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Posted by webyugioh - 16-12-09 18:01 - 4 comments - Read - Edit
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 | BAHAMUT's Crossword Puzzle of Boredom
incase you missed the NFO
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Posted by webyugioh - 29-09-09 22:30 - 47 comments - Read - Edit
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| | Yet another reason we all love Japan
"Cute Room"
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It's your money — spent it the way you like. Newly opened Akihabara business "Cute Room" is offering otaku a bevy of girlfriend-like services that include playing video games, watching DVDs and playing board games with young female (paid) friends.
The girls available are pictured above.
Cute Room does not appear to be offering sexual services. Below is a list of activities that are offered with the girls, along with prices:
QUOTE Play video games (Wii PS3 PSP XBOX DS) (30 minutes 1,000 yen)
Watch a DVD with the girl (30 minutes 1,000 yen)
Play board games or card games (20 minutes 800 yen)
Massage your hand (20 mins 1,200 yen)
Lie on her lap while she cleans your ear (20 mins 1,500 yen, 40 mins 3,000 yen)
Tsundere, dere dere slap (slapped both ways 1,000 yen)
Read a bedtime story (20 mins 1,000 yen)
Take a photo together (1,500 yen)
Handmade candy (2,000 yen)
Get a love letter (1,000 yen)
Get cellphone email from her (photo + message 500 yen for one, 1,200 yen for three)
Looking at you (with email movie 1 time 1000 yen, 2,500 yen for 3 times)
Bromide (300 yen each)
Exchange presents (need to call in advance, 1,500 yen)
One thousand yen is rough US$11.
Patrons select a room to spend with their paid girlfriend. Rooms include a big sister room, a maid's room, a classroom, a Japanese room or a Monster Hunter-style room. After selecting a room, customers pick out a costume for the girl. The price breakdown is ¥4,000 for 40 minutes, ¥5,500 for 60 minutes, ¥7,500 for 80 minutes and ¥2,000 for every 20 minutes after that.
Official Site
Kotaku Article
So, how much money would you spend? |
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Read 1,192 times - last 47 by comment |
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Posted by webyugioh - 20-09-09 01:22 - 4 comments - Read - Edit
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| | Urinal protocol vulnerability
| Ripped from here
Urinal protocol vulnerability
When a guy goes into the bathroom, which urinal does he pick? Most guys are familiar with the International Choice of Urinal Protocol. It’s discussed at length elsewhere, but the basic premise is that the first guy picks an end urinal, and every subsequent guy chooses the urinal which puts him furthest from anyone else peeing. At least one buffer urinal is required between any two guys or Awkwardness ensues.
Let’s take a look at the efficiency of this protocol at slotting everyone into acceptable urinals. For some numbers of urinals, this protocol leads to efficient placement. If there are five urinals, they fill up like this:
The first two guys take the end and the third guy takes the middle one. At this point, the urinals are jammed — no further guys can pee without Awkwardness. But it’s pretty efficient; over 50% of the urinals are used.
On the other hand, if there are seven urinals, they don’t fill up so efficiently:
There should be room for four guys to pee without Awkwardness, but because the third guy followed the protocol and chose the middle urinal, there are no options left for the fourth guy (he presumably pees in a stall or the sink).
For eight urinals, the protocol works better:
So a row of eight urinals has a better packing efficiency than a row of seven, and a row of five is better than either.
This leads us to a question: what is the general formula for the number of guys who will fill in N urinals if they all come in one at a time and follow the urinal protocol? One could write a simple recursive program to solve it, placing one guy at a time, but there’s also a closed-form expression. If f(n) is the number of guys who can use n urinals, f(n) for n>2 is given by:
The protocol is vulnerable to producing inefficient results for some urinal counts. Some numbers of urinals encourage efficient packing, and others encourage sparse packing. If you graph the packing efficiency (f(n)/n), you get this:
This means that some large numbers of urinals will pack efficiently (50%) and some inefficiently (33%). The ‘best’ number of urinals, corresponding to the peaks of the graph, are of the form:
The worst, on the other hand, are given by:
So, if you want people to pack efficiently into your urinals, there should be 3, 5, 9, 17, or 33 of them, and if you want to take advantage of the protocol to maximize awkwardness, there should be 4, 7, 13, or 25 of them.
