No.1240 
Original Fairy problems 
No.1240 Dmitri Turevski 
Solutions: (click to show/hide) 
white Kd3 Rg8
black Pg2e2 Kh3
hs#5.5 2 solutions (2+3) 

No.1240 
Original Fairy problems 
No.1240 Dmitri Turevski 
Solutions: (click to show/hide) 
white Kd3 Rg8
black Pg2e2 Kh3
hs#5.5 2 solutions (2+3) 
1...g2g1=Q 2.Rg8g4 Qg1d1 + 3.Kd3e3 Kh3h2 4.Ke3f2 e2e1=B + 5.Kf2f1 Kh2h1 {
}6.Rg4h4 + Be1*h4 # {(Indian)}
1...g2g1=R 2.Kd3e3 Rg1g4 3.Ke3f2 Rg4h4 4.Kf2g1 e2e1=S 5.Kg1h1 Se1f3 {
}6.Rg8g3 + Kh3*g3 # {(Herlin)
(C+ by Popeye 4.79)}

Terrific AUW with so few pieces, no fairy piece or condition but lot of moves, congrats Dmitri !
This is so beautiful !!
Apparently the first 5man!!
A terrific find! Either solution would make a good problem on its own. In the first solution 1…g1=Q puts a guard on e3, preventing 2.Ke3??. Then 4…e1=B must be delayed until after 4.Kf2. It is also neat how the bQ guards h5, so the check must be on h4. In the second solution the bR must unguard g1 as quickly as possible, which forces the black move order. Then Black must guard g1, but 4…e1=B? 5.Kh1 Bf2 6.Rg3+ Bxg3!
Very satisfying! Congratulations: this manymover discovery shows the control and flexibility coming from hs# aim.
AUw theme Record: P0576117
There’s at least one 3piece AUW: P1111864.
Change the stipulation to zd81 and you can remove both kings 🙂