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  • How to set webview scroll position within it on initial launch ?

    - by mob-king
    I have an issue with Webview to set the initial scroll position within it when first launched. By default it is left , I want it to scroll and show the view from center. Or is it possible to resize the view to fit the screen width initially, instead of scroll bars. I need this since the left part of the page I am loading is blank making user feel the page is not loaded.

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  • How to preserve sibling element position when one sibling is absolutely positioned?

    - by Casey
    In the snippet below, the child div is normally positioned until it is :hovered , when it becomes absolutely positioned. The reasoning behind this markup is to simulate a popup style in a limited environment where I can't use a <select> (among other limitations). When child is hovered, the sibling elements jump around, which is expected, as the contents of the block have changed. But how can I preserve their positioning? That is, what CSS can I add to prevent the siblings from jumping around when child is hovered. Javascript is also not allowed, so please no answers using JS. HTML: <div class="container"> <div class="child"> <span class="d4"></span> <label><input type="radio" name="radio" value="1"/>One</label> <label><input type="radio" name="radio" value="2"/>Two</label> </div> <input type="text" name="sibling"/> <button name="sibling2">Button</button> </div> CSS: .container, .child, button { display:inline-block; } .child { vertical-align: middle; width: 35px; height: 35px; } .child:hover { background: gray; position:absolute; width: 100px; height: auto; } .child:hover > .d4 { display: none; } .child label { display:none; } .child:hover label { display: inline-block; } .d4 { background-position: -411px -1px; width: 35px; height: 35px; background-image: url("https://i.imgur.com/zkgyBOi.png"); background-repeat: no-repeat; color: transparent; display: inline-block; } Here's a fiddle: http://jsfiddle.net/cpctZ/1/

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  • What variable dictates position of non-focused elements in the roundabout plugin?

    - by kristina childs
    Part of the problem here is that i'm not sure what the best language to use in order to find the solution. I search and searched so please forgive if this is already a thread somewhere. I'm using the roundabout plugin to cycle through 3 divs. Each div is 794px wide, which makes the roundabout-in-focus element 794 and the two not in focus 315.218px wide, positioned so half of each is hidden by the in-focus div. This is all well and good, however the total width of the display needs to stay within 1000px (ideally 980px, but i can fudge if need be.) Basically I want to make the non-focused divs be 3/4 hidden by the in-focus div but for the life of me can't figure out what variables i need to edit in order to do it. Unfortunately it's not one of the many easily-changed options like z-index and minScale. i tried minScale but it's clear this isn't going to work. the plugin outputs this code: <li class="roundabout-moveable-item" style="position: absolute; left: -57px; top: 205px; width: 319.982px; height: 149.513px; opacity: 0.7; z-index: 146; font-size: 5.6px;"> i need to find out what changes the left positioning so it's shifted closer to the center of the stage, like this: <li class="roundabout-moveable-item" style="position: absolute; left: 5px; top: 205px; width: 319.982px; height: 149.513px; opacity: 0.7; z-index: 146; font-size: 5.6px;"> i tried playing with the positioning functions of the plugin but all that did was shift everything in tandem left or right. any help is greatly appreciated. this site is going to be awesome once i figure out all this jquery stuff! here is a link to my .js file: http://avalon.eaw.com/scripts/jquery.roundabout2.js i've got an overflow:hidden on the to help guide the positioning of those no-focused items.

