Choosing random numbers and unique serveral

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• Random numbers and generate funds

  Hello.

  I would like to create a very simple equation random addition.

  I have: 3 boxes of dynamic text (r1_txt, r2_txt, a1_txt)

  I want the first two random numbers and 'a1_txt' to display the sum of the first two random numbers.

  Everything is perfect to the point of adding the two numbers and drop the answer in a1_text. What should I do for my last line of code?

  Thanks for your help!

  random numbers
  1.
  function randomNumbers (min:Number, max: Number) {}
  var Results:Number = Math.floor (Math.random () * max) + min;
  return results;
  }

  2.
  new_mc.addEventListener (MouseEvent.CLICK, showRandomnumber);

  3.
  function showRandomnumber(event:MouseEvent):void {}
  r1_txt. Text = randomNumbers (1,100)
  r2_txt. Text = randomNumbers (1,100)
  a1_txt. Text = Number (r1_txt.text) + Number (r2_txt.text);
  }

  Your showRandomnumber() function has a few problems.  TextFields working with channels, so whatever it is assigned to a textfield must be a string value.  You should actually get errors for the first two lines.

  function showRandomnumber(event:MouseEvent):void {}
  r1_txt. Text = String (randomNumbers (1,100));
  r2_txt. Text = String (randomNumbers (1,100));
  a1_txt. Text = String (Number (r1_txt.text) + Number (r2_txt.text));
  }

  If you change which is not fix things, then look to your textfields to the problem

• 3 display unique and random numbers on 6

  The following program gives me the results in the Panel out because I used the trace.  The problem is I want these numbers when I click on a button.  The numbers do not again when the button is clicked, for example not "3, 3, 2" but "3, 2, 5".»  Here is the program that selects random numbers (which, incidentally, are not mine): (I'll give gladly $15 to the animal charity of your choice for the one who can give me a hand with this problem, thank you!)

  var mynumbers:Array is ['1', '2', '3', '4', '5', '6'];.

  var mystuff:Array = [];

  var randomCount:Number = 3;

  var r: Number;

  for (var i = 0; i < randomCount; i ++)
  {

  r = Math.floor (Math.random () * mynumbers.length);

  Mystuff [mystuff. Length] = mynumbers.splice (r, 1);

  }

  trace (mystuff);

  and here is my button code
  

  generate_btn.addEventListener (MouseEvent.CLICK, coco);

  //3.;
  function coco(event:MouseEvent):void
  {
  }

  Oops, sorry.

  use:

  var mynumbers:Array is ['1', '2', '3', '4', '5', '6'];.

  var randomCount:Number = 3;

  function shuffle(a:Array) {}
  var p:int;
  var t: *;
  var ivar:int;
  for (ivar =. Length-1; Ivar > = 0; Ivar-) {}
  p = Math.Floor ((Ivar+1) * Math.Random ());
  t = a [ivar];
  a [ivar] = a [p];
  a [p] = t;
  }
  }

  and here is my button code
  

  generate_btn.addEventListener (MouseEvent.CLICK, coco);

  //3.;
  function coco(event:MouseEvent):void
  {

  shuffle(mynumbers);
  

  for (var i: int = 0; i<>

  {if(!this["tf_"+i])}

  This ["tf_" + i] = new TextField();

  }

  This ["tf_" + i] .text = mynumbers [i];

  addChild (this ["tf_" + i]);

  This ["tf_" + i] there = i * 40;

  }

  }

  Thanks for the giveaway.  That's very generous of you.

• I'm trying to generate a table of random numbers 30. After each 5 readings a new vi must open and indicate to the user than 5 readings were made. and continue with the generation of the table again.

  because I don't have a sensor now, I am currently generating a table of random numbers 30. After each 5 readings a warning should be given to the user 5 readngs are completed. This cycle must be repeated. the size of the table is 30.

  Please help me, waiting for response as soon as possible.

  Once I have the transducer, I'll take 30 analog samples and then after each 5 smaples this wraning will be displayed din a new VI

  Use a while loop with a delay time representing your sampling interval.

  Use is equal to the count Terminal to see if 4, then 4th iteration = 5th sample.

  Use a box structure. The real deal will only run on the 4th iteration.

  In the case of true place a Subvi with your message of your choice in the front panel. Go to the properties of the VI window and set ' open the front panel when it is called.

  The condition to closing of attention is not given to your description.

