Implicit Prime Factorisation

Правка en2, от Everule, 2022-10-16 07:12:08

Those of us who have been doing cp for a long time, have come to love the representation of numbers as an infinite prime power vector i.e.

$$$n = \prod p_i^{e_i} = [e_1, e_2, \ldots ] = v_n$$$

Where $$$n$$$ is a number and $$$p_i$$$ is the $$$i$$$ th prime number. This has many useful properties for problems, such as mapping

$$$n \times m \sim v_n + v_m$$$
$$$n / m \sim v_n - v_m$$$
$$$n \mid m \sim v_n \le v_m$$$
$$$\gcd(n, m) \sim \min(v_n, v_m)$$$
$$$lcm(n, m) \sim \max(v_n, v_m)$$$

All the above operations are done elementwise. However in some cases, the problem authors may forbid you from factorising the numbers, by perhaps increasing $$$n$$$ to around $$$10^{36}$$$ where there isn't any simple factorisation algorithm to do it in reasonable time. However using effective prime factorisation, we can use these properties freely on a set of numbers. If this set contained all integers, then it would require us to truly prime factorise the integer. However, we only need to use these operations on a subset of integers, making our task a lot simpler. We will make a finite array $$$q$$$ of size $$$t$$$ with the following properties.

$$$(1)$$$ Any element in $$$x \in S$$$ is representable by $$$q$$$, i.e. there exists $$$[f_1, f_2, \ldots f_t] = v'_n$$$ such that $$$x = \prod q_i^{f_i}$$$

$$$(2)$$$ For all the above operations, replacing $$$v_n$$$ with $$$v'_n$$$ should produce the same result.

Let us show, that if $$$q$$$ satisfies $$$(1)$$$ and $$$gcd(q_i, q_j) = 1$$$ for all $$$i < j$$$, then $$$q$$$ satisfies $$$(2)$$$.

Proofs

Let us now consider how we can find $$$q$$$ given $$$S$$$. Let us proceed element wise, and assume we have a valid solution, and we want to add another element.

Now let us add the next element $$$x \in S$$$. Go through $$$q$$$ until we spot an element that has a common factor with $$$x$$$. Let that element be $$$q_i$$$.

Let $$$P(q)$$$ be the set of elements produced by some product of elements in $$$q$$$. Notice that if $$$q$$$ has 2 elements that divide each other $$$a \mid b$$$, then $$$a, b$$$ is replaceable by $$$a, \frac{b}{a}$$$, and $$$P$$$ will not lose any elements. Let us do this recursively until neither divides the other, or an element becomes 1. Let us call this procedure reduce_pair.

(It maybe useful to look at the example implementation in parallel from here)

We run the procedure reduce_pair on the elements $$$q_i, x$$$, if $$$x$$$ becomes 1, we're done. If $$$q_i$$$ becomes 1, we remove it from the set and continue.

We then run an iterative algorithm on a list $$$L$$$, initialised with $$$[q_i, x]$$$. Let the elements before $$$L[ptr]$$$ be all co-prime. Initialise $$$ptr$$$ with 1. Run reduce_pair on all elements before $$$ptr$$$ with it. Then add their gcd to the end of the list if its not one, and divide both elements by it. This will make all elements upto $$$ptr + 1$$$ coprime. There can only be as many elements in this list as there are true primes in $$$q_i, x$$$. Therefore we can deduce that the size of the list will be at most $$$O(\log{\log{max(S)}})$$$.

Since this algorithm will be run $$$O(\log{x})$$$ times, the total algorithm will run in $$$O(n^2\log{A} + n\log{A} \times (\log{\log{A}})^2)$$$ Where $$$A = max(S)$$$, and we will get our array $$$q$$$.

Example implementation : 159590539

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en2 Английский Everule 2022-10-16 07:12:08 56
en1 Английский Everule 2022-10-16 06:58:29 4506 Initial revision (published)