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  • Java for loop with multiple incrementers

    - by user2517280
    Im writing a program which combines the RGB pixel values for 3 images, e.g. red pixel of image 1, green pixel of image 2 and blue pixel of image 3 and I want to then create a final image of it. Im using the code below, but this seems to be incrementing x2 and x3 whilst x1 is the same, i.e. not giving the right pixel value for same co-ordinate for each image. for (int x = 0; x < image.getWidth(); x++) { for (int x2 = 0; x2 < image2.getWidth(); x2++) { for (int x3 = 0; x3 < image3.getWidth(); x3++) { for (int y = 0; y < image.getHeight(); y++) { for (int y2 = 0; y2 < image2.getHeight(); y2++) { for (int y3 = 0; y3 < image3.getHeight(); y3++) { So I was wondering if anyone can tell me how to iterate through each of the 3 images on the same co-ordinate, so for example read 1, 1 of each image and record the red, green and blue value accordingly. Apologies if it doesnt make complete sense, its a bit hard to explain. I can iterate the values for one image fine but when I add in another, things start to go a bit wrong as obviously its quite a bit more complicated! I was thinking it might be easier to create an array and replace the according values in that just not sure how to do that effectively either. Thanks

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  • Help needed for writing a Set Based query for finding the highest marks obtained by the students

    - by priyanka.sarkar_2
    I have the below table declare @t table (id int identity, name varchar(50),sub1 int,sub2 int,sub3 int,sub4 int) insert into @t select 'name1',20,30,40,50 union all select 'name2',10,30,40,50 union all select 'name3',40,60,100,50 union all select 'name4',80,30,40,80 union all select 'name5',80,70,40,50 union all select 'name6',10,30,40,80 The desired output should be Id Name Sub1 Sub2 Sub3 Sub4 3 Name3 100 4 Name4 80 80 5 Name5 80 70 6 Name6 80 What I have done so far is ;with cteSub1 as ( select rn1 = dense_rank() over(order by sub1 desc),t.id,t.name,t.sub1 from @t t ) ,cteSub2 as ( select rn2 = dense_rank() over(order by sub2 desc),t.id,t.name,t.sub2 from @t t ) ,cteSub3 as ( select rn3 = dense_rank() over(order by sub3 desc),t.id,t.name,t.sub3 from @t t ) ,cteSub4 as ( select rn4 = dense_rank() over(order by sub4 desc),t.id,t.name,t.sub4 from @t t ) select x1.id,x2.id,x3.id,x4.id ,x1.sub1,x2.sub2,x3.sub3,x4.sub4 from (select c1.id,c1.sub1 from cteSub1 c1 where rn1 =1) as x1 full join (select c2.id,c2.sub2 from cteSub2 c2 where rn2 =1)x2 on x1.id = x2.id full join (select c3.id,c3.sub3 from cteSub3 c3 where rn3 =1)x3 on x1.id = x3.id full join (select c4.id,c4.sub4 from cteSub4 c4 where rn4 =1)x4 on x1.id = x4.id which is giving me the output as id id id id sub1 sub2 sub3 sub4 5 5 NULL NULL 80 70 NULL NULL 4 NULL NULL 4 80 NULL NULL 80 NULL NULL 3 NULL NULL NULL 100 NULL NULL NULL NULL 6 NULL NULL NULL 80 Help needed. Also how can I reduce the number of CTE's?

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  • Routing between 2 different subnets on 2 different interfaces in SonicOS

    - by Chris1499
    I'm having a bit of a problem allowing traffic between two of my subnets. Here's the structure I've built. The X0 interface has our windows server on it and it handles DHCP/DNS, etc. X1 has the WAN connection. The Sonicwall is handling DHCP on X2. The X3 interface is connected to a different vlan on the 48 port switch. The Sonicwall is handling DHCP on this network as well. So here's what i want to do. The network on X2 is for our guest wireless; i don't want it to be able to access any of the other networks, just the internet, so i that all blocked in the firewall. No issues there. The X3 network is going to be for programmable controllers, and needs to be able to access the X0 network where our computers are. This is where my problem is. I'm not able to get between the 192.168.2.xxx and the 192.168.1.xxx on interfaces X0 and X3 respectively. I have these rules set up in the firewall. The Lan Primary Subnet is the 192.168.2.0 on X0. So if i'm not mistaken, this will allow traffic between the two through the firewall. Now this is where I'm a little confused. Do i need to use NAT to get the traffic from X0 to go to X3 (and vice versa), or a static route, or both? Currently i have both, though i doubt they're done correctly (also in screenshot). I've tried to ping between the two without luck. Any advice, or if you see what's wrong with my setup, is much appreciated. If you need some more information, let me know. Thanks all! EDIT: So i found that i don't neither either NAT or a static route, that the setting in the firewall is enough. I can now ping from the 192.168.1.xxx network, however i can't access the server on the 192.168.2.xxx network. When i try to access i get "An error occured while reconnecting to Z: to server Microsoft Windows Network: The local device name is already in use. This connection has not been restored. What am i missing?

