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Lecture 8:

Hashing

David Evans

http://www.cs.virginia.edu/evans

Note: only 3

people (out of 4) have

voted that notes are

useful. I won’t make notes (regularly) until at least 10 people do.

CS588: Security and Privacy

University of Virginia

Computer Science

Picks random x

Bob

Alice

f (x)

Picks “odd”

or “even”

Alice wins

if x does

not match

Bob’s pick

“odd” or “even”

Checks

f (x) matches

value received

in step 1

x

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- One Way:
- For every integer x, easy to compute f(x)
- Given f (x), hard to find any information about x

- Collision Resistant:
- “Impossible” to find pair (x, y) where x y and f (x)=f (y)

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“dog”

“neanderthal”

“horse”

H (char s[]) = (s[0] – ‘a’) mod 10

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- Many-to-one: maps a large number of values to a small number of hash values
- Even distribution: for typical data sets, P(H(x) = n) = 1/N where N is the number of hash values and n = 0 .. N – 1.
- Efficient: H(x) is easy to compute.

How well does

H (char s[]) = (s[0] – ‘a’) mod 10

satisfy these properties?

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- One-way: for given h, it is hard to find x such that H(x) = h.
- Collision resistance:
Weak collision resistance: given x, it is hard to find y x such that H(y) = H(x).

Strong collision resistance: it is hard to find any x and y x such that H(y) = H(x).

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What goes wrong if f is

not one-way?

What goes wrong if f is not weak collision resistant?

What goes wrong if f is not strong collision resistant?

Picks random x

Bob

Alice

f (x)

Picks “odd”

or “even”

Alice wins

if x does

not match

Bob’s pick

“odd” or “even”

Checks

f (x) matches

value received

in step 1

x

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- Alice wants to send Bob and “I owe you” message.
- Bob should be able to show the message to a judge to compel Alice to pay up.
- Bob should not be able to make his own “I owe you” from Alice, or change the contents of the one she sent him.

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M

H(M)

Bob

Alice

M

H(M)

Hmmm...Bob can just make up M and H(M)!

Judge

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M

EKA[H(M)]

Bob

Alice

secret key KA

M

EKA[H(M)]

Can Bob cheat?

Shared secret KA

Can Alice cheat?

Yes, send Bob: M, junk.

Judge will think Bob cheated!

Judge

knows KA

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M

EKRA[H(M)]

Bob

Alice

knows KUA

{KUA, KRA}

M

EKRA[H(M)]

Why not just

use EKRA[M]?

Bob can verify H(M) by decrypting, but cannot forge M, EKRA[H(M)] pair without knowing KRA.

Known public-key encyrption algorithms are slow

Judge

knows KUA

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- Suppose we use: H (char s[]) = (s[0] – ‘a’) mod 10
- Alice sends Bob:
“I, Alice, owe Bob $2.”, EKRA[H (M)]

- Bob sends Judge:
“I, Alice, owe Bob $2000000.”, EKRA[H (M)]

- Judge validates
EKUA[EKRA[H (M)]] = H(“I, Alice, owe Bob $2000000.”)

and makes Alice pay.

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- Given x, it should be hard to find y x such that H(y) = H(x).
- Similar to a block cipher except no need for secret key:
- Changing any bit of x should change most of H(x).
- The mapping between x and H(x) should be confusing (complex and non-linear).

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- H(x) = DES (x, 0)
- Weak collision resistance?
- Given x, it should be hard to find y x such that H(y) = H(x).
- Yes – DES is one-to-one. (These is no such y.)

- A good hash function?
- No, its output is as big as the message!

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- Produce small number of bits (say 64) that depend on the whole message in a confusing, non-linear way.
- Have we seen anything like this?

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Pn

P2

P1

IV

...

DES

K

DES

DES

K

K

Cn

C2

C1

Use last ciphertext block as hash. Depends on all plaintext blocks.

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- Based on cipher block chaining
- No need for secret key or IV (just use 0)
- Don’t use DES
- Performance
- Better to use bigger blocks

- MD5 [Rivest92] – 512 bit blocks, produces 128-bit hash
- SHA [NIST95] – 512 bit blocks, 160-bit hash

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- 3DES is (probably) secure with 64-bit blocks, why do secure hash functions need at least 128 bit digests?
- 64 bits is fine for weak collision resistance, but we need strong collision resistance too.

