
Krypto Kakuro Puzzles by KrazyDadA Sample SolutionHere 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 9As 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 twosquare 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 foursquare word. Therefore E must be either 2 or 3, since a 4square 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 twosquare word with a onedigit clue, none of the component digits can be higher or equal to the clue digit. A related rule is that in a twosquare word with a twodigit 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 < ESince 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 twoletter 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 4letter 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 + 6Since 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 + ? + ? + Gand we have decoded some letters, we know this corresponds to 10 = 2 + ? + ? + GThe 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 = ? + 1We 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 + 1This 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 = _ + Aor 10 = _ + AAnd 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 = ? + ? + ? + 1and 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 + ? + ? + Ghas the possibilities 3 and 4. This means that the A in
DB = ? + Acan 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 + 1This 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,
Another vertical clue reads
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.
A horizontal clue reads 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 + 2Which 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 + 1Which 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 + 2So 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
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