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**Forcing Chains**

It was the first of a sequence of approaches called Forcing Chains. They’re built up of chains (or, more technically, Alternating Inference Chains), which are basic ON – OFF – ON – OFF outcomes. An ON or an OFF can start a chain. For example, when a candidate X is switched ‘on,’ all other candidates it can perceive are turned ‘off.’ If there is only one candidate left in the unit that it can see, it may turn on another candidate while it is in the ‘off state.

We look at the effects of a candidate being ON and then OFF in the strategy Forcing Chains, a group of candidates in a cell being ON (Cell Forcing Chains), or a number being ON in all occurrences in a unit (Unit Forcing Chains). We have a contradiction if the chains’ outcomes are the same – another candidate is always ON or OFF.

Digit Forcing Chains as a Family Before we get into particular instances, let’s take a look at the many sorts of attacks that make up this family. The beginning cell in this figure is on the left, and an eight has been picked. It’s either assaulting another digit (first two rows), a cell (middle two rows), or a unit (last two rows) (last two rows). The endpoints of the two chains are either ON or OFF in each assault.

If both ends of the Chain meet on the same digit, that digit can either be discarded or utilized as the answer to that cell if both ends are ON.

Similarly, by locating two ON or OFF digits, we may assault an entire cell. If OFF, the solution is the final remaining candidate – however, this only works if the cell has three candidates. On the other hand, if one of the two ON cells is ON, we know it’s the solution, and any other possibilities may be eliminated.

Finally, the unit will attack if the same number appears three times or more in that unit. You’ll see a trend presently. If we can discover two values that the chain ends say must be ON, one of them is the answer, and the remainder of X may be eliminated from the unit. If only three numbers remain and two are incorrect, the final choice is the solution.

**Figure 1:**

Load Example or: From the Beginning of the Digit Forcing Chain In this post, we’ll look at a difficult Sudoku with a clever set of Digit Forcing Chains with various outcomes. Figure 1 depicts the first.

In this technique, the ‘number’ is a single candidate; in this example, the 5 in J9. We’re looking into the effects of turning these five on or off. Following the blue Chain, we arrive at cell E1, where the absence of 5 results in the activation of 6. This Chain (the blue Chain) is as follows:

-5[J9]+5[H9]-5[H5]+5[E5]-5[E4]+8[E4]

-8[B4]+8[B1] -6[B1]+6 [E1]

If the 5 in J9 is switched on, we can follow a shorter chain to E9, where we can also turn on a 6. (the purple Chain) is the Chain for this:

+5[J9]-2[J9]+2[C9]-6[C9]+6[E9]

Hence, regardless of whether J9 is a five or not, we know that six will occur in row E in either E1 or E9, so the other 6s in E2 and E8 may be eliminated.

**Figure 2:**

Forcing Chain for the Second Digit: Example of a Load or: From the Beginning

The next stage is to find another Digit Forcing Chain; this time focused on the number 8 in E4. In the same way as the preceding example, the 8 in E2 will give us a five on either H5 (the blue Chain) or H9. Because of the either/or requirement, five is established in one of the two cells, allowing for the deletion of the remaining 5 in H2.

The blue Chain is -8[E4]+5[E4]-5[E5]+5[H5].

The purple Chain is +8[E4].

-8[B4]

+8[B1]\s-6[B1]

+6[E1]

-6[E9]+4\s{E9|A9}-4[H9]+5[H9] On E9|A9, there is a bothersome ALS that is tough to spot. The Chain enters cell E9 and switches off the 6, leaving just a 4/9 Naked Pair in A9 and H9. As a result, the 4 in H9 is removed, leaving us with the second fixed 5 in H9.

**Digit to Unit attack**

**Figure 3:**

Forcing Chain with the Third Digit Finally, we have one of the most typical forms of Digit Forcing Chain in this Sudoku, one that fixes a digit in another cell. Cell E6 is the starting point, and cell 8 is the digit we’re turning on or off. If the blue Chain is off, we follow it around the board to H2, where one is switched on. This is a link in a chain.

