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  • two-part dice pool mechanic

    - by bythenumbers
    I'm working on a dice mechanic/resolution system based off of the Ghost/Echo (hereafter shortened to G/E) tabletop RPG. Specifically, since G/E can be a little harsh with dealing out consequences and failure, I was hoping to soften the system and add a little more player control, as well as offer the chance for players to evolve their characters into something unique, right from creation. So, here's the mechanic: Players roll 2d12 against the two statistics for their character (each is a number from 2-11, and may be rolled above or below depending on the nature of the action attempted, rolling your stat exactly always fails). Depending on the success for that roll, they add dice to the pool rolled for a modified G/E style action. The acting player gets two dice anyhow, and I am debating offering a bonus die for each success, or a single bonus die for succeeding on both of the statistic-compared rolls. One the size of the dice pool is set, the entire pool is rolled, and the players are allowed to assign rolled dice to a goal and a danger. Assigned results are judged as follows: 1-4 means the attempted goal fails, or the danger comes true. 5-8 is a partial success at the goal, or partially avoiding the danger. 9-12 means the goal is achieved, or the danger avoided. My concerns are twofold: Firstly, that the two-stage action is too complicated, with two rolls to judge separately before anything can happen. Secondly, that the statistics involved go too far in softening the game. I've run some basic simulations, and the approximate statistics follow: 2 dice (up to) 3 dice (up to) 4 dice failure ~33% ~25% ~20% partial ~33% ~35% ~35% success ~33% ~40% ~45% I'd appreciate any advice that addresses my concerns or offers to refine my simulation (right now the first roll is statistically modeled as sign(1d12-1d12), where 0 is a success).

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  • Dice face value recognition

    - by Jakob Gade
    I’m trying to build a simple application that will recognize the values of two 6-sided dice. I’m looking for some general pointers, or maybe even an open source project. The two dice will be black and white, with white and black pips respectively. Their distance to the camera will always be the same, but their position on the playing surface will be random. (not the best example, the surface will be a different color and the shadows will be gone) I have no prior experience with developing this kind of recognition software, but I would assume the trick is to first isolate the faces by searching for the square profile with a dominating white or black color (the rest of the image, i.e. the table/playing surface, will in distinctly different colors), and then isolate the pips for the count. Shadows will be eliminated by top down lighting. I’m hoping the described scenario is so simple (read: common) it may even be used as an “introductory exercise” for developers working on OCR technologies or similar computer vision challenges.

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  • Calculating odds distribution with 6-sided dice

    - by Stephen
    I'm trying to calculate the odds distribution of a changing number of 6-sided die rolls. For example, 3d6 ranges from 3 to 18 as follows: 3:1, 4:3, 5:6, 6:10, 7:15, 8:21, 9:25, 10:27, 11:27, 12:25, 13:21, 14:15, 15:10, 16:6, 17:3, 18:1 I wrote this php program to calculate it: function distributionCalc($numberDice,$sides=6) { for ( $i=0; $i<pow($sides,$numberDice); $i++) { $sum=0; for ($j=0; $j<$numberDice; $j++) { $sum+=(1+(floor($i/pow($sides,$j))) % $sides); } $distribution[$sum]++; } return $distribution; } The inner $j for-loop uses the magic of the floor and modulus functions to create a base-6 counting sequence with the number of digits being the number of dice, so 3d6 would count as: 111,112,113,114,115,116,121,122,123,124,125,126,131,etc. The function takes the sum of each, so it would read as: 3,4,5,6,7,8,4,5,6,7,8,9,5,etc. It plows through all 3^6 possible results and adds 1 to the corresponding slot in the $distribution array between 3 and 18. Pretty straightforward. However, it only works until about 8d6, afterward i get server time-outs because it's now doing billions of calculations. But I don't think it's necessary because die probability follows a sweet bell-curve distribution. I'm wondering if there's a way to skip the number crunching and go straight to the curve itself. Is there a way to do this, so, for example, with 80d6 (80-480)? Can the distribution be projected without doing 6^80 calculations? I'm not a professional coder and probability is still new to me, so thanks for all the help! Stephen

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  • 3D Dice using Maya - for integration with iOS game app

