package matrix
import (
"math"
"github.com/astaxie/beego"
)
type Matrix4 struct {
Elements [16]float64 `json:"elements"`
}
type SQ struct {
//矩阵结构
M,N int //m是列数,n是行数
Data [][]float64
}
//矩阵定义
func (this*SQ)Set(m int,n int,data []float64) {
//m是列数,n是行数,data是矩阵数据(从左到右由上到下填充)
this.M=m
this.N=n
if len(data)!=this.M*this.N {
beego.Debug("矩阵定义失败")
return
}else {
k := 0
if this.M*this.N == len(data){
for i := 0; i < this.N; i++ {
var tmpArr []float64
for j := 0; j < this.M; j++ {
tmpArr = append(tmpArr, data[k])
k++
}
this.Data = append(this.Data, tmpArr)
}
}else {
beego.Debug("矩阵定义失败")
return
}
}
}
//a的列数和b的行数相等
//矩阵乘法
func Mul(a SQ,b SQ) [][]float64{
if a.M==b.M {
res := [][]float64{}
for i:=0;i<a.M;i++ {
t := []float64{}
for j:=0;j<b.M;j++ {
r := float64(0)
for k:=0;k<a.M;k++ {
r += a.Data[i][k]*b.Data[k][j]
}
t = append(t, r)
}
res = append(res,t)
}
return res
}else {
beego.Debug("两矩阵无法进行相乘运算")
return [][]float64{}
}
/*一个应用的例子
a := [][]int{
{1,2},
{3,4},
{5,6},
}
b := [][]int{
{1,2,3},
{3,4,1},
}
A := SQ{
2,3,
a,
}
B := SQ{
3,2,
b,
}
res := mul(A,B)
*/
}
//计算n阶行列式(N=n-1)
func Det(Matrix [][]float64,N int) float64 {
var T0,T1,T2,Cha int
var Num float64
var B [][]float64
if N>0 {
Cha=0
for i := 0; i < N; i++ {
var tmpArr []float64
for j := 0; j < N; j++ {
tmpArr = append(tmpArr, 0)
}
B = append(B, tmpArr)
}
Num=0
for T0=0;T0<=N;T0++{ //T0循环
for T1=1;T1<=N;T1++ { //T1循环
for T2=0;T2<=N-1;T2++ { //T2循环
if T2==T0 {
Cha = 1
}
B[T1-1][T2]=Matrix[T1][T2+Cha]
}//T2循环
Cha=0
}//T1循环
Num=Num+Matrix[0][T0]*Det(B,N-1)*math.Pow(-1,float64(T0))
}//T0循环
return Num
}else if N==0 {
return Matrix[0][0]
}
return 0
}
//矩阵求逆(N=n-1)
func Inverse(Matrix [][]float64,N int) (MatrixC [][]float64) {
var T0,T1,T2,T3 int
var B [][]float64
for i := 0; i < N; i++ {
var tmpArr []float64
for j := 0; j < N; j++ {
tmpArr = append(tmpArr, 0)
}
B = append(B, tmpArr)
}
Chay := 0
Chax := 0
var add float64
add = 1/Det(Matrix,N)
for T0=0;T0<=N;T0++{
for T3=0;T3<=N;T3++{
for T1=0;T1<=N-1;T1++{
if T1<T0 {
Chax = 0
}else {
Chax = 1
}
for T2=0;T2<=N-1;T2++{
if T2<T3 {
Chay = 0
}else{
Chay = 1
}
B[T1][T2]=Matrix[T1+Chax][T2+Chay]
}//T2循环
}//T1循环
Det(B,N-1)
MatrixC[T3][T0]=Det(B,N-1)*add*(math.Pow(-1, float64(T0+T3)))
}
}
return MatrixC
}