# 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
4 years, 11 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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).

- 4 years, 11 months ago

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.

Staff - 4 years, 11 months ago

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.

- 4 years, 11 months ago

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

$H + V + \frac{ 2 (n-H) + 2(n-V) - 4 } { 2} = H + V + (n-H) + (n-V) - 2 = 2n - 2$

Staff - 4 years, 11 months ago

O.k., great. Same idea, but more clearly stated than my exposition. :)

- 4 years, 11 months ago

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 $n = 2.$

- 4 years, 11 months ago

I like that quote. With induction, I guess you have to be careful where you start the journey. :)

- 4 years, 11 months ago

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.

Staff - 4 years, 11 months ago