CSE20 Lecture 5 4/12/11

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CSE20 Lecture 54/12/11

• CK Cheng
• UC San Diego

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Residual Numbers(NT-1 and Shaums Chapter 11)

• Introduction
• Definition
• Operations
• Range of numbers

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Introduction

• Applications communication, cryptography, and
  high performance signal processing
• Goal Simplify arithmetic operations (, -, x)
  when bit width n is huge, e.g. n 1000. Note no
  division is involved.
• Flow

Residual number (x1, x2, , xk) , -, x
operations for each xi under mi
Number x
Results
Chinese Remainder Theorem
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Definition

• Mod (Modular) operation.
• Given integer x and d (dgt 0), find q and r such
  that x qdr, 0lt r ltd,
• where q quotient, d divisor, and r remainder.
• We define xd r.
• Conversion to residual system
• Given moduli (m1, m2, , mk), where all mi are
  mutually prime, transform
• integer x to (r1, r2, , rk), where rixmi

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Definitions

• Mutually Prime Two integers a b are
• mutually (or relatively) prime if their greatest
• common divisor is 1.
• e.g. 3 8, 4 9, but not 6 9
• Residual number Given (m1, m2,,mk) where mis
  are mutually prime and a positive integer x
  ltMm1xm2xxmk
• (0 x ltM ) represent x as
• ( xm1, xm2,, xmk )

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Examples (xmiri)

• Given (m1, m2, m3) (3, 5, 7), convert
• x (r1, r2, r3).
• 0 (0, 0, 0) 030, 050, 07 0
• 2 (2, 2, 2) 232, 252, 272
• 21 (0, 1, 0) 2130 , 2151, 2170
• -2 (1, 3, 5) -231, -253, -275
• -3 (0, 2, 4) -330, -352, -374
• -21 (0, 4, 0) -2130, -215 4, -2170
• Hint 0lt ri lt mi

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Examples

• k 3 ( m1, m2, m3 ) ( 2, 3, 7 )
• M m1 x m2 x m3 2 x 3 x 7 42
• Given x30, ( xm1, xm2, xm3 )
• ( 302, 303, 307 ) ( 0, 0, 2 )
• Given y4, ( ym1, ym2, ym3 )
• ( 42, 43, 47 ) ( 0, 1, 4 )
• Given xy34, ((xy)m1,(xy)m2,(xy)m3 )
• ( 342, 343, 347 ) ( 0, 1, 6 )

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3. Modular Operations

• Theorem Given three integers x,y,d, we have
• (xy)d(xdyd)d.
• Proof
• Let x qxd rx, y qyd ry
• We have (xy)d (qxd rx qyd ry)d
• (rxry)d
• Therefore, (xy)d (xdyd)d

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3. Modular Operations

• Theorem Given three integers x,y,d, we have
• (xy)d(xd yd)d.
• Proof
• Let x qxd rx, y qyd ry
• We have (xy)d (qxd rx )(qyd ry)d
• (qxqyd2ryqxdrxqydrxry)d
• (rxry)d
• Therefore, (xy)d (xd yd)d

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3. Modular Operations

• What about division?
• Could we state the following equality?
• ((xd)/(yd))d (x/y)d
• Answer No! We have the following problems.
• yd can be zero.
• (xd)/(yd) or x/y can be fractional.

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Range of Numbers

• Theorem Given (m1, m2, , mk), where all mi are
  mutually prime, let Mm1xm2xxmk.
• For 0lt xlt M, the residual number
• (x1, x2, , xk) is distinct.
• Proof By contradiction, let 0lt yltx lt M.
• Suppose (x1, x2, , xk)(y1, y2, , yk) then
• x-y (0,0, ,0). However, for all numbers in the
  range of the interval, only 0 (0,0, ,0)
  because the mods mi are mutually prime.