Two identical strings: Difference between revisions
Line 1,190: | Line 1,190: | ||
<lang julia>function gentwoidenticalstringsinbase(base, maxnum, verbose=true) |
<lang julia>function gentwoidenticalstringsinbase(base, maxnum, verbose=true) |
||
dig = digits(maxnum; base) |
dig = digits(maxnum; base) |
||
len = length(dig) |
len, k = length(dig), len ÷ 2 |
||
k = len ÷ 2 |
|||
topnum = evalpoly(base, isodd(len) ? [base - 1 for i in 1:k] : dig[k+1:end]) |
topnum = evalpoly(base, isodd(len) ? [base - 1 for i in 1:k] : dig[k+1:end]) |
||
gen = [i * (base^k + 1) for i in 1:topnum] |
gen = [i * (base^k + 1) for i in 1:topnum] |
Revision as of 00:18, 5 April 2021
- Task
Find and display (here on this page) positive integers whose base 2 representation is the concatenation of two identical binary strings,
where n (in base ten) < 1,00010 (one thousand).
For each decimal number, show its decimal form and also its binary form.
8080 Assembly
<lang 8080asm> ;;; Print positive integers whose base-2 representation ;;; is the concatenation of two identical binary strings, ;;; for 1 < n < 1000 puts: equ 9 ; CP/M syscall to print a string org 100h lxi b,1 ; Counter loop: mov h,b ; HL = counter mov l,c call concat ; Get current concatenated number lxi d,1000 ; Reached the end yet? call cmp16 rnc ; Stop when >1000 push b ; Keep the counter push h ; And the concatenated number call hldec ; Print decimal value pop h ; Restore number call hlbin ; Print binary value lxi d,nl ; Print newline mvi c,puts call 5 pop b ; Restore counter inx b ; Increment counter jmp loop ;;; 16-bit compare HL to DE cmp16: mov a,h cmp d rnz mov a,l cmp e ret ;;; Concatenate HL with itself concat: push h ; Keep a copy of HL on the stack mov d,h ; DE = copy of HL mov e,l ctloop: mov a,d ; When DE=0, we are done ora e jz ctdone mov a,d ; Rotate DE left rar mov d,a mov a,e rar mov e,a dad h ; And rotate HL right (add to itself) jmp ctloop ctdone: pop d ; Retrieve old HL dad d ; Add to shifted version (same as OR) ret ;;; Print HL as a decimal value hldec: lxi d,outbuf push d ; Output pointer on the stack lxi b,-10 ; Divisor decdgt: lxi d,-1 ; Quotient div10: inx d ; Divide HL by 10 using trial subtraction dad b jc div10 mvi a,'0'+10 add l ; L contains remainder - 10 pop h ; Retrieve output pointer dcx h ; Store digit mov m,a push h xchg ; Continue with quotient mov a,h ; If any digits left ora l jnz decdgt ; Find the next digits pop d ; Otherwise, retrieve pointer mvi c,puts ; And print result using CP/M jmp 5 ;;; Print HL as a binary value hlbin: lxi d,outbuf ora a ; Zero the carry flag bindgt: mov a,h ; Rotate HL right rar mov h,a mov a,l rar mov l,a mvi a,0 ; A = '0' + carry flag (i.e. lowest bit) aci '0' dcx d ; Store digit stax d mov a,h ; Any more digits? ora l jnz bindgt ; If so, find next digits mvi c,puts ; Otherwise, print the result jmp 5 db '***********' outbuf: db 9,'$' nl: db 13,10,'$'</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
8086 Assembly
<lang asm>puts: equ 9 cpu 8086 org 100h section .text main: mov di,1 ; Counter .loop: mov ax,di call concat ; Concatenate current number to itself cmp ax,1000 jge .done ; Stop when >= 1000 mov si,ax ; Keep a copy of AX call pdec ; Print decimal value mov ax,si call pbin ; Print binary value mov bx,nl ; Print newline call pstr inc di ; Next number jmp .loop .done: ret ;;; Concatenate AX to itself concat: mov bx,ax ; Store a copy of AX in BP mov cx,ax ; Store a copy of AX in CX .loop: shl ax,1 ; Shift AX left shr cx,1 ; Shift CX right jnz .loop ; Keep going until CX is zero or ax,bx ; OR original AX with shifted AX ret ;;; Print AX as decimal pdec: mov bp,10 ; Divisor mov bx,outbuf ; Buffer pointer .loop: xor dx,dx div bp add dl,'0' ; Add '0' to remainder dec bx ; Store digit mov [bx],dl test ax,ax ; Any more digits? jnz .loop jmp pstr ; When done, print the result ;;; Print AX as binary pbin: mov bx,outbuf ; Buffer pointer .loop: shr ax,1 ; Shift AX mov dl,'0' ; ASCII 0 or 1 adc dl,0 dec bx mov [bx],dl ; Store digit test ax,ax jnz .loop pstr: mov ah,puts ; When done, print the result mov dx,bx int 21h ret section .data nl: db 13,10,'$' db '****************' outbuf: db 9,'$'</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
ALGOL 68
<lang algol68>BEGIN # show the decimal and binary representations of numbers that are of the concatenation of #
# two identical binary strings # # returns a binary representation of v # OP TOBINSTRING = ( INT v )STRING: IF v = 0 THEN "0" ELSE STRING result := ""; INT rest := v; WHILE rest > 0 DO IF ODD rest THEN "1" ELSE "0" FI +=: result; rest OVERAB 2 OD; result FI # TOBINSTRING # ; INT power of 2 := 1; FOR b WHILE IF b = power of 2 THEN power of 2 *:= 2 FI; INT cat value = ( b * power of 2 ) + b; cat value < 1000 DO print( ( whole( cat value, -4 ), ": ", TOBINSTRING cat value, newline ) ) OD
END</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
ALGOL W
<lang algolw>BEGIN
INTEGER PROCEDURE BITSP ( INTEGER VALUE BT ) ; BEGIN INTEGER BITN, BITRSL, BITIDX; BITN := BT; BITRSL := 0; BITIDX := 1; WHILE BITN > 0 DO BEGIN INTEGER BITNX; BITNX := BITN DIV 2; BITRSL := BITRSL + BITIDX*(BITN-BITNX*2); BITN := BITNX; BITIDX := BITIDX*10 END; BITRSL END BITSP ;
INTEGER PROCEDURE DPLBIT ( INTEGER VALUE DVAL ) ; BEGIN INTEGER DTEMP, DSHFT; DTEMP := DVAL; DSHFT := DVAL; WHILE DTEMP > 0 DO BEGIN DSHFT := DSHFT * 2; DTEMP := DTEMP DIV 2; END; DSHFT + DVAL END DPLBIT ;
BEGIN INTEGER N; N := 0; WHILE BEGIN N := N + 1; DPLBIT(N) < 1000 END DO WRITE( S_W := 0, I_W := 3, DPLBIT(N), ": ", I_W := 10, BITSP(DPLBIT(N)) ) END
END.</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
APL
<lang APL>↑(((⊂2∘⊥),⊂)(,⍨2∘⊥⍣¯1))¨⍳30</lang>
- Output:
3 1 1 10 1 0 1 0 15 1 1 1 1 36 1 0 0 1 0 0 45 1 0 1 1 0 1 54 1 1 0 1 1 0 63 1 1 1 1 1 1 136 1 0 0 0 1 0 0 0 153 1 0 0 1 1 0 0 1 170 1 0 1 0 1 0 1 0 187 1 0 1 1 1 0 1 1 204 1 1 0 0 1 1 0 0 221 1 1 0 1 1 1 0 1 238 1 1 1 0 1 1 1 0 255 1 1 1 1 1 1 1 1 528 1 0 0 0 0 1 0 0 0 0 561 1 0 0 0 1 1 0 0 0 1 594 1 0 0 1 0 1 0 0 1 0 627 1 0 0 1 1 1 0 0 1 1 660 1 0 1 0 0 1 0 1 0 0 693 1 0 1 0 1 1 0 1 0 1 726 1 0 1 1 0 1 0 1 1 0 759 1 0 1 1 1 1 0 1 1 1 792 1 1 0 0 0 1 1 0 0 0 825 1 1 0 0 1 1 1 0 0 1 858 1 1 0 1 0 1 1 0 1 0 891 1 1 0 1 1 1 1 0 1 1 924 1 1 1 0 0 1 1 1 0 0 957 1 1 1 0 1 1 1 1 0 1 990 1 1 1 1 0 1 1 1 1 0
