Decompiling 8051 binary, read from EEPROM - binary

I'm trying to decompile the firmware of a Logitech Freedom 2.4 Cordless Joystick. I've managed to get something of the EEPROM. (here)
The EEPROM that is used is the Microchip 25AA320, which is a 32Kbit SPI-EEPROM. The MCU is a nRF24E1G , that contains a 8051 MCU.
The ROM should be 4096 bytes, so I think that my reading program looped over it self 4 times.
I managed to extract a 4kB ROM (here), but the start of the file doesn't look clean.
I loaded both files into IDA Pro and Ghidra and selected the 8051 processor. They don't generate anything useful.
Could anyone help me decompiling this ROM?
I used this Arduino Sketch to dump the rom.
Together with this python script
## Author: Arpan Das
## Date: Fri Jan 11 12:16:59 2019 +0530
## URL: https://github.com/Cyberster/SPI-Based-EEPROM-Reader-Writer
## It listens to serial port and writes contents into a file
## requires pySerial to be installed
import sys
import serial
import time
start = time.time()
MEMORY_SIZE = 4096 # In bytes
serial_port = 'COM5'
baud_rate = 115200 # In arduino, Serial.begin(baud_rate)
write_to_file_path = "dump.rom"
output_file = open(write_to_file_path, "wb")
ser = serial.Serial(serial_port, baud_rate)
print("Press d for dump ROM else CTRL+C to exit.")
ch = sys.stdin.read(1)
if ch == 'd':
ser.write('d')
for i in range(MEMORY_SIZE/32): # i.e. MEMORY_SIZE / 32
# wait until arduino response with 'W' i.e. 1 byte of data write request
while (ser.read() != 'W'): continue
ser.write('G') # sends back write request granted signal
for j in range(32):
byte = ser.read(1);
output_file.write(byte);
print(str(MEMORY_SIZE - (i * 32)) + " bytes remaining.")
print '\nIt took', time.time()-start, ' seconds.'

This is what I did, the next part left is for you. My machine is a Win10 notebook, however I used unix tools because they are so capable.
First of all, I divided the 16KB dump into four 4KB parts. The first one was different from the other three. And the provided 4KB dump is different to all of these parts. I did not investigate this further, and simply took one of the other three parts that are all equal.
$ split -b 4K LogitechFreedom2.4CordlessJoystick.rom part
$ cmp partaa partab
partaa partab differ: byte 1, line 1
$ cmp partab partac
$ cmp partac partad
$ cmp dump.rom partaa
dump.rom partaa differ: byte 9, line 1
$ cmp dump.rom partab
dump.rom partab differ: byte 1, line 1
From the microcontroller's data sheet I learned that the EEPROM contents has a header of at least 3 bytes (chapter 10.2 at page 61).
These bytes are:
0b Version = 00, Reserved = 00, SPEED = 0.5MHz, XO_FREQ = 16MHz
03 Offset to start of user program = 3
0f Number of 256 bytes block = 15
The last entry seems to be off by one, because there seems to be code in the 16th block, too.
Anyway, these bytes look decent, so I cut the first 3 bytes.
$ dd if=partad of=rom.bin bs=1 skip=3
4093+0 records in
4093+0 records out
4093 bytes (4,1 kB, 4,0 KiB) copied, 0,0270132 s, 152 kB/s
$ dd if=partad of=head.bin bs=1 count=3
3+0 records in
3+0 records out
3 bytes copied, 0,0043809 s, 0,7 kB/s
$ od -Ax -t x1 rom.bin > rom.hex
$ od -Ax -t x1 head.bin > head.hex
The hex files are nice for loading them into an editor and look around.
I loaded the remaining 4093 bytes into a disassembler I once wrote and peeked around a bit. It looks promising, so I think you can go on without me now:
C0000: ljmp C0F54
C0003: setb 021H.2
reti
C000B: xch a,r5
inc r6
xrl a,r6
mov a,#0B2H
movc a,#a+pc
movx #r1,a
mov r7,a
setb 021H.2
reti
C0F54: mov psw,#000H
mov sp,#07BH
mov r0,#0FFH
mov #r0,#000H
djnz r0,C0F5C
ljmp C0C09

Related

OS dev and where the boot sector is in qemu

I found this article about writing a simple OS:
https://www.cs.bham.ac.uk/~exr/lectures/opsys/10_11/lectures/os-dev.pdf
In chapter 3 part 4 there it says that an "X" will be written to the boot sector at: 0x7c00 but, in qemu if you dump the memory you can see its not written to 0x7c00, its written to, 0x7fc8.
Why is this?
What bytes does qemu use for the boot sector?
https://gyazo.com/6dcfaeffea19dfbc6edffa22a1bf0c83
This is the code in the os-dev pdf:
mov ah , 0 x0e
mov bx , the_secret
add bx , 0 x7c00
mov al , [bx]
int 0 x10`
jmp $ ; Jump forever.
the_secret :
db "X "
; Padding and magic BIOS number.
times 510 -( $ - $$ ) db 0
dw 0 xaa55

Variable initialization in as8088

I'm currently writing a function that should basically just write characters from a string into variables.
When performing test prints my variables seem fine. But when I attempt to print the first variable assigned(inchar) outside of the function it returns a empty string, but the second variable (outchar) seems to return fine. Am I somehow overwriting the first variable?
This is my code:
_EXIT = 1
_READ = 3
_WRITE = 4
_STDOUT = 1
_STDIN = 1
_GETCHAR = 117
MAXBUFF = 100
.SECT .TEXT
start:
0: PUSH endpro2-prompt2
PUSH prompt2
PUSH _STDOUT
PUSH _WRITE
SYS
ADD SP,8
PUSH 4
PUSH buff
CALL getline
ADD SP,4
!!!!!!!!!
PUSH buff
CALL gettrans
ADD SP,4
ADD AX,1 !gives AX an intial value to start loop
1: CMP AX,0
JE 2f
PUSH endpro-prompt1
PUSH prompt1
PUSH _STDOUT
PUSH _WRITE
SYS
ADD SP,8
PUSH MAXBUFF
PUSH buff
CALL getline
ADD SP,2
!PUSH buff
!CALL translate
!ADD SP,4
JMP 1b
2: PUSH 0 ! exit with normal exit status
PUSH _EXIT
SYS
getline:
PUSH BX
PUSH CX
PUSH BP
MOV BP,SP
MOV BX,8(BP)
MOV CX,8(BP)
ADD CX,10(BP)
SUB CX,1
1: CMP CX,BX
JE 2f
PUSH _GETCHAR
SYS
ADD SP,2
CMPB AL,-1
JE 2f
MOVB (BX),AL
INC BX
CMPB AL,'\n'
JNE 1b
2: MOVB (BX),0
MOV AX, BX
SUB AX,8(BP)
POP BP
POP CX
POP BX
RET
gettrans:
PUSH BX
PUSH BP
MOV BP,SP
MOV BX,6(BP) !Store argument in BX
MOVB (inchar),BL ! move first char to inchar
1: INC BX
CMPB (BX),' '
JE 1b
MOVB (outchar),BL !Move char seperated by Space to outchar
MOV AX,1 !On success
POP BP
POP BX
RET
.SECT .BSS
buff:
.SPACE MAXBUFF
.SECT .DATA
prompt1:
.ASCII "Enter a line of text: "
endpro:
prompt2:
.ASCII "Enter 2 characters for translation: "
endpro2:
outchar:
.BYTE 0
inchar:
.BYTE 0
charct:
.BYTE 0
wordct:
.BYTE 0
linect:
.BYTE 0
inword:
.BYTE 0
This is the code used to test print
PUSH 1 ! print that byte
PUSH inchar
PUSH _STDOUT
PUSH _WRITE
SYS
ADD SP,8
CALL printnl !function that prints new line
PUSH 1 ! print that byte
PUSH outchar
PUSH _STDOUT
PUSH _WRITE
SYS
CALL printnl
ADD SP,8
There seem to be a number of as88 8088 simulator environments. But I noticed on many of the repositories of code this bug mentioned:
1. The assembler requires sections to be defined in the following order:
TEXT
DATA
BSS
After the first occurrences, remaining section directives may appear in any order.
