Draw Growing Stacks of Boxes

Question

The Task

In this challenge, your task is to draw an ASCII art representation of several stacks of boxes of increasing height. You are given as input the number of stacks, which is a positive integer. The first stack contains one box of size 2x2. The second stack contains 2 boxes of size 3x3. In general, the kth stack contains k boxes of size (k+1)x(k+1).

The borders of each box are drawn using the characters -|+, and their interior consists of whitespace. Adjacent boxes share their borders, and corners should always be drawn with +, even when they are part of a border of another box.

Examples

Output for 1:

++
++

Output for 2:

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

Output for 3:

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

Output for 5:

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

Rules and Scoring

The input can be received from STDIN, as a command line argument, or as a function argument. Output must go to STDOUT or closest equivalent. Any finite amount of trailing whitespace is allowed, as are preceding and trailing newlines, but there cannot be any extra preceding spaces.

This is code-golf, so the lowest byte count wins. Standard loopholes are disallowed.

Answer 1

CJam, 64 60 58 bytes

]ri:X{'|'}{I))*\I**XX*)Se[s}:L~:M.e>S'-LaI*~M}fI]zN*'}/'+*

Constructing each column at a time.

Try it online here

Answer 2

Java (407 349 chars)

A few chars thanks to @Zgarb and @Geobits

Code

void s(int q){int d,h,y,i,j,x,z,t=q*q+1;char b;for(i=0;i<t;i++){z=x=0;d=t-i;for(j=0;j<(q*q+q)/2+1;j++){b=' ';h=x*x+1;if(x==z){y=x+1;if((d<=h&d%(x==0?1:x)==(x==1?0:1))|(y<=q&d<=y*y+1&d%(y==0?1:y)==(y==1?0:1)))b='+';else if(d<=h|y<=q&d<=y*y+1)b='|';x++;z=1;}else{if(d<=h&d%(x==0?1:x)==(x==1?0:1))b='-';z++;}System.out.print(b);}System.out.println();}}

Not sure if this is optimal, but it's my first attempt, I will probably try to put it in a better golfing language later. Any suggestions are welcome!

Expanded

class StackingBlocks{
    public static void main(String[]a){
        int d,h,y,i,j,x,z,t,q=10;
        t=q*q+1;
        char b;
        for(i=0;i<t;i++){
            z=x=0;
            d=t-i;
            for(j=0;j<(q*q+q)/2+1;j++){
                b=' ';
                h=x*x+1;
                if(x==z){
                    y=x+1;
                    if((d<=h&d%(x==0?1:x)==(x==1?0:1))|(y<=q&d<=y*y+1&d%(y==0?1:y)==(y==1?0:1)))
                        b='+';
                    else if(d<=h|y<=q&d<=y*y+1)
                        b='|';
                    x++;
                    z=1;
                }else{
                    if(d<=h&d%(x==0?1:x)==(x==1?0:1))
                        b='-';
                    z++;
                }
                System.out.print(b);
            }
            System.out.println();
        }
    }
}

Check it out here.

Answer 3

Python 2, 144 128 bytes

n=input()
i=n*n
while-~i:j=x=1;l="";exec'y=i%j<1;z=i>j*j;l+=j*z*" "or"|+"[x|y]+" -"[y]*~-j;x=y^z>z;j+=1;'*n;print l+"|+"[x];i-=1

Bit twiddling. Bit twiddling everywhere.

Answer 4

Python, 188 bytes

Mathematically calculates the character at each x,y position. It was tricky making the +s print on both sides of each box as well as stopping the rightmost +s of what would be n+1th boxes.

n=input();l=1;c=0
for y in range(n*n,-1,-1):
 s=""
 for x in range((n*n+n)/2+1):k=((8*x+1)**.5+1)/2;i=int(k);b=y<=i**2;s+=" |-+"[((k==i)+2*((y%l+c)*(y%i+(k==n+1))<1))*b];l=i;c=b^1
 print s

Answer 5

C# - 304 bytes (function)

void b(int s){int h=s*s,w=h+s>>1,x,y,j;var c=new int[w+1,h+1];for(;s>0;s--){for(y=s*s-s;y>=0;y-=s){x=s*s-s>>1;for(j=0;j<s;){c[x+j,y]=c[x+j,y+s]=13;c[x,y+j]=c[x+s,y+j++]=92;}c[x,y]=c[x+s,y]=c[x+s,y+s]=c[x,y+s]=11;}}for(y=h;y>=0;y--){for(x=0;x<=w;x++)Console.Write((char)(32+c[x,y]));Console.WriteLine();}}

or 363 bytes (full code)

namespace System{class C{static void Main(string[]a){int s=int.Parse(a[0]),h=s*s,w=h+s>>1,x,y,j;var c=new int[w+1,h+1];for(;s>0;s--){for(y=s*s-s;y>=0;y-=s){x=s*s-s>>1;for(j=0;j<s;){c[x+j,y]=c[x+j,y+s]=13;c[x,y+j]=c[x+s,y+j++]=92;}c[x,y]=c[x+s,y]=c[x+s,y+s]=c[x,y+s]=11;}}for(y=h;y>=0;y--){for(x=0;x<=w;x++)Console.Write((char)(32+c[x,y]));Console.WriteLine();}}}}

