I wouldn't buy that argument.

Anyway, I request you (and everyone else) not to have multiple ids.

You can now check others' score as well in the score page - http://logicmastersindia.com/ISC2011/score.asp
Note that this just shows the complete list. Official scores (submissions before 4:30PM Saturday) will be published soon.

Note : The round is still running, hence the results are not public.

Hi Admin,

During the test time, In the last moment while submitting the answer for diagonal sudoku i typed 1 instead of typing 2. But after realizing that mistake, immediately i enterd a backspace and entered "2". After that i submitted the answer. But i did not notice the solutions after that last submission(was busy in trying out the other sudokus). Now in the site still i am seeing "1" in the place of "2".

Please consider this request if you have the mechanism for keep tracking of each and every action the user is doing.

Thanks,
Upendra

There is no mechanism to keep track of all the typing.
As you will see later, several players have done these kind of mistakes. I've to ask you (and others) to be very careful while entering the answer keys. There is nothing worse than solving a Sudoku correctly, but entering the solution wrongly.

Initially i did the typing mistake but after that with in 5 seconds i did a backspace and entered the right one which did not get reflected over there.

Anyway thanks for the quick response.

List of all Online qualifiers

List of all qualifiers

When will the national finals be held? Where? Would you conduct finals for ISC & IPC on same/consecutive date/s & venue?

sir,this is to bring to your kind information that i nikunj agarwal participated in INDIAN SUDOKU CHAMPIONSHIP and inspite of MY SCORE BEING 199 I AM NOT SELECTED AMONG THE TOP 15 PARTICIPANTS rather the guys scoring less than me-their score being around 187 are selected....May i know the reason for the same.........

11 chasing_dreams Sunil Singhal 198 Qualified
12 shivin garg shivin garg 187 Qualified
13 abhishekn abhishekn 185 Qualified
14 upendra upendra reddy 167 Qualified

Sure I can tell you the reason. Before that let me ask one question.
When did you start the "Online Round"? date and time?

date-25/04/2011 and time is around 11.00 a.m.

Perfect.
Do you still need an answer?
Check the rules here or the IB or the notes in results summary.

thanks...i didn't pay an attention to the word "OFFICIAL"
I only took a notice on -
"
LMI April Sudoku Test

Submissions will be allowed until 25-Apr-2011 16:00 IST (GMT +5:30)."

Hello everyone,

Thanks to the organizers for a nice set of puzzles. I greatly enjoyed many of them. Classic Sudoku 4 had a nice hurdle to clear, and I recall the Consecutive Sudoku and Killer being special as well.

Would someone be so kind to list a logical series of steps for the Product Frame puzzle? I spent a great deal of time on it and do not see any way to proceed.

Thanks again, and congratulations to the top performers.

Zason, the Product Frame was indeed tough (but it was the highest pointer as well) and it proved to be killer for many solvers.

I'm not sure if you didn't get the start or you get struck in the middle. But here are some steps to start with the puzzle.
If you are stuck at a certain point after solving, let me know so that I can explain from there.

First the obvious step, but it doesn't lead to much.


Next - 36 at right of Row 8 can't have 4. So it can't have 9 as well. So 36 will be (2*3*6)


Now, we have 12 at bottom of Col2 and 24 at left of Row9.
12 = (1*2*6) or (1*3*4)
24 = (1*3*8) or (1*4*6) or (2*3*4)

There are lot of common digits among them. To begin with R9C2 can be 1 or 2 or 3 or 4 or 6. But 3 / 4 / 6 are ruled immediately. So R9C2 can be 1 or 2.


On closer look, R9C2 can't be 1. [ If it were 1, 12 has to be 1*3*4 because 2 and 6 are already present in row8 towards the right. if 12 is 1*3*4, there is no way to satisfy 24 ]

So R9C2 = 2
I think this was a pretty difficult step.

Few obvious steps

Now 18 at bottom of Col9
18 = 1*2*9 [ 1 * 3 * 6 is rejected ]

Now the 56 at right of row1, it can't be 1*7*8
If it is 1*7*8, then (R2C5,R3C5) will be (1,8), and that means 1 or 8 has to be
appear at (R2C1,R2C2,R3C3) that is not possible since the product R2C1*R2C2*R3C3=84
So 56 = 2*4*7


That gives us lot of opening.



So 84 on left of row2 = 7*2*6 (because 4 is already at R2C6)



18 at top of col1 is 3*1*6






40 at left of col5 is 5*2*4





R5C8 can't be 3 [ if it is 3, R8C8 will be 6 and there is no way to make 12 at top of Col8]





In the last step, R4C9 can't be 7. If it is 7, R1C9 will be 4 and R4C8 will be 2 so there is no way to satisfy 12 at top of Col8 without 2 and 4.

Thank you very much Deb. This is the step I missed -- the 178 not being allowed because of what it does with the {18} in the 2nd row and 2nd box. This single step allowed me to finish the rest of the puzzle in about 5 more minutes. I should not have missed this, and the puzzle is much more clever than I had originally credited it as being.

Great contest, thanks again.

hi deb :

i could solve all those variants sudokus (after taking a print out and juggling with it for more than a day) except that V and X type sudoku,, can u plz guide me through the steps ? and
i took 3 hours to solve killer sudoku,, is there any strategy to solve it quicker ? would u mind sharing the tricks ?

thanks in advance ..!

Zason, Glad that you liked the Product Frame Sudoku (and others too). In this puzzle one has to focus on 2 given products at the same time (unlike easy Product/Sum frame puzzles where one has work out single product / sum). The step you missed (and the 12/24 in box7 to begin with) were difficult steps, and this puzzle turned out to be the differentiator at the top, which is probably not ideal from a competition point of view.

The Killer Sudoku was difficult for beginners and little tricky for experienced solvers. It will be tough to list all the steps. I'll just post the beginning steps, and some tricks that beginners may not be aware of.

First the 30 at R6C3 and then the 23 at R5C2 gives you lot of pencil marks and 7 at R6C3



Now the maximum sum of R4C2 and R4C3 is 9 (4+5). so 26 at R3C2 will be (4+5)+(8+9)



So (1,2,3) at (R4C1,R5C1,R6C1)



Now, this is the step beginners are likely to miss. Lets assume R6C1=X (X can be 1 or 2 or 3), so where does X appear in Box7.
The only place for X in box7 is R9C2.



Since the sum at R8C1 is 20, X has to be 3.

Similar tricks can be be used in top part of the grid later (note that the grid is symmetric except 2 cages)

One more In-Out trick that can be useful later is the 45 at R3C4.
So if R3C4 is A, where does A appear in Box5. The only place is R4C6. Similarly for B at R7C6.


I'm not going to list all the steps, but look at 24 at R4C8, 11 at R8C6, 10 at R3C6.

Also, check that you can find the value of R3C1 and R9C4 with simple addition/subtraction.

So, challenge yourself again to solve it in 20 minutes :-)

The XV sudoku was little different because there were only few X / V marks on the grid. So box 7 and box 3 can be filled a lot easily. [ For example, R7C2 can be 7 or 8. But it can't be 7 because then R8C3 will be 2 (and 2 is already present in column 3). So R7C2=8 ]

After that all you need to remember is 2 digits can't be consecutive if their sum is 5 or 10 9 (since no more X or V are marked in the grid).
It is really an easy puzzle and I ask you to try it again. if you still face issues, post here.

Gowrishankar: For XV sudoku, to add to Deb's points, you must realize that the inverse rule is also true, i.e., if there is no V between two cells, the sum cannot be 5, and similarly for X. Here are a few (more than a few, actually) steps to get you started. The red cells are all obtained by XV rules and the blue cells based on normal sudoku rules.

After this I suggest you proceed on your own. For example, if you look at the left bottom corner cell, it cannot be a 9 (..as then it violates the rule for X) and so on. Do let us know if you are still stuck somewhere.

hi deb,

the first few steps which u explained to solve killer one was suffice,it eases to solve the rest ,especially i dint know the technique by solving it this way,,

i dint get wot u said below :
"After that all you need to remember is 2 digits can't be consecutive if their sum is 5 or 10 9 (since no more X or V are marked in the grid). "( copied from ur EARLIER reply to my query)
my point is : 2+3=5 and 2&3 are consecutive ryt ? y cant de go in cells marked with V,,

hope u don mind to rectify my mistake in query...!

hi rai:
i could solve those cells given for X and V da, however ur inverse law i m not getting .

"if there is no V between two cells, the sum cannot be 5, and similarly for X" how ?

u mean t say in a box if ter is no V between any two cells ,then those two cells wont be occupied by digits adding to 5? i dint get ,,,,dont mind to elaborate....

consecutive is a bad word to use. I meant adjacent cells.

My point was this (and Rakesh was explaining the same thing)
If there is no v between 2 adjacent cells, the sum of 2 digits can't be 5. So 2 and 3 can't be placed in adjacent cells if there is no V. Similarly, 1 and 4
Same rule applies for X as well.