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Help the chess knight complete four clockwise laps on this racetrack, so that he lands on every square and never lands on the same square twice! The final square the knight lands on will be the same as its starting square.

In other words, find a closed knight's tour on this funnily-shaped board which crosses the checkered finish line from left to right four times.

, please, this can be done by hand.

enter image description here

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First, since we need to do four laps, then we need to make spaces for four routes since we cannot repeat squares. This means there should be four initial squares that would be the first square for each lap, and this would lead to four parallel routes. This would make it easier to find which square each route needs to go to next, while not blocking other routes. Also note that for the straight portion of the track, there is only one possibility for four parallel routes to go. So the options we have really are constrained on the corners.

This is one possible routes:

enter image description here

The problem with this is that each route is going back to itself when going to the next route, while what we want is that each route should go to another route.

So we need to find another possible routes that would lead to the routes cycling between themselves. Let's try another:

enter image description here

Oh no, apparently it still cycles back to itself.

Now, noticing that in each versions above, the method of turning is the same in all four corners, but they have different method. What if we combine them? Turns out that these two methods of turning is not enough. So I find the third way of turning, and by combining all three we have our final solution:

Final solution

enter image description here

Here the initial green leads to black, then black to light blue, and finally light blue to dark blue. Therefore, this completes the four laps touching all squares.

Commentary

In our final solution, the first turn is using the second method of turning, the second and third turn is using the first method of turning, and the fourth turn is using the third method of turning. Note that since the middle square in each turn is fixed (there is only one route that can use that), that leave us with only three routes to play with. Here I used three different methods. (In Retudin answer we can see the fourth one that I didn't use)

For example, the green route here is forced, since there is no other route that can reach that middle square in the turn.

enter image description here

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So I pretty much just went at it haphazardly and came up with this solution.

enter image description here

I went for 'fast' laps for the first three times around, but then found I had a substantial amount of open space in each corner, so I ended up spending a lot more time in the last lap, picking up the extra empty spaces. There was a bit of a ruckus right at the end of lap 3 though.

My answer is different from @justhalf's in that in their solution, they have almost each lap doing a zigzag in each turn, except lap two gets two and light blue gets none. I feel like there has to be more solutions than these two.

Now I'm looking at it, and noticing all of the straightaways consist of blocks that each lap hits. My usual method of knight's tour is to divy it up into 2x2 blocks and only hit each and every block once before hitting the first one again.

enter image description here

I feel like this could be explained a bit more, but I'm not sure how. Going to think on it a bit more.

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EDIT: oeps, the bottom right picture is o.c. not baab but caab
My answer is (currently) wrong , I end in the same position after a round, i.e. created four 1 round loops
There are 7 ways to do this. (if I did not miss anything)

There are three ways to get through a corner (see left column)
There are three types of route to get a closed loop (see right column)
Those can be rotated to get aaaa, caab, aabc, abca, bcaa, bcbc, cbcb
enter image description here

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    $\begingroup$ aaaa doesn't work. It will loop to itself, as shown in the first image on my answer. And cbcb and baab shouldn't work either. The goal is to make the configuration not the same after the fourth turn. Making them the same will not make four laps, but four independent routes. Also apparently there is a fourth way (my second picture) to make the turn. My solution used daac. But I'm sure there are other possibilities as well, akin to what you've done here. $\endgroup$
    – justhalf
    Aug 26 at 8:39
  • $\begingroup$ You are correct, I made 4 loops not 1. $\endgroup$
    – Retudin
    Aug 26 at 8:55

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