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
ABC
HOW TO REPORT n is.steady.square.below power.of.ten:
REPORT ( n * n ) mod power.of.ten = n
HOW TO MIGHT n BE A STEADY SQUARE BELOW power.of.ten:
IF n is.steady.square.below power.of.ten:
WRITE ( ( ( n << 1 ) ^ "^2 = " ) >> 10 ) ^ ( ( n * n ) >> 1 ) /
PUT 10 IN power.of.ten
PUT -10 IN n
FOR i IN { 0 .. 1000 }:
PUT n + 10 IN n
IF n = power.of.ten:
PUT power.of.ten * 10 IN power.of.ten
MIGHT ( n + 1 ) BE A STEADY SQUARE BELOW power.of.ten
MIGHT ( n + 5 ) BE A STEADY SQUARE BELOW power.of.ten
MIGHT ( n + 6 ) BE A STEADY SQUARE BELOW power.of.ten- 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
Action!
;;; find some steady squares - numbers whose squares end in the number
;;; e.g. 376^2 = 141 376
PROC Main()
CARD p ; the number to square, with the final digit replaced by 0
CARD n ; the number to square
CARD d10 ; 10^the number of digits in p, n
CARD s ; the square of n modulo d10
CARD f ; loop counter to choose 1, 5 or 6 as the final digit of q
CARD front ; the first 2 digits of n
CARD back ; the last two digits of n
d10 = 10
FOR p = 0 TO 10000 STEP 10 DO
IF p = d10 THEN
d10 ==* 10
FI
FOR f = 1 TO 3 DO
IF f = 1 THEN
n = p + 1
ELSEIF f = 2 THEN
n = p + 5
ELSE
n = p + 6
FI
IF n <= 255 THEN
s = ( n * n )
ELSE
front = n / 100
back = n MOD 100
s = ( back * back ) + ( 200 * ( ( front * back ) MOD 100 ) )
FI
s ==MOD d10
IF s = n THEN
Put(' ) PrintC( n )
FI
OD
OD
RETURN- Output:
1 5 6 25 76 376 625 9376
ALGOL 60
begin comment find steady squares - numbers whose square ends in the number
e.g.: 376^2 = 141 376 ;
integer powerOfTen, p;
powerOfTen := 10;
comment note the final digit must be 1, 5 or 6 ;
for p := 0 step 10 until 10 000 do begin
integer d;
if p = powerOfTen then begin
comment number of digits has increased ;
powerOfTen := powerOfTen * 10
end;
for d := 1, 5, 6 do begin
integer m, n, n2;
n := p + d;
n2 := n * n;
m := n2 - ( ( n2 % powerOfTen ) * powerOfTen ) ;
if m = n then outinteger( 1, n )
end
end
end- 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
Arturo
steady?: function [n][
mask: 1
d: n
while -> d <> 0 [
mask: mask * 10
d: d / 10
]
n = (n * n) % mask
]
loop 0..1000 'n [
loop [1 5 6] 'd [
r: d + 10 * n ; only check numbers that end with 1, 5, 6
if steady? r -> print ~"|r|^2 = |r*r|"
]
]
- 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 "$")) {
printf("%6d^2 = %12d\n",i,n)
count++
}
}
printf("\nSteady squares %d-%d: %d\n",start,stop,count)
exit(0)
}
- Output:
1^2 = 1
5^2 = 25
6^2 = 36
25^2 = 625
76^2 = 5776
376^2 = 141376
625^2 = 390625
9376^2 = 87909376
90625^2 = 8212890625
109376^2 = 11963109376
890625^2 = 793212890625
Steady squares 1-999999: 11
BASIC
Applesoft BASIC
100 HOME
110 FOR n = 1 TO 10000
120 m$ = STR$(n)
130 n2$ = STR$(n*n)
140 IF RIGHT$(n2$,LEN(m$)) = m$ THEN HTAB(5-LEN(m$)): PRINT m$;"^2 = ";N2$
150 NEXT n
160 END
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
BASIC256
for i = 1 to 9999
if isSteady(i) then
print rjust(i,4); "^2 = "; rjust(i * i,8)
end if
next i
end
function isSteady(n)
mask = 1
d = n
while d <> 0
mask *= 10
d = int(d / 10)
end while
return ((n * n) mod mask = n)
end function
- Output:
Same as FreeBASIC entry.
