Question

对于作业，我必须设计一个DPDA，要求用户输入转换函数并说明最终状态。一切似乎已经在工作，除了当我运行它并开始输入输入时，一个字符的第一个输入似乎推动了需要被推入堆栈的一个额外副本。同样的事情发生在堆栈的第一个弹出窗口，它似乎弹出一个额外的。所以从长远来看它仍然可以工作，但我不知道额外的char来自哪里。

这是负责输入和输出的Driver类的部分

DPDA dad = new DPDA(0, states, transitions, stack, finalStates2);

    System.out.print("Current status " + dad.getCurrentState() + ":" + dad.getStack().toString() + ", Enter input: ");
    String nextInput = input.next();
    while(!nextInput.equals(".")){

        dad.transition(nextInput);
        System.out.print("Current status " + dad.getCurrentState() + ":" + dad.getStack().toString()+ ", Enter input: ");
        nextInput = input.next();
    }
    if(dad.isFinalState()){
        System.out.println("String accepted");
    }else
        System.out.println("String not accepted");

这是DPDA.transition功能

public void transition(String input){

    for(int i = 0; i < transitions.size(); i++){
        if(transitions.get(i).getCurrentState() == this.getCurrentState() && 
                transitions.get(i).getInput().equals(input) &&
                transitions.get(i).getTopStack()==this.getStack().peek()){

            stack.pop();
            this.pushStringOnToStack(transitions.get(i).getPush());
            this.setCurrentState(transitions.get(i).getNextState());

        }
        else
            System.out.println("No transition available");
    }

}

这是pushStringOntoStack函数

public void pushStringOnToStack(String pushed){
        for(int i = pushed.length()-1; i >=0 ; i--){
            if(!pushed.equals(".")){
                stack.push(pushed.charAt(i));
            }
        }

    }

示例输入

4
ab
0 a $ 1 A$
1 a A 1 aA
1 a a 1 aa
1 b a 2 .
1 b A 3 .
2 b a 2 .
2 b A 3 .
-1
1 3 -1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
(Equivalent to)
number of states = 4
alphabet, {a, b}
transition δ (0,a,$) = (1,"A$")
transition δ (1,a,A) = (1,"aA")
transition δ (1,a,a) = (1,"aa")
transition δ (1,b,a) = (2, λ)
transition δ (1,b,A) = (3, λ)
transition δ (2,b,a) = (2, λ)
transition δ (2,b,A) = (3, λ)
end of transition function
final states: {1, 3}

示例输出

Current status 0:$, Enter input: a
Current status 1:A$, Enter input: a
Current status 1:aA$, Enter input: a
Current status 1:aaA$, Enter input: a
Current status 1:aaaA$, Enter input: b
Current status 2:aaA$, Enter input: b
Current status 2:aA$, Enter input: b
Current status 2:A$, Enter input: b
Current status 3:$, Enter input: .
String accepted.

我的DPDA堆栈会将额外的字符推到它上面，但仍能以某种方式工作？

0 个答案: