所 謂『相由心生』是說精神外顯的『形貌』從『用心方向』而來,這個『習焉不察』之內在『心相』,常可以用來分辨『行業』。一行有一行的規矩,百業有百業的訣竅,入了行,從了業,自然帶有某種『氣息』的吧!如何才能夠不著『相』?若可『無所住』而生其『心』,那麼既無『我心』何來『我相』的呢!!
那麼這個《子之武城》一事,是否有個『前言』對上『後語』,可分出『對錯好壞』的呢?也許有個『禮樂』之『理』和『禮樂』之『用』的差別,想那『子游』為武城宰,採用『禮樂』教化之道,孔夫子卻『莞爾』笑,豈有不『以子之言,擊子之語』的哩!然而夫子所謂『戲之』果真是說『割雞焉用牛刀?』是錯了嗎?恐是不樂見『禮樂』被當作了『名器』的吧!就像到了宋代的『存天理,去人欲』,導致『死生事小,失節事大』,終演成『禮教殺人』之憾事 !!於是
祇『能』這樣『用』,不『會』那樣『使』,終究難了『用大』之道 ── 無用而不通達 ── ,如如不動,應事而動,因事制宜。
正說著『以正治國』和『以奇用兵』,『為學之法』與『用學之法 』的不同,也須避免那『紙上談兵』之過。此事《孫子兵法》
地形‧第十
孫子曰:地形有通者、有掛者、有支者、有隘者、有險者、有遠者 。我可以往,彼可以來,曰通。通形者, 先居高陽,利糧道,以戰則利。可以往,難以返,曰掛。掛形者,敵無備,出而勝之,敵若有備,出而不勝,難以返,不利。我出而不利,彼出而不利,曰支 。支形 者,敵雖利我,我無出也,引而去之,令敵半出而擊之,利 。隘形者,我先居之,必盈之以待敵。若敵先居之,盈而勿從,不盈而從之。險形者,我先居之,必居高 陽以待敵;若敵先居之,引而去之,勿從也。遠形者,勢均,難以挑戰,戰而不利。凡此六者 ,地之道也,將之至任,不可不察也。
故兵有走者、有弛者、有陷者、有崩者、有亂者、有北者。凡此六者,非天之災,將之過也。夫勢均,以一擊十,曰走;卒強吏弱,曰馳;吏強卒弱,曰陷;大吏怒而不服,遇敵懟 而自戰,將不知其能,曰崩;將弱不嚴,教道不明,吏卒無常,陳兵縱橫,曰亂;將不能料敵,以少合衆,以弱擊強,兵無選鋒,曰北。凡此六者,敗之道也,將之 至任,不可不察也。
夫地形者,兵之助也。料敵制勝,計險厄遠近,上將之道也。知此而用戰者必勝,不知此而用戰者必敗。故戰道必勝,主曰無戰,必戰可也;戰道不勝,主曰必戰,無戰可也。故進不求名,退不避罪 ,唯民是保,而利合於主,國之寶也。
視卒如嬰兒,故可以與之赴深溪;視卒如愛子,故可與之俱死 。厚而不能使,愛而不能令,亂而不能治,譬若驕子,不可用也。
知吾卒之可以擊,而不知敵之不可擊,勝之半也;知敵之可擊,而不知吾卒之不可以擊,勝之半也;知敵之可擊,知吾卒之可以擊,而不知地形之不可以戰,勝之半也。故知兵者,動而不迷,舉而不窮。故曰:知彼知己,勝乃不殆;知天知地,勝乃可全。
講的好。
─── 《字詞網絡︰ WordNet 《六》 相 □ 而用 ○ !!》
假使說『形勢』比人強!那麼『形式』無所言耶?
就算『滿天變數』!!只要『彼此相關 』??
並非『無所拘束』乎 !!??
Specifying a transformation by three points
Given a set of three distinct points z1, z2, z3 on the Riemann sphere and a second set of distinct points w1, w2, w3, there exists precisely one Möbius transformation f(z) with f(zi) = wi for i = 1,2,3. (In other words: the action of the Möbius group on the Riemann sphere is sharply 3-transitive.) There are several ways to determine f(z) from the given sets of points.
Mapping first to 0, 1, ∞
It is easy to check that the Möbius transformation
with matrix
maps z1, z2, z3 to 0, 1, ∞, respectively. If one of the zi is ∞, then the proper formula for is obtained from the above one by first dividing all entries by zi and then taking the limit zi → ∞.
