# Semeai Mnemonic

Here is a mnemonic given in the the Nihon Kiin's __Handbook of Proverbs__?:

3=3, 4=5, 5=8, 6=12

What does this mean? Well, if you have 3 liberties in a large eye, then that is how many liberties it is worth in a capturing race. However, if you have 4 liberties, that provides 5 virtual liberties, because the opponent has to play 3 approach moves to yield atari once, and then another 2 approach moves after you have captured the first three stones. The bigger the large eye, the more virtual liberties it provides. 5 gives 8, and 6 gives 12, and so on.

Dieter: This mnemonic already features at the big eye liberties page. See the aliases. Perhaps you want to un-alias. I agree the mnemonic is currently hard to find.

Tamsin: Yes, I missed that. It also looks like there's already a term for virtual liberties, namely eye liberties. Perhaps we should simply delete my virtual liberties page or alias it to eye liberties. Being vain, though, I do think my term is a good one!

DougRidgway Also see capture metric liberty.

Hans: Here is an interesting aspect. If you add the trivial cases 1=1 and 2=2 you get 1=1, 2=2, 3=3, 4=5, 5=8, 6=12. The sequence of the numbers on the right side of the "="sign seems to be the beginning of the Fibonacci numbers except for 12 (should be 13, very sad!). This means something is wrong with Go mathematically:)

unkx80: Unfortunately, the Fibonacci sequence has the following recurrence relation

L(1) = 1 L(2) = 2 L(n) = L(n - 2) + L(n - 1) for n > 2

while the eye liberties has the following recurrence relation

L(1) = 1 L(2) = 2 L(n) = L(n - 1) + n - 2 for n > 2

so they are not the same. =)

Gronk: Not to mention that the eye liberty sequence terminates (the concept of eye liberties becomes meaningless if the eye is big enough that two eyes can be formed--typically around 8, depending on cutting points). Fibonacci's squence does not terminate, of course.

Shaydwyrm: The sequence is also complicated considerably by the existence of cutting points and the presence of the corner. These figures are only for the "perfect" eye shapes when found away from the corner.

Warfreak2: Given the above recurrence relation (which can be proved), the forward difference at each stage is linearly dependent on n, so the total number of liberties is quadratic in n: (n^2 - 3n + 6)/2, which is valid for n>=2. Probably no point remembering and using this formula though!

Fibonacci numbers haven't really got anything to do with this, that's just a coincidence.