基于指定字符集的字符左右匹配解码方法咨询

Hey there, let's work through this decoding problem step by step. First, let's unpack the rules and example you provided, then build out clear technical steps to crack the message.

First, Lay the Foundation: Map the Reference Character Set #

The key here is the ordered reference set: MONCLETABDFGHIJKPQRSUVWXYZ. We need to formalize the position and neighbor relationships of each character in this set, since the decoding relies on left/right neighbors.

First, create two core mappings (you can do this in any programming language, I'll use Python for examples):

# Define the reference set as an ordered string
ref_set = "MONCLETABDFGHIJKPQRSUVWXYZ"

# Create a character-to-index map and index-to-character map
char_to_idx = {char: idx for idx, char in enumerate(ref_set)}
idx_to_char = {idx: char for idx, char in enumerate(ref_set)}

# Precompute left (previous) and right (next) neighbors for each character
prev_neighbor = {}
next_neighbor = {}
for idx, char in enumerate(ref_set):
    # Handle edge cases: first character has no left neighbor, last has no right
    prev_neighbor[char] = ref_set[idx-1] if idx > 0 else None
    next_neighbor[char] = ref_set[idx+1] if idx < len(ref_set)-1 else None

With these mappings, we can instantly look up:

• The left neighbor of J is I (matches your example)
• The right neighbor of D is F (also matches your example)

Next, Test Plausible Decoding Rules #

Your description is a bit ambiguous, but based on the example, we can test a few logical rule sets to find the readable message. Let's start with the most likely candidates:

Rule 1: Decode each target character to its left neighbor in the reference set

This uses the first example (J → I). Let's run through the first few target pairs to see:

• JD → I (J's left neighbor) + B (D's left neighbor) → IB
• IR → H (I's left neighbor) + Q (R's left neighbor) → HQ
  If this doesn't produce readable words, move to the next rule.

Rule 2: Decode each target character to its right neighbor in the reference set

This uses the second example (D → F):

• JD → K (J's right neighbor) + F (D's right neighbor) → KF
• IR → J (I's right neighbor) + S (R's right neighbor) → JS

Rule 3: Pair-based decoding (since targets are split by spaces into 2-character groups)

It's possible each 2-character group maps to a single decoded character. For example:

• A group XY could mean "find the character in the reference set whose left neighbor is X and right neighbor is Y"
• Or, XY could mean "X is the right neighbor of the decoded character, Y is the left neighbor"

To test this, you'd loop through each 2-character group, check if there's a character in the reference set that fits the neighbor criteria, and collect those characters.

Final Step: Validate for Readable Output #

Since decoding should result in a coherent message, run each rule through the full target sequence:

target_sequence = "JD IR BW TQ DW SH UB XW AH NG AS XW CH UF BW FO DW SH UB BW TQ GW UH YG JV IE VM YE JD UO"
target_groups = target_sequence.split()

# Example implementation of Rule 1 (left neighbor decoding)
decoded_rule1 = []
for group in target_groups:
    for char in group:
        decoded_rule1.append(prev_neighbor[char])
print(''.join(decoded_rule1))

# Example implementation of Rule 2 (right neighbor decoding)
decoded_rule2 = []
for group in target_groups:
    for char in group:
        decoded_rule2.append(next_neighbor[char])
print(''.join(decoded_rule2))

Check the output of each rule—whichever produces a readable English string is the correct one. If none do, you may need to adjust the rule (e.g., wrap around the reference set for edge characters, or reverse the neighbor direction per group).

内容的提问来源于stack exchange，提问作者javax