If is similarly defined to map w1, w2, w3 to 0, 1, ∞, then the matrix which maps z1,2,3 to w1,2,3 becomes
The stabilizer of {0, 1, ∞} (as an unordered set) is a subgroup known as the anharmonic group.
縱使只一『式』 ,亦『勢』必將有所決矣︰
由於實難知能否破『迷惘』??!!
故而明指『任意』『三相異點』 之 的『投射』
真地存在呦☆
應該怎麼講
之所指哩★
如是再借
,求得 ,
豈非『任意』三點『投射』到『任意』三點呀◎
pi@raspberrypi:~ $ ipython3 Python 3.4.2 (default, Oct 19 2014, 13:31:11) Type "copyright", "credits" or "license" for more information. IPython 2.3.0 -- An enhanced Interactive Python. ? -> Introduction and overview of IPython's features. %quickref -> Quick reference. help -> Python's own help system. object? -> Details about 'object', use 'object??' for extra details. In [1]: from sympy import * In [2]: init_printing() In [3]: z1,z2,z3,w1,w2,w3 = symbols('z1,z2,z3,w1,w2,w3') In [4]: H1 = Matrix([[z2-z3,-z1*(z2-z3)],[z2-z1,-z3*(z2-z1)]]) In [5]: H1 Out[5]: ⎡z₂ - z₃ -z₁⋅(z₂ - z₃) ⎤ ⎢ ⎥ ⎣-z₁ + z₂ -z₃⋅(-z₁ + z₂)⎦ In [6]: H2 = Matrix([[w2-w3,-w1*(w2-w3)],[w2-w1,-w3*(w2-w1)]]) In [7]: H2 Out[7]: ⎡w₂ - w₃ -w₁⋅(w₂ - w₃) ⎤ ⎢ ⎥ ⎣-w₁ + w₂ -w₃⋅(-w₁ + w₂)⎦ In [8]: H2反矩陣 = H2.inv() In [9]: H2反矩陣 = Matrix([[H2反矩陣[0,0].simplify(), H2反矩陣[0,1].simplify()],[H2反矩陣[1,0].simplify(), H2反矩陣[1,1].simplify()]]) In [10]: H2反矩陣 Out[10]: ⎡ -w₃ -w₁ ⎤ ⎢─────────────────── ───────────────────⎥ ⎢(w₁ - w₃)⋅(w₂ - w₃) (w₁ - w₂)⋅(w₁ - w₃)⎥ ⎢ ⎥ ⎢ -1 -1 ⎥ ⎢─────────────────── ───────────────────⎥ ⎣(w₁ - w₃)⋅(w₂ - w₃) (w₁ - w₂)⋅(w₁ - w₃)⎦ In [11]: H = H2反矩陣 * H1 In [12]: H[0,0].simplify() Out[12]: w₁⋅(w₂ - w₃)⋅(z₁ - z₂) - w₃⋅(w₁ - w₂)⋅(z₂ - z₃) ─────────────────────────────────────────────── (w₁ - w₂)⋅(w₁ - w₃)⋅(w₂ - w₃) In [13]: H[0,1].simplify() Out[13]: -w₁⋅z₃⋅(w₂ - w₃)⋅(z₁ - z₂) + w₃⋅z₁⋅(w₁ - w₂)⋅(z₂ - z₃) ────────────────────────────────────────────────────── (w₁ - w₂)⋅(w₁ - w₃)⋅(w₂ - w₃) In [14]: H[1,0].simplify() Out[14]: (w₁ - w₂)⋅(-z₂ + z₃) + (w₂ - w₃)⋅(z₁ - z₂) ────────────────────────────────────────── (w₁ - w₂)⋅(w₁ - w₃)⋅(w₂ - w₃) In [15]: H[1,1].simplify() Out[15]: z₁⋅(w₁ - w₂)⋅(z₂ - z₃) - z₃⋅(w₂ - w₃)⋅(z₁ - z₂) ─────────────────────────────────────────────── (w₁ - w₂)⋅(w₁ - w₃)⋅(w₂ - w₃) In [16]: H[1,1].simplify() / H[1,0].simplify() Out[16]: z₁⋅(w₁ - w₂)⋅(z₂ - z₃) - z₃⋅(w₂ - w₃)⋅(z₁ - z₂) ─────────────────────────────────────────────── (w₁ - w₂)⋅(-z₂ + z₃) + (w₂ - w₃)⋅(z₁ - z₂) In [17]: