離散数理工学 第 2回 数え上げの基礎：漸化式の立て方
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- こうだい すわ
- 5 years ago
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1 :29 ( ) (2) / 45
2 ( ) 1 (10/6) ( ) (10/13) 2 (10/20) 3 ( ) (10/27) (11/3) 4 ( ) (11/10) 5 (11/17) 6 (11/24) 7 (12/1) 8 (12/8) ( ) (2) / 45
3 ( ) 9 (12/15) (12/22) 10 (1/5) 11 ( ) (1/12) 12 ( ) (1/19) 13 ( ) (1/26) 14 ( ) (2/2) (2/9) (2/16?) ( ) (2) / 45
4 ( ) (2) / 45
5 1 2 3 ( ) (2) / 45
6 (V, E) V E 2 V V 2 V = {1, 2, 3, 4, 5} E = {{1, 2}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {4, 5}} {2, 5} = {5, 2} ( ) ( ) (2) / 45
7 V = {1, 2, 3, 4, 5} E = {{1, 2}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {4, 5}} ( ) (2) / 45
8 G = (V, E) V G E G V G E G {u, v} E u, v v e v e u v u v V = {1, 2, 3, 4, 5} E = {{1, 2}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {4, 5}} 2, 3 {2, 3} {2, 3} ( ) (2) / 45
9 ( ) (2) / 45
10 ( ) (2) / 45
11 G = (V, E) G I V 2 u, v I {u, v} E ( ) (2) / 45
12 ( ) 22 ( ) (2) / 45
13 22 ( ) (2) / 45
14 G = (V, E) v V σ v {0, 1} σ v = 0 v σv = 1 v σ v = 1 v V = ( ) (2) / 45
15 P 1 P 2 P 3 P 4 P 5 P n ( ) (2) / 45
16 n ( ) (2) / 45
17 P 5 2 (A) = P 4 (B) = 2 P 3 { } P 5 = P 4 + P 3 ( ) (2) / 45
18 P 5 2 (A) = P 4 (B) = 2 P 3 { } P 5 = P 4 + P 3 ( ) (2) / 45
19 P 5 2 (A) = P 4 (B) = 2 P 3 { } P 5 = P 4 + P 3 ( ) (2) / 45
20 P 5 2 (A) = P 4 (B) = 2 P 3 { } P 5 = P 4 + P 3 ( ) (2) / 45
21 P 5 2 (A) = P 4 (B) = 2 P 3 { } P 5 = P 4 + P 3 ( ) (2) / 45
22 P 5 2 (A) = P 4 (B) = 2 P 3 { } P 5 = P 4 + P 3 ( ) (2) / 45
23 ( ) P n 2 ( n 3) (A) = P n 1 (B) = 2 P n 2 { } n 3 P n = P n 1 + P n 2 ( ) (2) / 45
24 ( ) P n 2 ( n 3) (A) = P n 1 (B) = 2 P n 2 { } n 3 P n = P n 1 + P n 2 ( ) (2) / 45
25 ( ) P n 2 ( n 3) (A) = P n 1 (B) = 2 P n 2 { } n 3 P n = P n 1 + P n 2 ( ) (2) / 45
26 a n = P n 2 (n = 1 ) a n = 3 (n = 2 ) a n 1 + a n 2 (n 3 ) ( ) (2) / 45
27 P n P 2 (G n ) G 1 G 2 G 3 G 4 G 5 G n ( ) (2) / 45
28 P n P 2 G n 2 (A) = (B) = 3 { }.... G k ( ) (2) / 45
29 P n P 2 G n 2 (A) = (B) = 3 { }.... G k ( ) (2) / 45
30 P n P 2 G n 2 (A) = (B) = 3 { }.... G k ( ) (2) / 45
31 P n P 2 (H n ) H 1 H 2 H 3 H 4 H 5 H n n 2 G n = H n + H n 1 ( ) (2) / 45
32 P n P 2 H n 2 (A) = (B) = 2 { }.... n 2 H n = G n 1 + H n 1 ( ) (2) / 45
33 P n P 2 H n 2 (A) = (B) = 2 { }.... n 2 H n = G n 1 + H n 1 ( ) (2) / 45
34 P n P 2 H n 2 (A) = (B) = 2 { } G n 1.. H n 1.. n 2 H n = G n 1 + H n 1 ( ) (2) / 45
