A Trip to Ripple Islands

Question

All aboard! All aboard! You have won an ALL-EXPENSES PAID TRIP to the magnificent Ripple Islands!
Now, the only price of admission is to solve this little puzzle. It is a Ripple Effect puzzle based off a map of the luxurious Ripple Islands!

The water, of course, doesn't need to be filled with numbers. That's a ridiculous proposition. You need only to fill each island segment of size N with the numbers 1 to N, such that any two identical numbers X in the same row or column have at least X cells between them. Deusovi explains the rules better here.

Anyway, good luck! To start you off, I've placed all the 5s onto the map already. I hope to see you on board.

If you're one of those people who don't like colours:

Answer 1

^{A note: so as not to make this answer too horribly big, I've reduced the size of each image. If there's an "s" or "m" or "t" before the ".png", you can strip it out to see the full-size image. Or just click the smaller image to open the full-size one.}

Final answer

Complete solution

1. Consider the bottom left island. The upper brown segment must contain 1 and 2, in some order. If the 1 is on the left, then the pink 2 must be at the corner of the L, so 2 can't appear anywhere in the lower brown segment. This is a contradiction, so we must have:

   

2. Now consider the island just above the really big one at the bottom. Assuming 1 is above 2 in the left-hand green segment, we know that the top of the lower brown segment must be 3 (clearly it can't be 2; and if it were 1, then 1 couldn't be anywhere in the pink segment). Then we can fill in the pink segment and the rest of the lower brown segment easily, and finally the 2 and 3 in the upper brown segment:

   

   But now 3 can't be anywhere on the right-hand green segment. Contradiction, so we can now fill in the left-hand green segment for sure, and thus the 2's in the pink and lower brown segments. Assuming a 3 at the top of the lower brown segment leads to a contradiction when 3 can't be anywhere in the upper brown segment:

   

   So we can now fill in most of the rest of this island:

   

Now we can finally start to make some cross-island deductions!

3. Consider the bottom right island. If the rightmost brown segment has 1 on the left, then the bottom of the pink L would have to be 4, so 4 couldn't be anywhere in the green segment. Contradiction, so 2 is on the left. This enables us to solve the pink L too, and we get:

   

   Now if the green segment has 4 on the right, then it must have 3 second from the left, so either 1 or 2 is on the left. That means the corner of the brown L must also be 1 or 2, leaving nowhere for 3 in the brown L. Contradiction, so the green segment must have 3 on the right.

   If the bottom pink segment has 1 on the left, we can fill in the upper brown segment starting from the 2. Then the second-to-left cell in the green segment must be 4 and the bottom right of the pink square must be 1, so the second-to-right cell in the green segment must be 2:

   

   But now the top right cell in the pink square can't be anything. Contradiction, so the bottom pink segment has 2 on the left, and we can fill in the rest of the bottom island quite quickly:

   

4. Consider the rightmost island. A cross-island deduction from the bottom one tells us that in the lower green segment, the 4 must be above the 3. Assume the 2 is next to the 4; then we can swiftly fill in the lower brown L as follows:

   

   But now 1,2,4 can't be in either of the two cells to either side of the corner in the pink segment, contradiction. So we can fill in the lower green segment and the lower brown L, and then we know where 2 must be in the pink segment, which enables us to fill in the rest:

   

Now we can use more cross-island deductions to complete the half-finished island from earlier:

5. Consider the rectangular island on the left. By cross-island deductions, we know the 3 in the brown segment must be in one of the two squares adjacent to the corner. If it's on the left, then the 3 in the pink L must be at the corner, leaving not enough room for the other two 3's on this island. Contradiction, so the 3 in the brown segment must be on the right, and we get:

   

6. Now consider the island on the left, second from the top, and the right-hand end of lower pink segment. By cross-island deduction, this can't be 3; nor can it be 2, because there must be a 2 in the green segment. So it must be 1, and we can swiftly fill in the rest of the island:

   

7. Now consider the top left island. Cross-island deductions from below enable us to fill in all the 3's: on the right for brown, on the right for green, at the bottom for pink. Then the corner of the green L can't be 2 (that would leave no possibilities for pink 2), so we can finish off the island:

   

8. After this, we can return to the rectangular island on the left. We now know that in the green L, 3 and 4 must both be on the bottom. But 2 can't be next to the corner (since then there'd be no place for 2 in the pink L), so 2 is at the top and 1 is below it. From this we can almost finish off the island:

   

9. Now consider the island in the middle of the second row. By cross-island deductions, the 3 in the lower brown L must be somewhere on the right, and thus so must the 3 in the upper brown L. If the upper brown L has a 3 at the top, then the green L has a 3 at the bottom, so the pink 3 can't go anywhere. Contradiction, so the upper brown L has a 3 at the corner. This means the pink 3 is at the bottom left, the green 3 is on the right, and the lower brown 3 is at the corner. By cross-island deductions from the left, we can fill in the rest:

   

Cross-island deductions now let us finish off two more islands:

10. Consider the top right island. By cross-island deductions from the left, the green 3 must be at the corner. So the pink 3 must be at the bottom, which leaves only one possibility for the 3 in the lower brown segment. Now the green 2 can't be at the bottom (that would leave nowhere for the 2 in the lower brown segment), so we can finish off the green segment:

    

    The brown L must have 3 on the right-hand side, so that leaves only one possibility for the pink 3. Also, 2 above 1 in the upper pink segment would leave no possibilities for the left cell in the brown L, so it must be 1 above 2. This enables us to complete the tiny island using cross-island deductions:

    

    And the rest is easy to deduce, yielding the final solution shown at the top.

Answer 2

Partial

Following on from @randal'thor:

1) By looking at the three at the bottom of the large island and the right of the small one we can place a three on the bottom island.

2) By using this three as well we can place in a couple more. The rightmost cannot be one tile lower as this would make it impossible to place a 3 in the green area.

3) A little trial and error with the placement of remaining 3's and 4's on this island yields this solution.

Using the 3's from the middle island also allows us to start placing 3's on the rightmost island too. (Also I missed a 3 at the bottom of the middle island in my images.)