, 0 ) ) ) FI;
last := next
OD;
print( ( newline ) )
END- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
APL
(⊂∘⍋⌷⊣)∪,∘.*⍨1+⍳4
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
AppleScript
Idiomatic
Uses an extravagantly long list, but gets the job done quickly and easily.
on task()
script o
property output : {}
end script
repeat (5 ^ 5) times
set end of o's output to missing value
end repeat
repeat with a from 2 to 5
repeat with b from 2 to 5
tell (a ^ b as integer) to set item it of o's output to it
tell (b ^ a as integer) to set item it of o's output to it
end repeat
end repeat
return o's output's integers
end task
task()
- Output:
{4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125}
Functional
Composing a solution from generic primitives, for speed of drafting, and ease of refactoring:
use framework "Foundation"
use scripting additions
------------------ DISTINCT POWER VALUES -----------------
-- distinctPowers :: [Int] -> [Int]
on distinctPowers(xs)
script powers
on |λ|(a, x)
script integerPower
on |λ|(b, y)
b's addObject:((x ^ y) as integer)
b
end |λ|
end script
foldl(integerPower, a, xs)
end |λ|
end script
sort(foldl(powers, ¬
current application's NSMutableSet's alloc's init(), xs)'s ¬
allObjects())
end distinctPowers
--------------------------- TEST -------------------------
on run
distinctPowers(enumFromTo(2, 5))
end run
------------------------- GENERIC ------------------------
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m ≤ n then
set lst to {}
repeat with i from m to n
set end of lst to i
end repeat
lst
else
{}
end if
end enumFromTo
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- sort :: Ord a => [a] -> [a]
on sort(xs)
((current application's NSArray's arrayWithArray:xs)'s ¬
sortedArrayUsingSelector:"compare:") as list
end sort
- Output:
{4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125}
AppleScriptObjC
Throwing together a solution using the most appropriate methods for efficiency and legibility.
use AppleScript version "2.4" -- Mac OS X 10.10 (Yosemite) or later.
use framework "Foundation"
on task()
set nums to {}
repeat with a from 2 to 5
repeat with b from 2 to 5
set end of nums to (a ^ b) as integer
set end of nums to (b ^ a) as integer
end repeat
end repeat
set nums to current application's class "NSSet"'s setWithArray:(nums)
set descriptor to current application's class "NSSortDescriptor"'s sortDescriptorWithKey:("self") ascending:(true)
return (nums's sortedArrayUsingDescriptors:({descriptor})) as list
end task
task()
- Output:
{4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125}
Arturo
print sort unique flatten map 2..5 'a [
map 2..5 'b -> a^b
]
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
BCPL
get "libhdr"
let pow(n, p) =
p=0 -> 1,
n * pow(n, p-1)
let sort(v, length) be
if length > 0
$( for i=0 to length-2
if v!i > v!(i+1)
$( let t = v!i
v!i := v!(i+1)
v!(i+1) := t
$)
sort(v, length-1)
$)
let start() be
$( let v = vec 15
let i = 0
for a = 2 to 5 for b = 2 to 5
$( v!i := pow(a,b)
i := i+1
$)
sort(v, 16)
for i = 0 to 15
if i=0 | v!i ~= v!(i-1) do writef("%N ", v!i)
wrch('*N')
$)- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
BQN
∧⍷⥊⋆⌜˜ 2+↕4
- Output:
⟨ 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125 ⟩
C
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int compare(const void *a, const void *b) {
int ia = *(int*)a;
int ib = *(int*)b;
return (ia>ib) - (ia<ib);
}
int main() {
int pows[16];
int a, b, i=0;
for (a=2; a<=5; a++)
for (b=2; b<=5; b++)
pows[i++] = pow(a, b);
qsort(pows, 16, sizeof(int), compare);
for (i=0; i<16; i++)
if (i==0 || pows[i] != pows[i-1])
printf("%d ", pows[i]);
printf("\n");
return 0;
}
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
C++
#include <iostream>
#include <set>
#include <cmath>
int main() {
std::set<int> values;
for (int a=2; a<=5; a++)
for (int b=2; b<=5; b++)
values.insert(std::pow(a, b));
for (int i : values)
std::cout << i << " ";
std::cout << std::endl;
return 0;
}
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
F#
set[for n in 2..5 do for g in 2..5->pown n g]|>Set.iter(printf "%d ")
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
Factor
USING: kernel math.functions math.ranges prettyprint sequences
sets sorting ;
2 5 [a,b] dup [ ^ ] cartesian-map concat members natural-sort .
