Truncatable primes
A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number; for example, the number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime. The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime. No zeroes are allowed in truncatable primes.
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
- C.f
Ada
<lang Ada> with Ada.Text_IO; use Ada.Text_IO; with Ada.Containers.Ordered_Sets;
procedure Truncatable_Primes is
package Natural_Set is new Ada.Containers.Ordered_Sets (Natural); use Natural_Set;
Primes : Set; function Is_Prime (N : Natural) return Boolean is Position : Cursor := First (Primes); begin while Has_Element (Position) loop if N mod Element (Position) = 0 then return False; end if; Position := Next (Position); end loop; return True; end Is_Prime;
function Is_Left_Trucatable_Prime (N : Positive) return Boolean is M : Natural := 1; begin while Contains (Primes, N mod (M * 10)) and (N / M) mod 10 > 0 loop M := M * 10; if N <= M then return True; end if; end loop; return False; end Is_Left_Trucatable_Prime;
function Is_Right_Trucatable_Prime (N : Positive) return Boolean is M : Natural := N; begin while Contains (Primes, M) and M mod 10 > 0 loop M := M / 10; if M <= 1 then return True; end if; end loop; return False; end Is_Right_Trucatable_Prime;
Position : Cursor;
begin
for N in 2..1_000_000 loop if Is_Prime (N) then Insert (Primes, N); end if; end loop; Position := Last (Primes); while Has_Element (Position) loop if Is_Left_Trucatable_Prime (Element (Position)) then Put_Line ("Largest LTP from 1..1000000:" & Integer'Image (Element (Position))); exit; end if; Previous (Position); end loop; Position := Last (Primes); while Has_Element (Position) loop if Is_Right_Trucatable_Prime (Element (Position)) then Put_Line ("Largest RTP from 1..1000000:" & Integer'Image (Element (Position))); exit; end if; Previous (Position); end loop;
end Truncatable_Primes; </lang> Sample output:
Largest LTP from 1..1000000: 998443 Largest RTP from 1..1000000: 739399
ALGOL 68
Note: This specimen retains the original C coding style.
<lang algol68>#!/usr/local/bin/a68g --script #
PROC is prime = (INT n)BOOL:(
[]BOOL is short prime=(FALSE, TRUE, TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, FALSE); IF n<=UPB is short prime THEN is short prime[n] # EXIT # ELSE IF ( NOT ODD n | TRUE | n MOD 3 = 0 ) THEN FALSE # EXIT # ELSE INT h := ENTIER sqrt(n)+3; FOR a FROM 7 BY 6 WHILE a<h DO IF ( n MOD a = 0 | TRUE | n MOD (a-2) = 0 ) THEN false exit FI OD; TRUE # EXIT # FI FI EXIT false exit: FALSE
);
PROC string to int = (STRING in a)INT:(
FILE f; STRING a := in a; associate(f, a); INT i; get(f, i); close(f); i
);
PROC is trunc prime = (INT in n, PROC(REF STRING)VOID trunc)BOOL: (
INT n := in n; STRING s := whole(n, 0); IF char in string("0", NIL, s) THEN FALSE # EXIT # ELSE WHILE is prime(n) DO s := whole(n, 0); trunc(s); IF UPB s = 0 THEN true exit FI; n := string to int(s) OD; FALSE EXIT true exit: TRUE FI
);
PROC get trunc prime = (INT in n, PROC(REF STRING)VOID trunc)VOID:(
FOR n FROM in n BY -1 TO 1 DO IF is trunc prime(n, trunc) THEN printf(($g(0)l$, n)); break FI OD; break: ~
);
main:(
INT limit = 1000000; printf(($g g(0) gl$,"Highest left- and right-truncatable primes under ",limit,":")); get trunc prime(limit, (REF STRING s)VOID: s := s[LWB s+1:]); get trunc prime(limit, (REF STRING s)VOID: s := s[:UPB s-1]); write("Press Enter"); read(newline)
)</lang> Output:
Highest left- and right-truncatable primes under 1000000: 998443 739399 Press Enter
AutoHotkey
<lang AutoHotkey>SetBatchLines, -1 MsgBox, % "Largest left-truncatable and right-truncatable primes less than one million:`n" . "Left:`t" LTP(10 ** 6) "`nRight:`t" RTP(10 ** 6)
LTP(n) { while n { n-- if (!Instr(n, "0") && IsPrime(n)) { Loop, % StrLen(n) if (!IsPrime(SubStr(n, A_Index))) continue, 2 break } } return, n }
RTP(n) { while n { n-- if (!IsPrime(SubStr(n, 1, 1))) n -= 10 ** (StrLen(n) - 1) if (!Instr(n, "0") && IsPrime(n)) { Loop, % StrLen(n) if (!IsPrime(SubStr(n, 1, A_Index))) continue, 2 break } } return, n }
IsPrime(n) { if (n < 2) return, 0 else if (n < 4) return, 1 else if (!Mod(n, 2)) return, 0 else if (n < 9) return 1 else if (!Mod(n, 3)) return, 0 else { r := Floor(Sqrt(n)) f := 5 while (f <= r) { if (!Mod(n, f)) return, 0 if (!Mod(n, (f + 2))) return, 0 f += 6 } return, 1 } }</lang> Output:
Largest left-truncatable and right-truncatable primes less than one million: Left: 998443 Right: 739399
Bracmat
Primality test: In an attempt to compute the result of taking a (not too big, 2^32 or 2^64, depending on word size) number to a fractional power, Bracmat computes the prime factors of the number and checks whether the powers of prime factors make the fractional power go away. If the number is prime, the output of the computation is the same as the input. <lang bracmat>( 1000001:?i & whl
' ( !i+-2:>0:?i & !i:?L & whl'(!L^1/2:#?^1/2&@(!L:% ?L)) & !L:~ )
& out$("left:" !i) & 1000001:?i & whl
' ( !i+-2:>0:?i & !i:?R & whl'(!R^1/2:#?^1/2&@(!R:?R %@)) & !R:~ )
& out$("right:" !i) )</lang> Output:
left: 998443 right: 739399
C
<lang C>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
- define MAX_PRIME 1000000
char *primes; int n_primes;
/* Sieve. If we were to handle 10^9 range, use bit field. Regardless,
* if a large amount of prime numbers need to be tested, sieve is fast. */
void init_primes() { int j; primes = malloc(sizeof(char) * MAX_PRIME); memset(primes, 1, MAX_PRIME); primes[0] = primes[1] = 0; int i = 2; while (i * i < MAX_PRIME) { for (j = i * 2; j < MAX_PRIME; j += i) primes[j] = 0; while (++i < MAX_PRIME && !primes[i]); } }
int left_trunc(int n) { int tens = 1; while (tens < n) tens *= 10;
while (n) { if (!primes[n]) return 0; tens /= 10; if (n < tens) return 0; n %= tens; } return 1; }
int right_trunc(int n) { while (n) { if (!primes[n]) return 0; n /= 10; } return 1; }
int main() { int n; int max_left = 0, max_right = 0; init_primes();
for (n = MAX_PRIME - 1; !max_left; n -= 2) if (left_trunc(n)) max_left = n;
for (n = MAX_PRIME - 1; !max_right; n -= 2) if (right_trunc(n)) max_right = n;
printf("Left: %d; right: %d\n", max_left, max_right); return 0; }</lang>output<lang>Left: 998443; right: 739399</lang>
Faster way of doing primality test for small numbers (1000000 isn't big), and generating truncatable primes bottom-up: <lang c>#include <stdio.h>
- define MAXN 1000000
int maxl, maxr;
int is_prime(int n) { int p; if (n % 3 == 0) return 0;
for (p = 6; p * p <= n; p += 6) if (!(n % (p + 1) && n % (p + 5))) return 0; return 1; }
void left(int n, int tens) { int i, nn;
if (n > maxl) maxl = n; if (n < MAXN / 10) for (tens *= 10, i = 1; i < 10; i++) if (is_prime(nn = i * tens + n)) left(nn, tens); }
void right(int n) { int i, nn; static int d[] = {1,3,7,9};
if (n > maxr) maxr = n; if (n < MAXN / 10) for (i = 1; i < 4; i++) if (is_prime(nn = n * 10 + d[i])) right(nn); }
int main(void) { left(3, 1); left(7, 1); right(3); right(5); right(7);
printf("%d %d\n", maxl, maxr);
return 0; }</lang>
- Output:
998443 739399
C#
<lang csharp>using System; using System.Diagnostics;
namespace RosettaCode {
internal class Program { public static bool IsPrime(int n) { if (n<2) return false; if (n<4) return true; if (n%2==0) return false; if (n<9) return true; if (n%3==0) return false; var r = (int) Math.Sqrt(n); var f = 6-1; while (f<=r) { if (n%f==0 ||n%(f+2)==0) return false; f += 6; } return true; }
private static bool IsRightTruncatable(int n) { for (;;) { n /= 10; if (n==0) return true; if (!IsPrime(n)) return false; } }
private static bool IsLeftTruncatable(int n) { string c = n.ToString(); if (c.Contains("0")) return false; for (int i = 1; i<c.Length; i++) if (!IsPrime(Convert.ToInt32(c.Substring(i)))) return false; return true; }
private static void Main() { var sb = new Stopwatch(); sb.Start(); int lt = 0, rt = 0; for (int i = 1000000; i>0; --i) { if (IsPrime(i)) { if (rt==0 && IsRightTruncatable(i)) rt = i; else if (lt==0 && IsLeftTruncatable(i)) lt = i; if (lt!=0 && rt!=0) break; } } sb.Stop(); Console.WriteLine("Largest truncable left is={0} & right={1}, calculated in {2} msec.", lt, rt, sb.ElapsedMilliseconds); } }
}</lang>
Largest truncable left is=998443 & right=739399, calculated in 16 msec.
Clojure
<lang Clojure>(use '[clojure.contrib.lazy-seqs :only [primes]])
(def prime?
(let [mem (ref #{})
primes (ref primes)]
(fn [n] (dosync (if (< n (first @primes))
(@mem n) (let [[mems ss] (split-with #(<= % n) @primes)] (ref-set primes ss) ((commute mem into mems) n)))))))
(defn drop-lefts [n]
(let [dropl #(if (< % 10) 0 (Integer. (subs (str %) 1)))] (->> (iterate dropl n)
(take-while pos? ,) next)))
(defn drop-rights [n]
(->> (iterate #(quot % 10) n) next (take-while pos? ,)))
(defn truncatable-left? [n]
(every? prime? (drop-lefts n)))
(defn truncatable-right? [n]
(every? prime? (drop-rights n)))
user> (->> (for [p primes :while (< p 1000000) :when (not-any? #{\0} (str p)) :let [l? (if (truncatable-left? p) p 0) r? (if (truncatable-right? p) p 0)] :when (or l? r?)]
[l? r?]) ((juxt #(apply max-key first %) #(apply max-key second %)) ,) ((juxt ffirst (comp second second)) ,) (map vector ["left truncatable: " "right truncatable: "] ,))
(["left truncatable: " 998443] ["right truncatable: " 739399])</lang>
CoffeeScript
<lang coffeescript># You could have symmetric algorithms for max right and left
- truncatable numbers, but they lend themselves to slightly
- different optimizations.
max_right_truncatable_number = (n, f) ->
# This algorithm only evaluates 37 numbers for primeness to # get the max right truncatable prime < 1000000. Its # optimization is that it prunes candidates for # the first n-1 digits before having to iterate through # the 10 possibilities for the last digit. if n < 10 candidate = n while candidate > 0 return candidate if f(candidate) candidate -= 1 else left = Math.floor n / 10 while left > 0 left = max_right_truncatable_number left, f right = 9 while right > 0 candidate = left * 10 + right return candidate if candidate <= n and f(candidate) right -= 1 left -= 1 throw Error "none found"
max_left_truncatable_number = (max, f) ->
# This is a pretty straightforward countdown. The first # optimization here would probably be to cache results of # calling f on small numbers. is_left_truncatable = (n) -> candidate = 0 power_of_ten = 1 while n > 0 r = n % 10 return false if r == 0 n = Math.floor n / 10 candidate = r * power_of_ten + candidate power_of_ten *= 10 return false unless f(candidate) true do -> n = max while n > 0 return n if is_left_truncatable n, f n -= 1 throw Error "none found"
is_prime = (n) ->
return false if n == 1 return true if n == 2 for d in [2..n] return false if n % d == 0 return true if d * d >= n
console.log "right", max_right_truncatable_number(999999, is_prime)
console.log "left", max_left_truncatable_number(999999, is_prime)
</lang>
output
<lang>
> coffee truncatable_prime.coffee
right 739399
left 998443
</lang>
D
<lang d>import std.stdio, std.math, std.string, std.conv, std.algorithm,
std.range;
bool isPrime(in int n) pure nothrow {
if (n <= 1) return false; foreach (immutable i; 2 .. cast(int)sqrt(real(n)) + 1) if (!(n % i)) return false; return true;
}
bool isTruncatablePrime(bool left)(in int n) pure {
immutable s = n.text; if (s.canFind('0')) return false; foreach (immutable i; 0 .. s.length) static if (left) { if (!s[i .. $].to!int.isPrime) return false; } else { if (!s[0 .. i + 1].to!int.isPrime) return false; } return true;
}
void main() {
enum n = 1_000_000; writeln("Largest left-truncatable prime in 2 .. ", n, ": ", iota(n, 1, -1).filter!(isTruncatablePrime!true).front); writeln("Largest right-truncatable prime in 2 .. ", n, ": ", iota(n, 1, -1).filter!(isTruncatablePrime!false).front);
}</lang>
- Output:
Largest left-truncatable prime in 2 .. 1000000: 998443 Largest right-truncatable prime in 2 .. 1000000: 739399
Fortran
<lang fortran>module primes_mod
implicit none logical, allocatable :: primes(:)
contains
subroutine Genprimes(parr)
logical, intent(in out) :: parr(:) integer :: i
! Prime sieve
parr = .true. parr (1) = .false. parr (4 : size(parr) : 2) = .false. do i = 3, int (sqrt (real (size(parr)))), 2 if (parr(i)) parr(i * i : size(parr) : i) = .false. end do
end subroutine
function is_rtp(candidate)
logical :: is_rtp integer, intent(in) :: candidate integer :: n
is_rtp = .true. n = candidate / 10 do while(n > 0) if(.not. primes(n)) then is_rtp = .false. return end if n = n / 10 end do
end function
function is_ltp(candidate)
logical :: is_ltp integer, intent(in) :: candidate integer :: i, n character(10) :: nstr
write(nstr, "(i10)") candidate is_ltp = .true. do i = len_trim(nstr)-1, 1, -1 n = mod(candidate, 10**i) if(.not. primes(n)) then is_ltp = .false. return end if end do
end function
end module primes_mod
program Truncatable_Primes
use primes_mod implicit none integer, parameter :: limit = 999999 integer :: i character(10) :: nstr
! Generate an array of prime flags up to limit of search
allocate(primes(limit)) call Genprimes(primes)
! Find left truncatable prime
do i = limit, 1, -1 write(nstr, "(i10)") i if(index(trim(nstr), "0") /= 0) cycle ! check for 0 in number if(is_ltp(i)) then write(*, "(a, i0)") "Largest left truncatable prime below 1000000 is ", i exit end if end do
! Find right truncatable prime
do i = limit, 1, -1 write(nstr, "(i10)") i if(index(trim(nstr), "0") /= 0) cycle ! check for 0 in number if(is_rtp(i)) then write(*, "(a, i0)") "Largest right truncatable prime below 1000000 is ", i exit end if end do
end program</lang> Output
Largest left truncatable prime below 1000000 is 998443 Largest right truncatable prime below 1000000 is 739399
Go
<lang go>package main
import "fmt"
func main() {
sieve(1e6) if !search(6, 1e6, "left", func(n, pot int) int { return n % pot }) { panic("997?") } if !search(6, 1e6, "right", func(n, _ int) int { return n / 10 }) { panic("7393?") }
}
var c []bool
func sieve(ss int) {
c = make([]bool, ss) c[1] = true for p := 2; ; { p2 := p * p if p2 >= ss { break } for i := p2; i < ss; i += p { c[i] = true } for { p++ if !c[p] { break } } }
}
func search(digits, pot int, s string, truncFunc func(n, pot int) int) bool {
n := pot - 1 pot /= 10
smaller:
for ; n >= pot; n -= 2 { for tn, tp := n, pot; tp > 0; tp /= 10 { if tn < tp || c[tn] { continue smaller } tn = truncFunc(tn, tp) } fmt.Println("max", s, "truncatable:", n) return true } if digits > 1 { return search(digits-1, pot, s, truncFunc) } return false
}</lang> Output:
max left truncatable: 998443 max right truncatable: 739399
Haskell
Using
from HackageDB
<lang haskell>import Data.Numbers.Primes(primes, isPrime) import Data.List import Control.Arrow
primes1e6 = reverse. filter (notElem '0'. show) $ takeWhile(<=1000000) primes
rightT, leftT :: Int -> Bool rightT = all isPrime. takeWhile(>0). drop 1. iterate (`div`10) leftT x = all isPrime. takeWhile(<x).map (x`mod`) $ iterate (*10) 10
main = do
let (ltp, rtp) = (head. filter leftT &&& head. filter rightT) primes1e6 putStrLn $ "Left truncatable " ++ show ltp putStrLn $ "Right truncatable " ++ show rtp</lang>
Output: <lang haskell>*Main> main Left truncatable 998443 Right truncatable 739399</lang>
Interpretation of the J contribution: <lang haskell>digits = [1..9] :: [Integer] smallPrimes = filter isPrime digits pow10 = iterate (*10) 1 mul10 = (pow10!!). length. show righT = (+) . (10 *) lefT = liftM2 (.) (+) ((*) . mul10)
primesTruncatable f = iterate (concatMap (filter isPrime.flip map digits. f)) smallPrimes</lang> Output: <lang haskell>*Main> maximum $ primesTruncatable righT !! 5 739399
- Main> maximum $ primesTruncatable lefT !! 5
998443</lang>
Icon and Unicon
<lang Icon>procedure main(arglist)
N := 0 < integer(\arglist[1]) | 1000000 # primes to generator 1 to ... (1M or 1st arglist) D := (0 < integer(\arglist[2]) | 10) / 2 # primes to display (10 or 2nd arglist) P := sieve(N) # from sieve task (modified) write("There are ",*P," prime numbers in the range 1 to ",N) if *P <= 2*D then every writes( "Primes: "|!sort(P)||" "|"\n" ) else every writes( "Primes: "|(L := sort(P))[1 to D]||" "|"... "|L[*L-D+1 to *L]||" "|"\n" ) largesttruncateable(P)
end
procedure largesttruncateable(P) #: find the largest left and right trucatable numbers in P local ltp,rtp
every x := sort(P)[*P to 1 by -1] do # largest to smallest if not find('0',x) then { /ltp := islefttrunc(P,x) /rtp := isrighttrunc(P,x) if \ltp & \rtp then break # until both found } write("Largest left truncatable prime = ", ltp) write("Largest right truncatable prime = ", rtp) return
end
procedure isrighttrunc(P,x) #: return integer x if x and all right truncations of x are in P or fails if x = 0 | (member(P,x) & isrighttrunc(P,x / 10)) then return x end
procedure islefttrunc(P,x) #: return integer x if x and all left truncations of x are in P or fails if *x = 0 | ( (x := integer(x)) & member(P,x) & islefttrunc(P,x[2:0]) ) then return x end</lang>
Sample output:
There are 78498 prime numbers in the range 1 to 1000000 Primes: 2 3 5 7 11 ... 999953 999959 999961 999979 999983 Largest left truncatable prime = 998443 Largest right truncatable prime = 739399
J
Truncatable primes may be constructed by starting with a set of one digit prime numbers and then repeatedly adding a non-zero digit (using the cartesian product of digit sequences) and, at each step, selecting the prime numbers which result.
In other words, given:
<lang j>selPrime=: #~ 1&p: seed=: selPrime digits=: 1+i.9 step=: selPrime@,@:(,&.":/&>)@{@;</lang>
The largest truncatable primes less than a million can be obtained by adding five digits to the prime seeds, then finding the largest value from the result:
<lang j> >./ digits&step^:5 seed NB. left truncatable 998443
>./ step&digits^:5 seed NB. right truncatable
739399</lang>
Java
<lang Java> import java.util.BitSet;
public class Main {
public static void main(String[] args){
final int MAX = 1000000;
//Sieve of Eratosthenes (using BitSet only for odd numbers) BitSet primeList = new BitSet(MAX>>1); primeList.set(0,primeList.size(),true);
int sqroot = (int) Math.sqrt(MAX); primeList.clear(0); for(int num = 3; num <= sqroot; num+=2) { if( primeList.get(num >> 1) ) { int inc = num << 1; for(int factor = num * num; factor < MAX; factor += inc) { //if( ((factor) & 1) == 1) //{ primeList.clear(factor >> 1); //} } } } //Sieve ends...
//Find Largest Truncatable Prime. (so we start from 1000000 - 1 int rightTrunc = -1, leftTrunc = -1; for(int prime = (MAX - 1) | 1; prime >= 3; prime -= 2) { if(primeList.get(prime>>1)) { //Already found Right Truncatable Prime? if(rightTrunc == -1) { int right = prime; while(right > 0 && primeList.get(right >> 1)) right /= 10; if(right == 0) rightTrunc = prime; }
//Already found Left Truncatable Prime? if(leftTrunc == -1 ) { //Left Truncation String left = Integer.toString(prime); if(!left.contains("0")) { while( left.length() > 0 ){ int iLeft = Integer.parseInt(left); if(!primeList.get( iLeft >> 1)) break; left = left.substring(1); } if(left.length() == 0) leftTrunc = prime; } } if(leftTrunc != -1 && rightTrunc != -1) //Found both? then Stop loop { break; } } } System.out.println("Left Truncatable : " + leftTrunc); System.out.println("Right Truncatable : " + rightTrunc); } } </lang> Output :
Left Truncatable : 998443 Right Truncatable : 739399
Lua
<lang lua>max_number = 1000000
numbers = {} for i = 2, max_number do
numbers[i] = i;
end
for i = 2, max_number do
for j = i+1, max_number do if numbers[j] ~= 0 and j % i == 0 then numbers[j] = 0 end end
end
max_prime_left, max_prime_right = 2, 2 for i = 2, max_number do
if numbers[i] ~= 0 then local is_prime = true local l = math.floor( i / 10 ) while l > 1 do if numbers[l] == 0 then is_prime = false break end l = math.floor( l / 10 ) end if is_prime then max_prime_left = i end is_prime = true local n = 10; while math.floor( i % 10 ) ~= 0 and n < max_number do if numbers[ math.floor( i % 10 ) ] ~= 0 then is_prime = false break end n = n * 10 end if is_prime then max_prime_right = i end end
end
print( "max_prime_left = ", max_prime_left ) print( "max_prime_right = ", max_prime_right )</lang>
Mathematica
<lang Mathematica>LeftTruncatablePrimeQ[n_] := Times @@ IntegerDigits[n] > 0 &&
And @@ PrimeQ /@ ToExpression /@ StringJoin /@ Rest[Most[NestList[Rest, #, Length[#]] &[Characters[ToString[n]]]]]
RightTruncatablePrimeQ[n_] := Times @@ IntegerDigits[n] > 0 &&
And @@ PrimeQ /@ ToExpression /@ StringJoin /@ Rest[Most[NestList[Most, #, Length[#]] &[Characters[ToString[n]]]]]</lang>
Example usage:
n = PrimePi[1000000]; While[Not[LeftTruncatablePrimeQ[Prime[n]]], n--]; Prime[n] -> 998443 n = PrimePi[1000000]; While[Not[RightTruncatablePrimeQ[Prime[n]]], n--]; Prime[n] -> 739399
MATLAB
largestTruncatablePrimes.m: <lang MATLAB>function largestTruncatablePrimes(boundary)
%Helper function for checking if a prime is left of right truncatable function [leftTruncatable,rightTruncatable] = isTruncatable(prime,checkLeftTruncatable,checkRightTruncatable)
numDigits = ceil(log10(prime)); %calculate the number of digits in the prime less one powersOfTen = 10.^(0:numDigits); %cache the needed powers of ten leftTruncated = mod(prime,powersOfTen); %generate a list of numbers by repeatedly left truncating the prime %leading zeros will cause duplicate entries thus it is possible to %detect leading zeros if we rotate the list to the left or right %and check for any equivalences with the original list hasLeadingZeros = any( circshift(leftTruncated,[0 1]) == leftTruncated ); if( hasLeadingZeros || not(checkLeftTruncatable) ) leftTruncatable = false; else %check if all of the left truncated numbers are prime leftTruncatable = all(isprime(leftTruncated(2:end))); end
if( checkRightTruncatable ) rightTruncated = (prime - leftTruncated) ./ powersOfTen; %generate a list of right truncated numbers rightTruncatable = all(isprime(rightTruncated(1:end-1))); %check if all the right truncated numbers are prime else rightTruncatable = false; end
end %isTruncatable()
nums = primes(boundary); %generate all primes <= boundary
%Flags that indicate if the largest left or right truncatable prime has not %been found leftTruncateNotFound = true; rightTruncateNotFound = true;
for prime = nums(end:-1:1) %Search through primes in reverse order
%Get if the prime is left and/or right truncatable, ignoring %checking for right truncatable if it has already been found [leftTruncatable,rightTruncatable] = isTruncatable(prime,leftTruncateNotFound,rightTruncateNotFound);
