Numbers in base 10 that are palindromic in bases 2, 4, and 16: Difference between revisions
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let start() be
for i =
if palindrome(i,2) & palindrome(i,4) & palindrome(i,16)
do writef("%N*N", i)</lang>
{{out}}
<pre>
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Revision as of 22:01, 24 June 2021
- Task
- Find numbers in base 10 that are palindromic in bases 2, 4, and 16, where n < 25000
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
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
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
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...
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.