1. The problem is to encode the string \texttt{BCAACBDAACSADDDAX} using Lempel-Ziv coding. 2. Lempel-Ziv coding is a dictionary-based compression algorithm that encodes sequences by referencing previously seen substrings. 3. We start with an empty dictionary and scan the input string from left to right, adding new substrings to the dictionary and outputting codes for the longest match found. 4. Step-by-step encoding: - Start with dictionary: empty - Read \texttt{B}, add \texttt{B} to dictionary as entry 1, output (0, B) - Read \texttt{C}, add \texttt{C} as entry 2, output (0, C) - Read \texttt{A}, add \texttt{A} as entry 3, output (0, A) - Read \texttt{A}, already in dictionary as 3, look ahead: Next char \texttt{C}, \texttt{AC} not in dictionary, add \texttt{AC} as 4, output (3, C) - Read \texttt{B}, add \texttt{B} as 1 already, look ahead: Next char \texttt{D}, \texttt{BD} not in dictionary, add \texttt{BD} as 5, output (1, D) - Read \texttt{A}, add \texttt{A} as 3 already, look ahead: Next char \texttt{A}, \texttt{AA} not in dictionary, add \texttt{AA} as 6, output (3, A) - Read \texttt{C}, add \texttt{C} as 2 already, look ahead: Next char \texttt{S}, \texttt{CS} not in dictionary, add \texttt{CS} as 7, output (2, S) - Read \texttt{A}, add \texttt{A} as 3 already, look ahead: Next char \texttt{D}, \texttt{AD} not in dictionary, add \texttt{AD} as 8, output (3, D) - Read \texttt{D}, add \texttt{D} as 9, output (0, D) - Read \texttt{D}, look ahead: Next char \texttt{D}, \texttt{DD} not in dictionary, add \texttt{DD} as 10, output (9, D) - Read \texttt{A}, add \texttt{A} as 3 already, look ahead: Next char \texttt{X}, \texttt{AX} not in dictionary, add \texttt{AX} as 11, output (3, X) 5. Final encoded output is the sequence of pairs: (0, B), (0, C), (0, A), (3, C), (1, D), (3, A), (2, S), (3, D), (0, D), (9, D), (3, X) This sequence represents the Lempel-Ziv encoding of the input string.