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how difficult is this one? http://www.calcudoku.org/forum/viewtopic.php?f=19&t=774 
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Author:  pnm [ Mon Feb 22, 2016 12:53 pm ] 
Post subject:  how difficult is this one? 
Hello, I got a question from a puzzler about the difficult Killer Sudoku of Feb 20th: "How come this puzzle is rated difficult? In my opinion it is easy." What do other people think? What are some of the step(s) that make this an easy puzzle? thanks, Patrick 
Author:  aquilae [ Tue Feb 23, 2016 11:45 am ] 
Post subject:  Re: how difficult is this one? 
I found this one to be more of a medium than an easy. This one required a few more types of techniques than easies usually do, or at least the way I did it, but this was one of the easiest killer sudoku puzzles labeled difficult that I've solved. These are the things I think make this puzzle easier, in order: (I'll be referring to the 9cell subdivisions as blocks and the groupings of a few cells that add up to something as boxes) Step 1: Right off the bat you can get b3 and a9 for free due to the fact that each column, row, and block must add up to 45. You can infer c8 and c9 quickly by similar logic. From there things line up almost perfectly so finding a single possible combination of numbers for each box the left three blocks, (or left 3 columns, whichever perspective you prefer), is very straightforward. Step 2: The 16 box and the 3cell 7 box only have one possible combination of numbers by default, which helps out a lot. The 16 box allows you to fill d8 and e8, which allows you to fill e1. Step 3: Finding a single combination for each box in the bottom three rows is pretty straightforward at this point. You start with the bottom row and work up. In the end you'll fill e6 and find that the only possible combination for h6 and i6 is 4 and 2. Step 4: That 29 box in the center of the puzzle has all but one of its cells in the ecolumn which of it takes up almost half. You already know all but one of the remaining other cells in that column and only two possibilities for that one cell. So you end up knowing that the 29 box has 8, 4, and 3 in the ecolumn, and also either 9 or 6. 9 would correspond to 5 in f2, and 6 would correspond to 8 in f2, but we know f2 can't be 8 because the 29 box already has an 8. Step 5: If you notice the top right block, 19+14+12 = 45, which means f1 must be equivalent to h3. Remember that h and i6 contain 4 and 2, which allows you to infer that g6 and h3 are equivalent by similar reasoning, and also allows you to find a single possible combination each for the 12 box and the 10 box in the middle right block. This gives you three equivalent cells in three different blocks, with only three possible numbers for them, which you can narrow down faster. After all that the rest is just cleanup. TLDR;  Free numbers (from stuff adding to 45)  Free combinations (i.e. the 3cell 7 and 2cell 16)  The number combinations of the big boxes are easily discovered, where in other puzzles boxes of 4 or more cells are usually the last ones you can narrow down.  3 equivalencies  All of these working together in just the right way to severely limit possibilities at each point 
Author:  case [ Wed Feb 24, 2016 5:54 pm ] 
Post subject:  Re: how difficult is this one? 
The procedure described by aquilae to solve the puzzle is pretty similar to the one I used. The only differences I can see are: 1) Once you have determined that the combination needed to fill h7 and i7 is 1 and 5, you know that the combination for h6 and i6 has to be 2 and 4, so you get that a bit sooner than at the end of step 3. 2) After completing step 4, you have most of the middle three blocks filled out, including the 2 in f5. You can now determine that the combination needed to fill f6 and g6 is 3 and 8. In step 5 it is reasoned that g6, h3 and f1 are equivalent. h3 cannot be 3 because of the 3 in b3, so you now know that the three equivalent boxes must be 8. 
Author:  nicow [ Sun Feb 28, 2016 1:43 am ] 
Post subject:  Re: how difficult is this one? 
I think it is easy, it is solvable by easy steps. 
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