university of hawaii ics141: discrete mathematics for ... · ics 141: discrete mathematics i –...

13-1 ICS 141: Discrete Mathematics I – Fall 2011

University of Hawaii University of Hawaii

ICS141: Discrete Mathematics for Computer Science I

Dept. Information & Computer Sci., University of Hawaii

Jan Stelovsky based on slides by Dr. Baek and Dr. Still

Originals by Dr. M. P. Frank and Dr. J.L. Gross Provided by McGraw-Hill


University of Hawaii University of Hawaii

Lecture 18a

Chapter 3. The Fundamentals 3.6 Applications of Integers Algorithms


University of Hawaii University of Hawaii Quiz

1.  What is the decimal expansion (1AF)16? 2.  What is the hexadecimal expansion of (287)10? 3.  What is the two’s complement of -7? 4.  Multiply (100)2 and (101)2 in binary system.

n  Hints n  162 = 256


University of Hawaii University of Hawaii Applications

n  Miscelaneous useful results n  Linear congruences n  Chinese Remainder Theorem n  Pseudoprimes

n  Fermat’s Little Theorem n  Public Key Cryptography

n  The Rivest-Shamir-Adleman (RSA) cryptosystem


University of Hawaii University of Hawaii Miscelaneous Results n  Theorem 1:

n  ∀a,b ∈ Z+: ∃s,t ∈Z: gcd(a,b) = sa + tb n  Lemma 1:

n  ∀a,b,c ∈ Z+: gcd(a,b)=1 ∧ a | bc → a|c n  Lemma 2:

n  If p is prime and p|a1a2…an (integers ai) then ∃i: p|ai.

n  Theorem 2: n  If ac ≡ bc (mod m) and gcd(c,m)=1, then

a ≡ b (mod m). (m∈Z+, a,b,c∈Z)


University of Hawaii University of Hawaii Theorem 1

n  Theorem 1: ∀a,b∈Z+: ∃s,t ∈Z such that gcd(a,b) = sa + tb

n  Proof: By induction over the value of the larger argument a.

n  Example: n  Express gcd(252, 198) = 18 as a linear

combination of 252 and 198.


University of Hawaii University of Hawaii Proof of Theorem 1 Theorem 1: ∀b,a∈Z+: ∃s,t ∈Z : gcd(a,b) = sa + tb Proof: (By induction over the value of the larger

argument a.) n  ByTheorem 0 gcd(a,b) = gcd(b,c) if c = a mod b, i.e.,

a = kb + c for some integer k, and thus c = a − kb. n  Now, since b<a and c<b, by inductive hypothesis,

we can assume that ∃u,v: gcd(b,c) = ub + vc. n  Substituting for c, this is ub+v(a−kb), which we can

regroup to get va + (u−vk)b. n  So now let s = v, and let t = u−vk, and we’re finished. n  The base case: s = 1 and t = 0.

This works for gcd(a,0), or if a=b originally. ■


University of Hawaii University of Hawaii Theorem 1: Example n  Express gcd(252, 198) = 18 as a linear combination of

252 and 198. n  252 = 1 ⋅ 198 + 54

198 = 3 ⋅ 54 + 36 54 = 1 ⋅ 36 + 18 36 = 2 ⋅ 18

n  18 = 54 – 1 ⋅ 36 = 54 – 1 ⋅ (198 – 3 ⋅ 54) = 4 ⋅ 54 – 1 ⋅ 198 = 4 ⋅ (252 – 1 ⋅ 198) – 1 ⋅ 198 = 4 ⋅ 252 – 5 ⋅ 198

n  Therefore, gcd(252, 198) = 18 = 4 ⋅ 252 – 5 ⋅ 198

Euclidean algorithm


University of Hawaii University of Hawaii Proof of Lemma 1 n  Lemma 1:

∀a,b,c ∈ Z+: gcd(a,b)=1 ∧ a|bc → a|c Proof:

n  Applying theorem 1, ∃s, t: sa + tb = 1. n  Multiplying through by c, we have that

sac + tbc = c. n  Since a|bc is given, we know that a|tbc,

and obviously a|sac. n  Thus (using the theorem on pp.202), it

follows that a|(sac+tbc); in other words, that a|c. ■


University of Hawaii University of Hawaii Proof of Lemma 2 n  Lemma 2: If p is prime and p|a1a2…an (integers

ai) then p|ai for some i.