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  • Sorting a list of numbers with modified cost

    - by David
    First, this was one of the four problems we had to solve in a project last year and I couldn’t find a suitable algorithm so we handle in a brute force solution. Problem: The numbers are in a list that is not sorted and supports only one type of operation. The operation is defined as follows: Given a position i and a position j the operation moves the number at position i to position j without altering the relative order of the other numbers. If i j, the positions of the numbers between positions j and i - 1 increment by 1, otherwise if i < j the positions of the numbers between positions i+1 and j decreases by 1. This operation requires i steps to find a number to move and j steps to locate the position to which you want to move it. Then the number of steps required to move a number of position i to position j is i+j. We need to design an algorithm that given a list of numbers, determine the optimal (in terms of cost) sequence of moves to rearrange the sequence. Attempts: Part of our investigation was around NP-Completeness, we make it a decision problem and try to find a suitable transformation to any of the problems listed in Garey and Johnson’s book: Computers and Intractability with no results. There is also no direct reference (from our point of view) to this kind of variation in Donald E. Knuth’s book: The art of Computer Programing Vol. 3 Sorting and Searching. We also analyzed algorithms to sort linked lists but none of them gives a good idea to find de optimal sequence of movements. Note that the idea is not to find an algorithm that orders the sequence, but one to tell me the optimal sequence of movements in terms of cost that organizes the sequence, you can make a copy and sort it to analyze the final position of the elements if you want, in fact we may assume that the list contains the numbers from 1 to n, so we know where we want to put each number, we are just concerned with minimizing the total cost of the steps. We tested several greedy approaches but all of them failed, divide and conquer sorting algorithms can’t be used because they swap with no cost portions of the list and our dynamic programing approaches had to consider many cases. The brute force recursive algorithm takes all the possible combinations of movements from i to j and then again all the possible moments of the rest of the element’s, at the end it returns the sequence with less total cost that sorted the list, as you can imagine the cost of this algorithm is brutal and makes it impracticable for more than 8 elements. Our observations: n movements is not necessarily cheaper than n+1 movements (unlike swaps in arrays that are O(1)). There are basically two ways of moving one element from position i to j: one is to move it directly and the other is to move other elements around i in a way that it reaches the position j. At most you make n-1 movements (the untouched element reaches its position alone). If it is the optimal sequence of movements then you didn’t move the same element twice.

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  • elisp: posn-at-point returns nil after goto-char. How to update the display before posn-at-point?

    - by Cheeso
    In emacs lisp, posn-at-point is documented as: posn-at-point is a built-in function in C source code. (posn-at-point &optional POS WINDOW) . Return position information for buffer POS in WINDOW. POS defaults to point in WINDOW; WINDOW defaults to the selected window. . Return nil if position is not visible in window. Otherwise, the return value is similar to that returned by event-start for a mouse click at the upper left corner of the glyph corresponding to the given buffer position: (WINDOW AREA-OR-POS (X . Y) TIMESTAMP OBJECT POS (COL . ROW) IMAGE (DX . DY) (WIDTH . HEIGHT)) The posn- functions access elements of such lists. ok, now I've got a function that looks something like this: (defun my-move-and-popup-menu () "move the point, then pop up a menu." (goto-char xxxx) (setq p (posn-at-point)) (my-popup-menu p ...) ) Basically, move the point, then retrieve the screen position at that point, and then popup a menu at that screen position. But I am finding that posn-at-point returns non-nil, only if the xxxx character position (the after position) is visible in the window, before the call to goto-char. It seems that the position is not actually updated until exit from the function. If goto-char goes a long way, more than one screenful, then the retrieved position is always nil, and my code doesn't know where to popup the menu. The reason I suggest that the position is not actually updated until exit from the function - when the menu successfully pops up, the cursor is clearly visible in its previous location while the popup menu is being displayed. When I dismiss the menu, the cursor moves to where I expected it to move, after the goto-char call. How can I get the position to be really updated, between goto-char and posn-at-point, so that posn-at-point will not return nil? In a Windows Forms application I would call Form.Update() or something similar to update the display in the middle of an event handler. What's the emacs version of that?

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  • jQuery: How to position one element relative to another?

