Autogram programs

Question

This sentence employs two a’s, two c’s, two d’s, twenty-eight e’s, five f’s, three g’s, eight h’s, eleven i’s, three l’s, two m’s, thirteen n’s, nine o’s, two p’s, five r’s, twenty-five s’s, twenty-three t’s, six v’s, ten w’s, two x’s, five y’s, and one z.

Such sentences are called autograms. Your task is to write a program or function that has a similar property: it takes as input a byte, and outputs an integer representing the number of times that byte appears in the program's source code. It should output 0 if its input doesn't appear in its source.

Your program must not read its own source code, directly or indirectly, but instead all the data it needs to give its output must be hard-coded, quine style. Your program must be at least one byte in length.

This is code-golf, so the shortest entry (in bytes) wins. I won't accept an answer, so feel free to compete for the shortest solution in your favourite language.

You are strongly encouraged to include an explanation in your post.

Leaderboards

Below are leaderboards both for overall score and per language:

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• N. Virgo: Sandbox. (It's been in there since 2014!)
• Uriel: Are functions allowed, or only full programs?
• user202729: Note: Answers similar to this one doesn't count as valid anymore.
• N. Virgo: FWIW, the proper quine definition is a bit stronger than I would really like for this challenge. For example, I would be quite happy to see a program that uses all the characters in some contiguous range, and just outputs a 1 if its input is in that range, even though this doesn't have "a section of the program which encodes a different part of the program". For that reason I'd have been willing to allow the retina example, though it would have been disappointing if it had put others off from posting, due to already just being 1 byte. (@user202729)
• N. Virgo: Because of my above comment, I've removed the statement that quine rules apply. (I've kept the part about not reading the source code.)

Answer 1

Octave, 44 bytes

@(x)sum([[39,'(())*,239==@[[]]msuxx']]==x)*2

Try it online!

The result of the function, when using each of the characters in the function as input:

' ( ) * , 2 3 9 = @ [ ] m s u x 
2 4 4 2 2 2 2 2 4 2 4 4 2 2 2 4 

All other input characters returns a zero.

The fact that I had to escape ' made this a whole lot longer. In order to account for the 4 parentheses, I just had a duplicate in the string. However, duplicating apostrophes takes apostrophes, so this just takes us further from the correct result. Therefore, I had to count the number of apostrophes with its ASCII-value, 39. This of course meant I had to check for the characters 3 and 9 too, making it all a lot longer.

• N. Virgo: This is the sort of thing I'm hoping to see more of. (Optimising for the problem, rather than just taking a standard quine and adding code to count the characters.)

Answer 2

Excel, 84 bytes

=IFERROR(MID("65496331125442343343233",FIND(A1,"""123456789,()=MINORFADEOMEN"),1),0)

Find() will look for the value in cell A1 within the string "123456789,()=MINORFADEOMEN (the """ at the beginning is to escape the character and will evaluate to just ").

Based on the result of Find(), the Mid() function will return the corresponding character from the string of numbers. This string was created through iteration until it stopped changing.

If the character in A1 is not found, Find() returns an error so the IfError() function escapes that to return 0 instead.

The OMEN at the end of the string being searched in the Find() function are duplicate letters so their position will never be returned but they were needed to adjust the character counts. Without them, there was an endless loop of changing counts. The letter arrangement is stylistic choice.

Answer 3

JavaScript (ES6), 70 bytes

Doesn't read the source of the function, but this is quite long. Takes input as a 1-character string.

i=>~(n='\34円=\'in(|)0257?:.>[]Odefx~'.indexOf(i))?'3733544333'[n]||2:0

Try it online!

• Yair Rand: You can shorten this a bunch by reworking a bit to lower the amount of distinct characters: i=>'36335633'[i='\36円=\'i|&()25.>[]Odefx~n'.indexOf(i)]||~i&&2 (8 bytes saved)

Answer 4

Retina, 1 byte

This wouldn't be valid if following strictly the quine rules, but the OP explicitely allowed it in a comment.

x

Try it online!

