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1

2 Euclid Euler Euler RSA ElGamal Diffie-Hellman Hamming Hamming code Hamming code

3 December 3, : Huffman S Sun, Sun Microsystems, Solaris Sun Microsystems, Inc.. UNIX UNIX System Laboratories, Inc..,,,.

4 ,., (Ethernet),,,.,,,,,.,,,,.,,,,,,.,, , A B, X, P, C, f : p(x) p(x) (P C), g : p(x) p(x) (C P ).,,,,., ,.,,,,, ASCII 1., f(x) =x +1. ASCII, Z/127Z, 1,,,., ASCII, 0x00 0x7F,, Z/127Z.

5 December 3, : ,, g(x) =x 1, f g =1., X, a X, b X, f(x) =ax + b, g(x) =a 1 x a 1 b,.,, a X, b X..,, ,., X X. A B f AB g AB f AB : X n (X ) k, g AB :(X ) k X n., f AB, B A, g AB B A.,, ,, Euler RSA,, , PGP, Secure Shell (SSH). PGP 1024 bits RSA, SSH Version 1 RSA. SSH Version 2, DSA. 2 HAL, IBM.

6 December 3, : Euclid a, b, gcd(a, b), Euclid.,, Euclid Euclid,. Algorithm (Euclidean Algorithm). a 0, a 1, gcd(a 0,a 1 ). 1. a j = a j+1 q j+1 + a j+2 q j+1 a j+2., q j+1, a j a j+1, a j a j+2 =0 a j+1 = gcd(a 0,a 1 ). a j+2 0.,. Theorem a, b, q, r, a = bq + r, gcd(a, b) = gcd(b, r)., Euclid., a, b, gcd(a, b) Euclid.,.,. Theorem a, b, a>b, O(log a). Proof., a j+2 < 1 2 a j.,. a j a j a j+2 < a j a j,., a j+1 > 1 2 a j., a j+2 = a j q j+1 a j+1 < (1 qj+1 2 )a j, q j+1 > 1, a j+2 < 1 2 a j., a j+2 < 1 2 a j., Euclid,. Theorem a, b, x, y, ax + by = gcd(a, b). x, y, a>b, O(log a) Z/mZ m., Z/mZ.

7 December 3, :33 6 Proposition a Z/mZ gcd(a, m) =1., a 1 Z/mZ O(log m). Proof. Theorem gcd(a, m) =1, ax + ym =1 x, y., ax 1 (mod m), x = a 1 in Z/mZ. gcd(a, m) > 1 ax + ym =1 x, y. Z/mZ (Z/mZ). Corollary p, Z/pZ. F p. Corollary ax b (mod m), gcd(a, m) =1,. gcd(a, m) = d>1, b d (d b ) d. Theorem (Fermat ). p. a, a p a (mod p). a p a p 1 1 (mod p). Proof. p a, a p, a p 0 a (mod p). p a., k, m [0,p 1], k m, ka ma (mod p)., (k m)a 0 (mod p)., {ka} p 1 k=0 {0,,p 1}., (p 1)!a p 1 (p 1)! (mod p), a p 1 1 (mod p) 3. Corollary a p n m (mod p 1), a m a n (mod p). Proof. n>m. p 1 n m, c, n = m + c(p 1)., a p 1 1 (mod p) c. Theorem ( ). {m i } n, x a i (mod m i )., M = m 1 m n,. O(log max m i ). Proof.,,. x, x, x x 0 (mod m i ). x x M = m 1 m n 0.,. M i = M/m i gcd(m i,m i )=1, N i M i 1 (mod m i ) N i., x = N a im i N i, x a i (mod m i ) Euler Euler Definition a, ϕ(a) ={a }. 3, F p, a p 1.

8 December 3, :33 7 p, ϕ(p) =p 1. Proposition a = p l1 1 plk k., d n ϕ(d) =n., ϕ(a) =p l1 1 (1 1 p 1 ) p l k k (1 1 p k ) Proof. a = p k, ϕ(p k )=p k p k 1., 1 p k 1 p, n N, np., n n [1,p k 1 1], p k 1 p k 1 1 p., ϕ(p k )=p(1 1 p )., a = mn, gcd(m, n) =1, ϕ(mn) =ϕ(m)ϕ(n). Euler. 1 mn 1, mn k. gcd(mn, k) = 1., j 1 = k (mod m), j 2 = k (mod n), (j 1,j 2 ), k,., k mn, j 1 m., gcd(j 1,m)=1. gcd(j 2,n)=1, j 1 ϕ(m), j 2 ϕ(n)., k ϕ(mn) (j 1,j 2 ) ϕ(m) ϕ(n).,,, ϕ(mn) =ϕ(m)ϕ(n)., f(n) = d n ϕ(d)., gcd(m, n) =1 f(nm) =f(n)f(m)., d mn d, d = d 1 d 2,(, d 1 m, d 2 n, gcd(d 1,d 2 )=1)., ϕ(d) =ϕ(d 1 )ϕ(d 2 ), f(mn) = ϕ(d) = ϕ(d 1 )ϕ(d 2 )= ϕ(d 1 ) ϕ(d 2 )=f(m)f(n) d mn d 1 m d 2 n d 1 m d 2 n., n = p α1 1 pαk k, f(n) =f(p α1 1 ) f(pαk k ), f(p α )=p α., f(p α )= α ϕ(p j )=1+ j=0 α (p j p j 1 )=p α j=1, f(p α )=p α., f(n) =n. Corollary n, p, q, ϕ(n). ϕ(n) p, q. Proof. ϕ(n) =(p 1)(q 1), p, q ϕ(n). n, p = 2, q = n/2, ϕ(n) = n/2 1, n., n = pq, ϕ(n) =(p 1)(q 1) = n +1 (p + q), n ϕ(n), p, q, x 2 2bx + n =0,b = n +1 ϕ(n).

