Basic strategies.
Example of “the use of maximums and minimums in the sum of cages”.
One year has elapsed since the original post was sent. Along this time I am sure this rule has been useful in many cases. However, just in case any calcudoker has still doubts on how to apply it I am including a recent puzzle which clarifies well its utility. The example below corresponds to the Sunday’s regular 9x9 (Jul 08, 2012).
Here is the puzzle:
We will apply the rule to determine the cage “3:” which “initially” could admit two combinations: the [1,3] or the [3,9] (the [2,6] not allowed due to the 6’s in rows 4 and 5). Graphic 2 below (the relevant cells shown in blue colour): The sum of the three leftmost columns must be 135 (= 3 x 45) so (naming, for clarity, s1 the sum of numbers in cage “3:”, s2 the sum of numbers in cage”3” and “a” the value of the single cell c7) we have:
s1 + s2 + a + 14 + 9 + 23 + 6 + 1 + 14 + 8 + 8 + 16 + 3 + 7 = 135, that is:
s1 + s2 + a = 26.
A first consequence, i.e., from this equation is that, being s1 even and s2 odd, “a” must be odd, that is, 3 or 5 (since 1, 7 and 9 are already present in column c). We apply the rule of maximums and minimums in this way: If the maximum possible value of s2 is 15 (case [6,9]) and the maximum possible value of “a” is 5, the minimum possible value of s1 should be 6 (26  15  5), consequently the cage “3:” cann’t be [1,3] and it must be [3,9] (with s1 = 12).
Now, s2 + a = 26  12 = 14. If a = 3 >>> s2 = 11 = [4,7], not allowed due to i6 = 7, then:
a = 5 and s2 = 9 = [3,6] with b6 = 6 and c6 = 3 (see Graphic 3 below):
We have defined the only possible combination [3,9] for “3:” and at the same time we have determined 4 additional cells (b6, c6, c7, d7). The official solution of the full puzzle:
