Basic Modular Arithmetics

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Basic Modular Arithmetics

a

Let m, n ∈ Z+.

m | n means m divides n; m is n’s divisor.

We call the numbers 0, 1, . . . , n − 1 the residue modulo n.

The greatest common divisor of m and n is denoted gcd(m, n).

The r in Theorem 49 (p. 391) is a primitive root of p.

We now prove the existence of primitive roots and then Theorem 49 (p. 391).

aCarl Friedrich Gauss.

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Basic Modular Arithmetics (concluded)

We use

a ≡ b mod n if n | (a − b).

So 25 ≡ 38 mod 13.

We use

a = b mod n

if n | (a − b) and 0 ≤ b < n; in other words, b is the remainder of a divided by n.

– So 25 = 12 mod 13.

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Euler’s

a

Totient or Phi Function

Let

Φ(n) = {m : 1 ≤ m < n, gcd(m, n) = 1}

be the set of all positive integers less than n that are prime to n.b

Φ(12) = {1, 5, 7, 11}.

Define Euler’s function of n to be φ(n) = |Φ(n)|.

φ(p) = p − 1 for prime p, and φ(1) = 1 by convention.

Euler’s function is not expected to be easy to compute without knowing n’s factorization.

aLeonhard Euler (1707–1783).

bZn is an alternative notation.

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    Q











I+Q/

HXOHUSKLQE 

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Two Properties of Euler’s Function

The inclusion-exclusion principlea can be used to prove the following.

Lemma 52 φ(n) = n Q

p|n(1 − 1p).

If n = pe11pe22 · · · pe`` is the prime factorization of n, then φ(n) = n

Y` i=1

µ

1 − 1 pi

.

Corollary 53 φ(mn) = φ(m) φ(n) if gcd(m, n) = 1.

aConsult any textbook on discrete mathematics.

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A Key Lemma

Lemma 54 P

m|n φ(m) = n.

Let Q`

i=1 pkii be the prime factorization of n and consider Y`

i=1

[ φ(1) + φ(pi) + · · · + φ(pkii) ]. (4)

Equation (4) equals n because φ(pki ) = pki − pk−1i by Lemma 52.

Expand Eq. (4) to yield X

k10≤k1,...,k0`≤k`

Y` i=1

φ(pki0i).

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The Proof (concluded)

By Corollary 53 (p. 404), Y`

i=1

φ(pki0i) = φ

à ` Y

i=1

pki0i

! .

So Eq. (4) becomes

X

k10≤k1,...,k`0≤k`

φ

à ` Y

i=1

pki0i

! .

Each Q`

i=1 pki0i is a unique divisor of n = Q`

i=1 pkii.

Equation (4) becomes

Xφ(m).

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The Density Attack for primes

Witnesses to compositeness

of n

All numbers < n

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The Density Attack for primes (continued)

1: for i = 1, 2, . . . , N do

2: Choose 1 ≤ m ≤ n randomly;

3: if m | n then

4: return “n is not a prime”;

5: end if

6: end for

7: return “n is (probably) a prime”;

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The Density Attack for primes (continued)

It works, but does it work well?

The ratio of numbers ≤ n relatively prime to n (the white area) is φ(n)/n.

When n = pq, where p and q are distinct primes, φ(n)

n = pq − p − q + 1

pq > 1 − 1

q 1 p.

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The Density Attack for primes (concluded)

So the ratio of numbers ≤ n not relatively prime to n (the grey area) is < (1/q) + (1/p).

The “density attack” has probability < 2/√

n of factoring n = pq when p ∼ q = O(√

n ).

The “density attack” to factor n = pq hence takes Ω(

n) steps on average when p ∼ q = O(√ n ).

– This running time is exponential: Ω(20.5 log2n).

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The Chinese Remainder Theorem

Let n = n1n2 · · · nk, where ni are pairwise relatively prime.

For any integers a1, a2, . . . , ak, the set of simultaneous equations

x = a1 mod n1, x = a2 mod n2,

...

x = ak mod nk,

has a unique solution modulo n for the unknown x.

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Fermat’s “Little” Theorem

a

Lemma 55 For all 0 < a < p, ap−1 = 1 mod p.

Consider aΦ(p) = {am mod p : m ∈ Φ(p)}.

aΦ(p) = Φ(p).

aΦ(p) ⊆ Φ(p) as a remainder must be between 0 and p − 1.

Suppose am = am0 mod p for m > m0, where m, m0 ∈ Φ(p).

That means a(m − m0) = 0 mod p, and p divides a or m − m0, which is impossible.

aPierre de Fermat (1601–1665).

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The Proof (concluded)

Multiply all the numbers in Φ(p) to yield (p − 1)!.

Multiply all the numbers in aΦ(p) to yield ap−1(p − 1)!.

As aΦ(p) = Φ(p), ap−1(p − 1)! = (p − 1)! mod p.

Finally, ap−1 = 1 mod p because p 6 |(p − 1)!.

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The Fermat-Euler Theorem

a

Corollary 56 For all a ∈ Φ(n), aφ(n) = 1 mod n.

The proof is similar to that of Lemma 55 (p. 412).

Consider aΦ(n) = {am mod n : m ∈ Φ(n)}.

aΦ(n) = Φ(n).

aΦ(n) ⊆ Φ(n) as a remainder must be between 0 and n − 1 and relatively prime to n.

Suppose am = am0 mod n for m0 < m < n, where m, m0 ∈ Φ(n).

That means a(m − m0) = 0 mod n, and n divides a or m − m0, which is impossible.

aProof by Mr. Wei-Cheng Cheng (R93922108, D95922011) on Novem- ber 24, 2004.

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The Proof (concluded)

a

Multiply all the numbers in Φ(n) to yield Q

m∈Φ(n) m.

Multiply all the numbers in aΦ(n) to yield aφ(n) Q

m∈Φ(n) m.

As aΦ(n) = Φ(n), Y

m∈Φ(n)

m = aφ(n)

 Y

m∈Φ(n)

m

 mod n.

Finally, aφ(n) = 1 mod n because n 6 | Q

m∈Φ(n) m.

aSome typographical errors corrected by Mr. Jung-Ying Chen (D95723006) on November 18, 2008.

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An Example

As 12 = 22 × 3,

φ(12) = 12 × µ

1 − 1 2

¶ µ

1 − 1 3

= 4.

In fact, Φ(12) = {1, 5, 7, 11}.

For example,

54 = 625 = 1 mod 12.

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Exponents

The exponent of m ∈ Φ(p) is the least k ∈ Z+ such that mk = 1 mod p.

Every residue s ∈ Φ(p) has an exponent.

1, s, s2, s3, . . . eventually repeats itself modulo p, say si = sj mod p, which means sj−i = 1 mod p.

If the exponent of m is k and m` = 1 mod p, then k|`.

Otherwise, ` = qk + a for 0 < a < k, and

m` = mqk+a = ma = 1 mod p, a contradiction.

Lemma 57 Any nonzero polynomial of degree k has at most k distinct roots modulo p.

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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) that have exponent k.

We already knew that R(k) = 0 for k 6 |(p − 1).

So X

k|(p−1)

R(k) = p − 1 as every number has an exponent.

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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. 417).

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 solutions of xk = 1 mod p.

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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?

Suppose ` < k and ` 6∈ Φ(k) with gcd(`, k) = d > 1.

Then

(s`)k/d = (sk)`/d = 1 mod p.

Therefore, s` has exponent at most k/d, which is less than k.

We conclude that

R(k) ≤ φ(k).

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Size of R(k) (concluded)

Because all p − 1 residues have an exponent, p − 1 = X

k|(p−1)

R(k) ≤ X

k|(p−1)

φ(k) = p − 1

by Lemma 54 (p. 405).

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. 391).

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A Few Calculations

Let p = 13.

From p. 414, 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.

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The Other Direction of Theorem 49 (p. 391)

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 6= 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.

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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. 414).

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. 417).

Note that k | φ(p) (p. 417).

As k ≤ φ(p), k < p − 1.

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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.

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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?

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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.

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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.

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An Algorithm for fsat Using sat

1: t := ²; {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

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Analysis

If sat can be solved in polynomial time, so can fsat.

There are ≤ n + 1 calls to the algorithm for sat.aBoolean 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. 219).

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.

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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 P

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.

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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;

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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?).

There are O(| x | + n2) calls to the algorithm for tsp (d).

So if tsp (d) can be solved in polynomial time, so can tsp.

Hence tsp (d) and tsp are equally hard (or easy).

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Randomized Computation

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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.

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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.

aRabin (1976); Solovay and Strassen (1977).

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“Four Most Important Randomized Algorithms”

a

1. Primality testing.b

2. Graph connectivity using random walks.c 3. Polynomial identity testing.d

4. Algorithms for approximate counting.e

aTrevisan (2006).

bRabin (1976); Solovay and Strassen (1977).

cAleliunas, Karp, Lipton, Lov´asz, and Rackoff (1979).

dSchwartz (1980); Zippel (1979).

eSinclair and Jerrum (1989).

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Bipartite Perfect Matching

We are given a bipartite graph G = (U, V, E).

U = {u1, u2, . . . , un}.

V = {v1, v2, . . . , vn}.

E ⊆ U × V .

We are asked if there is a perfect matching.

A permutation π of {1, 2, . . . , n} such that (ui, vπ(i)) ∈ E

for all i ∈ {1, 2, . . . , n}.

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A Perfect Matching

X

X

X

X

X

Y

Y

Y

Y

Y

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Symbolic Determinants

We are given a bipartite graph G.

Construct the n × n matrix AG whose (i, j)th entry AGij is a variable xij if (ui, vj) ∈ E and zero otherwise.

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Symbolic Determinants (concluded)

The determinant of AG is det(AG) = X

π

sgn(π) Yn i=1

AGi,π(i). (5) – π ranges over all permutations of n elements.

sgn(π) is 1 if π is the product of an even number of transpositions and −1 otherwise.

Equivalently, sgn(π) = 1 if the number of (i, j)s such that i < j and π(i) > π(j) is even.a

aContributed by Mr. Hwan-Jeu Yu (D95922028) on May 1, 2008.

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Determinant and Bipartite Perfect Matching

In P

π sgn(π)Qn

i=1 AGi,π(i), note the following:

– Each summand corresponds to a possible perfect matching π.

– All of these summands Qn

i=1 AGi,π(i) are different monomials and will not cancel.

It is essentially an exhaustive enumeration.

Proposition 58 (Edmonds (1967)) G has a perfect matching if and only if det(AG) is not identically zero.

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A Perfect Matching in a Bipartite Graph

X

X

X

X

X

Y

Y

Y

Y

Y

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The Perfect Matching in the Determinant

The matrix is

AG =









0 0 x13 x14 0

0 x22 0 0 0

x31 0 0 0 x35

x41 0 x43 x44 0

x51 0 0 0 x55









.

det(AG) = −x14x22x35x43x51 + x13x22x35x44x51 + x14x22x31x43x55 − x13x22x31x44x55, each denoting a perfect matching.

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How To Test If a Polynomial Is Identically Zero?

det(AG) is a polynomial in n2 variables.

There are exponentially many terms in det(AG).

Expanding the determinant polynomial is not feasible.

– Too many terms.

Observation: If det(AG) is identically zero, then it

remains zero if we substitute arbitrary integers for the variables x11, . . . , xnn.

But what is the likelihood of obtaining a zero when det(AG) is not identically zero?

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Number of Roots of a Polynomial

Lemma 59 (Schwartz (1980)) Let p(x1, x2, . . . , xm) 6≡ 0 be a polynomial in m variables each of degree at most d. Let M ∈ Z+. Then the number of m-tuples

(x1, x2, . . . , xm) ∈ {0, 1, . . . , M − 1}m such that p(x1, x2, . . . , xm) = 0 is

≤ mdMm−1.

By induction on m (consult the textbook).

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Density Attack

The density of roots in the domain is at most mdMm−1

Mm = md

M . (6)

So suppose p(x1, x2, . . . , xm) 6≡ 0.

Then a random

(x1, x2, . . . , xm) ∈ { 0, 1, . . . , M − 1 }m has a probability of ≤ md/M of being a root of p.

Note that M is under our control.

One can raise M to lower the error probability, e.g.

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Density Attack (concluded)

Here is a sampling algorithm to test if p(x1, x2, . . . , xm) 6≡ 0.

1: Choose i1, . . . , im from {0, 1, . . . , M − 1} randomly;

2: if p(i1, i2, . . . , im) 6= 0 then

3: return “p is not identically zero”;

4: else

5: return “p is (probably) identically zero”;

6: end if

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A Randomized Bipartite Perfect Matching Algorithm

a

We now return to the original problem of bipartite perfect matching.

1: Choose n2 integers i11, . . . , inn from {0, 1, . . . , 2n2 − 1}

randomly;

2: Calculate det(AG(i11, . . . , inn)) by Gaussian elimination;

3: if det(AG(i11, . . . , inn)) 6= 0 then

4: return “G has a perfect matching”;

5: else

6: return “G has no perfect matchings”;

7: end if

aLov´asz (1979). According to Paul Erd˝os, Lov´asz wrote his first sig- nificant paper “at the ripe old age of 17.”

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Analysis

If G has no perfect matchings, the algorithm will always be correct.

Suppose G has a perfect matching.

– The algorithm will answer incorrectly with

probability at most n2d/(2n2) = 0.5 with d = 1 in Eq. (6) on p. 447.

Run the algorithm independently k times and output

“G has no perfect matchings” if and only if they all say no.

– The error probability is now reduced to at most 2−k.

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L´oszl´o Lov´asz (1948–)

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Remarks

a

Note that we are calculating

prob[ algorithm answers “no” | G has no perfect matchings ], prob[ algorithm answers “yes” | G has a perfect matching ].

We are not calculatingb

prob[ G has no perfect matchings | algorithm answers “no” ], prob[ G has a perfect matching | algorithm answers “yes” ].

aThanks to a lively class discussion on May 1, 2008.

bNumerical Recipes in C (1988), “[As] we already remarked, statistics is not a branch of mathematics!”

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But How Large Can det(A

G

(i

11

, . . . , i

nn

)) Be?

It is at most

n! ¡

2n2¢n .

Stirling’s formula says n! ∼

2πn (n/e)n.

Hence

log2 det(AG(i11, . . . , inn)) = O(n log2 n) bits are sufficient for representing the determinant.

We skip the details about how to make sure that all intermediate results are of polynomial sizes.

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An Intriguing Question

a

Is there an (i11, . . . , inn) that will always give correct answers for the algorithm on p. 449?

A theorem on p. 544 shows that such a witness exists!

Whether it can be found efficiently is another question.

aThanks to a lively class discussion on November 24, 2004.

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Perfect Matching for General Graphs

Page 438 is about bipartite perfect matching

Now we are given a graph G = (V, E).

V = {v1, v2, . . . , v2n}.

We are asked if there is a perfect matching.

A permutation π of {1, 2, . . . , 2n} such that (vi, vπ(i)) ∈ E

for all vi ∈ V .

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The Tutte Matrix

a

Given a graph G = (V, E), construct the 2n × 2n Tutte matrix TG such that

TijG =







xij if (vi, vj) ∈ E and i < j,

−xij if (vi, vj) ∈ E and i > j, 0 othersie.

The Tutte matrix is a skew-symmetric symbolic matrix.

Similar to Proposition 58 (p. 442):

Proposition 60 G has a perfect matching if and only if det(TG) is not identically zero.

aWilliam Thomas Tutte (1917–2002).

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William Thomas Tutte (1917–2002)

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