For some reason, I'm spending some of my precious time trying to figure out Sudoku-solving-strategies on my own. I'm sure I could just Google these, but what the heck.
So, what follows could be a spoiler for those of you who are also going it alone
I figure I'm a "pretty good" Sudoku-er - I focus on only the hardest problems in whatever book I've grabbed. But sometimes they stump me. I recently had a conceptual breakthrough - still time before the spoiler...
This so-called "breakthrough" of mine is based on the idea that for each puzzle there is only one solution.
Given that "na" means "not applicable" and "(x)" means one of two possible locations,
Let's say you have this rough situation in two of the nine boxes (in the same row or column):
na na na
na na na
(91) 7 (91) <--- previously calculated that only 9 and 1 could go in these two locations
na na na
na na na
(91) na (91) <--- this is not a valid "possibilities" notation, because it would allow two solutions.
It turns out that this strategy applies a lot when I'm stuck. Is this a no-brainer that I was just slow to get to?
Spot false dilemmas now, ask me how!
(If you're not on the edge, you're taking up too much room.)
I must admit I don't like that sort of technique (you get similar cases in other types of logic problem - for some reason I don't think to use it as often in sudoku), because it assumes the existence of a unique solution, and intellectually I consider proving the solution is unique as part of the puzzle. But it's valid enough if you trust the setter.
There's a famous math problem which has a similar basis (at least it's famous among those who were calculus teaching assistants):
You have a sphere, and you drill a straight, circular, hole through the centre. (Along a diameter of the sphere.) The length of this hole is 10 centimetres. What is the volume of the remaining material?
Joined: Oct 14, 2002
I'm sure I'm falling for something here but don't we need to know the diameter of the drill bit?
Bert: no, the diameter of the hole is one of the unknowns.
Stephan: no, the diameter of the sphere is also one of the unknowns. And sorry for the unclarity, the hole does go all the way through the sphere. (So the diameter of the sphere is at least 10 centimetres but can be greater.)
Anyway, the point was that it is a fairly straightforward, but tedious, first-year calculus problem. (The volume is the result of a certain area being rotated around a central axis, so you would have to figure out what integral to do, and so forth. Details are left for the interested student.)
But every year there was one or two students who would say "Look, there's a unique answer to this, or you wouldn't have asked it. So it must be independent of the diameter of the hole. So let's set the diameter of the hole to be zero, then what we have is a sphere whose diameter is 10 centimetres and therefore the answer is just given by the formula for the volume of a sphere."
And it turns out, if you grind your way through the calculation correctly, that the answer is indeed independent of the diameter of the hole.
This always made for an interesting classroom discussion when the homework was graded and returned. Was that a legitimate assumption or not? (But we always gave those smart-asses full credit for showing good mathematical instincts.)
How can it be independent from the size of the drill? Like you said, for diameter zero, the solution equals the volume of the entire sphere. If we use a wider drill (possibly one that is so large that we grind away the entire sphere!) won't the solution be different from the volume of the original sphere? Doesn't that imply that the outcome *does* depend on the diameter of the hole (as one would intuitively think)?
Maybe I'm completely misunderstanding the question.
Joined: Oct 14, 2002
I am SO NOT going to dig out my old Calc text, but I think I get the drift and it's cool
So, if I'm understanding this correctly, you could start with a sphere of diameter 100 cm. In order to have a hole 10 cm high, the bit would have to be almost 100 cm. itself, so the resulting shape would be this thin donut-y thing (10 cm high)... is that the gist?
Stephan van Hulst wrote: If we use a wider drill (possibly one that is so large that we grind away the entire sphere!) won't the solution be different from the volume of the original sphere?
If you use a wider drill, then you need a bigger sphere so that the length of the hole is still 10 centimetres. (Remember, that's the only length you know.) Like Bert said, if the sphere is really big then you need to drill a really wide hole so that it's length is only 10 centimetres. Imagine drilling such a hole through the earth, for example -- you'd end up with a ring 12,800 kilometres around which was 10 centimetres high and thinner than a sheet of paper.
I suppose the question is whether you want to encourage the "trick" or not. If not, it would be easy enough to rephrase it "find the volume and prove that it is independent of the width of the drill bit" if you wanted to discourage it.
Or even just the "prove" bit, leaving room for some intuitive shortcuts. Just been googling, and found a neat proof of it without going to the trouble of calculating the volume.
Stephan van Hulst wrote:Oh geez... So the premise is "Drill a hole wide enough so that you're left with a ring that's 10cm high". I call shenanigans on the phrasing :P
Eh, I think that part was clear enough. But I don't think the phrasing here actually justifies the assumption that there's a single numeric answer. The answer could have been a formula based on the radius of the sphere, for example. And I think students need to be prepared to consider outside-the-box answers to questions, including "there are infinitely many solutions" and "there is no solution" for a given problem. Training them to assume there is one unique solution isn't good practice, because the real world isn't always like that. Still, I would be inclined to give full credit to students taking the shortcut, because it's good to consider such alternate approaches.
Mike Simmons wrote:But I don't think the phrasing here actually justifies the assumption that there's a single numeric answer. The answer could have been a formula based on the radius of the sphere, for example. And I think students need to be prepared to consider outside-the-box answers to questions, including "there are infinitely many solutions" and "there is no solution" for a given problem. Training them to assume there is one unique solution isn't good practice, because the real world isn't always like that. Still, I would be inclined to give full credit to students taking the shortcut, because it's good to consider such alternate approaches.
Yes, that was the gist of the classroom discussion which would take place. Bear in mind that first-year calculus problem sets basically consist entirely of calculations which result in a single correct number as the answer, so the assumption that there was a single correct number as the answer isn't exactly out-of-the-box. However identifying that as an assumption is in itself a problem-solving methodology; most students just take it for granted and are unable to articulate it as such.
Joined: Oct 14, 2002
Back to the original Sudoku idea, since I first started this thread I've done a few more puzzles, and this "unique solution" strategy is helping me crack a LOT of puzzles that had stumped me before. On the other hand, I have noticed that certain puzzles tend to be more tractable using a given strategy, so it's possible that I've just run into a collection of puzzles whose style is suited to analysis with this strategy.
Still, it feels like this is going to be an important tool in my Sudoku toolkit.
In high school I once started a Delphi project to make a Sudoku solver. I only added strategies to it that I used myself when I was solving a puzzle 'by hand', so I could see how far my strategies got me before I would have to 'guess' cells. My solver only ever managed to complete the simplest of puzzles