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  • Get to Know a Candidate (2 of 25): Merlin Miller–American Third Position Party

    - by Brian Lanham
    DISCLAIMER: This is not a post about “Romney” or “Obama”. This is not a post for whom I am voting. Information sourced for Wikipedia. Meet Merlin Miller of American Third Position Party In addition to being American Third Position Party nominee, Miller is an independent film maker.  He is a graduate of West Point and has commanded two units in the United States Army.  After military service he worked as an industrial engineering manager for Michelin Tire.  He learned about film making by earning an MFA from the University of Southern California. Mr. Miller is on the ballot in: CO, NJ, and TN. In the 2000’s, Miller began taking an increasingly paleoconservative political stance.  He claimed that Hollywood “ surreptitiously seeks to destroy our European-American heritage and our Christian-based traditional values, and replace them with values that debase these traditional values and elevate minorities as paragons of virtue and wisdom....Today’s motion pictures, in concert with other forms of mass media entertainment, are the greatest enemies to the well-being of our progeny and the future of our country.” Miller states that he "doesn't like" interracial marriage; however, he does not support outlawing interracial marriage, either.  Miller has denied being anti-Semitic, instead claiming that he merely opposes "favoritism" granted to Jews in the film industry.  He also opposes illegal immigration and what he refers to as "wide open borders" in the United States. The American Third Position Party (A3P) is a third positionist American political party which promotes white supremacy.  It was officially launched in January 2010 (although in November 2009 it filed papers to get on a ballot in California) partially to channel the right-wing populist resentment engendered by the financial crisis of 2007–2010 and the policies of the Obama administration and defines its principal mission as representing the political interests of white Americans. The party takes a strong stand against immigration and globalization, and strongly supports an anti-interventionist foreign policy. Although the party does not support labor unions, they do strongly support the labor rights of the American working class on a platform of placing American workers first over illegal immigrant workers and banning of overseas corporate relocation of American industry and technology Third Position or Third Alternative refers to a revolutionary nationalist political position that emphasizes its opposition to both communism and capitalism. Advocates of Third Position politics typically present themselves as "beyond left and right", instead claiming to syncretize radical ideas from both ends of the political spectrum Learn more about Merlin Miller and American Third Position Party on Wikipedia.

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  • Congratulations Nick Colebourn - Microsoft Certified Master

    - by Christian
    Congratulations to Nick Colebourn who was brave enough to take his MCM lab exam in Seattle during PASS last month (at very short notice!) and is now a Microsoft Certified Master in SQL Server! Nick’s momentous achievement is especially exciting for us as he’s now the 5th member of our team to achieve Microsoft’s highest technical qualification for SQL Server – Coeo now has more SQL Server MCM’s than any other Microsoft customer or partner in the WORLD! Thank you Nick, and congratulations; it’s well deserved and we’re all very proud of you!   Christian Bolton - MCA, MCM, MVP Technical Director http://coeo.com - SQL Server Consulting & Managed Services You can read more about the Certified Master program on Microsoft’s website here: http://bit.ly/aOFLxm

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  • Is Nick Clegg a man or a mouse?

    - by BizTalk Visionary
    Well we got the hung election so many of us wanted! I believe it really is time for electoral change. Why? Consider: the ConMen under Cameroon have polled 36% of the great British voting public – well those that got to vote!! That means 64% of us don’t want him as PM. So what gives him the right to govern? Well an ancient voting system ideal for two party politics. But for the last 30 years we’ve had multi-party politics and going forward we may see 4 or 5 parties stepping up. We have to set in place a system that makes this work! So what does that mean today: Nick has a golden chance to push forward the case and in fact the absolute right for the change. He needs to keep this in mind when he discusses coalition with both Labour and the ConMen. So the mouse approach: Decides it is only fair to side with the ‘biggest’ vote and team up with the ConMen. Chances of electoral change? Big fat zero. Chance of achieving any of his other targets. Big fat zero. Why? Simple (as the Meer Kat would say). Cameroon needs to become PM by hook or crook. Once PM he holds the whip hand. Labour will dump Brown and head off into Leadership race land, Clegg will be knocking on number 10, having meaningless meetings and seeing no reward. Finally while Labour is at 6‘s and 7’s  the ‘new’ PM will call a new election, gain the majority they need and dump luckless Nick!! So the man approach: Team up with Labour. As one of the conditions – Brown to go. Run referendum for PR. Get PR through then force Labour to have new election under PR. Nick now hero and should be in a much better place following a PR election!! The man bit is standing up to the media attack for supporting Labour. Come Nick – be a man for a better Britain!!

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  • Is Nick Clegg a man or a mouse?

    - by BizTalk Visionary
    Well we got the hung election so many of us wanted! I believe it really is time for electoral change. Why? Consider: the ConMen under Cameroon have polled 36% of the great British voting public – well those that got to vote!! That means 64% of us don’t want him as PM. So what gives him the right to govern? Well an ancient voting system ideal for two party politics. But for the last 30 years we’ve had multi-party politics and going forward we may see 4 or 5 parties stepping up. We have to set in place a system that makes this work! So what does that mean today: Nick has a golden chance to push forward the case and in fact the absolute right for the change. He needs to keep this in mind when he discusses coalition with both Labour and the ConMen. So the mouse approach: Decides it is only fair to side with the ‘biggest’ vote and team up with the ConMen. Chances of electoral change? Big fat zero. Chance of achieving any of his other targets. Big fat zero. Why? Simple (as the Meer Kat would say). Cameroon needs to become PM by hook or crook. Once PM he holds the whip hand. Labour will dump Brown and head off into Leadership race land, Glegg will be knocking on number 10, having meaningless meetings and seeing no reward. Finally while Labour is at 6‘s and 7’s  the ‘new’ PM will call a new election, gain the majority they need and dump luckless Nick!! So the man approach: Team up with Labour. As one of the conditions – Brown to go. Run referendum for PR. Get PR through then force Labour to have new election under PR. Nick now hero and should be in a much better place following a PR election!! The man bit is standing up to the media attack for supporting Labour. Come Nick – be a man for a better Britain!!

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  • 50 Years of LEDs: An Interview with Inventor Nick Holonyak [Video]

    - by Jason Fitzpatrick
    The man who powered on the first LED half a century ago is still around to talk about it; read on to watch an interview with LED inventor Nick Holonyak. The most fascinating thing about Holonyak’s journey to the invention of the LED was that he started off trying to build a laser and ended up inventing a super efficient light source: Holonyak got his PhD in 1954. In 1957, after a year at Bell Labs and a two year stint in the Army, he joined GE’s research lab in Syracuse, New York. GE was already exploring semiconductor applications and building the forerunners of modern diodes called thyristors and rectifiers. At a GE lab in Schenectady, the scientist Robert Hall was trying to build the first diode laser. Hall, Holonyak and others noticed that semiconductors emit radiation, including visible light, when electricity flows through them. Holonyak and Hall were trying to “turn them on,” and channel, focus and multiply the light. Hall was the first to succeed. He built the world’s first semiconductor laser. Without it, there would be no CD and DVD players today. “Nobody knew how to turn the semiconductor into the laser,” Holonyak says. “We arrived at the answer before anyone else.” But Hall’s laser emitted only invisible, infrared light. Holonyak spent more time in his lab, testing, cutting and polishing his hand-made semiconducting alloys. In the fall of 1962, he got first light. “People thought that alloys were rough and turgid and lumpy,” he says. “We knew damn well what happened and that we had a very powerful way of converting electrical current directly into light. We had the ultimate lamp.” How To Get a Better Wireless Signal and Reduce Wireless Network Interference How To Troubleshoot Internet Connection Problems 7 Ways To Free Up Hard Disk Space On Windows

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  • Why isn't this driver install working (sudo code)?

    - by Nick
    I have a soundcard that I'd like to use and I've been trying to install it and being a new Ubuntu user, I get about half way through this in the Terminal and it stops cooperating with me... See the link (soundcard hyperlink) but basically what I have here: I do the following and it works: sudo apt-get install subversion svn co https://line6linux.svn.sourceforge.net/svnroot/line6linux Change to the directory cd line6linux/driver/trunk Time to build from the source but first make sure you have the latest build and headers sudo apt-get install build-essential sudo apt-get install linux-headers Then after this point it says must specify file to install. Not sure how to do this or what it means. Then, running make gives the following output: ./set_revision.sh ./set_revision.sh: 9: test: https://line6linux.svn.sourceforge.net/svnroot/line6linux/driver/trunk: unexpected operator make -C /lib/modules/3.2.0-29-generic-pae/build CONFIG_LINE6_USB=m SUBDIRS=/home/nick/line6linux/driver/trunk modules make[1]: Entering directory /usr/src/linux-headers-3.2.0-29-generic-pae' CC [M] /home/nick/line6linux/driver/trunk/audio.o /home/nick/line6linux/driver/trunk/audio.c: In function ‘line6_init_audio’: /home/nick/line6linux/driver/trunk/audio.c:30:57: error: ‘THIS_MODULE’ undeclared (first use in this function) /home/nick/line6linux/driver/trunk/audio.c:30:57: note: each undeclared identifier is reported only once for each function it appears in make[2]: * [/home/nick/line6linux/driver/trunk/audio.o] Error 1 make[1]: * [module/home/nick/line6linux/driver/trunk] Error 2 make[1]: Leaving directory/usr/src/linux-headers-3.2.0-29-generic-pae' make: * [default] Error 2 This is in Ubuntu 12.04.1 LTS Another thing, semi related. Cut, copy, paste? Seems like it's different from program to program. I was in the terminal and hit Ctrl-C and then Ctrl-Shift-V in Firefox and it won't paste. But in terminal it will paste. I'm confused. Here is what it's giving me after I hit "Make": nick@NickUbuntu:~/line6linux/driver/trunk$ make ./set_revision.sh ./set_revision.sh: 9: test: https://line6linux.svn.sourceforge.net/svnroot/line6linux/driver/trunk: unexpected operator make -C /lib/modules/3.2.0-29-generic-pae/build CONFIG_LINE6_USB=m SUBDIRS=/home/nick/line6linux/driver/trunk modules make[1]: Entering directory /usr/src/linux-headers-3.2.0-29-generic-pae' CC [M] /home/nick/line6linux/driver/trunk/audio.o /home/nick/line6linux/driver/trunk/audio.c: In function ‘line6_init_audio’: /home/nick/line6linux/driver/trunk/audio.c:30:57: error: ‘THIS_MODULE’ undeclared (first use in this function) /home/nick/line6linux/driver/trunk/audio.c:30:57: note: each undeclared identifier is reported only once for each function it appears in make[2]: *** [/home/nick/line6linux/driver/trunk/audio.o] Error 1 make[1]: *** [_module_/home/nick/line6linux/driver/trunk] Error 2 make[1]: Leaving directory/usr/src/linux-headers-3.2.0-29-generic-pae' make: * [default] Error 2 Looks like these folks also had similar problems: http://ubuntuforums.org/showthread.php?t=1163608&page=3