    - by Anil
    My designer is building a 3D design of a dice using Maya. I want to integrate this in my iOS app so that the user can spin the dice and get a number. Then they play the game using that number. So, I have two questions: 1) How can I make the dice spin and stop at a random position so that a number is presented to the user? and 2) Once it stops spinning how can I detect the number that is displayed to the user (programmatically)? Many thanks. -Anil

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  • Fair dice over network w/o trusted 3rd party

    - by Kay
    Though it should be a pretty basic problem, I did not find a solution for it: How to play dice over a network without a trusted third party? The M players shall roll N dice, one player after another. No player may "cheat", i.e. change the outcome to his advantage, or "look into the future" before the next roll. Is that possible? I guess the solution would be something like public key crypto, where each player turns in an encrypted message. After all messages were collected you exchange the keys to decode the messages. Then the sha1(joined string of all decrypted messages) mod 6 + 1 is used to determine the die. The major problem I have: since the message [c/s]hould be anything, I don't know how to prevent tampering with the private keys. Esp. the last player to turn in his key could easily cheat (I guess). The game should even stay fair, if all players "conspire" against one player.

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  • Throw of a dice in Java

    - by Arkapravo
    The throw of a dice is a popular program in Java, public class Dice { /* This program simulates rolling a dice */ public static void main(String[] args) { int dice; // The number on the dice. dice = (int) (Math.random() * 6 + 1); System.out.println (dice); } } What I wish to do is make it repeat, 500 times. I have not been able to put this program into a loop of 500. I usually program in Python, thus I guess my Java has rusted ! Any help is most welcome !

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  • Generate (in R) a matrix of all possible outcomes for throwing n dice (ignoring order)

    - by Brani
    In cases where order does matter, it's rather easy to generate the matrix of all possible outcomes. One way for doing this is using expand.grid as shown here. What if it doesn't? If I'm right, the number of possible combinations is (S+N-1)!/S!(N-1)!, where S is the number of dice, each with N sides numbered 1 through N. (It is different from the well known combinations formula because it is possible for the same number to appear on more than one dice). For example, when throwing four six-sided dice, N=6 and S=4, so the number of possible combinations is (4+6-1)!/4!(6-1)! = 9!/4!x5! = 126. How can I generate a matrix of these 126 possible outcomes? Thank you.

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  • Generate a matrix of all possible outcomes for throwing n dice (ignoring order)

    - by Brani
    In cases where order does matter, it's rather easy to generate the matrix of all possible outcomes. One way for doing this is using expand.grid as shown here. What if it doesn't? If I'm right, the number of possible combinations is (S+N-1)!/S!(N-1)!, where S is the number of dice, each with N sides numbered 1 through N. (It is different from the well known combinations formula because it is possible for the same number to appear on more than one dice). For example, when throwing four six-sided dice, N=6 and S=4, so the number of possible combinations is (4+6-1)!/4!(6-1)! = 9!/4!x5! = 126. How can I generate a matrix of these 126 possible outcomes? Thank you.

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  • Combinatorics, probability, dice

    - by TarGz
    A friend of mine asked: if I have two dice and I throw both of them, what is the most frequent sum (of the two dice' numbers)? I wrote a small script: from random import randrange d = dict((i, 0) for i in range(2, 13)) for i in xrange(100000): d[randrange(1, 7) + randrange(1, 7)] += 1 print d Which prints: 2: 2770, 3: 5547, 4: 8379, 5: 10972, 6: 13911, 7: 16610, 8: 14010, 9: 11138, 10: 8372, 11: 5545, 12: 2746 The question I have, why is 11 more frequent than 12? In both cases there is only one way (or two, if you count reverse too) how to get such sum (5 + 6, 6 + 6), so I expected the same probability..?

