π’ Case 1 β f(n) is smaller than boundary
If
f(n) = O( np β Ξ΅ )
for some Ξ΅ > 0
T(n) = Ξ( np )
π Smaller work per level, leaves dominate
π‘ Case 2 β f(n) matches the boundary
If
f(n) = Ξ( np
Β· logk
n )
(k = number of log factors already in f(n))
T(n) = Ξ( np Β· logk+1 n )
π Each recursion level contributes equally β one extra log
π΄ Case 3 β f(n) is bigger than boundary
If
f(n) = Ξ©( np + Ξ΅ )
for some Ξ΅ > 0
T(n) = Ξ( f(n) )
π Root work dominates (regularity condition required)