I'd recommend in your code to move the BSS section after DATA in the event your as88 environment has a similar problem.
In your original code you had lines like this:
MOV (outchar),BX
[snip]
MOV (inchar),BX
You defined outchar and inchar as bytes. The 2 lines above move 2 bytes (16-bits) from the BX register to both one byte variables. This will cause the CPU to write the extra byte into the next variable in memory. You'd want to explicitly move a single byte. Something like this might have been more appropriate:
MOVB (outchar),BL
[snip]
MOVB (inchar),BL
As you will see this code still has a bug as I mention later in this answer. To clarify - the MOVB instruction will move a single byte from BL and place it into the variable.
When you do a SYS call for Write you need to pass the address of the buffer to print, not the data in the buffer. You had 2 lines like this:
PUSH (inchar)
[snip]
PUSH (outchar)
The parentheses say to take the value in the variables and push them on the stack. SYS WRITE requires the address of the characters to display. The code to push their addresses should look like:
PUSH inchar
[snip]
PUSH outchar
gettrans function has a serious flaw in handling the copy of a byte from one buffer to another. You have code that does this:
MOV BX,6(BP) !Store argument in BX
MOVB (inchar),BL ! move first char to inchar
1: INC BX
CMPB (BX),' '
JE 1b
MOVB (outchar),BL !Move char seperated by Space to outchar
MOV BX,6(BP) properly places that buffer address passed as an argument and puts it into BX. There appears to be a problem with the lines that look like:
MOVB (inchar),BL ! move first char to inchar
This isn't doing what the comment suggests it should. The line above moves the lower byte (BL) of the buffer address in BX to the variable inchar . You want to move the byte at the memory location pointed to by BX and put it into inchar. Unfortunately on the x86 you can't move the data from one memory operand to another directly. To get around this you will have to move the data from the buffer pointed to by BX into a temporary register (I'll choose CL) and then move that to the variable. The code could look like this:
MOVB CL, (BX)
MOVB (inchar),CL ! move first char to inchar
You then have to do the same for outchar so the fix in both places could look similar to this:
MOV BX,8(BP) !Store argument in BX
MOVB CL, (BX)
MOVB (inchar),CL ! move first char to inchar
1: INC BX
CMPB (BX),' '
JE 1b
MOVB CL, (BX)
MOVB (outchar),CL ! move second char to outchar
The instruction MOV (inchar),BX stores register BX to the memory location labelled inchar.
However, inchar has been defined as a .BYTE, but BX is a 16-bit register, (2 bytes,) so you are writing not only inchar but also outchar.
The only reason why it appears to work in the beginning is because the 8088 is a low-endian architecture, so the low-order byte of BX is being stored first, while the high-order byte follows.
So, try MOV (inchar),BL

NVVM_ERROR_INVALID_OPTION when using the CUDA kernel with Numbapro api

I want to execute a CUDA kernel in python using Numbapro API. I have this code:
import math
import numpy
from numbapro import jit, cuda, int32, float32
from matplotlib import pyplot
#cuda.jit('void(float32[:], float32[:], float32[:], float32[:], float32, float32, float32, int32)')
def calculate_velocity_field(X, Y, u_source, v_source, x_source, y_source, strength_source, N):
start = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
end = N
stride = cuda.gridDim.x * cuda.blockDim.x
for i in range(start, end, stride):
u_source[i] = strength_source/(2*math.pi) * (X[i]-x_source)/((X[i]-x_source)**2 + (Y[i]-y_source)**2)
v_source[i] = strength_source/(2*math.pi) * (Y[i]-x_source)/((X[i]-x_source)**2 + (Y[i]-y_source)**2)
def main():
N = 200 # number of points in each direction
x_start, x_end = -4.0, 4.0 # boundaries in the x-direction
y_start, y_end = -2.0, 2.0 # boundaries in the y-direction
x = numpy.linspace(x_start, x_end, N) # creates a 1D-array with the x-coordinates
y = numpy.linspace(y_start, y_end, N) # creates a 1D-array with the y-coordinates
X, Y = numpy.meshgrid(x, y) # generates a mesh grid
strength_source = 5.0 # source strength
x_source, y_source = -1.0, 0.0 # location of the source
start = timer()
#calculate grid dimensions
blockSize = 1024
gridSize = int(math.ceil(float(N)/blockSize))
#transfer memory to device
X_d = cuda.to_device(X)
Y_d = cuda.to_device(Y)
u_source_d = cuda.device_array_like(X)
v_source_d = cuda.device_array_like(Y)
#launch kernel
calculate_velocity_field[gridSize,blockSize](X_d,Y_d,u_source_d,v_source_d,x_source,y_source,strength_source,N)
#transfer memory to host
u_source = numpy.empty_like(X)
v_source = numpy.empty_like(Y)
u_source_d.to_host(u_source)
v_source_d.to_host(v_source)
elapsed_time = timer() - start
print("Exec time with GPU %f s" % elapsed_time)
if __name__ == "__main__":
main()
Is giving me this error:
NvvmError Traceback (most recent call last)
<ipython-input-17-85e4a6e56a14> in <module>()
----> 1 #cuda.jit('void(float32[:], float32[:], float32[:], float32[:], float32, float32, float32, int32)')
2 def calculate_velocity_field(X, Y, u_source, v_source, x_source, y_source, strength_source, N):
3 start = cuda.blockIdx.x * cuda.blockDim.x + cuda.threadIdx.x
4 end = N
5 stride = cuda.gridDim.x * cuda.blockDim.x
~/.anaconda3/lib/python3.4/site-packages/numba/cuda/decorators.py in kernel_jit(func)
89 # Force compilation for the current context
90 if bind:
---> 91 kernel.bind()
92
93 return kernel
~/.anaconda3/lib/python3.4/site-packages/numba/cuda/compiler.py in bind(self)
319 Force binding to current CUDA context
320 """
--> 321 self._func.get()
322
323 #property
~/.anaconda3/lib/python3.4/site-packages/numba/cuda/compiler.py in get(self)
254 cufunc = self.cache.get(device.id)
255 if cufunc is None:
--> 256 ptx = self.ptx.get()
257
258 # Link
~/.anaconda3/lib/python3.4/site-packages/numba/cuda/compiler.py in get(self)
226 arch = nvvm.get_arch_option(*cc)
227 ptx = nvvm.llvm_to_ptx(self.llvmir, opt=3, arch=arch,
--> 228 **self._extra_options)
229 self.cache[cc] = ptx
230 if config.DUMP_ASSEMBLY:
~/.anaconda3/lib/python3.4/site-packages/numba/cuda/cudadrv/nvvm.py in llvm_to_ptx(llvmir, **opts)
420 cu.add_module(llvmir.encode('utf8'))
421 cu.add_module(libdevice.get())
--> 422 ptx = cu.compile(**opts)
423 return ptx
424
~/.anaconda3/lib/python3.4/site-packages/numba/cuda/cudadrv/nvvm.py in compile(self, **options)
211 for x in opts])
212 err = self.driver.nvvmCompileProgram(self._handle, len(opts), c_opts)
--> 213 self._try_error(err, 'Failed to compile\n')
214
215 # get result
~/.anaconda3/lib/python3.4/site-packages/numba/cuda/cudadrv/nvvm.py in _try_error(self, err, msg)
229
230 def _try_error(self, err, msg):
--> 231 self.driver.check_error(err, "%s\n%s" % (msg, self.get_log()))
232
233 def get_log(self):
~/.anaconda3/lib/python3.4/site-packages/numba/cuda/cudadrv/nvvm.py in check_error(self, error, msg, exit)
118 sys.exit(1)
119 else:
--> 120 raise exc
121
122
NvvmError: Failed to compile
libnvvm : error: -arch=compute_52 is an unsupported option
NVVM_ERROR_INVALID_OPTION
I tried another numbapro examples and the same error ocurrs.
I don't know if it's a bug of Numbapro that doesn't support 5.2 compute capability or it's a problem of Nvidia NVVM... suggestions?
In theory it should be supported, but I don't know what is happening.
I'm using Linux with CUDA 7.0 and driver version 346.29
Finally I found a solution here
Solution 1:
conda update cudatoolkit
Fetching package metadata: ....
# All requested packages already installed.
# packages in environment at ~/.anaconda3:
#
cudatoolkit 6.0 p0
It looks like me updating the CUDA toolkit doesn't update to CUDA 7.0. A second solution can be done:
Solution 2
conda install -c numba cudatoolkit
Fetching package metadata: ......
Solving package specifications: .
Package plan for installation in environment ~/.anaconda3:
The following packages will be downloaded:
package | build
---------------------------|-----------------
cudatoolkit-7.0 | 1 190.8 MB
The following packages will be UPDATED:
cudatoolkit: 6.0-p0 --> 7.0-1
Proceed ([y]/n)? y
Before:
In [4]: check_cuda()
------------------------------libraries detection-------------------------------
Finding cublas
located at ~/.anaconda3/lib/libcublas.so.6.0.37
trying to open library... ok
Finding cusparse
located at ~/.anaconda3/lib/libcusparse.so.6.0.37
trying to open library... ok
Finding cufft
located at ~/.anaconda3/lib/libcufft.so.6.0.37
trying to open library... ok
Finding curand
located at ~/.anaconda3/lib/libcurand.so.6.0.37
trying to open library... ok
Finding nvvm
located at ~/.anaconda3/lib/libnvvm.so.2.0.0
trying to open library... ok
finding libdevice for compute_20... ok
finding libdevice for compute_30... ok
finding libdevice for compute_35... ok
-------------------------------hardware detection-------------------------------
Found 1 CUDA devices
id 0 b'GeForce GTX 970' [SUPPORTED]
compute capability: 5.2
pci device id: 0
pci bus id: 7
Summary:
1/1 devices are supported
PASSED
Out[4]: True
After:
In [6]: check_cuda()
------------------------------libraries detection-------------------------------
Finding cublas
located at ~/.anaconda3/lib/libcublas.so.7.0.28
trying to open library... ok
Finding cusparse
located at ~/.anaconda3/lib/libcusparse.so.7.0.28
trying to open library... ok
Finding cufft
located at ~/.anaconda3/lib/libcufft.so.7.0.35
trying to open library... ok
Finding curand
located at ~/.anaconda3/lib/libcurand.so.7.0.28
trying to open library... ok
Finding nvvm
located at ~/.anaconda3/lib/libnvvm.so.3.0.0
trying to open library... ok
finding libdevice for compute_20... ok
finding libdevice for compute_30... ok
finding libdevice for compute_35... ok
-------------------------------hardware detection-------------------------------
Found 1 CUDA devices
id 0 b'GeForce GTX 970' [SUPPORTED]
compute capability: 5.2
pci device id: 0
pci bus id: 7
Summary:
1/1 devices are supported
PASSED
Out[6]: True

Code Golf: Conway's Game of Life

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
The Challenge: Write the shortest program that implements John H. Conway's Game of Life cellular automaton. [link]
EDIT: After about a week of competition, I have selected a victor: pdehaan, for managing to beat the Matlab solution by one character with perl.
For those who haven't heard of Game of Life, you take a grid (ideally infinite) of square cells. Cells can be alive (filled) or dead (empty). We determine which cells are alive in the next step of time by applying the following rules:
Any live cell with fewer than two live neighbours dies, as if caused by under-population.
Any live cell with more than three live neighbours dies, as if by overcrowding.
Any live cell with two or three live neighbours lives on to the next generation.
Any dead cell with exactly three live neighbours becomes a live cell, as if by reproduction.
Your program will read in a 40x80 character ASCII text file specified as a command-line argument, as well as the number of iterations (N) to perform. Finally, it will output to an ASCII file out.txt the state of the system after N iterations.
Here is an example run with relevant files:
in.txt:
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
..................................XX............................................
..................................X.............................................
.......................................X........................................
................................XXXXXX.X........................................
................................X...............................................
.................................XX.XX...XX.....................................
..................................X.X....X.X....................................
..................................X.X......X....................................
...................................X.......XX...................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
Iterate 100 times:
Q:\>life in.txt 100
Resultant Output (out.txt)
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
..................................XX............................................
..................................X.X...........................................
....................................X...........................................
................................XXXXX.XX........................................
................................X.....X.........................................
.................................XX.XX...XX.....................................
..................................X.X....X.X....................................
..................................X.X......X....................................
...................................X.......XX...................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
................................................................................
The Rules:
You need to use file I/O to read/write the files.
You need to accept an input file and the number of iterations as arguments
You need to generate out.txt (overwrite if it exists) in the specified format
You don't need to deal with the edges of the board (wraparound, infinite grids .etc)
EDIT: You do need to have newlines in your output file.
The winner will be determined by character count.
Good luck!
Mathematica - 179 163 154 151 chars
a = {2, 2, 2};
s = Export["out.txt",
CellularAutomaton[{224, {2, {a, {2, 1, 2}, a}}, {1,1}},
(ReadList[#1, Byte, RecordLists → 2>1] - 46)/ 42, #2]〚#2〛
/. {0 → ".", 1 → "X"}, "Table"] &
Spaces added for readability
Invoke with
s["c:\life.txt", 100]
Animation:
You can also get a graph of the mean population over time:
A nice pattern for generating gliders from Wikipedia
AFAIK Mathematica uses a Cellular Automaton to generate random numbers using Rule 30.
MATLAB 7.8.0 (R2009a) - 174 171 161 150 138 131 128 124 characters
Function syntax: (124 characters)
Here's the easier-to-read version (with unnecessary newlines and whitespace added for better formatting):
function l(f,N),
b=char(importdata(f))>46;
for c=1:N,
b=~fix(filter2(ones(3),b)-b/2-3);
end;
dlmwrite('out.txt',char(b*42+46),'')
And here's how the program is run from the MATLAB Command Window:
l('in.txt',100)
Command syntax: (130 characters)
After a comment about calling functions with a command syntax, I dug a little deeper and found out that MATLAB functions can in fact be invoked with a command-line format (with some restrictions). You learn something new every day!
function l(f,N),
b=char(importdata(f))>46;
for c=1:eval(N),
b=~fix(filter2(ones(3),b)-b/2-3);
end;
dlmwrite('out.txt',char(b*42+46),'')
And here's how the program is run from the MATLAB Command Window:
l in.txt 100
Additional Challenge: Tweetable GIF maker - 136 characters
I thought for fun I'd see if I could dump the output to a GIF file instead of a text file, while still keeping the character count below 140 (i.e. "tweetable"). Here's the nicely-formatted code:
function l(f,N),
b=char(importdata(f))>46;
k=ones(3);
for c=1:N+1,
a(:,:,:,c)=kron(b,k);
b=~fix(filter2(k,b)-b/2-3);
end;
imwrite(~a,'out.gif')
Although IMWRITE is supposed to create a GIF that loops infinitely by default, my GIF is only looping once. Perhaps this is a bug that has been fixed in newer versions of MATLAB. So, to make the animation last longer and make the evolution steps easier to see, I left the frame delay at the default value (which seems to be around half a second). Here's the GIF output using the Gosper Glider Gun pattern:
Improvements
Update 1: Changed the matrix b from a logical (i.e. "boolean") type to a numerical one to get rid of a few conversions.
Update 2: Shortened the code for loading the file and used the function MAGIC as a trick to create the convolution kernel in fewer characters.
Update 3: Simplified the indexing logic, replaced ~~b+0 with b/42, and replaced 'same' with 's' as an argument to CONV2 (and it surprisingly still worked!).
Update 4: I guess I should have searched online first, since Loren from The MathWorks blogged about golfing and the Game of Life earlier this year. I incorporated some of the techniques discussed there, which required me to change b back to a logical matrix.
Update 5: A comment from Aslak Grinsted on the above mentioned blog post suggests an even shorter algorithm for both the logic and performing the convolution (using the function FILTER2), so I "incorporated" (read "copied") his suggestions. ;)
Update 6: Trimmed two characters from the initialization of b and reworked the logic in the loop to save 1 additional character.
Update 7: Eric Sampson pointed out in an e-mail that I could replace cell2mat with char, saving 4 characters. Thanks Eric!
