1

EP Revision 3/3/16

2

CODER (Encryption)

DECODER (Decryption)

Destroy enemy

Destroy enemy

17,4,23,12,8,………….

…. 34 22 34 12 . .

Destroy SadMan

I am confused

3

Data that can be read without any special measures is called plaintext. The method of disguising plaintext is called encryption. Encrypting plaintext results in ciphertext. The process of reverting ciphertext back to its original plaintext is called decryption. In conventional cryptography, (I.e. symmetric-key encryption), one key is used for both encryption and decryption. This project uses the Hill Cipher algorithm which was Developed by the mathematician Lester Hill in 1929.

Conventional cryptography

Destroy SADMAN

4,5,19,20,18,15,25,27,191,4,3,1,1

Destroy SADMAN

4

HILL CIPHER ALGORITHM

This encryption algorithm takes successive plaintext letters, pi, and substitutes for them m ciphertext letters, ci. The substitution is determined by m linear equations in which each character is assigned a numerical value (I.e. a = 0, b = 1, c = 2, .... , z = 25 26 possible values). For m = 3 , the system can be described as:

(k11 p1 + k12 p2 + k13 p3) = c1 (k21 p1 + k22 p2 + k23 p3) = c2 (k31 p1 + k32 p2 + k33 p3) = c3

This linear system can be expressed in matrix format:

=

3

2

1

3

2

1

333231

232221

131211

ccc

ppp

kkkkkkkkk

P1x3 K3x3 = C1x3 whereby K3x3 is called the key

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Encryption/decryption project

WRITE A C++ PROGRAM THAT WILL ENCRYPT AND DECRYPT A MESSAGE ENTERED AT THE KEYBOARD WITH WHITE SPACES ALLOWED. THE ENCRYPTION ALGORITHM TO ENCODE THE SENTENCE IS AS FOLLOWS (TRANSMITTER/CODER): 1) EACH LETTER IS FIRST CONVERTED TO A NUMBER: i.e. A = 0, B = 1, C = 2, …………., Z = 25, SPACE = 26 2) ARRANGE THE SEQUENCE OF NUMBERS INTO 3 x 1 MATRICES (i.e. COLUMN MATRICES P1, P2, P3, …….) 3) MULTIPLY EACH 3 x 1 MATRIX ABOVE WITH A 3 x 3 MATRIX, K, CALLED THE KEY. THE KEY MATRIX MUST BE INVERTIBLE

,......,, 332211 CKPCKPCKP ===

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4) THE ENCODED MATRICES ARE TRANSMITTED IN THE SEQUENCE (E1: ROW 1, ROW 2, ROW 3, E2: ROW 1, ROW 2, ROW 3, E3: ROW 1, ROW 2, ROW 3, ……) THE DECRYPTION ALGORITHM TO DECODE THE SENTENCE FOLLOWS (RECEIVER/DECODER): 1) ARRANGE THE SEQUENCE OF NUMBERS BACK INTO 3 x 1 MATRICES (i.e. COLUMN MATRICES P1, P2, P3, …….) 2) MULTIPLY EACH 3 x 1 MATRIX ABOVE WITH A 3 x 3 MATRIX, K-1, THE INVERSE OF THE KEY AS SHOWN: 3) THE RESULTING 3 x 1 DECODED MATRICES REPRESENT THE ORIGINAL 3 x 1 MATRICES W1, W2, W3 ……

,......,, 331

221

111 PCKPCKPCK === −−−

Encryption/decryption project

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4) CREATE THE SEQUENCE: (E1: ROW 1, ROW 2, ROW 3, E2: ROW 1, ROW 2, ROW 3, E3: ROW 1, ROW 2, ROW 3, ……) 5) EACH NUMBER IN THE SEQUENCE IS CONVERTED BACK TO A CHARACTER USING THE CONVERSION TABLE: i.e. A = 0, B = 1, C = 2, …………., Z = 25, SPACE = 26

FOR THIS PROJECT, THE KEY MUST BE GENERATED RANDOMLY, BE AN INVERTIBLE MATRIX AND IS TRANSMITTED ALONG WITH THE ENCRYPTED MESSAGE. MESSAGE TRANSMISSION: THE ENCRYPTED MESSAGE AND THE KEY ARE PLACED INTO A FILE. MESSAGE RECEPTION: THE FILE IS OPENED SO TO DECODE ITS CONTENT (KEY + MESSAGE).

