The use of maximums and minimums in the sum of cages

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: