Display a tally (in varying bases)

Question

Tallying is a simple counting system that works in base 5. There are various different tallying systems used around the world, but the one that is used in most English-speaking countries is perhaps the most simple - count units by marking vertical lines, then for each 5th mark put a horizontal line through the previous collection of four. This clusters the tally marks in groups of 5 (and makes them easier to quickly count).

You are going to write a program that displays tally marks up to a given value. But, tallying in only base 5 is boring! Therefore, your program should also be able to display tallies in different bases.

Input

The input will be either one or two non-negative integer values separated by a comma (e.g. 9 or 8,4). The first number is the value that should be displayed by the tally. The second value is the base of the tally. If the second value is not given, use base 5.

Output

The output will be the inputted value represented as ASCII art tally marks. Here are some examples you can test your program against - your output should match them exactly!

Input: 12 or 12,5

 | | | |   | | | |   | |
-+-+-+-+- -+-+-+-+-  | |
 | | | |   | | | |   | |

Input: 7,3

 | |   | |   |
-+-+- -+-+-  |
 | |   | |   |

Input: 4,2

 |   |
-+- -+-
 |   |

Input: 6,1 or 6,10 (notice the leading spaces)

 | | | | | |
 | | | | | |
 | | | | | |

Note also that base 1 is intended to be inconsistent - only vertical lines should be used.

If either of the inputted values is 0, there should be no output whatsoever (and your program should end gracefully).

Rules

• This is code-golf, so the shortest correct implementation (in bytes) wins.
• Input/output can be in any suitable medium (e.g. stdin/stdout, file...).
• Input can be in the form of multiple command-line arguments or separated by spaces, etc. if it is more suitable for your target language.
• Trailing newlines are allowed in the output. Trailing spaces are not. This rule only applies when there is an output (i.e. not when the inputted value is 0).
• Your code must default to base 5 when no base is input.

Answer 1

CJam 103 85 72 characters

Try it at http://cjam.aditsu.net/.

original

q","/(i:A\_,{~i}{;5}?:B_@*{(_{" |"*S"l"++AB/*AB%}{;MA}?\:J" |l""-+ "er\" |"*N+_J\+@2$+@J\+++"l"Ser}{;}?

Works by defining one set of tallys with spaces, lines and an l for spaces that should remain spaces. Then takes advantage of er (tranliteration) function to make the second line. Most inefficient part is dealing with the 1 and 0 special cases. Will edit as I improve it. Tip I took too long to realize: as the second input being 1 is the same as infinity or the first input +1, redefining it when it is equal to 1 saves alot of work.

most improved so far with comma delimited

l",":G/(i:A\_5a?~i:B_@*{(_" |":K*\{SG++AB/*AB%}{A}?\_KG+"-+ "er[\GSer_@@M]\K*N+*}{;}?

most improved so far with space delimited input

Naturally as CJam is really designed for space delimited input. Placing the input at 20 3 instead of 20,3 is a huge benefit.

ri:Aq_5s?S-i(_)A)?:B*{B(" |":K*STs++ABmd@@*_K"-+"er[\_@@M]\K*N+*TsSer}M?

Answer 2

Python 2 - 111 108 119 144 140 136 135 134 - Try it

Ok, let's give it a try:

i=input()
n,b=[(i,5),i][i>[]]
o=b-1
a=[n,n%b][b>1]*' |'
m=(b>1)*n/b
s=(' |'*o+'  ')*m+a
print(s+'\n'+('-+'*o+'- ')*m+a+'\n'+s)*(b*n>0)

Edit: I've overlooked that there should be no output if n==0 or b==0. This costs me 11 characters. :(

Edit: Ok, after fixing the second issue mentioned in the comments my solution basically converged to the one from BeetDemGuise.

Answer 3

Bash, 239 228 199 189 188

Here's my attempt, it could be golfed a lot.

Note: the second line does not subtract 5 from 2, it sets a default value if $2 is empty!

n=$1
b=${2-5}
((n<1&b<1))&&exit
while ((n>b&b>1));do
n=$[n-b]
x+=\
y+=-
for i in `seq $[b-1]`;{
x+='| '
y+=+-
}
x+=\
y+=\
done
for j in `seq $n`;{
x+=' |'
y+=' |'
}
echo -e "$x\n$y\n$x"

Answer 4

Python - 171 143

i=input();t,b=i if[0]<i else(i,5);b=[b,t+1][b==1];l,d,m,o=' |',t/b,t%b,b-1;r=(l*o+'  ')*d+l*m
if t*b>0:print r,'\n',('-+'*o+'- ')*d+l*m,'\n',r

The program is pretty straight-forward:

• Get the input and try to unpack into t,b. If that fails, simply assign the correct values.
• If the base was 1, change its value to something that can handle all vertical lines easily (t+1).
• Set some variables and create the bottom and top sections of the tallies.
• Print out the tallies if both t and b are non-zero.

EDIT 1: Use the input function instead of raw_input after some playing around.