    - by paul
    I have a hidden DIV which contains a toolbar-like menu. I have a number of DIVs which are enabled to show the menu DIV when the mouse hovers over them. Is there a built-in function which will move the menu DIV to the top right of the active (mouse hover) DIV? I'm looking for something like $(menu).position("topright", targetEl);

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  • jQuery animate slidding background image using mouse position wont stop... just keeps going

    - by simian
    I have a neat little jQuery script I am working on. Goal: parent div with overflow hidden holds a larger div with image. When I move the mouse to the left or right of the parent div, the div with image changes margin-left to move left or right. Problem is... if I move the mouse out of the parent div (left or right), the image keeps going. I need the image to stop when the mouse is not on the inside left or right edge of the parent div. Any ideas? <!DOCTYPE html> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <script type="text/javascript" src="http://code.jquery.com/jquery-latest.js"></script> <script type="text/javascript"> //<![CDATA[ jQuery.noConflict (); jQuery(document).ready(function(){ jQuery('#container').mousemove(function(e){ var parentWidth = jQuery('#container').width(); var parentWidthLeft = Math.round(.1 * jQuery('#container').width()); var parentWidthRight = Math.round(jQuery('#container').width() - parentWidthLeft); var parentHeight = jQuery('#container').height(); var parentOffset = jQuery('#container').offset(); var X = Math.round(e.pageX - parentOffset.left); var Y = Math.round(e.pageY - parentOffset.top); if (X<parentWidthLeft) { jQuery('#image').animate({'left': '-' + parentWidth }, 5000); } if (X>parentWidthRight) { jQuery('#image').animate({'left': parentWidth }, 5000); } if (X<=parentWidthRight && X>=parentWidthLeft) { jQuery('#image').stop(); } if (X<1) { jQuery('#image').stop(); } }); jQuery('#container').mouseleave(function(){ jQuery('#image').stop(); }); }); // ]]> </script> </head> <body> <div id="container" style="width: 500px; height: 500px; overflow: hidden; border: 10px solid #000; position: relative; margin: auto auto;"> <div id="image" style="position: absolute; left: 0; top: 0;"><img src="http://dump4free.com/imgdump/1/Carmen-Electra_1wallpaper1024x768.jpg" alt="" /></div> </div> </body> </html>

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  • Is the W3 standard a major factor when google decides SERP position?

    - by Camran
    I have a dynamic php website which index only has around 800 errors according to the w3 validator online. I have tried checking major websites like ebay, stackoverflow and others also, all with around 400 errors. So my first thought is, what good is that validator when it always displays errors? Secondly, will the errors affect my SERP ranking? ie, will me fixing these errors as good as I can increase my Google search position? Thanks

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  • Reverse subarray of an array with O(1)

    - by Babibu
    I have an idea how to implement sub array reverse with O(1), not including precalculation such as reading the input. I will have many reverse operations, and I can't use the trivial solution of O(N). Edit: To be more clear I want to build data structure behind the array with access layer that knows about reversing requests and inverts the indexing logic as necessary when someone wants to iterate over the array. Edit 2: The data structure will only be used for iterations I been reading this and this and even this questions but they aren't helping. There are 3 cases that need to be taking care of: Regular reverse operation Reverse that including reversed area Intersection between reverse and part of other reversed area in the array Here is my implementation for the first two parts, I will need your help with the last one. This is the rule class: class Rule { public int startingIndex; public int weight; } It is used in my basic data structure City: public class City { Rule rule; private static AtomicInteger _counter = new AtomicInteger(-1); public final int id = _counter.incrementAndGet(); @Override public String toString() { return "" + id; } } This is the main class: public class CitiesList implements Iterable<City>, Iterator<City> { private int position; private int direction = 1; private ArrayList<City> cities; private ArrayDeque<City> citiesQeque = new ArrayDeque<>(); private LinkedList<Rule> rulesQeque = new LinkedList<>(); public void init(ArrayList<City> cities) { this.cities = cities; } public void swap(int index1, int index2){ Rule rule = new Rule(); rule.weight = Math.abs(index2 - index1); cities.get(index1).rule = rule; cities.get(index2 + 1).rule = rule; } @Override public void remove() { throw new IllegalStateException("Not implemented"); } @Override public City next() { City city = cities.get(position); if (citiesQeque.peek() == city){ citiesQeque.pop(); changeDirection(); position += (city.rule.weight + 1) * direction; city = cities.get(position); } if(city.rule != null){ if(city.rule != rulesQeque.peekLast()){ rulesQeque.add(city.rule); position += city.rule.weight * direction; changeDirection(); citiesQeque.push(city); } else{ rulesQeque.removeLast(); position += direction; } } else{ position += direction; } return city; } private void changeDirection() { direction *= -1; } @Override public boolean hasNext() { return position < cities.size(); } @Override public Iterator<City> iterator() { position = 0; return this; } } And here is a sample program: public static void main(String[] args) { ArrayList<City> list = new ArrayList<>(); for(int i = 0 ; i < 20; i++){ list.add(new City()); } CitiesList citiesList = new CitiesList(); citiesList.init(list); for (City city : citiesList) { System.out.print(city + " "); } System.out.println("\n******************"); citiesList.swap(4, 8); for (City city : citiesList) { System.out.print(city + " "); } System.out.println("\n******************"); citiesList.swap(2, 15); for (City city : citiesList) { System.out.print(city + " "); } } How do I handle reverse intersections?