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  • Dice Emulation - ImageView

    - by Michelle Harris
    I am trying to emulate dice using ImageView. When I click the button, nothing seems to happen. I have hard coded this example to replace the image with imageView4 for debugging purposes (I was making sure the random wasn't fail). Can anyone point out what I am doing wrong? I am new to Java, Eclipse and Android so I'm sure I've probably made more than one mistake. Java: import java.util.Random; import android.app.Activity; import android.os.Bundle; import android.view.View; import android.widget.ArrayAdapter; import android.widget.ImageView; import android.widget.Spinner; import android.widget.Toast; public class Yahtzee4Activity extends Activity { /** Called when the activity is first created. */ @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); Spinner s = (Spinner) findViewById(R.id.spinner); ArrayAdapter adapter = ArrayAdapter.createFromResource( this, R.array.score_types, android.R.layout.simple_spinner_dropdown_item); adapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item); s.setAdapter(adapter); } public void onMyButtonClick(View view) { ImageView imageView1 = new ImageView(this); Random rand = new Random(); int rndInt = 4; //rand.nextInt(6) + 1; // n = the number of images, that start at index 1 String imgName = "die" + rndInt; int id = getResources().getIdentifier(imgName, "drawable", getPackageName()); imageView1.setImageResource(id); } } XML for the button: <Button android:id="@+id/button_roll" android:layout_width="wrap_content" android:layout_height="wrap_content" android:text="@string/roll" android:onClick="onMyButtonClick" />

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  • Why are these lines being skipped? (java)

    - by David
    Here's the relevant bit of the source code: class Dice { String name ; int x ; int[] sum ; ... public Dice (String name) { this.name = name ; this.x = 0 ; this.sum = new int[7] ; } ... public static void main (String[] arg) { Dice a1 = new Dice ("a1") ; printValues (a1) ; } public static void printDice (Dice Dice) { System.out.println (Dice.name) ; System.out.println ("value: "+Dice.x) ; printValues (Dice) ; } public static void printValues (Dice Dice) { for (int i = 0; i<Dice.sum.length; i++) System.out.println ("#of "+i+"'s: "+Dice.sum[i]) ; } } Here is the output: #of 0's: 0 #of 1's: 0 #of 2's: 0 #of 3's: 0 #of 4's: 0 #of 5's: 0 #of 6's: 0 Why didn't these two lines execute inside printDice: System.out.println (Dice.name) ; System.out.println ("value: "+Dice.x) ; if they had then i would expect to see "a1" and "Value: 0" printed at the top of the rows of #of's

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  • Algorithm for a dice problem

    - by vivekeviv
    I was thinking what should be the best algorithm for finding all the solutions of this puzzle. http://1cup1coffee.com/puzzle/endice/ Could backtracking be the an approach for solving this or can you suggest any other approach? Thanks

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  • recursion resulting in extra unwanted data

    - by spacerace
    I'm writing a module to handle dice rolling. Given x die of y sides, I'm trying to come up with a list of all potential roll combinations. This code assumes 3 die, each with 3 sides labeled 1, 2, and 3. (I realize I'm using "magic numbers" but this is just an attempt to simplify and get the base code working.) int[] set = { 1, 1, 1 }; list = diceroll.recurse(0,0, list, set); ... public ArrayList<Integer> recurse(int index, int i, ArrayList<Integer> list, int[] set){ if(index < 3){ // System.out.print("\n(looping on "+index+")\n"); for(int k=1;k<=3;k++){ // System.out.print("setting i"+index+" to "+k+" "); set[index] = k; dump(set); recurse(index+1, i, list, set); } } return list; } (dump() is a simple method to just display the contents of list[]. The variable i is not used at the moment.) What I'm attempting to do is increment a list[index] by one, stepping through the entire length of the list and incrementing as I go. This is my "best attempt" code. Here is the output: Bold output is what I'm looking for. I can't figure out how to get rid of the rest. (This is assuming three dice, each with 3 sides. Using recursion so I can scale it up to any x dice with y sides.) [1][1][1] [1][1][1] [1][1][1] [1][1][2] [1][1][3] [1][2][3] [1][2][1] [1][2][2] [1][2][3] [1][3][3] [1][3][1] [1][3][2] [1][3][3] [2][3][3] [2][1][3] [2][1][1] [2][1][2] [2][1][3] [2][2][3] [2][2][1] [2][2][2] [2][2][3] [2][3][3] [2][3][1] [2][3][2] [2][3][3] [3][3][3] [3][1][3] [3][1][1] [3][1][2] [3][1][3] [3][2][3] [3][2][1] [3][2][2] [3][2][3] [3][3][3] [3][3][1] [3][3][2] [3][3][3] I apologize for the formatting, best I could come up with. Any help would be greatly appreciated. (This method was actually stemmed to use the data for something quite trivial, but has turned into a personal challenge. :) edit: If there is another approach to solving this problem I'd be all ears, but I'd also like to solve my current problem and successfully use recursion for something useful.

