This is part of course on Orthogonal Array.
Suppose that you have given the following equation: $$\sum_{j=1}^{r}2^{k_j} = 2^n$$ Then, Proof that there is a partitions of $A_2^n$ into faces of dimensions $k_1, k_2, ..., k_r$.
For example: $A^3_2$ can be partitioned into 5 faces of dimensions 1,1,1,0,0. by the summation you see that this is true. The question is to prove that we can partitioned the set into r sets. Now, how to prove that? Do I need to show that the equation in the equality is hold for $0 \leq k_j \leq n$ or that is not enough
We can define the faces of dimension as following:
$A_q^n = |^{a_1,a_2, ...,a_n}_{l_1, l_2, ..., l_n}$ where $0 \leq a_i \leq q-1$. So, any codeword is membership of this set. let's see the example.
For example: for $A_2^3$ the faces of dimension:$|^{1,0}_{1,2}$ which is the codeword of first position = 1 and second position = 0. So we have $|^{1,0}_{1,2}=\{(101),(100)\}$
In this question, I'm looking to find intuitive idea of the proof rather than the proof.
Thank you!
Okay, here's a sketch for a proof, using infinite descent (or "inverse induction"): Let $(k_1,\ldots,k_r)$ be a counterexample which is minimal with respect to $r$. Under that condition, show that all $k_i$ must be distinct. From this you derive that $r=1$ and $k_1=n$ which contradicts the assumption. Done.