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) 2

Frobenius problem: Given relatively prime positive integers a 1 , . . . , a d , we call an in- teger n representable if there exist nonneg- ative integers m 1 , . . . , m d such that n = m 1 a 1 + · · · + m d a d . Find the largest integer (the Frobenius num- ber g ( a 1 , . . . , a d ) ) which is not representable. Consider the partition function ( m 1 , . . . , m d ) ∈ Z d � � ≥ 0 : p { a 1 ,...,a d } ( n ) := # m 1 a 1 + · · · + m d a d = n Frobenius problem: find the largest value for n such that p { a 1 ,...,a d } ( n ) = 0 . Geometri- cally, this partition function enumerates in- teger (“lattice”) points on the n -dilate of the polytope ( x 1 , . . . , x d ) ∈ R d : � � . x j ≥ 0 , x 1 a 1 + · · · + x d a d = 1 3

Some known results: • (Sylvester, 1884) g ( a 1 , a 2 ) = a 1 a 2 − a 1 − a 2 • (Erd¨ os, 1940’s, . . . ) n d − 1 � n d − 2 � p { a 1 ,...,a d } ( n ) = a 1 ··· a d ( d − 1)! + O • (Stanley, Wilf, 1970’s) n p { a 1 ,a 2 } ( n ) = a 1 a 2 + f ( n ) where f ( n ) is periodic in n with period a 1 a 2 . 4

Theorem (Tripathi, B-R) � a − 1 � � a − 1 � 2 n 1 n n p { a 1 ,a 2 } ( n ) = a 1 a 2 − − +1 . a 1 a 2 Here { x } = x − ⌊ x ⌋ denotes the fractional part of x , a − 1 1 a 1 ≡ 1 (mod a 2 ) , and a − 1 2 a 2 ≡ 1 (mod a 1 ) . “The proof is left as an exercise.” 5

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 − 1 ( ab − a − b + n ) − − + 1 a b � � � � = 2 − 1 b − 1 − 1+ n − 1+ n a + n ab − − a b � � − 1 = 1 − 1 If n = 0 use a to obtain a p { a,b } ( ab − a − b ) = � � � � 2 − 1 b − 1 1 − 1 1 − 1 a − − = 0 . a b ≤ 1 − 1 � m � If n > 0 note that a and hence a p { a,b } ( ab − a − b + n ) ≥ � � � � 2 − 1 b − 1 a + n 1 − 1 1 − 1 = n ab − − ab > 0 . a b 6

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 � � b − 1 ( ab − n ) p { a,b } ( ab − n ) = ab − n − a ab � � a − 1 ( ab − n ) − + 1 b − b − 1 n − a − 1 n � � � � = 2 − n ab − − a b ( ⋆ ) b − 1 n a − 1 n � � � � = − n ab + + a 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. 7

Extension: we call an integer n k -representable if p { a 1 ,...,a d } ( n ) = k , that is, n can be repre- sented in exactly k ways. Define g k ( a 1 , . . . , a d ) to be the largest k -representable integer. Theorem (B-R) g k ( a, b ) = ( k +1) ab − a − b This follows directly from Lemma p { a,b } ( n + ab ) = p { a,b } ( n ) + 1 Proof. � � b − 1 ( n + ab ) p { a,b } ( n + ab ) = n + ab − a ab � � a − 1 ( n + ab ) − + 1 b b − 1 n a − 1 n � � � � = n ab + 2 − − a b = p { a,b } ( n ) + 1 8

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 [ g k − 2 ( a, b ) + a + b , g k ( 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 . 9