Write a binary counter in quine

Question

Write two code fragments, which we will call s_zero and s_one.

Program (l, n) consists of l copies of s_zero and s_one corresponding with the digits of n in binary, padded with s_zero on the left.

For example, if s_zero = foo and s_one = bar then
Program (4, 0) = foofoofoofoo
Program (4, 1) = foofoofoobar
Program (4, 2) = foofoobarfoo
etc.

Program (l, n) must print the source of Program (l, (n + 1) mod (2^l)) to standard out. In the example above, foobarfoofoo must print foobarfoobar when executed.

Your score is the sum of the lengths of fragments s_zero and s_one

Answer 1

CJam, 29 + 29 = 58 bytes

The 0 code:

0{"_~"]]z(3\+2b(2b2>a\+z~[}_~

The 1 code:

1{"_~"]]z(3\+2b(2b2>a\+z~[}_~

Explanation

0                       " Push the current bit. ";
{"_~"                   " The basic quine structure. ";
    ]                   " Make an array of the current bit, the function and _~.
                          That is the code fragment itself. ";
    ]                   " Make an array of all code fragments in the stack. ";
    z(                  " Extract the array of the bits. ";
    3\+2b(2b2>          " Convert from base 2, decrease by one and convert back,
                          keeping the length. ";
    a\+z                " Put the bits back to the original position. ";
    ~                   " Dump the array of code fragments back into the stack. ";
    [                   " Mark the beginning position of the array of the next code fragment. ";
}_~

Answer 2

CJam, 47 + 47 = 94 bytes

The 0 code:

{`"_~"+]:+T2*0+:T)\AsSerS/_,(@2Y$#+2bU@->]zs}_~

The 1 code:

{`"_~"+]:+T2*1+:T)\AsSerS/_,(@2Y$#+2bU@->]zs}_~

Excuse the expletive.

I'm sure I can still shave off a few bytes there. I'll add an explanation once I decide that I can't be bothered to golf it any further.

Test it here.

Answer 3

Python 3.8 (pre-release), 264 250 120 + 120 = 240 bytes

Code 0: (120 bytes)

for i in(c:=[]),0:
 c+=0,;exec(s:="i or print(end=f'for i in(c:=[]),0:\\n c+={c[0]^all(c:=c[1:-1])},;exec(s:=%r)#'%s)")#

Try it online!

Code 1: (120 bytes)

for i in(c:=[]),0:
 c+=1,;exec(s:="i or print(end=f'for i in(c:=[]),0:\\n c+={c[0]^all(c:=c[1:-1])},;exec(s:=%r)#'%s)")#

Try it online!

Test Suite

Try it online!

How it works:

• c store the binary number of the program on the form of a list

The for loop execute twice :

• the first time, it assigns c to the correct value
• the second time, it prints the code replacing a char by 1 or 0 depending on c. The nth time c[0]^all(c:=c[1:-1]) is called, it returns the nth leftmost bit of c+1

the next for loop are ignored thanks to the comment #

The printing is a derivate of this quine : exec(s:="print('exec(s:=%r)'%s)")

Answer 4

CJam, 45 + 45 = 90 bytes

The 0 code:

{`]W=L0+:L,Ua*L2b)2b+L,W*>\f{X$!s/\s*"_~"}}_~

The 1 code:

{`]W=L1+:L,Ua*L2b)2b+L,W*>\f{X$!s/\s*"_~"}}_~

Explanation soon.

Try it online here

Answer 5

GolfScript, 37 + 37 = 74 bytes

0{`".~"+:q;]-2%1\{1$^.@&q@}%\;-1%~}.~

1{`".~"+:q;]-2%1\{1$^.@&q@}%\;-1%~}.~

Not quite as short as user23013's CJam solution, but I figured I'd post this anyway, if only to (marginally) increase the diversity of the languages used.

My solution not directly based on any of the existing solutions (and, indeed, I haven't examined them in detail, as I still don't read CJam very well), but they all feature variants of the same basic quine structure ({".~"}.~ in GolfScript, {"_~"}_~ in CJam). That's not really very surprising, since it seems to be one of the most efficient ways to write a quine with an arbitrary payload in these languages.

There are several parts of this code I don't really like, and I suspect it may be possible to golf this further, but I've spent too much time on this as it is.