我们得到了一个问题,要求使用字母qdn(四分之一,角钱,镍(创建有限状态机,该机器仅在增加40美分时接受。我有使用IF语句概念的基础知识。我想知道是否有更简单的方法需要更少的时间?
我已经尝试制作大量的if情况,但是使用该方法有很多步骤。
public class FSA2_rpf4961 {
public static void main(String[] args) {
//The program will read in a sequence of strings and test them against a
//FSM. Your strings may not contain blank spaces
System.out.println("Enter string to test or q to terminate");
Scanner in = new Scanner (System.in);
String testString = in.next();
while (!testString.equals("q"))
{
String testOutput = applyFSA2(testString);
System.out.println("For the test string "+testString+
", the FSM output is "+testOutput);
System.out.println("Enter next string to test or q to terminate:");
testString = in.next();
}
in.close();
}
public static String applyFSA(String s) {
String currentOut = "0"; // initial output in s0
String currentState = "s0"; // initial state
int i = 0;
while (i<s.length())
{
//quarter first
if (currentState.equals("s0") && s.charAt(i) == 'q')
{
currentState = "s1";
currentOut += 25; // collect output on move to s1
}
else if (currentState.equals("s1") && s.charAt(i) == 'd') {
currentState = "s2";
currentOut += 10;
}
else if (currentState.equals("s2") && s.charAt(i) == 'n') {
currentState = "s3";
currentOut += 5;
}
else if (currentState.equals("s1") && s.charAt(i) == 'n') {
currentState = "s4";
currentOut += 5;
}
else if (currentState.equals("s4") && s.charAt(i) == 'd') {
currentState = "s3";
currentOut += 10;
}
//dime first
else if (currentState.equals("s0") && s.charAt(i) == 'd')
{
currentState = "s5";
currentOut += 10; // d
}
我们需要它仅在增加40美分的情况下才能接受。这让我很困惑。
我将在这里一般说话,因为您当前的代码很奇怪,以至于几乎确定的破碎。我不太了解您的"输出"应该是什么。
a FSM是一组状态的定义以及输入如何导致状态过渡。看来您开始大致这样做,但是currentOut
被打破了。如果您的目标是获得一笔款项,那么您将打败练习的全部要点。总和是什么都没关系。重要的是您最终处于哪种状态 - 尤其是您是否处于"接受"状态,在这种情况下,字符串等于40美分的状态。
至于如何实现没有if
s的FSM,您通常可以将状态存储在数组中,而将您的状态名称为状态号。那时,理论上您可以摆脱所有您的if
s。(但是,在现实世界中,您可能仍然希望一个人忽略或拒绝(q | d | n(以外的其他字符。(
考虑这样的东西(伪代码(:
// Each array in `graph` represents a state.
// Each entry is the state number (ie: index into `graph`) to go to
// when you're in that state and see the corresponding char.
//
// BTW, the state graph makes a lot more sense when you consider nickels first.
// A nickel takes you to the "next" state, and dimes and quarters act like
// 2 and 5 nickels, respectively. When you do that, a pattern shows up.
graph = [
// n d q
//-----------
[ 1, 2, 5 ], // s0
[ 2, 3, 6 ], // s1
[ 3, 4, 7 ], // s2
[ 4, 5, 8 ], // s3
[ 5, 6, 9 ], // s4
[ 6, 7, 9 ], // s5
[ 7, 8, 9 ], // s6
[ 8, 9, 9 ], // s7
[ 9, 9, 9 ], // s8
[ 9, 9, 9 ] // s9 (fail state)
]
start = 0
accept = 8
fail = 9
// at this point, walking the graph is trivial.
state = start
for each char c in s:
index = "ndq".indexOf(c) // n->0, d->1, q->2, others -> -1
state = graph[state][index]
// Once the loop's done:
// if state == accept, you have exactly 40c.
// if state == fail, you have >40c. A FSM won't tell you how much,
// because FSMs can't count.
// any other state represents a known amount that's less than 40c.