Secret Key: stream ciphers & block ciphers

Stream CiphersIdea: try to simulate one-time pad• define a secret key (“seed”)• Using the seed generates a byte stream

(Keystream): i-th byte is function– only of the key (synchronous Stream cypher)

or– Of the key and first i-1 bytes of ciphertext

(asynchronous Stream cypher).• obtain ciphertext by using XOR of

textplain and keystream (bi-wise)

=

⊕

Synchronous Stream Cipher

Key

Ciphertext

Stream

Plaintext

encryption

Synchronous Stream Cipher

=

⊕

Key

Plaintext

Stream

Ciphertext

decryption

Cipher Streams in practice

• Many codes before 1940• Enigma - II world war (Germany)• A5 – GSM (encryption cell phone-

base station)• WEP - used in ethernet 802.11

(wireless)• RC-4 (Ron’s Code)

Example: RC-4

• RC: code proposed by Ron (Ron’s Code,Ron=Ronald Rivest)

• Considered safe: 1987 - 1994 kept secret,after ‘94 extensively studied

• Good for exporting (complain USrestrictions)

• Easy to program, fast• Very popular: Lotus Notes, SSL, Wep etc.

RC4: properties

• variable key lenght (byte)• synchronous• starting from the key it generates aapparently random permutation:

•Eventually the sequence will repeat•However long period > 10100 [in this way itsimulates one-time-pad]

• very fast: 1 byte of output requires 8-16instruction

RC-4 initialization1. j=02. S0=0, S1=1, …, S255=2553. Assume a key of 255 (bytes) k0,…,k255 (if the

key is shorter repeat)4. For i=0 to 255

j = (j + Si+ ki) mod 256exchange Si and Sj

In this way we obtain a permutation of 0, 1,…,255, the resulting permutation is afunction of the key

RC-4 Key-stream generation

Input: i,j, permutation of 0,1,…2551. i = (i+1) mod 2562. j = (j +Si) mod 2563. exchange Si and Sj

4. t = (Si + Sj) mod 2565. B = St

Output: BRecall: ciphertext EXOR of 1 byte of

plaintext and 1 byte of ciphertext

Real World Block Ciphers

• DES, 3-DES - (64 bit block, 56 bit key)• AES (Rijndael) (128-256 block)• RC-2• RC-5• IDEA ((64 bit block, 128 bit key)• Blowfish, Cast• Gost

ECB Mode Encryption(Electronic Code Book)

P1

Ek

C1

P2

Ek

C2

P3

Ek

C3

encrypt each plaintext block separately

Properties of ECB

• Simple and efficient• Parallel implementation possible• Does not conceal plaintext patterns• Active attacks are possible (plaintext can be easily manipulated by removing, repeating, or interchanging blocks).

ECB: plaintext repetitionsplaintext ciphertext ECB good cyphertext

CBC (Cipher Block Chaining) mode

m1

Ek

C1

m2

Ek

C2

m3

Ek

C3

seed

Previous ciphertext is XORed with current plaintextbefore encrypting current block.• Seed is used to start the process; it can be sent without encryption• Seed =0 safe in most but NOT all cases (eg assume the file withsalaries is sent once a month, with the same seed we can detectchanges in the salaries) therefore a random seed is better

CBC (Cipher Block Chaining): decryption

m1

Dk

C1

m2

Dk

C2

m3

Dk

C3

seed

ProblemIF a transmission error changes one bit of C(i-1) -THEN block mi changes in a predicatble wayBUT there are unpredictable changes in m (i-1);Solution: always use error detecting codes (for example CRC) to checkquality of transmissione

Properties of CBC

• Asynchronous stream cipher• Errors in one ciphertext block propagate• Conceals plaintext patterns• No parallel implementation known• Plaintext cannot be easily manipulated.• Standard in most systems: SSL, IPSec

etc.

OFB Mode(Output FeedBack)

An initialization vector s0 is use as a``seed'’ for a sequence of data blocks si

s0= seed

OFB modeDiscussion• If f is public (known to the adversary) then initial seed s0

must be encrypted (why?)• If f is a cryptographic funnction that depends on a secret

key then initial sees can be sent in the clear (why?)• Initial seed must be modified for EVERY new message -

even if is protected and unknown to the adversary (in factif the adv knows a pair message, initial seed then he canencode every message - why?)

