Todays puzzle is a Renban Diagonal. Rules of Classic sudoku apply. Additionally the 2 main diagonals also contain digits 1 to 9. Marked extra regions contain consecutive digits, not necessarily in sequential order.
It is indeed a great step forward for LMI as the Tapa Variation contest has moved to the LMI site from the oapc site. The 5th Tapa variation contest will be held on the weekend of 5th-6th February 2011.
The test will be of 75 minutes duration. The puzzles will be normal Tapa Variations. The IB will be uploaded by 31st Jan 2011.
Sorry, I got caught up in some personal works and could not devote much time to the blog. I hope to see that this kind of a break does not happen again. To continue our puzzle uploads, today's puzzle is a consecutive puzzle.
Rules of classic sudoku apply. All adjacent cells that contain consecutive numbers are marked by a thick line. Where the thick line does not exist, numbers cannot be consecutive.
Argio-Logic the Italian sudoku website will be hosting its Contest over the coming weekend. The sudoku types are listed below.
The contest starts on Thursday, 13th Jan 2011 at 9:00 PM GMT and ends on Monday, 17th Jan 2011 at 9:00 GMT. You can log in anytime during the weekend to take the test. Once you have started the contest you will have 150 minutes to complete the puzzles.
Today we will look at the solving technique for tackling the Consecutive Sudoku.
Whenever you start a consecutive sudoku, look for 1 and 9 as clues with a consecutive bar beside it. We can simply put 2 with 1 and 8 with nine since there is no other option available for these 2 digits. Also Another thing that can be borne in mind is if we have the numbers 2 and 8 in a box and they dont have a consecutive bar beside them, then the numbers 1 and 9 can never be in a cell with a consecutive bar in that box. ( this holds true if the consecutive bars are within the same box. If the consecutive bar leads to a different box then this does not apply.)
Take a look at the image below.
We have a 1 at R8C9 with a consecutive bar and a 9 at R7C4 with a consecutive bar beside them. We can straight-away put 2 and 8 respectively as 1 and 9 do not have any other consecutive numbers.
Another technique that is evident in this example. Look at the 8 and 2 in Box 3 and 6 respectively. In box 3, we have 8 at R1C8 and since for number 9, number 8 is the only consecutive number, 9 cannot be adjacent to 8 as there is no consecutive bar, and neither can 9 be in R2C7,R2C8 or R3C789 since all these cells have a consecutive bar, but 8 is already present in the box. Hence the only place for 9 is at R2C9.
Similarly, look at 2 in Box 6. Just like the above example, 1 cannot be in R5C789 and R6C789 since 2 is already present at R4C9. 1 also cannot be at R4C8 since there is no consecutive bar beside the 2. Hence the only place for 1 in Box 6 is R4C7.
Apart from these things always look for longer chains of consecutive numbers as they are usually the starting point of a puzzle.
Keeping this in mind let us attempt the consecutive sudoku that was displayed at this blog.
Take a look at box 2. we have a chain of 7 consecutive numbers and a chain of 2 consecutive numbers. This means that all 9 digits are part of some chain. To fulfil this criteria, the 2 digit chain has to be either 1 and 2 or 8 and 9. so the chains will be 1 and 2 and 3 to 9, or 1 to 7 and 8,9.
When we look at the 4 at R2C1 and 7 at R1C8, the chain of 3 to 9 cannot be possible. If 3 is at R3C6 then 4 has to be at R2C6 which is not possible and if 3 is at R3C4 then 7 has to be at R1C6 which is again not possible because of the 2 given digits at R1C8 and R2C1. hence the 7 digit chain is 1 to 7 and the 2 digit chain is 8 and 9. so we can safely put number 4 at R1C5. and since R3C6 is either 1 or a 7 R3C9 also has to be a 4.
Now in Box 3, the only place for the digit 8 is R2C9. If 8 were at R2C7 then R2C8 has to be 9 which doesnt allow for the 2 digit chain in Box 2. So we have 8 at R2C9 and 9 at R1C7 since there is no other place for the digit 9. and that also enables 9 at R2C4 and 8 at R1C4.
Now in R2C7 and R2C8, we can have a pair of 1,2 or 2,3 or 5,6. But 5 and 6 not possible because of the 7 digit chain in Box 2. So 5 and 6 have to be in Row 3 in box 3. Hence we have R3C7 as 6 and R3C8 as 5. Which also gives us 7 at R3C6 and the complete 7 digit chain in Box 2.
Since we have got a 3 at R2C5, R1C9 becomes a 3 and we have the chain of 1,2 at R2C7 and R2c8. since the 3 digits left at R3C1,R3C2 and R3C3 are 3,8,9 and there is a consecutive chain at R3C1 and R3C2. We can deduce that R3C3 is a 3. R3C1 is 9 and R3C2 is 8. Which also gives us R2C2 as 7 and R3C3 as 5 (naked single) for row 2. and similarly we have 1,2,6 for Row1 in box 1. so 1,2 chain is the consecutive chain and 6 comes in at R1C1.
After this the puzzle is easily solvable.
Hope this explanation was helpful and it gives you a better lead and approach to solve a consecutive sudoku.