if( leftTruncateNotFound && leftTruncatable ) %print out largest left truncatable prime display([num2str(prime) ' is the largest left truncatable prime <= ' num2str(boundary) '.']); leftTruncateNotFound = false; end
if( rightTruncateNotFound && rightTruncatable ) %print out largest right truncatable prime display([num2str(prime) ' is the largest right truncatable prime <= ' num2str(boundary) '.']); rightTruncateNotFound = false; end
%Terminate loop when the largest left and right truncatable primes have %been found if( not(leftTruncateNotFound || rightTruncateNotFound) ) break; end end
end </lang> Solution for n = 1,000,000: <lang MATLAB> >> largestTruncatablePrimes(1e6) 998443 is the largest left truncatable prime <= 1000000. 739399 is the largest right truncatable prime <= 1000000. </lang>
Nimrod
<lang nimrod>import sets, strutils, algorithm
proc primes(n): seq[int] =
result = @[] var multiples = initSet[int]() for i in 2..n: if i notin multiples: result.add i for j in countup(i*i, n, i): multiples.incl j
proc truncatablePrime(n): tuple[left: int, right: int] =
var primelist: seq[string] = @[] for x in primes(n): primelist.add($x) reverse primelist var primeset = toSet primelist for n in primelist: var alltruncs = initSet[string]() for i in 0..n.len: alltruncs.incl n[1..n.high] if alltruncs <= primeset: result.left = parseInt(n) break for n in primelist: var alltruncs = initSet[string]() for i in 0..n.len: alltruncs.incl n[0..i] if alltruncs <= primeset: result.right = parseInt(n) break
echo truncatablePrime(1000000)</lang> Output:
(left: 999961, right: 739399)
ooRexx
<lang ooRexx> -- find largest left- & right-truncatable primes < 1 million. -- an initial set of primes (not, at this time, we leave out 2 because -- we'll automatically skip the even numbers. No point in doing a needless -- test each time through primes = .array~of(3, 5, 7, 11)
-- check all of the odd numbers up to 1,000,000 loop j = 13 by 2 to 1000000
loop i = 1 to primes~size prime = primes[i] -- found an even prime divisor if j // prime == 0 then iterate j -- only check up to the square root if prime*prime > j then leave end -- we only get here if we don't find a divisor primes~append(j)
end
-- get a set of the primes that we can test more efficiently primeSet = .set~of(2) primeSet~putall(primes)
say 'The last prime is' primes[primes~last] "("primeSet~items 'primes under one million).'
say copies('-',66)
lastLeft = 0
-- we're going to use the array version to do these in order. We're still -- missing "2", but that's not going to be the largest loop prime over primes
-- values containing 0 can never work if prime~pos(0) \= 0 then iterate -- now start the truncations, checking against our set of -- known primes loop i = 1 for prime~length - 1 subprime = prime~right(i) -- not in our known set, this can't work if \primeset~hasIndex(subprime) then iterate prime end -- this, by definition, with be the largest left-trunc prime lastLeft = prime
end -- now look for right-trunc primes lastRight = 0 loop prime over primes
-- values containing 0 can never work if prime~pos(0) \= 0 then iterate -- now start the truncations, checking against our set of -- known primes loop i = 1 for prime~length - 1 subprime = prime~left(i) -- not in our known set, this can't work if \primeset~hasIndex(subprime) then iterate prime end -- this, by definition, with be the largest left-trunc prime lastRight = prime
end
say 'The largest left-truncatable prime is' lastLeft '(under one million).' say 'The largest right-truncatable prime is' lastRight '(under one million).'
</lang> Output:
The last prime is 999983 (78498 primes under one million). ------------------------------------------------------------------ The largest left-truncatable prime is 998443 (under one million). The largest right-truncatable prime is 739399 (under one million).
OpenEdge/Progress
<lang progress>FUNCTION isPrime RETURNS LOGICAL (
i_i AS INT
):
DEF VAR ii AS INT.
DO ii = 2 TO SQRT( i_i ):
IF i_i MODULO ii = 0 THEN RETURN FALSE.
END.
RETURN TRUE AND i_i > 1.
END FUNCTION. /* isPrime */
FUNCTION isLeftTruncatablePrime RETURNS LOGICAL (
i_i AS INT
):
DEF VAR ii AS INT. DEF VAR cc AS CHAR. DEF VAR lresult AS LOGICAL INITIAL TRUE. cc = STRING( i_i ).
DO WHILE cc > "": lresult = lresult AND isPrime( INTEGER( cc ) ). cc = SUBSTRING( cc, 2 ). END.
RETURN lresult.
END FUNCTION. /* isLeftTruncatablePrime */
FUNCTION isRightTruncatablePrime RETURNS LOGICAL (
i_i AS INT
):
DEF VAR ii AS INT. DEF VAR cc AS CHAR. DEF VAR lresult AS LOGICAL INITIAL TRUE. cc = STRING( i_i ).
DO WHILE cc > "": lresult = lresult AND isPrime( INTEGER( cc ) ). cc = SUBSTRING( cc, 1, LENGTH( cc ) - 1 ). END.
RETURN lresult.
END FUNCTION. /* isRightTruncatablePrime */
FUNCTION getHighestTruncatablePrimes RETURNS CHARACTER (
i_imax AS INTEGER
):
DEF VAR ii AS INT. DEF VAR ileft AS INT. DEF VAR iright AS INT.
