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Steady squares

From Rosetta Code
(Redirected from Steady Squares)
Steady squares 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.
Euler Project #284
Task


The 3-digit number 376 in the decimal numbering system is an example of numbers with the special property that its square ends with the same digits: 376*376 = 141376. Let's call a number with this property a steady square. Find steady squares under 10.000

AWK[edit]

 
# syntax: GAWK -f STEADY_SQUARES.AWK
BEGIN {
start = 1
stop = 999999
for (i=start; i<=stop; i++) {
n = i ^ 2
if (n ~ (i "$")) {
printf("%6d^2 = %12d\n",i,n)
count++
}
}
printf("\nSteady squares %d-%d: %d\n",start,stop,count)
exit(0)
}
 
Output:
     1^2 =            1
     5^2 =           25
     6^2 =           36
    25^2 =          625
    76^2 =         5776
   376^2 =       141376
   625^2 =       390625
  9376^2 =     87909376
 90625^2 =   8212890625
109376^2 =  11963109376
890625^2 = 793212890625

Steady squares 1-999999: 11

BASIC[edit]

ASIC[edit]

Compile with the Extended math option.

Translation of: C
 
REM Steady squares
FOR I = 1 TO 9999
N = I
GOSUB CheckIfSteady:
IF Steady <> 0 THEN
PRINT I;
PRINT " ^ 2 = ";
II& = I * I
PRINT II&
ENDIF
NEXT I
END
 
CheckIfSteady:
REM Result: Steady = 1 if N * N is steady; Steady = 0 otherwise.
Mask = 1
D = N
WHILE D <> 0
Mask = Mask * 10
D = D / 10
WEND
NNModMask& = N * N
NNModMask& = NNModMask& MOD Mask
IF NNModMask& = N THEN
Steady = 1
ELSE
Steady = 0
ENDIF
RETURN
 
Output:
     1 ^ 2 =            1
     5 ^ 2 =           25
     6 ^ 2 =           36
    25 ^ 2 =          625
    76 ^ 2 =         5776
   376 ^ 2 =       141376
   625 ^ 2 =       390625
  9376 ^ 2 =     87909376

FreeBASIC[edit]

function numdig( byval n as uinteger ) as uinteger
'number of decimal digits in n
dim as uinteger d=0
while n
d+=1
n\=10
wend
return d
end function
 
function is_steady_square( n as const uinteger ) as boolean
dim as integer n2 = n^2
if n2 mod 10^numdig(n) = n then return true else return false
end function
 
for i as uinteger = 1 to 10000
if is_steady_square(i) then print using "####^2 = ########";i;i^2
next i
Output:
   1^2 =        1
   5^2 =       25
   6^2 =       36
  25^2 =      625
  76^2 =     5776
 376^2 =   141376
 625^2 =   390625
9376^2 = 87909376

GW-BASIC[edit]

10 FOR N = 1 TO 10000
20 M$ = STR$(N)
30 M2#=N*N
40 M$ = RIGHT$(M$,LEN(M$)-1)
50 N2$ = STR$(M2#)
60 A = LEN(M$)
70 IF RIGHT$(N2$,A)= M$ THEN PRINT M$,N2$
80 NEXT N
Output:
1              1
5              25
6              36
25             625
76             5776
376            141376
625            390625
9376           87909376

Liberty BASIC[edit]

Translation of: C
Works with: Just BASIC version any
 
rem Steady squares
for i = 1 to 9999
if isSteady(i) then
print using("####",i); " ^ 2 = "; using("########", i * i)
end if
next i
end
 
function isSteady(n)
mask = 1
d = n
while d <> 0
mask = mask * 10
d = int(d / 10)
wend
isSteady = ((n * n) mod mask = n)
end function
 
Output:
   1 ^ 2 =        1
   5 ^ 2 =       25
   6 ^ 2 =       36
  25 ^ 2 =      625
  76 ^ 2 =     5776
 376 ^ 2 =   141376
 625 ^ 2 =   390625
9376 ^ 2 = 87909376

Tiny BASIC[edit]

Because TinyBASIC is limited to signed 16-bit integers, we need to perform the squaring by repeated addition and then take modulus. That makes for a pretty inefficient solution.

REM  N = THE NUMBER TO BE SQUARED
REM D = 10^THE NUMBER OF DIGITS IN N
REM M = THE SQUARE OF N, MODULO D
REM T = TEMP COPY OF N
 
LET N = 1
LET D = 10
10 IF N > 9 THEN LET D = 100
IF N > 99 THEN LET D = 1000
IF N > 999 THEN LET D = 10000
LET M = 0
LET T = N
20 LET M = M + N
LET M = M - (M/D)*D
LET T = T - 1
IF T > 0 THEN GOTO 20
rem PRINT N, " ", M
IF M = N THEN PRINT N
LET N = N + 1
IF N < 10000 THEN GOTO 10
END
Output:

1 5 6 25 76 376 625

9376

C[edit]

#include <stdio.h>
#include <stdbool.h>
 
bool steady(int n)
{
int mask = 1;
for (int d = n; d != 0; d /= 10)
mask *= 10;
return (n * n) % mask == n;
}
 
int main()
{
for (int i = 1; i < 10000; i++)
if (steady(i))
printf("%4d^2 = %8d\n", i, i * i);
return 0;
}
Output:
   1^2 =        1
   5^2 =       25
   6^2 =       36
  25^2 =      625
  76^2 =     5776
 376^2 =   141376
 625^2 =   390625
9376^2 = 87909376

CLU[edit]

n_digits = proc (n: int) returns (int)
i: int := 0
while n>0 do
i := i+1
n := n/10
end
return(i)
end n_digits
 
steady = proc (n: int) returns (bool)
sq: int := n ** 2
return (sq // 10**n_digits(n) = n)
end steady
 
start_up = proc ()
po: stream := stream$primary_output()
for i: int in int$from_to(1, 10000) do
if ~steady(i) then continue end
stream$putright(po, int$unparse(i), 4)
stream$puts(po, "^2 = ")
stream$putright(po, int$unparse(i**2), 8)
stream$putl(po, "")
end
end start_up
Output:
   1^2 =        1
   5^2 =       25
   6^2 =       36
  25^2 =      625
  76^2 =     5776
 376^2 =   141376
 625^2 =   390625
9376^2 = 87909376

Draco[edit]

proc nonrec steady(ulong n) bool:
ulong mask;
mask := 1;
while mask <= n do mask := mask * 10 od;
n*n % mask = n
corp
 
proc nonrec main() void:
word i;
for i from 1 upto 10000 do
if steady(i) then
writeln(i:4, "^2 = ", make(i,ulong)*i:8)
fi
od
corp
Output:
   1^2 =        1
   5^2 =       25
   6^2 =       36
  25^2 =      625
  76^2 =     5776
 376^2 =   141376
 625^2 =   390625
9376^2 = 87909376

F#[edit]

The Function[edit]

Implements No Search Required. large values may be produced using only integers.

 
// Steady Squares. Nigel Galloway: December 21st., 2021
let fN g=let n=List.fold2(fun z n g->z+n*g) 0L g (g|>List.rev) in (n,g)
let five,six=(5L,[|0L..9L|]),(6L,[|0L;9L;8L;7L;6L;5L;4L;3L;2L;1L|])
let stdySq(g0,N)=let rec fG n (g,l)=seq{let i=Array.item(int((n+g)%10L)) N in yield i; yield! (fG((n+g+2L*g0*i)/10L)(fN(i::l)))}
seq{yield g0; yield! fG(g0*g0/10L)(0L,[])}
 

Some Examples[edit]

 
stdySq six|>Seq.take 80|>Seq.rev|>Seq.iter(printf "%d");printfn ""
stdySq five|>Seq.take 80|>Seq.rev|>Seq.iter(printf "%d");printfn ""
 
Output:
61490109937833490419136188999442576576769103890995893380022607743740081787109376
38509890062166509580863811000557423423230896109004106619977392256259918212890625
Confirming Phix's example for 999 digits (in 11 thousands of sec).
 
stdySq six|>Seq.skip 920|>Seq.take 79|>Seq.rev|>Seq.iter(printf "%d");printfn "..."
 
Output:
7218745998663099139651109156359761242340631780203738180821664795072958006751247...
Real: 00:00:00.011
9999 digits
 
stdySq six|>Seq.skip 9920|>Seq.take 79|>Seq.rev|>Seq.iter(printf "%d");printfn "...";;
 
Output:
8908826164991254342660560818535016604238201034937718562215376152130910068662033...
Real: 00:00:00.330
If you have 57secs to spare then do 99999 digits, I leave it to the faithless to prove that this a Steady Square.
 
stdySq six|>Seq.skip 99920|>Seq.take 79|>Seq.rev|>Seq.iter(printf "%d");printfn "...";;
 
Output:
2755643458676224038154570844433833690960332159243668007360724907611570195135435...
Real: 00:00:57.520

Factor[edit]

Only checking numbers that end with 1, 5, and 6. See Talk:Steady_Squares for more details.

Works with: Factor version 0.99 2021-06-02
USING: formatting kernel math math.functions
math.functions.integer-logs prettyprint sequences
tools.memory.private ;
 
: steady? ( n -- ? )
[ sq ] [ integer-log10 1 + 10^ mod ] [ = ] tri ;
 
1000 <iota> { 1 5 6 } [
[ 10 * ] dip + dup steady?
[ dup sq commas "%4d^2 = %s\n" printf ] [ drop ] if
] cartesian-each
Output:
   1^2 = 1
   5^2 = 25
   6^2 = 36
  25^2 = 625
  76^2 = 5,776
 376^2 = 141,376
 625^2 = 390,625
9376^2 = 87,909,376

Fermat[edit]

Func Isstead( n ) = 
m:=n;
d:=1;
while m>0 do
d:=d*10;
m:=m\10;
od;
if n^2|d=n then Return(1) else Return(0) fi.;
 
for i = 1 to 9999 do
if Isstead(i) then !!(i,'^2 = ',i^2) fi;
od;
Output:
 1^2 =  1
 5^2 =  25
 6^2 =  36
 25^2 =  625
 76^2 =  5776
 376^2 =  141376
 625^2 =  390625
 9376^2 =  87909376

Go[edit]

Translation of: Wren
Library: Go-rcu
package main
 
import (
"fmt"
"rcu"
"strconv"
"strings"
)
 
func contains(list []int, s int) bool {
for _, e := range list {
if e == s {
return true
}
}
return false
}
 
func main() {
fmt.Println("Steady squares under 10,000:")
finalDigits := []int{1, 5, 6}
for i := 1; i < 10000; i++ {
if !contains(finalDigits, i%10) {
continue
}
sq := i * i
sqs := strconv.Itoa(sq)
is := strconv.Itoa(i)
if strings.HasSuffix(sqs, is) {
fmt.Printf("%5s -> %10s\n", rcu.Commatize(i), rcu.Commatize(sq))
}
}
}
Output:
Steady squares under 10,000:
    1 ->          1
    5 ->         25
    6 ->         36
   25 ->        625
   76 ->      5,776
  376 ->    141,376
  625 ->    390,625
9,376 -> 87,909,376

jq[edit]

Works with: jq

Works with gojq, the Go implementation of jq

# Input: an upper bound, or null for infinite
def steady_squares:
range(0; . // infinite)
| tostring as $i
| select( .*. | tostring | endswith($i));
 
10000
| steady_squares
Output:
0
1
5
6
25
76
376
625
9376

Julia[edit]

issteadysquare(n) = (s = "$n"; s == "$(n * n)"[end+1-length(s):end])
 
println(filter(issteadysquare, 1:10000)) # [1, 5, 6, 25, 76, 376, 625, 9376]
 

Haskell[edit]

import Control.Monad (join)
import Data.List (isSuffixOf)
 
--------------- NUMBERS WITH STEADY SQUARES --------------
 
p :: Int -> Bool
p = isSuffixOf . show <*> (show . join (*))
 
 
--------------------------- TEST -------------------------
main :: IO ()
main =
print $
takeWhile (< 10000) $ filter p [0 ..]
Output:
[0,1,5,6,25,76,376,625,9376]


Or, obtaining the squares by addition, rather than multiplication:

import Control.Monad (join)
import Data.Bifunctor (bimap)
import Data.List (isSuffixOf)
 
--------------- NUMBERS WITH STEADY SQUARES --------------
 
steadyPair :: Int -> Int -> [(Int, (String, String))]
steadyPair a b =
[ (a, ab)
| let ab = join bimap show (a, b),
uncurry isSuffixOf ab
]
 
--------------------------- TEST -------------------------
main :: IO ()
main =
putStrLn $
unlines
( uncurry ((<>) . (<> " -> ")) . snd
<$> takeWhile
((10000 >) . fst)
( concat $
zipWith
steadyPair
[0 ..]
(scanl (+) 0 [1, 3 ..])
)
)
Output:
0 -> 0
1 -> 1
5 -> 25
6 -> 36
25 -> 625
76 -> 5776
376 -> 141376
625 -> 390625
9376 -> 87909376

MAD[edit]

            NORMAL MODE IS INTEGER
VECTOR VALUES FMT = $I4,7H **2 = ,I8*$
THROUGH LOOP, FOR I=1, 1, I.G.10000
THROUGH POW, FOR MASK=1, 0, MASK.G.I
POW MASK = MASK*10
SQ = I*I
WHENEVER SQ-SQ/MASK*MASK.E.I
PRINT FORMAT FMT, I, SQ
END OF CONDITIONAL
LOOP CONTINUE
END OF PROGRAM
Output:
   1**2 =        1
   5**2 =       25
   6**2 =       36
  25**2 =      625
  76**2 =     5776
 376**2 =   141376
 625**2 =   390625
9376**2 = 87909376

Perl[edit]

#!/usr/bin/perl
 
use strict; # https://rosettacode.org/wiki/Steady_Squares
use warnings;
 
($_ ** 2) =~ /$_$/ and printf "%5d  %d\n", $_, $_ ** 2 for 1 .. 10000;
Output:
    1  1
    5  25
    6  36
   25  625
   76  5776
  376  141376
  625  390625
 9376  87909376

Phix[edit]

A number n ending in 2,3,4,7,8, or 9 will have a square ending in 4,9,6,9,4 or 1 respectively.
Further a number ending in k 0s will have a square ending in 2*k 0s, and hence always fail, so all possible candidates must end in 1, 5, or 6.
Further, the square of any k-digit number n will end in the same k-1 digits as the square of the number formed from the last k-1 digits of n,
in other words every successful 3-digit n must end with one of the previously successful answers (maybe zero padded), and so on for 4 digits, etc.
I stopped after 8 digits to avoid the need to fire up gmp. Finishes near-instantly, of course.

with javascript_semantics
sequence success = {1,5,6}  -- (as above)
atom p10 = 10
for digits=2 to 8 do
    for d=1 to 9 do
        for i=1 to length(success) do
            atom cand = d*p10+success[i]
            if remainder(cand*cand,p10*10)=cand then
                success &= cand
            end if
        end for
    end for
    p10 *= 10
end for
printf(1,"%d such numbers < 100,000,000 found:\n",length(success))
for i=1 to length(success) do
    atom si = success[i]
    printf(1,"%,11d^2 = %,21d\n",{si,si*si})
end for
Output:
15 such numbers < 100,000,000 found:
          1^2 =                     1
          5^2 =                    25
          6^2 =                    36
         25^2 =                   625
         76^2 =                 5,776
        376^2 =               141,376
        625^2 =               390,625
      9,376^2 =            87,909,376
     90,625^2 =         8,212,890,625
    109,376^2 =        11,963,109,376
    890,625^2 =       793,212,890,625
  2,890,625^2 =     8,355,712,890,625
  7,109,376^2 =    50,543,227,109,376
 12,890,625^2 =   166,168,212,890,625
 87,109,376^2 = 7,588,043,387,109,376

mpz (super fast to 1000 digits)[edit]

Obsessed with the idea the series could in fact be finite, I wheeled out gmp anyway... As per the talk page, it turns out that all steady squares (apart from 1) are in fact on a 5-chain and a 6-chain, which carry on forever. The following easily finishes in less than a second.

