Solution!

Logical reasoning

(For referencing cells, I’ll think about the grid is crammed out with numbers 1-49 like this)

1. The begin

There’s just one actual place to start out, and thats the one place you already know precisely how the trail leaves and enters the cell: the 2 empty corners. As there are simply two entry factors, the trail should begin like this:

Now we even have a bit extra info. Cell 44 should be shifting east, as if it was shifting west then the trail can be lower off from its path to the 49. Similarly, cell 14 should be shifting down, which means cell 6 is shifting east.

That means cell 44 should then transfer up, and cell 13 should transfer as much as attain 6 to keep away from consecutive east strikes:

2. One means path

Notice now that the paths alongside the sides solely have one entry/exit level, so the trail should proceed that means. To keep away from consecutive strikes, they have to additionally then reduce into the board, which in flip means cells 13 and 37 solely have a technique they can also go.

We additionally know two methods the trail cannot go to keep away from double east strikes, so we are able to block off that path.

3. No cell left behind

Consider cells 12 and 38. If cell 12 is reached by cell 19, then cell 5 is totally unreachable, so cell 12 should be reached by cell 5, which in flip extends the trail alongside the sting. Similarly for cell 38, it should go to 39 or threat isolating it.

Continuing these paths require an east transfer, which suggests we have now to increase the trail again into the grid too to keep away from consecutive east strikes.

4. Endpoints

We additionally know at this level that one arrow should be part of the again of the opposite arrow, whereas that arrow goes to the 49 and the tail extending again to the 1. Lets take a look at these endpoints really.

The endpoints are corners so even have two doable entries/exits – besides just one is used. If the 1 goes to 2, it should then go all the way down to 9 – however this leaves cell 3 remoted. Similar with the 49, entry from 48 leaves 47 remoted.

Hence 1 should go to eight, and 49 should be reached by 42. Now think about 8 extends down to fifteen, then having to chop into 16. This would go away 22 remoted, so 8 should go to 9 as a substitute (and identical backside proper, 42 should be reached by 41.

To keep away from isolating cells, 9 should go to 2 which should prolong to three and 10 – and 49 should path spherical equally:

5. Finishing the edges

Consider 20 and 30. If they prolong in the identical route, after which have to chop again into the grid, they go away cells 22 and 28 fully unreachable with no 2 transfer east, so 20 and 30 should lower into the grid, after which into 22 and 28 to keep away from isolating them.

This continues a loop spherical that completes the edges:

6. Connecting the dots, effectively, strains

Finally, we simply want to affix the strains appropriately and fill within the center.

So lets have a look at the center. We’ve really ‘created’ some corners, which we are able to use the logic from half 1 on:

Now have a look at 10. If it continues down, it creates a loop that has to go 2 east throughout the center to attach the edges. So 10 goes proper, and every thing falls into place: