Dunno about that, but he sure was a collosal dick.
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I’ll allow it but only because of The Tango 
Shhh…leave the weird Euros to their own devices. Come visit Toronto
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That is some really extreme mid 2000s comedy poster design right there.
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Huh, and here I was singing “Scotty doesn’t know” to myself just earlier this week…
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@TheChicken there are Chickens in Terranigma
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Atleast you didn’t use the hard r but still, cmon man
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Terra-ligma balls
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First step in hiding the past.
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What is the difference between a double cheese burger and a mcdouble? They both look like little versions of the double quarter pounder w/cheese.
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I asked the lady at the drive through and she told me “the McDouble is cheaper, but no one orders it .”
That was all she said.
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The difference is a slice of cheese. That’s literally it.
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God fucking bless America and our brilliance in coming up with new and unique food items.
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Wow! Thanks. I feel silly now bc I did look closely and didn’t notice 
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I’m literally 6 alcoholic drinks in and it didn’t take me that long to figure it out lmao.
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Once again here to shamelessly seek help for a question…
Mammad Javad’s Birthday Decorations
Let’s call him MJ for short. MJ’s birthdays are always disastrous, and there is no exception for this year as well. The date is near and Shaazzz’s cursed pals shall hold the best birthday party possible for MJ. The first thing to do is to light up the party room. Giving MJ’s interest in circles, Shaazzz pals decided to put n bulbs on a circle. But as Mehrdad was in charge of doing it again, not all bulbs were properly circuited. Some were emitting and some were not emitting any light; But MJ hates disorder and won’t stand some of his birthday lights to be off: He’ll kill Mehrdad in revenge.Each light has a switch, and pressing it will change the state of its respective light and the two lights adjacent to it. You should prove that there is a method to turn every light on by differing the state of the switches, and save Mehrdad’s life… :DDD
Note that there is no limit on the order and the count of switching operations.
I tried induction but it doesn’t seem to work with this one.
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Putting first thoughts into it, it should not be possible with at least n = 2. If
Some were emitting and some were not emitting any light
then you can only alternate between which of the two bulbs are emitting light with the given rules.
1: on, 2: off => 1: off, 2: on
…and reversed, no matter which switch is flipped.
However, if the rule
change the state of its respective light and the two lights adjacent to it
is interpreted as the other bulb counts twice (it is “left” and "right of the bulb you switch), it is also xor’d twice. In that sense the second bulb remains at the given state, making it possible for n=2.
However on n=3 it does not work for sure:
1: on, 2: on, 3: off => 1: off, 2: off, 3: on
…and reversed, no matter which switch is flipped.
I only am able to light all bulbs with n=4. And even if all other configurations with 1 or at least 4 bulbs can be all lighted up, it does not work with n=3 and, depending on the rule definition, n=2, so I’d say I’d refuse to prove that’d be possible. 
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Good point, I think the writer forgot to add a limit, n > 3, other than that Shaazzz is a quite a strong team 
(Shaazzz = Iran’s computer gold medalists)
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In that case let me look at my draft if I can conclude a proof for that case. 
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