BASIC
<lang BASIC>10 DEFINT A-Z: DIM B(15) 20 N=0 30 N=N+1 40 C=0: X=N 50 C=C+1 60 X=X\2 70 IF X>0 THEN 50 80 K=N+2^C*N 90 IF K>1000 THEN END 100 PRINT K, 110 FOR I=C*2 TO 1 STEP -1 120 B(I)=K AND 1 130 K=K\2 140 NEXT I 150 FOR I=1 TO C*2 160 PRINT USING "#";B(I); 170 NEXT I 180 PRINT 190 GOTO 30</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
Befunge
<lang befunge>1>:::>:#v_++:91+v >^ / >\vv**::<
^2\*2<>`#@_v v_v#!:<\+19\0.:< $ : ^ /2<
+v<>2%68*+\^ 1:^ $!, ^_^</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
C
<lang c>#include <stdio.h>
- include <stdint.h>
uint8_t bit_length(uint32_t n) {
uint8_t r; for (r=0; n; r++) n >>= 1; return r;
}
uint32_t concat_bits(uint32_t n) {
return (n << bit_length(n)) | n;
}
char *bits(uint32_t n) {
static char buf[33]; char *ptr = &buf[33]; *--ptr = 0; do { *--ptr = '0' + (n & 1); } while (n >>= 1); return ptr;
}
int main() {
uint32_t n, r; for (n=1; (r = concat_bits(n)) < 1000; n++) { printf("%d: %s\n", r, bits(r)); } return 0;
}</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
C#
<lang csharp>using System; using static System.Console; class Program { static void Main() { int c = 0, lmt = 1000;
for (int n = 1, p = 2, k; n <= lmt; n++) if ((k = n + n * (p += n >= p ? p : 0)) > lmt) break; else Console.Write("{0,3} ({1,-10}) {2}", k, Convert.ToString(k, 2), ++c % 5 == 0 ? "\n" : ""); Write("\nFound {0} numbers whose base 2 representation is the " + "concatenation of two identical binary strings.", c); } }</lang>
- Output:
Same as Visual Basic. NET
C++
<lang cpp>#include <iostream>
- include <string>
// Given the base 2 representation of a number n, transform it into the base 2 // representation of n + 1. void base2_increment(std::string& s) {
size_t z = s.rfind('0'); if (z != std::string::npos) { s[z] = '1'; size_t count = s.size() - (z + 1); s.replace(z + 1, count, count, '0'); } else { s.assign(s.size() + 1, '0'); s[0] = '1'; }
}
int main() {
std::cout << "Decimal\tBinary\n"; std::string s("1"); for (unsigned int n = 1; ; ++n) { unsigned int i = n + (n << s.size()); if (i >= 1000) break; std::cout << i << '\t' << s << s << '\n'; base2_increment(s); }
}</lang>
- Output:
Decimal Binary 3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
Cowgol
<lang cowgol>include "cowgol.coh";
sub bitLength(n: uint32): (l: uint8) is
l := 0; while n != 0 loop n := n >> 1; l := l + 1; end loop;
end sub;
sub concatBits(n: uint32): (r: uint32) is
r := (n << bitLength(n)) | n;
end sub;
sub printBits(n: uint32) is
var buf: uint8[33]; var ptr := &buf[32]; [ptr] := 0; loop ptr := @prev ptr; [ptr] := '0' + (n as uint8 & 1); n := n >> 1; if n == 0 then break; end if; end loop; print(ptr);
end sub;
var n: uint32 := 1; loop
var r := concatBits(n); if r > 1000 then break; end if; print_i32(r); print(": "); printBits(r); print_nl(); n := n + 1;
end loop;</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
Factor
<lang factor>USING: formatting kernel lists lists.lazy math math.parser sequences ;
1 lfrom [ >bin dup append bin> ] lmap-lazy [ 1000 < ] lwhile [ dup "%d %b\n" printf ] leach</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
FALSE
<lang FALSE>1[$$$[$][2/\2*\]#%|$1000>~][
$.": " 0\10\[$1&'0+\2/$][]#% [$][,]#% 1+
]#%%</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
FOCAL
<lang FOCAL>01.10 S N=0 01.20 S N=N+1 01.30 D 3 01.40 I (K-1000)1.5;Q 01.50 T %3,K,": " 01.60 D 4 01.70 G 1.2
02.10 S BC=0;S BT=N 02.20 S BC=BC+1 02.30 S BT=FITR(BT/2) 02.40 I (-BT)2.2
03.10 D 2;S I=BC;S BT=N 03.20 S BX=FITR(BT/2) 03.30 S I=I-1 03.40 S B(I)=BT-BX*2 03.50 S BT=BX 03.60 I (-I)3.2,3.2 03.70 F I=0,BC-1;S B(BC+I)=B(I) 03.80 S BC=BC*2;S K=0 03.90 F I=0,BC-1;S K=K*2+B(I)
04.10 F I=0,BC-1;D 4.3 04.20 T !;R 04.30 I (B(I))4.4,4.5,4.4 04.40 T "1" 04.50 T "0"</lang>
- Output:
= 3: 11 = 10: 1010 = 15: 1111 = 36: 100100 = 45: 101101 = 54: 110110 = 63: 111111 = 136: 10001000 = 153: 10011001 = 170: 10101010 = 187: 10111011 = 204: 11001100 = 221: 11011101 = 238: 11101110 = 255: 11111111 = 528: 1000010000 = 561: 1000110001 = 594: 1001010010 = 627: 1001110011 = 660: 1010010100 = 693: 1010110101 = 726: 1011010110 = 759: 1011110111 = 792: 1100011000 = 825: 1100111001 = 858: 1101011010 = 891: 1101111011 = 924: 1110011100 = 957: 1110111101 = 990: 1111011110
Forth
<lang forth>: concat-self
dup dup begin dup while 1 rshift swap 1 lshift swap repeat drop or
- print-bits
0 swap begin dup 1 and '0 + swap 1 rshift dup 0= until drop begin dup while emit repeat drop
- to1000
1 begin dup concat-self dup 1000 < while dup . 9 emit print-bits cr 1+ repeat 2drop
to1000 bye</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
Fortran
<lang fortran> program IdentStr
implicit none integer n, concat, bits n = 1 100 if (concat(n) .lt. 1000) then write (*,'(I3,2X,I11)') concat(n), bits(concat(n)) n = n + 1 goto 100 end if stop end
C Concatenate binary representation of number with itself
integer function concat(num) integer num, sl, sr sl = num sr = num 100 if (sr .gt. 0) then sl = sl * 2 sr = sr / 2 goto 100 end if concat = num + sl end
C Calculate binary representation of number
integer function bits(num) integer num, n, bx n = num bits = 0 bx = 1 100 if (n .gt. 0) then bits = bits + bx * mod(n,2) bx = bx * 10 n = n / 2 goto 100 end if end</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
FreeBASIC
<lang freebasic>dim as uinteger n=1, k=0 do
k = n + 2*n*2^int(log(n)/log(2)) if k<1000 then print k, bin(k) else end n=n+1
loop</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
Alternate
No log() function required. <lang freebasic>dim as uinteger n = 1, k = 0, p = 2 do
if n >= p then p = p + p k = n + n * p if k < 1000 then print k, bin(k) else end n = n + 1
loop</lang>
- Output:
Same as log() version.
Haskell
<lang haskell>import Control.Monad import Data.Bits import Text.Printf
-- Find the amount of bits required to represent a number nBits :: Int -> Int nBits = liftM2 (-) finiteBitSize countLeadingZeros
-- Concatenate the bits of a number to itself concatSelf :: Int -> Int concatSelf = (.|.) =<< ap shift nBits
-- Integers whose base-2 representation is the concatenation of -- two identical binary strings identStrInts :: [Int] identStrInts = map concatSelf [1..]
main :: IO () main = putStr $ unlines $ map (join $ printf "%d: %b") to1000
where to1000 = takeWhile (<= 1000) identStrInts</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
J
<lang J>(":,': ',":@#:)@(,~&.#:)"0 (>:i.30)</lang>
- Output:
3: 1 1 10: 1 0 1 0 15: 1 1 1 1 36: 1 0 0 1 0 0 45: 1 0 1 1 0 1 54: 1 1 0 1 1 0 63: 1 1 1 1 1 1 136: 1 0 0 0 1 0 0 0 153: 1 0 0 1 1 0 0 1 170: 1 0 1 0 1 0 1 0 187: 1 0 1 1 1 0 1 1 204: 1 1 0 0 1 1 0 0 221: 1 1 0 1 1 1 0 1 238: 1 1 1 0 1 1 1 0 255: 1 1 1 1 1 1 1 1 528: 1 0 0 0 0 1 0 0 0 0 561: 1 0 0 0 1 1 0 0 0 1 594: 1 0 0 1 0 1 0 0 1 0 627: 1 0 0 1 1 1 0 0 1 1 660: 1 0 1 0 0 1 0 1 0 0 693: 1 0 1 0 1 1 0 1 0 1 726: 1 0 1 1 0 1 0 1 1 0 759: 1 0 1 1 1 1 0 1 1 1 792: 1 1 0 0 0 1 1 0 0 0 825: 1 1 0 0 1 1 1 0 0 1 858: 1 1 0 1 0 1 1 0 1 0 891: 1 1 0 1 1 1 1 0 1 1 924: 1 1 1 0 0 1 1 1 0 0 957: 1 1 1 0 1 1 1 1 0 1 990: 1 1 1 1 0 1 1 1 1 0
Julia
filter version
<lang julia>function twoidenticalstringsinbase(base, maxnum, verbose=true)
found = Int[] for i in 1:maxnum dig = digits(i; base) k = length(dig) iseven(k) && dig[begin:begin+k÷2-1] == dig[begin+k÷2:end] && push!(found, i) end if verbose println("\nDecimal Base $base") for n in found println(rpad(n, 9), string(n, base=base)) end end return found
end
twoidenticalstringsinbase(2, 999) twoidenticalstringsinbase(16, 999)
</lang>
- Output:
Decimal Base 2 3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110 Decimal Base 16 17 11 34 22 51 33 68 44 85 55 102 66 119 77 136 88 153 99 170 aa 187 bb 204 cc 221 dd 238 ee 255 ff
generator version
<lang julia>function gentwoidenticalstringsinbase(base, maxnum, verbose=true)
dig = digits(maxnum; base) len, k = length(dig), len ÷ 2 topnum = evalpoly(base, isodd(len) ? [base - 1 for i in 1:k] : dig[k+1:end]) gen = [i * (base^k + 1) for i in 1:topnum] if verbose println("\nDecimal Base $base") for n in gen println(rpad(n, 9), string(n, base=base)) end end return gen
end
twoidenticalstringsinbase(2, 999) twoidenticalstringsinbase(16, 999)
</lang>
- Output:
Same as filter version
MAD
<lang MAD> NORMAL MODE IS INTEGER
INTERNAL FUNCTION(BT) ENTRY TO BITS. BITN = BT BITRSL = 0 BITIDX = 1
GETBIT WHENEVER BITN.G.0
BITNX = BITN/2 BITRSL = BITRSL + BITIDX*(BITN-BITNX*2) BITN = BITNX BITIDX = BITIDX*10 TRANSFER TO GETBIT END OF CONDITIONAL FUNCTION RETURN BITRSL END OF FUNCTION INTERNAL FUNCTION(DVAL) ENTRY TO DPLBIT. DTEMP = DVAL DSHFT = DVAL
DSTEP WHENEVER DTEMP.G.0
DSHFT = DSHFT * 2 DTEMP = DTEMP / 2 TRANSFER TO DSTEP END OF CONDITIONAL FUNCTION RETURN DSHFT + DVAL END OF FUNCTION THROUGH NUM, FOR N=1, 1, DPLBIT.(N).GE.1000
NUM PRINT FORMAT NFMT, DPLBIT.(N), BITS.(DPLBIT.(N))
VECTOR VALUES NFMT = $I3,2H: ,I10*$ END OF PROGRAM </lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
Pascal
<lang pascal>program IdenticalStrings; const
LIMIT = 1000;
var
n: Integer;
function BitLength(n: Integer): Integer;
var count: Integer; begin count := 0; while n > 0 do begin n := n shr 1; count := count + 1; end; BitLength := count; end;