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  • JSF - Random Number using Beans (JAVA)

    - by Alex Encore Tr
    I am trying to create a jsf application which, upon page refresh increments the hit counter and generates two random numbers. What should be displayed on the window may look something like this: On your On your roll x you have thrown x and x For this program I decided to create two Beans, one to hold the page refresh counter and one to generate a random number. Those look like this for the moment: CounterBean.java package diceroll; public class CounterBean { int count=0; public CounterBean() { } public void setCount(int count) { this.count=count; } public int getCount() { count++; return count; } } RandomNumberBean.java package diceroll; import java.util.Random; public class RandomNumberBean { int rand=0; Random r = new Random(); public RandomNumberBean() { rand = r.nextInt(6); } public void setNextInt(int rand) { this.rand=rand; } public int getNextInt() { return rand; } } I have then created an index.jsp to display the above message. <html> <%@ taglib uri="http://java.sun.com/jsf/core" prefix="f"%> <%@ taglib uri="http://java.sun.com/jsf/html" prefix="h"%> <f:view> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1"> <title>Roll the Dice</title> </head> <body> <h:form> <p> On your roll # <h:outputText value="#{CounterBean.count} " /> you have thrown <h:outputText value="#{RandomNumberBean.rand}" />and <h:outputText value="#{RandomNumberBean.rand} " /> </p> </h:form> </body> </f:view> </html> However, when I run the application, I get the following message: org.apache.jasper.el.JspPropertyNotFoundException: /index.jsp(14,20) '#{RandomNumberBean.rand}' Property 'rand' not found on type diceroll.RandomNumberBean Caused by: org.apache.jasper.el.JspPropertyNotFoundException - /index.jsp(14,20) '#{RandomNumberBean.rand}' Property 'rand' not found on type diceroll.RandomNumberBean I suppose there's a mistake with my faces-config.xml file, so I will post this here as well, see if somebody can provide some help: faces-config.xml <?xml version="1.0" encoding="UTF-8"?> <faces-config xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-facesconfig_2_0.xsd" version="2.0"> <managed-bean> <managed-bean-name>CounterBean</managed-bean-name> <managed-bean-class>diceroll.CounterBean</managed-bean-class> <managed-bean-scope>session</managed-bean-scope> </managed-bean> <managed-bean> <managed-bean-name>RandomNumberBean</managed-bean-name> <managed-bean-class>diceroll.RandomNumberBean</managed-bean-class> <managed-bean-scope>session</managed-bean-scope> </managed-bean> </faces-config>

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  • Will this ever result in a stack overflow error?

    - by David
    Will incrementing the instance variables of an object ever lead to a stack overflow error? For example: This method (java) will cause a stack overflow error: class StackOverflow { public static void StackOverflow (int x) { System.out.println (x) ; StackOverflow(x+1) ; } public static void main (String[]arg) { StackOverflow (0) ; } but will this?: (..... is a gap that i've put in to shorten the code. its long enough as it is.) import java.util.*; class Dice { String name ; int x ; int[] sum ; .... public Dice (String name) { this.name = name ; this.x = 0 ; this.sum = new int[7] ; } .... public static void main (String[] arg) { Dice a1 = new Dice ("a1") ; for (int i = 0; i<6000000; i++) { a1.roll () ; printDice(a1) ; } } .... public void roll () { this.x = randNum(1, this.sum.length) ; this.sum[x] ++ ; } public static int randNum (int a, int b) { Random random = new Random() ; int c = (b-a) ; int randomNumber = ((random.nextInt(c)) + a) ; return randomNumber ; } public static void printDice (Dice Dice) { System.out.println (Dice.name) ; System.out.println ("value: "+Dice.x) ; printValues (Dice) ; } public static void printValues (Dice Dice) { for (int i = 0; i<Dice.sum.length; i++) System.out.println ("#of "+i+"'s: "+Dice.sum[i]) ; } } The above doesn't currently cause a stack overflow error but could i get it too if i changed this line in main: for (int i = 0; i<6000000; i++) so that instead of 6 million something sufficiently high were there?

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  • null pointer exception comparing two strings in java.

    - by David
    I got this error message and I'm not quite sure whats wrong: Exception in thread "main" java.lang.NullPointerException at Risk.runTeams(Risk.java:384) at Risk.blobRunner(Risk.java:220) at Risk.genRunner(Risk.java:207) at Risk.main(Risk.java:176) Here is the relevant bits of code (i will draw attention to the line numbers within the error message via comments in the code as well as inputs i put into the program while its running where relevant) public class Risk { ... public static void main (String[]arg) { String CPUcolor = CPUcolor () ; genRunner (CPUcolor) ; //line 176 ... } ... public static void genRunner (String CPUcolor) // when this method runs i select 0 and run blob since its my only option. Theres nothing wrong with this method so long as i know, this is only significant because it takes me to blob runner and because another one of our relelvent line numbers apears. { String[] strats = new String[1] ; strats[0] = "0 - Blob" ; int s = chooseStrat (strats) ; if (s == 0) blobRunner (CPUcolor) ; // this is line 207 } ... public static void blobRunner (String CPUcolor) { System.out.println ("blob Runner") ; int turn = 0 ; boolean gameOver = false ; Dice other = new Dice ("other") ; Dice a1 = new Dice ("a1") ; Dice a2 = new Dice ("a2") ; Dice a3 = new Dice ("a3") ; Dice d1 = new Dice ("d1") ; Dice d2 = new Dice ("d2") ; space (5) ; Territory[] board = makeBoard() ; IdiceRoll (other) ; String[] colors = runTeams(CPUcolor) ; //this is line 220 Card[] deck = Card.createDeck () ; System.out.println (StratUtil.canTurnIn (deck)) ; while (gameOver == false) { idler (deck) ; board = assignTerri (board, colors) ; checkBoard (board, colors) ; } } ... public static String[] runTeams (String CPUcolor) { boolean z = false ; String[] a = new String[6] ; while (z == false) { a = assignTeams () ; printOrder (a) ; boolean CPU = false ; for (int i = 0; i<a.length; i++) { if (a[i].equals(CPUcolor)) CPU = true ; //this is line 384 } if (CPU==false) { System.out.println ("ERROR YOU NEED TO INCLUDE THE COLOR OF THE CPU IN THE TURN ORDER") ; runTeams (CPUcolor) ; } System.out.println ("is this turn order correct? (Y/N)") ; String s = getIns () ; while (!((s.equals ("y")) || (s.equals ("Y")) || (s.equals ("n")) || (s.equals ("N")))) { System.out.println ("try again") ; s = getIns () ; } if (s.equals ("y") || s.equals ("Y") ) z = true ; } return a ; } ... } // This } closes the class The reason i don't think i should be getting a Null:pointerException is because in this line: a[i].equals(CPUcolor) a at index i holds a string and CPUcolor is a string. Both at this point definatly have a value neither is null. Can anyone please tell me whats going wrong?

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  • 2 Shaders using the same vertex data

    - by Fonix
    So im having problems rendering using 2 different shaders. Im currently rendering shapes that represent dice, what i want is if the dice is selected by the user, it draws an outline by drawing the dice completely red and slightly scaled up, then render the proper dice over it. At the moment some of the dice, for some reason, render the wrong dice for the outline, but the right one for the proper foreground dice. Im wondering if they aren't getting their vertex data mixed up somehow. Im not sure if doing something like this is even allowed in openGL: glGenBuffers(1, &_vertexBuffer); glBindBuffer(GL_ARRAY_BUFFER, _vertexBuffer); glBufferData(GL_ARRAY_BUFFER, numVertices*sizeof(GLfloat), vertices, GL_STATIC_DRAW); glEnableVertexAttribArray(effect->vertCoord); glVertexAttribPointer(effect->vertCoord, 3, GL_FLOAT, GL_FALSE, 0, 0); glEnableVertexAttribArray(effect->toon_vertCoord); glVertexAttribPointer(effect->toon_vertCoord, 3, GL_FLOAT, GL_FALSE, 0, 0); im trying to bind the vertex data to 2 different shaders here when i load my first shader i have: vertCoord = glGetAttribLocation(TexAndLighting, "position"); and the other shader has: toon_vertCoord = glGetAttribLocation(Toon, "position"); if I use the shaders independently of each other they work fine, but when i try to render both one on top of the other they get the model mixed up some times. here is how my draw function looks: - (void) draw { [EAGLContext setCurrentContext:context]; glBindVertexArrayOES(_vertexArray); effect->modelViewMatrix = mvm; effect->numberColour = GLKVector4Make(numbers[colorSelected].r, numbers[colorSelected].g, numbers[colorSelected].b, 1); effect->faceColour = GLKVector4Make(faceColors[colorSelected].r, faceColors[colorSelected].g, faceColors[colorSelected].b, 1); if(selected){ [effect drawOutline]; //this function prepares the shader glDrawElements(GL_TRIANGLES, numIndices, GL_UNSIGNED_SHORT, 0); } [effect prepareToDraw]; //same with this one glDrawElements(GL_TRIANGLES, numIndices, GL_UNSIGNED_SHORT, 0); } this is what it looks like, as you can see most of the outlines are using the wrong dice, or none at all: links to full code: http://pastebin.com/yDKb3wrD Dice.mm //rendering stuff http://pastebin.com/eBK0pzrK Effects.mm //shader stuff http://pastebin.com/5LtDAk8J //my shaders, shouldn't be anything to do with them though TL;DR: trying to use 2 different shaders that use the same vertex data, but its getting the models mixed up when rendering using both at the same time, well thats what i think is going wrong, quite stumped actually.