DO ii = i_imax TO 1 BY -1 WHILE ileft = 0 OR iright = 0:
IF INDEX( STRING( ii ), "0" ) = 0 THEN DO: IF ileft = 0 AND isLeftTruncatablePrime( ii ) THEN ileft = ii. IF iright = 0 AND isRightTruncatablePrime( ii ) THEN iright = ii. END.
END.
RETURN SUBSTITUTE("Left: &1~nRight: &2", ileft, iright ).
END FUNCTION. /* getHighestTruncatablePrimes */
MESSAGE
getHighestTruncatablePrimes( 1000000 )
VIEW-AS ALERT-BOX.
</lang>
Output:
--------------------------- Message --------------------------- Left: 998443 Right: 739399 --------------------------- OK ---------------------------
PARI/GP
This version builds the truncatable primes with up to k digits in a straightforward fashion. Run time is about 15 milliseconds, almost all of which is I/O. <lang parigp>left(n)={ my(v=[2,3,5,7],u,t=1,out=0); for(i=1,n, t*=10; u=[]; for(j=1,#v, forstep(a=t,t*9,t, if(isprime(a+v[j]),u=concat(u,a+v[j])) ) ); out=v[#v]; v=vecsort(u) ); out }; right(n)={ my(v=[2,3,5,7],u,out=0); for(i=1,n, u=[]; for(j=1,#v, forstep(a=1,9,[2,4], if(isprime(10*v[j]+a),u=concat(u,10*v[j]+a)) ) ); out=v[#v]; v=u ); out }; [left(6),right(6)]</lang>
Perl
<lang perl>#!/usr/bin/perl use warnings; use strict;
use constant {
LEFT => 0, RIGHT => 1,
};
{ my @primes = (2, 3);
sub is_prime { my $n = shift; return if $n < 2;
for my $prime (@primes) { last if $prime >= $n; return unless $n % $prime; }
my $sqrt = sqrt $n; while ($primes[-1] < $sqrt) { my $new = 2 + $primes[-1]; $new += 2 until is_prime($new); push @primes, $new; return unless $n % $new; }
return 1; }
}
sub trunc {
my ($n, $side) = @_; substr $n, $side == LEFT ? 0 : -1, 1, q(); return $n;
}
sub is_tprime { # Absence of zeroes is tested outside the sub.
my ($n, $side) = @_; return (is_prime($n) and (1 == length $n or is_tprime(trunc($n, $side), $side)));
}
my $length = 6;
my @tprimes = ('9' x $length) x 2;
for my $side (LEFT, RIGHT) {
$tprimes[$side] -= 2 until -1 == index $tprimes[$side], '0' and is_tprime($tprimes[$side], $side);
}
print 'left ', join(', right ', @tprimes), "\n";</lang>
- Output:
left 998443, right 739399
Perl 6
<lang perl6>my @primes := 2, 3, 5, -> $p { ($p+2, $p+4 ... &prime)[*-1] } ... *; my @isprime = False,False; # 0 and 1 are not primes by definition sub prime($i) { @isprime[$i] //= ($i %% none @primes ...^ * > $_ given $i.sqrt.floor) }
sub ltp {
for 9...1 -> $a { for 9...1 -> $b { for 9...1 -> $c { for 9...1 -> $d { for 9...1 -> $e { for 9,7,3 -> $f { my @x := [\+] $f, $e, $d, $c, $b, $a Z* (1,10,100 ... *); return @x[*-1] if not grep {not prime $^n}, @x; } } } } } }
}
sub infix:<*+> ($a,$b) { $a * 10 + $b }
sub rtp {
for 7,5,3 {
for grep &prime, ($_ X*+ 9,7,3,1) { for grep &prime, ($_ X*+ 9,7,3,1) { for grep &prime, ($_ X*+ 9,7,3,1) { for grep &prime, ($_ X*+ 9,7,3,1) { for grep &prime, ($_ X*+ 9,7,3,1) { return $_; } } } } }
}
}
say "Highest ltp: ", ltp; say "Highest rtp: ", rtp;</lang>
- Output:
Highest ltp: 998443 Highest rtp: 739399
PicoLisp
<lang PicoLisp>(load "@lib/rsa.l") # Use the 'prime?' function from RSA package
(de truncatablePrime? (N Fun)
(for (L (chop N) L (Fun L)) (T (= "0" (car L))) (NIL (prime? (format L))) T ) )
(let (Left 1000000 Right 1000000)
(until (truncatablePrime? (dec 'Left) cdr)) (until (truncatablePrime? (dec 'Right) '((L) (cdr (rot L))))) (cons Left Right) )</lang>
Output:
-> (998443 . 739399)
PL/I
<lang PL/I> tp: procedure options (main);
declare primes(1000000) bit (1); declare max_primes fixed binary (31); declare (i, k) fixed binary (31);
max_primes = hbound(primes, 1); call sieve;
/* Now search for primes that are right-truncatable. */ call right_truncatable;
/* Now search for primes that are left-truncatable. */ call left_truncatable;
right_truncatable: procedure;
declare direction bit (1); declare (i, k) fixed binary (31);
test_truncatable:
do i = max_primes to 2 by -1; if primes(i) then /* it's a prime */ do; k = i/10; do while (k > 0); if ^primes(k) then iterate test_truncatable; k = k/10; end; put skip list (i || ' is right-truncatable'); return; end; end;
end right_truncatable;
left_truncatable: procedure;
declare direction bit (1); declare (i, k, d, e) fixed binary (31);
test_truncatable:
do i = max_primes to 2 by -1; if primes(i) then /* it's a prime */ do; k = i; do d = 100000 repeat d/10 until (d = 10); e = k/d; k = k - e*d; if e = 0 then iterate test_truncatable; if ^primes(k) then iterate test_truncatable; end; put skip list (i || ' is left-truncatable'); return; end; end;
end left_truncatable;
sieve: procedure;
declare (i, j) fixed binary (31);
primes = '1'b; primes(1) = '0'b;
do i = 2 to sqrt(max_primes); do j = i+i to max_primes by i; primes(j) = '0'b; end; end;
end sieve;
end tp; </lang>
739399 is right-truncatable 998443 is left-truncatable
PowerShell
<lang PowerShell>function IsPrime ( [int] $num ) {
$isprime = @{} 2..[math]::sqrt($num) | Where-Object { $isprime[$_] -eq $null } | ForEach-Object { $_ $isprime[$_] = $true for ( $i=$_*$_ ; $i -le $num; $i += $_ ) { $isprime[$i] = $false } } 2..$num | Where-Object { $isprime[$_] -eq $null }
}
function Truncatable ( [int] $num ) {