- Output:
{ 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125 }
FreeBASIC
redim arr(-1) as uinteger
dim as uinteger i
for a as uinteger = 2 to 5
for b as uinteger = 2 to 5
redim preserve arr(0 to ubound(arr)+1)
i = ubound(arr)
arr(i) = a^b
while arr(i-1)>arr(i) and i > 0
swap arr(i-1), arr(i)
i -= 1
wend
next b
next a
for i = 0 to ubound(arr)
if arr(i)<>arr(i-1) then print arr(i),
next iGo
package main
import (
"fmt"
"rcu"
"sort"
)
func main() {
var pows []int
for a := 2; a <= 5; a++ {
pow := a
for b := 2; b <= 5; b++ {
pow *= a
pows = append(pows, pow)
}
}
set := make(map[int]bool)
for _, e := range pows {
set[e] = true
}
pows = pows[:0]
for k := range set {
pows = append(pows, k)
}
sort.Ints(pows)
fmt.Println("Ordered distinct values of a ^ b for a in [2..5] and b in [2..5]:")
for i, pow := range pows {
fmt.Printf("%5s ", rcu.Commatize(pow))
if (i+1)%5 == 0 {
fmt.Println()
}
}
fmt.Println("\nFound", len(pows), "such numbers.")
}
- Output:
Ordered distinct values of a ^ b for a in [2..5] and b in [2..5]:
4 8 9 16 25
27 32 64 81 125
243 256 625 1,024 3,125
Found 15 such numbers.
Haskell
import qualified Data.Set as S
------------------ DISTINCT POWER NUMBERS ----------------
distinctPowerNumbers :: Int -> Int -> [Int]
distinctPowerNumbers a b =
(S.elems . S.fromList) $
(fmap (^) >>= (<*>)) [a .. b]
--------------------------- TEST -------------------------
main :: IO ()
main =
print $
distinctPowerNumbers 2 5
- Output:
[4,8,9,16,25,27,32,64,81,125,243,256,625,1024,3125]
or, as a one-off list comprehension:
import qualified Data.Set as S
main :: IO ()
main =
(print . S.elems . S.fromList) $
(\xs -> [x ^ y | x <- xs, y <- xs]) [2 .. 5]
or a liftA2 expression:
import Control.Applicative (liftA2)
import Control.Monad (join)
import qualified Data.Set as S
main :: IO ()
main =
(print . S.elems . S.fromList) $
join
(liftA2 (^))
[2 .. 5]
which can always be reduced (shedding imports) to the pattern:
import qualified Data.Set as S
main :: IO ()
main =
(print . S.elems . S.fromList) $
(\xs -> (^) <$> xs <*> xs)
[2 .. 5]
- Output:
[4,8,9,16,25,27,32,64,81,125,243,256,625,1024,3125]
J
~./:~;^/~2+i.4
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
jq
Works with gojq, the Go implementation of jq
For relatively small integers, such as involved in the specified task, the built-in function `pow/2` does the job:
[range(2;6) as $a | range(2;6) as $b | pow($a; $b)] | unique- Output:
[4,8,9,16,25,27,32,64,81,125,243,256,625,1024,3125]
However, if using gojq, then for unbounded precision, a special-purpose "power" function is needed:
def power($b): . as $in | reduce range(0;$b) as $i (1; . * $in);
[range(2;6) as $a | range(2;6) as $b | $a|power($b)] | unique- Output:
As above.
Julia
println(sort(unique([a^b for a in 2:5, b in 2:5])))
- Output:
[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]
Mathematica/Wolfram Language
Union @@ Table[a^b, {a, 2, 5}, {b, 2, 5}]
- Output:
{4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125}
Nim
import algorithm, math, sequtils, strutils, sugar
let list = collect(newSeq):
for a in 2..5:
for b in 2..5: a^b
echo sorted(list).deduplicate(true).join(" ")
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
Phix
with javascript_semantics function sqpn(integer n) return sq_power(n,{2,3,4,5}) end function sequence res = apply(true,sprintf,{{"%,5d"},unique(join(apply({2,3,4,5},sqpn),""))}) printf(1,"%d found:\n%s\n",{length(res),join_by(res,1,5," ")})
- Output:
15 found:
4 8 9 16 25
27 32 64 81 125
243 256 625 1,024 3,125
Perl
#!/usr/bin/perl -l
use strict; # https://rosettacode.org/wiki/Distinct_power_numbers
use warnings;
use List::Util qw( uniq );
print join ', ', sort { $a <=> $b } uniq map { my $e = $_; map $_ ** $e, 2 .. 5} 2 .. 5;
- Output:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
Python
from itertools import product
print(sorted(set(a**b for (a,b) in product(range(2,6), range(2,6)))))
- Output:
[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]
Or, for variation, generalizing a little in terms of starmap and pow:
'''Distinct power numbers'''
from itertools import product, starmap
# distinctPowerNumbers :: Int -> Int -> [Int]
def distinctPowerNumbers(a):
'''Sorted values of x^y where x, y <- [a..b]
'''
def go(b):
xs = range(a, 1 + b)
return sorted(set(
starmap(pow, product(xs, xs))
))
return go
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Distinct powers from integers [2..5]'''
print(
distinctPowerNumbers(2)(5)
)
# MAIN ---
if __name__ == '__main__':
main()
- Output:
[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]
R
This only takes one line.
unique(sort(rep(2:5, each = 4)^rep(2:5, times = 4)))- Output:
[1] 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
Quackery
[ [] swap
behead swap
witheach
[ 2dup = iff
drop done
dip join ]
join ] is unique ( [ --> [ )
[]
4 times
[ i 2 +
4 times
[ dup i 2 + **
rot join swap ]
drop ]
sort unique echo- Output:
[ 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125 ]
Raku
put squish sort [X**] (2..5) xx 2;
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
REXX
With this version of REXX, there's no need to sort the found numbers, or to eliminate duplicates.
/*REXX pgm finds and displays distinct power integers: a^b, where a and b are 2≤both≤5*/
parse arg lo hi cols . /*obtain optional arguments from the CL*/
if lo=='' | lo=="," then lo= 2 /*Not specified? Then use the default.*/
if hi=='' | hi=="," then hi= 5 /* " " " " " " */
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
w= 11 /*width of a number in any column. */
title= ' distinct power integers, a^b, where a and b are: ' lo "≤ both ≤" hi
say ' index │'center(title, 1 + cols*(w+1) )
say '───────┼'center("" , 1 + cols*(w+1), '─')
@.= .; $$= /*the default value for the @. array.*/
do a=lo to hi /*traipse through A values (LO──►HI).*/
do b=lo to hi /* " " B " " " */
x= a ** b; if @.x\==. then iterate /*Has it been found before? Then skip.*/
@.x= x; $$= $$ x /*assign power product; append to $$ */
end /*b*/
end /*a*/
$=; idx= 1 /*$$: a list of distinct power integers*/
do j=1 while words($$)>0; call getMin $$ /*obtain smallest number in the $$ list*/
$= $ right(commas(z), max(w, length(z) ) ) /*add a distinct power number ──► list.*/
