离散数学读书记录
离散数学读书记录
如果选课系统面子给够的话，我下学期大概有三门数学课。非常恐怖。
所以提前学一下离散，避免挂科。
用的书是 Discrete Mathematics and Its Applications 8th Edition
Chapter 1: The Foundations: Logic and Proofs
1.1 Propositional Logic
高中内容，略。
Exercises
T53
The \(n\) th statement in a list of 100 statements is "Exactly \(n\) of the statements in the list are false."
a) What conclusions can you draw from these statements?
b) Answer part (a) if the \(n\) th statement is "At least \(n\) of the statements in the list are false."
c) Aswer part(b) assuming that the list contains 99 statements.
Sol
a)
讨论：
-
假如有超过两个命题都是真的，那么天然矛盾。
-
假如没有任何一个命题是真的，也就是 100 条都是假的，那么第 100 条命题就是真的了，矛盾。
-
假如恰好有一个命题是真的，也就是 99 条都是假的，那么只有第 99 条命题是真的，其他都是假的。
综合，第 99 条是真的，别的都是假的。
b)
发现第 \(i\) 条命题是第 \(i+1,i+2,\dots,n\) 条命题的充分不必要条件。
这意味着存在一个唯一的 \(k\) ，满足前 \(k\) 条命题为真，而剩余的为假。
容易发现
\(k\) 仅能取 50 ，经验证无误。
c)
同理，
无整数解，矛盾。这一事实与命题本身可以判断真假又矛盾，是个悖论。
1.2 Applications of Propositional Logic
- 一些五三等习题集边角处的逻辑小谜题的出处 ( ﾟ∀。)
- 基础的逻辑门知识
- 搜索引擎与逻辑运算符
Exercises
T19
Each inhabitant of a remote village always tells the truth or always lies. A villager will give only a "Yes" or a "No" response to a question a tourist asks. Suppose you are a tourist visiting this area and come to a fork in the road. One branch leads to the ruins you want to visit; the other branch leads deep into the jungle. A village is standing at the fork in the road. What one question can you ask the village to determine which branch to take?
Sol
我觉得比较标准的一个答案是：
「如果我问你，左边这条路是不是可以到遗迹去，你会回答 Yes 吗？」
还有不讲道理的答案：
「如果你只会说谎，那左边不能走到遗迹去对吗，或者你不会说谎，那左边能走到遗迹去对吗？」
似乎也没有毛病。
T42
Solve this famous logic puzzle, attributed to Albert Einstein, and known as the zebra puzzle.
Five men with different nationalities and with different jobs live in consecutive houses on a street. These houses are painted different colors. The men have different pets and have different favorite drinks. Determine who owns a zebra and whose favorite drink in mineral water ( which is one of the favorite drinks ) given these clues:
The Englishman lives in the red house.
The Spaniard owns a dog.
The Japanese man is a painter.
The Italian drinks tea.
The Norwegian lives in the first house on the left.
The green house in immediately to the right of the white one.
The photographer breeds snails.
The diplomat lives in the yellow house.
Milk is drunk in the middle house.
The owner of the green house drinks coffee.
The Norwegian 's house is next to the blue one.
The violinist drinks orange juice.
The fox is in a house next to that of the physician
The horse is in a house next to that of the diplomat.
Sol
？？？什么破题
| ID | Nationality | Pet | Job | Favorite Drink | The color of house | The location of house |
|---|---|---|---|---|---|---|
| 1 | Englishman | Red | ||||
| 2 | Spaniard | Dog | ||||
| 3 | Japanese | Painter | ||||
| 4 | Italian | Tea | ||||
| 5 | Norwegian | 1 |
photographer breeds snails 那么 Pet 和 Job 已填写的都不可能对应这一关系。
diplomat lives in the yellow house 同样 Job 或 color 已填写的也不能对应这一关系。
violinist drinks orange juice 类似地，Job 或 Drink 已填写的不能对应这一关系。
总之，
photographer 可能在 ID 1 4 5
diplomat 可能在 ID 2 4 5
violinist 可能在 ID 1 2 5
physician 可能在 ID 1 2 4 5
类似地，
Green, White, Blue 只能在 ID 2 3 4
且 Blue 对应 Location 2
Green 的 Location 比 White 的大 1
设各ID的 location 是 \(x_1,x_2,x_3,x_4\) 那么，
铁尼玛无语，还推理个锤头，暴力得到答案：
| ID | Nationality | Pet | Job | Favorite Drink | The color of house | The location of house |
|---|---|---|---|---|---|---|
| 1 | Englishman | Snail | Photographer | Milk | Red | 3 |
| 2 | Spaniard | Dog | Violinist | Orange Juice | White | 4 |
| 3 | Japanese | Zebra | Painter | Coffee | Green | 5 |
| 4 | Italian | Horse | Physician | Tea | Blue | 2 |
| 5 | Norwegian | Fox | Diplomat | Mineral Water | Yellow | 1 |
没验证别的，我猜想应该只有这一个答案。
佛了。
1.3 Propositional Equivalences
-
Tautology: 重言式，恒为真值的复合命题。这个命名感觉有 You can say it again 的意思。
-
Contradiction: 矛盾式，恒为假值的符合命题。
-
Logical Equivalences: \(p\equiv q\) ( or \(p\Leftrightarrow q\) ) means \(p \leftrightarrow q\) is tautology
下面给出一些重要的 Logical Equivalences
Equivalence Name \(p\to q \equiv \neg p\or q\)
\(p\leftrightarrow q \equiv (\neg p\or q) \and (\neg q\or p)\)Conditional-disjunction equivalence \(p \and T \equiv p\)
\(p\or F \equiv p\)Identity laws \(p \or T \equiv T\)
\(p\and F \equiv F\)Domination laws \(p \and p\equiv p\)
\(p\or p \equiv p\)Idempotent laws \(\neg(\neg p) \equiv p\) Double negation law \(p\or q\equiv q\or p\)
\(p \and q \equiv q \and p\)Commutative laws \((p\or q)\or r\equiv q\or (p\or r)\)
\((p\and q)\and r\equiv q\and (p\and r)\)Associative laws \(p\or (q \and r) \equiv (p\or q) \and (p \or r)\)
\(p \and (q \or r) \equiv (p\and q) \or (p \and r)\)Distributive law \(\neg(p\and q) \equiv \neg p \or \neg q\)
\(\neg (p \or q) \equiv \neg p \and \neg q\)De Morgan's law \(p\or(p\and q) \equiv p\)
\(p\and(p\or q) \equiv p\)Absorption laws \(p\or \neg p \equiv T\)
\(p\and\neg p \equiv F\)Negation laws -
Satisfiability： 如果存在至少一种变量取值使得一个符合命题可以取真值，这个命题就是 satisfiable 的。具体讨论了八皇后问题和数独
Exercises
T43
Why are the duals of two equicalent compound propositions are also equivaalent, where these compound propositions contain only the operators \(\and,\or,\neg\) 。
Sol
一个复合命题的对偶应当是把命题中出现的 \(\or\) 和 \(\and\) 全部互相替换，\(T\) 和 \(F\) 互相替换得到的命题。
接下来说明等价命题的对偶也是等价命题。
首先看两个很自然的引理：
Lemma 1: \((p\equiv q) \Leftrightarrow (\neg p \equiv \neg q)\)
Lemma 2: 考虑 \(p=f(p_1,p_2,\dots),~~q=g(q_1,q_2,\dots)\) ，若 \(p\equiv q\) ，那么 \(f(\neg p_1,\neg p_2,\dots) \equiv g(\neg q_1,\neg q_2,\dots)\)
利用真值表，可以清楚地观察出这两个引理。
接着，首先考虑复合命题中的 \(\neg\) 运算符。为了方便说明，我们不断地对原命题使用 De Morgan's Law，使得所有 \(\neg\) 运算符都作用在了不可分的 atomic statement 上。这样可以简化后续的说明。
假设原命题可以划分成 \(p\equiv p_1\and p_2,~~q\equiv q_1\and q_2\) 的形式，那么
首先在 \(p \equiv q\) 的基础上用引理1，得到 \(\neg(p_1\and p_2) \equiv \neg(q_1\and q_2)\) ，
再用 De morgan's Law 有 \(\neg p_1 \or \neg p_2 \equiv \neg q_1 \or \neg q_2\)，
这时由引理2，就有 \(p_1\or p_2 \equiv q_1\or q_2\) ，完成了符号变换。
\(p\equiv p_1\or p_2,~~ q\equiv q_1\or q_2\) 或
\(p\equiv p_1 \and p_2,~~ q \equiv q_1\or q_2\) 或
\(p\equiv p_1 \or p_2,~~ q \equiv q_1\and q_2\) 也可以通过相似的流程完成符号变换。
我们按照运算优先级的逆序，不断地划分复合命题的结构，再重复这种 Lemma 1 -> De Morgan's Law -> Lemma 2 的循环，就可以把复合命题中的所有 \(\or,\and\) 完成转置。
说得不很清楚，但总之可以机械地通过这种方法，说明任意两个等价命题的对偶也等价。
T47, 48, 49
A collection of operators is called functionally complete if every compound propostion is logically equivalent to a compoud proposition involving only these logical operators.
