Wong Wai Ling, Lam Pui Ki 2011-3-22

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Identification number  clearly identify a person or a thing Check digit  an extra digit for the purpose of error detection 2

11-digit  sum of the first 10 digits is divided by 9  remainder = check digit Disadvantages  can’t detect mistake 1) replace 0 by 9 2) numbers interchange 3

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First letter  identify the country that issues the note  add up the numbers, add each single digit in the sum until a single digit remains  sum = 4+7+9+5 = 25, 2+5 = 7 5

Checksum  single digit obtained previously  refer to the particular country 6

12-digit 1 st : category 2 nd -6 th : manufacturer 7 th -11 th : identify the product 12 th : check digit  sum = 3 (1 st )+ 1 (2 nd )+ 3 (3 rd )+ … + 3 (11 th )+ 1 (12 th )  sum should end with a 0. (e.g. 10,20,30,…)  can detect all single position error and about 89% of other kinds of errors 7

  use a variation of the UPC scheme append check digits to the numbers assigned to banks Each bank has a 8-digit routing number  sum = 7 (1 st )+ 3 (2 nd )+ 9 (3 rd )+ … + 7 (7 th )+ 3 (8 th )  an extra 9 th digit (check digit) = last digit of the sum i.e. sum = 53 → check digit = 3  numbers 7 , 3 , 9 are called weights  can detect most transposition errors 8

13-digit  become worldwide standard  existing UPC numbers are converted to EAN by adding an extra “0” at the beginning  sum = 1 (1 st )+ 3 (2 nd )+ 1 (3 rd )+ … + 1 (11 th )+ 3 (12 th )+ 1 (13 th )  does not affect the check digit compared with UPC 9

Credit cards, libraries, blood banks, etc  Computation 1) add the digits in odd positions 2) double the result 3) add the numbers in odd positions that exceed 4 4) add the remaining digits (even positions)  check digit = the number needed to bring the total result to end with 0 10

 detect 100% single position error and 98% common error 11

10-digit e.g. a 1 - a 2 a 3 a 4 a 5 - a 6 a 7 a 8 a 9 - a 10 ↑ ↑ ↑ ↑ published publisher identification check digit  country sum = 10a 1 +9a 2 number +8a 3 +7a 4 +6a 5 +5a 6 +4a 7 +3a 8 +2a 9 +1a 10 is divisible by 11  however, the 10 th digit (check digit) could be 10  solution: use “X” instead of “10”  detect 100% single position error and transposition error  another type: 13-digit, same as EAN 12

  The ISBN 0-669-03925-4 is the result of a transposition of two adjacent digits not involving the first or last digit.

Determine the correct ISBN.

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 A series of dark and light spaces that represents characters Binary code  A system for representing data with two symbols Bar coding  Method for automated data collection 14

 A beam of light passes through the bars and spaces by scanning device   The differences in reflection intensities Dark bars reflect less; light spaces reflect more 15

  most often encountered, first used on grocery items in 1973 translate 12-digit number into bars 12-digit number  two five-digit numbers in between two single-digit 16

12-digit: 0-12345-67890-5  1 st number (0) : kinds of product  next 5 digits (12345): manufacturer number  next 5 digits (67890): product number  last digit (5) : check digit 17

    digits are represented by a space divided into 7 modules of equal width 2 long bars with one-module thickness in each end separated by a light space of one-module thickness the modules are named as guard bar patterns 18

   separate the manufacturer’s number and the product number not a part of identification number 5 modules: a light space, a long dark bar, a light space, a long dark bar, a light space 19

   one-module-thickness light space 0 one-module-thickness dark bar 1 code in the product number can be obtained from the code in the manufacturer’s number, vice versa 20

 replace 1 by 0, 0 by 1 Example  0111011 for manufacture’s number  = 1000100 for the product number manufacture’s number : odd number of 1’s product number : even number of 1’s Determine whether the barcode is scanned from LHS or RHS  scanning can be done correctly 21

1. Prove that the ISBN detects 100% of the single position errors.

2. The following is an actual identification number and bar code from a roll of wallpaper.

What appears to be wrong with them? Speculate on the reason for the apparent violation of the UPC format.

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Extra : Compare the advantages and the disadvantages of the algorithm methods stated in class. Give examples on the purpose of each method.

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Thank you!

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