with javascript_semantics
include mpfr.e
constant limit = 1000
sequence success = {"1","5","6"}    -- (as above)
sequence squared = {"1","25","36"}  -- (kiss)
integer count = 3
mpz ch5 = mpz_init(5),  -- the 5-chain
    ch6 = mpz_init(6),  -- the 6-chain
    p10 = mpz_init(10), 
    {d10,sqr,r10,t10,cand} = mpz_inits(5), ch
for digits=2 to limit-1 do
    for d=1 to 9 do
        ch = ch5
        mpz_mul_si(d10,p10,d)
        mpz_mul_si(t10,p10,10)
        for chain=5 to 6 do
            mpz_add(cand,d10,ch)
            mpz_mul(sqr,cand,cand)
            mpz_fdiv_r(r10,sqr,t10)
            if mpz_cmp(cand,r10)=0 then
                count += 1
                if digits<=12 or digits>=limit-3 then
                    success = append(success,shorten(mpz_get_str(cand)))
                    squared = append(squared,shorten(mpz_get_str(sqr)))
                end if
                mpz_set(ch,cand)
            end if
            ch = ch6
        end for
    end for
    mpz_mul_si(p10,p10,10)
end for
printf(1,"%d steady squares < 1e%d found:\n",{count,limit})
for i=1 to length(success) do
    printf(1,"%13s^2 = %25s\n",{success[i],squared[i]})
end for

No doubt you could significantly improve that by replacing the mul/div with mpz_powm_ui(r10, cand, 2, t10) and o/c not even trying to print any of the silly-length numbers.

Output:
1783 steady squares < 1e1000 found:
            1^2 =                         1
            5^2 =                        25
            6^2 =                        36
           25^2 =                       625
           76^2 =                      5776
          376^2 =                    141376
          625^2 =                    390625
         9376^2 =                  87909376
        90625^2 =                8212890625
       109376^2 =               11963109376
       890625^2 =              793212890625
      2890625^2 =             8355712890625
      7109376^2 =            50543227109376
     12890625^2 =           166168212890625
     87109376^2 =          7588043387109376
    212890625^2 =         45322418212890625
    787109376^2 =        619541169787109376
   1787109376^2 =       3193759921787109376
   8212890625^2 =      67451572418212890625
  18212890625^2 =     331709384918212890625
  81787109376^2 =    6689131260081787109376
 918212890625^2 =  843114912509918212890625
18745998663099139651...07743740081787109376 (997 digits)^2 = 35141246587691473110...07743740081787109376 (1,993 digits)
81254001336900860348...92256259918212890625 (997 digits)^2 = 66022127332570868008...92256259918212890625 (1,994 digits)
21874599866309913965...07743740081787109376 (998 digits)^2 = 47849811931116570591...07743740081787109376 (1,995 digits)
78125400133690086034...92256259918212890625 (998 digits)^2 = 61035781460491829128...92256259918212890625 (1,996 digits)
27812540013369008603...92256259918212890625 (999 digits)^2 = 77353738199525217326...92256259918212890625 (1,997 digits)
72187459986630991396...07743740081787109376 (999 digits)^2 = 52110293793214504525...07743740081787109376 (1,998 digits)

Note that should this produce two steady squares of the same length that begin with the same digit, the one that ends in 5 would be shown first, even if it is numerically after then one that ends in 6, not that there are any such < 1e1000. In other words add a flag that effectively swaps the ch = ch5 and ch = ch6 lines.

No Search Required using strings[edit]

with javascript_semantics
atom t0 = time()
constant limit = 9999
sequence fivesix = {"5","6"} -- (held backwards)
for chain=5 to 6 do
    string f56 = fivesix[chain-4]
    integer d0 = f56[1]-'0', d = 0, n = floor(d0*d0/10), 
            dn = iff(chain=6?10-n:n)
    f56 &= dn+'0'
    for digit=2 to limit-1 do
        n = floor((n+d+2*dn*d0)/10)
        d = 0
        for j=2 to digit do
            d += (f56[j]-'0')*(f56[-j+1]-'0')
        end for
        dn = remainder(n+d,10)
        if chain=6 and dn then dn=10-dn end if
        f56 &= dn+'0'
    end for
    fivesix[chain-4] = f56
end for
integer count = 1
printf(1,"%13s\n",{"1"})
for d=1 to limit do
    sequence r = {}
    for j=1 to 2 do
        string fj = fivesix[j]
        if fj[d]!='0' then
            count += 1
            if d<=12 then
                r &= {sprintf("%13s",{reverse(fj[1..d])})}
            elsif d=999 or d=9999 then
                r &= {sprintf("%s...%s (%d digits)",{reverse(fj[d-19..d]),reverse(fj[1..20]),d})}
            end if
        end if
    end for
    if length(r)>1 and r[2]<r[1] then r = reverse(r) end if
    for i=1 to length(r) do
        printf(1,"%s\n",{r[i]})
    end for
end for
printf(1,"%d steady squares < 1e%d found\n",{count,limit+1})
?elapsed(time()-t0)
Output:
            1
            5
            6
           25
           76
          376
          625
         9376
        90625
       109376
       890625
      2890625
      7109376
     12890625
     87109376
    212890625
    787109376
   1787109376
   8212890625
  18212890625
  81787109376
 918212890625
27812540013369008603...92256259918212890625 (999 digits)
72187459986630991396...07743740081787109376 (999 digits)
10911738350087456573...92256259918212890625 (9999 digits)
89088261649912543426...07743740081787109376 (9999 digits)
18069 steady squares < 1e10000 found
"4.0s"

Unfortunately it is not particularly fast, 7mins on a 10 year old i3 for 99,999 digits, with results that match F#. Then again, I suppose it is a near-perfect candidate for my (far future) plans to boost performance in version 2... Perhaps more for my future benefit than anyone else's, if we replace the for j=2 to digit do loop with:

        #ilASM{
            mov esi,[f56]
            mov edx,[digit]
            mov ecx,1
            shl esi,2
            sub edx,1
            xor eax,eax
            xor edi,edi
         @@:
            mov al,[esi+ecx]
            mov bl,[esi+edx]
            sub al,'0'
            sub bl,'0'
            mul bl
            add edi,eax
            xor eax,eax
            sub edx,1
            add ecx,1
            cmp edx,1
            jge @b
            mov [d],edi
            xor ebx,ebx
                }

we get the above plus

27556434586762240381...07743740081787109376 (99999 digits)
72443565413237759618...92256259918212890625 (99999 digits)
179886 steady squares < 1e100000 found
"12.2s"

Which is a whopping 35-fold speedup, so obviously all I need to do is make the compiler emit similarly efficient code...

Python[edit]

Procedural[edit]

print("working...")
print("Steady squares under 10.000 are:")
limit = 10000
 
for n in range(1,limit):
nstr = str(n)
nlen = len(nstr)
square = str(pow(n,2))
rn = square[-nlen:]
if nstr == rn:
print(str(n) + " " + str(square))
 
print("done...")
Output:
working...
Steady squares under 10.000 are:
1 1
5 25
6 36
25 625
76 5776
376 141376
625 390625
9376 87909376
done...

Functional[edit]

'''Steady squares'''
 
from itertools import count, takewhile
 
 
# isSteady :: Int -> Bool
def isSteady(x):
'''True if the square of x ends
with the digits of x itself.
'''

return isSuffixOf(
str(x)
)(
str(x ** 2)
)
 
 
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Roots of numbers with steady squares up to 10000
'''

xs = takewhile(
lambda x: 10000 > x,
(x for x in count(0) if isSteady(x))
)
for x in xs:
print(f'{x} -> {x ** 2}')
 
 
# ----------------------- GENERIC ------------------------
 
# isSuffixOf :: (Eq a) => [a] -> [a] -> Bool
def isSuffixOf(needle):
'''True if needle is a suffix of haystack.
'''

def go(haystack):
d = len(haystack) - len(needle)
return d >= 0 and (needle == haystack[d:])
return go
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
0 -> 0
1 -> 1
5 -> 25
6 -> 36
25 -> 625
76 -> 5776
376 -> 141376
625 -> 390625
9376 -> 87909376


Or, defining the squares as an additive accumulation:

'''Steady Squares'''
 
from itertools import accumulate, chain, count, takewhile
from operator import add
 
 
def main():
'''Numbers up to 10000 which have steady squares'''
print(
'\n'.join(
f'{a} -> {b}' for (a, b) in takewhile(
lambda ab: 10000 > ab[0],
enumerate(
accumulate(
chain([0], count(1, 2)),
add
)
)
) if isSuffixOf(str(a))(str(b))
)
)
 
 
# ----------------------- GENERIC ------------------------
 
# isSuffixOf :: (Eq a) => [a] -> [a] -> Bool
def isSuffixOf(needle):
'''True if needle is a suffix of haystack.
'''

def go(haystack):
d = len(haystack) - len(needle)
return d >= 0 and (needle == haystack[d:])
return go
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
0 -> 0
1 -> 1
5 -> 25
6 -> 36
25 -> 625
76 -> 5776
376 -> 141376
625 -> 390625
9376 -> 87909376

Raku[edit]

.say for ({$++²}*).kv.grep( {$^v.ends-with: $^k} )[1..10]
Output:
(1 1)
(5 25)
(6 36)
(25 625)
(76 5776)
(376 141376)
(625 390625)
(9376 87909376)
(90625 8212890625)
(109376 11963109376)

REXX[edit]

/* REXX */
Numeric Digits 50
Call time 'R'
n=1000000000
Say 'Steady squares below' n
Do i=1 To n
c=right(i,1)
If pos(c,'156')>0 Then Do
i2=i*i
If right(i2,length(i))=i Then
Say right(i,length(n)) i2
End
End
Say time('E')
Output:
Steady squares below 1000000000
         1 1
         5 25
         6 36
        25 625
        76 5776
       376 141376
       625 390625
      9376 87909376
     90625 8212890625
    109376 11963109376
    890625 793212890625
   2890625 8355712890625
   7109376 50543227109376
  12890625 166168212890625
  87109376 7588043387109376
 212890625 45322418212890625
 787109376 619541169787109376
468.422000

Ring[edit]

 
see "working..." +nl
see "Steady squatres under 10.000 are:" + nl
limit = 10000
 
for n = 1 to limit
nstr = string(n)
len = len(nstr)
square = pow(n,2)
rn = right(string(square),len)
if nstr = rn
see "" + n + " -> " + square + nl
ok
next
 
see "done..." +nl
 
Output:
working...
Steady numbers under 10.000 are:
1 -> 1
5 -> 25
6 -> 36
25 -> 625
76 -> 5776
376 -> 141376
625 -> 390625
9376 -> 87909376
done...

Wren[edit]

Library: Wren-fmt

Although it hardly matters for a small range such as this, one can cut down the numbers to be examined by observing that a steady square must end in 1, 5 or 6.

import "./fmt" for Fmt
 
System.print("Steady squares under 10,000:")
var finalDigits = [1, 5, 6]
for (i in 1..9999) {
if (!finalDigits.contains(i % 10)) continue
var sq = i * i
if (sq.toString.endsWith(i.toString)) Fmt.print("$,5d -> $,10d", i, sq)
}
Output:
Steady squares under 10,000:
    1 ->          1
    5 ->         25
    6 ->         36
   25 ->        625
   76 ->      5,776
  376 ->    141,376
  625 ->    390,625
9,376 -> 87,909,376

XPL0[edit]

int  N, P;
[for N:= 0 to 10000-1 do
[P:= 1;
repeat P:= P*10 until P>N;
if rem(N*N/P) = N then
[IntOut(0, N);
Text(0, "^^2 = ");
IntOut(0, N*N);
CrLf(0);
];
];
]
Output:
0^2 = 0
1^2 = 1
5^2 = 25
6^2 = 36
25^2 = 625
76^2 = 5776
376^2 = 141376
625^2 = 390625
9376^2 = 87909376

Yabasic[edit]

// Rosetta Code problem: http://rosettacode.org/wiki/Steady_squares
// by Galileo, 04/2022
 
for i = 1 to 10000
r$ = str$(i^2, "%9.f")
if i = val(right$(r$, len(str$(i)))) print i, "\t=", r$
next
Output:
1       =        1
5       =       25
6       =       36
25      =      625
76      =     5776
376     =   141376
625     =   390625
9376    = 87909376
---Program done, press RETURN---