Case 1: The two groups have the same mass, which means that X must be in the third group, then take any two balls from the third group, and then take any two balls from the previous two groups. If it is balanced, take one of the remaining two balls in the third group. If it is balanced, the rest of the third group is X. If it is unbalanced, it is of course X!
Case 2: The two groups have different qualities and use the same method. ..
(①, ②, ③ Weighing for three times)
Divide the ball into three groups with four in each group, such as: Group X (1, 2, 3, 4), Group Y (A, B, C, D) and Group Z (A, B, C, D).
(1) if X=Y, then q in Z.
Draw d from z, and add normal ball 1 call (a, B) (C, 1).
② If (a, B)=(C, 1), then q = d.
②if (A，B)& lt； (c, 1) is called a, b.
③ If A = B, then Q = C.
③ If A>, then Q = B.
③ if A<, then q = a.
②if (A, B)>(C, 1) is called a, b.
③ If A = B, then Q = C.
③ if A>, then q = a.
③ If A<, then Q = B.
(1) if X>Y and then q is in x or y.
Extract (3,4) from X, (d) from Y, X residual (1, 2) Y residual (a, b, c),
Swap (2) in X and (c) in Y, and then add the ordinary ball (d) to X,
Group x (1, c, d) and group y (a, b, 2) after recombination, and then called x, y.
② If X = Y, then q is in (3,4,d).
Because (1, 2,3,4) >: (a, b, c, d) (from weighing ①), Q = d (lighter than normal) or Q = (3,4) is the heavier one.
Weighing (3, 4).
③ If 3 = 4, then Q = D.
③if 3 & gt； 4, then Q = 3.
③if 3 & lt； 4, then Q = 4.
② If X>y and then Q are in (1, a, b).
It doesn't matter if 2 and C are exchanged, they are both normal balls, so Q = 1 (heavier than normal) or Q =(a, b) is the lighter one.
③ If a = b, then Q = 1
③ If a>, then Q = B.
③ if a<, then q = a.
② If X<y and then q are in (2, d).
2 and c determine the weights of x and y, so Q = 2 (heavier than normal) or Q = c (lighter than normal).
Compare 2 with an ordinary ball 1.
③ If 2 = 1, then q = c.
③if 2 & gt； 1 then Q = 2
③2 & lt； 1 impossible.
① If X<y is the same.