利用hadoop来解决“共同好友”的问题

假设A有好友B C D；B有好友A C D E；C有好友A B D E；D有好友A B C E;E有好友B C D。

A -> B C D

B -> A C D E

C -> A B D E

D -> A B C E

E -> B C D

则

对于A来说， A -> B C D

(A B) -> B C D

(A C) -> B C D

(A D) -> B C D

对于B来说， B -> A C D E

(A B) -> A C D E

(B C) -> A C D E

(B D) -> A C D E

(B E) -> A C D E

对于C来说， C -> A B D E

(A C) -> A B D E

(B C) -> A B D E

(C D) -> A B D E

(C E) -> A B D E

对于D来说， D -> A B C E

(A D) -> A B C E

(B D) -> A B C E

(C D) -> A B C E

(D E) -> A B C E

对于E来说， E -> B C D

(B E) -> B C D

(C E) -> B C D

(D E) -> B C D

汇总得到

(A B) -> (A C D E) (B C D)

(A C) -> (A B D E) (B C D)

(A D) -> (A B C E) (B C D)

(B C) -> (A B D E) (A C D E)

(B D) -> (A B C E) (A C D E)

(B E) -> (A C D E) (B C D)

(C D) -> (A B C E) (A B D E)

(C E) -> (A B D E) (B C D)

(D E) -> (A B C E) (B C D)

则共同好友的关系是

(A B) -> (C D)

(A C) -> (B D)

(A D) -> (B C)

(B C) -> (A D E)

(B D) -> (A C E)

(B E) -> (C D)

(C D) -> (A B E)

(C E) -> (B D)

(D E) -> (B C)