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工程数学(II--1).pdf

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工程 数学 II
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p §?? ? ? 2011c? ' a ?K ? k ? a S8 ? S8 Mobius ?“ 1‰ ‰ ‰′ ′ ′ 1 ? ?? a G=(V,E) ' a,XJ 3 V y' V = X[Y, X∩Y =;,? ?u?? l> e 2 E, ”: ?u X, ?u Y. ' aˇ~P G=(X,E,Y). XJ' a G l > e2E : u,v ,Kfu,vg∩X, fu,vg∩ Y .du ? ?a, !¥^ (x,y)5L ? ^l>. 2‰ ‰ ‰′ ′ ′ 2 ( ) ? M ?E ' a G=(X,E,Y) ,XJ M ¥?? l> e,e 0 ,§ o : p . >8 feg ‰· , ˇ ¥ 3 l>. M = ; · . du M ¥?? l> 3 : , ?– jMj ≤ minfjXj,jYjg. ‰ ‰ ‰′ ′ ′ 3 ( ) ? M ' a G , XJ M ·? jMj . 3IS ¥ 4×6 IS . 41 6 'OIP X =fx 1 ,x 2 ,x 3 ,x 4 g –9 Y =fy 1 ,y 2 ,y 3 ,y 4 ,y 5 ,y 6 g ¥ ×I£ f. y 1 y 2 y 3 y 4 y 5 y 6 x 1 x 2 x 3 x 4 × × × × × × × × ?K·3 ? – ? p ′ ? ‰′' a G = (X,E,Y), ¥ (x i ,y j )2 E ? ^ · ¥ (x i ,y j )?vkB\I£ ×. G ?Aup ′ { . G ·?K ). 4‰ ‰ ‰′ ′ ′ 4 (O O O · · ·? ? ?) M G = (X,E,Y) . P M =E\M. XJ G¥ 3{ ·? P =y 1 x 1 y 2 x 2 ···y p x p v: 1. x i 2X, y i 2Y; 2. P : y 1 ,x p · M ¥l>: ; 3. l> (y i ,x i )2M; 4. l>(x i ,y i+1 )2M. K? P ?u M O ·?. ( ·?) 5XJ 3’u M O ·? P,- M P = n (x i ,y i+1 ): (i=1,2,··· ,p?1) o , M P = n (y j ,x j ): (j =1,2,··· ,p) o ? 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G C X S,? jMj=jSj. ‰ ‰ ‰′ ′ ′ 10 ( I I I£ £ £? ? ?,\ \ \I I I£ £ £ ) M G=(X,E,Y) . X L ?X, Y L ?Y ? X L Y L ¥”: ?\I£”: . ? (x,y):x2X, y2Y I£?,XJ 1. x2X L , y62Y L ,? l> (x,y)2M; 2. x62X L , y2Y L ,? l> (x,y)2M. (x,y) I£?. ?§ \I£ ·: XJ (x,y) 1 ?a.,K y\I£: y(x); XJ (x,y) 1 ?a.,K x\I£: x(y). 11I£ { –£a : // \: 1) G= (X, E, Y); // 2) M: G 1) X ¥ · M ¥l>:”: \IP: x(*); 2) while( G ¥ 3 I£?(x,y) ) { (x,y) \I£; } e?·I£ { 1 ¢~. ¥o l> (x 3 ,y 3 ), (x 4 ,y 4 ), (x 5 ,y 5 ), (x 7 ,y 7 ) M,I£ ?iL?\I£ gS. “! # “! # “! # “! # “! # “! # “! # “! # “! # “! # “! # “! # “! # “! # y 1 y 2 y 3 y 4 y 5 y 6 y 7 x 1 x 2 x 3 x 4 x 5 x 6 x 7 H H H H H H H H H H H H H H @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ (?) (?) (?) 0 0 0 (x 1 ) 1 (x 6 ) 2 (y 4 ) 3 (x 4 ) 4 (y 5 ) 5 (x 5 ) 6 (x 5 ) 7 (x 5 ) 8 (y 3 ) 9 12I£ y(x): x \IP? (y,x)2M. I£ x(y): y \I£? (x,y)2M. (y ·“ I£ ?) I£ {( , Y ¥ \IP”: 8 P Y L , X¥ \IP”: 8 P X L . ‰ ‰ ‰′ ′ ′ 11 I£ {( ,XJ y2Y L ? y · M ¥: . K? y ?:. 3 a¥, y 1 ?:. XI£ ·? 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S CX, M ,?– jMj=jSj. y. ?( ' a {Xe: // {; // \: ' a G = (X, E, Y); // : G ; 1) ˉ G M; 2) while (true) { –(G, M) \N^I£ {; if(G ¥ 3?: y) { l y u£ ? M O ·? P; P ?U M; } else break; } 3) M ‰ ‰ ‰n n n 13 { E, O(jVj 2 jEj). y?: {¥ while 1g? L jVj. while N^I£ { # M. I£ { E, O(jVjjEj). # M E, ? LI£ {. l ( .y . 14 ' a l>?8 u CX” :?8. 14p L?K k n f8 A 1 ,···A n . ¥ A i ? Y. ?·? 3 n p fy 1 ,··· ,y n g ? y i 2 A i (1≤ i≤ n)? ·? ?3z 8 A i ¥ L,? z L U ?E ¥? ‰ ‰ ‰n n n 15 (? ? ?· · ·^ ^ ^ ) XJp L?Kk), K? ?? 1≤i 1 . ? – A i k ?S 2 , l A i i [···[A i k ?S 2 jA i i [···[A i k j≤jS 2 j1. 1 ? /: 3 A;§Q ??k4 , ??k4 . A + = x2X : 3 a2A,a4x du 3 ?u A 4 , jA + j: 1 (k =l) ?1 (l=k+1) 0 ? 43X =f1,2,··· ,ng,(X,\) Mobius…? X z:x4z4y μ(x,z)=δ(x,y), (x4y) X z:x\z,z\y μ(x,z)=δ(x,y), (x\y) - μ(1,y)=μ(y) X d\n μ(d)=0, (n>1) n=p ? . μ(1)+μ(p)=0; μ(p)=?1 n=p 2 μ(1)+μ(p)+μ(p 2 )=0; μ(p 2 )=0 n=p k ,(k≥2) , μ(p k )=0 44 n=pq, p,qp ?. μ(1)+μ(p)+μ(q)+μ(n)=0; μ(n)=1 n=pqr, p,q,rp ? μ(n)=?1 μ(12)=0 μ(n)·?¨…?: μ(mn)=μ(m)μ(n), (m?n) μ(n)= 8 >: 1 n=1 (?1) k n kp ? ?¨ 0 ? μ(d,n)=μ(1, n d )=μ( n d ) 45(2 X ,?) Mobius…? μ(A,B)=(?1) jBj?jAj y?: μ(A,A)=1. ( ??. k =jBj?jAj=1: μ(A,A)+μ(A,B)=0; μ(A,B)=?1=(?1) k k =jBj?jAj=2 μ(A,B)=1=(?1) 2 ? k 8B{. B =Atfx 1 ,x 2 ,··· ,x k g X C:A?C?B μ(A,C)=0 μ(A,B) = ? P C:A?C?B μ(A,C) = ? P C:A?C?B (?1) jCj?jAj = ? P k?1 i=0 k i (?1) i = (?1) k k k = (?1) jBj?jAj 46Mobius ?“ ‰n: (X,4) S8, μ Mobius…?, f :X 7!R. XJ g(x)= X z:z4x f(z) K f(x)= X y:y4x g(y)μ(y,x) y?: - h(x)= P y:y4x g(y)μ(y,x). h(x) = P y:y4x P z:z4y f(z)μ(y,x) = P y:y4x h μ(y,x) P z:z4y f(z) i = P y:y4x [μ(y,x) P z2X ζ(z,y)f(z)] = P y:y4x [ P z2X μ(y,x)ζ(z,y)f(z)] = P z2X h P y:y4x μ(y,x)ζ(z,y) i f(z) = P z2X δ(x,z)f(z) = f(x) 47(2 X ,?) Mobius ?“ g(L)= X K:K?L f(K), L?X K f(K)= X L:L?K g(L)μ(L,K)= X L:L?K g(L)(?1) jKj?jLj 48N‰ n - A 1 ,··· ,A m S f8. X =f1,2,...,mg K =fi 1 ,··· ,i p g?X - K= 8 < : s2S :s62 \ i2K A i ,s2 \ i62K A i 9 = ; = K=A 1 ∩···∩A i 1 ∩···∩A i p ∩···∩A m - f(K)=jKj - g(L)= X K:K?L f(K), K: g(L)=j \ i62L A i j, g(X\L)=j \ i2L A i j 49 Mobius ?“ f(K)= X L:L?K (?1) jKj?jLj g(L) f(X)= X L:L?X (?1) m?jLj j \ j62L A j j f(X)= X J:J?X (?1) jJj j \ j2J A j j jA 1 ∩···∩A m j= X J:J?X (?1) jJj j \ j2J A j j jA 1 ∩···∩A m j=jSj+ X J:J?X,J6=; (?1) jJj j \ j2J A j j jA 1 [···[A m j= X J:J?X,J6=; (?1) jJj?1 j \ j2J A j j jA 1 [A 2 j=jA 1 j+jA 2 j?jA 1 A 2 j jA 1 [A 2 [A 3 j=jA 1 j+jA 2 j+jA 3 j?jA 1 A 2 j?jA 1 A 3 j?jA 2 A 3 j+jA 1 A 2 A 3 j(X,\) Mobius ?“ X =f1,2,···ng XJ: g(n)= X d\n f(d) K f(n)= X d\n μ(d,n)g(d)= X d\n μ(n/d)g(d) 50X =f1,2,··· ,ng. - d\n. P S d n =fk2X :gcd(k,n)=dg X = G d\n S d n , n= X d\n jS d n j ~X: X =f1,2,··· ,12g, 12k 6 ˇf f1,2,3,4,6,12g S 1 12 = f1,5,7,11g S 2 12 = f2,10g S 3 12 = f3,9g S 4 12 = f4,8g S 6 12 = f6g S 12 12 = f12g X =S 1 12 tS 2 12 tS 3 12 tS 4 12 tS 6 12 tS 12 12 51 .…? φ(n) φ(n)=jS 1 n j=jfk :1≤k≤n,k?ngj jS d n j=jS 1 n/d j=φ(n/d) n= X d\n φ(n/d)= X d\n φ(d) Mobius ?“ φ(n)= X d\n μ(n/d)d= X d\n μ(d)n/d - n=p α 1 1 p α 2 2 ···p α k k d=p β 1 1 p β 2 2 ···p β k k (0≤β k ≤α k ) k 0≤β k ≤1 , μ(d)6=0. φ(n)=n?( n p 1 +···+ n p k )+( n p 1 p 2 +···)+(?1) k n p 1 ···p k φ(n)= n Y p\n1? 1 p ! 52
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