程序代數(Process Algebra)
Process Algebra 理論
提出者 | 理論名稱 | 縮寫 | 論文連結 | 簡介 |
---|---|---|---|---|
C. A. R. Hoare/Tony Hoare | Communicating Sequencing Process | CSP | Communicating Sequential Processes | 1978年C. A.R.Hoare提出的通訊順序程序 CSP,是面向分散式系統的程式設計語言 |
Robin Milner | Calculus of Communicating Systems | CCS | -- | 1973至1980年間發明了通訊系統演算CCS,是用於描述通訊併發系統的代數理論 |
J.A. Bergstra, J.W. Klop | Algebra of Communicating Processes with Abstraction | ACP | [ACP]http://dspace.library.uu.nl/handle/1874/12719) | Bergstra等人1984年提出的 ACP理論針對反應式、並行式和分散式系統,描述了兩個系統之間的互動行為 |
CSP基礎知識
- 原版教材PDF獲取,點我
注:目前已更新至2015版; - 中文版可參考周巢塵院士翻譯的《通訊順序程序》。但是年代比較久遠,是90年代的版本了。
第一章
1、對確定性程序,如何判斷兩個程序等價?
答: 確定性程序,需要判斷兩者alphabet(字母表)和traces(跡)是否相等。即:
①\(\alpha P=\alpha Q\)
②\(traces( P) =traces( Q )\)
2、\(traces(\mu X: A \cdot F(x)) = ?\)
答: \(traces(\mu X: A \cdot F(x)) = \{s|\exists n≥0,x \in A,s \le traces(F(x))^{n}\}\)
3、證明:
(下述兩道證明題均是採用數學歸納法證明)
(1)\(traces(RUN_{A}) = A^{*}.\)
(注:\(A^{*}\) means the set of sequences with elements in A)
(2)\(traces(VMS) = \cup_{n≥0} \{s| s≤< coin,choc >^{n},n≥0\}.\)
第二章
1、Let \(\alpha P = \{a,c\},\quad and \quad P = (a → c → P), \quad \alpha Q = \{b,c\}\quad and \quad Q = (c → b → Q).\)
(1)\(P || Q = ?\)
答:
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(2)Please prove that \(P|| Q \quad sat\quad 0 ≤ tr↓ a-tr↓ b ≤ 2.\)
答:
1.若 \(tr\) 未執行到迴圈階段,則 \(tr ↓ a = 1\) 或 \(0\), \(tr↓ b = 0\) 滿足不等式;
2.若 \(tr\) 執行到迴圈並恰好完成若干次迴圈,則由於每次迴圈 \(a\) 的個數 \(=\quad b\) 的個數,所以\(tr ↓ a − tr ↓ b = 1\)。
3.若 \(tr\) 執行到某次迴圈中,由於本次迴圈前滿足 \(tr ↓ a - tr ↓ b= 1\),
所以:
若執行 \(a → b → c → X\),則 \(tr↓ a − tr ↓ b =2\) 或 \(1\);
若執行 \(b → a → c → X\),則 \(tr ↓ a- tr ↓ b=0\) 或 \(1\);
綜上,\(0 ≤ tr↓ a- tr ↓ b≤ 2\)。
2、If P and Q never stop and if \(\alpha P \cap \alpha Q\) contains at most one element, then\((P || Q)\) never stops.
(1)請直觀解釋此結論的正確性。
答: 因為P和Q的字母表交集最多含有1個元素,所以不會觸發\((c → P)||(d → Q) = STOP \quad if c\ne d\)
(2)當 \(\alpha P \cap \alpha Q\) 含有 2 個或更多元素時,此結論不成立,舉例說明。
如\(\alpha P = \alpha Q = \{a, b\},\)
\(P = a \rightarrow b \rightarrow P;\)
\(Q = b \rightarrow a \rightarrow Q;\)
\(P || Q = STOP.\)
第三章
1、
(1)\(traces(P\sqcap Q) = ?\)
答: \(traces(P\sqcap Q)=traces(P) ∪ traces(Q)\)
(2)\(traces(P \square Q) = ?\)
答: \(traces(P \square Q)= traces(P)∪ traces(Q)\)
(3)\(refusals(P \sqcap Q) = ?\)
答: \(refusals(P\sqcap Q) = refusals(P) ∪ refusals(Q)\)
(4)\(refusals(P\square Q) = ?\)
答: \(refusals(P\square Q)=refusals(P) ∩ refusals(Q)\)
(5)令\(\alpha P = \alpha Q = \alpha P_{1} = \alpha Q_{1}= \{a,b,c\},\)
\(P_{1} = (a → b → STOP)\)
\(P_{2}= (b → c → STOP)\)
\(P = P_{1} \sqcap P_{2}\)
\(Q = P_{1}\square P_{2}\)
問:
①\(refusals(P) = ?\)
②\(refusals(Q) = ?\)
答:
\(refusals(P_{1}) = \{\{\},{b},{c},{b,c}\}\)
\(refusals(P_{2}) =\{\{\},{a},{c},{a,c}\}\)
\(refusals(P) = \{\{\},{a},{b},{c},{b,c},{a,c}\}\)
\(refusals(Q) =\{\{\},{c}\}\)
(6)\(refusals(P|| Q) = ?\)
答: \(refusals(P||Q)=\{X ∪ Y | X \in refusals(P) \wedge Y \in refusals(Q)\}\)
(7)\(refusals(P|||Q) = ?\)
答:\(refusals(P|||Q) =refusals(P\square Q) =refusals(P) \cap refusals(Q)\)
2.
