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.