Suppose we have $N$ integers $a_1,a_2,\dots,a_N$, Given $M$ more integers $b_1,b_2,\dots,b_M$($b_i$ is calculated from $a_1\dots a_n$ by some ways)
Now remove any $M$ numbers from $a_1,a_2,\dots,a_N, b_1,b_2,\dots,b_M$, I want to recover $a_1,a_2,\dots,a_N$
My question is, Can I find a way to calculate such $b_1,b_2,\dots,b_m$?
For example, suppose $M=1$, we can calculate $b_1$ as $ b_1=a_1\oplus a_2\oplus\dots\oplus a_N $ so if $a_i$ is missing ,we just need to XOR $b_1$ and left $a_i$.
For any $M$, my idea is to make $b_i$ as a linear combinations of $a_i$, that is $b_i = \sum_{j=1}^{N}k_{ij}a_j, 1\le i\le M$
Define A as a $(M+N)\times N$ matrix $ A = \left[ \begin{array}{cccc} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \vdots& \vdots& & \vdots\\ 0 & 0 & \dots & 1\\ k_{11}& k_{12} &\dots & k_{1N}\\ \vdots& \vdots & & \vdots\\ k_{M1}& k_{M2} &\dots& k_{MN} \\ \end{array} \right] $ The first $N$ rows form an identity matrix $I_N$
The problem is to find $k_{ij}$, such that remove any M rows of $A$, the left $N\times N$ matrix is still full rank.
I'm not sure whether we define $k_{ij}=i^{j-1}$ will work .