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Wednesday 8-Aug-2012

... by: Tarzan

Im the estranged, biettr dark cousin, who for most part, has been there but not really there, considering we grew up together. I was supposed to write here after reading a long time ago, so here i am, what with fucked up net connections, trips outta town for weddings, work, part time not really being part time, and generally trying to piece together a life.I think this is awesome, the stuff she's done, people she's got writing, innovation on her experiences, both Indian and what she's seen elsewhere, i think its pretty complete. Personally im not much of a blogger, i dont read much on this stuff either, but man, all this has got me interested .and i'm gonna keep coming back for more, coz i have a really vivid imagination, and im just gonna picture everything written and totally trip out on it ..great work Sheena ..keep it going, love u always ..

Sunday 29-May-2011

... by: sk


Hopefully the rest of my post is not filled with similar errors. I generally use one of two different (from yours) row/cell label groups. So, I tend to some embarrassing typos: The elimination is (7)b2, not b8. I apologize for the confusion.

Saturday 28-May-2011

... by: Andrew Stuart

Excellent and very interesting analysis again this week. I'm going to try and learn some of the lessons here to see what I can add to the AIC search options currently in the armory.

SK, I'm just trying to pick my way through your explanation and I'm stuck on the seemingly simple FXW(7)aj28,a1 => b8<>7. I assume that means Finned X-Wing. But I've mapped that out and the elimination is not in the finned box. Missing a trick here, I dont see how this imples b8<>7. send me a private email on if I'm being thick.

Saturday 28-May-2011

... by: fariadne

An Ariadne Lesson

When I give an "optimized" Ariadne solution using only the basic strategies 1-7 under Show Candidates, the reader has no idea of how much pain and suffering I endured in first finding an Ariadne-based solution. Once a solution is obtained, one can quickly deduce much more "optimized" Ariadne-based solutions. So it is a fair question to ask: Did you cheat and use Stuart's solution count button to get the solution? The answer is no.

So I decided to use this weeks sudoku to show how I first found a solution. Start by loading and saving the puzzle in the Stuart Solver. Then, as usual, turn off all strategies except the basic 1-7 under Show Candidates. You will be using the Clear and Re-Load red buttons many times during this demonstration.

Proceed as far as possible in the Solver. The Solver finds that cell B5 has the value 4. Also, at this point there are exactly 6 cells, A6, C3, E3, G8, H2 and H3 that are 2-candidate cells. So it seems reasonable to expect that some combination of these six cells could be used to fashion a path to the solution. (However, as we wilI sadly discover this is not the case, but on the other hand it will give an excellent example of the Ariadne technique.)

Since cells H3 and H2 share the same row I decided to use (H3, H2) as a starting point. The 4 values for (H3, H2) are (2,5), (2,7), (9,5), and (9,7). For (2,5) enter 2 for cell H3 and 5 for cell H2 and then proceed as far as possible in the Solver. The result will be inconclusive since we get neither a solution or a contradiction. Then clear, reload the puzzle and try (2, 7). This time the Solver finds a contradiction and so (2,7) is not a valid starting path. Similarly, (9,5) and (9,7) are inconclusive. Thus, at this time we can restrict our search for a valid starting path for (H3, H2) to (2,5), (9,5) and (9,7).

Next, extend the search for a valid starting path to (H3, H2, C3). There are 6 such ordered triples since (H3, H2) has 3 values and C3 has 2 possible values. By inspection, the ordered triples (9,5,9) and (9,7,9) are contradictions since that would result in column 3 having 2 cell values of 9. Check the other 4 using the Solver. They will all be inconclusive. Thus, at this time we can restrict our search for a valid starting path for (H3, H2, C3) to (2,5,8), (2,5,9), (9,5,8) and (9,7,8).

Next, extend the search for a valid starting path to (H3, H2, C3, E3). There are 8 4-tuples to consider. Five will be contradictions and the other 3 will be inconclusive.The 3 inconclusive 4-tuples are (2,5,8,7), (9,5,8,7) and (9,7,8,7). We have not solved the puzzle yet but we do know that the value for cell C3 is 8 and the value for cell E3 is 7. I suggest you load in the values for these two cells and save the resulting puzzle.

Next, extend the search for a valid starting path to (H3, H2, C3, E3, A6). There are 6 5-tuples to consider. Three will be contradictions and three will be inconclusive. The 3 inconclusive 5-tuples are (2,5,8,7,6), (9,5,8,7,6) and (9,7,8,7,9).

Next, extend the search for a valid starting path to (H3, H2, C3, E3, A6, G8). There are 6 6-tuples to check. Four will be contradictions and two will be inconclusive. The 2 inconclusive 5-tuples are (2,5,8,7,6,4) and (9,5,8,7,6,4). At this point in time we know that the value for cell G8 is 4. I suggest you load in this value for cell G8 and save the resulting puzzle.

We are almost done. A valid starting path for (H3, H2, C3, E3, A6, G8) is either (2,5,8,7,6,4) or (9,5,8,7,6,4). First enter the values (2,5,8,7,6,4) in the puzzle and proceed as far as possible. You will see quite a few cells with 2 possible values. After a few dead ends I found that using cell C4 would do the trick! The candidate values for C4 are 5 and 9. Enter 5 for cell C4 and proceed with the Solver to obtain the solution. Since we are assuming a unique solution we are done. Thus, assuming only the basic strategies 1-7, a valid path for the complete solution is (H3, H2, C3, E3, A6, G8, C4) = (2,5,8,7,6,4,5).

This certainly is one of the most lengthy solution paths I have used but, as mentioned earlier, it seemed a good bet that 6 2-candidate cells would suffice to find a path to the solution. Once locked in to that belief I would not let go. Usually, that is not a good idea. Like poker it is often better to fold and try to construct a different path!

