
Grouping efficiency  baskets, ingredients, and pies
Here is an interesting problem. I'm not sure how to approach it yet.
You have eight baskets. 1,2,3 etc.
Each basket is filled with different ingredients to different pies. Each pie takes exactly two ingredients.
You want to make seven types of pies. A,B,C etc.
But you can only pick 5 of the baskets to take home, or less.
You're only allowed to take two ingredients (e.g., A1, B2, or A) from any one basket for a pie.
Here are the ingredient contents for each basket. A1 or A2 would be one or the other half of the ingredients to make pie A. "A" means both the ingredients to make pie A  a combination of the both the two ingredients A1 and A2.
1  A2, B2
2  C,D,E,F
3  A2, B2
4  A1, B, C2
5  A2, B1, C, D1, E, F, G
6  A1, B, C2
7  A1, C1, F1
8  A, B, C2, E1
Assuming I'm some kind of efficiency nut ;), how do I figure out which five baskets to take home, to make *all seven pies*.
i) Do I need to take five baskets or can I get away with less?
ii) Assume there is an arbitrary advantage to taking baskets that have both ingredients to any one pie, does this change your answer?

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
Adam12 You have eight baskets. 1,2,3 etc.
Each basket is filled with different ingredients to different pies. Each pie takes exactly two ingredients.
You want to make seven types of pies. A,B,C etc.
You want to solve:
$\displaystyle \binom{n}{2}=\frac{n(n1)}{2}\ge 7$ for the least n.

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
adam12 1  a2, b2
2  c,d,e,f
3  a2, b2
4  a1, b, c2
5  a2, b1, c, d1, e, f, g
6  a1, b, c2
7  a1, c1, f1
8  a, b, c2, e1
I can see 3 baskets only...or did I miss something?
Code:
2 : c d e f
5 : a2 b1 c d1 e f g
6 : a1 b c2

Re: Grouping efficiency  baskets, ingredients, and pies
I forgot an important detail I'm afraid: You're only allowed to take two ingredients (e.g., A1, B2, or A) from any one basket for a pie.
Sorry! I'll fix the original

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
Plato You want to solve:
$\displaystyle \binom{n}{2}=\frac{n(n1)}{2}\ge 7$ for the least n.
What is this sort of problem and the solution approach called? (e.g., if I wanted to do more reading on them)

Re: Grouping efficiency  baskets, ingredients, and pies
VERY confusing:
Quote:
Originally Posted by
Adam12 > You want to make seven types of pies. A,B,C etc.
Why do you say 8 pies later?
> But you can only pick 5 of the baskets to take home, or less.
Do you mean "pick FROM 5 of the baskets"?
> You're only allowed to take two ingredients (e.g., A1, B2, or A)
> from any one basket for a pie.
Clarify that: is A alone because it represents 2 ingredients?
> 1  A2, B2
> 3  A2, B2
Why the duplicate baskets? Why not have 7 baskets (remove one of those)?
> Assuming I'm some kind of efficiency nut ;), how do I figure out
> which five baskets to take home, to make *all eight pies*.
You said 7 pies earlier! You can't take a basket home, right?
Do you mean "which 5 baskets to pick from"?

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
Adam12 What is this sort of problem and the solution approach called? (e.g., if I wanted to do more reading on them)
Well the answer I gave is to an entirely different problem.
I agree that the way you have put things makes no sense.
Here is a usual type problem in this category.
Suppose there are ten bins each containing a different ingredient.
In order to make seven different pies each containing exactly two different ingredients, what is the least number of bins we can choose?
This is what I thought you meant.
But now, I have no idea what you mean to ask.

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
Plato Well the answer I gave is to an entirely different problem.
I agree that the way you have put things makes no sense.
Here is a usual type problem in this category.
Suppose there are ten bins each containing a different ingredient.
In order to make seven different pies each containing exactly two different ingredients, what is the least number of bins we can choose?
This is what I thought you meant.
But now, I have no idea what you mean to ask.
I'm doing my best.
Hmm, your interpretation is close, except each bin contains specific ingredients (plural) which can be used for different pies.
It's not supposed to be realistic (although I can think of a real situation for it) though.
Quote:
Here is a usual type problem in this category.
What is the category called?

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
Wilmer VERY confusing:
It's seven, that was an error. (Fixed) AG
As for duplicate baskets  it's simply the way it is presented. If it makes no mathematical difference whether there are two baskets or one, I will go ahead and make it 7 baskets. I'm very inexperienced with this so I wanted to present the information as I have it.
Thanks for your patience.

Re: Grouping efficiency  baskets, ingredients, and pies
Now I notice 2 more duplicates:
4  A1, B, C2
6  A1, B, C2
Like, if A1 + A1 does not make up A (A1 + A2 being required)
then these duplicates make no sense; so there's only 6 baskets.
Like Plato, I give up...

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
Wilmer Now I notice 2 more duplicates:
4  A1, B, C2
6  A1, B, C2
Like, if A1 + A1 does not make up A (A1 + A2 being required)
then these duplicates make no sense; so there's only 6 baskets.
Like Plato, I give up...
In that case, I'll retract my compliment about patience  lol. And he didn't give up. Check the 2nd post of the thread. I believe it may be solved.
Make it 6 baskets if you like. It's a simple matter or organizing the question so that it makes sense to you.
The data is not negotiable. How you interpret it is a matter of mathematical reason, and it is up to you to do that if you want to solve the question.

Re: Grouping efficiency  baskets, ingredients, and pies
Well, not to leave you emptyhanded (no pun intended!),
we have 6 baskets; and you [PICK] comme suit (maximum of 2 per basket):
1: A2, B2 [NONE]
2: A1, B, C2 [C2]
3: A1, C1, F1 [C1]
4: C, D, E, F [D, E]
5: A, B, C2, E1 [A, B]
6: A2, B1, C, D1, E, F, G [F, G]
So now we can cook 7 pies, from an Apple pie to a Gooseberry pie (Rock)

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
Wilmer Well, not to leave you emptyhanded (no pun intended!),
we have 6 baskets; and you [PICK] comme suit (maximum of 2 per basket):
1: A2, B2 [NONE]
2: A1, B, C2 [C2]
3: A1, C1, F1 [C1]
4: C, D, E, F [D, E]
5: A, B, C2, E1 [A, B]
6: A2, B1, C, D1, E, F, G [F, G]
So now we can cook 7 pies, from an Apple pie to a Gooseberry pie (Rock)
I lol'd :)

Re: Grouping efficiency  baskets, ingredients, and pies
Quote:
Originally Posted by
Adam12 What is this sort of problem and the solution approach called? (e.g., if I wanted to do more reading on them)
asking again