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- It is hard to find any x and y x such that H(y) = H(x).
- Difference from weak:
- Attacker gets to choose both x and y, not just y.

- Scenario:
- Suppose Bob gets to write IOU message, send it to Alice, and she signs it.

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- Many-to-one: compresses
- Even distribution: P(H(x) = n) = 1/N
- Efficient: H(x) is easy to compute.
- One-way: given H(x), hard to find x
- Collision resistance:
Weak collision resistance: given x, it is hard to find y x such that H(y) = H(x).

Strong collision resistance: it is hard to find any x and y x such that H(y) = H(x).

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x

EKRA[H(x)]

Bob

Alice

knows KUA

{KUA, KRA}

y

EKRA[H(x)]

Bob picks x and y such that H(x) = H(y).

Judge

knows KUA

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Bob generates 210 different agreeable (to Alice) xi messages:

I, { Alice | Alice Hacker | Alice P. Hacker | Ms. A. Hacker }, { owe | agree to pay } Bob { the sum of | the amount of } { $2 | $2.00 | 2 dollars | two dollars } { by | before } { January 1st | 1 Jan | 1/1 | 1-1 } { 2006 | 2006 AD}.

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Bob generates 210 different agreeable (to Bob) yi messages:

I, { Alice | Alice Hacker | Alice P. Hacker | Ms. A. Hacker }, { owe | agree to pay } Bob { the sum of | the amount of } { $2 quadrillion | $2000000000000000 | 2 quadrillion dollars | two quadrillion dollars } { by | before } { January 1st | 1 Jan | 1/1 | 1-1 } { 2006 | 2006 AD}.

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- For each message xi and yi, Bob computes hxi = H(xi) and hyi = H(yi).
- If hxi = hyjfor some i and j, Bob sends Alice xi, gets EKRA[H(x)]back.
- Bob sends the judge yjand EKRA[H(xi)].
- Is this different from when Alice chooses x?

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- Hash function generate 64-bit digest (n = 264)
- Hash function is good (randomly distributed and diffuse)
- Chance a randomly chosen message maps to a given hash value: 1 in n = 2-64
- By hashing m good messages, chance that a randomly chosen bad message maps to one of the m different hash values: m * 2-64
- By hashing m good messages and m bad messages: m * m * 2-64(approximation)

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- m = 210
- 210 * 210 * 2-64 = 2-44 (still a pauper)
- Try m= 232
- 232 * 232 * 2-64 = 20 = 1 (yippee!)
- Flaw: some of the messages might hash to the same value, might need more than 232 to find match.

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What is the probability that two people in this room have the same birthday?

Text, Chapter 3.6

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Ways to assign k different birthdays without duplicates:

N = 365 * 364 * ... * (365 – k + 1)

= 365! / (365 – k)!

Ways to assign k different birthdays with possible duplicates:

D = 365 * 365 * ... * 365 = 365k

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Assuming real birthdays assigned randomly:

N/D = probability there are no duplicates

1 - N/D = probability there is a duplicate

= 1 – 365! / ((365 – k)!(365)k)

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n!

(n – k)! nk

P(n, k) = 1 –

Given k random selections from n possible values, P(n, k) gives the probability that there is at least 1 duplicate.

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P(no two match) = 1 – P(all are different)

P(2 chosen from N are different)

= 1 – 1/N

P(3 are all different)

= (1 – 1/N)(1 – 2/N)

P(n trials are all different)

= (1 – 1/N)(1 – 2/N) ... (1 – (n – 1)/N)

ln (P)

= ln (1 – 1/N) + ln (1 – 2/N) + ... ln (1 – (k – 1)/N)

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ln (P) = ln (1 – 1/N) + ... + ln (1 – (k – 1)/N)

For 0 < x < 1:ln (1 – x) x

ln (P) – (1/N + 2/N + ... + (n – 1)/N)

Gauss says:

1 + 2 + 3 + 4 + ... + (n – 1) + n = ½ n (n + 1)

So,

ln (P) ½ (k-1) k/N

Pe½ (k-1)k / N

Probability of match 1 –e½ (k-1)k / N

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P(n, k) > 1 – e-k*(k-1)/2n

- For n = 365, k = 20:
P(365, 20) > 1 – e-20*(19)/2*365

P(365, 20) > .4058

- For n = 264, k = 232: P (264, 232) > .39
- For n = 264, k = 233: P (264, 233) > .86
- For n = 264, k = 234: P (264, 234) > .9996

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- For n = 2128, k = 240: P (2128, 240) > 10-15
- If your guesses are random, need to try 240 inputs to have a 10-15 chance of finding a collision
- Assumes you hash function is perfect

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From http://www.freedom-to-tinker.com/archives/000664.html

#!/usr/bin/perl -w

use strict;

use Digest::MD5 qw(md5_hex);

# Create a stream of bytes from hex.

my @bytes1 = map {chr(hex($_))} qw(d1 31 dd 02 c5 e6 ee c4 69 3d 9a 06 98 af f9 5c 2f ca b5 87 12 46 7e ab 40 04 58 3e b8 fb 7f 89 55 ad 34 06 09 f4 b3 02 83 e4 88 83 25 71 41 5a 08 51 25 e8 f7 cd c9 9f d9 1d bd f2 80 37 3c 5b d8 82 3e 31 56 34 8f 5b ae 6d ac d4 36 c9 19 c6 dd 53 e2 b4 87 da 03 fd 02 39 63 06 d2 48 cd a0 e9 9f 33 42 0f 57 7e e8 ce 54 b6 70 80 a8 0d 1e c6 98 21 bc b6 a8 83 93 96 f9 65 2b 6f f7 2a 70);

my @bytes2 = map {chr(hex($_))} qw(d1 31 dd 02 c5 e6 ee c4 69 3d 9a 06 98 af f9 5c 2f ca b5 07 12 46 7e ab 40 04 58 3e b8 fb 7f 89 55 ad 34 06 09 f4 b3 02 83 e4 88 83 25 f1 41 5a 08 51 25 e8 f7 cd c9 9f d9 1d bd 72 80 37 3c 5b d8 82 3e 31 56 34 8f 5b ae 6d ac d4 36 c9 19 c6 dd 53 e2 34 87 da 03 fd 02 39 63 06 d2 48 cd a0 e9 9f 33 42 0f 57 7e e8 ce 54 b6 70 80 28 0d 1e c6 98 21 bc b6 a8 83 93 96 f9 65 ab 6f f7 2a 70);

# Print MD5 hashes

print md5_hex(@bytes1), "\n", md5_hex(@bytes2), "\n";

79054025255fb1a26e4bc422aef54eb4

79054025255fb1a26e4bc422aef54eb4

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- Collisions announced in SHA-0 at Crypto 2004
- No collisions yet found in SHA-1 (which replaced SHA-0 as a standard in 1994)
- NIST is nervous http://csrc.nist.gov/hash_standards_comments.pdf

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- “At the recent Crypto2004 conference, researchers announced that they had discovered a way to "break" a number of hash algorithms, including MD4, MD5, HAVAL-128, RIPEMD and the long superseded Federal Standard SHA-0 algorithm. The current Federal Information Processing Standard SHA-1 algorithm, which has been in effect since it replaced SHA-0 in 1994, was also analyzed, and a weakened variant was broken, but the full SHA-1 function was not broken and no collisions were found in SHA-1. The results presented so far on SHA-1 do not call its security into question. However, due to advances in technology, NIST plans to phase out of SHA-1 in favor of the larger and stronger hash functions (SHA-224, SHA-256, SHA-384 and SHA-512) by 2010.”

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We’ll cover SSL

after Spring Break…

but, this should make you nervous…

Wednesday 3:30

Chenxi Wang Seminar

“Defending against Large Scale Attacks on the Internet”

Thursday 9:30 (please arrive on time for class, not like usual!)

Chenxi Wang guest lecture

Using hashes to provide censorship-resistant publishing

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