No.1363 
Original Fairy problems 
Definitions: (click to show/hide)
No.1363 Claude Beaubestre 
Solution: (click to show/hide) 
White Qa8 Ba7 Ka6
Black Ke8 Bf2 Qh1
serZ(b8)13 (3+3) 

No.1363 
Original Fairy problems 
Definitions: (click to show/hide)
Stipulation serZ(b8)13: White play 13 serial moves in order to reach the square b8.
Madrasi: Units, other than Kings, are paralysed when they attack each other. Paralysed units cannot move, capture or give check, their only power being that of causing paralysis.
No.1363 Claude Beaubestre 
Solution: (click to show/hide) 
White Qa8 Ba7 Ka6
Black Ke8 Bf2 Qh1
serZ(b8)13 (3+3) 
1.Ka6b5 2.Kb5c4 3.Kc4d3 4.Kd3e2 5.Ke2f1 6.Kf1g1 7.Kg1h2 {
} 8.Kh2g3 9.Kg3f4 10.Kf4e5 11.Ke5d6 12.Kd6c7 13.Kc7b8 {z
(C+ by WinChloe and Popeye 4.79)}

what does it mean serZ(b8) ?
Z stands for ‘Zielfeld’, ‘field to be reached’. You have to get a piece to b8 in 13 consecutive moves, doesn’t matter which piece.
so, why not 5.Kxf2 and 6.Bb8 ?
Seems that was not correctly checked as Popeye also confirms the cooks.
So basically this?
BeginProblem
Stipulation serza213
Pieces
black Bh8 Pf7f5e4d3c2g6
white Kc1
EndProblem
1.Kc1d2 2.Kd2e3 3.Ke3f4 4.Kf4g5 5.Kg5h6 6.Kh6h7 7.Kh7g8 8.Kg8*f7 9.Kf7e6 10.Ke6d5 11.Kd5c4 12.Kc4b3 13.Kb3a2 z
Without a pawn.
serzb813
1.Kb5 2.Kc4 3.Kd3 4.Ke2 5.Kf1 6.Kхg1 7.Kf2 8.Ke3 9.Kd4 10.Kc5 11.Kb6 12.Ka7 13.Kb8 z
The “orthodox” length record for this stipulation with 6 units is 49 moves: http://pdb.dieschwalbe.de/search.jsp?expression=PROBID='P1113601‘. If you want it to be without captures, you can remove the pieces on f7 and h7, move the wK to h8 and get a serZf7 in 17 moves.
I think that original idea was to have a clear geometry and attractive use of madrasi effects.
So the following version seems nearer to author’s thoughts, though I was not able to resolve all issues).
White : Kd1 Qb2 Rf4
Black : Kf6 Qb4 Rc4 Pa2
serz[b1]16
The author’s concept can be save by little shifting of pieces:
serzb813 Madrasi (3+3)
B4k2/q7/K7/8/8/8/5Q2/7b
1.Kb5 2.Kc4 3.Kd3 4.Ke2 5.Kf1 6.Kg1 7.Kh2 8.Kg3 9.Kf4 10.Ke5 11.Kd6 12.Kc7 13.Kb8 z
If captures will be added then the solution length can be increased, for example:
serza255 Madrasi (3+10)
2b4B/5p2/1k4Qp/3n4/5rK1/r7/8/bqn5
1.Kh5 2.Kхh6 3.Kh7 4.Kg8 5.Kf8 6.Ke8 7.Kd8 8.Kхc8 9.Kd7 10.Kd6 11.Kхd5 12.Kd6 13.Ke7 14.Kf8 15.Kg8 16.Kh7 17.Kh6 18.Kg5 19.Kхf4 20.Kg5 21.Kh6 22.Kh7 23.Kg8 24.Kхf7 25.Ke6 26.Kd5 27.Kc4 28.Kb4 29.Kхa3 30.Kb4 31.Kc4 32.Kd5 33.Ke6 34.Kf7 35.Kg8 36.Kh7 37.Kh6 38.Kg5 39.Kf4 40.Ke3 41.Kd2 42.Kхc1 43.Kd2 (43.Kxb1??) 44.Ke3 45.Kf4 46.Kg5 47.Kh6 48.Kh7 49.Kg8 50.Kf7 51.Ke6 52.Kd5 53.Kc4 54.Kb3 55.Ka2 z
wK, like a “shuttle”, moves 5 times through right upper corner of the board.
Why so many captures? Just a simpler scheme (after Georgy):
White Qa7 Kf5 Rg3
Black Be8 Qg7 Pa6 Sc6 Rg6 Se4 Ke3
Stipulation serza853
Condition Madrasi
1.Kf5e6 2.Ke6d5 3.Kd5c4 4.Kc4b3 5.Kb3c2 6.Kc2d1 7.Kd1e1 8.Ke1f1 9.Kf1g2 10.Kg2h3 11.Kh3h4 12.Kh4h5 13.Kh5h6 14.Kh6h7 15.Kh7g8 16.Kg8f8 17.Kf8*e8 18.Ke8f8 19.Kf8g8 20.Kg8h7 21.Kh7h6 22.Kh6h5 23.Kh5h4 24.Kh4h3 25.Kh3g2 26.Kg2f1 27.Kf1e1 28.Ke1d1 29.Kd1c2 30.Kc2b3 31.Kb3c4 32.Kc4d5 33.Kd5*c6 34.Kc6d5 35.Kd5c4 36.Kc4b3 37.Kb3c2 38.Kc2d1 39.Kd1e1 40.Ke1f1 41.Kf1g2 42.Kg2h3 43.Kh3h4 44.Kh4h5 45.Kh5h6 46.Kh6h7 47.Kh7g8 48.Kg8f8 49.Kf8e8 50.Ke8d8 51.Kd8c8 52.Kc8b8 53.Kb8a8 z
Sorry, sorry, have missed one more condition under diagram – NoWhiteCapture! Corrected now.
Sorry, Julia, it’s my fault, I checked the prepublication badly.
5 captures/124 moves, but Madrasi is here a rather mechanical tool:
White Ka7 Rf7 Qb1
Black Qb7 Rc7 Ba6 Rc6 Pd6 Kf5 Sc4 Pf4 Ba3 Pe3 Pf3 Sc1
Stipulation serza1124
Condition Madrasi
20.Kxc641.Kxa660.Kxc482.Kxa3102.Kxc1 etc.
Thank you for your feedback ; they interested me a lot and learned. I propose the following problem with the same idea. But 2 pieces less and 1 more move.
White: Ka8, Ba7
Black: Kd7, Bg1
serZ(a7)14
Madrasi
Breton
1.Kb7 2.Ka6 3.Kb5 4.Kç4 5.Kd3 6.Ké2 7.Kf1 8.K×g1(×a7) 9.Kf2 10.Ké3 11.Kd4 12.Kç5 13.Kb6 14.Ka7