*Note:
*

*Cn is column number, reading left to right.*

*Rn is row number, reading top to bottom.*

If you search online, you can find a 1 – 18 solution that takes fewer steps than what I will show you here. However, that solution, as far as I can tell, does not follow a general strategy. This is a strategic solution to the puzzle, and an illustration of the merits of planning ahead.

The starting board:

First I notice that there are only two pairs of 2 – 8. I notice that the Head could easily be a 2 if I eliminate the 1. I notice that the tail is already an 8. And I notice that the other 8 – 2 is separated by 9s and 1s, and is easily eliminable.

The strategy is to effectively eliminate the 2 – 8 set by leaving a 2 at the Head and an 8 at the Tail.

There are a couple of ways to eliminate the 9s and 1s between the other 8 – 2 match. I notice that C1R3 begins the sequence 5 – 1 – 6. If I could eliminate the 1, that would leave 5 – 6, which I could use later, symmetrically, with the 4 – 5 in R1. So, let’s eliminate the 9s and 1s as follows:

Now, eliminate the 8 – 2 horizontal sequence between R1 and R2, then Check. This will leave an 8 – 2 horizontal sequence in the centre of the board, which should be eliminated.

Here are two blocked verticals. C5R2 – C5R3 – C5R4, and C9R2 – C9R3 – C9R4. If I could eliminate the 1 at C6R3, then I could unblock both verticals by eliminating the 7 – 3 horizontal match. (Note: it takes time to build your skills at thinking ahead like this, and keeping the numbers and patterns straight. Exercise patience.)

To make all this happen, I need more numbers. Since I have already made the game exponentially easier by effectively eliminating the 2 – 8 set, I’ll Check and survey.

Remain focused on the 1 at C6R3. This is your target. If you eliminate the 8 – 2 match in the centre, you expose another blocked vertical involving the 1 at C6R3. This can be unblocked as follows:

Now eliminate your target 1 at C6R3, as well as the 7 – 3 horizontal match it reveals.

Refocus your attention to symmetry. Notice that R1 contains 4 – 5 – 6 – 7, and in the jumble between R2 and R4 are 3 – 4 – 5 – 6. If you could remove those pesky 1s (and potentially some others), this would be a symmetrical sequence. Work toward that as follows.

Once you eliminate the 7 – 3 horizontal match, you’ll need to create a match with the 6 at C6R1. The match with C3R3 is blocked by the 5 at C1R3. However, that five can be eliminated horizontally if you eliminate the 6 – 1 – 4 sequence starting at C3R3.

To do so, you’ll need to eliminate the 1 at C4R3, thereby creating a horizontal match between C3R3 and C1R4. This is easily accomplished.

You’ve exposed the symmetric sequence 4 – 5 – 6 – 6 – 5 – 6. This illustrates the point that following a wise strategy often ends with good sequences. Be confident in your strategy.

Now eliminate the symmetric sequence you just revealed.

Things are looking good, and there are a few ways to proceed here. I chose to eliminate some 1s and 7s, to start revealing sequences of 4s – 5s – 6s. Start with the vertical 7 – 7 match at C3R6 – C3R7, which exposes a 1 – 1 vertical match.

Perhaps you have already noticed that your List starts with 2 – 3, and that if the 1 at C1R9 were eliminated, the List would end with 7 -8. This would set up a 2 – 3 – x – x – 7 – 8 sequence. It follows that there will be an even inventory of matching pairs between the 2 – 3 and the 7 – 8, which means there *might* be a way to eliminate them all.

Proceed with these considerations in mind.

You have established the 2 – 3 – x – x – 7 – 8 sequence, and you have also exposed another symmetric sequence beginning at C4R7, namely 1 – 4 – 5 – 6 – 7 – 3 – 4 – 5 – 6 – 1. Begin eliminating this latter sequence, but pause to notice the position of the remaining 1s.

If you continue with the horizontal sequence, eliminating the 1s, you will leave the remaining 1s stranded. However, you can eliminate one pair vertically and leave a horizontal match. The rest follows fairly obviously.