OpenLLM-France/wikimedia huggingface dataset:
# The Legend of Zelda/Underworld

Hyrule's underworld is made up of nine labyrinth dungeons. The entrance to some dungeons, particularly the earlier ones, are right out in the open, while some require a step or two to expose. Each labyrinth is in a particular shape, be it an object, symbol, or even a letter of the alphabet. Unlike the Overworld, the underworld of the second quest differs drastically from that of the first quest.

## General tips

* Most items that you can find in the dungeons can only be obtained when you defeat every monster in a room. The only monsters that do not count are bubbles or traps, which can not be beaten. Therefore, it is in your best interest to clear every room at least once if you are trying to maximize your collection.
* Doors which are not open or locked, but rather shut with a diamond pattern printed on the door, can only be opened either by defeating every monster in the room, or by pushing certain blocks in the room after all of the monsters are defeated.
* Blocks which must be pushed in order to open doors, or reveal staircases, can only be pushed once every monster in the room is defeated. The block to push is usually the first block from the left along the center (fourth) row of a dungeon room.
* If you unsure of how to proceed through a dungeon, be sure to bomb every wall that you encounter which does not contain a door. You will typically find passageways that lead further in to the dungeon. It is also not uncommon to find passages to rooms in the "eyes" or unmarked rooms on the dungeon map.
* In the second quest, it is also possible to walk through certain walls that are simply illusions. To do so, push against the wall until access to the room beyond is granted.
* When fighting against monsters, Link is capable of swinging his sword upwards through a block, and still hit a monster directly above the block on the opposite side. He can not do this in any other direction other than up.
* Another useful tactic when fighting monsters is using the ladder to position yourself safely on water where the monsters are incapable of reaching you. This is particularly effective against Darknuts, but completely ineffective against blue Wizzrobes who can still collide in to you while levitating over the water.

## 1st Quest

* Dungeon 1
* Dungeon 2
* Dungeon 3
* Dungeon 4
* Dungeon 5
* Dungeon 6
* Dungeon 7
* Dungeon 8
* Dungeon 9

## 2nd Quest

* Dungeon 1
* Dungeon 2
* Dungeon 3
* Dungeon 4
* Dungeon 5
* Dungeon 6
* Dungeon 7
* Dungeon 8
* Dungeon 9
# The Legend of Zelda/Underworld/Quest 1/Dungeon 3
## Objectives

* Obtain the Raft.
* Beat Manhandla.
* Obtain a Heart Container.
* Obtain the third Piece of Triforce.

## Dungeon walkthrough

No preparation is necessary before diving into this dungeon.
* The entrance to this dungeon is located at E-4 of section 8. Approach it from the area to the west.
* From the entrance, go left and clear this room and the above to obtain two keys.
* Move up to the next room above. The right wall can be bombed to expose a hidden passage that can serve as a significant shortcut to the end of the dungeon.
* Proceed left to the next room to obtain the Compass.
* From the compass room, enter the left room, and then the room below to locate a stairwell. Down these stairs, you will find the Raft.
* From the raft room, move up two rooms, and two rooms to the right and end up in a room with four doors.
* From here, the pathway above is optional if you wish to obtain information about the location of the White Sword. If you enter the room above, open the north door by pushing the left block. Enter the room to the left of the man to obtain a key.
* From the four door room, head right to next room and obtain the Map. You can bomb the right wall to expose a shortcut to the dungeon boss.
* Otherwise, from the map room, head down to the room below, clear it of the Darknuts, head right and defeat the Keeses and Zols, and open the door above.

## Boss

* The boss is a Manhandla. To defeat it, you must remove all four heads. You can do so by stabbing each one individually, but it's faster to place a well timed bomb that explodes on the center of the body, eliminating all four heads at one time. Since Manhandla moves unpredictably, it takes a little luck to drop the bomb at just the right place.
* Defeat it to earn a Heart Container. Continue on to the room above to obtain the third Piece of Triforce.

## Resupply

* One screen west then three screens north provide a healing Fairy.
* One screen north allows purchase of a Water of Life.
* Three screens north of the medicine woman allows purchasing Monster Bait.
* A Magical Shield can also be obtained two screens east and two screens north. Note that it is also available two screens east, but it is more expensive there.
# The Legend of Zelda/Overworld/Quest 2

Be careful that some of the hidden rooms will charge you for some R in order to fix their doors. Avoid them.

## Quest 2 overworld

* Get the Wooden Sword
* Get to the eastern Fairy Fountain
* Get some Bombs
** Use bombs to get 30 R (on the seashore)
** Push an Armos to get 30 R (in Green Forest)
** Push an Armos to get 10 R (in Green Forest)
* Buy a Blue Candle (for 60 R)
** Use the candle to get 10 R (in Green Forest)
** Use the candle to get 10 R (in Green Forest)
** Use the candle to get 30 R (in Green Forest)
** Use the candle to get 30 R (on Lake Hylia)
** Use the candle to get 30 R (on Lake Hylia)
* Buy a Magic Shield for 90 R (use the candle on Lake Hylia)
** Use the candle to get 10 R (on Lake Hylia)
* Get to the western Fairy Fountain
** Use the candle to get 100 R (in Brown Forest)
** Use the candle to get 10 R (in Brown Forest)
** Push a tombstone to get a Heart Container (in the Graveyard)
** Enter a crypt to get 30 R (in the Graveyard)
* Buy a Blue Ring (on the seashore)
* Get the Power Bracelet (on Death Mountain)
* Get the letter (use Power Bracelet on Death Mountain)
** Use bombs to get 30 R (on Death Mountain)
* Buy a Red Potion for 68 R (use Bombs on Death Mountain)

## Summary

This is all that can be collected before Dungeon 1, and it's clearly less equipment than in Quest 1:
* 4 Heart Containers instead of 3
* Wooden sword
* Magical Shield instead of small shield
* Blue Ring
* Bombs and Blue Candle
* Power Bracelet and Red Potion (difficult)
# The Legend of Zelda/Underworld/Quest 1/Dungeon 2
## Objectives

* Obtain the Magical Boomerang.
* Beat Dodongo.
* Obtain a Heart Container.
* Obtain the second Piece of Triforce.

## Walkthrough
### Prerequisites

None.

### Dungeon walkthrough

* The entrance to this dungeon is easily found at A-1 of section 2. Approach it from the area to the south.
* From the entrance, pick up one key from the room immediately to the right, and pick up another key in the room above and to the left of the entrance.
* From the entrance, go up once and right twice, to find the Compass. The north wall of the compass room can be bombed to reveal a hidden passage.
* Above the compass room, you can obtain the Map. The north wall of the map room can also be bombed.
* Above the map room, defeat the three Goriyas to obtain the Magical Boomerang. Be careful of the energy orbs fired from the statues in the four corners. The north wall can be bombed.
* From the magic boomerang room, the left room and the room above that one both contain keys. The room above the magic boomerang room has bombs. The north wall of that room can also be bombed.
* Three rooms above the boomerang room, a man gives you a hint about how to defeat the boss of the dungeon. The left wall of the man's room can be bombed.
* From the man's room, go left one room, and up one room to find the boss.

## Boss

* The boss is Dodongo. There are two ways to beat him. Either place two bombs in front of Dodongo so that he eats them, or place a bomb so that it blows up in his face before he eats it. If it blows up in his face, he will be stunned, leaving him vulnerable to a slash from the sword or to earn another set of bombs.
* Defeat him to earn a Heart Container. Continue on to the left room to obtain the second Piece of Triforce.

## Resupply

* A replacement shield is one screen south and one screen east, as is Monster Bait. You can buy health here.
# The Legend of Zelda/Secret passage guide

| Entrance | Left | Middle | Right |
| S4 B2 | S4 B5 | S1 B2 | S7 D3 |
| S7 D3 | S4 B2 | S4 B5 | S1 B2 |
| S1 B2 | S7 D3 | S4 B2 | S4 B5 |
| S4 B5 | S1 B2 | S7 D3 | S4 B2 |
# The Legend of Zelda/Underworld/Quest 1/Dungeon 4
## Objectives

* Obtain the Ladder.
* Defeat the 2 headed Gleeok.
* Obtain a Heart Container.
* Obtain the fourth Piece of Triforce.

### Prerequisites

* Raft.

### Advised Preparation

* Use the raft to launch off from the dock at D-4 in section one and collect a Heart Container.

### Dungeon Walkthrough

* To enter this dungeon, you must have obtained the Raft from dungeon three. Once you have it, you can launch the Raft from the dock at B-3 in section six.
* From the entrance, go left and clear this room to obtain a Key.
* After that, from the entrance, head up and open the right door to find the Compass. Remember that unless you have the Magical Sword, the Vires will split up in to two Keese.
* The compass room is one of the first dark rooms you encounter. If you have trouble going through these rooms, you can light them with the Blue Candle.
* From the compass room, return to the left, and head up. Obtain a key in this room, then head left again, and up to locate another Key.
* Head up once more, and you will reach a room whose north door is currently unreachable. Head right and defeat all of the Vires and Keese in the room to open the right door way. If you travel up from this room, you can see the Map, but you can not obtain it from this direction.
* When you enter the room with two individual blocks, defeat all of the Zols and Like-Likes, while taking care to avoid the Bubbles. When the room is cleared (except for the Bubbles) push the left block to the right to reveal a staircase. Head down the stairs to find the Ladder.
* Once you have obtained the Ladder, return to the room all the way on the left with the unreachable door. You may now use the Ladder to cross the gap and access the door. Enter it, and head right to obtain the Map.
* From the map room, you can return to the left, or you can bomb the top wall to reveal a pathway which leads to an unmarked room on the map. In it, you will find 10 Rupees, and every wall in that room can also be bombed.
* If you return to the left and head up, you will face another Manhandla which you must defeat to proceed up. Once you do, you will encounter a man who will reveal information about the waterfall.
* From the man's room, head right, and use the Ladder to cross the islands and collect a Key. Head right again to enter a trap room. Avoiding the traps, make your way to the bottom door.
* Here you must defeat all of the Vires and the resulting Keese that they become so that you can push the left block and continue on to the boss of the dungeon.

## Boss

* The boss is a two headed Gleeok. Each head must be stabbed until it starts to fly freely around the room. Once the head is free, it is still a threat, and can not be defeated. The only way to beat Gleeok is to destroy every one of the attached heads.
* Defeat it to earn a Heart Container. Continue on to the room above to obtain the fourth piece of the Triforce.

## Resupply

* Bombs are one screen to the west.
* A cheap Shield is one screen to the east.
* A woman selling Water of Life is two screens to the south, and one to the west.
# Topological Modules
## Table of Contents

* Constructions
* Banach spaces
* Hahn–Banach theorems
* Inductive limits and LF spaces
* Barrelled spaces
* Topological tensor products
* Differentiation
* Orthogonal projection
# Topological Modules/Barrelled spaces

Proposition (pointwise limit of continuous linear functions from a barrelled LCTVS into a Hausdorff TVS is continuous and linear):
Let $X$ be a barrelled LCTVS over a field $\mathbb {K}$, let $Y$ be a locally closed Hausdorff TVS over the same field, and suppose that a net $T_{\lambda }:X\to Y$ ($\lambda \in \Lambda$) of linear and continuous functions is given. Suppose further that $T:X\to Y$ is a function such that
    $\lim _{\lambda \in \Lambda }T_{\lambda }(x)=T(x)$.
Then $T$ is itself a linear and continuous functional.
Proof: First note that $T$ is linear, since whenever $\alpha \in \mathbb {K}$ and $v,w\in X$, we have
    $T(v+\alpha w)=\lim _{\lambda \in \Lambda }(T_{\lambda }(v)+\alpha T_{n}(w))=T(v)+\alpha T(w)$
since $Y$ is a Hausdorff space, where limits are well-defined, and by continuity of addition. Then note that $T$ is continuous, since for all $v\in X$ the set $\{T_{\lambda }|\lambda \in \Lambda \}$ is bounded, so that the Banach—Steinhaus theorem applies and the family $\{T_{\lambda }|\lambda \in \Lambda \}$ is uniformly bounded. Hence, suppose that $V\subseteq Y$ is a closed neighbourhood of the origin. By uniform boundedness, select $U\subseteq X$ to be an open neighbourhood of the origin so that
    $\forall \lambda \in \Lambda :T_{n}(U)\subseteq V$.
We conclude that $T(U)\subseteq V$, since closed sets contain their net limits. We conclude since $Y$ is locally closed, so that $V$ represents a generic neighbourhood. $\Box$
# Topological Modules/Hahn–Banach theorems

