香港身份证号码

For this example d = (1 0 0)
T
The error occurred at the 4th position. D is called a parity check matrix.
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Probability for Correct Sending
Suppose we are transmitting a message of 4 digits on a binary channel and q=0.9. Then the probability of correctly sent for No coding: q4=0.6561; Repetition code:
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X354670(?)
9(58)+8(33)+7(3)+6(5)+5(4)+4(6)+3(7)+2(0)+z = 902+z 被11整除，所以z=0。 我們可利用Modular arithmetic來簡化運算。 z = 9α8β7a6b5c4d3e2f ≡ 2α+3β+4a+5b5c4d3e2f (mod 11) 所以 z ≡ 2(58)+3(33)+4(3)+5(5) 5(4) 4(6) 3(7) 2(0) ≡ 2(3)+3(0)+12+252024210 ≡ 6+0+1+3+22+10=11 ≡ 0 (mod 11) 即 X354670(0) 是正確的香港身分證號碼。
a1 + a3 + a5 + a7 + a9 + a11 + a13 + 3(a2 + a4 + a6 + a8 + a10 + a12 ) ≡ 0 (mod 10)
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Parity Check: ISBN-13 Original ISBN-10: 0-471-61884-5 New ISBN-13: 978- 0-471-61884-?
1001 →(1001)G = (1001100) →100 0 100 = c
E Channel
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Examples of Single-Error Correcting Code
0 0 0 1 1 1 1 Let d=DcT, where D = 0 1 1 0 0 1 1 1 0 1 0 1 0 1
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No Encoding-Decoding System
Noise e
Message u=1001
Channel
Received Message v=0001
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Encoding-Decoding System
Noise e
Message u=1001
Encoder Eu=b=1001100
Channel
q12 + 12q11 p + 12 × 9 10 2 12 × 9 × 6 9 3 12 × 9 × 6 × 3 8 4 q p + q p + q p ≈ 0.8926 ; 2! 3! 4!
Hamming code: ≥ q7+7q6p ≈ 0.8503.
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Probability for Correct Sending Rate of a code is the ratio of the number of information digits with the length of the code. Rate of the repetition code is 1/3. Rate of the Hamming code is 4/7.
1× 0 + 2 × 4 + 3 × 7 + 4 × 1 + 5 × 6 + 6 × 1 + 7 × 8 + 8 × 8 + 9 × 4 = 225 ≡ 5 (mod 11)
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In general, assume the first 9 digits of an ISBN is a1a2…a9, where 0 ≤ ai ≤ 9 for 1 ≤ i ≤ 9, then the tenth digit (the check digit) is 9 1a1 + 2a2 + 3a3 + 4a4 + 5a5 + 6a6 + 7 a7 + 8a8 + 9a9 = ∑ iai
≡ a10 (mod 11),
i =1
where 0 ≤ a10 ≤ 10. If a10=10, then use X to instead of it.
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Parity Check: ISBN-13
Jan. 1, 2007, ISBN-13 will instead of ISBN-10 How do we convert ISBN-10 to ISBN-13?
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Encoded Message Digits in the original message : Information digits Digits added by the encoder : Redundancy digits Length of a code = the number of information digits + the number of redundancy digits
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Message We think of a message as a block of symbols from a finite alphabet. A commonly used alphabet is the set of two symbols 0 and 1. A possible message is 1001. This message is then transmitted over a communications channel that is subject to some amount of noise.
(9 + 8 + 4 + 1 + 1 + 8 + ?) + 3(7 + 0 + 7 + 6 + 8 + 4) = 30 + ?+ 3(32) ≡ ?+ 3(2) = ?+ 6 (mod 10)
? = 4
The new ISBN-13 is 978- 0-471-61884-4
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香港身份證號碼
X354670(?) 身份證號碼的「結構」，可以用 αβ abcdef (z)表示。 「α」可能是「空格」或是一個英文字母，「 β 」則 必定是英文字母。「 abcdef 」代表一個六位數字，而 「z」是作為檢碼之用，它的可能選擇是0, 1, 2, ... , 9, z 0, A(代表10)。這些代號的背後，都可配上一個編碼值。 透過編碼值，便可找出 9α+8β+7a+6b+5c+4d+3e+2f+z 的總和。該總和特別之處，是必須被11整除。利用這 特點，我們便能找出括號內的數字。試試看！
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Linear Codes
An (n, k) linear code C over a finite field F is a kdimensional vector subspace in F n. Elements of C is called codeword. If F={0, 1}, then C is called a binary code Any k×n matrix G whose rows are the basis vectors of C is called a generator matrix. This means that for each α∈C, there exists b=(b1,…,bk) such that α=bG. There exists (will show a proof later) an (nk)×n matrix H called parity check matrix of C such that
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Parity Check: ISBN-13 Drop the check digit from the existing ISBN10 Add the prefix 978 or 979 (usually is 978) Calculate the check digit using modulo 10: Suppose the 13 digits is a1a2…a13, where 0 ≤ ai ≤ 9 for 1 ≤ i ≤ 13, then it satisfies
α 或 β 的編碼值 的編碼值:
空格 A B C D E F G H 58 10 11 12 13 14 15 16 17 I J K L M N O P Q 18 19 20 21 22 23 24 25 26 R S T U V W X Y Z 27 28 29 30 31 32 33 34 35
被11整除所以我們可利用modulararithmetic11mod11mod2124202512即x3546700是正確的香港身分證號碼wethinksymbolsfromfinitealphabet
Error Correcting Code
Dr. W. C. Shiu Hong Kong Baptist University Department of Mathematics
Байду номын сангаас
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Examples of Single-Error Correcting Code Hamming (7,4) code: Given a matrix 1 0 0 0 0 1 1 0 1 0 0 1 0 1 G= 0 0 1 0 1 1 0 0 0 0 1 1 1 1 If u=(u1,u2,u3,u4), then we decode it as b=uG.
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Examples of single-error detecting code: 1. Repetition code: Each message is repeated once.
1001 →10011001 →10001001
E Channel
2. Even-parity check code: Add an extra digit to the message. It is a 0 if there are an even number of 1’s, otherwise a 1.
D E 1 E Channel
It can also correct some double errors and more errors.
100110011001→100010111001→ 100110011001 100110011001→100010001001→ 100010001000 100110011001→100010110101→ 100110011001
Message E-1(Dc)=1001
Decoder Dc=1001100
Received Message c=b+e=0001100
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Binary Symmetric Channel (BSC)
No memory. Receives and transmits two symbols 0 and 1. The BSC has the property that with probability q a transmitted digit will received correctly, and with probability p=1q it will not be. Boolean sum and product: + 0 1 0 0 1 1 1 0 × 0 1 0 0 0 1 0 1
1001 →10010 →10000
E Channel
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Examples of Single-Error Correcting Code Repetition code: Each message is repeated twice.
1001 100110011001 →100110001001 → →100110011001 →1001
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Examples: Let us vonsider the following example. I am a dog, so you have to worship me. Meet me at 8:00 c.m.
Parity Check: ISBN-10 (International Standard Book Number) 0-471-61884-5