The scripts provided here for computing these bounds are awk scripts, named reductionvectorscript10-17-10 and HFBoundsAndBettis10-17-2010. They are run as follows:

unix$ awk -f reductionvectorscript10-17-10 File1 File2 | awk -f HFBoundsAndBettis10-17-2010

File1 contains the matrix M. For 4 points, where 3 are collinear and 1 is off that line, the matrix M (and hence the lines in the file File1) would be as follows if we assume point 1 is the point off the line and points 2, 3 and 4 are the collinear points:

1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1Row 1 indicates a line goes through points 1 and 2, row 2 specifies a line goes through points 1 and 3, row 3 says a line goes through points 1 and 4, and row 4 says points 2, 3 and 4 are collinear. These rows can be given in any order.

File2 specifies the multiplicities m

2 3 4 5 3 4 5 2 3 3 3 3we have three fat points schemes, Z = 2p

Here is the output you would get if you run

awk -f reductionvectorscript10-17-10 File1 File2 | awk -f HFBoundsAndBettis10-17-2010

using the files as above:

unix$ awk -f reductionvectorscript10-17-10 File1 File2 | awk -f HFBoundsAndBettis10-17-2010 ******************************* Input vector: 12 9 6 4 2 1 Upper and lower Hilbert functions ARE equal. alpha lwrbnd= 6 lower bound on dim of I_t for t >= alpha-1: 0 3 8 15 23 33 44 57 alpha uppbnd= 6 upper bound on dim of I_t for t >= alpha-1: 0 3 8 15 23 33 44 57 Delta uprbnd= 1 2 3 4 5 6 4 3 2 2 1 1 Delta lwrbnd= 1 2 3 4 5 6 4 3 2 2 1 1 uprbnd= 1 3 6 10 15 21 25 28 30 32 33 34 lwrbnd= 1 3 6 10 15 21 25 28 30 32 33 34 The input is Bezout, so we now give upper and lower bounds on the Betti numbers. Note: the upper and lower bounds coincide. Upper bounds for the graded Betti numbers: nmbr of gens = 0 0 0 0 0 0 3 1 1 0 1 0 1 0 0 nmbr of syzs = 0 0 0 0 0 0 0 2 1 1 0 1 0 1 0 Lower bounds for the graded Betti numbers: nmbr of gens = 0 0 0 0 0 0 3 1 1 0 1 0 1 0 0 nmbr of syzs = 0 0 0 0 0 0 0 2 1 1 0 1 0 1 0 ******************************* Input vector: 11 8 6 4 3 2 Upper and lower Hilbert functions ARE equal. alpha lwrbnd= 6 lower bound on dim of I_t for t >= alpha-1: 0 1 6 13 22 32 44 alpha uppbnd= 6 upper bound on dim of I_t for t >= alpha-1: 0 1 6 13 22 32 44 Delta uprbnd= 1 2 3 4 5 6 6 3 2 1 1 Delta lwrbnd= 1 2 3 4 5 6 6 3 2 1 1 uprbnd= 1 3 6 10 15 21 27 30 32 33 34 lwrbnd= 1 3 6 10 15 21 27 30 32 33 34 The input is Bezout, so we now give upper and lower bounds on the Betti numbers. Note: the upper and lower bounds coincide. Upper bounds for the graded Betti numbers: nmbr of gens = 0 0 0 0 0 0 1 3 1 1 0 1 0 0 nmbr of syzs = 0 0 0 0 0 0 0 0 3 1 1 0 1 0 Lower bounds for the graded Betti numbers: nmbr of gens = 0 0 0 0 0 0 1 3 1 1 0 1 0 0 nmbr of syzs = 0 0 0 0 0 0 0 0 3 1 1 0 1 0 ******************************* Input vector: 9 6 4 3 2 Upper and lower Hilbert functions ARE equal. alpha lwrbnd= 5 lower bound on dim of I_t for t >= alpha-1: 0 1 6 13 21 31 alpha uppbnd= 5 upper bound on dim of I_t for t >= alpha-1: 0 1 6 13 21 31 Delta uprbnd= 1 2 3 4 5 5 2 1 1 Delta lwrbnd= 1 2 3 4 5 5 2 1 1 uprbnd= 1 3 6 10 15 20 22 23 24 lwrbnd= 1 3 6 10 15 20 22 23 24 The input is Bezout, so we now give upper and lower bounds on the Betti numbers. Note: the upper and lower bounds coincide. Upper bounds for the graded Betti numbers: nmbr of gens = 0 0 0 0 0 1 3 1 0 1 0 0 nmbr of syzs = 0 0 0 0 0 0 0 3 1 0 1 0 Lower bounds for the graded Betti numbers: nmbr of gens = 0 0 0 0 0 1 3 1 0 1 0 0 nmbr of syzs = 0 0 0 0 0 0 0 3 1 0 1 0