# Collect and sort square numbers in ascending order from three lists

Collect and sort square numbers in ascending order from three lists 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.

Collect and sort square numbers in ascending order from three lists.
list[1] = [3,4,34,25,9,12,36,56,36]
list[2] = [2,8,81,169,34,55,76,49,7]
list[3] = [75,121,75,144,35,16,46,35]

## 11l

V lists = [[3, 4, 34, 25, 9, 12, 36, 56, 36], [2, 8, 81, 169, 34, 55, 76, 49, 7], [75, 121, 75, 144, 35, 16, 46, 35]]
[Int] squares

F is_square(x)
R Int(sqrt(x)) ^ 2 == x

L(l) lists
L(x) l
I is_square(x)
squares.append(x)

squares.sort()
print(squares)
Output:
[4, 9, 16, 25, 36, 36, 49, 81, 121, 144, 169]

## ALGOL 68

Library: ALGOL 68-rows
Library: ALGOL 68-sort

Note, the sources of rows.incl.a68 and sort.incl.a68 are on separate pages on Rosetta Code, see the above links.

BEGIN # construct a list of the squares from some other lists                 #
# lists are represented by arrays in this sample                          #
PR read "rows.incl.a68" PR               # incldue row printing utilities #
PR read "sort.incl.a68" PR               # include row sorting  utilities #

# construct the lists of numbers required by the task #
[][]INT list = ( (  3,   4, 34,  25,  9, 12, 36, 56, 36 )
, (  2,   8, 81, 169, 34, 55, 76, 49,  7 )
, ( 75, 121, 75, 144, 35, 16, 46, 35     )
);
# find the elements of the lists that are squares                         #
INT r pos   := 0;
REF[]INT squares := HEAP[ 1 : 0 ]INT;          # start with an empty list #
FOR i FROM LWB list TO UPB list DO
FOR j FROM LWB list[ i ] TO UPB list[ i ] DO
IF   INT  n    = list[ i ][ j ];
REAL root = sqrt( n );
root = ENTIER root
THEN # have a square #
r pos +:= 1;
IF r pos > UPB squares THEN
# the squares array isn't big enough - make a larger one #
REF[]INT new squares
:= HEAP[ 1 : UPB squares + 10 ]INT;
new squares[ 1 : UPB squares ] := squares;
squares                        := new squares
FI;
squares[ r pos ] := n
FI
OD
OD;
squares QUICKSORT ELEMENTS( 1, r pos );
SHOW squares[ 1 : r pos ]
END
Output:
4 9 16 25 36 36 49 81 121 144 169

## Arturo

lists: [
[3,4,34,25,9,12,36,56,36]
[2,8,81,169,34,55,76,49,7]
[75,121,75,144,35,16,46,35]
]

squares: map 1..inc to :integer sqrt max flatten lists 'x -> x^2
square?: function [n]-> contains? squares n

print select sort fold.seed:[] lists [a,b][a++b] => square?
Output:
4 9 16 25 36 36 49 81 121 144 169

## AutoHotkey

list := [], t := [], result := []

list[1] := [3,4,34,25,9,12,36,56,36]
list[2] := [2,8,81,169,34,55,76,49,7]
list[3] := [75,121,75,144,35,16,46,35]

for i, l in list
for j, n in l
if ((s:=Sqrt(n)) = Floor(s))
t[n, j] := true

for n, obj in t
for i, v in obj
result.push(n)
Output:
[4, 9, 16, 25, 36, 36, 49, 81, 121, 144, 169]

## AWK

# syntax: GAWK -f COLLECT_AND_SORT_SQUARE_NUMBERS_IN_ASCENDING_ORDER_FROM_THREE_LISTS.AWK
BEGIN {
list[1] = "3,4,34,25,9,12,36,56,36"
list[2] = "2,8,81,169,34,55,76,49,7"
list[3] = "75,121,75,144,35,16,46,35"
for (i=1; i<=length(list); i++) {
n = split(list[i],list_arr,",")
for (j=1; j<=n; j++) {
if (is_square(list_arr[j])) {
sq_arr[i,j] = list_arr[j]
}
}
}
PROCINFO["sorted_in"] = "@val_num_asc"
for (i in sq_arr) {
printf("%d ",sq_arr[i])
}
printf("\n")
exit(0)
}
function is_square(n) {
return (int(sqrt(n))^2 == n)
}
Output:
4 9 16 25 36 36 49 81 121 144 169

## BQN

lists  ⟨[3,4,34,25,9,12,36,56,36]
[2,8,81,169,34,55,76,49,7]
[75,121,75,144,35,16,46,35]⟩

Output:
⟨ 4 9 16 25 36 36 49 81 121 144 169 ⟩

## Delphi

Works with: Delphi version 6.0

The program uses the Delphi standard "TList" component to collect the numbers. Since it normally holds pointers, the integers are recast as pointers. The TList components provides a quick sort routine that is customized by providing your own compare routine.

const List1: array [0..8] of integer = (3,4,34,25,9,12,36,56,36);
const List2: array [0..8] of integer = (2,8,81,169,34,55,76,49,7);
const List3: array [0..7] of integer = (75,121,75,144,35,16,46,35);

function Compare(P1,P2: pointer): integer;
{Compare for quick sort}
begin
Result:=Integer(P1)-Integer(P2);
end;

procedure SortedSquareNumbers(Memo: TMemo);
{Find and sort numbers that are perfect squares}
var I: integer;
var NumList: TList;
var S: string;

{Load only perfect squares from one of the list }
var I: integer;
begin
for I:=0 to High(L) do
if  Sqr(Sqrt(L[I]))=L[I] then
end;

begin
NumList:=TList.Create;
try
NumList.Sort(Compare);
S:='';
for I:=0 to NumList.Count-1 do
S:=S+' '+IntToStr(Integer(NumList[I]));
finally NumList.Free; end;
end;
Output:
4 9 16 25 36 36 49 55 76 81 121 144 169

## EasyLang

lists[][] = [ [ 3 4 34 25 9 12 36 56 36 ] [ 2 8 81 169 34 55 76 49 7 ] [ 75 121 75 144 35 16 46 35 ] ]
#
proc sort . d[] .
for i = 1 to len d[] - 1
for j = i + 1 to len d[]
if d[j] < d[i]
swap d[j] d[i]
.
.
.
.
func issqr x .
return if pow floor sqrt x 2 = x
.
for i to len lists[][]
for e in lists[i][]
if issqr e = 1
sqrs[] &= e
.
.
.
sort sqrs[]
print sqrs[]

## F#

// Collect and sort square numbers in ascending order from three lists. Nigel Galloway: December 21st., 2021
let fN g=g*g in printfn "%A" (([3;4;34;25;9;12;36;56;36]@[2;8;81;169;34;55;76;49;7]@[75;121;75;144;35;16;46;35])|>List.filter(fun n->(float>>sqrt>>int>>fN)n=n)|>List.sort)
Output:
[4; 9; 16; 25; 36; 36; 49; 81; 121; 144; 169]

## Factor

Works with: Factor version 0.99
USING: prettyprint project-euler.common sequences sorting ;

{ 3 4 34 25 9 12 36 56 36 }
{ 2 8 81 169 34 55 76 49 7 }
{ 75 121 75 144 35 16 46 35 }
3append [ perfect-square? ] filter sort .
Output:
{ 4 9 16 25 36 36 49 81 121 144 169 }

## FreeBASIC

function issquare( n as integer ) as boolean
if int(sqr(n))^2=n then return true else return false
end function

dim as integer i, j, nums(1 to 3, 1 to 9) = {{3,4,34,25,9,12,36,56,36},_
{2,8,81,169,34,55,76,49,7}, {75,121,75,144,35,16,46,35,-1}}, c, d, ii, jj      'pad the last list to make it the same length
redim as integer squares(-1)   'we don't know beforehand how many squares we'll find, so set up a placeholder

while true                     'keep going while there are still squares in the lists
c = 99999                  'sentinel value
for i = 1 to 3             'loop through lists
for j = 1 to 9
d = nums(i, j)
if issquare(d) then     'have we found a square?
if d < c then       'is it the smallest so far?
c = d
ii = i
jj = j
end if
end if
next j
next i
if c = 99999 then exit while    'if we didn't find a square, stop looking
redim preserve squares(0 to ubound(squares)+1)      'update our list
squares(ubound(squares)) = c
nums(ii,jj) = 13                'mod the original lists so we don't find that same square again
wend

for i = 0 to ubound(squares)        'output data
print squares(i);"  ";
next i
Output:
4   9   16   25   36   36   49   81   121   144   169

## Go

package main

import (
"fmt"
"math"
"sort"
)

func isSquare(n int) bool {
s := int(math.Sqrt(float64(n)))
return s*s == n
}

func main() {
lists := [][]int{
{3, 4, 34, 25, 9, 12, 36, 56, 36},
{2, 8, 81, 169, 34, 55, 76, 49, 7},
{75, 121, 75, 144, 35, 16, 46, 35},
}
var squares []int
for _, list := range lists {
for _, e := range list {
if isSquare(e) {
squares = append(squares, e)
}
}
}
sort.Ints(squares)
fmt.Println(squares)
}
Output:
[4 9 16 25 36 36 49 81 121 144 169]

import Data.List (sort)

------- PERFECT SQUARE SUBSET OF THREE LISTS SORTED ------

squaresSorted :: Integral a => [[a]] -> [a]
squaresSorted = sort . concatMap (filter isPerfectSquare)

isPerfectSquare :: Integral a => a -> Bool
isPerfectSquare = (==) <*> ((^ 2) . floor . sqrt . fromIntegral)

--------------------------- TEST -------------------------
main :: IO ()
main =
print \$
squaresSorted
[ [3, 4, 34, 25, 9, 12, 36, 56, 36],
[2, 8, 81, 169, 34, 55, 76, 49, 7],
[75, 121, 75, 144, 35, 16, 46, 35]
]
Output:
[4,9,16,25,36,36,49,81,121,144,169]

## J

There's many valid approaches here. Here's one of them:

list1=: 3 4 34 25 9 12 36 56 36
list2=: 2 8 81 169 34 55 76 49 7
list3=: 75 121 75 144 35 16 46 35

/:~ (#~ ] = <.&.%:) list1,list2,list3
4 9 16 25 36 36 49 81 121 144 169

## jq

Works with: jq

Works with gojq, the Go implementation of jq

def isSquare: sqrt | (. == floor);

def lists: [
[3,4,34,25,9,12,36,56,36],
[2,8,81,169,34,55,76,49,7],
[75,121,75,144,35,16,46,35]
];

[lists[][]] | unique | map(select(isSquare))
Output:
[4,9,16,25,36,49,81,121,144,169]

## Julia

vcat is list "addition" in Julia.

lists = [
[3,4,34,25,9,12,36,56,36],
[2,8,81,169,34,55,76,49,7],
[75,121,75,144,35,16,46,35]
]

squares = reduce(vcat, [[s for s in list if isqrt(s)^2 == s] for list in lists])
sort!(squares)
println(squares)

## Mathematica / Wolfram Language

list1 = {3, 4, 34, 25, 9, 12, 36, 56, 36};
list2 = {2, 8, 81, 169, 34, 55, 76, 49, 7};
list3 = {75, 121, 75, 144, 35, 16, 46, 35};
Sort@Select[Join[list1, list2, list3], IntegerQ[Sqrt[#]] &]
Output:

{4,9,16,25,36,36,49,81,121,144,169}

## Maxima

list[1]: [3,4,34,25,9,12,36,56,36]\$
list[2]: [2,8,81,169,34,55,76,49,7]\$
list[3]: [75,121,75,144,35,16,46,35]\$
block(makelist(sublist(list[i],lambda([x],integerp(sqrt(x)))),i,1,3),map(sort,%%));
Output:

[[4,9,25,36,36],[49,81,169],[16,121,144]]

## newLISP

(define (sqr? n) (= n (pow (round (sqrt n)) 2)))

(sort (filter sqr? (flat
'((3 4 34 25 9 12 36 56 36)
(2 8 81 169 34 55 76 49 7)
(75 121 75 144 35 16 46 35)))))
Output:
(4 9 16 25 36 36 49 81 121 144 169)

## Nim

import std/[algorithm, math, strutils, sugar]

func isSquare(n: Natural): bool =
let r = sqrt(n.toFloat).int
result = n == r * r

const
List1 = @[3, 4, 34, 25, 9, 12, 36, 56, 36]
List2 = @[2, 8, 81, 169, 34, 55, 76, 49, 7]
List3 = @[75, 121, 75, 144, 35, 16, 46, 35]

var squareList = collect:
for list in [List1, List2, List3]:
for n in list:
if n.isSquare:
n
squareList.sort()

echo squareList.join(" ")
Output:
4 9 16 25 36 36 49 81 121 144 169

## ooRexx

Parse Version v
Say v
l.1=.array~of(3,4,34,25,9,12,36,56,36)
l.2=.array~of(2,8,81,169,34,55,76,49,7)
l.3=.array~of(75,121,75,144,35,16,46,35)
st.0=0
Do li=1 To 3
Do e over l.li
If is_square(e) Then
Call store s
End
End
Call Show
Exit

is_square:
Parse Arg x
Do i=1 By 1 UNtil i**2>x
if i**2=x Then Return 1
End
Return 0

store:
do i=1 To st.0
/*
If st.i=e Then
Return
*/
If st.i>e Then Do
Do j=st.0 To i By -1
ja=j+1
st.ja=st.j
End
st.0=st.0+1
Leave i
End
End
st.i=e
If i>st.0 Then
st.0=i
Return

show:
ol='Ordered squares:'
Do i=1 To st.0
ol=ol st.i
End
Say ol
Exit
Output:
REXX-ooRexx_5.0.0(MT)_64-bit 6.05 8 Sep 2020
Ordered squares: 4 9 16 25 36 36 49 81 121 144 169

## Pascal

### Free Pascal

Program SortedSquares;
{\$mode ObjFPC}{\$H+}

Uses sysutils,fgl;

Type
tmap = specialize TFPGList<integer>;

Const List1: array Of integer = (3,4,34,25,9,12,36,56,36);
Const List2: array Of integer = (2,8,81,169,34,55,76,49,7);
Const List3: array Of integer = (75,121,75,144,35,16,46,35);

Var list : specialize TFPGList<integer>;

Procedure addtolist(arr : Array Of integer);
Var i : integer;
Begin
For i In arr Do
If ((frac(sqrt(i)) = 0) {check if a square, there will be no fraction}
And (list.indexof(i) = -1)) {make sure number isn't in list already}
End;

Function CompareInt(Const Item1,Item2: Integer): Integer;
Begin
result := item1 - item2;
End;

Var i : integer;
Begin
list := specialize TFPGList<integer>.create;
List.Sort(@CompareInt); {quick sort the list}
For i In list Do
write(i:4);
list.destroy;
End.
Output:
4   9  16  25  36  49  81 121 144 169

## Perl

Library: ntheory
#!/usr/bin/perl

use warnings;
use ntheory qw( is_square );
use List::Util qw( sum );

my @lists = (
[3,4,34,25,9,12,36,56,36],
[2,8,81,169,34,55,76,49,7],
[75,121,75,144,35,16,46,35]);

my \$sum = sum my @squares = grep is_square(\$_), sort { \$a <=> \$b } map @\$_, @lists;
print "sum \$sum  -  @squares\n";
Output:
sum 690  -  4 9 16 25 36 36 49 81 121 144 169

## Phix

with javascript_semantics
procedure squares(sequence lists)
sequence res = sort(join(lists)), sq = {1}
while res[\$]>sq[\$] do sq &= power(length(sq)+1,2) end while
res = filter(res,"in",sq)
end procedure

squares({{3,4,34,25,9,12,36,56,36},
{2,8,81,169,34,55,76,49,7},
{75,121,75,144,35,16,46,35}})

Alternatively, and a smidgen faster but not enough to be measurable here, you could replace that inner while loop with:

integer m = res[\$], s = 1, d = 1
while m>s do d += 2; s += d; sq &= s; end while
Output:

You could also apply res:=unique(res), anywhere/8 ways: new line in 3 places, or inline (two ways each) on either existing assignment, or as a replacement for the sort(), which would reduce the length by 1 (the duplicate 36) and yield a total of 654.

## Python

import math

print("working...")
list = [(3,4,34,25,9,12,36,56,36),(2,8,81,169,34,55,76,49,7),(75,121,75,144,35,16,46,35)]
Squares = []

def issquare(x):
for p in range(x):
if x == p*p:
return 1
for n in range(3):
for m in range(len(list[n])):
if issquare(list[n][m]):
Squares.append(list[n][m])

Squares.sort()
print(Squares)
print("done...")
Output:
working...
[4, 9, 16, 25, 36, 36, 49, 81, 121, 144, 169]
done...

## Quackery

sqrt+ is defined at Isqrt (integer square root) of X#Quackery.

[ sqrt+ nip 0 = ] is issquare ( n --> b )

[]
' [ [  3   4 34  25  9 12 36 56 36 ]
[  2   8 81 169 34 55 76 49  7 ]
[ 75 121 75 144 35 16 46 35 ] ]
witheach
[ witheach
[ dup issquare iff
join else drop ] ]
sort
echo
Output:
[ 4 9 16 25 36 36 49 81 121 144 169 ]

## Raku

my \$s = cache sort ( my \$l = ( cache flat

[3,4,34,25,9,12,36,56,36],
[2,8,81,169,34,55,76,49,7],
[75,121,75,144,35,16,46,35]

)).grep: * ∈ cache {\$++²}…*>\$l.max;

put ' ' ~ \$s.sum ~ '  ' ~ \$s.gist;
690  (4 9 16 25 36 36 49 81 121 144 169)

## REXX

Parse Version v
Say v
l.=0
Call mk_array 1,3,4,34,25,9,12,36,56,36
Call mk_array 2,2,8,81,169,34,55,76,49,7
Call mk_array 3,75,121,75,144,3,5,16,46,35
st.0=0
Do li=1 To 3
Do ii=1 To l.li.0
If is_square(l.li.ii) Then
Call store l.li.ii
End
End
Call Show
Exit

mk_array:
an=arg(1)
Do i=1 To arg()-1
l.an.i=arg(i+1)
End
l.an.0=arg()-1
Return

is_square:
Parse Arg x
Do i=1 By 1 Until i**2>x
if i**2=x Then Return 1
End
Return 0

store:
Parse Arg e
do i=1 To st.0
If st.i>e Then Do
Do j=st.0 To i By -1
ja=j+1
st.ja=st.j
End
st.0=st.0+1
Leave i
End
End
st.i=e
If i>st.0 Then
st.0=i
Return

show:
ol='Ordered squares:'
Do i=1 To st.0
ol=ol st.i
End
Say ol
Exit
Output:
REXX-Regina_3.9.4(MT) 5.00 25 Oct 2021
Ordered squares: 4 9 16 25 36 36 49 81 121 144 169

## Ring

see "working..." + nl
list = list(3)
list[1] = [3,4,34,25,9,12,36,56,36]
list[2] = [2,8,81,169,34,55,76,49,7]
list[3] = [75,121,75,144,35,16,46,35]
Primes = []

for n = 1 to 3
for m = 1 to len(list[n])
if issquare(list[n][m])
ok
next
next

Primes = sort(Primes)
showArray(Primes)

see nl + "done..." + nl

func issquare(x)
for n = 1 to sqrt(x)
if x = pow(n,2)
return 1
ok
next
return 0

func showArray(array)
txt = ""
see "["
for n = 1 to len(array)
txt = txt + array[n] + ","
next
txt = left(txt,len(txt)-1)
txt = txt + "]"
see txt
Output:
working...
[4,9,16,25,36,36,49,81,121,144,169]
done...

