Re: The difficult puzzle 6x6 of today ( 01NOV2011 )

arjen wrote:

My 'solution' was: Total of -4, -2 and -1 = 19 (6*21 - sum(+cages))

By 2-6 in -4 cage: 11 left over for sum '-2' and '-1' cage

Possible values:

1-3 / 3-4 (invalid because of +5 cage)

2-4 / 2-3 (invalid because of +3 cage)

3-5 / 1-2 (invalid because of +3 cage)

@clm: thanks for explaning an easier one.

@patrick: 'Show coordinates' doesnt work by solution view.

Welcome. Implicitly, in your solution, you manage a problem of “maximums and minimums in the sum of cages”, that sometimes may be reduced to the same problem with individual cells and cages, like in our case (this time I include a graphic so that other puzzlers may follow this topic in the future). The interesting thing, in the more compact equation,

**a3 + “4-” = f4 + 2** is that, once we know that “4-“ is the pair 15, we have

**a3 + 6 = f4 + 2**, that is,

**f4 = a3 + 4** and, since the maximum value for f4 is 6, and a3 can not be 1, it must be a3 = 2 and f4 = 6. We then quickly fill all those cells in blue and inmediately those in green (considering that c6 + d6 = 6 and then d6 = 4, c6 = 2) and those cells in brown (f5 + f6 = 7, so 25 for those). We finish the puzzle in a few minutes and I think that answers the initial jomapil’s question: “How is it possible to solve it in 5 minutes?", but we must be lucky to see where relationships like those may be present.