Hybrid argument for repeated but alternating sequence

Question

In the usual hybrid argument, it is shown that if two efficiently samplable distributions, $X$ and $Y$, are indistinguishable given a single sample, then they remain indistinguishable with any polynomial number of samples (say $N$).

This is proved by defining two sequences (For independent RVs): $X_0, X_1, ..., X_N$ and $Y_0, Y_1, ..., Y_N$, and the $N$ hybrids $H_k = X_0, X_1, ..., X_k, Y_{k+1}, Y_{k+2}, ..., Y_N $, are all indistinguishable — the usual proof creates these sequences and fills the "middle point" ($k$-th index) with the single (unknown) sample that we are given and uses the distinguisher on this created hybrid.

[EDIT: I edited the formulation of the question below to make everything more clear. I thank @Marc Ilunga and @ellipsoid for the feedback.]

My question is as follows: Let $R \in \{0,1\}^N$ be any possible $N$-bit string. Consider the hybrids $H_R = (Z_1,Z_2,...,Z_N)$ (recall $N$ is polynomial in some security parameter) where $Z_i=X$ if $R_i=0$ and $Z_i=Y$ if $R_i=1$ for all $i\in [N]$. There are $2^N$ possible hybrids, all of polynomial length $N$. From the classical hybrid argument formulation, I know that $H_{0^N}$ is indistinguishable from $H_{1^N}$ and indistinguishable from $H_{0^k || 1^{N-k}}$ for all $k$ ($||$ is bit-string concatenation). Now I have introduced these new alternating bit sequences where every position can be an $X$ or a $Y$. Are hybrids $H_R$ indistinguishable for all bit strings $R$ (i.e., any $H_R$ is indistinguishable from all other $H_{R'}$)?

Answer 1

Yes, fixing any $R$ and $R'$ the resulting distributions $H_R$ and $H_{R'}$ are indistinguishable. You can see (and prove) this similar to the "original" hybrid argument. Beginning with $H_R$ as $H_0$ replace/flip from left to right the random variables which are different from $H_R'$. More formally you can define $\Delta=R\oplus R'$ and at the 1st position $i$ of $\Delta$ which equals $1$ you replace Y by X or X by Y to get $H_1$. Proceed until the last (k-th) occurrence which leads to $H_k=H_{R'}$. A chain of hybrids where $H_i$ and $H_{i+1}$ are indistinguishable.