AIC with Groups

Grouped nodes were discussed on the Grouped X-Cycles page and it is very relevant to Alternating Inference Chains. Luckily, there’s nothing too scary about them although they may be harder to spot.

Note: Examples here will show up if Forcing Nets are turned off.

Rule 1 - Off-Chain Elimination

Grouped AIC : Load Example or : From the Start

Kicking off with a relatively short and 'continuous' loop. This means there is no start or end, nor any 'discontinuity' where two weak links join or two strong links join - and where we'd be looking for an elimination. Instead one can trace a route in either direction and the ON and OFF could be swapped. This type of loop allows us to make eliminations on all weak links, shown in yellow/red. One end or other of each link must be the highlighted candidate.

What makes this a 'Grouped' AIC chain is the double cell D3/F3. If you start with +8[D1] that potentially removes all other 8s in box 4. If that were so the only remaining 8 in column 3 is on B3. In order to make that link we combine D3 and F3 into a single pseudo-cell DF3. Note that if we use two or three cells in a group it has a specific orientation or alignment. It can only be useful in the direction the cells are aligned on. In this case it points along the column. The solver gives us:

AIC (w.Groups) Rule 1:
+4[B1]-4[D1]+8[D1]-8[D3|F3]+8[B3]-4[B3]+4[B1]
- Off-chain 8 taken off D2 - weak link: (8)D1 - to (8)D3|F3 =
- Off-chain 8 taken off F2 - weak link: (8)D1 - to (8)D3|F3 =
- Off-chain 7 taken off B3 - weak link between 8 and 4 in B3

Rule 2 - Strong Link Assertion

Grouped Cell AIC : Load Example or : From the Start

Rule 2 loops are 'discontinuous' in that they have a kink or point where the alternating weak/strong inferences break down. The thick blue line from 2[H5] to 2[H8] is very visible. A bit harder to see is the other link between 2 and 6 in the same cell H5. Rule 2 says we can assert that 2 must be the solution to H5.

But making this a 'Grouped' AIC is the pseudo-cell GH6. If A6 is not 6 then one/either 6s in G6 or H6 must be ON. The orientation coming in from the column means we must continue the loop by looking in the box. We can turn OFF 6s there including H5. The logic on those links works in reverse as well, box to column. The solver gives us:

AIC on 2 ((w.Groups) Discontinuous Alternating Nice Loop, length 8):
-2[H5]+2[H8]-2[A8]+6[A8]-6[A6]+6[G6|H6]-6[H5]+2[H5]
- Contradiction: When 2 is removed from H5 the chain implies it must be 2 - other candidates 6 can be removed

Rule 3 - Weak Link Elimination

Rule 3 Grouped Link example : Load Example or : From the Start

Finally we have the weak-link version of the 'discontinuous' loop where we can remove the candidate which forms the discontinuity. This is probably the most common type of elimination, followed by Rule 1 off-chains.

There is a very clear Grouped Cell in AC7. We can follow the solver's chain which comes in from the box and A3 with a -3. The 3s in AC7 are the only ones left in box 3 so we can combine them into a link cell if we follow their alignment down the column. This takes us to the -3 in J7.

To figure out we have a discontinuity we note we started with asserting J7 was 3 and the loop proves this assertion didn't work out. So the assertion was wrong and we can remove 3 from J7.

AIC on 3 ((w.Groups) Discontinuous Alternating Nice Loop, length 8):
+3[J7]-2[J7]+2[J9]-1[J9]+1[A9]-3[A9]+3[A7|C7]-3[J7]
- Contradiction: When J7 is set to 3 the chain implies it cannot be 3 - it can be removed

Go back to Alternating Inference Chains

Continue to AICs with ALSs

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... by: Rick

Sunday 27-Jul-2025

On the 20-7-2025 puzzle I was using the solver to compare with the route I took to solve the puzzle (as I had to resort to cell forcing chains at one point) and have come across an air where it jumps from a 4 in one cell to a pair of 9s in the same box. I can’t see how this is possible.

I have a screenshot but can’t see how to post it. Can anyone explain / help?

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... by: Robert

Saturday 23-Jan-2021

I've thought about this one quite a lot.

Specifically - is there any kind of sensible group other than the same value candidate appearing two or three times within the intersection of a row and a box, or within a column and a box? Answer: no. You can contemplate other sorts of groups, but you don't get any sort of inference you can't get beyond what you get with these sorts of groups, which consist of the same valued candidate two or three times.

And - is it ever convenient to consider a group that consists only of two candidates, even when that would be a "sub group" of one containing three candidates? That is, does it ever make sense to have the group exclude candidates it could otherwise include? Answer: yes. Suppose the candidate "8" appears in A1, A2, A3, B1, and C1, and nowhere else in the box. There are other occurrences of the candidate "8" in the first column, and in the first row. Then you can break the five candidates in the box into two groups, either (A1 A2 A3), and (B1 C1), or (A2 A3) and (A1 B1 C1) - it doesn't matter which way you do it. Then the two groups are strongly linked. If there are weak links between (A1 B1 C1) and another candidate in column 1 and between (A1 A2 A3) and another candidate in row 1, the chain can use these two groups, connected by a strong link. If you included A1 in both groups, you couldn't do this - the two groups could both be "on".

REPLY TO THIS POST

... by: smmcroberts

Tuesday 3-Dec-2019

In example 1, in cell B7, why isn't the bi-value link between candidates 1 and 3 a strong link?

Andrew Stuart writes:

It is a strong link but we can use it as a weak link for the sake of the chain. See Weak and Strong Links

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... by: Atomic

Saturday 7-Jul-2018

Are overlapping groups of different candidates allowed?
In other words, let's say you visit a block and find a vertical group of 1's, but the uppermost cell in that group also has 2 as a possibility. Now you revisit the block again in the chain and find a horizontal group of 2's... but the leftmost cell is shared with the 1's group. Will this still be a valid Grouped AIC?

Andrew Stuart writes:

Good question. I say no, you should not re-use any candidate in a chain. I extend that to using any other candidate in a cell or unit as part of a chain. It might work if you can see that the ON and OFF are complimentary, ie the links don't contradict each other. But the solver does not allow itself to compute this.

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