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Six. Six hyper-Chimeras suffice and are necessary on 20x20. (Six mere super-Chimeras would work likewise.) The problem amounts to placing as many hyper-Chimeras so all the squares are covered and they mutually protect one another. By definition in the last comment, hyper-Chimera is Angel n=4, controlling surrounding set of 9x9 squares from any starting square. 20x20 board is a to t files and 1 to 20 ranks. So, efficiently place hyper-Chimeras protecting themselves orthogonally at e5, e16, p5 and p16. That leaves corridors of ranks 10 and 11 and files j and k not yet fully covered and just out of those sitting four's reach. It takes two more hyper-Chimera centrally placed at either both j10 and k11 or else both j11 and k10. Strongish to say the least, enemy Angel n=19 reaches every square singly on the board any turn at will. We assume alternating turns whilst the hyper-Chimera spread out from a1, a2, a3, a4, a5 and a6. There would be no other solution of so few as 6 pieces only. They have to be e5, e16, p5, p16 plus what is on the central pair. Only the j10/11-k10/11 guarantee both full rank/file coverage of the gaps, self-preservation, and triumph over the Angel. /// The other question is how many moves to get there from the a-file positions? Assuming for follow-up related, similar but somewhat weaker super-Chimera, not hyper-Chimera, is more interesting. Any super-Chimera dangling unprotected en route is a loss, sacked at once by opposing Angel n=19; and five pieces left are unable ever to terminate endless deadlock since Angel n=19 here only needs 2 free squares to live forever.