@ 2011-02-14 12:02 PM (#3495 - in reply to #3492) (#3495) Top | |

Posts: 774 Country : India |
akash.doulani - 2011-02-14 11:38 AM One solution could be ...to start the test 30 minutes after you click on "Start" ...wasnt able to do anything in the last 30 minutes . someone please help me get over this serious problem. |

@ 2011-02-14 12:20 PM (#3496 - in reply to #3495) (#3496) Top | |

Country : India |
rakesh_rai - 2011-02-14 12:02 PM akash.doulani - 2011-02-14 11:38 AM One solution could be ...to start the test 30 minutes after you click on "Start" ...wasnt able to do anything in the last 30 minutes . someone please help me get over this serious problem. funniest answer :-) on a serious note, please help him if you can.... |

@ 2011-02-14 1:27 PM (#3497 - in reply to #3492) (#3497) Top | |

Posts: 460 Country : India |
akash.doulani - 2011-02-14 11:38 AM it was a great set of sudokus. started with 1st one but dont know how , i was stuck in that for a good 15 minutes and ultimately got it wrong. then did the 4 killers . then did the jigsaw diagonal. aroud 30 minutes were left and i started with disjoint diagonal . and as deb nicely put it "Akash continued his trademark not doing anything in last 30 minutes :-) " i dont know how i got it wrong . so wasnt able to do anything in the last 30 minutes . someone please help me get over this serious problem. Thanks Akash, I think what you should do is, if you get stuck in a puzzle or are wrong, put it aside and solve others and go back to the wrong one later with a fresh mind.. atleast that is what i do when i go wrong in a puzzle. If I start that wrong puzzle again immediately, the previous numbers keep popping into my mind and stop me from thinking clearly. Rishi |

@ 2011-02-14 1:29 PM (#3498 - in reply to #3494) (#3498) Top | |

Posts: 460 Country : India |
rakesh_rai - 2011-02-14 11:59 AM Is there a neat and elegant approach for solving the 150 pointer? Can someone share on the forum? I got stuck after a while...and had to resort to trial and error to complete it. I will share that Rakesh but I would love to see Motris and Rohan post their approach. I am not aware of Snyders exact time for that puzzle but I know Rohan cracked it in under 7 mins which made a mockery of the puzzle. It was awesome the way he solved it. @Rohan, maybe you can share your approach with ppl here. Rishi |

@ 2011-02-14 4:16 PM (#3499 - in reply to #3357) (#3499) Top | |

Posts: 739 Country : India |
First of all, nice set of puzzles Rishi. Got completely mixed up with the variants in the middle of the test, but managed to come out clean :D The 150 pointer was the Diagonal-Antiknight Sudoku. I started the test with this sudoku. I got a couple of numbers and was then stuck. A little guesswork made me progress with a couple of more numbers. By this time 4 minutes were gone. Then it struck me. Observe the given numbers. All 4s and 8s are symmetrically opposite. All 2s and 5s are symmetrically opposite. Similarly, 1s and 7s, 6s and 9s. Also, the 3s are symmetrically placed. Once you realise this, I think the puzzle is not very difficult because the entire grid will have these pairs in symmetrical opposite cells. Even after getting a few numbers, you will observe the pattern holds. Hence, I directly placed a 3 in the middle cell (R5C5). Then I solved the puzzle in about 3mins using this logic. Solving the toughest puzzle in 7 minutes does give you a boost in confidence :-) But I still couldn't complete all 12 :| If I found it, I'm sure many of the top solvers must have used this logic too :-) Rohan. |

@ 2011-02-14 5:43 PM (#3504 - in reply to #3499) (#3504) Top | |

Posts: 460 Country : India |
Rohan Rao - 2011-02-14 4:16 PM First of all, nice set of puzzles Rishi. Got completely mixed up with the variants in the middle of the test, but managed to come out clean :D The 150 pointer was the Diagonal-Antiknight Sudoku. I started the test with this sudoku. I got a couple of numbers and was then stuck. A little guesswork made me progress with a couple of more numbers. By this time 4 minutes were gone. Then it struck me. Observe the given numbers. All 4s and 8s are symmetrically opposite. All 2s and 5s are symmetrically opposite. Similarly, 1s and 7s, 6s and 9s. Also, the 3s are symmetrically placed. Once you realise this, I think the puzzle is not very difficult because the entire grid will have these pairs in symmetrical opposite cells. Even after getting a few numbers, you will observe the pattern holds. Hence, I directly placed a 3 in the middle cell (R5C5). Then I solved the puzzle in about 3mins using this logic. Solving the toughest puzzle in 7 minutes does give you a boost in confidence :-) But I still couldn't complete all 12 :| If I found it, I'm sure many of the top solvers must have used this logic too :-) Rohan. Thanks for the inputs Rohan.... You are right about the digits being symmetric. In a competitive environment it definitely helps in noticing such finer details which enable you to solve hard puzzles faster. At times in competitions where time is of essence, guesswork and T&E d o help us solve faster than actual logical methods. But for benefit of all I will explain below the intended path using the two constraints of Diagonal and Anti-Knight as was the intent of the puzzle. On a lighter note, the putting of 3 in R5C5 does not require the symmetry property and comes by default sudoku rules as we have 3 as a given in Box2,4,6 and 8. To begin with, we can place all the 3s in the given grid once we have the 3 in R5C5. I have pencil-marked all the other cells. 8 in box 7 will be in either R7C3 or R9C1 and no where else in the UR-BL diagonal. similarly, 4 will be in R1C9 or R3C7 for the same diagonal. the number 1 will be in either R2C9 or R1C9 in box 3 as per the given image below. Rishi (AKDiag1.png) Attachments ---------------- AKDiag1.png (33KB - 2 downloads) |

@ 2011-02-14 5:51 PM (#3505 - in reply to #3499) (#3505) Top | |

Posts: 460 Country : India |
Now let us look for the options of number 1 in the Tl-BR diagonal in the earlier image. We have the possible places of R1C1, R2C2, R7C7, R8C8 and R9C9. R9C9 is eliminated because of 1 being in either R1C9 or R2C9. in box 6 the possibilities of 1 are either R5C8, R6C7 or R6C8. This forms an L shape ( as explained in my blog the L Technique ) . Whichever cell 1 is in, it eliminates 1 from R8C8, as shown in image below. so that leaves only cells R1C1, R2C2 and R7C7 for the number 1 in UL-BR diagonal. (AKDiag2.png) Attachments ---------------- AKDiag2.png (45KB - 2 downloads) |

@ 2011-02-14 6:02 PM (#3506 - in reply to #3505) (#3506) Top | |

Posts: 460 Country : India |
Now let us look at possibilities of 1 in Row 4 and column 4. These have been marked by rectangular boxes. In R4, we have options of R4C2,R4C3 and R4C5. If 1 was in R4C2 or R4C3, it eliminates 1 from R5C3 since it is the same box and if it is in R4C5 then it is at a knights step from R5C3 and it eliminates 1 again. So 1 can be removed from R5C3 marked by a circle. Similarly the possibilities of 1 in Column 4 are R5C4, R6C4 and R9C4. marked by rectangles. If 1 was in R6C4 it eliminates 1 from R7C3, if 1 is in R5C4 or R9C4 it again eliminates 1 from R7C3 since both these cells are at a knights step from R7C3. So in col 4 whichever cell 1 is in, it eliminates 1 from R7C3 marked by a circle. So we have a revised options for 1 as in image 2. Now we have only R1C1, R2C2 and R7C7 for the digit 1. (AKDiag21.png) (AKDiag22.png) Attachments ---------------- AKDiag21.png (46KB - 1 downloads) AKDiag22.png (32KB - 1 downloads) |

@ 2011-02-14 6:11 PM (#3507 - in reply to #3506) (#3507) Top | |

Posts: 460 Country : India |
Now let us look at Col 3 for the remaining possibilities of number 1. We have R1C3, R2C3 and R4C3, marked by rectangles. If 1 was in R1C3 or R2C3, it eliminates 1 from R2C2 (marked by a circle ) as it is within same box, and if 1 is in R4C3 then also it eliminates 1 from R2C2 as it is at a knights step. so whichever cell in Col 3 the number 1 comes in, it is eliminated from R2C2. Now that leaves us only R1C1 and R7C7 as a possibility for 1 in the TL-BR diagonal, marked by rectangles. Now whichever cell in this diagonal 1 falls in, it cannot be at R7C1, marked by a circle so that eliminates 1 from R7C1 and we can safely put 6 at R7C1. (shown in the second image. (AKDiag3.png) (AKDiag31.png) Attachments ---------------- AKDiag3.png (46KB - 0 downloads) AKDiag31.png (32KB - 1 downloads) |

@ 2011-02-14 6:14 PM (#3508 - in reply to #3507) (#3508) Top | |

Posts: 460 Country : India |
Now that we have got a 6 at R7C1, by using the constraints we get all the 6s in the grid as shown below. Post this I think we can proceed normally and solve the entire grid. This was my interpretation of this puzzle and the reason I gave it such a high value. But in a competition the end justifies the means so you dont get marks for methods used but only for the correct solution. Hope this helps. Rishi (AKDiag4.png) Attachments ---------------- AKDiag4.png (30KB - 1 downloads) |

@ 2011-02-14 6:49 PM (#3509 - in reply to #3508) (#3509) Top | |

Posts: 774 Country : India |
purifire - 2011-02-14 6:14 PM Thanks Rishi for sharing your approach. I don't think my approach was much different from this. But, probably, I had got stuck somewhere where I could have used the anti-knight constraints to eliminate candidates, and had to guess for placing a 4 between two cells. I'll try to solve it again from the point you have left it here.This was my interpretation of this puzzle and the reason I gave it such a high value. But in a competition the end justifies the means so you dont get marks for methods used but only for the correct solution. Hope this helps. |

@ 2011-02-14 7:31 PM (#3510 - in reply to #3509) (#3510) Top | |

Posts: 774 Country : India |
Paper solvers had it simplified this time with a simpler answer entry system (by and large, with a few exceptions). Of course, there can be further improvements which can be discussed and sorted out. I am in the minority who are solving online. And there is one thing which I would like to mention, from an improvement perspective in future tests - the representation of the diagonal lines on the grid. This time we had five grids which had a diagonal constraint. The example on the site also had diagonal lines and was represented using gray lines. But in the test four out of the five grids had black (dashed) lines running diagonally. I am attaching an image which brings out the differences clearly. The pencilmarks in the gray one are clear, but the other one requires a lot of eye exercise while solving the sudoku - especially the numbers 1,3,5,7 and 9 - each of which interfere with one or more of the diagonal lines. We can surely avoid this by choosing slightly light colors for these lines. (diagonallines.png) Attachments ---------------- diagonallines.png (54KB - 1 downloads) |

@ 2011-02-14 10:40 PM (#3513 - in reply to #3504) (#3513) Top | |

Posts: 199 Country : United States |
I already sent Rishi my approach to the last puzzle. On things like Diagonals I often do minor coloring/trial steps to make eliminations. I certainly took longer than "7 minutes" but it was maybe 15 to do this: While I didn't instantly notice it, I caught on quickly that the puzzle had rotational digit symmetry (the 2/5 would be placed in symmetric spots, the 1/7, 6/9, 4/8 as well - 3 is self symmetric). Most of the solve involved considering the most limited digits (first 2s and 5s) and how certain forcing placements cause contradictions. I'll describe steps for "2" but same applies for "5" on other end After placing the 3's, 2 can be in three positions in box 4, but if it is in R5C2 it leaves no spot for a 2 in column 3, so a 2 is in R45C3, and in C2 in box 7. If the 2 is in R7C2 (pushing the resulting 2's), there is no possible place for a 2 on the LL-UR diagonal, so R8C2 = 2. Next, either there are two 2s at R4C3 and R3C4 or three 2s at R5C3, R4C4, R3C5. The latter case (more tempting to try) led to an immediate contradiction in the lower-right box as you can't put in 2's to avoid knight constraints or diagonal constraints. So 2 = R4C3 and R3C4. Then, 2 is in R6C56 but if in R6C5, then no way to put a 2 on the UL-LR diagonal. So 2 in R6C6. Then (describing 4, but 8 is symmetric), there must be a 4 somewhere in column 6. You know the upper-right has a 4 in either R3C7 or R1C9 and these form (from the anti-knight constraint) a perfect elimination of a 4 from R1C6. This is my favorite step. Anyway, other digits eliminate all but R9C6 for that 4. This gives several more 4's, and all the 2's as a result. Last necessary step comes from thinking about R5C7 spot. It contains a 7 or a 9. Whichever digit is in this cell must also be in R6C6 because of the eliminations in the center. But a 7 is already forced in C2 in R45, so you can't have the two 7s in the other spots, so this is a 9. From there it is just entering digits and copying the symmetric logic, but very interesting and challenging puzzle. I think the Anti-Knight Non Consecutive was my favorite, but this was a close second. Edited by motris 2011-02-14 10:44 PM |

