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, ? The combination of constraints (7) and (8) (Tab. 2) in this order gives the 2 following di-peptide constraints, ) ó?W (H+Q) W (H+Q) ó? C (H+Q)

, ? The combination of constraints (8) and (7) gives: H (C+W) ó? Q (C+W) Q (C+W) ó? H (C+W)

, ? The double combination of constraint (7) gives: W (C+W) ó? H (H+Q) C (C+W) ó? Q (H+Q) and the double combination of constraint (8) gives the reciprocal constraints

, ? The combination of constraints (2) and (10) in this order gives: Y (D+E) ó? SY Y (F+K+L+N+R+S) ó?

D. Terms, F. , and N. , are removed from the right part of the constraints as they are forbidden di-peptides (Tab. 3), i.e. DY = 0 The (ordered) combination of Constraints (10) and (2) gives the reciprocal constraints

, ? The combination of constraints (8) and (10) in this order gives (the reverse order gives the reciprocal constraints

, ? The combination of constraints (7) and (10) in this order gives (the reverse order gives the reciprocal constraints

. Hence, )) lead to a di-peptide constraint that divides into two di-peptide constraints; and four combinations)) lead to three di-peptide constraints. This increases the total count of " non-null " constraints by 14. Moreover, one of the 10 2 constraints derived from the amino-acid constraints, YY ó? YY, is removed from the constraint set, as it is redundant with the two null constraints excluding dipeptide YY both in the reference and in the overlapping frame (Tab. 3), This results in 10 2 -1 + 14 = 113 non-null di-peptide constraints in frame -2, as found by our graph algorithm (Fig. 3) applied to " heptons