Cover an unusual board with minimum chess rooks

Question

_{This is an entry for the 17^th Fortnightly Challenge.}

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On this unusual board, how many chess rooks minimum needed to attack all tiles, but they are not attacking each other ? Show us the formation.

Above are chess rooks attack pattern at blue tile.

To help them who can not edit picture use this numbering guide.

Answer 1

Here is a solution with

$5$ rooks $(6, 14, 26, 38, 50)$:

I don't think less can be possible since

$4$ are required to attack the ends of the horizontals which together cannot attack all other cells. See how Yellow is required to attack $(2,3,31,34)$ in the above solution.

Update

Here is code to search for solutions
- it is not generalised to any size, I just hard-coded the "rows" (horizontals, diagonals and anti-diagonals)
- it is zero based so all the numbers are one less that those shown in the question

N_CELLS = 57
ROWS = [{4,5,6,7,8,9,10,11},{19,20,21,22,23,24,25,26},{34,35,36,37,38,39,40,41},{49,50,51,52,53,54,55,56}, # Horizontals
        {50,45},{42,35,52,46,30},{27,37,43,15,20,54,47,31},{32,0,48,5,39,12,44,16,22,56,28},{33,1,41,7,13,17,24,29},{9,18,26,2,14},{11,3}, # Diagonals
        {0,4},{1,19,12,6,15},{34,8,2,13,16,21,27,30},{3,36,49,10,45,14,17,23,28,42,31},{32,38,43,46,18,51,25,29},{40,33,44,53,47},{48,55} # Anti-diagonals
       ]
ALL_CELLS = set(range(N_CELLS))

def makeAttackLookup():
    res = []
    for pos in range(N_CELLS):
        a = set()
        for r in ROWS:
            if pos in r:
                a |= r
        res.append(set(a))
    return res

ATTACK_LOOKUP = makeAttackLookup()

def iterSolutions(stop, curPos=[], curAttacks=set()):
    if len(curAttacks) == N_CELLS:
        yield curPos
    elif len(curPos) < stop:
        for pos in ALL_CELLS - curAttacks:
            if not curPos or pos > curPos[-1]:
                for solution in iterSolutions(stop, curPos + [pos], curAttacks | ATTACK_LOOKUP[pos]):
                    yield solution

Counting solutions (not collapsing into equivalence classes):

>>> c = 0
>>> for solution in iterSolutions(4): c+=1
...
>>> c
0
>>> for solution in iterSolutions(5): c+=1
...
>>> c
162

We can count how many there are up to symmetry by noting that

All solutions of $5$ must use exactly one of the top horizontal that is not an end: $\{6,7,8,9,10,11\}$
A solution using $6$ has a reflection in the vertical using $11$
Likewise for $7,10$ and $8,9$
There are $29$ solutions using $6$, $14$ using $7$ and $38$ using $8$
Total number of solutions up to symmetry is $29+14+38=81$ (half the number counted as expected by the fact there is only one reflection in the board).

Answer 2

I have a solution with 6 rooks: