Numbers in base 10 that are palindromic in bases 2, 4, and 16: Difference between revisions
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=={{header|APL}}== |
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{{works with|Dyalog APL}} |
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<lang apl>(⊢(/⍨)(2 4 16∧.((⊢≡⌽)(⊥⍣¯1))⊢)¨)0,⍳24999</lang> |
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{{out}} |
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<pre>0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845</pre> |
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=={{header|BASIC}}== |
=={{header|BASIC}}== |
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Revision as of 22:35, 24 June 2021
- Task
- Find numbers in base 10 that are palindromic in bases 2, 4, and 16, where n < 25,000
ALGOL 68
<lang algol68>BEGIN # show numbers in decimal that are palindromic in bases 2, 4 and 16 #
INT max number = 25 000; # maximum number to consider #
INT min base = 2; # smallest base needed #
INT max digits = BEGIN # number of digits max number has in the smallest base #
INT d := 1;
INT v := max number;
WHILE v >= min base DO
v OVERAB min base;
d PLUSAB 1
OD;
d
END;
# returns 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 : max digits ]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 numbers in decimal that are palendromic primes in bases 2, 4 and 16 #
FOR n FROM 0 TO max number DO
IF PALINDROMIC ( n DIGITS 16 ) THEN
IF PALINDROMIC ( n DIGITS 4 ) THEN
IF PALINDROMIC ( n DIGITS 2 ) THEN
print( ( " ", whole( n, 0 ) ) )
FI
FI
FI
OD
END</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
APL
<lang apl>(⊢(/⍨)(2 4 16∧.((⊢≡⌽)(⊥⍣¯1))⊢)¨)0,⍳24999</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
BASIC
<lang basic>10 DEFINT A-Z: DEFDBL R 20 FOR I=1 TO 25000 30 B=2: GOSUB 100: IF R<>I GOTO 70 40 B=4: GOSUB 100: IF R<>I GOTO 70 50 B=16: GOSUB 100: IF R<>I GOTO 70 60 PRINT I, 70 NEXT 80 END 100 R=0: N=I 110 IF N=0 THEN RETURN 120 R=R*B+N MOD B 130 N=N\B 140 GOTO 110</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
BCPL
<lang bcpl>get "libhdr" manifest $( MAXIMUM = 25000 $)
let reverse(n, base) = valof $( let r = 0
while n > 0
$( r := r*base + n rem base
n := n / base
$)
resultis r
$)
let palindrome(n, base) = n = reverse(n, base)
let start() be
for i = 0 to MAXIMUM
if palindrome(i,2) & palindrome(i,4) & palindrome(i,16)
do writef("%N*N", i)</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
C
<lang c>#include <stdio.h>
- define MAXIMUM 25000
int reverse(int n, int base) {
int r;
for (r = 0; n; n /= base)
r = r*base + n%base;
return r;
}
int palindrome(int n, int base) {
return n == reverse(n, base);
}
int main() {
int i, c = 0;
for (i = 0; i < MAXIMUM; i++) {
if (palindrome(i, 2) &&
palindrome(i, 4) &&
palindrome(i, 16)) {
printf("%5d%c", i, ++c % 12 ? ' ' : '\n');
}
}
printf("\n");
return 0;
}</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
Cowgol
<lang cowgol>include "cowgol.coh"; const MAXIMUM := 25000;
sub reverse(n: uint16, base: uint16): (r: uint16) is
r := 0;
while n != 0 loop
r := r * base + n % base;
n := n / base;
end loop;
end sub;
var i: uint16 := 0; var c: uint8 := 0; while i < MAXIMUM loop
if reverse(i,2) == i
and reverse(i,4) == i
and reverse(i,16) == i
then
c := c + 1;
print_i16(i);
if c == 15 then
print_nl();
c := 0;
else
print_char(' ');
end if;
end if;
i := i + 1;
end loop; print_nl();</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
Factor
<lang factor>USING: io kernel math.parser prettyprint sequences ;
25,000 <iota> [
{ 2 4 16 } [ >base ] with map [ dup reverse = ] all?
] filter [ pprint bl ] each nl</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
Go
<lang go>package main
import (
"fmt" "rcu" "strconv"
)
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("Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:")
var numbers []int
for i := int64(0); i < 25000; i++ {
b2 := strconv.FormatInt(i, 2)
if b2 == reverse(b2) {
b4 := strconv.FormatInt(i, 4)
if b4 == reverse(b4) {
b16 := strconv.FormatInt(i, 16)
if b16 == reverse(b16) {
numbers = append(numbers, int(i))
}
}
}
}
for i, n := range numbers {
fmt.Printf("%6s ", rcu.Commatize(n))
if (i+1)%10 == 0 {
fmt.Println()
}
}
fmt.Println("\n\nFound", len(numbers), "such numbers.")
}</lang>
- Output:
Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:
0 1 3 5 15 17 51 85 255 257
273 771 819 1,285 1,365 3,855 4,095 4,097 4,369 12,291
13,107 20,485 21,845
Found 23 such numbers.
Phix
with javascript_semantics function palindrome(string s) return s=reverse(s) end function function p2416(integer n) return palindrome(sprintf("%a",{{2,n}})) and palindrome(sprintf("%a",{{4,n}})) and palindrome(sprintf("%a",{{16,n}})) end function sequence res = apply(filter(tagset(25000,0),p2416),sprint) printf(1,"%d found: %s\n",{length(res),join(res)})
- Output:
23 found: 0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
PL/M
<lang plm>100H: /* CP/M CALLS */ BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS; EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT; PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
DECLARE MAXIMUM LITERALLY '25$000';
/* PRINT A NUMBER */ PRINT$NUMBER: PROCEDURE (N);
DECLARE S (7) BYTE INITIAL ('..... $');
DECLARE (N, P) ADDRESS, C BASED P BYTE;
P = .S(5);
DIGIT:
P = P - 1; C = N MOD 10 + '0'; N = N / 10; IF N > 0 THEN GO TO DIGIT; CALL PRINT(P);
END PRINT$NUMBER;
/* REVERSE NUMBER GIVEN BASE */ REVERSE: PROCEDURE (N, B) ADDRESS;
DECLARE (N, R) ADDRESS, B BYTE;
R = 0;
DO WHILE N > 0;
R = R*B + N MOD B;
N = N/B;
END;
RETURN R;
END REVERSE;
/* CHECK IF NUMBER IS PALINDROME */ PALIN: PROCEDURE (N, B) BYTE;
DECLARE N ADDRESS, B BYTE; RETURN N = REVERSE(N, B);
END PALIN;
DECLARE I ADDRESS, C BYTE; C = 0; DO I = 0 TO MAXIMUM;
IF PALIN(I,2) AND PALIN(I,4) AND PALIN(I,16) THEN DO;
CALL PRINT$NUMBER(I);
C = C + 1;
IF C = 15 THEN DO;
CALL PRINT(.(13,10,'$'));
C = 0;
END;
END;
END; CALL EXIT; EOF</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
Python
<lang python>def reverse(n, base):
r = 0
while n > 0:
r = r*base + n%base
n = n//base
return r
def palindrome(n, base):
return n == reverse(n, base)
cnt = 0 for i in range(25000):
if all(palindrome(i, base) for base in (2,4,16)):
cnt += 1
print("{:5}".format(i), end=" \n"[cnt % 12 == 0])
print()</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
Raku
<lang perl6>put "{+$_} such numbers:\n", .batch(10)».fmt('%5d').join("\n") given (^25000).grep: -> $n { all (2,4,16).map: { $n.base($_) eq $n.base($_).flip } }</lang>
- Output:
23 such numbers:
0 1 3 5 15 17 51 85 255 257
273 771 819 1285 1365 3855 4095 4097 4369 12291
13107 20485 21845
REXX
Programming note: the conversions of a decimal number to another base (radix) was ordered such that the fastest
base conversion was performed before the other conversions.
The use of REXX's BIFs to convert decimal numbers to binary and hexadecimal were used (instead of the base
function) because they are much faster).
This REXX version takes advantage that no even integers need be tested (except for the single exception: zero),
this makes the execution twice as fast.
<lang rexx>/*REXX pgm finds non─neg integers that are palindromes in base 2, 4, and 16, where N<25k*/
numeric digits 100 /*ensure enough dec. digs for large #'s*/
parse arg n cols . /*obtain optional argument from the CL.*/
if n== | n=="," then n = 25000 /*Not specified? Then use the default.*/
if cols== | cols=="," then cols= 10 /* " " " " " " */
w= 10 /*width of a number in any column. */
title= ' non-negative integers that are palindromes in base 2, 4, and 16, where N < ' ,
commas(n)
say ' index │'center(title, 1 + cols*(w+1) ) /*display the title for the output. */ say '───────┼'center("" , 1 + cols*(w+1), '─') /* " a sep " " " */ $= right(0, w+1) /*list of numbers found (so far). */ found= 1 /*# of finds (so far), the only even #.*/ idx= 1 /*set the IDX (index) to unity. */
do j=1 by 2 to n-1 /*find int palindromes in bases 2,4,16.*/
h= d2x(j) /*convert dec. # to hexadecimal. */
if h\==reverse(h) then iterate /*Hex number not palindromic? Skip.*/ /* ◄■■■■■■■■ a filter. */
b= x2b( d2x(j) ) + 0 /*convert dec. # to hex, then to binary*/
if b\==reverse(b) then iterate /*Binary number not palindromic? Skip.*/ /* ◄■■■■■■■■ a filter. */
q= base(j, 4) /*convert a decimal integer to base 4. */
if q\==reverse(q) then iterate /*Base 4 number not palindromic? Skip.*/ /* ◄■■■■■■■■ a filter. */
found= found + 1 /*bump number of found such numbers. */
$= $ right( commas(j), w) /*add the found number ───► $ 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) /*possible display residual output.*/ say '───────┴'center("" , 1 + cols*(w+1), '─') /*display the foot sep for output. */ say say 'Found ' found title 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 ? /*──────────────────────────────────────────────────────────────────────────────────────*/ base: procedure; parse arg #,t,,y; @= 0123456789abcdefghijklmnopqrstuvwxyz /*up to 36*/
@@= substr(@, 2); do while #>=t; y= substr(@, #//t + 1, 1)y; #= # % t
end; return substr(@, #+1, 1)y</lang>
- output when using the default inputs:
index │ non-negative integers that are palindromes in base 2, 4, and 16, where N < 25,000 ───────┼─────────────────────────────────────────────────────────────────────────────────────────────────────────────── 1 │ 0 1 3 5 15 17 51 85 255 257 11 │ 273 771 819 1,285 1,365 3,855 4,095 4,097 4,369 12,291 21 │ 13,107 20,485 21,845 ───────┴─────────────────────────────────────────────────────────────────────────────────────────────────────────────── Found 23 non-negative integers that are palindromes in base 2, 4, and 16, where N < 25,000
Ring
<lang ring> load "stdlib.ring" see "working..." + nl see "Numbers in base 10 that are palindromic in bases 2, 4, and 16:" + nl
row = 0 limit = 25000
for n = 1 to limit
base2 = decimaltobase(n,2)
base4 = decimaltobase(n,4)
base16 = hex(n)
bool = ispalindrome(base2) and ispalindrome(base4) and ispalindrome(base16)
if bool = 1
see "" + n + " "
row = row + 1
if row%5 = 0
see nl
ok
ok
next
see nl + "Found " + row + " numbers" + nl see "done..." + nl
func decimaltobase(nr,base)
decList = 0:15
baseList = ["0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"]
binList = []
binary = 0
remainder = 1
while(nr != 0)
remainder = nr % base
ind = find(decList,remainder)
rem = baseList[ind]
add(binList,rem)
nr = floor(nr/base)
end
binlist = reverse(binList)
binList = list2str(binList)
binList = substr(binList,nl,"")
return binList
</lang>
- Output:
working... Numbers in base 10 that are palindromic in bases 2, 4, and 16: 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845 Found 22 numbers done...
Seed7
<lang seed7>$ include "seed7_05.s7i";
const func boolean: palindrome (in string: input) is
return input = reverse(input);
const proc: main is func
local
var integer: n is 1;
begin
write("0 ");
for n range 1 to 24999 step 2 do
if palindrome(n radix 2) and palindrome(n radix 4) and palindrome(n radix 16) then
write(n <& " ");
end if;
end for;
end func;</lang>
- Output:
0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
Wren
<lang ecmascript>import "/fmt" for Conv, Fmt import "/seq" for Lst
System.print("Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:") var numbers = [] for (i in 0..24999) {
var b2 = Conv.itoa(i, 2)
if (b2 == b2[-1..0]) {
var b4 = Conv.itoa(i, 4)
if (b4 == b4[-1..0]) {
var b16 = Conv.itoa(i, 16)
if (b16 == b16[-1..0]) numbers.add(i)
}
}
} for (chunk in Lst.chunks(numbers, 8)) Fmt.print("$,6d", chunk) System.print("\nFound %(numbers.count) such numbers.")</lang>
- Output:
Numbers under 25,000 in base 10 which are palindromic in bases 2, 4 and 16:
0 1 3 5 15 17 51 85
255 257 273 771 819 1,285 1,365 3,855
4,095 4,097 4,369 12,291 13,107 20,485 21,845
Found 23 such numbers.
XPL0
<lang XPL0>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; [Count:= 0; for N:= 1 to 25000-1 do
if N = Reverse(N, 2) &
N = Reverse(N, 4) &
N = Reverse(N, 16) then
[IntOut(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. "); ]</lang>
- Output:
1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845 22 such numbers found.