Truncatable primes
You are encouraged to solve this task according to the task description, using any language you may know.
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.
- Examples
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
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
- Related tasks
- See also
- Truncatable Prime from MathWorld.]
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
Arturo
<lang arturo>leftTrunc [num]{ str $(toString num) ret #() loop $(range 0 $(size str)-1) [x]{ ret ret + $(slice str x $(size str)) } return $(map ret { toNumber & }) }
rightTrunc [num]{ str $(toString num) ret #() loop $(range $(size str) 1) [x]{ ret ret + $(slice str 0 x) } return $(map ret { toNumber & }) }
isTruncatablePrime [nums]{ all nums { isPrime & } }
MaxLeft 0 MaxRight 0
loop $(range 3 1000000 2) [n]{ if $(isTruncatablePrime $(leftTrunc n)) { MaxLeft n } if $(isTruncatablePrime $(rightTrunc n)) { MaxRight n } }
print "Max Left-Truncatable Prime found (<1000000): " + MaxLeft print "Max Right-Truncatable Prime found (<1000000): " + MaxRight</lang>
- Output:
Max Left-Truncatable Prime found (<1000000): 998443 Max Right-Truncatable Prime found (<1000000): 739399
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
AWK
<lang AWK>
- syntax: GAWK -f TRUNCATABLE_PRIMES.AWK
BEGIN {
limit = 1000000
for (i=1; i<=limit; i++) {
if (is_prime(i)) {
prime_count++
arr[i] = ""
if (truncate_left(i) == 1) {
max_left = max(max_left,i)
}
if (truncate_right(i) == 1) {
max_right = max(max_right,i)
}
}
}
printf("1-%d: %d primes\n",limit,prime_count)
printf("largest L truncatable: %d\n",max_left)
printf("largest R truncatable: %d\n",max_right)
exit(0)
} function is_prime(x, i) {
if (x <= 1) {
return(0)
}
for (i=2; i<=int(sqrt(x)); i++) {
if (x % i == 0) {
return(0)
}
}
return(1)
} function truncate_left(n) {
while (n != "") {
if (!(n in arr)) {
return(0)
}
n = substr(n,2)
}
return(1)
} function truncate_right(n) {
while (n != "") {
if (!(n in arr)) {
return(0)
}
n = substr(n,1,length(n)-1)
}
return(1)
} function max(x,y) { return((x > y) ? x : y) } </lang>
- Output:
1-1000000: 78498 primes largest L truncatable: 998443 largest R truncatable: 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; // 4790@3.6 using System.Collections.Generic; class truncatable_primes {
static void Main()
{
uint m = 1000000;
Console.Write("L " + L(m) + " R " + R(m) + " ");
var sw = System.Diagnostics.Stopwatch.StartNew();
for (int i = 1000; i > 0; i--) { L(m); R(m); }
Console.Write(sw.Elapsed); Console.Read();
}
static uint L(uint n)
{
n -= n & 1; n--;
for (uint d, d1 = 100; ; n -= 2)
{
while (n % 3 == 0 || n % 5 == 0 || n % 7 == 0) n -= 2;
if ((d = n % 10) == 3 || d == 7)
{
while (d1 < n && d < (d = n % d1) && isP(d)) d1 *= 10;
if (d1 > n && isP(n)) return n; d1 = 100;
}
}
}
static uint R(uint m)
{
var p = new List<uint>() { 2, 3, 5, 7 }; uint n = 20, np;
for (int i = 1; i < p.Count; n = 10 * p[i++])
{
if ((np = n + 1) >= m) break; if (isP(np)) p.Add(np);
if ((np = n + 3) >= m) break; if (isP(np)) p.Add(np);
if ((np = n + 7) >= m) break; if (isP(np)) p.Add(np);
if ((np = n + 9) >= m) break; if (isP(np)) p.Add(np);
}
return p[p.Count - 1];
}
static bool isP(uint n)
{
if (n < 7) return n == 2 || n == 3 || n == 5;
if ((n & 1) == 0 || n % 3 == 0 || n % 5 == 0) return false;
for (uint r = (uint)Math.Sqrt(n), d = 7; d <= r; d += 30)
if (n % (d + 00) == 0 || n % (d + 04) == 0 ||
n % (d + 06) == 0 || n % (d + 10) == 0 ||
