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Covering \( n \times n \) dots with continuous straight lines

Consider a \( n \times n \) grid of \( n^2 \) points. What is the minimum number of straight lines that are needed to cover all \(n^2 \) points if you were using a pen that could not be removed from the page?

Treat the points as a 0-dimensional object. They do not have length or width.
Treat the lines as a 1-dimensional object. They do not have width.

For \( n = 3 \), see Think outside the box sometimes.
For \( n = 4 \), see Inspired by Chester Robinson.

Note by Calvin Lin
1 year, 8 months ago

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It seems to require \((2n - 2)\) lines to cover a board of size \(n\times n\ (n > 1).\) The best solution for a \(5\times 5\) board uses \(8\) lines, and for \(6\times 6,\) I can find \(10\) lines. I cannot find a proof, and as the strong law of small numbers states, \[\text{"You can't tell by looking."}\]

More precisely it seems that it would require, at most and/or at least, \((2n - 2)\) lines. I have not been able to prove that this is insufficient for larger boards, or that it can be optimized for larger boards. Do you already have the answer Calvin? I have spent an hour on it and it drives me mad (I have spent the better part of the past 45 minutes trying to find an induction step). Caleb Townsend · 1 year, 8 months ago

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@Caleb Townsend For \( n = 2 \), you actually need 3 lines instead of 2. That is a "special case" that needs to be dealt with.

The answer is indeed \( 2n -2 \) for \( n \geq 3 \).

There is a straight-forward induction proof to show that \( 2n-2 \) lines are sufficient to cover a \(n \geq 3 \) board.

There is a "innovative" proof which shows that you need at least \( 2n - 2 \) lines to cover a \( n \geq 3 \) board. My idea for the proof arose from the induction step, and looking at what the construction looks like. Calvin Lin Staff · 1 year, 8 months ago

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@Calvin Lin Following up on your hint in the comment section of the \(n = 4\) question, suppose that a "legal" path through the points of the \(n\) by \(n\) grid has a total of \(k\) horizontal lines and vertical lines. Since there are \(2n\) rows and columns, there are \(2n - k\) rows and columns that aren't traversed by horizontal or vertical lines, respectively. Let there be \(h\) such horizontal lines and \(v\) such vertical lines, in which case \(2n - k = h + v\). The points these rows and columns have in common, i.e., their points of intersection, must then be traversed by a diagonal, (not necessarily a \(45^{\circ}\) diagonal).

If \(h = 0\) then the path has \(n\) horizontal lines, which would have to be connected by at least \(n - 1\) non-horizontal lines in order to have a continuous path. The path in this case would then have at least \(2n - 1\) lines, exceeding the already established ceiling of \(2n - 2\) lines. Similarly if \(v = 0\).

If \(h = 1\) then the path has \(n - 1\) horizontal lines, with a row of \(n\) points that will each have to have at least one distinct line passing though it to create a legal path. The path in this case would also have to then have at least \(2n - 1\) lines. Similarly if \(v = 1\).

Thus both \(h\) and \(v\) must be \(\ge 2.\) The convex shell of the set of intersection points discussed above will contain \(2h + 2(v - 2) = 2(h + v - 2) = 2(2n - k - 2)\) points, and any diagonal line, (\(45^{\circ}\) or otherwise), can only traverse a maximum of two of these points. Thus we need to have at least \(2n - k - 2\) lines on the path in order to traverse all the points on the grid in addition to the \(k\) horizontal and vertical lines we started with. This gives us a total of at least \((2n - k - 2) + k = 2n - 2\) lines on any legal path. Since we have already established a ceiling of \(2n - 2\) lines, we can conclude that the optimal solution path has \(2n - 2\) lines. Brian Charlesworth · 1 year, 8 months ago

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@Brian Charlesworth My solution is identical / similar to yours.

I let \(H\) be the number of horizontal lines, \(V \) be the number of vertical lines, \( D \) be the number of diagonal lines. It made it slightly easier to count / think about.

  1. If \( H = n \), then we will need \( n - 1 \) lines to connect them up, so a minimum of \( 2n - 1 \). Similarly if \( V = n \).
  2. Otherwise, there is a \( (n - H) \times ( n - V) \) subgrid of points that isn't covered. As pointed out, there are \( 2(n-H) + 2(n-V) - 4 \) points on the perimeter, and each diagonal line can cover at most 2 points. Hence, the total number of lines is t least

\[ H + V + \frac{ 2 (n-H) + 2(n-V) - 4 } { 2} = H + V + (n-H) + (n-V) - 2 = 2n - 2 \] Calvin Lin Staff · 1 year, 8 months ago

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@Calvin Lin I see; 4 months ago, I read in a textbook, "Induction is very powerful, but not so powerful that it can prove something that isn't true." Perhaps the reason I couldn't prove that the number of lines required is \(2n - 2\) is because I was assuming it was also true for \(n = 1\) and \(n = 2.\) Caleb Townsend · 1 year, 8 months ago

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@Caleb Townsend Read this wiki on Construction, which talks about the different possible approaches.

In this problem, because the solution for \( P_{n+1} \) does not necessarily have a subcase of \( P_{n} \), the forward induction and backward induction approaches will (most likely) not work. Instead, we have to resort to another approach.

In this case, we used the structure finder / structure avoider approach. The idea was that horizontal and vertical lines do great, but we need some diagonal lines to connect things up. However, diagonal lines do very poorly. So, we want to minimize the number of diagonal lines and maximize the number of horizontal / vertical lines, and that is where the push and pull of the problem comes in. Calvin Lin Staff · 1 year, 8 months ago

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@Caleb Townsend I like that quote. With induction, I guess you have to be careful where you start the journey. :) Brian Charlesworth · 1 year, 8 months ago

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@Calvin Lin O.k., great. Same idea, but more clearly stated than my exposition. :) Brian Charlesworth · 1 year, 8 months ago

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