function Concat(n: Integer): Integer;
begin Concat := n shl BitLength(n) or n; end;
procedure WriteBits(n: Integer);
var bit: Integer; begin bit := 1 shl (BitLength(n)-1); while bit > 0 do begin if (bit and n) <> 0 then Write('1') else Write('0'); bit := bit shr 1; end; end;
begin
n := 1; while Concat(n) < LIMIT do begin Write(Concat(n)); Write(': '); WriteBits(Concat(n)); WriteLn; n := n + 1; end;
end.</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
Python
<lang python>def bits(n):
"""Count the amount of bits required to represent n""" r = 0 while n: n >>= 1 r += 1 return r
def concat(n):
"""Concatenate the binary representation of n to itself""" return n << bits(n) | n
n = 1 while concat(n) <= 1000:
print("{0}: {0:b}".format(concat(n))) n += 1</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
Perl
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Two_identical_strings use warnings;
while( 1 )
{ my $binary = ( sprintf "%b", ++$- ) x 2; (my $decimal = oct "b$binary") >= 1000 and last; printf "%4d %s\n", $decimal, $binary; }</lang>
- Output:
3 11 10 1010 15 1111 36 100100 45 101101 54 110110 63 111111 136 10001000 153 10011001 170 10101010 187 10111011 204 11001100 221 11011101 238 11101110 255 11111111 528 1000010000 561 1000110001 594 1001010010 627 1001110011 660 1010010100 693 1010110101 726 1011010110 759 1011110111 792 1100011000 825 1100111001 858 1101011010 891 1101111011 924 1110011100 957 1110111101 990 1111011110
Phix
integer n = 1 sequence res = {} while true do string binary = sprintf("%b%b",n) integer decimal = to_number(binary,0,2) if decimal>1000 then exit end if res &= {sprintf("%-4d %-10s",{decimal,binary})} n += 1 end while printf(1,"Found %d numbers:\n%s\n",{n-1,join_by(res,5,6)})
- Output:
Found 30 numbers: 3 11 54 110110 187 10111011 528 1000010000 693 1010110101 858 1101011010 10 1010 63 111111 204 11001100 561 1000110001 726 1011010110 891 1101111011 15 1111 136 10001000 221 11011101 594 1001010010 759 1011110111 924 1110011100 36 100100 153 10011001 238 11101110 627 1001110011 792 1100011000 957 1110111101 45 101101 170 10101010 255 11111111 660 1010010100 825 1100111001 990 1111011110
Raku
<lang perl6>my @cat = (1..*).map: { :2([~] .base(2) xx 2) }; say "{+$_} matching numbers\n{.batch(5)».map({$_ ~ .base(2).fmt('(%s)')})».fmt('%15s').join: "\n"}\n"
given @cat[^(@cat.first: * > 1000, :k)];</lang>
- Output:
30 matching numbers 3(11) 10(1010) 15(1111) 36(100100) 45(101101) 54(110110) 63(111111) 136(10001000) 153(10011001) 170(10101010) 187(10111011) 204(11001100) 221(11011101) 238(11101110) 255(11111111) 528(1000010000) 561(1000110001) 594(1001010010) 627(1001110011) 660(1010010100) 693(1010110101) 726(1011010110) 759(1011110111) 792(1100011000) 825(1100111001) 858(1101011010) 891(1101111011) 924(1110011100) 957(1110111101) 990(1111011110)
REXX
version 1, sans formatting
<lang>/*REXX program finds/displays decimal numbers whose binary version is a doubled literal.*/ numeric digits 20 /*ensure 'nuff dec. digs for conversion*/
do #=1 for 1000-1; b= x2b( d2x(#) ) + 0 /*find binary values that can be split.*/ L= length(b); if L//2 then iterate /*get length of binary; if odd, skip. */ if left(b, L%2)\==right(b, L%2) then iterate /*Left half ≡ right half? No, skip it.*/ say right(#, 4)':' right(b, 12) /*display number in decimal and binary.*/ end /*#*/ /*stick a fork in it, we're all done. */</lang>
- output (shown at three-quarter size.)
