東京大学 情報理工学系研究科 コンピュータ科学専攻 2015年8月実施 専門科目II 問題2
Author
祭音Myyura (co-authored with GPT 5.6 SOL)
Description
Let N \mathbb N N be the set of all nonnegative integers. Let Q Q Q be a set of states defined by Q = N × N × N Q=\mathbb N\times\mathbb N\times\mathbb N Q = N × N × N , and let a transition relation ⟶ \longrightarrow ⟶ on Q Q Q be defined as follows.
( a , b , c ) ⟶ ( a − 1 , b − 1 , c + 2 ) (if a > 0 and b > 0 ) , ( a , b , c ) ⟶ ( a + 2 , b − 1 , c − 1 ) (if b > 0 and c > 0 ) , ( a , b , c ) ⟶ ( a − 1 , b + 2 , c − 1 ) (if c > 0 and a > 0 ) . ( † ) \begin{aligned}
(a,b,c)&\longrightarrow(a-1,b-1,c+2)
&&\text{(if }a>0\text{ and }b>0\text{)},\\
(a,b,c)&\longrightarrow(a+2,b-1,c-1)
&&\text{(if }b>0\text{ and }c>0\text{)},\\
(a,b,c)&\longrightarrow(a-1,b+2,c-1)
&&\text{(if }c>0\text{ and }a>0\text{)}.
\end{aligned}
\qquad(\dagger) ( a , b , c ) ( a , b , c ) ( a , b , c ) ⟶ ( a − 1 , b − 1 , c + 2 ) ⟶ ( a + 2 , b − 1 , c − 1 ) ⟶ ( a − 1 , b + 2 , c − 1 ) (if a > 0 and b > 0 ) , (if b > 0 and c > 0 ) , (if c > 0 and a > 0 ) . ( † )
Let ⟶ ∗ \longrightarrow^* ⟶ ∗ denote the reflexive transitive closure of ⟶ \longrightarrow ⟶ .
Answer the following questions.
(1) Enumerate all states q ∈ Q q\in Q q ∈ Q such that ( 1 , 2 , 3 ) ⟶ ∗ q (1,2,3)\longrightarrow^*q ( 1 , 2 , 3 ) ⟶ ∗ q , and draw a state transition graph.
(2) A state ( a , b , c ) (a,b,c) ( a , b , c ) is called a deadlock state if there exists no state q q q such that ( a , b , c ) ⟶ q (a,b,c)\longrightarrow q ( a , b , c ) ⟶ q .
Give a necessary and sufficient condition for a state ( a , b , c ) (a,b,c) ( a , b , c ) to be a deadlock state.
(3) Give a necessary and sufficient condition for a state ( a , b , c ) (a,b,c) ( a , b , c ) to have a deadlock state q q q such that ( a , b , c ) ⟶ ∗ q (a,b,c)\longrightarrow^*q ( a , b , c ) ⟶ ∗ q .
(4) Assume that, at each state ( a , b , c ) (a,b,c) ( a , b , c ) , one out of the three transitions defined in the above ( † ) (\dagger) ( † ) is chosen to take place, with the following probabilities.
( a , b , c ) ⟶ ( a − 1 , b − 1 , c + 2 ) with probability a b / ( a b + b c + c a ) , ( a , b , c ) ⟶ ( a + 2 , b − 1 , c − 1 ) with probability b c / ( a b + b c + c a ) , ( a , b , c ) ⟶ ( a − 1 , b + 2 , c − 1 ) with probability c a / ( a b + b c + c a ) . \begin{aligned}
(a,b,c)&\longrightarrow(a-1,b-1,c+2)
&&\text{with probability }ab/(ab+bc+ca),\\
(a,b,c)&\longrightarrow(a+2,b-1,c-1)
&&\text{with probability }bc/(ab+bc+ca),\\
(a,b,c)&\longrightarrow(a-1,b+2,c-1)
&&\text{with probability }ca/(ab+bc+ca).
\end{aligned} ( a , b , c ) ( a , b , c ) ( a , b , c ) ⟶ ( a − 1 , b − 1 , c + 2 ) ⟶ ( a + 2 , b − 1 , c − 1 ) ⟶ ( a − 1 , b + 2 , c − 1 ) with probability ab / ( ab + b c + c a ) , with probability b c / ( ab + b c + c a ) , with probability c a / ( ab + b c + c a ) .
Now let an initial state be ( 1 , 2 , 3 ) (1,2,3) ( 1 , 2 , 3 ) , and consider repeating the above probabilistic transitions for sufficiently many times. Compute the probability with which, after such transitions, the current state is either ( 1 , 2 , 3 ) (1,2,3) ( 1 , 2 , 3 ) , ( 3 , 1 , 2 ) (3,1,2) ( 3 , 1 , 2 ) or ( 2 , 3 , 1 ) (2,3,1) ( 2 , 3 , 1 ) .
题目描述
令 N \mathbb N N 为非负整数集,状态空间 Q = N 3 Q=\mathbb N^3 Q = N 3 。定义转移
( a , b , c ) → ( a − 1 , b − 1 , c + 2 ) ( a > 0 , b > 0 ) , ( a , b , c ) → ( a + 2 , b − 1 , c − 1 ) ( b > 0 , c > 0 ) , ( a , b , c ) → ( a − 1 , b + 2 , c − 1 ) ( c > 0 , a > 0 ) . \begin{aligned}
(a,b,c)&\to(a-1,b-1,c+2) &&(a>0,b>0),\\
(a,b,c)&\to(a+2,b-1,c-1) &&(b>0,c>0),\\
(a,b,c)&\to(a-1,b+2,c-1) &&(c>0,a>0).
\end{aligned} ( a , b , c ) ( a , b , c ) ( a , b , c ) → ( a − 1 , b − 1 , c + 2 ) → ( a + 2 , b − 1 , c − 1 ) → ( a − 1 , b + 2 , c − 1 ) ( a > 0 , b > 0 ) , ( b > 0 , c > 0 ) , ( c > 0 , a > 0 ) .
以 → ∗ \to^* → ∗ 表示自反传递闭包。
(1)列出所有满足 ( 1 , 2 , 3 ) → ∗ q (1,2,3)\to^*q ( 1 , 2 , 3 ) → ∗ q 的状态 q q q ,并画出状态转移图。
(2)若不存在 ( a , b , c ) → q (a,b,c)\to q ( a , b , c ) → q ,称 ( a , b , c ) (a,b,c) ( a , b , c ) 为死锁状态。给出死锁的充要条件。
(3)给出从 ( a , b , c ) (a,b,c) ( a , b , c ) 可达某个死锁状态的充要条件。
(4)在每个状态按权重 a b , b c , c a ab,bc,ca ab , b c , c a 选择上述三种可行转移,即相应概率为
a b / ( a b + b c + c a ) ab/(ab+bc+ca) ab / ( ab + b c + c a ) 、b c / ( a b + b c + c a ) bc/(ab+bc+ca) b c / ( ab + b c + c a ) 、c a / ( a b + b c + c a ) ca/(ab+bc+ca) c a / ( ab + b c + c a ) 。从 ( 1 , 2 , 3 ) (1,2,3) ( 1 , 2 , 3 ) 出发,求充分多次转移后,当前状态属于
{ ( 1 , 2 , 3 ) , ( 3 , 1 , 2 ) , ( 2 , 3 , 1 ) } \{(1,2,3),(3,1,2),(2,3,1)\} {( 1 , 2 , 3 ) , ( 3 , 1 , 2 ) , ( 2 , 3 , 1 )}
的概率。
Kai
(1)
全部可达状态(含初态自身)为
{ ( 1 , 2 , 3 ) , ( 3 , 1 , 2 ) , ( 2 , 3 , 1 ) , ( 0 , 4 , 2 ) , ( 4 , 2 , 0 ) , ( 2 , 0 , 4 ) , ( 0 , 1 , 5 ) , ( 1 , 5 , 0 ) , ( 5 , 0 , 1 ) } . \begin{aligned}
\{&(1,2,3),(3,1,2),(2,3,1),\\
&(0,4,2),(4,2,0),(2,0,4),\\
&(0,1,5),(1,5,0),(5,0,1)\}.
\end{aligned} { ( 1 , 2 , 3 ) , ( 3 , 1 , 2 ) , ( 2 , 3 , 1 ) , ( 0 , 4 , 2 ) , ( 4 , 2 , 0 ) , ( 2 , 0 , 4 ) , ( 0 , 1 , 5 ) , ( 1 , 5 , 0 ) , ( 5 , 0 , 1 )} .
下图的边标 1 , 2 , 3 1,2,3 1 , 2 , 3 对应题中的三种转移。图中九态均从初态可达,且对所有可行转移封闭,故没有其他可达状态。
(2)
只要至少两个坐标为正,这两个坐标对应的某一种转移就可执行。因此
( a , b , c ) 为死锁状态 ⟺ a , b , c 中至少两个为 0. \boxed{(a,b,c)\text{ 为死锁状态}\iff a,b,c\text{ 中至少两个为 }0.} ( a , b , c ) 为死锁状态 ⟺ a , b , c 中至少两个为 0.
(3)
答案为
a ≡ b ( m o d 3 ) 或 b ≡ c ( m o d 3 ) 或 c ≡ a ( m o d 3 ) . \boxed{a\equiv b\pmod3\quad\text{或}\quad
b\equiv c\pmod3\quad\text{或}\quad
c\equiv a\pmod3.} a ≡ b ( mod 3 ) 或 b ≡ c ( mod 3 ) 或 c ≡ a ( mod 3 ) .
每种转移都保持 a − b , b − c , c − a a-b,b-c,c-a a − b , b − c , c − a 的模 3 3 3 余数不变。若可达死锁,例如 ( s , 0 , 0 ) (s,0,0) ( s , 0 , 0 ) ,则原状态必有 b ≡ c ( m o d 3 ) b\equiv c\pmod3 b ≡ c ( mod 3 ) ;其余两类死锁同理,故该条件必要。
下面证明充分性。不妨设 a ≡ b ( m o d 3 ) a\equiv b\pmod3 a ≡ b ( mod 3 ) 。连续执行第一种转移,直到 a , b a,b a , b 中至少一个变为 0 0 0 。若二者同时为 0 0 0 ,已经到达死锁;否则得到 ( 0 , 3 k , C ) (0,3k,C) ( 0 , 3 k , C ) 或 ( 3 k , 0 , C ) (3k,0,C) ( 3 k , 0 , C ) 。若 C = 0 C=0 C = 0 也已死锁。若 C > 0 C>0 C > 0 ,则
( 0 , 3 k , C ) → ( 2 , 3 k − 1 , C − 1 ) → ( 1 , 3 k − 2 , C + 1 ) → ( 0 , 3 k − 3 , C + 3 ) . \begin{aligned}
(0,3k,C)&\to(2,3k-1,C-1)\\
&\to(1,3k-2,C+1)\\
&\to(0,3k-3,C+3).
\end{aligned} ( 0 , 3 k , C ) → ( 2 , 3 k − 1 , C − 1 ) → ( 1 , 3 k − 2 , C + 1 ) → ( 0 , 3 k − 3 , C + 3 ) .
这三步使 k k k 减少 1 1 1 。重复即可到达 ( 0 , 0 , C + 3 k ) (0,0,C+3k) ( 0 , 0 , C + 3 k ) 。另一种形状 ( 3 k , 0 , C ) (3k,0,C) ( 3 k , 0 , C ) 对称处理,故条件充分。
(4)
从初态可达的九个状态可按循环置换分成三类:
A = { ( 1 , 2 , 3 ) , ( 3 , 1 , 2 ) , ( 2 , 3 , 1 ) } , B = { ( 0 , 4 , 2 ) , ( 4 , 2 , 0 ) , ( 2 , 0 , 4 ) } , C = { ( 0 , 1 , 5 ) , ( 1 , 5 , 0 ) , ( 5 , 0 , 1 ) } . \begin{aligned}
A&=\{(1,2,3),(3,1,2),(2,3,1)\},\\
B&=\{(0,4,2),(4,2,0),(2,0,4)\},\\
C&=\{(0,1,5),(1,5,0),(5,0,1)\}.
\end{aligned} A B C = {( 1 , 2 , 3 ) , ( 3 , 1 , 2 ) , ( 2 , 3 , 1 )} , = {( 0 , 4 , 2 ) , ( 4 , 2 , 0 ) , ( 2 , 0 , 4 )} , = {( 0 , 1 , 5 ) , ( 1 , 5 , 0 ) , ( 5 , 0 , 1 )} .
从 A A A 出发,以 6 / 11 , 3 / 11 , 2 / 11 6/11,3/11,2/11 6/11 , 3/11 , 2/11 的概率分别进入 A , B , C A,B,C A , B , C ;从 B B B 必然进入 A A A ,从 C C C 必然进入 B B B 。故聚合链的转移矩阵为
P = ( 6 / 11 3 / 11 2 / 11 1 0 0 0 1 0 ) . P=
\begin{pmatrix}
6/11&3/11&2/11\\
1&0&0\\
0&1&0
\end{pmatrix}. P = 6/11 1 0 3/11 0 1 2/11 0 0 .
该链不可约,且 A A A 有自环,故分布收敛到唯一平稳分布。设其为 ( π A , π B , π C ) (\pi_A,\pi_B,\pi_C) ( π A , π B , π C ) ,则
π C = 2 11 π A , π B = 3 11 π A + π C = 5 11 π A . \pi_C=\frac2{11}\pi_A,\qquad
\pi_B=\frac3{11}\pi_A+\pi_C=\frac5{11}\pi_A. π C = 11 2 π A , π B = 11 3 π A + π C = 11 5 π A .
归一化后得到
( π A , π B , π C ) = ( 11 18 , 5 18 , 1 9 ) . (\pi_A,\pi_B,\pi_C)=\left(\frac{11}{18},\frac5{18},\frac19\right). ( π A , π B , π C ) = ( 18 11 , 18 5 , 9 1 ) .
题目所求即 11 / 18 \boxed{11/18} 11/18 。