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# Krypto Kakuro Puzzles by KrazyDad

## A Sample Solution

Here are the steps I use to solve puzzle 1 from book 1 of my "Challenging" Krypto Kakuro puzzles. It helps to start with a chart showing the possibilities decodings for each letter.

```A      0 1 2 3 4 5 6 7 8 9
B      0 1 2 3 4 5 6 7 8 9
C      0 1 2 3 4 5 6 7 8 9
D      0 1 2 3 4 5 6 7 8 9
E      0 1 2 3 4 5 6 7 8 9
F      0 1 2 3 4 5 6 7 8 9
G      0 1 2 3 4 5 6 7 8 9
H      0 1 2 3 4 5 6 7 8 9
I      0 1 2 3 4 5 6 7 8 9
J      0 1 2 3 4 5 6 7 8 9
```
As we eliminate possibilites, I will replace digits in the above chart with a space. On the puzzle sheet, I keep track of this stuff by drawing dots in the squares next to each letter on the lower right.

First we can eliminate a bunch of letters that can't have a value of 0.

The digits A,C,E,G appear in the puzzle, so they can't be zero. The digit D appears in front of a clue, so it can't be zero.

Moreover, the digit D appears in front of a clue DJ, which is for a two-square word. Therefore, D must be 1, since a 2 square word can't be higher than 17.

The digit E appears in front of a a clue EI for a four-square word. Therefore E must be either 2 or 3, since a 4-square word can't be higher than 30.

```A          2 3 4 5 6 7 8 9
B      0   2 3 4 5 6 7 8 9
C          2 3 4 5 6 7 8 9
D        1
E          2 3
F      0   2 3 4 5 6 7 8 9
G          2 3 4 5 6 7 8 9
H      0   2 3 4 5 6 7 8 9
I      0   2 3 4 5 6 7 8 9
J      0   2 3 4 5 6 7 8 9
```

A rule of Kakuro is that in a two-square word with a one-digit clue, none of the component digits can be higher or equal to the clue digit.

A related rule is that in a two-square word with a two-digit clue, none of the component digits can be lower or equal to the second digit of the clue.

This means that for the clue

```C = A + ?,   A must be smaller than C.
DB = ? + A, so A must be larger than B
DB = ? + E, so E must be larger than B

B < A < C
B < E
```
Since B is smaller than three other digits, it can't be 7, 8 or 9.

C can't have the value of the lowest value of A or B.

Since B is less than E, we can rule out all but the values of 0 and 2 for B.

Also, you'll notice that G is used as a clue for a two-letter word. So G can't be lower than 3.

```A            3 4 5 6 7 8
B      0   2
C              4 5 6 7 8 9
D        1
E          2 3
F      0   2 3 4 5 6 7 8 9
G            3 4 5 6 7 8 9
H      0   2 3 4 5 6 7 8 9
I      0   2 3 4 5 6 7 8 9
J      0   2 3 4 5 6 7 8 9
```

Okay, this next one is a bit tricky. There are two 4-letter clues that have a first letter of E.

```EG = ? + ? + ? + ?
EI = ? + ? + ? + ?
```
We already know that E must be 2 or 3. However, there is only one possible clue in which E would be 3.

```30 = 9 + 8 + 7 + 6
```
Since there are two unique clues that use the E, that means E must be 2. Having solved E, we can see that B must be 0.
```A            3 4 5 6 7 8
B      0
C              4 5 6 7 8 9
D        1
E          2
F            3 4 5 6 7 8 9
G            3 4 5 6 7 8 9
H            3 4 5 6 7 8 9
I            3 4 5 6 7 8 9
J            3 4 5 6 7 8 9
```

Since two letter clues can't be larger than 17, the second digit of a two letter clue can't be 8 or 9. This affects A and J
```A            3 4 5 6 7
J            3 4 5 6 7
```

Since we have a clue that reads

```DB = E + ? + ? + G
```
and we have decoded some letters, we know this corresponds to
```10 = 2 + ? + ? + G
```
The only four numbers that can add up to 10 are 1+2+3+4. This rules out anything higher than 4 for G.
```G            3 4
```

This next one is also a bit tricky.

There's a clue in the puzzle that reads

```G = ? + ?
```
Since we know G can only be 3 or 4, that means that one of the cells that add up to G must be a one.
```G = 1 + ?   OR G = ? + 1
```
We also know that the second cell in that equation is also the second cell in
```DD = ? + ?
```
and we know that DD = 11. Therefore, that cell can't be 1 (because that would produce the illegal combination (1 + 10) so the first cell must be 1, and we can fill that it in the solution puzzle.
```G = 1 + ?
```
Since the 1 is used in that position, we know that further down in the puzzle, where it says
```C = A + ?
```
That the ? can't be another 1, since that cell is in the same word that already contains a 1.

That means that C must be at least 2 greater than A.

```A < C + 1
```
This rules out 4 as a possible value for C.
```A            3 4 5 6 7
C                5 6 7 8 9
```

In the equation

DB = ? + E We've solved B and E, and can fill in the blank as an 8.

This means that in the equation

```DJ = ? + ?
```
The first blank is an 8, which means that J can't be 6, since that would produce 16 = 8 + 8, which is illegal (numbers can't repeat).
```A            3 4 5 6 7
B      0
C                5 6 7 8 9
D        1
E          2
F            3 4 5 6 7 8 9
G            3 4
H            3 4 5 6 7 8 9
I            3 4 5 6 7 8 9
J            3 4 5   7
```

Note that possible values for the blank in
```DJ = 8 + ?
```
are 5,6,7,9.

Now, for the clue DB = E + ? ? + G

we already know that DB is 10, and that the numbers 1,2,3,4 must each be used in the word (because the only combination of 4 unique digits that add up to 10 are 1,2,3 and 4.

We've already ruled out 1 for the first and last square (E and G).

Now we can rule out 1 for the third square as well, because this is part of

```DB = _ + A
```
or
```10 = _ + A
```
And if the blank in that equation is a 1, then A must be 9, but we already know that A isn't 9.

That means the second square in DB = E + ? + ? + G must be the 1.

Since that 1 is also the last square in

```EG = ? + ? + ? + 1
```
and we know that EG must be either 23 or 24, then the remaining three squares must add up to 22 or 23, which means that none of those squares can be less than 5, and one of those squares must be a 9 (based on the different combinations of 3 numbers that add up to 22 or 23).

At this point the third square in the expression

```DB = E + ? + ? + G
```
has the possibilities 3 and 4.

This means that the A in

```DB = ? + A
```
can only have the values of 6 or 7.

Because A < C + 1, we can further limit C as well, to 8 or 9.

```A                  6 7
B      0
C                      8 9
D        1
E          2
F            3 4 5 6 7 8 9
G            3 4
H            3 4 5 6 7 8 9
I            3 4 5 6 7 8 9
J            3 4 5   7
```

Limiting A to 6 or 7 limits the possibilites of each of the squares in the clue

```DA = ? + ?
```
to values of 7,8 or 9. The second square in that clue intersects a clue which reads
```EI = ? + ? + C + ?
```
Having narrowed down the last two squares of that clue to values of 8 or 9, this means that it's second square must have a value of 7.

Looking at all the cells of the puzzle, we have narrowed down the possiblities for each square to:
```(5,6,7,8,9), (8,9)
(5,6),(7),(8,9),(8,9)
(8),(5,6,7,9),(1),(2,3)
(2),(1),(3,4),(3,4)
(6,7),(2,3)
```
Now we can start narrowing down possible values for some of the sums, based on these combinations of digits.

One of the clues is

```EG = (5,6,7,8,9) + (5,6) + (5,6,7,9) + (1)
```
Possible values are
```23  = 8 + 5 + 9 + 1
23  = 7 + 6 + 9 + 1
23  = 9 + 6 + 7 + 1
24  = 8 + 6 + 9 + 1
```
This reduces the possibliies of each square to
```(7,8,9),(5,6),(7,9),(1)
```
Because DJ = (8) + (7,9)

we can reduce J to 5 or 7,

```DF = (8,9) + (1) + (3,4) + (2,3)
```
Its possible range of sums is 14 thru 17, which means F can be limited to 4,5,6,7.

```EI = (5,6) + (7) + (8,9) + (8,9)
```
This limits possible values of EI to 29 or 30. Since we know E is 2, I must be 9, and therefore C must be 8.
```A                  6 7
B      0
C                      8
D        1
E          2
F              4 5 6 7
G            3 4
H            3 4 5 6 7 8
I                        9
J                5   7
```

We can now fill in this clue:

```DD = (9) + ?
```
as
```11 = 9 + 2
```
Which gives us a value for G because G = 1 + 2.

If you've made it this far, you should be able to solve the rest of the puzzle fairly rapidly.

Filling out other squares in the puzzle, you will find that A must be 6.

The possibilities of the clue that sums to EG are now reduced to

```23 = (7,8,9) + (5) + (7,9) + (1)
```
which must be
```23 = 8 + 5 + 9 + 1
```
Which means the clue
```DJ = ? + ?
```
is
```17 = 8 + 9,
```
so J must be 7.

and the clue

```DF = C + ? + G + ?
```
must be
```14 = 8 + 1 + 3 + 2
```
So F must be 4, and H must be 5. At this point, the puzzle is solved, and the letters can be decoded as
```A                  6
B      0
C                      8
D        1
E          2
F              4
G            3
H                5
I                        9
J                    7
```