In this section, we discuss the parallel algorithm and the cyclic algorithm, respectively, for solving the variational inequality over the set of the common fixed points of finite strict pseudocontractions.
Let be a real Hilbert space and a strongly monotone and boundedly Lipschitzian operator. Let be a positive integer and a strict pseudocontraction for some such that We consider the problem of finding such that
Since is a nonempty closed convex subset of , (4.1) has a unique solution. Throughout this section, is an arbitrary fixed point, , is the Lipschitz constant of on , the fixed constant satisfies , and the sequence belongs to .
Firstly we consider the parallel algorithm. Take a positive sequence such that and let
By using Lemma 2.8, we assert that is a strict pseudocontraction with and holds. Thus VI(4.1) is equivalent to VI and we can use scheme (3.2) to solve VI(4.1). In fact, taking in the scheme (3.2), we get the socalled parallel algorithm
Using Lemma 2.8 and Thorem 3.2, the following conclusion can be deduced directly.
Theorem 4.1.
Suppose that and satisfy the same conditions as in Theorem 3.2. Then the sequence generated by the parallel algorithm (4.3) converges strongly to the unique solution of (4.1).
For each let
where the constant such that . Then we turn to defining the cyclic algorithm as follows:
Indeed, the algorithm above can be rewritten as
where , namely, is one of circularly. For convenience, we denote (4.6) as
We get the following result
Theorem 4.2.
If satisfies the following conditions:
(i);
(ii);
(iii), or
then the sequence generated by (4.6) converges strongly to the unique solution of .
Proof.
We break the proof process into six steps.

(1)
. We prove it by induction. Definitely . Suppose , that is,
We have from , (4.8), and Lemma 2.7 that
where Observing , we get
This together with (4.9) implies that
It suggests that . Therefore, for all . We can also prove that the sequences , , are all bounded.

(2)
By (4.6) and Lemma 2.7, we have
where Since satisfies (i)–(iii), using Lemma 2.2, we get

(3)
By (4.3) and , we have
Recursively,
By Lemma 2.6, is nonexpansive, we obtain
Adding all the expressions above, we get
Using this together with the conclusion of step (2), we obtain

(4)
. Assume that such that , we prove . By the conclusion of step (3), we get
Observe that, for each is some permutation of the mappings , since are finite, all the full permutation are , there must be some permutation that appears infinite times. Without loss of generality, suppose that this permutation is , we can take a subsequence such that
It is easy to prove that is nonexpansive. By Lemma 2.3, we get
Using Lemmas 2.6 and 2.9, we obtain

(5)
In fact, there exists a subsequence such that
Without loss of generality, we may further assume that Since is the solution of , we obtain

(6)
By (4.6), Lemmas 2.1(ii), and 2.7, we obtain
From the conclusion of step (5) and Lemma 2.2, we get