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
Gambas
Public Sub Main()
For i As Integer = 1 To 10000
If is_steady_square(i) Then Print Format$(i, "####"); "^2 = "; Format$(i ^ 2, "########")
Next
End
Function numdig(n As Integer) As Integer
Dim d As Integer = 0
While n
d += 1
n \= 10
Wend
Return d
End Function
Function is_steady_square(n As Integer) As Boolean
Dim n2 As Integer = n ^ 2
Return IIf(n2 Mod CInt(10 ^ numdig(n)) = n, True, False)
End Function
- Output:
Same as FreeBASIC entry.
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
MSX Basic
10 CLS
20 FOR N = 1 TO 10000
30 M$ = STR$(N)
40 M2# = N*N
50 M$ = RIGHT$(M$,LEN(M$)-1)
60 N2$ = STR$(M2#)
70 A = LEN(M$)
80 IF RIGHT$(N2$,A)= M$ THEN LOCATE 5-LEN(M$): PRINT M$;"^2 =";N2$
90 NEXT N
100 END
QBasic
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
FOR i = 1 TO 9999
IF isSteady(i) THEN
PRINT USING ("####^2 = ########"); i; i * i
END IF
NEXT i
- Output:
Same as FreeBASIC entry.
Run BASIC
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:
Same as FreeBASIC entry.
Tiny BASIC
Although TinyBASIC is limited to signed 16-bit integers, we can still handle the basic task by noting that we can split a four digit number into the first pair of digits and the seoncd pair. If we call these F and B say, then the number squared module 10000 is B*B + 200 * ( F*B MOD 100 ). Additionally, (as other samples have noted), we only need to consider numbers ending in 1, 5 or 6.
REM P = THE NUMBER TO SQUARE, WITH THE FINAL DIGIT REPLACED BY 0
REM X = THE FINAL DIGIT OF THE NUMBER, 1, 5 OR 6
REM N = THE NUMBER TO BE SQUARED
REM D = 10^THE NUMBER OF DIGITS IN N
REM S = THE SQUARE OF N, MODULO D
REM F = THE FRONT (FIRST TWO DIGITS) OF N
REM B = THE BACK (LAST TWO DIGITS) OF N
10 LET P=0
20 LET D=10
30 LET X=1
40 IF D=P THEN LET D=D*10
50 LET N=P+X
60 LET F=N/100
70 LET B=N-F*100
80 LET S=B*B+200*(F*B-F*B/100*100)
90 IF S-S/D*D=N THEN PRINT N
100 LET X=X+4
110 IF X=10 LET P=P+10
120 IF X=10 LET X=1
130 IF X=9 LET X=6
140 IF P<9990 THEN GOTO 40
150 END
- Output:
1
5
6
25
76
376
625
9376
True BASIC
FUNCTION issteady(n)
LET mask = 1
LET d = n
DO WHILE d <> 0
LET mask = mask*10
LET d = INT(d/10)
LOOP
LET res = MOD((n * n), mask)
IF res = n THEN
LET issteady = res
ELSE
LET issteady = 0
END IF
END FUNCTION
FOR i = 1 TO 9999
IF issteady(i) <> 0 THEN
PRINT USING("####"):i;
PRINT "^2 = ";
PRINT USING("########"): i*i
END IF
NEXT i
END
- Output:
Same as FreeBASIC entry.
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
C++
#include <iostream>
using namespace std;
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);
}
- Output:
Same as C entry.
