X-Cycles

X-Cycles is the first slightly more complex colouring strategy that we will discuss. There are three main differences between X-Cycles and other colouring strategies such as Singles's Chains, 3D Medusa and XY-Chains:

Our main goal is to create a cycle, instead of chain. In other words, we try to link the last candidate with the first candidate at some point when creating the chain.
We introduce strong and weak links between candidates. A detailed explanation of this concept can be seen below.
The way we assign colours and link candidates is slightly different. For this method (and Alternating Inference Chains) the green colour means ON (i.e. we assume or it follows that the candidate will be placed in the cell) and the red colour means OFF (i.e. we assume or it follows that the candidate will not be placed in the cell). In comparison, the colours in the methods mentioned above are independent of the placement of candidates - we do not know which colour means ON and which means OFF.

X-Cycles naturally extend to Alternating Inference Chains. Both methods have one minor difference in the way we create chains, but are, otherwise the same. This is why we will carefully explain the new concepts here and we will apply them later in one of the most powerful methods - AIC. An interesting fact is that we can represent X-Wing, Swordfish and Jellyfish with an X-Cycle.

Strong vs Weak Links

In previous methods, we explicitly used the ⇔ sign to denote that a link works both ways. For example, suppose we have a row with exactly two candidates, e.g. A2 and A5 have a candidate 9, without any other such candidates in the same row. If we colour (9, A2), this means that we can colour (9, A5) . The opposite works as well. If we colour (9, A5) then we can colour (9, A2). We denote this relationship with a solid line between candidates and we write ⇔. This way of linking is said to be strong. With a strong link the colours are not all that important, as the implication works both ways, independently of which colour denotes the ON candidate.

Let us now introduce the new concept of a weak link. For the rest of this article a green colour means ON and a red colour means OFF. Suppose we have the same example as above, but there is an additional candidate in row A, say (9, A3). Now, if we colour (9, A2), then we can still colour (9, A5), since we have assumed that green means ON and so we can turn OFF any other candidates in the row. So we have (9, A2) ⇒ (9, A5). The opposite, however, is not true. If we assume (9, A5) is OFF, then it is not necessary for (9, A2) to be ON. The reason is because we have an additional (9, A3) candidate in the row. Summarizing, a weak link is one, where an ON candidate turns OFF another candidate, but the opposite is not true, due to an additional candidate in the same house.

Let us clarify these new concepts with an example. The Xs on the figure denote an arbitrary candidate. Suppose we start with (X, D3) and go clockwise. The candidate is ON, so looking at row D, we can create weak links with all other candidates X. In this case we, have chosen (X, D3) ⇒ (X, D7). Notice that the implication is written and works only one way. If we colour (X, D7), any of the Xs on D3, D6 or D9 could be ON, so the reverse implication is not correct. Thus we have a weak link and we mark it with a dashed line.

Let us continue with the chain. There are only two X candidates in column 7, so we can create a strong link. In other words, (X, D7) ⇔ (X, H7). Next, we note that (X, H7) is ON, so we can create a weak link with any other candidate X in row H. In this case we have (X, H7) ⇒ (X, H5). The rest of the links are strong and so the final chain is:

(X, D3) ⇒ (X, D7) ⇔ (X, H7) ⇒ (X, H5) ⇔ (X, E5) ⇔ (X, E2) ⇔ (X, D3)

Note, that we have finished the chained where we started, i.e. we have created a cycle (or a loop). There are two elimination rules which we will look at next. In the first, there are no "errors" in the loop, i.e. all colours are consistent. In the second, when we follow the chain, the first candidate turns out to be both ON and OFF. Let us look at some examples.

X-Cycles Rule 1 - Perfect Loops

Load in Solver

In this first example, the X-Cycle, with candidate 7, is aligned in the same way as in the example above. So instead of focusing on the chain, let us explain the logic behind the first rule. A cycle is said to be perfect if the candidate we start with keeps the same colour in the last step of creating the chain. In this case we start with (7, D3) and go clockwise. Following the chain we arrive at (7, E2), which is OFF and the final link confirms that (7, D3) is ON.

Suppose, now, that we start with (7, D3). We can follow the same chain counter-clockwise with swapped colours. The links will remain the same. This means that in D3 we can either place a 7 (Case 1) or not (Case 2). Both cases are valid and we do not know which one is correct. What we can observe in both cases, though, is the following: if there exists a candidate which sees two colours from the chain, we can eliminate it. We explain why this is the case in the next paragraph.

Let us apply rule 1 for this board. On this screenshot the starting cell is marked in dark blue and all cells with eliminations, are marked in light-blue. Let us carefully observe , for example, (7, H1). This candidate sees both (7, H5) and (7, H7). From our observation above, either H5 or H7 will be a 7. Hence, we can eliminate (7, H1).

This is true for all light-blue cells. Both (7, H2) and (7, H9) see (7, H5) and (7, H7). Also, both (7, D6) and (7, D9) see (7, D7) and (7, D3).

To summarize rule 1: If we obtain a perfect cycle, we can eliminate all off-chain candidates X which see two differently coloured candidates X from the chain.

X-Cycles Rule 2 - Imperfect Loops

Load in Solver

Rule 2 is much more common than rule 1. We start with a candidate and colour it. We obtain a loop, but in the last step it turns out that the initial colour was wrong, i.e. the last link implies that the starting candidate has the reverse colour. Let us see an example of this. Starting with (2, D8) ⇔ (2, F9), we create a chain, where the last cell is (2, H8). If we follow the last link, we see that (2, H8) ⇔ (2, D8). Now, this is strange. We start with a colour and the chain implies that the colour is wrong. This does not mean that our logic (or method) is wrong, but rather, that the initial colour we started with is wrong. Hence, for this specific example, (2, D8) is the wrong colour for this candidate. Hence the correct colour is ON and we can simply place the 2 in D8.

To summarize rule 2. If we create a chain, where we arrive at a contradicting colour for the first candidate, this means that the initial colour is wrong. In this first example, the initial colour is OFF, so we can place the candidate in the cell. In the next example we will see the reverse case, where we start with an ON candidate.

This is a nice board with 6 different X-Cycles in the solution. The illustrated cycle starts with (3, B1) and continues clockwise. We see that the final cell is (3, C3), which we can weakly link to the starting (3, B1). We notice that the colours are not consistent, so the initial colour for B1 is wrong. In other words, we can eliminate the candidate 3 from B1.

Load in Solver