This document contains code to encode and decode Japanese characters using a two-byte encoding scheme that requires only 10 fingers. It subtracts 0x3000 from character values in the Hiragana Unicode block to fit them into a single byte. It defines a Fingers codec class with encode and decode methods that take a string as input, add flag and value bytes to a buffer, and return the encoded bytes or decoded string. Sample usage encodes and decodes a test string, printing the results.

1. I can count to 1023
Anthony Shaw

2. 1
2
4
8
16
32
64
128
256
512

3. 1
4
16
32
64
128
256
512
21

5. >>> ord('a')
97
>>> ord('あ')
12354
>>> o = ord('あ')
>>> f'{o:08b}'
'11000001000010'
Oh No! We need 14 fingers for
this encoding

6. 0 1 2 3 4 5 6 7 8 9 A B C D E F
U+304
x
ぁ あ ぃ い ぅ う ぇ え ぉ お か が き ぎ く
U+305
x
ぐ け げ こ ご さ ざ し じ す ず せ ぜ そ ぞ た
U+306
x
だ ち ぢ っ つ づ て で と ど な に ぬ ね の は
U+307
x
ば ぱ ひ び ぴ ふ ぶ ぷ へ べ ぺ ほ ぼ ぽ ま み
U+308
x
む め も ゃ や ゅ ゆ ょ よ ら り る れ ろ ゎ わ
ゟ = 12447
ぁ = 12354
Hiragana Unicode Block

7. Unicode UCS code blocks for Japanese
Characters
0x3000-0x303F
Japanese-style
punctuation
0x3040-0x309F Hiragana
0x30A0-0x30FF Katakana
Therefore we can subtract 0x3000 from the
value to fit it within a single byte

8. 1
2
4
8
16
32
64
128
Flag 1
Flag 2?

9. import codecs
class Fingers(codecs.Codec):
def encode(self, input: str, errors='strict'):
buffer = bytearray()
for c in input:
if ord(c) & 0x3000:
flag = 1
v = ord(c) - 0x3000
else:
flag = 0
v = ord(c)
buffer.extend([flag, v]) # Add 2 byte to buffer
return bytes(buffer)

10. class Fingers(codecs.Codec):
def decode(self, input: bytes, errors: str = "strict"):
buffer = bytearray(input)
result = ''
while buffer:
flag = buffer.pop(0) # First Byte
v = buffer.pop(0) # Second Byte
if flag:
v += 0x3000
result += chr(v)
return result, len(input)
codec = Fingers()
output = codec.encode("Hello とうきょう")
print(f"Output is {output}")
print(codec.decode(output))