• Extension: it can be modified in such a way that only k bitsare used to compute the ciphertext (k-OFB)

Properties of OFB

• Synchronous stream cipher• Errors in ciphertext do not propagate• Pre-processing is possible• Conceals plaintext patterns• No parallel implementation known• Active attacks by manipulating plaintext

are possible

CTR (Counter Mode)

seed

Ek

C1

seed +1

Ek

C2

seed +2

Ek

C3

Similar to OFB•There are problems in repeated use of same seed (like OFB)•CTR vs OFB: using CTR you can decrypt the message starting fromblock i for any i (i.e. You do not need to decrypt from the first blockas in OFB)

m1 m2 m3

AES Proposed Modes

• CTR (Counter) mode (OFB modification):Parallel implementation, offline pre-processing, provable security, simple andefficient

• OCB (Offset Codebook) mode - parallelimplementation, offline preprocessing,provable security (under specificassumptions), authenticity

Strengthening a GivenCipher

• Design multiple key lengths – AES• Whitening - the DESX idea• Iterated ciphers – Triple DES (3-

DES), triple IDEA and so on

Triple Cipher - DiagramP

Ek1

C

Ek2

Ek3

Iterated Ciphers

• Plaintext undergoes encryption repeatedlyby underlying cipher

• Ideally, each stage uses a different key• In practice triple cipher is usually

C= Ek1(Ek2(Ek1(P))) [EEE mode] orC= Ek1(Dk2(Ek1(P))) [EDE mode]EDE is more common in practice

Two or Three Keys

• Sometimes only two keys are used in 3-DES• Identical key must be at beginning and end• Legal advantage (export license) due to

smaller overall key size• Used as a KEK in the BPI protocol which

secures the DOCSIS cable modemstandard

Adverary’s goal

• Final goal: find the secret key• Partial goals:

– Reduee the no of possible keys– Detect patterns in the text– Decode part of the text– Modify the cipertext obtaining a plausible text

(even without breaking the cipher; evenwithout knowing which modifications)

Repeated coding

• To increase the robustness performmultiple encryption. How many times?2,3, 678?

• In practice triple cipherC= Ek1(Ek2(Ek1(P))) [EEE mode ] orC= Ek1(Dk2(Ek1(P))) [EDE mode]EDE more used

Double DES: man in the middleattack

Cipher twice with two different keys? NOMan in th emiddle attack. Requirements

– Known plaintext and ciphertext– 2k+1 encryption and decryption (2 keys of k bit)– |k|2|k| memory space– Idea: try all possible encryptions of the plaintext and all

possible decryption of the ciphertext. Check for a pairof keys that transform the palintext in the ciphertext.

– Note: the method can be applied to all block codes

Triple encodingP

Ek1

C

Ek2

Ek3

Triple encoding and CBCM1

Ek1

Ek2

M2

Ek1

C1

Mn

Ek1

0000000

... . . . .....

Ek3

Ek2

C2

Ek3

In the picture: External CBC: code (using triple encoding) eachblock ; then concatenate

Other possibility: Internal CBC (the concatenation depends onthe level of encoding)

Group and Fields

AESAdvanced Encryption Standard

Review - Groups

Def (group): A set G with a binary operation +(addition) is called a commutative group if

1 ∀ a,b∈G, a+b∈G2 ∀ a,b,c∈G, (a+b)+c=a+(b+c)3 ∀ a,b∈G, a+b=b+a4 ∃ 0∈G, ∀ a∈G, a+0=a5 ∀ a∈G, ∃ -a∈G, a+(-a)=0

+,0, and -aare only notations!

Sub-groups

• Let (G, +) be a group, (H,+) is a sub-groupof (G,+) if it is a group, and H⊆G.

• Claim: Let (G, +) be a finite group, and H⊆G. If H is closed under +, then (H,+) is asub-group of (G,+).

