［Royale with Cheese］”And you know what they call a Quarter Pounder with Cheese in Paris?” “They don't call it a Quarter Pounder with cheese?” “No man, they got the metric system. They wouldn't know what the f*** a Quarter Pounder is”（＊）
TLDR：大鄉里出中環系列。萬物皆數！足三兩應該係香港特有
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1. 好似寫過，但寫多次。應該好多人都有留意，Quarter Pounder 中文係「足三兩」。點解？因為四份一磅，就真係三兩咁重！係咪好神奇呢BB！這樣的知識你也知道，你真厲害。
2. 唔使識整GPU唔使識光刻機，中間個圖寫住。3斤係4磅，而1斤16兩（唔識唱「半斤八兩」冇所謂，個圖左邊寫住呀！），咁即係 4磅等於48兩。兩邊除16，4份1磅就即係3兩，所以Quarter Pounder即係足三兩！係咪巧神奇呢！！！！數學之美呀！
3. 不過，再聰明或煩膠少少嘅你，應該會知道，一切嘅關鍵，係嚟自邊個最初中西文化交流，將3斤變成4磅？1斤16兩你大約明白，但3斤變成4磅，fusion 嘢喎，好arbitrary，邊度走出嚟的？
4. 原來係，光緒年代嘅香港法律！1884年，開咗埠了。所以咪有呢個 華洋雜交嘅3斤變成4磅．
5. 咁大家應該知，東亞其他國家嘅斤呀，尺呀，同香港都唔同。但我仲有一個疑問，咁其他地方係咪叫足三兩？答案明顯唔會，只有香港（又，澳門呢？都應該咁叫）呢條式先成立。
6. 咁其他地方點叫Quarter Pounder？就真係考起喎！日本，當然係叫返片假名，西洋事物嘛。韓國？唔知，冇去過麥當勞。台灣呢？出奇地我去台灣成百次，都好似未去過麥當勞，但目測，台灣係從來都冇Quarter Pounder嘅！「你可以睇到美國有幾愛台灣」。而同樣地，大陸我去過麥當勞，好似唔覺有Quarter Pounder．
7. 又，原來不經不覺已經冇咗「Double足三兩」。只係得返le Big Mac．This city is dying
8. 你睇，人地去中環街市就同文青女影相，我就研究埋啲乜撚足三兩，證明我真係溝唔到女嘅毒撚。「不過好彩我有錢」（認真，其實唔係，我低人工時先最多女。潘驢鄧小閒，唔一定要有錢或靚仔嘅。咁你睇我個肚腩點都唔似驢，所以，咪剩返嗰兩味。）
（＊） https://ytube.io/3L3c
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一週年！比別人知得多。subscribe now（https://bityl.co/4Y0h）。Ivan Patreon，港美市場評點，專題號外，每日一圖，好文推介。每星期6篇，月費100，已經1900人訂！ 畀年費仲有85折，20/40年費VIP 送本人著作一本。
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arbitrary數學 在 [試題] 96下顏嗣鈞離散數學期中考- 看板NTU-Exam 的推薦與評價
課程名稱︰離散數學
課程性質︰必修
課程教師︰顏嗣鈞
開課學院：電資學院
開課系所︰電機系
考試日期（年月日）︰2008/04/21
考試時限（分鐘）：120分鐘
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題：
（以下邏輯語句，以'V'表示'or'、'^'表示'and'、'~'表示'negation'，與這門課使用
的符號略有不同，此外，以下以A代表for any，以E代表there exist。）
1. (15 pts) True or false? No explanations needed. Score= {Right-1/2Wrong};
so do not make uninformed guesses. (Please mark O for 'true' and X for
'false'.)
(a) ExAyP(x,y)→AyExP(x,y) is a valid formula.
(b) The sentence 'Today is Thursday.' is a proposition.
(c) P(d)^(~Ex(R(x)^~Q(x)))‧(~Ex(Q(x)^P(x)))├ ~R(d).
(d) (p_1→q_1)^(p_2→q_2)├ (p_1 V p_2→q_1 V q_2) can be proven using
Natural Deduction of propositional logic.
(e) (~q^(~p→q))→~p is a tautology.
(f) AxP(x) V AyQ(y) and Ax(P(x) V Q(x)) are logically equivalent.
(g) ExEyP(x,y) and EyExP(x,y) are logically equivalent.
(h) ExAyEzP(x,y,z) and EzAyExP(x,y,z) are logically equivalent.
(i) A formula ψ in propositional logic is satisfiable if ψ always evaluates
to True for every truth assignment.
(j) There exists a set A such that A is a subset of 2^A. (Note: 2^A, the power set of
A, ={B| B is a subset of A})
(k) If A is countably infinite, so is 2^A
(l) Every infinite set contains a countably infinite subset.
(m) {ψ, {{a}}} is the power set of some set.
(n) If A is countably infinite and B is an arbitrary set, then A∩B is either
finite of countably infinite.
(o) The union of infinitely many countably infinite sets is countably
infinite.
2. (10 pts) Use the following predicates
D(x): x is a dog
C(x): x is a dog catcher
T(x): x is a town
L(x,y), x lives in y
B(x,y), x has bitten y
to obtain a first-order formula corresponding to the following:
At least one town has a dog catcher who has been bitten by none of the dogs
in town.
3. (10 pts) Write a propositional formula to express the following statement:
If Clifton does not live in France, then he does not speak French.
Clifton does not drive a Datsun.
If Clifton lives in France, then he rides a bicycle.
Either Clifton speaks French, or he drives a Datsun.
Hence, Clifton rides a bicycle.
(Note: you need to identify the propositions first.)
4. (10 pts) Prove that for an arbitrary infinite set A, there is no
one-to-one correspondence between A and 2^A. (Hint: proof by contradiction.
Assuming f to be such a one-to-one correspondence, define S={x| x belongs to A
^x does not belong to f(x)}…)
5. (10 pts) Prove formally that if for any i≧1, A_i is countably infinite,
then
∪A_i=A_1∪A_2∪…∪A_i∪...
i≧1
is also countably infinite.
6. (15 pts) Prove that the following program is totally correct. (Do not
forget to show that the program always terminates.)
(Hint: For partial correctness, use loop invariant { z=x(y-n+1)‧n≧1 }.)
{y>0}
z:=x;
n:=y;
while n>1 do
begin
z:=z+x;
n:=n-1;
end
{z=x*y}
7. (15 pts) Calculate (in detail) the following weakest preconditions (WPs):
(Recall that WP(S,{Q}) represents the "weakest condition" P such that P{S}Q.)
(a) WP(if x<0 then x:=x+2 else y:=x+3, {x>0^y<0}).
(b) WP(if odd(x) then x:=x+1, {x=10}). (odd(x) means x is an odd number.)
(c) WP(if y<0 then x:=y-v, {x=y-v^y>0})
8. (15 pts) Answer the following two questions:
(a) (7 pts) Use the truth table method to prove ((p→r)^(q→r))╞ ((p V q)→
r)
(b) (8 pts) Use natural deduction (see the following table) to prove
((p→r)^(q→r))├ ((p V q)→r).
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◆ From: 220.133.169.66
※ 編輯: boggart0803 來自: 220.133.169.66 (06/18 10:44)
※ 編輯: boggart0803 來自: 220.133.169.66 (06/18 10:45)
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