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
Dart
bool steady(int n) {
int mask = 1;
for (int d = n; d != 0; d ~/= 10) mask *= 10;
return (n * n) % mask == n;
}
void main() {
for (int i = 1; i < 10000; i++) if (steady(i)) print('$i^2 = ${i * 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
Delphi
{This routine is normally part of a separate library, but it is included here for clarity}
procedure GetDigits(N: integer; var IA: TIntegerDynArray);
{Get an array of the integers in a number}
{Numbers returned from least to most significant}
var T,I,DC: integer;
begin
DC:=Trunc(Log10(N))+1;
SetLength(IA,DC);
for I:=0 to DC-1 do
begin
T:=N mod 10;
N:=N div 10;
IA[I]:=T;
end;
end;
function IsSteadySquare(N: integer): boolean;
{compare digits of N and N^2 and see if they matchs}
var Dig1,Dig2: TIntegerDynArray;
var I: integer;
begin
Result:=False;
{Get digits}
GetDigits(N,Dig1);
GetDigits(N*N,Dig2);
{Compare digits}
for I:=0 to High(Dig1) do
if Dig1[I]<>Dig2[I] then exit;
Result:=True
end;
procedure ShowSteadySquares(Memo: TMemo);
var I: integer;
begin
for I:=1 to 10000-1 do
if IsSteadySquare(I) then
Memo.Lines.Add(Format('%6.0n^2 = %10.0n',[I+0.0,I*I+0.0]));
end;
- 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
9,376^2 = 87,909,376
Elapsed Time: 11.168 ms.
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
EasyLang
func steady n .
mask = 1
d = n
while d > 0
mask *= 10
d = d div 10
.
return if n * n mod mask = n
.
for i = 1 to 9999
if steady i = 1
print i & " * " & i & " = " & i * i
.
.
- Output:
1 * 1 = 1 5 * 5 = 25 6 * 6 = 36 25 * 25 = 625 76 * 76 = 5776 376 * 376 = 141376 625 * 625 = 390625 9376 * 9376 = 87909376
Euler
The original Euler implementations didn't have loops built-in, however they can be constructed using labels and gotos.
As procedures can take literal procedures as parameters, procedures to provide a loops can be defined. This is used here to define a "while" loop procedure.
Note that text between ` and ' is a procedure literal. The text "new while; while <- `...'" defines a new variable and assigned the while-loop procedure to it. When called, the text of the condition and loop body must be enclosed in ` and ' to make them procedures - otherwise they would not be evaluated each time as requried.
All Euler variables are declared using "new var" (only one variable per "new"), labels must be declared with "label" and procedure parametrers are declared with "formal" (again, only one label or parameter per "label" or "formal" declaration).
Everything is Euler is an expression (apart from new/label/formal) and returns a value (although the value of a "goto" can't be used), so the "else" part of an "if" is not optional, hence the "else 0"s appearing in the code below.
begin
new maxNumber; new powerOfTen; new lastDigit; new n;
new while;
while <- ` formal condition; formal loopBody;
begin
label again;
again: if condition then begin loopBody; goto again end else 0
end
'
;
maxNumber <- 10 000;
powerOfTen <- 10;
lastDigit <- ( 1, 5, 6 );
n <- -10;
while( ` [ n <- n + 10 ] <= maxNumber '
, ` begin
new d;
if n = powerOfTen then powerOfTen <- powerOfTen * 10
else 0;
d <- 0;
while( ` [ d <- d + 1 ] <= length lastDigit '
, ` begin
new nd; new n2;
nd <- n + lastDigit[ d ];
n2 <- nd * nd;
if n2 mod powerOfTen = nd then out nd else 0
end
'
)
end
'
)
end
$
- Output:
NUMBER 1
NUMBER 5
NUMBER 6
NUMBER 25
NUMBER 76
NUMBER 376
NUMBER 625
NUMBER 9376
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
Fe
(= steadySquares
(fn (maxNumber) ; max number to consider
(let powerOfTen 10) ; 10^(the number of digits in n
(let lastDigit (list 1 5 6)); a steady square must end with 1, 5 or 6
(let lastResult (cons 0 nil)); latest steady square start with a dummy 0
(let result lastResult) ; list of steady squares - dummy leading 0
(let n -10) ; multiple of 10 to consider
(while (do (= n (+ n 10)) ; find steady squares up to maxNumber
(<= n maxNumber)
)
(if (is n powerOfTen)
(= powerOfTen (* powerOfTen 10)); number of digits has increased
)
(let d lastDigit)
(while (not (atom d)) ; try n + each possible lastDigit
(let nd (+ n (car d)))
(let n2 (* nd nd))
; Fe doesn't have a mod operator, integer division or a way
; to truncate a float to an integer, so we calculate
; n2 mod powerOfTen using repeated subtraction - but see
; FizzBuzz for an example of doing it with a C function
(let n2%p10 n2)
(let mDivisor (* powerOfTen powerOfTen))
(while (<= powerOfTen mDivisor)
(while (<= mDivisor n2%p10)
(= n2%p10 (- n2%p10 mDivisor))
)
(= mDivisor (/ mDivisor 10))
)
(if (is nd n2%p10)
(do (setcdr lastResult (cons nd nil))
(= lastResult (cdr lastResult))
)
)
(= d (cdr d))
)
)
(cdr result) ; return the list of steady squares without the dummy 0
)
)
(print (steadySquares 10000))
- Output:
(1 5 6 25 76 376 625 9376)
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
Haxe
class Main // steady squares
{
static inline var MAX_NUMBER = 10000;
static function main()
{
var powerOfTen = 10;
var pad = ' ';
var lastDigit = [ 1, 5, 6 ]; // possible final digits
var n = -10;
for ( n10 in 0...Math.floor( MAX_NUMBER / 10 ) + 1 ) {
n += 10;
if( n == powerOfTen ) { // the number of digits just increased
powerOfTen *= 10;
pad = pad.substr( 1 );
}
for( d in 0...lastDigit.length ){
var nd = n + lastDigit[ d ];
var n2 = nd * nd;
if( n2 % powerOfTen == nd ){ // have a steady square
Sys.println( '$pad$nd^2 = $n2' );
}
}
}
}
}
- 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
\
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).
For example:
bigsteady=: {{
Y=. 1+s=.0x
whilst. Y < y do.
s=. (#~ issteady) ,(Y*i.10)+/s
Y=. Y*10
end.
s #~ y>s
}}
Example use:
$bigsteady 1e20
37
$bigsteady 1e30
54
q:54
2 3 3 3
9 6$bigsteady 1e30
0 1 5 6 25 76
376 625 9376 90625 109376 890625
2890625 7109376 12890625 87109376 212890625 787109376
1787109376 8212890625 18212890625 81787109376 918212890625 9918212890625
40081787109376 59918212890625 259918212890625 740081787109376 3740081787109376 6259918212890625
43740081787109376 56259918212890625 256259918212890625 743740081787109376 2256259918212890625 7743740081787109376
92256259918212890625 392256259918212890625 607743740081787109376 2607743740081787109376 7392256259918212890625 22607743740081787109376
77392256259918212890625 977392256259918212890625 9977392256259918212890625 19977392256259918212890625 80022607743740081787109376 380022607743740081787109376
619977392256259918212890625 3380022607743740081787109376 6619977392256259918212890625 93380022607743740081787109376 106619977392256259918212890625 893380022607743740081787109376
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
Works with jaq, the Rust 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]
Lua
Uses the fact the final digit can only be 1, 5 or 6 (see the talk page).
do --[[ 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
local function possibleSteadySuare ( n, p10 )
if ( n * n ) % p10 == n then
io.write( string.format( "%4d", n ), "^2: ", n * n, "\n" )
end
end
local powerOfTen = 10
-- note the final digit must be 1, 5 or 6
for p = 0,10000,10 do
if p == powerOfTen then
-- number of digits has increased
powerOfTen = powerOfTen * 10
end
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
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
Miranda
main :: [sys_message]
main = [Stdout (lay (map showsteady (takewhile (< 10000) steadies)))]
where showsteady n = shownum n ++ "^2 = " ++ shownum (n^2)
steadies :: [num]
steadies = filter steady [1..]
steady :: num->bool
steady n = n = n^2 mod 10^numdigits n
numdigits :: num->num
numdigits n = 1, if n<10
= 1 + numdigits (n div 10), otherwise- 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
Nim
import std/[algorithm, strutils]
var ends = @["1", "5", "6"]
var steady = @[1, 5, 6]
while ends.len != 0:
var newEnds: seq[string]
for e in ends:
for d in '1'..'9':
let s = d & e
let n = parseInt(s)
if n >= 10_000: break
if ($(n * n)).endsWith(s):
steady.add n
newEnds.add s
ends = newEnds
echo "Steady squares under 10_000: "
for n in sorted(steady):
echo n, "² = ", n * n
- Output:
Steady squares under 10_000: 1² = 1 5² = 25 6² = 36 25² = 625 76² = 5776 376² = 141376 625² = 390625 9376² = 87909376
Oberon-07
(* find some steady squares - numbers whose squares end in the number *)
(* e.g. 376^2 = 141 376 *)
MODULE SteadySquares;
IMPORT Out;
CONST maxNumber = 10000;
VAR n, powerOfTen :INTEGER;
PROCEDURE PossibleSteadySquare( r: INTEGER );
VAR r2: INTEGER;
BEGIN
r2 := r * r;
IF ( r2 MOD powerOfTen ) = r THEN
Out.Int( r, 6 );Out.String( "^2 = " );Out.Int( r2, 1 );Out.Ln
END
END PossibleSteadySquare;
BEGIN
powerOfTen := 10;
FOR n := 0 TO maxNumber BY 10 DO
IF n = powerOfTen THEN
(* the number of digits has increased *)
powerOfTen := powerOfTen * 10
END;
PossibleSteadySquare( n + 1 );
PossibleSteadySquare( n + 5 );
PossibleSteadySquare( n + 6 )
END
END SteadySquares.
- Output:
1^2 = 1
5^2 = 25
6^2 = 36
25^2 = 625
76^2 = 5776
376^2 = 141376
625^2 = 390625
9376^2 = 87909376
Perl
#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Steady_Squares
use warnings;
($_ ** 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/0
const maxnumber = 10000;
var p10, n, d, nd, n2;
begin
p10 := 10;
n := 0;
while n <= maxnumber do begin
if n = p10 then p10 := p10 * 10;
d := 0;
while d < 6 do begin
if d = 5 then d := 6;
if d = 1 then d := 5;
if d = 0 then d := 1;
nd := n + d;
n2 := nd * nd;
n2 := n2 - ( ( n2 / p10 ) * p10 );
if n2 = nd then ! nd
end;
n := n + 10
end
end.
- Output:
1
5
6
25
76
376
625
9376
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 8-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$LAST LITERALLY '3'; /* SUBSCRIPT OF LAST DIGIT PAIR */
DECLARE DEC$LEN LITERALLY '4'; /* LENGTH OF AN 8-DIGIT BCD NUMBER */
DECLARE DEC$8 LITERALLY '( DEC$LEN )BYTE'; /* TYPE DECLARATION OF AN */
/* 8-DIGIT BCD NUMBER - 4 BYTES */
PR$DEC: PROCEDURE( A$PTR ); /* PRINT AN UNSIGNED BCD NUMBER */
DECLARE A$PTR ADDRESS;
DECLARE A BASED A$PTR DEC$8;
DECLARE ( D, ZERO$CHAR, I, V ) BYTE;
ZERO$CHAR = ' ';
DO I = 0 TO DEC$LAST - 1;
V = A( I );
D = SHR( V AND 0F0H, 4 );
IF D = 0 THEN CALL PR$CHAR( ZERO$CHAR );
ELSE CALL PR$CHAR( D + ( ZERO$CHAR := '0' ) );
D = V AND 0FH;
IF D = 0 THEN CALL PR$CHAR( ZERO$CHAR );
