Game of the glasses on the windowsill

Question

The windowsill above the sink is where my wife and I place our dirty wine glasses. And while both of us love each other, neither of us love loading the dishwasher. As a result, these dirty glasses accumulate en masse on the windowsill until there is physically no room left to place another.

For a while, it was simply up to chance who would wind up having to do this chore. But over a period of months, without either of us having spoken of the matter aloud, my wife and I found ourselves embroiled in an intensely competitive game of strategy, complete with rules of fair play. The first such rule was that whosoever found themself unable to place another glass on the windowsill must concede defeat, and load the dishwasher with all the glasses. The second such rule was that any glasses already placed would under no circumstances be disturbed until the washing. The third such rule was that we would rigidly adhere to a previous habit of ours, that of myself retiring my emptied glass of wine after dinner, and of my wife nursing her drink until later in the evening. We would dirty no other glasses during the day.

As such, I conspired a strategy to escape the chore entirely: When the windowsill was empty, I would place a glass in its center, and then precisely mirror each of her subsequent placements. I won each and every game.

My wife wasn't a fan of this strategy, but she wasn't about to admit defeat. Her riposte took the form of her feigning adoration for a truly dreadful potted plant she had found and purchased at the market, and then placing it, conspicuously off-center, right there on our windowsill gameboard. Her motive was obvious, but for its creativity I deemed it would be unsporting on my part to call her foul. Thus, a silent addendum to the rules: The plant would never be moved. (Nor would it be watered, but that was the result of an unrelated dispute between us.)

Alas, I must now seek counsel from greater puzzlers so that I might escape an equal division of labor. How do I defeat my wife in this new game, what should be my strategy? The dimensions of the windowsill, glasses, and plant are as follows:

• The windowsill is 82 inches wide (why yes, it is very large and impressive)
• The glasses are identical, each 4 inches in diameter
• The potted plant is 8 inches in diameter, placed 51 inches from the left of the windowsill

To be clear, intervals are continuous, not discrete: We are allowed to place glasses freely, not merely in a whole number of inches from the right or left.

Answer 1

Here is my evaluation:

Let's take the glass as the length unit. The windowsil has size 82/4 = 20.5 glasses, the empty spaces are 51/4 = 12.75 and 23/4 = 5.75 glasses wide.

If you think of the capacity of a space as an integer number of glasses, you can see that regardless of how much extra space you have, a space of capacity N can be split in two spaces of total capacity N-1 or N-2. It is N-1 or N-2 depending on the exact placement of the glass.

How to do N-1 or N-2? The available space is $N + \delta$ where $0 \leq \delta < 1$. To split N into A and B, where A+B = N-1, you can place the glass to leave space $A+\delta/2$ and $B+\delta/2$. To do A+B = N-2, you can place the glass to leave space $A+1/2+\delta/2$ and $B+1/2+\delta/2$. In both cases the spaces plus the glass total to $N+\delta$.

You cannot do A+B = N-3 because you would have to distribute 2 extra units between the spaces, making at least one of them gain one unit in size. The capacity would then be larger than we want.

This game is known as Kayles. One version has the form of a row of bowling pins where each player knocks down one or two adjacent pins. You can see how it relates to the glasses on the windowsill.

The solution to this problem uses the Sprague-Grundy theorem stating that such a problem is equivalent to playing Nim. Each space translates to a heap in Nim. The size of the heap is the nimber of the space.

The nimber for capacity 5 is 4 and the nimber for capacity 12 is also 4. That means that the position with the flower pot is equivalent to playing nim with two heaps of 4 token. Your wife might as well have place the flower pot in the dead center of the windowsill ("for esthetic reasons" of course). If you place the first glass, your wife can ensure to place the last glass.

Conclusion

There is no escape loading the dishwasher. If your wife loaded the dishwasher, she can make sure she places the last glass and force you to do it next.
There was that other puzzling web site where a woman was asking about where to put a flower pot on the windowsill to force her lazy husband to load the dishwasher. Not sure it is related...

And yet...

You are asking how to "escape an equal division of labor". You could do that by always being the one loading the dishwasher. For this, all you need to do is intentionally lose the game when your wife starts.

Easy, right? Not quite! What if your loving wife won't let you do all the work and tries to prevent you from losing. Can you guarantee to lose the game when you start?

That is called the misère version of the game. You want your opponent to play the last possible move. The misère Kayles was studied by William Sibert who published a solution. I didn't understand the solution, so I wrote a brute-force program to decide whether you could guarantee not to do the last move.

It happens that the misère version is losing for the starting position of 12 + 5.
So basically the one who loads the dishwasher can decide who does it next.

Therefore, once you had the task to of filling the diswasher, your wife will start the game in a losing position and you can force her to be the last to put a glass on the windowsill. Do the dishes, rince and repeat.

So you can escape an equal division of labor, maybe not the way you intended.