Suduro is a new puzzle that blends the rules and challenges of Kakuro and Sudoku. As far as I know, I (Dan Tow) am the inventor of this game. In Suduro, you solve a Kakurolike puzzle in the first stage, in order to create a Sudoku puzzle, which you then solve in the second stage using the usual strategies for solving Sudoku puzzles. However, there is no rule against beginning to solve the Sudoku puzzle when you have just a partial solution to the Kakurolike puzzle, and this can be a good or even necessary strategy for completing stage 1. The rules are best illustrated by example, so here is the first Suduro puzzle I built, as an illustration:
12 
11 
8 









14 
6 
27 









15 
28 
10 
11 
18 
12 
18 
21 
6 
4 
31 
10 


11 
0 
0 
0 
0 
9 
0 
0 
0 
2 


12 
1 
4 
0 
2 
0 
0 
0 
5 
0 


8 
0 
7 
0 
0 
0 
0 
1 
0 
0 


10 
2 
0 
1 
0 
4 
0 
3 
0 
0 


16 
5 
0 
0 
0 
0 
2 
0 
9 
0 


21 
0 
0 
6 
0 
0 
0 
0 
7 
8 


13 
0 
0 
5 
0 
8 
0 
0 
0 
0 


26 
3 
6 
0 
9 
0 
0 
0 
8 
0 


14 
0 
1 
0 
7 
0 
4 
0 
2 
0 
The gray and white squares must be filled in with the numbers 19, just as for Sudoku. Just as Sudoku requires, each row, column, and 3by3 subgrid (as outlined by the thicker black lines) uses each of the numbers 19 exactly once. However, unlike Sudoku, Suduro does not provide starting numbers for the problem solution by filling in roughly onethird of the grid squares in advance. Instead, Suduro shows some of the grid squares colored gray, and provides clues to the sums of groups of the gray squares, just as Kakuro provides sums of rows and columns of squares. In Suduro, the numbers in the black squares along the left edge of the puzzle are the sums of the gray squares in each row. The numbers in the black squares along the top edge of the puzzle are the sums of each column of gray squares. Finally, the numbers in the small 3by3 grid to the upper left of the main grid are the sums of the gray squares within each 3by3 subgrid inside the main grid. Examples will make this clearer, so I’ll illustrate the beginnings of the solution of this Suduro, with row and column labels added so I can refer to the rows and columns clearly.
12 
11 
8 









14 
6 
27 
1 
2 
3 
4 
5 
6 
7 
8 
9 
15 
28 
10 
11 
18 
12 
18 
21 
6 
4 
31 
10 

a 
11 
0 
0 
0 
0 
9 
0 
0 
0 
2 

b 
12 
1 
4 
0 
2 
0 
0 
0 
5 
0 

c 
8 
0 
7 
0 
0 
0 
0 
1 
0 
0 

d 
10 
2 
0 
1 
0 
4 
0 
3 
0 
0 

e 
16 
5 
0 
0 
0 
0 
2 
0 
9 
0 

f 
21 
0 
0 
6 
0 
0 
0 
0 
7 
8 

g 
13 
0 
0 
5 
0 
8 
0 
0 
0 
0 

h 
26 
3 
6 
0 
9 
0 
0 
0 
8 
0 

i 
14 
0 
1 
0 
7 
0 
4 
0 
2 
0 
Consider column 5: The three gray squares at a5, d5, and g5 must add up to 21, according to the clue on the top row. They must also be different numbers between 1 and 9, since they must end up as part of a valid Sudoku grid. The largest legal sum for two of these numbers is 17 (8+9). That makes the smallest possible value for any of the three numbers 4 because a number less than 4 for any of these numbers would require that the other two numbers add up to more than 17. In the row d, however, the only legal combination of 4 numbers (for the 4 gray squares in row d) adding up to 10 is 1, 2, 3, and 4, in some order. Since d5 belongs to both row d and column 5, it must be 4, since only 4 works in both row d and column 5. (Rows, columns, and subgrids having unusually high or unusually low sums for the given number of gray squares provide especially useful clues.) Since d5 is 4, and the sum of the gray squares for the center subgrid (gray squares d5 and e6) is 6 (according to the clue in the center of the upperright 3by3 grid), e6 must be 2. Since e6 and i6, in column 6, add up to 6, according to the toprow clue, i6 must be 4. So far, the solution looks like:
12 
11 
8 









14 
6 
27 
1 
2 
3 
4 
5 
6 
7 
8 
9 
15 
28 
10 
11 
18 
12 
18 
21 
6 
4 
31 
10 

a 
11 
0 
0 
0 
0 
9 
0 
0 
0 
2 

b 
12 
1 
4 
0 
2 
0 
0 
0 
5 
0 

c 
8 
0 
7 
0 
0 
0 
0 
1 
0 
0 

d 
10 
2 
0 
1 
0 
4 
0 
3 
0 
0 

e 
16 
5 
0 
0 
0 
0 
2 
0 
9 
0 

f 
21 
0 
0 
6 
0 
0 
0 
0 
7 
8 

g 
13 
0 
0 
5 
0 
8 
0 
0 
0 
0 

h 
26 
3 
6 
0 
9 
0 
0 
0 
8 
0 

i 
14 
0 
1 
0 
7 
0 
4 
0 
2 
0 
The remaining gray squares in the
bottommiddle subgrid (g5, h4, and i4) must add up to 24 (284), according to
the clue in the upperleft 3by3 grid. Only the numbers 7, 8, and 9 (in some
order) legally sum to 24, so none of these squares is less than 7. This minimum
number, 7, is also the maximum number allowable in row i, since i2+i8 cannot
sum less than 3 (1+2). Therefore, i4=7, and i2 and i8 are 1 and 2, in some
order. The maximum value possible in any of the 3 gray squares (summing to 8,
according to the clue) in the upperright subgrid is 5, since the other two
squares must add up to at least 3 (1+2). Now examine column 8: If we assume
both the maximum value of b8 (5), and the maximum value, 2, of i8, the other 3
gray squares (e8, f8, and h8) in column 8 must add up to 24 (
If you’ve never solved a Kakuro puzzle, this line of reasoning may be hard to follow, and might seem like it would take a long time to carry out, but if you are an experienced Kakuro puzzler, it will seem familiar, except for the minor extra twist provided by the subgridsum clues, added to the clues for rows and columns. Just as for Kakuro, practice will lead to much faster puzzlesolving, as this sort of reasoning becomes secondnature.
At this point in the solution process, you’re over the hump for stage1, and the rest of the stage1 solution for the gray squares is relatively straightforward. The partial solution so far is:
12 
11 
8 









14 
6 
27 
1 
2 
3 
4 
5 
6 
7 
8 
9 
15 
28 
10 
11 
18 
12 
18 
21 
6 
4 
31 
10 

a 
11 
0 
0 
0 
0 
9 
0 
0 
0 
2 

b 
12 
1 
4 
0 
2 
0 
0 
0 
5 
0 

c 
8 
0 
7 
0 
0 
0 
0 
1 
0 
0 

d 
10 
2 
0 
1 
0 
4 
0 
3 
0 
0 

e 
16 
5 
0 
0 
0 
0 
2 
0 
9 
0 

f 
21 
0 
0 
6 
0 
0 
0 
0 
7 
8 

g 
13 
0 
0 
5 
0 
8 
0 
0 
0 
0 

h 
26 
3 
6 
0 
9 
0 
0 
0 
8 
0 

i 
14 
0 
1 
0 
7 
0 
4 
0 
2 
0 
We already know that c7 is 1 or 2, based on the sum of the topright subgrid being 8, but c7 cannot be 2, since that would require that d7 also be 2 for the column 7 to sum to 4 (according to the toprow clue), violating the distinctvalues requirement of every column, row, and subgrid. Therefore, c7 must be 1. You can now fill in a whole series of values just by searching for rows, columns, and subgrids having only one remaining unsolved gray square, and subtracting the solved values from the clues. For example, the upperright subgrid has just a9 unsolved, and a9+1+5=8, so a9=2. Following this method, (with “C,” “R,” and “S” indicating use of a column, row, or subgrid clue, respectively), you can solve for a9=2 (S), f9=8 (C), i2=1 (R), c2=7 (R), h8=8 (S), d7=3 (C), a5=9 (R), b4=2 (S), g5=8 (C), g3=5 (R), and h4=9 (C&S). The partial stage1 solution now looks like:
12 
11 
8 









