I am trying to understand amplitude amplification and I am able to find two formulations (almost identical).
(A) No ancilla
Usually, and what I do understand quite well, is that you start with the equal superposition state $$ |\psi_0\rangle = \frac{1}{\sqrt{N}}\sum_i |i\rangle $$ or otherwise by projecting to the "marked" subspace (say its only one state) $|m\rangle$ and its orthogonal completion $|k\rangle$ such that $\langle k | m \rangle = 0$, \begin{align} |\psi_0\rangle &= \frac{\sqrt{N}-1}{\sqrt{N}}|k\rangle + \frac{1}{\sqrt{N}}|m\rangle \\ &= \cos(\phi/2)|k\rangle + \sin(\phi/2)|m\rangle \end{align}
and then you apply two projections, the "oracle" encoding the Boolean function that does the phase inversion: $$ U_f = (\mathbb{1}-2|m\rangle\langle m|) $$ followed by the diffusion operator that projects back onto $|\psi_0\rangle$ and reads $$ V = (2|\psi_0\rangle \langle \psi_0|-\mathbb{1}). $$ The resulting state after $d$ iterations is $(VU_f)^d|\psi_0\rangle$ and it has achieved to amplify the amplitude of the state essentially by projecting it onto $|m\rangle$.
(B) With ancilla
However, in a very similar construction (see this paper for example) I read that one can directly start with the state in $n+1$ qubits (before we only had $n$) \begin{align} |\psi_0\rangle &= \mathcal{A}|0\rangle^n|0\rangle \\ &=\cos(\phi)|k\rangle\otimes|0\rangle + \sin(\phi)|m\rangle \otimes|1\rangle \\ &=\cos(\phi)|k\rangle |0\rangle + \sin(\phi)|m\rangle |1\rangle \end{align} (not sure why the angle discrepancy!) for some operator $\mathcal{A}$ that does the job of projecting into the $\{|k\rangle, |m\rangle \}$ space, and then apply $$ Q = \mathcal{A}S_0 \mathcal{A}^{-1}S_\chi $$ where $S_0 = (\mathbb{1}-2|0\rangle^{n+1}\langle 0|^{n+1})$ and $S_\chi= (\mathbb{1}^{\otimes n} \otimes Z)$. It is clear that $Z$ acting on the last qubit it will change its sign (sends $|0\rangle \to |0\rangle$ and $|1\rangle$ to $-|1\rangle$). Thus, we start from \begin{align} |\psi_0 \rangle = \cos(\phi)|k\rangle |0\rangle + \sin(\phi) |m\rangle |1\rangle \end{align} and applying $S_\chi$ changes is this to \begin{align} |\psi_0 \rangle = \cos(\phi)|k\rangle |0\rangle -\sin(\phi) |m\rangle |1\rangle \end{align} Then applying $\mathcal{A}^{-1}$ takes this back to (what???) and finally $S_0$ it projects this onto $|m\rangle$.
Overall we have $$ Q^d|\psi_0\rangle = \cos((2m+1)\phi)|k\rangle + \sin((2m+1)\phi)|m\rangle $$ which amplifies a lot the second summand.
Both (A) and (B) do the same. So the question is what is the difference really? Especially for (B) I do not see how the operator $\mathcal{A}^{-1}$ is applied.
Why do we need the extra ancilla state if the former approach does the same job?