ELSE CALL PR$CHAR( D + ( ZERO$CHAR := '0' ) );
END;
V = A( DEC$LAST );
D = SHR( V AND 0F0H, 4 );
IF D = 0 THEN CALL PR$CHAR( ZERO$CHAR );
ELSE CALL PR$CHAR( D + '0' );
D = V AND 0FH;
CALL PR$CHAR( D + '0' );
END PR$DEC ;
INIT$DEC: PROCEDURE( A$PTR ); /* SETS THE BCD VALUE IN A TO 0 */
DECLARE A$PTR ADDRESS;
DECLARE A BASED A$PTR DEC$8;
DECLARE I BYTE;
DO I = 0 TO DEC$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$8;
DECLARE ( I, P, D1, D2 ) BYTE;
DECLARE V ADDRESS;
V = B;
P = DEC$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$8, B BASED B$PTR DEC$8;
DECLARE I BYTE;
DO I = 0 TO DEC$LAST;
A( I ) = B( I );
END;
END MOV$DEC ;
ADD$DEC: PROCEDURE( A$PTR, B$PTR ); /* 8-DIGIT BCD ADDITION RESULT IN A */
DECLARE ( A$PTR, B$PTR ) ADDRESS;
DECLARE A BASED A$PTR DEC$8, B BASED B$PTR DEC$8;
DECLARE ( A0, A1, A2, A3 ) BYTE;
DECLARE ( B0, B1, B2, B3 ) BYTE;
/* SEPARATE THE DIGIT PAIRS */
A0 = A( 0 ); A1 = A( 1 ); A2 = A( 2 ); A3 = A( 3 );
B0 = B( 0 ); B1 = B( 1 ); B2 = B( 2 ); B3 = B( 3 );
/* DO THE ADDITIONS */
A3 = DEC( A3 + B3 );
A2 = DEC( A2 PLUS B2 );
A1 = DEC( A1 PLUS B1 );
A0 = DEC( A0 PLUS B0 );
/* RETURN THE RESULT */
A( 0 ) = A0; A( 1 ) = A1; A( 2 ) = A2; A( 3 ) = A3;
END ADD$DEC;
/* 8-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$8, ACCUMULATOR DEC$8;
CALL MOV$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$DEC ;
BIN$DEC4: PROCEDURE( A$PTR )ADDRESS; /* CONVERT A 4-DIGIT BCD NUMBER TO */
DECLARE A$PTR ADDRESS; /* BINARY */
DECLARE A BASED A$PTR DEC$8;
DECLARE ( D, V, RESULT ) ADDRESS, I BYTE;
RESULT = 0;
DO I = DEC$LAST - 1 TO DEC$LAST;
V = A( I );
D = SHR( V AND 0F0H, 4 );
RESULT = ( RESULT * 10 ) + D;
D = V AND 0FH;
RESULT = ( RESULT * 10 ) + D;
END;
RETURN RESULT;
END BIN$DEC4 ;
/* TASK */
DECLARE ( P, Q, F, POWER$OF$10 ) ADDRESS;
DECLARE SQ DEC$8;
POWER$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$DEC( .SQ, Q );
/* CONVERT THE LAST 4 DIGITS OF THE DECIMAL NUMBER TO BINARY */
/* AND COMPARE THE MODULI */
IF BIN$DEC4( .SQ ) MOD POWER$OF$10 = Q THEN DO;
/* FOUND ANOTHER STEADY SQUARE */
IF Q < 10 THEN CALL PR$CHAR( ' ' );
IF Q < 100 THEN CALL PR$CHAR( ' ' );
IF Q < 1000 THEN CALL PR$CHAR( ' ' );
CALL PR$NUMBER( Q );
CALL PR$STRING( .'**2: $' );
CALL PR$DEC( .SQ );
CALL PR$NL;
END;
END;
END;
EOF- 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
Prolog
works with swi-prolog
steadySquare([], _, []).
steadySquare([N| NTail], Modulo, SteadyList):-
Modulo =< N,!,
Modulo1 is Modulo * 10,
steadySquare([N| NTail], Modulo1, SteadyList).
steadySquare([N| NTail], Modulo, [N| SteadyList]):-
N ^ 2 mod Modulo =:= N,!,
steadySquare(NTail, Modulo, SteadyList).
steadySquare([_| NTail], Modulo, SteadyList):-
steadySquare(NTail, Modulo, SteadyList).
candidateList(Limit, List):-
candidateList(5, Limit, List0),
append([0,1], List0, List).
candidateList(N, Limit, [N, N1| Tail]):-
N < Limit,!,
N1 is N + 1,
N10 is N + 10,
candidateList(N10, Limit, Tail).
candidateList(_, _, []).
showList(_, []):-!.
showList(FrmStr, [StSquare| Tail]):-
Sqr is StSquare ^ 2,
format(FrmStr, [StSquare, Sqr]),
showList(FrmStr, Tail).
do:- candidateList(1000000, List),
steadySquare(List, 1, SteadySquareList),
last(SteadySquareList, LastStSqr),
LastLen is 1 + floor(log10(LastStSqr)),
LastSqrLen is 1 + floor(log10(LastStSqr ^ 2)),
swritef(FrmStr, '~|~t~d~%d+ ~|~t~d~%d+~n', [LastLen, LastSqrLen]),
showList(FrmStr, SteadySquareList).
- Output:
?- time(do).
0 0
1 1
5 25
6 36
25 625
76 5776
376 141376
625 390625
9376 87909376
90625 8212890625
109376 11963109376
890625 793212890625
% 900,133 inferences, 0.235 CPU in 0.235 seconds (100% CPU, 3834748 Lips)
true.
PROMAL
As with the Tiny BASIC sample, 16-bit languages must avoid overflow when solving this task. Uses long multiplication modulo the appropriate power of ten and the fact that the final digit must be 1, 5 or 6 (see the Discussion page).
PROGRAM steadySquares
INCLUDE LIBRARY
WORD p ; the number to square, with the final digit replaced by 0
WORD n ; the number to square
WORD d10 ; 10^the number of digits in p, n
WORD s ; the square of n modulo d10
WORD f ; loop counter to choose 1, 5 or 6 as the final digit of n
WORD front ; the first two digits of n
WORD back ; the last two digits of n
BEGIN
d10 = 10
p = 0
WHILE p < 10000
IF p = d10
d10 = d10 * 10
FOR f = 1 TO 3
CHOOSE f
1
n = p + 1
2
n = p + 5
ELSE
n = p + 6
IF n <= 255
s = ( n * n )
ELSE
front = n / 100
back = n % 100
s = ( back * back ) + ( 200 * ( ( front * back ) % 100 ) )
s = s % d10
IF s = n
OUTPUT " #W", n
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)
Quackery
[ 1 swap
[ 10 / dup while
dip 1+ again ]
drop ] is digitcount ( n --> n )
[ dup 2 **
over digitcount
10 swap **
mod = ] is steady ( n --> b )
[]
10000 times
[ i^ steady if [ i^ join ] ]
echo- Output:
[ 0 1 5 6 25 76 376 625 9376 ]
Refal
$ENTRY Go {
= <FindSteady 1 9999>;
};
FindSteady {
s.N s.Max, <Compare s.N s.Max>: '+' = ;
s.N s.Max, <Steady <Symb s.N>>: F = <FindSteady <+ s.N 1> s.Max>;
s.N s.Max = <ShowSteady s.N> <FindSteady <+ s.N 1> s.Max>;
};
ShowSteady {
s.N = <Prout <Symb s.N> '^2 = ' <* s.N s.N>>;
};
Steady {
e.N, <Symb <* <Numb e.N> <Numb e.N>>>: e.X e.N = T;
e.N = F;
};- 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
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...
RPL
Brute force approach
Numbers are converted into strings to make comparison easier.
≪ { 1 } 5 ROT FOR n
n →STR n SQ →STR
DUP SIZE 3 PICK SIZE - 1 + OVER SIZE SUB
IF == THEN n + END
NEXT
≫ ‘STEDY’ STO
10000 STEDY
- Output:
1: { 1 5 6 25 76 376 625 9376 }
Slight optimization
Steady numbers must end with 5 or 6, but there's no double-digits-or-more ones ending with 1: if ##b1 was the base 10 representation of such a number, the 2nd digit from the right shall be b (because of steadiness) and 2b (because of the elevation to square), which means b=0. By recurrence, all digits shall be zero.