• Examples• Lagrange theorem: if G is finite and (H,+)

is a sub-group of (G,+) then |H| divides |G|

Order of Elements

• Let an denote a+…+a (n times)• We say that a is of order n if an = 0, and

for any m<n, am≠0• Examples• Euler theorem: In the multiplicative group

of Zm, every element is of order at most φ(m).

Cyclic Groups• Claim: let G be a group and a be an

element of order n. The set <a>={1, a,…,an-1} is a sub-group of G.• a is called the generator of <a>.• If G is generated by a, then G is

called cyclic, and a is called aprimitive element of G.

• Theorem: for any prime p, themultiplicative group of Zp is cyclic

GroupZ set of integers (positive and negative) ;Zn integer modulo n (0,,2,3,…,n-1); Z*n =(1,2,3,…,n-1);- Z and addition (0 identity; -a inverse of a) is a group- Zn and addition addizione is a group (0 identity; -a inverse of a)- Zn and multiplication is NOT a group (inverse exist only for 1 and -1)- Set of rational numbers and multiplication is a group- Z*n [a mod n] and multiplication IS NOT ALWAYS a group

- n=6 then {1,2,3,4,5} is not close (2*3= 0 mod 6)- n prime then it is a group

- Zn* [a mod n] and multiplication if MCD(a,n) = 1 is a group ( 1 is identity- And if as + nt = 1 mod n then s is inverse of a

- n =15 then {1,2,4,7,8,11,13,14}- n=5 {1,2,3,4} (in fact all numbers are prim ewith 5)

Review - RingsDef (ring): A set F with two binaryoperations + (addition) and · (multiplication) is called a commutative ring with identity if

6 ∀ a,b∈F, a·b∈F7 ∀ a,b,c∈F, (a·b)·c=a·(b·c)8 ∀ a,b∈F, a·b=b·a9 ∃ 1∈F, ∀ a∈F, a·1=a10 ∀ a,b,c∈F,a·(b+c)=a·b+a·c

1 ∀ a,b∈F, a+b∈F2 ∀ a,b,c∈F, (a+b)+c=a+(b+c)3 ∀ a,b∈F, a+b=b+a4 ∃ 0∈F, ∀ a∈F, a+0=a5 ∀ a∈F, ∃ -a∈F, a+(-a)=0

+,·,0, 1 and-a are only notations!

Review - FieldsDef (field): A set F with two binaryoperations + (addition) and · (multiplication) is called a field if

6 ∀ a,b∈F, a·b∈F7 ∀ a,b,c∈F, (a·b)·c=a·(b·c)8 ∀ a,b∈F, a·b=b·a9 ∃ 1∈F, ∀ a∈F, a·1=a10 ∀ a,b,c∈F,a·(b+c)=a·b+a·c

1 ∀ a,b∈F, a+b∈F2 ∀ a,b,c∈F, (a+b)+c=a+(b+c)3 ∀ a,b∈F, a+b=b+a4 ∃ 0∈F, ∀ a∈F, a+0=a5 ∀ a∈F, ∃ -a∈F, a+(-a)=0

11 ∀ a≠0∈F, ∃ a-1∈F, a·a-1=1

+,·,0, 1,-a and a-1 are

only notations!

Review - Fields

A field is a commutative ring with identity where eachnon-zero element has a multiplicative inverse

∀ a≠0∈F, ∃ a-1∈F, a·a-1=1

Equivalently, (F,+) is a commutative (additive) group,and (F \ {0}, ·) is a commutative (multiplicative) group.

ExamplesZn with addition and multiplication is a ring but not always a field• n=15 NO ( {1,2,3,4,….,15} is not a group with resepct ot

multiplication)• n=5 Yes ( {1,2,3,4} is a gropu w.r.t. multiplication)

Polynomials over Fields

Let f(x)= an·xn + an-1·xn-1 + an-2·xn-2 + … + a1·x + a0be a polynomial of degree n in one variable x over a fieldF (namely an, an-1,…, a1, a0 ∈ F).

Theorem: The equation f(x)=0 has at most n solutions in F.

