CSE20 Lecture 6: Number Systems 5. Residual Numbers (cont)

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CSE20 Lecture 6 Number Systems5. Residual
Numbers (cont) 6. Cryptography

• CK Cheng
• UC San Diego

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Residual Numbers(NT-1 and Shaums Chapter 11)

• Introduction
• Definition
• Operations
• Inverse Conversion

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Inverse Conversion
Number x
Mod Operation
Residual number (x1, x2, , xk) , -, x
operations for each xi under mi
Moduli (m1, m2, , mk)
Results
Chinese Remainder Theorem
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Chinese Remainder Theorem

• Given a residual number (r1, r2, , rk) with
  moduli (m1, m2, , mk), where all mi are mutually
  prime, set M m1m2 mk, and MiM/mi.
• 1. Find Si that (MiSi)mi 1 (Si an inverse of
  Mi in mod mi)
• 2. The corresponding number
• x (?i1,k(Mi Si ri))M.

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Example

• Given (m1,m2,m3)(2,3,7), M23742, we have
• M1m2m33721 (M1S1)m1(21S1)21
• M2m1m32714 (M2S2)m2(14S2)31
• M3m1m2236 (M3S3)m3(6S3)71
• Thus, (S1, S2, S3) (1,2,6)
• For a residual number (0,2,1)
• x(M1S1r1 M2S2r2 M3S3r3)M
• (2110 1422 661 )42
• ( 0 56 36 )42 9242 8

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Example

• For a residual number (1,2,5)
• x(M1S1r1 M2S2r2 M3S3r3)M
• (2111 1422 665)42
• (21 56 180)42
• 25742 5

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Example iClicker

• Given (m1,m2,m3)(2,3,5), M23530, we have
• M1m2m33515 (M1S1)m1(15S1)21
• M2m1m32510 (M2S2)m2(10S2)31
• M3m1m2236 (M3S3)m3(6S3)51
• Thus, (S1, S2, S3) is
• A. (1, 1, 1)
• B. (1, 2, 1)
• C. (2, 1, 2)
• D. None of the above

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Example iClicker

• Given (m1,m2,m3)(2,3,5), M23530, we have
• M1m2m33515 (M1S1)m1(15S1)21
• M2m1m32510 (M2S2)m2(10S2)31
• M3m1m2236 (M3S3)m3(6S3)51
• For a residual number (x1,x2,x3)(1,2,3), the
  corresponding number x is
• A. 5
• B. 19
• C. 23
• D. None of the above

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Proof of Chinese Remainder Theorem

• Let A ?i1,k(Mi Si ri), we show that
• 1. Amv rv and 2. xAM is unique.
• 1. Amv ?i1,k(Mi Si ri) mv
• S(MiSiri) mvmv (MvSvrv)mv
• (MvSv)mv rvmv mv rvmv rv
• 2. Proof was shown in lecture 5.

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6. Cryptography

1. Introduction
2. RSA Protocol
3. Remarks

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6.1 Cryptography Introduction

• Application of residual number systems
• Number theory (skip)
• Show the basic concept and process
• Many variations

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6.2 RSA Protocol
Private M

• Function P(X)XeN is public.
• Function S(X)XdN is secret.
• Message M is private, but P(M) is observed by
  all.
• Desired feature S(P(M))M.
• Example (e,N)(7,55), (d,N)(23,55)
• M12 gt P(12)1275523 gt S(23)23235512
• M8 gt P(8)87552 gt S(2)22355?

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6.2 RSA Protocol

• Npq where p q are primes and kept secret.
• e is mutually prime to f(N)(p-1)(q-1)
• d is the inverse of e mod f(N), i.e. (ed)f(N)1
• Theorem S(P(M))P(S(M))M for 0ltMltN
• Note that S(P(M))MedN
• Theorem Mf(N)N1 for 0ltMltN
• Assumption p q are hard to find. Consequently,
  it is difficult to derive d.

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6.2 RSA Protocol

• Npq where p q are primes and kept secret.
• e is mutually prime to f(N)(p-1)(q-1)
• d is the inverse of e mod f(N), i.e. (ed)f(N)1
• Example Npq3x1133, f(N)(3-1)(11-1)20
• Let e3, then d7 (3x7201).
• M9 gt P(9)93333 gt S(3)3733?

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6.3 Remark

• Residual number system is used in cryptography.
• RSA protocol uses public key for coding P(X) and
  secret key to decode S(X).
• Use wide words (gt1000 bits) so that the solution
  is computationally expensive without the
  knowledge of the function S(X).