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## PowerPoint Slideshow about ' Cryptographic Protocols' - camille-banks

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SKEY

- SKEY relies on one-way function
- Alice enters a random number R to the computer
- Computer computes f(R), f(f(R)), f(f(f(R))), and so on, about 100 times
- Called X1, x2, …, x100
- The computer prints out the list of x1 to x100 to Alice. It also computes x101 and store in DB associated with Alice’s name, and removes x1 to x100 from the system

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SKEY (cont.)

- Alice first enter her name and x100. The computer calculates f(x100) and compares with x101
- Then the computer replaces x101 with x100. Alice also erases x100 from her list

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SKID2

- SKID2 and SKID3 are symmetric identification protocol that uses MAC to provide security
- In SKID2, assume that Alice and Bob share a secret key K

Alice Bob: RA

Bob Alice: RB, HK(RA, RB, IDB)

RA and RB are random numbers generated by Alice and Bob, respectively

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SKID3

- Provide mutual authentication between Alice and Bob

Alice Bob: RA

Bob Alice: RB, HK(RA, RB, IDB)

Alice Bob: HK(RB, IDA)

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Encrypted Key Exchange Protocol

- Alice and Bob share a common password P. Using this protocol, they can authenticate to each other and generate a common session key K

A B: A, EP(K’)

B A: EP(EK’ (K))

A B: EK(RA)

B A: EK(RA, RB)

A B: EK(RB)

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Problems of Online Key Generation

- A shared key has been used for various purposes:
- As authentication token
- As a key for cryptographic operation e.g. symmetric encryption or keyed-hash function.
- However, a number of message passes must be performed in order to generate a new session key.
- The more frequent the messages are passed, the higher chance it can be attacked
- Offline key distribution is preferred.
- It also reduces frequency of secret key update process.

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Rubin’s Approach

- A client shares K with a bank.
- The client generates a token T, where

T = {fifty-dollars-book-Bob’s-store}K

- The client sends T to the bank to authenticate herself to the bank.
- The bank decrypts T to receive the information and verify the client.
- The value of T changes in every transaction depending on purchase details. However, the collision might occur.

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Li et al.’s Approach

- A client and a bank share a long-term secret S and initial token Tinit.
- The client generates a token Tnew and sends it to authenticate herself to the bank, where

Tnew = h(Tcur, S)

3. The bank verifies Tnew from {Tinit, S}.

- Security of the system is based on the length of T and S and security of hash function.

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Kungpisdan et al.’s Approach

- A client and a bank share a long-term secret CCI. Also they shares a secret Y which has to be updated periodically.
- The client generates Yi, i = 1,…,n.

Yi = h(i-bit-shift-of-(CCI,Y))

3. The client sends {Yi, i} to the bank. i is randomly chosen.

- Yi is then used as a key for encryption and MAC.
- As Yi is not sent in order, it is difficult to retrieve both Y and CCI.

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Limited-Use Key Generation

- Alice shares KAB with Bob.
- DK is distributed between them.

KAB, DK

2. Alice generates the set of Ki, i=1,…,m

from KAB and DK

{K1,…,Km}

3. Alice selects two preference keys from

the set of Ki.

Select KMid1and KMid2

SIK, DK

4. Calculate SIK from KMid1and KMid2.

5. Calculate SKj, j=1,…,n, from SIK and DK.

{SK1,…,SKn}

6. To update SKj, Alice returns to step 3.

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Limited-Use Key Generation (cont.)

K1 = h(DK, KAB)

K2 = h(DK, K1)

…

Km = h(DK, Km-1)

KAB, DK

{K1,…,Km}

- Generate a random number r
- KMid1 = middle key of {K1,…,Kr}
- KMid2 = middle key of {K1,…,KMid1}

Select KMid1and KMid2

SIK = h(KMid1, KMid2)

SIK, DK

SK1 = h(SIK, DK)

SK2 = h(SIK, SK1)

…

SKn = h(SIK, SKn-1)

{SK1,…,SKn}

SK1, r

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Session Key Generation

K1 = h(DK, KAB)

K2 = h(DK, K1)

…

Km = h(DK, Km-1)

SK1 = h(SIK, DK)

SK2 = h(SIK, SK1)

…

SKn = h(SIK, SKn-1)

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Secret Splitting

- Sometimes we need to keep our information secret
- You could tell company’s secret to the most trusted employee, but what if he/she defects to the competition?
- Secret Splitting: take a message and divide it up into pieces. Each piece by itself means nothing, but put them together and the message appears

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Secret Splitting – 2 people

- Trent generates a random-bit string R, the same length as the message M.
- Trent XORs M with R to generate S.

M R = S

- Trent gives R to Alice and S to Bob
- To construct the message, Alice and Bob has to XOR their pieces together:

R S = M

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Secret Splitting – 4 people

- Trent generates 3 random strings, R, S, and T, the same length as the message M
- Trent XORs M with the three strings to generate P

M R S T = U

- Trent gives R to Alice, S to Bob, T to Carol, and U to Dave
- Alice, Bob, Carol, and Dave get together and compute

R S T U = M

- What happens if Carol is fired, and Trent is not around?

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Secret Sharing

- What happens if any one of the people who holds secret is not around?
- Threshold scheme: take any message and divide it into n pieces, called shadows or shares, such that any m of them can be used to reconstruct the message
- This is called an (m, n)-threshold scheme

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(m, n)-Threshold Scheme

- Choose a prime p, which is larger then the number of possible shadows and larger than the largest possible secret.
- To share a secret , generate an arbitrary polynomial of degree m-1.
- If you want to create a (3, n)-threshold scheme, generate a quadratic polynomial:

(ax2 + bx + M) mod p

- a and b are chosen randomly. They are kept secret and are discarded after the shadows are handed out. M is the message. p must be made public

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(m, n)-Threshold Scheme (cont.)

- The shadows are obtained by evaluating the polynomial at n different points:

ki = F(xi)

- Any three shadows can be used to create three equations

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(m, n)-Threshold Scheme (cont.)

- For example, M = 11. We want to construct (3, 5)-threshold scheme
- Generate a quadratic equation (a =7, b = 8, chosen randomly)

F(x) = (7x2 + 8x + 11) mod 13

- The five shadows are:
- K1 = F(1) = 7 + 8 + 1 0 (mod 13)
- K2 = F(2) = 28 + 16 + 1 3 (mod 13)
- K3 = F(3) = 63 + 24 + 1 7 (mod 13)
- K4 = F(4) = 112 + 32 + 1 12 (mod 13)
- K5 = F(5) = 175 + 40 + 1 5 (mod 13)

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(m, n)-Threshold Scheme (cont.)

- To reconstruct M from 3 out of the shadows, k2, k3, and k5 solve the set of linear equations:

a * 22 + b * 2 + M 3 (mod 13)

a * 32 + b * 3 + M 7 (mod 13)

a * 52 + b * 5 + M 5 (mod 13)

- The solution is a = 7, b = 8, and M = 11. So, M is recovered.

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Key Escrow

- Alice creates her private/public-key pair. She splits the private key into several public and private pieces
- She sends a public piece and corresponding private piece to each trustee in an encrypted form. She also sends the public key to KDC
- Each trustee perform calculation on the received information to confirm that it is correct. Each trustee stores the private piece somewhere secure and sends the public piece to KDC
- KDC performs the calculation on the public pieces and the public key. If everything is correct, it signs the public key and returns the signed public key to Alice

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Fair DH (5 trustees)

- In basic DH, a group of users share a prime p, and a generator g. Alice’s private key is s, and her public key is t = gs mod p
- Alice chooses five intergers s1, s2, s3, s4, and s5, each less than p-1.

Alice’s private key is

s = (s1 + s2 + s3 + s4 + s5) mod p-1

Alice’s public key is: t = gs mod p

Alice also computes

ti = gsi mod p, for i = 1 to 5.

Alice’s public key shares are ti and private key share are si

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Fair DH(cont.)

2. Alice sends a private key piece and corresponding public key piece to each trustee.

- Send s1 and t1 to trustee 1, and send t to KDC
- Each trustee verifies that

ti = gsi mod p

If so, the trustee signs ti and sends it to KDC. The trustee stores si in a secure place.

- After receiving all five public pieces, KDC verifies that

t = (t1 * t2 * t3 * t4 * t5) mod p

If so, KDC approves the public key.

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