9 December 3, :33 8 Theorem (Euler ). gcd(a, m) =1 a ϕ(m) 1 (mod m). Proof. p, m = p α. α =1, Fermat. α.,, a pα 1 p α 2 =1+p α 1 b b. p, a pα p α 1 1 p., m., Euler, a ϕ(m) 1 (mod p α ). Corollary n. d, e, de 1 n p, p 1., a de a (mod n). Proof., n = p 1 p k, p i,., a de a (mod p i )., p i a., p i a, de 1=(p i 1)c c., Fermat a pi 1 1 (mod p i ), c, , p, q = p f., q F q, F q F p f 4. F q,. Proposition a F q q 1. Proof. a q 1 =1. F q (q 1 ), a., q 1 F q a, F q q 1,., a q 1 =1., d a., a d =1 d, d q 1, a r = a q 1 bd =1 r<d., d. Theorem , g F q, gcd(j, q 1)=1, g j. ϕ(q 1). F q, q 1. Proof. a F q d, d q 1., a, a2,,a d =1. d gcd(j, d) =1 ϕ(d) a j. {a j } d j=1 xd =1., x d =1 d., d., gcd(j, d) =d > 1, a j., d/d, j/d, (a j ) d/d =(a d ) j/d =1. gcd(j, d) =1, a j d., a j d, (a d ) j = a d =1, (a d ) gcd(j,d) = a d =1., a d., a j gcd(j, d) =1, d., d, d ϕ(d)., d (q 1), d,., Euler d q 1 ϕ(d) =q 1, F q d q 1 d., q 1 ϕ(q 1). 4 F q F p f.

10 December 3, :33 9 Definition G, b, y G., b x = y, x Z b y. Example G = F 19 =(Z/19Z)., 2 F 19 7 F 19 6., 2 F , q 1, b y F q. Algorithm b F q, y F q, bx = y x Z., q 1= p pα., x mod p α., x p, x mod q q 1 p, b p r p,j = b j(q 1)/p, j =0, 1,,p 1,. 2. p, x x 0 + x 1 p + + x α 1 p α 1 (mod p α ), x i. (a) x 0, y (q 1)/p. y q 1 =1, y (q 1)/p = b x(q 1)/p = b x0(q 1)/p = r p,x0., y (q 1)/p, {r p,j }, j x 0. (b) y y 1 = y/b x0, (y 1 ) (q 1)/p2., (y 1 ) (q 1)/p = b x(q 1)/p x0 =1,, (y 1 ) (q 1)/p2 = b (x x0)(q 1)/p2 = b x1(q 1)/p = r p,x1,, j x 1. (c), α 1. Remark ,, q 1., {r p,j }, (y i ) (q 1)/pi+1. Example (F 37 ), = , r 2,0 =1,r 2,1 = 1., 2 36/3 26 (mod 37), {r 3,j } = {1, 26, 10}., 28=2 x, p =2, x 0 +2x 1 x mod /2 1 (mod 37), x 0 =0,28 36/4 1 (mod 37), x 1 =1., x 2 (mod 4). p =3, x 0 +3x 1 = x mod 9, 28 36/3 26 (mod 37) x 0 =1,28 36/9 10 (mod 37), x 1 =2., x 7 (mod 9)., x =34.

11 December 3, : , Z/mZ, f : Z/mZ Z/mZ, f(x) =ax + b.. Example A Z, 0 25., Z/26Z. Example A Z, 0 26., Z/27Z. Example A Z,?, 0 27., Z/28Z., f(x) =ax + b, gcd(a, m) =1 f 1 (x) =a 1 x a 1 b. a gcd(a, m) 1, c 1, c 2 Z/mZ, f(c 1 )=f(c 2 )., gcd(a, m) =1. Example Example 1.3.1, f(x) =3x +24., ABCDEFGHIJKLMNOPQRSTUVWXYZ YBEHKNQTWZCFILORUXADGJMPSV. Example Example 1.3.3, f(x) =23x +19. ABCDEFGHIJKLMNOPQRSTUVWXYZ? TOJE?WRMHCZUPKFAXSNID VQLGBY ,, a, b.,.,, E T

12 December 3, :33 11.,,,. Example Example 1.3.4,, K, D., E, T, a 1 = a, a 1 b = b, 10a + b = 4 (mod 26), 3a + b = 19 (mod 26).,, 7a = 11 (mod 26), a = 9 (mod 26), b =4 10a = 18 (mod 26). Example Example 1.3.5, B,?,,, E, a + b = 26 (mod 28), 27a + b = 4 (mod 28). a, b, 2a = 22 (mod 28), a = 11 (mod 14), a =11 25 (mod 28), b =15 1 (mod 28),.,, I, T, 8a + b = 18 (mod 28), a =11, b = RSA,, RSA. RSA,,., N, k, Z/(N k )Z. l (k <l), Z/(N l )Z., f : Z/(N k )Z Z/(N l )Z.

13 December 3, :33 12, p, q, n = pq N k <n<n l., e, e p, q, ϕ(n) =(p 1)(q 1)., A,. K E,A =(n A,e A ). K D,A =(n A,d A )., d A e 1 A (mod ϕ(n A))., f(p ) P ea (mod n A ) A., A, g(c) C d A (mod n A )., d A e A 1 (mod ϕ(n A )), d A e A (p 1), (q 1), n A = pq, Corollary , P eada P (mod n A )., A B, B,, B. B,. Remark , n = pq, (n, d) e., RSA RSA, p, q,.. p 1, q 1 GCD,., n = pq., p, q. p>q, n = pq =( p+q 2 )2 ( p q 2 )2, s =(q p)/2., t =(p + q)/2 n, t 2 n., n, n<t t, t 2 n, t =(p + q)/2 s =(q p)/ ,. RSA. A B A,. 1. A,,. 2. B A,.,., RSA PGP,

14 December 3, :33 13, md5 5, md5, Diffie-Hellman g k, g a g ak., ElGamal q, F q., g F q., A a = a A F q,., g a F q. P A, B k, (g k Pg ak ) A., a g ak. A (g k ) a, Pg ak g ak = P Diffie-Hellman, q g F q 6,. Diffie-Hellman,,.,,. a F q, g a F q,, a. A, B,,, g a(a)a(b)., a(a) A. A B g a(a)a(b), g a (B) a(a)., A,B, A,B,. Diffie-Hellman, Sun Microsystems Solaris 2.x NIS+. 5 md5 digest string,,,,,. 6 g F q.

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16 ,,.,.,.,, ASCII 0x00 0x7F 128, 7 bits.,, 0/1.,, 10 bits 1., (start bit) (stop bit),, 7 bits, 1 bit., 1 bit. 2, 7 bits,, stop bit 1.,, 10 bits 1,.,, 10 bits 1,.,, 8 bits,, MNR,.,,, ,.,,,.,..,,., RAID. RAID 5, n, n 1,,. RAID bits 9 bits. 2,.