-8[E6]+8[E4]-8[B4]+8[B1]

+6[B2] -6[B1]

-6[D2]+4[D2]-4[H2]+1[H2]

Because of the existence of 8 in E6, 1 in H2 must be present. The second purple chain is +8[E6]-3[E6]+3[H6]-1[H6]+1[H2].

So, regardless of whether E6 is eight or not, the chains show that H2 must be 1, solving the puzzle.

**Second Example**

The Digit Forcing Chains are aimed towards a certain unit: Example of a Load or: Starting at the Beginning Two chains may end up resting on the same digit in the same row, column, or box if they are carefully traversed. Starting on G9, the example works its way around to the 1s on row D. When the five on G9 is ON, we receive a one on D7. We receive a one on D1 if the 5 in G9 is turned off. As a result, there can’t be any more 1s in that row. At this point in the problem, eliminating three other 1s is a bit of a hit. The solver will give you the following results:

**Digit Forcing Chain in Row**:

1s in Row 4 are fixed on [D1, D7] due to G9.

-5[G9]+5[F9]

-5[F1]+5[C1] -4[C1]+4[B1]-1[B1] -5[F1]+5[C1] -4[C1]+4[B1]-1[B1]

+5[G9] +1[D1]

-5[G7]

+2[G7]

-2[D7]

+1[D7]

We may take one away from D2, one away from D8, and one away from D9.

Forcing Chain from Second Digit to Unit: Load Example or: From the Beginning

But the story does not end there. The following step makes use of the new chain link created by removing those 1s. With two 1s on row D, we create a new conjugate pair and add another link to the larger Chain. This allows us to repair the 2s in column 7, resulting in a 2 in H7 elimination.

**Digit Forcing Chain in Column**:

2s in Col 7 are fixed on [D7, G7] due to G9.

-5[G9]+5[F9]

-5[F1]+5[C1]

+4[B1] -4[C1]

-1[B1]

+1[D1]

-1[D7]

+5[G9] +2[D7]

-5[G7]

We may subtract two from H7 by using +2[G7].

**Cell Forcing Chains**

A Strong Link exists when two Cells in an area are the only Cells in that region that have a certain candidate: If the candidate isn’t the first cell’s answer, it must be the second cell’s solution, and vice versa. So, for example, in a “Bi-Value” Cell, if one of the choices is not the answer, the other contender must be.

Strong Links, on the other hand, are only considered in Chaining Strategies as links connecting a Cell where we declare the candidate is not the solution to a Cell where it must be the solution.

A Weak Link connects a Cell where the candidate is “ON” to another cell called “OFF.” Starting from each candidate in a given cell, we create chains alternating Weak Links and Strong Links in the Cell Forcing Chain approach. The candidate “ON” is the starting point for all chains. Type 1 (always “ON”) FORCING CELL: All chains meet in a Cell with the same candidate “ON.” This contender must be the answer for that cell since whatever the solution for the Chain’s first Cell is, it is always “ON.”

**Cell ****Forcing Type 2**

(always “OFF”): In a Cell with the same candidate “OFF,” all chains meet. Therefore, this candidate cannot be the solution for that cell since it is always “OFF” regardless of the Chain’s beginning Cell solution.

The blue Chain indicates that candidate 1 is not the solution in A2 if it is the solution in F4.

The pink Chain means that candidate one cannot be the solution in A2 if candidate 3 is the solution in F4.

Candidate 1 cannot answer A2, regardless of the F4 solution. So it’s possible to get rid of it in A2.

**Cell Forcing Type 3**

(all chains meet in a Cell with some candidates “ON”) Because one of the chains specifies the answer for all of its Nodes, and because we’ve established all feasible chains, the candidates “ON” in the last cell are the sole options. As a result, none of the other choices in that cell can be the answer.

**Cell ****Forcing ****Type 4**

(some candidates “ON” in a region): all chains turn on the same candidate in the same region (Row, Column, or Square). Because one of the chains identifies the solution for all of its Nodes, and because we’ve created all of the potential chains, the candidate can’t be the solution in any other Cell in that region.

If candidate 5 is the answer in C8, the pink Chain indicates that candidate five must be the solution in either H4 or H5.

If candidate 9 is the answer in C8, the blue Chain suggests that candidate five must be the solution in G5 or H5.

Candidate 5 cannot be the answer in G4 regardless of the C8 solution. It’s possible to get rid of it in G4.

**Digit Forcing Chains**

A Strong Link exists when two Cells in an area are the only Cells in that region that include a given candidate: If the candidate isn’t the first cell’s response, it must be the second cell’s solution, and vice versa. For example, in a “Bi-Value” Cell, if one of the choices is not the answer, the other contender must be. Strong Ties, on the other hand, are only considered in Chaining Strategies as links between Cells where we claim the candidate is not the solution (we declare the candidate is in the “OFF” state) and Cells where it must be the solution (we say the candidate is in the “ON” state).

A Weak Link connects a Cell where the candidate is “ON” to another called “OFF.”

Starting with a single candidate in a given cell, we establish two chains alternating Weak Links and Strong Links in the Digit Forcing Chain approach. The candidate “OFF” starts the first Chain, while “ON” starts the second.

**Type 1 Forcing Digit**

(always “ON”): the two chains meet in a Cell with the same candidate “ON.” This candidate must answer that cell since it is always “ON” regardless of the candidate’s initial state in the Chain’s first Cell.

If candidate 9 is “ON” in G8, the blue Chain will result in candidate 9 becoming “ON” in E7.

If candidate 9 is “OFF” at G8 and we follow the pink Chain, we will finally arrive at E7, where candidate 9 is “ON.”

Candidate 9 is the solution in E7 in all hypotheses (candidate nine being or not being the solution in G8).

**Type 2 Forcing Digit**

(always “OFF”): the two chains meet in a Cell with the same candidate “OFF.” This candidate cannot answer that cell since it is always “OFF” regardless of the candidate’s initial state in the Chain’s first Cell.

Candidate 2 is set “OFF” in F6 by following the blue Chain if candidate 7 is the answer in D6 (i.e., is “ON” in D6).

If candidate 7 is not the answer in D6 (i.e., is “OFF” in D6), then-candidate two is put “OFF” in F6 following the pink Chain.

Candidate 2 cannot be the solution in F6 in either of the assumptions (candidate seven being or not being the solution in D6).

**Type 3** **Forcing Digit **

(two “ON” candidates): the two chains meet in a Cell with two “ON” candidates. Because one of the chains specifies the answer for all of its Nodes, and because there are only two potential chains, the two choices “ON” in the last cell are the sole options. As a result, none of the other choices in that cell can be the answer.

**Type 4 Forcing Digit**

(two candidates “ON” in a region): the two chains turn the same candidate “ON” twice in the same region (Row, Column, or Square). Because one of the chains identifies the solution for all of its Nodes, and because there are only two chains feasible, the candidate cannot be the solution in any other Cell of that region.

If candidate 7 is the answer in E5 (i.e., is “ON” in E5), then-candidate five is put “ON” in C4 by following the blue Chain.

If candidate 7 is not the answer in E5 (i.e., is “OFF” in E5), then-candidate five is put “ON” in A6 by following the pink Chain.

Candidate 5 must answer either C4 or A6in all hypotheses. This means that option five cannot be the answer in any cell where C4 and A6 are present, such as Square “2,” eliminating candidate 5 in A4.

**Sudoku Solving Techniques**

**Forcing Chain**

Coloring and Forcing Chain are pretty similar. However, instead of looking at a digit that can be a candidate for precisely two cells in a row, column, or 3×3 box, we’ll look at a cell with exactly two digits as candidates.