In retina a single-line program counts the occurrences of that regex in the input. This challenge is solved by any single ASCII character except `, ., +, *, ?, [, (, ), ^, $, \ and newline.

Answer 5

C# (Visual C# Compiler), (削除) 88 (削除ここまで) 57 bytes

x=>"\\\"(())==>>??..::3300CCaaiinoossttx".Contains(x)?3:0

Try it online!

Returns 3 if the string contains the passed char, otherwise returns 0. The string contains each character of the code atleast once and exactly the amount required to have the char 3 times in the code.

-31 bytes thanks to Kevin Cruijssen

• Kevin Cruijssen: Nice approach! +1 from me. You can save some bytes by removing the parenthesis around (a)=> and taking the input as string instead of char so the .ToString() is no longer necessary (and a+"" could have been used as well, for code-golf challenges you'll never need .ToString()). Also, since .Contains already contains an a, you might make the input-variable a different unused character (like x or q) to have 3 for each character instead of 4. EDIT: Never mind the last part, I see " is used 4 times as well now.
• Hyarus: @KevinCruijssen Thanks for the advice. But is string valid as input? Considering OP asked for a byte as input?
• Kevin Cruijssen: As long as it's a single character string I'm pretty sure it is. Most languages can only input string-values anyway, and some that have characters still choose to input as string (like the Java, JavaScript, and Japt for example. Input and output formats are usually pretty flexible. But you could always ask OP in a comment if you're still doubting. :)
• Hyarus: @KevinCruijssen sounds reasonable. And if we escape one of the " we can even get down to the desired 3 per char
• N. Virgo: Single character strings as input are fine by me.

Answer 6

Haskell, 66 bytes

The first two versions are essentially a quine which filters the required characters and then takes the length:

q c=length.filter(==c)$id<>show$"q c=length.filter(==c)$id<>show$"

Try it online or test with all characters! ^*

---

Alternative without (<>), 72 bytes

q c=length.filter(==c)$(++)<*>show$"q c=length.filter(==c)$(++)<*>show$"

Try it online or test with all characters!

---

Non-quine alternative, (削除) 87 (削除ここまで) 86 bytes

Might still be improved, but I'm happy that I got it down to only three different groups.

I like this one the best, even though it counts the most bytes. It's computes the character/byte count by as a sum of 2,3 and 7 (note how some characters are in multiple groups):

u t=sum[fst e|e<-[(2," ()-237<=dflnst|"),(3," ()[\\]`mstu"),(7,"\",ee")],t`elem`snd e]

Try it online or test with all characters!

---

^{* imports (<>) because TIO's GHC version is 8.0.2}

Answer 7

Python 2, (削除) 54 (削除ここまで) (削除) 52 (削除ここまで) 32 bytes

^{-20 bytes thanks to ovs}

("''+.23cnotu()"*2+'"*'*3).count

Try it online!

• Stewie Griffin: How come you don't have either lambda or input?
• Neil: @StewieGriffin Python automatically binds a method reference, so "xxx".count is equivalent to lambda c:"xxx".count(c).
• Rod: @StewieGriffin and you use both in the same way ->f="xxx".count and f=lambda c:"xxx".count(c) will be called as f('a')

Answer 8

Husk, (削除) 11 10 (削除ここまで) 8 bytes

Thanks Leo for -2 bytes!

#sD"#sD"

Try it online!

Explanation

This makes sure that it only uses ASCII (because show would mess it up) and that each character is contained twice:

#sD"#sD" -- character as input, eg. '"'
 "#sD" -- string literal (note the redundant '"'): "#sD"
 D -- double: "#sD#sD"
 s -- show: "\"#sD#sD\""
# -- number of occurences: 2

Standard quine extension, 11 bytes

#hS+s"#hS+s

Try it online!

