如何从距离矩阵计算原始矢量？
我对矢量和矩阵有一个小问题。
假设向量V = {v1，v2，...，vn}。我生成一个n×n距离矩阵M，定义如下：
M_ij = | v_i - v_j |这样我，j属于[1，n]。
也就是说，方阵中的每个元素M_ij是V中两个元素的绝对距离。
例如，我有一个向量V = {1,3,3,5}，距离矩阵将是
M = [
0 2 2 4;
2 0 0 2;
2 0 0 2;
4 2 2 0; ]
看起来很简单。现在来问题了。给定这样的矩阵M，如何获得初始V？
谢谢。
基于这个问题的一些答案，似乎答案并不是唯一的。所以，现在假设所有初始向量已经归一化为0均值和1个方差。问题是：给定这样一个对称距离矩阵M，如何确定初始归一化向量？
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碉罕城爸
肉簧咸缮
# Assume v[1] is 0 v[1] = 0 # e is value of first non-zero vector element e = 0 # ei is index of first non-zero vector element ei = 0 for i = 2...n: # if all vector elements have been 0 so far if e == 0: # get the current distance from element 1 and its index # this new element may still be 0 e = d[1,i] ei = i v[i] = e elseif d[1,i] == d[ei,i] + v[ei]: # v[i] <= v[1] # v[i] is to the left of v[1] (assuming v[ei] > v[1]) v[i] = -d[1,i] else: # some other case; v[i] is to the right of v[1] v[i] = d[1,i]恋卡
Let v_1 = 0, and let l_k = | v_k | for kin [2,n] and p_k the parity of v_k Let p_1 = 1 for(int i = 2; i < n; i++) if( | l_i - l_(i+1) | != M_i(i+1) ) p_(i+1) = - p_i else p_(i+1) = p_iLet d = Sqrt(v_1^2 + v_2^2 + ... + v_n^2) Vector = {0, v_1 / d, v_2 / d, ... , v_n / d} or {0, -v_1 / d, -v_2 / d, ... , -v_n / d}