A scip model for solving the puzzle at https://www.puzzleprime.com/brain-teasers/deduction/the-ping-pong-puzzle/

the-ping-pong-puzzle.zpl

# A scip model for solving the puzzle at 
# https://www.puzzleprime.com/brain-teasers/deduction/the-ping-pong-puzzle/


set players  := { 'A','B','C' };
param NM := 21; 	                # 21 = (10+15+17)/2     number of matches
set matches  := { 1..NM };

set m3p := matches*players*players*players;
set m2p := matches*players*players;

# two play at a time, one doesn't
# x[m,p1,p2,p3] = 1 means that p1 and p2 were the players of the m-th game,
#                         with p3 watching
var x[m3p] binary;

# who won the m-th match between p1 and p2?
#	1 means the first player (p1) won
var w[matches] binary;


# players must be different
subto t1:forall <m,p1,p2,p3> in m3p
	with (p1==p2 or p2==p3 or p1==p3) do x[m,p1,p2,p3]==0;
	 
	 
# the winner of each match stays on the table and
# will play the next match with the third player
subto t2:forall <m,p1,p2,p3> in m3p with m < NM do
	x[m+1,p1,p2,p3] ==
		# either p1 won the previous match as first player
		x[m,p1,p3,p2]*w[m] +
		# or p1 won the previous match as second player
		x[m,p3,p1,p2]*(1-w[m]);
	 

# A played 10 matches in total
subto nma: sum <m,p1,p2> in m2p: (x[m,'A',p1,p2] + x[m,p1,'A',p2]) == 10;
# B played 15 matches in total
subto nmb: sum <m,p1,p2> in m2p: (x[m,'B',p1,p2] + x[m,p1,'B',p2]) == 15;
# C played 17 matches in total
subto nmc: sum <m,p1,p2> in m2p: (x[m,'C',p1,p2] + x[m,p1,'C',p2]) == 17;