n % (d + 12) == 0 || n % (d + 16) == 0 ||
n % (d + 22) == 0 || n % (d + 24) == 0) return false;
return true;
}
}</lang>
Output: L 998443 R 739399 24 μs
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>
Common Lisp
<lang lsip> (defun start ()
(format t "Largest right-truncatable ~a~%" (max-right-truncatable)) (format t "Largest left-truncatable ~a~%" (max-left-truncatable)))
(defun max-right-truncatable ()
(loop for el in (6-digits-R-truncatables)
maximizing el into max
finally (return max)))
(defun 6-digits-R-truncatables (&optional (lst '(2 3 5 7)) (n 5))
(if (zerop n) lst (6-digits-R-truncatables (R-trunc lst) (- n 1))))
(defun R-trunc (lst)
(remove-if (lambda (x) (not (primep x)))
(loop for el in lst append (mapcar (lambda (x) (+ (* 10 el) x)) '(1 3 7 9)))))
(defun max-left-truncatable ()
(loop for el in (6-digits-L-truncatables)
maximizing el into max
finally (return max)))
(defun 6-digits-L-truncatables (&optional (lst '(3 7)) (n 5))
(if (zerop n) lst (6-digits-L-truncatables (L-trunc lst (- 6 n)) (- n 1))))
(defun L-trunc (lst n)
(remove-if (lambda (x) (not (primep x)))
(loop for el in lst append (mapcar (lambda (x) (+ (* (expt 10 n) x) el)) '(1 2 3 4 5 6 7 8 9)))))
(defun primep (n)
(primep-aux n 2))
(defun primep-aux (n d)
(cond ((> d (sqrt n)) t)
((zerop (rem n d)) nil)
(t (primep-aux n (+ d 1)))))
</lang>
- Output:
Largest right-truncatable 739399 Largest left-truncatable 998443
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
EchoLisp
<lang lisp>
- does p include a 0 in its decimal representation ?
(define (nozero? n) (= -1 (string-index (number->string n) "0")))
- right truncate
- p and successive quotients by 10 (integer division) must be primes
(define (right-trunc p) (unless (zero? p) (and (prime? p) (right-trunc (quotient p 10))))) (remember 'right-trunc)
- left truncate
- p and successive modulo by 10, 100, .. must be prime
(define (left-trunc p (mod 1000000)) (unless (< mod 1) (and (prime? p) (nozero? p) (left-trunc (modulo p mod) (/ mod 10)))))
- start from 999999. stop on first found
(define (fact-trunc trunc) (for ((p (in-range 999999 100000 -1))) #:break (when (trunc p) (writeln p) #t))) </lang> Output: <lang lisp> (fact-trunc left-trunc) 998443 (fact-trunc right-trunc) 739399 </lang>
Eiffel
<lang Eiffel> class APPLICATION
create make
feature
make do io.put_string ("Largest right truncatable prime: " + find_right_truncatable_primes.out) io.new_line io.put_string ("Largest left truncatable prime: " + find_left_truncatable_primes.out) end
find_right_truncatable_primes: INTEGER -- Largest right truncatable prime below 1000000. local i, maybe_prime: INTEGER found, is_one: BOOLEAN do from i := 999999 until found loop is_one := True from maybe_prime := i until not is_one or maybe_prime.out.count = 1 loop if maybe_prime.out.has ('0') or maybe_prime.out.has ('2') or maybe_prime.out.has ('4') or maybe_prime.out.has ('6') or maybe_prime.out.has ('8') then is_one := False else if not is_prime (maybe_prime) then is_one := False elseif is_prime (maybe_prime) and maybe_prime.out.count > 1 then maybe_prime := truncate_right (maybe_prime) end end end if is_one then found := True Result := i end i := i - 2 end ensure Result_is_smaller: Result < 1000000 end
