Counter counter

Question

In typography, a counter is the area of a letter that is entirely or partially enclosed by a letter form or a symbol. A closed counter is a counter that is entirely enclosed by a letter form or symbol. You must write a program takes a string as input and prints the total number of closed counters in the text.

Your input:

• May be a command line input, or from STDIN, but you must specify which.

• Will consist entirely of the printable ASCII characters, meaning all ASCII values between 32 and 126 inclusive. This does include spaces. More information.

Now, this does vary slightly between fonts. For example, the font you're reading this in considers 'g' to have one closed counter, whereas the google font has 'g' with two closed counters. So that this is not an issue, here are the official number of closed counters per character.

All the symbols with no closed counters:

 !"'()*+,-./12357:;<=>?CEFGHIJKLMNSTUVWXYZ[\]^_`cfhijklmnrstuvwxyz{|}~

Note that this includes space.

Here are all the symbols with one closed counter:

#0469@ADOPQRabdegopq

And here are all the symbols with 2 closed counters:

$%&8B

And last but not least, here are some sample inputs and outputs.

Programming Puzzles and Code-Golf should print 13

4 8 15 16 23 42 should print 5

All your base are belong to us should print 12

Standard loopholes apply should print 12

Shortest answer in bytes is the winner! should print 8

Answer 1

Python 3, 63

print(sum(map(input().count,"#0469@ADOPQRabdegopq$%&8B$%&8B")))

A straightforward approach. Iterates over each character with a closed counter, summing the number of occurrences, doing so twice for characters with two closed counters. It would be the same length to instead write

"#0469@ADOPQRabdegopq"+"$%&8B"*2

Python 3 is needed to avoid raw_input.

Answer 2

CJam, 41 39 37 34 bytes

"$%&8Badopq#0469@Rbeg"_A<eu+qfe=1b

Thanks to @jimmy23013 for golfing off 3 bytes!

Try it online.

How it works

"$%&8Badopq#0469@Rbeg"             e# Push that string.
                      _A<          e# Retrieve the first 10 characters.
                         eu+       e# Convert to uppercase and append.
                                   e# This pushes "$%&8Badopq#0469@Rbeg$%&8BADOPQ".
                            q      e# Read from STDIN.
                             fe=   e# Count the occurrences of each character. 
                                1b e# Base 1 conversion (sum).

Answer 3

Pyth, 31 bytes

sm@tjC"cúÁ-ÈN%³rØ|­"3Cdz

Demonstration.

Note that the code may not be displayed properly due to the use of non-ASCII characters. The correct code is at the link.

I made a lookup table of the output desired for each input character, rotated it by 32 to make use of Pyth's modular indexing, stuck a 1 at the beginning, and interpreted it as a base 3 number, giving the number 2229617581140564569750295263480330834137283757. I then converted this number to base 256 and converted it to a string, which is the string used in the answer.

Answer 4

sed, 51

With golfing help from @manatwork and @TobySpeight:

s/[$%&8B]/oo/g
s/[^#0469@ADOPQRabdegopq]//g
s/./1/g

Input from STDIN. With this meta-question in mind, the output is in unary:

$ echo 'Programming Puzzles and Code-Golf
4 8 15 16 23 42
All your base are belong to us
Standard loopholes apply
Shortest answer in bytes is the winner!' | sed -f countercounter.sed
1111111111111
11111
111111111111
111111111111
11111111
$ 

Answer 5

Perl, 41

$_=y/#0469@ADOPQRabdegopq//+y/$%&8B//*2

41 characters +1 for the -p flag.

This uses y/// to count the characters.

echo 'Programming Puzzles and Code-Golf' | perl -pe'$_=y/#0469@ADOPQRabdegopq//+y/$%&8B//*2'

Answer 6

GNU APL, 39 bytes

+/⌈20÷⍨26-'$%&8B#0469@ADOPQRabdegopq'⍳⍞

Try it online in GNU APL.js.

How it works

                                      ⍞ Read input.
          '$%&8B#0469@ADOPQRabdegopq'⍳  Compute the indexes of the input characters
                                        in this string. Indexes are 1-based.
                                        26 == 'not found'
       26-                              Subtract each index from 26.
   20÷⍨                                 Divide those differences by 20.
  ⌈                                     Round up to the nearest integer.
+/                                      Add the results.

Answer 7

JavaScript, 86

I/O via popup. Run the snippet in any drecent browser to test.

for(c of prompt(t=0))c='$%&8B#0469@ADOPQRabdegopq'.indexOf(c),~c?t+=1+(c<5):0;alert(t)

Answer 8

K, 54 43 42 37 bytes

+//(30#"$%&8B#0469@ADOPQRabdegopq"=)'

Cut off 5 bytes thanks to @JohnE!

