A1 tells you that E5 is an Alternator.
E5’s first and third statements cannot both be true, because it would make a triangle of Alternators (E5, C3, and either A5 or E1). This would violate A1’s second statement. Therefore, E5 is a False-start-Alternator; so its second statement is true and E1 is a Liar.
E1 and E5 are both lying when they claim C3 is a Liar and and Alternator, respectively. It must be a Truthteller.
C3 tells us that there is only one Liar in each Diagonal, so A5 can not be a Liar. Also, A5 can not be an Alternator, because that would make E5’s first statement true, so A5 must be a Truthteller.
From C3, one of B2 or D4 must be a Liar, and the other must be an Alternator or a Truthteller.
From C3’s second statement, either
* A) one of B2 & D4 is an Alternator and one of D2 & B4 is, also.
* B) neither of B2 & D4 is an Alternator and both of D2 & B4 are Alternators
However, if we choose A, then from E1, we know that B2 would have to be an Alternator and D4 the Liar, but that leaves you unable to make A5’s second statement true. (Remember that A5 does not consider different flavors of Alternator to be different species.) So selection B, above, must be correct: D2 & B4 are both Alternators.
B4 and D2 must start with opposite true/false state, but we don’t know which is which. B4 can’t be determined, but consider the second statement by D2. We know from A5 that there are three Liars in row 2, but there is no way to put them into the row without either two of them being adjacent or putting one in E2, which is adjacent to E1. Therefore, D2 must be lying with his second statement, implying his third statement is true: D1 is a Truthteller.
So now we know that B4 is a false-first-Alternator, telling us that B2 is not a Truthteller, and we already know that B2 is not an Alternator, so B2 is a Liar.
We also know from C3 that D4 is not a Liar, and from E1 that it is not an Alternator, so D4 must be a Truthteller.
From D4, we know that B1 is either a Truthteller or a true-first Alternator. If B1 is a Truthteller, this makes a second Truthteller who is adjacent to C1. In order to make B4’s third statement false, we would have to put another Truthteller in C2. However, that would make B2’s first statement true, which we can’t allow. Therefore, B1 can not be a Truthteller; he must be an true-first Alternator.
Similarly, C2 can not be a Truthteller or it would make B4’s third statement true, and it can’t be an Alternator or it would make B2’s second statement true. C2 must be a Liar.
From D4, B1’s first statement is true, so C1 is not a Truthteller. However, D1 claims to be adjacent to more Truthtellers than A1 is, so E2 must be a Truthteller.
From A5’s first statement, A2 must be a Liar.
From A2’s second statement, there must be at least as many Alternators as Liars. We can see four Liars already.
* From A2’s first statement, one of A3 & A4 must be a Liar.
* From E2, one of D3 & D5 must be a Liar.
* From B2’s second statement, at least one of C1, C4, & C5 must be a Liar.
So there are seven Liars, at least.
Now consider where it is impossible to put Alternators. Any square that can see both a true-first-Alternator and a false-first-Alternator can not support either. These squares are marked with X’s.
E3 and E4 are both prevented from being Alternators, by the above logic. However they also can not both be Truthtellers, because that would make C2’s statement true, so one of those two must be a Liar, as well.
This means that there are at least 8 Liars on the board.
In order to get eight Alternators, we will need every available space.
C1 and D3 are both Alternators (both false-first).
One of A3 & A4 can be an Alternator, but not both.
One of C4 & C5 can be an Alternator, but not both.
Of those two sets, if A3 is, then C5 cannot be, and if A4 is, then C4 cannot be.
So, either A3 & C4 are both Alternators (both true-first) or A4 & C5 both are.
Also, B3 and B5 both must be Truthtellers, or again there would be too many Liars.
A3 and A4 must be Alternator and Liar, so B3 tells us that E3 is a Truthteller.
From B5, we know that the combination of Alternators discussed on the previous page must be A4 & C5, leaving A3 and C4 as Liars.
Finally, E4 must be a Liar to prevent three Truthtellers from appearing in a bunch.