A bit more of general ideas

Правка en7, от adamant, 2022-07-03 16:04:31

Hi everyone!

Here's another collection of little tricks and general ideas that might make your life better (or maybe worse).

1. Evaluating polynomial modulo small prime $$$p$$$. Given a polynomial $$$q(x)$$$, you may evaluate it modulo $$$p$$$ in all possible arguments. To do this, compute $$$q(0)$$$ separately and use chirp Z-transform to compute $$$q(g^0), q(g^1), \dots, q(g^{p-2})$$$, where $$$g$$$ is a primitive root modulo $$$p$$$.

This method can be used to solve 1054H - Epic Convolution.

2. Generalized Euler theorem. Let $$$a$$$ be a number, not necessarily co-prime with $$$m$$$, and $$$k > \log_2 m$$$. Then

$$$ a^k \equiv a^{\phi(m) + k \mod \phi(m)} \pmod m, $$$

where $$$\phi(m)$$$ is Euler's totient. This follows from the Chinese remainder theorem, as it trivially holds for $$$m=p^d$$$.

This fact can be used in 906D - Power Tower.

3. Range add/range sum in 2D. Fenwick tree, generally, allows for range sum/point add queries.

Let $$$s_{xy}$$$ be a sum on $$$[1,x] \times [1,y]$$$. If we add $$$c$$$ on $$$[a, +\infty) \times [b, +\infty)$$$, the sum $$$s_{xy}$$$ would change as

$$$ s_{xy} \mapsto s_{xy} + (x-a+1)(y-b+1)c, $$$

for $$$x \geq a$$$ and $$$y \geq b$$$. To track these changes, we may represent $$$s_{xy}$$$ as

$$$ s_{xy} = s_{xy}^{(0)}+ x \cdot s_{xy}^{(x)} + y \cdot s_{xy}^{(y)} + xy \cdot s_{xy}^{(xy)}, $$$

which allows us to split the addition of $$$c$$$ on $$$[a,+\infty) \times [b,+\infty)$$$ into additions in $$$(a;b)$$$:

$$$\begin{align} s_{xy}^{(0)} &\mapsto s_{xy}^{(0)} + (a-1)(b-1)c, \\ s_{xy}^{(x)} &\mapsto s_{xy}^{(x)} - (b-1)c, \\ s_{xy}^{(y)} &\mapsto s_{xy}^{(y)} - (a-1)c, \\ s_{xy}^{(xy)} &\mapsto s_{xy}^{(xy)} + c. \end{align}$$$
code

The solution generalizes 1-dimensional Fenwick tree range updates idea from Petr blog from 2013.

4. DP on convex subsets. You want to compute something related to convex subsets of a given set of points in 2D space.


You sort points over bottom-left point $$$O$$$, then over point $$$B$$$ and go through all pairs $$$(A, C)$$$ with two pointers

This can be done with dynamic programming, which generally goes as follows:

  1. Iterate over possible bottom left point $$$O$$$ of the convex subset;
  2. Ignore points below it and sort points above it by angle that they form with $$$O$$$;
  3. Iterate over possible point $$$B$$$ to be the "last" in the convex subset. It may only be preceded by a point that was sorted before it and succeeded by a points that was sorted after it when the points were sorted around $$$O$$$;
  4. Sort considered points around $$$B$$$, separately in "yellow" and "green" areas (see picture);
  5. Iterate over possible point $$$C$$$ which will succeed $$$B$$$ in the convex subset;
  6. Set of points that may precede $$$B$$$ with a next point $$$C$$$ form a contiguous prefix of points before $$$B$$$;
  7. The second pointer $$$A$$$ to the end of the prefix is maintained;
  8. Eventually, for every $$$B$$$, all valid pairs of $$$A$$$ and $$$C$$$ are iterated with two pointers.

This allows to consider in $$$O(n^3)$$$ all the convex subsets of a given set of points, assuming that sorting around every point $$$B$$$ was computed beforehand in $$$O(n^2 \log n)$$$ and is now used to avoid actual second sorting of points around $$$B$$$.

5. Knapsack on segments. You're given $$$a_1, \dots, a_n$$$ and need to answer $$$q$$$ queries. Each query is whether $$$a_l, a_{l+1}, \dots, a_r$$$ has a subset of sum $$$w$$$. This can be done with dynamic programming $$$L[r][w]$$$ being the right-most $$$l$$$ such that $$$a_l, \dots, a_r$$$ has a subset with sum $$$w$$$:

$$$ L[r][w] = \max(L[r-1][w], L[r-1][w-a_r]). $$$

6. Data structure with co-primality info. There is a structure that supports following queries:

  1. Add/remove element $$$x$$$ from the set, all prime divisors of $$$x$$$ are known;
  2. Count the number of elements in the structure that are co-prime with $$$x$$$.

Without loss of generality, we may assume that the numbers are square-free.

Let $$$w(x)$$$ be the number of distinct prime divisors of $$$x$$$ and $$$N_x$$$ be the amount of numbers divisible by $$$x$$$ in the structure. When $$$x$$$ is added or removed from the structure, you need to update $$$2^{w(x)}$$$ values of $$$N_x$$$. Now, having $$$N_x$$$, how to count numbers co-prime with $$$x$$$?

$$$ A_x = \sum\limits_{d | x} (-1)^{w(d)} N_d = \sum\limits_{d | x} \mu(d) N_d, $$$

where $$$\mu(d)$$$ is the Möbius function. This formula essentially uses inclusion-exclusion principle, as $$$N_d$$$ counts numbers divisible by $$$d$$$ and we need to count numbers that are not divisible by any divisor of $$$x$$$.

The method was used in 102354B - Yet Another Convolution.

7. Generalized inclusion-exclusion. Let $$$A_1, \dots, A_n$$$ be some subsets of a larger set $$$S$$$. Let $$$\overline{A_i} = S \setminus A_i$$$.

With the inclusion-exclusion principle, we count the number of points from $$$S$$$ that lie in neither of $$$A_i$$$:

$$$ \left|\bigcap\limits_{i=1}^n \overline{A_i}\right| = \sum\limits_{m=0}^n (-1)^m \sum\limits_{|X|=m} \left|\bigcap\limits_{i \in X} A_i\right|, $$$

assuming the empty intersection to be the full set $$$S$$$. We may split the formula half-way as

$$$ \left|\bigcup\limits_{|Y|=r} \left( \bigcap\limits_{i \in Y} A_i \bigcap\limits_{j \in Y} \overline{A_j} \right)\right| = \sum\limits_{m=r}^n (-1)^{m-r} \binom{m}{r} \sum\limits_{|X|=m} \left|\bigcap\limits_{i \in X} A_i\right|. $$$

This way, we're able to count the number of points from $$$S$$$ that lie in exactly $$$r$$$ set among $$$A_1, \dots, A_n$$$.

Explanation lies in the fact that for a fixed $$$Y$$$, we may use PIE directly:

$$$ \left|\bigcap\limits_{i \in Y} A_i \bigcap\limits_{j \in Y} \overline{A_j}\right| = \sum\limits_{m=r}^n (-1)^{m-r} \sum\limits_{\substack{|X| = m \\ Y \subset X}} \left|\bigcap\limits_{i \in X} A_i\right|, $$$

then if summing up over all possible $$$Y$$$, each set $$$X$$$ will always have $$$(-1)^{m-r}$$$ coefficient and will occur for $$$\binom{m}{r}$$$ sets $$$Y$$$.

8. Finding roots of polynomials over $$$\mathbb Z_p$$$. You're given $$$q(x)$$$. You want to find all $$$a \in \mathbb Z_p$$$, such that $$$q(a)=0$$$.

