An introduction by Michael Nüsken

An introduction by Michael Nüsken

Universität Paderborn FB17 Mathematik & Informatik AG Algorithmische Mathematik

Mathematics for James Bond & Co.or

How to conceal Her Majesty‘s secret?

or

FB17 - Mathematik & Informatik

Michael NüskenApril 19, 2023

Cryptography:Mathematics (not only) for

James Bond & Co.

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KALLEKOKALOLLOLEKOKALOLLOLEKOKALOLLOLE

Secret talking




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Secret talking

JAMES

BOND

JOJAMOMESOSBOBONONDOD

JOJAMOMESOSBOBONONDOD




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Secret talking

• Encrypt: difficult to learn• Decrypt: easy• Break: easy, even without knowledge




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Caesar‘s Encryption

D

ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZ




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DE

ABCDEFGHIJKLMNOPQRSTUVWXYZ




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DEEFGHIJKLMNOPQRSTUVWXYZ




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DEFGHIJKLMNOPQRSTUVWXYZABC




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D


EFGHIJKLMNOPQRSTUVWXYZABC

JAMES

Encrypt:




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D



JAMES

Encrypt:

M




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D



JAMES

Encrypt:

MD




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D



JAMES

Encrypt:

MDP




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D



JAMES

Encrypt:

MDPH




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D



JAMES

Encrypt:

MDPHV




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D



JAMES

Encrypt:

MDPHV

VHFUHWSECRET

Decrypt:




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D



Ring




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Caesar‘s Basic Situation

I want to write toCleopatra ...




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But Brutus must not known what ...




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?




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!





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Now I can write to

Cleopatra and Brutus

will not know

what ...

I hope Caesar‘s writing soon!

Ha ha ha!If they knew how easy ...




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Modern Times‘ Cryptography

Giovanni Batista Porta (1563)De Furtivis Literarum Notis

Encryption methods are

• refined,

• mechanized and

• remain symmetric.

Enigma (invented 1918)Cipher machine of the German Forces


• refined,

• mechanized


• refined,




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An example: words as keys

Modern Times‘ Cryptography

CROOK

TOHERMAJESTYTHEQUEEN

CROOK

VFVSBORXSCVPHVOSLSSX

The secret

keyword

CROOKCROOKThe original message




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One-Time-Pad

• Random sequence instead of words as key.

• Absolutely secure!• Provably.• Therein unique.• Problem: large key

lengths.

Coat hanger of a Stasi spywith hidden One-Time-Pad

(From: Spiegel Spezial 1/1990)




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Modern Times‘ Situation

Symmetric keys!




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Why letters? With numbers we could

calculate ...

Mathematics 1

Numbers instead of letters





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Mathematics 1



Advantage: Numbers can be

• added 17 + 10 = 27

• multiplied 10 · 5 = 50


0 5 10 151-1-5 20 25 Z

There you are ...




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Mathematics 1

Hum, we rolled up letters ...




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0 5 10 151-1-5

We begin with the number

axis.

Mathematics 1

„Rolling up“:




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0 5 10 151-1-5

-5

-4

-3

-2-1 0 1

2

3

4

5

6

7

8

9

10

11121314

15

16

17

18

19

20

21

22

23

2425 26

275051 52

Mathematics 1

„Rolling up“:

53

Z26

This is a ring!




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35

50 = 24

Mathematics 1

-5

-4

-3

-2-1 0 1

2

3

4

5

6

7

8

9

10

11121314

15

16

17

18

19

20

21

22

23

2425 26 275051 52 53

Z 26

In the ring Z26 we can:

!Calculating in rings is easy!

• Multiply 10 · 5 = ?

• Add 17 + 10 = ?

27 = 1

• Add 17 + 10 = 1

• Multiply 10 · 5 = 24




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Mathematics 1

-5

-4

-3

-2-1 0 1

2

3

4

5

6

7

8

9

10

11121314

15

16

17

18

19

20

21

22

23

2425 26 275051 52 53

Z 26

In the ring Z26 we can:

Mathematicians call this:

Calculating „modulo 26“.

• Add 17 + 10 = 1

• Multiply 10 · 5 = 24




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Cryptography until 1950

Bletchley Park Manson

Alan M. Turing (1912-1954)

All broken!

