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
AWK
<lang AWK>
- syntax: GAWK -f STEADY_SQUARES.AWK
BEGIN {
start = 1 stop = 999999 for (i=start; i<=stop; i++) { n = i ^ 2 if (n ~ (i "$")) { printf("%6d^2 = %12d\n",i,n) count++ } } printf("\nSteady squares %d-%d: %d\n",start,stop,count) exit(0)
} </lang>
- Output:
1^2 = 1 5^2 = 25 6^2 = 36 25^2 = 625 76^2 = 5776 376^2 = 141376 625^2 = 390625 9376^2 = 87909376 90625^2 = 8212890625 109376^2 = 11963109376 890625^2 = 793212890625 Steady squares 1-999999: 11
BASIC
ASIC
Compile with the Extended math option.
<lang basic> REM Steady squares FOR I = 1 TO 9999
N = I GOSUB CheckIfSteady: IF Steady <> 0 THEN PRINT I; PRINT " ^ 2 = "; II& = I * I PRINT II& ENDIF
NEXT I END
CheckIfSteady: REM Result: Steady = 1 if N * N is steady; Steady = 0 otherwise. Mask = 1 D = N WHILE D <> 0
Mask = Mask * 10 D = D / 10
WEND NNModMask& = N * N NNModMask& = NNModMask& MOD Mask IF NNModMask& = N THEN
Steady = 1
ELSE
Steady = 0
ENDIF RETURN </lang>
- Output:
1 ^ 2 = 1 5 ^ 2 = 25 6 ^ 2 = 36 25 ^ 2 = 625 76 ^ 2 = 5776 376 ^ 2 = 141376 625 ^ 2 = 390625 9376 ^ 2 = 87909376
FreeBASIC
<lang freebasic>function numdig( byval n as uinteger ) as uinteger
'number of decimal digits in n dim as uinteger d=0 while n d+=1 n\=10 wend return d
end function
function is_steady_square( n as const uinteger ) as boolean
dim as integer n2 = n^2 if n2 mod 10^numdig(n) = n then return true else return false
end function
for i as uinteger = 1 to 10000
if is_steady_square(i) then print using "####^2 = ########";i;i^2
next i</lang>
- Output:
1^2 = 1 5^2 = 25 6^2 = 36 25^2 = 625 76^2 = 5776 376^2 = 141376 625^2 = 390625 9376^2 = 87909376
GW-BASIC
<lang gwbasic>10 FOR N = 1 TO 10000 20 M$ = 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</lang>
- Output:
1 1 5 25 6 36 25 625 76 5776 376 141376 625 390625 9376 87909376
Liberty BASIC
<lang lb> 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 </lang>
- 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.
<lang tinybasic>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</lang>
- Output:
1 5 6 25 76 376 625
9376
C
<lang 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;
}</lang>
- 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
<lang 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</lang>
- 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
<lang 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</lang>
- 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. <lang fsharp> // 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,[])}
</lang>
Some Examples
<lang fsharp> stdySq six|>Seq.take 80|>Seq.rev|>Seq.iter(printf "%d");printfn "" stdySq five|>Seq.take 80|>Seq.rev|>Seq.iter(printf "%d");printfn "" </lang>
- Output:
61490109937833490419136188999442576576769103890995893380022607743740081787109376 38509890062166509580863811000557423423230896109004106619977392256259918212890625
- Confirming Phix's example for 999 digits (in 11 thousands of sec).
<lang fsharp> stdySq six|>Seq.skip 920|>Seq.take 79|>Seq.rev|>Seq.iter(printf "%d");printfn "..." </lang>
- Output:
7218745998663099139651109156359761242340631780203738180821664795072958006751247... Real: 00:00:00.011
- 9999 digits
<lang fsharp> stdySq six|>Seq.skip 9920|>Seq.take 79|>Seq.rev|>Seq.iter(printf "%d");printfn "...";; </lang>
- 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.
<lang fsharp> stdySq six|>Seq.skip 99920|>Seq.take 79|>Seq.rev|>Seq.iter(printf "%d");printfn "...";; </lang>
- 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.
<lang factor>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</lang>
- 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
<lang 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;</lang>
- 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
<lang 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)) } }
}</lang>
- 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
<lang 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 ..]</lang>
- Output:
[0,1,5,6,25,76,376,625,9376]
or retaining the string pair when the test succeeds:
<lang haskell>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 <>)</lang>
- 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:
<lang haskell>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 ..]) ) )</lang>
- Output:
0 -> 0 1 -> 1 5 -> 25 6 -> 36 25 -> 625 76 -> 5776 376 -> 141376 625 -> 390625 9376 -> 87909376
JavaScript
Procedural
<lang javascript>// 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, ' '));</lang>
- 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
<lang javascript>(() => {
"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();
})();</lang>
- 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 <lang 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</lang>
- Output:
0 1 5 6 25 76 376 625 9376
Julia
<lang 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] </lang>
MAD
<lang 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</lang>
- 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
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Steady_Squares use warnings;
($_ ** 2) =~ /$_$/ and printf "%5d %d\n", $_, $_ ** 2 for 1 .. 10000;</lang>
- 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...
Python
Procedural
<lang python>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...")</lang>
- 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
<lang python>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()</lang>
- 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:
<lang python>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()</lang>
- Output:
0 -> 0 1 -> 1 5 -> 25 6 -> 36 25 -> 625 76 -> 5776 376 -> 141376 625 -> 390625 9376 -> 87909376
Raku
<lang perl6>.say for ({$++²}…*).kv.grep( {$^v.ends-with: $^k} )[1..10]</lang>
- Output:
(1 1) (5 25) (6 36) (25 625) (76 5776) (376 141376) (625 390625) (9376 87909376) (90625 8212890625) (109376 11963109376)
REXX
<lang 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')</lang>
- 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
<lang 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 </lang>
- 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
<lang javascript>// 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, ' '));
</lang>
- 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. <lang ecmascript>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 -> $,10d", i, sq)
}</lang>
- Output:
Steady squares under 10,000: 1 -> 1 5 -> 25 6 -> 36 25 -> 625 76 -> 5,776 376 -> 141,376 625 -> 390,625 9,376 -> 87,909,376
XPL0
<lang XPL0>int N, P; [for N:= 0 to 10000-1 do
[P:= 1; repeat P:= P*10 until P>N; if rem(N*N/P) = N then [IntOut(0, N); Text(0, "^^2 = "); IntOut(0, N*N); CrLf(0); ]; ];
]</lang>
- Output:
0^2 = 0 1^2 = 1 5^2 = 25 6^2 = 36 25^2 = 625 76^2 = 5776 376^2 = 141376 625^2 = 390625 9376^2 = 87909376
Yabasic
<lang Yabasic>// Rosetta Code problem: http://rosettacode.org/wiki/Steady_squares // by Galileo, 04/2022
for i = 1 to 10000
r$ = str$(i^2, "%9.f") if i = val(right$(r$, len(str$(i)))) print i, "\t=", r$
next</lang>
- Output:
1 = 1 5 = 25 6 = 36 25 = 625 76 = 5776 376 = 141376 625 = 390625 9376 = 87909376 ---Program done, press RETURN---