This is done in two steps. First, you compute

$$$ h(x) = \gcd(q(x), x^{p}-x) $$$

to get rid of non-linear or repeated linear factors of $$$q(x)$$$, as

$$$ x^p - x \equiv \prod\limits_{a=0}^{p-1} (x - a) \pmod p. $$$

Second, you pick random $$$a$$$ and compute

$$$ \gcd(h(x), (x+a)^{\frac{p-1}{2}}-1). $$$

This will filter roots of $$$h(x)$$$ by whether they're quadratic residues if $$$a$$$ is added to them or not.

Quadratic residues make up $$$\frac{p-1}{2}$$$ of numbers in $$$\mathbb Z_p$$$ and are distributed uniformly, so you'll have at least $$$\frac{1}{2}$$$ chance to get non-trivial divisor of $$$h(x)$$$. This is particularly useful when you want to solve e.g. $$$x^2 \equiv a \pmod p$$$, which can be done in $$$O(\log p)$$$ with this algorithm.

The method is called Berlekamp–Rabin algorithm and can be generalized to find all factors of $$$q(x)$$$ over $$$\mathbb Z_p$$$ (see this comment).

9. Matching divisible by $$$m$$$. You're given a weighted bipartite graph and you need to check if there exists a perfect matching that sums up to the number that is divisible by $$$m$$$. In other words, whether there exists a permutation $$$\sigma_1, \dots, \sigma_n$$$ such that

$$$ A_{1 \sigma_1} + \dots + A_{n \sigma_n} \equiv 0 \pmod m. $$$

For this, we introduce matrices $$$R^{(0)}, \dots, R^{(m-1)}$$$ such that

$$$ R^{(k)}_{ij} = x_{ij} \omega^{k A_{ij}}, $$$

where $$$A_{ij}$$$ is weight between $$$i$$$ in the first part and $$$j$$$ in the second part, $$$x_{ij}$$$ is a random number when there is an edge between $$$i$$$ and $$$j$$$ or $$$0$$$ otherwise, and $$$\omega$$$ is a root of unity of degree $$$m$$$. The determinants of such matrices is then

$$$ \det R^{(k)} = \sum\limits_{\sigma \in S_n} \left( (-1)^{N(\sigma)} \prod\limits_{i=1}^n x_{i \sigma_i}\right) (\omega^k)^{\sum\limits_{i=1}^n A_{i \sigma_i}}, $$$

where $$$N(\sigma)$$$ is a parity of $$$\sigma$$$. If you sum them up, you get

$$$ \sum\limits_{k=0}^{m-1} \det R^{(k)} = \sum\limits_{\sigma \in S_n} \left( (-1)^{N(\sigma)} \prod\limits_{i=1}^n x_{i \sigma_i}\right) \sum\limits_{k=0}^{m-1} (\omega^k)^{\sum\limits_{i=1}^n A_{i \sigma_i}}. $$$

But at the same time,

$$$ \sum\limits_{k=0}^{m-1} \omega^{kx} = \begin{cases} m &, x \equiv 0 \pmod m,\\ 0&, x \not\equiv 0 \pmod m. \end{cases} $$$

Thus, a summand near $$$\sigma_1, \dots, \sigma_n$$$ will be non-zero only if $$$A_{1\sigma_1} + \dots + A_{n \sigma_n}$$$ sums up to the number divisible by $$$m$$$.

Therefore, the property can be checked in $$$O(mn^3)$$$.

The method was used in CSAcademy — Divisible Matching.

Теги tutorial, i love tags

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en23 Английский adamant 2022-07-03 20:43:38 82
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en9 Английский adamant 2022-07-03 16:22:40 0 (published)
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en2 Английский adamant 2022-07-03 05:02:53 2774 Tiny change: 'e with $m$ and $k > ' -> 'e with $m$, and $k > '
en1 Английский adamant 2022-07-03 02:59:55 412 Initial revision (saved to drafts)