≪ { 1 } 0 ROT FOR n
5 6 FOR j
n j + →STR LAST SQ →STR
DUP SIZE 3 PICK SIZE - 1 + OVER SIZE SUB
IF == THEN n j + + END
NEXT
10 STEP
≫ ‘STEDY’ STO
Ruby
p (0..10_000).select{|n| (n*n).to_s.end_with? n.to_s }
- Output:
[0, 1, 5, 6, 25, 76, 376, 625, 9376]
Rust
fn is_steady_square(n: u32) -> bool {
fn iter(n: u32, m: u32) -> u64 {
if m <= n {iter(n, 10 * m)}
else {m as u64}
}
let nl = n as u64;
nl * nl % iter(n, 1) == nl
}
fn main() {
let start = std::time::Instant::now();
let limit = 10_000_000;
let list5 = (5..=limit).step_by(10)
.flat_map(|n|[n, n+1])
.filter(|&n| is_steady_square(n));
let list: Vec<u32> = (0..2).into_iter().chain(list5).collect();
let max = *list.iter().last().unwrap();
let maxl = max as u64;
let duration = start.elapsed().as_millis();
let max_len = max.to_string().len();
let sqr_len = (maxl*maxl).to_string().len();
println!("{:>max_len$} {:>sqr_len$}", "num", "square");
for n in list {
let nl = n as u64;
println!("{:max_len$} {:sqr_len$}", n, nl*nl);
}
println!("time(ms): {duration}");
}
- Output:
num square
0 0
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
time(ms): 50
Scala
ready for Scala3
def steadySquares(cand: Seq[Int], modulo: Long, acc: Seq[Int]): Seq[Int] = {
if (cand.isEmpty) return acc
val num = cand.head
val numl = num.toLong
val modulo1 = if (modulo > numl) modulo else 10 * modulo
val acc1 = if (numl * numl % modulo1 != numl) acc
else acc :+ num
steadySquares(cand.tail, modulo1, acc1)
}
val limit = 1_000_000
val candidates = Seq(0, 1) ++ (5 to limit by 10).flatMap(n => Seq(n, n + 1))
val list = steadySquares(candidates, 1, Seq())
@main def main = {
val start = System.currentTimeMillis
val max = list.last.toLong
val Seq(maxLen, sqrLen) = Seq(max, max * max).map(_.toString.length)
for (steadySquare <- list) {
val stSqr = steadySquare.toLong
val sqr = stSqr * stSqr
println("%%%dd %%%dd".format(maxLen, sqrLen).format(stSqr, sqr))
}
val duration = System.currentTimeMillis - start
println(s"time(ms): $duration")
}
- Output:
0 0
1 1
5 25
6 36
25 625
76 5776
376 141376
625 390625
9376 87909376
90625 8212890625
109376 11963109376
890625 793212890625
time(ms): 7
SETL
program steady_squares;
loop for n in [1..10000] | steady n do
print(str n + "^2 = " + str (n**2));
end loop;
op steady(n);
return n**2 mod 10**#str n = n;
end op;
end 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
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
VTL-2
110 P=0
120 D=10
130 X=1
140 D=D=P*9*D+D
150 N=P+X
160 F=N/100
170 B=N-(F*100)
180 S=F*B-(F*B/100*100)*200+(B*B)
190 #=S-(S/D*D)=N=0*220
200 ?=N
210 $=32
220 X=X+4
230 #=X=10=0*260
240 P=P+10
250 X=1
260 #=X=9=0*280
270 X=6
280 #=P<9990*140- Output:
1 5 6 25 76 376 625 9376
Wren
Although it hardly matters for a small range such as this, one can cut down the numbers to be examined by observing that a steady square must end in 1, 5 or 6.
import "./fmt" for Fmt
System.print("Steady squares under 10,000:")
var finalDigits = [1, 5, 6]
for (i in 1..9999) {
if (!finalDigits.contains(i % 10)) continue
var sq = i * i
if (sq.toString.endsWith(i.toString)) Fmt.print("$,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
Categories:
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