35 P n P 2 H n 2 (A) = (B) = 2 { } G n 1.. H n 1.. n 2 H n = G n 1 + H n 1 ( ) (2) / 45
36 P n P 2 b n = G n c n = H n b n = c n = { 3 (n = 1 ) c n + c n 1 (n 2 ) { 2 (n = 1 ) b n 1 + c n 1 (n 2 ) ( ) (2) / 45
37 1 2 3 ( ) (2) / 45
38 A 1: def fnct(n) 2: print "a" 3: if n > 2 4: fnct(n-1) 5: fnct(n-2) 6: end 7: end fnct(n) a ( ) (2) / 45
39 n a n a n a n a ( ) (2) / 45
40 A 1: def fnct(n) 2: print "a" 3: if n > 2 4: fnct(n-1) 5: fnct(n-2) 6: end 7: end f n = fnct(n) a ( ) (2) / 45
41 A 1: def fnct(n) 2: print "a" 3: if n > 2 4: fnct(n-1) 5: fnct(n-2) 6: end 7: end 2 n 1 a 4 5 ( ) (2) / 45
42 A 1: def fnct(n) 2: print "a" 3: if n > 2 4: fnct(n-1) 5: fnct(n-2) 6: end 7: end f n = { 1 (n 2 ) 1 + f n 1 + f n 2 (n 3 ) ( ) (2) / 45
43 1: def gcd(a, b) # precondition: a >= b 2: print "G" 3: if b == 0 4: return a 5: else 6: gcd(b, a % b) 7: end 8: end ( ) a % b = a b ( a mod b ) gcd(a, b) G ( ) ( ) (2) / 45
44 (1) a b G ( ) (2) / 45
45 (2) a b G ( ) (2) / 45
46 1: def gcd(a, b) # precondition: a >= b 2: print "G" 3: if b == 0 4: return a 5: else 6: gcd(b, a % b) 7: end 8: end g n = max {gcd(a, b) G } a 1,b n g n = b n g n ( ) (2) / 45
47 a, b 1 a b a a mod b 2 a = bq + r ( 0 r < b) a mod b = r a b q 1 a a b r < b 2 2 a a a b > r = a bq a b < a = a r 2 ( ) n n n 2 = n 2 ( ) (2) / 45
48 a, b 1 a b a a mod b 2 a = bq + r ( 0 r < b) a mod b = r a b q 1 a a b r < b 2 2 a a a b > r = a bq a b < a = a r 2 ( ) n n n 2 = n 2 ( ) (2) / 45
49 a, b 1 a b a a mod b 2 a = bq + r ( 0 r < b) a mod b = r a b q 1 a a b r < b 2 2 a a a b > r = a bq a b < a = a r 2 ( ) n n n 2 = n 2 ( ) (2) / 45
50 a, b 1 a b a a mod b 2 a = bq + r ( 0 r < b) a mod b = r a b q 1 a a b r < b 2 2 a a a b > r = a bq a b < a = a r 2 ( ) n n n 2 = n 2 ( ) (2) / 45
51 a, b 1 a b a a mod b 2 a = bq + r ( 0 r < b) a mod b = r a b q 1 a a b r < b 2 2 a a a b > r = a bq a b < a = a r 2 ( ) n n n 2 = n 2 ( ) (2) / 45
52 1: def gcd(a, b) # precondition: a >= b 2: print "G" 3: if b == 0 4: return a 5: else 6: gcd(b, a % b) 7: end 8: end g n = gcd(a, b) G a, b... ( b = n) ( ) (2) / 45
53 (1) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G a mod b = 0 g n = 2 ( gcd(b, a mod b) ) a mod b 0 ( ) (2) / 45
54 (1) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G a mod b = 0 g n = 2 ( gcd(b, a mod b) ) a mod b 0 ( ) (2) / 45
55 (1) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G a mod b = 0 g n = 2 ( gcd(b, a mod b) ) a mod b 0 ( ) (2) / 45
