Polytopes, lattice points, and a problem of Frobenius Matthias Beck - - PDF document

Polytopes, lattice points, and a problem of Frobenius Matthias Beck SUNY Binghamton Sinai Robins Temple University www.math.binghamton.edu/matthias

“If you think it’s simple, then you have mis- understood the problem” Bjarne Strustrup (lecture at Temple University, 11/25/97)

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Frobenius problem: Given relatively prime positive integers a1, . . . , ad , we call an in- teger n representable if there exist nonneg- ative integers m1, . . . , md such that n = m1a1 + · · · + mdad . Find the largest integer (the Frobenius num- ber g(a1, . . . , ad) ) which is not representable. Consider the partition function p{a1,...,ad}(n) := #

• (m1, . . . , md) ∈ Zd

≥0 :

m1a1 + · · · + mdad = n

• Frobenius problem: find the largest value for

n such that p{a1,...,ad}(n) = 0 . Geometri- cally, this partition function enumerates in- teger (“lattice”) points on the n -dilate of the polytope

• (x1, . . . , xd) ∈ Rd :

xj ≥ 0, x1a1 + · · · + xdad = 1

• .

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Some known results:

• (Sylvester, 1884)

g(a1, a2) = a1a2 − a1 − a2

• (Erd¨
• s, 1940’s, . . . )

p{a1,...,ad}(n) =

nd−1 a1···ad(d−1)!+O

• nd−2
• (Stanley, Wilf, 1970’s)

p{a1,a2}(n) =

n a1a2 + f(n)

where f(n) is periodic in n with period a1a2 .

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Theorem (Tripathi, B-R) p{a1,a2}(n) =

n a1a2−

• a−1

2 n

a1

• −
• a−1

1 n

a2

• +1 .

Here {x} = x − ⌊x⌋ denotes the fractional part of x , a−1

1 a1 ≡ 1 (mod a2) ,

and a−1

2 a2 ≡ 1 (mod a1) .

“The proof is left as an exercise.”

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Corollary (Sylvester) g(a, b) = ab − a − b Proof. p{a,b}(ab − a − b + n) = ab−a−b+n

ab

−

• b−1(ab−a−b+n)

a

• −
• a−1(ab−a−b+n)

b

• + 1

= 2 − 1

b − 1 a + n ab −

• −1+n

a

• −
• −1+n

b

• If n = 0 use
• −1

a

• = 1 − 1

a to obtain

p{a,b}(ab − a − b) = 2 − 1

b − 1 a −

• 1 − 1

a

• −
• 1 − 1

b

• = 0 .

If n > 0 note that m

a

• ≤ 1 − 1

a and hence

p{a,b}(ab − a − b + n) ≥ 2 − 1

b − 1 a + n ab −

• 1 − 1

a

• −
• 1 − 1

b

• = n

ab > 0 .

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Corollary (Sylvester) Exactly half of the in- tegers between 1 and (a − 1)(b − 1) are rep- resentable.

• Proof. If n ∈ [1, ab − 1] is not a multiple of

a or b then p{a,b}(ab − n) = ab−n

ab

−

• b−1(ab−n)

a

• −
• a−1(ab−n)

b

• + 1

= 2 − n

ab −

• −b−1n

a

• −
• −a−1n

b

• (⋆)

= − n

ab +

• b−1n

a

• +
• a−1n

b

• = 1 − p{a,b}(n) .

(⋆) follows from {−x} = 1 − {x} if x ∈ Z . Hence for n between 1 and ab − 1 and not divisible by a or b , exactly one of n and ab − n is not representable. There are ab − a − b + 1 = (a − 1)(b − 1) such integers.

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Extension: we call an integer n k-representable if p{a1,...,ad}(n) = k , that is, n can be repre- sented in exactly k ways. Define gk(a1, . . . , ad) to be the largest k-representable integer. Theorem (B-R) gk(a, b) = (k+1)ab−a−b This follows directly from Lemma p{a,b}(n + ab) = p{a,b}(n) + 1 Proof. p{a,b}(n + ab) = n+ab

ab

−

• b−1(n+ab)

a

• −
• a−1(n+ab)

b

• + 1

= n

ab + 2 −

• b−1n

a

• −
• a−1n

b

• = p{a,b}(n) + 1

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More exercises:

• Given k ≥ 2 , the smallest k-representable

integer is ab(k − 1) .

• Given k ≥ 2 , the smallest interval con-

taining all k-representable integers is [ gk−2(a, b) + a + b , gk(a, b) ] .

• There are ab − 1 uniquely representable
• integers. Given k ≥ 2 , there are exactly

ab k-representable integers.

• Extend all of this to d > 2 .

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