Suko-kulink (Sudoku, Suko and Slitherlink tri-brid)

Question

Rules:

1. The lower part of the puzzle is a sudoku. Basic sudoku rules applied.
2. The upper and right of the puzzle are suko puzzles.
3. The numbers in circle and to the selected square has the same number when solving the slitherlink. Note: The circled square is in yellow. Only the yellow square with NO circle or pointed arrow are the clues from sudoku to slitherlink.
4. After solving the sukos and sudoku, the squares in yellow together with the numbers for suko are clues for solving slitherlink.

Rules for each puzzle:

Sudoku: Each column, row and 3x3 region must contain the numbers 1-9 once, and only once.

Suko:

Quadrant totals: These are the numbers given in the circles inside the grid itself.

Layout totals: There will always be three different colours for the cells in each puzzle. The numbers above the grid tell you what these should add to.

Slitherlink: Draw a single loop around the grid so that all the clues are satisfied. The clues tell you how many of its four sides are part of the loop.

Good luck!

Answer 1

Okay, I believe I have a solution. Resolving the sudoku and suko puzzles produces this grid:

Unfortunately, I believe there is an ambiguity in the third suko puzzle whereby the rows could read either 527/649/381 or 563/249/781, with 3-7 and 2-6 being interchangeable pairs.

To understand how to solve a suko puzzle, here is an example using the one level with the top row of the sudoku. The aim is to use the numbers 1-9 once each to fill the 9 spaces so that the sum of each sub-square of 4 digits equals the number they surround, while the sum of numbers of the same colour equals the number indicated to the side...

1. We know that the green squares sum to 17. These occupy 3 spaces around the 20 - the remaining blue space around the 20 must therefore be a 3 (since 17 + 3 = 20). Since the blues must sum to 6 the other two blue spaces must be occupied by 1 and 2.

2. This means that the two greens adjacent to the 13 must sum to 10. The third green must therefore be 7 in order for the greens to total 17 altogether. The remaining two green spaces must be occupied by 6 and 4 in order to sum to 10.

3. The orange spaces must therefore be some permutation of 5, 8 and 9. Consider if the centre space was 6. We would then need two of 5, 8 and 9 to sum to 12 (the remainder from 25 - 7 - 6); however, this is impossible. Thus the middle space must instead be 4. This then requires two of 5, 8 and 9 to sum to 14 (the remainder from 25 - 7 - 4); this must be the 5 and the 9 - the 8 therefore occupies bottom-left.

4. Considering the bottom left sub-square, this needs to total 19. With a 4 and an 8 already, we require 7 from the remaining two numbers. From the options available this can only be 2 and 5. Thus the suko puzzle is fully resolved!

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If we then interpret the rules as per my comment on the question, for any square in a suko puzzle which contains a circle, that circled number must be input into all of the other circles it is chained to, and in the yellow square at the end of its arrowhead. (I realise that can be confusing.) The resulting slitherlink for us to solve is then:

This is actually quite tricky, but there is a logical path to its solution, albeit one which requires a lot of x-steps-ahead thinking:

Step 1:

(i) First the usual slitherlink 'ins' involving 3's.
(ii) To satisfy the 2 on the bottom row and both 3's, the loop must path beneath the 2.
(iii) We can also add a few short sections of path elsewhere in the grid. (Bottom-left: that segment must be filled to avoid forming a loop restricted by the first 1 on the bottom row. Top-left: Segment below 2 must be filled regardless of which of the remaining sides of the 3 is correct. Top-right: Segment on edge above the 2 must be filled due to restrictions form the 1's below.)

Step 2:

Both 2's on the right must be passed along 2 sides forming the NE-SW diagonal. This helps force two sides of the neighbouring 3, then the 2 to its NW, and then spaces continue to fall until the top-right section is complete.

Step 3:

(i) Now for some conditional thinking. The segment to the left of the 2 in row 8 must be part of the loop regardless of which of the other sides of the 2 is filled. Furthermore, if we try to pass the 2 in row 8 on its right-hand side we eventually reach a scenario where the first 1 in row 7 needs 2 lines - a contradiction.
(ii) Thus, this 2 must be passed to the south. This has knock-on effects up most of the left-side of the grid.

Step 4:

(i) Now consider what would happen if the first 2 in row 2 was passed to the north. We would end up forming a closed loop - not allowed.
(ii) Thus it must be passed on the left instead.

Step 5:

(i) What about the 3 in row 6? If we pass it on the right we contrive a situation where the 2 in the first column has a loose end that cannot be satisfied.
(ii) Instead, we must pass it to the south.

Step 6:

(i) Consider next the second 2 in row 5. If we pass it to the left, the knock-on logic produces a scenario where the 2 to its right cannot be satisfied.
(ii) Instead it must be passed to the south.
(iii) Can it also be passed to the north? Not without creating another illegal situation...
(iv) Thus it must be passed on the right.

Step 7 (last one):

(i) Finally, note that the first 2 in row 4 must have 4 connections, including one either above or to the left of the top 2 in column 5. If it goes above the 2, we force a situation which leaves a stray segment below it.
(ii) Thus we pass it on the left instead, and there is only one way to complete the grid without breaking the loop.