Exponents and Primitive Roots
• From Fermat’s “little” theorem, all exponents divide p − 1.
• A primitive root of p is thus a number with exponent p − 1.
• Let R(k) denote the total number of residues in Φ(p) = {1, 2, . . . , p − 1} that have exponent k.
• We already knew that R(k) = 0 for k ̸ |(p − 1).
• So ∑
k|(p−1)
R(k) = p − 1 as every number has an exponent.
Size of R(k)
• Any a ∈ Φ(p) of exponent k satisfies xk = 1 mod p.
• Hence there are at most k residues of exponent k, i.e., R(k) ≤ k, by Lemma 57 (p. 458).
• Let s be a residue of exponent k.
• 1, s, s2, . . . , sk−1 are distinct modulo p.
– Otherwise, si = sj mod p with i < j.
– Then sj−i = 1 mod p with j − i < k, a contradiction.
• As all these k distinct numbers satisfy xk = 1 mod p, they comprise all the solutions of xk = 1 mod p.
Size of R(k) (continued)
• But do all of them have exponent k (i.e., R(k) = k)?
• And if not (i.e., R(k) < k), how many of them do?
• Pick sℓ, where ℓ < k.
• Suppose ℓ ̸∈ Φ(k) with gcd(ℓ, k) = d > 1.
• Then
(sℓ)k/d = (sk)ℓ/d = 1 mod p.
• Therefore, sℓ has exponent at most k/d < k.
• We conclude that
R(k) ≤ ϕ(k).
Size of R(k) (concluded)
• Because all p − 1 residues have an exponent, p − 1 = ∑
k|(p−1)
R(k) ≤ ∑
k|(p−1)
ϕ(k) = p − 1
by Lemma 54 (p. 445).
• Hence
R(k) =
ϕ(k) when k|(p − 1) 0 otherwise
• In particular, R(p − 1) = ϕ(p − 1) > 0, and p has at least one primitive root.
• This proves one direction of Theorem 49 (p. 431).
A Few Calculations
• Let p = 13.
• From p. 455, we know ϕ(p − 1) = 4.
• Hence R(12) = 4.
• Indeed, there are 4 primitive roots of p.
• As
Φ(p − 1) = {1, 5, 7, 11}, the primitive roots are
g1, g5, g7, g11 for any primitive root g.
The Other Direction of Theorem 49 (p. 431)
• We show p is a prime if there is a number r such that 1. rp−1 = 1 mod p, and
2. r(p−1)/q ̸= 1 mod p for all prime divisors q of p − 1.
• Suppose p is not a prime.
• We proceed to show that no primitive roots exist.
• Suppose rp−1 = 1 mod p (note gcd(r, p) = 1).
• We will show that the 2nd condition must be violated.
The Proof (continued)
• So we proceed to show r(p−1)/q = 1 mod p for some prime divisor q of p − 1.
• rϕ(p) = 1 mod p by the Fermat-Euler theorem (p. 455).
• Because p is not a prime, ϕ(p) < p − 1.
• Let k be the smallest integer such that rk = 1 mod p.
• With the 1st condition, it is easy to show that k | (p − 1) (similar to p. 458).
• Note that k | ϕ(p) (p. 458).
• As k ≤ ϕ(p), k < p − 1.
The Proof (concluded)
• Let q be a prime divisor of (p − 1)/k > 1.
• Then k|(p − 1)/q.
• By the definition of k,
r(p−1)/q = 1 mod p.
• But this violates the 2nd condition.
Function Problems
• Decision problems are yes/no problems (sat, tsp (d), etc.).
• Function problems require a solution (a satisfying truth assignment, a best tsp tour, etc.).
• Optimization problems are clearly function problems.
• What is the relation between function and decision problems?
• Which one is harder?
Function Problems Cannot Be Easier than Decision Problems
• If we know how to generate a solution, we can solve the corresponding decision problem.
– If you can find a satisfying truth assignment efficiently, then sat is in P.
– If you can find the best tsp tour efficiently, then tsp (d) is in P.
• But decision problems can be as hard as the corresponding function problems.
fsat
• fsat is this function problem:
– Let ϕ(x1, x2, . . . , xn) be a boolean expression.
– If ϕ is satisfiable, then return a satisfying truth assignment.
– Otherwise, return “no.”
• We next show that if sat ∈ P, then fsat has a polynomial-time algorithm.
• sat is a subroutine (black box) that returns “yes” or
“no” on the satisfiability of the input.
An Algorithm for fsat Using sat
1: t := ϵ; {Truth assignment.}
2: if ϕ ∈ sat then
3: for i = 1, 2, . . . , n do
4: if ϕ[ xi = true ] ∈ sat then 5: t := t ∪ { xi = true};
6: ϕ := ϕ[ xi = true ];
7: else
8: t := t ∪ { xi = false};
9: ϕ := ϕ[ xi = false ];
10: end if 11: end for 12: return t;
13: else
14: return “no”;
15: end if
Analysis
• If sat can be solved in polynomial time, so can fsat.
– There are ≤ n + 1 calls to the algorithm for sat.a – Boolean expressions shorter than ϕ are used in each
call to the algorithm for sat.
• Hence sat and fsat are equally hard (or easy).
• Note that this reduction from fsat to sat is not a Karp reduction (recall p. 247).
• Instead, it calls sat multiple times as a subroutine and moves on sat’s outputs.
aContributed by Ms. Eva Ou (R93922132) on November 24, 2004.
tsp and tsp (d) Revisited
• We are given n cities 1, 2, . . . , n and integer distances dij = dji between any two cities i and j.
• tsp (d) asks if there is a tour with a total distance at most B.
• tsp asks for a tour with the shortest total distance.
– The shortest total distance is at most ∑
i,j dij.
∗ Recall that the input string contains d11, . . . , dnn.
∗ Thus the shortest total distance is less than 2| x | in magnitude, where x is the input (why?).
• We next show that if tsp (d) ∈ P, then tsp has a polynomial-time algorithm.
An Algorithm for tsp Using tsp (d)
1: Perform a binary search over interval [ 0, 2| x | ] by calling tsp (d) to obtain the shortest distance, C;
2: for i, j = 1, 2, . . . , n do
3: Call tsp (d) with B = C and dij = C + 1;
4: if “no” then
5: Restore dij to old value; {Edge [ i, j ] is critical.}
6: end if
7: end for
8: return the tour with edges whose dij ≤ C;
Analysis
• An edge that is not on any optimal tour will be eliminated, with its dij set to C + 1.
• An edge which is not on all remaining optimal tours will also be eliminated.
• So the algorithm ends with n edges which are not eliminated (why?).
• This is true even if there are multiple optimal tours!a
aThanks to a lively class discussion on November 12, 2013.
Analysis (concluded)
• There are O(| x | + n2) calls to the algorithm for tsp (d).
• Each call has an input length of O(| x |).
• So if tsp (d) can be solved in polynomial time, so can tsp.
• Hence tsp (d) and tsp are equally hard (or easy).
Randomized Computation
I know that half my advertising works, I just don’t know which half.
— John Wanamaker
I know that half my advertising is a waste of money, I just don’t know which half!
— McGraw-Hill ad.
Randomized Algorithms
a• Randomized algorithms flip unbiased coins.
• There are important problems for which there are no known efficient deterministic algorithms but for which very efficient randomized algorithms exist.
– Extraction of square roots, for instance.
• There are problems where randomization is necessary.
– Secure protocols.
• Randomized version can be more efficient.
– Parallel algorithm for maximal independent set.b
aRabin (1976); Solovay and Strassen (1977).
b“Maximal” (a local maximum) not “maximum” (a global maximum).