Steady squares
- 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
Action!
... which is based on the Tiny BASIC sample with optimisations, including that the final digit of the squares must be 1, 5 or 6 (see the discussion page).
;;; find some steady squares - numbers whose squares end in the number
;;; e.g. 376^2 = 141 376
;;; returns ( value * multiplier ) MOD modulus using repeated addition
CARD FUNC multiplyMod( CARD value, multiplier, modulus )
CARD result, i
result = 0
FOR i = 1 TO multiplier / 2 DO
result = ( result + value + value ) MOD modulus
OD
IF multiplier MOD 2 = 1 THEN
result = ( result + value ) MOD modulus
FI
RETURN( result )
PROC Main()
CARD p ; the number to square, with the final digit replaced by 0
CARD q ; the number to square
CARD d ; 10^the number of digits in p, q
CARD s ; the square of q modulo d
CARD f ; loop counter to choose 1, 5 or 6 as the final digit of q
d = 10
FOR p = 0 TO 10000 STEP 10 DO
IF p = d THEN
d ==* 10
FI
FOR f = 1 TO 3 DO
IF f = 1 THEN
q = p + 1
ELSEIF f = 2 THEN
q = p + 5
ELSE
q = p + 6
FI
IF q <= 255 THEN
s = ( q * q ) MOD d
ELSEIF q < 1000 THEN
s = multiplyMod( q, q, d )
ELSE
s = multiplyMod( q, q MOD 100, d )
s ==+ ( 100 * ( multiplyMod( q, q / 100, d ) MOD 100 ) )
FI
IF s = q THEN
PrintCE( q )
FI
OD
OD
RETURN
- Output:
1 5 6 25 76 376 625 9376
ALGOL 68
Using the observation that the final digit must be 1, 5 or 6 (See Talk page).
BEGIN # find some steady squares - numbers whose squares end in the number #
# e.g. 376^2 = 141 376 #
INT max number = 10 000; # maximum number we will consider #
INT power of ten := 10;
[]INT last digit = ( 1, 5, 6 );
FOR n FROM 0 BY 10 TO max number DO
IF n = power of ten THEN
# the number of digits just increased #
power of ten *:= 10
FI;
FOR d FROM LWB last digit TO UPB last digit DO
INT nd = n + last digit[ d ];
INT n2 = nd * nd;
IF n2 MOD power of ten = nd THEN
# have a steady square #
print( ( whole( nd, -5 ), "^2 = ", whole( n2, 0 ), newline ) )
FI
OD
OD
END
- 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
ALGOL W
begin % find steady squares - numbers whose square ends in the number %
% e.g.: 376^2 = 141 376 %
% checks wheher n^2 mod p10 = n, i.e. n is a steady square and displays %
% a message if it is %
procedure possibleSteadySuare ( integer value n, p10 ) ;
if ( n * n ) rem p10 = n then write( i_w := 4, s_w := 0, n, "^2: ", i_w := 1, n * n );
integer powerOfTen;
powerOfTen := 10;
% note the final digit must be 1, 5 or 6 %
for p := 0 step 10 until 10000 do begin;
if p = powerOfTen then begin
% number of digits have increased %
powerOfTen := powerOfTen * 10
end if_p_eq_powerOfTen ;
possibleSteadySuare( p + 1, powerOfTen );
possibleSteadySuare( p + 5, powerOfTen );
possibleSteadySuare( p + 6, powerOfTen )
end
end.
- 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
AWK
# syntax: GAWK -f STEADY_SQUARES.AWK
BEGIN {
start = 1
stop = 999999
for (i=start; i<=stop; i++) {
n = i ^ 2
if (n ~ (i "$")) {
printf("%6d^2 = %12d\n",i,n)
count++
}
}
printf("\nSteady squares %d-%d: %d\n",start,stop,count)
exit(0)
}
- Output:
1^2 = 1 5^2 = 25 6^2 = 36 25^2 = 625 76^2 = 5776 376^2 = 141376 625^2 = 390625 9376^2 = 87909376 90625^2 = 8212890625 109376^2 = 11963109376 890625^2 = 793212890625 Steady squares 1-999999: 11
BASIC
ASIC
Compile with the Extended math option.
