# Write self-validating code

Write code that takes a string as input, and outputs a truthy or falsey value depending on whether or not the string follows these rules:

If you stack each character on top of each other, convert to binary and sums each column, then all the sums should be identical. You may assume the input string contains only printable ASCII-characters (code points 32 - 126).

As an example:

The input O5vy_+~ should return a truthy value, since its binary representation is:

1001111  | O
0110101  | 5
1110110  | v
1111001  | y
1011111  | _
0101011  | +
1111110  | ~
-------
5555555  <- Sum of bits in each column. Should give a truthy value.


The input PPCG should return a falsey value, since its binary representation is:

1010000  | P
1010000  | P
1000011  | C
1000111  | G
-------
4020122  <- Should give a falsey value


The twist is: Your code should return a truthy value if it's used as input to your function / program. I.e. the code must adhere to the same rules as above (your code can contain characters that are not ASCII 32-126).

Your program/function only needs to handle printable ASCII as input. If your code contains something else, 8 bit, 16 bit encoding, Unicode, a custom made character set (or something else), then the binary representation of it should adhere to the same rules, but your code doesn't need to handle it as input.

This is , so standard rules apply.

• How long will the input string be? Can we assume the sum will always be 7 digits long? – Okx Feb 25 '17 at 21:53
• Also, if our program uses characters other than ASCII characters, what happens? – Okx Feb 25 '17 at 21:58
• I guess that "then the binary representation of it should adhere to the same rules" should explicitly exclude the clause "only needs to handle printable ASCII as input" (otherwise one could write code with just one byte that maps to non-printable ASCII). – Jonathan Allan Feb 25 '17 at 22:07
• @Okx you may assume the input string is less than 1kB. The input will only be printable ASCII that can be represented using 7 bits, so yes: There will always be 7 integer (not necessarily digits) sums. – Stewie Griffin Feb 25 '17 at 22:35
• @StewieGriffin That's not a very good clarification. If I have a non-ASCII answer, and you try and input the program into the program, and it doesn't work because it only supports ASCII, what happens? – Okx Feb 25 '17 at 22:52

## JavaScript (ES6), 123122120 110 bytes

S=>[...S].map(S=>R.map((_GSSSSSSVWWW,V)=>R[V]-=S.charCodeAt()>>V&1),R=[_=3^3,_,_,_,_,_,_])&&!R.some(S=>S^R[_])


Below is a hexdump with bit sums.

Addr. | Dump                                            | #6 #5 #4 #3 #2 #1 #0
------+-------------------------------------------------+---------------------
00-0F | 53 3D 3E 5B 2E 2E 2E 53 5D 2E 6D 61 70 28 53 3D |  8 11  9 11  9  9  9
10-1F | 3E 52 2E 6D 61 70 28 28 5F 47 53 53 53 53 53 53 | 20 18 19 17 14 20 19
20-2F | 56 57 57 57 2C 56 29 3D 3E 52 5B 56 5D 2D 3D 53 | 30 24 32 25 26 30 29
30-3F | 2E 63 68 61 72 43 6F 64 65 41 74 28 29 3E 3E 56 | 41 37 37 32 34 38 36
40-4F | 26 31 29 2C 52 3D 5B 5F 3D 33 5E 33 2C 5F 2C 5F | 47 47 48 43 44 47 46
50-5F | 2C 5F 2C 5F 2C 5F 2C 5F 5D 29 26 26 21 52 2E 73 | 54 57 55 54 56 56 54
60-6D | 6F 6D 65 28 53 3D 3E 53 5E 52 5B 5F 5D 29       | 64 64 64 64 64 64 64


### Demo

let f =

S=>[...S].map(S=>R.map((_GSSSSSSVWWW,V)=>R[V]-=S.charCodeAt()>>V&1),R=[_=3^3,_,_,_,_,_,_])&&!R.some(S=>S^R[_])

console.log(f("O5vy_+~"))
console.log(f("Hello World!"))
console.log(f(S=>[...S].map(S=>R.map((_GSSSSSSVWWW,V)=>R[V]-=S.charCodeAt()>>V&1),R=[_=3^3,_,_,_,_,_,_])&&!R.some(S=>S^R[_])))

# MATL, 10 9 bytes

BXs&=?I&]


Input is a string enclosed with single quotes (if the input contains single qoutes, escape them by duplicating).

Output is 3 as truthy and nothing (empty output) as falsy.

Try it online!

The code in binary is as follows:

B     1 0 0 0 0 1 0
X     1 0 1 1 0 0 0
s     1 1 1 0 0 1 1
&     0 1 0 0 1 1 0
=     0 1 1 1 1 0 1
?     0 1 1 1 1 1 1
I     1 0 0 1 0 0 1
&     0 1 0 0 1 1 0
]     1 0 1 1 1 0 1

Sum   5 5 5 5 5 5 5


### Explanation

B      % Input string (implicit). Convert each char to its ASCII code, and
% then to binary. This gives a binary matrix, with each char of the
% input corresponding to a row
Xs     % Sum of each column. Gives a row vector
&=     % All pairwise equality comparisons
?      % If all are true
I    %    Push 3
&    %    Specify that the next function, namely implicit display, will
%    take one input, instead of the whole stack which is the default
]      % End
% Display (implicit)


# Jelly, 11 10 bytes

OBUSE&889.


Try It Online! Or see tests and self-input (the code is all printable ASCII, which have the same values in Jelly's code page, as seen below).

Char -> Hex -> Decimal -> Binary
O       0x4F   79         0b1001111
B       0x42   66         0b1000010
U       0x55   85         0b1010101
S       0x53   83         0b1010011
E       0x45   69         0b1000101
&       0x26   38         0b0100110
8       0x38   56         0b0111000
8       0x38   56         0b0111000
9       0x39   57         0b0111001
.       0x2E   46         0b0101110
-------
5555555


### How?

OBUSE&889. - Main link: string
O          - cast to ordinals
B         - convert to binary
U        - upend (reverses each to prepare for vectorised sum)
S       - sum (vectorises)
E      - all equal? (yields 1 if all bit-sums are equal and 0 if not)
889. - 889.0
&     - bitwise and (1 & 889.0 is 1; and 0 & 889.0 is 0)

• Your code looks much like Dennis's. – Erik the Outgolfer Feb 26 '17 at 10:54
• Yeah, I saw after. – Jonathan Allan Feb 26 '17 at 10:55

# Jelly, 11 10 bytes

OBUSE$*8?8  Uses no no-ops or comments. Try it online! ### Binary breakdown O 1 0 0 1 1 1 1 B 1 0 0 0 0 1 0 U 1 0 1 0 1 0 1 S 1 0 1 0 0 1 1 E 1 0 0 0 1 0 1$  0 1 0 0 1 0 0
*  0 1 0 1 0 1 0
8  0 1 1 1 0 0 0
?  0 1 1 1 1 1 1
8  0 1 1 1 0 0 0
————————————————
∑  5 5 5 5 5 5 5


### How it works

OBUSE*8?8 Main link. Argument: s (string) O Ordinal; map all characters in s to their code points. B Binary; convert each code point to base 2. U Upend; reverse each binary array to right-align the digits. 8? If 8 is non-zero (it is): SE          Sum the corresponding digits and test the the sums for equality.
Else (never happens):
*  8      Raise all binary digits to the eighth power.

• Is Jelly TC using only printable ASCII? – Pavel Feb 26 '17 at 1:25
• I don't think so. – Dennis Feb 26 '17 at 2:08
• I like that this code starts with OBUSE because it sounds like ABUSE. – Esolanging Fruit Feb 27 '17 at 4:24

# Mathematica, 88 bytes

Total@IntegerDigits[ToCharacterCode@#,2,7]~MatchQ~{"?";a_ ..}&


Contains many unprintable characters between the quotes. Has 49 of each bit.

Here's the hexdump:

0000-0010:  54 6f 74 61-6c 40 49 6e-74 65 67 65-72 44 69 67  Total@In tegerDig
0000-0020:  69 74 73 5b-54 6f 43 68-61 72 61 63-74 65 72 43  its[ToCh aracterC
0000-0030:  6f 64 65 40-23 2c 32 2c-37 5d 7e 4d-61 74 63 68  ode@#,2, 7]~Match
0000-0040:  51 7e 7b 22-3f 1f 1f 1f-1f 1f 1f 1f-1f 1f 1f 1f  Q~{"?... ........
0000-0050:  1f 1f 1f 1f-1f 1a 1a 1a-1a 18 18 18-18 18 10 22  ........ ......."
0000-0058:  3b 61 5f 20-2e 2e 7d 26                          ;a_...}&


# Octave, 53 52 bytes

Making a complete rewrite helped me golf the code 5 bytes, but I had to add more no-ops, making it a net-save of only 1 byte.

@(_)~diff(sum(de2bi(+_)))%RRPPPVVVW?????????________


I can't add a TIO-link, since none of the online interpreters have implemented the communication toolbox necessary for de2bi. Changing it to dec2bin instead would cost 4 bytes (2 for working code, and two no-ops).

