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.

Task

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

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 # sorting and printing 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;
    QUICKSORT squares FROMELEMENT 1 TOELEMENT 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

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]

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]

jq

Works with: jq

Works with gojq, the Go implementation of jq

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

The Task

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}			

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

Perl

Library: ntheory

#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Add_and_sort_square_numbers_in_ascending_order_from_three_lists
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)
    printf(1,"Added:%,d, Sorted:%V\n",{sum(res),res})
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:

Added:690, Sorted:{4,9,16,25,36,36,49,81,121,144,169}

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...

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 'Added - Sorted';
put ' ' ~ $s.sum ~ '  ' ~ $s.gist;

Added - Sorted
 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

load "stdlib.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])
           add(Primes,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...

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