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  • Multiple range product in Python

    - by Tyr
    Is there a better way to do this: perms = product(range(1,7),range(1,7),range(1,7)) so that I can choose how many ranges I use? I want it to be equivalent to this, but scalable. def dice(num) if num == 1: perms = ((i,) for i in range(1,7)) elif num == 2: perms = product(range(1,7),range(1,7)) elif num == 3: perms = product(range(1,7),range(1,7),range(1,7)) #... and so on but I know there has to be a better way. I'm using it for counting dice outcomes. The actual code def dice(selection= lambda d: d[2]): perms = itertools.product(range(1,7),range(1,7),range(1,7)) return collections.Counter(((selection(sorted(i)) for i in perms))) where I can call it with a variety of selectors, like sum(d[0:2]) for the sum of the lowest 2 dice or d[1] to get the middle dice.

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  • Agile Testing Days 2012 – My First Conference!

    - by Chris George
    I’d like to give you a bit of background first… so please bear with me! In 1996, whilst studying for my final year of my degree, I applied for a job as a C++ Developer at a small software house in Hertfordshire  After bodging up the technical part of the interview I didn’t get the job, but was offered a position as a QA Engineer instead. The role sounded intriguing and the pay was pretty good so in the absence of anything else I took it. Here began my career in the world of software testing! Back then, testing/QA was often an afterthought, something that was bolted on to the development process and very much a second class citizen. Test automation was rare, and tools were basic or non-existent! The internet was just starting to take off, and whilst there might have been testing communities and resources, we were certainly not exposed to any of them. After 8 years I moved to another small company, and again didn’t find myself exposed to any of the changes that were happening in the industry. It wasn’t until I joined Red Gate in 2008 that my view of testing and software development as a whole started to expand. But it took a further 4 years for my view of testing to be totally blown open, and so the story really begins… In May 2012 I was fortunate to land the role of Head of Test Engineering. Soon after, I received an email with details for the “Agile Testi However, in my new role, I decided that it was time to bite the bullet and at least go to one conference. Perhaps I could get some new ideas to supplement and support some of the ideas I already had.ng Days” conference in Potsdam, Germany. I looked over the suggested programme and some of the talks peeked my interest. For numerous reasons I’d shied away from attending conferences in the past, one of the main ones being that I didn’t see much benefit in attending loads of talks when I could just read about stuff like that on the internet. So, on the 18th November 2012, myself and three other Red Gaters boarded a plane at Heathrow bound for Potsdam, Germany to attend Agile Testing Days 2012. Tutorial Day – “Software Testing Reloaded” We chose to do the tutorials on the 19th, I chose the one titled “Software Testing Reloaded – So you wanna actually DO something? We’ve got just the workshop for you. Now with even less powerpoint!”. With such a concise and serious title I just had to see what it was about! I nervously entered the room to be greeted by tables, chairs etc all over the place, not set out and frankly in one hell of a mess! There were a few people in there playing a game with dice. Okaaaay… this is going to be a long day! Actually the dice game was an exercise in deduction and simplification… I found it very interesting and is certainly something I’ll be using at work as a training exercise! (I won’t explain the game here cause I don’t want to let the cat out of the bag…) The tutorial consisted of several games, exploring different aspects of testing. They were all practical yet required a fair amount of thin king. Matt Heusser and Pete Walen were running the tutorial, and presented it in a very relaxed and light-hearted manner. It was really my first experience of working in small teams with testers from very different backgrounds, and it was really enjoyable. Matt & Pete were very approachable and offered advice where required whilst still making you work for the answers! One of the tasks was to devise several strategies for testing some electronic dice. The premise was that a Vegas casino wanted to use the dice to appeal to the twenty-somethings interested in tech, but needed assurance that they were as reliable and random as traditional dice. This was a very interesting and challenging exercise that forced us to challenge various assumptions, determine/clarify requirements but most of all it was frustrating because the dice made a very very irritating beeping noise. Multiple that by at least 12 dice and I was dreaming about them all that night!! Some of the main takeaways that were brilliantly demonstrated through the games were not to make assumptions, challenge requirements, and have fun testing! The tutorial lasted the whole day, but to be honest the day went very quickly! My introduction into the conference experience started very well indeed, and I would talk to both Matt and Pete several times during the 4 days. Days 1,2 & 3 will be coming soon…  

