




Post an idea here... Discussion... Thursday 24Apr2014... by: cialis online canadattcxmAnQS cialis online canada discount cialis online viagra canada onlineSaturday 14Dec2013Tuesday 10Dec2013Sunday 8Dec2013Sunday 8Dec2013Thursday 5Dec2013Tuesday 3Dec2013Monday 2Dec2013Monday 18Feb2013... by: CadetThis has been going on for over a year. The problem comes and goes. I tried on many mhiecnas pretty sure the problem is Yahoo. Not seem to care much about it. You have not answered my question.Friday 1Feb2013... by: SotirSE=9.0SSTS to UP=29 1) AUR(34)AF79: =>A79<>3 2) FXW(5): GHJ4=A4A9=J9,=>J56<>5 3) (1)ABC5=(14)J5=(49)J3=J1(9=78)DE1(8=5)E2(5=6)E5(6=1)D4.=>A4<>1 4) (8)C3=C2(1=19)CBD2(4)D2=(4)D3,=>D3<>8 5) Kraken : (1)J456,=>J6<>2 (1)J4(1=6)D4(6=5)E5(5=8)E2C2=(82)C3=(2)C6  (14)J5=(48)J3=(82)C3=(2)C6  (1)J6 6) (3=5)A4B56=(53)B7=(3)C9,=>C5<>3 7) (7=1)J6F6=F3A3=A56(1=7)C5,=>BC6<>7 8) (8)C3=(84)J3=(41)J5=J4(1=6)D4(6=5)E5(5=8)E2,=>C2<>8. STE Wednesday 26Dec2012... by: raySorry ! typing error. It should be the 6_tuple (J7,H9,G8,J4,D5,D6) = (4,2,1,3,8,9).Tuesday 25Dec2012... by: rayOops! The puzzle is " Second Flush".Tuesday 25Dec2012... by: rayThe "Golden Flush" puzzle can be solved only basic strategies and Fariadne's method. Using thebilocation cells (J7,H9,G8,D5,D6) to form a progressive 5tuple by forcing the cell values and retaining only the "inconclusive results" progressive ntuple values . When the 5tuple(J7,H9,G8, D5,D6) = (4,2,1,3,8,9) a solutions is found. Using this solution, the following other solutions can be found: 2cell B4A7 = (8,6) and one of the 3cell F1F2F3 = (2,6,4) when applying only basic strategies. Sunday 25Nov2012... by: Steve Pnot too bad 2.21Monday 6Aug2012... by: fariadneSorry, I meant June 26.Monday 6Aug2012... by: fariadneUpdate:In regard to my comments on June 24. The upper bound is indeed at least 3 as Puzzle #66 shows! As is to be expected, Puzzle #66 is extremely difficult. Monday 2Jul2012... by: LerenNot quite sure what all the fuss about these so called "hardest" puzzles is. # 28 is completely trivial when the double Exocet structure at the start (38 eliminations) is cleared. # 49 is harder but solves similarly to puzzles like Fata Morgana or Tungston rod with an Exocet/Abi loop structure at the start (16 eliminations). The latest Inkala puzzle has a couple of Rank 0 logic structures at the start so technically is a bit weaker than Escargot but takes a bit longer to solve .The "most difficult" puzzle for my solver at the moment is the following, called "Second Flush" 12..3....4....1.2...52..1..5..4..2......6..7......3..8.5....9....9.7..3......8..6 This is closely followed by many (ie many 000's) of others, including the following "Golden Nugget", a classic "most difficult" puzzle. .......39.....1..5..3.5.8....8.9...6.7...2...1..4.......9.8..5..2....6..4..7..... Enjoy!! Saturday 30Jun2012... by: fariandeThe New Inkala PuzzleEmploying the algorithm found on the Puzzle #56 site, we find a solution path for the new Inkala Puzzle cited in the comments by Andrew. As is always the case, we only use the basic strategies 16. Load and save the puzzle in the Solver and turn off all strategies other than the basic 16 under Show Candidates. Then proceed as far as possible in the Solver. The Solver gets stuck before it fills in any cells. At this point in time, I think that cells D4 and E4 lying in Column 4 and Box 5 are good cells to begin a solution path since both these cells have the same set of 4 candidate values and these values are widely distributed within the aforementioned Column and Box cells. So we will start with cell D4 having the two candidate values 2, 3, 8 and 9. Inserting 2 for D4 and proceeding with the Solver yields an inconclusive result. Similarly, 3, 8 and 9 also yield inconclusive results. Thus, all four candidates get carried to the next step. Next we add cell E4 to the mix and search for a valid starting path of the form (D4, E4). Since E4 has the same four candidates as D4, the 16 candidates for a (D4, E4) starting path are (2,2), (2,3), (2,8), (2,9), (3,2), (3,3), (3,8), (3,9), (8,2), (8,3), (8,8), (8,9), (9,2), (9,3), (9,8) and (9,9). Of course, the four 2tuples (2,2), (3,3), (8,8) and (9,9) are immediate contradictions leaving us with only twelve 2tuples to check. Inserting (2,3) for (D4, E4) and proceeding with the Solver yields an inconclusive result. Inserting (2,8) for (D4, E4) and proceeding with the Solver yields the complete solution and we are done . Thus, (D4, E4) = (2,8) is a 2cell solution path using only the strategies 16. Using the solution just found it is easy to show that there are no 1cell solution paths using only the strategies 16. The 2cell solution path (D4, E4) = (2,8) we found is not optimal. It takes some tedious work but it is not difficult to show that there are four optimal 2cell solution paths and they are (B2, J4) = (4,3), (E3, D3) = (9,4), (F5, E4) = (6,8) and (G2, J4 ) = ( 2,3) in which the first component of each path is a 3candidate cell and the second component is a 4candidate cell giving a product of 3x4=12 instead of 4x4=16. Friday 29Jun2012... by: GouesOh, sorry, didn't see it being there already :DFriday 29Jun2012... by: Andrew StuartI have changed my tune on potential for Level3 puzzples. I've now checked all the level 2 unsolvables. David Filmer is the hands down winner with these two puzzles:
So our weak conjecture, fariande, might be defeated. The first puzzle is tantalizingly close to a level 3. You can load Arto Inkala's puzzle from this link Tuesday 26Jun2012... by: fariandeAndrew,The data cited in your article makes it a reasonable conjecture that the least upper bound is 2. In all likelihood we will never know its true value. I personally tend to think that it is larger and is probably 3. My only reason for thinking this is that there are far too many cases in which 2 wellchosen cell assignments are required. A weak reason for sure! Monday 25Jun2012... by: Andrew StuartHi All. Too much to write so I've created an article. Love to here your views either here or on that page.Sunday 24Jun2012... by: Andrew StuartHi AllI have some great stats and nice charts to put up. I'll do a write up tomorrow. I've been looking at how inserting a number (or two) effects the difficulty of a puzzle along the lines of fariande's idea. More tomorrow :) Sunday 24Jun2012... by: Andrew StuartInteresting discussion. I shall have to add this test and see if it's possible to find an upper bound over 2.Saturday 23Jun2012... by: fariandeOptimal Solutions Using Only The BasicStrategies 16 In Stuart’s Solver I find the information given below very interesting as it shows that all of Stuart’s unsolvable puzzles to date are only one or two wellchosen cell assignments away from being solvable by using only the basic strategies 16 in Stuart’s Solver. The same is true for some of the more famous sudokus such as the Golden Nugget, the Easter Monster, Al Escargot and Arto Inkala’s Sudoku of 8/23/2010. In fact, I have yet to see an example in which more than two cells are required. So is 2 the least upper bound for the number of cells required? If not what is the least upper bound? My gut feeling is that the only way to find the least upper bound is to test every sudoku. For the moment however, I would be happy to see an example in which more than two cells are required. As usual I am assuming that Stuart’s Solver is being used with all strategies turned off except the basic 16 under Show Candidates. The table below shows that only