T47: Show that \(\neg,\or,\and\) form a functionally complete collection of logical operators.
T48: Show that \(\neg\) and \(\and\) form a functionally complete collection of logical operators.
T49: Show that \(\neg\) and \(\or\) form a functionally complete collection of logical operators.
Sol
T47: 根据 Conditional-disjunction equivalence 和 Negation laws 可以替换掉 \(\to, \leftrightarrow,T,F\) ，于是结论成立。
T48: 在T47的基础上逆用 De Morgan's Law ，即 \(p\or q \equiv \neg(\neg(p\and q))\) 即可。
T49: 同理根据 \(p\and q \equiv \neg(\neg(p\or q))\) 即可
T55
We now present a group of exercises that involve the logical operators NAND and NOR.
The proposition \(p\) NAND \(q\) is ture when either \(p\) or \(q\) or both are false; and it is false when both \(p\) and \(q\) are true.
The proposition \(p\) NOR \(q\) is ture when both \(p\) and \(q\) are false; and it is false otherwise.
The proposition \(p\) NAND \(q\) and \(p\) NOR \(q\) are denoted by \(p~|~q\) and \(p\downarrow q\) respectively.
T50-54 passed
T55: Find a compound proposition logically equivalent to \(p\to q\) using only the logically operator \(\downarrow\)
Sol
顺着做T50-54可以得到
- \(p\downarrow p \equiv \neg p\)
- \((p \downarrow q) \downarrow (p \downarrow q) \equiv p \or q\)
又因为 \(p\to q \equiv \neg p\or q\)
所以有 \(p \to q \equiv ((p \downarrow p) \downarrow q) \downarrow ((p \downarrow p) \downarrow q)\)
挺奇妙的。
T59
How many different truth table of compound propositions are there that involve the propositional variables \(p\) and \(q\) ?
Sol
要说明这个问题其实是有点难度的。假如我们知道逻辑的这套体系非常完备，就可以应用简单的数学知识，得到是 16 种。（考虑一个真值表，固定 \(p,q\) 两列不动，在 compound proposition 那一列填值，可以从 0 0 0 0 写到 1 1 1 1，共16种）。
但是现在我们未必知道体系是完备的，所以应该需要举例子，就非常麻烦。
我太懒，答案略。
1.4 Predicates and Quantifiers
-
A statement of the form \(P(x_1,x_2,\dots,x_n)\) is the value of the propositional function \(P\) at the \(n\)-tuple \((x_1,x_2,\dots,x_n)\), and \(P\) is also called an \(n\)-place predicate or and \(n\)-ary predocate
举例来讲，可以令 \(Q(x,y)\) 代表 \(x+y=3\) ，那么 \(Q(1,2)\) 就是真值，\(Q(1,1)\) 就是假值。
做上一节习题的时候不小心用到这个玩意了，，，
-
Quantifiers 即量词。考虑把某个 propositional function 变成一个 proposition ，除了代入具体值意外，还可以通过量词。也就是高中那种内容：\(\forall x, P(x)\) 和 \(\exist x, P(x)\) 。还额外附加了一种 \(\exist!x,P(x)\) （或写成 \(\exist_1x,P(x)\) ）表示唯一存在。
优先级方面呢，\(\forall x P(x) \or Q(x)\) 实际上的意义是 \((\forall x P(x)) \or Q(x)\) 而不是 \(\forall x (P(x)\or Q(x))\)
-
Logic Programming 给了一个逻辑式编程的小例子。
Exercises
梦回高中，略。
1.5 Nested Quantifiers
-
提示了一些含有多个逻辑量词的命题，确实是超出高中范围了，不过学微积分的时候也教过了。
\(\forall x~\exist y ,~~x+y=0\) 就理解成 \(\forall x Q(x)\)， where \(Q(x)\) is \(\exist y P(x,y)\), where \(P(x,y)\) is \(x+y =0\)
另一个经典的例子是，\(\lim_{x\to x_0} f(x) = C\) 的定义即为 \(\forall \varepsilon >0,~\exist \delta >0,\forall x \in \mathring{U}_\delta(x_0),|f(x)-C|<\varepsilon\)
否定是需要把量词逐个改掉
Exercises
T17
Express each of these sstem specifications using predicates, quantifiers, and logical connectives, if necessary.
a) Every user has access to exactly one mailbox
b) There is a process that continues to run during all error conditions only if the kernel is working correctly.
c) All users on the campus network can access all websites whose url has a .edu extension.
d) There are exactly two systems that monitor every remote server.