Friday 27-May-2011

... by: sk

If I were to WAG 2 candidates, given the almost avoidable sets with candidates 15, and considering the givens with candidates 67 - and knowing that this puzzle stumped andrew's solver - a reasonable wag would be to assume that to avoid unavoidable sets, candidates(67) all form such sets, but have at least one given in each. Thus, hazarding (67) f45 would be a WAG for speed. Luckily, this works fine by fariadne's protocol.

Note, this is still nothing better than a slightly educated WAG.

Wednesday 25-May-2011

... by: sk

This puzzle is very close to a reverse BUG on candidates(15). Therefor, one would expect to get perhaps some mileage from uniqueness patterns using(15). I have included more explanation to hopefully expose how solving one such as this can be much about pattern combination.

0) Start 23 one single; HT(289)ach3; FXW(7)aj28,a1 => b8<>7; end sssts; MSF(7) ajBox7,r128 => e2<>7
1) HP(14)d8f9 = (4)e89 - e1 = HP(45-51)d13 = (1)d8 => d8 <>3689
also: HT(145)d138 = (4)d46 - f46 = (4-1)f9 = (1)d8

2) This step is really the only truly difficult one. For those not familiar with complex Quasi sorts of patterns embedded in AIC type structures, I hope the following explanation helps: This step is based primarily upon the relationship created by the UR(15)ac59 => at least one of the following must be true:
[(5)c4, (6)ac9,(3)c9]. Adding (6)a6=(6)c4 within box 2 to the mix allows one to group these animals into a Mutant Quasi XWing(6)Box(2),ac9 = (3)a9. The kraken [(1)column 2] that uses this pattern below is one of the milder kraken types, as one of the kraken animals directly forbids the eventual target and it reduces to Identity Loop (8)g6 = (1)g2

(1)a2 - b123 = b8 - (1=4)d8 - gh8 = (4)g9 - NP(49)g68 = (8)g6
(1)b2 - HP(36)be2 = (6)a2 - QXW(NP(69)a6ac5,(6)ac9)) = UR(15)ac59 = (3)c9 - (3)j9 = NQ(2549)j9g789 - (49=8)g6

=> g2<>8; 5 singles; LC(5)b123; LC(9)j456; LC(7)ef3; Kite(5) => j9<>5; LC(5)j45;

3) The next two steps beat up the same relationship of a partial XY Chain, amongst other things.
NT(157)agh2 = (6)a2 - (6=9)a6 - (9=3)j6 - (3=2)h5 - h12 = (2-1)g1 = (1)g2 => b2<>1

4) To avoid the UR(15)bd13 at least one of these must be true [(1)f3,(1)b8,(5)b2]. One can group the (1)'s together using the almost Loop (A Quasi Xwing, which does not look like such an animal) [(1)f3 = b8 - d8 = f9 LOOP] = (5)b2. Using Almost Loops like this is much like using an ALS in a chain. (They are logically indistinguishable). Additionally, I nest together a simpler deduction that also targets(6)a8.: the pattern also embeds a two string kite(1).

[(1)f3 = b8 - d8 = f9 LOOP] = (5)b2 - NT(157)agh2 = (6)a2 - (6=9)a6 - (9=3)j6 - (3=2)h5 - (2=9)h3 - (9=8)c3 - a3 = (8)a8 => a8<>16, c2<>1; 2 Singles; LC(1); XYZ Wing(36)e2, (56)b3. (356)b2 => a2<>6; HP(36)be52; LC(6); LC(5)

5) Death blossom like: XWing(3)be28 = (3-6)h8 = (6)e8 - (6=3)e2 => e1579<>3; one single;

6) Can structure this as an ALS XZ rc(6) or an APE, but this seems simpler:
(6)h8 = e8 - (6=2)e9 - (2=3)j9 => h8<>3; XWing(3);

7) (5)d3 = d1 - (5=1)b1 - b8 = (1)d8 => d3<>1; ste

Wednesday 25-May-2011

... by: fariadne

There are no one-cell Ariadne solutions using only the basic strategies 1-7 under Show candidates but there are two-cell Ariadne solutions. One such solution is shown below.

Load the puzzle in the Stuart solver, turn off all strategies except 1-7 under Show candidates and proceed as far as possible in the Solver. The candidate values for cell C3 are 8 and 9. Enter 8 for the cell value and proceed as far as possible in the Solver. The candidate values for cell F4 are 2, 4 and 6. Enter 6 for the cell value and proceed with the Solver. You will get the complete solution.

Note: It was a struggle for me to first find any Ariadne solution and secondly to use that solution to find the "optimized" two-cell Ariadne solution given above.

Wednesday 25-May-2011

... by: NtG

Talk about a candidate dense puzzle!

Tuesday 24-May-2011

... by: Andrew Stuart

Hi everyone. Very embarrassed to see a blank puzzle for the first days of this week. My graphics card blew last Friday and was only repaired today, so I have been offline and unable to see problems. Humblest apologies. This week's Unsolvable is an asymmetrical one - just to be different.

Tuesday 24-May-2011

... by: James

Thanks for the replies. Glad to know it's not my problem. Hope Andrew catches this before the week runs out.

Monday 23-May-2011

... by: L÷wenzahn

Sorry, this week the grid is blank again, and also the printable version does not work, it's even worse than the first time.
Last week all was perfectly working.

Monday 23-May-2011

... by: sk

I have the same issue.

Sunday 22-May-2011

... by: James

I can't see any givens in the grid. Is it my computer?

Each week a new 'unsolvable' will be published and the previous will be accessible here from this archive section. If you like very tough puzzles, these are for you.
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Page created on 03-May-2011, last modified on 03-May-2011.
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