56 (1) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G a mod b = 0 g n = 2 ( gcd(b, a mod b) ) a mod b 0 ( ) (2) / 45
57 (1) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G a mod b = 0 g n = 2 ( gcd(b, a mod b) ) a mod b 0 ( ) (2) / 45
58 (1) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G a mod b = 0 g n = 2 ( gcd(b, a mod b) ) a mod b 0 ( ) (2) / 45
59 (2) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G = 2 + gcd(a mod b, b mod (a mod b)) G 2 + max a 1,b b/2 {gcd(a, b ) G } = 2 + g b/2 = 2 + g n/2 b mod (a mod b) b 2 n 1 g n 2 + g n/2 ( ) (2) / 45
60 (2) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G = 2 + gcd(a mod b, b mod (a mod b)) G 2 + max a 1,b b/2 {gcd(a, b ) G } = 2 + g b/2 = 2 + g n/2 b mod (a mod b) b 2 n 1 g n 2 + g n/2 ( ) (2) / 45
61 (2) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G = 2 + gcd(a mod b, b mod (a mod b)) G 2 + max a 1,b b/2 {gcd(a, b ) G } = 2 + g b/2 = 2 + g n/2 b mod (a mod b) b 2 n 1 g n 2 + g n/2 ( ) (2) / 45
62 (2) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G = 2 + gcd(a mod b, b mod (a mod b)) G 2 + max a 1,b b/2 {gcd(a, b ) G } = 2 + g b/2 = 2 + g n/2 b mod (a mod b) b 2 n 1 g n 2 + g n/2 ( ) (2) / 45
63 (2) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G = 2 + gcd(a mod b, b mod (a mod b)) G 2 + max a 1,b b/2 {gcd(a, b ) G } = 2 + g b/2 = 2 + g n/2 b mod (a mod b) b 2 n 1 g n 2 + g n/2 ( ) (2) / 45
64 (2) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G = 2 + gcd(a mod b, b mod (a mod b)) G 2 + max a 1,b b/2 {gcd(a, b ) G } = 2 + g b/2 = 2 + g n/2 b mod (a mod b) b 2 n 1 g n 2 + g n/2 ( ) (2) / 45
65 (2) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G = 2 + gcd(a mod b, b mod (a mod b)) G 2 + max a 1,b b/2 {gcd(a, b ) G } = 2 + g b/2 = 2 + g n/2 b mod (a mod b) b 2 n 1 g n 2 + g n/2 ( ) (2) / 45
66 (2) g n = gcd(a, b) G = 1 + gcd(b, a mod b) G = 2 + gcd(a mod b, b mod (a mod b)) G 2 + max a 1,b b/2 {gcd(a, b ) G } = 2 + g b/2 = 2 + g n/2 b mod (a mod b) b 2 n 1 g n 2 + g n/2 ( ) (2) / 45
67 ( ) ( ) { = 1 n = 0 g n 2 + g n/2 n 1 ( ) (2) / 45
68 1: def collatz(n) 2: print n 3: if n % 2 == 0 4: collatz(n/2) 5: else 6: collatz(3*n+1) 7: end 8: end ( ) n collatz(n) 1 n (Oliveira e Silva 10) ( ) (2) / 45
69 1 2 3 ( ) (2) / 45
70 ( ) (2) / 45
71 ( ) (2) / 45
72 ( ) TA OK OK ( ) (2) / 45
73 1 2 3 ( ) (2) / 45
離散数理工学 第 2回 数え上げの基礎：漸化式の立て方
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