(1)\(divergences(Chaos) = ?\)
答: \(divergences(Chaos) = A^*\)
(2)\(divergences(X: B → P(X)) = ?\)
答: \(\{⟨x⟩\smallfrown s | x \in B \wedge s \in divergences(P(x))\}\)
(3)\(divergences(P \sqcap Q) = ?\)
答: \(divergences(P) ∪ divergences(Q)\)
(4)\(divergences(P\square Q) = ?\)
答: \(divergences(P) ∪ divergences(Q)\)
(5)\(divergences(P∥Q) = ?\)
答: \(\{s \smallfrown t|t \in (\alpha P ∪ \alpha Q) ^{*} \wedge
((s \upharpoonright\alpha P \in divergences( P )\wedge s \upharpoonright \alpha Q \in traces(Q)) ∨
(s \upharpoonright \alpha P \in traces(P) \wedge s\upharpoonright \alpha Q \in divergences(Q))\}\)
(6)\(divergences(P|||Q) = ?\)
答: \(\{u | \exists s, t • u \quad interleaves (s, t) \wedge ((s \in divergences(P) \wedge t \in traces(Q)) ∨ (s \in traces(P) \wedge t \in divergences(Q)))\}\)
3.
(1)\(failures(P) = ?\)
答: \(failures(P) =\{(s, X)| s \in traces(P) \wedge X \in refusals(P/s)\}\)
(2)P 與 Q 的定義如上述第三章的 1、(3)所定義:
問:\(failures(P) = ?\) \(failures(Q) = ?\)
(3)
①\(failures(P \sqcap Q) = ?\)
答: \(failures(P \sqcap Q) =failures(P)\cup failures(Q)\)
②\(failures(X: B → P(X)) = ?\)
答: \(\{(<>, X)| X \subseteq (\alpha P − B)\} ∪ \{(⟨x⟩ \smallfrown s, X)| x \in B \wedge (s, X) \in failures(P(x))\}\)
③\(failures(P ∥ Q) = ?\)
答: \(failures(P||Q) = \{(s, X \cup Y )|s \in (\alpha P ∪ \alpha Q) ^{*} \wedge (s \upharpoonright \alpha P, X) \in failures(P) \wedge (s\upharpoonright \alpha Q, Y ) \in failures(Q)\} \cup \{(s, X)|s \in divergences(P||Q)\}\)
④\(failures(P \square Q) = ?\)
答: \(\{(s, X)|(s, X) \in failures(P) ∩ failures(Q)) \vee (s \ne <>\wedge (s, X) \in failures(P) \cup failures(Q))\} \cup \{(s, X)| s \in divergences(P \square Q)\}\)
⑤\(failures(P|||Q) = ?\)
答: \(\{(s, X)| ∃t, u• (t, X) \in failures(P) \wedge (u, X) \in failures(Q) \} ∪ \{(s, X)| s \in divergences(P|||Q)\}\)
4.對非確定性程序,如何判斷兩個程序等價?
答:對非確定性程序而言,使用traces已經無法區分(如,第三章的 1、(3)所定義的兩程序\(P\)和\(Q\):\(\alpha P=\alpha Q\),且\(traces(P)=traces(Q )\));進一步引入\(refusals\),但是用\(refusals\)來判斷,具有侷限性。最終,通過\(alphabet\)、\(divergences\)和\(failures\)綜合判斷。
即:
①\(\alpha P=\alpha Q\)
②\(divergences(P)=divergences(Q)\)
③\(failures(P)=failures(Q)\)
CSP: Operational Semantics
1、如何從 CSP 通訊的操作語義角度理解 CSP 併發定義中要求公共事件須同步?
答:
A和B之間存在通訊的管道,可以傳送某種型別的訊息,B在接收到A的訊息之前,並不清楚A傳送的內容,只知道型別;
只有在A傳送的同時,B同步接收,雙方才可以通訊,因此公共事件須同步。
2、從 CSP 的操作語義的角度定義:
(1)\(failures(P) = ?\)
答: \(failures(P) ={}_{df}\{s,X|\exists P_{1},P_{2}\cdot P\stackrel{s}{ \implies}P1\wedge P_{1}\xrightarrow {*}P_2\wedge stable(P_2)\wedge \forall c\in X\cdot \lnot (P_2\rightarrow)\}\)
(2)\(divergences(P) = ?\)
答: \(divergences(P) = {}_{df}\{s|\exists P_{1}\cdot P\stackrel{s}{ \implies}{s} P_{1}\wedge \uparrow P_{1}\}\)
CCS: Bisimulation
1.CCS 中 Strong Bisimulation 是如何定義的?
A binary relation \(S \subseteq P × P\) over agents is a strong bisimulation if \((P, Q) \in S\) implies, for all \(\alpha \in Act\),
(1) Whenever \(P \xrightarrow {\alpha }P'\) then, for some \(Q'\) , \(Q\xrightarrow {\alpha}Q'\) and \((P' ,Q' ) \in S\)
(2) Whenever \(Q \xrightarrow {\alpha } Q'\) then, for some \(P'\) , \(P \xrightarrow {\alpha }P'\) and \((P', Q') \in S\)
Denoted by \(P \sim Q\).
2.CCS 中 Weak Bisimulation 是如何定義的?
A binary relation \(S \subseteq P × P\) over agents is a weak bisimulation if \((P, Q) \in S\) implies, for all \(\alpha \in Act\),
(1) Whenever \(P \xrightarrow {\alpha } P'\) then, for some \(Q'\) , \(Q \stackrel{ \hat\alpha }{ \implies}Q'\) and \((P' ,Q' ) \in S\)
(2) Whenever \(Q \xrightarrow {\alpha } Q'\) then, for some \(P'\) , \(P \stackrel{ \hat\alpha }{ \implies} P'\) and \((P', Q') \in S\)
Denoted by \(P \approx Q\).