[bonanova]

Here's a G-rated generalization of Rookie's condom conundrum.
Take a look, if only to understand the problem being posed.

Now to introduce a modicum of professionalism, we'll talk about surgical gloves, instead.
d doctors wish to examine p patients p <= d] in a manner that every male over 50 is aware of.
No doctor wishes to wear a glove contaminated by a patient.
No patient wishes to be exposed to a glove worn by another doctor.
As in the "cc" puzzle, they may be turned inside out and/or worn on top of another glove.
There are three cases, increasing in complexity:
d=p=2. Well, Rookie's puzzle gives this answer.
p=1; d=2n+1 n is an integer]
any p <= d
Edit: If all dp combinations of doctor/patient examinations are performed, then ...
What is the minimum number of surgical gloves required in each case?
examinations and surfaces. Edited April 12, 2009 by bonanova
Added condition shown in red ...

[Izzy]

1. Give a glove to each doctor, let each doctor do one patient. Two gloves used.

2. Well, if there's only one patient, I assume you only need one glove?

Unless you want every doctor to inspect every patient. *shrugs*

[Guest]

Here's a G-rated generalization of Rookie's condom conundrum.
Take a look, if only to understand the problem being posed.

Now to introduce a modicum of professionalism, we'll talk about surgical gloves, instead.
d doctors wish to examine p patients p <= d] in a manner that every male over 50 is aware of.
No doctor wishes to wear a glove contaminated by a patient.
No patient wishes to be exposed to a glove worn by another doctor.
As in the "cc" puzzle, they may be turned inside out and/or worn on top of another glove.
There are three cases, increasing in complexity:
d=p=2. Well, Rookie's puzzle gives this answer.
p=1; d=2n+1 n is an integer]
any p <= d
What is the minimum number of surgical gloves required in each case?examinations and surfaces.

Clarification needed:

No doctor wishes to wear a glove contaminated by a patient.
Does this mean having the same side (inside or out) which is contaminated by a patient touch the hand? In other words - the doctor can still wear a glove when the outer side has been contaminated, correct?

No patient wishes to be exposed to a glove worn by another doctor.
Same question, basically. The patient doesn't care who wore the glove - as long as the side used for his examination is untouched (by doctors and patients). Is this correct?

[bonanova]

Yes - to both of the clarifying questions.

The approach to use as regards your idea is suggested in the spoiler.

Also note the omission and correction to the OP.

All patients are examined by all doctors.

That's an important part of the puzzle.

[bonanova]

1. Give a glove to each doctor, let each doctor do one patient. Two gloves used.

2. Well, if there's only one patient, I assume you only need one glove?

Unless you want every doctor to inspect every patient. *shrugs*

Thanks Izzy, there's a reason why your answer makes sense ... besides your intellect, of course.

I left out a crucial part of the puzzle; namely that all patients are examined by all doctors.

See Edit in OP.

[plasmid]

You don't need any surgical gloves at all. Non-sterile exam gloves will do just fine for that exam.

Total number of gloves used is n+1.

Take one glove and designate it the "bum glove". Take 2n of the doctors and put them in pairs that will each share a glove - the first one wears the pair's glove with the bum glove on top for the exam, the second doc takes the pair's glove and turns it inside out so his hand is facing the unused surface before putting the bum glove over it and examining. After the 2n docs have gone, the +1 doc can do his exam with the (still unused) inner surface of the bum glove.

You could use a similar approach to cover the subset of case 3 where there are an even number of both doctors and patients. Pair them off (pairs of doctors and pairs of patients) that will each share a glove. Each doctor uses his/her surface of the exam glove, and covers it with the appropriate surface of the patient's glove for the exam that's about to be performed. Total number of gloves used would be:

patients/2 + doctors/2

If both the number of docs and the number of patients are odd, then remove one doctor and one patient from the group and handle the remaining even number of doctors and patients as above. The odd ones out also get to split a glove between the two of them: the doctor uses the inside surface of the odd glove out and covers it with the other patients' gloves for all of their exams. The odd patient out uses the other surface for all of his exams with the other doctors wearing their gloves on the inside. The odd pair out can do their exam with just the odd glove out and nothing in between. Total gloves is still:

(patients + doctors) / 2

That still leaves the question open: if there are an odd number of patients and even doctors (or vice versa), can you do it with fewer than (patients + doctors + 1) / 2 gloves? And also, did you have something different in mind, 'cause the OP makes the p <= d stipulation for case 3 that wouldn't seem to make any difference with this set of solutions.

[Guest]

Even if wearing the same glove as another docter is far more appealing than sharing a condom, I'm assuming that doctors do not want to share gloves - or else the puzzle would be too straight forward.

I haven't found any proof of concept yet, but it seems to me that since we are always able to protect at least 1 side of a glove, and a used side can never be used for another thing, the number of d+p = number of glove-sides required = g*2 (round up).

or...

Gloves required = (d+p) / 2 - rounded up

I'm still trying to find proof of this - or to the contrary of course

Edited April 12, 2009 by uhre

[Guest]

Arh well, initial thoughts can be dangerous...

While I still haven't found proof of any concept, at least I could verify that my previous claim was incorrect.

My claim was that for any SGR(d,p) [surgical Glove Requirement given d doctors and p patients ] the following would be true:

SGR(d,p) = (d+p+1) / 2 [integer division = cut decimals / floor]

While this works for all d+p = 4, and d+p = 5, it does not work for all d+p = 6. In particular:

SGR(5,1) = 3

SGR(4,2) = 4

SGR(3,3) = 4

SGR(2,4) = 4

SGR(1,5) = 3

(I know the OP said for any p <= d, but it seems to work just as well for p > d)

Anyway, I'm sure there is some sort of valid explanation, but it eludes me for now.

[Guest]

I need a clarification please:

If a glove is used on a patient then its outer surface is considered to be contaminated. And now if we put on another fresh glove over this contaminated glove then will the inner-surface of the new glove considered to be contaminated?

[Izzy]

Heh, figured you wanted every doctor to do every patient anyway, but was lazy. Was also too nice to suggest the doctors shouldn't use gloves at all.

Anyway..

I'm going to start with p=d, and just call them n. Number of surgical gloves used is after the comma.

n = 0, 0

n = 1, 1

n = 2, 2

n = 3, 5 (?)

n = 4, 8 (?)

n = 5, 13 (?)

Basically, I'm thinking n²/2 = g

But if there are less patients... Umm..

d = doctor, p = patient, g = glove

dp - (d-p)/2 = g

And g always rounds up to the nearest whole number.

[Guest]

I don't unerstand why this won't work.

Take a glove. Take a Doctor and Take a patient. Let the doctor use the glove on the patient. Have him take it off. Call another doctor to put on the same glove. Have him examine the same patient. Again, call another doctor to examine the same patient with the same glove and so--on until the last doctor.

Now all doctor moves to next patient. A new glove is taken out. Its used on the patient by all the doctors. Now all doctors move to next patient. But this time the previously used two gloves are put one on another such that contaminated surface are made the inside, thus leaving a composite glove with both surface clean. All docs use this glove on the third patient. For the forth a new (third) glove is used and for the fith third glove and the composite gloves are combined again. So I have this answer

No of patient: 1 2 3 4 5 6 7 ...

No. of gloves: 1 2 2 3 3 4 4 ....

I think I am wrong somewhere since, My answer is irrespective of no. of doctors.

[Izzy]

Because it's insanely difficult to take of those plastic glove thing without getting them inside out?

[Prof. Templeton]

Since we are talking about magical math gloves and the OP says nothing about wearing gloves or not wearing gloves that touched another doctor's hand, I agree with the-genius' solution above which only depends on the number of patients. This solution wouldn't work for prophylactics, however - ewww! I'm not sure, but I think bn may have intended a solution that considers if a glove comes in contact with a doctors hand that side of the glove is also considered contaminated. Clarification?

[bonanova]

I don't unerstand why this won't work.

Take a glove. Take a Doctor and Take a patient. Let the doctor use the glove on the patient. Have him take it off. Call another doctor to put on the same glove. Have him examine the same patient. Again, call another doctor to examine the same patient with the same glove and so--on until the last doctor.

Now all doctor moves to next patient. A new glove is taken out. Its used on the patient by all the doctors. Now all doctors move to next patient. But this time the previously used two gloves are put one on another such that contaminated surface are made the inside, thus leaving a composite glove with both surface clean. All docs use this glove on the third patient. For the forth a new (third) glove is used and for the fith third glove and the composite gloves are combined again. So I have this answer

No of patient: 1 2 3 4 5 6 7 ...

No. of gloves: 1 2 2 3 3 4 4 ....

I think I am wrong somewhere since, My answer is irrespective of no. of doctors.

No patient wishes to be exposed to a glove worn by another doctor.

[Joe's Student]

No patient wishes to be exposed to a glove worn by another doctor.

Irrespective of whether or not the part that touches the patient hasn't been touched by another doctor? If the inside has been touched by another doctor, I'm taking it that that fulfills the above condition? I wanted to be clear in my head.

What if a doctor wears two gloves. Then the next doctor wears the top glove to examine the same patient? Is that allowed as it wasn't actually touched by the first doctor.

Edited April 13, 2009 by Joe's Student

[Prof. Templeton]

No patient wishes to be exposed to a glove worn by another doctor.

The doctors are wearing 2 gloves on a hand every time they examine a patient, so the examination glove never contacts the doctors hand. Are the doctors wearing the gloves or are the gloves wearing gloves? Silly question, but you get what I mean. Is a glove considered "worn" even if there is another glove between it and the doctor? If it is, then the case 2 solutions of n+1 for 2n+1 doctors aren't valid since both the inner glove and outer glove have been "worn". Right?

[Guest]

Here's how I read it:

If a glove has been touched on either surface directly by a doctor's hand, it cannot be used by another doctor for the same patient.

If any single glove surface has been touched directly by a patient, it cannot be touched by a doctor or another patient.

The only thing that is unclear is - if two surfaces touch where one of them has been touched by a doctor, should we consider the second surface "dirty"?

In any case,

Worst case: d + (p/2) rounding up.

Each doctor gets a glove, and each pair of patients gets a glove. When all doctors have seen a patient, invert the glove and see the next patient, and so on.

Better case: as has been shown already, (d+p)/2 rounding up. Each pair of doctors gets a glove and each pair of patients gets a glove. Doctors can invert their gloves between each member of the pair the same way the patients do.

The question is whether this can be improved (as using 2 gloves for 3 doctors to see 1 patient, etc)

[Guest]

bonanova, you said: No, patient wishes to be exposed to the glove worn by another doctor. Does this mean the patient won't like to be exposed to the glove worn by a doctor even if the outside surface of the glove is fresh? If he won't deny in this case, then I would like to clarify my above slolution.

When a doctor after examinig a patient takes out his glove to hand it to another doctor to examinie the same patient, he takes the glove out without turning it insideout, so that even after taking the glove out, the outside surface remains contaminated (by the patient) and the inside surface remains fresh (touched only by the doctor's hand).