@ 2011-02-14 10:59 PM (#3514 - in reply to #3513) (#3514) Top | |

Posts: 460 Country : India |
motris - 2011-02-14 10:40 PM I already sent Rishi my approach to the last puzzle. On things like Diagonals I often do minor coloring/trial steps to make eliminations. I certainly took longer than "7 minutes" but it was maybe 15 to do this: While I didn't instantly notice it, I caught on quickly that the puzzle had rotational digit symmetry (the 2/5 would be placed in symmetric spots, the 1/7, 6/9, 4/8 as well - 3 is self symmetric). Most of the solve involved considering the most limited digits (first 2s and 5s) and how certain forcing placements cause contradictions. I'll describe steps for "2" but same applies for "5" on other end After placing the 3's, 2 can be in three positions in box 4, but if it is in R5C2 it leaves no spot for a 2 in column 3, so a 2 is in R45C3, and in C2 in box 7. If the 2 is in R7C2 (pushing the resulting 2's), there is no possible place for a 2 on the LL-UR diagonal, so R8C2 = 2. Next, either there are two 2s at R4C3 and R3C4 or three 2s at R5C3, R4C4, R3C5. The latter case (more tempting to try) led to an immediate contradiction in the lower-right box as you can't put in 2's to avoid knight constraints or diagonal constraints. So 2 = R4C3 and R3C4. Then, 2 is in R6C56 but if in R6C5, then no way to put a 2 on the UL-LR diagonal. So 2 in R6C6. Then (describing 4, but 8 is symmetric), there must be a 4 somewhere in column 6. You know the upper-right has a 4 in either R3C7 or R1C9 and these form (from the anti-knight constraint) a perfect elimination of a 4 from R1C6. This is my favorite step. Anyway, other digits eliminate all but R9C6 for that 4. This gives several more 4's, and all the 2's as a result. Last necessary step comes from thinking about R5C7 spot. It contains a 7 or a 9. Whichever digit is in this cell must also be in R6C6 because of the eliminations in the center. But a 7 is already forced in C2 in R45, so you can't have the two 7s in the other spots, so this is a 9. From there it is just entering digits and copying the symmetric logic, but very interesting and challenging puzzle. I think the Anti-Knight Non Consecutive was my favorite, but this was a close second. You did indeed send me your methodology Thomas and in hind sight your approach appears much better than what I have put up. The only reason I mentioned in the forum for you and Rohan to post your methods was that others would be genuinely interested in knowing how you approached the puzzle. :) Rishi |

@ 2011-02-14 11:09 PM (#3515 - in reply to #3510) (#3515) Top | |

Posts: 460 Country : India |
rakesh_rai - 2011-02-14 7:31 PM Paper solvers had it simplified this time with a simpler answer entry system (by and large, with a few exceptions). Of course, there can be further improvements which can be discussed and sorted out. I am in the minority who are solving online. And there is one thing which I would like to mention, from an improvement perspective in future tests - the representation of the diagonal lines on the grid. This time we had five grids which had a diagonal constraint. The example on the site also had diagonal lines and was represented using gray lines. But in the test four out of the five grids had black (dashed) lines running diagonally. I am attaching an image which brings out the differences clearly. The pencilmarks in the gray one are clear, but the other one requires a lot of eye exercise while solving the sudoku - especially the numbers 1,3,5,7 and 9 - each of which interfere with one or more of the diagonal lines. We can surely avoid this by choosing slightly light colors for these lines. Thanks Rakesh for bring this up.. I have received a lot of comments on the use of lines as diagonals and will avoid it in the future. Rishi |

@ 2011-02-15 2:48 AM (#3516 - in reply to #3357) (#3516) Top | |

Posts: 170 Country : Germany |
Thanks for the great puzzles. I'm not usually much into Sudoku, but these were great fun. (I did poorly on the test, but solved the rest later.) By the way, the Anti Knight Disjoint is solvable without the "disjoint" constraint. Yes, finding this out was an accident. Cheers Rob |

@ 2011-02-15 8:42 AM (#3517 - in reply to #3516) (#3517) Top | |

Posts: 10 Country : Canada |
I didn't see the symmetry of the 150 point puzzle but I was able to solve it logically. For instance you can see from the first diagram of Rishi's, that R2C8 is a 5 because the anti-knight moves cancel out the possibilities for 5 being in any other spot on the diagonal. Edited by WaterlooMathie 2011-02-15 8:43 AM |

@ 2011-02-21 12:11 AM (#3561 - in reply to #3357) (#3561) Top | |

Posts: 4 Country : Poland |
Hi, Is there anywhere password to get the PB? regards |

@ 2011-02-21 4:44 AM (#3564 - in reply to #3561) (#3564) Top | |

Country : India |
pafcio123 - 2011-02-21 12:11 AM Please login in the Double Delight page and you should see the password.Hi, Is there anywhere password to get the PB? regards |