(Following are two comments, posted this way because I ("r.e.s.") cannot post comments directly.)
Comment on the answer by "Deedlit":
He does this by showing that
TREE(3)
must be very large - specifically, he constructs a sequence of more thann(4)
rooted trees labeled from{1,2,3}
such thatTi
has at mosti
vertices, for noi,j
withi<j
do we haveTi
homeomorphically embeddable intoTj
, and each tree contains either a 2 label or a 3 label. We can obviously continue this with tree(n(4)) trees with all labels 1.
That's not quite right. His first tree T1
uses label 3
(so this label cannot be used later at all), followed by more than n(4)
trees using labels 1,2
-- not, as you wrote, using labels 2,3
. It's because of the way these latter {1,2}
-labelled trees are constructed, that they can nevertheless be followed by a long sequence of trees using only label 1
. (I show the beginning of his sequence in my other comment below, using bracket expressions in which the bracket-types (),[],{}
correspond to his labels 1,2,3
respectively.)
Comment on the answer by "Feldman Denis":
[
TREE(3)
is] the length of the longest sequence(T2,T3,T4,…,Tn)
of labeled trees such thatTk
has at mostk
nodes labeleda
orb
, andTi
is not a subtree ofTj
fori<j
.
Rather than "is not a subtree of", that should be "is not homeomorphically embedded in", which is a very much more stringent requirement. (There might not even exist a longest such sequence in the less-stringent case. A similar situation occurs for Friedman's n()
function -- in that case, the relation "is not a subsequence of" is more stringent than "is not a substring of" -- there being no longest sequence in the latter case.) With this correction, and by starting with T2
, the length of the resulting sequence will of course be TREE(3) - 1
.
BTW, a convenient representation of TREE(3)
uses nested bracket expressions (well-formed in the usual way with pairs of matching brackets) involving only three bracket-types -- say (),[],{}
-- each rooted tree being uniquely represented by a nest of such brackets (up to isomorphism with respect to sibling order). TREE(3)
is then the length of a longest sequence (T1,T2,T3,T4,…,Tn)
of nests such that each Tk
has at most k
bracket pairs and for no i<j
is Ti
embedded in Tj
. Here X
is embedded in Y
means that X
can be obtained from Y
by erasing zero or more matching bracket-pairs.
(Note that, because T1
must be some single bracket pair which cannot then appear in any later nest in an embedding-free sequence, an equivalent definition is obtained by assuming T1={}
, so TREE(3)
is one greater than the length of a longest embedding-free sequence (T2,T3,T4,…,Tn)
(starting with index 2) of nests formed as before but using only two bracket types (),[]
.)
Another thing to note is that TREE(3)
assumes rooted trees with unordered siblings, so, for example, the nests ([]())
and (()[])
are not regarded as distinct. Some authors have treated wqo's for rooted trees with ordered siblings, with corresponding "longest sequence" results.
To illustrate the use of bracket expressions, here is a representation of the initial tree sequence used by Friedman to prove the lower bound mentioned by the OP:
T1 {}
T2 [[]]
T3 [()()]
T4 [((()))]
T5 ([][][][])
T6 ([][][](()))
T7 ([][](()()()))
T8 ([][](()(())))
T9 ([][](((((()))))))
T10 ([][]((((())))))
T11 ([][](((()))))
T12 ([][]((())))
T13 ([][](()))
T14 ([][]())
...
NB: It should be noted that the article linked by the OP does not treat Friedman's TREE
function, but a rather different function TR
. The confusion may be partly due to the fact that "TR
" is also what Friedman called the TREE
function before he changed it to the latter name in a follow-up article to the one mentioned in Deedlit's posting.