Dynamic programming--optimal binary search tree

Posted 2020-07-22

key_probability = [-1, 0.15, 0.1, 0.05, 0.1, 0.2]
virtual_key_probability = [0.05, 0.1, 0.05, 0.05, 0.05, 0.1]

expect_cost = [[0] *10 for i in range(10)]
probability_sum =  [[0] *10 for i in range(10)]
root = [[0] *10 for i in range(10)]

def optimal_bst():
 global expect_cost
 global probability_sum
 for i in range(1, 6):
  expect_cost[i][i-1] = virtual_key_probability[i-1]
  probability_sum[i][i-1] = virtual_key_probability[i-1]
 
 for step in range(0, 5):
  for i in range(1, 6):
   this_time_end = i + step
   if this_time_end > 5:
    break
   probability_sum[i][this_time_end] = probability_sum[i][this_time_end-1] + key_probability[this_time_end] + virtual_key_probability[this_time_end]
   # print i,this_time_end, probability_sum[i][this_time_end]
   print "item", expect_cost[i][i-1], expect_cost[i+1][this_time_end], probability_sum[i][this_time_end]
   print "value", expect_cost[i][i-1]+expect_cost[i+1][this_time_end]+probability_sum[i][this_time_end]
   expect_cost_min = expect_cost[i][i-1] + expect_cost[i+1][this_time_end] + probability_sum[i][this_time_end]
   print "expect", expect_cost_min
   root[i][this_time_end] = i
   # print i, this_time_end, expect_cost_min
   for j in range(i, this_time_end+1): 
    cur_cost = expect_cost[i][j-1] + expect_cost[j+1][this_time_end] +  probability_sum[i][this_time_end]
    if cur_cost < expect_cost_min:
     expect_cost_min = cur_cost
     root[i][this_time_end] =j

   
   expect_cost[i][this_time_end] = expect_cost_min
   # print i, this_time_end, expect_cost[i][this_time_end]

def find_root(begin,end,r):
 if begin >= end:
  return
 k = root[begin][end]
 print begin, end, k
 find_root(begin,k-1,r)
 find_root(k+1,end,r)

optimal_bst()

# find_root(1,5)
for i in range(1,6):
 print root[i]
# print root
print "value",expect_cost[1][5]