Sol
-
a) \(\forall u,\exist_1 m, A(u,m)\), where \(A(u,m)\) means user \(u\) has access to mailbox \(m\)
-
b) \(\exist p, \forall e, R(p) \and Cond(e) \to R(kernel)\),
where \(R(x)\) means \(x\) is working / running correctly, and \(Cond(x)\) means under the \(x\) condition
-
c) \(\forall u \in Campus ~~\forall w (\)
w.url.find(".edu") != string::npos\(\to H(u,w))\),where \(H(u,w)\) means the user \(u\) has access to the website \(w\)
-
d) \(\forall r,\exist_2s, ~M(s,r)\), where \(M(s,r)\) means \(s\) monitors \(r\)
俺自己发明的符号！难道不能用吗（
上网了找答案，说的是
\[\exist x\exist y(x\not=y\and\forall z ((\forall s M(z,s)) \leftrightarrow (z=x\or z=y))) \]where \(M(z,s)\) means thar system \(z\) monitors remote sever \(s\)
意外地好难啊orz
T48
Show that \(\forall x P(x)\or \forall x Q(x) \equiv \forall x\forall y(P(x) \or Q(y))\) , where all quantifiers have the same nonempty domain.
Sol
如果 \(\forall x P(x) \or \forall x Q(x)\) 是真值，讨论可能的情况：
- \(\forall x P(x)\) 为真，而 \(\forall y (T\or Q(y))\) 为真，于是 \(\forall y(P(x) \or Q(y))\) 为真，从而右式为真。
- \(\forall x Q(x)\) 为真，而 \(\forall x (P(x) \or T)\) 为真，于是 \(\forall y(P(x) \or Q(y))\) 为真，从而右式为真。
如果左式是假值，那么 \(\forall x P(x)\) 与 \(\forall x Q(x)\) 均为假，从而 \(\exist x, \neg P(x),~~\exist y,\neg Q(y)\) ，于是 \(\exist x \exist y \neg(P(x)\or Q(y))\) 。这恰好是右式的否定，从而右式也是假值。
综合，\(\forall x P(x)\or \forall x Q(x) \equiv \forall x\forall y(P(x) \or Q(y))\)
T49
a) Show that \(\forall x P(x) \and \exist x Q(x)\) is logically equivalent to \(\forall x \exist y (P(x)\and Q(y))\) , where all quantifiers have the same nonempty domain.
b) Show that $\forall x P(x) \or \exist x Q(x) $ is equivalent to \(\forall x \exist y (P(x)\or Q(y))\) , where all quantifiers have the same nonempty domain.
Sol
陷在 logically equivalent 和 equivalent 区别的泥淖里畏首畏尾。
往前翻书试图区别未果，
google 之倒是有 logically equivalent 和 martial equivalent 的区别：
好像也不太一样（
所以到底是怎么样呢？
看答案反推，发现 logically equivalent 只需要前真后即真，equivalent 还需要附加一条后真前即真。
-
a) 如果 \(\forall x P(x) \and \exist x Q(x)\) 是真值，就需要 \(\forall x P(x)\) 为真且 \(\exist x Q(x)\) 为真。由后者可以得到 \(\exist y (T \and Q(y))\) 为真。再由前者就得到 \(\forall x \exist y (P(x)\and Q(y))\) 为真。于是做完了。
-
b)
如果 \(\forall x P(x) \or \exist x Q(x)\) 是真值，讨论可能的情况：
- \(\forall x P(x)\) 为真，而有 \(\exist y (T \or Q(y))\) 为真，从而 \(\forall x \exist y (P(x) \or Q(y))\) 为真。
- \(\exist x Q(x)\)为真，而有 \(\forall x (P(x)\or T)\) 为真，从而 \(\forall x \exist y (P(x)\or Q(y))\) 为真。
相反地，考虑 \(\forall x \exist y (P(x) \or Q(y))\) 为真的情况：
- 若 \(\forall x P(x)\) 为真，此式显然为真；
- 否则，即 \(\exist x_0 \neg P(x_0)\) ，在这种情况下此式仍然需要为真，那么就必然有 \(\exist y Q(y)\) 为真。
综合，两式等价。
T50, 51, 52
明天再做吧orz
