One tough outlaw

Paul · Site Admin Joined: 01 Jan 1970 Posts: 88

Here's the highest-rated puzzle in the new batch I've just generated. Most outlaws aren't rated all that high, because they just need one or two techniques that SDXP doesn't support (so which don't count towards their rating). This one needs 1,477's worth of SDXP solving steps and a series of non-SDXP steps as well. Scanraid's solver (the Boss, in my opinion) had to use just about every trick in the book, including a couple of ALSs and some grouped inference chains.

Anyone fancy a go? Smile

GreenLantern · Joined: 23 Nov 2006 Posts: 95

Fun puzzle, Paul! Thanks for sharing. Here is how I solved it:

1. 2-String Kite in 2's: r4c35,r69c6 => r9c3<>2

Paul · Site Admin Joined: 01 Jan 1970 Posts: 88

Hi GreenLantern - I have to admit that when I said "Anyone fancy a go?" I did mean you! And only you could describe this mind-boggling grid as a "fun puzzle"! Smile

I hadn't heard of a 2-string kite before, but Googled it and discovered it's a neat way of spotting a discontinuous nice loop as a static pattern.

The 2-string kite consists of two conjugate pairs (on the same candidate), one in a row, the other in a column, with two of the squares (one from each pair) in the same box. The candidate can be eliminated from any square that's related to the other two squares in the pattern (i.e. the ones that aren't in the same box as each other).

Here's the puzzle at the start, before the kite elimination:

GreenLantern · Joined: 23 Nov 2006 Posts: 95

I consider 2-String Kites and Skyscrapers to be simple forms of Multicoloring. For example, if there was no '2' in r6c5 in the starting candidates grid, I would have described the Step #1 deduction as a Simple Coloring in 2's. I believe SDXP would describe that as a Chain of Conjugate Pairs.

Step #2 is written in Eureka AIC (Alternating Inference Chains) notation. I am now using AICs in addition to Nice Loops because I believe they are an excellent way of notating certain patterns like XY-chains. The AIC I wrote is actually 2 AICs in one. You can think of it as an extended XY-chain.

If we leave the last two nodes off of Step #2, we have the following AIC:

(6=2)r7c2-(2=7)r6c2-(7=5)r5c2-(5=2)r4c3-(2=6)r4c5

This is how you notate an XY-chain. '6' is the unlinked digit and r7c2 and r4c5 are the endpoints of the chain. The way you read an AIC is to first look at the endpoints. Either r7c2=6 or r4c5=6 since AICs are bidirectional like Nice Loops. Therefore, any cell that can see both endpoints can have '6' eliminated from its candidate list.

Reading left to right, r7c2=2 => (r7c2<>6 and r6c2<>2) => r6c2=7 => r5c2<>7 => r5c2=5 => r4c3<>5 => r4c3=2 => r4c5<>2 => r4c5=6. So, r7c2<>6 => r4c5=6. If we followed the AIC from right to left, then we would conclude that r4c5<>6 => r7c2=6.

Now, let's look at the whole AIC in Step #2:

(6=2)r7c2-(2=7)r6c2-(7=5)r5c2-(5=2)r4c3-(2=6)r4c5-r4c7=r5c9

It is understood that when you leave off the parenthesis before a node that the digit involved will be the same as in the previous node. So, r4c5=6 => r4c7<>6 => r5c9=6. So overall, r7c2<>6 => (r4c5=6 AND r5c9=6). This is an XY-chain which allows one to conclude r7c59<>6.