Proof (by induction): n  If n=1, this is immediate since p|a0 → p|a0.

Suppose the lemma is true for all n<k and p|a1…ak. n  If p|m where m=a1…ak-1 then we have it inductively. n  Otherwise, we have p|mak but ¬(p|m).

Since m is not a multiple of p, and p has no factors, m has no common factors with p, thus gcd(m,p)=1. So by applying Lemma 1, p|ak. ■


University of Hawaii University of Hawaii Theorem 2 n  Theorem 2: Let m∈Z+ and a,b,c∈Z.

If ac ≡ bc (mod m) and gcd(c,m)=1, then a ≡ b (mod m). Proof:

n  Since ac ≡ bc (mod m), this means m | ac−bc. n  Factoring the right side, we get m | c(a − b).

Since gcd(c,m)=1, lemma 1 implies that m | a−b, in other words, that a ≡ b (mod m). ■

n  Examples n  20 ≡ 8 (mod 3) i.e. 5 ⋅ 4 ≡ 2 ⋅ 4 (mod 3)

Since gcd(4, 3) = 1, 5 ≡ 2 (mod 3) n  14 ≡ 8 (mod 6) but 7 ≡ 4 (mod 6) (as gcd(2,6) ≠ 1)


University of Hawaii University of Hawaii Linear Congruences, Inverses n  A congruence of the form ax ≡ b (mod m) is

called a linear congruence. (m∈Z+, a,b∈Z, and x: variable) n  To solve the congruence is to find the x’s that

satisfy it. n  An inverse of a, modulo m is any integer

a-1 such that a-1a ≡ 1 (mod m). n  If we can find such an a-1, notice that we can

then solve ax ≡ b (mod m) by multiplying through by it, giving a-1ax ≡ a-1b (mod m), thus 1·x ≡ a-1b (mod m), thus x ≡ a-1b (mod m).


University of Hawaii University of Hawaii Theorem 3

n  Theorem 3: If gcd(a,m)=1 (i.e. a and m are relatively prime) and m > 1, then a has a inverse a-1 unique modulo m. Proof:

n  By theorem 1, ∃s,t: sa + tm = 1, so sa + tm ≡ 1 (mod m). n  Since tm ≡ 0 (mod m), sa ≡ 1 (mod m).

Thus s is an inverse of a (mod m). n  Theorem 2 guarantees that if ra ≡ sa ≡ 1 then r ≡ s,

thus this inverse is unique modulo m. (All inverses of a are in the same congruence class as s.) ■


University of Hawaii University of Hawaii Example n  Find an inverse of 3 modulo 7

n  Since gcd(3, 7) = 1, by Theorem 3 there exists an inverse of 3 modulo 7.

n  7 = 2·3 + 1 n  From the above equation, –2·3 + 1·7 = 1 n  Therefore, –2 is an inverse of 3 modulo 7

n  Note that every integer congruent to –2 modulo 7 is also an inverse of 3, such as 5, –9, 12, and so on.)


University of Hawaii University of Hawaii Example n  What are the solutions of the linear

congruence 3x ≡ 4 (mod 7)? n  –2 is an inverse of 3 modulo 7 (previous slide) n  Multiply both side by –2: –2·3x ≡ –2·4 (mod 7) n  –6·x ≡ x ≡ –8 ≡ 6 (mod 7) n  Therefore, the solutions to the congruence are

the integers x such that x ≡ 6 (mod 7), i.e. 6, 13, 20, 27,… and –1, –8, –15,…

n  e.g. 3·13 = 39 ≡ 4 (mod 7)


University of Hawaii University of Hawaii An Application of Primes!

n  When you visit a secure web site (https:… address, indicated by padlock icon in IE, key icon in Netscape), the browser and web site may be using a technology called RSA encryption.

n  This public-key cryptography scheme involves exchanging public keys containing the product pq of two random large primes p and q (a private key) which must be kept secret by a given party.

n  So, the security of your day-to-day web transactions depends critically on the fact that all known factoring algorithms are intractable!


