6.451 Principles of Digital Communication II Wednesday, March 9, 2005 MIT, Spring 2005 Handout #12. Problem Set 5 Solutions

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1 6.51 Priciples of Digital Commuicatio II Weesay, March 9, 2005 MIT, Sprig 2005 Haout #12 Problem Set 5 Solutios Problem 5.1 (Eucliea ivisio algorithm). (a) For the set F[x] of polyomials over ay fiel F, show that the istributive law hols: (f 1 (x)+ f 2 (x))h(x) = f 1 (x)h(x)+ f 2 (x)h(x). Usig the associative a commutative laws a the rules of polyomial arithmetic, we have ( )( ) (f (f 1i + f 2i )x i h j x j 1 (x)+ f 2 (x)) h(x) = = (f 1i + f 2i )h j x i+j a i j i j f (f 1i h j + f 2i h j )x i+j 1 (x)h(x)+ f 2 (x)h(x) =. Fially, (f 1i + f 2i )h j = f 1i h j + f 2i h j by the istributive law over F. (b) Use the istributive law to show that for ay give f(x) a g(x) i F[x], there is a uique q(x) a r(x) with eg r(x) < eg g(x) such that f(x) = q(x)g(x)+ r(x). Suppose that f(x) ca be writte i two ways: f(x) = q(x)g(x)+ r(x) = q (x)g(x)+ r (x) where eg r(x) < eg g(x) a eg r (x) < eg g(x). Usig the istributive law, we have i j (q(x) q (x))g(x)+(r(x) r (x)) = 0. (1) If q(x) = q (x), the (q(x) q (x))g(x) = 0, so (1) implies r(x) = r (x). If q(x) = q (x), the (q(x) q (x))g(x) 0 a has egree eg g(x), whereas r(x) r (x) has egree < eg g(x), so (1) caot hol. Thus the quotiet q(x) a remaier r(x) are uique. Problem 5.2 (uique factorizatio of the itegers). Followig the proof of Theorem 7.7, prove uique factorizatio for the itegers Z. We follow the statemet a proof of Theorem 7.7, replacig statemets about polyomials by correspoig statemets about itegers: Theorem 7.0 (Uique factorizatio of itegers) Every positive iteger Z with magitue 2 may be writte i the form k = p i, i=1 where each p i, 1 i k, is a prime iteger. This factorizatio is uique, up to the orer of the factors. 1

2 Proof. If is a prime, the = is the esire uique factorizatio. If is ot a prime, the ca be factore ito the prouct ab of two otrivial factors each less tha, which i tur ca be factore, a so forth. Sice magitues ecrease with each factorizatio, this process ca oly termiate i a prime factorizatio. Now we ee to prove uiqueess. Thus assume hypothetically that the theorem is false a let be the smallest iteger that has more tha oe such factorizatio, = a 1 a k = b 1 b j ; j, k 1, (2) where a 1,...,a k a b 1,...,b j are prime itegers. We will show that this implies a iteger smaller tha with o-uique factorizatio, a this cotraictio will prove the theorem. Now a 1 caot appear o the right sie of (2), else it coul be factore out for a immeiate cotraictio. Similarly, b 1 caot appear o the left. Without loss of geerality, assume b 1 a 1. By the Eucliea ivisio algorithm, a 1 = qb 1 + r. Sice a 1 is prime, r =0 a 0 <r <b 1 a 1.Now r has a prime factorizatio r = r 1 r,where b1 is ot a ivisor of ay of the r i, sice it has greater magitue. Substitutig ito (2), we have (qb 1 + r 1 r )a 2 a k = b 1 b j, or, efiig = r 1 r a 2 a k a rearragig terms, = r 1 r a 2 a k = b 1 (b 2 b j qa 2 a k ). Now is positive, because it is a prouct of positive itegers; it is less tha, sice r< a 1 ; a it has two ifferet factorizatios, with b 1 a factor i oe but ot a ivisor of ay of the factors i the other; cotraictio. Problem 5. (fiig irreucible polyomials). (a) Fi all prime polyomials i F 2 [x] of egrees a 5. [Hit: There are three prime polyomials i F 2 [x] of egree a six of egree 5.] We ca immeiately elimiate all polyomials which have the egree-1 factor x (i.e., whose costat term is 0) or the egree-1 factor x +1 (i.e., whichhaveaeveumber of ozero coefficiets). This elimiates of the caiate polyomials. We the ee to sieve out oly multiples of the egree-2 prime polyomial x 2 + x + 1 a the two egree- prime polyomials, x + x + 1 a its reverse, x + x This leaves four egree- polyomials. Oe of these is (x 2 + x +1) 2 = x + x The remaiig three are prime: 2 x + x +1,x + x +1,x + x + x + x +1. Similarly, this leaves eight egree-5 polyomials. Two of these are multiples of the egree- 2 prime polyomial with oe of the two egree- prime polyomials, amely (x 2 + x +1) 5 5 (x + x +1) = x + x + 1 a its reverse, x + x + 1. The remaiig six are prime: x + x 2 +1,x + x + x + x +1,x 5 + x + x + x +1, 5 5 a their reverses x 5 + x +1,x + x + x + x 2 +1,x + x + x + x +1. 2

3 (b) Show that x 16 + x factors ito the prouct of the prime polyomials whose egrees ivie, a x 2 + x factors ito the prouct of the prime polyomials whose egrees ivie 5. The prime polyomials whose egrees ivie are x, x + 1,x 2 + x + 1 a the three egree- prime polyomials above. Straightforwar polyomial multiplicatio shows that 2 x(x +1)(x 2 + x +1)(x + x +1)(x + x + x + x +1)(x + x +1) = x + x. (Note that (x 2 +x+1)(x +x +1)(x +x+1) = x 10 +x 5 +1 a (x+1)(x +x +x 2 +x+1) = x 5 +1.) Similarly, the prime polyomials whose egrees ivie 5 are x, x + 1 a the six egree-5 prime polyomials above. Agai, straightforwar polyomial multiplicatio shows that their prouct is x 2 + x. 16 Problem 5. (The ozero elemets of F g(x) form a abelia group uer multiplicatio). Let g(x) be a prime polyomial of egree m, a r(x),s(x),t(x) polyomials i F g(x). (a) Prove the istributive law, i.e., (r(x)+ s(x)) t(x) = r(x) t(x)+ s(x) t(x). [Hit: Express each prouct as a remaier usig the Eucliea ivisio algorithm.] By the istributive law for oriary polyomials, we have (r(x)+ s(x))t(x) = r(x)t(x)+ s(x)t(x). Followig the hit, write r(x)t(x) = q 1 (x)g(x)+ r 1 (x),s(x)t(x) = q 2 (x)g(x)+ r 2 (x), a (r(x) + s(x))t(x) = q (x)g(x) + r (x), where eg r i (x) < eg g(x) for i =1, 2,. The r(x) t(x) = r 1 (x),s(x) t(x) = r 2 (x), a (r(x) + s(x)) t(x) = r (x). Now from the equatio above, which implies q (x)g(x)+ r (x) = q 1 (x)g(x)+ r 1 (x)+ q 2 (x)g(x)+ r 2 (x). 0=(q (x) q 1 (x) q 2 (x))g(x)+(r (x) r 1 (x) r 2 (x)) Sice 0 = 0q(x) + 0 a such a ecompositio is uique, we have r (x) = r 1 (x)+ r 2 (x). (b) For r(x) =0, show that r(x) s(x) = r(x) t(x) if s(x) = t(x). The equatio r(x) s(x) = r(x) t(x) implies r(x) (s(x) t(x)) = 0; but sice g(x) is irreucible, this implies either r(x) =0 or s(x) = t(x). (c) For r(x) 0, show that as s(x) rus through all ozero polyomials i F g(x), the prouct r(x) s(x) also rus through all ozero polyomials i F g(x). By part (b), the proucts r(x) s(x) are all ozero a are all istict as as s(x) rus m through the F 1 ozero polyomials i F g(x), so they must be all of the F m 1 ozero polyomials i F g(x).

4 () Show from this that r(x) 0 hasamo-g(x) multiplicative iverse i F g(x) ; i.e., that r(x) s(x) =1 for some s(x) F g(x). By part (c), the proucts r(x) s(x) iclue every ozero polyomial i F g(x), icluig 1. Therefore, give r(x) =0 F g(x), there exists a uique s(x) =0 F g(x) such that r(x) s(x) =1; i.e., such that s(x) is the multiplicative iverse of r(x) i F g(x). Sice the multiplicatio operatio is associative a commutative a has ietity 1, it follows that the ozero elemets of F g(x) form a abelia group uer multiplicatio. Problem 5.5 (Costructio of F 2 ) (a) Usig a irreucible polyomial of egree 5 (see Problem 5.), costruct a fiite fiel F 2 with 2 elemets. We ca costruct F 2 usig ay of the 6 irreucible polyomials of egree 5 fou i Problem 5.. Usig g(x) = x 5 + x 2 + 1, the fiel F 2 is efie as the set of all 2 biary polyomials of egree or less uer polyomial arithmetic moulo g(x). (b) Show that aitio i F 2 ca be performe by vector aitio of 5-tuples over F 2. The sum of two polyomials of egree or less is obtaie by a compoetwise sum of their coefficiets, whether moulo g(x) orot. (c) Fi a primitive elemet F 2. Express every ozero elemet of F 2 as a istict power of. Show how to perform multiplicatio a ivisio of ozero elemets i F 2 usig this log table. The set F 2 of ozero elemets of F 2 is the set of roots of the equatio x 1 =1 i F2; i.e., every β F 2 satisfies β 1 = 1, so the multiplicative orer of every elemet must ivie 1, which is prime. There is oe elemet of multiplicative orer 1, amely 1. The remaiig 0 elemets must therefore have multiplicative orer 1; i.e., there are 0 primitive elemets i F 2. Therefore = x must be primitive. We compute its powers, reucig x 5 to x as ecessary: = x, 2 2 = x, = x, = x, 5 = x 2 +1, 6 = x + x, 2 7 = x + x, 8 = x + x 2 +1, 9 = x + x + x, 10 = x +1, 11 2 = x + x +1, 12 = x + x 2 + x, 1 2 = x + x + x,

5 1 = x + x + x 2 +1, 15 2 = x + x + x + x +1, 16 = x + x + x +1, 17 = x + x +1, 18 = x +1, 19 = x 2 + x, 20 2 = x + x, 21 = x + x, 22 = x + x 2 +1, 2 2 = x + x + x +1, 2 = x + x + x 2 + x, 25 = x + x +1, 26 2 = x + x + x +1, 27 = x + x +1, 28 = x + x 2 + x, 29 = x +1, 0 = x + x, 1 =1. The prouct of i a j is i+j. The quotiet of i ivie by j is i j. I both cases the expoets are compute moulo 1, sice 1 =1. () Discuss the rules for multiplicatio a ivisio i F 2 whe oe of the fiel elemets ivolve is the zero elemet, 0 F 2. The prouct of 0 with ay fiel elemet is 0. Divisio by 0 is ot efie; i.e., it is illegal (as with the real or complex fiel). Problem 5.6 (Seco ozero weight of a MDS coe) Show that the umber of coewors of weight +1 i a (, k, ) liear MDS coe over Fq is ( )( ( ) ) +1 N +1 = (q 2 1) (q 1), +1 where the first term i paretheses represets the umber of coewors with weight i ay subset of +1 cooriates, a the seco term represets the umber of coewors with weight equal to. Cosier ay subset of + 1 = k + 2 cooriates. Take two of these cooriates a combie them with the remaiig k 2 cooriates to form a iformatio set. Fix the compoets i the k 2 cooriates to zero, a let the remaiig two cooriates rufreelythrough F q. These q 2 iformatio set combiatios must correspo to q 2 5

6 coewors. (I fact, we may view this subset of coewors as a shortee ( + 1, 2,) MDS coe.) Oe of these coewors must be the all-zero coewor, sice the coe is liear. The remaiig q 2 1 coewors must have weight or + 1. Sice there are q 1coewors of weight with support i ay subset of cooriate positios, the umber of ( coewors of weight whose support is i ay subset of + 1 cooriate positios is ) +1 (q 1) (the umber of coewors of weight i a ( + 1, 2,) MDS coe). So the umber of coewors of weight +1 i ay + 1 cooriate positios is ( ( ) ) +1 (q 2 1) (q 1). ( ) Sice there are istict subsets of + 1 cooriate positios, the give expressio +1 for N +1 follows. Note that ( ( ) ) +1 (q 2 1) (q 1) =(q +1)(q 1) ( +1)(q 1) = (q )(q 1). This implies that if >,the q, sice otherwise N +1 woul become egative. I other wors, there exists o (, k, = k + 1) MDS coe over F q with q< <. For example, there exist o biary MDS coes other tha the (,, 1), (, 1, 2) a (, 1,) coes (a (, 0, ), if you like). More geerally, whe q + 2, there exist forbie values of, amely q Similarly, by cosierig shortee coes of legths +2,+,...,, we ca compute N+2,N +,...,N. Problem 5.7 (N a N +1 for certai MDS coes) (a) Compute the umber of coewors of weights 2 a i a (, 1, 2) SPC coe over F 2. ( ) ( ) We have N 2 =(q 1) 2 = ; i.e., there is a weight-2 coewor for every cooriate 2 ( ) pair. The N =(q )(q 1) = 0. This is cosistet with the efiitio of a SPC coe as the set of all eve-weight -tuples. (b) Compute the umber of coewors of weights 2 a i a (, 1, 2) liear coe over F. ( ) ( ) Here we have N 2 =(q 1) 2 =2 ; i.e., there are two weight-2 coewors for every 2 ( ) ( ) cooriate pair. The N =(q )(q 1) =2. For example, the (, 2, 2) RS coe over F has geerators (111, 012) a coewors {000, 111, 222, 012, 120, 201, 021, 102, 210}, with N 2 =6 a N = 2. Thusigeeral a zero-sum coe over a fiel larger tha F 2 has o-weight coewors. 6

7 (c) Compute the umber of coewors of weights a i a (, 2, ) liear coe over F. ( ) Here we have N =(q 1) = 2() = 8, so all o-zero coewors have weight. (Verificatio: N = 0 because q =.) The (, 2, ) liear coe over F give i the itrouctio of Chapter 8 (or i the ext problem) is a example of such a coe, calle a tetracoe. Problem 5.8 ( Doubly extee RS coes) (a) Cosier the followig mappig from (F q ) k to (F q ) q+1. Let (f 0,f 1,...,f k 1 ) be ay k-tuple over F q, a efie the polyomial f(z) = f 0 + f 1 z + + f k1 z k 1 of egree less tha k. Map (f 0,f 1,...,f k 1 ) to the (q +1)-tuple ({f(β j ),β j F q },f k 1 ) i.e., to the RS coewor correspoig to f(z), plus a aitioal compoet equal to f k 1. Show that the q k (q +1)-tuples geerate by this mappig as the polyomial f(z) rages over all q k polyomials over F q of egree less tha k form a liear ( = q + 1,k, = k + 1) MDS coe over F q. [Hit: f(z) has egree less tha k 1 if a oly if fk 1 =0.] The coe evietly has legth = q + 1. It is liear because the sum of coewors correspoig to f(z) a f (z) is the coewor correspoig to f(z)+ f (z), aother polyomial of egree less tha k. Its imesio is k because o polyomial other tha the zero polyomial maps to the zero (q + 1)-tuple. To prove that the miimum weight of ay ozero coewor is = k + 1, use the hit a cosier the two possible cases for f k 1 : If f k 1 0, the eg f(z) = k 1. By the fuametal theorem of algebra, the RS coewor correspoig to f(z) has at most k 1 zeroes. Moreover, the f k 1 compoet is ozero. Thus the umber of ozero compoets i the coe (q +1)- tuple is at least q (k 1) + 1 = k +1. If f k 1 =0 a f(z) 0,the eg f(x) k 2. By the fuametal theorem of algebra, the RS coewor correspoig to f(z) has at most k 2 zeroes, so the umber of ozero compoets i the coe (q+1)-tuple is at least q (k 2) = k+1. (b) Costruct a (, 2, ) liear coe over F. Verify that all ozero wors have weight. The geerators of a extee RS (, 2, ) tetracoe over F are (1110, 0121), a the coe is {0000, 1110, 2220, 0121, 1201, 2011, 0212, 1022, 2102}, with N =8 a N =0 (as show i Problem 5.7(c); compare the zero-sum (, 2, 2) coe of Problem 5.7(b)). 7