These calculations suggest a few other hacks. Guys: if you enter a bathroom with an awkward number of vacant urinals in a row, rather than taking one of the end ones, you can take one a third of the way down the line. This will break the awkward row into two optimal rows, turning a worst-case scenario into a best-case one. On the other hand, say you want to create awkwardness. If the bathroom has an unawkward number of urinals, you can pick one a third of the way in, transforming an optimal row into two awkward rows.
And, of course, if you want to make things really awkward, I suggest printing out this article and trying to explain it to the guy peeing next to you.
Discussion question: This is obviously a male-specific issue. Can you think of any female-specific experiences that could benefit from some mathematical analysis, experiences which — being a dude — I might be unfamiliar with? Alignments of periods with sequences of holidays? The patterns to those playground clapping rhymes? Whatever it is that goes on at slumber parties? Post your suggestions in the comments!
Edit: The protocol may not be international, but I’m calling it that anyway for acronym reasons. |
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Posted by webyugioh - 22-04-09 21:58 - 2 comments - Read - Edit
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| | A Date Idea Analyzed
Idea for the XKCD Blog
| Here is yet something else I found entertaining whilst procrastinating:
From the XKCD blog:
QUOTE A Date Idea Analyzed I don’t do conventions very often, but I recently went to ConBust out in Northampton, MA, while visiting some friends. While I was there, I had a guy propose something fascinating to me. I can’t remember the guy’s name, so if he or one of his friends sees this, post your info in the comments. (Edit: it was a dude by name of Thom Howe.) The guy Thom had an idea for a date. He wanted to rent a cherry picker, drive it to her door, and pick her up in it.  Then, he’d drive to the beach, and get there at just the right time to watch the sun set.  Once the sun had set, he’d activate the cherry picker, they’d be lifted up above the beach …  … and they’d watch the sun set again. Clearly, this is an excellent idea, and any girl would be lucky to see Thom at her door. But is it plausible? How fast and how high does the cherry picker have to go? I tried to work out the answer for him there at the table, but there was a line of people and there wasn’t time. But when I got home, I remembered it again, and I’ve worked out the solution. Here’s the situation:  By the time the earth has rotated through angle theta, the cherry picker will have to have climbed to height h. After t seconds, theta in radians is:  The height of the lift above the center of the earth is:  So the height above the surface (sea level) is:  Substituting everything so far we get this expression for the height the lift needs to reach t seconds after sunset to stay even with the sun.  Now, an actual cherry picker has a maximum lift rate (I Googled some random cherry picker specs, and 0.3 m/s is a normal enough top lift rate.) We’ll call that rate v, so the actual height of the lift will be this:  Substituting that in and solving for v, we get this:  (That’s arcsecant, not arcsecond). This equation tells us how fast the lift has to go to get from the ground to height h in time for the sunset1. But we can also get the answer by just trying a few different heights. We plug it in to Google Calculator^2: 2*pi*6 meters/(day*arcsec(6 meters/(radius of earth)+1))and find that h=6 meters gives about the right speed. So, given a standard cherry picker, he’ll get his second sunset when they’re about six meters up, 20 seconds later. You might notice that I’m ignoring the fact that he’s not starting at sea level — he’s a couple meters above it. This is actually pretty significant, since the sunset line accelerates upward, and it brings down his second-sunset height quite a bit. If he got a faster lift, or used an elevator, the correction would become less necessary. Extra credit3 for anyone who wants to derive the expression for the height of the second sunset given the lift speed and height of first sunset. For now, I recommend he dig a hole in the sand and park the lift in it, so their eyes are about at sea level4.  1 Ideally, we’d solve for h, but it’s inside the arcsec and that looks like it’s probably hard. Do one of you wizards with Maple or Mathematica wanna find the result? 2 If you work in one of the physical sciences and don’t use Google Calculator for all your evaluatin’, you’re missing out. I wish there were a command-line version so I could more easily look/scroll through my history. I know Google Calculator is largely a frontend to the unix tool units, but it’s better than units and available everywhere. 3 Redeemable for regular credit, which is not redeemable for anything. 4 I suggest a day when there aren’t many waves.
I found it amusing.
Tell me this is not a great idea. |
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