Python 3 infinite recursion; see n say

Question

I am trying to make a program in python that takes an index value and outputs a string based on the input value; then I need it to respond in english. For example, index 0 = value 0 One zero., index 1 = value 10 One one. One zero., index 2 = value 1110 Three ones. One zero. , index 3 = value 3110 One three. Two ones. One zero., etc.

initialIndexGlobal = "0"

nextIndexGlobal = ""

string = initialIndexGlobal

def stringHead():
    initialIndexGlobal[0]

def StringTail():
    initialIndexGlobal[1:]

def checkValueIterative(string):
    nextIndex = ""
    string = string + "4"
    initialValue = string[0]
    counter = 0

    for w in string:
        if w == initialValue:
            counter = counter + 1
        else:
            secondDigit = str(initialValue)
            firstDigit = str(counter)
            #print(firstDigit + secondDigit)
            global nextIndexGlobal
            nextIndexGlobal = nextIndexGlobal + str(firstDigit) + str(secondDigit)
            string = string[(counter):]
            checkValueIterative(string)
            break
    return(nextIndex)

def checkValueRecursion(index):
    if index == 0:
        print(initialIndexGlobal)
    else:
        checkValueIterative(initialIndexGlobal)
        global initialIndexGlobal
        global nextIndexGlobal
        initialIndexGlobal = nextIndexGlobal
        nextIndexGlobal = ""
        checkValueRecursion(index-1)

def runLen(string, index):
    if string == '':
        return 0
    if stringHead()==index:
        return 1 + runLen(StringTail, index)
    else:
        return 0


def say(string):
    if string == "":
        return
    else:
        if ( string[ 0 ] )== "0" and runLen( string,string[ 0 ]==3 ):
            print( "three zeros" )
            return say( string[ 3:1 ] )

        elif (string[ 0 ] == "0" and runLen( string,string[ 0 ]==2 ) ):
            print( "two zeros" )
            return say( string[ 2:1 ] )

        elif ( string[ 0 ]=="0" ):
            print( "one zero" )
            return say( string )

        if (string[ 0 ] == "1" and runLen( string,string[ 0 ]==3 ) ):
            print( "three ones" )

            return say( string[ 3:1 ] )
        elif (string[ 0 ] == "1" and runLen( string,string[ 0 ]==2 ) ):
            print( "two ones" )
            return say( string[ 2:1 ] )

        elif (string[ 0 ] == "1" ):
            print( "one one" )
            return say( string )

        if (string[ 0 ] == "2" and runLen( string,string[ 0 ]==3 ) ):
            print( "three twos" )
            return say( string[ 3:1 ] )

        elif (string[ 0 ] == "2" and runLen( string,string[ 0 ]==2 ) ):
            print( "two twos" )
            return say( string[ 2:1 ] )

        elif (string[ 0 ]=="2" ):
            print( "one two" )
            return say( string )

        if (string[ 0 ]=="3"and runLen( string,string[ 0 ]==3 ) ):
            print( "three threes" )
            return say( string[ 3:1 ] )

        elif (string[ 0 ]=="3"and runLen( string,string[ 0 ]==2 ) ):
            print( "two Threes" )
            return say( string[ 2:1 ] )

        elif (string[ 0 ]=="3" ):
            print( "one three" )
            return say( string )



#index = input("Enter Index value here: ")
checkValueRecursion(10)
runLen(string, initialIndexGlobal)
say(string)
input()

This is what I have so far. It calculates the index value correctly, but it does not repeat it in english; instead it re-curses until it crashes. How would I go about fixing this?

First, that final block of ifs and elifs is a nightmare. Rewrite that with some spacing and nest it so it's easier to understand. Second, you've got (at least) one significant typo, the last return has the variable misspelled. — joemaller
– joemaller, Commented Oct 1, 2013 at 19:45
I fixed the spacing of the ifs and elifs and I fixed the typos. What do you mean nesting? — user2835532
– user2835532, Commented Oct 1, 2013 at 20:14
within the say(string) function, you are saying return say(X) so it just calls itself again. You want to return X... (although I'm not convinced that X of string[2:1] etc. is defined correctly... What are you trying to get with the [3:1] slicing? — beroe
– beroe, Commented Oct 1, 2013 at 21:39
I think that we're just trying to cut out that part of the string so that the function can read it. — user2835532
– user2835532, Commented Oct 1, 2013 at 22:35

user2836773 · Accepted Answer · 2013-10-01 23:02:44Z

Ok, we figured it out. We completely removed the old say() completely and rewrote it. Thanks for the responses.

initialIndexGlobal = "0"
nextIndexGlobal = ""
stringComposition = ""

def numberToLetter(string):
    numtotext = ""
    for w in string:
        if w == "0":
            numtotext = numtotext + "zero "
        elif w == "1":
            numtotext = numtotext + "one "
        elif w == "2":
            numtotext = numtotext + "two "
        else:
            numtotext = numtotext + "three "
    return numtotext

def checkValueIterative(string):
    nextIndex = ""
    string = string + "4"
    initialValue = string[0]
    counter = 0
    for w in string:
        if w == initialValue:
            counter = counter + 1
        else:
            secondDigit = str(initialValue)
            firstDigit = str(counter)
            #print(firstDigit + secondDigit)
            global nextIndexGlobal
            nextIndexGlobal = nextIndexGlobal + str(firstDigit) + str(secondDigit)
            string = string[(counter):]
            checkValueIterative(string)
            break
    return(nextIndex)

def checkValueRecursion(index):
    if index == 0:
        return
    else:
        checkValueIterative(initialIndexGlobal)
        global initialIndexGlobal
        global nextIndexGlobal
        initialIndexGlobal = nextIndexGlobal
        nextIndexGlobal = ""
        checkValueRecursion(index-1)

def main():
    indexValue=int(input("What is the index value?"))
    checkValueRecursion(indexValue)
    print(str(indexValue) + " : " + str(initialIndexGlobal) + " : " + str(numberToLetter(initialIndexGlobal)))
main()

cwallenpoole · Accepted Answer · 2013-10-01 21:46:41Z

0

say will loop infinitely on the string '0'

say("0")
# runLen("0",False) and runLen("0",True) won't return 2 or 3
# We move on to the line:
    if string[ 0 ]=="0": #this passes because it is true.
        print('one zero')
        say(string)

As a note, you have a lot of runLen( string,string[ 0 ]==3 ), I think you really mean runLen( string,string[ 0 ] ) ==3. Also, string = initialIndexGlobal should probably be removed.

answered Oct 1, 2013 at 21:46

cwallenpoole

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