Palindromic primes in base 16: Difference between revisions
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Found 13 palindromic primes in base 16 |
Found 13 palindromic primes in base 16 |
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done... |
done... |
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</pre> |
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=={{header|Ruby}}== |
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<syntaxhighlight lang="ruby">res = Prime.each(500).filter_map do |pr| |
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str = pr.to_s(16) |
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str if str == str.reverse |
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end |
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puts res.join(", ")</syntaxhighlight> |
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{{out}} |
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<pre>2, 3, 5, 7, b, d, 11, 101, 151, 161, 191, 1b1, 1c1 |
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</pre> |
</pre> |
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Revision as of 18:07, 15 April 2023
- Task
Find palindromic primes n in base 16, where n < 50010
11l
F is_prime(a)
I a == 2
R 1B
I a < 2 | a % 2 == 0
R 0B
L(i) (3 .. Int(sqrt(a))).step(2)
I a % i == 0
R 0B
R 1B
L(n) 500
V h = hex(n)
I h == reversed(h) & is_prime(n)
print(h, end' ‘ ’)
print()
- Output:
2 3 5 7 B D 11 101 151 161 191 1B1 1C1
Action!
INCLUDE "H6:SIEVE.ACT"
BYTE FUNC Palindrome(CHAR ARRAY s)
BYTE l,r
l=1 r=s(0)
WHILE l<r
DO
IF s(l)#s(r) THEN RETURN (0) FI
l==+1 r==-1
OD
RETURN (1)
PROC IntToHex(INT i CHAR ARRAY hex)
CHAR ARRAY digits="0123456789ABCDEF"
BYTE d
hex(0)=0
WHILE i#0
DO
d=i MOD 16
hex(0)==+1
hex(hex(0))=digits(d+1)
i==/16
OD
RETURN
BYTE Func IsPalindromicPrime(INT i CHAR ARRAY hex
BYTE ARRAY primes)
BYTE d
INT rev,tmp
IF primes(i)=0 THEN
RETURN (0)
FI
IntToHex(i,hex)
IF Palindrome(hex) THEN
RETURN (1)
FI
RETURN (0)
PROC Main()
DEFINE MAX="499"
BYTE ARRAY primes(MAX+1)
INT i,count=[0]
CHAR ARRAY hex(5)
Put(125) PutE() ;clear the screen
Sieve(primes,MAX+1)
FOR i=2 TO MAX
DO
IF IsPalindromicPrime(i,hex,primes) THEN
Print(hex) Put(32)
count==+1
FI
OD
PrintF("%EThere are %I palindromic primes",count)
RETURN
- Output:
Screenshot from Atari 8-bit computer
2 3 5 7 B D 11 101 151 161 191 1B1 1C1 There are 13 palindromic primes
ALGOL 68
BEGIN # find primes that are palendromic in base 16 #
# sieve the primes to 499 #
PR read "primes.incl.a68" PR
[]BOOL prime = PRIMESIEVE 499;
# returns an array of the digits of n in the specified base #
PRIO DIGITS = 9;
OP DIGITS = ( INT n, INT base )[]INT:
IF INT v := ABS n;
v < base
THEN v # single dogit #
ELSE # multiple digits #
[ 1 : 10 ]INT result;
INT d pos := UPB result + 1;
INT v := ABS n;
WHILE v > 0 DO
result[ d pos -:= 1 ] := v MOD base;
v OVERAB base
OD;
result[ d pos : UPB result ]
FI # DIGITS # ;
# returns TRUE if the digits in d form a palindrome, FALSE otherwise #
OP PALINDROMIC = ( []INT d )BOOL:
BEGIN
INT left := LWB d, right := UPB d;
BOOL is palindromic := TRUE;
WHILE left < right AND is palindromic DO
is palindromic := d[ left ] = d[ right ];
left +:= 1;
right -:= 1
OD;
is palindromic
END;
# print the palendromic primes in the base 16 #
STRING base digits = "0123456789ABCDEF";
FOR n TO UPB prime DO
IF prime[ n ] THEN
# have a prime #
IF []INT d = n DIGITS 16;
PALINDROMIC d
THEN
# the prime is palindromic in base 16 #
print( ( " " ) );
FOR c FROM LWB d TO UPB d DO print( ( base digits[ d[ c ] + 1 ] ) ) OD
FI
FI
OD
END
- Output:
2 3 5 7 B D 11 101 151 161 191 1B1 1C1
AppleScript
on isPrime(n)
if ((n < 4) or (n is 5)) then return (n > 1)
if ((n mod 2 = 0) or (n mod 3 = 0) or (n mod 5 = 0)) then return false
repeat with i from 7 to (n ^ 0.5) div 1 by 30
if ((n mod i = 0) or (n mod (i + 4) = 0) or (n mod (i + 6) = 0) or ¬
(n mod (i + 10) = 0) or (n mod (i + 12) = 0) or (n mod (i + 16) = 0) or ¬
(n mod (i + 22) = 0) or (n mod (i + 24) = 0)) then return false
end repeat
return true
end isPrime
on task()
set digits to "0123456789ABCDEF"'s characters
set output to {"2"} -- Take "2" as read.
repeat with n from 3 to 499 by 2 -- All other primes are odd.
if (isPrime(n)) then
-- Only the number's hex digit /values/ are needed for testing.
set vals to {}
repeat until (n = 0)
set vals's beginning to n mod 16
set n to n div 16
end repeat
-- If they're palindromic, build a text representation and append this to the output.
if (vals = vals's reverse) then
set hex to digits's item ((vals's beginning) + 1)
repeat with i from 2 to (count vals)
set hex to hex & digits's item ((vals's item i) + 1)
end repeat
set output's end to hex
end if
end if
end repeat
return output
end task
task()
- Output:
{"2", "3", "5", "7", "B", "D", "11", "101", "151", "161", "191", "1B1", "1C1"}
Arturo
palindrome?: function [a][
and? -> prime? a
-> equal? digits.base:16 a reverse digits.base:16 a
]
print map select 1..500 => palindrome? 'x -> upper as.hex x
- Output:
2 3 5 7 B D 11 101 151 161 191 1B1 1C1
AWK
# syntax: GAWK -f PALINDROMIC_PRIMES_IN_BASE_16.AWK
BEGIN {
start = 1
stop = 499
for (i=start; i<=stop; i++) {
hex = sprintf("%X",i)
if (is_prime(i) && hex == reverse(hex)) {
printf("%4s%1s",hex,++count%10?"":"\n")
}
}
printf("\nPalindromic 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 reverse(str, i,rts) {
for (i=length(str); i>=1; i--) {
rts = rts substr(str,i,1)
}
return(rts)
}
- Output:
2 3 5 7 B D 11 101 151 161 191 1B1 1C1 Palindromic primes 1-499: 13
C++
See C++ solution for task 'Palindromic Primes'.
F#
This task uses Extensible Prime Generator (F#)
let rec fN g=[yield g%16; if g>15 then yield! fN(g/16)]
primes32()|>Seq.takeWhile((>)500)|>Seq.filter(fun g->let g=fN g in List.rev g=g)|>Seq.iter(printf "%0x "); printfn ""
- Output:
2 3 5 7 b d 11 101 151 161 191 1b1 1c1
Factor
USING: kernel math.parser math.primes prettyprint sequences
sequences.extras ;
500 primes-upto [ >hex ] [ dup reverse = ] map-filter .
- Output:
V{ "2" "3" "5" "7" "b" "d" "11" "101" "151" "161" "191" "1b1" "1c1" }
FreeBASIC
Function isprime(num As Ulongint) As Boolean
For i As Integer = 2 To Sqr(num)
If (num Mod i = 0) Then Return False
Next i
Return True
End Function
Function reverse(Byval text As String) As String
Dim As String text2 = text
Dim As Integer x, lt = Len(text)
For x = 0 To lt Shr 1 - 1
Swap text2[x], text2[lt - x - 1]
Next x
Return text2
End Function
Dim As Integer inicio = 2, final = 499, cont = 0
For i As Integer = inicio To final
Dim As String hexi = Str(Hex(i))
If isprime(i) = True And hexi = reverse(hexi) Then
cont += 1
Print Hex(i); " ";
End If
Next i
Print !"\n\nEncontrados"; cont; " primos palindrómicos entre " & inicio & " y " & final
Sleep
- Output:
2 3 5 7 B D 11 101 151 161 191 1B1 1C1 Encontrados 13 primos palindrómicos entre 2 y 499
Go
package main
import (
"fmt"
"rcu"
"strconv"
"strings"
)
func reverse(s string) string {
chars := []rune(s)
for i, j := 0, len(chars)-1; i < j; i, j = i+1, j-1 {
chars[i], chars[j] = chars[j], chars[i]
}
return string(chars)
}
func main() {
fmt.Println("Primes < 500 which are palindromic in base 16:")
primes := rcu.Primes(500)
count := 0
for _, p := range primes {
hp := strconv.FormatInt(int64(p), 16)
if hp == reverse(hp) {
fmt.Printf("%3s ", strings.ToUpper(hp))
count++
if count%5 == 0 {
fmt.Println()
}
}
}
fmt.Println("\n\nFound", count, "such primes.")
}
- Output:
Primes < 500 which are palindromic in base 16: 2 3 5 7 B D 11 101 151 161 191 1B1 1C1 Found 13 such primes.
jq
Works with gojq, the Go implementation of jq
This entry uses a generator that produces an unbounded stream of arrays of the form [dec, hex], where `dec` is the palindromic prime as a JSON number, and `hex` is the JSON string corresponding to its hexadecimal representation.
For a suitable implementation of `is_prime`, see e.g. Erdős-primes#jq.
# '''Preliminaries'''
def emit_until(cond; stream): label $out | stream | if cond then break $out else . end;
# decimal number to exploded hex array
def exploded_hex:
def stream:
recurse(if . > 0 then ./16|floor else empty end) | . % 16 ;
if . == 0 then [48]
else [stream] | reverse | .[1:]
| map(if . < 10 then 48 + . else . + 87 end)
end;
The Task
# Output: a stream of [decimal, hexadecimal] values
def palindromic_primes_in_base_16:
(2, (range(3; infinite; 2) | select(is_prime)))
| exploded_hex as $hex
|select( $hex | (. == reverse))
| [., ($hex|implode)] ;
emit_until(.[0] >= 500; palindromic_primes_in_base_16)
- Output:
[2,"2"] [3,"3"] [5,"5"] [7,"7"] [11,"b"] [13,"d"] [17,"11"] [257,"101"] [337,"151"] [353,"161"] [401,"191"] [433,"1b1"] [449,"1c1"]
Julia
using Primes
ispal(n, base) = begin dig = digits(n, base=base); dig == reverse(dig) end
palprimes(N, base=16) = [string(i, base=16) for i in primes(N) if ispal(i, base)]
foreach(s -> print(s, " "), palprimes(500, 16)) # 2 3 5 7 b d 11 101 151 161 191 1b1 1c1
Mathematica /Wolfram Language
Giving the base 10 numbers and the base 16 numbers:
Select[Range[499], PrimeQ[#] \[And] PalindromeQ[IntegerDigits[#, 16]] &]
BaseForm[%, 16]
- Output:
{2, 3, 5, 7, 11, 13, 17, 257, 337, 353, 401, 433, 449} {Subscript[2, 16],Subscript[3, 16],Subscript[5, 16],Subscript[7, 16],Subscript[b, 16],Subscript[d, 16],Subscript[11, 16],Subscript[101, 16],Subscript[151, 16],Subscript[161, 16],Subscript[191, 16],Subscript[1b1, 16],Subscript[1c1, 16]}
Nim
import strformat, strutils
func isPalindromic(s: string): bool =
for i in 1..s.len:
if s[i-1] != s[^i]: return false
result = true
func isPrime(n: Natural): bool =
if n < 2: return false
if n mod 2 == 0: return n == 2
if n mod 3 == 0: return n == 3
var d = 5
while d * d <= n:
if n mod d == 0: return false
inc d, 2
if n mod d == 0: return false
inc d, 4
return true
var list: seq[string]
for n in 0..<500:
let h = &"{n:x}"
if h.isPalindromic and n.isPrime: list.add h
echo "Found ", list.len, " palindromic primes in base 16:"
echo list.join(" ")
- Output:
Found 13 palindromic primes in base 16: 2 3 5 7 b d 11 101 151 161 191 1b1 1c1
Perl
(1 x $_ ) !~ /^(11+)\1+$/ # test if prime
and $h = sprintf "%x", $_ # convert to hex
and $h eq reverse $h # palindromic?