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  • Accessing the same service more than twice in the nick of time

    - by PointedC
    I have an application that will access interface service A which is to run from windows startup. This service is used by program B and my application functions on B's presence after getting a pointer to A. The scenario is translated as follows, public interface A{} ///my program public class MyProgram { public MyProgram() { ProgramB.DoA(); } public A GetA(){} } public class ProgramB { void DoA(){} } The translated source is not true, but that seems to be what I am looking for. In order to eliminate the overhead of allocating and realocating dynamic accesses to the same service used by other processes, would you please provide an actual solution to the problem ?(I am all out of any idea now)

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  • Miller-rabin exception number?

    - by nightcracker
    Hey everyone. This question is about the number 169716931325235658326303. According to http://www.alpertron.com.ar/ECM.HTM it is prime. According to my miller-rabin implementation in python with 7 repetitions is is composite. With 50 repetitions it is still composite. With 5000 repetitions it is STILL composite. I thought, this might be a problem of my implementation. So I tried GNU MP bignum library, which has a miller-rabin primality test built-in. I tested with 1000000 repetitions. Still composite. This is my implementation of the miller-rabin primality test: def isprime(n, precision=7): if n == 1 or n % 2 == 0: return False elif n < 1: raise ValueError("Out of bounds, first argument must be > 0") d = n - 1 s = 0 while d % 2 == 0: d //= 2 s += 1 for repeat in range(precision): a = random.randrange(2, n - 2) x = pow(a, d, n) if x == 1 or x == n - 1: continue for r in range(s - 1): x = pow(x, 2, n) if x == 1: return False if x == n - 1: break else: return False return True And the code for the GMP test: #include <gmp.h> #include <stdio.h> int main(int argc, char* argv[]) { mpz_t test; mpz_init_set_str(test, "169716931325235658326303", 10); printf("%d\n", mpz_probab_prime_p(test, 1000000)); mpz_clear(test); return 0; } As far as I know there are no "exceptions" (which return false positives for any amount of repetitions) to the miller-rabin primality test. Have I stumpled upon one? Is my computer broken? Is the Elliptic Curve Method wrong? What is happening here? EDIT I found the issue, which is http://www.alpertron.com.ar/ECM.HTM. I trusted this applet, I'll contact the author his applet's implementation of the ECM is faulty for this number. Thanks. EDIT2 Hah, the shame! In the end it was something that went wrong with copy/pasting on my side. NOR the applet NOR the miller-rabin algorithm NOR my implementation NOR gmp's implementation of it is wrong, the only thing that's wrong is me. I'm sorry.

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  • The Latest Dish

    - by Oracle Staff
    Black Eyed Peas to Headline at Appreciation Event If you're coming to OpenWorld to fill up on the latest in IT solutions, be sure to save room for dessert. At the Oracle OpenWorld Appreciation Event, you'll be savoring the music of the world's hottest funk pop band, Black Eyed Peas, plus superstar rock legends Don Henley, of the Eagles, and Steve Miller. Save the date now: When: Wednesday, September 22, 8 p.m-12 a.m. Where: Treasure Island, San Francisco OpenWorld's annual thank-you event will be our most spectacular yet. Treasure Island, in the center of scenic San Francisco Bay, will once again serve as a rockin' oasis for Oracle customers and partners as they groove to the beat and enjoy delicious food, drinks, and festivities. Get all the details here.

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  • How to get the Jabber ID for a Multi User Chat nick

    - by Kutzi
    I'm trying to get the Jabber ID for a nick in a multi user chat, but the following code returns only null: class JabberMUCMessageListenerAdapter implements PacketListener { private final MultiUserChat muc; public JabberMUCMessageListenerAdapter(MultiUserChat muc) { this.muc = muc; } @Override public void processPacket(Packet p) { if (p instanceof Message) { final Message msg = (Message) p; String jid = muc.getOccupant(msg.getFrom()).getJid(); // returns null ... } } } Does anyone know, what I'm doing wrong?

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  • storing and retrieving socket

    - by Trevor Newhook
    From what I can understand, once I create a socket, I can then create an array to store it with userArray[socket.nickname]=socket; I can then send a message to it with: io.sockets.socket(userArray[data.to]).emit('private message', tstamp(), socket.nickname, message); The basic logic is to store a copy of each socket in an object, identified by nickname. When I want to send a message to that socket, I use the copy of the socket, and send the message via io.sockets.socket(id).emit(). The entire server code is below: io.sockets.on('connection', function (socket) { socket.on('user message', function (msg) { socket.broadcast.emit('user message', tstamp(), socket.nickname, msg); updateLog('user message', socket.nickname, msg); }); socket.on('private message', function(data) { socket.get(data.nickname, function (err, name) { console.log('Chat message by ', name); }); updateLog('private message', socket.nickname, data.message); message=data.message; io.sockets.socket(userArray[data.to]).emit('private message', tstamp(), socket.nickname, message); }); socket.on('get log', function () { updateLog(); // Ensure old entries are cleared out before sending it. io.sockets.emit('chat log', log); }); socket.on('nickname', function (nick, fn) { var i = 1; var orignick = nick; while (nicknames[nick]) { nick = orignick+i; i++; } fn(nick); nicknames[nick] = socket.nickname = nick; userArray[socket.nickname]=socket; socket.set('nickname', nick, function () { socket.emit('ready'); }); socket.broadcast.emit('announcement', nick + ' connected'); // io.sockets.socket(userArray[nick]).emit('newID', 'Your name is: ' + nick, '. Your ID is: '+ userArray[nick]); io.sockets.emit('nicknames', nicknames); });

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  • Linux Unable to Write to Directory Despite Permissions

    - by Nick Q.
    I'm trying to give myself permissions to /var/www/ however for some reason I am unable to do so. Currently what I'm facing is this: nick@server1:/var$ ls -l drwxrwxr-x 5 root wwwusers 232 Mar 15 19:31 www nick@server1:/var$ groups nick wwwusers nick@server1:/var$ mkdir www/trying mkdir: cannot create directory `www/trying': Permission denied I am running Ubuntu 10.04 LTS on a VPS and am used to running unix on my own machine so I may be doing something absolutely stupid, but I would like to be able to have the group wwwusers be able to write to www.

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  • Linux Unable to Write to Directory Despite Permissions

    - by Nick Q.
    I'm trying to give myself permissions to /var/www/ however for some reason I am unable to do so. Currently what I'm facing is this: nick@server1:/var$ ls -l drwxrwxr-x 5 root wwwusers 232 Mar 15 19:31 www nick@server1:/var$ groups nick wwwusers nick@server1:/var$ mkdir www/trying mkdir: cannot create directory `www/trying': Permission denied I am running Ubuntu 10.04 LTS on a VPS and am used to running unix on my own machine so I may be doing something absolutely stupid, but I would like to be able to have the group wwwusers be able to write to www.

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  • Still confused about JavaScript's 'this'.

    - by Nick Lowman
    I've been reading through quite a few articles on the 'this' keyword when using JavaScript objects and I'm still somewhat confused. I'm quite happy writing object orientated Javascript and I get around the 'this' issue by referring the full object path but I don't like the fact I still find 'this' confusing. I found a good answer here which helped me but I'm still not 100% sure. So, onto the example. The following script is linked from test.html with <script src="js/test.js"></script> if (!nick) { var nick = {}; } nick.lowman = function(){ var helloA = 'Hello A'; console.log('1.',this, this.helloA); var init = function(){ var helloB = 'Hello B'; console.log('2.',this, this.helloB); } return { init: init } }(); nick.lowman.init(); What kind of expected to see was 1. Object {} nick.lowman, 'Hello A' 2. Object {} init, 'Hello B' But what I get is this? 1. Window test.html, undefined 2. Object {} init, undefined I think I understand some of what's happening there but I would mind if someone out there explains it to me. Also, I'm not entirely sure why the first 'console.log' is being called at all? If I remove the call to the init function //nick.lowman.init() firebug still outputs 1. Window test.html, undefined. Why is that? Why does nick.lowman() get called by the window object when the html page loads? Many thanks

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  • A Nondeterministic Engine written in VB.NET 2010

