我在第三次迭代中卡住的這些 while 循環做錯了什么？

Question

嘗試在 python 3.9 中實現 FSA，

編輯：

就上下文而言，這是我作為一個學習編碼的人必須做的第一個重大項目。 我遲到了 5 周才被扔進一門大學課程，基本上被告知要盡力趕上。 我很難弄清楚循環機制在做什么，並且 class 與變量結構應該看起來像。 FSA 有問題“在此識別器中，起始 state 是 S1，而 S7 是唯一接受 state。（不接受空字符串）。如果上述 FSA 接受，您的程序應要求用戶輸入字符串並打印 'True'它，否則為“False”。打印結果和“再見”消息后，程序應立即停止。”

我的表格顯示了 FSA 正在做什么：表格

我一直在努力保留我所有工作的 github。

當它檢查輸入的數字時，它應該根據所需的步驟對它們進行測試，並根據它是否滿足要求來向后或向前移動。 對於此示例，att 應該作為唯一成功通過測試。

我一直無法讓循環正常工作。 我可以讓前兩個發生，但它要么卡在第三個，要么根本不運行。

att = 'bbaccb'


class rec:
    a = 'a'
    b = 'b'
    c = 'c'
    i = 0
    x = 1
    cd = str(att[i])

    def fx():
        rec.x = rec.x + 1
        print(rec.i, rec.x, rec.cd)

    def fi():
        rec.i = rec.i + 1
        print(rec.i, rec.x, rec.cd)

    def bx():
        rec.x = rec.x - 1
        print(rec.i, rec.x, rec.cd)

    def sx():
        rec.x = rec.x + 0
        print(rec.i, rec.x, rec.cd)

    def fail():
        print('False','\n','Goodbye')
        quit()

    def pas():
        print('True\n', 'Goodbye\n')
        quit()

    def forward1():
        rec.fx()
        rec.fi()

    def stepback():
        rec.bx()
        rec.fi()

    def stepback2():
        rec.x = rec.x - 2
        rec.fi()

    def standstill():
        rec.sx()
        rec.fi()
    
    def steps(object): 
        while rec.x > 0:
            print(rec.i, rec.x, rec.cd)
            while rec.x == 1:
                print(rec.i, rec.x, rec.cd)
                if rec.cd == rec.b:
                    rec.forward1()
                if rec.cd == rec.c:
                    rec.forward1()
            else:
                rec.fail()
            
                while rec.x == 2:
                    print(rec.i, rec.x, rec.cd)
                    if rec.cd == rec.a:
                        rec.stepback()
                    if rec.cd == rec.b:
                        rec.forward1()
                else:
                    rec.fail()
                    while rec.x == 3:
                        print(rec.i, rec.x, rec.cd)
                        if rec.cd == rec.a:
                            rec.forward1()
                    else:
                        rec.fail()
                        
                        while rec.x == 4:
                            print(rec.i, rec.x, rec.cd)
                            if rec.cd == rec.a:
                                rec.standstill()
                            if rec.cd == rec.b:
                                rec.stepback()
                            if rec.cd == rec.c:
                                rec.forward1()
                        else:
                            rec.fail()
                            while rec.x == 5:
                                print(rec.i, rec.x, rec.cd)
                                if rec.cd == rec.b:
                                    rec.stepback2()
                                if rec.cd == rec.c:
                                    rec.forward1()
                            else:
                                rec.fail()
                                while rec.x == 6:
                                    print(rec.i, rec.x, rec.cd)
                                    if rec.cd == rec.a:
                                        rec.stepback()
                                    if rec.cd == rec.b:
                                        rec.forward()
                                    if rec.cd == rec.c:
                                        rec.standstill()
                                else:
                                    rec.fail()
                                    while rec.x == 7:
                                            print(rec.i, rec.x, rec.cd)
                                            pas()
        else:
            rec.fail()

Answer 1

我坐下來仔細看了看，然后想出了一個解決方案。 我不確定它是否正確，但它確實運行了我認為是此 FSA 的“密碼”。

#==============================================================================
#Variables
#==============================================================================
# Each accepted digit
a = 'a'
b = 'b'
c = 'c'
i = 0
# Placement in attempted string

# Inputed array from the user
arr = input(str('Please enter the string to be recognized: '))

#testing
#arr = 'bbaccb'
#==============================================================================
"""
In this recognizer, the starting state is S1, and S7 is the only accepting state
(empty strings are not accepted). Your program should ask the user to enter a
string and print "True" if the FSA accepts it, "False" otherwise. After printing
the result (and a 'Goodbye' message), the program should immediately stop.
"""

"""
Step 1:
    If arr[i] is an a, fail.
    If b, go to next step with next digit.
    If c, check digit 2
"""
def step1(object):
    global i
    #print(arr[i])
    if arr[i] == a:
        print(false())
    if arr[i] == b:
        forward(object)
        step2(object)
    if arr[i] == c:
        i = i + 1
        step1(object)
    else:
        print(false())
"""
Step 2:
    If arr[i] is an a, return to step 1 with next digit.
    If b, go to next step with next digit.
    If c, fail.
"""
def step2(object):
    global i
    #print(arr[i])
    if arr[i] == a:
        forward(object)
        step1(object)
    if arr[i] == b:
        forward(object)
        step3(object)
    if arr[i] == c:
        print(false())
    else:
        print(false())
"""
Step 3:
    If arr[i] is an a, go to next step with next digit.
    If b, fail.
    If c, fail.
"""
def step3(object):
    global i
    #print(arr[i])
    if arr[i] == a:
        forward(object)
        step4(object)
    if arr[i] == b or c:
        print(false())
    else:
        print(false())
"""
Step 4:
    If arr[i] is an a, check the next digit.
    If b, return to step 3 with the next digit.
    If c, go to next step with next digit.
"""
def step4(object):
    global i
    #print(arr[i])
    if arr[i] == a:
        forward(object)
        step4(object)
    if arr[i] == b:
        forward(object)
        step3(object)
    if arr[i] == c:
        forward(object)
        step5(object)
    else:
        print(false())
"""
Step 5:
    If arr[i] is an a, fail.
    If b, return to step 3 with the next digit.
    If c, go to next step with next digit.
"""
def step5(object):
    global i
    #print(arr[i])
    if arr[i] == a:
        print(fail())
    if arr[i] == b:
        forward(object)
        step3(object)
    if arr[i] == c:
        forward(object)
        step6(object)
    else:
        print(false())
"""
Step 6:
    If arr[i] is an a, return to step 5 with next digit.
    If b, complete task.
    If c, check next digit.
"""
def step6(object):
    global i
    #print(arr[i])
    if arr[i] == a:
        forward(object)
        step5(object)
    if arr[i] == b:
        print(true())
    if arr[i] == c:
        step6(object)
    else:
        print(false())

#If the program compleates the task, then it's 'True'.
def true():
    #print(arr[i])
    print('True.')
    print('Goodbye.')
    quit()

# Otherwise, the arr is 'False'.
def false():
    #print(arr[i])
    print('False.')
    print('Goodbye.')
    quit()

# Move to next item in string
def forward(object):
    global i
    i = i + 1
    #return i
#==============================================================================
# To Do
#==============================================================================
print(step1(arr))