In common with a number of other Magnum Opus puzzles, it is recommended that "pencil mark" numbers be entered into the cells of a KAKURO puzzle
to show the numbers which might potentially be the final number for each cell. The solving strategies are then applied to gradually remove these "pencil marks" until
there is only one left. This last remaining number in any cell becomes the solution number for that cell.
When specific cells need to be referred to in this discussion, they will be designated as C4-R3 which means the 4th cell in row 3.
Fundamental Strategy.
The first step is to "cast out" pencil marks for digits which cannot legally be used in forming the required summations. For example, consider
C3-R1 and C4-R1. The numbers in these two cells must add up to 4, and this can only be ahieved if the numbers are 1 and 3 (2s are not permitted, as that would give us
two digits the same in a solution line). The following table lists the digits which can be cast out for all possible digit sums for cases having 2, 3 or 4 digits.
Note that there are examples of situations where there will be no digits to cast out. For example if the sum is 14 for a three digit case, then no digits are
available for casting out. The example puzzle has a total of three such cases.
2 Digits | 3 Digits | 4 Digits |
3 | : | __3456789 | 6 | : | ___456789 | 10 | : | ____56789 |
4 | : | _2_456789 | 7 | : | __3_56789 | 11 | : | ___4_6789 |
5 | : | ____56789 | 8 | : | _____6789 | 12 | : | ______789 |
6 | : | __3__6789 | 9 | : | ______789 | 13 | : | _______89 |
7 | : | ______789 | 10 | : | _______89 | 14 | : | ________9 |
8 | : | ___4___89 | 11 | : | ________9 | 26 | : | 1________ |
9 | : | ________9 | 19 | : | 1________ | 27 | : | 12_______ |
10 | : | ____5____ | 20 | : | 12_______ | 28 | : | 123______ |
11 | : | 1________ | 21 | : | 123______ | 29 | : | 123__6___ |
12 | : | 12___6___ | 22 | : | 1234_____ | 30 | : | 12345____ |
13 | : | 123______ | 23 | : | 12345_7__ | | | |
14 | : | 1234__7__ | 24 | : | 123456___ | | | |
15 | : | 12345____ | | | | | | |
16 | : | 123456_8_ | | | | | | |
17 | : | 1234567__ | | | | | | |
The right hand graphic above shows the situation after all available cast outs have been completed.
Further Strategies.
Having reached this point, a number of opportunities present themselves:-
Look at C4-R1. The 1 or the 3 will need either a 4 or a 6 from C4-R2 to add up to 7. The remaining numbers in C4-R2 can be removed.
The 6 in C1-R3 would require a 9 in C1-R4 to add up to 15. Since there is no 9 available, the 6 can be removed.
The 6, 7 and 8 in C1-R4 require either 3, 2 or 1 to add up to 9. The other digits in C2-R4 may be removed.
Finally, as you have no doubt noted, C3-R1 contains a single digit 1 which becomes the solution for that cell. Having converted it to a
solution digit, 1s may be removed from C4-R1, C3-R2 and C3-R3.
The right hand graphic shows the puzzle after all of the above steps have been completed.
Further developments.
The following steps are now more or less obvious:-
C4-R1 contains a single 3 which can be converted to a solution digit.
The 6 in C4-R2 can be removed, and the 4 converted to a solution digit to give the required sum of 7.
The 4s in C2-R2 and C3-R2 can be removed, leaving a single 2 in C3-R2 which can be converted to a solution digit.
The 2s in C2-R2 and C3-R3 can be removed, leaving a single 4 in C3-R3 which can be converted to a solution digit.
The 4 in C2-R3 can be removed.
Finally a solution digit of 8 can be placed in C2-R2 as this is required to give a sum of 14, and the 8 can be removed from C2-R3.
The right hand graphic shows the puzzle after all of the above steps have been completed.
Back to Fundamentals.
At this point we can recognise that C1-R3 and C2-R3 represent a two digit configuration which must sum to 12, while C2-R3 and C2-R4 represent a
two digit configuration which must sum to 6, and so we can resume casting out using the tables previously discussed. The right hand graphic shows the situation
after this has been done. From here on, completion of the solution is a simple matter indeed.
Solvers with a mathematical bent may prefer to generate a set of eight simultaneous equations,
and to solve those to arrive at a solution to the puzzle. Whatever works for you is OK.
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