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CS 312: Algorithm Analysis
Lecture #3: Algorithms for
Modular Arithmetic,
Modular Exponentiation
Slides by: Eric Ringger, with contributions from Mike Jones, Eric Mercer, Sean Warnick
Announcements
 HW #1 Due Now
 Always start of class
 Always show work
 FERPA protects your student record
 Need waiver to return graded work without
cover sheet
Objectives
 Add the Max Rule to your asymptotic
analysis toolbox
 Review modular arithmetic
 Discuss and analyze algorithms for:
 modular arithmetic
 modular exponentiation
Max. rule
 Another useful rule for Asymptotic
analysis.
O( f(n) + g(n) ) = O( max( f(n), g(n) ) )
 Examples:
Goal for Ch. 1
 Appreciate the role of theoretical analysis in
the security of RSA.
 Requires: Solve, analyze, and use (!) two
important and related problems:
 Factoring: Given a number N, express it as a
product of its prime numbers
 Primality Testing: Given a number N, determine
whether it is prime
 Which one is harder?
Algorithms for Integer Arithmetic
 Computing Device:
 Binary operations are constant time
 Arithmetic operations on arbitrary length integers
may require more time
 For an integer 𝑥, we talk about its
representation in bits:
 𝑛 = log 2 𝑥
 Pad length of 𝑥 to the next power of 2 (using 0s)
if necessary.
Algorithms for Integer Arithmetic
 Addition
 Multiplication
 Division
Algorithms for Integer Arithmetic
 Addition: Θ(𝑛)
 Multiplication: Θ(𝑛2 )
 Division: Θ(𝑛2 )
Modular Arithmetic
Congruency
An important distinction
 Congruency
 Equality, using the modulus operator
Properties
 Associativity:
 Commutativity:
 Distributivity:
Substitution Rule
Useful Consequence
xy  (x mod z)y
(mod z)
xy mod z = (x mod z)y mod z
 Example:
Modular Addition
Modular Multiplication
Goal: Modular Exponentiation
 We need to compute
xy mod N
for values of x, y, and N that are several
hundred bits long.
 Can we do so quickly?
Sequential Exponentiation
Describe a simple algorithm for doing exponentiation:
function seqexp (x, y)
Input: An n-bit integer x and a non-negative integer
exponent y (arbitrarily large)
Output: xy
if y=0: return 1
r=x
for i = 1 to y-1 do
r=r⋅x
return r
Analysis of Sequential
Exponentiation
function seqexp (x, y)
Input: An n-bit integer x and a non-negative
integer exponent y (arbitrarily large)
Output: xy
if y=0: return 1
r=x
for i = 1 to y-1 do
r=rx
return r
Modular Exponentiation, Take I
function modexp (x, y, N)
Input: Two n-bit integers x and N, a non-negative
integer exponent y (arbitrarily large)
Output: xy mod N
if y=0: return 1
r = x mod N
for i = 1 to y-1 do
r = (r ⋅ x) mod N
return r
Modular Exponentiation, Take I
function modexp (x, y, N)
Input: Two n-bit integers x and N, a non-negative
integer exponent y (arbitrarily large)
Output: xy mod N
if y=0: return 1
r = x mod N
for i = 1 to y-1 do
r = (r ⋅ x) mod N
return r
New Ideas
 Represent y (the exponent) in binary
 Then break down xy into factors using the
non-zero bits of y
 Also: compute the factors using repeated
squaring
 Reduce factors using substitution rule
Modular Exponentiation, Take II
function modexp (x, y, N)
Input: Two n-bit integers x and N, a non-negative
integer exponent y (arbitrarily large)
Output: xy mod N
Recursive call
if y=0: return 1
z = modexp(x, floor(y/2), N)
if y is even:
return z2 mod N
else:
return x ⋅ z2 mod N
Right shift
Multiplication
Analysis of Modular Exponentiation




Each multiplication is Q(n2)
Each modular reduction is Q(n2)
There are log(y)=m of them
Thus, modular exponentiation is in
Q(n2 log y) = Q(n2 m)
function modexp (x, y, N)
if y=0: return 1
z = modexp(x, floor(y/2), N)
if y is even:
return z2 mod N
else:
return x z2 mod N
Modular Exponentiation (II),
Iterative Formulation
function modexp (x, y, N)
Input: Two n-bit integers x and N, a non-negative
integer exponent y (arbitrarily large)
Output: xy mod N
if y = 0: return 1
i = y; r = 1; z = x mod N
while i > 0
if i is odd: r = r ⋅ z mod N
z = z2 mod N
i = floor(i/2)
return r
Modular Exponentiation
 xy mod N
 Key Insights:
1. Exponent y can be represented in binary
2. Problem can be factored into one factor per
binary digit
3. Each factor can be reduced mod N
(substitution rule)
Example
We’re employing
same insights and a little more
cleverness than the
algorithm.
Example #2
function modexp (x, y, N)
Input: Two n-bit integers x and N, an integer
exponent y (arbitrarily large)
Output: xy mod N
if y = 0: return 1
i = y; r = 1; z = x mod N
while i > 0
if i is odd: r = r z mod N
z = z2 mod N
i = floor(i/2)
return r
310 mod10
x  3, y  10, N  10
i  10, r  1, z  3mod10  3
z  32 mod10  9
i5
r  1 9 mod10  9
z  92 mod10  81mod10  1
i2
z  12 mod10  1
i 1
r  9 1mod10  9
z 1
i0
return 9
Strictly tracing the
algorithm.
Example #2
function modexp (x, y, N)
Input: Two n-bit integers x and N, an integer
exponent y (arbitrarily large)
Output: xy mod N
if y = 0: return 1
i = y; r = 1; z = x mod N
while i > 0
if i is odd: r = r z mod N
z = z2 mod N
i = floor(i/2)
return r
310 mod10
x  3, y  10, N  10
i  10, r  1, z  3mod10  3
z  32 mod10  9
i5
r  1 9 mod10  9
z  92 mod10  81mod10  1
i2
z  12 mod10  1
i 1
r  9 1mod10  9
z 1
i0
return 9
Example
20
3 mod 10
Needed: two volunteers:
Volunteer A: use our final modexp() to compute it.
Volunteer B: compute 320 then reduce mod 10
Efficiency
 The key point is that xy mod N is easy
 modexp is in Q(n2 log y)
 In fact, it requires about 1.5 log2 y multiplications for typical y
 seqexp required y-1 multiplications
 When x, y, and N are 200 digit numbers
 Assume 1 multiplication of two 200 digit numbers takes 0.001
seconds
 modexp typically takes about 1 second
 seqexp would require 10179 times the Age of the Universe!
 Only works when y is an integer.
Assignment
 Read: Section 1.4
 HW #2:
 Problem 1.25 using modexp,
 Then redo 1.25 but replace 125 with 126 for
the exponent
 Implement modular exponentiation now as a
step toward finishing Project #1
Next
 Primality Testing