EDIT 2: Thanks to Falko for pointing out a small bug with my non-zero checking. Now our code is basically identical, less some variable names and some small logic.

EDIT 3: Thanks to how Python compares sequences and different types, we can compare i to a list to get a shorter version of our try...except block.

Here is the ungolfed version:

i=input()

# If True, `i` must be a list
if [0]<i:
    t,b=i
# Otherwise, we know its a number (since `list` comes after `int` lexicographically.)
else:
    b=5
    t=i

b = b if b==1 else t+1
l=' |'
d=t/b
m=t%b
o=b-1

r=(l*o+'  ')*d+l*m
if t and b:
    print r,'\n',('-+'*o+'- ')*d+l*m,'\n',r

Answer 5

C - 193

I'm so sorry. Dealing with the special case for 1 was a bit of a bad hack so I guess this could be golfed more with a better approach. Also, this code includes a new line at the beginning of the output, so if this is not allowed, please let me know.

Of course, very ugly-looking defines always help :)

Character count includes only necessary spaces and new lines.

#define P printf(
#define L P" |")
#define A P"\n");for(i=0;i<a;)b==1?i++,L:i++&&i%b==0?P
i;
main(a,b)
{
    scanf("%d,%d",&a,&b)<2?b=5:!b?a=0:a;
    if(a){
    A"  "):L;
    A"- "):a%b&&i>a/b*b?L:P"-+");
    A"  "):L;}
}

Answer 6

Java, 343

class C{public static void main(String[]a){long n=new Long(a[0])+1,b=a.length>1?new Long(a[1]):5;if(b>0){if(b<2)b=(int)2e9;int i;for(i=1;i<n;i++)p(i%b>0?" |":"  ");p("\n");for(i=1;i<n-n%b;i++)p(i%b>0?"-+":"- ");if(n>b)p("- ");for(i=1;i<n%b;i++)p(" |");p("\n");for(i=1;i<n;i++)p(i%b>0?" |":"  ");}}static void p(String s){System.out.print(s);}}

Less golfed:

class C {
  public static void main(String[] a) {
    long n=new Long(a[0])+1, b=a.length>1 ? new Long(a[1]) : 5;
    if(b>0) {
      if(b<2) b=(int)2e9; // if the base is 1, pretend the base is 2 billion
      int i;
      for(i=1;i<n;i++) p(i%b>0 ? " |" : "  ");
      p("\n");
      for(i=1;i<n-n%b;i++) p(i%b>0 ? "-+" : "- ");
      if(n>b) p("- ");
      for(i=1;i<n%b;i++) p(" |");
      p("\n");
      for(i=1;i<n;i++) p(i%b>0 ? " |" : "  ");
    }
  }
  static void p(String s) {
    System.out.print(s);
  }
}

Answer 7

Perl - 167 165 156

my($n,$b)=($ARGV[0]=~/(\d+)(?:,(\d+))?/,5);print$b>1?map{join(" ",($_ x($b-1))x int($1/$b)," | "x($1%$b))."\n"}(" | ","-+-"," | "):join" ",("---")x$1if$1*$b

ungolfed

my($n,$b) = ($ARGV[0] =~ /(\d+)(?:,(\d+))?/, 5);
print($b>1 ?
    map{ 
        join(" ",($_ x ($b-1)) x int($1/$b)," | " x ($1%$b))."\n"
    } (" | ","-+-"," | ") :
    join " ", ("---") x $1
) if $1 * $b

Answer 8

C# 271bytes

Not the shortest, I couldn't golf the input reading due to it needing to accept 0 as an input.

using C=System.Console;class P{static void Main(){var L=C.ReadLine().Split(',');int t=int.Parse(L[0]),f=L.Length>1?int.Parse(L[1]):5,r=3,i;for(var R="";t*f>0&r-->0;C.WriteLine(R.TrimEnd()))for(R="",i=0;i<t;R+=r==1&i++<t-t%f?(i%f<1?"- ":"-|"):i%f<1?"  ":" |")f+=f<2?t:0;}}

Formatted code:

using C=System.Console;

class P
{
    static void Main()
    {
        var L=C.ReadLine().Split(',');
        int t=int.Parse(L[0]),f=L.Length>1?int.Parse(L[1]):5,r=3,i;

        for(var R="";t*f>0&r-->0;C.WriteLine(R.TrimEnd()))
            for(R="",i=0;i<t;R+=r==1&i++<t-t%f?(i%f<1?"- ":"-|"):i%f<1?"  ":" |")
                f+=f<2?t:0;
    }
}

Answer 9

Lua - 219 203 bytes

I went for the make d copies of b copies of "|", then add r copies of "|" at the end. I feel like maybe I should have gone with the 'tally up' the "|"s to the string one at a time.

l=' |'s=string.rep _,_,a,b=io.read():find'(%d+)%D*(%d*)'b=tonumber(b)or 5 d=(a-a%b)/b f=b>1 and s(s(l,b-1)..'  ',d)g=b>1 and s(s('-+',b-1)..'- ',d)r=b>1 and a%b or a e=s(l,r)..'\n'print(f..e..g..e..f..e)

ungolfed:

l=' |'          --the base string
s=string.rep    --string.rep will be used a lot, so best shorten it