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  • In this program(Java) I'm trying to make a dice roller. How do I make it so it rolls a bunch of times and adds the rolls?

    - by Mac
    import java.util.Random; public class dice { private int times; private int roll; private int side; Random roller = new Random(); public void setTimes(int sides) { times = sides; } public void setSides(int die) { side = die; } public int getRoll() //this is where the "rolling" happens { int total = 0; int c = 0; while (c <= times) { c = c + 1; int rol = 0; roll = roller.nextInt(side) + 1; rol = rol + roll; total = rol; } return total; } } If you need the GUIWindow and the main, just ask

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  • Code to generate random numbers in C++

    - by user1678927
    Basically I have to write a program to generate random numbers to simulate the rolling of a pair of dice. This program should be constructed in multiple files. The main function should be in one file, the other functions should be in a second source file, and their prototypes should be in a header file. First I write a short function that returns a random value between 1 and 6 to simulate the rolling of a single 6-sided die.Second, i write a function that pretends to roll a pair of dice by calling this function twice. My program starts by asking the user how many rolls should be made. Then I write a function to simulate rolling the dice this many times, keeping a count of exactly how many times the values 2,3,4,5,6,7,8,9,10,11,12(each number is the sum of a pair of dice) occur in an array. Later I write a function to display a small bar chart using these counts that ideally would look something like below for a sample of 144 rolls, where the number of asterisks printed corresponds to the count: 2 3 4 5 6 7 8 9 10 11 12 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Next, to see how well the random number generator is doing, I write a function to compute the average value rolled. Compare this to the ideal average of 7. Also, print out a small table showing the counts of each roll made by the program, the ideal count based on the frequencies above given the total number of rolls, and the difference between these values in separate columns. This is my incomplete code so far: "Compiler visual studio 2010" int rolling(){ //Function that returns a random value between 1 and 6 rand(unsigned(time(NULL))); int dice = 1 + (rand() %6); return dice; } int roll_dice(int num1,int num2){ //it calls 'rolling function' twice int result1,result2; num1 = rolling(); num2 = rolling(); result1 = num1; result2 = num2; return result1,result2; } int main(void){ int times,i,sum,n1,n2; int c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11;//counters for each sum printf("Please enter how many times you want to roll the dice.\n") scanf_s("%i",&times); I pretend to use counters to count each sum and store the number(the count) in an array. I know i need a loop (for) and some conditional statements (if) but m main problem is to get the values from roll_dice and store them in n1 and n2 so then i can sum them up and store the sum in 'sum'.