four of the Stuart Puzzles do not have a 1cell solution path and each of these four have a 2cell solution path. For the 1cell solution path puzzles, the first number inside the curly brackets is the number of cell candidates for the chosen cell when the Solver gets stuck and the second number is the number of 1cell solution paths. The given solution is optimal in the sense that no other cell used for a 1cell solution path has a smaller number of cell candidates. For the 2cell solution path puzzles, the number inside the curly brackets is the product of the number of cell candidates for the two chosen cells when the Solver gets stuck. The solution is optimal in the sense that no other pair of cells used for a 2cell solution path has a smaller product. Puzzle #1: D2=9 {2,18} Puzzle #2: J4=3 {2,16} Puzzle #3: G4=5 {4,3}} Puzzle #4: (C3,F4)=(8,6) {2x3=6} Puzzle #5: D8=1 {4,3} Puzzle #6: B1=1 {3,3} Puzzle #7: A2=9 {2,13} Puzzle #8: B2=7 {3,21} Puzzle #9: D9=9 {2,12} Puzzle #10: G7=5 {3,12} Puzzle #11: A3=7 {2,26} Puzzle #12: A2=6 {2,13} Puzzle #13: G7=4 {2,7} Puzzle #14: A3=1 {3,7} Puzzle #15: H3=4 {4,2} Puzzle #16: A5=5 {2,10} Puzzle #17: E1=3 {3,3} Puzzle #18: G2=8 {2,16} Puzzle #19: J6=3 {2,7} Puzzle #20: D4=7 {2,7} Puzzle #21: H9=1 {2,16} Puzzle #22: B7=9 {2,17} Puzzle #23: C2=5 {3,5} Puzzle #24: B6=5 {3,5} Puzzle #25: A9=4 {2,16} Puzzle #26: B2=1 {3,6} Puzzle #27: A2=5 {3,15} Puzzle #28: (C1,E8)=(3,9) {3x4=12} Puzzle #29: (B6,H1)=(9,9) {3x2=6} Puzzle #30: D2=6 {2,14} Puzzle #31: E6=7 {2,13} Puzzle #32: E1=6 {3,6} Puzzle #33: A8=8 {3,4} Puzzle #34: G3=9 {3,2 Puzzle #35: C4=6 {2,11} Puzzle #36: E1=4 {3,6} Puzzle #37: E2=7 {3,6} Puzzle #38: E4=5 {4,4} Puzzle #39: A1=5 {4,4} Puzzle #40: C6=3 {2,14 Puzzle #41: J2=5 {3,10} Puzzle #44: F5=8 {2,25} Puzzle #45: B5=7 {3,6} Puzzle #46: F5=8 {3,7} Puzzle #47: A9=8 {3,3} Puzzle #48: C3=8 {3,2} Puzzle #49:(A2,D3)=(4,8) {3x5=15} Puzzle #50: G1=1 {3,10} Puzzle #51: C3=3 {4,6} Puzzle #52: A2=6 {2,11} Puzzle #53: F3=9 {3,9} Puzzle #54: B6=2 {2,6} Puzzle #55: G3=8 {4,1} Puzzle #56: E9=3 {3,9} Puzzle #57: H4=9 {3,5} Puzzle #58: G4=4 {3,6} Puzzle #59: F5=8 {5,1} Puzzle#60: B6=6 {4,2} Puzzle #61: F3=5 {2, 17} Saturday 23Jun2012... by: fariandeTechnically we use the algorithm found on the Puzzle #56 site to finds a solution path using only the strategies 16 although the solution in this case is so simple the algorithm is really not needed.Load and save the puzzle in the Solver and turn off all strategies other than the basic 16 under Show Candidates. Then proceed as far as possible in the Solver. The Solver fills in the correct cell values for the three cells E8, F1 and F8 before getting stuck. At this time, Row F is certainly the best place to start looking for a solution path since the four unknown cells in that row split into 15 and 34 pairs. We start with the 2candidate cell F3 having the candidates 1 and 5. Inserting 1 for F3 and proceeding with the Solver yields an inconclusive result. However, inserting 5 for F3 and proceeding with the Solver yields the complete solution and we are done. Using the solution just found it is easy to show that there are exactly seventeen 1cell solution paths using only the basic strategies 16. They are A3=1, A4=3, B6=5, C5=1, D1=9, D2=1, D4=6, E2=8, E5=5, E9=6, F3=5, F6=1, G2=5, G5=6, H5=3, J1=8 and J4=1. I consider this to be the easiest of the 61 puzzles. There are nine 2candidate cells when the Solver gets stuck and five of these cells give a 1cell solution path. They are D4=6, E2=8, E5=5, F3=5 and F6=1. Saturday 23Jun2012... by: JC Van HAy+6D4 is a backdoorwhile +1D4>{+5F6 and [+1F3>+1A56>(9B5=(91)A5=(14)J5=4J34D3=(49)D2=9B2)>+6B3]}>+7B6>contradiction using only singles 
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