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The links you provide give both the same error
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The problem is that part of the rules that fall:
The creature lands on its belly, unless it avoids suffering damage from falling.
You suggest giving the PC an advantage over the attack and no damage when it should really be down to earth after suffering damage.
The question then is whether the negation of the harm is fair or not. The problem with falling rules is that they do not consider what you are landing. If you fall to about 30 meters and you end up in a deep pond, your damage must be reduced or canceled. For this example, it is about how you land in the water. A clean dive or a pin fall would be the perfect way to land, a belly flop would be the worst.
Landing on an opponent would follow a similar logic. Unless they fall on something hard, like a Golem, the body of their opponents should be usable to break their fall a little.
I suggest that the PC launches an acrobatic test to land properly on its opponent. Have DC determined by the distance fell. If they fail, their opponent and themselves suffer damage and do not receive their attack.
Keep in mind the distance between the attacker and the target. If they are 15 feet high in the air on a branch and their opponent is 6 feet tall, it is only 9 feet.
If they fall too far, you may want to consider a rollover (perhaps athleticism) to accurately jump on their opponent.
I've tried to start with something like below. Note that traversal etc. it's not included, it's the part for storage and modifications. It seems like it could be a lot more stylish in some places, but we do not know exactly how. I'm mainly trying to improve how the code almost repeats itself for accessors and installers for left and right subtrees.
import org.apache.commons.lang3.ArrayUtils;
public class Tree {
private int() treeArray;
private int depth;
public Tree(int treeDepth) {
this.depth = treeDepth;
this.treeArray = new int((int) Math.pow(2, treeDepth));
treeArray(0) = Integer.MIN_VALUE;
}
public int returnRoot() {
return treeArray(1);
}
public int() getTreeArray() {
return treeArray;
}
public int getDepth() {
return depth;
}
public void setTreeArray(int() treeArray) {
this.treeArray = treeArray;
}
public Tree returnLeftSubTree() {
if (depth == 0) return null;
Tree ret = new Tree(depth - 1);
for (int i = 1; i < depth; i ++) {
for (int j = 0; j < depth - 1; j ++) {
ret.getTreeArray()((int) Math.pow(2, i - 1) + j)
= treeArray((int) Math.pow(2, i) + j);
}
}
return ret;
}
public Tree returnRightSubTree() {
if (depth == 0) return null;
Tree ret = new Tree(depth - 1);
for (int i = 1; i < depth; i ++) {
for (int j = 0; j < depth - 1; j ++) {
ret.getTreeArray()((int) Math.pow(2, i - 1) + j)
= treeArray((int) (Math.pow(2, i) + Math.pow(2, i - 1)) + j);
}
}
return ret;
}
public void setLeftSubTree(Tree leftSub) {
if (leftSub.getDepth() >= depth) {
this.treeArray = ArrayUtils.addAll(
this.treeArray,
new int((int) (Math.pow(2, leftSub.getDepth() + 1) - Math.pow(2, depth))));
this.depth = leftSub.getDepth() + 1;
}
for (int i = 1; i < depth; i ++) {
for (int j = 0; j < depth - 1; j ++) {
treeArray((int) Math.pow(2, i) + j) =
leftSub.getTreeArray()((int) Math.pow(2, i - 1) + j);
}
}
}
public void setRightSubTree(Tree rightSub) {
if (rightSub.getDepth() >= depth) {
this.treeArray = ArrayUtils.addAll(
this.treeArray,
new int((int) (Math.pow(2, rightSub.getDepth() + 1) - Math.pow(2, depth))));
this.depth = rightSub.getDepth() + 1;
}
for (int i = 1; i < depth; i ++) {
for (int j = 0; j < depth - 1; j ++) {
treeArray((int) (Math.pow(2, i) + Math.pow(2, i - 1)) + j)
= rightSub.getTreeArray()((int) Math.pow(2, i - 1) + j);
}
}
}
}
If we have a related graph, weighted and not directed. Can any one provide an example where DFS will not return the MST? No matter what summit we are leaving?
I need help to understand a theorem and its proof from a script. He says, "There is a sequence of $ n $ insert and delete operations in a tree (2,3) that requires $ Omega ($n log n$) $ many splitting and merging operations. "
In fact, I do not know at all what such a sequence looks like. I've tried performing some insert and delete operations on a tree (2,3) and I've got different results on split and merge operations, but I think that there is a "special" sequence to maximize the splitting and merging operations.
I would be very grateful if anyone can help me with this question, so I can try to see how the proof of the theorem can be made.
Greetings
I immerse myself in the data structures and the PATRICIA tree, in particular.
It is written almost in all resources that if the PAT tree is binary (for example, alphabet is {0, 1}), then there are n-1 internal nodes (where n is the number of elements), but the problem is that I can not find proof or find it on the web.
How can we show that the number of internal nodes is n-1 in this case?
Thank you!