REM Steady squares
FOR I = 1 TO 9999
N = I
GOSUB CheckIfSteady:
IF Steady <> 0 THEN
PRINT I;
PRINT " ^ 2 = ";
II& = I * I
PRINT II&
ENDIF
NEXT I
END
CheckIfSteady:
REM Result: Steady = 1 if N * N is steady; Steady = 0 otherwise.
Mask = 1
D = N
WHILE D <> 0
Mask = Mask * 10
D = D / 10
WEND
NNModMask& = N * N
NNModMask& = NNModMask& MOD Mask
IF NNModMask& = N THEN
Steady = 1
ELSE
Steady = 0
ENDIF
RETURN
- Output:
1 ^ 2 = 1 5 ^ 2 = 25 6 ^ 2 = 36 25 ^ 2 = 625 76 ^ 2 = 5776 376 ^ 2 = 141376 625 ^ 2 = 390625 9376 ^ 2 = 87909376
Chipmunk Basic
10 rem Steady squares
20 for n = 1 to 10000
30 m$ = str$(n)
40 n2$ = str$(n*n)
50 if right$(n2$,len(m$)) = m$ then print m$,n2$
60 next n
70 end
- Output:
1 1 5 25 6 36 25 625 76 5776 376 141376 625 390625 9376 87909376
FreeBASIC
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
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
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
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
Yabasic
// 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---
C
#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
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
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#
The Function
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
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
Only checking numbers that end with 1, 5, and 6. See Talk:Steady_Squares for more details.
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
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
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
Haskell
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 retaining the string pair when the test succeeds:
import Control.Monad (join)
import Data.Bifunctor (bimap)
import Data.List (isSuffixOf)
---------------------- STEADY NUMBERS --------------------
steadyPair :: Int -> [(String, String)]
steadyPair n =
[ (s, s2)
| let (s, s2) = join bimap show (n, n * n),
s `isSuffixOf` s2
]
--------------------------- TEST -------------------------
main :: IO ()
main =
( \xs ->
let (w, w2) = join bimap length (last xs)
in mapM_
( putStrLn . uncurry ((<>) . (<> " -> "))
. bimap
(justifyRight w ' ')
(justifyRight w2 ' ')
)
xs
)
$ [0 .. 10000] >>= steadyPair
------------------------- GENERIC ------------------------
justifyRight :: Int -> Char -> String -> String
justifyRight n c = (drop . length) <*> (replicate n c <>)
- Output:
0 -> 0 1 -> 1 5 -> 25 6 -> 36 25 -> 625 76 -> 5776 376 -> 141376 625 -> 390625 9376 -> 87909376
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
J
Implementation:
issteady=: {{ Y-:N{.":*:y[N=.-#Y=.":y }}"0
issteady=: (": -: -@#@": {. ":@*:)"0 NB. tacit alternative
Task example:
I.issteady i.1e4
0 1 5 6 25 76 376 625 9376
Note that for larger values we would want to take advantage of the suffix characteristics of these numbers (a multi-digit steady square would have a suffix which is a steady square).
JavaScript
Procedural
// Steady squares
function steady(n) {
// Result: true if n * n is steady; false otherwise.
var mask = 1;
for (var d = n; d != 0; d = Math.floor(d / 10))
mask *= 10;
return (n * n) % mask == n;
}
for (var i = 1; i < 10000; i++)
if (steady(i))
console.log(i.toString().padStart(4, ' ') + "^2 = " +
(i * i).toString().padStart(8, ' '));
- 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
Functional
(() => {
"use strict";
// ----------------- STEADY SQUARES ------------------
// isSteady :: Int -> Bool
const isSteady = n =>
Boolean(steadyPair(n).length);
// steadyPair :: Int -> [(String, String)]
const steadyPair = n => {
// An empty list if n is not steady, otherwise a
// list containing a tuple of (n, n^2) strings.
