Code Golf: Triforce

Question

This is inspired by/taken from this thread: http://www.allegro.cc/forums/thread/603383 The Problem Assume the user gives you a numeric input ranging from 1 to 7. Input should be taken from the console, arguments are less desirable. When the input is 1, print the following: ``` *********** ********* ******* ***** *** * ``` Values greater than one should generate multiples of the pattern, ending with the one above, but stacked symmetrically. For example, 3 should print the following: ``` *********** *********** *********** ********* ********* ********* ******* ******* ******* ***** ***** ***** *** *** *** * * * *********** *********** ********* ********* ******* ******* ***** ***** *** *** * * *********** ********* ******* ***** *** * ``` Bonus points if you print the reverse as well. ``` *********** *********** ********* ********* ******* ******* ***** ***** *** *** * * *********** ********* ******* ***** *** * * *** ***** ******* ********* *********** * * *** *** ***** ***** ******* *…

Accepted Answer

Assembler, 165 bytes assembled

Build Instructions

- Download A86 from here

- Add a reference to the A86 executable into your DOS search path

- Paste the code below into a text file (example: `triforce.asm`)

- Invoke the assembler: `a86 triforce.asm`

- This will create a .COM file called `triforce.com`

- Type `triforce` to run

This was developed using the standard WinXP DOS box (Start->Programs->Accessories->Command Prompt). It should work with other DOS emulators.

Assemble using A86 and requires WinXP DOS box to run the .COM file it produces. Press 'q' to exit, keys 1-7 to draw the output.

```
  l20:mov ah,7
      int 21h
      cmp al,'q'
      je ret
      sub al,'0'
      cmp al,1
      jb l20
      cmp al,7
      ja l20
      mov [l0-1],al
      mov byte ptr [l7+2],6
      jmp $+2
      mov ah,2
      mov ch,0
      mov bh,3
   l0:mov bl,1
   l1:mov dh,0
   l3:cmp dh,ch
      je l2
      mov dl,32
      int 21h
      inc dh
      jmp l3
      ret
   l2:mov dh,bh
   l6:mov cl,12
   l5:mov dl,42
      cmp cl,bl
      ja l4
      mov dl,32
      cmp dh,1
      je l21
   l4:int 21h
      dec cl
      jnz l5
  l21:dec dh
      jnz l6
      mov dl,10
      int 21h
      mov dl,13
      int 21h
  l10:inc ch
   l9:add bl,2
   l7:cmp ch,6
      jne l1
  l13:add byte ptr [l7+2],6
  l11:dec bh
  l12:cmp bh,0
      jne l0
      xor byte ptr [l0+1],10
      xor byte ptr [l9+1],40
      xor byte ptr [l10+1],8
      xor byte ptr [l13+1],40
      sub byte ptr [l7+2],12
      mov dh,[l0-1]
      inc dh
      xor [l12+2],dh
      xor byte ptr [l11+1],8
      xor byte ptr [l1+1],1
      inc bh
      cmp byte ptr [l0+1],11
      je l0
      jmp l20
```

It uses lots of self-modifying code to do the triforce and its mirror, it even modifies the self-modifying code.

Code Golf: Triforce

Solution

Problem