Remark: The theorem does not hold over rings with identity. For example, in Z24 the equation 6·x = 0 has six solutions (0,4,8,12,16,20).

Polynomial RemaindersLet f(x)= an·xn + an-1·xn-1 + an-2·xn-2 + … + a1·x + a0

g(x)= bm·xm + bm-1·xm-1 + bm-2·xm-2 + … + b1·x + b0be two polynomials over F such that m < n (or m=n).

Theorem: There is a unique polynomial r(x) of degree < mover F such that f(x) = h(x) · g(x) + r(x).

Remark: r(x) is called the remainder of f(x) modulo g(x).

Finite FieldsDef (finite field): A field (F,+,·) is called a finite field if the

set F is finite.

Example: Zp denotes {0,1,...,p-1}. We define + and · as additionand multiplication modulo p, respectively.

One can prove that (Zp,+,·) is a field iff p is prime.

Q.: Are there any finite fields except (Zp,+,·) ?

Galois Fields GF(pk)

Évariste Galois (1811-1832)

Theorem: For every prime power pk (k=1,2,…) there is aunique finite field containing pk elements. These fields aredenoted by GF(pk).There are no finite fields with other cardinalities.

Polynomials over Finite FieldsPolynomial equations and factorizations in finitefields can be different than over the rationals.

Examples from an XMAPLE session:

Irreducible PolynomialsA polynomial is irreducible in GF(p) if it does not factor overGF(p). Otherwise it is reducible.

Examples:

The same polynomial is reducible in Z5 but irreducible in Z2.

Implementing GF(p^k) arithmetic

Theorem: Let f(x) be an irreducible polynomialof degree k over Zp.

The finite field GF(pk) can be realized as the set

of degree k-1 polynomials over Zp, with additionand multiplication done modulo f(x).

Example: Implementing GF(2^k)

By the theorem the finite field GF(25) can be realized as

the set of degree 4 polynomials over Z2, with additionand multiplication done modulo the irreducible polynomialf(x)=x5+x4+x3+x+1.

The coefficients of polynomials over Z2 are 0 or 1.So a degree k polynomial can be written down by k+1 bits.For example, with k=4:

x3+x+1 (0,1,0,1,1)

x4+ x3+x+1 (1,1,0,1,1)

Implementing GF(2^k)

Addition: bit-wise XOR (since 1+1=0)

x3+x+1 (0,1,0,1,1) + x4+ x3+x (1,1,0,1,0)------------------------------- x4 +1 (1,0,0,0,1)

Multiplication: Polynomial multiplication, and then remainder modulo the defining polynomial f(x):

Implementing GF(2^k)

For small size finite field, a lookup table is the most efficientmethod for implementing multiplication.

(1,1,0,1,1) *(0,1,0,1,1)

= (1,1,0,0,1)

Implementing GF(25) in XMAPLE

Irreducible polynomial

More GF(25) Operations in XMAPLEAddition: b+c

test primitive element

e <--inverse of a Multiplication: a*e

Loop forfinding primitiveelements

Back to Symmetric BlockCiphers

out in

DES AES

Historic NoteDES (data encryption standard) is a symmetric block cipherusing 64 bit blocks and a 56 bit key.

Developed at IBM, approved by the US goverment (1976)as a standard. Size of key (56 bits) was apparently small enough to allow the NSA (US national security agency) tobreak it exhaustively even back in 70’s.

In the 90’s it became clear that DES is too weak for contemporary hardware & algorithmics. (Best attack, Matsui“linear attack”, requires only 243 known plaintext/ciphertextpairs.)

Historic Note (cont.)The US government NIST (national inst. of standards and technology) announced a call for an advanced encryption standard in 1997.

This was an international open competition.Overall, 15 proposals were made and evaluated, and 6 were finalists. Out of those, a proposal namedRijndael, by Daemen and Rijmen (two Belgians) was chosen in February 2001.

AES - Advanced Encryption Standard

• Symmetric block cipher• Key lengthes: 128, 192, or 256 bits• Approved US standard (2001)

AES Design Rationale

• Resistance to all known attacks.

