Substring primes: Difference between revisions
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=={{header|Picat}}== |
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===Checking via substrings=== |
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<lang Picat>% get all the substrings of a number |
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subs(N) = findall(S, (append(_Pre,S,_Post,N.to_string), S.len > 0) ). |
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go => |
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Ps = [], |
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foreach(Prime in primes(500)) |
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(foreach(N in subs(Prime) prime(N) end -> Ps := Ps ++ [Prime] ; true) |
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end, |
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println(Ps).</lang> |
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{{out}} |
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<pre>[2,3,5,7,23,37,53,73,373]</pre> |
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===Checking via predicate=== |
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{{trans|AWK}} |
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Here we must use cut (<code>!</code>) to ensure that the test does not continue after a satisfied test. This is a "red cut" (i.e. removing it would change the logic of the program) and these should be avoided if possible. |
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<lang Picat>t(N,false) :- |
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not prime(N),!. |
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t(N,true) :- |
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N < 10,!. |
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t(N,false) :- |
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not prime(N mod 100), !. |
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t(N,false) :- |
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not prime(N mod 10),!. |
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t(N,false) :- |
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not prime(N // 10),!. |
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t(N,true) :- |
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N < 100,!. |
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t(N,false) :- |
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not prime(N // 100),!. |
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t(N,false) :- |
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not prime((N mod 100) // 10),!. |
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t(_N,true). |
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go2 => |
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println(findall(N,(member(N,1..500),t(N,Status),Status==true))).</lang> |
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{{out}} |
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<pre>[2,3,5,7,23,37,53,73,373]</pre> |
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=={{header|Phix}}== |
=={{header|Phix}}== |
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Revision as of 22:19, 13 May 2022
- Task
Find all primes in which all substrings (in base ten) are also primes.
This can be achieved by filtering all primes below 500 (there are 95 of them), but there are better ways.
- Advanced
Solve by testing at most 15 numbers for primality. Show a list of all numbers tested that were not prime.
11l
<lang 11l>F is_prime(n)
I n == 2
R 1B
I n < 2 | n % 2 == 0
R 0B
L(i) (3 .. Int(sqrt(n))).step(2)
I n % i == 0
R 0B
R 1B
V results = [2, 3, 5, 7] V odigits = [3, 7] [Int] discarded V tests = 4
V i = 0 L i < results.len
V r = results[i]
i++
L(od) odigits
I (r % 10) != od
V n = r * 10 + od
I is_prime(n)
results.append(n)
E
discarded.append(n)
tests++
print(‘There are ’results.len‘ primes where all substrings are also primes, namely:’) print(results) print("\nThe following numbers were also tested for primality but found to be composite:") print(discarded) print("\nTotal number of primality tests = "tests)</lang>
- Output:
There are 9 primes where all substrings are also primes, namely: [2, 3, 5, 7, 23, 37, 53, 73, 373] The following numbers were also tested for primality but found to be composite: [27, 57, 237, 537, 737, 3737] Total number of primality tests = 15
Action!
<lang Action!>INCLUDE "H6:SIEVE.ACT"
BYTE FUNC IsSubstringPrime(INT x BYTE ARRAY primes)
CHAR ARRAY s(4),tmp(4) INT len,start,sub
IF primes(x)=0 THEN RETURN (0) FI
StrI(x,s)
FOR len=1 TO s(0)-1
DO
FOR start=1 TO s(0)-len+1
DO
SCopyS(tmp,s,start,start+len-1)
sub=ValI(tmp)
IF primes(sub)=0 THEN
RETURN (0)
FI
OD
OD
RETURN (1)
PROC Main()
DEFINE MAX="499" BYTE ARRAY primes(MAX+1) INT i
Put(125) PutE() ;clear the screen
Sieve(primes,MAX+1)
FOR i=2 TO MAX
DO
IF IsSubstringPrime(i,primes) THEN
PrintIE(i)
FI
OD
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
2 3 5 7 23 37 53 73 373
ALGOL 68
<lang algol68>BEGIN # find primes where all substrings of the digits are prime #
# find the primes of interest #
PR read "primes.incl.a68" PR
[]BOOL prime = PRIMESIEVE 500;
FOR p TO UPB prime DO
IF prime[ p ] THEN
INT d := 10;
BOOL is substring := TRUE;
WHILE is substring AND d <= UPB prime DO
INT n := p;
WHILE is substring AND n > 0 DO
is substring := prime[ n MOD d ];
n OVERAB 10
OD;
d *:= 10
OD;
IF is substring THEN print( ( " ", whole( p, 0 ) ) ) FI
FI
OD
END</lang>
- Output:
2 3 5 7 23 37 53 73 373
ALGOL W
starts with a hardcoded list of 1 digit primes ( 2, 3, 5, 7 ) and constructs the remaining members of the sequence (in order) using the observations that the final digit must be prime and can't be 2 or 5 or the number wouldn't be prime. Additionally, the final digit pair cannot be 33 or 77 as these are divisible by 11. <lang algolw>begin % find primes where every substring of the digits is also priome %
% sets p( 1 :: n ) to a sieve of primes up to n %
procedure Eratosthenes ( logical array p( * ) ; integer value n ) ;
begin
p( 1 ) := false; p( 2 ) := true;
for i := 3 step 2 until n do p( i ) := true;
for i := 4 step 2 until n do p( i ) := false;
for i := 3 step 2 until truncate( sqrt( n ) ) do begin
integer ii; ii := i + i;
if p( i ) then for s := i * i step ii until n do p( s ) := false
end for_i ;
end Eratosthenes ;
% it can be shown that all the required primes are under 1000, however we will %
% not assume this, so we will allow for 4 digit numbers %
integer MAX_NUMBER, MAX_SUBSTRING;
MAX_NUMBER := 10000;
MAX_SUBSTRING := 100; % assume there will be at most 100 such primes %
begin
logical array prime( 1 :: MAX_NUMBER );
integer array sPrime( 1 :: MAX_SUBSTRING );
integer tCount, sCount, sPos;
% adds a substring prime to the list %
procedure addPrime ( integer value p ) ;
begin
sCount := sCount + 1;
sPrime( sCount ) := p;
writeon( i_w := 1, s_w := 0, " ", p )
end addPrime ;
% sieve the primes to MAX_NUMBER %
Eratosthenes( prime, MAX_NUMBER );
% clearly, the 1 digit primes are all substring primes %
sCount := 0;
for i := 1 until MAX_SUBSTRING do sPrime( i ) := 0;
for i := 2, 3, 5, 7 do addPrime( i );
% the subsequent primes can only have 3 or 7 as a final digit as they must end %
% with a prime digit and 2 and 5 would mean the number was divisible by 2 or 5 %
% as all substrings on the prime must also be prime, 33 and 77 are not possible %
% final digit pairs %
sPos := 1;
while sPrime( sPos ) not = 0 do begin
integer n3, n7;
n3 := ( sPrime( sPos ) * 10 ) + 3;
n7 := ( sPrime( sPos ) * 10 ) + 7;
if sPrime( sPos ) rem 10 not = 3 and prime( n3 ) then addPrime( n3 );
if sPrime( sPos ) rem 10 not = 7 and prime( n7 ) then addPrime( n7 );
sPos := sPos + 1
end while_sPrime_sPos_ne_0 ;
write( i_w := 1, s_w := 0, "Found ", sCount, " substring primes" )
end
end.</lang>
- Output:
2 3 5 7 23 37 53 73 373 Found 9 substring primes
AWK
<lang AWK>
- syntax: GAWK -f SUBSTRING_PRIMES.AWK
- converted from FreeBASIC
BEGIN {
start = 1
stop = 500
for (i=start; i<=stop; i++) {
if (is_substring_prime(i)) {
printf("%d ",i)
count++
}
}
printf("\nSubString Primes %d-%d: %d\n",start,stop,count)
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 is_substring_prime(n) {
if (!is_prime(i)) { return(0) }
if (n < 10) { return(1) }
if (!is_prime(n%100)) { return(0) }
if (!is_prime(n%10)) { return(0) }
if (!is_prime(int(n/10))) { return(0) }
if (n < 100) { return(1) }
if (!is_prime(int(n/100))) { return(0) }
if (!is_prime(int((n%100)/10))) { return(0) }
return(1)
} </lang>
- Output:
2 3 5 7 23 37 53 73 373 SubString Primes 1-500: 9
BASIC
BASIC256
<lang freebasic>function isPrime(v)
if v < 2 then return False
if v mod 2 = 0 then return v = 2
if v mod 3 = 0 then return v = 3
d = 5
while d * d <= v
if v mod d = 0 then return False else d += 2
end while
return True
end function
function isSubstringPrime (n)
if not isPrime(n) then return False if n < 10 then return True if not isPrime(n mod 100) then return False if not isPrime(n mod 10) then return False if not isPrime(n \ 10) then return False if n < 100 then return True if not isPrime(n \ 100) then return False if not isPrime((n mod 100) \ 10) then return False return True
end function
for i = 1 to 500
if isSubstringPrime(i) then print i; " ";
next i end</lang>
- Output:
Igual que la entrada de FreeBASIC.
