User blog comment:LittlePeng9/Random Turing machines/@comment-25418284-20130609012459/@comment-1605058-20130609080827

Here we can see output of specific 2-state 4-color machine. If I'm not mistaken, by flipping halt transition machine head will land on rightmost 3. By flipping whole machine, it will land on leftmost 3. Also, replace symbol 2 with x, 3 with ( and 1 with something else. Now, instead of halting, make machine transit to state 0 in this machine:

0 x x r 0

0 ( _ r 1

0 _ x l 3

1 * ) r 2

2 * ( r 0

3 * * l 3

3 _ y l 7

3 ( [ l 4

3 x [ r 6

4 ) ] l 4

4 _ y r 5

5 * * r 5

5 x y r 5

5 _ ] r 6

6 * * r 6

6 x y r 6

6 _ ] r 0

7 _ x r 7

7 y _ r 7

7 [ ( r 7

7 ] ) r 7

7 x _ r M1

This machine will replace x((((...((x with x ( ... )) )) ) which is proper input for Buchholz hydra machine with 683 nested w brackets. M1 is state 1 in Buchholz hydra machine. So with total of 168 states we have machine outputing over BH(683) symbols. \(\Sigma(168,7)>\text{BH}(683)\). By adding 3 states we can extend x(((...((x by 3 ('s, thus adding one more w pair of brackets. So I can give bound \(\Sigma(3x+168,7)>\text{BH}(683+x)\). Using my color reduction technique we can get \(\Sigma(75x+4200)>\text{BH}(683+x)\). This best bound I can think of.