The New Sudoku Players' Forum

[Crazy Girl]

Came across Turkey's Qualifying puzzles to the World Sudoku Championships in Lucca here.

The First 2 puzzles are regular sudoku, (despite looks, not SudokuX). In the third puzzle all neighbouring numbers with a difference of 4 are marked with a red circle on the side between them. In the final puzzle one has to use the numbers found in the above puzzles and add clues until the puzzle is solvable.

Here is the 4th puzzle with the clues A-L (anyone free to independently confirm the clues A-L)

Code: Select all

 A7 .  .  | B6 . C3 | .  .  .
 .  .  .  | .  .  . | .  D8 .
 .  .  .  | .  .  . | .  .  .
----------+---------+--------
 .  .  .  | 3E .  . | F5 .  .
 .  .  G8 | .  .  . | .  .  .
 .  H4  . | .  I8 . | .  .  .
----------+---------+--------
 J9 .  .  | K5 .  . | L3 .  .
 .  .  *  | .  .  . | .  .  .
 .  .  .  | .  *  . | .  .  .

two cells marked with (*) must also be clues.

Is there a logical way to find the minimum number of clues for which this puzzle can be solved, or is it abit of 'Trial and Error' involved

[Wolfgang]

Strange problem, pure T&E. You already have 7x7 possibilities for the 2 cells and are left then with 49 16-clues with millions of possible solutions. Also with a program it is not easy to find a sudoku with the minimum number of clues then. Or did i miss some additional constraint ?

[Crazy Girl]

No additional constraint, the 4 puzzles are marked on a points system, for the fourth puzzle, where c is the number of clues the points awarded are

100 - square of (c-27) if c>=27

100 + square of (27-c) if c<=27.

so obviously to score highly one needs to get the minimum number of clues for the puzzle.

If you make the puzzle symmetrical, then there would be 28 clues, but one of the cells can be determined from the information currently present, so that makes 27 clues, a coincidence

[udosuk]

I suspect the 2 extra shaded cells in the 4th puzzle must be identical, so 7 choices there... Afterwards you have 14 clues, and theoretically 3 more extra clues could give you a valid puzzle... So the max total points you could score is 60+140+100+100+(27-17)^2 = 500, right?

[Crazy Girl]

I suspect the 2 extra shaded cells in the 4th puzzle must be identical, so 7 choices there... Afterwards you have 14 clues, and theoretically 3 more extra clues could give you a valid puzzle... So the max total points you could score is 60+140+100+100+(27-17)^2 = 500, right?

Yes, but how you find the 3 extra clues to give one 17 clues, and be able to solve it is a mystery to me.

[udosuk]

Yes, but how you find the 3 extra clues to give one 17 clues, and be able to solve it is a mystery to me.

This is mostly likely to be a programming problem... I think gfoyle & co. would be able to find the perfect solution...

[Ocean]

Here is the 4th puzzle with the clues A-L (anyone free to independently confirm the clues A-L)

Code: Select all

 A7 .  .  | B6 . C3 | .  .  .
 .  .  .  | .  .  . | .  D8 .
 .  .  .  | .  .  . | .  .  .
----------+---------+--------
 .  .  .  | 3E .  . | F5 .  .
 .  .  G8 | .  .  . | .  .  .
 .  H4  . | .  I8 . | .  .  .
----------+---------+--------
 J9 .  .  | K5 .  . | L3 .  .
 .  .  *  | .  .  . | .  .  .
 .  .  .  | .  *  . | .  .  .

two cells marked with (*) must also be clues.

Is there a logical way to find the minimum number of clues for which this puzzle can be solved, or is it abit of 'Trial and Error' involved

Try this:

Code: Select all

 A7 .  .  | B6 . C3 | .  .  .
 .  .  .  | .  .  . | .  D8 X4
 .  .  .  | .  .  . | .  .  .
----------+---------+--------
 .  .  .  | 3E .  . | F5 .  .
 .  Y1 G8 | .  .  . | .  .  .
 .  H4  . | .  I8 . | .  .  .
----------+---------+--------
 J9 .  .  | K5 .  . | L3 .  .
 Z2 .  *7 | .  .  . | .  .  .
 .  .  .  | .  *1 . | .  .  .