• Speed and code compactness.

• Simplicity.

AES Specifications• Input & output block length: 128 bits.

• State: 128 bits, arranged in a 4-by-4 matrix of bytes.

A3,3A3,2A3,1A3,0

A2,3A2,2A2,1A2,0

A1,3A1,2A1,1A1,0

A0,3A0,2A0,1A0,0 Each byte is viewedas an element in GF(28)

Input/Output: A0,0, A1,0, A2,0, A3,0, A0,1,…

AES Specifications• Key length: 128, 196, 256 bits.

Cipher Key Layout: n = 128, 196, 256 bits, arranged in a 4-by-n/32 matrix of bytes.

K3,3

K2,3

K1,3

K0,3

K3,4

K2,4

K1,4

K0,4

K3,5K3,2K3,1K3,0

K2,5K2,2K2,1K2,0

K1,5K1,2K1,1K1,0

K0,5K0,2K0,1K0,0

Initial layout: K0,0, K1,0, K2,0, K3,0, K0,1,…

AES Specifications

• High level code:• AES(State,Key)

– KeyExpansion(Key,ExpandKey)– AddRoundKey(State,ExpandKey[0])– For (i=1; i<R; i++)

Round(State,ExpandKey[i]);– FinalRound(State,ExpandKey[R]);

Encryption: Carried out in rounds

input block (128 bits)

output block (128 bits)

Secret key (128 bits)

Rounds in AES128 bits AES uses 10 rounds, no shortcutsknown for 6 rounds• The secret key is expanded from 128 bits to 10 round keys, 128 bits each.• Each round changes the state, then XORS the round key. (For longer keys, addOne round for every extra 32 bits)

Each rounds complicates things a little. Overall it seems infeasible to invert without the secret key (but easy given the key).

AES Specifications: One Round

A3,3A3,2A3,1A3,0

A2,3A2,2A2,1A2,0

A1,3A1,2A1,1A1,0

A0,3A0,2A0,1A0,0

Transform the state by applying:

1. Substitution.2. Shift rows3. Mix columns

4. XOR round key

Substitution operates on every Byteseparately: Ai,j <-- Ai,j

-1 (multiplicative inverse in GF(28)which is highly non linear.)

Substitution (S-Box)

If Ai,j =0, don’t change Ai,j .

Clearly, the substitution is invertible.

Cyclic Shift of Rows

A3,0A3,3A3,2A3,1

A2,1A2,0A2,3A2,2

A1,2A1,1A1,0A1,3

A0,3A0,2A0,1A0,0 no shift shift 1 position shift 2 positions shift 3 positions

Clearly, the shift is invertible.

Mixing Columns Every state column is considered as a Polynomial over GF(28)

Multiply with an invertible polynomial03 x3 + 01x2 + 01x + 02 (mod x4 + 1)Inv = 0B x3 + 0D x2 +09 x + 0E

Round: Subbytes(State) ShiftRows(State) MixColumns(State) AddRoundKey(State,ExpandedKey[i])

Key Expansion

• Generate a “different key” per round• Need a 4 x 4 matrix of values (over

GF(28)) per round• Based upon a non-linear transformation of

the original key.• Details available:• The Design of Rijndael, Joan Daemen and

Vincent Rijmen, Springer

Breaking AESBreaking 1 or 2 rounds is easy.

It is not known how to break 5 rounds.

Breaking the full 10 rounds AES efficiently (say 1 year on existing hardware, or in less than 2128 operations) is considered impossible ! (a good, tough challenge…)

Exercises1. Evaluate error propagation in CBC e OFB:

• Show how an adevrsary can modify a block as he/sheprefers assuming that the remaining part of themessage is modified

• Discuss the security of this and techniques foravoiding such attacks

2. CBC and OFB use and initial seed that must be known toboth the sender and the receiver• Assume that the initial seed is sent in the clear (so

it is known to the adversary). Show how theadversary is able to modify part of the messahe.Conclusion: either the initial seed is fixed inadvance or it muts be encypted and sent before thmessage

• Break OFB if you use the same key and the sameinitial seed mote than once