17 December 3, : , F 2. n, n 1, {a i } n 1.,, a a n 1 + a n =0 a n. {a i } n 1.,, a i., a i = a a i 1 + a i a n.,,,.. 2.2, (code)., Definition A = {a 1,,a n } alphabet, a i. A word, A, a = a i1 a ik, a =(a i1,,a ik ). (S, P) source, S alphabet word, P S P : S [0, 1], word s i S. A A. a A, len(a) a (word length) n-ary code n alphabet, A = {a 1,,a n } alphabet A C, source (S, P ), f : S C encoding function., a C codeword. C size C., A = {0, 1} binary code. Example A = {a, b, c, d}, S = {a, b, c, d}, f(a) =0,f(b) = 10, f(c) = 11, f(d) = 110,. Definition a C, len(a) block code, len(a) C (code length). Example , A, S = A, S 7. a C, len(a) =7 block code Hamming block code C,.

18 December 3, :33 17 Definition v 1, v 2 C, d(v 1,v 2 )= {(v 1 ) i (v 2 ) i }. C Hamming distance, C. C minimun distance, d(c) = min d(v, w) v,w C. r, size M, minimun distance t (r, M, t)-code.. Definition C t-error-detecting, codeword t, alphabet alphabet, codeword. exactly t-error-detecting, t + 1-error-detecting t-error-detecting code. C t-error-correcting, codeword t, alphabet alphabet, codeword. exactly t-error-correcting, t + 1-error-correcting t-error-correcting code. Theorem t-error-correcting t-error-detecting., C exactly t-error-detecting, d(c) =t +1. Proof.. C t-error-detecting., v 1, v 2 C, d(v 1,v 2 ) t, v 1 t alphabet, v 2 C. t-error-detecting., d(c) t +1., exactly t-error-detecting, v C, v t +1 alphabet, codeword w C. d(v, w) =t +1, d(c) =t +1. d(c) =t +1, v C, t alphabet C., C exactly t-error-detecting. Theorem C exactly t-error-correcting, d(c) =2t +1 2t +2. Proof. d(c) =2t +1 2t +2., v A w C t alphabet., 0<d(v, w) t., w w C, d(v,w ) >d(v, w)., v A codeword w., d(v, w ) t, d(w, w ) d(v,w)+d(v,w ) t + t =2t<d(C), C minimun Hamming distance 2t +1 2t +2., C t-error-correcting., d(c) =2t +1, w, w C, d(w, w )=2t +1., w v t +1 alphabet, w,

19 December 3, :33 18 w alphabet., d(w, v) =t +1, d(w,v)=2t +1 (t +1)=t,, codeword,. d(c) =2t +2, v codeword, w, w,., C exactly t-error-correcting. C exactly t-error-correcting, d(w, w ) 2t codewords w, w C., w alphabet, w w alphabet,., d(c) > 2t., d(c) > 2t +2, t + 1-error-correcting, exactly t-error-correcting, 2t <d(c) 2t +2. Definition Hamming distance codeword error correcting, minimum distance decoding. Corollary d(c) =d C exactly [(d 1)/2]-error-correcting. Definition Block code C minimum Hamming distance d(c) =t. Block code C, x A len(c), w C, d(w, x) t., C (perfect),, codeword, Hamming distance t. Theorem (Sphere packing condition). C q alphabet, (n, M, d)-code., C, d =2t +1, d,. t ( ) n V q (n, t) = (q 1) k, k k=0 MV q (n, t) =q n Proof. d. d =2t +2, a, b C, d(a, b) =2t +2., word x, d(a, x) =d(b, x) =t +1. C., d =2t +1. C, word x, a C,., x S t (a) ={x : word,d(x, a) t} A n = a C S t (a), disjoint union (2.2.1)., a C word x d(a, x) =k, a, x ( n ) k (q 1) k., a k, q 1., S t (a) = t k=0 ( ) n (q 1) k k

20 December 3, :33 19, (2.2.1) q n A n = a C S t (a) = M V q (n, t). Definition Block code C 1 C 2 equivalent, C 1 C 2 alphabet,, size, σ S n (n alphabet ), alphabet {π i } n, c 1 c n C 1 π 1 (c σ(1) ) π n (c σ(n) ) C 2., C 1 C 2 scalar multiple equivalent, C 1, C 2 alphabet F q, π i α i F q, π i (s) =α i s. 2.3,, (linear code) Definition q p q = p f, V (n, q) F q n., L linear code, L V (n, q)., dim L = k, L [n, k]-code., L minimun distance d, [n, k, d]-code., V (n, q), a L F q n (a 1,,a n )., [n, k]-code M M = q k., linear code block code. Definition L linear code, d L Hamming distance, d(v, 0) = w(v) w : C Z, C Hamming weight. x, y L, x y intersection x y. x y =(x 1 y 1,,x n y n ) Lemma V (n, q) Linear code L. 1. x, y V (n, q), d(x, y) =w(x y). 2. x, y V (n, 2), d(x, y) =w(x)+w(y) 2w(x y). Theorem Linear code L, d(l) =w(l). d(l) L minimum Hamming distance. Proof. c, d L codeword, c d L codework., d(l) = min d(c, d) = min w(c d) = min w(c) =w(l). c d L c d L c 0 L

21 December 3, : M(k n, F q ) F q k n. Definition L [n, k]-code. L V (n, q), G M(k n, F q ), L = {xg : x V (k,q)}. G L generator matrix. Example V (4, 2) G = [4, 3]-code, ] ] [x 1 x 2 x = [x 1 + x 3 x 1 + x 2 x 2 + x 3 x Linear code generator matrix G,, G =(I k A). Generator matrix (I k A), generator matrix standard form. V (n, q). Definition x =(x i ), y =(y i ) V (n, q), x y x y =,. L [n, k]-code. L dual code, n x i y i L = {x V (n, q) :x c = 0 for all c L}.. Proposition L [n, k]-code, G generator matrix, 1. L generator matrix G t. 2. L [n, n k]-code. 3. (L ) = L. Remark Q, R, C X, X V V, V V = {0}., F q,.

22 December 3, :33 21 Example V (4, 2) [4, 2]-code L = {0000, 1100, 0011, 1111} L = L. L = L linear code self-dual., V (4, 2) L, e 1 + e 2, e 3 + e 4. 0 V, V e 1 e 2, e 3 e 4, F 2 e 2 = e 2, e 4 = e 4, V = V., dim(l) + dim(l )=n, G L generator matrix G =(I k A) standard form., H =( A t I n k ), ( ) GH t A =(I k A) = A + A =0 I n k. H L generator matrix. H L partity check matrix., x L xh t =0. Parity check matrix, H =(A I m ), standard form. Example V (n, 2) [n, n 1]-code 1 G =(I n 1 A), A =. 1., (x 1,,x n 1 ), (x 1,,x n,x x n 1 ).,, x x n 1,(x 1,,x n ) parity. parity chech matrix H,, x L. H =( A t I 1 )= [ ] 1 1 (x x n 1 )+x n =0

23 December 3, : Theorem L [n, k, d]-code. L linear [n, k]-code, w(l) =d., r H r, d = r +1. Proof. H k 1,,k n., c L, c = c 1 c n, ch t =0 c 1 k c n k n =0., w(c) c i 0, c i1,,c iw(c), c i1 k i1 + + c iw(c) k iw(c) =0., k 1,,k n w(c)., r<d. d<w(c) k i1,,k id, c = c 1 c n w(c) =d, c i1 k i1 + + c id k id =0., c L, w(c) =d<w(c), minimum Hamming distance., r = d 1. Corollary Linear [n, k]-code L parity check matrix H, r H r. d(l) =r +1,, L [n, k, r +1]-code, exactly r-error-detecting, exactly [r/2]-error-correcting. Example Example , H k i = [1], d =2., exactly 1-error-detecting,.,, codeword. Remark Corollary r rankh., H 0, r =0, rankh 0., H rankh, r rankh. x V (n, q),,,. Parity check matrix H V (n, q) L, x V (n, q) L, x L x =0 V (n, q)/l. Definition x V (n, q), xh t syndrome., syndrome, minimum distance decoding,. minimun distance decoding, x, c, c L, x c weight., c L, x c, V (n, q)/l coset x + L, x + L weight c L,.,.

24 December 3, :33 23 Theorem L parity check matrix., L minimun distance decoding, x, codeword c = x a., a x + L weight. a syndrome word weight.,., {c i } m = L,. 0 c 1 c m a 1 c 1 + a 1 c m + a 1. a s c 1 + a s c m + a s L. V (n, q), word, weight, a 1, a 1 + c i.,, 0 weight, a j + L., V (n, q)/l,, coset weight. a j coset leader., V (n, q)., x, x a j + c i, x a j = c i codeword. syndrome decoding. syndrome decoding., c L, x = c + e., e., e coset leader a j, syndrome decoding process, c = x a j = x e,,., e coset leader c i + a j, x a j x e = x (c i + a j )=c,. Example L V (4, 2) [4, 2]-code, generator matrix G [ ] G =, , L cosets, [0000] = {0000, 0100, 1101, 1001}, [1000] = {1000, 1100, 0101, 0001}, [0010] = {0010, 0110, 1111, 1011}, [1010] = {1010, 1110, 0111, 0011} , G standard form, parity check matrix H, [ ] [ ] G =, H = , coset leader syndrome,

25 December 3, :33 24, x = 1110, coset leader syndorme = 0000 th = 1000 th = 0010 th = 1010 th 1110 H t =11, syndrome coset leader 1010, x = , d(l) =1. codeword., exactly 1-error-detecting, Hamming code,, Hamming code F q r V (r, q), V 1 = V (r, q), c 1 V 1 c 1 0. c 2 V 2 = V 1 {αc 1 : α 0}, c 2 V 2 c 2 0.,., {c i } n., n,, V (r, q) = q r, {αc : α 0} = q 1 n = qr 1 q 1. {c i } n H, H parity check matrix L H q (r), Hamming code., H r n matrix, generator matrix G, (n r) n matrix, H q (r) [n, k]-code (k = n r).,. Theorem Hamming code H q (r) minimun weight d(h q (r)) = 3., H q (r) c i, scaler multiple equivalent., H q (r). Proof., H q (r). d =3, t =1, d =2t +1., V q (r, t) =1+ qr 1 q 1 (q 1) = qr., size M H q (r) V (r, q) k, M = q k., MV q (r, 1) = q r., H q (r) sphere packing condition,.

26 December 3, :33 25 d(h q (r)) = 3., c k, {c i } k 1, H,. d(h q (r)) 3., {c i } n,., e 1, e 2, e 1, e 2, e 1 + e 2., d(h q (r))=3.,, H q (r) parity check matrix H,. H q (r) parity check matrix H 1 H 2, scaler multiple equivalent. Hamming code L 1, L 2., x L 1 xh1 t =0., H 1 H 2, σ S n, H 1 {c 1 i }n H 2 {c 2 i }n α 1 i c 1 σ(i) = c2 i., x =[x 1,,x n ], σ x =[α 1 x σ(1),,α n x σ(n) ], 0=xH t 1 =(σ x)h t 2 =0., L 1 L 2 scaler multiple equivalent. Corollary Hamming code H q (r) q, r, 1-error-correcting code. Example H 2 (3) parity check matrix H 1 : H 1 = H 3 (3) parity check matrix H 2 : H 2 = Hamming code Hamming code H q (r),. H H q (r) parity check matrix. x i., c H q (r), x = c+αe i. α F q., syndrome xh t = αe i H t. e i H t H i. α, H 0 1., syndrome 0 α., syndrome s, 0 α, α 1 s H, i, x α 1 e i codeword. Example H 2 parity check matrix H 3 (3)., y = decode. [ ] [ ] [ ] H2 t = = =2 e 7, y 7 2, codeword c =

27 December 3, : , [n, k]-code, minimum Hamming weight d.. Theorem (Gilbert-Varhamov Bound). F q [n, k]-code minimum Hamming weight. q k q n < ). (q 1) i d 2 i=0 ( n 1 i Proof. Parity check matrix H d 1.. F q n k., n k F q., i, d 2.,, N i =(i 1, ) +(i 1, ) + +(i 1 d 2, ) ( ) ( ) ( ) i 1 i 1 i 1 = (q 1) + (q 1) (q 1) d 2, 1 2 d 2, q n k N i , n = i. q n k N n 1 > ,, [13, 8, 4]-code linear code., 2-error-detecting, 1-error-correcting code., SIMM (Serial Inline Memory Module) 3. SIMM,, 3 Dual Inline Memory Module (DIMM).

