(I) GotoBALS, kkimur/charpoly.html 2
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- かずゆき こうじょう
- 9 years ago
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1 sdmp Maple - (Ver.2) ( ) September 27,
2 (I) GotoBALS, kkimur/charpoly.html 2
3 (II) Nehalem CPU GotoBLAS Intel CPU Nehalem CPU, GotoBLAS, Hyper-Thread technology off 3
4 Windows BIOS off Mac Hyper-Thread technology off Mac OS root, terminal, nvram SMT=0 OFF 4
5 Maple,, 4, sdmp rpearcea/ 4,? 5
6 4 (1) (2) (3)Cylindrical Algebraic Decomposition Projection (4)Dixon (5), 6
7 Dixon f(x, y) = x 2 + y 2 3, g(x, y) = xy 1 f(x, y) g(x, y) f(α, y) g(α, y) x α = ( α 1 ) y 1 1 y 3 + 3y = 0 x 1 α 7
8 y = λ, v = x 1, λ 1 1 λ 3 + 3λ v = 0 λ 1 1 λ 3 + 3λ = 0, y,, 3 8
9 (1) (2) 1 (2+3x)(14+15x) x 4+5x (6+7x)(14+15x) 11+12x+13x 3 (2+3x)(16+17x) 6 + 7x (2 + 3x)(4 + 5x) 8(2 + 3x)( x) (9 + 10x)( x + 13x 3 ) (3) a b(4 + 5x) 8c ad( x + 13x 3 ) a = x, b = x, c = x, d = x, 9
10
11 1., O(2 n ), 2. -,???, 3.fraction-free Gauss, O(n 3 ), 4.,???, mod 5.Berkowitz, Fadeev, O(n 4 ), 10
12 fraction-free Gauss n n C, ĉ ( 1) 0,0 = 1, ĉ (0) i,j = C i,j, ĉ (k) i,j = (ĉ(k 1) k,k ĉ (k 1) i,j ĉ (k 1) k,j ĉ (k 1) i,k )/ĉ (k 2) k 1,k 1 k + 1 i n, k + 1 j n, k = 1,, n 1, ĉ n,n (n 1), 0,, Jacobi,, 11
13 fraction-free Gauss n n C, τ 0,0 ( 1) = 1, τ i,j (0) = C i,j, τ i,j (k) = (τ k,k (k 1) τ i,j (k 1) τ k,j (k 1) τ i,k (k 1) )/τ k 1,k 1 (k 2) k + 1 i n, k + 1 j n, k = 1,, n 1, τ (n 1) n,n, 0,, Jacobi,, 12
14 Berkowitz (I) A = T A = A r S R a r+1,r a r+1,r RS a r+1,r RA r 2 r S RA r 3 r S 1 RA r 1 r S RA r 2 r S a r+1,r+1 13
15 Berkowitz (II) [procedure Berkowitz(A)] if dim A=1 then return [1, A 1,1 ] else r (dim A) 1 Decompose A = A r S R a r+1,r+1 Compute T A [X r+1,, X 1 ] Berkowitz(A r ) [X r+2,, X 1 ] T A [X r+1,, X 1 ] return [X r+2,, X 1 ] 14
![R a r+1,r+1 Compute T A [X r+1,, X 1 ] Berkowitz(A r )](/docs-images/49/20185101/images/page_15.jpg "[X r+2,, X 1 ] T A [X r+1,, X 1 ] return [X r+2,, X 1 ]")
16 ? 1., , 15
17 Resultant( ) 1.Collins subresultant
18 f = a 2 x 2 + a 1 x + a 0,g = b 1 x + b 0 Sylvester res x (f, g) = a 2 a 1 a 0 b 1 b b 1 b 0 m n (m + n) (m + n) 17
19 f = a 2 x 2 + a 1 x + a 0,g = b 1 x + b 0 Bezout res x (f, g) = b 0 b 1 b 1 a 0 b 0 a 2 b 1 a 1 m n (m > n) m m 18
20 1., 2. - Quantifier Elimination Q.E., Cylindrical Algebraic Decomposition C.A.D., Resultant( ) 19
21 a 1 a 2 a 3 a 4 b 1 b 2 b 3 b 4 c 1 c 2 c 3 c 4 d 1 d 2 d 3 d 4 = +a 1 (b 2 (c 3 d 4 d 3 c 4 ) c 2 (b 3 d 4 d 3 b 4 ) +d 2 (b 3 c 4 c 3 b 4 )) b 1 (a 2 (c 3 d 4 d 3 c 4 ) c 2 (a 3 d 4 d 3 a 4 ) + d 2 (a 3 c 4 c 3 a 4 )) +c 1 (a 2 (b 3 d 4 d 3 b 4 ) b 2 (a 3 d 4 d 3 a 4 ) +d 2 (a 3 b 4 b 3 a 4 )) d 1 (a 2 (b 3 c 4 c 3 b 4 ) b 2 (a 3 c 4 c 3 a 4 ) + c 2 (a 3 b 4 b 3 a 4 )) (bottom up),! 20
22 - L 1 = a 4,4 a 5,5 a 4,5 a 5,4, L 2 = a 4,3 a 5,5 a 4,5 a 5,3, L 3 = a 4,3 a 5,4 a 4,4 a 5,3, L 4 = a 4,2 a 5,5 a 4,5 a 5,2, L 5 = a 4,2 a 5,4 a 4,4 a 5,2, L 6 = a 4,2 a 5,3 a 4,3 a 5,2, L 7 = a 4,1 a 5,2 a 4,2 a 5,1, L 8 = a 4,1 a 5,5 a 4,5 a 5,1, L 9 = a 4,1 a 5,4 a 4,4 a 5,1, L 10 = a 4,1 a 5,3 a 4,3 a 5,1 det(a) = +(a 1,1 a 2,2 a 1,2 a 2,1 )(a 3,3 L 1 a 3,4 L 2 + a 3,5 L 3 ) (a 1,1 a 2,3 a 1,3 a 2,1 )(a 3,2 L 1 a 3,4 L 4 + a 3,5 L 5 ) +(a 1,1 a 2,4 a 1,4 a 2,1 )(a 3,2 L 2 a 3,3 L 4 + a 3,5 L 6 ) (a 1,1 a 2,5 a 1,5 a 2,1 )(a 3,2 L 3 a 3,3 L 5 + a 3,4 L 6 ) +(a 1,2 a 2,3 a 1,3 a 2,2 )(a 3,1 L 1 a 3,4 L 8 + a 3,5 L 9 ) (a 1,2 a 2,4 a 1,4 a 2,2 )(a 3,1 L 2 a 3,3 L 8 + a 3,5 L 10 ) +(a 1,2 a 2,5 a 1,5 a 2,2 )(a 3,1 L 3 a 3,3 L 9 + a 3,4 L 10 ) +(a 1,3 a 2,4 a 1,4 a 2,3 )(a 3,1 L 4 a 3,2 L 8 + a 3,5 L 7 ) (a 1,3 a 2,5 a 1,5 a 2,3 )(a 3,1 L 5 a 3,2 L 9 + a 3,4 L 7 ) +(a 1,4 a 2,5 a 1,5 a 2,4 )(a 3,1 L 6 a 3,2 L 10 + a 3,3 L 7 ) 21
23 =, Lisp Lisp Hash Hash,? 22
24 index,
25 w[k], index, (2 3 4), c[i][j] = i j 1 n=5; i=3; u=1; b1=n-1; for(k=i;k>0;k--){ for(b2=b1;b2>=w[k];b2--){ u=u+c[b2][k]; } b1=b2-1; } 24
26 geobucket,, 25
27 , 1 heap (sdmp ) (Singular ) (f 1 + f 2 ) (g 1 + g 2 ) = f 1 g 1 + f 1 g 2 + f 2 g 1 + f 2 g 2 26
28 ?, f = 5x 2 y + 6x + 7y + 8 (5, [2, 1]) (6, [1, 0]) (7, [0, 1]) (8, [0, 0]) 27
29 (heap ) X j, Y j,, m n (X 1 + X X m )(Y 1 + Y Y n ) = X 1 (Y 1 + Y Y n ) + X 2 (Y 1 + Y Y n ) + X m (Y 1 + Y Y n ) 28
30 (X 1 Y 1 + X 1 Y X 1 Y n ) + (X 2 Y 1 + X 2 Y X 2 Y n ) + (X m Y 1 + X m Y X m Y n ) Y 1, Y 2,, Y n,, X j,, m 29
31 , =m, 1,, =1, 2, 2 root,, 2, root 2, 2,, 2, 30
32 sdmp, 31
33 heap u = ( a 1 b 1 c 1 ) v = b 2 b 3 c 2 c 3 a 2 a 3 c 2 c 3 a 2 a 3 b 2 b 3 a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 b 2 b 3 c 2 c 3 b 1 a 2 a 3 c 2 c 3 + c 1 a 2 a 3 b 2 b 3 = u v 32
34 u[1] v[1], u[2] v[2], u[3] v[3], u[1] v[1], u[2] v[2], u[3] v[3],, 1 33
35 sdmp Maple, kkimur/susemi etc.html 1.sdmp 2. 1,, 34
36 sdmp sdmp, exponent vector 1 word (pack ) f = 5x 2 y + 6x + 7y + 8 (5, [2, 1]) (6, [1, 0]) (7, [0, 1]) (8, [0, 0]) 35
37 32bit, pack=2 16bit 16bit 2 1,, exponent vector 36
38 sdmp 1 + x 2 + 2x 2 + 2x 4 + 4x = 0,, 37
39 a 1 + a 2 x a 3 + a 4 x a 5 + a 6 x a 7 + a 8 x generic position ( ) 38
40 , O(n 2 ) 39
41 , Linear Assignment Problem 40
42 Hungarian algorithm ,, Hungarian algorithm, O(n 3 ) bound cf. Hungarian algorithm, maple, 41
43 1., kkimur/susemi etc.html 4. kkimur/susemi interpolation.html,, 42
1 Abstract 2 3 n a ax 2 + bx + c = 0 (a 0) (1) ( x + b ) 2 = b2 4ac 2a 4a 2 D = b 2 4ac > 0 (1) 2 D = 0 D < 0 x + b 2a = ± b2 4ac 2a b ± b 2
1 Abstract n 1 1.1 a ax + bx + c = 0 (a 0) (1) ( x + b ) = b 4ac a 4a D = b 4ac > 0 (1) D = 0 D < 0 x + b a = ± b 4ac a b ± b 4ac a b a b ± 4ac b i a D (1) ax + bx + c D 0 () () (015 8 1 ) 1. D = b 4ac
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2007 -1-1 1 1 1 1 12 31 2 2 3 4 -2-5 6 CPU 3 Windows98 1 -3-2. 3. -4-4 2 5 1 1 1 -5- 50000 50000 50000 50000 50000 50000 50000 50000 50000 50000-6- -7-1 Windows 2 -8-1 2 3 4 - - 100,000 200,000 500,000
