(1) use t1(p) = a1+b1,tk (p) = bk+max {tk-1(p), a1.

(2)T2(P)=max{a+b+d, a+c+d}, T2(P')= max {c+d+b, c+a+b}, and discussed by classification. With the new definition, the sizes of T2 (p) and T2(p') can be compared.

(3) According to the new definition, we can draw a conclusion.

Answer:

Solution:

( 1)T 1(P)=2+5=7，T2(P)= 1+max{T 1(P)，2+4}= 1+max{7，6 } = 8；

(2)T2(P)=max{a+b+d，a+c+d}，T2(P′)= max { c+d+b，c+a+b}。

When m=a, T2(P')=max{c+d+b, c+a+b}=c+d+b,

∵a+b+d≤c+d+b, and a+c+d≤c+b+d, ∴ T2 (P) ≤ T2 (P');

When m=d, T2(P')=max{c+d+b, c+a+b}=c+a+b,

∵a+b+d≤c+a+b, and a+c+d≤c+a+d, ∴ T2 (P) ≤ T2 (P');

Whether m=a and m=d, ∴ t2 (p) ≤ t2 (p';

(3) Number pairs (4,6), (1 1,1), (16,1), (/kloc-0. T3(P)42，T4(P)=50，T5(P)=52。