Ruby 1.9 - 189 178 159 155 153 chars
f,n=$*
c=IO.read f
n.to_i.times{i=0;c=c.chars.map{|v|i+=1
v<?.?v:('...X'+v)[[83,2,-79].map{|j|c[i-j,3]}.to_s.count ?X]||?.}*''}
File.new('out.txt',?w)<<c
Edit:
Handles newlines with 4 chars less.
Can remove 7 more (v<?.?v:) if you allow it to clobber newlines when the live cells reach the edges.
perl, 127 129 135 chars
Managed to strip off a couple more characters...
$/=pop;#b=split'',<>;map{$n=-1;#b=map{++$n;/
/?$_:($t=grep/X/,#b[map{$n+$_,$n-$_}1,80..82])==3|$t+/X/==3?X:'.'}#b}1..$/;print#b
Python - 282 chars
might as well get the ball rolling...
import sys
_,I,N=sys.argv;R=range(3e3);B=open(I).read();B=set(k for k in R if'A'<B[k])
for k in R*int(N):
if k<1:b,B=B,set()
c=sum(len(set((k+o,k-o))&b)for o in(1,80,81,82))
if(c==3)+(c==2)*(k in b):B.add(k)
open('out.txt','w').write(''.join('.X\n'[(k in B)-(k%81<1)]for k in R))
Python 2.x - 210/234 characters
Okay, the 210-character code is kind of cheating.
#coding:l1
exec'xÚ=ŽA\nÂ#E÷sŠº1­ƒÆscS‰ØL™Æª··­âî¿GÈÿÜ´1iÖ½;Sçu.~H®J×Þ-‰­Ñ%ª.wê,šÖ§J®d꘲>cÉZË¢V䀻Eîa¿,vKAËÀå̃<»Gce‚ÿ‡ábUt¹)G%£êŠ…óbÒüíÚ¯GÔ/n×Xši&ć:})äðtÏÄJÎòDˆÐÿG¶'.decode('zip')
You probably won't be able to copy and paste this code and get it to work. It's supposed to be Latin-1 (ISO-8859-1), but I think it got perverted into Windows-1252 somewhere along the way. Additionally, your browser may swallow some of the non-ASCII characters.
So if it doesn't work, you can generate the file from plain-old 7-bit characters:
s = """
23 63 6F 64 69 6E 67 3A 6C 31 0A 65 78 65 63 27 78 DA 3D 8E 41 5C 6E C2
40 0C 45 F7 73 8A BA 31 13 AD 83 15 11 11 C6 73 08 63 17 05 53 89 D8 4C
99 C6 AA B7 B7 AD E2 EE BF 47 C8 FF DC B4 31 69 D6 BD 3B 53 E7 75 2E 7E
48 AE 4A D7 DE 90 8F 2D 89 AD D1 25 AA 2E 77 16 EA 2C 9A D6 A7 4A AE 64
EA 98 B2 3E 63 C9 5A CB A2 56 10 0F E4 03 80 BB 45 16 0B EE 04 61 BF 2C
76 0B 4B 41 CB C0 E5 CC 83 03 3C 1E BB 47 63 65 82 FF 87 E1 62 55 1C 74
B9 29 47 25 A3 EA 03 0F 8A 07 85 F3 62 D2 FC ED DA AF 11 47 D4 2F 6E D7
58 9A 69 26 C4 87 3A 7D 29 E4 F0 04 74 CF C4 4A 16 CE F2 1B 44 88 1F D0
FF 47 B6 27 2E 64 65 63 6F 64 65 28 27 7A 69 70 27 29
"""
with open('life.py', 'wb') as f:
f.write(''.join(chr(int(i, 16)) for i in s.split()))
The result of this is a valid 210-character Python source file. All I've done here is used zip compression on the original Python source code. The real cheat is that I'm using non-ASCII characters in the resultant string. It's still valid code, it's just cumbersome.
The noncompressed version weighs in at 234 characters, which is still respectable, I think.
import sys
f,f,n=sys.argv
e=open(f).readlines()
p=range
for v in p(int(n)):e=[''.join('.X'[8+16*(e[t][i]!='.')>>sum(n!='.'for v in e[t-1:t+2]for n in v[i-1:i+2])&1]for i in p(80))for t in p(40)]
open('out.txt','w').write('\n'.join(e))
Sorry about the horizontal scroll, but all newlines in the above are required, and I've counted them as one character each.
I wouldn't try to read the golfed code. The variable names are chosen randomly to achieve the best compression. Yes, I'm serious. A better-formatted and commented version follows:
# get command-line arguments: infile and count
import sys
ignored, infile, count = sys.argv
# read the input into a list (each input line is a string in the list)
data = open(infile).readlines()
# loop the number of times requested on the command line
for loop in range(int(count)):
# this monstrosity applies the rules for each iteration, replacing
# the cell data with the next generation
data = [''.join(
# choose the next generation's cell from '.' for
# dead, or 'X' for alive
'.X'[
# here, we build a simple bitmask that implements
# the generational rules. A bit from this integer
# will be chosen by the count of live cells in
# the 3x3 grid surrounding the current cell.
#
# if the current cell is dead, this bitmask will
# be 8 (0b0000001000). Since only bit 3 is set,
# the next-generation cell will only be alive if
# there are exactly 3 living neighbors in this
# generation.
#
# if the current cell is alive, the bitmask will
# be 24 (8 + 16, 0b0000011000). Since both bits
# 3 and 4 are set, this cell will survive if there
# are either 3 or 4 living cells in its neighborhood,
# including itself
8 + 16 * (data[y][x] != '.')
# shift the relevant bit into position
>>
# by the count of living cells in the 3x3 grid
sum(character != '.' # booleans will convert to 0 or 1
for row in data[y - 1 : y + 2]
for character in row[x - 1 : x + 2]
)
# select the relevant bit
& 1
]
# for each column and row
for x in range(80)
)
for y in range(40)
]
# write the results out
open('out.txt','w').write('\n'.join(data))
Sorry, Pythonistas, for the C-ish bracket formatting, but I was trying to make it clear what each bracket was closing.
Haskell - 284 272 232 chars
import System
main=do f:n:_<-getArgs;s<-readFile f;writeFile"out.txt"$t s$read n
p '\n'_='\n'
p 'X'2='X'
p _ 3='X'
p _ _='.'
t r 0=r
t r n=t[p(r!!m)$sum[1|d<-1:[80..82],s<-[1,-1],-m<=d*s,m+d*s<3240,'X'==r!!(m+d*s)]|m<-[0..3239]]$n-1
F#, 496
I could reduce this a lot, but I like this as it's still in the ballpark and pretty readable.
open System.IO
let mutable a:_[,]=null
let N y x=
[-1,-1;-1,0;-1,1;0,-1;0,1;1,-1;1,0;1,1]
|>Seq.sumBy(fun(i,j)->try if a.[y+i,x+j]='X' then 1 else 0 with _->0)
[<EntryPoint>]
let M(r)=
let b=File.ReadAllLines(r.[0])
a<-Array2D.init 40 80(fun y x->b.[y].[x])
for i=1 to int r.[1] do
a<-Array2D.init 40 80(fun y x->
match N y x with|3->'X'|2 when a.[y,x]='X'->'X'|_->'.')
File.WriteAllLines("out.txt",Array.init 40(fun y->
System.String(Array.init 80(fun x->a.[y,x]))))
0
EDIT
428
By request, here's my next stab:
open System
let mutable a,k=null,Array2D.init 40 80
[<EntryPoint>]
let M r=
a<-k(fun y x->IO.File.ReadAllLines(r.[0]).[y].[x])
for i=1 to int r.[1] do a<-k(fun y x->match Seq.sumBy(fun(i,j)->try if a.[y+i,x+j]='X'then 1 else 0 with _->0)[-1,-1;-1,0;-1,1;0,-1;0,1;1,-1;1,0;1,1]with|3->'X'|2 when a.[y,x]='X'->'X'|_->'.')
