python - backtracking - solution hint for generating schedule -

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i'm trying solve following problem:

there 3 hours of math, 2 of phisics , 3 of informatics. need generate possible schedules day must have @ least 1 hour of these subjects , @ maximum 3 of these (it math-math-math or math-physics-informatics)

the schedule 5 days. came following solution:

let generate(k, o) 'generator' function k-th day o hours, o between 1 , 3

for every day k , hour o pick subject still in our left. keep track of having array counter[i] = takencount, 1, 2, 3 corresponding math, physics , informatics, , takencount means how many hours of subject used.

and configuration use matrix v[k][o] gives me configuration day k, hours o

this code i've tried far

h = ['', 'm', 'f', 'i'] counter = [0, 0, 0, 0] limit = [0, 3, 2, 3] v = [['' x in range(4)] x in range(6)]  def generate(k, o):   if k == 6:     if counter[1] == limit[1] , counter[2] == limit[2] , counter[3] == limit[3]:       showconfig()   else:     in range(1, 4, 1):       if counter[i] != limit[i]:         v[k][o] = h[i]         counter[i] += 1          generate(k + 1, 1)         generate(k + 1, 2)         generate(k + 1, 3)          v[k][o] = ''         counter[i] -= 1  def showconfig():   print '-- configuration --'   print counter    in range(1, 6, 1):     print 'day ', i,     success = false      j in range(1, 4, 1):       print v[i][j],    print '-- end --'  generate(1, 1) generate(1, 2) generate(1, 3)

it's written in python think that's not problem, because it's pretty easy read

the problem here never have hours inside v[i][2] , v[i][3], v[i][1], , condition

counter[1] == 3, counter[2] == 2 , counter[3] == 3

never gets true , not know why.

also after generate following day, k + 1, hours 1, 2 , 3, need reset current v[k][o] to '' , counter[i] counter[i] - 1

this not homework or simillar. i'm asking if solution , if is, did wrong in code? because can't seem find out

output example:

( (m), (m, f), (m, f, i), (i), (i) ) ... ( (m, i), (f), (i), (m, f, i), (m) ) ( (m, f), (i), (i), (m, f, i), (m) )

where m math, f physics , informatics, , each paranthesis day in week, 5 days in 1 week

here want:

from itertools import permutations def possible_permutations(*args):     length_args = len(args)     args_list, permutations_list = [], []     arg in args:         args_list.append(arg)         permutations_list.append(tuple(permutations(args_list)))     # clean ugly list of touples of touples list of touples.     perms = []     perm in permutations_list:         p in perm:             perms.append(p)     return perms  def schedules(days, *classes):     possible_perms = possible_permutations(*classes)     perm in permutations(possible_perms, days):         print(perm) schedules(5, 'math', 'science', 'english')

old answer

it not entirly clear me asking, belive solves question.

from itertools import permutations  classes = {'math':3, 'physics':2, 'informatics':3} def possible_schedules(**classes_hours):     k, v = [k k, v in classes_hours.items()], [v k, v in classes_hours.items()]     k_perms, v_perms = permutations(k), permutations(v)     perms = zip(k_perms, v_perms)     k_perm, v_perm in perms:         perm = tuple(zip(k_perm, v_perm))         print(perm)  possible_schedules(**classes)


output:

(('physics', 2), ('math', 3), ('informatics', 3)) (('physics', 2), ('informatics', 3), ('math', 3)) (('math', 3), ('physics', 2), ('informatics', 3)) (('math', 3), ('informatics', 3), ('physics', 2)) (('informatics', 3), ('physics', 2), ('math', 3)) (('informatics', 3), ('math', 3), ('physics', 2))

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