Digit Forcing Chains

This is the start of a family of strategies called Forcing Chains. As the name implies they are made from chains - or formally - Alternating Inference Chains which are simple ON - OFF- ON - OFF consequences. Chains can start with an ON or an OFF. When a candidate X it turned 'on' it immediately turns 'off' all other candidates it can see. When a candidate is in the 'off' state it might turn on another candidate if there is only one left in the unit it can see.

In the strategy Forcing Chains we look at the consequences of a candidate being first ON and then OFF, or a group of candidates in a cell being ON (Cell Forcing Chains) or a number being ON in all instances in a unit (Unit Forcing Chains). If the consequences of the chains are identical - that another candidate is always ON or always OFF - then we can assert something about those candidates or immediate neighbours.

Before we look at specific examples, its worth going over the different types of attack in this family. In this diagram the starting cell is on the left - and an 8 is chosen. It is attacking either another digit (first two rows), a cell (middle two rows) or a unit (last two rows). In each attack the ends of the two chains are either ON or OFF.

If the two ends of the chain meet on the same digit, the we can remove that digit if the ends are OFF, or it's the solution to that cell if the chain ends are both ON.

Likewise, we can attack a whole cell by finding two ON or OFF digits. If OFF, then the last remaining candidate is the solution - but this only works if there are 3 candidates in the whole cell. If ON, then we know that one of those two ON cells is the solution, so any other candidates can be removed.

And finally, the unit attack, based on the same number appearing three or more times in that unit. You will see the pattern now. If we can find two numbers that the chain ends say must be ON, then one of them is the solution and the rest of that number X can be removed from the unit. If there are three numbers left and we identify two numbers that are off, then the last candidate is the solution.

The 'digit' in this strategy is a single candidate - in this case the 2 in B1. We are looking at the consequences of this 2 being ON or OFF. Following the purple chain where 2 is ON we get to the cell H9 where the consequence of +2[B1] is to turn 2 in H9 ON as well. The purple chain is:
+2[B1]-2[B6]+1[B6]-1[C4] +1[E4]-1[E9]+5[E9]-5[D9] +7[D9]-7[H9]+2[H9]

Now, if the 2 in B1 were OFF we can trace another shorter chain (blue) to H9 were we also find the 2 can also be turned ON. The chain for this is
-2[B1]+2[C3]-2[H3]+2[H9]
So whether B1 is 2 or not, we know 2 will appear in H9 so the other candidate 7 can be removed and we get a solved cell.

Type 1 is pretty rare compared to other types but has more punch.

This difficult puzzle has three in a row Digit Forcing Chains of type 2.

The first one is centred on 5 in H4. It finds that either way the 5 behaves it will disallow 6 in E2. Unlike type one this eliminates a single candidate - chipping away at the numbers left. But it is much more common.

So if -5[H5] we get this chain to -6[E2]:
-5[H4]+5[J5]-5[J2] +5[E2]-6[E2]
If +5[H5] we get the purple chain to -6[E2]:
+5[H4]-6[H4]+6[H6] -6[D6]+6[D1|D2]-6[E2]
No fancy links.

To finish this Sudoku off there are three Type 3 Digit Forcing Chains and then a couple of 3D Medusas. This first one looks like it hops around a lot but it is just taking advantage of a links between candidates in some cells.

Starting with the shorter blue chain where 4 is OFF in B3 we arrive at F3 and must turn 7 ON as it is the last 7 in the row after -7[F5. The chain is
-4[B5]+9[B5]-9[E5]+7[E5] -7[F5]+7[F3]
The longer purple chain follows turning 4 OFF and it finds its way to F3 but wants to turn 8 ON instead. The chain is +4[B5]-4[C4]+4[C7]-2[C7] +2[D7]-1[D7]+1[F7] -8[F7]+8[F3].
So, in conclusion whether B5 is ON or OFF a 7 or a 8 fits in F3 - other candidates can be removed.

Let us take a peek at the second Digit Forcing Chain that comes immediately after the one above.

B1 is being favoured and the focus is on candidate 1. Following the chains we arrive at F7 where 1 and 8 are ON and possible solutions.

Type 3. Whether B1 is 1 or not the chains imply F7 must be [1 or 8] - other candidates 6/9 can be removed
-1[B1]+1[B2]-1[F2]+1[F7]

+1[B1]-8[B1]+8[B4]-9[B4] +9[B5]-9[E5]+7[E5]-7[F5] +7[F3]-8[F3]+8[F7]