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  • Hot-hot-hot for jobs in Java and mobile software

    - by hinkmond
    It's hot-hot-hot! The market for Java and mobile developers keeps growing hotter, and hotter--so says the latest Dice survey. See: Dice survey says Java & Mobile are tops Here's a quote: The market for mobile developers is expanding faster than the talent pool can adapt, a Dice survey indicates. Software developers in general—as well as Java, mobile software and Microsoft .Net developers in particular—are in short supply today. Those fields represent four of the top five most difficult positions IT managers are looking to fill... ... The New York/New Jersey metro area led the country with 8,871 positions listed... So, if you are looking to get into software development get crackin' in learning Java mobile programming and move to NY or NJ. Let's go Mets! Hinkmond

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  • constructing a recursive function returning an array

    - by Admiral Kunkka
    I'm developing a function that has a random chance to loop through itself and put it's results in one array for me to use later in my PHP class. Is there a better way to do this more organized, specifically case 5. The array becomes sloppy if it rolls 5, after 5, after 5 looking unpleasant. private function dice($sides) { return mt_rand(1, $sides); } private function randomLoot($dice) { switch($dice) { case 1: $array[] = "A fancy mug."; break; case 2: $array[] = "A buckler."; break; case 3: $array[] = "A sword."; break; case 4: $array[] = "A jewel."; break; case 5: $array[] = "A treasure chest with contents:"; $count = $this->dice(3); $i = 1; while($i <= $count) { $array[] = $this->randomLoot($this->dice(5)); $i++; } break; } return $array; }

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  • Calculate the number of ways to roll a certain number

    - by helloworld
    I'm a high school Computer Science student, and today I was given a problem to: Program Description: There is a belief among dice players that in throwing three dice a ten is easier to get than a nine. Can you write a program that proves or disproves this belief? Have the computer compute all the possible ways three dice can be thrown: 1 + 1 + 1, 1 + 1 + 2, 1 + 1 + 3, etc. Add up each of these possibilities and see how many give nine as the result and how many give ten. If more give ten, then the belief is proven. I quickly worked out a brute force solution, as such int sum,tens,nines; tens=nines=0; for(int i=1;i<=6;i++){ for(int j=1;j<=6;j++){ for(int k=1;k<=6;k++){ sum=i+j+k; //Ternary operators are fun! tens+=((sum==10)?1:0); nines+=((sum==9)?1:0); } } } System.out.println("There are "+tens+" ways to roll a 10"); System.out.println("There are "+nines+" ways to roll a 9"); Which works just fine, and a brute force solution is what the teacher wanted us to do. However, it doesn't scale, and I am trying to find a way to make an algorithm that can calculate the number of ways to roll n dice to get a specific number. Therefore, I started generating the number of ways to get each sum with n dice. With 1 die, there is obviously 1 solution for each. I then calculated, through brute force, the combinations with 2 and 3 dice. These are for two: There are 1 ways to roll a 2 There are 2 ways to roll a 3 There are 3 ways to roll a 4 There are 4 ways to roll a 5 There are 5 ways to roll a 6 There are 6 ways to roll a 7 There are 5 ways to roll a 8 There are 4 ways to roll a 9 There are 3 ways to roll a 10 There are 2 ways to roll a 11 There are 1 ways to roll a 12 Which looks straightforward enough; it can be calculated with a simple linear absolute value function. But then things start getting trickier. With 3: There are 1 ways to roll a 3 There are 3 ways to roll a 4 There are 6 ways to roll a 5 There are 10 ways to roll a 6 There are 15 ways to roll a 7 There are 21 ways to roll a 8 There are 25 ways to roll a 9 There are 27 ways to roll a 10 There are 27 ways to roll a 11 There are 25 ways to roll a 12 There are 21 ways to roll a 13 There are 15 ways to roll a 14 There are 10 ways to roll a 15 There are 6 ways to roll a 16 There are 3 ways to roll a 17 There are 1 ways to roll a 18 So I look at that, and I think: Cool, Triangular numbers! However, then I notice those pesky 25s and 27s. So it's obviously not triangular numbers, but still some polynomial expansion, since it's symmetric. So I take to Google, and I come across this page that goes into some detail about how to do this with math. It is fairly easy(albeit long) to find this using repeated derivatives or expansion, but it would be much harder to program that for me. I didn't quite understand the second and third answers, since I have never encountered that notation or those concepts in my math studies before. Could someone please explain how I could write a program to do this, or explain the solutions given on that page, for my own understanding of combinatorics? EDIT: I'm looking for a mathematical way to solve this, that gives an exact theoretical number, not by simulating dice

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