onigame ([info]onigame) wrote,
@ 2009-06-12 12:06:00
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Just a Small Sudoku


Who needs the complexity of 9 numbers? Fill in each cell with a number from 1 to 6 so that no number appears more than once in any of the rows, columns, and 2x2 boxes.

Puzzle 3 in the "too hard to be published" series.

This one is small enough to be quite tractable to trial-and-error... but if you can find a logical path to the solution, consider yourself good.

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[info]cyrebjr
2009-06-12 11:56 pm UTC (link)
I don't know what a logic-without-guessing path looks like for this one, but I found the answer relatively quickly after guessing the upper-left corner. quotient By the way, have you considered easier versions of these THTBP puzzles?

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[info]onigame
2009-06-13 01:06 am UTC (link)
Yes, there are easier versions of these puzzles ... they are being published, of course. :-)

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[info]lhopitalified
2009-06-13 06:10 pm UTC (link)
I never quite understood the "difference" between trial-and-error and logic for these puzzles. Isn't it simply the depth of recursion? Filling in single-candidate squares can be thought of as trying all possible numbers and taking the only one that does not cause an immediate contradiction, which must be the solution then, by exclusion. (assuming the rest of the board is correct)

That said, I still have yet to develop a good intuition for where/when/what to guess to speed up my solving.

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[info]zundevil
2009-06-14 07:48 am UTC (link)
I may not be the right guy to offer any insight here, but I will say that...well, I didn't hardly get anywhere before I used T-n-E. In fact, I'm still not done...although I figured out one square...I think R1C3.

I think you're right about your definition there; 1-ply doesn't count as T-n-E. Then again, at that level it's simply applying the first principles, so something about it seems ok. I'll defer to Wei-Hwa.

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[info]onigame
2009-06-14 09:43 am UTC (link)
I just typed a five-paragraph reply on this topic with a broken hand, and then lost it all when I hit backspace. Remind me again in about three weeks, when my hand will be healed and I'll probably feel less disgusted.

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[info]motris
2009-06-14 06:44 pm UTC (link)
To me a hallmark of trial-and-error techniques is that the solver cannot always say that there is only one answer to the puzzle. Admittedly, a trial and error strategy in sudoku sometimes proves this when your guess fails, but how many times do you erase after the guess works to show the other path doesn't? My guess is none. A logical path may have a trial-and-error like step with a simple conclusion - this is how I would view the Y-wing deduction for example - but you can "prove" in written form that there is just the single answer.

As the person who has done the second most "udoku"/"deficit sudoku" in the world after [info]onigame, I'll try to fill-in for our injured author. I find accounting for the "missing numbers" in these puzzles is rather useful and in this case you can model the 3x3 squares as a set of double-valued cells that also have 1-6 in the rows and columns. The top-center square contains a 2 and some other number that is "missing" from that box. The lower-right square, from its bivalue cells, is missing (1 or 2) and (4 or 5) for example. In the grid itself, you can make the most progress in either/or choices with the 1's and 4's that lead to x-wing type eliminations. Specifically, you can say there are two 1s in opposite corners of R23C26 and R15C45 as well as two 4s in opposite corners of R14C15 and R26C36.

Now here is my sort of "simple trial-and-error" discovery of a logical solution. Given the contingencies on 4's and 1's, consider if the top-center box is also missing a 4 from placing a 4 at R2C6. This results in pushing a 1 to R3C6 and a 1 to R1C4 as 24 are already missing from that box, meaning 1 is missing in the top-right. You'll see that the 24 pair missing in the top-center box is also missing in the middle-left and bottom-right boxes. Here is how "missing number" mapping helps you here. The problem comes from having no place in the left-most column to exclude a 1. It cannot be the top-left box as the top-right one already misses a 1. It cannot be the middle-left as that is missing 2 and 4. The bottom-left has a 1 in it. So, you cannot solve a grid with 4 at R2C6 as you'd necessarily place 2 1's in either C1 or C2.

So, put the 4 in at R2C3 and at R6C6. Now, notice that R2C6 can only be 1 or 6. If R1C5 is a 1, then there is no 1 in the top-center box which means R2C4 is a 6. This leave no possible digits for R2C6. So, R1C4 is a 1 and from here I think you can get to the end satisfactorily.

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