Posts for MahaTmA

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Some searching leads me to the same conclusion, although I did guess it by the part about the placenta fusing with internal organs.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Randil wrote:
It would be cool to see how high we can go with this :)
11= 11 + 0 12 = 11+1 13=11+2 14=13+1 15=13+2 16=11+5 17=17+0 18=17+1 19=17+2 20=13+7=17+3 21=19+2 22=19+3 23=23+0 24=23+1 25=23+2 This is, of course, only accepting that 0 is a prime. this is even though D(0)=NU{0} and not just {0,1} which it should be
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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I have one of the earliest implentations of the techonology, the Microsoft Sidewinder [something] gamepad, and I must say, that if the dead-zones aren't completely out in the weather (E.g. you must hold it on it's side to gain full movement in on direction), the concept is actually quite effective. Direction of blows etc. could be controlled by the thumbstick or tiltfunction, and we would have the first beat-em-up where facing your opponent (Is always either too difficult or too easy due to collision detection of various qualitys) is no longer a prerequisite to winning.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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http://en.wikipedia.org/wiki/Knapsack_problem
The knapsack problem is often solved using dynamic programming, though no polynomial-time algorithm is known for the general problem.
E.g. no general formulae has yet been found
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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That's because the problem has, up until now anyhow, been unsolveable. Mathematician's are getting close, but if they can't come up with some clever trick such as the one for Fermat's last Theorem, they will never solve it. Exhaustive methods just aint doing the trick here. For the record, Fermat's last theorem: Prove, that X^n + Y^n = Z^n has no solutions for n>2
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Post the answer to that in binary code, and I'll set you up for a Nobel prize.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Math gets your brain a-swinging, and I like that.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Feel the same way. ]a;b[ is the interval which is between a and b, but is never actually a or b, like you said :)
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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f(x) doesn't have to be a linear function. f'(c) is a number which corresponds to the inclination constant of the tangent in (c,f(c)). If the definitions for f are true, you can pick a c so that those two are equal. A rough translation of what it's called in Danish (Since I'm not really too stiff in academical English) would be the "mean value sentence." It's a proven mathematical sentence, so that step is valid. To clarify, think of a "curved" function. Pick two points on that function's graph and put a line between them. Now raise/lower the line until it coincides with a tangent for f. There's your proof in informal form.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Fascinating... My guess would be it takes into account how "deep" a page is, and lets this be expressed by the length of the stalk. It then colours the stalks by which colour is dominant on the page, letting the size of 1 stalk denote how many connections can be made from that site (e.g. links)
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Here's a funny wee thing, try to find the fault: Let f(x) be continuos in [a;b] and differentiable in ]a;b[ Let us assume that f(x) isn't constant in the interval [a;b], this means that there exist numbers x1 and x2 within [a;b] , so that: x1 < x2 => f(x1) < f(x2) or x1 < x2 => f(x1) > f(x2) Assume the first is true, and: x1 < x2 => f(x1) < f(x2) (1) We only look at this, since the proof for the other sentence follows the same form. It is then known, that there exists a number c in [x1,x2] so that: f'(c) = (f(x2)-f(x1))/(x2-x1) As a cause of (1), the numerator and denominator are both positive, so the fraction is also positive. Ergo: f'(c) > 0 (2) Define the function g as: g(x)=(f(x)-f(c))^(1/3) , a=<x=<b Ergo: g(c) = 0 (3) From the definition of g comes: g³(x) = f(x)-f(c) <=> f(x) = g³(x)+f(c) We differentiate the last expression, and get (since f(c) is a constant): f'(x)= 3*g²(x)*g'(x) + 0 = 3*g²(x)*g'(x) (4) since g(c)=0 according to (3), (4) gives that f'(c)=0, which is inconsistent with (2). This means that our assumption (1) is wrong, and thus: If f(x) is continuos in [a;b] and differentiable in ]a;b[, f(x) is constant in [a;b] Sure does make the life of a lot of people easier.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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I do believe it's widely accepted that it isn't. For instance: Natural numbers can be divisors. 0 can't, since otherwise it would be a divisor to every other natural numer, much like 1 is. Also, it would make 1 the only prime in the world (A prime being defined as a natural number with only 2 divisors)
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Gorash got it, and yes, the N vs Z does make a difference, since for N, there is no solution as 0 is not included in N.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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MahaTmA wrote:
Here's an interesting problem to keep them gears cranking: 14 natural numbers are written down onto a piece of paper. Whichever number is crossed, the remaining numbers can be grouped into 3 groups with the same sum. Is there a conclusive answer as to which numbers had been written down, and if so, which are they?
