Bicliques for permutations So I'm trying to understand the complexity of the biclique attack on Grostl and I have some questions:

1) The paper introduces sliced bicliques where on Grostl one constructs a biclique of dimension 1 and accordingly, we get 4 $Q's$ at state#5 and 4 $P's$ at state#8. What I understood is that each pair of $Q's$ should hash to the same value at the matching point (state#13), similarly for the $P's$. My question is how is this possible if we have 4 different $Q's$? My understanding is that can happen if we consider only the parts at state#13 that are not affected by the difference (white bytes), is my undarstanding correct?

2) The complexity of the attack should be equal to #bicliques(cost of biclique construction + cost of computing forward and backwards to the matching point + cost of checking for a full match if partial matching took place), right?

3) Cost of biclique construction is the $2^{70}$ for the super Sbox solutions and consequently using the free 48 bytes in the $Q's$, we can have $2^{384}$ bicliques without the need to construct new one. However I cannot understand the $2\times(8 + 16 + 2 + 7 + 56 + 8) = 194$ Sboxes for both the forward and backward recomputation. I understand that we're supposed to recompute the white bytes that affects the 4 leftmost columns white bytes at the matching states only but howcome there are 6 terms in the above computation althought we only have 2 rounds in each direction after the biclique? and how are the 194 Sboxes equal to $2^{-3}$ call of the permutation if the 10-rounds permutation has 640 Sbox lookups in Total?

4) What about the complexity of checking for the full match if we get a partial match?

Thanks in advance.
A sliced biclique over $g$ is two set of states ${Q_{i,j}}$ and ${P_{i,j}}$ such that for every pair $(i,j)$
$$
P_{i,j} = Q_{i,j}.
$$
The states are chosen so that the difference between $Q_{i,j}$ for fixed $i$ is small, and the same must hold for $P_{i,j}$ and fixed $j$.

$Q$ and $P$ map to the same hash value $H$, but it is not state #13, but rather
$$
H = \#1\oplus\#13.
$$
You use this equation to filter out incompatible pairs $(i,j)$. The entire trick is to treat some $Q$ as the same, and similarly with $P$ to reduce the computations.

So to answer your other questions

2) Right

3) These are the numbers of recomputed S-boxes in each round from 1 to 6th. There are $6*64 = 384$ S-boxes in total, so there must be $2^{-2}$ calls, not $2^{-3}$.