II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K
II. () 7 F 7 = { 0,, 2, 3, 4, 5, 6 }., F 7 a, b F 7, a b, F 7,. (a) a, b,,. (b) 7., 4 5 = 20 = 2 7 + 6, 4 5 = 6 F 7., F 7,., 0 a F 7, ab = F 7 b F 7. (2) 7, 6 F 6 = { 0,, 2, 3, 4, 5 },,., F 6., 0 0 a F
x V x x V x, x V x = x + = x +(x+x )=(x +x)+x = +x = x x = x x = x =x =(+)x =x +x = x +x x = x ( )x = x =x =(+( ))x =x +( )x = x +( )x ( )x = x x x R
V (I) () (4) (II) () (4) V K vector space V vector K scalor K C K R (I) x, y V x + y V () (x + y)+z = x +(y + z) (2) x + y = y + x (3) V x V x + = x (4) x V x + x = x V x x (II) x V, α K αx V () (α + β)x
A11 (1993,1994) 29 A12 (1994) 29 A13 Trefethen and Bau Numerical Linear Algebra (1997) 29 A14 (1999) 30 A15 (2003) 30 A16 (2004) 30 A17 (2007) 30 A18
2013 8 29y, 2016 10 29 1 2 2 Jordan 3 21 3 3 Jordan (1) 3 31 Jordan 4 32 Jordan 4 33 Jordan 6 34 Jordan 8 35 9 4 Jordan (2) 10 41 x 11 42 x 12 43 16 44 19 441 19 442 20 443 25 45 25 5 Jordan 26 A 26 A1
4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx
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0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9
1-1. 1, 2, 3, 4, 5, 6, 7,, 100,, 1000, n, m m m n n 0 n, m m n 1-2. 0 m n m n 0 2 = 1.41421356 π = 3.141516 1-3. 1 0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c),
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy
Appendix A BASIC BASIC Beginner s All-purpose Symbolic Instruction Code FORTRAN COBOL C JAVA PASCAL (NEC N88-BASIC Windows BASIC (1) (2) ( ) BASIC BAS
Appendix A BASIC BASIC Beginner s All-purpose Symbolic Instruction Code FORTRAN COBOL C JAVA PASCAL (NEC N88-BASIC Windows BASIC (1 (2 ( BASIC BASIC download TUTORIAL.PDF http://hp.vector.co.jp/authors/va008683/
行列代数2010A
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5 / / $\mathrm{p}$ $\mathrm{r}$ 8 7 double 4 22 / [10][14][15] 23 P double 1 $\mathrm{m}\mathrm{p}\mathrm{f}\mathrm{u}\mathrm{n}/\mathrm{a
![5 / / $\mathrm{p}$ $\mathrm{r}$ 8 7 double 4 22 / [10][14][15] 23 P double 1 $\mathrm{m}\mathrm{p}\mathrm{f}\mathrm{u}\mathrm{n}/\mathrm{a](/thumbs/91/105326127.jpg "5 / / $\mathrm{p}$ $\mathrm{r}$ 8 7 double 4 22 / [10][14][15] 23 P double 1 $\mathrm{m}\mathrm{p}\mathrm{f}\mathrm{u}\mathrm{n}/\mathrm{a")
double $\mathrm{j}\mathrm{s}\mathrm{t}$ $\mathrm{q}$ 1505 2006 1-13 1 / (Kinji Kimura) Japan Science and Technology Agency Faculty of Science Rikkyo University 1 / / 6 1 2 3 4 5 Kronecker 6 2 21 $\mathrm{p}$
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1729 1 2016 10 28 1 1729 1111 1111 1729 (1887 1920) (1877 1947) 1729 (, 2016:17) 12 3 1728 9 3 729 1729 = 12 3 + 1 3 = 10 3 + 9 3 (1) 1 1 2 1729 1729 19 13 7 = 1729 = 12 3 + 1 3 = 10 3 + 9 3 13 7 = 91
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