KAKURO MANIA
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Kakuro example Let's solve a puzzle together. Generally, we first look for cells with minimum possible alternatives. | ||
| Step 1. Very helpful is to discover unique combinations in the puzzle like the two intersecting sequences highlighted with blue and red on the picture. The only 2-cell combination with sum 3 is 3 = 1+2. Analogously, 4 = 1+3. So, the value of the intersecting cell must be 1. | |
| Step 2. Now, take a look at the intersection on the next picture. Again we have a unique combination at the down clue (10 = 1+2+3+4), so the maximal possible value for each cell in this sequence is 4. Although there are many possible combinations for the across clue (13 = 4+9 = 5+8 = 6+7), the minimal possible value for each cell in that sequence is 4. Therefore, for the intersecting cell, there is no other possible value except 4. | 
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| Step 3. The next highlighted clues show a similar situation - intersection between unique combination sequences (23 = 6+8+9 and 16 = 1+2+3+4+6), where the only possible value for the intersecting cell is 6. | |
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Step 4. Now, let's analyze the blue highlighted sequence. As we mentioned, the only possible combination here is 16 = 1+2+3+4+6. So, we have 3 empty cells - the second, the fourth and the fifth - and 3 missing values - 1, 2 and 4. If we look for the right cell for the 4, we will exclude the second one, because the 4 already exists in the down sequence. We will also exclude the fourth one, because 4 doesn't appear in the possible cell values of the down clue (11 = 1+2+3+5). Therefore, the 4 must be placed in the fifth cell of the sequence. Afterwards, the 2 already appears in the down sequence along the fourth cell, so we will place it in the second cell. | 
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| Step 5. Look at the red highlighted down sequence. As we mentioned, the only possible combination here is 11 = 1+2+3+5. The remaining empty cells must be filled with 3 and 5. But 5 can't appear in the blue highlighted across sequence, because the entire sum of the sequence is 5. | |
| Step 6. At last, the missing 1 and 3 in the highlighted sequence can fill the empty cells in only one way, thus making the rest of the solution obvious. | 
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| Congratulations! We have a solved puzzle. | |
| Copyright © 2012 Boyan Kolev, Sofia, Bulgaria | ||