Explanation

#hS+s"#hS+s -- character as input, eg. '"'
 "#hS+s -- string literal: "#hS+s"
 S+ -- join with itself: ("#hS+s"++)
 s -- | and itself shown: "\"#hS+s\""
 -- : "#hS+s\"#hS+s\""
 h -- init: "#hS+s\"#hS+s"
# -- number of occurences in string: 1

• Leo: You can manage the quotes more easily by doubling the string before showing it: Try it online!

Answer 9

Java 10, (削除) 164 (削除ここまで) (削除) 81 (削除ここまで) 57 bytes

q->"q-->>\\\"..ccoonttaaiiss(())??33::00".contains(q)?3:0

Port of @Hyarus' C# answer, so make sure to upvote him!

Explanation:

Try it online.

q-> // Method with String parameter and integer return-type
 "q->\\\"..ccoonttaaiiss(())??33::00".contains(q)?
 // If the string above contains the input character
 3 // Return 3
 : // Else:
 0 // Return 0

---

Old quine 164-bytes answer:

c->{var s="c->{var s=%c%s%1$c;return s.format(s,34,s).replaceAll(%1$c[^%1$c+c+']',%1$c%1$c).length();}";return s.format(s,34,s).replaceAll("[^"+c+']',"").length();}

Explanation:

Try it online.

c->{ // Method with char parameter and integer return-type
 var s="c->{var s=%c%s%1$c;return s.format(s,34,s).replaceAll(%1$c[^%1$c+c+']',%1$c%1$c).length();}";
 // Unformatted source code
 return s.format(s,34,s) // Create the formatted source code (the quine),
 .replaceAll("[^"+c+']',"") // remove all characters not equal to the input,
 .length();} // and return the length

quine-part:

• The String s contains the unformatted source code.
• %s is used to input this String into itself with the s.format(...).
• %c, %1$c and the 34 are used to format the double-quotes.
• s.format(s,34,s) puts it all together

Challenge part:

• .replaceAll("[^"+c+']',"") removes all characters except those equal to the input.
• .length() then takes the length of this String.

---

NOTE: .split(c).length (with String input instead of char) may seem shorter, but has two problems:

1. The first character gives an incorrectly result, so if c (with c-> as leading part) is input, it will incorrectly return one character too few. This can be fixed by adding +(c==99?1:0) in both the source code and unformatted source code String (and changing .split(c) to .split(c+""), but then we still have the following problem:
2. If a regex character (i.e. $) is input, the .split will interpret it as a regex, giving an incorrect result.

Answer 10

Haskell, 58 bytes

f c=sum[2|x<-succ '!':"f c=sum[2|x<-succ '!':,x==c]",x==c]

Try it online! or verify the solution.

Same byte count alternatives:

f c=sum[2|x<-tail$show"f c=sum[2|x<-tail$show,x==c]",x==c]
f c=sum[1|x<-id<>show$"f c=sum[1|x<-id<>show,ドルx==c]",x==c]

---

Haskell, 90 bytes

sum.($zip"1234\",ドル.[\\]opu+()iklmsz652"$[4,4,11,3,3,3,16,5,3,3,3,3,3,3,3]++[2,2..]).lookup

Try it online! or verify the solution. Uses the fact that sum Nothing = 0 and e.g. sum (Just 4) = 4.

Answer 11

Smalltalk, 112 (削除) 132 (削除ここまで) bytes

Smalltalk isn't exactly known for its golfing suitability :-)

Method defined in class Character (tested in VA Smalltalk and Squeak, should work in other dialects such as VisualWorks and Pharo as well):

a^(0to:6),#(10 18)at:(#('' ',[]^|+68cForu' '#0adefln' '1it' '()' ':s' ' ' '''')findFirst:[:s|s includes:self])+1

Characters occurring in the source are grouped by occurrence count. The groups are tested for the first one which contains the receiver, and the matching occurrence count is returned.