28 December 3, :33 27 ECC., SIMM,., SIMM.,, ECC SIMM., ECC Error Correcting Code, [39, 32, 4]-code linear code, 32.

29

30 ,,, , 0 1., binary sequencial data. binary 0 1, sequencial., ,, , ,,,. run length archive.,.,.

31 December 3, : , (S, P )., S {x i } M, x i P (x i ). binary sequencial data, S = {0, 1},.,. Definition A =(S, p), S = M,, M M 1 H b (A) =H b (p(x 1 ),,p(x M )) = p(x k )log b p(x k )= p(x k )log b p(x k )., b>1., 0log0=0.,. k=1 Theorem H, 0 p i 1, p 1 =1, m H b (p 1,,p M )= p i log b p i. 1. H b (p 1,,p n ), p i. 2. H b (1/n,, 1/n) <H b (1/(n +1),, 1/(n + 1)) n N. 3. b i N, b i = n,. k=1 k=1 H b ( 1 n,, 1 n )=H b( b 1 n,, b k k n )+ b i n H b( 1,, 1 ) b i b i,,,,. 1.,. 2. S {B i } k, B i = b i., b i = n., B i. P (B i )=B i /n., x j B k, P (x j )= n j=1 P (x j B i )P (B i )= 1 b k b k n = 1 n., P (x j B i ), B i x j, 0 x j B i, P(x j B i )= 1 x j B i b i

32 December 3, :33 31.,,,., = H b ( b 1 n,, b k n ), =. = k P (B i ) B i k P (B i )H b ( 1,, 1 ) b i b i Example , S. 1. S =2. H 2 ( 1 2, 1 2 )= 1 2 log log 2 2=1. 2. S =3, H 3 ( 1 3, 1 3, 1 3 )=1 3 log log log 3 3=1, H 2 ( 1 3, 1 3, 1 3 )=1 3 log log log 2 3 = log (bits). b =2, H bit., S =3,, Definition X S = {x i } n. P (X = x i)=p(x i ), n 1 H b (X) = p(x i )log b p(x i )., X S 1 = {x i } n, Y S 2 = {y j } m j=1, P (X = x i,y = y j )=p(x i,y j ), H b (X, Y )= i,j p(x i,y j )log b 1 p(x i,y j ). Example S = {x i } n, X P (X = x i)= 1 n.,. H b (X) =H b ( 1 n,, 1 n )= n, S = {x, y}, P (X = x) =1,P (X = y) =0,. H b (X) =0 1 n log b n =log b n

33 December 3, : Theorem P = {p i } n, Q = {q i} n, 0 p i,q i 1, p i = q i =1., n p i log b 1 p i n p i log b 1 q i, p i = q i. Remark n =. Theorem X {x i } n.,. 0 H b (X) log b n Theorem X, Y,. H b (X, Y ) H b (X)+H b (Y ), H. 3.3, (S, p). alphabet A = {a i } r, C A. Definition (S, p) C f : S C,, C (avarage length), AveLen(C) =., S = {s i } n. n p(s i )len(f(s i )) Example S = {a, b, c, d}, p(a) =p(b) =2/17, p(c) =9/17, p(d) =4/17. (C 1,f 1 ), (C 2,f 2 ) C 1 = {0, 11, 100, 101}, C 2 = {00, 10, 11, 01010}, f 1 (a) =11, f 2 (a) = 01010, f 1 (b) =0, f 2 (b) =00, f 1 (c) = 100, f 2 (c) =10, f 1 (d) =10, f 2 (d) =11,. 2 AveLen(C 1,f 1 )= = 41 17, 2 AveLen(C 2,f 2 )= = 40 17,

34 December 3, :33 33 Definition C, c 1 c k, d 1 d j C codeword, k = j c i = d i. C, C codeword, codeword. Example S = {a, b, c}, C = {0, 01, 001}, f(a) =0,f(b) = 01, f(c) = 001, 001 ab c.. S = {a, b, c}, C = {1, 01, 001}, f(a) =1,f(b) = 01, f(c) = 001,. S = {a, b, c}, C = {0, 01, 011}, f(a) =0,f(b) = 01, f(c) = 011, 01,,. S = {a, b, c, d}, C = {0, 10, 110, 1110}, f(a) =0,f(b) = 10, f(c) = 110, f(d) = 1110,,

35 December 3, :33 34 Theorem (Kraft ). C r alphabet, {l 1,,l n }., n k=1 1 r lk 1., {l 1,,l n } r, r alphabet C, C = n, {l 1,,l n }. Proof. C = {c i } n, len(c i)=l i., Kurft., L = max{l i }, c i = x i x li. c i C, C, x = x 1 x li y li+1 y L C., L word C, n L r L li = r L 1., L word r L, L r L 1 r li.,. r li, Kraft., α j {l i } j. Step 1 Step 2 Step 3 r L α 1 r, alphabet A α 1, codeword. 2 codeword x = x 1 x 2 x 1 1 codeword, r 2 α 1 r.,. α 2 r 2 α 1 r 3 codeword x = x 1 x 2 x 3, x 1 1 codeword, x 1 x 2 2 codeword, r 3 r(α 1 r α 2 ).,. α 3 <r 3 α 1 r 2 α 2 r, α 1 r n 1 + α 2 r n α n r n. r n, r + + α n r n 1, Kruft. α 1,. Theorem (McMillan s Theorem). C, Kraft.

