选择一组对，以最小化组的均方根。

Question

简化问题

我有~40个电阻(所有的值都是+-5%)，我需要选择其中的12个，使它们尽可能相似。

解决方案:我列出它们的顺序，并采取与最小均方根连续12个。

实际问题

我有~40个电阻(所有的值都是+-5%)，我必须选择12对对，使它们的电阻尽可能相似。

备注

这对(R1，R2)的电阻是R1+R2。我并不真正关心编程语言，但假设我正在寻找C++或Python的解决方案，这是我最熟悉的两种语言。

Answer 1

这给出了相当好的结果(在MATLAB中)

a = ones(40,1) + rand(40,1)*0.1-0.05; % The resistors
vec = zeros(40,2);        % Initialize matrix
indices = zeros(40,2);    % Initialize matrix
a = sort(a);              % Sort vector of resistors
for ii = 1:length(a)
  vec(ii,:) = [a(ii) a(ii)];    % Assign resistor values to row ii of vec
  indices(ii,:) = [ii,ii];      % Corresponding resistor number (index)
  for jj = 1:length(a)
    if sum(abs((a(ii)+a(jj))-2*mean(a))) < abs(sum(vec(ii,:))-2*mean(a))
      vec(ii,:) =  [a(ii) a(jj)];    % Check if the new set is better than the
      indices(ii,:) = [ii, jj];      % previous, and update vec and indices if true.
    end
  end
end

[x, idx] = sort(sum(vec')');   % Sort the sum of the pairs 
final_list = indices(idx);     % The indices of the sorted pairs

这就是我绘制它的结果：

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Answer 2

这不是最优的，但应该会带来一些不错的结果。它的速度非常快，所以如果你需要从10000个电阻中选择1000对.

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define GROUPS 12
#define N 40

int compare (const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}

int main ()
{
    // generate random numbers 
    float *values = (float *)malloc(sizeof(float) * N);
    srand(time(0));
    for (int i = 0; i < N; i++)
        values[i] = 950 + rand()%101;

    qsort(values, N, sizeof(float), compare);

    // find "best" pairing
    float bestrms = -1;
    int beststart = -1;
    float bestmean = -1;
    for (int start = 0; start <= N - 2 * GROUPS; start++)
    {
        float sum = 0;
        for (int i = start; i < start + 2 * GROUPS; i++)
            sum += values[i];

        float mean = sum / GROUPS;

        float square = 0;
        for (int i = 0; i < GROUPS; i++)
        {
            int x = start + 2 * GROUPS - 1 - i;
            float first = values[start + i];
            // in a sorted sequence of 24 resistors, always pair 1st with 24th, 2nd with 23rd, etc
            float second = values[start + 2 * GROUPS - 1 - i];
            float err = mean - (first + second);
            square += err * err;
        }

        float rms = sqrt(square/GROUPS);       

        if (bestrms == -1 || rms < bestrms)
        {
            bestrms = rms;
            beststart = start;
            bestmean = mean;
        }
    }

    for (int i = 0; i < GROUPS; i++)
    {             
        float first = values[beststart + i];
        float second = values[beststart + 2 * GROUPS - 1 - i];
        float err = bestmean - (first + second);
        printf("(%f, %f) %f %f\n", first, second, first + second, err);
    }
    printf("mean %f rms %f\n", bestmean, bestrms);
    free(values);
}

Answer 3

排序，然后配对1与2，3与4，5与6，等等。找出每对之间的差异，然后重新排序，选择差异最小的12。