Start by looking at the Sudoku puzzle below:

The choices in the yellow cell are exactly two digits 1 and 4. So let’s trace the pathways of these two scenarios:

- If the yellow cell is 1, the green cell with a common column must be 7. The red cell that shares the same 3×3 box as the green cell must be number three. Therefore, the pink cell that shares the same column as the red cell and the same 3×3 box as the yellow cell must be 4. The pink cell can no longer be four since it is in the same column as the blue cell.
- If the yellow cell has a value of 4, the blue cell in the same row as the yellow cell cannot have a value of 4.

The blue cell cannot be 4 in either circumstance. As a result, the numeral four may be securely ruled out of the blue cell as a contender.

This approach, like coloring, can be used to validate a digit for a cell. For example, the following problem is for you to solve:

There are just two possibilities in the yellow cell: 1 and 4. Let’s trace the pathways of these two scenarios:

- If the yellow cell has a value of 1, the blue cell in the same column must have a value of 2.
- If the yellow cell is 4, the green cell that shares the same 3×3 box as the yellow cell cannot be 4, and the two green cells in the sixth column create a Naked Pair pattern with only the digits 2 and 9 as choices. As a result, the red cell that shares the same column as these green cells cannot be two but must instead be 1.

The blue cell must be 2 in both cases. As a result, we can establish that the blue cell is number two.

Please make sure you understand the Introduction to Chains before diving into Forcing Chains or Nets.

Any sequence that leads to a contradiction or verity and hence compels anything is known as a forcing chain (any Discontinuous Nice Loop or any AIC is a Forcing Chain by that definition). Multiple Forcing Chains are made up of chains that do not lead to a contradiction on their own. Instead, all chains can establish either a truth or a contradiction, resulting in a forcing.

Multiple chains have the same result in a variety. If one of the chains’ premises must be true, the outcome must also be true. The following are some examples of premises that can give integrity:

- Chains starting with all candidates in a cell (at least one must be true, and at least one must be false)
- Chains starting with all candidates in a cell (at least one must be true, and at least one must be false)
- Chains that begin with all of a UR’s extra digits (currently not implemented in HoDoKu)

All chains begin with the same premise but lead to contradictory conclusions in a contradiction. Examples of inconsistencies include:

- Chains demonstrating that a cell can’t hold a digit
- Chains removing all instances of a digit from a house
- Chains inserting multiple digits into a cell
- Chains inserting multiple digits into a house

The possibilities of Multiple Forcing Chains are nearly limitless.

Multiple Forcing Chains have practically infinite possibilities.

The sheer number of links in a Forcing Chain or Net might render the picture entirely meaningless in hookup. As a result, only one Chain may be displayed (see Using the Hint System in the Users Manual).

The example presents a straightforward Forcing Chain Verity that establishes that r1c7 must be four (the first image shows all links at once, the other three images show the three chains).

Chain Verity Forcing => r1c7=4

r4c4=1 r8c4=6 r6c8=6 r1c8=3 r1c7=4

r4c6=1 r4c7=3 r1c7=4

r4c7=1 r5c8=5 r1c8=3 r1c7=4

**Meaning:** It can be proven that r1c7 is 4 for every conceivable location of digit 1 in row 4. As a result, r1c7 must be 4.

There is a Forcing Chain Contradiction for every Forcing Chain Verity.

Chain Contradiction Forced in r4 => r1c7=4

r1c7<>4 r1c7=3 r1c8=5 r5c8=1 r8c4=1 r4c4<>1

r1c7<>4 r1c7=3 r4c6=3 r4c6<>1

r1c7<>4 r1c7=3 r1c8=5 r5c8=1 r4c7<>1

**Meaning:** Row 4 does not include digit one if r1c7 is not 4. The premise must be incorrect because this isn’t feasible; therefore, four may go in r1c7.

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Set as Nodes in Forcing Chains