DFA或确定性有限自动机是一种有限状态机，如果达到最终状态，则该字符串会接受字符串(在某些特定条件下)，否则将其拒绝。
问题：给定“0′和” 1′‘01’或‘10’别人拒绝的字符串。不论是否接受，也要打印状态图。由于在DFA中没有内存的概念，因此我们一次只能检查一个字符，从第0个字符开始。此问题的输入集为{0，1} 。对于输入集中的每个字符，DFA的每个状态都会重定向到另一个有效状态。
DFA机器：对于以上问题陈述，我们必须首先构建DFA机器。 DFA机器类似于具有各种状态和转换的流程图。与上述问题相对应的DFA机器如下所示， Q3和Q4为最终状态： 
例子：
Input: 010101
Output:
State transitions are q0->q1->q3->q4
->q3->q4->q3->YES
Explanation : 010101 ends with "01".
Input: 0100
Output:
State transitions are q0->q1->q3->q4->q1->NO
Explanation : 0100 ends with "00",
which is not equal to any of "01" or "10".
算法：
- Define the minimum number of states required to make the state diagram. Use functions to define various states.
- List all the valid transitions. Each state must have a transition for every valid symbol.
- Define the final states by applying the base condition.
- Define all the state transitions using state function calls.
- Define a returning condition for the end of the string.
对于给定的DFA计算机：
- Q0, Q1, Q2, Q3, Q4 are defined as the number of states.
- 0 and 1 are valid symbols. Each state has transitions for 0 and 1.
- Q3 and Q4 are defined as the final states.
- Suppose at state Q0, if 0 comes, the function call is made to Q1. So, if 1 comes, the function call is made to Q2.
- If the program reaches the end of the string, the output is made according to the state, the program is at.
执行：
C++
// CPP Program to DFA that accepts string ending
// with 01 or 10.
#include
using namespace std;
// Various states of DFA machine are defined
// using functions.
void q1(string, int);
void q2(string, int);
void q3(string, int);
void q4(string, int);
// End position is checked using the string
// length value.
// q0 is the starting state.
// q1 and q2 are intermediate states.
// q3 and q4 are final states.
void q1(string s, int i)
{
cout << "q1->";
if (i == s.length()) {
cout << "NO \n";
return;
}
// state transitions
// 0 takes to q1, 1 takes to q3
if (s[i] == '0')
q1(s, i + 1);
else
q3(s, i + 1);
}
void q2(string s, int i)
{
cout << "q2->";
if (i == s.length()) {
cout << "NO \n";
return;
}
// state transitions
// 0 takes to q4, 1 takes to q2
if (s[i] == '0')
q4(s, i + 1);
else
q2(s, i + 1);
}
void q3(string s, int i)
{
cout << "q3->";
if (i == s.length()) {
cout << "YES \n";
return;
}
// state transitions
// 0 takes to q4, 1 takes to q2
if (s[i] == '0')
q4(s, i + 1);
else
q2(s, i + 1);
}
void q4(string s, int i)
{
cout << "q4->";
if (i == s.length()) {
cout << "YES \n";
return;
}
// state transitions
// 0 takes to q1, 1 takes to q3
if (s[i] == '0')
q1(s, i + 1);
else
q3(s, i + 1);
}
void q0(string s, int i)
{
cout << "q0->";
if (i == s.length()) {
cout << "NO \n";
return;
}
// state transitions
// 0 takes to q1, 1 takes to q2
if (s[i] == '0')
q1(s, i + 1);
else
q2(s, i + 1);
}
// Driver Code
int main()
{
string s = "010101";
// all state transitions are printed.
// if string is accpetable, YES is printed.
// else NO is printed
cout << "State transitions are ";
q0(s, 0);
} Java
// Java Program to DFA that accepts string ending
// with 01 or 10.
class GFG
{
// End position is checked using the string
// length value.
// q0 is the starting state.
// q1 and q2 are intermediate states.
// q3 and q4 are final states.
static void q1(String s, int i)
{
System.out.print("q1->");
if (i == s.length())
{
System.out.println("NO");
return;
}
// state transitions
// 0 takes to q1, 1 takes to q3
if (s.charAt(i) == '0')
q1(s, i + 1);
else
q3(s, i + 1);
}
static void q2(String s, int i)
{
System.out.print("q2->");
if (i == s.length())
{
System.out.println("NO ");
return;
}
// state transitions
// 0 takes to q4, 1 takes to q2
if (s.charAt(i) == '0')
q4(s, i + 1);
else
q2(s, i + 1);
}
static void q3(String s, int i)
{
System.out.print("q3->");
if (i == s.length())
{
System.out.println("YES");
return;
}
// state transitions
// 0 takes to q4, 1 takes to q2
if (s.charAt(i) == '0')
q4(s, i + 1);
else
q2(s, i + 1);
}
static void q4(String s, int i)
{
System.out.print("q4->");
if (i == s.length())
{
System.out.println("YES");
return;
}
// state transitions
// 0 takes to q1, 1 takes to q3
if (s.charAt(i) == '0')
q1(s, i + 1);
else
q3(s, i + 1);
}
static void q0(String s, int i)
{
System.out.print("q0->");
if (i == s.length())
{
System.out.println("NO");
return;
}
// state transitions
// 0 takes to q1, 1 takes to q2
if (s.charAt(i) == '0')
q1(s, i + 1);
else
q2(s, i + 1);
}
// Driver Code
public static void main (String[] args)
{
String s = "010101";
// all state transitions are printed.
// if string is accpetable, YES is printed.
// else NO is printed
System.out.print("State transitions are ");
q0(s, 0);
}
}
// This code is contributed by AnkitRai01Python3
# Python3 Program to DFA that accepts string ending
# with 01 or 10.
# End position is checked using the string
# length value.
# q0 is the starting state.
# q1 and q2 are intermediate states.
# q3 and q4 are final states.
def q1(s, i) :
print("q1->", end="");
if (i == len(s)) :
print("NO");
return;
# state transitions
# 0 takes to q1, 1 takes to q3
if (s[i] == '0') :
q1(s, i + 1);
else :
q3(s, i + 1);
def q2(s, i) :
print("q2->", end = "");
if (i == len(s)) :
print("NO");
return;
# state transitions
# 0 takes to q4, 1 takes to q2
if (s[i] == '0') :
q4(s, i + 1);
else :
q2(s, i + 1);
def q3(s, i) :
print("q3->", end = "");
if (i == len(s)) :
print("YES");
return;
# state transitions
# 0 takes to q4, 1 takes to q2
if (s[i] == '0') :
q4(s, i + 1);
else :
q2(s, i + 1);
def q4(s, i) :
print("q4->", end = "");
if (i == len(s)) :
print("YES");
return;
# state transitions
# 0 takes to q1, 1 takes to q3
if (s[i] == '0') :
q1(s, i + 1);
else :
q3(s, i + 1);
def q0( s, i) :
print("q0->", end = "");
if (i == len(s)) :
print("NO");
return;
# state transitions
# 0 takes to q1, 1 takes to q2
if (s[i] == '0') :
q1(s, i + 1);
else :
q2(s, i + 1);
# Driver Code
if __name__ == "__main__" :
s = "010101";
# all state transitions are printed.
# if string is accpetable, YES is printed.
# else NO is printed
print("State transitions are", end = " ");
q0(s, 0);
# This code is contributed by AnkitRai01C#
// C# Program to DFA that accepts string ending
// with 01 or 10.
using System;
class GFG
{
// End position is checked using the string
// length value.
// q0 is the starting state.
// q1 and q2 are intermediate states.
// q3 and q4 are final states.
static void q1(string s, int i)
{
Console.Write("q1->");
if (i == s.Length )
{
Console.WriteLine("NO");
return;
}
// state transitions
// 0 takes to q1, 1 takes to q3
if (s[i] == '0')
q1(s, i + 1);
else
q3(s, i + 1);
}
static void q2(string s, int i)
{
Console.Write("q2->");
if (i == s.Length)
{
Console.WriteLine("NO ");
return;
}
// state transitions
// 0 takes to q4, 1 takes to q2
if (s[i] == '0')
q4(s, i + 1);
else
q2(s, i + 1);
}
static void q3(string s, int i)
{
Console.Write("q3->");
if (i == s.Length)
{
Console.WriteLine("YES");
return;
}
// state transitions
// 0 takes to q4, 1 takes to q2
if (s[i] == '0')
q4(s, i + 1);
else
q2(s, i + 1);
}
static void q4(string s, int i)
{
Console.Write("q4->");
if (i == s.Length)
{
Console.WriteLine("YES");
return;
}
// state transitions
// 0 takes to q1, 1 takes to q3
if (s[i] == '0')
q1(s, i + 1);
else
q3(s, i + 1);
}
static void q0(string s, int i)
{
Console.Write("q0->");
if (i == s.Length)
{
Console.WriteLine("NO");
return;
}
// state transitions
// 0 takes to q1, 1 takes to q2
if (s[i] == '0')
q1(s, i + 1);
else
q2(s, i + 1);
}
// Driver Code
public static void Main()
{
string s = "010101";
// all state transitions are printed.
// if string is accpetable, YES is printed.
// else NO is printed
Console.Write("State transitions are ");
q0(s, 0);
}
}
// This code is contributed by AnkitRai01输出：
State transitions are q0->q1->q3->q4->q3->q4->q3->YES
复杂度： O(n)，其中长度为n的字符串需要遍历n个状态。
一个问题陈述可能有多个DFA。