Basic strategies (perhaps trivial but may be useful to someone).

Fifth rule: Addition of corners, border, inner area. And why the numbers in the corners of a 4x4 must be the same as those in the inner area (this property may be used in the timed puzzles).

Let’s call “b” the sum of all numbers in the border (the “addition value” of the border). Let’s call “C” the sum of the four corners. Let’s call “I” the addition of all numbers in the inner area (the “value” of the inner area). Let’s call “p” the size of the puzzle and let’s call “s” the sum of the numbers of a line (row or column) (as we know this sum is 10, 15, 21, 28, 36, …, for puzzles 4x4, 5x5, 6x6, 7x7, 8x8, …).

If we sum all the numbers in the top and bottom rows and in the leftmost and rightmost columns, that total will be 4s (four times a line) and it will be equal to the “border”

plus the four corners since we have counted the corners twice (the four corners are duplicated in this count). So:

b + C = 4s

b + I = ps (this second equation because the border

plus the inner area constitute the full puzzle and the sum of all numbers in the puzzle equals p multiplied by s (the “value” of a line). Subtracting these two equations:

I - C = (p - 4)s

I = C + (p - 4)s or also C = I - (p - 4)s. Sometimes, in a 6x6 (p = 6, s= 21), the corners are given so: I = C + (6 – 4)s = C + 2s = C + 42 (or C = I - 42). When the corners are given, the inner area may be deduced easily, but in other cases, not so intuitive, the inner area is given (or easily calculated) and then we deduce C (if in this case one of the corners is known, for instance, we determine the sum of the other three and then we can analyze the combinations for the “virtual triangle” they constitute).

Now let’s see the special case of the 4x4’s: I = C + (4 - 4)s = C + 0 = C,

the sum of the corners must be equal to the sum of the numbers in the inner area (the square 2x2 inside). But not only this. Further, the numbers in the corners and in the inner area must be the same. The demonstration is easy. Suppose that we have two equal numbers inside this “heart”, for instance, two 1’s. As these 1’s occupy (“erase”) columns B and C and rows 2 and 3, the other two 1’s must go to two (opposite) corners. Let’s see all the possibilities:

Nrs. inside the “heart” Corners

1123 11 23 (the rest 5 [for a total of 7] is only made with the 2-3,

1-4 producing three 1’s in the corners)

1124 11 24 (only combination for a sum of 6, 3-3 is not possible

because 3-3 in the other two corners force two 3’s

inside the “heart” against the hypothesis)

1134 11 34 (only combination for a sum of 7)

2213 22 13 (obviously four 2’s in the corners are not possible)

2214 22 14 (a sum of 5 with 2-3 is not possible, three 2’s in the

corners)

2234 22 34 (only combination for a sum of 7)

3312 33 12

3314 33 14 (a sum of 5 with 2-3 impossible, three 3’s in the corners)

3324 33 24 (obviously four 3’s in the corners are not possible)

4412 44 12

4413 44 13 (2-2 is not possible since the other two 2’s would go the

“heart” against the hypothesis)

4423 44 23 (1-4 makes three 4’s in the corners).

Now, three equal numbers inside the “heart” are impossible (two equal numbers contiguous). Finally, if the four numbers inside the inner area are different, 1-2-3-4 (I = 10), the four numbers in the corners must also be 1-2-3-4 (C = 10), different numbers, because if two of them were equal, that would force the repeated number to go twice to the “heart” against the hypothesis.

An easy example (practical):

The June, 20 4x4 difficult (Puzzle Id: 299530).The 3 in D1 forces a 3 in the “heart” but in B2 is not possible (it would require two consecutive 1’s to complete the cage “1-“); in C2 is not possible because 3 is not a factor of 8; in C3 is not possible because the 1 in A4 makes impossible a sum of 3 in C4-D4, so a 3 must go to B3 being the other 3 [prime factor of the “72x”] necessarily in A2; the rest is very quick (4 in B4, 2 in A3, etc.).

Looking to the general expression: I = C + (p-4)s, we observe that if “C” increases “I” also increases and vice versa; obviously if both increase simultaneously the “walls” in the border must decrease and vice versa. Let’s call “w” the sum of all numbers in the four walls (the border without the cornes):

w = b - C = 4s - C - C = 4s - 2C (in the case of a 6x6, for instance, w = 84 - 2C).

In the case of a 6x6, C could range from a minimum of 6 (1, 1, 2, 2) to a maximum of 22 (6, 6, 5, 5) then I [= C + (6-4)s = C + 2s = C+ 42] could range from 48 to 64; the value of the four “walls”, w, could range from 72 (84 - 12) to 40 (84 - 44).