Cooking rota

[wildeabandon]

Say you have three people, and sometimes they all eat together, and sometimes they eat together in each of the pairs. Say you want to keep track of whose turn it is to cook in each of these three configurations. Say you'd quite like some kind of physical method of doing this so it can be incremented in the kitchen. I'm thinking it must be possible with jars and beads, or some kind of abacus like thing, but I can't entirely pinpoint how best to do it. If it's something that it's easy to make a second copy of for keeping track of whose turn it is to buy dinner in the same groupings that would be awesome.

Any ideas?

Comments

[robert_jones]

It seems to me that we don't want four separate rotas, one for the three of us and one for each of three pairs.

Let's say between the three of us, cooking has gone C, J, S

Then S isn't here, and it's your turn to cook. When next we three eat together, it will be my turn to cook, then yours, then Ramesh's.

If you, like it will go C, J, S, [C], J, C, S and not C, J, S, [C], C, J, S, which would be silly. Similarly if we had several turn of pairwise cooking, then we would fit pairwise into the three-way rota however was appropriate, e.g. C, J, S, [C], [J], C, J, [C], S, J, C, etc.

[robert_jones]

So I think you could do it with a triangle, with C, J and S on the three sides. Then we rotate the triangle by 120° for three-way cooking and reflect it along the appropriate axis for two-way cooking.

[wildeabandon]

That's equivalent to the three jars suggested below, isn't it? So now we just have to decide which is easier to implement.

[robert_jones]

No, it's not equivalent. Taking x, y, z to mean C has x beans, J has y and S has z, and allowing alphabetical order to break ties, the sequence C, J, S, [C], [J], C, J, [C], S, J would go

3, 3, 3
5, 2, 2
4, 4, 1
3, 3, 3
4, 2, 3
3, 3, 3
5, 2, 2
4, 4, 1
5, 3, 1
4, 2, 3
3, 4, 2

Which would make it S's turn to cook, when it should be yours.

[robert_jones]

Or more simply, the sequence, C, [J] goes

3, 3, 3
5, 2, 2
4, 3, 2

So as between you and me, I have to cook twice in a row!

[ciphergoth]

Every time you cook a meal, move a bean from every jar for a person you cooked for into your jar.

[wildeabandon]

So it's always the turn of the person with fewest beans in their jar to cook? Yes, that sounds sensible. I think it would give slightly too much value the additional difficulty in cooking for an extra person, except Ramesh being veggie means that it often is twice as difficult.

Thanks.

[meihua]

This is beautifully simple. Thank you.

[robert_jones]

It doesn't matter how beautifully simple it is, when it gives the wrong answer!

[meihua]

I saw who posted it and assumed it was correct. I'll look forward to seeing the discussion.

[wildeabandon]

Does it give the wrong answer to the question as I stated it? (I agree that your solution is a better one for the question as I should have stated it)

[robert_jones]

Also, how can you call a solution involving three jars and at least six beads simpler than one involving one triangle?

[meihua]

Because the triangle solution as described didn't make any sense to me, and also because it doesn't seem to handle the situation where it just so happens that someone cooks a few times in a row, just because that's how it works out.

[andrewducker]

I'd ignore the configurations, as otherwise you end up in situations where someone would have to cook twice in succession (which doesn't seem fair) - and it's not much harder to cook for three than for two.

In which case a simple "Take it in turns to cook, and if you missed your turn then it's your turn next" rule would probably work best, and be fair 95% of the time.

[robert_jones]

The problem is, this would work well if the three people all lived together. In fact (although Sebastian didn't say so), one of them is only there 30% of the time, or so. It doesn't seem fair to treat him as 'missing his turn' on the large number of occasions when he isn't here!

[andrewducker]

Being away 70% of the time would affect things a lot. It'd basically mean that he'd cook every time he was there, which would be unfair.

I suppose that what you actually want is to keep the ratio of "times cooked/"times eaten" the same between people.

Which basically means keeping track of both numbers (via a whiteboard and mental arithmetic). Or the beads method, which should be fine most of the time.

I do worry that any mechanical method will have intermittent problems - not least at times when The Method says it's your turn to cook, but real life makes it tricky.

[robert_jones]

not least at times when The Method says it's your turn to cook, but real life makes it tricky

That's a problem with having a rota, not with having a mechanical system for implementing it.

[meihua]

After reading all of the comments, may I just say that you are all lovely? *smiles*

[meihua]

(so I'm femme, sue me.)

[fleurione]

how about a calendar that has two initials marked on it for each day. seems the simplest option ;)

[fleurione]

ahh i have rethought about this and realised i misunderstood the original question :)

the beans in a jar thing works, but instead of having a jar for each person, have a jar for each pair of people? i can't quite work out how, but it seems it would show more easily.. something.. hm

[(Anonymous)]

If you found me in the heaven, will you marry me?
Will we have wedding?

[khalinche]

To extend the bean/jar system further, you could have a different coloured bean for each person. This would make things more messy and analogue, which may be contrary to the precision you seem to require here, but it might do better for establishing the balance of cooking between any two of the people in the triangle as well as all three.

Say C had pinto beans, J had kidney beans and S had black beans. Every time S cooked for J and C he would put a black bean in each of their jars, but if C was just cooking for J he would put a pinto bean in J's jar and not in S's. That would give you a visual indicator of how many dinners had been cooked for who and by who (whom?): if, say, C had cooked J dinner twenty days out of thirty in the month, J's jar would be embarrassingly full of pinto beans and he would see that he owed C some dinners, without being too worried about S.

The optimum would be a nicely speckled mix of beans in each jar, after which you could soak them up and make delicious bean chili.

[stumonkey]

points system. When you cook, you get a point and a tick on a bit of paper. The person who's turn it is is the person with the least points. In case of tie, paper scissors stone for it.
Alternatively you can adopt a more relaxed attitude and it's not fixed turns, but you can see if you're shirking your duty with the same points system

[stumonkey]

note: the problem I can see with beans in jars is that you have to actually count them, you can't look at a glance (which you can with a piece of paper / whiteboard)