I try to learn rust and I feel like a dummy at every step. I've followed the tutorial on creating lists at Rust and I'm trying to apply the knowledge to implement a simple BST. The code seems pretty awful though: – / Any suggestions for improving the situation?
pub struct Tree {
root: Link,
}
type Link = Option>>;
pub struct Node {
elem: T,
left: Link,
right: Link
}
impl Tree {
pub fn new() -> Self {
Tree {root: None}
}
pub fn add(&mut self, elem: T) {
let mut current;
match self.root {
None => {
self.root = Some(Box::new(Node{elem: elem, left: None, right: None}));
return;
},
Some(_) => {
current = self.root.as_mut();
}
}
loop {
if elem < current.as_ref().unwrap().elem {
if current.as_ref().unwrap().left.is_none() {
current.unwrap().left = Some(Box::new(Node{elem: elem, left: None, right: None}));
break;
} else {
current = current.unwrap().left.as_mut();
}
} else {
if current.as_ref().unwrap().right.is_none() {
current.unwrap().right = Some(Box::new(Node{elem: elem, left: None, right: None}));
break;
} else {
current = current.unwrap().right.as_mut();
}
}
}
}
}
#(cfg(test))
mod test {
use super::Tree;
#(test)
fn basics() {
let mut tree = Tree::new();
tree.add(5);
tree.add(3);
tree.add(2);
tree.add(4);
assert_eq!(tree.root.as_ref().unwrap().elem, 5);
assert_eq!(tree.root.as_ref().unwrap().left.as_ref().unwrap().elem, 3);
assert_eq!(tree.root.as_ref().unwrap().left.as_ref().unwrap().left.as_ref().unwrap().elem, 2);
assert_eq!(tree.root.as_ref().unwrap().left.as_ref().unwrap().right.as_ref().unwrap().elem, 4);
}
}
I want to find the minimum number of tree cuts so that each pair of trees in a sequence of trees alternates strictly decreasing and strictly increasing. Example: In (2, 3, 5, 7), the minimum number of tree cuts is 2 – a possible final solution is (2, 1, 5, 4).
My search pattern is a graph where each node is a possible configuration of all tree heights and each edge is a tree cut (= a decrease in the height of a tree). In this model, a possible path from the initial node to the objective node in the above example would be (2,3,5,7) – (2,1,5,7) – (2,1,5, 4). I used a width search first to find the node of the goal. Since BFS does not cross the already traversed nodes, the part of the graph that I cross during the search is actually a tree data structure.
The only improvement I could think for this algorithm was to use a priority queue ordering to explore the possible nodes in ascending order, starting with the number of cuts (as is already done by the traditional BFS system). ) and by the number of strictly increasing nodes. triplets. This ensures that an objective node with the minimum number of N cuts will be the first node of all nodes with N cuts to be evaluated and that the search can finish a little faster.
The time required to run this algorithm increases exponentially with the number of trees. Is there another algorithm / idea that could be used to speed it up?
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I'm trying to create a binary tree using JS. I am not totally convinced that what I have done is good enough, but I have not found a better solution. Data structures like this one are somewhat new to me, which is why I would like to hear your comments on the code.
class Tree {
constructor(initial = null) {
this.root = initial === null ? null : this._createNode(initial);
}
_createNode(value, children = null) {
let left = null;
let right = null;
if (typeof children === "number") {
left = children;
} else if (Array.isArray(null)) {
(left, right) = children.sort((a, b) => a - b);
}
return {
value,
left,
right
};
}
_getNewPlace(value) {
let currentNode = this.root;
while (currentNode !== null) {
if (currentNode.value > value) {
if (currentNode.left === null) {
currentNode.left = this._createNode(value);
return;
} else {
currentNode = currentNode.left;
}
} else {
if (currentNode.right === null) {
currentNode.right = this._createNode(value);
return;
} else {
currentNode = currentNode.right;
}
}
}
}
getRoot() {
return this.root;
}
addNode(item) {
if (this.root === null) {
this.root = this._createNode(item);
} else {
this._getNewPlace(item);
}
return this;
}
}
module.exports = {
Tree
};
And tests:
const { Tree } = require("./index");
describe("Binary Tree tests", () => {
it("should create an empty tree", () => {
const t = new Tree();
expect(t.getRoot()).toBe(null);
});
it("should create a root node first", () => {
const t = new Tree();
t.addNode(1);
expect(t.getRoot()).toMatchObject({
value: 1,
left: null,
right: null
});
});
it("should create a root node from constuctor, when value is provided", () => {
const t = new Tree(1);
expect(t.getRoot()).toMatchObject({
value: 1,
left: null,
right: null
});
});
it("should create an additional node if there is a root", () => {
const t = new Tree(1);
t.addNode(2);
expect(t.getRoot()).toMatchObject({
value: 1,
left: null,
right: {
value: 2,
left: null,
right: null
}
});
});
});
And finally, the code on CodeSandbox: https://codesandbox.io/s/binary-tree-search-xq000
Thanks in advance!
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