and print "$h " # much rejoicing
for 1..500;
- Output:
1 2 3 5 7 b d 11 101 151 161 191 1b1 1c1
Phix
with javascript_semantics function palindrome(string s) return s=reverse(s) end function sequence res = filter(apply(true,sprintf,{{"%x"},get_primes_le(500)}),palindrome) printf(1,"found %d: %s\n",{length(res),join(res)})
- Output:
found 13: 2 3 5 7 B D 11 101 151 161 191 1B1 1C1
Quackery
See Palindromic primes#Quackery for rest of code. This is a trivial modification.
16 base put
500 times
[ i^ isprime if
[ i^ digits palindromic if
[ i^ echo sp ] ] ]
base release
- Output:
2 3 5 7 B D 11 101 151 161 191 1B1 1C1
Raku
Trivial modification of Palindromic primes task.
say "{+$_} matching numbers:\n{.batch(10)».fmt('%3X').join: "\n"}"
given (^500).grep: { .is-prime and .base(16) eq .base(16).flip };
- Output:
13 matching numbers: 2 3 5 7 B D 11 101 151 161 191 1B1 1C1
REXX
/*REXX program finds and displays hexadecimal palindromic primes for all N < 500. */
parse arg hi cols . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi= 500 /*Not specified? Then use the default.*/
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
call genP /*build array of semaphores for primes.*/
w= 8 /*max width of a number in any column. */
title= ' palindromic primes in base 16 that are < ' hi
if cols>0 then say ' index │'center(title, 1 + cols*(w+1) )
if cols>0 then say '───────┼'center("" , 1 + cols*(w+1), '─')
finds= 0; idx= 1 /*define # of palindromic primes & idx.*/
$= /*hex palindromic primes list (so far).*/
do j=1 for hi; if \!.j then iterate /*J (decimal) not prime? Then skip.*/ /* ◄■■■■■■■■ a filter. */
x= d2x(j); if x\==reverse(x) then iterate /*Hex value not palindromic? " " */ /* ◄■■■■■■■■ a filter. */
finds= finds + 1 /*bump the number of palindromic primes*/
if cols<0 then iterate /*Build the list (to be shown later)? */
$= $ right( lowerHex(x), w) /*use a lowercase version of the hex #.*/
if finds//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) /*possible display residual output.*/
if cols>0 then say '───────┴'center("" , 1 + cols*(w+1), '─')
say
say 'Found ' finds title
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
lowerHex: return translate( arg(1), 'abcdef', "ABCDEF") /*convert hex chars──►lowercase.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: !.= 0; hip= max(hi, copies(9,length(hi))) /*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; sq.#= @.# **2 /*number of primes so far; prime². */
/* [↓] generate more primes ≤ high.*/
do j=@.#+2 by 2 to hip /*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 sq.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; sq.#= j*j; !.j= 1 /*bump # of Ps; assign next P; P²; P# */
end /*j*/; return
- output when using the default inputs:
index │ palindromic primes in base 16 that are < 500 ───────┼─────────────────────────────────────────────────────────────────────────────────────────── 1 │ 2 3 5 7 b d 11 101 151 161 11 │ 191 1b1 1c1 ───────┴─────────────────────────────────────────────────────────────────────────────────────────── Found 13 palindromic primes in base 16 that are < 500
Ring
load "stdlib.ring"
see "working..." + nl
see "Palindromic primes in base 16:" + nl
row = 0
limit = 500
for n = 1 to limit
hex = hex(n)
if ispalindrome(hex) and isprime(n)
see "" + upper(hex) + " "
row = row + 1
if row%5 = 0
see nl
ok
ok
next
see nl + "Found " + row + " palindromic primes in base 16" + nl
see "done..." + nl
- Output:
working... Palindromic primes in base 16: 2 3 5 7 B D 11 101 151 161 191 1B1 1C1 Found 13 palindromic primes in base 16 done...
Ruby
res = Prime.each(500).filter_map do |pr|
str = pr.to_s(16)
str if str == str.reverse
end
puts res.join(", ")
- Output:
2, 3, 5, 7, b, d, 11, 101, 151, 161, 191, 1b1, 1c1
Rust
See Rust solution for task 'Palindromic Primes'.
Seed7
$ include "seed7_05.s7i";
const func boolean: isPrime (in integer: number) is func
result
var boolean: prime is FALSE;
local
var integer: upTo is 0;
var integer: testNum is 3;
begin
if number = 2 then
prime := TRUE;
elsif odd(number) and number > 2 then
upTo := sqrt(number);
while number rem testNum <> 0 and testNum <= upTo do
testNum +:= 2;
end while;
prime := testNum > upTo;
end if;
end func;
const proc: main is func
local
var integer: n is 0;
var string: hex is "";
begin
for n range 2 to 499 do
if isPrime(n) then
hex := n radix 16;
if hex = reverse(hex) then
write(hex <& " ");
end if;
end if;
end for;
end func;
- Output:
2 3 5 7 b d 11 101 151 161 191 1b1 1c1
Sidef
func palindromic_primes(upto, base = 10) {
var list = []
for (var p = 2; p <= upto; p = p.next_palindrome(base)) {
list << p if p.is_prime
}
return list
}
var list = palindromic_primes(500, 16)
list.each {|p|
say "#{'%3s' % p}_10 = #{'%3s' % p.base(16)}_16"
}
- Output:
2_10 = 2_16 3_10 = 3_16 5_10 = 5_16 7_10 = 7_16 11_10 = b_16 13_10 = d_16 17_10 = 11_16 257_10 = 101_16 337_10 = 151_16 353_10 = 161_16 401_10 = 191_16 433_10 = 1b1_16 449_10 = 1c1_16
Wren
import "/math" for Int
import "/fmt" for Conv, Fmt
System.print("Primes < 500 which are palindromic in base 16:")
var primes = Int.primeSieve(500)
var count = 0
for (p in primes) {
var hp = Conv.Itoa(p, 16)
if (hp == hp[-1..0]) {
Fmt.write("$3s ", hp)
count = count + 1
if (count % 5 == 0) System.print()
}
}
System.print("\n\nFound %(count) such primes.")
- Output:
Primes < 500 which are palindromic in base 16: 2 3 5 7 B D 11 101 151 161 191 1B1 1C1 Found 13 such primes.
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;
];
func Reverse(N, Base); \Reverse order of digits in N for given Base
int N, Base, M;
[M:= 0;
repeat N:= N/Base;
M:= M*Base + rem(0);
until N=0;
return M;
];
int Count, N;
[SetHexDigits(1);
Count:= 0;
for N:= 0 to 500-1 do
if IsPrime(N) & N=Reverse(N, 16) then
[HexOut(0, N);
Count:= Count+1;
if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\);
];
CrLf(0);
IntOut(0, Count);
Text(0, " such numbers found.
");
]
- Output:
2 3 5 7 B D 11 101 151 161 191 1B1 1C1 13 such numbers found.