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Blitz++ 矩陣相乘(張量運算) 示例

// 整理?by?RobinKin

// Blitz++?張量計算的示例

/* ****************************************************************************

?*?matmult.cpp?????Blitz++?tensor?notation?example

?*****************************************************************************

?*?This?example?illustrates?the?tensor-like?notation?provided?by?Blitz++.

? */

#include? < blitz / array.h >

#include? < iostream >

using ? namespace ?blitz;

int ?main()

{

???? // ?Create?two?4x4?arrays.??We?want?them?to?look?like?matrices,?so

???? // ?we'll?make?the?valid?index?range?1..4?(rather?than?0..3?which?is

???? // ?the?default).

????Range?r( 1 , 4 );

????Array < float , 2 > ?A(r,r),?B(r,r);

???? // ?The?first?will?be?a?Hilbert?matrix:

???? //

???? // ?a???=???1

???? // ??ij???-----

???? // ???????i+j-1

???? //

???? // ?Blitz++?provides?a?set?of?types?{?firstIndex,?secondIndex,?

?}

???? // ?which?act?as?placeholders?for?indices.??These?can?be?used?directly

???? // ?in?expressions.??For?example,?we?can?fill?out?the?A?matrix?like?this:

????firstIndex?i;???? // ?Placeholder?for?the?first?index

????secondIndex?j;??? // ?Placeholder?for?the?second?index

????A? = ? 1.0 ? / ?(i + j - 1 );

????cout? << ? " A?=? " ? << ?A? << ?endl;

???? // ?A?=?4?x?4

???? // ?????????1???????0.5??0.333333??????0.25

???? // ???????0.5??0.333333??????0.25???????0.2

???? // ??0.333333??????0.25???????0.2??0.166667

???? // ??????0.25???????0.2??0.166667??0.142857

???? // ?Now?the?A?matrix?has?each?element?equal?to?a_ij?=?1/(i+j-1).

???? // ?The?matrix?B?will?be?the?permutation?matrix

???? //

???? // ?[?0?0?0?1?]

???? // ?[?0?0?1?0?]

???? // ?[?0?1?0?0?]

???? // ?[?1?0?0?0?]

???? //

???? // ?Here?are?two?ways?of?filling?out?B:

????B? = ?(i? == ?( 5 - j));????????? // ?Using?an?equation?--?a?bit?cryptic

????cout? << ? " B?=? " ? << ?B? << ?endl;

???? // ?B?=?4?x?4

???? // ?????????0?????????0?????????0?????????1

???? // ?????????0?????????0?????????1?????????0

???? // ?????????0?????????1?????????0?????????0

???? // ?????????1?????????0?????????0?????????0

????B? = ? 0 ,? 0 ,? 0 ,? 1 ,??????????? // ?Using?an?initializer?list

???????? 0 ,? 0 ,? 1 ,? 0 ,???????????

???????? 0 ,? 1 ,? 0 ,? 0 ,

???????? 1 ,? 0 ,? 0 ,? 0 ;

????cout? << ? " B?=? " ? << ?B? << ?endl;

???? // ?Now?some?examples?of?tensor-like?notation.

????Array < float , 3 > ?C(r,r,r);?? // ?A?three-dimensional?array:?1..4,?1..4,?1..4

????thirdIndex?k;????????????? // ?Placeholder?for?the?third?index

???? // ?This?expression?will?set

???? //

???? // ?c????=?a???*?b

???? // ??ijk????ik????kj

? // ???C?=?A(i,k)?*?B(k,j);

???cout? << ? " C?=? " ? << ?C? << ?endl;

???? // ?In?real?tensor?notation,?the?repeated?k?index?would?imply?a

???? // ?contraction?(or?summation)?along?k.??In?Blitz++,?you?must?explicitly

???? // ?indicate?contractions?using?the?sum(expr,?index)?function:

????Array < float , 2 > ?D(r,r);

????D? = ?sum(A(i,k)? * ?B(k,j),?k); // 指標收縮,?計算矩陣積

???? // ?The?above?expression?computes?the?matrix?product?of?A?and?B.

????cout? << ? " D?=? " ? << ?D? << ?endl;

???? // ?D?=?4?x?4

???? // ??????0.25??0.333333???????0.5?????????1

???? // ???????0.2??????0.25??0.333333???????0.5

???? // ??0.166667???????0.2??????0.25??0.333333

???? // ??0.142857??0.166667???????0.2??????0.25

???? // ?Indices?like?i,j,k?can?be?used?in?any?order?in?an?expression.

???? // ?For?example,?the?following?computes?a?kronecker?product?of?A?and?B,

???? // ?but?permutes?the?indices?along?the?way:

????Array < float , 4 > ?E(r,r,r,r);???? // ?A?four-dimensional?array

????fourthIndex?l;???????????????? // ?Placeholder?for?the?fourth?index

????E? = ?A(l,j)? * ?B(k,i); // 指標輪換

// cout?<<?"E?=?"?<<?E?<<?endl;

???? // ?Now?let's?fill?out?a?two-dimensional?array?with?a?radially?symmetric

???? // ?decaying?sinusoid.

???? int ?N? = ? 64 ;??????????????????? // ?Size?of?array:?N?x?N

????Array < float , 2 > ?F(N,N);

???? float ?midpoint? = ?(N - 1 ) / 2 .;

???? int ?cycles? = ? 3 ;

???? float ?omega? = ? 2.0 ? * ?M_PI? * ?cycles? / ? double (N);

???? float ?tau? = ? - ? 10.0 ? / ?N;

????F? = ?cos(omega? * ?sqrt(pow2(i - midpoint)? + ?pow2(j - midpoint)))

???????? * ?exp(tau? * ?sqrt(pow2(i - midpoint)? + ?pow2(j - midpoint)));

// cout?<<?"F?=?"?<<?F?<<?endl;

???? return ? 0 ;

}

// 輸出

A? = ? 4 ?x? 4 [????????? 1 ??????? 0.5 ?? 0.333333 ?????? 0.25 ????????? 0.5 ?? 0.333333 ?????? 0.25 ??????? 0.2 ???? 0.333333 ?????? 0.25 ??????? 0.2 ?? 0.166667 ???????? 0.25 ??????? 0.2 ?? 0.166667 ?? 0.142857 ?]

B? = ? 4 ?x? 4

[????????? 0 ????????? 0 ????????? 0 ????????? 1 ?

?????????? 0 ????????? 0 ????????? 1 ????????? 0 ?

?????????? 0 ????????? 1 ????????? 0 ????????? 0 ?

?????????? 1 ????????? 0 ????????? 0 ????????? 0 ?]

B? = ? 4 ?x? 4

[????????? 0 ????????? 0 ????????? 0 ????????? 1 ?

?????????? 0 ????????? 0 ????????? 1 ????????? 0 ?

?????????? 0 ????????? 1 ????????? 0 ????????? 0 ?

?????????? 1 ????????? 0 ????????? 0 ????????? 0 ?]

D? = ? 4 ?x? 4

[?????? 0.25 ?? 0.333333 ??????? 0.5 ????????? 1 ?

???????? 0.2 ?????? 0.25 ?? 0.333333 ??????? 0.5 ?

??? 0.166667 ??????? 0.2 ?????? 0.25 ?? 0.333333 ?

??? 0.142857 ?? 0.166667 ??????? 0.2 ?????? 0.25 ?]

posted on 2006-07-01 13:24 楊粼波閱讀(1067) 評論(0) 編輯收藏引用

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Blitz++ 矩陣相乘(張量運算) 示例