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才: 能力
学: 学问
识: 見识 (世界/人生观, 对社会、 历史、 人生的感悟)
Tuesday, 2 April 2013
Logic: Monkey's Dilemma
西遊记 (The Journey to the West)
On the way to the West, the team came to a Y-junction, either go Left or Right.
The Monkey King (孙悟空 ) went out reconnaissance, only came back few days later.
The Pig (猪八戒) suspects this fellow is a monster in disguise of the monkey, may not tell the truth.
The smart Monk (三藏) asks only 2 questions to detect his identity :
Q1. Is 'Left' direction correct ?
Q2. If I ask you Q1, you would say "Yes", would-you?
The true Monkey will answer "Yes/Yes" or "No/No", regardless the correct direction.
Why ?
Case 1. If the Correct direction =Left
Monkey: (Q1, Q2) = (Y,Y)
Monster (tell lie) = (N,Y)
Case 2. If the Correct direction =Right
Monkey : =(N,N)
Monster (tell lie) =(Y, N)
On the way to the West, the team came to a Y-junction, either go Left or Right.
The Monkey King (孙悟空 ) went out reconnaissance, only came back few days later.
The Pig (猪八戒) suspects this fellow is a monster in disguise of the monkey, may not tell the truth.
The smart Monk (三藏) asks only 2 questions to detect his identity :
Q1. Is 'Left' direction correct ?
Q2. If I ask you Q1, you would say "Yes", would-you?
The true Monkey will answer "Yes/Yes" or "No/No", regardless the correct direction.
Why ?
Case 1. If the Correct direction =Left
Monkey: (Q1, Q2) = (Y,Y)
Monster (tell lie) = (N,Y)
Case 2. If the Correct direction =Right
Monkey : =(N,N)
Monster (tell lie) =(Y, N)
Mathematician Pope
The ‘Number’ Pope Sylvester II @ 999 AD
“The just man lives by faith, but it is good that he should combine Science with his faith.”
The mathematician Pope introduced ‘1,2,3...9’ to replace Roman numerals in Europe.
He was inspired by Alcuin’s (Abbot of St. Martin in Tours, France) Math book “Problems to Sharpen the Young” (Propositiones ad Acuendos Juvenes)
“The just man lives by faith, but it is good that he should combine Science with his faith.”
The mathematician Pope introduced ‘1,2,3...9’ to replace Roman numerals in Europe.
He was inspired by Alcuin’s (Abbot of St. Martin in Tours, France) Math book “Problems to Sharpen the Young” (Propositiones ad Acuendos Juvenes)
Mediant
Discovered by a French watchmaker Achille Brocot (1817–1878) http://en.wikipedia.org/wiki/Achille_Brocot
Mediant = $latex \frac {a+c}{b+d}$
$latex \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d} $
a= 132/456 = 0.07
b= 258/338 = 0.76
find c between a, b ?
$Latex c = \frac {132+258} {456+338}
= \frac{390}{794} =0.49$
Mediant = $latex \frac {a+c}{b+d}$
$latex \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d} $
a= 132/456 = 0.07
b= 258/338 = 0.76
find c between a, b ?
$Latex c = \frac {132+258} {456+338}
= \frac{390}{794} =0.49$
Combinatoric in Accounting
Prove:
$latex \displaystyle\sum_{n=2}^{n}{_n}C_r = 2^n-1-n$
$latex \displaystyle\sum_{n=2}^{n}{_n}C_r 2^r= 3^n-1-2n $
Note:
$latex {_4}C_2=\frac {4.3}{2!}$
Proof:
1.
$latex \displaystyle\sum_{n=2}^{n}{_n}C_r $ =(1+1)ⁿ -1- $latex \displaystyle {_n}C_1 =2^n-1-n$
2.
$latex \displaystyle\sum_{n=2}^{n}{_n}C_r 2^{r}$=(1+2)ⁿ -1 - $latex \displaystyle {_n}C_1 .2^{1}= 3^n-1-2n$
An accounting transaction = Debit p accounts + Credit q accounts (p, q ≥ 1)
In a company with total n accounts,
Prove: there are $latex 3^n -2^{n+1} + 2$ transactions.
Proof:
Let the Set of all accounts T = {a1, a2, ..., an}
aj = account with value '+' (credit), or '-' (debit), or 0 (nil)
1. Trivial transaction: To= {0, 0,.....0} = 1 way
2. Choose r accounts from n = $latex \displaystyle {_n}C_r $
3. Slot '+' or '-' in these r accounts = $latex 2^r $ways
exclude 2 impossible all '+', '-' transactions = $latex 2^r- 2$ ways
4.
