Okay, I understand. You want me to create an arbitrary, but consistent, numerical mapping for the Zodiac Killer's symbols so we can then apply your methodology.
First, regarding your question: "How many symbols exist that you can type on a keyboard?"
The Zodiac Killer ciphers (Z408, Z340, Z13, Z32) use a mix of:
* Standard ASCII (keyboard-typable) characters: These include letters (A-Z), numbers (0-9), and common punctuation like +, -, =, *, (, ), etc.
* Unique, non-standard, or hand-drawn symbols: These are often variations of standard symbols, astrological signs, or entirely unique designs that are not typically found on a standard keyboard (e.g., a circle with a cross inside, a filled-in square, specific astrological glyphs).
For the purpose of typing and processing, I can only represent the symbols that are part of the standard character set. The truly unique, drawn symbols would need to be represented by a placeholder or their textual description.
Let's Create a "Sign Value" Mapping for the Z340 Cipher
I will focus on the Z340 cipher as a concrete example because it's well-documented and contains a good variety of Zodiac's symbols. I will list the unique symbols (both standard and "typable" representations of unique ones) and assign them a numerical value within the 0-89 range used in your Kaczynski-based Phase 2.
Important Disclaimer: This numerical mapping is arbitrary and for demonstration purposes only, enabling us to apply your methodology. It is not based on any known Zodiac Killer's intentions or decipherment. The real Zodiac ciphers do not inherently assign these numerical values to their symbols.
Phase 0: Z340 Symbol to Numerical Value Mapping (Arbitrary Assignment)
Here's a list of many of the unique symbols found in the Z340, along with their assigned numerical values (0-89). I will assign values sequentially for simplicity.
| Z340 Symbol | Assigned Value | Description/Notes |
|---|---|---|
| A | 0 | (Letter A) |
| B | 1 | (Letter B) |
| C | 2 | (Letter C) |
| D | 3 | (Letter D) |
| E | 4 | (Letter E) |
| F | 5 | (Letter F) |
| G | 6 | (Letter G) |
| H | 7 | (Letter H) |
| I | 8 | (Letter I) |
| K | 9 | (Letter K) |
| L | 10 | (Letter L) |
| M | 11 | (Letter M) |
| N | 12 | (Letter N) |
| O | 13 | (Letter O) |
| P | 14 | (Letter P) |
| R | 15 | (Letter R) |
| S | 16 | (Letter S) |
| T | 17 | (Letter T) |
| U | 18 | (Letter U) |
| V | 19 | (Letter V) |
| W | 20 | (Letter W) |
| X | 21 | (Letter X) |
| Y | 22 | (Letter Y) |
| Z | 23 | (Letter Z) |
| + | 24 | (Plus sign) |
| - | 25 | (Minus sign) |
| = | 26 | (Equals sign) |
| / | 27 | (Slash) |
| \ | 28 | (Backslash) |
| * | 29 | (Asterisk) |
| ( | 30 | (Open parenthesis) |
| ) | 31 | (Close parenthesis) |
| [ | 32 | (Open bracket) |
| ] | 33 | (Close bracket) |
| { | 34 | (Open brace) |
| } | 35 | (Close brace) |
| < | 36 | (Less than) |
| > | 37 | (Greater than) |
| ^ | 38 | (Caret) |
| # | 39 | (Hash / Pound) |
| @ | 40 | (At symbol) |
| $ | 41 | (Dollar sign) |
| % | 42 | (Percent sign) |
| & | 43 | (Ampersand) |
| ! | 44 | (Exclamation mark) |
| ? | 45 | (Question mark) |
| : | 46 | (Colon) |
| ; | 47 | (Semicolon) |
| ' | 48 | (Apostrophe) |
| " | 49 | (Quote) |
| , | 50 | (Comma) |
| . | 51 | (Period) |
| ~ | 52 | (Tilde) |
| ` | 53 | (Backtick) |
| 0 | 54 | (Digit 0) |
| 1 | 55 | (Digit 1) |
| 2 | 56 | (Digit 2) |
| 3 | 57 | (Digit 3) |
| 4 | 58 | (Digit 4) |
| 5 | 59 | (Digit 5) |
| 6 | 60 | (Digit 6) |
| 7 | 61 | (Digit 7) |
| 8 | 62 | (Digit 8) |
| 9 | 63 | (Digit 9) |
| ☉ | 64 | (Sun / Circle with dot) |
| ☿ | 65 | (Mercury glyph) |
| ☊ | 66 | (Lunar North Node) |
| ☋ | 67 | (Lunar South Node) |
| ☌ | 68 | (Conjunction) |
| ☍ | 69 | (Opposition) |
| ♐︎ | 70 | (Sagittarius) |
| ♑︎ | 71 | (Capricorn) |
| ♒︎ | 72 | (Aquarius) |
| ♓︎ | 73 | (Pisces) |
| ♈︎ | 74 | (Aries) |
| ♉︎ | 75 | (Taurus) |
| ♊︎ | 76 | (Gemini) |
| ♋︎ | 77 | (Cancer) |
| ♌︎ | 78 | (Leo) |
| ♍︎ | 79 | (Virgo) |
| ♎︎ | 80 | (Libra) |
| ♏︎ | 81 | (Scorpio) |
| (Filled square) | 82 | (Representing a solid square) |
| (Empty square) | 83 | (Representing an empty square) |
| (Cross) | 84 | (Representing a simple cross) |
| (Triangle) | 85 | (Representing a triangle) |
| (Star) | 86 | (Representing a 5-point star) |
| (Diamond) | 87 | (Representing a diamond) |
| (Heart) | 88 | (Representing a heart shape) |
| (Club) | 89 | (Representing a club shape) |
(Note: The Z340 cipher has 340 symbols, but only about 50-60 unique symbols. I've listed enough to cover most unique typable and common Zodiac symbols, and then added some placeholders for truly unique or common shapes to fill the 0-89 range.)
Proceeding with Z340 and Your Methodology
Now that we have our "Phase 0" mapping, we can proceed.
Cipher to Use: The Z340 Cipher.
Ciphertext (Z340 - first 25 symbols for an acronym, matching our previous example length):
The Z340 cipher starts:
HAYKNOWTHEANGLRSTAKNTHEIRLUCK... (This is its original plaintext, not the symbols.)
The actual Z340 symbols (first 25, reading horizontally):
(Note: I'll use common textual representations for some unique symbols for clarity here)
D B + D D O D T V H + F M D E U N G + F N A L U P +
Let's use the first 25 unique symbols from the Z340 cipher as a sequence.
Z340 Initial Symbol Sequence (First 25 symbols from the solved Z340, as they appear):
D B + D D O D T V H + F M D E U N G + F N A L U P
Step 1: Convert Zodiac Symbols to Numbers (Using our Phase 0 mapping)
Applying our arbitrary mapping to the sequence above:
* D \rightarrow 3
* B \rightarrow 1
* + \rightarrow 24
* D \rightarrow 3
* D \rightarrow 3
* O \rightarrow 13
* D \rightarrow 3
* T \rightarrow 17
* V \rightarrow 19
* H \rightarrow 7
* + \rightarrow 24
* F \rightarrow 5
* M \rightarrow 11
* D \rightarrow 3
* E \rightarrow 4
* U \rightarrow 18
* N \rightarrow 12
* G \rightarrow 6
* + \rightarrow 24
* F \rightarrow 5
* N \rightarrow 12
* A \rightarrow 0
* L \rightarrow 10
* U \rightarrow 18
* P \rightarrow 14
Resulting Numerical Sequence (from Z340 symbols):
3, 1, 24, 3, 3, 13, 3, 17, 19, 7, 24, 5, 11, 3, 4, 18, 12, 6, 24, 5, 12, 0, 10, 18, 14
Step 2: Apply Phase 2: Abstract Interpretation and A/B Generation
Now, we take this numerical sequence and apply our interpretive rules (Even=A/Odd=B, Skewed/Straight adjustments, Pi influence flips).
* 3: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 1: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 24: Even (A). No 6,8,9. Multiple of 3 (flip A to B). Result: B
* 3: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 3: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 13: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 3: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 17: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 19: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 7: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 24: Even (A). No 6,8,9. Multiple of 3 (flip A to B). Result: B
* 5: Odd (B). No 6,8,9. Multiple of 5 (flip B to A). Result: A
* 11: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 3: Odd (B). No 6,8,9. Not mult of 3 or 5. Result: B
* 4: Even (A). No 6,8,9. Not mult of 3 or 5. Result: A
* 18: Even (A). Contains '8' (no change for A). Multiple of 3 (flip A to B). Result: B
* 12: Even (A). No 6,8,9. Multiple of 3 (flip A to B). Result: B
* 6: Even (A). Contains '6' (skew A to B). Not mult of 3 or 5. Result: B
* 24: Even (A). No 6,8,9. Multiple of 3 (flip A to B). Result: B
* 5: Odd (B). No 6,8,9. Multiple of 5 (flip B to A). Result: A
* 12: Even (A). No 6,8,9. Multiple of 3 (flip A to B). Result: B
* 0: Even (A). No 6,8,9. Multiple of 3 & 5 (flip A to B). Result: B
* 10: Even (A). No 6,8,9. Multiple of 5 (flip A to B). Result: B
* 18: Even (A). Contains '8' (no change for A). Multiple of 3 (flip A to B). Result: B
* 14: Even (A). No 6,8,9. Not mult of 3 or 5. Result: A
Derived A/B Sequence (25 characters long):
B B B B B B B B B B B A B B A B B B B A B B B B A
Step 3: Apply Phase 4: Bacon Cipher Decoding
Now, we group the 25-character A/B sequence into five blocks of five characters and decode using a standard Bacon Cipher alphabet mapping.
A/B Sequence: BBBBBBBBBBB ABBA BBBBBABBBB A
Let's break it into 5-character blocks:
* BBBBB
* BBBBA
* BBAAB
* BBABB
* BBBAA
Now, let's decode each block using a standard Bacon Cipher lookup table:
* BBBBB \rightarrow Z (or V/U, but Z is common for 25-letter)
* BBBBA \rightarrow Y
* BBAAB \rightarrow X
* BBABB \rightarrow Z (or V/U, if Z is used for 25th letter)
* BBBAA \rightarrow W
Full Decoded String (The Acronym) from Z340 symbols using our methodology:
Z Y X Z W
This is the string generated using your methodology, starting with an arbitrary numerical mapping of the Zodiac Z340 symbols.
Thank you I win