if j//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
if $\=='' then say center(idx, 7)"│" substr($, 2) /*possible display residual output.*/
say '───────┴'center("" , 1 + cols*(w+1), '─')
say
say 'Found ' commas(j-1) title
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
getMin: parse arg z .; p= 1; #= words($$) /*assume min; # words in $$.*/
do m=2 for #-1; a= word($$, m); if a>=z then iterate; z= a; p= m
end /*m*/; $$= delword($$, p, 1); return /*delete the smallest number.*/
- output when using the default inputs:
index │ distinct power integers, a^b, where a and b are: 2 ≤ both ≤ 5 ───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── 1 │ 4 8 9 16 25 27 32 64 81 125 11 │ 243 256 625 1,024 3,125 ───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── Found 15 distinct power integers, a^b, where a and b are: 2 ≤ both ≤ 5
- output when using the inputs of: 0 5
index │ distinct power integers, a^b, where a and b are: 0 ≤ both ≤ 5 ───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── 1 │ 0 1 2 3 4 5 8 9 16 25 11 │ 27 32 64 81 125 243 256 625 1,024 3,125 ───────┴─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
- output when using the inputs of: 0 9
index │ distinct power integers, a^b, where a and b are: 0 ≤ both ≤ 9 ───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── 1 │ 0 1 2 3 4 5 6 7 8 9 11 │ 16 25 27 32 36 49 64 81 125 128 21 │ 216 243 256 343 512 625 729 1,024 1,296 2,187 31 │ 2,401 3,125 4,096 6,561 7,776 15,625 16,384 16,807 19,683 32,768 41 │ 46,656 59,049 65,536 78,125 117,649 262,144 279,936 390,625 531,441 823,543 51 │ 1,679,616 1,953,125 2,097,152 4,782,969 5,764,801 10,077,696 16,777,216 40,353,607 43,046,721 134,217,728 61 │ 387,420,489 ───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────── Found 61 distinct power integers, a^b, where a and b are: 0 ≤ both ≤ 9
Ring
load "stdlib.ring"
see "working..." + nl
see "Distinct powers are:" + nl
row = 0
distPow = []
for n = 2 to 5
for m = 2 to 5
sum = pow(n,m)
add(distPow,sum)
next
next
distPow = sort(distPow)
for n = len(distPow) to 2 step -1
if distPow[n] = distPow[n-1]
del(distPow,n-1)
ok
next
for n = 1 to len(distPow)
row++
see "" + distPow[n] + " "
if row%5 = 0
see nl
ok
next
see "Found " + row + " numbers" + nl
see "done..." + nl- Output:
working... Distinct powers are: 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125 Found 15 numbers done...
Sidef
[2..5]*2 -> cartesian.map_2d {|a,b| a**b }.sort.uniq.say
Alternative solution:
2..5 ~X** 2..5 -> sort.uniq.say
- Output:
[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]
Wren
import "/seq" for Lst
import "/fmt" for Fmt
var pows = []
for (a in 2..5) {
var pow = a
for (b in 2..5) {
pow = pow * a
pows.add(pow)
}
}
pows = Lst.distinct(pows).sort()
System.print("Ordered distinct values of a ^ b for a in [2..5] and b in [2..5]:")
for (chunk in Lst.chunks(pows, 5)) Fmt.print("$,5d", chunk)
System.print("\nFound %(pows.count) such numbers.")- Output:
Ordered distinct values of a ^ b for a in [2..5] and b in [2..5]:
4 8 9 16 25
27 32 64 81 125
243 256 625 1,024 3,125
Found 15 such numbers.
XPL0
int A, B, N, Last, Next;
[Last:= 0;
loop [Next:= -1>>1; \infinity
for A:= 2 to 5 do \find smallest Next
for B:= 2 to 5 do \ that's > Last
[N:= fix(Pow(float(A), float(B)));
if N>Last & N<Next then Next:= N;
];
if Next = -1>>1 then quit;
IntOut(0, Next); ChOut(0, ^ );
Last:= Next;
];
]- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125