Distinct power numbers: Difference between revisions
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<pre>[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]</pre> |
<pre>[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]</pre> |
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This only takes one line. |
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<lang R>unique(sort(rep(2:5, each = 4)^rep(2:5, times = 4)))</lang> |
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<pre> [1] 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125</pre> |
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=={{header|Raku}}== |
=={{header|Raku}}== |
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Revision as of 15:27, 9 October 2021
- 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
ALGOL 68
<lang algol68>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</lang>
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
APL
<lang APL>(⊂∘⍋⌷⊣)∪,∘.*⍨1+⍳4</lang>
- 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. <lang applescript>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()</lang>
- Output:
<lang applescript>{4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125}</lang>
Functional
Composing a solution from generic primitives, for speed of drafting, and ease of refactoring:
<lang applescript>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</lang>
- 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. <lang applescript>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()</lang>
- Output:
<lang applescript>{4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125}</lang>
BCPL
<lang 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')
$)</lang>
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
BQN
<lang bqn>∧⍷⥊⋆⌜˜ 2+↕4</lang>
- Output:
⟨ 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125 ⟩
C
<lang 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;
}</lang>
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
C++
<lang cpp>#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;
}</lang>
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
F#
<lang fsharp> set[for n in 2..5 do for g in 2..5->pown n g]|>Set.iter(printf "%d ") </lang>
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
Factor
<lang factor>USING: kernel math.functions math.ranges prettyprint sequences sets sorting ;
2 5 [a,b] dup [ ^ ] cartesian-map concat members natural-sort .</lang>
- Output:
{ 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125 }
Go
<lang go>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.")
}</lang>
- 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
<lang 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</lang>
- Output:
[4,8,9,16,25,27,32,64,81,125,243,256,625,1024,3125]
or, as a one-off list comprehension:
<lang haskell>import qualified Data.Set as S
main :: IO () main =
(print . S.elems . S.fromList) $ (\xs -> [x ^ y | x <- xs, y <- xs]) [2 .. 5]</lang>
or a liftA2 expression: <lang haskell>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]</lang>
which can always be reduced (shedding imports) to the pattern: <lang haskell>import qualified Data.Set as S
main :: IO () main =
(print . S.elems . S.fromList) $
(\xs -> (^) <$> xs <*> xs)
[2 .. 5]</lang>
- Output:
[4,8,9,16,25,27,32,64,81,125,243,256,625,1024,3125]
J
<lang j>~./:~;^/~2+i.4</lang>
- 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:
<lang jq>[range(2;6) as $a | range(2;6) as $b | pow($a; $b)] | unique</lang>
- Output:
[4,8,9,16,25,27,32,64,81,125,243,256,625,1024,3125]
However, if using gojq, then for unbounded precision, a special-purpose "power" function is needed:<lang jq>def power($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</lang>
- Output:
As above.
Julia
<lang julia>println(sort(unique([a^b for a in 2:5, b in 2:5])))</lang>
- Output:
[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]
Mathematica/Wolfram Language
<lang Mathematica>Union @@ Table[a^b, {a, 2, 5}, {b, 2, 5}]</lang>
- Output:
{4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125}
Nim
<lang 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(" ")</lang>
- 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
<lang 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;</lang>
- Output:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
Python
<lang python>from itertools import product print(sorted(set(a**b for (a,b) in product(range(2,6), range(2,6)))))</lang>
- 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:
<lang python>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()</lang>
- Output:
[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]
R
This only takes one line. <lang R>unique(sort(rep(2:5, each = 4)^rep(2:5, times = 4)))</lang>
- Output:
[1] 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125
Raku
<lang perl6>put squish sort [X**] (2..5) xx 2;</lang>
- 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. <lang rexx>/*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.*/</lang>
- 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
<lang 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 </lang>
- 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
<lang ruby>[2..5]*2 -> cartesian.map_2d {|a,b| a**b }.sort.uniq.say</lang>
Alternative solution: <lang ruby>2..5 ~X** 2..5 -> sort.uniq.say</lang>
- Output:
[4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]
Wren
<lang ecmascript>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.")</lang>
- 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
<lang 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;
];
]</lang>
- Output:
4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125