Thread: ISBN Code Checking...!

Since I can't understand why the position of the X or ? in the ISBN values, makes a difference, the whole if(th1 > th2) block of code, is a mystery to me.

Here's some pseudo code that I'd suggest studying:

Code:

define X 10

This is all in main():
   integers isbn[10]={2,4,6,3,7,8,1,X,9,'?'}; //2463781X95
   integers i, valid=0, sum=0, mysteryNum=0, trial;

   for i=0;i<10;i++
      if(isbn[i]==10)
         print "%c ",'X';
      else if isbn[i]=='?'
         print"%c ",'?';
      else
         print "%d ",isbn[i];
   

   if isbn[0]==X 
      valid=0;
      print "\n Invalid Code: The ISBN first value, can't be an X.\n";
   

   this is the else to the if directly above, and it is the heart of the   
   program:

      for i=0,sum=0;i<10;i++   //multiply each digit by it's index +1, but
         if isbn[i] != '?'         //exclude the mystery number, if present
            sum += isbn[i] * (i+1);
         
         else 
            mysteryNum=i+1; //we have a mystery number at i, but it could be at index zero, (which we don't want, so +1
         
      
      print "\nOriginal ISBN sum: %d   Original sum mod 11: %d\n\n",sum, sum%11;

      if sum % 11==0  //we have a valid ISBN number
         if mysteryNum //with a mystery number
            mysteryNum=0; //so the mystery number if zero, gives us a valid ISBN number

         valid=1;  //and the ISBN number is valid
             
      else we need to test the mystery number  // ISBN number is invalid - test the mystery array[mysteryNum-1], for values. 
         for i=1;i<11;i++ { //i=0 has already been tested, and failed
            trial = sum + (i*mysteryNum);
            
            /* for debug only (I made an error down below, and this is
            a good way to show how you can find and fix errors)
            if i==2 
               printf("trial: %d  sum: %d  mysteryNum: %d\n",trial,sum, mysteryNum); getchar();
             
            */
            if trial % 11==0 
               valid=1;
               isbn[mysteryNum-1]=i;
               sum = trial;
               break;
            
         end of for
      end of else 
      if sum 
         print "\nFinal ISBN sum: %d   Final sum mod 11: %d\n\n",sum, sum % 11
         if valid
            print "The ISBN code is valid\n"
         else
            print "The ISBN code is invalid\n"
      
         for i=0;i<10;i++ 
            if isbn[i]==10
               isbn[i]='X';
            if isbn[i]=='?' OR isbn[i]=='X'
               print "%c ",isbn[i];
            else
               print "%d ",isbn[i];
         end of for
      end of if
   end of else
   
   printf("\n");
   return 0;
end of program

Originally Posted by Adak

Since I can't understand why the position of the X or ? in the ISBN values, makes a difference, the whole if(th1 > th2) block of code, is a mystery to me.

Here's some pseudo code that I'd suggest studying:

Code:

define X 10

This is all in main():
   integers isbn[10]={2,4,6,3,7,8,1,X,9,'?'}; //2463781X95
   integers i, valid=0, sum=0, mysteryNum=0, trial;

   for i=0;i<10;i++
      if(isbn[i]==10)
         print "%c ",'X';
      else if isbn[i]=='?'
         print"%c ",'?';
      else
         print "%d ",isbn[i];
   

   if isbn[0]==X 
      valid=0;
      print "\n Invalid Code: The ISBN first value, can't be an X.\n";
   

   this is the else to the if directly above, and it is the heart of the   
   program:

      for i=0,sum=0;i<10;i++   //multiply each digit by it's index +1, but
         if isbn[i] != '?'         //exclude the mystery number, if present
            sum += isbn[i] * (i+1);
         
         else 
            mysteryNum=i+1; //we have a mystery number at i, but it could be at index zero, (which we don't want, so +1
         
      
      print "\nOriginal ISBN sum: %d   Original sum mod 11: %d\n\n",sum, sum%11;

      if sum % 11==0  //we have a valid ISBN number
         if mysteryNum //with a mystery number
            mysteryNum=0; //so the mystery number if zero, gives us a valid ISBN number

         valid=1;  //and the ISBN number is valid
             
      else we need to test the mystery number  // ISBN number is invalid - test the mystery array[mysteryNum-1], for values. 
         for i=1;i<11;i++ { //i=0 has already been tested, and failed
            trial = sum + (i*mysteryNum);
            
            /* for debug only (I made an error down below, and this is
            a good way to show how you can find and fix errors)
            if i==2 
               printf("trial: %d  sum: %d  mysteryNum: %d\n",trial,sum, mysteryNum); getchar();
             
            */
            if trial % 11==0 
               valid=1;
               isbn[mysteryNum-1]=i;
               sum = trial;
               break;
            
         end of for
      end of else 
      if sum 
         print "\nFinal ISBN sum: %d   Final sum mod 11: %d\n\n",sum, sum % 11
         if valid
            print "The ISBN code is valid\n"
         else
            print "The ISBN code is invalid\n"
      
         for i=0;i<10;i++ 
            if isbn[i]==10
               isbn[i]='X';
            if isbn[i]=='?' OR isbn[i]=='X'
               print "%c ",isbn[i];
            else
               print "%d ",isbn[i];
         end of for
      end of if
   end of else
   
   printf("\n");
   return 0;
end of program

I think the position of X and ? do matters... the X could be anywhere except at the front... and also ? could be any where even at the front... So while Multiplying we have to skip the ? and also have to remember it's position so that the MystNumb we have found can be multiplied with the position and added to the sum of rest 9 numbers. and the new sum will be divisible by 11.

so what about this ex. I copied it from a books ISBN Code....
8121?2605X

now the number that I have hided from you is 9. So in place of ? it would be 9. We need to find the numb via C.

And very much thank you for your response.

Originally Posted by Tipu Sultan

the X could be anywhere except at the front...

As Tim and I have already told you that's wrong. The X is only valid as the last digit (check digit).

From http://www.isbn.org/standards/home/i...ml/usmapp.htm:

...
3. Construction of an international standard book number
An international standard book number consists of ten digits[4] made up of the following parts
...
Footnote 4: The digits are the arabic numerals 0-9; in the case of the check digit only, an X can sometimes occur (see 2.4)

Bye, Andreas

Originally Posted by AndiPersti

As Tim and I have already told you that's wrong. The X is only valid as the last digit (check digit).

From http://www.isbn.org/standards/home/i...ml/usmapp.htm:


Bye, Andreas

THANKS A LOT... that reduces my work a lot more... now we just need to deal with 9 numbers and accordingly no need to deal with the position of the X... Thanks again....