36 December 3, : Huffman Definition C r alphabet, C = {c 1,,c n }, p(c i )=p i., MinAveLen r (p 1,,p n )., {p i } n, MinAveLen r(p 1,,p n ). Definition optimal encoding scheme, {p i } n, MinAveLen r(p i,,p n )= AveLen(c 1,,c n )= p i len(c i )., Huffman.,,. Example , 0.20, 0.18, 0.13, 0.10, 0.06, 0.05, 0.03, 0.01., 0.24, 0.20, 0.18, 0.13, 0.10, 0.06, (0.05, 0.03, 0.01) 0.24, 0.20, 0.18, (0.13, 0.10, 0.09, 0.06) 0.38, 0.24, 0.20, 0.18, 0.38 (0), 0.24(1), 0.20(2), 0.18(3) 0.24(1), 0.20(2), 0.18(3), (0.13(00), 0.10(01), 0.09 (02), 0.06(03)) 0.24(1), 0.20(2), 0.18(3), 0.13(00), 0.10(01), 0.06(02), (0.05(020), 0.03(021), 0.01(022)).,, p i,.,. reduction, alphabet. Algorithm (Huffman Encoding). P =(p 1,,p n ) r alphabet. 1., n r, C =(0,,n 1). 2. n>r, (a) P, p 1 p 2 p n. (b) s n (mod r 1), 2 s r 1 2. (c) Q =(p 1,,p n s,q), q = p n s p n. (d) Q,, D =(c 1,,c n s,d). 1 r = 2 (binary encoding), s =2. 2 s r, s = r.

37 December 3, :33 36 (e) C =(c 1,,c n s,d0,d1,,d(s 1)).,, MinAveLen r (p 1,,p n ). Lemma P =(p 1,,p n ), Q =(p 1,,p n s,q), q = p n s p n., MinAveLen r (p 1,,p n ) = MinAveLen r (p 1,,p n s,q)+q. Proof. D =(c 1,,c n s,d), C =(c 1,,c n s,d0,d1,,d(s 1)), n s AveLen(D) = p i len(c i )+qlen(d), n s AveLen(C) = p i len(c i )+ n s = p i len(c i )+ n p i (len(d i )+1) i=n s+1 n i=n s+1 q(len(d i ) + 1) = AveLen(D)+q., D (p 1,,p n s,q), optimal, MinAveLen r (p 1,,p n ) AveLen(C) = AveLen(D)+q = MinAveLen r (p 1,,p n s,q). C optimal, n s MinAveLen r (p 1,,p n s,q) AveLen(D) = p i len(c i )+q(len(d)+1) = n p i len(c i ) q = AveLen(C) q = MinAveLen r (p 1,,p n ) q,, MinAveLen r (p 1,,p n ) MinAveLen r (p 1,,p n s,q)+q., MinAveLen r (p 1,,p n ) = MinAveLen r (p 1,,p n s,q)+q Lemma P =(p 1,,p n ), p 1 p 2 p n., r alphabet optimal encoding scheme C =(c 1,,c n ), s codewords L, d0,d1,,d(s 1)., q = p n s p n, MinAveLen r (p 1,,p n ) = MinAveLen r (p 1,,p n s,q)+q.

38 December 3, :33 37 Proof. P optimal r alphabet, n l n, C =(c 1,,c n ). C optimal, n AveLen(C) = p i len(c i ) = MinAveLen r (p 1,,p n )., C, C,. C p i >p j len(c i ) len(c j )., len(c i ) > len(c j ) c i c j,,., p 1 p n, l 1 l n., k codewords L., l n k <l n k+1 = = l n = L., k s., s codewords L word. Kraft, n K = r li 1., K r L + r (L 1) 1, Kraft, l n = L l n 1=L 1,.,, l 1,,l n 1,l n 1,., 1+r L + r (L 1) <K 1. r L, r L r +1<r L K r L., r L K = r L r + α, α {2, 3,,r} (3.4.1). r L K,. 1. r 1 r K 1, (3.4.1), r L K α (mod r) r L K = n i 1 r L li, u, r u 1 (mod r) 1, r L K n (mod r) 1.

39 December 3, : , α n (mod r) 1. 2 α r s n (mod r) 1, α = s., r L K = r L r + s. 4. r r L r, r L K s (mod r). 5. L l i =0 i>n k, n n k r L K = r L li = r L li + k., r L K k (mod r). 6. s k (mod r)., 2 s r, k 0, k = s, k>r. k 0, r s, k s., k s., s codewords, L. c C len(c) =L codeword, c = dx., d0,d1,,d(s 1) c x 0, 1,,s 1., C, k s, C codeword L., C optimal., L codewords k, s., C codewords, C s. Proof of Algorithm. n. n r, optimal encoding scheme. Q,, Q <n, D optimal., D, AveLen(D) = MinAveLen r (p 1,,p n s,q) = MinAveLen r (p 1,,p n ) q., D, C, AveLen(C) = AveLen(D)+q = MinAveLen r (p 1,,p n )., C optimal. Remark r =2, encoding scheme., s =2. 3.5, (S, P ), Shannon (Noiseless coding theorem). Lemma C =(c 1,,c n ) P =(p 1,,p n )., H r (p 1,,p n ) AveLen(c 1,,c n ), len(c i ) = log r 1/p i.

40 December 3, :33 39 Proof. C, Kraft, n 1 r len(ci) 1., q i =1/r len(ci), Entropy, H r (p 1,,p n )= = n p i log r 1 p i n 1 p i log r n p i log r r len(ci) = = AveLen(C),, p i = q i. q i n p i len(c i ) Theorem (Noiseless coding theorem (Shannon)). P = (p 1,,p n ), H r (p 1,,p n ) MinAveLen r (p 1,,p n ) <H r (p 1,,p n )+1. Proof. C =(c 1,,c n ) MinAveLen r (p 1,,p n ). l i = len(c i ), C Kraft.,,,, l i < log r 1 p i p i <r li 1=, Kraft., n p i < n r li log r 1 p i l i (3.5.1)., C (3.5.1) l i len(c i )=l i, 1 l i < log r +1 p i.,.,. AveLen(C) = n p i l i < log r 1 p i l i < log r 1 p i + 1 (3.5.2) n p i (log r 1 p 1 +1)=H r (p 1,,p n )+1

41 December 3, :33 40,. (S, P), (S n,p n ). S n S n word. P n x = x 1 x n, x i C P n (x) =P (x 1 ) P(x n ).,. Theorem H r (P n )=nh r (P ). Outline of Proof., {x i }, H r (X, Y ) H r (X)+H r (Y ) p(x i )p(y j )=p(x i,y j ).,. Theorem P, H r (P ) 1 n MinAveLen(P n ) H r (P )+ 1 n., (S, P), n optimal encoding, H r (P ).