$declen = [math]::abs($num).ToString().Length $primes = @() $ltprimes = @{} $rtprimes = @{} 1..$declen | ForEach-Object { $ltprimes[$_]=@{}; $rtprimes[$_]=@{} } IsPrime $num | ForEach-Object { $lastltprime = 2 $lastrtprime = 2 } { $curprim = $_ $curdeclen = $curprim.ToString().Length $primes += $curprim if( $curdeclen -eq 1 ) { $ltprimes[1][$curprim] = $true $rtprimes[1][$curprim] = $true $lastltprime = $curprim $lastrtprime = $curprim } else { $curmod = $curprim % [math]::pow(10,$curdeclen - 1) $curdiv = [math]::floor($curprim / 10) if( $ltprimes[$curdeclen - 1][[int]$curmod] ) { $ltprimes[$curdeclen][$curprim] = $true $lastltprime = $curprim } if( $rtprimes[$curdeclen - 1][[int]$curdiv] ) { $rtprimes[$curdeclen][$curprim] = $true $lastrtprime = $curprim } } if( ( $ltprimes[$curdeclen - 2].Keys.count -gt 0 ) -and ( $ltprimes[$curdeclen - 1].Keys.count -gt 0 ) ) { $ltprimes[$curdeclen -2] = @{} } if( ( $rtprimes[$curdeclen - 2].Keys.count -gt 0 ) -and ( $rtprimes[$curdeclen - 1].Keys.count -gt 0 ) ) { $rtprimes[$curdeclen -2] = @{} } } { "Largest Left Truncatable Prime: $lastltprime" "Largest Right Truncatable Prime: $lastrtprime" }
}</lang>
PureBasic
<lang PureBasic>#MaxLim = 999999
Procedure is_Prime(n)
If n<=1 : ProcedureReturn #False ElseIf n<4 : ProcedureReturn #True ElseIf n%2=0: ProcedureReturn #False ElseIf n<9 : ProcedureReturn #True ElseIf n%3=0: ProcedureReturn #False Else Protected r=Round(Sqr(n),#PB_Round_Down) Protected f=5 While f<=r If n%f=0 Or n%(f+2)=0 ProcedureReturn #False EndIf f+6 Wend EndIf ProcedureReturn #True
EndProcedure
Procedure TruncateLeft(n)
Protected s.s=Str(n), l=Len(s)-1 If Not FindString(s,"0",1) While l>0 s=Right(s,l) If Not is_Prime(Val(s)) ProcedureReturn #False EndIf l-1 Wend ProcedureReturn #True EndIf
EndProcedure
Procedure TruncateRight(a)
Repeat a/10 If Not a Break ElseIf Not is_Prime(a) Or a%10=0 ProcedureReturn #False EndIf ForEver ProcedureReturn #True
EndProcedure
i=#MaxLim Repeat
If is_Prime(i) If Not truncateleft And TruncateLeft(i) truncateleft=i EndIf If Not truncateright And TruncateRight(i) truncateright=i EndIf EndIf If truncateleft And truncateright Break Else i-2 EndIf
Until i<=0
x.s="Largest TruncateLeft= "+Str(truncateleft) y.s="Largest TruncateRight= "+Str(truncateright)
MessageRequester("Truncatable primes",x+#CRLF$+y)</lang>
Python
<lang python>maxprime = 1000000
def primes(n):
multiples = set() prime = [] for i in range(2, n+1): if i not in multiples: prime.append(i) multiples.update(set(range(i*i, n+1, i))) return prime
def truncatableprime(n):
'Return a longest left and right truncatable primes below n' primelist = [str(x) for x in primes(n)[::-1]] primeset = set(primelist) for n in primelist: # n = 'abc'; [n[i:] for i in range(len(n))] -> ['abc', 'bc', 'c'] alltruncs = set(n[i:] for i in range(len(n))) if alltruncs.issubset(primeset): truncateleft = int(n) break for n in primelist: # n = 'abc'; [n[:i+1] for i in range(len(n))] -> ['a', 'ab', 'abc'] alltruncs = set([n[:i+1] for i in range(len(n))]) if alltruncs.issubset(primeset): truncateright = int(n) break return truncateleft, truncateright
print(truncatableprime(maxprime))</lang>
Sample Output
(998443, 739399)
Racket
<lang racket>
- lang racket
(require math/number-theory)
(define (truncate-right n)
(quotient n 10))
(define (truncate-left n)
(define s (number->string n)) (string->number (substring s 1 (string-length s))))
(define (contains-zero? n)
(member #\0 (string->list (number->string n))))
(define (truncatable? truncate n)
(and (prime? n) (not (contains-zero? n)) (or (< n 10) (truncatable? truncate (truncate n)))))
- largest left truncatable prime
(for/first ([n (in-range 1000000 1 -1)]
#:when (truncatable? truncate-left n)) n)
- largest right truncatable prime
(for/first ([n (in-range 1000000 1 -1)]
#:when (truncatable? truncate-right n)) n)
- Output
998443 739399 </lang>
REXX
Extra code was added to the prime number generator as this is the section of the REXX program that consumes the vast majority of the time. <lang REXX>/*REXX pgm finds largest left- & right-truncatable primes ≤1m (or arg1).*/ parse arg high .; if high== then high=1000000 /*assume 1 million.*/ !.=0; Lp=0; Rp=0; w=length(high) /*placeholders for primes, Lp, Rp*/ @.1=2; @.2=3; @.3=5; @.4=7; @.5=11; @.6=13; @.7=17 /*some low primes. */ !.2=1; !.3=1; !.5=1; !.7=1; !.11=1; !.13=1; !.17=1 /*low prime flags. */
- =7; s.#=@.#**2 /*number of primes so far, prime²*/
/*───────────────────────────────────────generate more primes ≤ high. */
do j=@.#+2 by 2 to high /*only find odd primes from here.*/ if j//3 ==0 then iterate /*is J divisible by three? */ if right(j,1)==5 then iterate /*is the right-most digit a "5" ?*/ if j//7 ==0 then iterate /*is J divisible by seven? */ if j//11 ==0 then iterate /*is J divisible by eleven? */ if j//13 ==0 then iterate /*is J divisible by thirteen? */ /*[↑] above five lines saves time*/ do k=7 while s.k<=j /*divide by known odd primes. */ if j//@.k==0 then iterate j /*Is J divisible by X? ¬ prime.*/ end /*k*/ #=#+1 /*bump number of primes found. */ @.#=j; s.#=j*j /*assign to sparse array; prime².*/ !.j=1 /*indicate that J is a prime.*/ end /*j*/
/*─────────────────────────────────────find largest left truncatable P. */
do L=# by -1 for #; if pos(0,@.L)\==0 then iterate do k=1 for length(@.L)-1; _=right(@.L,k) /*L truncate #.*/ if \!._ then iterate L /*Truncated # ¬prime? Skip it.*/ end /*k*/ leave /*leave the DO loop, we found one*/ end /*L*/
/*─────────────────────────────────────find largest right truncatable P.*/
do R=# by -1 for #; if pos(0,@.R)\==0 then iterate do k=1 for length(@.R)-1; _=left(@.R,k) /*R truncate #.*/ if \!._ then iterate R /*Truncated # ¬prime? Skip it.*/ end /*k*/ leave /*leave the DO loop, we found one*/ end /*R*/
/*───────────────────────────────────────show largest left/right trunc P*/ say 'The last prime found is ' @.# " (there are" # 'primes ≤' high")." say copies('─',70) /*show a separator line. */ say 'The largest left-truncatable prime ≤' high " is " right(@.L,w) say 'The largest right-truncatable prime ≤' high " is " right(@.R,w)
/*stick a fork in it, we're done.*/</lang>
output
The last prime found is 999983 (there are 78498 primes ≤ 1000000). ────────────────────────────────────────────────────────────────────── The largest left-truncatable prime ≤ 1000000 is 998443 The largest right-truncatable prime ≤ 1000000 is 739399
Ruby
<lang ruby>def left_truncatable?(n)
truncatable?(n) {|i| i.to_s[1..-1].to_i}
end
def right_truncatable?(n)
truncatable?(n) {|i| i/10}
end
def truncatable?(n, &trunc_func)
return false if n.to_s.include? "0" loop do n = trunc_func.call(n) return true if n.zero? return false unless Prime.prime?(n) end
end
require 'prime' primes = Prime.each(1_000_000).to_a.reverse
p primes.detect {|p| left_truncatable? p} p primes.detect {|p| right_truncatable? p}</lang>
returns
998443 739399
An Alternative Approach
Setting BASE to 10 and MAX to 6 in the Ruby example here Produces:
The largest left truncatable prime less than 1000000 in base 10 is 998443
Tcl
<lang tcl>package require Tcl 8.5
- Optimized version of the Sieve-of-Eratosthenes task solution
proc sieve n {
set primes [list] if {$n < 2} {return $primes} set nums [dict create] for {set i 2} {$i <= $n} {incr i} { dict set nums $i "" } set next 2 set limit [expr {sqrt($n)}] while {$next <= $limit} { for {set i $next} {$i <= $n} {incr i $next} {dict unset nums $i} lappend primes $next
dict for {next -} $nums break
} return [concat $primes [dict keys $nums]]
}
proc isLeftTruncatable n {
global isPrime while {[string length $n] > 0} {
if {![info exist isPrime($n)]} { return false } set n [string range $n 1 end]
} return true
} proc isRightTruncatable n {
global isPrime while {[string length $n] > 0} {
if {![info exist isPrime($n)]} { return false } set n [string range $n 0 end-1]
} return true
}
- Demo code
set limit 1000000 puts "calculating primes up to $limit" set primes [sieve $limit] puts "search space contains [llength $primes] members" foreach p $primes {
set isPrime($p) "yes"
} set primes [lreverse $primes]
puts "searching for largest left-truncatable prime" foreach p $primes {
if {[isLeftTruncatable $p]} {
puts FOUND:$p break
}
}
puts "searching for largest right-truncatable prime" foreach p $primes {
if {[isRightTruncatable $p]} {
puts FOUND:$p break
}
}</lang> Output:
calculating primes up to 1000000 search space contains 78498 members searching for largest left-truncatable prime FOUND:998443 searching for largest right-truncatable prime FOUND:739399
XPL0
<lang XPL0>code CrLf=9, IntOut=11;
func Prime(P); \Return true if P is a prime number int P; \(1 is not prime, but 2 is, etc.) int I; [if P<=1 then return false; \negative numbers are not prime for I:= 2 to sqrt(P) do
if rem(P/I) = 0 then return false;
return true; ];
func RightTrunc(N); \Return largest right-truncatable prime < one million int N; int M; [for N:= 1_000_000-1 downto 2 do
[M:= N; loop [if not Prime(M) then quit; M:= M/10; if rem(0) = 0 then quit; \no zeros allowed if M=0 then return N; ]; ];
];
func LeftTrunc(N); \Return largest left-truncatable prime < one million int N; int M, P; [for N:= 1_000_000-1 downto 2 do
[M:= N; P:=100_000; loop [if not Prime(M) then quit; M:= rem(M/P); P:= P/10; if M<P then quit; \no zeros allowed if M=0 then return N; ]; ];
];
[IntOut(0, LeftTrunc); CrLf(0);
IntOut(0, RightTrunc); CrLf(0);
]</lang>
Output:
998443 739399
zkl
Using Extensible prime generator#zkl and a one meg bucket of bytes, construct a yes/no lookup table for all primes <= one million (<80,000). <lang zkl>const million=0d1_000_000;
var pTable=Data(million+1,Int); // actually bytes (0).pump(million+1,pTable,T(Void,0)); // initialize to 0 primes:=Utils.Generator(Import("sieve").postponed_sieve); while((p:=primes.next())<million){ pTable[p]=1; }
fcn rightTrunc(n){
while(n){ if(not pTable[n]) return(False); n/=10; } True
} fcn leftTrunc(n){ // 999,907 is not allowed
ns:=n.toString(); if (ns.holds("0")) return(False); while(ns){ if(not pTable[ns]) return(False); ns=ns[1,*]; } True
}
[million..0,-1].filter1(rightTrunc):
"%,d is a right truncatable prime".fmt(_).println();
[million..0,-1].filter1(leftTrunc):
"%,d is a left truncatable prime".fmt(_).println();</lang>
- Output:
739,399 is a right truncatable prime 998,443 is a left truncatable prime