  Consider that rather than usign a Subvi to do this, you can use the "dialog box one/two/three button" or "display message" live in the palette "user interface and dialogue."

  Please try it out and send your own VI. Do not provide us with a working solution.

  Kind regards

• How can I generate multiple unique random numbers?

  Hello

  I am trying to generate multiple random numbers between a given set of numbers (1-52) and do not have the same number generated twice within this group. I can compare the numbers of last and next with the function to compare, but how would I go about comparison of all numbers generated without using a huge list of shift registers...

  Any help/ideas are welcome and appreciated.

  Jason

  Solved my problem. IM passing the random number through a registry to offset to each case and build a table every time. I then searches the table for the new random number. If the number is not found I get a value of-1, another thing is an index value of 0-everything. If a comparator greater than (-1) indicates that the same number is in the table and then I can raise this matter until the same number is not found.

  Kind regards

  Jason

• I want just a bunch of random numbers (about 120 of them) from the list, select 'all' and get the computer to rearrange in CNC

  I try both notepad and wordpad. I'm in Winows XP 2003.  I want just a bunch of random numbers (about 120 of them) from the list, select "all" and reorganize in order digital computer. I can't understand how I've done it before. Google says select the balls feature, but all that is put a point in front of a certain number. There is no arrow down to select. No A - Z. No 1,2,3. what Miss me? is the Notepad or wordpad not the place to do that?

  Use Excel.

• Random generation of numbers and their use in mathematics

  I m doing a project and had an idea to do a math game thing where the numbers are generated randomly in both slots. For example, it is between 1 and 100 and the other the same and then add them up. The player will have to add them together and then type them in the reply box and if it's correct or not. I need help with the coding, as I'm new to CS5.

  I worked with a more ancient and outdated Flash version a little more than a year. So I know a little things General and some of the things that must be accomplished. However, ActionScript is fairly new, and I tried to use mathematics. Random because it seems that everyone uses in their tutorials. Yet, I couldn't make it work when playing with it. After you have created the random number, as I said before, I would like to be able to add/subtract/multiply/divide them to another number for an answer, but not to display the answer until a button is hit with the response users. Any help would be greatly appreciated.

  Thank you.

  If any other information is necessary, I can give that info.

  Assuming that your implementation of the stage via the Flash IDE, you can do 4 text boxes and give them instance names leftNumber, answerNumber and mathSymbol, rightNumber. Also create a button with an instance name someButton.

  Make sure that the first 3 fields (leftNumber, mathSymbol, rightNumber are dynamic and the answerNumber field is entered).

  Set your document class hand (file-> setting Actionscript-> Document class-> tpye in 'Hand'

  Now create a new as3 at the same level as your FLA and name it Main.as

  Here is an example of code that I did very briefly to your main class.

  Obviously, this has been done short hand and needs some changes, but it should help you get started (sorry for the [lack] of formatting).

  package

  {

  import flash.display.MovieClip;

  import flash.events.MouseEvent;

  SerializableAttribute public class Main extends MovieClip

  {

  private const HIGHEST_NUMBER:Number = 100;

  private var randomLeft:Number;

  private var randomRight:Number;

  private var randomOperator:String;

  private var correctAnswer:Number;

  public void Main (): void {}

  create random numbers

  randomLeft = Math.floor (Math.random () * HIGHEST_NUMBER) + 1;

  randomRight = Math.floor (Math.random () * HIGHEST_NUMBER) + 1;

  var operatorSwitch:Number = Math.floor (Math.random () * 3) + 1;

  If (operatorSwitch == 1) {randomOperator = '+';}          }

  Else if (operatorSwitch == 2) {randomOperator = "-";}          }

  else {randomOperator = 'x';          }

  correctAnswer = solveProblem();

  setupDisplay();

  This ['someButton'] .addEventListener (MouseEvent.CLICK, checkAnswer, false, 0, true);

  }

  private void solveProblem (): number {}

  If (randomOperator == '+') {return randomLeft + randomRight;}          }

  Else if (randomOperator == '-') {return randomLeft - randomRight;}          }

  else {return randomLeft * randomRight;}          }

  }

  private function setupDisplay (): void {}

  This ["leftNumber"] .text = String (randomLeft);

  This ["rightNumber"] .text = String (randomRight);

  This ["mathSymbol"] .text = String (randomOperator);

  This ["answerNumber"] .text = '? ';

  }

  private void checkAnswer(evt:MouseEvent):void {}

  If (Number (this ["answerNumber"] .text) == correctAnswer) {}

  trace ("YOU SMART if");

  } else {}

  trace ("TRY AGAIN");

  }

  }

  }

  }

• unique random numbers

  How to get the unique random numbers each time in AS3

  I searched but not able to get satisfactory results

  use:

  var tempArray:Array =]

  for (var i: uint = 0; i<>

  tempArray.push (i);

  }

  Shuffle (tempArray);

  You can now browse tempArray for random integers from 0 to moveXml.movie.length () - 1

  function shuffle(a:Array) {}
  var p:int;
  var t: *;
  var ivar:int;
  for (ivar =. Length-1; Ivar > = 0; Ivar-) {}
  p = Math.Floor ((Ivar+1) * Math.Random ());
  t = a [ivar];
  a [ivar] = a [p];
  a [p] = t;
  }
  }

• Random numbers to send SMS that I did not send.

  So, my phone has recently more than 70 messages sent to a large random number and text is long and in another language like Chinese or something. I never sent anything to a number like that. Also, I received an e-mail saying that my Apple ID was signed to another device which I have no gift too. Just today, my boyfriend's best friend got a text saying to click on a link and I wouldn't show pictures but in don't even have her number, I need my phone at all and he got the text of a number, we were not familiar with. He had my Facebook profile with her picture. Help, please!

  You will need change the password of your Apple ID, as well as all the other accounts where you use the same password, preferably to something unique for each account, since each of them using the same password for multiple accounts can endanger. If you think that your ID Apple has been compromised - Apple supports

• Random numbers appearing in messages

  I am running Firefox 4.0.1 on 64-bit Windows 7 and every time that I post on a forum or send a Gmail account, it adds random numbers at the end of my entry. I tried using Chrome and it doesn't do this, what's the problem?

  You paste content or add a signature?

  Start Firefox in Firefox to solve the issues in Safe Mode to check if one of the extensions of the origin of the problem (switch to the DEFAULT theme: Firefox (Tools) > Add-ons > appearance/themes).

  • Makes no changes on the start safe mode window.
  • https://support.Mozilla.com/kb/safe+mode
  • https://support.Mozilla.com/kb/troubleshooting+extensions+and+themes

• Medium-sized random numbers

  So what I need to do, is to generate a series of random numbers. I've got that.

  However, the tricky part is that I need to average the random number and the previous 3 numbers random and exit on a waveform graph.

  Any help or pointers would be great. Thank you

  Simple as pie. He called a VI means Pt by PT. set your buffer to 4 and it will be an average intensity absorbed with the last 3.

  

• How the random numbers will be generated

  The "dice" in LabVIEW function is ised to generate random numbers between 0 and 1. If I create a [100000] array with random numbers between 0 and 100, the appearance of 1 to 100 is the same (about 1000 times each), but the appearance of the 0 is only 500.

  So my question is, on the basis of which will be generated random numbers?

  Mitu salvation,

  It is not a problem of the RNG, it's a problem of your function rounded!

  "To U8" allows you to convert the entire random DBL. ToU8 rounds up to the next integer. So all the number of 0-0.5 will get rounded to 0, but all the numbers from 0.5 to 1.5 will get rounded to 1. If you have twice the range of a number rounded to 1 at the beach of rounding to zero - the same goes for your "end of range" with rounding to 100. 100 should be also less likelihood (in your VI) to appear...

  To get the same probability, you should (explicitly) roundpupils before converting in U8!

  BTW. You can also search the forum to get the same answer by searching for "random"...

• Division of random numbers, see if the result is an integer.

  Simple and yet impossible to find an easy way around it.

  I have 2 RANDOM numbers, I'm dividing them, and I want to see if the result is an integer or not. Basically, I need to know if the divison of random numbers 2 is an integer.

  Who wants KUDOS?

  /SPAW

  Use Quotient and remains, check if the remainder is 0.

• How LabVIEW can generate random numbers according to any pdf or histogram?

  I would like to know if there is a VI that generates a set of random numbers according to any probability density, the probability mass function or the graph histogram (this without explicitly knowing the function).

  Thank you.

  Acoustic wrote:

  I would like to know if there is a VI that generates a set of random numbers according to any probability density, the probability mass function or the graph histogram (this without explicitly knowing the function).

