英文:
Finding two Vectors
问题
我有一个大小为100x1的向量S。使用S创建子向量A和B，例如A=S(1:50)和B=S(5:55)。注意A和B中元素的偏移。生成一个结果向量R=a*A + b*B。
现在问题是，已知a、b、R和A与B之间的偏移，如何计算A和B？
英文:
I have a vector S of size say 100x1. Sub-vectors A and B are created using S, say A=S(1:50) and B=S(5:55). Note the shift in the elements in A and B. A resultant vector is generated R=a*A + b*B.
Now the problem is a, b, R and the shift between A and B are given. How to calculate A and B?
答案1
得分: 3
你有一个欠定的方程系统，所以无法得到精确的结果。但Matlab完全能够估计这种方程系统的解决方案：
% 定义参数
start = 10;
offset = 2;
len = 20;
a = 5;
b = 25;
% 从随机数组S创建A和B
S = randi(30,1,100);
A = S(start:start+len-1);
B = S(start+offset:start+offset+len-1);
% 计算R
R = a*A + b*B;
% 创建我们的线性方程系统
M = cat(2,diag(ones(len,1)*a),zeros(len,offset)) + cat(2,zeros(len,offset),diag(ones(len,1)*b));
% 解决方程系统
S_theta = (M\R.').';
% 比较S_theta和S
S_theta(offset+1:end) - B
结果：
请注意，如果满足以下条件，结果会更好：
- 偏移量小（方程系统中未知数较少）
- a和b之间的差异大。
当我们执行：
M\R.'
Matlab解决方程系统 Mx = R，由于M是欠定的（我们有len个方程和len+offset个未知数），Matlab使用QR Solver来找到最小范数残差解决方案，类似于Ax = B类型的问题。
英文:
You have an underdetermined system of equation. So you cannot get the exact result. But matlab is totally able to estimate the solution of such a system of equation:
% Define the parameters
start = 10;
offset = 2;
len = 20;
a = 5;
b = 25;
% Create A and B from a random array S.
S = randi(30,1,100);
A = S(start:start+len-1);
B = S(start+offset:start+offset+len-1);
% We compute R
R = a*A+b*B;
% We create our system of linear equation:
M = cat(2,diag(ones(len,1)*a),zeros(len,offset))+cat(2,zeros(len,offset),diag(ones(len,1)*b));
% We solve the system
S_theta = (M\R.').'
Let's compare S_theta to S:
S_theta(offset+1:end)-B
Result:
Noticed that the result will be better if:
- The offset is small (less unknows in the system of equation)
- The difference between a and b is big.
When we do:
M\R.'
Matlab solve the system Mx = R, since M is underdetermined (we have len equations for len+offset unknows) then matlab use a QR Solver that find minimum-norm-residual solution to Ax = B type of problem.
答案2
得分: 0
您基本上有一个具有变量数量等于numel(R)+shift和等式数量等于numel(R)的线性系统。我认为不可能得到您想要的结果，因为未知变量始终多于等式数目：
% 第一组变量
R(1) = a*S(1)+b*S(5); % 一个等式，两个未知数
R(5) = a*S(5)+b*S(9); % 两个等式，三个未知数
R(9) = a*S(9)+b*S(13); % 三个等式，四个未知数
% 第二组变量
R(2) = a*S(2)+b*S(6); % 四个等式，六个未知数
R(6) = a*S(6)+b*S(10); % 五个等式，七个未知数
% 依此类推
请注意，这些变量和等式之间的数量关系可能会导致无法唯一解决系统。
英文:
You essentially have a linear system with a number of variables equaling numel(R)+shift and a number of equations equaling numel(R). I don't think it's possible to get what you want, since there are always more unknown variables than you have equations:
% First chain of variables
R(1) = a*S(1)+b*S(5); %One equation, two unknowns
R(5) = a*S(5)+b*S(9); %Two equations, three unknowns
R(9) = a*S(9)+b*S(13); %Three equations, four unknowns
% Second chain of variables
R(2) = a*S(2)+b*S(6); %Four equations, six unknowns
R(6) = a*S(6)+b*S(10); %Five equations, seven unknowns
% etc.
答案3
得分: -2
在字符串S的前面（或后面）添加shift个零，将前面的部分乘以a，后面的部分乘以b，然后将它们相加。这将问题简化为子串搜索。你可以采用蛮力搜索或使用更复杂的方法；请参考以下链接：
https://en.wikipedia.org/wiki/String-searching_algorithm
英文:
Slap shift zeroes on the front (resp. back) of S, multiply the former by a and the latter by b, and add them together. This reduces the problem to a substring search. You can either brute force that or use something more sophisticated; see
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