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affine cipher

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Section Seven

Fall 2003Chris Christensen

Class notesAffine Ciphers

Cryptography

Combining the two processes of the Caesar cipher and the multiplicative cipher results in a transformation of the form. This is called an affine cipher.

affine (adjective): via French, from Latin ad "to, at," and finis "boundary, border," of unknown prior origin. Although the original meaning of the Latin adjective affinis was "neighboring, adjacent," the most common meaning eventually became "related by marriage." The notion of kinship or affinity was carried into French, and from there to English. In mathematics, the affine transformations include translation [and] stretching [multiplication] .

Affine transformations maintain a "kinship" between the original object and the transformed object. For example, in the plane, a pair of parallel lines is transformed into a pair of parallel lines.

The Number of Possible Keys

Now we have two parts to the key an additive part b (the shift) and a multiplicative part m (the decimation interval). As there are for Caesar ciphers, there are 26 choices for b. For each of those choices, there are (as there are for multiplicative ciphers) 12 possible choices for m. Therefore, there are possible affine ciphers. Of course, one of these does nothing the cipher with b = 0 and m = 1 yields C = P, the plaintext alphabet. This would obviously not be a good choice for encryption.

The 312 affine ciphers include the 26 Caesar ciphers (the affine ciphers with m = 1; C = P + b) and the multiplicative ciphers (the affine ciphers with b = 0, C = mP).

Here is an example of an affine cipher with additive key 5 and multiplicative key 7.

Affine cipher

Multiplicative key = 7 and additive key = 5

a112L

b219S

c326Z

d47G

e514N

f621U

g72B

h89I

i916P

j1023W

k114D

l1211K

m1318R

n1425Y

o156F

p1613M

q1720T

r181A

s198H

t2015O

u2122V

v223C

w2310J

x2417Q

y2524X

z265E

Cryptanalysis

Recognition and Cryptanalysis

Just like the multiplicative cipher, an affine cipher with decimates the alphabet but if there is also a shift. We can recognize an affine cipher from its frequency chart if we can spot a decimation and a shift.

Here is a frequency chart transformed by the affine cipher with additive key 5 and multiplicative key 7.

Frequencies

Additive key = 5 and Multiplicative key = 7

A11111111

B11

C1

D

E

F1111111

G1111

H111111

I1111

J11

K1111

L1111111

M111

N1111111111111

O111111111

P1111111

Q

R111

S1

T

U111

V111

W

X11

Y11111111

Z111

If , unlike the multiplicative cipher, ciphertext Z will not correspond to plaintext z. It is much more difficult to determine the decimation interval. Notice that in this case cipher E corresponds to plaintext z. Counting backwards from E in intervals of 7 we find

E (low)

X (low)

Q (low)

J (low)

C(low)

V(not high)

which from b = 5 and m = 7.

Another Cryptanalysis by Frequency Analysis

If we knew or had reason to suspect that we were dealing with an affine cipher, we might use the following method of cryptanalysis.

Recall that for a Caesar cipher C = P + b we need only one plaintext-ciphertext letter correspondence to determine the shift b and for a multiplicative cipher C = mP we need only one plaintext-ciphertext letter correspondence to determine the multiplier m. On way of cryptanalyzing those ciphers is to assume that the most common ciphertext letter corresponded to the plaintext e. (If that turned out to be an incorrect choice, we assumed that another high frequency plaintext letter corresponded to e, and we continued this process until the correct key was determined.)

For an affine cipher, we need to determine two keys b and m. We need two ciphertext-plaintext correspondences to do that.

Example 7.1

Here is a ciphertext

OINRF HORXH ONAPF VHLHM NZOFU OINAN GRLZI PYNJL HOINM KVBLY GMKVB SFLAG LAALY BNRNY OVHNG SXPO

and here is a frequency analysis of the ciphertext

A11111

B111

C

D

E

F1111

G1111

H111111

I1111

J1

K11

L1111111

M111

N1111111111

O11111111

P111

Q

R1111

S11

T

U1

V1111

W

X11

Y1111

Z11

Let us assume that an affine cipher was used. We need two plaintext-ciphertext correspondences. We might begin by assuming the most frequent ciphertext letter N corresponds to plaintext e and the second most frequent ciphertext letter O corresponds to plaintext t.

This would give us two equations to solve modulo 26.

If ciphertext N (C = 14) corresponds to plaintext e (P = 5), .

If ciphertext O (C = 15) corresponds to plaintext t (P = 20), .

We have a system of equations to solve modulo 26.

Here is a solution. We begin by subtracting the first congruence from the second.

m = 7. Now substitute this into one of the congruences, say the first.

b = 5.

If we were correct in our assumptions that ciphertext letter N corresponds to plaintext letter e and ciphertext letter O corresponds to plaintext letter t, then the affine cipher has additive key 5 and multiplicative key 7.

Use the key given above to decipher the ciphertext.

Brute Force

Although it would not be pleasant to do by hand (however, you might be willing to do that if the security of the free world depended upon it), it would not be hard for a computer to print out the 312 plaintext messages which result in this ciphertext. Then, we just need to find the one that makes sense.

Another Approach

We could search through the ciphertext for an enciphered version of the. Here are the 312 affine ciphers of the.

