Experiment 105

“Affine Hill ciphers of size 2 to 5, solved from every known letter, not only whole blocks”

0015 and 0044 solved Hill matrices from whole crib blocks and used the partial blocks only as checks, which left the 4x4 affine case open (it needs five whole blocks). Written as decryption, p = B c + w, every known plaintext letter gives one linear equation in one row of B, because the carved block is always known. So each row of an n x n affine Hill is solved from all the known letters in that row position, partial blocks included, for n = 2 to 5.

revertedstatus
a new rootparents
2 mintime, all phases
0candidates
0/24known letters
unclocked
1

The experiment method

What the method does with the 24 known letters.

A Hill ciphera Hill cipher A cipher that turns letters into numbers, takes them two, three or four at a time, and mixes each group with a fixed table of numbers. cuts the messagethe message The 97 letters of English that Jim Sanborn wrote and enciphered. into blocks of a few letters and mixes each block with a small table of numbers. Experiments 15 and 44 solved the tablethe table The right-hand half of the sculpture, 26 rows of the alphabet each shifted one place further, starting from the word KRYPTOS. from blocks where all letters are known. Turning the method around, every single known letter gives one equation, even in a block that is only partly known. With that, even the 4-letter version with an extra shift, which had too few whole blocks before, can be solved exactly.

Words in the margin are defined on the glossary page.

1.1

The method at work

The carved letters go in at the top; the reading comes out at the bottom.

OBKRUOXOGHULBSOLIFBBWEASTNORTHEASTOTWTQSJQSSEKZZWATJKLUDIAWINFBBERLINCLOCKWGDKZXTJCDIGKUHUAUEKCARIN: THE COPPER, ALL 97 LETTERS, THE 24 KNOWN PLACES IN ACCENTa 4x4 Hill table with a shift: the copper in blocks of 4; each known letter is one equation for one row of the tableOBKRUOXOGHULBSOLIFBBWEASTNORTHEASTOTWTQSJQSSEKZZWATJKLUDIAWINFBBERLINCLOCKWGDKZXTJCDIGKUHUAUEKCARplace 22 is the second letter of block 21-24, WFLR = [22, 5, 11, 17]: row 2 times those numbers, plus its shift, must give E = 4ochre: the 7 known letters in a second position, 7 equations for row 2's five unknowns; the other rows get theirs the same waysettings by table size, every alignment of the blocks and four letter numberingstableK4 alivedeadopencoppers alivecoppers open2x2 plain0800-00-02x2 affine0800-00-03x3 plain01200-00-03x3 affine01200-00-04x4 plain01600-00-04x4 affine01600-00-05x5 plain02000-10-05x5 affine06140-010-20open: too few equations to decideOUT: NO READING; NO HILL TABLE OF SIZE 2 TO 5 FITSEASTNORTHEASTBERLINCLOCK
the cipher

A Hill cipher mixes each block of a few letters with a small table of numbers, sometimes plus a shift. Earlier runs solved it only from blocks whose letters were all known, too few for the 4-letter version with a shift.

the method

Every single known letter gives one equation for one row of the table, even in a partly known block. Solved mod 2 and mod 13, each row either has a solution or does not.

the bound

Plain and affine tables of size 2 to 4 are dead at every alignment and numbering. At size 5 the plain table is dead; the affine one is mostly open, too few equations, as on the scrambled coppers.

2

Best candidate against the known letters

This experiment produced no candidate: the search ruled everything out before a full text was built. Its product is the bounded negative in section 7.

3

Walkthrough, by hand

Redo the method on a few letters with pencil and paper.

  1. Cut the copper into blocks of 4 from place 1: OBKR, UOXO, and so on; place 22 is the second letter of block 6 (places 21-24, WFLR).
  2. Number the letters, A = 0 ... Z = 25.
  3. Place 22 holds E = 4, the second letter of its block, so row 2 of the table times the block's four numbers, plus shift 2, equals 4 mod 26.
  4. Every known letter in a block's second position gives such an equation for row 2: five unknowns, about six equations.
  5. Solve each row; if all rows have solutions and the table can be undone, the setting fits.
the worksheet: the first steps by hand, on squared paper
4X4 AFFINE HILL, BLOCKS FROM PLACE 1, LETTERS A = 0: EQUATIONS FOR ROW 2 (THE SECOND LETTER OF EACH BLOCK)block placescarved blocknumbersknown second letter21-24WFLR22 5 11 17E = 425-28VQQP21 16 16 15N = 1329-32RNGK17 13 6 10H = 733-36SSOT18 18 14 19T = 1965-68YPVT24 15 21 19R = 1769-72TMZF19 12 25 5C = 273-76PKWG15 10 22 6K = 10Row 2 has five unknowns: four tableentries and the shift. Each line isone equation: a x the first number +b x the second + c x the third + d xthe fourth + shift = the knownletter, mod 26.Seven equations, five unknowns: twochecks.
4

Worked example, generated

The same steps run by the code on the real letters.

