René J. Millour
ESP72
h#9 11 black invisible pieces


Solution
1.Ixh6 Bf3 2.Ixg6 Kg4 3.Ixf6 Kf4 4.Ixe6 Ke4 5.Ixd6 Kd4 6.Ixc6 Kc4 7.Ixg2 Kb4 8.Ixh1 Ka4 9.g5 Bh5 #

The given position shows that 5 pieces were captured by the wPs. The 11 other black pieces are present, but invisible.
The wK uses the path Kh4-g4-f4-e4-d4-c4-b4-a4, implying that there is no static Invisible on g5-f5-e5-d5-c5-b5: for example an Invisible is supposed to be at b5, with Kd4 it cannot be S, with Kc4 not K Q B P, with Kb4 not R, in other words the K's march proves that this Invisible cannot exist.

1.Ixh6 [Ih6 not K (self-check) ant not R (no R-move)] 1... Bf3 [Ih6 not Q (self-check)].
2.Ixg6 [This I is new because Ih6 (not Q,R) cannot Ih6xg6] 2... Kg4 [Ih6 not S / Ig6 not Q,R (self-check) because no other I can interfere on g5 (see the preliminary remark above, for example I(S)g5: later with Ke4 this S is invalidated), but Ig6 may be K (coming from f5 before 2...Kg4)].
3.Ixf6 [This I is new: not I(Kg6)xf6 (self-check) / If6 not K (self-check)] 3... Kf4 [Ih6 not B, Ih6 is P / Ig6 not S / If6 not Q,R].
4.Ixe6 [New I / Ie6 not K (would have stayed in check on 7 or 5)] 4... Ke4 [Ig6 not B / If6 not S / Ie6 not Q,R].
5.Ixd6 [New I / Id6 not K (would have stayed in check on 7 or 5)] 5... Kd4 [If6 not B, If6 is P / Ie6 not S / Id6 not Q,R].
6.Ixc6 [New I / Ic6 not K (self-check)] 6... Kc4 [Ie6 not B, Ie6 is P / Id6 not S / Ic6 not Q,R].
7.Ixg2 [New I because Ig6 (not Q,R) cannot Ig6xg2, and Bf3 prevents I(Bc6)xg2. What kind of piece on g2? Not a K (self-check). For I(Ph3)xg2, the P would not come from g7 because 3...Kf4 has established that Ih6 is a P, necessarily coming from g7 by 1.I(Pg7)xh6. This Ph3 would come from h7 after hxgxh in order to allow the march of the wP from h2 to h6, or (if Ig6 is K) from f7 after fxgxh. Two captures seem possible as Bc1 and Pa2 are missing. For that, Pa2 has to promote, but Pa7 also has to promote to be captured by a wP. There is no victim left for a black or white axb (remember, 14 + 2 captures by h/fxgxh = 16, and 0 + 11 Invisibles + 5 captures by wPs = 16), colliding Ps, thus Ig2 is not P. For 7.I(B)xg2, the B at h3 or f1 would be incompatible with 2...Kg4 or 6...Kc4 [of course not 6.I(Bf1)xg2 Kc4 7.Ixc6 Kb4 etc, because Ic6 can be R], thus Ig2 is not B. But Ig2 may be S as I(Se1)xg2 is valid. I(Rf2/c2/b2/a2)xg2 is also valid. The sole square where a Q is compatible with the K's march is b2, and I(Qb2)xg2 is possible. This means Ig2 is S, R or Q] 7... Kb4 [Id6 not B, Id6 is P / Ic6 not S].
8.Ixh1 [Not I(Sf2/g3)xh1 (S incompatible with 4...Ke4). This means the sole piece able to capture on h1 is Ig2. And Ig2xh1 being a diagonal move, Ig2 is not S and not R. After the precise moves 7-8.Ib2xg2xh1, Ih1 is Q. Now we are sure that the g-file can be opened, and that it has really been opened. Therefore, Ig6 is not K (self-check by Rg1), Ig6 is P] 8... Ka4 [Ic6 not B, Ic6 is P].
An Invisible is Q on h1 and 6 Invisibles are Ps having just made g7xh6, f7xg6, e7xf6, d7xe6, c7xd6 and b7xc6. Therefore, the 4 last Invisibles are necessarily Bf8, Bc8, Ke8 and Qd8 because, with bPs on g7-f7-e7-d7-c7-b7, these 4 pieces never could leave their home-squares. Hence the continuation: 9.g5 Bh5 #.
It can be added that the Q on h1 is necessarily the promoted bPa (remember, bPh cannot promote for hxgxh reasons). The wPa could sacrifice to allow the promotion, and the pieces captured by the wPs were 2 Ss, 2 Rs and Ph, this one disappearing on g after hxg taking the Bc1. The g-file can also be opened by 8.Ig2xh2, but no dual here: in fact no mate because, necessarily Q or R, Ih2 guards h5. Why not 1.Ixh6 Kg4 2.Ixg6, White cannot 2...Bf3 and must play 2...Kf4 and Ig6 can be R. Why not 1.Ixg6 Kg4 2.Ixh6, once more Bh5 is locked, White must play 2...Kf4 and Ih6 can be S. Why not 1.Ixg6 Bf3 2.Ixh6, because Ig6 can be R and 2.I(Rg6)xh6+ Kg4 is possible.
ESP72
Final position, showing the 11 black Invisibles