42 ,,.,.,, CD-R 640 MB , R, Fourier.,,., f : R C, Fourier F(f)(ξ) = f(ξ) = 1 f(x)e xξi dx 2π.,, Fourier F (f)(x) = ˇf(x) = 1 f(ξ)e xξi dξ 2π R R, FF =1, F F =1, F(f) L 2 = f L 2, F(f g) =F(f)F(g)., f g (f g)(x) = R f(x t)g(t) dt, f g,,. f : S 1 C, 2π, f Fourier, f(k) = 1 2π S 1 f(x)e kxi dx, f(x) = 1 2π n= f(k)e kxi

43 December 3, : Fourier f(k), f 2kπ., F, R Fourier, F ({ f})(x) = 1 2π n= f(k)e kxi FF =1, F F =1, F(f)(k) 2 = f 2 L 2, n= F(f g) =F(f)F(g)., Fourier S 1 Fourier S 1 R Fourier,, S 1, R R n 1., G Fourier., G n G = {a 0,a 1,,a n 1 }. Z/nZ, G = Z/nZ., G C(G), { 1 x = y, δ x (y) = 0 x y {δ k } n 1 k=0 n., C(G) n 1 n 1 f 2 = f(k) 2 = f(k), f(k) norm, C(G) Hilbert space, { 1 e 2πi n n k=0 k=0 kx } n 1 k=0 C(G) 2., C(G) f(x) = 1 n 1 c k e 2πi n kx n k=0. G Fourier., G Fourier F, Fourier F F(f)(k) = f(k) = 1 n 1 f(x)e 2πi n kx f C(G), n x=0 F (f)(x) = ˇf(x) = 1 n 1 f(k)e 2πi n kx f C(Ĝ) n 1,. 2, G,. k=0

44 December 3, :33 43., Fourier., Z/nZ, G, R, R Z/nZ., [0, 2π] a k =2kπ/n., C(G) f C(G), C(R) 2π, f C(S 1 ) {a k }., 2π 1/n C(G). Fourier, Fourier , T =2π 3., 1/n, n Hz.,. (2 n 1 1) 2 n 1 1, n., Hz, 16 bits.. 4.2, R 2π Fourier. R 2π S 1, C(S 1 ) R 2π, S 1 f C(S 1 ), Fourier F(f) F(f)(k) = f(k) = 1 2π S 1 f(x)e kxi dx., F(f) k Z., {c(k)} k=, Fourier F (c) F (c)(x) =č(x)= 1 2π k=.,. Theorem f, g C(S 1 ), 1. F F(f) =f, c(k)e kxi 3,,,, 2π. 4 Fourier,,.

45 December 3, : Parseval n 1 k=0 F(f)(k) 2 = f 2 L 2 (S 1), 3. F(f g)(k) =F(f)(k)F(g)(k)., (f g)(x) = S 1 f(x y)g(y) dy.,. Theorem {(2π) 1/2 e 2πi n kx } n 1 k=0 C(S 1 ). f,g = f(k), g(k) dx S Fourier, f C(S 1 ), Fourier F(f) F(f)(k) f Fourier, F(f)(k) Fourier f Fourier., f Fourier,, F(f) Fourier S n (f,x) = 1 2π k n. f, S n (f,x) = 1 f(y)e kyi dye kxi = 1 2π S 2π 1., = 1 2π k n S 1 f(y) k n f(k)e kxi k n S 1 f(y)e k(x y)i dy e k(x y)i dy = 1 S1 f(y) e n(x y)i e (n+1)(x y)i dy 2π 1 e (x y)i e n(x y)i e (n+1)(x y)i = e (n+1/2)(x y)i e (n+1/2)(x y)i sin((n +1/2)(x y)) = 1 e (x y)i e (x y)i/2 e (x y)i/2 sin((x y)/2),. Theorem Fourier S n (f,x) S n (f,x) = f(y)k n (x y) dy =(f K n )(x) S 1.,. K n Dirichlet. K n (x) = 1 sin((n +1/2)x) 2π sin(x/2) 4.3, 1/n.

46 December 3, : ,. Theorem {n 1/2 e 2πi n kx } n 1 k=0 C(G) n 1 f,g = f(k), g(k) k=0. Proof. ω = e 2πi n., fk (x) =ω kx, n 1 n 1 f j,f k = ω jl ω kl = l=0 l=0 ω (j k)l., f j,f k = nδ jk., F. Theorem f, g C(G), 1. F F(f) =f, 2. Parseval n 1 k=0 F(f)(k) 2 = n 1 x=0 f(x) 2, 3. F(f g)(k) =F(f)(k)F(g)(k)., (f g)(x) = n 1 y=0 f(x y)g(y). Proof. 1. Fourier F Fourier F F(f)(k) = 1 n 1 f(x)ω kx, n x=0 F (f)(x) = 1 n 1 f(k)ω kx n k=0., nf F(f)(y) = n 1 k=0 n 1 n 1 = k=0 x=0 n 1 n 1 f(k)ω ky = f(x)ω ky ω kx k=0 x=0 f(x)ω k(y x)., { n 1 n y = x, ω k(x y) = 0 y x k=0

47 December 3, :33 46,,. 2. n 1 n 1 nf F(f)(y) = f(x)ω k(y x) = nf(y) k=0 x=0.,. 3. n 1 n F(f)(k) 2 = f(x)ω kx 2 = n 1 n x=0 n 1 n 1 F(f)(k) 2 = k=0 x=0 y=0 f(x)f(y)ω kx ω ky n 1 n 1 x=0 y=0 n 1 f(x)f(y) k=0 n 1 n 1 nf(f g)(k) = f(x y)g(y)ω kx x=0 y=0 n 1 n 1 = f(x y)ω k(x y) x=0 y=0 ω k(y x) = n f(x) 2 n 1 n 1 g(y)ω ky =( f(z)ω kz )( g(y)ω ky )=F(f)(k)F(g)(k). z=0 y=0 Fourier,. Proposition f C(G), ReF(f)(k) =ReF(f)(n k), ImF(f)(k) = ImF(f)(n k)., π,., f C(S 1 )., f Fourier f(k) = 1 2π S 1 f(x)e kxi dx (4.3.1)., Fourier N S N (f,x) = 1 2π k N., f,. f(k)e kxi (4.3.2)