Place numbers in the grid such that each row, column and 3x3 box contain the numbers 1 to 9. If the absolute difference between two digits in neighbouring cells equals 1, then they're separated by a white dot. If the digit in a cell is half of the digit in a neighbouring cell, then they're separated by a black dot. The dot staying between '1' and '2' can have any of these dots.
Today we will look at a solving technique for Non-Consecutive Sudoku. The images below are for explanatory purposes only and not an actual puzzle.
Let us take a look at box 1 in the above example. We have pencil marks of 1,2 at R1C1. 1,2,3 at R1C2 and 1,2,3 at R3C2. In a Non-Consecutive sudoku, whenever we have consecutive numbers as pencil marks in a cell, then we can simply remove them from the adjacent cells. For example, if R1C2 was 1, then R1C1 would become 2, and if R1C2 was 2 then R1C1 would be 1. Since either scenario is not allowed in a Non-Consecutive sudoku, we can safely eliminate 1 and 2 from R1C2.
Since we have 3 in R1C2, we can again eliminate 2 from R1C1. so we have 1 at R1C1 and 2 at R3C2.
Now let us look at box 6. We have options of 1,2,7 at R4C7,R5C7 and R6C7. since 1 and 2 are consecutive numbers they will always be separated by the third number. So 7 will always be in R5C7.
On 8th/9th January 2011, LogicMastersIndia will be holding a Puzzle Contest named Puzzle Jackpot. This contest is created by Serkan Yurekli of Turkey. He is the editor of Akil Oyunlari puzzle magazine and they have published several books including logic puzzles.
The test is of 2 hours duration and can be started anytime during the weekend of 8th and 9th January 2011. Your timer will start once you click on the "Start Puzzle Jackpot" button on the LMI website. You will also have a 5 minute grace period for answer submission. A Timer will be on once you start the test. Once your time has run out you will not be allowed to enter any answers.
The puzzle types are :
Skyscraper and Variations
Snake and Variations
First Seen Snake
Tapa and Variations
Knapp Daneben Tapa
Easy As Tapa
The IB can be downloaded here. The Puzzle booklet of 16 pages has been uploaded. You can download the puzzle booklet here. Further discussions on this contest is at the LMI Website
Today we will look at the solving technique for Magic Square Sudoku. Now we know that all the 9 centers of the 3x3 boxes add upto 15 row wise, column wise and diagonally. Always remember that the middle cell ( R5C5 ) will always be a 5. The four corners will always be even numbers and the remaining 4 centers will be odd numbers.
At the corners of the center dots, 2 and 8 will always be diagonally opposite to each other and the numbers 4 and 6 will be diagonally opposite to each other. These two digonal sums will add upto 15 since the number 5 is always at the center at R5C5.
Let us take yesterdays Magic Square Sudoku for example. I have updated the pencil marks for the centers.
Now let us look at R2C8. Since the only even number that can fit there is 8, we also get the number 2 at R8C2, so that the diagonal sum is 15. That leaves us with options 4 and 6 for R2C2 and R8C8. Since number 4 already exists in Box 1, we can safely place number 6 in R2C2 and number 4 in R8C8. so we have placed 5 numbers till now and we have the diagonals in place. Diagonals here meaning the centers.
Now we can fit the odd numbers as per the sums of each row of centers. We have 6 and 8 as the two centers in R2. So in order to have a sum of 15 we can safely place the number 1 in R2C5, 7 in R5C2, 3 in R5C8 and 9 in R8C5. Once all these 9 digits placed, the sudoku can be solved like any other Classic Sudoku.
This coming weekend the Italian website www.argio-logic.net will be hosting the Hybrids competition number 3.
The contest starts at 9:00 PM ( GMT ) on Thursday 6th Jan 2011 and ends at 9:00 AM ( GMT ) on Monday 10th Jan. The contest is an online contest and all solvings have to be done online on their website.
Today we will look at a solving technique for tackling the Anti-Knight Sudoku. To demonstrate that, I am using the Antiknight-Non-consecutive puzzle that I had displayed earlier on my blog.
I have updated the pencil marks for the grid. Now let us look at the possibilities for number 8 in Box 5. When we look at it, we see that the possibilities for number 8 forms an L shape, marked by the orange lines.
Now, let us draw a reverse L shape extending to 2 cells on either side from the junction point of the original L shape. This new reverse L shape is marked by green lines in the next image.
Since the number 8 has to be in one of the 3 cells in the original orange colored L, it can NEVER exist at the ends of the green L (R2C4 and R4C2 ). Hence the possibility of number 8 can be removed from these 2 cells.
Now you may wonder why this holds true, so let me go further and explain. Let us take R4C2 for example. now if 8 was in R4C4 or R4C5, then it is in the same row and automatically 8 gets eliminated from R4C2. And if 8 was in R5C4 then it is eliminated again from R4C2 since it is at a knights step. so irrespective of where 8 comes in the original orange L shape, it cannot be at R4C2. similarly it also holds true for R2C4. So we can safely remove 8 from R4C2 and place a 9 there.
This is always true for any L shape, however please remember that this L shape rule is true only and only if the other cells in that box (box 5 in this example ) do not contain a possibility of the number 8.