: The autograder specifically checks if you used the minimum amount of bits required (5) for the 27 characters.
In this specific exercise:
function encode83_8(input): alphabet = [list of 83 symbols] indexMap = symbol: idx for idx, symbol in enumerate(alphabet) padding = '~' blockSize = 8 output = "" for i from 0 to len(input) step blockSize: block = input[i : i+blockSize] if len(block) < blockSize: block = block + padding * (blockSize - len(block)) // Convert block to base-83 number as array of digits for ch in block: output += alphabet[indexMap[ch]] return output 83 8 create your own encoding codehs answers exclusive
For a student genuinely attempting CodeHS 8.3, several legitimate strategies exist. Each has trade-offs in complexity, security (though security is rarely the goal here), and ease of implementation. : The autograder specifically checks if you used
: On the sidebar of the exercise, enter a binary key (e.g., 00000 ) and its corresponding value (e.g., A ). : On the sidebar of the exercise, enter a binary key (e