    - by neil chen
    When I'm reading SICP (Structure and Interpretation of Computer Programs) recently, I'm very interested in the concept of an "Nondeterministic Algorithm". According to wikipedia:  In computer science, a nondeterministic algorithm is an algorithm with one or more choice points where multiple different continuations are possible, without any specification of which one will be taken. For example, here is an puzzle came from the SICP: Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment housethat contains only five floors. Baker does not live on the top floor. Cooper does not live onthe bottom floor. Fletcher does not live on either the top or the bottom floor. Miller lives ona higher floor than does Cooper. Smith does not live on a floor adjacent to Fletcher's.Fletcher does not live on a floor adjacent to Cooper's. Where does everyone live? After reading this I decided to build a simple nondeterministic calculation engine with .NET. The rough idea is that we can use an iterator to track each set of possible values of the parameters, and then we implement some logic inside the engine to automate the statemachine, so that we can try one combination of the values, then test it, and then move to the next. We also used a backtracking algorithm to go back when we are running out of choices at some point. Following is the core code of the engine itself: Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--Public Class NonDeterministicEngine Private _paramDict As New List(Of Tuple(Of String, IEnumerator)) 'Private _predicateDict As New List(Of Tuple(Of Func(Of Object, Boolean), IEnumerable(Of String))) Private _predicateDict As New List(Of Tuple(Of Object, IList(Of String))) Public Sub AddParam(ByVal name As String, ByVal values As IEnumerable) _paramDict.Add(New Tuple(Of String, IEnumerator)(name, values.GetEnumerator())) End Sub Public Sub AddRequire(ByVal predicate As Func(Of Object, Boolean), ByVal paramNames As IList(Of String)) CheckParamCount(1, paramNames) _predicateDict.Add(New Tuple(Of Object, IList(Of String))(predicate, paramNames)) End Sub Public Sub AddRequire(ByVal predicate As Func(Of Object, Object, Boolean), ByVal paramNames As IList(Of String)) CheckParamCount(2, paramNames) _predicateDict.Add(New Tuple(Of Object, IList(Of String))(predicate, paramNames)) End Sub Public Sub AddRequire(ByVal predicate As Func(Of Object, Object, Object, Boolean), ByVal paramNames As IList(Of String)) CheckParamCount(3, paramNames) _predicateDict.Add(New Tuple(Of Object, IList(Of String))(predicate, paramNames)) End Sub Public Sub AddRequire(ByVal predicate As Func(Of Object, Object, Object, Object, Boolean), ByVal paramNames As IList(Of String)) CheckParamCount(4, paramNames) _predicateDict.Add(New Tuple(Of Object, IList(Of String))(predicate, paramNames)) End Sub Public Sub AddRequire(ByVal predicate As Func(Of Object, Object, Object, Object, Object, Boolean), ByVal paramNames As IList(Of String)) CheckParamCount(5, paramNames) _predicateDict.Add(New Tuple(Of Object, IList(Of String))(predicate, paramNames)) End Sub Public Sub AddRequire(ByVal predicate As Func(Of Object, Object, Object, Object, Object, Object, Boolean), ByVal paramNames As IList(Of String)) CheckParamCount(6, paramNames) _predicateDict.Add(New Tuple(Of Object, IList(Of String))(predicate, paramNames)) End Sub Public Sub AddRequire(ByVal predicate As Func(Of Object, Object, Object, Object, Object, Object, Object, Boolean), ByVal paramNames As IList(Of String)) CheckParamCount(7, paramNames) _predicateDict.Add(New Tuple(Of Object, IList(Of String))(predicate, paramNames)) End Sub Public Sub AddRequire(ByVal predicate As Func(Of Object, Object, Object, Object, Object, Object, Object, Object, Boolean), ByVal paramNames As IList(Of String)) CheckParamCount(8, paramNames) _predicateDict.Add(New Tuple(Of Object, IList(Of String))(predicate, paramNames)) End Sub Sub CheckParamCount(ByVal count As Integer, ByVal paramNames As IList(Of String)) If paramNames.Count <> count Then Throw New Exception("Parameter count does not match.") End If End Sub Public Property IterationOver As Boolean Private _firstTime As Boolean = True Public ReadOnly Property Current As Dictionary(Of String, Object) Get If IterationOver Then Return Nothing Else Dim _nextResult = New Dictionary(Of String, Object) For Each item In _paramDict Dim iter = item.Item2 _nextResult.Add(item.Item1, iter.Current) Next Return _nextResult End If End Get End Property Function MoveNext() As Boolean If IterationOver Then Return False End If If _firstTime Then For Each item In _paramDict Dim iter = item.Item2 iter.MoveNext() Next _firstTime = False Return True Else Dim canMoveNext = False Dim iterIndex = _paramDict.Count - 1 canMoveNext = _paramDict(iterIndex).Item2.MoveNext If canMoveNext Then Return True End If Do While Not canMoveNext iterIndex = iterIndex - 1 If iterIndex = -1 Then Return False IterationOver = True End If canMoveNext = _paramDict(iterIndex).Item2.MoveNext If canMoveNext Then For i = iterIndex + 1 To _paramDict.Count - 1 Dim iter = _paramDict(i).Item2 iter.Reset() iter.MoveNext() Next Return True End If Loop End If End Function Function GetNextResult() As Dictionary(Of String, Object) While MoveNext() Dim result = Current If Satisfy(result) Then Return result End If End While Return Nothing End Function Function Satisfy(ByVal result As Dictionary(Of String, Object)) As Boolean For Each item In _predicateDict Dim pred = item.Item1 Select Case item.Item2.Count Case 1 Dim p1 = DirectCast(pred, Func(Of Object, Boolean)) Dim v1 = result(item.Item2(0)) If Not p1(v1) Then Return False End If Case 2 Dim p2 = DirectCast(pred, Func(Of Object, Object, Boolean)) Dim v1 = result(item.Item2(0)) Dim v2 = result(item.Item2(1)) If Not p2(v1, v2) Then Return False End If Case 3 Dim p3 = DirectCast(pred, Func(Of Object, Object, Object, Boolean)) Dim v1 = result(item.Item2(0)) Dim v2 = result(item.Item2(1)) Dim v3 = result(item.Item2(2)) If Not p3(v1, v2, v3) Then Return False End If Case 4 Dim p4 = DirectCast(pred, Func(Of Object, Object, Object, Object, Boolean)) Dim v1 = result(item.Item2(0)) Dim v2 = result(item.Item2(1)) Dim v3 = result(item.Item2(2)) Dim v4 = result(item.Item2(3)) If Not p4(v1, v2, v3, v4) Then Return False End If Case 5 Dim p5 = DirectCast(pred, Func(Of Object, Object, Object, Object, Object, Boolean)) Dim v1 = result(item.Item2(0)) Dim v2 = result(item.Item2(1)) Dim v3 = result(item.Item2(2)) Dim v4 = result(item.Item2(3)) Dim v5 = result(item.Item2(4)) If Not p5(v1, v2, v3, v4, v5) Then Return False End If Case 6 Dim p6 = DirectCast(pred, Func(Of Object, Object, Object, Object, Object, Object, Boolean)) Dim v1 = result(item.Item2(0)) Dim v2 = result(item.Item2(1)) Dim v3 = result(item.Item2(2)) Dim v4 = result(item.Item2(3)) Dim v5 = result(item.Item2(4)) Dim v6 = result(item.Item2(5)) If Not p6(v1, v2, v3, v4, v5, v6) Then Return False End If Case 7 Dim p7 = DirectCast(pred, Func(Of Object, Object, Object, Object, Object, Object, Object, Boolean)) Dim v1 = result(item.Item2(0)) Dim v2 = result(item.Item2(1)) Dim v3 = result(item.Item2(2)) Dim v4 = result(item.Item2(3)) Dim v5 = result(item.Item2(4)) Dim v6 = result(item.Item2(5)) Dim v7 = result(item.Item2(6)) If Not p7(v1, v2, v3, v4, v5, v6, v7) Then Return False End If Case 8 Dim p8 = DirectCast(pred, Func(Of Object, Object, Object, Object, Object, Object, Object, Object, Boolean)) Dim v1 = result(item.Item2(0)) Dim v2 = result(item.Item2(1)) Dim v3 = result(item.Item2(2)) Dim v4 = result(item.Item2(3)) Dim v5 = result(item.Item2(4)) Dim v6 = result(item.Item2(5)) Dim v7 = result(item.Item2(6)) Dim v8 = result(item.Item2(7)) If Not p8(v1, v2, v3, v4, v5, v6, v7, v8) Then Return False End If Case Else Throw New NotSupportedException End Select Next Return True End FunctionEnd Class    And now we can use the engine to solve the problem we mentioned above:   Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--Sub Test2() Dim engine = New NonDeterministicEngine() engine.AddParam("baker", {1, 2, 3, 4, 5}) engine.AddParam("cooper", {1, 2, 3, 4, 5}) engine.AddParam("fletcher", {1, 2, 3, 4, 5}) engine.AddParam("miller", {1, 2, 3, 4, 5}) engine.AddParam("smith", {1, 2, 3, 4, 5}) engine.AddRequire(Function(baker) As Boolean Return baker <> 5 End Function, {"baker"}) engine.AddRequire(Function(cooper) As Boolean Return cooper <> 1 End Function, {"cooper"}) engine.AddRequire(Function(fletcher) As Boolean Return fletcher <> 1 And fletcher <> 5 End Function, {"fletcher"}) engine.AddRequire(Function(miller, cooper) As Boolean 'Return miller = cooper + 1 Return miller > cooper End Function, {"miller", "cooper"}) engine.AddRequire(Function(smith, fletcher) As Boolean Return smith <> fletcher + 1 And smith <> fletcher - 1 End Function, {"smith", "fletcher"}) engine.AddRequire(Function(fletcher, cooper) As Boolean Return fletcher <> cooper + 1 And fletcher <> cooper - 1 End Function, {"fletcher", "cooper"}) engine.AddRequire(Function(a, b, c, d, e) As Boolean Return a <> b And a <> c And a <> d And a <> e And b <> c And b <> d And b <> e And c <> d And c <> e And d <> e End Function, {"baker", "cooper", "fletcher", "miller", "smith"}) Dim result = engine.GetNextResult() While Not result Is Nothing Console.WriteLine(String.Format("baker: {0}, cooper: {1}, fletcher: {2}, miller: {3}, smith: {4}", result("baker"), result("cooper"), result("fletcher"), result("miller"), result("smith"))) result = engine.GetNextResult() End While Console.WriteLine("Calculation ended.")End Sub   Also, this engine can solve the classic 8 queens puzzle and find out all 92 results for me.   Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--Sub Test3() ' The 8-Queens problem. Dim engine = New NonDeterministicEngine() ' Let's assume that a - h represents the queens in row 1 to 8, then we just need to find out the column number for each of them. engine.AddParam("a", {1, 2, 3, 4, 5, 6, 7, 8}) engine.AddParam("b", {1, 2, 3, 4, 5, 6, 7, 8}) engine.AddParam("c", {1, 2, 3, 4, 5, 6, 7, 8}) engine.AddParam("d", {1, 2, 3, 4, 5, 6, 7, 8}) engine.AddParam("e", {1, 2, 3, 4, 5, 6, 7, 8}) engine.AddParam("f", {1, 2, 3, 4, 5, 6, 7, 8}) engine.AddParam("g", {1, 2, 3, 4, 5, 6, 7, 8}) engine.AddParam("h", {1, 2, 3, 4, 5, 6, 7, 8}) Dim NotInTheSameDiagonalLine = Function(cols As IList) As Boolean For i = 0 To cols.Count - 2 For j = i + 1 To cols.Count - 1 If j - i = Math.Abs(cols(j) - cols(i)) Then Return False End If Next Next Return True End Function engine.AddRequire(Function(a, b, c, d, e, f, g, h) As Boolean Return a <> b AndAlso a <> c AndAlso a <> d AndAlso a <> e AndAlso a <> f AndAlso a <> g AndAlso a <> h AndAlso b <> c AndAlso b <> d AndAlso b <> e AndAlso b <> f AndAlso b <> g AndAlso b <> h AndAlso c <> d AndAlso c <> e AndAlso c <> f AndAlso c <> g AndAlso c <> h AndAlso d <> e AndAlso d <> f AndAlso d <> g AndAlso d <> h AndAlso e <> f AndAlso e <> g AndAlso e <> h AndAlso f <> g AndAlso f <> h AndAlso g <> h AndAlso NotInTheSameDiagonalLine({a, b, c, d, e, f, g, h}) End Function, {"a", "b", "c", "d", "e", "f", "g", "h"}) Dim result = engine.GetNextResult() While Not result Is Nothing Console.WriteLine("(1,{0}), (2,{1}), (3,{2}), (4,{3}), (5,{4}), (6,{5}), (7,{6}), (8,{7})", result("a"), result("b"), result("c"), result("d"), result("e"), result("f"), result("g"), result("h")) result = engine.GetNextResult() End While Console.WriteLine("Calculation ended.")End Sub (Chinese version of the post: http://www.cnblogs.com/RChen/archive/2010/05/17/1737587.html) Cheers,  

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  • Can the public ssh key from my local machine be used to access two different users on a remote serve

    - by Nick
    I have an new ubuntu (hardy 8.04) server, it has two users, User1 and User2. User1 is listed in sudoers. I appended my public ssh key (my local machine's public key local/Users/nick/.ssh/id_rsa.pub) to authorized_keys in remote_server/home/user1/.ssh/authorized_keys, changed the permissions on user1/.ssh/ to 700 and user1/.ssh/authorized_keys to 600 and both file and folder are owned my User1. Then added I User1 to sshd_config (AllowUsers User1). This works and I can login into User1 debug1: Offering public key: /Users/nick/.ssh/id_rsa debug1: Server accepts key: pkalg ssh-rsa blen 277 debug1: Authentication succeeded (publickey). debug1: channel 0: new [client-session] debug1: Entering interactive session. Last login: Mon Mar 15 09:51:01 2010 from ..*.* I then copied the authorized_keys file remote_server/home/user1/.ssh/authorized_keys to remote_server/home/user2/.shh/authorized_keys and changed the permissions and ownership and added User2 to AllowUsers in sshd_config (AllowUsers User1 User2). Now when I try to login to User2 it will not authenticate the same public key. debug1: Offering public key: /Users/nick/.ssh/id_rsa debug1: Authentications that can continue: publickey debug1: Trying private key: /Users/nick/.ssh/identity debug1: Trying private key: /Users/nick/.ssh/id_dsa debug1: No more authentication methods to try. Permission denied (publickey). Am I missing something fundamental about the way ssh works? Thanks in advance, Nick

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  • Active directory integration not working properly with winbind and samba

    - by tubaguy50035
    I'm trying to get my linux box to use active directory authentication. I believe I have almost everything setup correctly. I'm able to issue wbinfo -g and wbinfo -u and see all the groups and users respectively. Brief intro to my setup: The username I use on my linux box to do admin things is nick. My active directory username is nwalke. They have two different passwords. I am able to log in to the box with nick and that user's password and I'm also able to login as nwalke with nwalke's password. The curious bit: Upon creating the active directory user's home directory, I run a script that requires root access. This is to setup some system wide things like a samba share for them. When I log in as nwalke, I enter my nwalke password and it succeeds. I'm then greeted with [sudo] password for nick:. If I enter my nwalke password here, it says Sorry, try again.. If I enter nick's password, it says Sorry, user nick is not allowed to execute scriptname as root. If I do groups as nwalke, I see that magically my user has been given the group nick. Now, I accidentally thought that nick had a UID of 100, not 1000. So originally in my smb.conf I had idmap uid 1000-10000. The only thing I can think of, is that I logged in with nwalke while that was still set and now I'm just being presented with a UID of 1000 forcing linux to think I'm nick. I'm not really sure where to go from here. Like I said, I'm fairly certain active directory is communicating with my server properly, but something must not be mapped right on the linux side. Any thoughts? Here is my smb.conf: [global] security = ads netbios name = hostname realm = COMPANY.COM password server = adshost.company.com workgroup = COMPANY idmap uid = 10000-90000 idmap gid = 10000-90000 winbind separator = + winbind enum users = no winbind enum groups = no winbind use default domain = yes template homedir = /home/%D/%U template shell = /bin/bash client use spnego = yes domain master = no load printers = no printing = bsd printcap name = /dev/null disable spoolss = yes Let me know if more information about something is required.

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  • It is possible to record a data that have a straight row in mysql based on date or sequence?

