Guide: If a space is filled in green, it is confirmed to be green. Same goes with red spaces. If a space contains a green “G” or a red “R”, those spaces are not necessarily green or red. They are just marked to make situations easier to explain.

Start by clicking the space with the star: We are given the hints: Wedge 1 is the only wedge that is all green. No loop contains three consecutive green or red spaces. So click on the remaining two spaces in wedge 1: The next clues are: Considered together, the four wedges labeled with factors of eight have two green spaces and two red spaces in each loop. The outer loop contains fewer green spaces than the center loop.

We can figure out now how many green spaces are in each loop. There are eight spaces per loop, and we can’t have three consecutive spaces of the same color in the same loop. That puts the amount of green spaces in each loop in between three and five. The outer loop contains one less green space than the middle loop, so the only possible combination of green spaces is the outer loop with four green spaces, middle loop with five, and center loop with five. (Also, keep in mind that with five green spaces in a row, there cannot be two consecutive red spaces.)

Let’s work with the middle and center loops for the moment. The only green space is in wedge 1 and no red spaces are known. The last remaining clue we have requires us to look at wedges 1, 2, 4, and 8, the four factors of eight. For each loop, we have two green spaces and two red spaces for each of those wedges.

Let us now assume that the space in the center loop in wedge 8 is green. Because wedges 1 and 8 contain green spaces in the center loop, wedges 2 and 4 must contain red spaces in the center loop: Now we can fill in wedges 3 and 5 with green spaces. In order to avoid three consecutive green spaces, wedge 7 contains a red space, and then wedge 6 contains a green space to finish with five green spaces and three red spaces: Note: All of this can also be done with the middle loop because they both contain 5 green spaces. I will only show the center loop to save time and space.

Next, let’s assume wedge 2 contains a green space. Wedges 4 and 8 are now red: In order to avoid three consecutive green spaces, wedge 3 contains a red space. BUT, wedge 4 already has a red space. This solution does NOT work, so wedge 2 must contain a red space in the center loop. (The middle loop for wedge 2 is also red for the exact same reason).

Lastly, we can set wedge 4 to contain a green space and wedge 8 to have a red space: We can fill in the remaining spaces in the inner loop: Let’s compare the two completed center loops:  Which spaces are the same color in both situations? Wedges 3 and 6 in the center loop: Remember, wedge 1 is the only all-green wedge. So the spaces in the outer loop and in wedges 3 and 6 are red: Our next step is to do the exact same process as we just did with the center loop, but instead for the outer loop. Let’s set wedge 8 as green and wedges 2 and 4 as red: However, this does NOT work because we now have three consecutive red spaces. So the space in the outer loop and in wedge 8 is red. Let’s make wedge 2 green instead: This loop can now be completed: Lastly, let’s make wedge 4 green and fill in the rest of the loop: There are two common green spaces when comparing the outer loops from the past two scenarios: There’s the last clue! Two wedges contain red spaces in the center loop and the outer loop. This doesn’t help us right away, but we will need it.

Back to the center loop, let’s pretend wedge 8 is green and wedge 4 is red: There needs to be a red space in wedge 5: This allows us to now fill in the rest of the middle loop using the same logic used as in the center loop (keeping in mind the factors of eight rule here): There are two spaces left in the outer loop. Looking carefully, there has to be one green space and one red space, and there are no restrictions on which one is which.

So we are left with two unique solutions:  We’re not done yet. Back to the center loop, let’s imagine wedge 4 is green and 8 is red. We can fill the entire center loop: Putting a red space in wedge 7, we can also complete the middle loop: And then the last two spaces can be either red or green:  So we are left with 4 unique solutions:    Which one could it be? We haven’t used the last hint yet: Two wedges contain red spaces in the center loop and the outer loop.

Which solution fits the hint? Only the one on the bottom right! (Wedges 2 and 8)

FINAL SOLUTION: 