University of Hawaii University of Hawaii Public Key Cryptography

n  In private key cryptosystems, the same secret “key” string is used to both encode and decode messages. n  This raises the problem of how to securely communicate the key strings.

n  In public key cryptosystems, there are two complementary keys instead. n  One key decrypts the messages that the other one encrypts.

n  This means that one key (the public key) can be made public, while the other (the private key) can be kept secret from everyone. n  Messages to the owner can be encrypted by anyone using the public

key, but can only be decrypted by the owner using the private key. n  Like having a private lock-box with a slot for messages.

n  Or, the owner can encrypt a message with their private key, and then anyone can decrypt it, and know that only the owner could have encrypted it.

n  This is the basis of digital signature systems. n  The most famous public-key cryptosystem is RSA.

n  It is based entirely on number theory!


University of Hawaii University of Hawaii Rivest-Shamir-Adleman (RSA) n  Choose a pair p, q of large random prime

numbers with about the same number of bits n  Let n = pq

n  Choose exponent e that is relatively prime to (p−1)(q−1) and 1 < e <(p−1)(q−1)

n  Compute d, the inverse of e modulo (p−1)(q−1).

n  The public key consists of: n, and e. n  The private key consists of: n, and d.


University of Hawaii University of Hawaii RSA Encryption n  To encrypt a message encoded as an integer:

n  Translate each letter into an integer and group them to form larger integers, each representing a block of letters. Each block is encrypted using the mapping C = Me mod n.

n  Example: RSA encryption of the message STOP with p = 43, q = 59, and e = 13 n  n = 43 x 59 = 2537 n  gcd(e, (p–1)(q–1)) = gcd(13, 42·58) = 1 n  STOP -> 1819 1415 n  181913 mod 2537 = 2081; 141513 mod 2537 = 2182 n  Encrypted message: 2081 2182


University of Hawaii University of Hawaii RSA Decryption n  To decrypt the encoded message C,

n  Compute M = Cd mod n n  Recall that d is an inverse of e modulo (p−1)(q−1).

n  Example: RSA decryption of the message 0981 0461 encrypted with p = 43, q = 59, and e = 13 n  n = 43 x 59 = 2537; d = 937 n  0981937 mod 2537 = 0704 n  0461937 mod 2537 = 1115 n  Decrypted message: 0704 1115 n  Translation back to English letters: HELP


University of Hawaii University of Hawaii Why RSA Works Theorem (Correctness of RSA): (Me)d ≡ M (mod n). Proof: n  By the definition of d, we know that de ≡ 1 [mod (p−1)(q−1)].

n  Thus by the definition of modular congruence, ∃k: de = 1 + k(p−1)(q−1).

n  So, the result of decryption is Cd ≡ (Me)d = Mde = M1+k(p−1)(q−1) (mod n)

n  Assuming that M is not divisible by either p or q, n  Which is nearly always the case when p and q are very large, n  Fermat’s Little Theorem tells us that Mp−1≡1 (mod p)

and Mq−1≡1 (mod q) n  Thus, we have that the following two congruences hold:

n  First: Cd ≡ M·(Mp−1)k(q−1) ≡ M·1k(q−1) ≡ M (mod p) n  Second: Cd ≡ M·(Mq−1)k(p−1) ≡ M·1k(p−1) ≡ M (mod q)

n  And since gcd(p,q)=1, we can use the Chinese Remainder Theorem to show that therefore Cd≡M (mod pq): n  If Cd≡M (mod pq) then ∃s: Cd=spq+M, so Cd≡M (mod p) and

(mod q). Thus M is a solution to these two congruences, so (by CRT) it’s the only solution.■


University of Hawaii University of Hawaii Uniqueness of Prime Factorizations

“The prime factorization of any positive integer n is unique.” Proof: Suppose that the positive integer n can be written as

the product of two different ways, i.e. n = p1…ps = q1…qt are equal products of two nondecreasing sequences of primes. Assume (without loss of generality) that all primes in common have already been divided out, so that ∀ij: pi ≠ qj. But since p1…ps = q1…qt, we have that p1|q1…qt, since p1·(p2…ps) = q1…qt. Then applying lemma 2, ∃j: p1|qj. Since qj is prime, it has no divisors other than itself and 1, so it must be that pi=qj. This contradicts the assumption ∀ij: pi ≠ qj. Consequently, the two lists must have been identical to begin with! ■

The “hard” part of proving the Fundamental Theorem of Arithmetic.


