# [PHP] Easiest way to get the Instagram id of the given username without an access token

Details:

1) Easiest way to get the instagram id of the given username.
3) Only a few lines of code.
4) PHP example available on github.

How it works?:
2) Send a get request to:
3) The returned JSON contains the user id.
Structure of the returned JSON:

Code:

Usage:
Send a get request with the parameter “username”

Credits:
http://stackoverflow.com/a/38342137/5334285

# Prototyping with Axure RP – Goaaal

Goaaal-足球新闻APP原型-北大用户体验研究2016

【简介】
Goaaal 是一款手机上的足球新闻 APP。在 APP 上可以阅读最新的足球新闻,查

【APP图片】

【逻辑流程图】

【视频】

# Simple Pokemon Go Bot

GITHUB: https://github.com/anakornk/Pokemon-Go-Catcher

How it works?
1) This bot will warp to the specified location (manually set the location in the config.properties file)
2) Will loot pokestops and try to catch pokemon nearby.

Please note: the bot does not recycle items and pokemon.

# Pokemon Go Sniper Java

FOR EDUCATIONAL PURPOSES ONLY

GITHUB: https://github.com/anakornk/PokemonGoSniperJava

How it works? (Sniping without getting softbans)
1) Warp to the specified location
2) encounter the pokemon
3) warp back
4) catch the pokemon (throw pokeballs)

# Java 飞机大战 / Space Shooter

GITHUB: https://github.com/anakornk/Java-SpaceShooter
Java 飞机大战（DISTANCE,UTL）

enemyInfo.java – 记录关于敌人飞船的信息 如生命值，图片

bulletInfo.java – 记录关于各种子弹的信息 如伤害力，图片

constant.java – 关于游戏界面的信息 如屏幕大小

# What is an algorithm for enclosing the maximum number of points in a 2-D plane with a fixed-radius circle?

What is an algorithm for enclosing the maximum number of points in a 2-D plane with a fixed-radius circ… by @chhung6

There is an $O(N^2 \log N)$ deterministic algorithm. This is usually (if not always) sufficient for competitive programming contests / online judges (e.g., POJ 1981 Circle and Points).

## Idea:

If there exists a circle C that encloses M points, then by slightly moving C it will touch at least two points. [1]

So, essentially we just need to check all circles which touch (at least) 2 points:

1. For each pair of points in the given set, construct a circle with radius R that touches both points. For each point pair there are at most two such circles.
2. For each constructed circle, check for each of the given points is inside it.
3. Return the circle that encloses the maximum number of points.

The runtime of this algorithm is $O(N^3)$ because there are at most $O(N^2)$ such circles in step 1, and the linear scan at step 2 takes $O(N)$ time.

## Runtime Improvement:

So, how can we achieve $O(N^2 \log N)$?

The idea is as follows:

Pick an arbitrary point P from the given set. We rotate a circle C with radius R using P as the "axis of rotation" (by convention, in the counter-clockwise direction), i.e. we keep C to touch P any time during the rotation. While C is being rotated, we maintain a counter to count the number of points C encloses.

Note that this counter only changes when some point Q enters (or leaves) the region of the circle C. Our goal is to come up with an algorithm that will increment (or decrement) this counter, whenever some other point QP enters (or leaves) the region of C.

The state of the (rotating) circle C can be described by a single parameter $\theta$, where $(r, \theta)$ are the polar coordinates of the center of the circle C, if we pick P as the fixed point of the polar coordinate system. With this system, rotating C means increasing $\theta$.

(Figure from Wikipedia: Points in a polar coordinate system)

For each other point Q (≠ P), we can actually compute the range of $\theta$  for which C covers Q. Put more formally, C encloses Q whenever (iff) $\theta \in [\alpha, \beta]$. [2]

So, up to this point, the original problem has been reduced to:

What is an optimal value of $\theta$ that lies in the most number of $[\alpha, \beta]$ intervals?

The reduced problem can be solved with a pretty standard $O(N \log N)$ algorithm.[3] This reduced problem has to be solved N times (one for each point P), hence the time complexity $O(N^2 \log N)$.

## Remarks:

The above algorithms assume that a point is considered being enclosed in a circle even if it is on the circumference. If not, we could still tweak them a bit, for example, by changing the intervals of $\theta$ from being close-close to close-open. (Hmm, probably – didn't think too much to verify, but you get the idea.)

[1]: Unless all pairwise distances are > 2R, where the answer to the maximum number of points will be 1. This can be checked in $O(N^2)$ time.
[2]: We ignore Q's with distance > 2R from P.
[3]: Probably need to be careful when handling these cyclic intervals.

What is an algorithm for enclosing the maximum number of points in a 2-D plane with a fixed-radius circle?

# Pokemon Go Bangkok Gym Analysis

Thank you Pokémon Go Thailand FanClub for sharing.

# XSS Vulnerabilities on Joomla v3.3.3

Sent a report to the joomla! strike tream but have never received a response.
Seems like the vuln is not critical.
To trigger these xss(es) vuln you will need access to the administrative panel. Since the vuln(s) are basically the same but are in different parameters I’ll be showing an example to just one.
1) Go to the Global Configuration page:
2) Type in the following payload in “Site Name”

" onclick=alert(1) test="

3) Save
4)  On the admin page, if you click on the site logo or the site name, or on your main public page when you click on the site name, javascript will be executed.

<header class="header">
<div class="container-logo">
</div>
<a class="brand pull-left" href="/joomla">
<div class="header-search pull-right">