“…The remaining unions of three paths can be packed into K 0 6 as follows: 1, 6, 3, 5, 6, 2, 2), (3, 3, 1, 2, 5, 4, 4), (5, 5, 1, 4, 6, 6), P 9 ∪ P 6 ∪ P 6 → (1, 1, 2, 6, 5, 3, 4, 2, 2), (3, 3, 2, 5, 4, 4), (5, 5, 1, 4, 6, 6), P 7 ∪ P 6 ∪ P 6 → (1, 1, 6, 3, 4, 2, 2), (3, 3, 2, 5, 4, 4), (5, 5, 1, 4, 6, 6), P 9 ∪ P 5 ∪ P 5 → (1, 1, 5, 4, 6, 5, 3, 2, 2), (3, 3, 1, 4, 4), (5, 5, 2, 6, 6), P 8 ∪ P 6 ∪ P 5 → (1, 1, 6, 5, 3, 4, 2, 2), (3, 3, 1, 5, 4, 4), (5, 5, 2, 6, 6), P 7 ∪ P 7 ∪ P 5 → (1, 1, 6, 3, 4, 2, 2), (3, 3, 1, 5, 6, 4, 4), (5, 5, 2, 6, 6), P 8 ∪ P 7 ∪ P 4 → (1, 1, 5, 2, 4, 6, 2, 2), (3, 3, 1, 6, 3, 4, 4), (5,5,6,6), P 9 ∪ P 6 ∪ P 4 → (1, 1, 6, 2, 5, 4, 3, 2, 2), (3, 3, 1, 2, 4, 4), (5,5,6,6), P 10 ∪ P 5 ∪ P 4 → (1, 1, 6, 2, 5, 1, 4, 3, 2, 2), (3,3,5,4,4), (5,5,6,6), P 11 ∪ P 4 ∪ P 4 → (1, 1, 6, 2, 3, 1, 2, 5, 4, 2, 2), (3,3,4,4), (5,5,6,6), P 6 ∪ P 6 ∪ P 6 → (1, 1, 3, 6, 2, 2), (3, 3, 2, 5, 4, 4), (5, 5, 1, 4, 6, 6), P 8 ∪ P 5 ∪ P 5 → (1, 1, 5, 6, 4, 3, 2, 2), (3, 3, 1, 4, 4), (5, 5, 2, 6, 6), P 7 ∪ P 6 ∪ P 5 → (1, 1, 6, 5, 3, 2, 2), (3, 3, 1, 5, 4, 4), (5, 5, 2, 6, 6), P 7 ∪ P 7 ∪ P 4 → (1, 1, 6, 4, 5, 2, 2), (3, 3, 4, 1, 2, 4, 4), (5,5,6,6), (1, 1, 6, 3, 4, 5, 2, 2), (3, 3, 2, 1, 4, 4), (5,5,6,…”