intro to crypto and mod n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Intro to Crypto and Mod PowerPoint Presentation
Download Presentation
Intro to Crypto and Mod

Loading in 2 Seconds...

play fullscreen
1 / 31

Intro to Crypto and Mod - PowerPoint PPT Presentation


  • 56 Views
  • Uploaded on

Intro to Crypto and Mod. Supplementary Notes. Prepared by Raymond Wong. Presented by Raymond Wong. e.g.1 (Page 4). E.g., m = 21 n = 9 21 can be expressed as 9 x 2 + 3 (i.e., nq + r). r is defined to be 21 mod 9. q = 2 r = 3. 21 mod 9 is equal to 3.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Intro to Crypto and Mod' - maude


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
intro to crypto and mod

Intro to Crypto and Mod

Supplementary Notes

Prepared by Raymond Wong

Presented by Raymond Wong

e g 1 page 4
e.g.1 (Page 4)
  • E.g., m = 21 n = 9 21 can be expressed as 9 x 2 + 3 (i.e., nq + r)

r is defined to be 21 mod 9

q = 2

r = 3

21 mod 9 is equal to 3

0  r < n

e g 2 page 9
e.g.2 (Page 9)
  • Illustration of [(-m) mod n] = n – [m mod n]
  • E.g., m = 4 n = 5E.g., m = 9 n = 5

4 mod 5 = 4

-4 mod 5

= 5 – (4 mod 5)

= 5 – 4 = 1

9 mod 5 = 4

-9 mod 5

= 5 – (9 mod 5)

= 5 – 4 = 1

e g 3 page 10
e.g.3 (Page 10)

2 x 8 mod 9 = 16 mod 9 = 7

(2 mod 9) (8 mod 9) = 2 x 8 = 16

Conclusion: 2 x 8 mod 9  (2 mod 9) (8 mod 9)

(2 + 8) mod 9 = 1

(2 mod 9) + (8 mod 9) = 2 + 8 = 10

Conclusion: (2 + 8) mod 9  (2 mod 9) + (8 mod 9)

e g 4 page 11
e.g.4 (Page 11)
  • Illustration of Lemma 2.2
  • E.g., i = 1 n = 5

1 mod 5 = 1

(1 + (-1)x5) mod 5 = -4 mod 5 = 1

(1 + 5) mod 5 = 6 mod 5 = 1

(1 + (-2)x5) mod 5 = -9 mod 5 = 1

(1 + 2x5) mod 5 = 11 mod 5 = 1

(1 + (-3)x5) mod 5 = -14 mod 5 = 1

(1 + 3x5) mod 5 = 16 mod 5 = 1

(1 + k.5) mod 5 = 1

e g 5 page 11
e.g.5 (Page 11)
  • Illustration of Lemma 2.2
  • E.g., i = 11 n = 5

11 mod 5 = 1

(11 + (-1)x5) mod 5 = 6 mod 5 = 1

(11 + 5) mod 5 = 16 mod 5 = 1

(11 + 2x5) mod 5 = 21 mod 5 = 1

(11 + (-2)x5) mod 5 = 1 mod 5 = 1

(11 + 3x5) mod 5 = 26 mod 5 = 1

(11 + (-3)x5) mod 5 = -4 mod 5 = 1

(11 + k.5) mod 5 = 1

e g 6 page 11
e.g.6 (Page 11)
  • Prove that 11 mod 5 = (11 + k.5) mod 5for all integers k

Let r = 11 mod 5

By Euclid’s Division Theorem, we can write 11 = 5q + r

where q and r are two unique integers and 0  r < 5

Consider 11 + k.5

= (5q + r) + k.5

= 5q + r + k.5

= 5q + k.5 + r

= 5(q + k) + r

By the definition of Euclid’s division theorem, (11 + k.5) mod 5 = r

e g 7 page 12
e.g.7 (Page 12)
  • Illustration of Lemma 2.3
  • E.g.,

(2 + 8) mod 9 = (2 + (8 mod 9)) mod 9

= ((2 mod 9) + 8) mod 9

= ((2 mod 9) + (8 mod 9)) mod 9

(2.8) mod 9 = (2 . (8 mod 9)) mod 9

= ((2 mod 9) . 8) mod 9

= ((2 mod 9) . (8 mod 9)) mod 9

e g 8 page 12
e.g.8 (Page 12)

2 x 8 mod 9 = 16 mod 9 = 7

((2 mod 9) (8 mod 9)) mod 9 = 2 x 8 mod 9 = 16 mod 9 = 7

Conclusion: 2 x 8 mod 9 = ((2 mod 9) (8 mod 9)) mod 9

(2 + 8) mod 9 = 1

((2 mod 9) + (8 mod 9)) mod 9 = (2 + 8) mod 9 = 10 mod 9 = 1

Conclusion: (2 + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9

e g 8
e.g.8

Claim: (20 + 17) mod 9 = ((20 mod 9) + (17 mod 9)) mod 9

(2 + 8) mod 9 = 1

((2 mod 9) + (8 mod 9)) mod 9 = (2 + 8) mod 9 = 10 mod 9 = 1

Conclusion: (2 + 8) mod 9 = ((2 mod 9) + (8 mod 9)) mod 9

e g 81
e.g.8

Lemma 2.2 (Y + 9k) mod 9 = Y mod 9

Claim: (20 + 17) mod 9 = ((20 mod 9) + (17 mod 9)) mod 9

Why is it correct?

By Euclid’s Division Theorem, we can write 20 as follows. 20 = 9q1 + r1 where q1 and r1 are some unique integers.

= 9q1 + (20 mod 9)

By Euclid’s Division Theorem, we can write 17 as follows. 17 = 9q2 + r2 where q2 and r2 are some unique integers.

= 9q2 + (17 mod 9)

Consider (20 + 17) mod 9

= {[9q1 + (20 mod 9)] + [9q2 + (17 mod 9)]} mod 9

= {9q1 + (20 mod 9) + 9q2 + (17 mod 9)} mod 9

= [ (20 mod 9) + (17 mod 9) + 9q1 + 9q2] mod 9

= [ (20 mod 9) + (17 mod 9) + 9(q1 + q2)] mod 9

= [ (20 mod 9) + (17 mod 9)] mod 9

(by Lemma 2.2)

e g 9 page 14
e.g.9 (Page 14)
  • E.g. 0 +5 2 = 2
  • E.g., 1.52 = 2

(0 + 2) mod 5 = 2 mod 5 = 2

(1.2) mod 5 = 2 mod 5 = 2

e g 10 page 15
e.g.10 (Page 15)
  • Illustration of Theorem 2.4
    • Commutative law
    • Associative law
    • Distributive law

2 x9 8 = 8 x9 2

2 +9 8 = 8 +9 2

2 x9 (8 x9 1) = (2 x9 8) x9 1

2 +9 (8 +9 1) = (2 +9 8) +9 1

2 x9 (8 +9 1) = (2 x9 8) +9 (2 x9 1)

e g 11 page 24

encryption

decryption

I love UST.

kfjEfklje$3

e.g.11 (Page 24)

password

password

I love UST.

e g 11

encryption

decryption

I love UST.

kfjEfklje$3

e.g.11

keroro

keroro

I love UST.

e g 111

encryption

decryption

I love UST.

kfjEfklje$3

kfjEfklje$3

e.g.11

keroro

keroro

I love UST.

Undecipherable

(cannot be decrypted easily)

attacker

sender

receiver

keroro

keroro

e g 112

I love UST.

e.g.11

encryption

0

I love UST.

slide18

J mpwf VTU.

z

b

c

d

e

f

g

h

i

j

k

a

m

o

p

q

r

s

t

u

v

w

x

y

l

n

encryption

1

I love UST.

slide19

K nqxg WUV.

a

c

d

e

f

g

h

i

j

k

l

b

n

p

q

r

s

t

u

v

w

x

y

z

m

o

encryption

2

I love UST.

slide20

L oryh XVW.

b

d

e

f

g

h

i

j

k

l

m

c

o

q

r

s

t

u

v

w

x

y

z

a

n

p

encryption

3

I love UST.

e g 12 page 32

11

e.g.12 (Page 32)

Encryption function

= 5 .12 x

Multiplication modn

encryption

a = 5

n = 12

Is there any

division modn ?

7

Encrypted value

= 5 .12 7

= 5 . 7 mod 12

= 35 mod 12

= 11

e g 13 page 32
e.g.13 (Page 32)
  • Examples that an inverse function exists

T

S

a

1

2

b

f

3

c

d

4

S

T

A function!

a

1

2

b

f-1

3

c

d

4

e g 14 page 32
e.g.14 (Page 32)
  • Examples that an inverse function does not exist

T

S

a

1

2

b

f

3

c

4

S

T

Not a function!

a

1

2

b

No such f-1

3

c

4

e g 15 page 34

Encrypted

Encrypted

0

6

0

3

e.g.15 (Page 34)

Case (a): a = 4, n = 12

f4, 12 (x) = 4.x mod 12

The inverse of f4, 12 does not exist

S

T

0

0

1

1

The receiver cannot determine the original number.

2

2

3

3

4

4

f4, 12

5

5

6

6

sender

receiver

7

7

8

8

9

9

10

10

sender

receiver

11

11

e g 16 page 35

Encrypted

Encrypted

6

6

6

2

e.g.16 (Page 35)

Case (b): a = 3, n = 12

f3, 12 (x) = 3.x mod 12

The inverse of f3, 12 does not exist

S

T

0

0

1

1

The receiver cannot determine the original number.

2

2

3

3

4

4

f3, 12

5

5

6

6

sender

receiver

7

7

8

8

9

9

10

10

sender

receiver

11

11

e g 17 page 36

Encrypted

Encrypted

5

1

11

7

e.g.17 (Page 36)

Case (c): a = 5, n = 12

f5, 12 (x) = 5.x mod 12

The inverse of f5, 12 exists

S

T

0

0

1

1

The receiver can uniquely determine the original number.

2

2

3

3

4

4

f5, 12

5

5

6

6

sender

receiver

7

7

8

8

9

9

10

10

sender

receiver

11

11

e g 18 page 41

Encrypted

Encrypted

Encrypted

1

5

5

1

5

e.g.18 (Page 41)
  • Private-key cryptosystems

Encrypt this message with this private-key

Decrypt this message with the same private-key

sender

receiver

private-key (e.g., 2)

private-key (e.g., 2)

It should be kept privately at the sender’s side.

It should be kept privately at the receiver’s side.

e g 19 page 41

Encrypted

Encrypted

Encrypted

5

1

5

5

1

e.g.19 (Page 41)
  • Public-key cryptosystems

Encrypt this message with a public-key

Decrypt this message with a secret-key

It has some relationships with the public key.

sender

receiver

public key (e.g., 2)

secret-key (e.g., 4)

It can be kept publicly.

It should be kept privately at the receiver’s side.

e g 19 page 411

Public Key Directory

Encrypted

Encrypted

Encrypted

5

1

5

5

1

e.g.19 (Page 41)

This directory is accessible to the public.

Raymond

2

Peter

7

  • Public-key cryptosystems

Encrypt this message with a public-key

Decrypt this message with a secret-key

Peter

Raymond

sender

receiver

public key (e.g., 2)

secret-key (e.g., 4)

It can be kept publicly.

It should be kept privately at the receiver’s side.

e g 20 page 44
e.g.20 (Page 44)
  • An ideal key pair (public key and secret key)
    • Given a public key,
      • it is difficult for the adversary to deduce the secret key
e g 20

Encrypted

Encrypted

Encrypted

338

167

338

167

338

e.g.20
  • Public-key cryptosystems

rev (1000 – M) = rev (1000 – 167)

= rev (833)

= 338

1000 – rev(C) = 1000 – rev (338)

= 1000 – 833

= 167

sender

receiver

public key

secret-key

It is not secure. Why?