Field Guide

Part 0 simple, five-minute solve

Use KRYPTOS, read the back-side helper letter at the same row and column, apply the one-bit gate from position, and compute P = (C - R) mod 26.

Fast sanity check: position 22 has ciphertext F (5). With gate-adjusted shift R=1, plaintext is (5-1) mod 26 = 4 = E, the first letter of EAST.

Ground rules

A step only belongs in this guide if a solver can execute it on-site using one of three sources: a physical witness visible on the sculpture, a clue witness from the public clue system (Morse slabs, K1–K3 misspellings, Sanborn’s confirmed anchors), or simple arithmetic (writing, counting, and calculator work).

Every step must answer three questions: What do I look at? What exact thing do I do next? and Why does the sculpture or clue tell me to do that?

The answer (so you know where you’re headed)

THE COMPASS ROSE IS HERE X
EAST NORTHEAST
THIS IS YOUR POSITION X
COMMISSION
BERLIN CLOCK
WHICH IS NORTHEAST OF HERE X

The X’s are word separators. The message tells you: the compass rose marks your spot, your orientation is east-northeast, and the Berlin Clock is northeast of where you stand.

The four-step on-site procedure

Standing at the sculpture (or with photographs of both sides), K4 decodes in four steps. The whole point of the May 2026 reframe: there is no algorithm to invent. The keystream is the back of the copper.

Step 1: Use the KRYPTOS keyword

The same keyword used to decrypt K1, K2, and K3. Build the keyed alphabet by writing KRYPTOS first, then the remaining letters of the alphabet in order, skipping any letter already used:

K R Y P T O S A B C D E F G H I J L M N Q U V W X Z

Step 2: Read the keystream off the back of the sculpture

For each K4 cipher position, walk around to the tableau side of the copper screen at the same row and column. The character there is the helper letter T for that position. Read all 97 helper letters in order:

positions  1-4    cap:    W X Z K                              (4 cells)
positions  5-35   Y row:  Y X Z K R Y P T O S A B C D E F G H I J L M N Q U V W X Z K R  (31 cells)
positions 36-66   Z row:  Z Z K R Y P T O S A B C D E F G H I J L M N Q U V W X Z K R Y  (31 cells)
positions 67-97   footer: _ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z A B C D  (31 cells)

That is the entire 97-character physical keystream. The footer is a leading blank, the standard A–Z alphabet, and a partial wrap; the physical reason is that the tableau ends after the Z row and the footer shelf carries a standard alphabet.

Step 3: Apply the one-bit gate

At each position, add a one-bit gate (1 or 0) to the shift. The gate value is fixed by the position: you can compute it from the ciphertext letter and its lane, or from the grid coordinates (col31, tier mod 3, seam-corner), see Part 6C for the exact rule. This is a solved step: the gate is fixed by position and fully verifiable across all 97 cells.

Step 4: Apply Quagmire III subtraction

For each position, compute:

R = column_shift(T) + gate
P = ( stdpos(C) − R ) mod 26

where column_shift(T) is the standard Quagmire III column offset for the helper letter T against the KRYPTOS-keyed tableau, and gate is the one-bit gate value from Step 3. The plaintext letter P appears.

For solvers who do not have physical access, reconstructing K4 from the cipher transcript and public clues alone, the same R values are derivable algorithmically. That deeper-view derivation is documented in Parts 1–10 below.

Deeper view, remote-solver derivation

The four-step procedure above works at the sculpture. For solvers reconstructing K4 remotely from public clues and the cipher transcript alone, the helper-card values, the shift table, and the gate can all be derived algorithmically. The following ten parts document that remote-solver path in full.

Parts 1–5 lay out the physical and visual elements a solver would consult. Parts 6–10 carry the full arithmetic derivation through to the worked examples and the master constants.

This derivation path produces identical R values to the four-step on-site procedure at all 97 positions. The canonical helper-card values published with the May 2026 update appear on the verification page; both paths verify against the same published R-grid.

Rev 16 summary link: the Y master template rule and Y->Z1/Z2 transform reductions are consolidated in Appendix, Y master template.

Part 1: What the solver sees

Before any math starts, orient yourself to the five physical elements that matter.

1. The cipher side

The curved copper surface with cut-out letters. Find OBKR near the bottom-right, this is where K4 starts. Also find YAHR, four slightly raised letters that serve as a directional clue (explained in Part 6).

2. The tableau side

On the reverse of the copper sheet: a large grid of letters, a Vigenère square built on the KRYPTOS alphabet. Find the 26 shifted rows, the bottom three rows labeled X, Y, and Z (your helper rows), and the footer row with its leading blank space.

3. The entrance slab (Morse code)

A stone slab near the entrance carries Morse-coded phrases. The ones that matter for the solve:

• T IS YOUR POSITION, Look up T in the KRYPTOS alphabet: K=0, R=1, Y=2, P=3, T=4. The number 4 is the tail formula multiplier used to compute f values for tail letters: f = (4 × kpos + 1) mod 26. It also highlights T itself (f(T) = 13, the alphabet midpoint).
• RQ. These two letters appear together as a Morse pair. They share the same f value: f(R) = f(Q) = 3. This confirms the 3-family seed from the compass.
• SOS, Brackets the f-value anchor: f(S) = 12, f(O) = 0, f(S) = 12. O = 0 is the hinge/reset point for the entire row-value system. S = 12 is one step below T = 13.
• VIRTUALLY INVISIBLE, Describes the key structure: the f and g values are hidden in plain sight within the sculpture’s clue system, not engraved as obvious numbers.
• DIGETAL (misspelled on purpose). The correct spelling is DIGITAL. The E→I swap follows the misspelling bridge rule: f(I) = f(E) + 19 mod 26. This is the fourth misspelling in the system (joining IQLUSION, UNDERGRUUND, DESPARATLY).

4. The compass rose

A stone compass rose set into the courtyard ground. Its needle points to 67.5° (east-northeast) and 247.5° (the opposite bearing).

A standard 16-point compass divides the full 360° circle into 16 equal sectors. Each sector spans 360° ÷ 16 = 22.5°. The compass point number is the bearing divided by this sector width:

• 67.5° ÷ 22.5° = 3 (the 3rd compass point: east-northeast)
• 247.5° ÷ 22.5° = 11 (the 11th compass point: west-southwest)

These two numbers, 3 and 11, are the family seeds for the row key.

5. The solved sections K1, K2, K3

The first three Kryptos passages contain deliberate misspellings that encode the +19 bridge rule:

• K1: IQLUSION, should be ILLUSION. Q was used where L belongs. The bridge rule says: f(correct) = f(wrong) + 19. So f(L) = f(Q) + 19 = 3 + 19 = 22.
• K2: UNDERGRUUND, should be UNDERGROUND. U was used where O belongs. f(O) = f(U) + 19 = 7 + 19 = 26 mod 26 = 0. This independently confirms O = 0.
• K3: DESPARATLY, should be DESPERATELY. A was used where E belongs. f(E) = f(A) + 19 = 4 + 19 = 23.

The offset 19 is itself on-site: it is the KRYPTOS-alphabet position of N (K=0, R=1, … N=19). N is the “bridge row” that connects the misspelling chain to the row-value system.

Part 2: Where to start

Step 1: Find OBKR

Go to the cipher side. Look for the row ending with ? O B K R (the question mark is a visible character on the sculpture). This is the cap row.

Step 2: Start at O

Put your finger on the O in OBKR. This is position 1, the first letter of K4. The plaintext itself confirms this: “THE COMPASS ROSE IS HERE” begins at O.

Step 3: Number the path

From that O, continue reading cut-out letters in normal reading order (left to right, then down to the next line). Number each letter 1 through 97.

O B K R U O X O G H U L B S O L I F B B W F L R V Q Q P R N G K S S O T W T Q S J Q S S E K Z Z W A T J K L U D I A W I N F B N Y P V T T M Z F P K W G D K Z X T J C D I G K U H U A U E K C A R

The path breaks into four copper rows: positions 1–4 (the cap row tail), 5–35, 36–66, and 67–97 (each 31 letters wide).

Part 3: How to line up the tableau

Use the tableau in same-row, no-sideways-shift alignment. When you look at position 5 on the cipher side, look at the same physical spot on the tableau side, in the same row. Do not slide the tableau left or right.

When aligned correctly, the helper letters come from:

Positions 1–4:   X-row tail  → W X Z K
Positions 5–35:  Full Y row  (31 letters)
Positions 36–66: Full Z row  (31 letters, "Z1 pass")
Positions 67–97: Footer row  (31 letters, "Z2 pass" converted into KALPHA state)

There is also one “predecessor slot” just before position 1, the letter V from the X row. Think of it as the doorstep you step over before entering.

Part 4: Building the helper packet

Take a piece of paper and copy out the helper letters for each pass:

Positions 1–4 (X pass)

W  X  Z  K

Positions 5–35 (Y pass)

Y X Z K R Y P T O S A B C D E F G H I J L M N Q U V W X Z K R

That is 31 letters, the full Y row wraps around the seam.

Positions 36–66 (Z1 pass)

Z Z K R Y P T O S A B C D E F G H I J L M N Q U V W X Z K R Y

Positions 67–97 (Z2 pass, footer basis)

_ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z A B C D

The visible letters are the standard alphabet footer, not another read of the Z row. For the control lookup, the footer converts into Z2’s KALPHA-state basis by the carried one-step rule described in Part 5.

Part 5: Why there are two Z passes

K3’s plaintext describes Howard Carter peering through a breach: “LOWER PART OF THE DOORWAY WAS REMOVED… TINY BREACH IN THE UPPER LEFT HAND CORNER… CAN YOU SEE ANYTHING.” K2 mentions information “TRANSMITTED UNDERGROUND TO AN UNKNOWN LOCATION.”

In cipher terms, the lower half of the keystream is covered by two passes. The first pass (Z1, positions 36–66) reads the full Z row off the back of the sculpture. The second pass (Z2, positions 67–97) reads the standard-alphabet footer row beneath it, not a re-read of the Z row. For lookup, that visible footer converts into KALPHA state: effective_Z2_key = KALPHA[(stdpos(letter) - 1) mod 26], with the leading blank carrying Z. The two passes use linked control cards: the Z2 values are the Z1 values plus a set of delta corrections from a 3×3 control card (g_Z2 = g_Z1 + Z2_delta).

Part 6: Building the row value (f) and control value (g)

6A. The row value (f)

Every letter in the KRYPTOS alphabet has a row value called f. For each position, you apply f to the ciphertext letter (C) to determine which row of the tableau to enter.

The f values are built from three on-site seeds:

• Compass seed 3, the compass rose points to 67.5°; on a 16-point compass (360° ÷ 16 = 22.5° per point), that is point 3
• Compass seed 11, the opposite bearing 247.5° ÷ 22.5° = point 11
• Bridge offset 19, the KRYPTOS-alphabet position of N (kpos = 19), encoded in the K1/K2/K3 misspellings

The four letters YAHR (raised on the cipher side) split the alphabet into families: Y and A belong to the 11-side, H and R to the 3-side. The tail formula from “T IS YOUR POSITION” uses kpos(T) = 4 as its multiplier.

How each f value is built, worked examples

Every one of the 26 f values comes from one of five routes, all anchored in the seeds above. Here is one worked example of each route; you can check every value against the full table underneath.

Route 1. The KRYPTOS keyword head (K R Y P T O S). These seven letters carry the seed f values that anchor the whole table: K=8, R=3, Y=16, P=11, T=13, O=0, S=12. Four are pinned directly by clues: O=0 is the hinge confirmed by the UNDERGRUUND misspelling, R=3 and P=11 are the two compass seeds (67.5° and 247.5°), and T=13 is the alphabet midpoint named by “T IS YOUR POSITION.” Every other letter is derived from this head.

Route 2. The tail formula (the seven highest letters N Q U V W X Z, all with kpos ≥ 19). Each uses f = (4 × kpos + 1) mod 26; the multiplier 4 is kpos(T) from “T IS YOUR POSITION.”

Example, X:  kpos(X) = 24
   f(X) = (4 × 24 + 1) mod 26 = 97 mod 26 = 19  ✓

Route 3. The misspelling bridge (+19) (derives L, E, I). Each deliberate misspelling pairs a wrong letter with the correct one; the rule is f(correct) = f(wrong) + 19 mod 26.

Example, L, from K1’s IQLUSION (Q written where L belongs):
   f(L) = f(Q) + 19 = 3 + 19 = 22  ✓     (f(Q) = 3 comes from Route 2)

Route 4, Family chains (+19 steps from a compass seed; derives A, J, D, F, B, H, M). The 11-family steps out from the P=11 seed; the 3-family from the R=3 seed.