14 
6 
27 
1 
2 
3 
4 
5 
6 
7 
8 
9 
15 
28 
10 
11 
18 
12 
18 
21 
6 
4 
31 
10 

a 
11 
0 
0 
0 
0 
9 
0 
0 
0 
2 

b 
12 
1 
4 
0 
2 
0 
0 
0 
5 
0 

c 
8 
0 
7 
0 
0 
0 
0 
1 
0 
0 

d 
10 
2 
0 
1 
0 
4 
0 
3 
0 
0 

e 
16 
5 
0 
0 
0 
0 
2 
0 
9 
0 

f 
21 
0 
0 
6 
0 
0 
0 
0 
7 
8 

g 
13 
0 
0 
5 
0 
8 
0 
0 
0 
0 

h 
26 
3 
6 
0 
9 
0 
0 
0 
8 
0 

i 
14 
0 
1 
0 
7 
0 
4 
0 
2 
0 
For the sum in column 8 to work out legally, e8 and f8 must be 7 and 9, in some order, but if e8 is 7, then the row clue for row e requires that e1 also be 7, which violates the distinctvalues rule, so e8 must be 9, and f8 must be 7. Filling in singleremaining gray squares for rows e and f, then column 3, then row d, places values f3=6, e1=5, d3=1, and d1=2. We’re now almost finished with stage1, with the partial solution:
12 
11 
8 









14 
6 
27 
1 
2 
3 
4 
5 
6 
7 
8 
9 
15 
28 
10 
11 
18 
12 
18 
21 
6 
4 
31 
10 

a 
11 
0 
0 
0 
0 
9 
0 
0 
0 
2 

b 
12 
1 
4 
0 
2 
0 
0 
0 
5 
0 

c 
8 
0 
7 
0 
0 
0 
0 
1 
0 
0 

d 
10 
2 
0 
1 
0 
4 
0 
3 
0 
0 

e 
16 
5 
0 
0 
0 
0 
2 
0 
9 
0 

f 
21 
0 
0 
6 
0 
0 
0 
0 
7 
8 

g 
13 
0 
0 
5 
0 
8 
0 
0 
0 
0 

h 
26 
3 
6 
0 
9 
0 
0 
0 
8 
0 

i 
14 
0 
1 
0 
7 
0 
4 
0 
2 
0 
Just 4 gray squares remain unsolved, so even trialanderror would suffice from here, but logic can still help. Since the squares b1 and h1 must sum to 4, to leave a sum of 11 for the column1 gray squares, h1 can only be 1 or 3. However, the lower left subgrid already has a 1, so h1 must be 3. Setting h1=3 leads to h2=6, b1=1, and b2=4. This completes the stage1 solution for the gray squares, which leaves an ordinary Sudoku puzzle still to be solved. If you already solve Sudoku puzzles, you’ll find stage2, completion of the Sudoku puzzle, to be familiar, using the skills you have already mastered. If you are not experienced with Sudoku, there are lots of books and online materials to teach detailed Sudoku strategy, but you will find that you can get an awfully long way just by applying one simple rule, repeatedly and systematically: Considering one number at a time, check each subgrid that is missing that number to see whether that number is already restricted to just a single legal square within that subgrid. Sometimes, you cannot restrict a number to a single square, but you can place it in a single row or column, and this is also useful, allowing better restrictions on that number in the neighboring subgrids. Here is the initial stage2 Sudoku, after stage 1 is complete:

1 
2 
3 
4 
5 
6 
7 
8 
9 
a 
0 
0 
0 
0 
9 
0 
0 
0 
2 
b 
1 
4 
0 
2 
0 
0 
0 
5 
0 
c 
0 
7 
0 
0 
0 
0 
1 
0 
0 
d 
2 
0 
1 
0 
4 
0 
3 
0 
0 
e 
5 
0 
0 
0 
0 
2 
0 
9 
0 
f 
0 
0 
6 
0 
0 
0 
0 
7 
8 
g 
0 
0 
5 
0 
8 
0 
0 
0 
0 
h 
3 
6 
0 
9 
0 
0 
0 
8 
0 
i 
0 
1 
0 
7 
0 
4 
0 
2 
0 
(The Kakurolike clues to the graysquares sums no longer matter, since all the gray squares are filled in correctly, so I’ve left them off.) Just to start you off, the 1s at d3 and c7, together with the alreadyfilled in gray squares in the middleleft subgrid, leave square e9 as the only legal location for the 1 in that subgrid. With a 1 at e9, in turn, and the 1 at c7, only g8 remains as a legal location for the 1 in the bottomright subgrid. Simple repetition of this question, “Is there just one legal location remaining for value n in this subgrid?” very systematically for each number n and each subgrid, repeatedly, will fill in a surprisingly large fraction of even the hardest Sudoku. The trick here is to be very systematic, and to revisit subgrids after more values are filled in, since the more values get filled in, the more likely it is that just one legal square remains for a given value. In the case of this particular Sudoku, this approach alone suffices to solve the entire problem, giving the complete solution:











6 
5 
3 
1 
9 
8 
7 
4 
2 

1 
4 
9 
2 
7 
3 
8 
5 
6 

8 
7 
2 
4 
6 
5 
1 
3 
9 

2 
9 
1 
8 
4 
7 
3 
6 
5 

5 
8 
7 
6 
3 
2 
4 
9 
1 

4 
3 
6 
5 
1 
9 
2 
7 
8 

7 
2 
5 
3 
8 
6 
9 
1 
4 

3 
6 
4 
9 
2 
1 
5 
8 
7 

9 
1 
8 
7 
5 
4 
6 
2 
3 
There’s just one more trick I want to illustrate, because it is so useful, although it didn’t happen to be needed for that first Suduro: I call this the “partial sums” trick. Often it is possible to know both the sum of 2 squares and the sum of 3 squares, which include those 2 squares, allowing you to deduce the third square. For example, let’s say you had the following partial Suduro:
10 
13 
14 









14 
6 
27 
1 
2 
3 
4 
5 
6 
7 
8 
9 
15 
28 
10 
11 
18 
10 
18 
21 
8 
11 
31 
9 

a 
11 
0 
0 
0 
0 

0 




b 
14 
1 
4 
0 

0 
0 
0 

0 

c 
12 
0 

0 


0 

0 
0 
We know from the left subgrid clue that b1+b2=10, and we know from the rowb clue that b1+b2+b4=14, so b4 must be 4 (1410). The partial solution now looks like:
10 
13 
14 









14 
6 
27 
1 
2 
3 
4 
5 
6 
7 
8 
9 
15 
28 
10 
11 
18 
10 
18 
21 
8 
11 
31 
9 

a 
11 
0 
0 
0 
0 

0 




b 
14 
1 
4 
0 
4 
0 
0 
0 

0 

c 
12 
0 

0 


0 

0 
0 
This combination of row or column clues with subgrid clues is frequently useful. Even where we cannot use the trick at the beginning of a puzzle, it is often useful after some of the squares are solved. For example, with the 4 filled in at b4, there are two ways to solve for c7: From the middle subgrid clue, and knowing that b4=4, we know that c4+c5=9 (134), and from the rowc clue, we know that c4+c5+c7=12, so c7=3 (129). We also could deduce even from the beginning of the puzzle that c7=3 since a7+a9=11 (from the rowa clue) and a7+a9+c7=14 (from the right subgrid clue).
Here’s more depth for real Suduro scientists:
Happy problem solving!