48 December 3, :33 47 Assumption N>0, S N (f,x) =f(x)., f N +1., k>n f(k) =0, f(k) G =2πZ/NZ g C(G) Fourier., g(x) = 1 N 1 N., G 2πZ/NZ,., (4.3.2),., (4.3.3), (4.3.4), k=0 x = 2πl, l =0,,N 1 N f(x) = 1 2π k N f(k)e kxi (4.3.3) f(k)e kxi (4.3.4) N 1 k=0., f(k)e kxi = Ng(x) = 2πf(x), x = 2π l, l =0,,N 1 N., (4.3.4) (4.3.5), f(x) = 1 f(k)e kxi 2π.,, f( 2π N N l)= 2π g(2π l), l =0,,N 1 (4.3.5) N = 1 2πN = 1 N = = f(x) = N 1 l=0 N 1 l=0 N 1 l=0 k N N 1 k N l=0 N 1 f( 2π N f( 2π N l) 1 N k N l=0 f( 2π N l) 1 N K n (x) = 1 N g( 2π 2π l)e N lki e kxi N l)eik(x 2π N l) k N e ik(x 2π N l) sin((x 2π N l)(n +1/2)) sin((x 2π N l)/2) sin(x(n +1/2)) sin(x/2) (4.3.6) f( 2π N l)k N(x 2π N l)=(f K N )(x) (4.3.7)., Fourier.

49 December 3, :33 48 Remark , f, (4.3.3).,. Theorem (Shannon ). f C(S 1 ) Fourier N +1, f N. f, Fourier N/2.,. Corollary f C(S 1 ), f N, N/2. N, N/2 Nykist., Hz, Hz , KHz, 48 KHz.,,,,.,,,,, Fourier,,.,. 100 Hz.,,,, Fourier,.,,,,. Time Division Multiple Access (TDMA),.,, Fourier. Code Division Multiple Access (CDMA), G = Z/N/Z, f(k) = N 1 x=0 5, 20Hz Hz. f(x)ω kx

50 December 3, :33 49, f(k), N, N 1., N f(k), N 2, N(N 1).,., N =2 m, x k. x = x m 1 2 m x x 0 2 0, a = a m 1 2 m a a , f(x) =f(x m 1,,x 0 ), f(k) = f(k m 1,,k 0 ).,,, f(k m 1,,k 0 )= kx x 0 (k 0 +2k m 1 k m 1 ) +2x 1 (k 0 +2k m 2 k m 2 ) + +2 m 1 x m 1 k 0, (mod 2 m ) 1 x 0=0 K j = k j k j 1 x m 1=1., f 1 (k 0,x m 2,,x 0 )= f 2 (k 0,k 1,x m 3,,x 0 )=..=.. f m (k 0,,k m 1 )= f(x m 1,,x 0 )ω 2m 1 x m 1K0 ω x0km 1 1 x m 1=0 1 x m 2=0 f(x m 1,,x 0 )ω 2m 1 x m 1K 0, f(k 0,x m 2,,x 0 )ω 2m 2 x m 2K 1, 1 f(k 0,,k m 2,x 0 )ω x0km 1 x 0=0 = N f(k m 1,,k 0 ).,, f j (k 0,,k j 1,,x 0 ),,, N, 2N, N, m log 2 N, 2N log 2 N, N log 2 N. N = p 1 p l, N (p 1,,p l ) 6. N = a 0 + a 1 p 1 + a 1 p 1 p a l+1 p 1 p l 6,, j, p j =2., p j.

51 December 3, :33 50 N =8=2 3,. f(0) ω 0 ω 0 ω 0 ω 0 ω 0 ω 0 ω 0 ω 0 f(0) f(4) ω 0 ω 4 ω 0 ω 4 ω 0 ω 4 ω 0 ω 4 f(1) f(2) ω 0 ω 2 ω 4 ω 6 ω 0 ω 2 ω 4 ω 6 f(2) f(6) f(1) = ω 0 ω 6 ω 4 ω 2 ω 0 ω 6 ω 4 ω 2 f(3) ω 0 ω 1 ω 2 ω 3 ω 4 ω 5 ω 6 ω 7 f(4) f(5) ω 0 ω 5 ω 2 ω 7 ω 4 ω 1 ω 6 ω 3 f(5) f(3) ω 0 ω 3 ω 6 ω 1 ω 4 ω 7 ω 2 ω 5 f(6) f(7) ω 0 ω 7 ω 6 ω 5 ω 4 ω 3 ω 2 ω 1 f(7) ω 0 ω ω 0 0 ω ω 0 ω ω 0 0 ω ω 0 ω ω 0 0 ω = 0 0 ω 0 ω ω 2 0 ω ω 0 ω ω 0 0 ω ω 0 ω ω 0 0 ω ω 0 ω ω 0 0 ω ω 0 ω ω 2 0 ω 6 ω ω f(0) 0 ω ω f(1) 0 0 ω ω 0 0 f(2) ω ω 0 f(3) ω ω f(4) 0 ω ω f(5) 0 0 ω ω 6 0 f(6) ω ω 7 f(7) 4.3.5, f(x), g(x), f(x)g(x). 0, f, g N,, N f(x) = a k x k =(a 0,,a N ), k=0 N g(x) = b k x k =(b 0,,b N ) k=0, (fg)(x) = 2N k=0 j=0 k a j b k j x k., a k = b k =0,N<k 2N, f(x) =a =(a 0,,a 2N ), g(x) =b =(b 0,,b 2N )

52 December 3, :33 51, fg k c k., c k = 2N j=0 a j b 2N j c k =(a b)(k)., Fourier, ĉ(k) = (a b)(k) =â(k) b(k),, c k = F (â b)(k).

53

54 53 [1] N. Koblitz,. 1997, Springer-Tokyo. [2] S. Garfinkel, PGP, 1996, O Reilly Japan. [3] S. Roman, Coding and Information Theory, 1992, Springer-Verlag. [4] T.W. Körner,, 1996,.

one way two way (talk back) (... ) C.E.Shannon 1948 A Mathematical theory of communication. 1 ( ) 0 ( ) 1

one way two way (talk back) (... ) C.E.Shannon 1948 A Mathematical theory of communication. 1 ( ) 0 ( ) 1 1 1.1 1.2 one way two way (talk back) (... ) 1.3 0 C.E.Shannon 1948 A Mathematical theory of communication. 1 ( ) 0 ( ) 1 ( (coding theory)) 2 2.1 (convolution code) (block code), 3 3.1 Q q Q n Q n 1 Q

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(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y [ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)

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