I found no way to avoid any of the 27 no-ops. All function names and parentheses are between either below 64, or higher than 96, meaning all "necessary" characters have a 1 in the 6th position (from the right, 2^5). I had a solution with only 23 no-ops, but the code itself was longer. The actual code is 25 bytes, and has the following column sum when counting the bits of the binary equivalent:

15   22    6   15   10    9   13


There are 22 bits in the 6th position from the right (2^5), and only 6 bits in the 4th position from the right (2^3). That means, we have to add at least 16 bytes, to get the 6 up to 22. Now, the comment character % adds a bit to the 6th position, increasing it to 23. All printable ASCII-characters needs at least one of the two top bits to be 1. Therefore, adding 17 bytes will give us at least 27 bits in each of the two "top spots" (2^6 and 2^5). Now, we have 27 bits in the top two spots, and 22 in the rest. In order to get to an equilibrium, we have to add 10 bytes, to get to an even 32 bits in each position.

An explanation of the new code (52 bytes):

@(_)~diff(sum(de2bi(+_)))
@(_)      % An anonymous function that take a variable _ as input
% We use underscore, instead of a character, since it has the
% most suitable binary represetation
de2bi(+_)    % Convert the input string to a binary matrix
sum(de2bi(+_))   % Take the sum of each column
diff(sum(de2bi(+_)))  % And calculate the difference between each sum
~diff(sum(de2bi(+_)))  % Negate the result, meaning 0 becomes true,
% and everything else becomes false


A vector containing only 1s (true) is evaluated to true in Octave, and a vector containing at least one zero is evaluated to false in Octave.

An explanation of the old code (53 bytes):

@(_)!((_=sum(de2bi(+_)))-_(1))%RRRFVVVVVVVVV_____????

@(_)      % An anonymous function that take a variable _ as input
% We use underscore, instead of a character, since it has the
% most suitable binary represetation
!     % Negate the result, meaning 0 becomes true, and everything else becomes false
de2bi(+_)         % Convert the input string to a binary matrix
sum(de2bi(+_))        % Take the sum of each column
(_=sum(de2bi(+_)))       % Assign the result to a new variable, also called _
% It's not a problem that we use the same variable name, due
% to the order of evaluation
((_=sum(de2bi(+_)))-_(1)) % Subtract the first element of the new variable _
% If all elements of the new variable _ are identical, then this
% should give us a vector containing only zeros,
% otherwise, at least one element should be non-zero
!((_=sum(de2bi(+_)))-_(1))  % And finally, we negate this.


A vector containing only 1s (true) is evaluated to true in Octave, and a vector containing at least one zero is evaluated to false in Octave.

## JavaScript (ES6), 139111 107 bytes

f=
S=>![...""+1E6].some((____________ABQWWWWWWWWW,P)=>P*=R^(R^=R,[...S].map(Q=>R+=Q.charCodeAt()>>P&1),R),R=0)
<textarea oninput=o.textContent=f(this.value) style=width:100% rows=10>S=>![...""+1E6].some((____________ABQWWWWWWWWW,P)=>P*=R^(R^=R,[...S].map(Q=>R+=Q.charCodeAt()>>P&1),R),R=0)</textarea><div id=o>true

Contains 81 63 61 of each bit.

# Scala, 149 bytes

_.map(C=>("0"*7++(BigInt(C)toString 2))takeRight 7 map(_-48)).transpose.map(_.sum).toSet.size==1//______________________________


### Usage:

val f:(String=>Any)=_.map(C=>("0"*7++(BigInt(C)toString 2))takeRight 7 map(_-48)).transpose.map(_.sum).toSet.size==1//______________________________
println(f("string here")


### Hexdump:

00000000  5f 2e 6d 61 70 28 43 3d  3e 28 22 30 22 2a 37 2b  |_.map(C=>("0"*7+|
00000010  2b 28 42 69 67 49 6e 74  28 43 29 74 6f 53 74 72  |+(BigInt(C)toStr|
00000020  69 6e 67 20 32 29 29 74  61 6b 65 52 69 67 68 74  |ing 2))takeRight|
00000030  20 37 20 6d 61 70 28 5f  2d 34 38 29 29 2e 74 72  | 7 map(_-48)).tr|
00000040  61 6e 73 70 6f 73 65 2e  6d 61 70 28 5f 2e 73 75  |anspose.map(_.su|
00000050  6d 29 2e 74 6f 53 65 74  2e 73 69 7a 65 3d 3d 31  |m).toSet.size==1|
00000060  2f 2f 5f 5f 5f 5f 5f 5f  5f 5f 5f 5f 5f 5f 5f 5f  |//______________|
00000070  5f 5f 5f 5f 5f 5f 5f 5f  5f 5f 5f 5f 5f 5f 5f 5f  |________________|
00000080  1f 1f 1f 1f 1e 1e 1e 1e  16 16 16 16 16 12 12 10  |................|
00000090  10 10 10 10 10                                    |.....|


### Ungolfed:

string =>
string.map(char =>
(
"0" * 7 ++ BigInt(char).toString(2)
).takeRight(7).map(n=>n-48)
).transpose
.map(bits=>bits.sum)
.toSet
.size == 1
//______________________________


### Explanation:

string =>                      //create an anonymous function with a parameter string
string.map(char =>           //map each char in the string to
(
"0" * 7                  //a string of 7 zeroes
++                       //concatenated with
BigInt(char).toString(2) //the ascii value as a binary string
).takeRight(7)             //the last 7 items from this sequence
.map(n=>n-48)              //where each digit is mapped to its numerical value
).transpose                  //transpose, so the colums become rows and vice-versa
.map(bits=>bits.sum)         //maps the bits in each column to their sum
.toSet                       //and convert the sequence of sums to a set
.size == 1                   //which has 1 element of the sums are the same
//______________________________


# J, 45 bytes

[:(*/@:={.)[:+/2 #.inv 3 u:]NB.____UUUUUUUUDD


Try it online! Includes test cases for most submissions submitted, along with the source code.

_R _S=mod _S 2:_R(div _S 2)
_Z _S|_V:_W<-take 7.foldl1(zipWith(+))$_R.fromEnum<$>_S=all(==_V)_W
--________


Try it online! Usage: _Z "some string" returns either True or False.

There are some unprintable chars in the comment on the last line, so here is a string of the program using escaped chars:

"_R _S=mod _S 2:_R(div _S 2)\n_Z _S|_V:_W<-take 7.foldl1(zipWith(+))$_R.fromEnum<$>_S=all(==_V)_W\n--___\US\US\US\ETB\DC3\DC3\DC3\DC3\DC3\DC3\DC2\DC2_____"


Each bit occurs 68 times.

The shortest code I came up with was 82 bytes:

b n=mod n 2:b(div n 2)


However the sums of the bits for this code are [33,28,41,48,20,79,46], so 79 - 20 = 59 no-ops plus 2 bytes for starting a comment would additionally be needed, totalling in 143 bytes.

While rearranging the program I found that using upper case letters as variable names helps to level the sums because they don't have the bit in the 6th position set. Because Haskell does not allow variable names to start with an upper case letter they need to be prepended with _, which also does not set the 6th bit.

In doing so I ended up with the above solution which has 97 bytes before adding the no-ops and the bist sum to [50,47,56,56,48,68,60], so (68 - 47) = 21, so only 21 bytes need to be added in the comment.

# PHP, 9593 91 bytes

I am so happy that PHP function names are case insensitive!

FOR(ZZSSSSQ__*;$W=ORD($argn[$T++]);)FOR($V=7;$V--;)$R[$V]+=$W>>$V&1;PRINT MIN($R)==MAX(\$R);


where the * must be replaced with ASCII 151 (0x97). (PHP would complain about any control character in the code - apart from \r and \n, but I need something with bit 4 set, so I added 128.)

+1 byte for pure printable ASCII: Use _7 instead.

Run with echo '<input>' | php -nR '<code>' or test it online. Output is 1 for truthy, empty for falsy.

# Python 2, 117 bytes

All "spaces" are tabs to reduce number of 0x20 bits.

def Y(S):
O=map(sorted,zip(*['{:07b}'.format(ord(W))for   W   in  S]))
return  O[1:]==O[:-1]#V_____________


Contains 66 of each bit. (There is no '%07b' as explained in this issue.)

Hex dump:

00000000: 64 65 66 09 59 28 53 29 3a 0a 09 4f 3d 6d 61 70  def.Y(S):..O=map
00000010: 28 73 6f 72 74 65 64 2c 7a 69 70 28 2a 5b 27 7b  (sorted,zip(*['{
00000020: 3a 30 37 62 7d 27 2e 66 6f 72 6d 61 74 28 6f 72  :07b}'.format(or
00000030: 64 28 57 29 29 66 6f 72 09 57 09 69 6e 09 53 5d  d(W))for.W.in.S]
00000040: 29 29 0a 09 72 65 74 75 72 6e 09 4f 5b 31 3a 5d  ))..return.O[1:]
00000050: 3d 3d 4f 5b 3a 2d 31 5d 23 56 5f 5f 5f 5f 5f 5f  ==O[:-1]#V______
00000060: 5f 5f 5f 5f 5f 5f 5f 16 16 16 16 16 16 16 16 16  _______.........
00000070: 16 16 14 14 10                                   .....

• If you read the description of the bug report... "resolution: not a bug". – mbomb007 Feb 27 '17 at 22:53