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
version 1, with formatting
<lang rexx>/*REXX program finds/displays decimal numbers whose binary version is a doubled literal.*/ numeric digits 100 /*ensure hangling of larger integers. */ parse arg hi cols . /*obtain optional argument from the CL.*/ if hi== | hi=="," then hi= 1000 /* " " " " " " */ if cols== | cols=="," then cols= 4 /* " " " " " " */ w= 20 /*width of a number in any column. */
@dnbn= ' decimal integers whose binary version is a doubled binary literal, N < ' , commas(hi)
if cols>0 then say ' index │'center(@dnbn, 1 + cols*(w+1) ) if cols>0 then say '───────┼'center("" , 1 + cols*(w+1), '─')
- = 0; idx= 1 /*initialize # of integers and index. */
$= /*a list of nice primes (so far). */
do j=1 for hi-1; b= x2b( d2x(j) ) + 0 /*find binary values that can be split.*/ L= length(b); h= L % 2 /*obtain length of the binary value. */ if L//2 then iterate /*Can binary version be split? No, skip*/ if left(b, h)\==right(b, h) then iterate /*Left half match right half? " " */ #= # + 1 /*bump the number of integers found. */ if cols==0 then iterate /*Build the list (to be shown later)? */ c= commas(j) || '(' || b")" /*maybe add commas, add binary version.*/ $= $ right(c, max(w, length(c) ) ) /*add a nice prime ──► list, allow big#*/ if #//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 say 'Found ' commas(#) @dnbn 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 ?</lang>
- output when using the default inputs:
index │ decimal integers whose binary version is a doubled binary literal, N < 1,000 ───────┼───────────────────────────────────────────────────────────────────────────────────── 1 │ 3(11) 10(1010) 15(1111) 36(100100) 5 │ 45(101101) 54(110110) 63(111111) 136(10001000) 9 │ 153(10011001) 170(10101010) 187(10111011) 204(11001100) 13 │ 221(11011101) 238(11101110) 255(11111111) 528(1000010000) 17 │ 561(1000110001) 594(1001010010) 627(1001110011) 660(1010010100) 21 │ 693(1010110101) 726(1011010110) 759(1011110111) 792(1100011000) 25 │ 825(1100111001) 858(1101011010) 891(1101111011) 924(1110011100) 29 │ 957(1110111101) 990(1111011110) Found 30 decimal integers whose binary version is a doubled binary literal, N < 1,000
Ring
<lang ring>load "stdlib.ring"
decList = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] baseList = ["0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"]
see "working..." + nl see "Numbers whose base 2 representation is the juxtaposition of two identical strings:" + nl
row = 0 limit1 = 1000
for n = 1 to limit1
bin = decimaltobase(n,2) ln = len(bin) if ln & 1 = 0 if left(bin,ln/2) = right(bin,ln/2) row++ see sfl(n, 3) + " (" + sfrs(bin, 10) + ") " if row % 5 = 0 see nl ok ok ok
next
? nl + "Found " + row + " numbers whose base 2 representation is the juxtaposition of two identical strings" ? "done..."
func decimaltobase(nr,base)
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
- a very plain string formatter, intended to even up columnar outputs
def sfrs x, y
l = len(x) x += " " if l > y y = l ok return substr(x, 1, y)
- a very plain string formatter, intended to even up columnar outputs
def sfl x, y
s = string(x) l = len(s) if l > y y = l ok return substr(" ", 11 - y + l) + s</lang>
- Output:
working... Numbers whose base 2 representation is the juxtaposition of two identical strings: 3 (11 ) 10 (1010 ) 15 (1111 ) 36 (100100 ) 45 (101101 ) 54 (110110 ) 63 (111111 ) 136 (10001000 ) 153 (10011001 ) 170 (10101010 ) 187 (10111011 ) 204 (11001100 ) 221 (11011101 ) 238 (11101110 ) 255 (11111111 ) 528 (1000010000) 561 (1000110001) 594 (1001010010) 627 (1001110011) 660 (1010010100) 693 (1010110101) 726 (1011010110) 759 (1011110111) 792 (1100011000) 825 (1100111001) 858 (1101011010) 891 (1101111011) 924 (1110011100) 957 (1110111101) 990 (1111011110) Found 30 numbers whose base 2 representation is the juxtaposition of two identical strings done...
