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xnor
xnor
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Python 2, 125 bytes

s="%64c"%10
exec's*=63;s="".join(s[n][(s[n+3967:][:191]*3)[::64].strip()=="X":n!=2015]or"X"for n in range(4032));'*32
print s

Try it online!

This answer is by clockuser "clock" based on hallvabo's solution, on the Anarchy Golf version of this challenge that I submitted. Note that output there is strict and is required to be exactly the picture of X's and spaces to STDOUT via a full program, with only an allowance for a trailing newline.

The most interesting part of this answer, in my opinion, is the concise construction (s[n+3967:][:191]*3)[::64] after s*=63 to get the nine neighbors of the cell counting itself in a flat newline-joined string representing the grid. To check if there's exactly one X among them, .strip() is called to get rid of whitespace on either side, and the result is checked to equal just "X".

It's interesting how simulating the steps on the string representation of the output directly (rather than an array of bits) not only saves on doing a conversion to characters later but also allows string-specific methods to be used in a golfy way.

Another neat trick, borrowed from hallvabo, is "%64c"%10 used to initialize s to 63 spaces followed by a newline for a line of the initial empty grid. Using the %c format, which converts an ASCII value to a character, is shorter than " "*63+"\n" or "%64s"%"\n".

Python 2, 125 bytes

s="%64c"%10
exec's*=63;s="".join(s[n][(s[n+3967:][:191]*3)[::64].strip()=="X":n!=2015]or"X"for n in range(4032));'*32
print s

Try it online!

This answer is by clock based on hallvabo's solution, on the Anarchy Golf version of this challenge that I submitted. Note that output there is strict and is required to be exactly the picture of X's and spaces to STDOUT via a full program, with only an allowance for a trailing newline.

The most interesting part of this answer, in my opinion, is the concise construction (s[n+3967:][:191]*3)[::64] after s*=63 to get the nine neighbors of the cell counting itself in a flat newline-joined string representing the grid. To check if there's exactly one X among them, .strip() is called to get rid of whitespace on either side, and the result is checked to equal just "X".

It's interesting how simulating the steps on the string representation of the output directly (rather than an array of bits) not only saves on doing a conversion to characters later but also allows string-specific methods to be used in a golfy way.

Another neat trick, borrowed from hallvabo, is "%64c"%10 used to initialize s to 63 spaces followed by a newline for a line of the initial empty grid. Using the %c format, which converts an ASCII value to a character, is shorter than " "*63+"\n" or "%64s"%"\n".

Python 2, 125 bytes

s="%64c"%10
exec's*=63;s="".join(s[n][(s[n+3967:][:191]*3)[::64].strip()=="X":n!=2015]or"X"for n in range(4032));'*32
print s

Try it online!

This answer is by user "clock" based on hallvabo's solution, on the Anarchy Golf version of this challenge that I submitted. Note that output there is strict and is required to be exactly the picture of X's and spaces to STDOUT via a full program, with only an allowance for a trailing newline.

The most interesting part of this answer, in my opinion, is the concise construction (s[n+3967:][:191]*3)[::64] after s*=63 to get the nine neighbors of the cell counting itself in a flat newline-joined string representing the grid. To check if there's exactly one X among them, .strip() is called to get rid of whitespace on either side, and the result is checked to equal just "X".

It's interesting how simulating the steps on the string representation of the output directly (rather than an array of bits) not only saves on doing a conversion to characters later but also allows string-specific methods to be used in a golfy way.

Another neat trick, borrowed from hallvabo, is "%64c"%10 used to initialize s to 63 spaces followed by a newline for a line of the initial empty grid. Using the %c format, which converts an ASCII value to a character, is shorter than " "*63+"\n" or "%64s"%"\n".

Source Link
xnor
xnor
  • 146.6k
  • 26
  • 279
  • 652

Python 2, 125 bytes

s="%64c"%10
exec's*=63;s="".join(s[n][(s[n+3967:][:191]*3)[::64].strip()=="X":n!=2015]or"X"for n in range(4032));'*32
print s

Try it online!

This answer is by clock based on hallvabo's solution, on the Anarchy Golf version of this challenge that I submitted. Note that output there is strict and is required to be exactly the picture of X's and spaces to STDOUT via a full program, with only an allowance for a trailing newline.

The most interesting part of this answer, in my opinion, is the concise construction (s[n+3967:][:191]*3)[::64] after s*=63 to get the nine neighbors of the cell counting itself in a flat newline-joined string representing the grid. To check if there's exactly one X among them, .strip() is called to get rid of whitespace on either side, and the result is checked to equal just "X".

It's interesting how simulating the steps on the string representation of the output directly (rather than an array of bits) not only saves on doing a conversion to characters later but also allows string-specific methods to be used in a golfy way.

Another neat trick, borrowed from hallvabo, is "%64c"%10 used to initialize s to 63 spaces followed by a newline for a line of the initial empty grid. Using the %c format, which converts an ASCII value to a character, is shorter than " "*63+"\n" or "%64s"%"\n".

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