A technique I've been using a lot of recently, I'm not sure has a name (I haven't seen it here, will look at the other links posted recently). Basically, a good real-life analogy is airspace. If a digit in one set of three boxes is only in one row/column, then it can be eliminated from the same row/column in the other two boxes that it is adjacent to. Also, if two boxes are restricted to the same two rows/columns, then the third must be in the remaining row/column.
That sort of thing.
I like working in "triplets" (each box can be divided up into three triplets in two directions: row/column).
Techniques for fast solving
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by Dusty Chalk » Tue Aug 23, 2005 5:46 pm
Last edited by Dusty Chalk on Tue Aug 23, 2005 7:46 pm, edited 1 time in total.
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by cwtl » Wed Aug 24, 2005 1:14 am
[quote="george-no1"][quote="King Voodoo"]So far I've never had to use the techniques with quads, triples, x-wings and stuff like that.[/quote]
If you've never used triples, which puzzles are you doing? Even when you use advanced tactics, you can't avoid using triples in most sudokus from Times difficult/Pappocom moderate and above.
G :)[/quote]
I'm new at this - what are X-wings, quads & triples? Is there a glossary of terminology somewhere?
Thanks
If you've never used triples, which puzzles are you doing? Even when you use advanced tactics, you can't avoid using triples in most sudokus from Times difficult/Pappocom moderate and above.
G :)[/quote]
I'm new at this - what are X-wings, quads & triples? Is there a glossary of terminology somewhere?
Thanks
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by Hakosuuji » Mon Sep 12, 2005 3:33 am
I have developed a technique that I call "Bars and Stacks". It saves a lot of time used for scanning and hatching.
This is how it works:
Draw 2 grids, one with the intersections of the rows and boxes (the bars), the other with the intersections of columns and boxes (the stacks).
For each value, you must determine the status it has in each bar and stack. You can use colors for these statuses, and write them in the grid. The statuses are:
1. Filled
2. Blocked
3. Required
4. Undetermined
Place the givens into these grids, with status "filled"
These givens cannot appear in the remaining bars and stacks, which belong to the same group. Each bar and stack has 4 peers that can be blocked for the givens. Write the numbers down with status "blocked"
Each time you write a block, check the remaining 2 peers. e.g. if you block a value because it is given in another bar in the same row, you must check the 2 bars of the same box. All "follow-up" logic goes round the corner like this.
If one of the 2 remaining peers is also blocked for that value, the last unblocked bar or stack MUST contain that value. Write it down with the status "required"
Once you have determined that a certain bar or stack requires a certain value, that value can then be blocked from it's remaining peers. You only need to check 2 of them because the other 2 were already blocked.
When filling the grid, keep an eye on fully loaded bars and stacks. Only 3 values fit into it, so when the sum of both "filled" and "required" values is 3, the thing is fully loaded, and all remaining values can be blocked. This will allow you to possibly detect more requirements in its peers.
Each cell belongs to one bar and one stack. When a bar with a required value intersects with a stack that requires the same value, the cell at the intersection MUST contain that value.
More deductions can be made: when 2 cells in a bar or stack are filled, and a 3rd value is required, you know the location for that 3rd value.
Use the "Bars & Stacks" technique together with other solving techniques. I originally used it as a secondary method, but since it can eliminate a lot of possibilities in a relatively short time, I now use it as the primary method. It is also very useful in revealing those important subsets at the intersection of row/column and box.
Maybe I have developed something that is already known by a different name. Please tell me where to find more documentation on the subject when that is the case.
This is how it works:
Draw 2 grids, one with the intersections of the rows and boxes (the bars), the other with the intersections of columns and boxes (the stacks).
For each value, you must determine the status it has in each bar and stack. You can use colors for these statuses, and write them in the grid. The statuses are:
1. Filled
2. Blocked
3. Required
4. Undetermined
Place the givens into these grids, with status "filled"
These givens cannot appear in the remaining bars and stacks, which belong to the same group. Each bar and stack has 4 peers that can be blocked for the givens. Write the numbers down with status "blocked"
Each time you write a block, check the remaining 2 peers. e.g. if you block a value because it is given in another bar in the same row, you must check the 2 bars of the same box. All "follow-up" logic goes round the corner like this.
