Bay 12 Games Forum

A minecart cannon would be a useful weapon against Titans.

Plus everything described here:

http://myweb.tiscali.co.uk/matthaywood/main/Warwagons.htm

Megabeasts are incredibly clever and masters of disguise...

... this is not the giant Roc you are looking for... (Waves wing in a strange gesture)...

Throwing without hands, this reminds me one old game...

Guybrush Threepwood: How can you see without eyeballs?
Murray: How can you walk around without a brain? Some things no-one can answer

Forget the Alt- cheat, you cannot really swim in this game, which is good and shows how realistic the game is. The problem, you see, lies in the fact that upon trying to approach the water, you must first cross half of the distance. Once you do that, you still have the other half remaining, and this new half consists of two yet smaller sub-halves, and both of them too, and so on, so on, and with each half overcame new and new appear like heads of a hydra. Therefore, time slows down, DF gums up with all the infinity-approaching number of super-precise iterations... and you will never make it. A guy called Zenon would explain better than I can.

It's a bit like falling into a black hole. You should try it sometimes.

Oh... I read "arsenic booze". Now that would be something...

Deities of birth, new life e.t.c.

It's not that creepy really

(also common)

What is creepy on the other hand, is when they praise pestilence and wish you find death soon, meaning it with well intent and sincerity.

Well, I have heard equal share of nonsense from real worshippers of certain (mostly monotheistic) religions. It does not strike me so odd - in cults, reason and critical thinking must go down the drain.

Just brace for Dwarf Fortress - OpenTTD edition.

0.9999.... (Q further in the text) is equal to 1 - we can prove it by proving that negation leads to contradiction.

Let's assume that Q != 1. That means a real number P must exist that falls between Q and 1, i.e. Q < P < 1 [1].

We know that P > Q. We will try to construct it's decimal representation, because any real number has a decimal representation (though transcendent real numbers have infinite non-periodic ones). The first digit must be 0 followed by decimal point - 1 and greater would mean P > 1 and -1 and smaller would mean P < Q, which would contradict the original assumption [1]. Lets find the second number.  Clearly, it must be 9 - any other value would mean that P<Q (example - 0.7 < 0.999... ). This will be our first induction step [I1].

Then we will show that for n-th (n is a natural number) digit in decimal representation of Q, that if this number is 9, the following number must be 9 also, formally:

( d[n] = 9 ) => ( d[n+1] = 9 ) [I2]

Let's have a number S that equals 0.9999....9, and the number of nines past the decimal point is equal to n+1. Clearly, S < P. If our sought number Q have anything else on n+1 th position than 9, it's clear that immediately Q < S, and for every possible combination of the remaining digits in Q's decimal representation, Q will be always smaller than P, which is once again, contradiction, because for any digit on m-th position, where m>n, m-th digit in the representation of Q can never be higher than P's 9.

Now we use mathematical induction using [I1] (first number after decimal point must be 9) and [I2] (the next number must be always 9 too) to prove that our sought number Q must be expressed only by 0.9999...., and therefore is in fact equal to P.

P = Q, but that contradicts P < Q < 1, that means that we have proven by contradiction that no such real number Q that would fall between P and 1 exists. If no such real Q exists, then P must be equal to 1, i.e. 0.9999.... = 1

QED

Note - I don't know if this is entirely correct, I am out of school for some time.