Gah, just reviewed this. The numbers needn't be natural, they just need to be whole (e.g. from Z, not N. - 0 and negatives are included)
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Here's an interesting problem to keep them gears cranking: 14 natural numbers are written down onto a piece of paper. Whichever number is crossed, the remaining numbers can be grouped into 3 groups with the same sum. Is there a conclusive answer as to which numbers had been written down, and if so, which are they?
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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http://en.wikipedia.org/wiki/Halting_problem What this means, quite basically, is; "You can never know if a computation will finish, and, if so, what the result is, before it actually finishes." Since the scripts we listed never will finish (one script will act on the other's input and vice versa), there is never a sitation in which Z>Y or Z<Y, because every time we ask the question "is Z>Y?", another Y (or Z) will have come out which is large than the current Y (or Z), rendering any inquiry into the numbers useless. Thus, it is impossible to determine who wins and who loses. The outcome of such an inquiry, however, will be "you both won AND lost."
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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I shall make sure you never win! If(Z>Y){ Y = Z; Y++;} Now our comps will go into a moebius loop, in which we will never know the outcome, thus rendering any competition null and void. Since you can only know the outcome of an operation when (if) it ends, noone wins and we both win. at the same time. My conclusion is this: Pogo sticks make legs funny.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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todd wrote:
The recent discussion reminds me of another interesting problem I saw a few years ago. Write the largest natural number you can. Any notation is acceptable as long as: 1) It is easily understood (this is subjective and depends on your readership, but I think posters in this thread will be reasonable) 2) The representation of your number (and any function definitions, explanations, etc.) must not exceed a certain number of characters. I don't remember the character limit on the original problem, but let's make it 500 characters.
f(x)=(x^x)! n=(f◦f◦f◦f◦f◦f◦f......f◦f◦f)(9) Type out the finite series (f◦f◦f..f◦f◦f) as many times as can be done. Now, the quest for optimization is at hand. EDIT: didn't read previous posts, so an optimization would be: f(x)=(x↑↑↑↑↑x)!! This competition could go on forever :)
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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dood did post that he got it to work, see the first page, second post from the bottom. I am concerned however, that he didn't post a replay...
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Don't let this die, please! This must be the holy grail for ABAHB speedruns. Could someone do the new-found glitch, then post a demo of it in action at least?
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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That's more like it! I belive I am the one who requested a Zangief movie all that time ago, and my dreams came true with this one. Plenty of variety and lots of things I don't remember ever pulling off IRL. A definitive "yes" vote, if only for Zangief and the ex-pres dancing in the end. Priceless.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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88 posts and a usernumber <5? :) Anyhow, good timeattack on this classic, albeit a little too short imo.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Torus Trooper Level 1 to 25 in one stretch, no deaths on "extreme" mode. So far, I've managed to get to level 4 in one stretch with deaths.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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Truncated wrote:
>After having met the salesman the second time, and completing "world one", you die once, causing you to backtrack. What is the reason for this? That death (first after salesman) doesn't cause him to backtrack. He starts in the same position as he died. There is a death later around 3:30 which causes trackback, I'm guessing it's just for invulnerability and new health reasons (to not die later in an even worse place).
I think the death I meant was that one, but it seems to be intentional, although it is not clear as to how. Anyhow, my vote is Yep.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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After having met the salesman the second time, and completing "world one", you die once, causing you to backtrack. What is the reason for this? Also, is the side from which the merchant enters fixed, or is it dependant on the direction you are facing? If it's the latter, I see a second or two of improvement by manipulating from where he enters. Very entertaining run and a great addition from a newcomer :) Hope to see more of your material :)
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
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