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 nest.

Présentation au sujet: "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 nest."— Transcription de la présentation:

1 Integer Caratheodory theorems

2 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 nest pas indep, la comb lin permet deliminer un de ses elements.

3 Hilbert bases : generators with nice integrality properties integer vectors of the cone with integer coefficients: Caratheodory bounds 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 …

4 multiprocessor scheduling C fixed integer (typically small) INPUT: l 1,…,l C job lengths, n 1,…, n C coeffs,m,D IN 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

5 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)

6 Cutting Stock C integer. Given n, l IN C, D IN M = n where the columns 0 of M are solutions of integer l T x D, x integer minimize the sum of NP-hard ; Is it in NP ? Yes ! (Eisenbrand 05)

7 Trilling: le problème de limprimeur C modèles D poses imprimées sur le même film demandes: n i 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

8 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

9 New Caratheodory Thm 1: X + d, n:= |X| 2 and b int.cone (X) Then B X, |B| log 2 (b i +1): b int.cone (B) Proof: If |X| > log 2 (b i +1), then X has two disjoint subsets with equal sums. Thm 2: X + d, n:= |X| 2 and b int.cone (X) Then B X, |B| d log(2nM+1) b int.cone (B) M:=max | | in X; =d 1+ + d log 2 (M+1) d log d