Enigma

Turing Bombe

NCR Bombe (Dayton,USA)




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Cryptography Today

• Euro cheque cards, ATMs




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Cryptography Today

• Euro cheque cards, ATMs• Interbank money transfer




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Cryptography Today

• Euro cheque cards, ATMs• Interbank money transfer• Satellite communication, PayTV




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Cryptography Today

• Euro cheque cards, ATMs• Interbank money transfer• Satellite communication, PayTV• Telephone, mobile phone




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Cryptography Today

• Euro cheque cards, ATMs• Interbank money transfer• Satellite communication, PayTV• Telephone, mobile phone• Internet shopping




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Cryptography Today

• Euro cheque cards, ATMs• Interbank money transfer• Satellite communication, PayTV• Telephone, mobile phone• Internet shopping• and much more ...




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Cryptography Today

A lot of the methods applied here use symmetric keys!

Is this necessary?

• Euro cheque cards, ATMs• Interbank money transfer• Satellite communication, PayTV• Telephone, mobile phone• Internet shopping• and much more ...




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Is Symmetry Inevitable?

Let‘s reconsider the situation: Where is the

symmetry?




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Let‘s reconsider the situation:

Hey, the situation is indeed not symmetric!




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Answer of modern cryptography:

No, there is another

way!

• 1970-74: the British Secret Service CESG

• 1976: Diffie & Hellman

• 1978: Rivest, Shamir & Adleman:

This answers has already been given by




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Encryption is easy.Breaking is difficult.

Using a one-way function:

cipher textsmessages

Decryption is difficult!???Decryption is easy given the trapdoor!

Using a one-way function with trapdoor:We can do without symmetry?! And how?




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1 times234

Exponentiation is repeated multiplication:

Z35

0 1 23

4

5

6

7

8

9

10

11

12

13

1415

1617181920

21

22

23

24

25

26

27

28

29

30

31

3233

34

56 ·4

Oops, now it goes round!

RSA . . . ?Mathematics 2




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Mathematics 2


Z35

0 1 23

4

5

6

7

8

9

10

11

12

13

1415

1617181920

21

22

23

24

25

26

27

28

29

30

31

3233

34

·46 times

·51 times23456

Repeated „·5“ also

goes round!




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Mathematics 2


·46 times

6 times ·5

·21 times2345678910 times1112Z35

0 1 23

4

5

6

7

8

9

10

11

12

13

1415

1617181920

21

22

23

24

25

26

27

28

29

30

31

3233

34

I notice: After 24 steps it

always goes round in Z35.




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Mathematics 2

Z35

0 1 23

4

5

6

7

8

9

10

11

12

1314

151617181920

2122

23

24

25

26

27

28

29

3031

3233 34

Sure, Bond. Mathematic

s tells us:

Sorry? Q, please explain!?

Since 35 = 5 · 7, exponentiation in the ring Z35 has repetition frequency L = 4 · 6 = 24.




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Hello?

RSA by Example




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RSA by Example

Moneypenny, listen! I absolutely need the

secret information on Fort Knox. But

Goldfinger must not get it.




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RSA by Example

Yes sure, but yesterday Q has

declared all secret keys invalid ...




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RSA by Example

Yes, I know! Listen, Q has forseen this and has explained RSA to

me:Tell me first the length

of the text.




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RSA by Example

Hum, OK, but you take the

responsibility.It is only one letter.




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RSA by Example

Hum, now I need two prime numbers, say: p = 5 and q = 7. Then

the ring‘s size is N = p · q = 35 which is large

enough and therepetition frequency isL = (p-1) · (q-1) = 4 · 6

= 24.




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RSA by Example

Further I need two numbers e and d, such

thate·d = 1 + multiple of L.

I have L = 24 and e should be random, say e

= 5. And now d = 5 should be fine, yes,

indeed: 5·5 = 1 + 24.




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p = 5, q = 7.N = 35, L = 24.e = 5, d = 5.

RSA by Example

OK, I‘ve got it. My public key is N = 35 and e = 5. So

please multiply the number x of the letter in the ring Z35 five times with itself and tell me y = x^e, that is in our case y =

x^5 = x·x·x·x·x.




