Need help in analyzing the time complexity of this algorithm

Question

There is an algorithm which requires us to print all subsequences of the prime factors of a number. For example, if the number is 6, we will print:

{}, {2}, {3}, {2 3}

Now the given constraint is that the prime factor of a number will not exceed 13.

The complexity is clearly O(pow(2, prime factors of the number))

Now, my doubt is: Since we know that the number of primes less than or equal to 13 is just 6. This makes the worst time complexity O(pow(2, 6)) which is clearly O(1). So, can we call the algorithm a constant time algo, since we already know all the constant values which are possible candidates for the time complexity?

But there are infinitely many numbers whose prime factors are all less than or equal to 13. — r3mainer
– r3mainer, Commented Aug 22, 2018 at 17:39

btilly · Accepted Answer · 2018-08-22 19:07:52Z

1

Not so fast.

If k = 2^n₂ * 3^n₃ * 5^n₅ * 7^n₇ * 11^n₁₁ * 13^n₁₃ then the number of subsequences is (n₂+1) * (n₃+1) * (n₅+1) * (n₇+1) * (n₁₁+1) * (n₁₃+1) and the average length of those sequences is (n₂ + n₃ + n₅ + n₇ + n₁₁ + n₁₃)/2 It is easy to put the bound that each of those terms is O(log(k)). And from that, you can put an upper limit of O(log(k)⁷) on the size of the output. (If p^n_p ≈ k^1/6 for each p then all 6 terms are simultaneously of that scale.)

If the number does not start factored, you have to add in the difficulty of factoring the number first.

answered Aug 22, 2018 at 19:07

btilly

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2 Comments

Leandro Caniglia Over a year ago

Shouldn't the average length of those sequences be the sum of the n_i divided by 6 rather than 2? (there are 6 primes here). This of course doesn't change the conclusions of your answer; it is just side observation.

btilly Over a year ago

@LeandroCaniglia No, because that's the average exponent for a single prime across all sequences, and the exponents evenly from 0 to n_i. That averaging is where the "/2" comes from.

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