Code Golf: Hourglass

Asked 2009-11-05T21:43:15.583

The challenge

The shortest code by character count to output an hourglass according to user input.

Input is composed of two numbers: First number is a greater than 1 integer that represents the height of the bulbs, second number is a percentage (0 - 100) of the hourglass' capacity.

The hourglass' height is made by adding more lines to the hourglass' bulbs, so size 2 (the minimal accepted size) would be:

_____
\   /
 \ /
 / \
/___\

Size 3 will add more lines making the bulbs be able to fit more 'sand'.

Sand will be drawn using the character x. The top bulb will contain N percent 'sand' while the bottom bulb will contain (100 - N) percent sand, where N is the second variable.

'Capacity' is measured by the amount of spaces () the hourglass contains. Where percentage is not exact, it should be rounded up.

Sand is drawn from outside in, giving the right side precedence in case percentage result is even.

Test cases

Input:
    3 71%
Output:
    _______
    \x  xx/
     \xxx/
      \x/
      / \
     /   \
    /__xx_\

Input:
    5 52%
Output:
    ___________
    \         /
     \xx   xx/
      \xxxxx/
       \xxx/
        \x/
        / \
       /   \
      /     \
     /  xxx  \
    /xxxxxxxxx\

Input:
    6 75%
Output:
     _____________
     \x         x/
      \xxxxxxxxx/
       \xxxxxxx/
        \xxxxx/
         \xxx/
          \x/
          / \
         /   \
        /     \
       /       \
      /         \
     /_xxxxxxxxx_\

Code count includes input/output (i.e full program).

language-agnostic code-golf rosetta-stone

Edit Report

3 Answers

C/C++, a dismal 945 characters...

Takes input as parameters: a.out 5 52%

#include<stdio.h>
#include<memory.h>
#include<stdlib.h>
#define p printf

int h,c,*l,i,w,j,*q,k;const char*
 z;int main(int argc,char**argv)
  {h=atoi(argv[1]);c=(h*h*atoi(
   argv[2])+99)/100;l=new int[
    h*3];for(q=l,i=0,w=1;i<h;
     i++,c=(c-w)&~((c-w)>>31
      ),w+=2)if(c>=w){*q++=
       0;*q++ =0;* q++=w;}
        else {*q++=(c+1)/
         2;*q++=w-c;*q++
          =c/2;}p("_");
           for(i=0;i<h
            ;i ++)p (
             "__");p
              ("\n"
               );q
                =
               l+h
              *3-1;
             for (i=
            --h;i>=0;
           i--){p("%*"
          "s\\",h-i,"")
         ; z= "x\0 \0x";
        for(k=0;k<3;k++,q
       --,z+=2)for(j=0;j<*
      q;j++)p(z);q-=0;p("/"
     "\n");}q=l;for(i=0;i<=h
    ;i++){z =i==h? "_\0x\0_":
   " \0x\0 ";p("%*s/",h-i,"");
  for(k=0;k<3;k++,q++,z+=2)for(
 j=0;j<*q;j++)p(z);p("\\\n") ;}}

...and the decrypted version of this for us mere humans:

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

#define p printf

int h, c, *l, i, w, j, *q, k;
const char *z;

int main(int argc, char** argv)
{
    h = atoi(argv [1]);
    c = (h*h*atoi(argv[2])+99)/100;
    l = new int[h*3];
    for (q = l,i = 0,w = 1; i<h; i++,c = (c-w)&~((c-w)>>31),w += 2) {
        if (c>=w) {
            *q++ = 0;
            *q++ = 0;
            *q++ = w;
        } else {
            *q++ = (c+1)/2;
            *q++ = w-c;
            *q++ = c/2;
        }
    }
    p("_");
    for (i = 0; i<h; i++) {
        p("__");
    }
    p("\n");
    q = l+h*3-1;
    for (i = --h; i>=0; i--) {
        p("%*s\\",h-i,"");
        z = "x\0 \0x";
        for (k = 0; k<3; k++,q--,z += 2) {
            for (j = 0; j<*q; j++) {
                p(z);


                        
                        
                            answered 2009-11-05T23:58:17.790


            
            
                
                    
                        
                            
                            
                        
                        14
                        
                            
                            
                        
                        
                    
                    
                        
                        Python, 213 char

N,p=map(int,raw_input()[:-1].split())
S=N*N-N*N*(100-p)/100
_,e,x,b,f,n=C='_ x\/\n'
o=""
r=1
while N:N-=1;z=C[N>0];s=min(S,r);S-=s;t=r-s;v=s/2;w=s-v;r+=2;o=n+e*N+b+x*v+e*t+x*w+f+o+n+e*N+f+z*w+x*t+z*v+b
print _*r+o


                        
                        
                            answered 2009-11-06T18:40:30.547
                            
                        
                        
                            
                                
                                
                            
                        
                        
                            
                            
                            
                                
                                
                            
                            
                        
                    
                
            
            
            
                
                    
                        
                            
                            
                        
                        1
                        
                            
                            
                        
                        
                    
                    
                        
                        Python - 272 chars

X,p=map(int,raw_input()[:-1].split())
k=X*X;j=k*(100-p)/100
n,u,x,f,b,s='\n_x/\ '
S=list(x*k+s*j).pop;T=list(s*k+u*(2*X-j-1)+x*j).pop
A=B=""
for y in range(X):
 r=S();q=T()
 for i in range(X-y-1):r=S()+r+S();q+=T();q=T()+q
 A+=n+s*y+b+r+f;B=n+s*y+f+q+b+B
print u+u*2*X+A+B


                        
                        
                            answered 2009-11-06T08:44:58.287



        
        
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