[Tutorial] An interesting counting problem related to square product

Правка en18, от SPyofcode, 2021-10-29 19:50:27

The statement:

Given two integer $$$n, k, p$$$, $$$(1 \leq k \leq n < p)$$$.

Count the number of array $$$a[]$$$ of size $$$k$$$ that satisfied

  • $$$1 \leq a_1 < a_2 < \dots < a_k \leq n$$$
  • $$$a_i \times a_j$$$ is perfect square $$$\forall 1 \leq i < j \leq n$$$

Since the number can be big, output it under modulo $$$p$$$.




Yet you can submit the problem for $$$k = 3$$$ here.




Solution for k = 1

The answer just simply be $$$n$$$




Solution for k = 2


Algorithm

We need to count the number of pair $$$(a, b)$$$ that $$$1 \leq a < b \leq n$$$ and $$$a \times b$$$ is perfect square.

Every positive integer $$$x$$$ can be represent uniquely as $$$x = u \times p^2$$$ for some positive integer $$$u, p$$$ and $$$u$$$ as small as possible ($$$u$$$ is squarefree number).

Let represent $$$x = u \times p^2$$$ and $$$y = v \times q^2$$$ (still, minimum $$$u$$$, $$$v$$$ ofcourse).

We can easily proove that $$$x \times y$$$ is a perfect square if and if only $$$u = v$$$.

So for a fixed squarefree number $$$u$$$. You just need to count the number of ways to choose $$$p^2$$$.

The answer will be the sum of such ways for each fixed $$$u$$$.


Implementation

Implementation using factorization
Implementation 1
Implementation 2
Implementation related to Möbius function

Complexity

So about the complexity....

For the implementation using factorization, it is $$$O(n \log n)$$$.

Hint 1
Hint 2
Proof
Bonus

For the 2 implementations below, the complexity is linear.

Hint 1
Hint 2
Hint 3
Hint 4
Proof

For the last implementation, the complexity is Linear

Hint 1
Hint 2
Proof



Solution for general k

Using the same logic above, we can easily solve the problem.

But for a fixed $$$u$$$, we can improve the counting using combinatorics.

Yet nCk is a well known problem therefore I wont describe it here.

O(n) solution



A better solution for k = 2


Idea

In the above approach, we fix $$$u$$$ as a squarefree and count $$$p^2$$$.

But what if I fix $$$p^2$$$ to count $$$u$$$ instead ?

Yet you can see that the first loop now is $$$O(\sqrt{n})$$$, but it will still $$$O(n)$$$ total because of the second loop

Swap for loop implementation

Approach

Let $$$f(n)$$$ is the number of pair $$$(a, b)$$$ that $$$1 \leq a < b \leq n$$$ and $$$(a, b, n)$$$ is a three-term geometric progression.

Let $$$g(n)$$$ is the number of pair $$$(a, b)$$$ that $$$1 \leq a \leq b \leq n$$$ and $$$(a, b, n)$$$ is a three-term geometric progression.

Let $$$F(n) = \overset{n}{\underset{p=1}{\Large \Sigma}} f(p)$$$.

But why do we need these functions anyway

So it is no hard to prove that $$$g(n) = f(n) + 1$$$.

This interesting sequence $$$g(n)$$$ is A000188, having many properties, such as

  • Number of solutions to $$$x^2 \equiv 0 \pmod n$$$.
  • Square root of largest square dividing $$$n$$$.
  • Max $$$gcd \left(d, \frac{n}{d}\right)$$$ for all divisor $$$d$$$.

Well, to make the problem whole easier, I gonna skip all the proofs to use this property (still, you can use the link in the sequence for references).

$$$g(n) = \underset{d^2 | n}{\Large \Sigma} \phi(d)$$$.

From this property, we can solve the problem in $$$O(\sqrt{n})$$$.

Hint 1
Hint 2
Hint 3
Hint 4
Solution

Yet this paper also takes you to something similar.


Implementation

O(sqrt n log log sqrt n) solution
O(sqrt) solution



A better solution for general k


Algorithm

As what clyring decribed here

Let $$$f_k(n)$$$ is the number of set $$$(a_1, a_2, \dots, a_k, n)$$$ that $$$1 \leq a_1 < a_2 < \dots < a_k \leq n$$$ and $$$(a_1, a_2, \dots, a_k, n)$$$ is a $$$(k+1)$$$-term geometric progression.

Let $$$g_k(n)$$$ is the number of set $$$(a_1, a_2, \dots, a_k, n)$$$ that $$$1 \leq a_1 \leq a_2 \leq \dots \leq a_k \leq n$$$ and $$$(a_1, a_2, \dots, a_k, n)$$$ is a $$$(k+1)$$$-term geometric progression.

Let $$$F_k(n) = \overset{n}{\underset{p=1}{\Large \Sigma}} f_k(p)$$$.

Let $$$s_k(n)$$$ is the number of way to choose $$$p^2$$$ among those $$$k$$$ numbers when you fix squarefree $$$u$$$ (though we are doing in reverse)

The formula

Implementation

O(sqrt n log sqrt n)

Complexity

The complexity is $$$O(\sqrt{n} \log \sqrt{n})$$$

Hint 1
Hint 2
Hint 3
Proof



Tasks

Solved A: Can we also use phi function or something similar to solve for $$$k = 3$$$ in $$$O(\sqrt{n})$$$ ?

Solved B: Can we also use phi function or something similar to solve for general $$$k$$$ in $$$O(\sqrt{n})$$$ ?

C: Can we also solve the problem where there can be duplicate: $$$a_i \leq a_j (\forall\ i < j)$$$ and no longer $$$a_i < a_j (\forall\ i < j)$$$ ?

D: Can we solve the problem where there is no restriction between $$$k, n, p$$$ ?

E: Can we solve for negative integers, whereas $$$-n \leq a_1 < a_2 < \dots < a_k \leq n$$$

F: Can we solve for a specific range, whereas $$$L \leq a_1 < a_2 < \dots < a_k \leq R$$$

G: Can we solve for cube product $$$a_i \times a_j \times a_k$$$ effectively ?

H: Can we solve if it is given $$$n$$$ and queries for $$$k$$$ ?

I: Can we solve if it is given $$$k$$$ and queries for $$$n$$$ ?




Contribution

  • Yurushia for pointing out the linear complexity of squarefree sieve.

  • clyring for fixing typos, and the approach for tasks A, B, C, D, E, G

Теги combinatorics

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