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N = 35, e = 5.

RSA by Example

You are expecting a lot of me, Bond! But I calculate ...

p = 5, q = 7.N = 35, L = 24.e = 5, d = 5.




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p = 5, q = 7.N = 35, L = 24.e = 5, d = 5.

N = 35, e = 5.

RSA by Example

Ha, Bond certainly thinks I cannot do that!

Ha! Fine:N = 35, e = 5.

The letter is an E and x = 4 is the corresponding

number in the ring. Thus

y = x^e = 4·4·4·4·4 = 9.




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N = 35, e= 5.x = 4:y = x^e = 4·4·4·4·4 = 9.

N = 35, e = 5.

RSA by Example

OK, y = 9.

p = 5, q = 7.N = 35, L = 24.e = 5, d = 5.




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N = 35, e= 5.x = 4:y = x^e = 4·4·4·4·4 = 9.

p = 5, q = 7.N = 35, L = 24.e = 5, d = 5.

y = 9.

N = 35, e = 5.

RSA by Example

Thanks, Moneypenny. You have been a great

help.




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N = 35, e= 5.x = 4:y = x^e = 4·4·4·4·4 = 9.

p = 5, q = 7.N = 35, L = 24.e = 5, d = 5.

y = 9.

N = 35, e = 5.

RSA by Example

Now, let‘s see: N = 35, d = 5 and y = 9. Thus the result

is z = y^d = 9·9·9·9·9 = 4.

Hence the letter is an E!




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N = 35, e= 5.x = 4:y = x^e = 4·4·4·4·4 = 9.

p = 5, q = 7.N = 35, L = 24.e = 5, d = 5.

y = 9:z = y^d = 9·9·9·9·9 = 4.

y = 9.

N = 35, e = 5.

RSA by Example




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p, q prime.N = p·q,L = (p-1)·(q-1).e·d = 1 + L·t.

In the ring ZN:z = y^d.message = z.

In the ring ZN:x = message,y = x^e.

RSA in General

y

N, e




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RSA Works

y

N, eYes, Bond. Mathematics

tells us:

z = y^d = x^(e·d) = x^(1+L·t) = x.

Q, does this always work?




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Attacking RSA

y

N, e

Dmnt ...What the hell can I do with these

numbers?

Well, decomposing N is hard ...




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y^d y

RSA Trapdoor Function

Encryption is easy: N and e suffice.Breaking is hard.

Here, N and e define the RSA trapdoor function:

Decryption is hard!???Decryption is easy given the trapdoor d!

And where is the trapdoor function?

cipher textsy

messagesx

x x^e in the ring ZN




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Basic Situation in RSA

BLAST!

Two keys!

A public keyfor encryption

A secret key for decryption (the

trapdoor)




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Modern Cryptography

• Modern symmetric methods– common keys– high security– very fast

• RSA-like methods (Public-Key)– no shared keys– very high security– not that fast

• Real world: Both strategies combined• SigG: e.g. RSA for signatures




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Summary

• Symmetric methods– Disadvantage: symmetric keys

• Rings– Calculating in rings– Exponentiation– Repetition frequency

• RSA– Trapdoor function– Advantage: public keys




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Hints

• RSA: Do it yourself– Three tasks for James Bond ...– We meet in D/E3.301.

• Visual Cryptography– ...




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Hints

• RSA: Do it yourself– Three tasks for James Bond ...– We meet in D/E3.301.

• Visual Cryptography– You will be picked up ...

• Game addition chains– How can exponentiation be

fast?




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Time Schedule

• 1000 D2Welcome

• 1100

RSA: Bond‘s first task

• Ca.1300

Lunch• 1400

RSA: Two further tasks

• 1530 D2Farewell

• In parallel:• Visual

Cryptography• Game addition

chains




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RSA In General

y

N, e




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Do it yourself

• We are at your assistance:– Jamshid Shokrollahi– Martin Otto– Olaf Müller– Preda Mihăilescu– Tanja Lange– Michael Nöcker– Michael Nüsken– Prof. Joachim von zur Gathen

• OK, then: Help Bond with his first task … The BMW Z8 with built-in laptop awaits you.

An introduction by Michael Nüsken

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