Older version:

f:+/(#&"#0469@ADOPQRabdegopq$%&8B$%&8B"=)'

Original:

f:+/{2-(5="$%&8B"?x;20="#0469@ADOPQRabdegopq"?x;0)?0}'

Answer 9

C, 127 bytes

n;f(char*i){char*o="#0469@ADOPQRabdegopq",*t="$%&8B";for(;*i;i++)if(strchr(o,*i))n++;else if(strchr(t,*i))n+=2;printf("%d",n);}

Pretty straightforward. Ungolfed version:

int num = 0;
void f(char* input)
{
    char *one="#0469@ADOPQRabdegopq";
    char *two="$%&8B";

    for(;*input;input++)
        if(strchr(one, *input))     //If current character is found in first array
            num ++;
        else if(strchr(two, *input))//If cuurent character is found in second array
            num += 2;

    printf("%d", num);
}

Test it here

If function arguments are not allowed, then the stdin version takes up to 141 bytes:

n;f(){char*o="#0469@ADOPQRabdegopq",*t="$%&8B",i[99],*p;gets(i);for(p=i;*p;p++)if(strchr(o,*p))n++;else if(strchr(t,*p))n+=2;printf("%d",n);}

Note that the above version assumes that the input is atmost 98 characters long.

Test it here

Command-line arguments version (143 bytes):

n;main(c,v)char**v;{char*o="#0469@ADOPQRabdegopq",*t="$%&8B",*p=v[1];for(;*p;p++)if(strchr(o,*p))n++;else if(strchr(t,*p))n+=2;printf("%d",n);}

Test it here

Answer 10

Python 2, 96 90 75 67+2 = 69 Bytes

Can't think of any other way to do this... is what I would have thought until I saw xnor's solution. I'll post what I had anyways.

Thanks to FryAmTheEggman for saving 6 bytes

Alright, now I'm happy with this.

Thanks to xnor for the find trick, saving 4 bytes.

Added two bytes since input needs to be enclosed in quotes.

print sum('#0469@ADOPQRabdegopq$%&8B'.find(x)/20+1for x in input())

Answer 11

Java, 162

class C{public static void main(String[]a){System.out.print(-a[0].length()+a[0].replaceAll("[#0469@ADOPQRabdegopq]","..").replaceAll("[$%&8B]","...").length());}}

Well if it has to be a full program... It's just a one-liner that matches characters and replaces them with a longer string. Then, returns the difference in length from the original. Unfortunately, java doesn't really have anything to just count the number of matches.

Here it is with line breaks:

class C{
    public static void main(String[]a){
        System.out.print(
                -a[0].length() +
                a[0].replaceAll("[#0469@ADOPQRabdegopq]","..")
                .replaceAll("[$%&8B]","...")
                .length()
                        );
    }
}

Answer 12

Pyth - 35 bytes

Uses the obvious method of in first + *2 in second. Thanks @FryTheEggman.

s/Lz+"#0469@ADOPQRabdegopq"*2"$%&8B

Try it here online.

Answer 13

Javascript, 114 95 bytes

alert(prompt().replace(/[#046@ADOPQRabdegopq]/g,9).replace(/[$%&8B]/g,99).match‌​(/9/g).length)

Thanks to Ismael Miguel for helping me golf this.

Answer 14

Ruby, 59 bytes

a=gets;p a.count('#0469@ADOPQRabdegopq')+2*a.count('$%&8B')

Input from command line or stdin. Shortest so far using a non-esoteric language.

Update: chilemagic beat me

Answer 15

Retina, 44 bytes

1

[#0469@ADOPQRabdegopq]
1
[$%&8B]
11
[^1]
<empty line>

Gives output in unary.

Each line should go to its own file or you can use the -s flag. E.g.:

> echo "pp&cg"|retina -s counter
11111

The pairs of lines (pattern - substitute pairs) do the following substitution steps:

• Remove 1's
• Substitute 1-counter letters with 1
• Substitute 2-counter letters with 11
• Remove everything but the 1's

Answer 16

J, 43

As a function:

   f=:[:+/[:,30$'$%&8B#0469@ADOPQRabdegopq'=/]
   f 'text goes here'
6

46 bytes (command line)

As a standalone command-line program:

echo+/,30$'$%&8B#0469@ADOPQRabdegopq'=/>{:ARGV

Save the above line as counter2.ijs and call from the command line:

$ jconsole counter2.ijs 'Programming Puzzles and Code Golf'
13

Answer 17

Julia, 77 74 bytes

t=readline();length(join(t∩"#0469@ADOPQRabdegopq")*join(t∩"\$%&8B")^2)

This reads text from STDIN and prints the result to STDOUT.

Ungolfed + explanation:

# Read text from STDIN
t = readline()

# Print (implied) to STDOUT the length of the intersection of t with the
# 1-closed counter list joined with the duplicated intersection of t with
# the 2-closed counter list
length(join(t ∩ "#0469@ADOPQRabdegopq") * join(t ∩ "\$%&8B")^2)

Example:

julia> t=readline();length(join(t∩"#0469@ADOPQRabdegopq")*join(t∩"\$%&8B")^2)
Programming Puzzles and Code Golf
13

Answer 18

rs, 56 bytes

_/
[#0469@ADOPQRabdegopq]/_
[$%&8B]/__
[^_]/
(_+)/(^^\1)

Live demo.

Answer 19

GNU APL, 37 characters

+/,⍞∘.=30⍴'$%&8B#0469@ADOPQRabdegopq'

construct a character vector which contains 2-counter characters twice (30⍴)

compare each input char with every character in the vector (∘.=)

sum up ravelled matches (+/,)

Answer 20

Javascript 159, 130 Bytes

function c(b){return b.split("").map(function(a){return-1!="$%&8B".indexOf(a)?2:-1!="#0469@ADOPQRabdegopq".indexOf(a)?1:0}).reduce(function(a,b){return a+b})};

unminified:

function c(b) {
    return b.split("").map(function(a) {
        return -1 != "$%&8B".indexOf(a) ? 2 : -1 != "#0469@ADOPQRabdegopq".indexOf(a) ? 1 : 0
    }).reduce(function(a, b) {
        return a + b
    })
};

With the help of @edc65:

function c(b){return b.split('').reduce(function(t,a){return t+(~"$%&8B".indexOf(a)?2:~"#0469@ADOPQRabdegopq".indexOf(a)?1:0)},0)}

Answer 21

Haskell, 117

a="#0469@ADOPQRabdegopq"
b="$%&8B"
c n[]=n
c n(x:s)
 |e a=f 1
 |e b=f 2
 |True=f 0
 where
 e y=x`elem`y
 f y=c(n+y)s

c is a function c :: Int -> String -> Int which takes a counter and a string and goes through the string one letter at a time checking if the current letter is a member of the 1 point array or the 2 point array and calls itself for the rest of the string after incrementing the counter the appropriate amount.

Call with counter = 0 in ghci:

ghci> c 0 "All your base are belong to us"
12

Answer 22

C#, 157

void Main(string s){int v=0;v=s.ToCharArray().Count(c=>{return "#0469@ADOPQRabdegopq".Contains(c)||"$%&8B".Contains(c)?++v is int:false;});Console.Write(v);}

Ungolfed:

void Main(string s)
{
    int v = 0;

    v = s.ToCharArray()
    .Count(c => 
    {
        return "#0469@ADOPQRabdegopq".Contains(c) || "$%&8B".Contains(c) ? ++v is int:false;
    });

    Console.Write(v);
}

Converting the string into a char array, then seeing if each char is in either counter. If it is in the second one, I just have it incrementing the counter again.

Answer 23

Erlang, 103 bytes

This is a complete program that runs using escript. The first line of the file must be blank (adding 1 byte).

main([L])->io:write(c(L,"#0469@ADOPQRabdegopq")+2*c(L,"$%&8B")).
c(L,S)->length([X||X<-L,S--[X]/=S]).

Sample run:

$ escript closed.erl 'Shortest answer in bytes is the winner!'
8$

Answer 24

C, 99 bytes

n;f(char*i){for(char*o="#0469@ADOPQRabdegopq$%&8B",*t=o+20;*i;)n+=2-!strchr(o,*i)-!strchr(t,*i++);}

Explanation

I further golfed the answer of Cool Guy; it got too long to be a comment. Instead of the if/else, I took advantage of ! converting a pointer to bool. I also made o include t so that I could add "is in o" and "is in t" for the total number of counters.

Expanded code

#include <string.h>
int num = 0;

void f(char* input)
{
    char *one = "#0469@ADOPQRabdegopq"  "$%&8B";
    char *two = strchr(one, '$');

    while (*input) {
        num += 2 - !strchr(one, *input) - !strchr(two, *input));
        ++input;
    }
}

Output is in num, which must be cleared before each call.

Test program and results

#include <stdio.h>
int main() {
    const char* a[] = {
        "Programming Puzzles and Code-Golf",
        "4 8 15 16 23 42",
        "All your base are belong to us",
        "Standard loopholes apply",
        "Shortest answer in bytes is the winner!",
        NULL
    };
    for (const char** p = a;  *p;  ++p) {
        n=0;f(*p);
        printf("%3d: %s\n", n, *p);
    }
    return 0;
}

 13: Programming Puzzles and Code-Golf
  5: 4 8 15 16 23 42
 12: All your base are belong to us
 12: Standard loopholes apply
  8: Shortest answer in bytes is the winner!
 37: n;f(char*i){for(char*o="#0469@ADOPQRabdegopq$%&8B",*t=o+20;*i;)n+=2-!strchr(o,*i)-!strchr(t,*i++);}

The code itself contains 37 counters by its own metric.