## RPL

≪ + + OBJ→ → n
≪ { } 1 n START
IF DUP √ FP THEN DROP ELSE + END NEXT
SORT
≫ ≫ 'GETSQ' STO
{ 3 4 34 25 9 12 36 56 36 } { 2 8 81 169 34 55 76 49 7 } { 75 121 75 144 35 16 46 35 } GETSQ

Output:

1: { 4 9 16 25 36 36 49 81 121 144 169 }

## Ruby

class Integer
def square?
isqrt = Integer.sqrt(self)
isqrt*isqrt == self
end
end

list = [3,4,34,25,9,12,36,56,36].chain(
[2,8,81,169,34,55,76,49,7],
[75,121,75,144,35,16,46,35])

p list.select(&:square?).sort
Output:
[4, 9, 16, 25, 36, 36, 49, 81, 121, 144, 169]

## Rust

fn main( ) {
let list1 : Vec<i32> = vec![ 3 , 3 , 34 , 25 , 9 , 12 , 36 , 56 , 36] ;
let list2 : Vec<i32> = vec![2, 8 , 81 , 169 , 34 , 55 , 76 , 49 , 7] ;
let list3 : Vec<i32> = vec![ 75 , 121 , 75 , 144 , 35 , 16 , 46 , 35 ] ;
let mut all_numbers : Vec<f32> = Vec::new( ) ;
list1.iter( ).for_each( | n | all_numbers.push( *n as f32 ) ) ;
list2.iter( ).for_each( | n | all_numbers.push( *n as f32)) ;
list3.iter( ).for_each( | n | all_numbers.push( *n as f32) ) ;
let squares : Vec<&f32> = all_numbers.iter( ).filter( | i | {
let root = i.sqrt( ) ;
root == root.floor( )
}).collect( ) ;
let mut found : Vec<i32> = squares.iter( ).map( | n | **n as i32 ).
collect( )  ;
found.sort( ) ;
println!("{:?}" , found ) ;
}
Output:
[9, 16, 25, 36, 36, 49, 81, 121, 144, 169]

## Sidef

var lists = [
[3,4,34,25,9,12,36,56,36],
[2,8,81,169,34,55,76,49,7],
[75,121,75,144,35,16,46,35],
]

say lists.flat.grep{.is_square}.sort
Output:
[4, 9, 16, 25, 36, 36, 49, 81, 121, 144, 169]

## Wren

Library: Wren-math
import "./math" for Int

var lists = [
[3,4,34,25,9,12,36,56,36],
[2,8,81,169,34,55,76,49,7],
[75,121,75,144,35,16,46,35]
]

var squares = []
for (list in lists) {
for (e in list) if (Int.isSquare(e)) squares.add(e)
}
squares.sort()
System.print(squares)
Output:
[4, 9, 16, 25, 36, 36, 49, 81, 121, 144, 169]

## XPL0

The task description seems to dictate how a program is supposed to solve this, rather than merely provide a procedure that determines the solution. XPL0 doesn't have a built-in list capability like Python. A simple substitute is to treat the lists as arrays. A complication is that list[3] is shorter than the others, which is adjusted here. XPL0 also doesn't have a built-in sort routine (although one is provided in its library). Displaying a small number of sorted items is easily done with a few lines of code.

int List(3+1), Min, I, J, S, MI, MJ;
def Inf = -1>>1;
[List(1):= [3,4,34,25,9,12,36,56,36];
List(2):= [2,8,81,169,34,55,76,49,7];
List(3):= [75,121,75,144,35,16,46,35,Inf];
loop   [Min:= Inf;
for I:= 1 to 3 do
for J:= 0 to 8 do
[S:= sqrt(List(I,J));
if S*S = List(I,J) then
if List(I,J) <= Min then
[Min:= List(I,J);  MI:= I;  MJ:= J];
];
if Min = Inf then quit;
IntOut(0, Min);  ChOut(0, ^ );
List(MI, MJ):= Inf;
];
]
Output:
4 9 16 25 36 36 49 81 121 144 169