const
s = `${n}`,
s2 = `${n ** 2}`;
return s2.endsWith(s) ? [
[s, s2]
] : [];
};
// ---------------------- TESTS ----------------------
const main = () => {
const
range = enumFromTo(0)(1E4),
pairs = range.flatMap(steadyPair),
[w, w2] = pairs[pairs.length - 1]
.map(x => x.length);
return [
range.filter(isSteady).join(", "),
pairs.map(([n, n2]) => {
const
steady = n.padStart(w, " "),
square = n2.padStart(w2, " ");
return `${steady} -> ${square}`;
})
.join("\n")
]
.join("\n\n");
};
// --------------------- GENERIC ---------------------
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = m =>
n => Array.from({
length: 1 + n - m
}, (_, i) => m + i);
// MAIN ---
return main();
})();
- Output:
0, 1, 5, 6, 25, 76, 376, 625, 9376 0 -> 0 1 -> 1 5 -> 25 6 -> 36 25 -> 625 76 -> 5776 376 -> 141376 625 -> 390625 9376 -> 87909376
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
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]
MAD
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
#!/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
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)
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
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...
PL/M
... under CP/M (or an emulator)
Although integers are restricted to unsigned 16-bit, 8080 PL/M also allows the use of BCD values. This sample uses 4-digit BCD to find the steady squares. As with other samples, only numbers ending in 1, 5 or 6 are considered (see the Discussion page).
100H: /* FIND SOME STEADY SQUARES - NUMBERS WHOSE SQUARES END IN THE NUMBER */
/* E.G. 376^2 = 141$376 */
/* CP/M SYSTEM CALL AND I/O ROUTINES */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
/* BCD ARITHMETIC */
DECLARE DEC$LAST LITERALLY '1'; /* SUBSCRIPT OF LAST DIGIT PAIR */
DECLARE DEC$LEN LITERALLY '2'; /* LENGTH OF A 4-DIGIT BCD NUMBER */
DECLARE DEC$4 LITERALLY '( DEC$LEN )BYTE'; /* TYPE DECLARATION OF A */
/* 4-DIGIT BCD NUMBER - 2 BYTES */
INIT$DEC: PROCEDURE( A$PTR ); /* SETS THE BCD VALUE IN A TO 0 */
DECLARE A$PTR ADDRESS;
DECLARE A BASED A$PTR ( 0 )BYTE;
DECLARE I BYTE;
DO I = 0 TO DEC$LAST;
A( I ) = 0;
END;
END INIT$DEC ;
SET$DEC: PROCEDURE( A$PTR, B ); /* SETS THE BCD VALUE IN A TO B */
DECLARE ( A$PTR, B ) ADDRESS;
DECLARE A BASED A$PTR DEC$4;
DECLARE ( I, P, D1, D2 ) BYTE;
DECLARE V ADDRESS;
V = B;
P = DEC$LAST;
DO I = 0 TO DEC$LAST;
IF V = 0
THEN A( P ) = 0;
ELSE DO;
D1 = V MOD 10;
D2 = ( V := V / 10 ) MOD 10;
A( P ) = SHL( D2, 4 ) OR D1;
V = V / 10;
END;
P = P - 1;
END;
END SET$DEC ;
MOV$DEC: PROCEDURE( A$PTR, B$PTR ); /* ASSIGN THE BCD VALUE IN B TO A */
DECLARE ( A$PTR, B$PTR ) ADDRESS;
DECLARE A BASED A$PTR DEC$4, B BASED B$PTR DEC$4;
DECLARE I BYTE;
DO I = 0 TO DEC$LAST;
A( I ) = B( I );
END;
END MOV$DEC ;
ADD$DEC: PROCEDURE( A$PTR, B$PTR ); /* 4-DIGIT BCD ADDITION RESULT IN A */
DECLARE ( A$PTR, B$PTR ) ADDRESS;
DECLARE A BASED A$PTR DEC$4, B BASED B$PTR DEC$4;
DECLARE ( A0, A1 ) BYTE;
DECLARE ( B0, B1 ) BYTE;
/* SEPARATE THE DIGIT PAIRS */
A0 = A( 0 ); A1 = A( 1 );
B0 = B( 0 ); B1 = B( 1 );
/* DO THE ADDITIONS */
A1 = DEC( A1 + B1 );
A0 = DEC( A0 PLUS B0 );
/* RETURN THE RESULT */
A( 0 ) = A0; A( 1 ) = A1;
END ADD$DEC;
/* 4-DIGIT BCD MULTIPLICATION BY AN UNSIGNED INTEGER VIA ETHIOPIAN */
/* MULTIPLICATION, RESULT IN A */
MUL$DEC: PROCEDURE( A$PTR, B );
DECLARE ( A$PTR, B ) ADDRESS;
DECLARE V ADDRESS, R DEC$4, ACCUMULATOR DEC$4;
CALL MOV$DEC( .R, A$PTR );
V = B;
CALL INIT$DEC( .ACCUMULATOR );
DO WHILE( V > 0 );
IF ( V AND 1 ) = 1 THEN DO;
CALL ADD$DEC( .ACCUMULATOR, .R );
END;
V = SHR( V, 1 );
CALL ADD$DEC( .R, .R );
END;
CALL MOV$DEC( A$PTR, .ACCUMULATOR );
END MUL$DEC ;
BIN$DEC: PROCEDURE( A$PTR )ADDRESS; /* CONVERT A 4-DIGIT BCD NUMBER TO */
DECLARE A$PTR ADDRESS; /* BINARY */
DECLARE A BASED A$PTR DEC$4;
DECLARE ( D, V, RESULT ) ADDRESS, I BYTE;
RESULT = 0;
DO I = 0 TO DEC$LAST;
V = A( I );
D = SHR( V AND 0F0H, 4 );
RESULT = ( RESULT * 10 ) + D;
D = V AND 0FH;
RESULT = ( RESULT * 10 ) + D;
END;
RETURN RESULT;
END BIN$DEC ;
/* TASK */
DECLARE ( P, Q, F, POWER$OF$10 ) ADDRESS;
DECLARE SQ DEC$4;
POWER$OF$10 = 10;
DO P = 0 TO 10$000 BY 10;
IF P = POWER$OF$10 THEN DO;
/* REACHED THE CURRENT POWER OF TEN - THE NUMBERS NOW HAVE ANOTHER */
/* DIGIT */
POWER$OF$10 = POWER$OF$10 * 10;
END;
DO F = 0 TO 2;
DO CASE F;
/* 0 */ Q = P + 1;
/* 1 */ Q = P + 5;
/* 2 */ Q = P + 6;
END;
CALL SET$DEC( .SQ, Q );
CALL MUL$DEC( .SQ, Q );
/* CONVERT THE DECIMAL NUMBER TO BINARY */
IF BIN$DEC( .SQ ) MOD POWER$OF$10 = Q MOD POWER$OF$10 THEN DO;
/* FOUND ANOTHER STEADY SQUARE */
CALL PR$CHAR( ' ' );
CALL PR$NUMBER( Q );
END;
END;
END;
EOF
- Output:
1 5 6 25 76 376 625 9376
PROMAL
As pointed out by the Tiny BASIC sample, 16-bit languages must avoid overflow. This sample is based on the Tiny BASIC sample but applies a number of optimisations, including the fact that the final digit must be 1, 5 or 6 (see the Discussion page).
PROGRAM steadySquares
INCLUDE LIBRARY
; returns ( value * multiplier ) % modulus using repeated addition
FUNC WORD multiplyMod
ARG WORD value
ARG WORD multiplier
ARG WORD modulus
WORD result
WORD i
BEGIN
result = 0
FOR i = 1 TO multiplier / 2
result = ( result + value + value ) % modulus
IF multiplier % 2 = 1
result = ( result + value ) % modulus
RETURN result
END
WORD p ; the number to square, with the final digit replaced by 0
WORD q ; the number to square
WORD d ; 10^the number of digits in p, q
WORD s ; the square of q modulo d
WORD n ; loop counter for numbers without the final digit
WORD f ; loop counter to choose 1, 5 or 6 as the final digit of q
BEGIN
d = 10
p = 0
FOR n = 0 TO 1000
IF p = d
d = d * 10
FOR f = 1 TO 3
CHOOSE f
1
q = p + 1
2
q = p + 5
ELSE
q = p + 6
IF q <= 255
s = ( q * q ) % d
ELSE IF q < 1000
s = multiplyMod( q, q, d )
ELSE
s = multiplyMod( q, q % 100, d )
s = s + ( 100 * ( multiplyMod( q, q / 100, d ) % 100 ) )
IF s = q
OUTPUT "#4W#C", q
p = p + 10
END
- Output:
1 5 6 25 76 376 625 9376
Python
Procedural
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
'''Steady squares'''
from itertools import chain
# steadyPair :: Int -> [(String, String)]
def steadyPair(x):
'''An empty list if x^2 is not suffixed, in decimal,
by the decimal digits of x. Otherwise a list
containing a tuple of the decimal strings of (x, x^2)
'''
s, s2 = str(x), str(x**2)
return [(s, s2)] if s2.endswith(s) else []
# ------------------------ TESTS -------------------------
# main :: IO ()
def main():
'''Roots of numbers with steady squares up to 10000
'''
ns = range(1, 1 + 10000)
xs = concatMap(steadyPair)(ns)
w, w2 = (len(x) for x in xs[-1])
print([n for n in ns if steadyPair(n)])
print()
print(
'\n'.join([
f'{s.rjust(w, " ")} -> {s2.rjust(w2, " ")}'
for (s, s2) in xs
])
)
# ----------------------- GENERIC ------------------------
# concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
'''A concatenated list over which a function has been
mapped.
The list monad can be derived by using a function f
which wraps its output in a list, (using an empty
list to represent computational failure).
'''
def go(xs):
return list(chain.from_iterable(map(f, xs)))
return go
# MAIN ---
if __name__ == '__main__':
main()
- Output:
[1, 5, 6, 25, 76, 376, 625, 9376] 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 str(b).endswith(str(a))
)
)
# 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
.say for ({$++²}…*).kv.grep( {$^v.ends-with: $^k} )[1..10]
- Output:
(1 1) (5 25) (6 36) (25 625) (76 5776) (376 141376) (625 390625) (9376 87909376) (90625 8212890625) (109376 11963109376)
REXX
/* REXX */
Numeric Digits 50
Call time 'R'
n=1000000000
Say 'Steady squares below' n
Do i=1 To n
c=right(i,1)
If pos(c,'156')>0 Then Do
i2=i*i
If right(i2,length(i))=i Then
Say right(i,length(n)) i2
End
End
Say time('E')
- Output:
Steady squares below 1000000000 1 1 5 25 6 36 25 625 76 5776 376 141376 625 390625 9376 87909376 90625 8212890625 109376 11963109376 890625 793212890625 2890625 8355712890625 7109376 50543227109376 12890625 166168212890625 87109376 7588043387109376 212890625 45322418212890625 787109376 619541169787109376 468.422000
Ring
see "working..." +nl
see "Steady squatres under 10.000 are:" + nl
limit = 10000
for n = 1 to limit
nstr = string(n)
len = len(nstr)
square = pow(n,2)
rn = right(string(square),len)
if nstr = rn
see "" + n + " -> " + square + nl
ok
next
see "done..." +nl
- Output:
working... Steady numbers under 10.000 are: 1 -> 1 5 -> 25 6 -> 36 25 -> 625 76 -> 5776 376 -> 141376 625 -> 390625 9376 -> 87909376 done...
TypeScript
// Steady squares
function steady(n: number): bool {
// Result: true if n * n is steady; false otherwise.
var mask = 1;
for (var d = n; d != 0; d = Math.floor(d / 10))
mask *= 10;
return (n * n) % mask == n;
}
for (var i = 1; i < 10000; i++)
if (steady(i))
console.log(i.toString().padStart(4, ' ') + "^2 = " +
(i * i).toString().padStart(8, ' '));
- Output:
1^2 = 1 5^2 = 25 6^2 = 36 25^2 = 625 76^2 = 5776 376^2 = 141376 625^2 = 390625 9376^2 = 87909376
Wren
Although it hardly matters for a small range such as this, one can cut down the numbers to be examined by observing that a steady square must end in 1, 5 or 6.
import "./fmt" for Fmt
System.print("Steady squares under 10,000:")
var finalDigits = [1, 5, 6]
for (i in 1..9999) {
if (!finalDigits.contains(i % 10)) continue
var sq = i * i
if (sq.toString.endsWith(i.toString)) Fmt.print("$,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
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