_,_,a,b=io.read():find'(%d+)%D*(%d*)' --reads a,b I'm probably way of the mark with this one

b=tonumber(b)or 5

d=(a-a%b)/b -- shorter than math.floor(a/b), d equal the vertical mark

f=b>1 and s(s(l,b-1)..'  ',d) or '' --creates d multiples of b multiples of "|" more or less
g=b>1 and s(s('-+',b-1)..'- ',d)or''--same as above but with the middle "-+-"

r=b>1 and a%b or a --idk maybe i should set r before d(a- a%b )/b

e=s(l,r)..'\n'  -- makes the remainder string, notice that if b==1  then e will output all the "|" 

print(f..e..g..e..f..e) -- here's where the real magic happens!

Sample:

c:\Programming\AnarchyGolfMine>lua test.lua
13,5
 | | | |   | | | |   | | |
-+-+-+-+- -+-+-+-+-  | | |
 | | | |   | | | |   | | |


c:\Programming\AnarchyGolfMine>lua test.lua
6,2
 |   |   |
-+- -+- -+-
 |   |   |


c:\Programming\AnarchyGolfMine>lua test.lua
18,1
 | | | | | | | | | | | | | | | | | |
 | | | | | | | | | | | | | | | | | |
 | | | | | | | | | | | | | | | | | |

Answer 10

JavaScript (193)

This might be overly complex.

s=prompt().split(",");a=+s[0];b=s[1];b=b?+b:5;o=b>1;v=" | ";q=w="";for(g=~~(a/b),c=o?a-g:a;g+1;g--,q+=" ",w+=" ")for(i=o;c&&i<b;i++)c--,q+=v,w+=g&&o?"-+-":v;if(a&&b)console.log(q+'\n'+w+'\n'+q)

---

Commented version

s=prompt().split(",");
a=+s[0];
b=s[1];
b=b?+b:5;   // convert b to int and default to 5
o=b>1;      // special handling for b0 and b1
v=" | ";
q=w="";
// calculate number of lines and number of groups
for(g=~~(a/b),c=o?a-g:a;g+1;g--,q+=" ",w+=" ")
    for(i=o;c&&i<b;i++)
        c--,  // decrease line count
        q+=v,
        w+=g&&o?"-+-":v; // use " | " for last group and "-+-" for others
if(a&&b) // handle b0
    console.log(q+'\n'+w+'\n'+q)

Answer 11

Python - 127 123 122

Just sneaking in with a slightly shorter python version.

edit: Fixed 0 not printing nothing and rejigger, ended up the same length

k=input()
k=i,j=((k,5),k)[k>[]]
for m in[' |','-+',' |']*all(k):
 print(m*(j-1)+m[0]+' ')*(i/j*(j>1))+' |'*(i%(j+(j<2)*i))

Answer 12

Python 2, 134 126 123 114 bytes

lambda i,j=5,a=" |":"\n".join(("",(a*~-j+"  ","-+"*~-j+"- ")[x%2]*(i/j))[j>1]+a*(i,i%j)[j>1]for x in(0,1,2)if i*j)

Try it online!

Old question I know but fun to have a go at anyway. Chance to try out a few tricks I've learned since joining.

Answer 13

C (207 characters)

The newline right before exit is for legibility only.

#define P printf(
#define t(v)for(a=c;b<=a;a-=b)a-c&&P" "),P v+1),q(b,v);q(a," |");P"\n");
q(n,c){while(n--)P c);}a;b;c;main(){scanf("%d,%d",&c,&b)<2?b=5:0;b&&c||
exit();b==1?b=32767:0;t("| ")t("+-")t("| ")}

scanf usage shamelessy stolen from Allbeert. Notice that this solution does not compile with gcc as it tries to enforce a nonexistant prototype for exit. Compile with a working C compiler like tcc. Also, this function may or may not work on 64 bit platforms. Use with care.

Here is the original ungolfed implementation this is based on:

#include <stdio.h>
#include <stdlib.h>

static void
tally_line(int base, int count, const char *str)
{
     int follower = 0, i;

     /* full tallies first */
     for (; count >= base; count -= base) {
          if (follower++)
               putchar(' ');

          /* only print second character */
          printf(str + 1);

          for (i = 0; i < base; i++)
               printf(str);
     }

     /* partial tally */
     for (i = 0; i < count; i++)
          printf(" |");

     /* newline */
     puts("");
}

extern int
main(int argc, char **argv)
{
     int base, count;

     /* do away with program name */
     count = atoi(*++argv);

     base = argc - 3 ? 5 : atoi(*++argv);

     /* remove 0 later */
     base | count || exit(0);

     /* a crossed-out tally never appears for large numbers */
     if (base == 1)
          base = 32767;

     tally_line(base, count, "| ");
     tally_line(base, count, "+-");
     tally_line(base, count, "| ");

     return (EXIT_SUCCESS);
}