1. We claim that the other two element edges in g(t) are also included in triangles from g(t). (Proof of claim.) To see this, consider one of the other element edges in g(t), say (y 1;is in a triangle
2. z t ) as Y -edge. Thus, (z 1 , z t ) must be in a triple together with (z 1 , z 2 ), contradicting the hypothesis that (z 1 , z 2 ) is undamaged. (End of claim.) Hence the 3(1?6?)q undamaged element edges can be divided into groups of three that correspond to (1 ? 6?)q undamaged gadgets. Then the corresponding (1 ? 6?)q triples in instance I form a matching. We are now ready for the main result of this section. Theorem 2.6. WWI-3 is APX-hard even when all vectors in V X ? V Y ? V Z are 0-1 vectors with exactly two nonzero entries per vector. Proof. When we apply the reduction to a perfect instance I of MAX-3DM-3, Lemma 4.2 yields c OP T (I) = q + 12s for the resulting instance I of WWI-3. A (1 + )-approximation algorithm for WWI-3 would imply that we can compute, in polynomial time, a solution of I with objective value at most equal to (1 + ) c OP T (I) ? , and this matching can be found in polynomial time. Hence, a PTAS for WWI-3 would imply a PTAS for any perfect instance of 3-bounded MAX-3DM. 5. Binary inputs and fixed p;} must be in M . This implies, in turn, that (x t , u t ) and (z t , u t ) must be in the same triple, which can only contain