principle of inclusion

When the number of cases of an event is difficult to count directly, it is solved by dividing it into smaller and simpler counts problem partitioning. When the partial events are exclusive, they simply add up, but even if they do not, the inclusion-division principle can be used if the common part can be found.

$|A\cup B|=|A|+|B|-|A\cap B|$

• https://ja.wikipedia.org/wiki/包除原理

• binary feature.

• When you want to find 000 but it is difficult to find it directly, calculate *** - (1** + 1 + 1) + (11 + 11 + *11) - (111)

• In particular, if the size f(k) of the set is determined by the number k of 1s

  • ${\sum }_k-1^k\left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k)$

• https://twitter.com/opt_cp/status/1298569230439157762

• $|{\bigcap }_{i\in V}{A}_i|={\sum }_{S\subseteq V}(-1{)}^{|S|}|{\bigcap }_{i\in S}{A}_i^{\mathsf{c}}|$

• Instead of thinking |A or B| = |A| + |B| - |A and B|, it is easier to think |¬A and ¬B| = |U| - |A| - |B| + |A and B| [src https://twitter.com/Euglenese/status/1277087618606329856?s= 20] ABC172E

  • $|A\cup B|=|A|+|B|-|A\cap B|$
  • $|A^{\mathsf{c}}\cap B^{\mathsf{c}}|=|U|-|A|-|B|+|A\cap B|$

• Name the events properly and transform the equation; be aware of which events can be easily calculated.

  • Consider the case where N=3 this time. Let #() denote the number of elements in the event.

  • Pi = (the event that Ai=Bi), we want to find #(¬P1 ∧ ¬P2 ∧ ¬P3), and #(P1) or #(P1 ∧ P2) can be easily computed. #(¬P1 ∧ ¬P2) is not so easy to deal with, so it is easy to use de Morgan to convert it to #(¬(P1 ∨ P2 ∨ P3)) and think of it as the inclusive divisor principle.

  • https://twitter.com/oinori1197/status/1276894798209679360

• The Inclusion-Decomposition Principle Expand the range of solvable counts. tsutaj

• https://compro.tsutaj.com//archive/181015_incexc.pdf

• If small, full search

• Symmetry results in combination

  • O(2^N) becomes O(N)

• O(3^N) when using the inclusion principle for each subset

  • Why? → Sum of the number of subsets of a subset.
  • This is a heavy I
    • fast Möbius transformation can make this O(N2^N)

relevance - binomial coefficient - Moebius transform - perfect permutation

• https://twitter.com/sugarknri/status/1290212350696415233
  • Euler’s constant https://twitter.com/kaitou_ryaku/status/1239816691523051520

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