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  • Move a sphere along the swipe?

    - by gameOne
    I am trying to get a sphere curl based on the swipe. I know this has been asked many times, but still it's yearning to be answered. I have managed to add force on the direction of the swipe and it works near perfect. I also have all the swipe positions stored in a list. Now I would like to know how can the curl be achieved. I believe the the curve in the swipe can be calculated by the Vector dot product If theta is 0, then there is no need to add the swipe. If it is not, then add the curl. Maybe this condition is redundant if I managed to find how to curl the sphere along the swipe position The code that adds the force to sphere based on the swipe direction is as below: using UnityEngine; using System.Collections; using System.Collections.Generic; public class SwipeControl : MonoBehaviour { //First establish some variables private Vector3 fp; //First finger position private Vector3 lp; //Last finger position private Vector3 ip; //some intermediate finger position private float dragDistance; //Distance needed for a swipe to register public float power; private Vector3 footballPos; private bool canShoot = true; private float factor = 40f; private List<Vector3> touchPositions = new List<Vector3>(); void Start(){ dragDistance = Screen.height*20/100; Physics.gravity = new Vector3(0, -20, 0); footballPos = transform.position; } // Update is called once per frame void Update() { //Examine the touch inputs foreach (Touch touch in Input.touches) { /*if (touch.phase == TouchPhase.Began) { fp = touch.position; lp = touch.position; }*/ if (touch.phase == TouchPhase.Moved) { touchPositions.Add(touch.position); } if (touch.phase == TouchPhase.Ended) { fp = touchPositions[0]; lp = touchPositions[touchPositions.Count-1]; ip = touchPositions[touchPositions.Count/2]; //First check if it's actually a drag if (Mathf.Abs(lp.x - fp.x) > dragDistance || Mathf.Abs(lp.y - fp.y) > dragDistance) { //It's a drag //Now check what direction the drag was //First check which axis 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) && canShoot) //If the movement was to the right) { //Right move float x = (lp.x - fp.x) / Screen.height * factor; rigidbody.AddForce((new Vector3(x,10,16))*power); Debug.Log("right "+(lp.x-fp.x));//MOVE RIGHT CODE HERE canShoot = false; //rigidbody.AddForce((new Vector3((lp.x-fp.x)/30,10,16))*power); StartCoroutine(ReturnBall()); } else { //Left move float x = (lp.x - fp.x) / Screen.height * factor; rigidbody.AddForce((new Vector3(x,10,16))*power); Debug.Log("left "+(lp.x-fp.x));//MOVE LEFT CODE HERE canShoot = false; //rigidbody.AddForce(new Vector3((lp.x-fp.x)/30,10,16)*power); StartCoroutine(ReturnBall()); } } else { //the vertical movement is greater than the horizontal movement if (lp.y>fp.y) //If the movement was up { //Up move float y = (lp.y-fp.y)/Screen.height*factor; float x = (lp.x - fp.x) / Screen.height * factor; rigidbody.AddForce((new Vector3(x,y,16))*power); Debug.Log("up "+(lp.x-fp.x));//MOVE UP CODE HERE canShoot = false; //rigidbody.AddForce(new Vector3((lp.x-fp.x)/30,10,16)*power); StartCoroutine(ReturnBall()); } else { //Down move Debug.Log("down "+lp+" "+fp);//MOVE DOWN CODE HERE } } } else { //It's a tap Debug.Log("none");//TAP CODE HERE } } } } IEnumerator ReturnBall() { yield return new WaitForSeconds(5.0f); rigidbody.velocity = Vector3.zero; rigidbody.angularVelocity = Vector3.zero; transform.position = footballPos; canShoot =true; isKicked = false; } }

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  • Sorting Algorithms

    - by MarkPearl
    General Every time I go back to university I find myself wading through sorting algorithms and their implementation in C++. Up to now I haven’t really appreciated their true value. However as I discovered this last week with Dictionaries in C# – having a knowledge of some basic programming principles can greatly improve the performance of a system and make one think twice about how to tackle a problem. I’m going to cover briefly in this post the following: Selection Sort Insertion Sort Shellsort Quicksort Mergesort Heapsort (not complete) Selection Sort Array based selection sort is a simple approach to sorting an unsorted array. Simply put, it repeats two basic steps to achieve a sorted collection. It starts with a collection of data and repeatedly parses it, each time sorting out one element and reducing the size of the next iteration of parsed data by one. So the first iteration would go something like this… Go through the entire array of data and find the lowest value Place the value at the front of the array The second iteration would go something like this… Go through the array from position two (position one has already been sorted with the smallest value) and find the next lowest value in the array. Place the value at the second position in the array This process would be completed until the entire array had been sorted. A positive about selection sort is that it does not make many item movements. In fact, in a worst case scenario every items is only moved once. Selection sort is however a comparison intensive sort. If you had 10 items in a collection, just to parse the collection you would have 10+9+8+7+6+5+4+3+2=54 comparisons to sort regardless of how sorted the collection was to start with. If you think about it, if you applied selection sort to a collection already sorted, you would still perform relatively the same number of iterations as if it was not sorted at all. Many of the following algorithms try and reduce the number of comparisons if the list is already sorted – leaving one with a best case and worst case scenario for comparisons. Likewise different approaches have different levels of item movement. Depending on what is more expensive, one may give priority to one approach compared to another based on what is more expensive, a comparison or a item move. Insertion Sort Insertion sort tries to reduce the number of key comparisons it performs compared to selection sort by not “doing anything” if things are sorted. Assume you had an collection of numbers in the following order… 10 18 25 30 23 17 45 35 There are 8 elements in the list. If we were to start at the front of the list – 10 18 25 & 30 are already sorted. Element 5 (23) however is smaller than element 4 (30) and so needs to be repositioned. We do this by copying the value at element 5 to a temporary holder, and then begin shifting the elements before it up one. So… Element 5 would be copied to a temporary holder 10 18 25 30 23 17 45 35 – T 23 Element 4 would shift to Element 5 10 18 25 30 30 17 45 35 – T 23 Element 3 would shift to Element 4 10 18 25 25 30 17 45 35 – T 23 Element 2 (18) is smaller than the temporary holder so we put the temporary holder value into Element 3. 10 18 23 25 30 17 45 35 – T 23   We now have a sorted list up to element 6. And so we would repeat the same process by moving element 6 to a temporary value and then shifting everything up by one from element 2 to element 5. As you can see, one major setback for this technique is the shifting values up one – this is because up to now we have been considering the collection to be an array. If however the collection was a linked list, we would not need to shift values up, but merely remove the link from the unsorted value and “reinsert” it in a sorted position. Which would reduce the number of transactions performed on the collection. So.. Insertion sort seems to perform better than selection sort – however an implementation is slightly more complicated. This is typical with most sorting algorithms – generally, greater performance leads to greater complexity. Also, insertion sort performs better if a collection of data is already sorted. If for instance you were handed a sorted collection of size n, then only n number of comparisons would need to be performed to verify that it is sorted. It’s important to note that insertion sort (array based) performs a number item moves – every time an item is “out of place” several items before it get shifted up. Shellsort – Diminishing Increment Sort So up to now we have covered Selection Sort & Insertion Sort. Selection Sort makes many comparisons and insertion sort (with an array) has the potential of making many item movements. Shellsort is an approach that takes the normal insertion sort and tries to reduce the number of item movements. In Shellsort, elements in a collection are viewed as sub-collections of a particular size. Each sub-collection is sorted so that the elements that are far apart move closer to their final position. Suppose we had a collection of 15 elements… 10 20 15 45 36 48 7 60 18 50 2 19 43 30 55 First we may view the collection as 7 sub-collections and sort each sublist, lets say at intervals of 7 10 60 55 – 20 18 – 15 50 – 45 2 – 36 19 – 48 43 – 7 30 10 55 60 – 18 20 – 15 50 – 2 45 – 19 36 – 43 48 – 7 30 (Sorted) We then sort each sublist at a smaller inter – lets say 4 10 55 60 18 – 20 15 50 2 – 45 19 36 43 – 48 7 30 10 18 55 60 – 2 15 20 50 – 19 36 43 45 – 7 30 48 (Sorted) We then sort elements at a distance of 1 (i.e. we apply a normal insertion sort) 10 18 55 60 2 15 20 50 19 36 43 45 7 30 48 2 7 10 15 18 19 20 30 36 43 45 48 50 55 (Sorted) The important thing with shellsort is deciding on the increment sequence of each sub-collection. From what I can tell, there isn’t any definitive method and depending on the order of your elements, different increment sequences may perform better than others. There are however certain increment sequences that you may want to avoid. An even based increment sequence (e.g. 2 4 8 16 32 …) should typically be avoided because it does not allow for even elements to be compared with odd elements until the final sort phase – which in a way would negate many of the benefits of using sub-collections. The performance on the number of comparisons and item movements of Shellsort is hard to determine, however it is considered to be considerably better than the normal insertion sort. Quicksort Quicksort uses a divide and conquer approach to sort a collection of items. The collection is divided into two sub-collections – and the two sub-collections are sorted and combined into one list in such a way that the combined list is sorted. The algorithm is in general pseudo code below… Divide the collection into two sub-collections Quicksort the lower sub-collection Quicksort the upper sub-collection Combine the lower & upper sub-collection together As hinted at above, quicksort uses recursion in its implementation. The real trick with quicksort is to get the lower and upper sub-collections to be of equal size. The size of a sub-collection is determined by what value the pivot is. Once a pivot is determined, one would partition to sub-collections and then repeat the process on each sub collection until you reach the base case. With quicksort, the work is done when dividing the sub-collections into lower & upper collections. The actual combining of the lower & upper sub-collections at the end is relatively simple since every element in the lower sub-collection is smaller than the smallest element in the upper sub-collection. Mergesort With quicksort, the average-case complexity was O(nlog2n) however the worst case complexity was still O(N*N). Mergesort improves on quicksort by always having a complexity of O(nlog2n) regardless of the best or worst case. So how does it do this? Mergesort makes use of the divide and conquer approach to partition a collection into two sub-collections. It then sorts each sub-collection and combines the sorted sub-collections into one sorted collection. The general algorithm for mergesort is as follows… Divide the collection into two sub-collections Mergesort the first sub-collection Mergesort the second sub-collection Merge the first sub-collection and the second sub-collection As you can see.. it still pretty much looks like quicksort – so lets see where it differs… Firstly, mergesort differs from quicksort in how it partitions the sub-collections. Instead of having a pivot – merge sort partitions each sub-collection based on size so that the first and second sub-collection of relatively the same size. This dividing keeps getting repeated until the sub-collections are the size of a single element. If a sub-collection is one element in size – it is now sorted! So the trick is how do we put all these sub-collections together so that they maintain their sorted order. Sorted sub-collections are merged into a sorted collection by comparing the elements of the sub-collection and then adjusting the sorted collection. Lets have a look at a few examples… Assume 2 sub-collections with 1 element each 10 & 20 Compare the first element of the first sub-collection with the first element of the second sub-collection. Take the smallest of the two and place it as the first element in the sorted collection. In this scenario 10 is smaller than 20 so 10 is taken from sub-collection 1 leaving that sub-collection empty, which means by default the next smallest element is in sub-collection 2 (20). So the sorted collection would be 10 20 Lets assume 2 sub-collections with 2 elements each 10 20 & 15 19 So… again we would Compare 10 with 15 – 10 is the winner so we add it to our sorted collection (10) leaving us with 20 & 15 19 Compare 20 with 15 – 15 is the winner so we add it to our sorted collection (10 15) leaving us with 20 & 19 Compare 20 with 19 – 19 is the winner so we add it to our sorted collection (10 15 19) leaving us with 20 & _ 20 is by default the winner so our sorted collection is 10 15 19 20. Make sense? Heapsort (still needs to be completed) So by now I am tired of sorting algorithms and trying to remember why they were so important. I think every year I go through this stuff I wonder to myself why are we made to learn about selection sort and insertion sort if they are so bad – why didn’t we just skip to Mergesort & Quicksort. I guess the only explanation I have for this is that sometimes you learn things so that you can implement them in future – and other times you learn things so that you know it isn’t the best way of implementing things and that you don’t need to implement it in future. Anyhow… luckily this is going to be the last one of my sorts for today. The first step in heapsort is to convert a collection of data into a heap. After the data is converted into a heap, sorting begins… So what is the definition of a heap? If we have to convert a collection of data into a heap, how do we know when it is a heap and when it is not? The definition of a heap is as follows: A heap is a list in which each element contains a key, such that the key in the element at position k in the list is at least as large as the key in the element at position 2k +1 (if it exists) and 2k + 2 (if it exists). Does that make sense? At first glance I’m thinking what the heck??? But then after re-reading my notes I see that we are doing something different – up to now we have really looked at data as an array or sequential collection of data that we need to sort – a heap represents data in a slightly different way – although the data is stored in a sequential collection, for a sequential collection of data to be in a valid heap – it is “semi sorted”. Let me try and explain a bit further with an example… Example 1 of Potential Heap Data Assume we had a collection of numbers as follows 1[1] 2[2] 3[3] 4[4] 5[5] 6[6] For this to be a valid heap element with value of 1 at position [1] needs to be greater or equal to the element at position [3] (2k +1) and position [4] (2k +2). So in the above example, the collection of numbers is not in a valid heap. Example 2 of Potential Heap Data Lets look at another collection of numbers as follows 6[1] 5[2] 4[3] 3[4] 2[5] 1[6] Is this a valid heap? Well… element with the value 6 at position 1 must be greater or equal to the element at position [3] and position [4]. Is 6 > 4 and 6 > 3? Yes it is. Lets look at element 5 as position 2. It must be greater than the values at [4] & [5]. Is 5 > 3 and 5 > 2? Yes it is. If you continued to examine this second collection of data you would find that it is in a valid heap based on the definition of a heap.

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  • Making pascal's triangle with mpz_t's

    - by SDLFunTimes
    Hey, I'm trying to convert a function I wrote to generate an array of longs that respresents Pascal's triangles into a function that returns an array of mpz_t's. However with the following code: mpz_t* make_triangle(int rows, int* count) { //compute triangle size using 1 + 2 + 3 + ... n = n(n + 1) / 2 *count = (rows * (rows + 1)) / 2; mpz_t* triangle = malloc((*count) * sizeof(mpz_t)); //fill in first two rows mpz_t one; mpz_init(one); mpz_set_si(one, 1); triangle[0] = one; triangle[1] = one; triangle[2] = one; int nums_to_fill = 1; int position = 3; int last_row_pos; int r, i; for(r = 3; r <= rows; r++) { //left most side triangle[position] = one; position++; //inner numbers mpz_t new_num; mpz_init(new_num); last_row_pos = ((r - 1) * (r - 2)) / 2; for(i = 0; i < nums_to_fill; i++) { mpz_add(new_num, triangle[last_row_pos + i], triangle[last_row_pos + i + 1]); triangle[position] = new_num; mpz_clear(new_num); position++; } nums_to_fill++; //right most side triangle[position] = one; position++; } return triangle; } I'm getting errors saying: incompatible types in assignment for all lines where a position in the triangle is being set (i.e.: triangle[position] = one;). Does anyone know what I might be doing wrong?

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  • COCOS2D - Animation Stops During Move

    - by maiko
    Hey guys! I am trying to run a "Walk" style animation on my main game sprite. The animations work fine, and my sprite is hooked up to my joystick all fine and dandy. However, I think where I setup the call for my walk animations are wrong. Because everytime the sprite is moving, the animation stops. I know putting the animation in such a weak if statement is probably bad, but please tell me how I could get my sprite to animate properly while it is being moved by the joystick. The sprite faces the right direction, so I can tell the action's first frame is being called, however, it doesn't animate until I stop touching my joystick. Here is How I call the action: //WALK LEFT if (joypadCap.position.x <= 69 /*&& joypadCap.position.y < && joypadCap.position.y 40 */ ) { [tjSprite runAction:walkLeft]; }; //WALK RIGHT if ( joypadCap.position.x = 71 /* && joypadCap.position.y < 100 && joypadCap.position.y 40 */) { [tjSprite runAction:walkRight]; }; THIS: is the how the joystick controls the character: CGPoint newLocation = ccp(tjSprite.position.x - distance/8 * cosf(touchAngle), tjSprite.position.y - distance/8 * sinf(touchAngle)); tjSprite.position = newLocation; Please help. Any alternative ways to call the characters walk animation would be greatly appreciated!

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  • Positioning decorated series of div tags on screen using offset, DOM JQUERY RELATED

    - by Calibre2010
    Hi, I am using JQuery to position a series of div tags which basically use a class inside of the tag which decorates the divs as bars. So the div is a green box based on its css specifications to the glass. I have a list of STARTING postions, a list of left coordiantes- for the starting points I wish to position my DIV say 556, 560, 600 these automatically are generated as left positions in a list I have a list of ENDING postions, a list of left coordiantes- for the ending points I wish to position my DIV say 570, 590, 610 these automatically are generated as left positions in a list now for each start and end position, the bar(green box) i want to be drawn with its appropriate width as follows. so say f is the offset or position of the start and ff the offset or position of the end : Below draws the green box based on only one start and end position LEFT. if (f.left != 0) { $("#test").html($("<div>d</div>")).css({ position: 'absolute', left: (f.left) + "px", top: (f.top + 35) + "px", width: (ff.left - f.left) + 25 + "px" }).addClass("option1"); } I am looking to loop through the list of positons in the list and draw multiple green boxs based on the positions on the screen. The above code draws just one green box from the last offset position.

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