PureBasic
<lang PureBasic>Procedure isPrime(v.i)
If v <= 1 : ProcedureReturn #False
ElseIf v < 4 : ProcedureReturn #True
ElseIf v % 2 = 0 : ProcedureReturn #False
ElseIf v < 9 : ProcedureReturn #True
ElseIf v % 3 = 0 : ProcedureReturn #False
Else
Protected r = Round(Sqr(v), #PB_Round_Down)
Protected f = 5
While f <= r
If v % f = 0 Or v % (f + 2) = 0
ProcedureReturn #False
EndIf
f + 6
Wend
EndIf
ProcedureReturn #True
EndProcedure
Procedure isSubstringPrime (n)
If Not isPrime(n) : ProcedureReturn #False ElseIf n < 10 : ProcedureReturn #True ElseIf Not isPrime(n % 100) : ProcedureReturn #False ElseIf Not isPrime(n % 10) : ProcedureReturn #False ElseIf Not isPrime(n / 10) : ProcedureReturn #False ElseIf n < 100 : ProcedureReturn #True ElseIf Not isPrime(n / 100) : ProcedureReturn #False ElseIf Not isPrime((n % 100)/10) : ProcedureReturn #False EndIf ProcedureReturn #True
EndProcedure
OpenConsole() For i.i = 1 To 500
If isSubstringPrime(i) : Print(Str(i) + " ") : EndIf
Next i Input() CloseConsole()</lang>
- Output:
Igual que la entrada de FreeBASIC.
Yabasic
<lang freebasic>sub isPrime(v)
if v < 2 then return False : fi
if mod(v, 2) = 0 then return v = 2 : fi
if mod(v, 3) = 0 then return v = 3 : fi
d = 5
while d * d <= v
if mod(v, d) = 0 then return False else d = d + 2 : fi
wend
return True
end sub
sub isSubstringPrime (n)
if not isPrime(n) then return False : fi if n < 10 then return True : fi if not isPrime(mod(n, 100)) then return False : fi if not isPrime(mod(n, 10)) then return False : fi if not isPrime(int(n / 10)) then return False : fi if n < 100 then return True : fi if not isPrime(int(n / 100)) then return False : fi if not isPrime(int(mod(n, 100))/10) then return False : fi return True
end sub
for i = 1 to 500
if isSubstringPrime(i) then print i, " "; : fi
next i end</lang>
- Output:
Igual que la entrada de FreeBASIC.
C++
<lang cpp>#include <iostream>
- include <vector>
std::vector<bool> prime_sieve(size_t limit) {
std::vector<bool> sieve(limit, true);
if (limit > 0)
sieve[0] = false;
if (limit > 1)
sieve[1] = false;
for (size_t i = 4; i < limit; i += 2)
sieve[i] = false;
for (size_t p = 3; ; p += 2) {
size_t q = p * p;
if (q >= limit)
break;
if (sieve[p]) {
size_t inc = 2 * p;
for (; q < limit; q += inc)
sieve[q] = false;
}
}
return sieve;
}
bool substring_prime(const std::vector<bool>& sieve, unsigned int n) {
for (; n != 0; n /= 10) {
if (!sieve[n])
return false;
for (unsigned int p = 10; p < n; p *= 10) {
if (!sieve[n % p])
return false;
}
}
return true;
}
int main() {
const unsigned int limit = 500;
std::vector<bool> sieve = prime_sieve(limit);
for (unsigned int i = 2; i < limit; ++i) {
if (substring_prime(sieve, i))
std::cout << i << '\n';
}
return 0;
}</lang>
- Output:
2 3 5 7 23 37 53 73 373
Factor
For fun, a translation of FreeBASIC.
<lang factor>USING: combinators kernel math math.primes prettyprint sequences ;
- ssp? ( n -- ? )
{
{ [ n prime? not ] [ f ] }
{ [ n 10 < ] [ t ] }
{ [ n 100 mod prime? not ] [ f ] }
{ [ n 10 mod prime? not ] [ f ] }
{ [ n 10 /i prime? not ] [ f ] }
{ [ n 100 < ] [ t ] }
{ [ n 100 /i prime? not ] [ f ] }
{ [ n 100 mod 10 /i prime? not ] [ f ] }
[ t ]
} cond ;
500 <iota> [ ssp? ] filter .</lang>
- Output:
V{ 2 3 5 7 23 37 53 73 373 }
FreeBASIC
Since this is limited to one, two, or three-digit numbers I will be a bit cheeky. <lang freebasic>#include "isprime.bas"
function is_ssp(n as uinteger) as boolean
if not isprime(n) then return false if n < 10 then return true if not isprime(n mod 100) then return false if not isprime(n mod 10) then return false if not isprime(n\10) then return false if n < 100 then return true if not isprime(n\100) then return false if not isprime( (n mod 100)\10 ) then return false return true
end function
for i as uinteger = 1 to 500
if is_ssp(i) then print i;" ";
next i print</lang>
- Output:
2 3 5 7 23 37 53 73 373
Go
Using a limit
<lang go>package main
import (
"fmt" "rcu"
)
func main() {
primes := rcu.Primes(499)
var sprimes []int
for _, p := range primes {
digits := rcu.Digits(p, 10)
var b1 = true
for _, d := range digits {
if !rcu.IsPrime(d) {
b1 = false
break
}
}
if b1 {
if len(digits) < 3 {
sprimes = append(sprimes, p)
} else {
b2 := rcu.IsPrime(digits[0]*10 + digits[1])
b3 := rcu.IsPrime(digits[1]*10 + digits[2])
if b2 && b3 {
sprimes = append(sprimes, p)
}
}
}
}
fmt.Println("Found", len(sprimes), "primes < 500 where all substrings are also primes, namely:")
fmt.Println(sprimes)
}</lang>
- Output:
Found 9 primes < 500 where all substrings are also primes, namely: [2 3 5 7 23 37 53 73 373]
Advanced
<lang go>package main
import (
"fmt" "rcu"
)
func main() {
results := []int{2, 3, 5, 7} // number must begin with a prime digit
odigits := []int{3, 7} // other digits must be 3 or 7
var discarded []int
tests := 4 // i.e. to obtain initial results in the first place
// check 2 digit numbers or greater
// note that 'results' is a moving feast. If the loop eventually terminates that's all there are.
for i := 0; i < len(results); i++ {
r := results[i]
for _, od := range odigits {
// the last digit of r and od must be different otherwise number would be divisible by 11
if (r % 10) != od {
n := r*10 + od
if rcu.IsPrime(n) {
results = append(results, n)
} else {
discarded = append(discarded, n)
}
tests++
}
}
}
fmt.Println("There are", len(results), "primes where all substrings are also primes, namely:")
fmt.Println(results)
fmt.Println("\nThe following numbers were also tested for primality but found to be composite:")
fmt.Println(discarded)
fmt.Println("\nTotal number of primality tests =", tests)
}</lang>
- Output:
There are 9 primes where all substrings are also primes, namely: [2 3 5 7 23 37 53 73 373] The following numbers were also tested for primality but found to be composite: [27 57 237 537 737 3737] Total number of primality tests = 15
jq
Works with gojq, the Go implementation of jq
In the following, we verify that there are only nine "substring primes" as defined for this task..
See e.g. Erdős-primes#jq for a suitable implementation of `is_prime`. <lang jq>def emit_until(cond; stream): label $out | stream | if cond then break $out else . end;
def primes:
2, (range(3;infinite;2) | select(is_prime));
def is_substring(checkPrime):
def isp: if . == "" then true else tonumber|is_prime end;
(if checkPrime then is_prime else true end)
and (tostring
| . as $s
| all(range(0;length) as $i | range($i; length+1) as $j | [$i,$j];
$s[.[0]:.[1]]|isp ));
- Output an array of the substring primes less than or equal to `.`
def substring_primes:
. as $n
| reduce emit_until(. > $n; primes) as $p ( null;
if $p | is_substring(false)
then . += [$p]
else .
end );
- Input: an array of the substring primes less than or equal to 373.
- Output: any other substring primes.
