clarocca wrote:I understand the explanation provided in the answer explanations. Yet, are there any tips for how you can know when you've written out every possibility? I think I would sit there second guessing whether or not I've realized every possibility and waste a great amount of time.
Hello clarocca,
That's a good question. Sometimes you can know easily you've exhausted every possibility, e.g., if you have a group of A and B, and a group of 1, 2, and 3, and you know that there are only six pairs you can make (A1, A2, A3, B1, B2, B3). Sometimes it's harder, as in the present game.
With the m's and s's, one chokepoint is that the fifth load must be an m, "mulch". So given that there are three m's, you can do a "count" of sorts: there's one configuration is possible if all three m's slop over to fill spaces 5, 6, and 7 (so that there's only one changeover); and if you "walk" the block of 3 m's over one space, and then another space, you have m's in 4-5-6 and 3-4-5 respectively (yielding two changeovers).
Then after that, you see what configurations there are when there's no more three-block of m's. There's always the m in space 5, so if it's left alone with no m on either side, the only place the other 2 m's can be is on the far left, together as a block in 1-2. Otherwise, there'd be too many changeovers.
All that's left is if the m in space 5 has an m on either side, If it's in space 4, then the remaining m has to be at either end, 1 or 7, or there'd be too many changeovers. Similarly, if the m is in space 6, then again, the only place the last m can be is in space 1, to avoid more than three cleanings. (The last m can't be in space 7, because we already used up the scenario with m's in 5-6-7.)
So if you do things in that systematic way, it helps track things.
Hope this helps,
David