Old method:

a^('''''((((())))):::[[[[]]]]^^^^^0000066666aaaacccccdddddeefffFFFFFiiilllnnnnnrrrrrsssTTTTTuuuu'includes:self)ifTrue:[6]ifFalse:[0]

Every character that does appear in the method appears exactly 6 times (by being repeated in the string constant), so the method just checks whether the receiver is contained in the string and returns 6 if it is, 0 otherwise.

After defining the method as above, you can validate it using

| code |
code := Character sourceCodeAt: #a.
((0 to: 255) collect: [:b | b asCharacter]) reject: [:c | c a = (code occurrencesOf: c)]

The result should be empty.

• Martin Ender: Welcome to PPCG!

Answer 12

JavaScript, 31 bytes

f=c=>~-`f=${f}`.split(c).length

Try it online

• N. Virgo: I think this reads its own source code, doesn't it? That's explicitly banned, in bold, in the question.
• Shaggy: @Nathaniel, see here.
• N. Virgo: Hmm. Well honestly I think that's stupid, but I can't really disagree with meta consensus I suppose.
• user202729: @Nathaniel For all programs that make eval easy, they also make quine easy.
• N. Virgo: @user202729 I don't think that follows, does it? You can't do the same trick in Python, for example. This is a reverse eval, not an eval

Answer 13

Python 2, 57 bytes

lambda c,s='lambda c,s=%r:(s%%s).count(c)':(s%s).count(c)

Try it online!

Answer 14

05AB1E, 17 bytes

2"D34çýs¢"D34çýs¢

Try it online!

Answer 15

Haskell, 96 bytes

n"n"=15;n"="=14;n" "=2;n";"=13;n"\\"=3;n"\""=25;n"0"=2;n"1"=4;n"2"=4;n"3"=4;n"4"=5;n"5"=5;n n3=0

Try it online!

Haskell, 109 bytes

n 'n'=15;n '='=14;n ' '=14;n ';'=13;n '\\'=3;n '\''=25;n '0'=2;n '1'=5;n '2'=4;n '3'=4;n '4'=5;n '5'=5;n n3=0

Try it online!

Haskell, 122 bytes

n 'n'=21
n '='=14
n ' '=14
n '\n'=12
n '\\'=4
n '\''=32
n '0'=2
n '1'=5
n '2'=5
n '3'=3
n '4'=5
n '5'=5
n nnnnnn4'''''''=0

Try it online!

Explanations

These answers are not terribly sophisticated. It is a series of declarations, one for each character present in the program. At the end we have a catch all that returns 0 for characters not present in the program.

I use a couple of tricks to minimize the number of characters necessary in the program and from there I fiddled with things until the numbers turned out just right. You can see that I've padded the variable name in the last declaration, in all 3 of them. The difference between the 3 programs is whether I chose to use a new line or ; for line breaks and whether I chose to take Chars as input or Strings. The ; approach doesn't seem inherently superior to the others it just gets luck and ends up shorter, however it does seem that using Strings is a better idea than Chars because Chars require spaces after the function name an Strings do not.

• ASCII-only: :| what happened to your spelling in the explanation

Answer 16

Whitespace, 140 bytes

[S S S N
_Push_0][S N
S _Duplicate][S N
S _Duplicate][T N
T S _Read_STDIN_as_character][T T T _Retrieve][S S S T S S T N
_Push_9][T S S T _Subtract][S N
S _Duplicate][N
T S S T N
_If_0_Jump_to_Label_TAB][S S S T N
_Push_1][T S S T _Subtract][S N
S _Duplicate][N
T S S N
_If_0_Jump_to_Label_NEWLINE][S S S T S T T S N
_Push_22][T S S T _Subtract][N
T S T N
_If_0_Jump_to_Label_SPACE][N
S T N
_Jump_to_Label_PRINT][N
S S S T N
_Create_Label_TAB][S S S T S S T S T N
_Push_37][N
S T N
_Jump_to_Label_PRINT][N
S S S N
_Create_Label_NEWLINE][S S S T S S S S T N
_Push_33][N
S T N
_Jump_to_Label_PRINT][N
S S T N
_Create_Label_SPACE][S S S T S S S T T S N
_Push_70][N
S S N
_Create_Label_PRINT][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

70 spaces, 37 tabs, and 33 new-lines used.

