This notebook addresses the case where there is one simple zero over each of the four $2$-torsion points, and one double zero descends to one of the points in $\{(0,0), (1,0)\}$ and the other double zero descends to one of the points in $\{(0,1), (1,1)\}$.
We recall that in $\mathcal{H}(2,2,1^4)$, $d_{opt} = 18$ and by the arrangement above, there is a $2$-torsion point in the boundary of each horizontal cylinder that has exactly one simple zero above it.
import re
#This loads all of the 1-cylinder diagrams in H(2,1,1) formatted as python lists
H2_1_1_cyl_diags = [[[[0, 2, 1, 6, 4, 5, 3], [0, 6, 5, 2, 1, 4, 3]]],
[[[0, 2, 5, 1, 3, 6, 4], [0, 2, 6, 5, 1, 3, 4]]],
[[[0, 2, 5, 3, 1, 6, 4], [0, 2, 6, 3, 1, 5, 4]]],
[[[0, 6, 2, 3, 4, 5, 1], [0, 2, 4, 6, 5, 3, 1]]],
[[[0, 6, 4, 1, 2, 3, 5], [0, 1, 3, 6, 5, 4, 2]]]]
#This loads all of the functions for processing cylinder diagrams
%run ./ST5_fcns/cyl_diag_fcns.ipynb
H2_1_1_vertex_data = strat_odd_sc(H2_1_1_cyl_diags)
Among partitions of $36$ into seven odd numbers the minimum of the maximum numbers among all partitions is $2d_{opt}/7 = 6$, which is $6$ in this case.
Solving $36 - 2(6) - 2t_0 \geq 0$ implies that the largest value of $t_0$ is $12$.
Since the situation is symmetric, the same argument applies for $s_0$ and $t_0$ reversed.
In summary:
$$6 \leq s_0, t_0 \leq 12$$
#This loads all of the standard partition functions needed for nearly every case
#It also loads the partition evaluate function
%run ./ST5_fcns/partition_functions.ipynb
if True:
create_sc_partition_file((), part_length = 7, t0_range = range(6,13), d_opt = 18,
filename_root = 'ST5_data//H_2_2_1t4//2_branch_point//partitions//H2_1_1_part')
#Load the partitions
if True:
with open('ST5_data//H_2_2_1t4//2_branch_point//partitions//H2_1_1_part', 'r') as file:
H2_1_1_part = eval_part(file.read())
#This loads all of the align_list functions needed for nearly every case
#This includes the align_list evaluate function
%run ./ST5_fcns/align_list_fcns.ipynb
if True:
for t in range(6,13):
align_list_write_file(H2_1_1_part, H2_1_1_vertex_data,
'ST5_data//H_2_2_1t4//2_branch_point//align_list//H_2_1_1_align_list',
t0 = t)
#This loads all of the visible_align_list functions needed for nearly every case
%run ./ST5_fcns/align_list_visible_fcns.ipynb
if True:
generate_all_align_list_visible_files(range(6,13), range(6,13), 18,
'ST5_data//H_2_2_1t4//2_branch_point//',
'align_list//H_2_1_1_align_list_',
'align_list_visible//H_2_1_1_align_list_')
#At this point, none of the files above need to be loaded for this to run.
#Once the files above are generated, the file sizes are small enough that these can be loaded on the fly.
#This loads all of the visible_align_list functions needed for nearly every case
%run ./ST5_fcns/combine_align_list_visible_fcns.ipynb
if True:
combine_align_list_visible_write_file(s_range = range(6,13),
s_filename_root = 'align_list_visible//H_2_1_1_align_list_',
t_range = range(6,13),
t_filename_root = 'align_list_visible//H_2_1_1_align_list_',
d_opt = 18,
root_dir = 'ST5_data//H_2_2_1t4//2_branch_point//')
#This loads all of the vertical permutation check functions needed for nearly every case
#This function can be run once admissible_list is written and without needing to load any other file
%run ./ST5_fcns/vert_perm_check_fcns.ipynb
if True:
vert_perm_check_file(18, 'ST5_data//H_2_2_1t4//2_branch_point//')
#This checks that all (both) of the vertical permutations have the correct length of 2*d_opt
if True:
all_vert_perm_check_file(18, 'ST5_data//H_2_2_1t4//2_branch_point//')
%run ./ST5_fcns/slope_test_fcn.ipynb
if True:
final_list = slope_test(18, 'ST5_data//H_2_2_1t4//2_branch_point//')