- Comment: if there are any others, they would have to be constructed
- from the numbers in the input array, as by assumption it includes
- all substring primes less than 100.
def verify:
. as $sp | range(0;length) as $i | range(0;length) as $j | ([$sp[$i, $j]] | map(tostring) | add | tonumber) as $candidate | if $candidate | IN($sp[]) then empty elif $candidate | is_substring(true) then $candidate else empty end;
500 | substring_primes | "Verifying that the following are the only substring primes:",
., "...", ( [verify] as $extra | if $extra == [] then "done" else $extra end )</lang>
- Output:
Verifying that the following are the only substring primes: [2,3,5,7,23,37,53,73,373] ... done
Julia
<lang julia>using Primes
const pmask = primesmask(1, 1000)
function isA085823(n, base = 10, sieve = pmask)
dig = digits(n; base=base)
for i in 1:length(dig), j in i:length(dig)
k = evalpoly(base, dig[i:j])
(k == 0 || !sieve[k]) && return false
end
return true
end
println(filter(isA085823, 1:1000))
</lang>
- Output:
[2, 3, 5, 7, 23, 37, 53, 73, 373]
Advanced task
<lang julia>using Primes
const nt, nons = [0], Int[]
counted_primetest(n) = (nt[1] += 1; b = isprime(n); !b && push!(nons, n); b)
- start with 1 digit primes
const results = [2, 3, 5, 7]
- check 2 digit candidates
for n in results, i in [3, 7]
if n != i
candidate = n * 10 + i
candidate < 100 && counted_primetest(candidate) && push!(results, candidate)
end
end
- check 3 digit candidates
for n in results, i in [3, 7]
if 10 < n < 100 && n % 10 != i
candidate = n * 10 + i
counted_primetest(candidate) && push!(results, candidate)
end
end
println("Results: $results.\nThe function isprime() was called $(nt[1]) times.") println("Discarded candidates: ", nons)
- Because 237, 537, and 737 are already excluded, we cannot generate any larger candidates from 373.
</lang>
- Output:
Results: [2, 3, 5, 7, 23, 37, 53, 73, 373]. The function isprime() was called 10 times. Discarded candidates: [27, 57, 237, 537, 737]
Mathematica/Wolfram Language
<lang Mathematica>Select[Range[500], PrimeQ[#] && AllTrue[Subsequences[IntegerDigits[#], {1, \[Infinity]}], FromDigits /* PrimeQ] &]</lang>
- Output:
{2, 3, 5, 7, 23, 37, 53, 73, 373}
Nim
The algorithm we use solves the advanced task as it finds the substring primes with only 11 primality tests. Note that, if we limit to numbers with at most three digits, 10 tests are sufficient. As we don’t use this limitation, we need one more test to detect than 3737 is not prime.
<lang Nim>import sequtils, strutils
type
Digit = 0..9 DigitSeq = seq[Digit]
func isOddPrime(n: Positive): bool =
## Check if "n" is an odd prime. assert n > 10 var d = 3 while d * d <= n: if n mod d == 0: return false inc d, 2 return true
func toInt(s: DigitSeq): int =
## Convert a sequence of digits to an int. for d in s: result = 10 * result + d
var result = @[2, 3, 5, 7]
var list: seq[DigitSeq] = result.mapIt(@[Digit it])
var primeTestCount = 0
while list.len != 0:
var newList: seq[DigitSeq]
for dseq in list:
for d in [Digit 3, 7]:
if dseq[^1] != d: # New digit must be different of last digit.
inc primeTestCount
let newDseq = dseq & d
let candidate = newDseq.toInt
if candidate.isOddPrime:
newList.add newDseq
result.add candidate
list = move(newList)
echo "List of substring primes: ", result.join(" ") echo "Number of primality tests: ", primeTestCount</lang>
- Output:
List of substring primes: 2 3 5 7 23 37 53 73 373 Number of primality tests: 11
Perl
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Substring_primes use warnings;
my %prime;
LOOP: for (2 .. 500 )
{
my %substrings = ();
/.+(?{ $prime{$&} or $substrings{$&}++ })(*FAIL)/;
for my $try ( sort { $a <=> $b } keys %substrings )
{
$try < 2 and next LOOP;
$prime{$try} || (1 x $try) !~ /^(11+)\1+$/ ? $prime{$try}++ : next LOOP;
}
}
printf " %d" x %prime . "\n", sort {$a <=> $b} keys %prime;</lang>
- Output:
2 3 5 7 23 37 53 73 373
Picat
Checking via substrings
<lang Picat>% get all the substrings of a number subs(N) = findall(S, (append(_Pre,S,_Post,N.to_string), S.len > 0) ).
go =>
Ps = [], foreach(Prime in primes(500)) (foreach(N in subs(Prime) prime(N) end -> Ps := Ps ++ [Prime] ; true) end, println(Ps).</lang>
- Output:
[2,3,5,7,23,37,53,73,373]
Checking via predicate
Here we must use cut (!) to ensure that the test does not continue after a satisfied test. This is a "red cut" (i.e. removing it would change the logic of the program) and these should be avoided if possible.
<lang Picat>t(N,false) :-
not prime(N),!.
t(N,true) :-
N < 10,!.
t(N,false) :-
not prime(N mod 100), !.
t(N,false) :-
not prime(N mod 10),!.
t(N,false) :-
not prime(N // 10),!.
t(N,true) :-
N < 100,!.
t(N,false) :-
not prime(N // 100),!.
t(N,false) :-
not prime((N mod 100) // 10),!.
t(_N,true).
go2 =>
println(findall(N,(member(N,1..500),t(N,Status),Status==true))).</lang>
- Output:
[2,3,5,7,23,37,53,73,373]
Phix
This tests a total of just 15 numbers for primality.
with javascript_semantics --sequence tested = {} --function a085823(integer p=0) -- sequence res={} function a085823(sequence res={}, tested={}, integer p=0) for i=(p!=0)+1 to 4 do integer t = get_prime(i) if t!=remainder(p,10) and (p=0 or t!=5) then t += p*10 if is_prime(t) then -- {res,tested} = a085823(res&t,tested,t) {res,tested} = a085823(deep_copy(res)&t,deep_copy(tested),t) -- [1] -- res &= t -- res &= a085823(t) else tested &= t end if end if end for return {res,tested} -- return res end function sequence {res,tested} = a085823() -- sort() if you prefer... --sequence res = a085823() -- sort() if you prefer... printf(1,"There are %d such A085823 primes: %V\n",{length(res),res}) printf(1,"%d innocent bystanders falsly accused of being prime (%d tests in total): %V\n", {length(tested),length(tested)+length(res),tested})
[1]It is possible that the compiler could be improved to avoid the need for those deep_copy().
The other seven commented out lines show a different way to avoid any use of deep_copy().
- Output:
There are 9 such A085823 primes: {2,23,3,37,373,5,53,7,73}
6 innocent bystanders falsly accused of being prime (15 tests in total): {237,27,3737,537,57,737}
Raku
<lang perl6>my @p = (^10).grep: *.is-prime;
say gather while @p {
.take for @p;
@p = ( @p X~ <3 7> ).grep: { .is-prime and .substr(*-2,2).is-prime }
}</lang>
- Output:
(2 3 5 7 23 37 53 73 373)
Stretch Goal
<lang perl6>my $prime-tests = 0; my @non-primes; sub spy-prime ($n) {
$prime-tests++; my $is-p = $n.is-prime;
push @non-primes, $n unless $is-p; return $is-p;
}
my @p = <2 3 5 7>;
say gather while @p {
.take for @p;
@p = ( @p X~ <3 7> ).grep: { !.ends-with(33|77) and .&spy-prime };
} .say for :$prime-tests, :@non-primes;</lang>
- Output:
(2 3 5 7 23 37 53 73 373) prime-tests => 11 non-primes => [27 57 237 537 737 3737]
REXX
<lang rexx>/*REXX program finds/shows decimal primes where all substrings are also prime, N < 500.*/ parse arg n cols . /*obtain optional argument from the CL.*/ if n== | n=="," then n= 500 /*Not specified? Then use the default.*/ if cols== | cols=="," then cols= 10 /* " " " " " " */ call genP /*build array of semaphores for primes.*/ w= 10 /*width of a number in any column. */ title= ' primes (base ten) where all substrings are also primes, where N < ' commas(n) say ' index │'center(title, 1 + cols*(w+1) ) /*display the title of the output. */ say '───────┼'center("" , 1 + cols*(w+1), '─') /* " " separator " " " */ found= 0; idx= 1 /*define # substring primes found; IDX.*/ $= /*a list of substring primes (so far). */
do j=1 for #; p= @.j; p2= substr(p, 2) /*search for primes that fit criteria. */
if verify(p, 014689, 'M')>0 then iterate /*does it contain any of these digits? */
if verify(p2, 25 , 'M')>0 then iterate /* " P2 " " " " " */
L= length(p) /*obtain the length of the P prime.*/
do k=1 for L-1 /*test for primality for all substrings*/
do m=k+1 to L; y= substr(p, k, m-1) /*extract a substring from the P prime.*/
if \!.y then iterate j /*does substring of P not prime? Skip.*/
end /*m*/
end /*k*/
found= found + 1 /*bump the number of substring primes. */
$= $ right( commas(p), w) /*add a substring prime ──► $ list. */
if found//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
if $\== then say center(idx, 7)'│' substr($, 2) /*display any residual numbers.*/ say '───────┴'center("" , 1 + cols*(w+1), '─') /* " a foot separator. */ say; say 'Found ' words($) title /* " the summary. */ exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ? /*──────────────────────────────────────────────────────────────────────────────────────*/ genP: !.= 0 /*placeholders for primes (semaphores).*/
@.1=2; @.2=3; @.3=5; @.4=7; @.5=11 /*define some low primes. */
!.2=1; !.3=1; !.5=1; !.7=1; !.11=1 /* " " " " flags. */
#=5; s.#= @.# **2 /*number of primes so far; prime². */
/* [↓] generate more primes ≤ high.*/
do j=@.#+2 by 2 to n-1 /*find odd primes from here on. */
parse var j -1 _; if _==5 then iterate /*J ÷ by 5? (right digit)*/
if j//3==0 then iterate; if j//7==0 then iterate /*" " " 3?; J ÷ by 7? */
do k=5 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 numbers ≤ √ J */
#= #+1; @.#= j; s.#= j*j; !.j= 1 /*bump # of Ps; assign next P; P²; P# */
end /*j*/; return</lang>
- output when using the default inputs:
index │ primes (base ten) where all substrings are also primes, where N < 500 ───────┼─────────────────────────────────────────────────────────────────────────────────────────────────────────────── 1 │ 2 3 5 7 23 37 53 73 373 ───────┴─────────────────────────────────────────────────────────────────────────────────────────────────────────────── Found 9 primes (base ten) where all substrings are also primes, where N < 500
Ring
<lang ring> load "stdlib.ring"
see "working..." + nl see "Numbers in which all substrings are primes:" + nl
row = 0 limit1 = 500
for n = 1 to limit1
flag = 1
strn = string(n)
for m = 1 to len(strn)
for p = 1 to len(strn)
temp = substr(strn,m,p)
if temp != ""
if isprime(number(temp))
flag = 1
else
flag = 0
exit 2
ok
ok
next
next
if flag = 1
see "" + n + " "
ok
next
see nl + "Found " + row + " numbers in which all substrings are primes" + nl see "done..." + nl </lang>
- Output:
working... Numbers in which all substrings are primes: 2 3 5 7 23 37 53 73 373 Found 9 numbers in which all substrings are primes done...
Sidef
Generic solution for any base >= 2. <lang ruby>func split_at_indices(array, indices) {
var parts = [] var i = 0
for j in (indices) {
parts << array.slice(i, j)
i = j+1
}
parts
}
func consecutive_partitions(array, callback) {
for k in (0..array.len) {
combinations(array.len, k, {|*indices|
var t = split_at_indices(array, indices)
if (t.sum_by{.len} == array.len) {
callback(t)
}
})
}
}
func is_substring_prime(digits, base) {
for k in (^digits) {
digits.first(k+1).digits2num(base).is_prime || return false
}
consecutive_partitions(digits, {|part|
part.all { .digits2num(base).is_prime } || return false
})
return true
}
func generate_from_prefix(p, base, digits) {
var seq = [p]
for d in (digits) {
var t = [d, p...]
if (is_prime(t.digits2num(base)) && is_substring_prime(t, base)) {
seq << __FUNC__(t, base, digits)...
}
}
return seq
}
func substring_primes(base) { # finite sequence for each base >= 2
var prime_digits = (base-1 -> primes) # prime digits < base
prime_digits.map {|p| generate_from_prefix([p], base, prime_digits)... }\
.map {|t| digits2num(t, base) }\
.sort
}
for base in (2..20) {
say "base = #{base}: #{substring_primes(base)}"
}</lang>
- Output:
base = 2: [] base = 3: [2] base = 4: [2, 3, 11] base = 5: [2, 3, 13, 17, 67] base = 6: [2, 3, 5, 17, 23] base = 7: [2, 3, 5, 17, 19, 23, 37] base = 8: [2, 3, 5, 7, 19, 23, 29, 31, 43, 47, 59, 61, 157, 239, 251, 349, 379, 479, 491] base = 9: [2, 3, 5, 7, 23, 29, 47] base = 10: [2, 3, 5, 7, 23, 37, 53, 73, 373] base = 11: [2, 3, 5, 7, 29, 79] base = 12: [2, 3, 5, 7, 11, 29, 31, 41, 43, 47, 67, 71, 89, 137, 139, 359, 499, 503, 521, 569, 571, 809, 857, 859, 6043] base = 13: [2, 3, 5, 7, 11, 29, 31, 37, 41, 67, 379] base = 14: [2, 3, 5, 7, 11, 13, 31, 41, 47, 53, 73, 83, 101, 103, 109, 157, 167, 193, 439, 661, 1033, 2203] base = 15: [2, 3, 5, 7, 11, 13, 37, 41, 43, 47, 107, 167, 197, 557, 617, 647] base = 16: [2, 3, 5, 7, 11, 13, 37, 43, 53, 59, 61, 83, 179, 181, 211, 691, 947, 3389] base = 17: [2, 3, 5, 7, 11, 13, 37, 41, 47, 53, 223, 631] base = 18: [2, 3, 5, 7, 11, 13, 17, 41, 43, 47, 53, 59, 61, 67, 71, 97, 101, 103, 107, 131, 137, 139, 211, 239, 241, 251, 311, 313, 317, 751, 787, 859, 1069, 1103, 1109, 1213, 1223, 1283, 1289, 1759, 1831, 1861, 1871, 1931, 1933, 2371, 3803, 4349, 4523, 5639, 5647, 15467, 19867, 34807] base = 19: [2, 3, 5, 7, 11, 13, 17, 41, 43, 59, 97, 211] base = 20: [2, 3, 5, 7, 11, 13, 17, 19, 43, 47, 53, 59, 67, 71, 73, 79, 103, 107, 113, 151, 157, 223, 227, 233, 239, 263, 271, 277, 347, 353, 359, 383, 397, 1063, 1423, 1427, 1433, 1439, 1471, 1583, 1597, 3023, 4663, 4783, 5273, 5279, 7673, 28663]
Wren
Using a limit
<lang ecmascript>import "/math" for Int var primes = Int.primeSieve(499) var sprimes = [] for (p in primes) {
var digits = Int.digits(p)
var b1 = digits.all { |d| Int.isPrime(d) }
if (b1) {
if (digits.count < 3) {
sprimes.add(p)
} else {
var b2 = Int.isPrime(digits[0] * 10 + digits[1])
var b3 = Int.isPrime(digits[1] * 10 + digits[2])
if (b2 && b3) sprimes.add(p)
}
}
} System.print("Found %(sprimes.count) primes < 500 where all substrings are also primes, namely:") System.print(sprimes)</lang>
- Output:
Found 9 primes < 500 where all substrings are also primes, namely: [2, 3, 5, 7, 23, 37, 53, 73, 373]
Advanced
This follows the logic in the OEIS A085823 comments. <lang ecmascript>import "/math" for Int
var results = [2, 3, 5, 7] // number must begin with a prime digit var odigits = [3, 7] // other digits must be 3 or 7 as there would be a composite substring otherwise var discarded = [] var tests = 4 // i.e. to obtain initial results in the first place
// check 2 digit numbers or greater // note that 'results' is a moving feast. If the loop eventually terminates that's all there are. for (r in results) {
for (od in odigits) {
// the last digit of r and od must be different otherwise number would be divisible by 11
if ((r % 10) != od) {
var n = r * 10 + od
if (Int.isPrime(n)) results.add(n) else discarded.add(n)
tests = tests + 1
}
}
}
System.print("There are %(results.count) primes where all substrings are also primes, namely:") System.print(results) System.print("\nThe following numbers were also tested for primality but found to be composite:") System.print(discarded) System.print("\nTotal number of primality tests = %(tests)")</lang>
- Output:
There are 9 primes where all substrings are also primes, namely: [2, 3, 5, 7, 23, 37, 53, 73, 373] The following numbers were also tested for primality but found to be composite: [27, 57, 237, 537, 737, 3737] Total number of primality tests = 15
XPL0
<lang XPL0>func IsPrime(N); \Return 'true' if N is a prime number int N, I; [if N <= 1 then return false; for I:= 2 to sqrt(N) do
if rem(N/I) = 0 then return false;
return true; ];
int Digit, Huns, Tens, Ones, N; [Digit:= [0, 2, 3, 5, 7]; \leading zeros are ok for Huns:= 0 to 4 do
for Tens:= 0 to 4 do
if Huns+Tens=0 or IsPrime(Digit(Huns)*10+Digit(Tens)) then
for Ones:= 1 to 4 do \can't end in 0
[N:= Digit(Huns)*100 + Digit(Tens)*10 + Digit(Ones);
if N<500 & IsPrime(N) & IsPrime(Digit(Tens)*10+Digit(Ones)) then
[IntOut(0, N); ChOut(0, ^ )];
];
]</lang>
- Output:
2 3 5 7 23 37 53 73 373