Encryption/decryption project

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Letters are converted to numbers A = 0, b = 1, c = 2, …………., z = 25, space = 26 Create 3 x 1 column matrices p1, p2, p3, … Generate a random 3 x 3 matrix k (k-1 must exist) Generate encoded matrices kpi = ci Transmit encoded matrices and key

Algorithm

Encryption:

Decryption:

Receive encoded matrices ci Receive key k Calculate inverse k-1 Re-create the original words k-1 ci = pi Each number is converted back to letters

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Matrices are denoted by upper case letters (i.e. A) Each Entry of the matrix is called an Element Each element is given a position. aij. i = row#, j = column# The size of a matrix is given by n x m (n = Number of rows, m = Number of columns) The number of elements in a matrix is (n) . (m) Rectangular matrix: n != m Square matrix: n = m A column vector is an n x 1 matrix A row vector is an 1 x m matrix Two matrices are equal if all corresponding elements are equal An identity matrix ( I ) is a square matrix for which the diagonal elements are equal to 1 and all other elements are equal to zero Two square matrices A, B are inverse of each other if AB = BA = I The inverse of a matrix An is a matrix Bn such that An Bn = Bn An = In. We define A as A-1. As such A A-1 = A-1 A = I

Matrix Algebra Basics

10

A of inverse thebe tosaid isA then I A A AA if

R|A|A whereby oft determinan theas defined is |A| det(A)

tionMultiplicaScalar

size) same theof b and(A n SubtractioMatrix

size) same theof b and(A Addition Matrix

and size same theof are BA, if

1-1- ==

∈=

→∀=

→∀−=−

→∀+=+

∀==

ijijmxn

ijijijmxnmxn

ijmxnmxn

ijijijmxnmxn

KaKA

baBA

baBA

baBA

ijij

Matrix Algebra Basics

11

( ) ( ) ( )

[ ] Adjoint)(

Inverse)(Cofactor)1(

tDeterminan)det(

Transpose

tionMultiplicaMatrix

1

1 1 1

→=

→=

→+−=

→=

→== →=

→

−

= = =∑∑∑

TCAAdj

AAAdjA

MijjiCij

AA

abBTAaA

ba

nxmjiijmxnij

m

i

r

j

n

kkjik

(Amn Bnr = Cmr) I = IT

I is symmetric

Matrix Algebra Basics

12

=

333231

232221

131211

333231

232221

131211

333231

232221

131211

ccccccccc

bbbbbbbbb

xaaaaaaaaa

][][][ rxmrxnnxm =

3 x 3 3 x 3

m = 3 n = 3

n = 3 r = 3

3 x 3

m = 3 r = 3

Matrix Multiplication

13

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a11b11

C11 Matrix Multiplication

14

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a11b11+ a12b21

C11 Matrix Multiplication

15

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a11b11+ a12b21 + a13b31

C11 Matrix Multiplication

16

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a21b12

C22 Matrix Multiplication

17

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a21b12+ a22b22

C22 Matrix Multiplication

18

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a21b12+ a22b22 + a23b32

C22 Matrix Multiplication

19

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a31b13

C33 Matrix Multiplication

20

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a31b13+ a32b23

C33 Matrix Multiplication

21

333231

232221

131211

333231

232221

131211

bbbbbbbbb

xaaaaaaaaa

a31b13+ a32b23 + a33b33 ∑∑∑= = =

m

i

r

j

n

kkjikba

1 1 1

C33 Matrix Multiplication

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Matrix Multiplication #include <iostream> using namespace std; void main() { int i, j, k, count = 0, total = 0; int A[ ] [ 3 ] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; int B[ 3 ] [ 3 ] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; for (i = 0; i < 3; i++) { for (j = 0; j < 3; j++){ for (k = 0; k < 3; k++) total += A[i][k] * B[k][j]; cout << total << ' '; ++count; if( count % 3 == 0) cout << endl; total = 0; } } return; }

∑∑∑= = =

3

1

3

1

3

1i j kkjikba

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Determinant of a 2x2

Example:

11211121121)8()17()3()5(|| =

−

−⇒−=−=A

|A| is called the determinant of A and |A|-1 is called the reciprocal of |A| whereby |A| |A|-1 = 1:

=

31785

A

||1|| 1

AA =−

=

2221

1211

aaaa

A 21211211|| aaaaA −=if then

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Determinant of An with n > 2 Laplacian Expansion

∑=

=

++=

=

3

111

131312121111

333231

232221

131211

jJj CAA

CACACAA

AAAAAAAAA

A

Expand along the first row

3x3 SQUARE MATRIX

*

*

MijjiCij +−= )1(

Cofactors Minor

25 3231

1211

23

333231

232221

131211

AA

AAM

AAAAAAAAA

A

=

=

i= 2 j = 3

3231

121123

AAAA

M =

Finding the Minor

26

=−=

=

+23

3223

333231

232221

131211

)1( MC

AAAAAAAAA

A

3231

1211

AAAA

−

Finding the Cofactor

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EXAMPLE

−−=

434110433

A

Find |A|:

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Determinant of A3

=

333231

232221

131211

aaaaaaaaa

A

Find |A|:

Procedure:

=

32

22

12

31

21

11

333231

232221

131211

aaa

aaa

aaaaaaaaa

A

|A| = a11a22a33 + a12a23a31 + a13a21a32 – a31a22a13 – a32a23a11 – a33a21a12

Another Approach

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EXAMPLE Find |A|:

Procedure:

−−=

434110433

A

−−−−=

313

403

434110433

A

|A| = -12 -12 + 0 -16 +9 – 0 = -31

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[ ]

COFACTORMijjiCij

ADJOINTTCAAdj

INVERSEA

AAdjA

→+−=

→=

→=−

)1(

)(

)(1

MINOR

0≠A

Inverse of An

Matrix of co-factors

Adjoint

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EXAMPLE

−−=

434110433

A

Find A-1:

Adjoint

Matrix of co-factors

Determinant Inverse

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Inverse of A2

|A|-1 Example:

Note the decimal numbers (Accuracy ISSUE)

|A| = (5)(3) – (17)(8) = -121

−

−=

−

−

−

=−

041.014.0066.0025.0

12151783

1A

A A-1 = A-1 A = I If A and A-1 are inverse:

|A| is called the determinant of A and |A|-1 is called the reciprocal of |A|

=

31785

A

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Accuracy Issue Example:

−

−=−

125.0042.0062.0188.01K

=

=

924

9236

P

K

=

⇒=

129171

924

9236

xCPK

=

−

−⇒=−

9.815.24

129171

125.0042.0062.0188.01 xPCK

WTRANSMITTED

WRECEIVED

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Accuracy Issue

Example:

=⇒

=

9.815.24

924

RECEIVEDDTRANSMITTE WW

Accuracy Issue: Because the inverse of a matrix will result with values of decimal format, the decryption will result with some errors in the decoding (i.e. a b) due to the decimal portion. Possible Solutions: Select only determinants with |A| = 1. The inverse is the adjoint divided by |A| (i.e. divide by 1) Round decimal numbers (i.e. if decimal portion ≥ 0.5, round up otherwise round down) Add 0.5 to each element – 2.1 + 0.5 = 2.6 Round to 2 (integer division) 2.6 + 0.5 = 3.1 Round to 3 (integer division)

⇒

924

9.815.24

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EXAMPLE 1

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FULL EXAMPLE

Encrypt/Decrypt Message: DESTROY SADMAN

Use matrix K below for the key:

−−=

434110433

KK is invertible:

−−−

−−=−

0968.00968.01290.0968.09032.01290.0

2258.07742.00323.01K

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FULL EXAMPLE

Encrypt/Decrypt Message: DESTROY SADMAN

1. Assign each letter of the alphabet a number: A = 1, B = 2, C = 3, ……….X = 24, Y = 25, Z = 26, White Space = 27

DESTROY SADMAN 4,5,19,20,18,15,25,27,19,1,4,13,1,12

=

192725

3W

=

1954

1W

=

2712

15W

2. Create 3 x 1 column matrices

=

1341

4W

=

151820

2W

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FULL EXAMPLE

Encrypt/Decrypt Message: DESTROY SADMAN

=

+++++−−

=

−−=

1072449

)761516()1950(

)761512(

1954

434110433

1KW

3. Encode each 3 x 1 column matrix using the key

Do the same for KWi = Ei Where i = 2,3,4,5

4. Transmit 49, 24, 107, ….. E1

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FULL EXAMPLE

Encrypt/Decrypt Message: DESTROY SADMAN

5. Group the incoming numbers into 3 x 1 column vectors

1072449

……….

6. Decode each 3 x 1 column matrix by multiplying by the inverse key

=

−−

=−

1954

1072449

...1290.0

....1290.0

....0323.01

1EK DES

40

EXAMPLE 2 Acknowledgments: 18, 15, 8, 9, 20, 19,15,4,9,1

41

Step One (Encryption): Enter a message:

input =

inputnum =

“example”

example

{7

ASCII a-97

b-98

c-99

d-100

e-101

|-94

101-94=7

{7, 26, 3, 15, 18, 14, 7}

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Step Two (Encryption): inputnum = {7, 26, 3, 15, 18, 14, 7} 0, 0

7

26

3

15

18

14

7

0

0

Generated Key 11

4

-8

2

1

-1

-8

-3

6

43

Step Three (Encryption):

11

4

-8

2

1

-1

-8

-3

6

7

26

3

105

45

-64

encrypted = {105, 45, -64 {105, 45, -64, 89, 36, -54, 77, 28, -56}

e x a m p l e filler

44

Step One (Decryption): encrypted = {105, 45, -64, 89, 36, -54, 77, 28, -56}

3

0

4

-4

2

-5

2

1

3

105

45

-64

7

26

3

Inverse of key matrix

decrypted = {7, 26, 3 , 15, 18, 14, 7, 0, 0}

45

Step Two (Decryption): decrypted = {7, 26, 3, 15, 18, 14, 7, 0, 0}

0? ASCII a-97

b-98

c-99

d-100

e-101

NO?

7+94=101

e Printed on screen: x a m p l e

26+94=120 0?

YES?

Stop printing