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Given a number $n$ and an $n\times n$ grid. Consider a connected path from the lower left corner to the upper right corner. Here, a path means a series of adjacent blocks that do not revisit any previous block. How many blocks need to be eliminated in the grid such that there is one and only one path from the lower left corner to the upper right corner?

Please note that the path can go down and go left, it does not need to be the "shortest".

For example, the following is not allowed (red for path 1, yellow for path 2, orange for common):

enter image description here

My guess is $1$ for $n=2$, $2$ for $n=3$, $4$ for $n=4$, and $2n-5$ for $n\ge 5$. However, I cannot prove them as a lower bound.

The Construction is as follows (for $n=2,3,4,5$) enter image description here

All these paths are going up and right; maybe there are other routes possible, but I cannot prove that it is not optimal.

P.S. This bound is verified for $2\le n\le 7$ by a user on Stack Exchange. The link is here.

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  • $\begingroup$ The symmetric difference of any two paths is a loop. So, no loop is allowed. $\endgroup$ Commented Aug 30 at 4:03
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    $\begingroup$ @AlapanDas Not really. Loops are allowed. You can see in the case when $n=4$ or $n=5$. As long as this loop does not intersect any path, it is allowed. $\endgroup$ Commented Aug 30 at 4:27
  • $\begingroup$ I see. Thanks for correcting me. $\endgroup$ Commented Aug 30 at 6:18

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An upper bound is given by $2n-5$ for all $n\geq 5$. You first construct a zig-zag wall of thickness two using $2n-4$ blocks. But you can safely delete one. A picture is worth a thousand words. In the following pictures, black boxes are the wall and the deleted box is painted blue.

For $n$ even, (showing $n=12$ case)

n=12

For $n$ odd, (showing $n=13$ case)

n=13

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    $\begingroup$ This confirms the conjectured formula as an upper bound, but not as a lower bound $\endgroup$ Commented Sep 3 at 9:22
  • $\begingroup$ Oh, you are right. I'll correct my answer. $\endgroup$ Commented Sep 3 at 12:01
  • $\begingroup$ Thank you so much for your answer! Yes, my construction for $n\ge 5$ is constructed like this. My construction is just to extend the horizontal bar and vertical bar for $n=5$. However, I am looking for a lower bound though... Sorry for the confusion. $\endgroup$ Commented Sep 3 at 17:08

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