4x4 affine Hill, blocks from place 1, letters A = 0: equations for row 2 (the second letter of each block)

block places  carved block  numbers          known second letter
  21-24              WFLR  [22, 5, 11, 17]  E = 4
  25-28              VQQP  [21, 16, 16, 15] N = 13
  29-32              RNGK  [17, 13, 6, 10]  H = 7
  33-36              SSOT  [18, 18, 14, 19] T = 19
  65-68              YPVT  [24, 15, 21, 19] R = 17
  69-72              TMZF  [19, 12, 25, 5]  C = 2
  73-76              PKWG  [15, 10, 22, 6]  K = 10

row 2 has five unknowns (four table entries and a shift); each line above is one equation
5

Hypothesis, for a cryptanalyst

Family, key space, the constraint under test, and the clues it reads.

Family 8 open item 'the 4x4 affine' (families.md); the HILL column of the tableau (C21, Bauer 2016-17). Model: blocks of n carved letters c (n = 2-5), message block p = B c + w mod 26, B invertible mod 26, w any vector (w = 0 is plain Hill). Row r of (B, w) has n + 1 unknowns; each block whose r-th message letter is known gives one equation. Solved exactly mod 2 and mod 13 (all solutions enumerated when a row is underdetermined, up to a cap) and joined by the Chinese remainder theorem; a setting survives if every row has a solution and some choice makes B invertible. Block alignments 0 to n-1 (the last partial block left plain); numberings: standard or KRYPTOS order, 0- or 1-based. Ten shuffled coppers.

6

Prediction

Written before the run; the result decides it.

No setting with at least one check survives for n = 2-4, plain or affine; n = 5 affine has no checks (6 unknowns per row, about 5 equations) and is reported as untestable. Zero candidates.

7

Method

  1. For each n, alignment and numbering, group the known letters by block and row position.
  2. For each row, solve the equations mod 2 and mod 13.
  3. Join the rows, require B invertible mod 26, and count the spare equations (checks).
  4. Decrypt survivors.
  5. Repeat on ten shuffled coppers.
8

Result and bounds

The harness's record: candidates, gates, loss, and the negative result this experiment leaves behind.

0candidates returned
0passed every gate
0/24known letters, best
–lowest loss
nonechampion before

2x2 plain: K4 {'alive': 0, 'dead': 8, 'open': 0}, controls alive [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]; 2x2 affine: K4 {'alive': 0, 'dead': 8, 'open': 0}, controls alive [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]; 3x3 plain: K4 {'alive': 0, 'dead': 12, 'open': 0}, controls alive [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]; 3x3 affine: K4 {'alive': 0, 'dead': 12, 'open': 0}, controls alive [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]; 4x4 plain: K4 {'alive': 0, 'dead': 16, 'open': 0}, controls alive [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]; 4x4 affine: K4 {'alive': 0, 'dead': 16, 'open': 0}, controls alive [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]; 5x5 plain: K4 {'alive': 0, 'dead': 20, 'open': 0}, controls alive [0, 0, 0, 0, 0, 0, 0, 0, 0, 1]; 5x5 affine: K4 {'alive': 0, 'dead': 6, 'open': 14}, controls alive [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]. K4 alive with checks: 0.

What was searched, as the harness recorded it.

family
affine and plain Hill, sizes 2-5, every alignment, four numberings, solved row by row from every known letter
by size
2x2 plain
k4
alive
0
dead
8
open
0
controls alive
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
controls open
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
2x2 affine
k4
alive
0
dead
8
open
0
controls alive
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
controls open
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
3x3 plain
k4
alive
0
dead
12
open
0
controls alive
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
controls open
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
3x3 affine
k4
alive
0
dead
12
open
0
controls alive
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
controls open
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
4x4 plain
k4
alive
0
dead
16
open
0
controls alive
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
controls open
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
4x4 affine
k4
alive
0
dead
16
open
0
controls alive
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
controls open
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
5x5 plain
k4
alive
0
dead
20
open
0
controls alive
0, 0, 0, 0, 0, 0, 0, 0, 0, 1
controls open
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
5x5 affine
k4
alive
0
dead
6
open
14
controls alive
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
controls open
12, 16, 10, 12, 20, 16, 10, 14, 12, 20
k4 alive
none
controls
10 shuffled coppers (seed 105), identical test

How this experiment sits among the others is on the lineage page; its family's open hypotheses are on the families page.

Rendered from the record on 2026-10-18 12:26 UTC; updated after every experiment.
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