IO.File.WriteAllLines("out.txt",Array.init 40(fun y->String(Array.init 80(fun x->a.[y,x]))))
0
That's a 14% reduction with some basic golfing. I can't help but feel that I'm losing by using a 2D-array/array-of-strings rather than a 1D array, but don't feel like doing that transform now. Note how I elegantly read the file 3200 times to initialize my array :)
Ruby 1.8: 178 175 chars
f,n=$*;b=IO.read f
n.to_i.times{s=b.dup
s.size.times{|i|t=([82,1,-80].map{|o|b[i-o,3]||''}*'').count 'X'
s[i]=t==3||b[i]-t==?T??X:?.if s[i]>13};b=s}
File.new('out.txt','w')<<b
Newlines are significant (although all can be replaced w/ semicolons.)
Edit: fixed the newline issue, and trimmed 3 chars.
Java, 441... 346
Update 1 Removed inner if and more ugliness
Update 2 Fixed a bug and gained a character
Update 3 Using lots more memory and arrays while ignoring some boundaries issues. Probably a few chars could be saved.
Update 4 Saved a few chars. Thanks to BalusC.
Update 5 A few minor changes to go below 400 and make it just that extra bit uglier.
Update 6 Now things are so hardcoded may as well read in the exact amount in one go. Plus a few more savings.
Update 7 Chain the writing to the file to save a char. Plus a few odd bits.
Just playing around with BalusC's solution. Limited reputation means I couldnt add anything as a comment to his.
class M{public static void main(String[]a)throws Exception{int t=3240,j=t,i=new Integer(a[1])*t+t;char[]b=new char[i+t],p={1,80,81,82};for(new java.io.FileReader(a[0]).read(b,t,t);j<i;){char c=b[j],l=0;for(int n:p)l+=b[j+n]/88+b[j-n]/88;b[j+++t]=c>10?(l==3|l+c==90?88:'.'):c;}new java.io.FileWriter("out.txt").append(new String(b,j,t)).close();}}
More readable(?) version:
class M{
public static void main(String[]a)throws Exception{
int t=3240,j=t,i=new Integer(a[1])*t+t;
char[]b=new char[i+t],p={1,80,81,82};
for(new java.io.FileReader(a[0]).read(b,t,t);j<i;){
char c=b[j],l=0;
for(int n:p)l+=b[j+n]/88+b[j-n]/88;
b[j+++t]=c>10?(l==3|l+c==90?88:'.'):c;
}
new java.io.FileWriter("out.txt").append(new String(b,j,t)).close();
}
}
Scala - 467 364 339 chars
object G{def main(a:Array[String]){val l=io.Source.fromFile(new java.io.File(a(0)))getLines("\n")map(_.toSeq)toSeq
val f=new java.io.FileWriter("out.txt")
f.write((1 to a(1).toInt).foldLeft(l){(t,_)=>(for(y<-0 to 39)yield(for(x<-0 to 79)yield{if(x%79==0|y%39==0)'.'else{val m=t(y-1)
val p=t(y+1);val s=Seq(m(x-1),m(x),m(x+1),t(y)(x-1),t(y)(x+1),p(x-1),p(x),p(x+1)).count('X'==_)
if(s==3|(s==2&t(y)(x)=='X'))'X'else'.'}})toSeq)toSeq}map(_.mkString)mkString("\n"))
f.close}}
I think there is much room for improvement...
[Edit] Yes, it is:
object G{def main(a:Array[String]){var l=io.Source.fromFile(new java.io.File(a(0))).mkString
val f=new java.io.FileWriter("out.txt")
var i=a(1).toInt
while(i>0){l=l.zipWithIndex.map{case(c,n)=>if(c=='\n')'\n'else{val s=Seq(-83,-82,-81,-1,1,81,82,83).map(_+n).filter(k=>k>=0&k<l.size).count(l(_)=='X')
if(s==3|(s==2&c=='X'))'X'else'.'}}.mkString
i-=1}
f.write(l)
f.close}}
[Edit] And I have the feeling there is still more to squeeze out...
object G{def main(a:Array[String]){val f=new java.io.FileWriter("out.txt")
f.write(((1 to a(1).toInt):\(io.Source.fromFile(new java.io.File(a(0))).mkString)){(_,m)=>m.zipWithIndex.map{case(c,n)=>
val s=Seq(-83,-82,-81,-1,1,81,82,83)count(k=>k+n>=0&k+n<m.size&&m(k+n)=='X')
if(c=='\n')c else if(s==3|s==2&c=='X')'X'else'.'}.mkString})
f.close}}
The following solution uses my own custom domain-specific programming language which I have called NULL:
3499538
In case you are wondering how this works: My language consists of only of one statment per program. The statement represents a StackOverflow thread ID belonging to a code golf thread. My compiler compiles this into a program that looks for the best javascript solution (with the SO API), downloads it and runs it in a web browser.
Runtime could be better for new threads (it may take some time for the first upvoted Javascript answer to appear), but on the upside it requires only very little coding skills.
Javascript/Node.js - 233 236 characters
a=process.argv
f=require('fs')
m=46
t=f.readFileSync(a[2])
while(a[3]--)t=[].map.call(t,function(c,i){for(n=g=0;e=[-82,-81,-80,-1,1,80,81,82][g++];)t[i+e]>m&&n++
return c<m?c:c==m&&n==3||c>m&&n>1&&n<4?88:m})
f.writeFile('out.txt',t)
C - 300
Just wondered how much smaller and uglier my java solution could go in C. Reduces to 300 including the newlines for the preprocessor bits. Leaves freeing the memory to the OS! Could save ~20 by assuming the OS will close and flush the file too.
#include<stdio.h>
#include<stdlib.h>
#define A(N)j[-N]/88+j[N]/88
int main(int l,char**a){
int t=3240,i=atoi(a[2])*t+t;
char*b=malloc(i+t),*j;
FILE*f;
fread(j=b+t,1,t,fopen(a[1],"r"));
for(;j-b-i;j++[t]=*j>10?l==3|l+*j==90?88:46:10)
l=A(1)+A(80)+A(81)+A(82);
fwrite(j,1,t,f=fopen("out.txt","w"));
fclose(f);
}
MUMPS: 314 chars
L(F,N,R=40,C=80)
N (F,N,R,C)
O F:"RS" U F D C F
.F I=1:1:R R L F J=1:1:C S G(0,I,J)=($E(L,J)="X")
F A=0:1:N-1 F I=1:1:R F J=1:1:C D S G(A+1,I,J)=$S(X=2:G(A,I,J),X=3:1,1:0)
.S X=0 F i=-1:1:1 F j=-1:1:1 I i!j S X=X+$G(G(A,I+i,J+j))
S F="OUT.TXT" O F:"WNS" U F D C F
.F I=1:1:R F J=1:1:C W $S(G(N,I,J):"X",1:".") W:J=C !
Q
Java, 556 532 517 496 472 433 428 420 418 381 chars
Update 1: replaced 1st StringBuffer by Appendable and 2nd by char[]. Saved 24 chars.
Update 2: found a shorter way to read file into char[]. Saved 15 chars.
Update 3: replaced one if/else by ?: and merged char[] and int declarations. Saved 21 chars.
Update 4: replaced (int)f.length() and c.length by s. Saved 24 chars.
Update 5: made improvements as per hints of Molehill. Major one was hardcoding the char length so that I could get rid of File. Saved 39 chars.
Update 6: minor refactoring. Saved 6 chars.
Update 7: replaced Integer#valueOf() by new Integer() and refactored for loop. Saved 8 chars.
Update 8: Improved neighbour calculation. Saved 2 chars.
Update 9: Optimized file reading since file length is already hardcoded. Saved 37 chars.
import java.io.*;class L{public static void main(String[]a)throws Exception{int i=new Integer(a[1]),j,l,s=3240;int[]p={-82,-81,-80,-1,1,80,81,82};char[]o,c=new char[s];for(new FileReader(a[0]).read(c);i-->0;c=o)for(o=new char[j=s];j-->0;){l=0;for(int n:p)l+=n+j>-1&n+j<s?c[n+j]/88:0;o[j]=c[j]>13?l==3|l+c[j]==90?88:'.':10;}Writer w=new FileWriter("out.txt");w.write(c);w.close();}}
More readable version:
import java.io.*;
class L{
public static void main(String[]a)throws Exception{
int i=new Integer(a[1]),j,l,s=3240;
int[]p={-82,-81,-80,-1,1,80,81,82};
char[]o,c=new char[s];
for(new FileReader(a[0]).read(c);i-->0;c=o)for(o=new char[j=s];j-->0;){
l=0;for(int n:p)l+=n+j>-1&n+j<s?c[n+j]/88:0;
o[j]=c[j]>10?l==3|l+c[j]==90?88:'.':10;
}
Writer w=new FileWriter("out.txt");w.write(c);w.close();
}
}
Closing after writing is absoletely mandatory, else the file is left empty. It would otherwise have saved another 21 chars.
Further I could also save one more char when I use 46 instead of '.', but both javac and Eclipse jerks with a compilation error Possible loss of precision. Weird stuff.
Note: this expects an input file with \n newlines, not \r\n as Windows by default uses!
PHP - 365 328 322 Characters.
list(,$n,$l) = $_SERVER["argv"];
$f = file( $n );
for($j=0;$j<$l;$j++){
foreach($f as $k=>$v){
$a[$k]="";
for($i=0;$i < strlen( $v );$i++ ){
$t = 0;
for($m=-1;$m<2;$m++){
for($h=-1;$h<2;$h++){
$t+=ord($f[$k + $m][$i + $h]);
}
}
$t-=ord($v[$i]);
$a[$k] .= ( $t == 494 || ($t == 452 && ord($v[$i])==88)) ? "X" : "." ;
}
}
$f = $a;
}
file_put_contents("out.txt", implode("\n", $a ));
I'm sure this can be improved upon but I was curious what it would look like in PHP. Maybe this will inspire someone who has a little more code-golf experience.
Updated use list() instead of $var = $_SERVER["argv"] for both args. Nice one Don
Updated += and -= this one made me /facepalm heh cant believe i missed it
Updated file output to use file_put_contents() another good catch by Don
Updated removed initialization of vars $q and $w they were not being used
R 340 chars
cgc<-function(i="in.txt",x=100){
require(simecol)
z<-file("in.txt", "rb")
y<-matrix(data=NA,nrow=40,ncol=80)
for(i in seq(40)){
for(j in seq(80)){
y[i,j]<-ifelse(readChar(z,1) == "X",1,0)
}
readChar(z,3)
}
close(z)
init(conway) <- y
times(conway)<-1:x
o<-as.data.frame(out(sim(conway))[[100]])
write.table(o, "out.txt", sep="", row.names=FALSE, col.names=FALSE)
}
cgc()
I feel it's slightly cheating to have an add in package that does the actual automata for you, but I'm going with it cos I still had to thrash around with matricies and stuff to read in the file with 'X' instead of 1.
This is my first 'code golf', interesting....
c++ - 492 454 386
my first code golf ;)
#include<fstream>
#define B(i,j)(b[i][j]=='X')
int main(int i,char**v){for(int n=0;n<atoi(v[2]);++n){std::ifstream f(v[1]);v[1]="out.txt";char b[40][83];for(i=0;i<40;++i)f.getline(b[i],83);std::ofstream g("out.txt");g<<b[0]<<'\n';for(i=1;i<39;++i){g<<'.';for(int j=1;j<79;++j){int k=B(i-1,j)+B(i+1,j)+B(i,j-1)+B(i,j+1)+B(i-1,j-1)+B(i+1,j+1)+B(i+1,j-1)+B(i-1,j+1);(B(i,j)&&(k<2||k>3))?g<<'.':(!B(i,j)&&k==3)?g<<'X':g<<b[i][j];}g<<".\n";}g<<b[0]<<'\n';}}
A somewhat revised version, replacing some of the logic with a table lookup+a few other minor tricks:
#include<fstream>
#define B(x,y)(b[i+x][j+y]=='X')
int main(int i,char**v){for(int n=0;n<atoi(v[2]);++n){std::ifstream f(v[1]);*v="out.txt";char b[40][83], O[]="...X.....";for(i=0;i<40;++i)f>>b[i];std::ofstream g(*v);g<<b[0]<<'\n';for(i=1;i<39;++i){g<<'.';for(int j=1;j<79;++j){O[2]=b[i][j];g<<O[B(-1,0)+B(1,0)+B(0,-1)+B(0,1)+B(-1,-1)+B(1,1)+B(1,-1)+B(-1,1)];}g<<".\n";}g<<b[0]<<'\n';}}
Perl – 214 chars
What, no perl entries yet?
$i=pop;#c=<>;#c=map{$r=$_;$u='';for(0..79)
{$K=$_-1;$R=$r-1;$u.=((&N.(&N^"\0\W\0").&N)=~y/X//
|(substr$c[$r],$_,1)eq'X')==3?'X':'.';}$u}keys#c for(1..$i);
sub N{substr$c[$R++],$K,3}open P,'>','out.txt';$,=$/;print P#c
Run with: conway.pl infile #times
Another Java attempt, 361 chars
class L{public static void main(final String[]a)throws Exception{new java.io.RandomAccessFile("out.txt","rw"){{int e=88,p[]={-1,1,-80,80,-81,81,-82,82},s=3240,l=0,i=new Byte(a[1])*s+s,c;char[]b=new char[s];for(new java.io.FileReader(a[0]).read(b);i>0;seek(l=++l%s),i--){c=b[l];for(int n:p)c+=l+n>=0&l+n<s?b[l+n]/e:0;write(c>13?(c==49|(c|1)==91?e:46):10);}}};}}
And a little more readable
class L {
public static void main(final String[]a) throws Exception {
new java.io.RandomAccessFile("out.txt","rw"){{
int e=88, p[]={-1,1,-80,80,-81,81,-82,82},s=3240,l=0,i=new Byte(a[1])*s+s,c;
char[] b = new char[s];
for (new java.io.FileReader(a[0]).read(b);i>0;seek(l=++l%s),i--) {
c=b[l];
for (int n:p)
c+=l+n>=0&l+n<s?b[l+n]/e:0;
write(c>13?(c==49|(c|1)==91?e:46):10);
}
}};
}
}
Very similar to Molehill’s version. I've tried to use a different FileWriter and to count the cell's neighbors without an additional variable.
Unfortunately, RandomAccessFile is a pretty long name and it is required that you pass an file access mode.
RUST - 469 characters
Don't know if I should post this here, (this post is 3 years old) but anyway, my try on this, in rust (0.9):
use std::io::fs::File;fn main(){
let mut c=File::open(&Path::new(std::os::args()[1])).read_to_end();
for _ in range(0,from_str::<int>(std::os::args()[2]).unwrap()){
let mut b=c.clone();for y in range(0,40){for x in range(0,80){let mut s=0;
for z in range(x-1,x+2){for t in range(y-1,y+2){
if z>=0&&t>=0&&z<80&&t<40&&(x !=z||y !=t)&&c[t*81+z]==88u8{s +=1;}}}
b[y*81+x]=if s==3||(s==2&&c[y*81+x]==88u8){88u8} else {46u8};}}c = b;}
File::create(&Path::new("out.txt")).write(c);}
For people interested, here is the code before some agressive golfing:
use std::io::fs::File;
fn main() {
let f = std::os::args()[1];
let mut c = File::open(&Path::new(f)).read_to_end();
let n = from_str::<int>(std::os::args()[2]).unwrap();
for _ in range(0,n)
{
let mut new = c.clone();
for y in range(0,40) {
for x in range(0,80) {
let mut sum = 0;
for xx in range(x-1,x+2){
for yy in range(y-1,y+2) {
if xx >= 0 && yy >= 0 && xx <80 && yy <40 && (x != xx || y != yy) && c[yy*81+xx] == 88u8
{ sum = sum + 1; }
}
}
new[y*81+x] = if sum == 3 || (sum == 2 && c[y*81+x] == 88u8) {88u8} else {46u8};
}
}
c = new;
}
File::create(&Path::new("out.txt")).write(c);
}
ét voilà
you may want to use this html file. no file input, but a textarea that does the job!
there is also some html and initiation and vars. the main routine has only 235 characters.
It's hand-minified JS.
