搜索资源列表
originalsimpleM111.m
- 原始单纯形法（大M法，无需给出初始基变量）。输入：C是n维行向量，A是m*n的系数矩阵，b是m维列向量 输出：x最优解（如果有的话），fval最优值，flag解的状态说明，interation求解时的循环次数-The original simplex method (big M method, without giving the initial basic variable). Input: C is the n-dimensional row vector, A is the coef
cubesum
- Suppose there is a X x Y x Z 3D matrix A of numbers having coordinates (i, j, k) where 0 ≤ i < X, 0 ≤ j < Y, 0 ≤ k < Z. Now another X x Y x Z matrix B is defined from A such that the (i, j, k) element of B is the sum of all the the numbers i
lisanshuxushiyan3
- 以偶对的形式输入一个无向简单图的边，建立该图的邻接矩阵，判断图是否连通（A）。并计算任意两个结点间的距离（B）。对不连通的图输出其各个连通支（C）。-Even on the form to input an undirected graph edge, the establishment of the adjacency matrix to determine whether the connectivity graph (A). And calculate any distance betwe
dist
- 开车从起始点Ａ到目的地Ｂ的路线有多条。给你一张描述待选路线的表（n*n的矩阵A），让你找出行车距离最短的路线。表中表示了任意两个路口的连通情况，以及距离。矩阵元素a(i,j)=0表示路口i,j不连通，a(i,j)!=0表示路口i,j的行车距离。其中起始点Ａ在路口1，目的地Ｂ在路口n 。完成源程序DIST.CPP中Dijkstra函数的编写。-A drive to the destination from a starting point a number of B' s line. Giv
lbg
- 本程序可以将一个给定线性系统转换成龙伯格能控/能观II型。此线性系统由A,B,C,D四个矩阵描述-This program can transform a given linear system controllability Jackie Berg/observability II type. This linear system is composed of A, B, C, D four matrix descr iption
givenhousehold
- 利用该程序可计算任意对称矩阵M，给出区间(a,b)上的特征值个数-Calculated using this program any symmetric matrix M, gives the interval (a, b) on the eigenvalues
amaterial5
- 复合材料属性矩阵A，B,D,As的确定，通过这些矩阵可以计算材料刚度矩阵-form material matrix of composite material
assign3_m100295cs
- a) On sample artificial, 8X8 / 16X16 images, take the DFT, DCT, WT, & HT. Print the image & transform matrix side by side b) Repeat the above on real images of size 256X256, and display the transform coefficients as 8-bit intensity images along
Gauss_pivot.m
- Method Gaussian Elimination without pivoting for Linear Systems Solve Ax = b using Gaussian elimination without pivoting Inputs : A is the n-by-n coefficient matrix b is the n-by-k right hand side matrix Outputs : x is the n-by-k
linproj
- 模型是结构体类型的线性投影Y = linproj(X, model) 其中W.model为线性投影矩阵， Y = model.W *X + model.b b为偏差值- Descr iption: Y = linproj(X, model) linearly projects data in X such that Y = model.W *X+ model.b out_data = linproj(in_data, model) projects
MatrixProcessing
- 有两个方阵A、B，编程利用数组，求矩阵A与矩阵B的和矩阵C，并找出和矩阵C主对角线最大元素及其位置。 要求： 1）用自定义函数input实现数组的输入； 2）用自定义函数add实现矩阵的加运算； 3）用自定义函数diag_max实现查找； 4）在主函数输出和矩阵、输出查找结果。 -There are two square A, B, using an array of programming, find the matrix A and matrix B and matr
BytmianFangzhen
- B样条的方阵生成图形，利用MATLAB生成B样条的源码-B spline matrix generated graphics, the use of MATLAB source code generated B-spline
juzhen
- B样条的定义，并通过软件生成输出矩阵的数值-B-spline is defined and the output matrix generated by the numerical software
gaosisaier
- 迭代法求解线性方程组，首先将方程组AX=B中未知数X给定，计算出矩阵B。然后把矩阵B带入，利用雅各比迭代法反求X-Iterative method for solving linear equations, the first of equations AX = B will be in the unknown X given to calculate the matrix B. Then the matrix B into the use of Jacobi iterative method
satu
- int i, j, k,b,n,s,t,r=1 float sum, c float a[12][12] //pengisian matrix printf("Masukkan ordo matriks : ") scanf(" d",&n) printf("masukkan banyak persamaan : ") scanf(" d",&b) for(i=0 i<n i++){ for(j=0 j&
satu
- int i, j, k,b,n,s,t,r=1 float sum, c float a[12][12] //pengisian matrix printf("Masukkan ordo matriks : ") scanf(" d",&n) printf("masukkan banyak persamaan : ") scanf(" d",&b) for(i=0 i<n i++){ for(j=0 j&
konversi-bilangan
- int i, j, k,b,n,s,t,r=1 float sum, c float a[12][12] //pengisian matrix printf("Masukkan ordo matriks : ") scanf(" d",&n) printf("masukkan banyak persamaan : ") scanf(" d",&b) for(i=0 i<n i++){ for(j=0 j&
satu
- int i, j, k,b,n,s,t,r=1 float sum, c float a[12][12] //pengisian matrix printf("Masukkan ordo matriks : ") scanf(" d",&n) printf("masukkan banyak persamaan : ") scanf(" d",&b) for(i=0 i<n i++){ for(j=0 j&
STRASSEN
- 矩阵乘法是线性代数中最常见的运算之一，它在数值计算中有广泛的应用。若A和B是2个n×n的矩阵，则它们的乘积C=AB同样是一个n×n的矩阵-Matrix multiplication is linear algebra is the most common operation, it is one of the numerical calculation is widely used. If A and B is 2 n* n matrix, then their product C = AB i
RM
- 计算两组向量之间的旋转矩阵。输入参数为在两个不同坐标系中的同名向量（模为1），返回参数是一个3X3的旋转矩阵。-The function of RM realize the rotation between two sets of vectors, whose magnitude are 1, defined in two different refrence frames. Input parameters are two sets of vectors r and b, return val