http://www.sw.it.aoyama.ac.jp/2013/Math1/lecture5.html
© 2005-13 Martin J. Dürst 青山学院大学
Let's understand nCm = n!/ (m!·(n-m)!) directly!
Permutations: Different ways of ordering n different items
Number of permutations: n!
nCm is
the number of permutations of n items (n!)
divided by the number of permutations of the excluded items
((n-m)!)
divided by the number of permutations of the selected items (m!)
k-Permutation (or sequence without repetition): number of differently ordered k items taken from n different items.
nPm = n·(n-1)·(n-2)·...·(n-m+2)·(n-m+1) = ∏ni=n-m+1 i = n! / (n-m)!
| disjunction | conjunction | negation | ||
| or | and | not | ||
| precedence | low | middle | high | |
| A | B | A ∨ B | A ∧ B | ¬B |
|---|---|---|---|---|
| F | F | F | F | T |
| F | T | T | F | F |
| T | F | T | F | |
| T | T | T | T | |
Hint: Compare with laws for set operations; use truth tables to verify
(A ∨ ¬B) ∧ B = A ∧ B ∨ ¬B ∧ B = (A ∧ B) ∨ F = A ∧ B
¬(A ∨ ¬B) = ¬A ∧ ¬¬B = ¬A ∧ B
Application: Proof of absorption law from other laws
A ∧ (A ∨ B) = (A∨F) ∧ (A∨B) = A ∨ F∧B = A ∨ F = A
When looking at the laws of logical operations, we see the following:
If we exchange all instances of ∧ and ∨, and T and F, we get another law.
Examples:
T ∧ A = A; dual: F ∨
A = A
(¬A∨B) ∧ C =
C∧¬A ∨ B∧C; dual: ¬A∧B ∨ C =
(C∨¬A) ∧ (B∨C)
This is true in general. It can be proved using the truth tables for ∧ and ∨.
This is called the duality principle.
It is very useful for remembering the laws of logical operations.
Assume we are given a truth table (boolean function) such as the following:
| A | B | C | ? |
| F | F | F | F |
| F | F | T | T |
| F | T | F | F |
| F | T | T | T |
| T | F | F | T |
| T | F | T | F |
| T | T | F | T |
| T | T | T | F |
Can you find a logical formula corresponding to this truth table?
Is there a way to find a logical formula for every truth table (boolean function)?
加法標準形 (選言標準形、disjunctive normal form): 変数の (否定の) 積の和
乗法標準形 (連言標準形、conjunctive normal form): 変数の (否定の) 和の積
標準形の性質:
加法標準形の場合 [乗法標準形の場合は [] 内 (双対原理使用)]
式が正しい理由:
| A | B | C | ? | 加法標準形の項 | 乗法標準形の項 |
| F | F | F | T | ¬A ∧ ¬B ∧ ¬C | |
| F | F | T | T | ¬A ∧ ¬B ∧ C | |
| F | T | F | F | - | A ∨ ¬B ∨ C |
| F | T | T | F | - | A ∨ ¬B ∨ ¬C |
| T | F | F | F | - | ¬A ∨ B ∨ C |
| T | F | T | T | A ∧ ¬B ∧ C | |
| T | T | F | F | - | ¬A ∨ ¬B ∨ C |
| T | T | T | T | A ∧ B ∧ C |
加法標準形: ¬A∧¬B∧¬C ∨ ¬A∧¬B∧C ∨ A∧¬B∧C ∨ A∧B∧C
乗法標準形: (A∨¬B∨C) ∧ (A∨¬B∨¬C) ∧ (¬A∨B∨C) ∧ (¬A∨¬B∨C)
二つは基本的に同じが、使う｢道具」(式、図) が違う。
カルノー図表は標準形の構造 ((否定) の積の和等) を保持
違う構造でもっと単純化できる例も存在
例: A∧B∧C ∨ A∧¬B∧C ⇒ A∧C∧ (B ∨ ¬B) ⇒ A∧C
以前のスライドの式全体: A∧B∧C ∨ A∧¬B∧C ∨ ¬A∧¬B∧C ∨ ¬A∧¬B∧¬C ⇒ A∧C ∨ ¬A∧¬B
注意: 複数の単純化の道で、(式の構成が) 異なる結果が可能
| A=F B=F |
A=T B=F |
A=T B=T |
A=F B=T |
|
|---|---|---|---|---|
| C=F D=F |
T | T | F | T |
| C=T D=F |
F | T | T | F |
| C=T D=T |
F | T | T | F |
| C=F D=T |
F | F | F | T |
提出: 来週の木曜日 (11月14日)、22:00 (厳守)、Moodle
にて。形式は例に準拠
(プレーンテキスト, メモ帳など)。ファイル名は
solution5.txt。
| A = F B = F |
A
= F B = T |
A
= T B = F |
A
= T B = T |
簡単な式 |
| F | F | F | F | F |
| F | F | F | T | A ∧ B |
| F | F | T | F | ? |
| ? | ? | ? | ? | ? |
| A | B | A ∧ B |
| F | F | F |
| F | T | F |
| T | F | F |
| T | T | T |