Counterfeit Coin Problem Variant - Two Counterfeits
Yes, 4 weightings is possible. Even more, this is still true even if it is not known whether the combined weights of the 2 counterfeits is heavier, lighter, or same as that of 2 normal coins
Notation
First, let's introduce some notation.
Coins are labelled 1 through 8. H, L, and n denotes the heavy counterfeit, the light counterfeit, and a normal coin, respectively.
Weightings are denoted, for instance, 12-34 for weighting coins 1 and 2 against 3 and 4. The result is denoted 12>34, 12=34, or 12<34 if 12 is heavier, weights the same as, and lighter than 34, respectively.
1234:H means H is among coins 1, 2, 3, and 4. Similarly, 1234:L means L is among coins 1, 2, 3, and 4.
Due to the highly symmetric nature of the problem. A lot of without-loss-of-generality assumptions will be made. As such, they will not be called out.
Algorithm
Begin by 12-34 and 56-78.
Case 1: Double unbalanced (12>34, 56>78)
In this case, we know that either 12:H, 78:L or 56:H, 34:L. Do 13-57 next.
If 13>57 , then either 1:H, 7:L or 1:H, 8:L or 2:H, 7:L. These can be distinguished by 28-nn.
If 13=57, then a simple 2-8 to distinguish 2:H, 8:L and 6:H, 4:L
Case 2: Balanced-unbalanced (12>34, 56=78)
In this case, we know that 12:H and/or 34:L. Do 1-2 next.
If 1>2, then 1:H, 234:L. A simple 2-3 resolves that.
If 1=2, then either 3:H, 4:L or 4:H, 3;L. So 3-4.
Case 3: Double balanced (12=34, 56=78)
This means both H and L is within the same pair. Do 135-246.
If 135>246, then either 1:H, 2:L, 3:H, 4:L, or 5:H, 6:L. Do 1-3 to distinguish.
If 135=246, then either 7:H, 8:L or 8:H, 7:L. Do 7-8 to distinguish.