Let T1 = $latex \displaystyle\sum_{n=2}^{n}{_n}C_r .(2^r-2)$
Total transaction = To + T1
= $latex \displaystyle 1+ \sum_{n=2}^{n}{_n}C_r .(2^r-2)$
{Apply previous results}
= $latex \displaystyle 1+ \sum_{n=2}^{n}{_n}C_r .2^r -2.\sum_{n=2}^{n}{_n}C_r$
= $latex 1+3^n-1-2n -2(2^n-1-n) $
= $latex 3^n -2^{n+1} + 2$ [QED]
$latex \displaystyle\sum_{n=2}^{n}{_n}C_r = 2^n-1-n$
$latex \displaystyle\sum_{n=2}^{n}{_n}C_r 2^r= 3^n-1-2n $
Note:
$latex {_4}C_2=\frac {4.3}{2!}$
Proof:
1.
$latex \displaystyle\sum_{n=2}^{n}{_n}C_r $ =(1+1)ⁿ -1- $latex \displaystyle {_n}C_1 =2^n-1-n$
2.
$latex \displaystyle\sum_{n=2}^{n}{_n}C_r 2^{r}$=(1+2)ⁿ -1 - $latex \displaystyle {_n}C_1 .2^{1}= 3^n-1-2n$
An accounting transaction = Debit p accounts + Credit q accounts (p, q ≥ 1)
In a company with total n accounts,
Prove: there are $latex 3^n -2^{n+1} + 2$ transactions.
Proof:
Let the Set of all accounts T = {a1, a2, ..., an}
aj = account with value '+' (credit), or '-' (debit), or 0 (nil)
1. Trivial transaction: To= {0, 0,.....0} = 1 way
2. Choose r accounts from n = $latex \displaystyle {_n}C_r $
3. Slot '+' or '-' in these r accounts = $latex 2^r $ways
exclude 2 impossible all '+', '-' transactions = $latex 2^r- 2$ ways
4.
Let T1 = $latex \displaystyle\sum_{n=2}^{n}{_n}C_r .(2^r-2)$
Total transaction = To + T1
= $latex \displaystyle 1+ \sum_{n=2}^{n}{_n}C_r .(2^r-2)$
{Apply previous results}
= $latex \displaystyle 1+ \sum_{n=2}^{n}{_n}C_r .2^r -2.\sum_{n=2}^{n}{_n}C_r$
= $latex 1+3^n-1-2n -2(2^n-1-n) $
= $latex 3^n -2^{n+1} + 2$ [QED]
Relationship-Mapping-Inverse (RMI)
Relationship-Mapping-Inverse (RMI)
(invented by Prof Xu Lizhi 徐利治 中国数学家 http://baike.baidu.com/view/6383.htm)
Find Z = a*b
By RMI Technique:
Let f Homomorphism: f(a*b) = f(a)+f(b)
Let f = log
log: R+ --> R
=> log (a*b) = log a + log b
1. Calculate log a (=X), log b (=Y)
2. X+Y = log (a*b)
3. Find Inverse log (a*b)
4. ANSWER: Z = a*b
Prove:
$latex \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= 2$
1. Take f = log for Mapping:
$latex \log\sqrt{2}^{\sqrt{2}^{\sqrt{2}}} $
$latex = \sqrt{2}\log\sqrt{2}^{\sqrt{2}}$
$latex = \sqrt{2}\sqrt{2}\log\sqrt{2} $
$latex = 2\log\sqrt{2} $
$latex = \log (\sqrt{2})^2 $
$latex = \log 2$
2. Inverse of log (bijective):
$latex \log \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= \log 2$
$latex \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= 2$
(invented by Prof Xu Lizhi 徐利治 中国数学家 http://baike.baidu.com/view/6383.htm)
Find Z = a*b
By RMI Technique:
Let f Homomorphism: f(a*b) = f(a)+f(b)
Let f = log
log: R+ --> R
=> log (a*b) = log a + log b
1. Calculate log a (=X), log b (=Y)
2. X+Y = log (a*b)
3. Find Inverse log (a*b)
4. ANSWER: Z = a*b
Prove:
$latex \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= 2$
1. Take f = log for Mapping:
$latex \log\sqrt{2}^{\sqrt{2}^{\sqrt{2}}} $
$latex = \sqrt{2}\log\sqrt{2}^{\sqrt{2}}$
$latex = \sqrt{2}\sqrt{2}\log\sqrt{2} $
$latex = 2\log\sqrt{2} $
$latex = \log (\sqrt{2})^2 $
$latex = \log 2$
2. Inverse of log (bijective):
$latex \log \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= \log 2$
$latex \sqrt{2}^{\sqrt{2}^{\sqrt{2}}}= 2$
Quantum Physics = Math
CN Yang (杨振宁) Yang-Mills Equation in Quantum Physics and ‘Fibre Bundle’ in Differential Geometry by SS Chern (陈省身) are equivalent, however both men worked independently for 30 years!
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