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Integer Caratheodory theorems. Linear Caratheodory. Given A={a 1 ,…, a n } IR d . For all v cone(A) there exists B A, |B| ≤ d st v cone(B) Proof : Si A n’est pas indep, la comb lin permet d’eliminer un de ses elements.
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Linear Caratheodory Given A={a1,…, an} IRd. For all v cone(A) there exists B A, |B| ≤ d st v cone(B) Proof : Si A n’est pas indep, la comb lin permet d’eliminer un de ses elements.
Hilbert bases : generators with nice integrality properties integer vectors of the cone with integer coefficients: Caratheodorybounds depending on n. Cook, Fonlupt, Schrijver (1984) : 2d – 1 S. (1990) : 2d – 2, d for d=3 d in combopt cases, and stronger conjectures Bruns, Gubeladze (1999): counterexamples …
multiprocessor scheduling C fixed integer (typically small) INPUT:l1,…,lC job lengths, n1,…, nC coeffs,m,DIN QUESTION:Is there a schedule with m machines, in D time ? McCormick, Smallwood, Spieksma ‘93: ‘C=2’ P C arbitrary: NP-hard C fixed : open ‘bin packing with condensed input’
Example C=3 D=30 l=(15, 10, 6) n=(1, 2, 4) 2 0 0 1 1 0 1 0 3 0 1 0 2 2 0 0 5 0 2 1 4 Min = ? 2 or 3 ? 3 (If 2, we need a solution with a gap of 1)
Cutting Stock C integer. Given n, lINC , D IN M = n where the columns ≥ 0 of M are solutions of integer lTx ≤ D, x integer minimize the sum of NP-hard ; Is it in NP ? Yes ! (Eisenbrand ’05)
Trilling: le problème de l’imprimeur C modèles D poses imprimées sur le même film demandes: ni du modèle i (i=1, …, C) . Trouver le nombre d de films différents avec au plus D poses sur chacun la multiplicité j (j=1, …, d) des films, tq les demandes sont satisfaites et 10 000 d + (1 + … + d ) est min ‘m’
Example C=5 D=6 (Si C, D fix: polynomial !) n=(29976, 35121, 20749, 75286, 90959) Minimiser 10 000 fois le nombre de films + nombre de tirages 1 0 29976 1 0 35121 0 3 20749 2 0 75286 2 3 90959 37643 6917 1 0 0 3 0 2 2 1 3 0 31790 11707
New Caratheodory Thm 1: X +d , n:= |X| ≥ 2 and bint.cone (X) Then B X, |B| ≤ log2 (bi +1): bint.cone (B) Proof: If |X| > log2(bi +1), then X has two disjoint subsets with equal sums. Thm 2: X +d , n:= |X| ≥ 2 and bint.cone (X) Then B X, |B| ≤d log(2nM+1) bint.cone (B) M:=max | | in X;=d1+ + d log2 (M+1) d log d