find_left_truncatable_primes: INTEGER -- Largest left truncatable prime below 1000000. local i, maybe_prime: INTEGER found, is_one: BOOLEAN do from i := 999999 until found loop is_one := True from maybe_prime := i until not is_one or maybe_prime.out.count = 1 loop if not is_prime (maybe_prime) then is_one := False elseif is_prime (maybe_prime) and maybe_prime.out.count > 1 then if maybe_prime.out.at (2) = '0' then is_one := False else maybe_prime := truncate_left (maybe_prime) end end end if is_one then found := True Result := i end i := i - 2 end ensure Result_is_smaller: Result < 1000000 end
feature {NONE}
is_prime (n: INTEGER): BOOLEAN --Is 'n' a prime number? require positiv_input: n > 0 local i: INTEGER max: REAL_64 math: DOUBLE_MATH do create math if n = 2 then Result := True elseif n <= 1 or n \\ 2 = 0 then Result := False else Result := True max := math.sqrt (n) from i := 3 until i > max loop if n \\ i = 0 then Result := False end i := i + 2 end end end
truncate_left (n: INTEGER): INTEGER -- 'n' truncated by one digit from the left side. require truncatable: n.out.count > 1 local st: STRING do st := n.out st.remove_head (1) Result := st.to_integer ensure Result_truncated: Result.out.count = n.out.count - 1 end
truncate_right (n: INTEGER): INTEGER -- 'n' truncated by one digit from the right side. require truncatable: n.out.count > 1 local st: STRING do st := n.out st.remove_tail (1) Result := st.to_integer ensure Result_truncated: Result.out.count = n.out.count - 1 end
end </lang>
- Output:
Largest right truncatable prime: 739399 Largest left truncatable prime: 999431
Elena
ELENA 4.x : <lang elena>import system'calendar; import extensions;
const MAXN = 1000000;
extension mathOp {
isPrime()
{
int n := cast int(self);
if (n < 2) { ^ false };
if (n < 4) { ^ true };
if (n mod:2 == 0) { ^ false };
if (n < 9) { ^ true };
if (n mod:3 == 0) { ^ false };
int r := n.sqrt();
int f := 5;
while (f <= r)
{
if ((n.mod(f) == 0) || (n.mod(f + 2) == 0))
{ ^ false };
f := f + 6
};
^ true
}
isRightTruncatable()
{
int n := self;
while (n != 0)
{
ifnot (n.isPrime())
{ ^ false };
n := n / 10
};
^ true
}
isLeftTruncatable()
{
int n := self;
int tens := 1;
while (tens < n)
{ tens := tens * 10 };
while (n != 0)
{
ifnot (n.isPrime())
{ ^ false };
tens := tens / 10;
n := n - (n / tens * tens)
};
^ true
}
}
program() {
var n := MAXN; var max_lt := 0; var max_rt := 0;
while (max_lt == 0 || max_rt == 0)
{
if(n.toString().indexOf("0") == -1)
{
if ((max_lt == 0) && (n.isLeftTruncatable()))
{
max_lt := n
};
if ((max_rt == 0) && (n.isRightTruncatable()))
{
max_rt := n
}
};
n := n - 1
};
console.printLine("Largest truncable left is ",max_lt);
console.printLine("Largest truncable right is ",max_rt);
}</lang>
- Output:
Largest truncable left is 998443 Largest truncable right is 739399
Elixir
<lang elixir>defmodule Prime do
defp left_truncatable?(n, prime) do
func = fn i when i<=9 -> 0
i -> to_string(i) |> String.slice(1..-1) |> String.to_integer end
truncatable?(n, prime, func)
end
defp right_truncatable?(n, prime) do
truncatable?(n, prime, fn i -> div(i, 10) end)
end
defp truncatable?(n, prime, trunc_func) do
if to_string(n) |> String.match?(~r/0/),
do: false,
else: trunc_loop(trunc_func.(n), prime, trunc_func)
end
defp trunc_loop(0, _prime, _trunc_func), do: true
defp trunc_loop(n, prime, trunc_func) do
if elem(prime,n), do: trunc_loop(trunc_func.(n), prime, trunc_func), else: false
end
def eratosthenes(limit) do # descending order
Enum.to_list(2..limit) |> sieve(:math.sqrt(limit), [])
end
defp sieve([h|_]=list, max, sieved) when h>max, do: Enum.reverse(list, sieved)
defp sieve([h | t], max, sieved) do