University of Hawaii University of Hawaii Chinese Remainder Theorem n  Theorem: (Chinese remainder theorem.)

Let m1,…,mn > 0 be pairwise relatively prime and ai ,…,an arbitrary integers. Then the equations system x ≡ ai (mod mi) (for i=1,..,n) has a unique solution modulo m = m1 m2···mn. Proof:

n  Let Mi = m/mi. (Thus gcd(mi, Mi)=1.) n  So by Theorem 3, ∃yi=Mi such that yiMi≡1 (mod mi). n  Now let x = ∑i aiyiMi = a1y1M1 + a2y2M2 +···+ anynMn. n  Since mi|Mk for k≠i, Mk≡0 (mod mi), so

x≡aiyiMi≡ai (mod mi). Thus, the congruences hold. (Uniqueness is an exercise.) □


University of Hawaii University of Hawaii Computer Arithmetic with Large Integers

n  By Chinese Remainder Theorem, an integer a where 0≤a<m=∏mi, gcd(mi,mj≠i)=1, can be represented by a’s residues mod mi: (a mod m1, a mod m2, …, a mod mn)

n  To perform arithmetic with large integers represented in this way, n  Simply perform operations on the separate residues!

n  Each of these might be done in a single machine operation.

n  The operations may be easily parallelized on a vector machine.

n  Works so long as m > the desired result.


University of Hawaii University of Hawaii Computer Arithmetic Example n  For example, the following numbers are relatively prime:

m1 = 225−1 = 33,554,431 = 31 · 601 · 1,801 m2 = 227−1 = 134,217,727 = 7 · 73 · 262,657 m3 = 228−1 = 268,435,455 = 3 · 5 · 29 · 43 · 113 · 127 m4 = 229−1 = 536,870,911 = 233 · 1,103 · 2,089 m5 = 231−1 = 2,147,483,647 (prime)

n  Thus, we can uniquely represent all numbers up to m = ∏mi ≈ 1.4×1042 ≈ 2139.5 by their residues ri modulo these five mi. n  E.g., 1030 = (r1 = 20,900,945; r2 = 18,304,504; r3 = 65,829,085;

r4 = 516,865,185; r5 = 1,234,980,730) n  To add two such numbers in this representation,

n  Just add the residues using machine-native 32-bit integers. n  Take the result mod 2k−1:

n  If result is ≥ the appropriate 2k−1 value, subtract out 2k−1 n  or just take the low k bits and add 1.

n  Note: No carries are needed between the different pieces!


University of Hawaii University of Hawaii Pseudoprimes n  Ancient Chinese mathematicians noticed that whenever

n is prime, 2n−1≡1 (mod n). n  Some also claimed that the converse was true.

n  However, it turns out that the converse is not true! n  If 2n−1≡1 (mod n), it doesn’t follow that n is prime.

n  For example, 341=11·31, but 2340≡1 (mod 341). n  Composites n with this property are called

pseudoprimes. n  More generally, if bn−1≡1 (mod n) and n is composite,

then n is called a pseudoprime to the base b.


University of Hawaii University of Hawaii Carmichael Numbers n  These are sort of the “ultimate pseudoprimes.” n  A Carmichael number is a composite n such that

bn−1≡1 (mod n) for all b relatively prime to n. n  The smallest few are 561, 1105, 1729, 2465,

2821, 6601, 8911, 10585, 15841, 29341. n  Well, so what? Who cares?

n  Exercise for the student: Do some research and find me a useful & interesting application of Carmichael numbers.


University of Hawaii University of Hawaii Fermat’s Little Theorem n  Fermat generalized the ancient observation that

2p−1≡1 (mod p) for primes p to the following more general theorem:

n  Theorem: (Fermat’s Little Theorem.) n  If p is prime and a is an integer not divisible by p,

then ap−1≡1 (mod p). n  Furthermore, for every integer a we have

ap ≡ a (mod p). n  Example (Exponentiation MOD a Prime)

n  Find 2301 mod 5: By FLT, 24 ≡ 1 (mod 5). Hence, 2300 = (24)75 ≡ 1 (mod 5). Therefore, 2301=(2300)·2 ≡ 1·2 (mod 5)≡2 (mod 5)

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