If one of the 2 remaining peers is also blocked for that value, the last unblocked bar or stack MUST contain that value. Write it down with the status "required"
Once you have determined that a certain bar or stack requires a certain value, that value can then be blocked from it's remaining peers. You only need to check 2 of them because the other 2 were already blocked.
When filling the grid, keep an eye on fully loaded bars and stacks. Only 3 values fit into it, so when the sum of both "filled" and "required" values is 3, the thing is fully loaded, and all remaining values can be blocked. This will allow you to possibly detect more requirements in its peers.
Each cell belongs to one bar and one stack. When a bar with a required value intersects with a stack that requires the same value, the cell at the intersection MUST contain that value.
More deductions can be made: when 2 cells in a bar or stack are filled, and a 3rd value is required, you know the location for that 3rd value.
Use the "Bars & Stacks" technique together with other solving techniques. I originally used it as a secondary method, but since it can eliminate a lot of possibilities in a relatively short time, I now use it as the primary method. It is also very useful in revealing those important subsets at the intersection of row/column and box.
Maybe I have developed something that is already known by a different name. Please tell me where to find more documentation on the subject when that is the case.
- Hakosuuji
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by PhatFingers » Thu Sep 15, 2005 9:22 pm
When I'm solving a newspaper sudoku, I scan the chutes for any easy-pickins... mostly I'm looking for chutes with either two instances of the same number, or one instance of a number in one box and a different internal row or column in line with that chute that blocks an instance of that number, so I can focus on the third box. In that third box, the above conditions will have narrowed the focus to only three cells that can hold your number.
Cross-check those three cells with the rows or columns orthogonal (crossways) to your chute, and you can often eliminate two possibilities and write in the number.
If you can eliminate one possibility from the three cells, pretend there's a 3x3 grid inside each cell numbered 1-9 and place a dot in each of the two cells in the position of the imaginary numbers for the one number you're addressing. A dot in the top left corner, for example, signifies a one. I use the dots exclusively to show that I've narrowed a number to two possibilities within a box. I may write in other types of pencil marks later for other issues.
The dots are useful in a variety of ways.
* Eliminate one at any future point and you can write in the other without having to remember or resolve why.
* If two cells in a box share the same two dots, you can eliminate those two numbers as possibilities from any of the other cells.
* If you solve all but two cells of a row (or column) later in the game, then you can quickly scan the two intersecting colums (or rows) for a box having two dots matching either of your missing values, both in your column (or row) and treat the dots the same as if you had encountered the number itself, immediately resolving both cells by elimination.
Often that technique is enough to solve an easy or medium puzzle, and is a nice quick prep for the harder ones.
Cross-check those three cells with the rows or columns orthogonal (crossways) to your chute, and you can often eliminate two possibilities and write in the number.
If you can eliminate one possibility from the three cells, pretend there's a 3x3 grid inside each cell numbered 1-9 and place a dot in each of the two cells in the position of the imaginary numbers for the one number you're addressing. A dot in the top left corner, for example, signifies a one. I use the dots exclusively to show that I've narrowed a number to two possibilities within a box. I may write in other types of pencil marks later for other issues.
The dots are useful in a variety of ways.
* Eliminate one at any future point and you can write in the other without having to remember or resolve why.
* If two cells in a box share the same two dots, you can eliminate those two numbers as possibilities from any of the other cells.
* If you solve all but two cells of a row (or column) later in the game, then you can quickly scan the two intersecting colums (or rows) for a box having two dots matching either of your missing values, both in your column (or row) and treat the dots the same as if you had encountered the number itself, immediately resolving both cells by elimination.
Often that technique is enough to solve an easy or medium puzzle, and is a nice quick prep for the harder ones.
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- Joined: 12 September 2005
20 posts
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