Theorem (geometrical Hahn–Banach theorem):
Let $V$ be a real topological vector space, and let $U\subseteq V$ be open and convex so that $0\notin V$. Then there exists a hyperplane $W\leq V$ not intersecting $U$.
Proof: The set of all vector subspaces of $V$ that do not intersect is inductive and also nonempty (because of the zero subspace). Hence, by Zorn's lemma, pick a maximal vector subspace $W\leq V$ that does not intersect $U$. Claim that $W$ is a hyperplane. If not, $V/W$ has dimension $\geq 2$. Now the canonical map $p:V\to V/W$ is open, so that $U':=p(U)$ is an open, convex subset of $V/W$. We consider the cone
    $C:=\bigcup _{\lambda >0}\lambda U'$
and note that it has a nonzero boundary point; for otherwise $C$ would be clopen in $V/W\setminus \{0\}$ which is path-connected (indeed by assumption $\operatorname {dim} V/W\geq 2$, so that for any two points $x,y\in V/W$ we find a 2-dimensional plane containing both, and by using a "corner point" $z$ when $x,y$ do lie on a line through the origin, we may connect them in $V/W\setminus \{0\}$, because a segment in a TVS yields a continuous path by continuity of addition and scalar multiplication), so that $C=V/W\setminus \{0\}$, which is impossible because for any $x\neq 0$ in $V/W$, we then have $x\in \mu U'$, $-x\in \lambda U'$ for $\mu ,\lambda >0$, so that $0\in U'$ by convexity, a contradiction. Hence, let $w\in \partial C$. Then the line $L$ generated by $w$ does not intersect $C$ and hence not $U'$, and $p^{-1}(L)$ is a larger subspace of $V$ that does not intersect $U$ than $W$ in contradiction to the maximality of the latter. $\Box$
# Topological Modules/Orthogonal projection

Theorem (Von Neumann ergodic theorem):
Let $H$ be Hilbert space, and let $U:H\to H$ be a unitary operator. Further, let the orthogonal projection onto the space $W:=\{x\in H|Ux=x\}$ be given by $P_{W}:H\to H$. Then
    $\lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}U^{k}=P_{W}$,
where the limit is taken with respect to the operator norm on $B(H)$, the space of bounded operators on $H$. Moreover, the inequality
    $\left\|{\frac {1}{n}}\sum _{k=1}^{n}U^{k}-P_{W}\right\|\leq {\frac {2}{n}}$
is a valid estimate for the convergence rate.
Proof: Suppose first that $x\in H$ and $z\in W$. Then
    ${\begin{aligned}\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-P_{W}(x),z\rangle &=\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-P_{W}(x),z\rangle +{\frac {n}{n}}\langle P_{W}(x)-x,z\rangle \\&=\langle {\frac {1}{n}}\sum _{k=1}^{n}U^{k}x-x,z\rangle \\&={\frac {1}{n}}\sum _{k=1}^{n}\left(\langle U^{k}x,z\rangle -\langle x,z\rangle \right)\\&{\overset {z\in W}{=}}{\frac {1}{n}}\sum _{k=1}^{n}\left(\overbrace {\langle U^{k}x,U^{k}z\rangle } ^{=\langle x,z\rangle }-\langle x,z\rangle \right)=0.\end{aligned}}$
Further, if we set
    $y_{n}:={\frac {1}{n}}\sum _{k=1}^{n}U^{k}x$,
we obtain
    ${\begin{aligned}\|Uy_{n}-y_{n}\|^{2}&=\langle y_{n}-Uy_{n},y_{n}-Uy_{n}\rangle \\&={\frac {1}{n^{2}}}\langle Ux-U^{n+1}x,Ux-U^{n+1}x\rangle \\&={\frac {1}{n^{2}}}\left(\langle Ux,Ux\rangle -\langle U^{n+1}x,Ux\rangle -\langle Ux,U^{n+1}x\rangle +\langle U^{n+1}x,U^{n+1}x\rangle \right)\\&{\overset {\text{Cauchy‒Schwarz}}{\leq }}{\frac {4\|x\|^{2}}{n^{2}}}.\end{aligned}}$
If now the sequence $(y_{n})_{n\in \mathbb {N} }$ is convergent, we see that its limit is indeed contained within $W$. From the respective former consideration, we may hence infer that the sequence $(y_{n})_{n\in \mathbb {N} }$ does in fact converge to $P_{W}(x)$. We are thus reduced to proving the convergence of the sequence in operator norm. Since $H$ is Hilbert space, proving that $\left({\frac {1}{n}}\sum _{k=1}^{n}U^{k}\right)_{n\in \mathbb {N} }$ is a Cauchy sequence will be sufficient. But since
    ${\begin{aligned}\|y_{n}-y_{mn}\|&=\left\|y_{n}-\sum _{k=1}^{m}{\frac {1}{m}}U^{nk}y_{n}\right\|\\&\leq \sum _{k=1}^{m}{\frac {1}{m}}\|y_{n}-U^{nk}y_{n}\|\\&\leq {\frac {2\|x\|}{n}}\end{aligned}}$
for $m\geq 1$ this is the case; the gaps are closed using that
    ${\begin{aligned}\|y_{n}-y_{m}\|&=\left\|{\frac {m-n}{mn}}\sum _{k=1}^{n}U^{k}x-{\frac {1}{m}}\sum _{k=n+1}^{m}U^{k}x\right\|\\&\leq {\frac {m-n}{mn}}\|x\|+{\frac {m-n}{m}}\|x\|.\end{aligned}}$
Taking $m\to \infty$ in the next to last computation yields the desired rate of convergence. These computations also reveal the underlying cause of convergence: The sequence becomes more and more uniform, since applying $U^{m}$ to it does not change it by a large amount. $\Box$
# Topics in Abstract Algebra/Non-commutative rings

A ring is not necessarily commutative but is assumed to have the multiplicative identity.
Proposition.
$R$
$R\to R$
Theorem (Levitzky).
$R$
# Topological Modules/Constructions

Definition (quotient topological module):
Let $M$ be a topological module over the topological ring $R$, and let $N\leq M$ a submodule with the subspace topology. Then the module $M/N$ together with the quotient topology, ie. the final topology induced by the quotient map $q:M\to M/N$, is called the quotient module of $M$.
Proposition (quotient map of topological quotient is open):
Let $M$ be a topological module and $N\leq M$ a submodule. Then the map $q:M\to M/N$ is open.
Proof: Let $U\subseteq M$ be any open set. We have
    $p^{-1}(p(U))=\bigcup _{n\in \mathbb {N} }U+n$
which is open as the union of open sets. $\Box$
Proposition (quotient topological module is topological module):
Let $M$ be a topological module and $N\leq M$ a submodule. Then the quotient module $M/N$ is a topological module with the subspace topology.
Proof: $\Box$
# Topological Modules/Banach spaces

Definition (Banach space):
A Banach space is a complete normed space.
TODO: Links
Proposition (series criterion for Banach spaces):
Let $X$ be a normed space with norm $\|\cdot \|$. Then $X$ is a Banach space if and only if
    $\sum _{n=1}^{\infty }\|x_{n}\|<\infty$ implies that $\lim _{N\to \infty }\sum _{n=1}^{N}x_{n}$ exists in $X$,
whenever $(x_{n})_{n\in \mathbb {N} }$ is a sequence in $X$.
Proof: Suppose first that $X$ is a Banach space. Then suppose that $\sum _{n=1}^{\infty }\|x_{n}\|$ converges, where $(x_{n})_{n\in \mathbb {N} }$ is a sequence in $X$. Then set $S_{N}:=\sum _{n=1}^{N}x_{n}$; we claim that $(S_{N})_{N\in \mathbb {N} }$ is a Cauchy sequence. Indeed, for $M>0$ sufficiently large, we have
    $N\geq M\Rightarrow \|S_{M}-S_{N}\|=\left\|\sum _{n=M+1}^{N}x_{n}\right\|\leq \sum _{n=M+1}^{N}\|x_{n}\|\leq \sum _{n=M+1}^{\infty }\|x_{n}\|<\epsilon$.
Hence, $(S_{N})_{N\in \mathbb {N} }$ also converges, because $X$ is a Banach space.
Now suppose that for all sequences $(x_{n})_{n\in \mathbb {N} }$ the implication
    $\sum _{n=1}^{\infty }\|x_{n}\|<\infty \Rightarrow \lim _{N\to \infty }\sum _{n=1}^{N}x_{n}$
holds. Let then $(y_{n})_{n\in \mathbb {N} }$ be a Cauchy sequence in $X$. By the Cauchy property, choose, for all $k\in \mathbb {N}$, a number $N_{k}\in \mathbb {N}$ such that $\|y_{m}-y_{n}\|<1/2^{k}$ whenever $m,n\geq N_{k}$. We may assume that $N_{1}\leq N_{2}\leq \cdots \leq N_{k}\leq \cdots$, ie. $(N_{k})_{k\in \mathbb {N} }$ is an ascending sequence of natural numbers. Then define $x_{1}:=y_{N_{1}}$ and for $k\geq 2$ set $x_{k}:=y_{N_{k}}-y_{N_{k-1}}$. Then
    $\sum _{j=1}^{k}x_{j}=y_{N_{k}}$.
Moreover,
    $\sum _{j=1}^{k}\|x_{j}\|\leq \|y_{N_{1}}\|+\sum _{j=2}^{k}2^{-(k-1)}$,
so that
    $\sum _{j=1}^{\infty }\|x_{j}\|$
converges as a monotonely increasing, bounded sequence. By the assumption, the sequence $(S_{k})_{k\in \mathbb {N} }$ converges, where
    $S_{k}=\sum _{j=1}^{k}x_{j}=y_{N_{k}}$.
Thus, $(y_{n})_{n\in \mathbb {N} }$ is a Cauchy sequence that has a convergent subsequence and is hence convergent. $\Box$
# AP Biology/Control of Gene Expression

Gene expression is regulated by the pathways of transcription, translation, and eventual modification of proteins that are formed.
Gene expression starts at the DNA level with the transcription of the DNA to form mRNA by creating the complement of the DNA using RNA polymerase. In order to initiate transcription, one needs transcription factors which are specific to regions of the DNA. Transcription factors can increase the binding of RNA polymerase. Transcription factors can be formed from the cell or can be made of complexes, such as a hormone and hormone receptor which will bind to a region in the DNA called the hormone response element. Once RNA polymerase binds, then the next level of regulation is translation.
In prokaryotes, translation can be affected by the feedback of transcription on translation such as in the trp operon.
In eukaryotes, translation is influenced by the post-transcriptional modifications and the presence of the necessary factors for translation.
# AP Biology/Energy and Metabolism

Contents | Preface | Diagrams | Resources | Contributors | Edit Navigation Bar

## Key Phrases

* energetics/bioenergetics
* heterotroph - autotroph (stanley miller and harold urey)
* enzymes (organic catalysts)
* Reactions (exer-/ender- gonic)
* activation energy
* ATP
* Aerobic/anaerobic
* Cellular respiration
* Fermentation
* Glycolysis
* Krebs Cycle (Citric Acid Cycle)
* Oxidative Phosphorylation
from the Cell Biology wikibook
wikipedia: cellular respiration
# AP Biology/Contributors

Contents | Preface | Diagrams | Resources | Contributors | Edit Navigation Bar
Did you fix a spelling error? Write an article? Create a pretty image? If you answered yes to any of these, or made any other tiny contribution to the AP Biology wikibook, please sign your name here:
    cnelson - Haha... I'm the first contributor. I started this out of a desire to learn biology, but not wanting to go out and buy a review book, as I felt it was inadequate for the price. I have since taken the AP Biology test, and must say I was well prepared. Waiting for my score now.
    rroepke - I found this page while looking for study guides to AP Bio. I have written some of the things in here. I am currently taking AP Bio, and taking the exam in May.
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    Werdan7
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    Blurpeace