Bar Crawl Planning + Math = Happy Drunks!
January 4, 2015 2:49 PM   Subscribe

I am currently planning a bar crawl and could use some help. There are 6 groups and 6 bars. Each group will start at a different bar (Stop 1). Each group will then meet up with a different group at Stop 2, Stop 3, etc., totaling 6 stops. Each group will need to meet each other group one time. Ideally, no group would repeat a bar, ie. each group will meet each other group and hit all 6 designated bars. I've been trying to manually chart this and I'm pulling my hair out. Is there a way to figure this out with math or some other voodoo that I'm not aware of?
posted by tryniti to Grab Bag (14 answers total) 4 users marked this as a favorite
Well, your optimal case (each group goes to each bar once and meets each other group once) can't be done (assuming you don't allow some sort of bye round, where a group goes to a bar but doesn't meet another group).

To meet each other group, a group would need to do five different stops (six minus one for the fact that the group doesn't need to meet itself).

To go to each bar once, a group would need to make six stops, which means they either have to meet some other group twice, or have a stop in which they don't meet any group.
posted by firechicago at 2:56 PM on January 4, 2015

The sixth stop could be everyone at the same bar?

Realistically though ui can see this falling apart after the second bar or so... Could be more stress than fun!
posted by KateViolet at 2:58 PM on January 4, 2015

In reply to firechicago, as I understand it, that first Stop would involve each group at a different bar (the 'bye round'), and only meet up with the other groups at the other five stops. It's theoretically possible.
posted by Paragon at 2:59 PM on January 4, 2015

Ah, sorry, it's not possible as stated. If no-one goes to a bar more than once, each bar can only be visited by three different pairs (e.g. AB, CD, and EF). But because each group starts in a bar on their own, there will always be one other group that has no-one to visit that bar with.
posted by Paragon at 3:19 PM on January 4, 2015 [2 favorites]

It doesn't need to work. Set a number of switch times. At the time, it's each group's job to meet up with a group they haven't met yet (of their choice) at a place they haven't been yet. Negotiate in real time on Twitter or something.

It's going to fall apart after a round or two anyway, so might as well give them a way to just call all-in at one place without forgetting anyone.
posted by ctmf at 3:26 PM on January 4, 2015

By fall apart, I mean some group is going to meet an external group of hot guys or girls and want to stay a while longer. This will screw up any pre-scripted rotation scheme.
posted by ctmf at 3:31 PM on January 4, 2015 [2 favorites]

Response by poster: To address a few things:

Paragon is correct, the first bar is the "bye", so to speak. Each group will meet each other group once after, for a total of 6 bars. I actually already have the schedule of which groups meets which at what time figure out, it's the locations that are tripping me up.

And for those saying it won't work, this is the fifth one of these I've done, and they work great. There are challenges that each group must complete with a member of each other group, so there's incentive to stick to the plan. A schedule is a must so that we don't end up flooding one bar with 200+ people at one time if 4 groups decide to show up at one time.

The difference this year is that I only have 6 bars to work with, and I'd prefer that people don't have to repeat a bar if possible. Again, the groupings and schedule are already figured out, but if there is an easier way to figure out the actual combinations of groups/bars without just brute-forcing it, I'm all ears.
posted by tryniti at 3:44 PM on January 4, 2015 [1 favorite]

(Sorry to spam with replies)

Something in your initial conditions will have to give to make this work, because the conditions are mutually contradictory. The closest configuration I can find looks something like this:


(Bars A-F; Groups 1-6)

In this configuration everyone begins with another group rather than on their own (because of the issue above). Everyone sees every other group once, and visits all bars except one.

If you want everyone to visit every bar, then add a new leg:
In this final bar they'll reunite with a group they had previously seen.
posted by Paragon at 3:47 PM on January 4, 2015 [3 favorites]

I don't know if it's immediately relevant, but a Sestina (a form of poem) has features that are at least somewhat reminiscent:

6 repeated units (an end-word) that must appear in six different locations without overlap (each end of a line for a sestet), each position only once. The pattern is a folding one, with the last end-word of one stanza the first end-word of the next, and ensues as follows:

Stanza 1:--------Stanza 2:----------Stanza 3:----------Stanza 4:-----------Stanza 5:-----------Stanza 6:

I don't know just how relevant this is, but my tuition tells me it is in some way. Hope it's helpful.
posted by Gowellja at 4:00 PM on January 4, 2015 [1 favorite]

Paragon, there's a mistake in your 5-bar solution. Group 2 visits bar A twice and misses two bars, while group 5 visits bar E twice and misses two bars.

I wasn't able to find a solution for five bars, but I found a way for every group to visit four bars without seeing another group more than once. (Of course, each group misses one other group entirely.)
A  B  C  D 
12 56 -- 34
46 13 25 --
35 -- 14 26
-- 24 36 15
Tryniti, I'm curious what the structure was like in past years. Did you accept some overlap, since there were more bars to visit? Or did the numbers somehow work out that everyone could visit every bar and see every group once?
posted by Lirp at 5:48 PM on January 4, 2015 [1 favorite]

Response by poster: Lirp - in past years, we've had more bars than groups, so not every group visited every bar, but there weren't repeats. This is a new city, and we could only nail down 6 bars, hence my dilemma.

I kept plugging things into a spreadsheet and got it worked out so that every group meets every other group once, and each group has to repeat one bar. So, each group will only visit 5 of the 6 bars and will have to visit a bar twice, but it's the best I can come up with.

Each person will be trading items with their drinking challenge partner from other groups, with a goal to collect all the items by the end of the night. So, the "group meets every other group once" part is more critical than repeating bars.
posted by tryniti at 5:59 PM on January 4, 2015

Fun problem!

Here's a way to get the pairs to work out correctly. Label the six groups 1, 2, 3, 4, 5, 6

Stop 2: 12 34 56
Stop 3: 13 25 46
Stop 4: 14 26 35
Stop 5: 15 24 36
Stop 6: 16 23 45

Every group meets every other group exactly once!

As for organizing the bars, that's a bit trickier. I'll come back to this one tomorrow...
posted by math at 9:08 PM on January 4, 2015

If you change this around, think of each bar checking off each team instead:

Each bar checks off one team at the first stop. Then every stop after that involves 3 bars and six teams, so each bar checks off either 0 or two teams in each step. A single bar can't get to an even number of teams (6) when starting with 1 and adding 0 or 2 each time. It worked in the past because every bar would end with an odd number of teams because at least one team had skipped it.
posted by soelo at 8:59 AM on January 5, 2015 [1 favorite]

Nthing this falling apart a little ways in, but it is a cool idea!

Maybe make a last bar/fall back bar that everyone can circle around to at a certain time.
posted by PlutoniumX at 10:48 AM on January 7, 2015

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