Where can I find a similar view from a hotel room like Costa Rica overlooking Whale Tail National Park?

This is "called the lattice house of Puntarenas in Costa Rica. Absolutely beautiful. In the background is the" whale tail "national park But I have already visited Costa Rice before. Where can I find a similar view in

  1. Hawaii

  2. Japan

  3. Australia

  4. New Zealand?

enter description of image here

Head & Tail Games Collection Earn from 200% (earn $ 5-7 / day)

I collect a webist containing the head and the tail.

Notes:

1. Search the story

2. If winning 199% – 250%, the story loses 3-4 times -> play. If you win 300-400, the story loses 6-7 times -> play.

3. Change IP

4. Change the site bet

5. Play again

List of sites:

http://www.0145-2367.com

http://www.1goldgame.com

http://www.1gold-game.com

http://www.1goldgame.info

http://www.2goldgame.info

http://www.300game.info

http://www.7double.com

http://www.bestheadtailgame.com

http://www.bet2gold.com

http://www.betgold.info

http://www.betlr.us

http://www.dice-roller.net

http://www.eggame.info

http://www.evenorodd.com

http://www.gamblinglr.com

http://www.gangsterbet.com

http://www.getgold.us

http://www.getlr.us

http://www.gold-gambler.com

http://www.gugold.com

http://www.headtail.us

http://www.headxtail.tk

http://www.libertyreservegame.com

http://www.lrboxes.com

http://www.lrgame.us

http://www.lucky2bet.com

http://www.luckygold.info

http://www.perfectmoneygame.com

http://www.lrgame.us

http://www.v-slot.com

Good luck

java – Determines if a chain has a mirrored head and tail

Problem statement:

Given a string, look for a mirror image chain (backward) at the beginning and end of the given chain. In other words, zero or more characters at the very beginning of the given string and at the very end of the string in reverse order (possibly overlapping). For example, the string "abXYZba" has the mirror end "ab".

mirrorEnds ("abXYZba") → "ab"
mirrorEnds ("abca") → "a"
mirrorEnds ("aba") → "aba"

Here is my solution to the problem in java:

  public String mirrorEnds(String string) {
      final int len = string.length();
      final int half = len / 2;
      String result = "";

      for (int i = 0; i < half; i++) {
        if (string.charAt(i) != string.charAt(len -1 -i)) {
            break;      
        } else {
          result += string.substring(i, i + 1);
        }
      }
      return result.length() == half ? string : result;
    }

Is it safe to say that in terms of time complexity, the solution is already optimal? Any other review comments are also welcome.

p.s. - please ignore indentation!

What are the long tail and short tail keywords?

What are the long tail and short tail keywords?

unit – Snake tail in Snake game is not working properly

I make a simple snake game in Unity 2D. And the snake tail doesn't follow just behind the snake head as it should.

This is what is happening: https://imgur.com/a/zG6hbd7

Here is my code:

using System.Collections;

using System.Collections.Generic;
using UnityEngine;
using System.Linq;

Public class SnakeMovement class: MonoBehavior
{

private Vector3 fp;   //First touch position
private Vector3 lp;   //Last touch position

private float dragDistance;  //minimum distance for a swipe to be registered
private float speed = 13f;

private Rigidbody2D rb;
private Vector2 moveVelocity;

public GameObject tail;

public List tailPositions;

void Start() {
    dragDistance = Screen.height * 5 / 100; //dragDistance is 5% height of the screen
    rb = GetComponent();
}

void Update() {

    if (Input.touchCount == 1) // user is touching the screen with a single touch
    {
        Touch touch = Input.GetTouch(0); // get the touch
        if (touch.phase == TouchPhase.Began) //check for the first touch
        {
            fp = touch.position;
            lp = touch.position;
        }
        else if (touch.phase == TouchPhase.Moved) // update the last position based on where they moved
        {
            lp = touch.position;
        }
        else if (touch.phase == TouchPhase.Ended) //check if the finger is removed from the screen
        {
            lp = touch.position;  //last touch position. Ommitted if you use list

            //Check if drag distance is greater than ?% of the screen height
            if (Mathf.Abs(lp.x - fp.x) > dragDistance || Mathf.Abs(lp.y - fp.y) > dragDistance)
            {//It's a drag
             //check if the drag is vertical or horizontal

                if (Mathf.Abs(lp.x - fp.x) > Mathf.Abs(lp.y - fp.y)) {   //If the horizontal movement is greater than the vertical movement...
                    if ((lp.x > fp.x)) {   //Right swipe
                        if(moveVelocity != Vector2.left) {
                            moveVelocity = Vector2.right;
                        }
                    }
                    else {   //Left swipe
                        if(moveVelocity != Vector2.right) {
                            moveVelocity = Vector2.left;
                        }
                    }
                }
                else {   //the vertical movement is greater than the horizontal movement
                    if (lp.y > fp.y) {   //Up swipe
                        if(moveVelocity != Vector2.down) {
                            moveVelocity = Vector2.up;
                        }
                    }
                    else {   //Down swipe
                        if(moveVelocity != Vector2.up) {
                            moveVelocity = Vector2.down;
                        }
                    }
                }

            }
        }
    }

}

void FixedUpdate() {
    Vector3 lastPos = transform.position;

    rb.MovePosition(rb.position + moveVelocity * speed * Time.deltaTime);

    if (tailPositions.Count >= 1) {
        tailPositions.Last().position = lastPos;
        tailPositions.Insert(0, tailPositions.Last ());
        tailPositions.RemoveAt(tailPositions.Count - 1);
    }
}

void OnTriggerEnter2D(Collider2D trigger) {
    Vector2 spawnPos = new Vector2(5, 5);

    if (trigger.tag == "Food") {
        GameObject newTail = Instantiate(tail, spawnPos, Quaternion.identity) as GameObject;
        newTail.transform.parent = GameObject.Find("Tail Holder").transform;
        tailPositions.Add(newTail.transform);
    }
}

}

Thank you very much for taking the time to help me.

functional programming – What is the difference between recursion and tail recursion?

The difference is exactly where the recursive call is called: if it is called on the "tail" of the function, it is called the recursive tail.

The function's "tail" is your last call. This is the last calculation / calculation performed by the function, and right after, no processing is done before returning its value.

For example, considering this function to calculate the factorial of a number, in F #:

let rec fatorial n : int64 = if n <= 1L then 1L else n * fatorial (n - 1L)

This function calculates the factorial of any number entered. Unless this number is very high. Because in this function, with each recursive call, the stack increases and a very large number can cause a stack overflow. Imagine your execution:

fatorial(5) -> 5 * fatorial(5 - 1) ->
5 * fatorial(4) -> 5 * 4 * fatorial(4 - 1) ->
5 * 4 * fatorial(3) -> 5 * 4 * 3 * fatorial(3 - 1) ->
5 * 4 * 3 * fatorial(2) -> 5 * 4 * 3 * 2 * fatorial(2 - 1) ->
5 * 4 * 3 * 2 * fatorial(1) -> 5 * 4 * 3 * 2 * 1 ->
120

If you observe, on each recursive call, the number of functions called increases, because the program can only calculate the result of the last function called, then calculate the result of those who called it. Thus, the stack overflows.

How to avoid this? Use of recursive queue calls. Functional language compilers often transform recursive queue calls into loops, as this is perfectly possible. Why not do it directly? Because it would lose the qualities and benefits of functional programming.

I will illustrate a factorial function in F # using recursive tail calls:

let fatorial n =
  let rec _fatorial n acc : int64 =
    if n <= 1L then acc else _fatorial (n - 1L) (acc * r)
  _fatorial n 1L

Note that in this case, the recursive function is NOT fatorial, and yes _fatorial. I said _fatorial inside fatorial so we can call it with a single argument, and the recursive function uses an accumulator.

The main difference is that in the recursive tail function, the tail call is the recursive call, not * as in the first case. If you observe the flow of the call, it will run as follows:

fatorial(5)       ->
_fatorial(5, 1)   -> _fatorial(5 - 1, 1 * 5)  ->
_fatorial(4, 5)   -> _fatorial(4 - 1, 5 * 4)  ->
_fatorial(3, 20)  -> _fatorial(3 - 1, 20 * 3) ->
_fatorial(2, 60)  -> _fatorial(2 - 1, 60 * 2) ->
_fatorial(1, 120) -> 120

As you can see, at each step, the number of calls neither increases nor decreases. As soon as the recursive function is called, only it is called at the end, without further calculations.

When a compiler ready to do so sees a recursive call at the tail, it automatically transforms it into a loop during optimizations. With that, you don't lose the benefits or elegance of functional programming, but you also don't run the risk of encountering a stack overflow.

Using a reflector, I can see that the recursive function code would imperatively look like this (in C #):

internal static long _fatorial@8(long n, long acc)
{
  while (n > 1L)
  {
    long arg_1F_0 = n - 1L;
    acc *= n;
    n = arg_1F_0;
  }
  return acc;
}

public static long fatorial(long n)
{
  return Fatorial._fatorial(n, 1L);
}

The compiler actually turns its recursive function into a loop. On the other hand, the function which does not use the cause of recursion remains intact.

A good way to find out whether or not your function uses tail recursion is to try to simulate it in Clojure. Since Clojure has no native tail recursion, you should use the recur, which will throw an exception if it is not used on the tail.

; Causa uma exceção pois a chamada de cauda é *
(defn fatorial (n)
  (if (<= n 1)
      1
      (* n (recur (dec n)))))

; Funciona pois a chamada de cauda é recur
(defn fatorial
  ((n) (fatorial n 1))
  ((n acc) (if (<= n 1)
               acc
               (recur (dec n) (* acc n)))))

scala – What better way to map function value: foreach tail recursion?

I want to map functions from my function library (Map called chain) to enter a channel str. It also works (Twitter.removeRT, …) are regular expressions that replace the input substrings str. I think it is best to save these regular expression functions in Map as in my example.

So the code:

Tail recursion variant

  def filterTwitter2 (str: String): String = {

    @scala.annotation.tailrec
    def recFilter(str: String, chain: Map(String, (String) => String)): String = {
      chain.headOption match {
        case Some(v) =>
          val filteredString = v._2(str)
          recFilter(filteredString, chain.tail)
        case None => str
      }
    }

    val chain = Map(String,(String) => String)(
      "f1"-> Twitter.removeRT,
      "f2"-> Twitter.removeNickName,
      "f3"-> Twitter.removeURL,
      "f6"-> Emoticons.removePunctRepetitions,
      "f7"-> Emoticons.removeHorizontalEmoticons,
      "f9"-> Emoticons.normalizeEmoticons,
      "f10"-> Beautify.removeCharRepetitions,
      "f12"-> Beautify.removeNSpaces
    )

    recFilter(str, chain)
  }

foreach variant

  def filterTwitter (str: String): String = {
    var tmp = str
    val chain = Map(String,(String) => String)(
      "f1"-> Twitter.removeRT,
      "f2"-> Twitter.removeNickName,
      "f3"-> Twitter.removeURL,
      "f6"-> Emoticons.removePunctRepetitions,
      "f7"-> Emoticons.removeHorizontalEmoticons,
      "f9"-> Emoticons.normalizeEmoticons,
      "f10"-> Beautify.removeCharRepetitions,
      "f12"-> Beautify.removeNSpaces
    )
    chain.foreach {
      case (name, func) => tmp = func(tmp)
    }
    tmp
  }

So the questions:

  1. Is it okay to save functions in the card? What could be better for this?
  2. What's better: tail recursion variant or variant with foreach?
  3. Maybe there is a better solution to this problem?

probability – Exercise on the tail $ sigma $ -algebra

Let $ (X_n) _n $ be a sequence of real random variables, $ T_k = sigma (X_n) _ {n ge k} $ and $ T = bigcap_n T_n $ tail $ sigma $-algebra.
Prove it $ limsup_n frac {X_1 + … + X_n} {n} $ East $ T $-measurable.

I can use it

Lemma ($ (a_n) $, $ (b_n) $ are real sequences)

Yes $ lim_n a_n = 0 $, then $ limsup_n (a_n + b_n) = limsup_nb_n $.

My attempt

To take $ m ge 1 $, we have: $ limsup_n frac {X_1 + … + X_n} {n} = lim_ {k gt m} sup_ {n ge k} ( frac {X_1 + … + X_m} {m} frac {m} {n} + frac {X_ {m + 1} + … + X_n} {n}) = limsup_n frac {X_ {m + 1} + … + X_n} {n} $ because $ frac {X_1 + … + X_m} {m} frac {m} {n} at $ 0 as $ n to + infty $.

Since $ limsup_n frac {X_ {m + 1} + … + X_n} {n} $ East $ T_m $– measurable, it follows that $ limsup_n frac {X_1 + … + X_n} {n} $ East $ T_m $– measurable for each $ m ge 1 $, it is therefore $ T $-measurable.

Is it correct? Thanks in advance

Encoding of the tail operation in untyped lambda calculation

Pierce's 5.2.8 exercise proposes to code lists with a folding operation. I've learned that this encoding is also known as Scott encoding. I was entitled to most encodings:

$ text {nil} = lambda c. lambda n. n $

$ text {cons} = lambda h. lambda t. lambda c. lambda n. vs ; h ; (t ; c ; n) $

$ text {isnil} = lambda l. l ; ( lambda h. lambda t. text {fls}) ; text {tru} $

$ text {head} = lambda l. l ; ( lambda h. lambda t. h) text {fls} $

but the one for the queue failed, I guess because I did not think about the case, nothing. However, the book's solution has left me puzzled:

$ text {tail} = lambda l. text {fst} (l ; ( lambda x. lambda p. text {pair} ( text {snd} ; p) ; ( text {cons} ; x ; ( text { snd} ; p))) ; ( text {pair} ; text {nil} ; text {nil})) $

Can any one explain the strategy behind this coding?

5th dnd – Can a changeling turn into races with a tail / horn?

I've read that a changeling can turn into any creature they've seen, unless it's bigger / smaller than their size and that They can not change themselves into something that has a different body shape than the one they have. So, no additional members or others. So my question is: can a Changeling be transformed into Tiefling for example? Since they have horns and a tail. I think the horn will not be a problem, but maybe the tail would do it. Idk, that's why I ask