[68AttendGem]

I'm working on a logic problem. Maybe there are some math whizzes among us.

There are two parties happening on a Friday night. The two parties happen to be going on in opposing sides of a duplex house. Each attendee is given a hand stamp upon entry into each party, which allows the attendee to leave and return freely. Each party has it's own hand-stamp, and party attendees are only stamped once.

We don't know how many people were at these parties, but we know how we can classify them.

Here are some variables to consider (these are categories of attendees):

for PARTY #1:

CRASHER_1 = an attendee who crashed the party -- no hand-stamp present (i.e., they sneaked in)
FLOATER_1 = an attendee of both parties (both hand-stamps present)
LOYALIST_1 = an attendee with only party #1's hand-stamp
DEFECTOR_1 = an attendee with only THE OTHER PARTY'S hand-stamp (how would this happen?)

for PARTY #2:

CRASHER_2 = an attendee who crashed the party -- no hand-stamp present (i.e., they sneaked in)
FLOATER_2 = an attendee of both parties (both hand-stamps present)
LOYALIST_2 = an attendee with only party #2's hand-stamp
DEFECTOR_2 = an attendee with only THE OTHER PARTY'S hand-stamp (how would this happen?)

How do we calculate the net, overall direction of traffic from one party to the other? In other words, which party was more popular? Are any of the variables USELESS?

Good luck!

[LillyPlay]

The party that had naked chicks was more popular. Duhh.

[feannigvogten]

The party without the logic problem creator apparently rocked [thumbup]

However on a more serious note... I don't believe there's enough data to work the variables into any finite answer. IMO

Short of writing you an algebraic equation to express how I cannot find an answer, I don't know what else to say.

[evarekataVame]

Well it's quite simple really. Both parties suck as there are only 4 people attending so 100% of people leave and go to the cracking party across the road which doesn't require stamps to get in.

[Squeernemergo]

They were equally lame.

[Bwvapays]

Party 1 should suicide bomb Party 2

[CealialactBek]

Just looking at face value, but they both appear to have the same traffic as the variables do not differ.

[WhileKelf]

Crasher1 would be defector2
Crasher2 would be defector1
Floater 1 = floater 2
Loyalist 1 attended only party 1
Loyalist 2 attended only party 2

So, a total of five people, three of which visited both barties, but its impossible to say which party was more popular, as there is no timeline.

[viawbambutt]

Crasher1 would be defector2
Crasher2 would be defector1
Floater 1 = floater 2
Loyalist 1 attended only party 1
Loyalist 2 attended only party 2

So, a total of five people, three of which visited both barties, but its impossible to say which party was more popular, as there is no timeline.

I doubt he meant 5 people, it would just end up with 5 types of people, each of which have an unknown value.

Crasher and defectors would NOT be the same thing of course, a defector has a stamp for party A but went to party B (or the other way around) and Crashers have no stamp at all.

Floaters are useless in determining which one was more popular as they were present on both. Assuming you want to see the results of people actually staying at a party and not how much income from the stamps they generated it should be something like this:

Total1 = Crasher1 + Defector1 + Loyalist1
Total2 = Crasher2 + Defector2 + Loyalist2

Again, you could add up the floaters to find out how much of a % lead one party had over the other, but other than that floater1=floater2 so you can omit it if you just want to figure out which is more popular.

As for the direction of traffic, it's only really the defectors you want to look at, all the other variables (Crasher + Loyalist) are static for both parties.

That's if I understood it correctly

[WhileKelf]

A defector has a stamp for one party, but goes to the other party as well, but without the second stamp - only way for that would be if they crashed the second party.
That's why I have each representing two of the descriptions.

Least ways, that's the way I interpreted it. I'm sure Will will come in and correct everyone, though

[unmalryAlalry]

Least ways, that's the way I interpreted it. I'm sure Will will come in and correct everyone, though

Indeed, then DM will come in and post about something Will posted about, and then Lazer will come in and post about DM.

then the odd love triangle starts again [rofl]