Example, A then J, stepping out from the 11-family seed P:
   f(A) = f(P) + 19 = 11 + 19 = 30 mod 26 = 4  ✓
   f(J) = f(A) + 19 =  4 + 19 =          23  ✓

Route 5, Closure rules (the last two letters, C and G). Each is locked by a unique algebraic closure, the only letter in the whole alphabet satisfying its rule. C uses the zero-reset closure f(C) + kpos(C) = 26:

Example, C:  kpos(C) = 9
   f(C) = 26 − 9 = 17  ✓     (C is the only letter where f + kpos ≡ 0 mod 26)

G is locked the same way by a dual closure (it simultaneously equals the difference card and zero-locks with the Z2 reveal card). The full closure-rule derivations for C and G are on the verification page.

6B. The control value (g) and the control card

The control value g depends on which pass you are in (X, Y, Z1, or Z2) and which helper letter you read from the tableau. Each pass has its own g lookup.

The heart of the Z-pass g values is a 3×3 control card. The card has three rows and three columns:

The row names come from K3’s plaintext (“CAN YOU SEE ANYTHING” = LOOK; “I INSERTED THE CANDLE” = PLACE; K2’s “TRANSMITTED UNDERGROUND” = SEND). The column names (EDGE, MIDDLE, OUT) describe the letter’s structural position within each row.

For helper letters not on the control card, the Z1 scaffold default is g = 13 for letters whose kpos mod 8 falls in {0, 2, 3, 4}. Three exceptions (T, D, N) use card values instead.

6C. The one-bit gate

After computing r = (f + g) mod 26, apply the one-bit gate, a +1 or +0 correction. The gate value at each position is fixed and verifiable, and is most directly obtained from the position itself: from the ciphertext letter and its lane, or from the three grid coordinates below. The result is always 0 or 1.

The gate value is fully determined by position via three structural coordinates:

• col31, the position’s column within the 31-character copper row (the packet phase).
• tier mod 3, the carrier cycle induced by the 31/14 wrap beat (31 mod 14 = 3).
• seam_corner, a binary flag for lanes {14, 1, 2}, the seam handoff zone where the 31-column rows wrap into the 14-lane grid.

At most positions the packet phase and carrier cycle determine the gate. At the seam-corner lanes, the override resolves all remaining conflicts exactly. The result is always 0 or 1, no lookup table required.

6D. The final subtraction

Now you have everything for one position. The complete arithmetic:

Step 1: r = ( f(C) + g(T) ) mod 26
Step 2: R = r + gate
Step 3: P = ( stdpos(C) − R ) mod 26

Here C is the ciphertext letter and T is the helper letter from the active tableau row. f is applied to C (the ciphertext); g is applied to T (the helper).

For cipher and plaintext numbers, use the standard alphabet (A=0, B=1, … Z=25). If the result is negative, add 26. The answer is always between 0 and 25.

Part 7: Fully worked examples

One example from each pass, showing every step of the field procedure.

Example 1: Position 1 (X pass)

Example 2: Position 14 (Y pass)

Example 3: Position 48 (Z1 pass)

For this position, D is in the LOOK-MIDDLE class of the Z1 control mapping.

Example 4: Position 82 (Z2 pass)

Part 8: Why every clue matters

The clues are not decoration. They are the instruction manual. Every major clue does a specific job in the solve:

Part 9: Starter worksheet

Copy this layout onto paper. Fill in each row as you solve. The first five positions are pre-filled to get you started:

Column legend

Part 10: The master constants

The entire K4 system is built from eight numbers, all found on-site:

Everything traces back to these.

Cipher classification

K4 is a Quagmire III variant with a physical keystream and a one-bit gate. The keyword is KRYPTOS, the same one used for K1, K2, and K3. The keystream is the tableau side of the copper screen: at each K4 ciphertext position, the helper letter is the character on the back of the sculpture at the same row and column. The KRYPTOS-keyed tableau printed on the sculpture is the lookup square. The sculpture itself is the key.

The computation is the standard Quagmire III shift composition: r = (f(C) + g_state(T)) mod 26, where f is the row-identity card applied to the ciphertext letter and g_state is the column shift for the helper letter under whichever of the four helper cards (g_X, g_Y, g_Z1, g_Z2) covers this position’s 31-cell pass.

The one-bit gate applies as a final additive correction: R = r + gate, with the gate value fixed by position (Part 6C). Then P = (stdpos(C) − R) mod 26 recovers the plaintext at all 97 positions.