This puzzle is a follow-up for Does The Rubik’s Cube In This Painting Have A Solved State?
What is the most complicated illegal “partial state” possible for a Rubik’s Cube?
To be more specific: a partial state is a configuration of 27 colours for three adjacent faces, such as shown below. A partial state is legal if you can call the colours for all stickers on the three unseen faces and solve the resulting cube. Solving means all faces are a single colour, but not necessarily the standard permutation of colours (therefore White is not necessarily opposite Yellow etc). If this is not possible then the partial state is illegal.
Given an illegal partial state, the complexity is the minimum number of stickers you must consider to prove it is indeed illegal. In the diagram below we need only consider the five green edge stickers to prove illegality because … here’s a knowledge bomb from Puzzling Stack Exchange … every colour in a Rubik’s Cube has exactly four edge pieces 😊 Therefore the complexity is 5.
Your task is to find an illegal partial state with the highest possible complexity as defined above.