PR 2023 R1 - Classics (13^{th} - 19^{th} Jan) Score •Discuss

SM 2023 R1 - Standard & Odd Even (20^{th} - 26^{th} Jan) Score •Discuss

SM 2023 R1 - Standard & Odd Even (20

@ 2011-07-19 11:41 AM (#5240 - in reply to #5239) (#5240) Top | |

Posts: 89 Country : India |
I would second david's point on the fact that puzzles should be sequenced as per points, but i loved the time based scoring system, knowing your strengths and selecting the right strategy should be rewarded |

@ 2011-12-08 9:10 AM (#6160 - in reply to #5092) (#6160) Top | |

Posts: 1 |
can any body help out for how to start the "crossnumber sudoku and number 5 still alive"? Edit : Removed a link not related to sudoku / puzzles Edited by debmohanty 2011-12-08 9:15 AM |

@ 2011-12-09 4:11 AM (#6163 - in reply to #6160) (#6163) Top | |

Posts: 66 Country : Hungary |
atc121212 - 2011-12-08 9:10 AM can any body help out for how to start the "crossnumber sudoku and number 5 still alive"? Edit : Removed a link not related to sudoku / puzzles Crossnumber: There are just 4 3-long and 4-long numbers. Not too hard to find their places. Four instance in the 1st row the 3-long number can be just one from the list. After the 3-long number in the last row also unique. You can put the other two 3-long number easily. Now you can find the places of 4-long numbers. Number 5: A bit harder (it is 78 points) Four instance in the left-middle box there are 3 cages. The sum 5+5+15 or 5+15+15 But the second isnt good, because the total of remained cells would be 10, and a 6 is given. So it is 1+3+6, but R3C1=3 Hence the correct sum 5+5+15 so this 3 cages contain (2+3) (1+4) (7+8) and the 2 remained cells (5,9) You can consider the other side as well, the 2 remained cells (1,5) You could write down numbers, but these are important informations. If you look at the 5th column you will see 4 cages. The total of this cages 40 or 35 or 30 etc. So 40. And R5C5=5. The number pairs in this column (14) (23) (69) (78) and just one cell can be 9. (R3C5) Now you can find the 9 and 6 in the right-middle box. (5,9) in the left-middle box and (1,5) in the right-middle box. Moreover all 9s unique. I hope this will be enough. |