I thought about cost for this puzzle in terms of how far each glyph was from an access point. After placing the input and output glyphs, there are 6 free hexes at access+1. 2 of these need to have triplex, 1 needs to have a debonder, and 1 needs to have either end of duplication. This leaves 2 hexes which can contain the calcifier or remaining ends of the debonder and duplication. This means one of these must be at least 2 away from an access point, and the choice of which glyph to put at access+2 leads to very different solutions. Of these options I found the elemental end of duplication to be the easiest. Placing half the debonder or the salt end of duplication at access+2 requires bonding products to a larger structure with a lot of temporary bonds and debonds, while the calc at access+2 or further means the earth-water reagent permanently takes up access points and forces more complicated maneuvering around it.