3D Medusa is the fourth chaining method we will look at. While X-Cycles extends Singles's Chains via the way
we create links, 3D Medusa considers links between different candidate numbers. In other words,
both X-Cycles and Single’s chains work on a single candidate, while 3D Medusa works on multiple candidates.
In this method we only consider strong links, as described in the X-Cycles article, and not weak links. We can connect candidates, which are the same, by using the rule described in Single’s Chains. We can also links two different candidates only if they are the only two candidates in a cell. To clarify this, let us consider a cell which has only two candidates 1 and 5. If 1 is ON then 5 is clearly OFF and vice versa. The link is strong because there are only two candidates in the cell and each colour implies the other (i.e. works both ways).
Let us jump straight into the rules. There are 6 ways we can eliminate, which makes this method quite powerful. You may recognise some of the rules from previous chaining strategies.
This rule follows a similar logic to Singles's Chains Rule 2 (which works for 3D Medusa as well).
If there are two candidates with the same colour in the same cell then this must be the
OFF colour, i.e. both candidates can not be the value for a given cell.
Rule 2 is the same as in Single’s Chains. If there are two candidates X,
see each other (share a house) and they have the same colour, then that colour is OFF,
as a house can not have the same two values. Let us quickly see this in an example.
Looking closely at row A, we notice that there are two 5 candidates which are red. This means that red is the OFF colour and so we can eliminate all red candidates and place all green candidates
If there are no contradictions as per rules 1 and 2, the chain is said to be ”nice” (which is important for loops
in X-Cycles and Alternating Inference Chains).
This means that we cannot deduce the correct colour from the chain, but we can eliminate from outside of the chain.
This rule is the same as rule 4 in Singles's Chains and follows the same idea as rule 3. If an uncoloured candidate X
“sees” (shares a house with) two differently coloured candidates X, then we can eliminate the uncoloured candidate.
The reason is that the uncoloured candidate sees both colours and one of the colours will correspond to ON (i.e. will be placed).
Let us see this in an example.
There are no contradictions in the chain, presented in this board, but let us look more closely at candidate 9 in cell F7. This candidate sees (9, H7) and (9, D9). Either colour can be the ON colour, but since (9, F7) sees both, it cannot be the correct value for F7 and we can eliminate it.
The same holds for (9, H7), which sees again (9, H7) and (9, D9).
Rule 5 is similar to rule 4, but instead of looking for two coloured candidates in the same house,
one of them will be in the same cell as the uncoloured candidate. Let us clearly state the rule:
If an uncoloured candidate X sees a coloured candidate X (both share a house) and the uncoloured
candidate is in a cell, where another candidate is coloured differently, then we can eliminate the uncoloured candidate.
Let us see this in an example.
Consider the chain on this board. Looking at D7, suppose that the green 5 is ON, in which case we can eliminate the candidate 2. If the the green 5 is OFF then (2, D9) is ON and we can again eliminate the candidate 2 on D7. The same holds for candidate 6 in E7 which has a red 5 in the cell and sees a green 6 in the same square.
This rule is fairly straightforward, but may be harder to spot. We consider a cell, which
is not in the chain and determine the correct colours within the chain. The rule states that
if there exists a cell, which is not in the chain, and every candidate in that cell is seen by
candidates in the chain of the same colour, then the given colour is OFF. Sounds complicated so
let us clarify the rule with a quick example.
Let us consider the cell A5 in the following board. The candidate 2 in this cell sees (2, B4) and the candidate 7 sees (7, A7). If the red colour was ON then there would be no possible candidates for A5, which is a contradiction. In other words, A5 is “emptied” by the red colour and so the colour is OFF. Hence, we can eliminate all red candidates and place the green candidates.