On the Subject of ASCII Mazes

Why is it called ASCII?

The screen on the bottom of the module can be scrolled through to obtain 12 characters which are delimited by �. Use the following table^* to convert these characters to binary. Call this 96-bit binary value the main code.

NUL 00000000

SOH 00000001

STX 00000010

ETX 00000011

EOT 00000100

ENQ 00000101

ACK 00000110

BEL 00000111

BS 00001000

HT 00001001

LF 00001010

VT 00001011

FF 00001100

CR 00001101

SO 00001110

SI 00001111

DLE 00010000

DC1 00010001

DC2 00010010

DC3 00010011

DC4 00010100

NAK 00010101

SYN 00010110

ETB 00010111

CAN 00011000

EM 00011001

SUB 00011010

ESC 00011011

FS 00011100

GS 00011101

RS 00011110

US 00011111

00100000

! 00100001

" 00100010

# 00100011

$ 00100100

% 00100101

& 00100110

' 00100111

( 00101000

) 00101001

* 00101010

+ 00101011

, 00101100

- 00101101

. 00101110

/ 00101111

0 00110000

1 00110001

2 00110010

3 00110011

4 00110100

5 00110101

6 00110110

7 00110111

8 00111000

9 00111001

: 00111010

; 00111011

< 00111100

= 00111101

> 00111110

? 00111111

@ 01000000

A 01000001

B 01000010

C 01000011

D 01000100

E 01000101

F 01000110

G 01000111

H 01001000

I 01001001

J 01001010

K 01001011

L 01001100

M 01001101

N 01001110

O 01001111

P 01010000

Q 01010001

R 01010010

S 01010011

T 01010100

U 01010101

V 01010110

W 01010111

X 01011000

Y 01011001

Z 01011010

[ 01011011

\ 01011100

] 01011101

^ 01011110

_ 01011111

` 01100000

a 01100001

b 01100010

c 01100011

d 01100100

e 01100101

f 01100110

g 01100111

h 01101000

i 01101001

j 01101010

k 01101011

l 01101100

m 01101101

n 01101110

o 01101111

p 01110000

q 01110001

r 01110010

s 01110011

t 01110100

u 01110101

v 01110110

w 01110111

x 01111000

y 01111001

z 01111010

{ 01111011

| 01111100

} 01111101

~ 01111110

DEL 01111111

Ç 10000000

ü 10000001

é 10000010

â 10000011

ä 10000100

à 10000101

å 10000110

ç 10000111

ê 10001000

ë 10001001

è 10001010

ï 10001011

î 10001100

ì 10001101

Ä 10001110

Å 10001111

É 10010000

æ 10010001

Æ 10010010

ô 10010011

ö 10010100

ò 10010101

û 10010110

ù 10010111

ÿ 10011000

Ö 10011001

Ü 10011010

ø 10011011

£ 10011100

Ø 10011101

× 10011110

ƒ 10011111

á 10100000

í 10100001

ó 10100010

ú 10100011

ñ 10100100

Ñ 10100101

ª 10100110

º 10100111

¿ 10101000

® 10101001

¬ 10101010

½ 10101011

¼ 10101100

¡ 10101101

« 10101110

» 10101111

░ 10110000

▒ 10110001

▓ 10110010

│ 10110011

┤ 10110100

Á 10110101

 10110110

À 10110111

© 10111000

╣ 10111001

║ 10111010

╗ 10111011

╝ 10111100

¢ 10111101

¥ 10111110

┐ 10111111

└ 11000000

┴ 11000001

┬ 11000010

├ 11000011

─ 11000100

┼ 11000101

ã 11000110

à 11000111

╚ 11001000

╔ 11001001

╩ 11001010

╦ 11001011

╠ 11001100

═ 11001101

╬ 11001110

¤ 11001111

ð 11010000

Ð 11010001

Ê 11010010

Ë 11010011

È 11010100

ı 11010101

Í 11010110

Î 11010111

Ï 11011000

┘ 11011001

┌ 11011010

█ 11011011

▄ 11011100

¦ 11011101

Ì 11011110

▀ 11011111

Ó 11100000

ß 11100001

Ô 11100010

Ò 11100011

õ 11100100

Õ 11100101

µ 11100110

þ 11100111

Þ 11101000

Ú 11101001

Û 11101010

Ù 11101011

ý 11101100

Ý 11101101

¯ 11101110

´ 11101111

≡ 11110000

± 11110001

‗ 11110010

¾ 11110011

¶ 11110100

§ 11110101

÷ 11110110

¸ 11110111

° 11111000

¨ 11111001

· 11111010

¹ 11111011

³ 11111100

² 11111101

■ 11111110

nbsp 11111111

^* This character encoding is DOS Code Page 850.

Disarming the module

• Take the 7th to 48th bits from the main code. Group these into 7 rows of 6 bits. Number the rows 1–7.
• Use the color of the top-right LED in the following table to determine which rows need to be horizontally flipped and flip them. Then use the same table to determine whether or not to reverse the order of the rows. The result specifies the vertical walls in a 7×7 maze.

  LED color	black	blue	green	cyan	red	magenta	yellow	white
  Flip	none	even	none	even	all	odd	all	odd
  Reverse	no	no	yes	yes	no	no	yes	yes

• Take the 49th to 90th bits from the main code. Group these into 6 rows of 7 bits. Number the rows 1–6.
• Use the same table above with the color of the bottom-left LED to determine which rows need horizontal flipping and whether to reverse their order. The result specifies the horizontal walls of the same maze.
• Interweave the lines in such a way that the 7×7 maze is reconstructed. Note that the first vertical wall of each row is horizontally in between the first and second horizontal wall columns; similarly, the first horizontal wall of each column is between the first two vertical wall rows.
• Call the first 3 bits of the main code a and the next 3 bits b. Use the color of the top-left LED in the following table to obtain your starting position in the maze. The first value is the 0-based x-coordinate (column), the second is the 0-based y-coordinate (row). A tilde (~) means to negate the three bits.

  black	blue	green	cyan	red	magenta	yellow	white
  (a, b)	(b, a)	(a, ~b)	(b, ~a)	(~a, b)	(~b, a)	(~a, ~b)	(~b, ~a)

• Use the last six bits of the main code and the color of the bottom-right LED in the same way to obtain your goal position in the maze.
• Navigate the maze using the arrow buttons on the module. When you are on the goal position, press the green button with a target symbol to submit. The module is disarmed if the position is correct. Otherwise, a strike is incurred.
• The red button with an exclamation mark can be used to reset your position in the maze to the starting position.