Wong Wai Ling, Lam Pui Ki 2011-3-22 1. Identification number clearly identify a person or a thing...

Wong Wai Ling, Lam Pui Ki2011-3-22

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Identification numberclearly identify a person or a thing

Check digitan extra digit for the purpose of error detection

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11-digitsum of the first 10 digits is divided by 9remainder = check digit

Disadvantagescan’t detect mistake1) replace 0 by 92) numbers interchange

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First letteridentify the country that issues the noteadd up the numbers, add each single digit in the sum until a single digit remainssum = 4+7+9+5 = 25, 2+5 = 7

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Checksumsingle digit obtained previouslyrefer to the particular country

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12-digit1st: category2nd-6th: manufacturer7th-11th: identify the product12th: check digitsum = 3(1st)+1(2nd)+3(3rd)+ … +3(11th)+1(12th)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

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use a variation of the UPC scheme append check digits to the numbers assigned to banksEach bank has a 8-digit routing number sum = 7(1st)+3(2nd)+9(3rd)+ … +7(7th)+3(8th) an extra 9th 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

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13-digitbecome worldwide standardexisting UPC numbers are converted to EAN by adding an extra “0” at the beginningsum = 1(1st)+3(2nd)+1(3rd)+ … +1(11th)+3(12th)+1(13th)does not affect the check digit compared with UPC

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Credit cards, libraries, blood banks, etcComputation1) add the digits in odd positions2) double the result3) add the numbers in odd positions that exceed 44) add the remaining digits (even positions)check digit = the number needed to bring the total result to end with 0

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detect 100% single position error and 98% common error

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10-digite.g. a1 - a2a3a4a5 - a6a7a8a9 - a10

↑ ↑ ↑ ↑

sum = 10a1+9a2+8a3+7a4+6a5+5a6+4a7+3a8+2a9+1a10

is divisible by 11however, the 10th digit (check digit) could be 10solution: use “X” instead of “10”detect 100% single position error and transposition erroranother type: 13-digit, same as EAN

publishedcountry

publisher identificationnumber

check digit

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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 symbolsBar coding Method for automated data collection

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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

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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

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12-digit: 0-12345-67890-51st number (0) : kinds of productnext 5 digits (12345): manufacturer numbernext 5 digits (67890): product numberlast digit (5) : check digit

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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

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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

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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

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replace 1 by 0, 0 by 1Example 0111011 for manufacture’s number = 1000100 for the product number manufacture’s number : odd number of 1’s product number : even number of 1’sDetermine whether the barcode is scanned from LHS or

RHS scanning can be done correctly

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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.(write at least two)

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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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