Enciphered the

Multiplicative key

Additive key

ACJ1513

ACR2315

ADJ2125

AEF1725

AGB1911

AOL17

AQH319

ASD55

AUZ717

AWV93

AXP2522

AYR1115

BDK1514

BDS2316

BEK210

BFG170

BHC1912

BPM18

BRI320

BTE56

BVA718

BXW94

BYQ2523

BZS1116

CAT1117

CEL1515

CET2317

CFL211

CGH171

CID1913

CQN19

CSJ321

CUF57

CWB719

CYX95

CZR2524

DAS2525

DBU1118

DFM1516

DFU2318

DGM212

DHI172

DJE1914

DRO110

DTK322

DVG58

DXC720

DZY96

EAZ97

EBT250

ECU1119

EGN1517

EGU2319

EHN213

EIJ173

EKF1915

ESP111

EUL323

EWH59

EYD721

FBA98

FCU251

FDW1120

FHO1518

FHW2320

FIO214

FJK174

FLG1916

FTQ112

FVM324

FXI510

FZE722

GAF723

GCB99

GDV252

GEX1121

GIP1519

GIX2321

GJP215

GKL175

GMH1917

GUR113

GWN325

GYI511

HBG724

HDC910

HEW253

HFY1122

HJQ1520

HJY2322

HKQ216

HLM176

HNI1918

HUS114

HXO30

HZK512

IAL513

ICH725

IED911

IFX254

IGZ1123

IKR1521

IKZ2323

ILR217

IMN177

IOJ1919

IWT115

IYP31

JBM514

JDI70

JFE912

JGY255

JHA1124

JLA2324

JLS1522

JMS218

JNO178

JPK1920

JXU116

JZQ32

KAR33

KCN515

KEJ71

KGF913

KHZ256

KIB1125

KMB2325

KMT1523

KNT219

KOP179

KQL1921

KYV117

LBS34

LDO516

LFK72

LHG914

LIA257

LJC110

LNC230

LNU1524

LOU2110

LPQ1710

LRM1922

LZW118

MAX119

MCT35

MEP517

MGL73

MIH915

MJB258

MKD111

MOD231

MOV1525

MPV2111

MQR1711

MSN1923

NBY120

NDU36

NFQ518

NHM74

NJI916

NKC259

NLE112

NPE232

NPW150

NQW2112

NRS1712

NTO1924

OCZ121

OEU37

OGR519

OIN75

OKJ917

OLD2510

OMF113

OQF233

OQX151

ORX2113

OST1713

OUP1925

PDA122

PFW38

PHS520

PJO76

PLK918

PME2511

PNG114

PRG234

PRY152

PSY2114

PTU1714

PVQ190

QEB123

QGX39

QIT521

QKP77

QML919

QNF2512

QOH115

QSH235

QSZ153

QTZ2115

QUV1715

QWR191

RFC124

RHY310

RJU522

RLQ78

RNM920

ROG2513

RPI116

RTA154

RTI236

RUA2116

RVW1716

RXS192

SGD125

SIZ311

SKV523

SMR79

SON921

SPH2514

SQJ117

SUB155

SUJ237

SVB2117

SWX1717

SYT193

THE10

TJA312

TLW524

TNS710

TPO922

TQI2515

TRK118

TVC156

TVK238

TWC2118

TXY1718

TZU194

UAV195

UIF11

UKB313

UMX525

UOT711

UQP923

URJ2516

USL119

UWD157

UWL239

UXD2119

UYZ1719

VBW196

VJG12

VLC314

VNY50

VPU712

VRQ924

VSK2517

VTM1110

VXE158

VXM2310

VYE2120

VZA1720

WAB1721

WCX197

WKH13

WMD315

WOZ51

WQV713

WSR925

WTL2518

WUN1111

WYF159

WYN2311

WZF2121

XAG2122

XBC1722

XDY198

XLI14

XNE316

XPA52

XRW714

XTS90

XUM2519

XVO1112

XZG1510

XZO2312

YAH1511

YAP2313

YBH2123

YCD1723

YEZ199

YMJ15

YOF317

YQB53

YSX715

YUT91

YVN2520

YWP1113

ZBI1512

ZBQ2314

ZCI2124

ZDE1724

ZFA1910

ZNK16

ZPG318

ZRC54

ZTY716

ZVU92

ZWO2521

ZXQ1114

It could be an unpleasant experience to compare one-by-one all of the trigraphs of the ciphertext against the table, but we luck out. The very first trigraph OIN appears in the table as the enciphered version of the when the multiplicative key is 7 and the additive key is 5.

Exercises

1. Construct a plaintext-ciphertext correspondence for an affine cipher with multiplicative key 11 and additive key 16.

2. Encipher the following message using an affine cipher with multiplicative key 11 and additive key 16.

The Russians and Germans also solved PURPLE.

3. Use frequency analysis to cryptanalyze the following ciphertext:

EJURB IOJMR XGEMH HUBXW TWJZM QEJUR BISUS BWQEI QVGZW NBCIV GZMJU YWUSB JWMQS WHWEB JIZGE MHHUB IOWZW JMBWM BXGJN YWUSB JWMQ

4. Search through the ciphertext in exercise 3 for an enciphered version of the. Then determine the multiplicative and additive keys and recover the plaintext.

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