  If you have a given arbitrary distribution, you can simulate random data of thesholding mapping in the integral standard service with a simple random number 0... 1.

  This old post of mine should show you the General procedure and more details. (Is there even an former example). Of course, make sure that the desired function is all positive, of course.

  

  

• I receive a failure Audit Event Id 532 in the event of safety in numbers of Web servers.

  Hello

  I'm a domain administrator has recently left his job and his account has been disabled. Since I have disabled his account I get Failure Audit Event Id 532 in the event of safety in numbers of Web servers.

  Original event ID Title: Kerberos 532

  The event Id error on the Web server:

  Event type: Failure Audit
  Event source: security
  Event category: opening/closing session
  Event ID: 532
  Date: 10/07/2012
  Time: 14:38:12
  User: NT AUTHORITY\SYSTEM
  Computer: SERVERWEB2
  Description:
  Connection failure:
  Reason: The specified user account has expired
  User name:
  Domain:
  Logon type: 3
  Logon process: Authz
  Authentication package: Kerberos
  Workstation name: SERVERWEB2
  The name of the user calling: SERVERWEB2$
  Caller domain: DOMAIN name
  Caller logon ID: (0x0, 0x3E7)
  Calling process ID: 2532
  Transited Services: -.
  Source network address: -.
  Source port: -.

  At the same time, I get a DNS error in Netlogon.log on the same server:

  07/10 14:38:12 [SESSION] I_NetLogonGetAuthData called: (null) DOMAIN name (flags, 0x1)
  07/10 14:38:12 [MISC] DsGetDcName function called: Dom: DNS. DOMAIN.NAME Acct: (null) flags: DS RET_DNS
  07/10 14:38:12 [MISC] NetpDcGetName: DNS. DOMAIN.NAME using updated information in cache
  07/10 14:38:12 [MISC] DsGetDcName function returns 0: Dom: NOM_DOMAINE Acct: (null) flags: DS RET_DNS

  At the same time I get 4769 Failure Audit event IDs in the event of security in Active Directory:

  Log name: security
  Source: Microsoft-Windows-security-auditing
  Date: 10/07/2012 14:38:12
  Event ID: 4769
  Task category: Ticket to Service Kerberos Operations
  Level: Information
  Keywords: Audit failure
  User: n/a
  Computer: ActiveDirectory2.DNS.DOMAIN.NAME
  Description:
  A Kerberos service ticket has been requested.

  Account information:
  Account name: * address email is removed from the privacy *
  Account domain: DNS. DOMAIN.NAME
  Logon GUID: {00000000-0000-0000-0000-000000000000}

  Service Information:
  Service name: host/serverweb2.dns.domain.name
  Service ID: NULL SID

  Network information:
  Customer's address: 192.168.101.11
  Client port: 1681

  Additional information:
  Ticket options: 0 x 40810000
  Ticket encryption type: 0xffffffff
  Error code: 0 x 12
  Transited Services: -.

  This event is generated whenever access is requested to a resource such as a computer or a Windows service.  The name service indicates the resource to which access has been requested.

  This event can be correlated with the Windows login events by comparing fields GUID for session opening in each event.  The logon event occurs on the machine that was consulted, which is often a different machine than the domain controller that issued the service ticket.

  Options of ticket, the types of encryption and failure codes are defined in RFC 4120.
  The event XML:
  http://schemas.Microsoft.com/win/2004/08/events/event">
   
     
      4769
      0
      0
      14337
      0
      0 x 8010000000000000
     
      859551364
     
     
      Security
      ActiveDirectory2.dns.domain.name
     
   
   
      E-mail address is removed from the privacy *.
      DNS.domain.Name
      Host/serverweb2. DNS.domain.Name
      S 1-0-0
      0 x 40810000
      0xFFFFFFFF
      192.168.101.11
      1681
      0x12
      {00000000-0000-0000-0000-000000000000}
      -
   
  

  What I have so far:

  1. If I activate the user account of the former employee, it connects are deleted.

  2. deleted and joined the server from the domian, always I had questions.

  Any ideas please.

  Sikora

  For more information, see Help and Support Center at http://go.microsoft.com/fwlink/events.asp.

  Hi sarathchelika,

  You must post your question to the TechNet forums because it caters to an audience of it professionals.

  To do this, you must refer to the below mentioned link.

  http://social.technet.Microsoft.com/forums/en-us/categories/

  Hope this helps!