def all_NaN(matrix):
for row in matrix:
for x in row:
if not isNaN(x):
# NaN is not equal to itself, so we have an actual number here
return False
return True
def jap_score(m, sol):
if all_NaN(m):
raise ValueError("Matrix is all NaNs.")
total = 0
for i in range(len(m)):
for j in range(len(m[0])):
mm = m[i][j]
ss = sol[i][j]
# watch out for NaNs
if isNaN(mm):
# skip this element, so we don't have NaN scores for subproblems
continue
total += mm * ss
if total != total: # NaN
raise ValueError(
"Score is NaN for matrix {0} and solution {1}.".format(m, sol))
return total
def get_second_bests(normalized_matrix):
"""
Takes a normalized_matrix in which every row has a maximum value of zero.
Returns a list of the maximum non-zero value in each row.
"""
if max([max(row) for row in normalized_matrix]) > 0:
raise ValueError("Matrix must have maximum value of zero.")
result = []
for row in normalized_matrix:
non_zeros = [i for i in filter(lambda x: x != 0, row)]
if non_zeros == []: # all values in the row are zero; i.e. the same
result.append(0)
else:
non_NaNs = [i for i in filter(lambda x: not isNaN(x), non_zeros)]
if non_NaNs == []:
result.append(NaN)
else:
result.append(max(non_NaNs))
return result
def nchildren(self, v=0):
v = int(v)
assert not isNaN(v) # JS wont fail but make NaN
return v
def age(self, v=0):
v = float(v)
assert not isNaN(v) # JS wont fail but make NaN
return v
def yell(self, v):
v = int(v)
assert not isNaN(v) # JS wont fail but make NaN
return {'action': 'yell', 'value': v}
def jap(m):
"""
Takes input m (the n*m matrix where element (i,j) tells how fit person i is for job j).
Returns an n*m matrix where element (i,j) is 1 if person i is assigned to job j and 0 otherwise.
"""
original_m = [[m[i][j] for j in range(len(m[0]))] for i in range(len(m))]
# print("original matrix:",original_m)
# Note that we assume WLOG n <= m, so we assign each person to one job.
# If there are more people than jobs, then note the symmetry of the problem, so just transpose m, solve, and then transpose the result.
need_to_transpose = len(m) > len(m[0])
if need_to_transpose:
m = transpose(m)
N = len(m)
M = len(m[0])
pairs = [] # to be list of tuples (set of tuples could also work)
#if all_NaN(m): # if it's somehow already all NaN
#raise ValueError("Input matrix is all NaNs.")
limiter = 0
limit = 10**4
while not all_NaN(m): # m will be all NaNs when we are done assigning jobs
limiter += 1
if limiter > limit:
raise RuntimeError("While loop ran too many times.")
# print("this remnant of original matrix:",m)
for r in range(N): # for each row
row_without_NaN = [i for i in filter(lambda x: not isNaN(x), m[r])]
if row_without_NaN == []:
# print("skipping normalizing row {0} due to all NaN".format(r))
continue
best = max(row_without_NaN) # best fit value for this person
# print("best val:",best)
m[r] = [
i - best for i in m[r]
] # subtract best value from everything in the row; everything will end up non-positive
# print("normalized row:",m[r])
# print("THIS SHOULD NOT BE SKIPPED IN ANY CALL")
# method: find the rows with the greatest gap between the best and the second-best columns; these must be given highest priority
# keep in mind that a row which is all NaN will have been skipped already, so if second_bests is all NaN then there is one zero in the row
# if there is just one zero in one row, then that pair should be part of the solution! this is most often the form of the base case
best_pair = None
best_pair_found = False
second_bests = [
i for i in filter(lambda x: True, get_second_bests(m))
] # not isNaN(x)
if second_bests == []:
# print("breaking because second_bests is empty")
break
elif all_NaN([second_bests]):
singleton_rows = [
i for i in filter(
lambda row: isNaN(second_bests[row]) and not all_NaN(
[m[row]]), range(N))
]
# print("row indices with one non-NaN value:",singleton_rows)
# print("breaking because second_bests is all NaN")
best_pair = (singleton_rows[0], m[singleton_rows[0]].index(0)
) # just take the first one for now
best_pair_found = True
if not best_pair_found:
lowest_second_best = min(
[i for i in filter(lambda x: not isNaN(x), second_bests)])
# print("lowest_second_best:",lowest_second_best)
#if isNaN(lowest_second_best): # this was causing problems because we needed to remove NaN before calling min()
#break
important_row_indices = [
i for i in filter(
lambda r: second_bests[r] == lowest_second_best, range(N))
] # prioritize the rows with the worst second-bests
# print("next priority row indices (lowest_second_best = {0}): {1}".format(lowest_second_best,important_row_indices))
if len(important_row_indices) == 0:
raise IndexError(
"No rows selected for priority with matrix {0}.".format(m))
best_score = -Inf
# best_pair = None # already initialized
for important_row_index in important_row_indices:
important_row = m[important_row_index]
# print("this important row:",important_row)
for zero_index in filter(lambda x: important_row[x] == 0,
range(M)):
# print("this zero index for row {0}: {1}".format(important_row,zero_index))
this_eliminated_matrix = eliminate(m, important_row_index,
zero_index)
# print("eliminating row and col to produce matrix",this_eliminated_matrix)
if all_NaN(this_eliminated_matrix):
best_pair = (important_row_index, zero_index)
# print("breaking because elimination produced all NaNs, adding this pair to solution")
break
# print("making recursive call\n")
subsolution = jap(
this_eliminated_matrix
) # recursive call; watch out for treachery
# print("next subsolution:",subsolution)
# print("scoring matrix",this_eliminated_matrix)
score = jap_score(this_eliminated_matrix, subsolution)
if score > best_score:
# print("subsolution accepted")
best_score = score
best_pair = (important_row_index, zero_index)
else:
pass # print("subsolution rejected")
# print("best score found for subproblems:",best_score)
pairs.append(best_pair)
m = eliminate(m, best_pair[0], best_pair[1])
result = [[1 if (i, j) in pairs else 0 for j in range(M)]
for i in range(N)] # turn the ordered pairs into sparse matrix
# take out rows and columns by replacing them with NaN, then repeating procedure until whole matrix is NaN
# DON'T ever actually remove rows or columns! This will make indices much harder to work with.
# transpose back if we transposed before solving
if need_to_transpose:
result = transpose(result)
# if not is_fully_assigned(result): # note that subproblem results will not be all assigned, so don't actually check for this
# pass # raise ValueError("Result {0} has left some possible pairs unassigned.".format(result))
# print("returning result",result,"\n")
return result
def age(self, v=0):
v = float(v)
assert not isNaN(v) # JS wont fail but make NaN
return v
def yell(self, v):
v = int(v)
assert not isNaN(v) # JS wont fail but make NaN
return {'action': 'yell', 'value': v}
def nchildren(self, v=0):
v = int(v)
assert not isNaN(v) # JS wont fail but make NaN
return v