    - by user1987816
    I want to get the data that have a straight Sell more than 3 times, it is possible in mysql? If not, how to get it right? I'm need it on mysql or php. my database:- +----------+---------------------+--------+ | Username | Date | Action | +----------+---------------------+--------+ | Adam | 2014-08-20 22:30:20 | Sell | | Adam | 2014-08-20 22:30:20 | Sell | | Adam | 2014-08-20 22:30:20 | Sell | | Adam | 2014-08-20 22:30:20 | Buy | | Adam | 2014-08-20 22:30:20 | Buy | | Adam | 2014-08-20 22:30:20 | Sell | | Adam | 2014-08-20 22:30:20 | Sell | | Adam | 2014-08-20 22:30:20 | Sell | | Adam | 2014-08-20 22:30:20 | Sell | | Nick | 2014-08-20 22:30:20 | Sell | | Nick | 2014-08-20 22:30:20 | Sell | | Nick | 2014-08-20 22:30:20 | Sell | | Nick | 2014-08-20 22:30:20 | Sell | | Nick | 2014-08-20 22:30:20 | Buy | +----------+---------------------+--------+ From the table above, I need to list out all data that have a straight sell more then 3 times. RESULT +----------+---------------------+--------+-------------+ | Username | Date | Action | Straight 3+ | +----------+---------------------+--------+-------------+ | Adam | 2014-08-20 22:30:20 | Sell | 3 | | Adam | 2014-08-20 22:30:20 | Sell | 4 | | Nick | 2014-08-20 22:30:20 | Sell | 4 | +----------+---------------------+--------+-------------+

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  • How properly perform passing operation result to View

    - by atomAltera
    I'm developing web site on self made MVC engine. I have actionController that handles operations like register, login, post submit and etc. actionController receives operation name and parameters. Of course it mast handle errors such user with same nick already exists or password is to short about which action handler have to notify user. The question is which is the best way to organize errors, such that View could easily get localized user notification message. I see two ways First one: define error constants like ERR_NICK_BUSY = '1' ERR_NICK_INVALID = '2' ... and localization map local[ERR_NICK_BUSY] = 'User with the same nick already registered' local[ERR_NICK_INVALID ] = 'Nick, you entered is invalid' ... And second one: define abstract constants like ERR_FIELD_BUSY = '1' ERR_FIELD_INVALID = '2' ... and pass them with field name. In this case localization looks like local['nick_'+ERR_FIELD_BUSY] = 'User with the same nick already registered' ... I don't like both this methods. Can you advise something else?

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  • Using the public ssh key from local machine to access two remote users [closed]

    - by Nick
    I have an new Ubuntu (Hardy 8.04) server; it has two users, Alice and Bob. Alice is listed in sudoers. I appended my public ssh key (my local machine's public key local/Users/nick/.ssh/id_rsa.pub) to authorized_keys in remote_server/home/Alice/.ssh/authorized_keys, changed the permissions on Alice/.ssh/ to 700 and Alice/.ssh/authorized_keys to 600, and both the file and folder are owned my Alice. Then added I Alice to sshd_config (AllowUsers Alice). This works and I can login into Alice: ssh -v [email protected] ... debug1: Offering public key: /Users/nick/.ssh/id_rsa debug1: Server accepts key: pkalg ssh-rsa blen 277 debug1: Authentication succeeded (publickey). debug1: channel 0: new [client-session] debug1: Entering interactive session. Last login: Mon Mar 15 09:51:01 2010 from 123.456.789.00 I then copied the authorized_keys file remote_server/home/Alice/.ssh/authorized_keys to remote_server/home/Bob/.shh/authorized_keys and changed the permissions and ownership and added Bob to AllowUsers in sshd_config (AllowUsers Alice Bob). Now when I try to login to Bob it will not authenticate the same public key. ssh -v [email protected] ... debug1: Offering public key: /Users/nick/.ssh/id_rsa debug1: Authentications that can continue: publickey debug1: Trying private key: /Users/nick/.ssh/identity debug1: Trying private key: /Users/nick/.ssh/id_dsa debug1: No more authentication methods to try. Permission denied (publickey). Am I missing something fundamental about the way ssh works?

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  • Can the same ssh key be used to access two different users on the same server?

    - by Nick
    I have an new ubuntu (hardy 8.04) server, it has two users, User1 and User2. User1 is listed in sudoers. I appended my public ssh key to authorized_keys in /home/user1/.ssh/authorized_keys, changed the permissions on user1/.ssh/ to 700 and user1/.ssh/authorized_keys to 600 and both file and folder are owned my User1. Then added I User1 to sshd_config (AllowUsers User1). This works and I can login into User1 debug1: Offering public key: /Users/nick/.ssh/id_rsa debug1: Server accepts key: pkalg ssh-rsa blen 277 debug1: Authentication succeeded (publickey). debug1: channel 0: new [client-session] debug1: Entering interactive session. Last login: Mon Mar 15 09:51:01 2010 from 86.141.61.197 I then copied the authorized_keys file to /home/user2/.shh/ and changed the permissions and ownership and added User2 to AllowUsers in sshd_config (AllowUsers User1 User2). Now when I try to login to User2 it will not authenticate the same public key. debug1: Offering public key: /Users/nick/.ssh/id_rsa debug1: Authentications that can continue: publickey debug1: Trying private key: /Users/nick/.ssh/identity debug1: Trying private key: /Users/nick/.ssh/id_dsa debug1: No more authentication methods to try. Permission denied (publickey). Am I missing something fundamental about the way ssh works? Thanks in advance, Nick

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