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1 Massachusetts Istitute of Techology Departmet of Electrical Egieerig ad Computer Sciece 6.02 Solutios to Chapter 5 Updated: February 16, 2012 Please sed iformatio about errors or omissios to hari; questios are best asked o piazza. 1. The Hammig distace of a -bit word is the sum of the Hammig distaces for every sigle bit of the -bit word. Listig all of the possible combiatios of x, y, ad z for a sigle bit (ote: due to symmetry, oly four of the possible combiatios are relevat): x y z HD(x, y) + HD(y, z) HD(x, z) We d that the triagle iequality holds for each combiatio. Sice the Hammig distace satises the triagle iequality for every bit, it satises the triagle iequality for the etire -bit word. 2. (23, 15, 3) = umber of data bits + umber of parity bits = = 23 k = umber of data bits = 15 d = 3. See sectio 5.5 for a discussio of why d = (a) It is ot possible to obtai a rate lower tha 1 /3. I order to achieve this rate, each codeword will cosist of a sigle data bit ad 2 parity bits. If we icrease the umber of data bits i each codeword, we obtai a rate that is larger tha 1 /3; ad i order to decrease the rate, we would eed to have a codeword that cosisted of less tha 1 data bit, which is ot possible. (b) We would like to see whether rectagular codes exist whose rates are 1/2, 2/3,..., l 1 /l. We kow that the rate of a rectagular code with r rows ad c colums is rc = l 1 rc + r + c l rc lr lc = 0 rc lr lc + l 2 = l 2 (r l)(c l) = l 2 rc rc+r+c. If we set r l = l ad c l = l, the equatio is satised, which meas that a rectagular code with parameters (4l 2 + 4l, 4l, 3) has rate l /l+1 (there may be other solutios for particular values of l, e.g., whe l is ot a prime umber). This is a iterestig observatio because although the umber of parity bits i a rectagular code grows at least as fast as the square-root of the umber of message bits (l), it is still possible to achieve high code rates of the form l /l+1.

2 4. (a) (5, 2) = umber of data bits + umber of parity bits = = 5 k = umber of data bits = 2 (b) E 0 = (D 0 + P 0 ) mod 2 E 1 = (D 0 + D 1 + P 1 ) mod 2 E 2 = (D 1 + P 2 ) mod 2 (c) The sydrome table is: E 2 E 1 E 0 Corrective Actio 000 o errors 001 sigle error (P 0 has a error, ip to correct) 010 sigle error (P 1 has a error, ip to correct) 011 sigle error (D 0 has a error, ip to correct) 100 sigle error (P 2 has a error, ip to correct) 101 multiple errors (uable to correct) 110 sigle error (D 1 has a error, ip to correct) 111 multiple errors (uable to correct) Observe that the umber of sydrome table etries correspodig to the o error or sigle correctable error case is + 1 = 6. (d) E 0 = 1, E 1 = 1, E 2 = 0 which correspods to a sigle error i D 0 from table i the solutio of part (c). The receiver ca the ip D 0 to obtai the corrected codeword (e) No such code is possible. I order to have sigle-bit error correctio, the boud 2 k + 1 must be satised (2 k = = 5). 5. (a) (i) Code rate = k / = 1 /2 (ii) 3. All miimum weight, o-zero codewords are D 1 D 2 D 3 P 1 P 2 P 3 = ,010110,001011, (iii) Hammig distace = 3. Sice the code is a liear block code, Theorem 6.5 applies ad states that the miimum Hammig distace is equal to the weight of the o-zero codeword with smallest weight. 6. The table is: E 3 E 2 E 1 Error Patter 000 o errors 001 sigle error (P 1 has a error, ip to correct) 010 sigle error (P 2 has a error, ip to correct) 011 sigle error (D 2 has a error, ip to correct) 100 sigle error (P 3 has a error, ip to correct) 101 sigle error (D 1 has a error, ip to correct) 110 sigle error (D 3 has a error, ip to correct) 111 multiple errors (uable to correct) 7. The additio of the extra parity bit icreases the miimum Hammig distace from 3 to 4, but the extra parity bit has o eect o the error correctio capability. As a result, the code

3 ca detect up to 3 bit errors while it ca oly correct 1 bit error. A simple example why the error correctio capability is ot icreased ca be foud whe cosiderig the ew sydrome E 1 E 2 E 3 E 4, where E 1, E 2, ad E 3 are deed i the problem 5 ad E 4 = 3 i=1 D ip i + P 4. Assume that the calculated sydrome is This sydrome is geerated for the followig bit errors: D 1 D 2, D 3 P 4, ad P 2 P 3. As a result, the code ca detect ay three bit errors, but it caot correct them because of the multiple correctio possibilities. To show that the miimum Hammig distace is 4, cosider errors i the bits D 1, D 2, D 3, ad P 4. The sydrome for this case will be 0000 idicatig that o error occurred. Ay other combiatio of 4 bit errors will be detected by the code sice the calculated sydrome will be o-zero. I additio, ay combiatio of 1, 2, or 3 bit errors will also result i a o-zero sydrome. 8. (a) The code is a liear block code because the sum of ay two codewords is aother codeword. The rate is k / = 2 /3. (b) The code is a liear block code because the sum of ay two codewords is aother codeword. The rate is k / = 2 /3. (c) The code is ot a liear block code because the sum of 111 ad 100 is 011, which is ot a codeword. (d) The code is a liear block code because the sum of ay two codewords is aother codeword. The rate is k / = 2 /5. (e) The code is a liear block code sice the sum of with is equal to I this case, = 5 but k = 0 so the code rate is 0 (i.e., the receiver already kows what is set so o iformatio is trasferred). 9. A (,k) block code ca represet i its parity bits at most 2 k patters that must cover all of the error cases we wish to correct, as well as the oe case with o errors. Whe the miimum Hammig distace is 2t+1, the code ca correct up to t errors. The umber of ways i which the trasmissio ca experiece 0, 1, 2,..., t errors is equal to 1+ ( ( 1) + ( 2) + + t). This umber must ot exceed 2 k because the maximum umber of expressible sydromes is 2 k, which proves the assertio. 10. We ca verify if a specic code exists if the boud k (for sigle error correctio or derivatios of this boud for a larger umber of error corrects) is satised. (a) YES. 2 = 1, + 1 = 32 2 k = = 2 5 = 32. There are eough parity bits to esure that a miimum Hammig distace of 3 is possible. (b) NO. 2 = 1, + 1 = 33 2 k = = 2 5 = 32. There are o liear block codes that ca correct a sigle error, so (32, 27, 3) is ot a possible code. (c) YES. The simple parity code (addig all the bits i the message together so that each codeword has a eve umber of oes) is a (43, 42, 2) code; i geeral, simple parity is a ( + 1,, 2) code for ay 1. (d) YES. (27, 18, 3) are the parameters. First, ote that 2 = 1, + 1 = 28 2 k = = 2 9, so there seem to be eough parity bits to costruct such a code. The rectagular code with parameters r = 6, c = 3 gives us = rc + r + c = 27 ad k = rc = 18, ad we kow that ay rectagular code with r, c > 1 has Hammig distace 3. (e) Very similar to a PSet problem; see PSet solutios after the due date!

4 11. The (15,11) code ca be costructed as follows: idex biary idex (15,11) code p 1 p 2 d 1 p 3 d 2 d 3 d 4 p d 5 d 6 d 7 d 8 d 9 d 10 d 11 The above costructio shows that there are four parity bits (or equatios) where 7 message bits cotribute to each parity bit. For example, the biary idex idicates that the parity check equatio for p 1 is p 1 = d 1 + d 2 + d 4 + d 5 + d 7 + d 9 + d 11, which cotais 7 message bits. The rest of the parity check equatios yield similar results. 12. See the solutio for problem 1 for a proof of Theorem 6.2. Theorem 6.3 may be established as follows. A maximum likelihood decoder maximizes the quatity P(r c); i.e., it ds c so that the probability that r was received give that c was set is maximized. Cosider ay codeword c. For a BSC with error probability p e, if r ad c have a Hammig distace of d, the P(r c) = p d e(1 p e ) N d, where N is the legth of the received codeword (ad also the legth of each valid codeword). It's more coveiet to take the logarithm of this coditioal probaility, also termed the log-likelihood: 1 log P(r c) = d log p e + (N d) log(1 p e ) = d log p e 1 p e + N log(1 p e ). (1) If p e < 1/2, the pe 1 p e < 1 ad the log term is egative (otherwise, it's o-egative). As a result, miimizig the log likelihood boils dow to miimizig d, because the secod term o the RHS of Eq. (1) is a costat. This completes the proof of Theorem By deitio, the sum of ay two codewords i a liear block code is also a codeword. We will rst cosider the case whe we have all eve weight codewords. Sice the sum of ay two eve weight codewords is also a eve weight codeword, it is possible to have a liear block code that cosists of oly eve weight codewords. Now cosider the case where we have all odd weight codewords. Sice the sum of ay of these codewords must also be a codeword ad the sum of two odd weight codewords is a eve codeword, all possible liear combiatios of these codewords will result i a equal umber of eve ad odd weight codewords. This shows that ay liear block code must either have oly eve weight codewords, or have a equal umber of eve ad odd weight codewords. 14. First assig 0 = red ad 1 = blue. Oce everyoe has formed a lie, start from the back of the lie (i.e., the tallest perso) ad have that perso say the sum (modulo 2) of all hat colors i frot of him or her (i.e., the overall parity). The ext-tallest perso i lie will the take the umber (or color i this case) that the perso behid them said ad add that to the sum of all hat colors i from of him or her, all additios beig doe modulo 2. The result of this sum is the color of their hat, which they yell out to the ext perso i lie. This cotiues util everyoe has yelled out the colors of their hats. The oly perso that may yell out a 1 The base of the logarithm does't matter to us at this stage, but traditioally the log likelihood is deed as the atural logarithm (base e).

5 icorrect hat color is the tallest perso i lie who made the rst aoucemet (because that perso aouced the overall parity), which would give the team a score N 1.

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