Snobol
<lang snobol> define("bits(n)") :(bits_end) bits bits = gt(n,0) remdr(n,2) bits :f(return)
n = n / 2 :(bits)
bits_end
define("concat(n)m") :(concat_end)
concat concat = n
m = n
c_loop m = gt(m,0) m / 2 :f(c_done)
concat = concat * 2 :(c_loop)
c_done concat = concat + n :(return) concat_end
n = 0
loop n = n + 1
m = concat(n) output = lt(m,1000) m ": " bits(m) :s(loop)
end</lang>
- Output:
3: 11 10: 1010 15: 1111 36: 100100 45: 101101 54: 110110 63: 111111 136: 10001000 153: 10011001 170: 10101010 187: 10111011 204: 11001100 221: 11011101 238: 11101110 255: 11111111 528: 1000010000 561: 1000110001 594: 1001010010 627: 1001110011 660: 1010010100 693: 1010110101 726: 1011010110 759: 1011110111 792: 1100011000 825: 1100111001 858: 1101011010 891: 1101111011 924: 1110011100 957: 1110111101 990: 1111011110
Visual Basic .NET
Based on the Alternate version.
<lang vbnet>Imports System.Console Module Module1
Sub Main() Dim p, c, k, lmt as integer : p = 2 : lmt = 1000 For n As Integer = 1 to lmt p += If(n >= p, p, 0) : k = n + n * p If k > lmt Then Exit For Else c += 1 Write("{0,3} ({1,-10}) {2}", k, Convert.ToString( k, 2), If(c Mod 5 = 0, vbLf, "")) Next : WriteLine(vbLf + "Found {0} numbers whose base 2 representation is the concatenation of two identical binary strings.", c) End Sub
End Module</lang>
- Output:
3 (11 ) 10 (1010 ) 15 (1111 ) 36 (100100 ) 45 (101101 ) 54 (110110 ) 63 (111111 ) 136 (10001000 ) 153 (10011001 ) 170 (10101010 ) 187 (10111011 ) 204 (11001100 ) 221 (11011101 ) 238 (11101110 ) 255 (11111111 ) 528 (1000010000) 561 (1000110001) 594 (1001010010) 627 (1001110011) 660 (1010010100) 693 (1010110101) 726 (1011010110) 759 (1011110111) 792 (1100011000) 825 (1100111001) 858 (1101011010) 891 (1101111011) 924 (1110011100) 957 (1110111101) 990 (1111011110) Found 30 numbers whose base 2 representation is the concatenation of two identical binary strings.
Wren
<lang ecmascript>import "/fmt" for Conv, Fmt
var i = 1 while(true) {
var b2 = Conv.itoa(i, 2) b2 = b2 + b2 var d = Conv.atoi(b2, 2) if (d >= 1000) break Fmt.print("$3d : $s", d, b2) i = i + 1
} System.print("\nFound %(i-1) numbers.")</lang>
- Output:
3 : 11 10 : 1010 15 : 1111 36 : 100100 45 : 101101 54 : 110110 63 : 111111 136 : 10001000 153 : 10011001 170 : 10101010 187 : 10111011 204 : 11001100 221 : 11011101 238 : 11101110 255 : 11111111 528 : 1000010000 561 : 1000110001 594 : 1001010010 627 : 1001110011 660 : 1010010100 693 : 1010110101 726 : 1011010110 759 : 1011110111 792 : 1100011000 825 : 1100111001 858 : 1101011010 891 : 1101111011 924 : 1110011100 957 : 1110111101 990 : 1111011110 Found 30 numbers.