TLA+ 中的咖啡罐问题:无法表达任务
Coffee Can Problem in TLA+ : cannot express a task
我正在尝试在 TLA+ 中为 David Gries’ Coffee Can Problem 建模,但我卡在了这一部分:
“关于最后剩下的豆子的颜色,作为罐头中最初黑白豆数量的函数,你怎么说?”
我不知道该如何处理。你能提供一些建议或提示吗? (也欢迎提供方法论)
这是我在 TLA+ 中的代码:
------------------------------ MODULE CanBean ------------------------------
EXTENDS Naturals, FiniteSets, Sequences
\* filter <- 2 | max_can <- 5
CONSTANT filter, max_can
VARIABLES picked, Can, Whites, Blacks
vars == <<picked, Can, Whites, Blacks>>
IsBlack(i) == i % filter = 0
IsWhite(i) == i % filter /= 0
GetWhite(a,b) == IF IsWhite(a) THEN a ELSE b
GetBlack(a,b) == IF IsBlack(a) THEN a ELSE b
AreBothWhite(a,b) == IsWhite(a) /\ IsWhite(b)
Pick == /\ picked = <<>>
/\ \E a,b \in Can : a/= b /\ picked' = <<a,b>>
/\ UNCHANGED <<Whites, Blacks, Can>>
Process == /\ picked /= <<>>
/\ LET a == picked[1] b == picked[2] IN
/\ \/ AreBothWhite(a,b) /\ Whites' = Whites \ {a,b} /\ \E m \in Nat \cap Can : Blacks' = Blacks \cup { filter * m }
\/ ~AreBothWhite(a,b) /\ Blacks' = Blacks \ { GetBlack(a,b) } /\ UNCHANGED Whites
/\ picked' = <<>>
/\ Can' = Blacks' \cup Whites'
Terminating == /\ Cardinality(Can) = 1
/\ UNCHANGED vars
TypeInvariantOK == /\ \A n \in Can : n \in Nat
/\ LET length == Len(picked)
IN length = 2 \/ length = 0
Init == /\ picked = <<>>
/\ Can = 1..max_can
/\ Blacks = { n \in Can : n % filter = 0 }
/\ Whites = { n \in Can : n % filter /= 0 }
Next == Pick \/ Process \/ Terminating
=============================================================================
\* Modification History
\* Last modified Sun Feb 21 14:53:34 CET 2021
\* Created Sat Feb 20 19:50:01 CET 2021
我写了一个简化的 TLA+ 规范,你可能会觉得有用:
---------------------------- MODULE CoffeeCan -------------------------------
EXTENDS Naturals
VARIABLES can
Can == [black : Nat, white : Nat]
\* Initialize can so it contains at least one bean.
Init == can \in {c \in Can : c.black + c.white >= 1}
BeanCount == can.black + can.white
PickSameColorBlack ==
/\ BeanCount > 1
/\ can.black >= 2
/\ can' = [can EXCEPT !.black = @ - 1]
PickSameColorWhite ==
/\ BeanCount > 1
/\ can.white >= 2
/\ can' = [can EXCEPT !.black = @ + 1, !.white = @ - 2]
PickDifferentColor ==
/\ BeanCount > 1
/\ can.black >= 1
/\ can.white >= 1
/\ can' = [can EXCEPT !.black = @ - 1]
Termination ==
/\ BeanCount = 1
/\ UNCHANGED can
Next ==
\/ PickSameColorWhite
\/ PickSameColorBlack
\/ PickDifferentColor
\/ Termination
MonotonicDecrease == [][BeanCount > 1 => BeanCount' < BeanCount]_<<can>>
EventuallyTerminates == <>(ENABLED Termination)
Spec ==
/\ Init
/\ [][Next]_<<can>>
/\ WF_<<can>>(Next)
THEOREM Spec =>
/\ MonotonicDecrease
/\ EventuallyTerminates
=============================================================================
您可以通过覆盖 Nat
的定义来对其进行模型检查。
正如您所注意到的,从这个规范中很容易看出 bean 的数量单调减少(这可以用 MonotonicDecrease
时间 属性 来验证)因此该过程必须在有限数量的步骤。第二个问题似乎涉及概率,因此不太适合TLA+。 TLC 确实有能力 simulate basic random systems in TLA+, but this is too limited to directly encode the answer to question two as a system invariant. There's a formal specification language called PRISM 创建来处理概率系统,尽管它的语言比 TLA+ 更不符合人体工程学。
编辑:我为这个问题写了一个 PRISM 规范,发现了一些非常有趣的东西——我们根本不需要处理概率!这是规范 - min() 和 max() 看似多余的用法是因为 PRISM 对不使变量超出范围的更新非常挑剔,即使该更新被采用的概率为 0:
dtmc
const int MAX_BEAN = 20;
formula total = black + white;
formula p_two_black = (black/total)*(max(0,black-1)/(total-1));
formula p_two_white = (white/total)*(max(0,white-1)/(total-1));
formula p_different = 1.0 - p_two_black - p_two_white;
init
total >= 1 & total <= MAX_BEAN
endinit
module CoffeeCan
black : [0..(2*MAX_BEAN)];
white : [0..MAX_BEAN];
[] total > 1 ->
p_two_black : (black' = max(0, black - 1))
+ p_two_white : (black' = min(2*MAX_BEAN, black + 1)) & (white' = max(0, white - 2))
+ p_different : (black' = max(0, black - 1));
[] total = 1 -> true;
endmodule
插入一些测试初始值后,我使用 PRISM 检查了以下公式,它基本上是在问“我们以单个白豆终止的概率是多少?”:
P=?[F black = 0 & white = 1]
你知道我发现了什么吗?当您从偶数个白豆开始时,概率为零,而当您从奇数个白豆开始时,概率为 1!我们可以将此断言编码为 PRISM 将验证为真的属性:
P>=1 [mod(white, 2) = 1 => (F black = 0 & white = 1)]
P>=1 [mod(white, 2) = 0 => (F black = 1 & white = 0)]
回想起来这是显而易见的,因为白豆的数量只会减少两个。因此,如果您从偶数个白豆开始,您将以零个白豆结束,如果您以奇数个白豆开始,您将以一个白豆结束。您可以将此时间 属性 添加到上述 TLA+ 模型中以进行检查:
WhiteBeanTermination ==
IF can.white % 2 = 0
THEN <>(can.black = 1 /\ can.white = 0)
ELSE <>(can.black = 0 /\ can.white = 1)
我正在尝试在 TLA+ 中为 David Gries’ Coffee Can Problem 建模,但我卡在了这一部分:
“关于最后剩下的豆子的颜色,作为罐头中最初黑白豆数量的函数,你怎么说?”
我不知道该如何处理。你能提供一些建议或提示吗? (也欢迎提供方法论)
这是我在 TLA+ 中的代码:
------------------------------ MODULE CanBean ------------------------------
EXTENDS Naturals, FiniteSets, Sequences
\* filter <- 2 | max_can <- 5
CONSTANT filter, max_can
VARIABLES picked, Can, Whites, Blacks
vars == <<picked, Can, Whites, Blacks>>
IsBlack(i) == i % filter = 0
IsWhite(i) == i % filter /= 0
GetWhite(a,b) == IF IsWhite(a) THEN a ELSE b
GetBlack(a,b) == IF IsBlack(a) THEN a ELSE b
AreBothWhite(a,b) == IsWhite(a) /\ IsWhite(b)
Pick == /\ picked = <<>>
/\ \E a,b \in Can : a/= b /\ picked' = <<a,b>>
/\ UNCHANGED <<Whites, Blacks, Can>>
Process == /\ picked /= <<>>
/\ LET a == picked[1] b == picked[2] IN
/\ \/ AreBothWhite(a,b) /\ Whites' = Whites \ {a,b} /\ \E m \in Nat \cap Can : Blacks' = Blacks \cup { filter * m }
\/ ~AreBothWhite(a,b) /\ Blacks' = Blacks \ { GetBlack(a,b) } /\ UNCHANGED Whites
/\ picked' = <<>>
/\ Can' = Blacks' \cup Whites'
Terminating == /\ Cardinality(Can) = 1
/\ UNCHANGED vars
TypeInvariantOK == /\ \A n \in Can : n \in Nat
/\ LET length == Len(picked)
IN length = 2 \/ length = 0
Init == /\ picked = <<>>
/\ Can = 1..max_can
/\ Blacks = { n \in Can : n % filter = 0 }
/\ Whites = { n \in Can : n % filter /= 0 }
Next == Pick \/ Process \/ Terminating
=============================================================================
\* Modification History
\* Last modified Sun Feb 21 14:53:34 CET 2021
\* Created Sat Feb 20 19:50:01 CET 2021
我写了一个简化的 TLA+ 规范,你可能会觉得有用:
---------------------------- MODULE CoffeeCan -------------------------------
EXTENDS Naturals
VARIABLES can
Can == [black : Nat, white : Nat]
\* Initialize can so it contains at least one bean.
Init == can \in {c \in Can : c.black + c.white >= 1}
BeanCount == can.black + can.white
PickSameColorBlack ==
/\ BeanCount > 1
/\ can.black >= 2
/\ can' = [can EXCEPT !.black = @ - 1]
PickSameColorWhite ==
/\ BeanCount > 1
/\ can.white >= 2
/\ can' = [can EXCEPT !.black = @ + 1, !.white = @ - 2]
PickDifferentColor ==
/\ BeanCount > 1
/\ can.black >= 1
/\ can.white >= 1
/\ can' = [can EXCEPT !.black = @ - 1]
Termination ==
/\ BeanCount = 1
/\ UNCHANGED can
Next ==
\/ PickSameColorWhite
\/ PickSameColorBlack
\/ PickDifferentColor
\/ Termination
MonotonicDecrease == [][BeanCount > 1 => BeanCount' < BeanCount]_<<can>>
EventuallyTerminates == <>(ENABLED Termination)
Spec ==
/\ Init
/\ [][Next]_<<can>>
/\ WF_<<can>>(Next)
THEOREM Spec =>
/\ MonotonicDecrease
/\ EventuallyTerminates
=============================================================================
您可以通过覆盖 Nat
的定义来对其进行模型检查。
正如您所注意到的,从这个规范中很容易看出 bean 的数量单调减少(这可以用 MonotonicDecrease
时间 属性 来验证)因此该过程必须在有限数量的步骤。第二个问题似乎涉及概率,因此不太适合TLA+。 TLC 确实有能力 simulate basic random systems in TLA+, but this is too limited to directly encode the answer to question two as a system invariant. There's a formal specification language called PRISM 创建来处理概率系统,尽管它的语言比 TLA+ 更不符合人体工程学。
编辑:我为这个问题写了一个 PRISM 规范,发现了一些非常有趣的东西——我们根本不需要处理概率!这是规范 - min() 和 max() 看似多余的用法是因为 PRISM 对不使变量超出范围的更新非常挑剔,即使该更新被采用的概率为 0:
dtmc
const int MAX_BEAN = 20;
formula total = black + white;
formula p_two_black = (black/total)*(max(0,black-1)/(total-1));
formula p_two_white = (white/total)*(max(0,white-1)/(total-1));
formula p_different = 1.0 - p_two_black - p_two_white;
init
total >= 1 & total <= MAX_BEAN
endinit
module CoffeeCan
black : [0..(2*MAX_BEAN)];
white : [0..MAX_BEAN];
[] total > 1 ->
p_two_black : (black' = max(0, black - 1))
+ p_two_white : (black' = min(2*MAX_BEAN, black + 1)) & (white' = max(0, white - 2))
+ p_different : (black' = max(0, black - 1));
[] total = 1 -> true;
endmodule
插入一些测试初始值后,我使用 PRISM 检查了以下公式,它基本上是在问“我们以单个白豆终止的概率是多少?”:
P=?[F black = 0 & white = 1]
你知道我发现了什么吗?当您从偶数个白豆开始时,概率为零,而当您从奇数个白豆开始时,概率为 1!我们可以将此断言编码为 PRISM 将验证为真的属性:
P>=1 [mod(white, 2) = 1 => (F black = 0 & white = 1)]
P>=1 [mod(white, 2) = 0 => (F black = 1 & white = 0)]
回想起来这是显而易见的,因为白豆的数量只会减少两个。因此,如果您从偶数个白豆开始,您将以零个白豆结束,如果您以奇数个白豆开始,您将以一个白豆结束。您可以将此时间 属性 添加到上述 TLA+ 模型中以进行检查:
WhiteBeanTermination ==
IF can.white % 2 = 0
THEN <>(can.black = 1 /\ can.white = 0)
ELSE <>(can.black = 0 /\ can.white = 1)