Solved

# palindrome with recursion

Posted on 2004-10-27
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I am trying to write a palindrome function  that will tell me if a number is a palindrome. The tricky part about this, I think, is that I want to call the following two recursive functions from within the palindrome() to do the work.

-----------------------
unsigned digitCounter(int number)
{
if (number != 0)
{
return digitCounter(number / 10) + 1;
}
else
{
return 0;
}
}
---------------------
int power(int x, unsigned n)
{
if (n == 0)
{
//anchor case
return 1;
}
else
{      //inductive case (n > 0)
return x * power(x, n - 1);
}
}
---------------------
Here is what I have for the palindrome() so far, but its not quite right.

bool palindrome(int number)
{
int firstDigit, lastDigit;
if (number == 1)
{
return true;
}
else

firstDigit = number / (10 ^ numDigits - 1);
//gives last digit
lastDigit = number % 10;

if (firstDigit != lastDigit)
{
return false;
}
else
{
palindrome(number % 10) / 10 && lastDigit - 2;
}
}
0
Question by:jandhb

LVL 9

Expert Comment

bool palindrome(int number)
{
int firstDigit, lastDigit;
/* calculate numDigits here. the function looks correct */

if (numDigits == 1)
{
return true;
}
else

firstDigit = number / (10 ^ numDigits - 1);
lastDigit = number % 10;

if (firstDigit != lastDigit)
{
return false;
}
else
{
int newNumber = number - firstDigit*(10 ^ numDigits - 1); //remove first digit
newNumber /= 10; //remove last digit
palindrome(newNumber);
}
}
0

LVL 39

Expert Comment

>> I want to call the following two recursive functions from within the palindrome()

But you didn't...

bool palindrome(int number)
{
int firstDigit, lastDigit;

int numDigits = digitCounter(number);

if (numDigits == 1)
{
return true;
}
else
{
int dec = 10;
for (int i = 0; i < numDigits-1; ++i)
dec *= 10;
firstDigit = number / dec;
lastDigit = number % 10;

if (firstDigit != lastDigit)
{
return false;
}
else
{
int newNumber = number - dec; //remove first digit
newNumber /= 10; //remove last digit
return palindrome(newNumber);
}
}
}

Regards, Alex
0

LVL 12

Expert Comment

Hi jandhb,
Do you need to process your number as an integer? It's a lot easier when you convert it into text first:

bool palindrome(int number){
char ptext[12]; /* 10 digits, sign and trailing \0 */

sprintf(ptext,"%d",number);

return is_palindrome(ptext,0,strlen(ptext)-1);
}

...I'm sure you can easily write the is_palindrome() function yourself.

Cheers!

Stefan
0

LVL 1

Author Comment

Do you need to process your number as an integer?

The number I will be passing to this function will be an integer. In your examples where are you calling the two other functions from within the palindrome()? These recursive functions need to be called to do the work. Maybe I'm missing it. Here again are the two other functions....

unsigned digitCounter(int number)
{
if (number != 0)
{
return digitCounter(number / 10) + 1;
}
else
{
return 0;
}
}
----------------------------------------------
int power(int x, unsigned n)
{
if (n == 0)
{
//anchor case
return 1;
}
else
{      //inductive case (n > 0)
return x * power(x, n - 1);
}
}
----------------------------------------

0

LVL 39

Accepted Solution

>>>> Maybe I'm missing it.

Yes, i used digitCounter in the post above. I didn't use power(), but made a while loop calculating the power of 10 myself. I removed some errors and used both  functions now:

unsigned digitCounter(int number)
{
if (number != 0)
{
return digitCounter(number / 10) + 1;
}
else
{
return 0;
}
}

int power(int x, unsigned n)
{
if (n == 0)
{
//anchor case
return 1;
}
else
{     //inductive case (n > 0)
return x * power(x, n - 1);
}
}

bool palindrome(int number)
{
int firstDigit, lastDigit;

int numDigits = digitCounter(number);

if (numDigits <= 1)
{
return true;
}
else
{
int dec = power(10, numDigits-1);
firstDigit = number / dec;
lastDigit = number % 10;

if (firstDigit != lastDigit)
{
return false;
}
else
{
int newNumber = number - (firstDigit*dec); //remove first digit
newNumber /= 10; //remove last digit
return palindrome(newNumber);
}
}
}

Regards, Alex
0

LVL 1

Author Comment

Alex, thank you very much for your help here.
0

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