B(T’)= B(T) + p(x)[d(b) − d(x)] − p(b)[d(b) − d(x)] Now taking [d(b) − d(x)] as common B(T’) = B(T) − (p(b) − p(x))(d(b) − d(x)) It means that B(T’) <= B(T) as (p(b) − p(x))(d(b) − d(x)) is positive because both (p(b) − p(x)) and (d(b) − d(x)) are positive
B(T’)= B(T) + p(x)[d(b) − d(x)] − p(b)[d(b) − d(x)] Now taking [d(b) − d(x)] as common B(T’) = B(T) − (p(b) − p(x))(d(b) − d(x)) It means that B(T’) <= B(T) as (p(b) − p(x))(d(b) − d(x)) is positive because both (p(b) − p(x)) and (d(b) − d(x)) are positive
Hufman_tree.docx