list = for x <- t, rem(x,h)>0, do: x
sieve(list, max, [h | sieved])
end
defp prime_table(_, [], list), do: [false, false | list]
defp prime_table(n, [n|t], list), do: prime_table(n-1, t, [true|list])
defp prime_table(n, prime, list), do: prime_table(n-1, prime, [false|list])
def task(limit \\ 1000000) do
prime = eratosthenes(limit)
prime_tuple = prime_table(limit, prime, []) |> List.to_tuple
left = Enum.find(prime, fn n -> left_truncatable?(n, prime_tuple) end)
IO.puts "Largest left-truncatable prime : #{left}"
right = Enum.find(prime, fn n -> right_truncatable?(n, prime_tuple) end)
IO.puts "Largest right-truncatable prime: #{right}"
end
end
Prime.task</lang>
- Output:
Largest left-truncatable prime : 998443 Largest right-truncatable prime: 739399
Factor
<lang>USING: formatting fry grouping.extras kernel literals math math.parser math.primes sequences ; IN: rosetta-code.truncatable-primes
CONSTANT: primes $[ 1,000,000 primes-upto reverse ]
- number>digits ( n -- B{} ) number>string string>digits ;
- no-zeros? ( seq -- ? ) [ zero? not ] all? ;
- all-prime? ( seq -- ? ) [ prime? ] all? ;
- truncate ( seq quot -- seq' ) call( seq -- seq' )
[ 10 digits>integer ] map ;
- truncate-right ( seq -- seq' ) [ head-clump ] truncate ;
- truncate-left ( seq -- seq' ) [ tail-clump ] truncate ;
- truncatable-prime? ( n quot -- ? ) [ number>digits ] dip
'[ @ all-prime? ] [ no-zeros? ] bi and ; inline
- right-truncatable-prime? ( n -- ? ) [ truncate-right ]
truncatable-prime? ;
- left-truncatable-prime? ( n -- ? ) [ truncate-left ]
truncatable-prime? ;
- find-truncatable-primes ( -- ltp rtp )
primes [ [ left-truncatable-prime? ] find nip ]
[ [ right-truncatable-prime? ] find nip ] bi ;
- main ( -- ) find-truncatable-primes
"Left: %d\nRight: %d\n" printf ;
MAIN: main</lang>
- Output:
Left: 998443 Right: 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
FreeBASIC
Version 1
<lang freebasic>' FB 1.05.0 Win64
Function isPrime(n As Integer) As Boolean
If n Mod 2 = 0 Then Return n = 2 If n Mod 3 = 0 Then Return n = 3 Dim d As Integer = 5 While d * d <= n If n Mod d = 0 Then Return False d += 2 If n Mod d = 0 Then Return False d += 4 Wend Return True
End Function
Dim As UInteger i, j, p, pow, lMax = 2, rMax = 2 Dim s As String
' largest left truncatable prime less than 1000000 ' It can't end with 1, 4, 6, 8 or 9 as these numbers are not prime ' Nor can it end in 2 if it has more than one digit as such a number would divide by 2 For i = 3 To 999997 Step 2
s = Str(i) If Instr(s, "0") > 1 Then Continue For cannot contain 0 j = s[Len(s) - 1] - 48 If j = 1 OrElse j = 9 Then Continue For p = i pow = 10 ^ (Len(s) - 1) While pow > 1 If Not isPrime(p) Then Continue For p Mod= pow pow \= 10 Wend lMax = i
Next
' largest right truncatable prime less than 1000000 ' It can't begin with 1, 4, 6, 8 or 9 as these numbers are not prime For i = 3 To 799999 Step 2
s = Str(i) If Instr(s, "0") > 1 Then Continue For cannot contain 0 j = s[0] - 48 If j = 1 OrElse j = 4 OrElse j = 6 Then Continue For p = i While p > 0 If Not isPrime(p) Then Continue For p \= 10 Wend rMax = i
Next
Print "Largest left truncatable prime : "; lMax Print "Largest right truncatable prime : "; rMax Print Print "Press any key to quit" Sleep</lang>
- Output:
Largest left truncatable prime : 998443 Largest right truncatable prime : 739399
Version 2
Construct primes using previous found primes. <lang freebasic>' version 10-12-2016 ' compile with: fbc -s console
Dim Shared As Byte isPrime()
Sub sieve(m As UInteger)
Dim As Integer i, j ReDim isPrime(m)
For i = 4 To m Step 2
isPrime(i) = 1
Next
For i = 3 To Sqr(m) Step 2
If isPrime(i) = 0 Then
For j = i * i To m Step i * 2
isPrime(j) = 1
Next
End If
Next
End Sub
' ------=< MAIN >=------
- Define max 1000000 'upto 2^30 max for 32bit OS
Dim As UInteger a(), lt_prime(5000), rt_prime(100) Dim As UInteger i, j, j1, p1, p2, left_max, right_max
sieve(max)
' left truncatable primes ' if odd and ends with 3 or 7, never ends 1 or 9 (no prime ' never ends on a 2 or 5 and starts with 1 to 9 lt_prime(1) = 3 : lt_prime(2) = 7 p1 = 1 : p2 = 2
Do
For i = 1 To 9
j = Val( Str(i) + Str(lt_prime(p1)) )
If j > max Then Exit Do
If isPrime(j) = 0 Then ' if prime then add to the list
p2 += 1
lt_prime(p2) = j
If Left_max < j Then left_max = j
End If
Next
p1 += 1
Loop Until p1 > p2 ' no more numbers to process
' right truncatable prime ' start with 2, 3, 5 or 7 and end with 1, 3, 7 or 9 rt_prime(1) = 2 : rt_prime(2) = 3 : rt_prime(3) = 5 : rt_prime(4) = 7 p1 = 1 : p2 = 4 Dim As UInteger end_num(1 To 4) => {1, 3, 7, 9}
Do
j1 = rt_prime(p1) * 10
If j1 > max Then Exit Do
For i = 1 To 4
j = j1 + End_num(i)
If isprime(j) = 0 Then ' if prime then add to the list
p2 += 1
rt_prime(p2) = j
' If right_max < j Then right_max = j
End If
Next
p1 += 1
Loop Until p1 > p2 ' no more numbers to process ' the last one added is the biggest right_max = rt_prime(p2)
Print Print "The biggest left truncatable prime below"; max; " is "; left_max Print "The biggest right truncatable prime below"; max; " is "; right_max
' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
The biggest left truncatable prime below 1000000 is 998443 The biggest 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 (combine all possibilities of a truncatable prime digit sequence with each digit) 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>
Here, selPrime discards non-prime numbers from a list, so seed is the list 2 3 5 7.
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>
Note that we are using the same combining function and same basic procedure in both cases. The difference is which side of the number we add arbitrary digits to, for each step.
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 && right % 2 != 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
Julia
There are several features of Julia that make solving this task easy. Julia has excellent built-in support for prime generation and testing. The built-in mathematical functions prevpow and divrem are quite handy for implementing isltruncprime. <lang Julia> function isltruncprime{T<:Integer}(n::T, base::T=10)
isprime(n) || return false
p = n
f = prevpow(base, p)
while 1 < f
(d, p) = divrem(p, f)
isprime(p) || return false
d != 0 || return false
f = div(f, base)
end
return true
end
function isrtruncprime{T<:Integer}(n::T, base::T=10)
isprime(n) || return false
p = n
while base < p
p = div(p, base)
isprime(p) || return false
end
return true
end
hi = 10^6
for i in reverse(primes(hi))
isltruncprime(i) || continue
println("The largest left truncatable prime ≤ ", hi, " is ", i, ".")
break
end
for i in reverse(primes(hi))
isrtruncprime(i) || continue
println("The largest right truncatable prime ≤ ", hi, " is ", i, ".")
break
end </lang>
- Output:
The largest left truncatable prime ≤ 1000000 is 998443. The largest right truncatable prime ≤ 1000000 is 739399.
Kotlin
<lang scala>// version 1.0.5-2
fun isPrime(n: Int) : Boolean {
if (n < 2) return false
if (n % 2 == 0) return n == 2
if (n % 3 == 0) return n == 3
var d : Int = 5
while (d * d <= n) {
if (n % d == 0) return false
d += 2
if (n % d == 0) return false
d += 4
}
return true
}
fun main(args: Array<String>) {
var j: Char
var p: Int
var pow: Int
var lMax: Int = 2
var rMax: Int = 2
var s: String
// calculate maximum left truncatable prime less than 1 million
loop@ for( i in 3..999997 step 2) {
s = i.toString()
if ('0' in s) continue
j = s[s.length - 1]
if (j == '1' || j == '9') continue
p = i
pow = 1
for (k in 1..s.length - 1) pow *= 10
while(pow > 1) {
if (!isPrime(p)) continue@loop
p %= pow
pow /= 10
}
lMax = i
}
// calculate maximum right truncatable prime less than 1 million
loop@ for( i in 3..799999 step 2) {
s = i.toString()
if ('0' in s) continue
j = s[0]
if (j == '1' || j == '4' || j == '6') continue
p = i
while(p > 0) {
if (!isPrime(p)) continue@loop
p /= 10
}
rMax = i
}
println("Largest left truncatable prime : " + lMax.toString())
println("Largest right truncatable prime : " + rMax.toString())
}</lang>
- Output:
Largest left truncatable prime : 998443 Largest right truncatable prime : 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>
Maple
<lang Maple> MaxTruncatablePrime := proc({left::truefalse:=FAIL, right::truefalse:=FAIL}, $) local i, j, c, p, b, n, sdprimes, dir; local tprimes := table();
if left = true and right = true then
error "invalid input";
elif right = true then
dir := "right";
else
dir := "left";
end if;
b := 10;
n := 6;
sdprimes := select(isprime, [seq(1..b-1)]);
for p in sdprimes do
if assigned(tprimes[p]) then
next;
end if;
i := ilog[b](p)+1;
j := 1;
while p < b^n do
if dir = "left" then
c := j*b^i + p;
else
c := p*b + j;
end if;
if j >= b or c > b^n then # we have tried all 1 digit extensions of p, add p to tprimes and move back 1 digit
tprimes[p] := p;
if i = 1 then # if we are at the first digit, go to the next 1 digit prime
break;
end if;
i := i - 1;
j := 1;
if dir = "left" then
p := p - iquo(p, b^i)*b^i;
else
p := iquo(p, b);
end if;
elif assigned(tprimes[c]) then
j := j + 1;
elif isprime(c) then
p := c;
i := i + 1;
j := 1;
else
j := j+1;
end if;
end do;
end do;
return max(indices(tprimes, 'nolist'));
end proc;</lang>
> MaxTruncatablePrime(right); MaxTruncatablePrime(left);
739399
998443
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>
Nim
<lang nim>import sets, strutils, algorithm
proc primes(n: int64): seq[int64] =
result = @[]
var multiples = initSet[int64]()
for i in 2..n:
if i notin multiples:
result.add i
for j in countup(i*i, n, i.int):
multiples.incl j
proc truncatablePrime(n: int64): tuple[left: int64, right: int64] =
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-1:
alltruncs.incl n[i..n.high]
if alltruncs <= primeset:
result.left = parseInt(n)
break
for n in primelist:
var alltruncs = initSet[string]()
for i in 0..n.len-1:
alltruncs.incl n[0..i]
if alltruncs <= primeset:
result.right = parseInt(n)
break
echo truncatablePrime(1000000'i64)</lang> Output:
(left: 998443, 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
Typically with Perl we'll look for a CPAN module to make our life easier. This basically just follows the task rules:
<lang perl>use ntheory ":all"; sub isltrunc {
my $n = shift; return (is_prime($n) && $n !~ /0/ && ($n < 10 || isltrunc(substr($n,1))));
} sub isrtrunc {
my $n = shift; return (is_prime($n) && $n !~ /0/ && ($n < 10 || isrtrunc(substr($n,0,-1))));
} for (reverse @{primes(1e6)}) {
if (isltrunc($_)) { print "ltrunc: $_\n"; last; }
} for (reverse @{primes(1e6)}) {
if (isrtrunc($_)) { print "rtrunc: $_\n"; last; }
}</lang>
- Output:
ltrunc: 998443 rtrunc: 739399
We can be a little more Perlish and build up n-digit lists then select the last one: <lang perl>use ntheory ":all";
my @lprimes = my @rprimes = (2,3,5,7);
@lprimes = sort { $a <=> $b }
map { my $p=$_; map { is_prime($_.$p) ? $_.$p : () } 1..9 } @lprimes
for 2..6;
@rprimes = sort { $a <=> $b }
map { my $p=$_; map { is_prime($p.$_) ? $p.$_ : () } 1..9 } @rprimes
for 2..6;
print "ltrunc: $lprimes[-1]\nrtrunc: $rprimes[-1]\n";</lang>
Or we can do everything ourselves: <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>constant ltp = $[2, 3, 5, 7], -> @ltp {
$[ grep { .&is-prime }, ((1..9) X~ @ltp) ]
} ... *;
constant rtp = $[2, 3, 5, 7], -> @rtp {
$[ grep { .&is-prime }, (@rtp X~ (1..9)) ]
} ... *;
say "Highest ltp = ", ltp[5][*-1]; say "Highest rtp = ", rtp[5][*-1];</lang>
- Output:
Highest ltp: 998443 Highest rtp: 739399
Phix
A slightly different approach. Works up to N=8, quite fast - 10^8 in 5s with ~90% of time spent creating the basic sieve and ~10% propagation and final scan. <lang Phix>constant N = 6, limit = power(10,N) -- standard sieve: enum L,R -- (with primes[i] as mini bit-field) sequence primes = repeat(L+R, limit) primes[1] = 0 for i=2 to floor(sqrt(limit)) do
if primes[i] then
for k=i*i to limit by i do
primes[k] = 0
end for
end if
end for
-- propagate non-truncateables up the prime table: for p=1 to N-1 do
integer p10 = power(10,p) -- ie 10, 100, .. 100_000
for i=p10+1 to p10*10-1 by 2 do -- to 99, 999, .. 999_999
if primes[i] then
integer l = remainder(i,p10),
r = floor(i/10)
integer pi = and_bits(primes[l],L)+and_bits(primes[r],R)
if pi and find('0',sprint(i)) then pi = 0 end if
primes[i] = pi
end if
end for
end for
integer maxl=0, maxr=0
for i=limit-1 to 1 by -2 do
integer pi = primes[i]
if pi then
if maxl=0 and and_bits(pi,L) then maxl = i end if
if maxr=0 and and_bits(pi,R) then maxr = i end if
if maxl!=0 and maxr!=0 then exit end if
end if
end for ?{maxl,maxr}</lang>
- Output:
{998443,739399}
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 computation time. <lang REXX>/*REXX program finds largest left─ and right─truncatable primes ≤ 1m (or argument 1).*/ parse arg high .; if high== then high=1000000 /*Not specified? Then use 1m*/ !.=0; w=length(high) /*placeholders for primes; max width. */ @.1=2; @.2=3; @.3=5; @.4=7; @.5=11; @.6=13; @.7=17 /*define some low primes. */ !.2=1; !.3=1; !.5=1; !.7=1; !.11=1; !.13=1; !.17=1 /*set some low prime flags. */
- =7; s.#=@.#**2 /*number of primes so far; prime². */
/* [↓] generate more primes ≤ high.*/
do j=@.#+2 by 2 for max(0, high%2-@.#%2-1) /*only find odd primes from here on out*/
if j// 3==0 then iterate /*is J divisible by three? */
parse var j -1 _; if _==5 then iterate /* " " " " five? (right digit)*/
if j// 7==0 then iterate /* " " " " seven? */
if j//11==0 then iterate /* " " " " eleven? */
if j//13==0 then iterate /* " " " " thirteen? */
/* [↑] the above five lines saves time*/
do k=7 while s.k<=j /* [↓] divide by the known odd primes.*/
if j//@.k==0 then iterate j /*Is J ÷ X? Then not prime. ___ */
end /*k*/ /* [↑] only process up to the √ J */
#=#+1 /*bump the number of primes found. */
@.#=j; s.#=j*j; !.j=1 /*assign next prime; prime²; prime #.*/
end /*j*/
/* [↓] find largest left truncatable P*/
do L=# by -1 for #; digs=length(@.L) /*search from top end; get the length.*/
do k=1 for digs; _=right(@.L, k) /*validate all left truncatable primes.*/
if \!._ then iterate L /*Truncated number not prime? Skip it.*/
end /*k*/
leave /*egress, found left truncatable prime.*/
end /*L*/
/* [↓] find largest right truncated P.*/
do R=# by -1 for #; digs=length(@.R) /*search from top end; get the length.*/
do k=1 for digs; _=left(@.R, k) /*validate all right truncatable primes*/
if \!._ then iterate R /*Truncated number not prime? Skip it.*/
end /*k*/
leave /*egress, found right truncatable prime*/
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 for the output.*/ 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 all done. */</lang>
output when using the default input:
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
Ring
<lang ring>
- Project : Truncatable primes
for n = 1000000 to 1 step -1
flag = 1
flag2 = 1
strn = string(n)
for nr = 1 to len(strn)
if strn[nr] = "0"
flag2 = 0
ok
next
if flag2 = 1
for m = 1 to len(strn)
strp = right(strn, m)
if isprime(number(strp))
else
flag = 0
exit
ok
next
if flag = 1
nend = n
exit
ok
ok
next see "Largest left truncatable prime : " + nend + nl
for n = 1000000 to 1 step -1
flag = 1
strn = string(n)
for m = 1 to len(strn)
strp = left(strn, len(strn) - m + 1)
if isprime(number(strp))
else
flag = 0
exit
ok
next
if flag = 1
nend = n
exit
ok
next see "Largest right truncatable prime : " + nend + nl
func isprime num
if (num <= 1) return 0 ok
if (num % 2 = 0 and num != 2) return 0 ok
for i = 3 to floor(num / 2) -1 step 2
if (num % i = 0) return 0 ok
next
return 1
</lang> Output:
Largest left truncatable prime : 998443 Largest right truncatable prime : 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
Scala
This example uses lazily evaluated lists. The functions to determine if a number is a truncatable prime construct a list of truncated numbers and check that all the elements in the list are prime. <lang scala>object TruncatablePrimes {
def main(args: Array[String]): Unit = {
val max = 1000000
println(
s"""|ltPrime: ${ltPrimes.takeWhile(_ <= max).last}
|rtPrime: ${rtPrimes.takeWhile(_ <= max).last}
|""".stripMargin)
}
def ltPrimes: LazyList[Int] = 2 #:: LazyList.from(3, 2).filter(isLeftTruncPrime)
def rtPrimes: LazyList[Int] = 2 #:: LazyList.from(3, 2).filter(isRightTruncPrime)
def isPrime(num: Int): Boolean = (num > 1) && !LazyList.range(3, math.sqrt(num).toInt + 1, 2).exists(num%_ == 0)
def isLeftTruncPrime(num: Int): Boolean = !num.toString.contains('0') && Iterator.unfold(num.toString){str => if(str.nonEmpty) Some((str.toInt, str.tail)) else None}.forall(isPrime)
def isRightTruncPrime(num: Int): Boolean = !num.toString.exists(_.asDigit%2 == 0) && Iterator.unfold(num.toString){str => if(str.nonEmpty) Some((str.toInt, str.init)) else None}.forall(isPrime)
}</lang>
- Output:
ltPrime: 998443 rtPrime: 739399
Sidef
<lang ruby>func t_prime(n, left=true) {
var p = %w(2 3 5 7);
var f = (
left ? { '1'..'9' ~X+ p }
: { p ~X+ '1'..'9' }
)
n.times {
p = f().grep{ .to_i.is_prime }
}
p.map{.to_i}.max
}
say t_prime(5, left: true) say t_prime(5, left: false)</lang>
- Output:
998443 739399
Tcl
<lang tcl>package require Tcl 8.5
- Optimized version of the Sieve-of-Eratosthenes task solution
proc sieve n {
set primes [list]
if {$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
VBScript
<lang vb> start_time = Now
lt = 0 rt = 0
For h = 1 To 1000000 If IsLeftTruncatable(h) And h > lt Then lt = h End If If IsRightTruncatable(h) And h > rt Then rt = h End If Next
end_time = now
WScript.StdOut.WriteLine "Largest LTP from 1..1000000: " & lt WScript.StdOut.WriteLine "Largest RTP from 1..1000000: " & rt WScript.StdOut.WriteLine "Elapse Time(seconds) : " & DateDiff("s",start_time,end_time)
'------------ Function IsLeftTruncatable(n) IsLeftTruncatable = False c = 0 For i = Len(n) To 1 Step -1 If InStr(1,n,"0") > 0 Then Exit For End If If IsPrime(Right(n,i)) Then c = c + 1 End If Next If c = Len(n) Then IsLeftTruncatable = True End If End Function
Function IsRightTruncatable(n) IsRightTruncatable = False c = 0 For i = Len(n) To 1 Step -1 If InStr(1,n,"0") > 0 Then Exit For End If If IsPrime(Left(n,i)) Then c = c + 1 End If Next If c = Len(n) Then IsRightTruncatable = True End If End Function
Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function </lang>
- Output:
Largest LTP from 1..1000000: 998443 Largest RTP from 1..1000000: 739399 Elapse Time(seconds) : 49
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).fill(0); // actually bytes, all zero 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
}</lang> <lang zkl>[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
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