Usually I use the Create Labels in the order NSSN, NSSSN, NSSTN, NSSSSN, NSSSTN, NSSTSN, NSSTTN, etc. But because printing a number where the binary S=0/T=1 is used affects the number I need to output, I used the labels NSSN, NSSSN, NSSTN, and NSSSTN instead, which gave the perfect amount of spaces/tabs to be printed with the binary numbers SSSTSSSSTN (33; amount of new-lines), SSSTSSTSTN (37; amount of tabs), and SSSTSSSTTSN (70; amount of spaces).

Explanation in pseudo-code:

Character c = STDIN-input as character
If c is a tab:
 Print 37
Else if c is a new-line:
 Print 33
Else if c is a space:
 Print 70
Else
 Print 0

Example runs:

Input: space

Command Explanation Stack Heap STDIN STDOUT STDERR
SSSN Push 0 [0]
SNS Duplicate top (0) [0,0]
SNS Duplicate top (0) [0,0,0]
TNTS Read STDIN as character [0,0] {0:32} \n
TTT Retrieve [0,32] {0:32}
SSSTSSTN Push 9 [0,32,9] {0:32}
TSST Subtract top two (32-9) [0,23] {0:32}
SNS Duplicate top (23) [0,23,23] {0:32}
NTSSTN If 0: Jump to Label_TAB [0,23] {0:32}
SSSTN Push 1 [0,23,1] {0:32}
TSST Subtract top two (23-1) [0,22] {0:32}
SNS Duplicate top (22) [0,22,22] {0:32}
NTSSN If 0: Jump to Label_NEWLINE [0,22] {0:32}
SSSTSTTSN Push 22 [0,22,22] {0:32}
TSST Subtract top two (22-22) [0,0] {0:32}
NTSTN If 0: Jump to Label_SPACE [0] {0:32}
NSSTN Create Label_SPACE [0] {0:32}
SSSTSSSTTSN Push 70 [0,70] {0:32}
NSTN Jump to Label_PRINT [0,70] {0:32}
NSSN Create Label_PRINT [0,70] {0:32}
TNST Print as integer [0] {0:32} 70
 error

Program stops with an error: No exit defined.
Try it online (with raw spaces, tabs, and new-lines only).

Input: tab

STDIN will be \t (9) instead, in which case it will be 0 at the first If 0 check, goes to LABEL_TAB/NSSSTN, and will push and print 37 instead.

Try it online (with raw spaces, tabs, and new-lines only).

Input: new-line

STDIN will be \n (10) instead, in which case it will be 0 at the second If 0 check, goes to Label_NEWLINE/NSSSN, and will push and print 33 instead.

Try it online (with raw spaces, tabs, and new-lines only).

Input: anything else

Any other input-character will do NSTN (Jump to Label_PRINT) after the third If 0 check, printing the 0 that was still on the stack (which we've duplicated at the very beginning).

Try it online (with raw spaces, tabs, and new-lines only).

• qwr: This is a great language for this challenge.

Answer 17

Japt, 27 bytes

\Ua"" a a a " ÄU\\\"a "aU Ä
 " ÄU\\\"a " // Given this string literal,
 aU // find the last index of the input
 Ä // and add +1.
\Ua"" a a a // Do nothing useful, but make the results match.

Longer than the existing Japt answer, but uses a different approach.
Has the inherent limitation that all chars need to occur a unique number of times.

Takes input as a string.

Try it online!

Answer 18

Perl, 130 bytes

+print+0+((0)x40,6,6,0,3,43,0,0,0,22,12,6,3,5,2,4,0,1,0,0,0,1,0,1,(0)x28,1,0,1,(0)x6,1,(0)x4,1,(0)x4,1,1,1,0,2,0,1,0,0,0,5)[ord<>]

Has no newline or other whitespace. Reads a line from the standard output, but only cares about its first bytes, then prints the number of times that byte occurs in its own source code in decimal, without a newline.

The program is straightforward. Most of the source code is occupied by a literal table that gives the answer for the each possible byte. Trailing zeros are omitted, and adjacent zeros are run-time compressed, but there's no special trick other than that. The rest of the program simply reads the input and looks up the answer in the table.

For example, the part 22, 12, 6, 3, 5, 2, 4, 0, 1, 0 in the source code gives the frequency of digits, so there are 22 zeroes, 12 ones, 6 twos etc in the source code. As a result, if you enter 0 to the standard input of the program, the program will print 22.

Answer 19

C (gcc), 1033 bytes

#include <stdio.h>
int main(int argc,char *argv[]){int r=0*14811;switch((int)argv[1][0]){case' ':r=6;break;case'"':r=3;break;case'#':r=2;break;case'%':r=2;break;case'\'':r=101;break;case'(':r=5;break;case')':r=5;break;case'*':r=5*1*1;break;case'.':r=2;break;case':':r=51;break;case';':r=103;break;case'<':r=2;break;case'=':r=52;break;case'>':r=2;break;case'[':r=4;break;case'\\':r=3;break;case']':r=4;break;case'0':r=11;break;case'1':r=20;break;case'2':r=20;break;case'3':r=9;break;case'4':r=7;break;case'5':r=12;break;case'6':r=3;break;case'7':r=2;break;case'8':r=5;break;case'9':r=2;break;case'a':r=106;break;case'b':r=51;break;case'c':r=55;break;case'd':r=4;break;case'e':r=102;break;case'f':r=2;break;case'g':r=4;break;case'h':r=4;break;case'i':r=10;break;case'k':r=51;break;case'l':r=2;break;case'm':r=2;break;case'n':r=8;break;case'o':r=2;break;case'p':r=2;break;case'r':r=108;break;case's':r=53;break;case't':r=8;break;case'u':r=2;break;case'v':r=3;break;case'w':r=2;break;case'{':r=3;break;case'}':r=3;break;}printf("%d",r);}

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This is by NO means a golfed answer, but it was fun to try to accomplish this challenge in a language which I am not familiar with. It wasn't a particularly difficult challenge until it came time to find the occurrences of the digits, now THAT was a challenge. Had to do a little creative balancing :)

Answer 20

C (gcc), 192 bytes

F(J){J=J-70?J-40?J-41?J-74?J-'{'?J-'}'?J-39?J-48?J-49?J-50?J-51?J-52?J-53?J-54?J-55?J-56?J-57?J-47?J-61?J-63?J-45?J-58?J-59?0:1:23:23:23:1:2:3:3:4:4:14:14:10:10:15:6:4:2:2:25:1:1:1;}//84332211

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Probably possible to golf down further. Uses a comment at the end as a 'scratch space' to add extra digits. When I have to change a digit from X to Y, I change one of the Y's in the scratch to an X to compensate. Other than that this is just a function that takes an integer, using the assignment trick to return a value based on a large ternary conditional.

• LambdaBeta: I can save 1 byte by moving the numeric constant to after the zero as 0*84332211 like in the other C answer, but I can also save bytes by refactoring that to be smaller, so I'm not going to bother just yet.
• qwr: Can you save bytes by using a range check for the less common numbers? that is if we assme 4-9 are rare characters, we can handle them all at once with a conditional check.
• LambdaBeta: Almost certainly. We could probably remove even more by picking clever names for things so that certain symbols never appear or so that the bit-patterns can be used (e.g. J&1? to match all odd characters). I may further golf this if I find the time to do so.

Answer 21

x86 .COM, 17 bytes, controversial

0120 BF2001 MOV DI,0120 (120 be the current address)
0123 B91100 MOV CX,0011
0126 AE SCASB
0127 7502 JNZ 012B
0129 FEC4 INC AH
012B E2F9 LOOP 0126
012D C1E808 SHR AX,8
0130 C3 RET

36 bytes

0100 BF???? MOV DI,(an copy of this code)
0103 B91200 MOV CX,0012
0106 AE SCASB
0107 7503 JNZ 010C
0109 80C402 ADD AH,02
010C E2F8 LOOP 0106
010E C1E808 SHR AX,8
0111 C3 RET

• Razvan Socol: Isn't this reading it's own code?
• l4m2: @RazvanSocol I think it's like the JavaScript solution
• ASCII-only: @RazvanSocol well it doesn't require saving in a file so it's okay. a lot of fungeoids and JS also do this
• qwr: I argue you are reading code that is executed, which counts as "reading own source code". Now you could argue having a string copy of the program, as most solutions here do, is also "reading source code", but the string isn't executed.

Answer 22

APL (Dyalog Classic), 30 bytes

⊢⊢⊢11-11-11-'''''''1''⊢-⍳⍳0'⍳⊢

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Explanation

In APL, single quotes within strings are escaped by doubling, so '''''''1''⊢-⍳⍳0' is the string '''1'⊢-⍳⍳0, which contain every character used in the program.

APL arrays are by default 1-indexed, and the index-of function, interestingly, returns 1 + max index if the element is not found.

So, by using index-of on the string and the input returns

Input Index Count
' 1 10
1 4 7
⊢ 6 5
- 7 4
⍳ 8 3
0 10 1
<other> 11 0

As one can see, 11 - index gives the count of the character in the program. So, the basic algorithm is

11-'''''''1''⊢-⍳⍳0'⍳⊢

The rest is bloating the character counts to allow them to fit nicely into the slots.

Answer 23

Japt, 14 bytes

Qi"QiUè2 " 2èU

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Explanation

 :Implicit input of character string U
 "QiUè2 " :String literal
Qi :Append a quotation mark
 2 :Repeat twice
 èU :Count the occurrences of U

Answer 24

V, 12 bytes

ñéÑ~"qpxøÖa

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Hexdump:

00000000: f1e9 d17e 2271 7078 f8d6 1261 ...~"qpx...a

Answer 25

Jelly, 16 bytes

";)w8a2";)""w8a2

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Every character appears exactly twice.

";)w8a2";)""w8a2
";)w8a2";)"" the string ;)w8a2""
 w8a2 return 2 if input in string

Answer 26

x86, (削除) 42 (削除ここまで) 40 bytes

Here I use the same strategy as others: Create a string copy of the program with unique bytes, then return 2 in al if the input al is in the string. If we allow ourselves to read code that is actually run, we get l4m2's solution.

I got to make use of a cool string instruction scasb. To the best of my knowledge, there are no duplicate bytes, but this is something I'd easily screw up. Loading the string address takes 5 bytes but I'm not aware of any shorter solution (64-bit lea of rip with offset takes 6 bytes).

-2 by jumping backwards to avoid using 02 twice.

.section .text
.globl main
main:
 mov 0ドルxff, %eax
start:
 push 20ドル # program length
 pop %ecx # counter
 mov $str, %edi # load string
loop:
 scasb # if (al == *(edi++)) 
 jne loop1 
 mov 2,ドル %al # ret 2
end: ret 
loop1:
 loop loop # do while (--counter)
 xor %eax, %eax # ret 0
 jmp end
str: .byte 0x6a,0x14,0x59,0xbf,0xf4,0x83,0x04,0x08 
 .byte 0xae,0x75,0x03,0xb0,0x02,0xc3,0xe2,0xf8
 .byte 0x31,0xc0,0xeb,0xf9 

Hexdump (of binary file format elf32-i386, as obj file unfortunately has 00 bytes for str address):

000003e0 6a 14 59 bf f4 83 04 08 ae 75 03 b0 02 c3 e2 f8 |j.Y......u......|
000003f0 31 c0 eb f9 6a 14 59 bf f4 83 04 08 ae 75 03 b0 |1...j.Y......u..|
00000400 02 c3 e2 f8 31 c0 eb f9 |....1...|

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x86, 256 bytes

Boring answer that is the equivalent of a giant comment. Input in cl, immediately returns 1 in al. I'll make an actual answer when I have the free time.

00000039 b0 01 c3 00 02 03 04 05 06 07 08 09 0a 0b 0c 0d |................|
00000049 0e 0f 10 11 12 13 14 15 16 17 18 19 1a 1b 1c 1d |................|
00000059 1e 1f 20 21 22 23 24 25 26 27 28 29 2a 2b 2c 2d |.. !"#$%&'()*+,-|
00000069 2e 2f 30 31 32 33 34 35 36 37 38 39 3a 3b 3c 3d |./0123456789:;<=|
00000079 3e 3f 40 41 42 43 44 45 46 47 48 49 4a 4b 4c 4d |>?@ABCDEFGHIJKLM|
00000089 4e 4f 50 51 52 53 54 55 56 57 58 59 5a 5b 5c 5d |NOPQRSTUVWXYZ[\]|
00000099 5e 5f 60 61 62 63 64 65 66 67 68 69 6a 6b 6c 6d |^_`abcdefghijklm|
000000a9 6e 6f 70 71 72 73 74 75 76 77 78 79 7a 7b 7c 7d |nopqrstuvwxyz{|}|
000000b9 7e 7f 80 81 82 83 84 85 86 87 88 89 8a 8b 8c 8d |~...............|
000000c9 8e 8f 90 91 92 93 94 95 96 97 98 99 9a 9b 9c 9d |................|
000000d9 9e 9f a0 a1 a2 a3 a4 a5 a6 a7 a8 a9 aa ab ac ad |................|
000000e9 ae af b1 b2 b3 b4 b5 b6 b7 b8 b9 ba bb bc bd be |................|
000000f9 bf c0 c1 c2 c4 c5 c6 c7 c8 c9 ca cb cc cd ce cf |................|
00000109 d0 d1 d2 d3 d4 d5 d6 d7 d8 d9 da db dc dd de df |................|
00000119 e0 e1 e2 e3 e4 e5 e6 e7 e8 e9 ea eb ec ed ee ef |................|
00000129 f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 fa fb fc fd fe ff |................|

Answer 27

R, 135 bytes

Inspired by this Python answer.

Previous versions were broken. Thanks to @Giuseppe for pointing out that paste was not required it saved 18 bytes or so. lengths(regmatches(z,gregexpr(x,z)))is from this answer.

function(x,z=rep(c("afilo2679=:","hmpu15'","nstxz","cgr","e","()",'"',","),c(2:5,7,9,15,16)))sum(lengths(regmatches(z,gregexpr(x,z))))

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Answer 28

Pari/GP, 36 bytes

f=(x)->#[x|c<-Vec(Str("f=",f)),x==c]

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Answer 29

Ruby, 48 bytes

->x{%q[->x{%q[].count(x.chr)*2}].count(x.chr)*2}

%q[str] is a more convenient way to write a string literal than "str" because it can be nested inside itself without any escaping. So I just put the entire code except for the copy inside it, then double the count.

Answer 30

Jelly, (削除) 12 (削除ここまで) 10 bytes

";Ṿċƈ";Ṿċƈ

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Reads a character from STDIN.

12 bytes

Ḣ"Ḣċ;Ṿ¤";Ṿ¤ċ

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Using input from command line argument.

Also 12 bytes:

ẇ"Ṿẇ¤Ḥ"Ṿ¤Ḥ""

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