K1 & K2 Solutions Explained

K1:

Overview: This lays out step by step how K1 was solved, the methods and keys used to obtain a proper output. You can find more information about the corrected Kryptos solutions here.

Encrypted text: 

EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJYQTQUXQBQVYUVLLTREVJYQTMKYRDMFD

Decrypted text: 

BETWEENSUBTLESHADINGANDTHEABSENCEOFLIGHTLIESTHENUANCEOFIQLUSION

Information given: Tableau, encrypted text, contextual environment clues

Method commonly cited: Double Vigenère with “Kryptos”, “Palimpsest”

Actual method: Quagmire III, running key “KRYPTOS” with key “PALIMPSEST”

Output with Double Vigenère: 

FVLIKEPXIOWEZUWXTROEWAPHRZARDYIUUUPRLVJPNDDVLHABHBPLYPTFIDBVERA

Output with Quagmire 3: 

BETWEENSUBTLESHADINGANDTHEABSENCEOFLIGHTLIESTHENUANCEOFIQLUSION

Steps to Solve Kryptos K1 (Quagmire III Cipher):

Step 1: Understand the Cipher Basics
 Quagmire III uses two keys:

  • Alphabet Key (used to rearrange the alphabet): “KRYPTOS”
  • Running Key (used repeatedly to decode each letter): “PALIMPSEST”

Step 2: Create a Keyed Alphabet
First, we write down the alphabet (A-Z). But instead of starting with “A”, we use the Alphabet Key first. Like this:

KRYPTOSABCDEFGHIJLMNQUVWXZ

Notice we put letters from “KRYPTOS” first, and then added the remaining letters from the alphabet in order (skipping any we’ve already used). This is the text we get when we look at the ‘tableau’, that sort of spiral of text you see the word “KRYPTOS” in over and over.

Step 3: Use the Running Key
The Running Key (“PALIMPSEST”) repeats itself to match the length of the encrypted message. For example, if your cipher text is 10 letters long, “PALIMPSEST” repeats as “PALIMPSESTPALIMP…” until it matches the cipher’s length exactly.

Step 4: Decoding Each Letter (the math)
 To decode, use this simple subtraction method for each cipher letter:

  • Find the position of the current Running Key letter in your special keyed alphabet. Call this position the “shift number.”
  • Find the position of your cipher (encrypted) letter in that same keyed alphabet.
  • Subtract the “shift number” from the cipher letter’s position. If this number is negative, just add 26 to wrap around.
  • The result gives you the position of the decoded letter in the keyed alphabet.

Here’s an example with the first letter of Kryptos K1:

Cipher Letter: E
Running Key Letter: P

  • Special Alphabet: KRYPTOSABCDEFGHIJLMNQUVWXZ
  • Position of P = 3 (count from 0: K=0, R=1, Y=2, P=3…)
  • Position of E = 11 (K=0, R=1, Y=2, P=3, T=4, O=5, S=6, A=7, B=8, C=9, D=10, E=11…)

Now subtract:
11 (cipher letter E) – 3 (key letter P) = 8

Position 8 in your special alphabet is the decoded letter: B

You repeat this for every letter in your cipher. If you follow these steps carefully for every letter, you’ll end up with the verified, known plaintext solution:

K2:

Encrypted text: 

VFPJUDEEHZWETZYVGWHKKQETGFQJNCE GGWHKK?DQMCPFQZDQMMIAGPFXHQRLG TIMVMZJANQLVKQEDAGDVFRPJUNGEUNA QZGZLECGYUXUEENJTBJLBQCRTBJDFHRR YIZETKZEMVDUFKSJHKFWHKUWQLSZFTI HHDDDUVH?DWKBFUFPWNTDFIYCUQZERE EVLDKFEZMOQQJLTTUGSYQPFEUNLAVIDX FLGGTEZ?FKZBSFDQVGOGIPUFXHHDRKF FHQNTGPUAECNUVPDJMQCLQUMUNEDFQ ELZZVRRGKFFVOEEXBDMVPNFQXEZLGRE DNQFMPNZGLFLPMRJQYALMGNUVPDXVKP DQUMEBEDMHDAFMJGZNUPLGEWJLLAETG

Decrypted text:

ITWASTOTALLYINVISIBLEHOWSTHATPOSSIBLETHEYUSEDTHEEARTHSMAGNETICFIELDXTHEINFORMATIONWASGATHEREDANDTRANSMITTEDUNDERGRUUNDTOANUNKNOWNLOCATIONXDOESLANGLEYKNOWABOUTTHISTHEYSHOULDITSBURIEDOUTTHERESOMEWHEREXWHOKNOWSTHEEXACTLOCATIONONLYWWTHISWASHISLASTMESSAGEXTHIRTYEIGHTDEGREESFIFTYSEVENMINUTESSIXPOINTFIVESECONDSNORTHSEVENTYSEVENDEGREESEIGHTMINUTESFORTYFOURSECONDSWESTXLAYER2

Information given: Tableau, encrypted text, contextual environment clues, clues from K1

Method commonly cited: Double Vigenère with “Kryptos”, “Abscissa”

Actual method: Quagmire3, alphabet key “KRYPTOS” with key “Abscissa”

Output with Double Vigenère: 

QHJAZXDPXJLCWNHKUKLYZTSSUFOHQAKPL... (random, nonsensical text)

Output with Quagmire 3:  

ITWASTOTALLYINVISIBLEHOWSTHATPOSSIBLE....

Steps to solve:

  1. Construct a keyed alphabet by writing the Alphabet Key (“KRYPTOS”) followed by the remaining letters of the alphabet in normal order, skipping duplicates.
  2. Use the Running Key (“ABSCISSA”) repeatedly, expanding it to match the length of the ciphertext.
  3. Decrypt each cipher letter individually by:
    • Finding the position of the corresponding letter from the Running Key in the special keyed alphabet (this position becomes your numeric shift).
    • Finding the cipher letter’s position in the keyed alphabet.
    • Subtracting the numeric shift from the cipher letter’s position (add 26 if negative).
    • The resulting number gives the plaintext letter’s position in the keyed alphabet.
  4. Continue this process for each ciphertext character until the entire message is decoded.

Verifiable solution:

[To run, create a python file, then in the cmd, run that python file, these are as simple as possible so that even someone who does not understand code can go through them easily. Or visit this github page.]

import tkinter as tk
from tkinter import ttk, messagebox

# Build a keyed alphabet (no repeats)
def build_keyed_alphabet(key):
    key = key.upper()
    alphabet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
    keyed_alpha = ''
    for char in key + alphabet:
        if char not in keyed_alpha:
            keyed_alpha += char
    return keyed_alpha

# Quagmire III decryption
def quagmire3_decrypt(cipher, alphabet_key, running_key):
    cipher = cipher.upper().replace(' ', '')
    alphabet = build_keyed_alphabet(alphabet_key)
    running_key = running_key.upper()
    full_key = (running_key * ((len(cipher) // len(running_key)) + 1))[:len(cipher)]
    
    plaintext = ""
    key_index = 0
    for char in cipher:
        if char.isalpha():
            k_char = full_key[key_index]
            key_index += 1
            row = alphabet.find(k_char)
            col = alphabet.find(char)
            decrypted_char = alphabet[(col - row) % 26]
            plaintext += decrypted_char
        else:
            plaintext += char
    return plaintext


# GUI Application
class QuagmireCipherSolver:
    def __init__(self, root):
        self.root = root
        self.root.title("Quagmire III Cipher Solver")

        ttk.Label(root, text="Cipher Text:").grid(row=0, column=0, sticky=tk.W)
        self.cipher_entry = ttk.Entry(root, width=60)
        self.cipher_entry.grid(row=0, column=1, padx=5, pady=5)

        ttk.Label(root, text="Alphabet Key:").grid(row=1, column=0, sticky=tk.W)
        self.alpha_key_entry = ttk.Entry(root, width=60)
        self.alpha_key_entry.grid(row=1, column=1, padx=5, pady=5)

        ttk.Label(root, text="Running Key:").grid(row=2, column=0, sticky=tk.W)
        self.run_key_entry = ttk.Entry(root, width=60)
        self.run_key_entry.grid(row=2, column=1, padx=5, pady=5)

        self.solve_button = ttk.Button(root, text="Solve", command=self.solve_cipher)
        self.solve_button.grid(row=3, column=1, pady=10)

        ttk.Label(root, text="Decrypted Text:").grid(row=4, column=0, sticky=tk.W)
        self.result_text = tk.Text(root, height=15, width=80)
        self.result_text.grid(row=5, column=0, columnspan=2, padx=5, pady=5)

        # Default test values
        self.cipher_entry.insert(0, "EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJYQTQUXQBQVYUVLLTREVJYQTMKYRDMFD")
        self.alpha_key_entry.insert(0, "KRYPTOS")
        self.run_key_entry.insert(0, "PALIMPSEST")

    def solve_cipher(self):
        cipher = self.cipher_entry.get()
        alpha_key = self.alpha_key_entry.get()
        run_key = self.run_key_entry.get()

        if not cipher or not alpha_key or not run_key:
            messagebox.showerror("Input Error", "All fields must be filled in.")
            return

        plaintext = quagmire3_decrypt(cipher, alpha_key, run_key)

        self.result_text.delete('1.0', tk.END)
        self.result_text.insert(tk.END, plaintext)

if __name__ == '__main__':
    root = tk.Tk()
    app = QuagmireCipherSolver(root)
    root.mainloop()