4 Syntax coloring

## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)

puts (6 * 9).to_s(13)


## Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))


Output:

@  @   @@
@  @  @  @
@@@@@    @
@   @@
@  @@@@


## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)


## Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))


Output:

@  @   @@
@  @  @  @
@@@@@    @
@   @@
@  @@@@


## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)


## Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))


Output:

@  @   @@
@  @  @  @
@@@@@    @
@   @@
@  @@@@

3 added 5 characters in body

## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)


## Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))


Output:

@  @   @@
@  @  @  @
@@@@@    @
@   @@
@  @@@@


## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)


## Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))


Output:

@  @   @@
@  @  @  @
@@@@@    @
@   @@
@  @@@@


## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)


## Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))


Output:

@  @   @@
@  @  @  @
@@@@@    @
@   @@
@  @@@@

2 added 955 characters in body

## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)


## Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))


Output:

@  @   @@
@  @  @  @
@@@@@    @
@   @@
@  @@@@


## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)


## Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)


## Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
(2**17)**0 * 0b11100000000000000 +
(2**17)**1 * 0b00100000000000000 +
(2**17)**2 * 0b00100000000000000 +
(2**17)**3 * 0b11111000000000000 +
(2**17)**4 * 0b00100000000000000 +
(2**17)**5 * 0b00000000000000000 +
(2**17)**6 * 0b01001000000000000 +
(2**17)**7 * 0b10011000000000000 +
(2**17)**8 * 0b10011000000000000 +
(2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
print("".join(" @"[f(x, y)] for x in range(10)))


Output:

@  @   @@
@  @  @  @
@@@@@    @
@   @@
@  @@@@

1