The New Sudoku Players' Forum

[Jean-Christophe]

Another variation on a theme. This time a “real” jigsaw puzzle.



The rules :
All rules for “vanilla” & killer sudoku apply. No number may be repeated within a cage.
Each cage has a color out of : green, blue, pink, yellow.
No cage may “touch” another cage of the same color, not even at the corners.
Some cages are given in the grid with their sums and colors.
Below the grid, you’ll find a list of cages with their shape, number of cells, sum and color.
You should place all the cages. They may be turned to any orientation, but cannot be flipped, not upside-down.
The cages layout has 180° rotational symmetry (AKA slider symmetry).
It can be solved by logic since I designed it by logic.
Enjoy

[udosuk]

Is the original version flawed?

[Jean-Christophe]

If you mean the one I posted yesterday but removed 5 mins later, yes, this one was flawed. I just hoped nobody saw it, but you must be continuously connected to this forum ;-)

I triple checked the one displayed here, which is OK (AFAIK)

[udosuk]

I just hoped nobody saw it, but you must be continuously connected to this forum ;-)

I missed a lot of action in this forum... But just caught that 5-min window you posted that flawed version... (BTW I also saw it on another forum and your site...) So it was just sheer good luck (or in your case - gravely bad luck )...

I still have the pic of the flawed version... Must be very valuable now... Anyone wants a peek?

[Jean-Christophe]

Here are some hints.

Hint 1:

The cages layout has 180° rotational symmetry (AKA slider symmetry)

I can't imagine solving this puzzle without this bit of information. Think of the implications, you should be able to place one cage.

Hint 2:

You should be able to find out the color of R5C7. Notice they aren't any cage with a single cell. This should give you the color of another cell. Then see what works with the color(s) and hint 1, which should give you the color of another cell

Hint 3:
Find out the color of R6C7. Then use the same logic as in hint 2


Hint 4:
Find out the color of R7C6. Then use the same color logic as before

Hint 5:
Look at the pieces. Some are laaaarge. Several have the same color and can't touch each other. You should be able to place some large cages

To be continued...

[emm]

I did the jigsaw part OK but am stumped on the puzzle apart from some exclusions. I'm waiting patiently for a hint.

[Jean-Christophe]

Here is my walkthrough in tiny text you may view by copy-paste to a text editor like notepad

Step 1
The cages layout has 180° rotational symmetry. Thus the cage over R5C5 must have rotational symmetry, with an odd number of cells.
There are two possible shapes : 3 cells in a row (Y 20/3), or 5 cells Z shaped (Y 27/5)
It can't be Y 20/3, because we would be left with only one B 14/3 and only one Y 27/5 which cannot be placed symmetrically in the rest of the puzzle.
Thus it must be Y 27/5 which orientation is like in the list
-> R4C45+R5C5+R6C56 = Y 27/5

Step 2
R5C7 = P (because R4C6 = B, R6C6 = Y, R5C8 = G)
R5C6 must also be P since there is no cage with a single cell.
R5C67 cannot be P 7/2 because symetrically R5C34 cannot be Y 6/2
Thus R5C67 belongs to a larger P cage : either P/3 (L shaped) or P 31/6
R6C7 <> P -> R4C7 = P

Step 3
R6C7 = B -> R6C8 = B
R6C78 is either B 21/3 or B/4, all with an L shape
-> R7C7 = B
R7C6 <> B -> R7C6 = G -> R8C6 = G

Step 4
We have not less than 5 different green cages to place
The given cages are restring their possible locations
-> R234C123 <> G, R456C789 <> G
R78C6 is part of a green cage
Placing the large G 21/6
Trying horizontal orientation like in the list
Not in R789C1234 because it would leave a single cell in R9C1.
Not in R789C2345 because it would touch another G cage in R78C6. And also because symetrically, the P 31/6 can't touch R4C7 which is P.
Not in R789C6789 because it would leave two empty cells in R8C89 and symetrically in R2C12 next to P 31/6. None of them can be P 7/2.
Trying horizontal orientation turned 180°
G 21/6 not in R123C5678, because P 31/6 can't be placed symetrically
There is no other way to place G 21/6 horizontaly
Trying vertically with the long side on the left
Not in R6789C123 because symetrically the P 31/6 can't touch R5C7 which is P
The only place left is in R5678C234
-> R5C34+R6C23+R78C2 = G 21/6
-> R234C8+R4C7+R5C67 = P 31/6

Step 5
We still have to place 4 different green cages
Two of the green cages have a T shape and must be placed symetrically since they are the only 2 ones with a T shape.
A T shaped cage cannot go in N3, because it can't go symetrically in N7
For the same reason, it cannot go in N9 & N1
Thus the 2 T shaped green cage must go in N2, the other in N8 which must include R78C6
-> R23C4 is part of a green cage
R23C3 <> G -> R123C4+R2C5 = G/4 -> R789C6+R8C5 = G/4

Step 6
G 17/4 can't go in N9 -> in N3
G 17/4 can't be placed vertically because it would leave a single empty cell in R4C9
-> R1C789+R2C7 = G 17/4
-> R8C3+R9C123 = Y 18/4
R678C1 = B 14/3, R234C9 = Y 20/3
R8C4+R9C45 = B 21/3, R1C56+R2C6 = P 20/3 (one of the two P 20/3)
R6C78+C78C7 = B/4, R23C3+R4C23 = B/4
R1C1+R2C12 = P 20/3 (the other one), R1C23 = Y 6/2
R8C89+R9C9 = G 21/3, R9C78 = P 7/2

Step 7
We have now placed all the cages and know the sum of all but 4 of them :
The 2 T shaped G/4 : either 15 or 21
The 2 L shaped B/4 : either 17 or 21
Outies of N8 -> R6C4+R7C3 = 24+21 + (15|21) - 45 = 15|21
Max of 2 cells = 17 -> G 15/4 in N8, G 21/4 in N2
R6C4+R7C3 = 15 = {69|78}, R7C45 = 9
45 on N2 -> R3C56 = 4 = {13}, R3C7+R4C6 = 12 = {48|57}

Step 8
Outies of C123 -> R56C4 = 20+6 + (17|21) + 14+14+21+14+15+18 - 135 = 4|8
Since R6C4 = {6789} -> R56C4 <> 4 = 8
-> B 21/4 in N14, B 17/4 in N69
R56C4 = 8 = [17|26] -> R7C3 = {89}
45 on N5 -> R45C6 = 10 = [46|73|82] -> R3C7 = {458}

Step 9
45 on N1 -> R4C123 = 16, R23C3+R3C12 = 19
45 on N9 -> R6C789 = 8 = {134|125} = {1...}, R7C789+R8C7 = 17
45 on R6789 -> R6C56-R5C34 = 14
Min R5C34 = {12} = 3 -> Min R6C56 = 14+3 = 17 = {89} = Max R6C56
-> R5C34 = {12}, R6C56 = {89}
-> R45C6 = [46|73] = {(6|7)...} forms a complex naked pair on {67} with R6C4 -> not elsewhere in N5

Step 10
Cage 21/6 = {123456} -> R6C23+R78C2 = {3456} (BTW -> R45C2 <> {3456})
45 on N4 R5C3+R6C123 = 15
R5C3 = {12} -> R6C123 = 13|14 = {346|347|356}
-> 2 of R6 locked in R6C789 = {125} -> R5C89 <> {57}
-> R5C7 = 7 (hidden single in R5)
45 on R5 -> R5C56 = 9 = [36], R4C6 = 4, R3C567 = [138], R6C4 = 7, R5C34 = [21], R7C345 = [836]
R5C12 = {59}, R5C89 = {48}, R4C9 = 6, R4C78 = {39}, R23C8 = {24}, R23C9 = {59}
R5C89 = [84], R67C9 = {12}, R7C8 = 5, R6C7 = 5
R89C9 = {78}, R8C8 = 6
R1C89 = [73], R12C7 = {16}, R78C7 = {29}, R6C89 = [12], R7C9 = 1, R9C78 = [43], R4C78 = [39]
Cage 14/3 in R3C12+R4C1 = {167} -> R4C123 = [187], R23C3 = [15], R1C23 = [24]
...


Edited to fix a typo

[emm]

Thanks for that - I needed a hint.

You might like to correct one typo - Step 5 & 7 G 21/4 is in N2 not N3.

There was a lot of thinking in this one - thanks for the puzzle.