<!DOCTYPE html>
<html><body><textarea id="t" style="width:600px;height:600px;font-family:Courier">
</textarea></body><script type="text/javascript">var o,c,m=new Array(3200),
k=new Array(3200),y,v,l,p;o=document.getElementById("t");for(y=0;y<3200;y++)
{m[y]=Math.random()<0.5;}setInterval(function(){p="";for(y=0;y<3200;y++){c=0;
for(v=-1;v<2;v+=2){c+=m[y-1*v]?1:0;for(l=79;l<82;l++)c+=m[y-l*v]?1:0;}
k[y]=c==3||m[y]&&c==2;}p="";for(y=0;y<3200;y++){p+=(y>0&&y%80==0)?"\n":"";
m[y]=k[y];p+=(m[y]?"O":"-");}o.innerHTML=p;},100);</script></html>
One of the classic patterns
***
..*
.*
My avatar was created using my version of the Game of Life using this pattern and rule(note that it is not 23/3):
#D Thanks to my daughter Natalie
#D Try at cell size of 1
#R 8/1
#P -29 -29
.*********************************************************
*.*******************************************************.*
**.*****************************************************.**
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****************************.*.****************************
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****************************.*.****************************
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**.*****************************************************.**
*.*******************************************************.*
.*********************************************************
IMHO - as I learned Conway's Game of Life the trick wasn't writing short code, but code that could do complex life forms quickly. Using the classic pattern above and a wrapped world of 594,441 cells the best I could ever do was around 1,000 generations / sec.
Another simple pattern
**********
.
................*
.................**
................**.......**********
And gliders
........................*...........
......................*.*...........
............**......**............**
...........*...*....**............**
**........*.....*...**..............
**........*...*.**....*.*...........
..........*.....*.......*...........
...........*...*....................
............**......................

Code Golf: Triforce

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
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.
*********** ***********
********* *********
******* *******
***** *****
*** ***
* *
***********
*********
*******
*****
***
*
*
***
*****
*******
*********
***********
* *
*** ***
***** *****
******* *******
********* *********
*********** ***********
Can we try and keep it to one answer per language, that we all improve on?
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.
GolfScript - 43 chars
~:!6*,{:^' '
*'*'12*' '
^6%.+)*+
-12>!^
6/-*
n}
/
~:!6*,{:^' '*'*'12*' '^6%.+)*+-12>!^6/-*n}/
48 Chars for the bonus
~:!6*,.-1%+{
:^' '*'*'12
*' '^6%.+
)*+-12>
!^6/-
*n}
/
~:!6*,.-1%+{:^' '*'*'12*' '^6%.+)*+-12>!^6/-*n}/
Python - 77 Chars
n=input()
for k in range(6*n):print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6)
n=input()
for k in range(6*n):j=1+k%6*2;print' '*k+('*'*(12-j)+' '*j)*(n-k/6)
89 Chars for the bonus
n=input();R=range(6*n)
for k in R+R[::-1]:print' '*k+('*'*11+' '*11)[k%6*2:][:12]*(n-k/6)
114 Chars Version just using string replacements
u,v=' *';s=(v*11+u)*input()
while s.strip():print s;s=u+s.replace(*((v*2+u,u*3),(v*1+u*10,v*11))[' * 'in s])[:-2]
Unk Chars all in one statement, should work w/ 2.x and 3.x. The enumerate() is to allow the single input() to work for both places you need to use it.
print ('\n'.join('\n'.join(((' '*(6*n))+' '.join(('%s%s%s'%(' '*(5-x),'*'*(2*x+1),' '*(5-x)) for m in range(i + 1)))) for x in range(5,-1,-1)) for n, i in enumerate(range(int(input())-1,-1,-1))))
Yet Another Method
def f(n): print '\n'.join(' '*6*(n-r)+(' '*(5-l)+'*'*(l*2+1)+' '*(5-l)+' ')*r for r in xrange(1, n+1) for l in xrange(6))
f(input())
Ruby - 74 Chars
(6*n=gets.to_i).times{|k|puts' '*k+('*'*(11-(j=k%6*2))+' '*(j+1))*(n-k/6)}
COBOL - 385 Chars
$ cobc -free -x triforce.cob && echo 7| ./triforce
PROGRAM-ID.P.DATA DIVISION.WORKING-STORAGE SECTION.
1 N PIC 9.
1 M PIC 99.
1 value '0100***********'.
2 I PIC 99.
2 K PIC 99.
2 V PIC X(22).
2 W PIC X(99).
PROCEDURE DIVISION.ACCEPT N
COMPUTE M=N*6
PERFORM M TIMES
DISPLAY W(1:K)NO ADVANCING
PERFORM N TIMES
DISPLAY V(I:12)NO ADVANCING
END-PERFORM
DISPLAY ''
ADD 2 TO I
IF I = 13 MOVE 1 TO I ADD -1 TO N END-IF
ADD 1 TO K
END-PERFORM.
K could be returned to outside the group level. An initial value of zero for a numeric with no VALUE clause is compiler-implementation dependent, as is an initial value of space for an alpha-numeric field (W has been cured of this, at no extra character cost). Moving K back would save two characters. -free is compiler-dependant as well, so I'm probably being over-picky.
sed, 117 chars
s/$/76543210/
s/(.).*\1//
s/./*********** /gp
:
s/\*(\**)\*/ \1 /gp
t
:c
s/\* {11}\*/ ************/
tc
s/\* / /p
t
Usage: $ echo 7 | sed -rf this.sed
First attempt; improvements could probably be made...
Ruby 1.9 - 84 characters :
v=gets.to_i
v.times{|x|6.times{|i|puts' '*6*x+(' '*i+'*'*(11-2*i)+' '*i+' ')*(v-x)}}
Perl - 72 chars
die map$"x$_.("*"x(12-($l=1+$_%6*2)).$"x$l)x($n-int$_/6).$/,0..6*($n=<>)
78 chars
map{$l=$_%6*2;print$"x$_,("*"x(11-$l).$"x$l.$")x($n-int$_/6),$/}0..6*($n=<>)-1
87 chars
$n=<>;map{$i=int$_/6;$l=$_%6*2;print$"x$_,("*"x(11-$l).$"x$l.$")x($n-$i),$/}(0..6*$n-1)
97 chars
$n=<>;map{$i=int$_/6;$l=$_%6;print$"x(6*$i),($"x$l."*"x(11-2*$l).$"x$l.$")x($n-$i),$/}(0..6*$n-1)
108 chars
$n=<>;map{$i=int$_/6;$l=$_%6;print ""." "x(6*$i),(" "x$l."*"x(11-2*$l)." "x$l." ")x($n-$i),"\n";}(0..6*$n-1)
Powershell, 78 characters
0..(6*($n=read-host)-1)|%{" "*$_+("*"*(12-($k=1+$_%6*2))+" "*$k)*(.4+$n-$_/6)}
Bonus, 92 characters
$a=0..(6*($n=read-host)-1)|%{" "*$_+("*"*(12-($k=1+$_%6*2))+" "*$k)*(.4+$n-$_/6)}
$a
$a|sort
The output is stored in an array of strings, $a, and the reverse is created by sorting the array. We could, of course, just reverse the array, but it would be more characters to type :)
Haskell - 131 138 142 143 Chars
(⊗)=replicate
z o=[concat$(6*n+m)⊗' ':(o-n)⊗((11-m-m)⊗'*'++(1+m+m)⊗' ')|n<-[0..o-1],m<-[0..5]]
main=getLine>>=mapM_ putStrLn.z.read
This one is longer (146 148 chars) at present, but an interesting, alternate line of attack:
(⊗)=replicate
a↑b|a>b=' ';_↑_='*'
z o=[map(k↑)$concat$(6*n)⊗' ':(o-n)⊗"abcdefedcba "|n<-[0..o-1],k<-"abcdef"]
main=getLine>>=mapM_ putStrLn.z.read
FORTRAN - 97 Chars
Got rid of the #define and saved 8 bytes thanks to implict loops!
$ f95 triforce.f95 -o triforce && echo 7 | ./triforce
READ*,N
DO K=0,N*6
M=2*MOD(K,6)
PRINT*,(' ',I=1,K),(('*',I=M,10),(' ',I=0,M),J=K/6+1,N)
ENDDO
END
125 bytes for the bonus
READ*,N
DO L=1,N*12
K=L+5
If(L>N*6)K=N*12-L+6
M=2*MOD(K,6)
PRINT"(99A)",(32,I=7,K),((42,I=M,10),(32,I=0,M),J=K/6,N)
ENDDO
END
FORTRAN - 108 Chars
#define R REPEAT
READ*,N
DO I=0,6*N
J=MOD(I,6)*2
PRINT*,R(' ',I)//R(R('*',11-J)//R(' ',J+1),N-I/6)
ENDDO
END
JavaScript 1.8 - SpiderMonkey - 118 chars
N=readline()
function f(n,c)n>0?(c||' ')+f(n-1,c):''
for(i=0;i<N*6;i++)print(f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1)))
w/ bonus - 151 chars
N=readline()
function f(n,c)n>0?(c||' ')+f(n-1,c):''
function l(i)print(f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1)))
for(i=0;i<N*6;i++)l(i)
for(;i--;)l(i)
Usage: js thisfile.js
JavaScript - In Browser - 154 characters
N=prompt()
function f(n,c){return n>0?(c||' ')+f(n-1,c):''}
s='<pre>'
for(i=0;i<N*6;i++)s+=f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1))+'\n'
document.write(s)
The non-obfuscated version (before optimizations by gnarf):
var N = prompt();
var S = ' ';
function fill(c, n) {
for (ret=''; n--;)
ret += c;
return ret;
}
var str = '<pre>';
for (i=0; i<N*6; i++) {
str += fill(S, i);
for (j=0; j<N-i/6; j++)
str += fill('*', 11-i%6*2) + fill(S, i%6*2+1);
str += '\n';
}
document.write(str);
Here's a different algorithm that uses replace() to go from one line to the next of each line of a triangle row:
161 characters
N=readline()
function f(n,c){return n>0?(c||' ')+f(n-1,c):''}l=0
for(i=N;i>0;){r=f(i--,f(11,'*')+' ');for(j=6;j--;){print(f(l++)+r)
r=r.replace(/\*\* /g,' ')}}
F#, 184 181 167 151 147 143 142 133 chars
let N,r=int(stdin.ReadLine()),String.replicate
for l in[0..N*6-1]do printfn"%s%s"(r l" ")(r(N-l/6)((r(11-l%6*2)"*")+(r(l%6*2+1)" ")))
Bonus, 215 212 198 166 162 158 157 148 chars
let N,r=int(stdin.ReadLine()),String.replicate
for l in[0..N*6-1]#[N*6-1..-1..0]do printfn"%s%s"(r l" ")(r(N-l/6)((r(11-l%6*2)"*")+(r(l%6*2+1)" ")))
C - 120 Chars
main(w,i,x,y){w=getchar()%8*12;for(i=0;i<w*w/2;)y=i/w,x=i++%w,putchar(x>w-2?10:x<y|w-x-1<y|(x-y)%12>=11-2*(y%6)?32:42);}
Note that this solution prints some trailing spaces (which is okay, right?). It also relies on relational operators having higher precedence than bitwise OR, saving two characters.
124 Chars
main(n,i,k){n=getchar()&7;for(k=0;k<6*n;k++,putchar(10))for(i=-k-1;++i<12*n-2*k-1;putchar(32+10*(i>=0&&(11-i%12>2*k%12))));}
C - 177 183 Chars
#define P(I,C)for(m=0;m<I;m++)putchar(C)
main(t,c,r,o,m){scanf("%d",&t);for(c=t;c>0;c--)for(r=6;r>0;r--){P((t-c)*6+6-r,32);for(o=0;o<c;o++){P(r*2-1,42);P(13-r*2,32);}puts("");}}
C - 222 243 Chars (With Bonus Points)
#define P(I,C)for(m=0;m<I;m++)putchar(C)
main(t,c,r,o,m){scanf("%d",&t);for(c=t-1;-c<2+t;c-=1+!c)for(r=c<0?1:6;c<0?r<7:r>0;r+=c<0?1:-1){P((t-abs(c+1))*6+6-r,32);for(o=0;o<abs(c+1);o++){P(r*2-1,42);P(13-r*2,32);}puts("");}}
This is my first Code Golf submission as well!
Written in C
Bonus points (492 chars):
p(char *t, int c, int s){int i=0;for(;i<s;i++)printf(" ");for(i=0;i<c;i++)printf("%s",t);printf("\n");}main(int a, char **v){int i=0;int k;int c=atoi(v[1]);for(;i<c;i++){p("*********** ",c-i,i);p(" ********* ",c-i,i);p(" ******* ",c-i,i);p(" ***** ",c-i,i);p(" *** ",c-i,i);p(" * ",c-i,i);}for(i=0;i<c;i++){k=c-i-1;p(" * ",1+i,k);p(" *** ",1+i,k);p(" ***** ",1+i,k);p(" ******* ",1+i,k);p(" ********* ",1+i,k);p("*********** ",i+1,k);}}
Without bonus points (322 chars):
p(char *t, int c, int s){int i=0;for(;i<s;i++)printf(" ");for(i=0;i<c;i++)printf("%s",t);printf("\n");}main(int a, char **v){int i=0;int k;int c=atoi(v[1]);for(;i<c;i++){p("*********** ",c-i,i);p(" ********* ",c-i,i);p(" ******* ",c-i,i);p(" ***** ",c-i,i);p(" *** ",c-i,i);p(" * ",c-i,i);}}
First time posting, too!
Lua, 121 chars
R,N,S=string.rep,io.read'*n',' 'for i=0,N-1 do for j=0,5 do X=R(S,j)print(R(S,6*i)..R(X..R('*',11-2*j)..X..S,N-i))end end
123
R,N,S=string.rep,io.read'*n',' 'for i=0,N-1 do for j=0,5 do print(R(S,6*i)..R(R(S,j)..R('*',11-2*j)..R(S,j)..S,N-i))end end
PHP, 153
<?php $i=fgets(STDIN);function r($n,$c=' '){return$n>0?$c.r($n-1,$c):'';}for($l=0;$l<$i*6;){$z=$l%6*2;echo r($l).r($i-$l++/6,r(11-$z,'*').r($z+1))."\n";}
with Bonus, 210
<?php $i=fgets(STDIN);function r($n,$c=' '){return$n>0?$c.r($n-1,$c):'';}$o=array();for($l=0;$l<$i*6;){$z=$l%6*2;$o[]=r($l).r($i-$l++/6,r(11-$z,'*').r($z+1));}print join("\n",array_merge($o,array_reverse($o)));
dc 105 chars
123 129 132 139 141
[rdPr1-d0<P]sP?sn
0sk[1lk6%2*+sj32lkd0<Plnlk6/-si
[[*]12lj-d0<P32ljd0<Pli1-dsi0<I]dsIx
10Plk1+dskln6*>K]dsKx
Mathematica, 46 characters
The answer prints sideways.
TableForm#{Table["*",{l,#},{l},{j,6},{2j-1}]}&
HyperTalk - 272 chars
function triforce n
put"******" into a
put n*6 into h
repeat with y=0 to h-1
put" " after s
put char 1 to y of s after t
repeat n-y div 6
get y mod 6*2
put char 1 to 11-it of (a&a)&&char 1 to it of s after t
end repeat
put return after t
end repeat
return t
end triforce
Indentation is neither needed nor counted (HyperCard automatically adds it).
Miscellanea:
Since there is no notion of console or way to access console arguments in HyperCard 2.2 (that I know of), a function is given instead. It can be invoked with:
on mouseUp
ask "Triforce: "
put triforce(it) into card field 1
end mouseUp
To use this, a card field would be created and set to a fixed-width font. Using HyperCard's answer command would display a dialog with the text, but it doesn't work because:
The answer dialog font (Chicago) is not fixed-width.
The answer command refuses to display long text (even triforce(2) is too long).
Common Lisp, 150 characters:
(defun f(n o)(unless(= n 0)(dotimes(x 6)(format t"~v#{~a~:*~}~-1:*~v#{~?~2:*~}~%"
o" "n"~11#: "(list(- 11(* 2 x))#\*)))(f(1- n)(+ 6 o))))
77 char alternative python solution based on gnibbler's:
n=input()
k=0
exec"print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6);k+=1;"*6*n
Amazingly the bonus came out exactly the same also (101 chars, oh well)
n=input()
l=1
k=0
s="print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6);k+=l;"*6*n
exec s+'l=-1;k-=1;'+s