chrisl0lz chrisl0lz - 6 months ago 330
Python Question

Primitive Calculator - Dynamic Approach

I'm having some trouble getting the correct solution for the following problem:


Your goal is given a positive integer n, find the minimum number of
operations needed to obtain the number n starting from the number 1.


More specifically the test case I have in the comments below.

# Failed case #3/16: (Wrong answer)
# got: 15 expected: 14
# Input:
# 96234
#
# Your output:
# 15
# 1 2 4 5 10 11 22 66 198 594 1782 5346 16038 16039 32078 96234
# Correct output:
# 14
# 1 3 9 10 11 22 66 198 594 1782 5346 16038 16039 32078 96234
# (Time used: 0.10/5.50, memory used: 8601600/134217728.)


def optimal_sequence(n):
sequence = []

while n >= 1:
sequence.append(n)

if n % 3 == 0:
n = n // 3
optimal_sequence(n)

elif n % 2 == 0:
n = n // 2
optimal_sequence(n)

else:
n = n - 1
optimal_sequence(n)

return reversed(sequence)

input = sys.stdin.read()
n = int(input)
sequence = list(optimal_sequence(n))
print(len(sequence) - 1)
for x in sequence:
print(x, end=' ')


It looks like I should be outputting 9 where I'm outputting 4 & 5 but I'm not sure why this isn't the case. What's the best way to troubleshoot this problem?

Answer

You are doing a greedy approach. Whey you have n == 10 you check and see it's divisible by 2 so you assume that's the best step, which is wrong in this case.

What you need to do is proper dynamic programming. v[x] will hold the minimum number of steps to get to result x.

def solve(n):
  v = [0]*(n+1)  # so that v[n] is there
  v[1] = 1  # length of the sequence to 1 is 1
  for i in range(1,n+1):
    if not v[i]: continue
    if v[i+1] == 0 or v[i+1] > v[i] + 1: v[i+1] = v[i] + 1
    # Similar for i*2 and i*3

  solution = []
  while n > 1:
    solution.append(n)
    if v[n-1] == v[n] - 1: n = n-1
    if n%2 == 0 and v[n//2] == v[n] -1: n = n//2
    # Likewise for n//3
  solution.append(1)
  return reverse(solution)
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