I tried to avoid if statements. Ungolfed:

namespace N
{
    public class Explained
    {
        static void boxes(string[] args)
        {
            int size = int.Parse(args[0]);
            int height = size * size + 1;
            int width = size * (size + 1) / 2 + 1;
            var canvas = new int[width, height];
            for (; size > 0; size--)
                drawboxes(size, canvas);

            for (int y = height - 1; y >= 0; y--)
            {
                for (int x = 0; x < width; x++)
                    Console.Write((char)(32 + canvas[x, y]));
                Console.WriteLine();
            }
        }

        static void drawboxes(int size, int[,] canvas)
        {
            int x = size * (size - 1) / 2;
            for (int i = size - 1; i >= 0; i--)
            {
                drawbox(x, i * size, size, canvas);
            }
        }

        static void drawbox(int x, int y, int size, int[,] canvas)
        {
            for (int i = 0; i < size; i++)
            {
                canvas[x + i, y] = 13; // +32 = '-'
                canvas[x + i, y + size] = 13;
                canvas[x, y + i] = 92; // +32 = '|'
                canvas[x + size, y + i] = 92;
            }
            canvas[x, y] = 11; // +32 = '+'
            canvas[x + size, y] = 11;
            canvas[x + size, y + size] = 11;
            canvas[x, y + size] = 11;
        }
    }
}

Answer 6

Ruby (205 bytes)

Takes the number as a command line arguments. It starts with a fail leading newlines, but that's allowed.

n=$*[0].to_i
m=n+1
f=m.times.inject(:+)+1
c=((" "*f+p=?+)*n*m).split p
y=0
1.upto(n){|b|(b*b+1).times{|x|d=x%b==0;r=c[x]
d&&b.times{|g|r[y+g]=?-}
r[y]=d||r[y]==p ?p:?|
r[y+b]=d ?p:?|}
y+=b}
puts c.reverse

Answer 7

JavaScript (ES6), 293 bytes

(n,o='',r,f,t,u,b,c,e=n*n+1,i)=>{for(t=0;e>t;t++){for(c=b=0,d=e-t,u=0;(n*n+n)/2+1>u;u++)i=" ",r=b*b+1,b==c?(f=b+1,d<=r&d%(0==b?1:b)==(1==b?0:1)|n>=f&d<=f*f+1&d%(0==f?1:f)==(1==f?0:1)?i="+":d<=r|n>=f&d<=f*f+1&&(i="|"),b++,c=1):(d<=r&d%(0==b?1:b)==(1==b?0:1)&&(i="-"),c++),o+=i;o+="\n"}return o}

I ran this in Firefox. Ignore the " the console adds between the strings. This is mostly ES5 stuff but I'll try to golf this more.

Ungolfed / ES5

function box(n, o, r, f, t, u, b, c, e, i) {
  if (o === undefined) o = "";
  if (e === undefined) e = n * n + 1;
  return (function() {
    for (t = 0; e > t; t++) {
      for (c = b = 0, d = e - t, u = 0;
        (n * n + n) / 2 + 1 > u; u++) i = " ", r = b * b + 1, b == c ? (f = b + 1, d <= r & d % (0 == b ? 1 : b) == (1 == b ? 0 : 1) | n >= f & d <= f * f + 1 & d % (0 == f ? 1 : f) == (1 == f ? 0 : 1) ? i = "+" : d <= r | n >= f & d <= f * f + 1 && (i = "|"), b++, c = 1) : (d <= r & d % (0 == b ? 1 : b) == (1 == b ? 0 : 1) && (i = "-"), c++), o += i;
      o += "\n";
    }
    return o;
  })();
}

document.getElementById('g').onclick = function(){ document.getElementById('o').innerHTML = box(+document.getElementById('v').value) };

<input id="v"><button id="g">Run</button><pre id="o"></pre>

Answer 8

Python 2, 294 290

I got it working, but I still need to golf it more. I'm so happy, though, that was tough (for me, at least)!

I'll probably add an explanation later, unless it's immediately clear to someone...? I kind of doubt it.

Try it here

n=input()
w=n*n+n+2>>1
a=eval(`[[' ']*w]*-~n**2`)
r=range
j=[i*i+i>>1for i in r(n+1)]
p=0
for i in r(w):
 if i in j:
    p+=p<n
    for k in r(p*p+1):a[~k][i]='+'if k%p<1or' '<a[~k][i-1]<'.'else'|'
 else:
    for k in r(p*p+1):a[~k][i]=' 'if k%p else'-'
print'\n'.join(''.join(i)for i in a)

Answer 9

Python - 243 bytes

Generates all the columns, replacing the overlaps on columns except the first one. Then it pads with spaces, transposes, and prints.

Q=input()
Y=[]
for i in range(Q):
    f="+"+i*"-"+"+";x=map(list,zip(*([f]+["|"+" "*i+"|"]*i)*(i+1)+[f]))
    if i:y=Y.pop();x[0][-len(y):]=y
    Y+=x
print"\n".join("".join(i)for i in zip(*["".join(j[::-1]).ljust(Q*Q+1," ")for j in Y])[::-1])

I am considering translating to Pyth, but I will need a replacement for the pad function.