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  • how to print a char od struct()

    - by make
    Hi Could someone please tell us to print a char when receiving data as a truct? here is an EXP: ... struct rcv{ int x1; float x2; char x3; }; rcv data_rcv; ... if (recv(socket, &data_rcv, sizeof(data_rcv), 0) < 0) printf("recv() failed"); ... printf("x1 = %d\n", data_rcv.x1); printf("x2 = %f\n", data_rcv.x2); printf("x3 = %s\n", data_rcv.x3); // it doesn't print anything, why? ... Thanks for your replies-

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  • how to print a char from struct

    - by make
    Hi Could someone please tell us to print a char when receiving data as a struct? Here is an example: ... struct rcv{ int x1; float x2; char *x3; }; rcv data_rcv; ... if (recv(socket, &data_rcv, sizeof(data_rcv), 0) < 0) printf("recv() failed"); ... printf("x1 = %d\n", data_rcv.x1); printf("x2 = %f\n", data_rcv.x2); printf("x3 = %s\n", data_rcv.x3); // it doesn't print anything, why? ... Thanks for your replies-

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  • Oracle FLEXCUBE delivers 'Bank-in-a-Box' with Oracle Database Appliance

    - by margaret hamburger
    Another great example of how Oracle Database Appliance simplifies the deployment of high availability database solutions making it easy for Oracle Partners and ISVs to deliver value added solutions to customers on a simple, reliable and affordable database platform. Oracle FLEXCUBE Universal Banking recently announced that it runs on Oracle Database Appliance X3-2 to deliver mid-size banks a compelling banking-in-a-box solution. With this certification, banks can benefit from a low-IT-footprint, high-performance, full-scale banking technology that is engineered to support end-to-end business requirements. In a recent performance test of Oracle FLEXCUBE Universal Banking on Oracle Database Appliance X3-2, the system managed more than 2.6 million online transactions in 60 minutes. This equated to roughly 744 transactions per second with an average response time of 156 milliseconds for 98 percent of the transactions. Likewise, the solution completed end-of-month batch processing for 10 million customer accounts in 123 minutes during the performance test.  Learn more about Oracle Database Appliance Solution-in-a-Box.

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  • Engineered Systems Announcements at OOW

    - by Cinzia Mascanzoni
    Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Calibri","sans-serif"; mso-bidi-font-family:"Times New Roman";} See below the  announcements of new Exa systems at OOW Oracle Exadata X3 Database In-Memory Machine and  Oracle Exalogic Elastic Cloud X3-2 announcements.

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  • Database Insider - October 2012 issue

    - by Javier Puerta
    The October issue of the Database Insider newsletter is now available. (Full newsletter here) NEWS   Newly Launched Oracle Exadata X3 Redefines Extreme Performance At Oracle OpenWorld 2012, Oracle announced the general availability of Oracle Exadata Database Machine X3, a complete package of servers, storage, networking, and software that is massively scalable, secure, and fully redundant—and ideally suited for the varied and unpredictable workloads of cloud computing. Read More WEBCASTS What Are Oracle Users Doing to Improve Availability and Disaster Recovery? The Independent Oracle Users Group (IOUG) surveyed more than 350 data managers and professionals regarding planned and unplanned downtime, database high availability, and disaster recovery solutions. Download the report and watch the Webcast today.

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  • Multiple WAN interfaces on SonicWall TZ 100?

    - by Chad Decker
    I'm using a SonicWall TZ 100 with a basic configuration of X0 for the LAN and X1 for the WAN. The WAN uses DHCP to obtain its routable IP address. I want to obtain a second routable IP from my ISP. I'm in luck because my cable company will provide me with an additional dynamic IP for $5/mo. How do I bind this IP to my SonicWall? My additional dynamic IP will not be consecutive to the original one. It won't even be on the same class C. I think what I want to do is to use one of the empty ports/interfaces (X2, X3, or X4), tell that interface to use DHCP, and then add that interface to the WAN "zone". I can't figure out how to do this though. Here's what I've tried so far: (1) I've looked in Network Interfaces. I see X0 and X1 but the other unused interfaces don't show up. I don't see an "Add" button to add the new interfaces. (2) I've looked in Network Zones. I see that X0, X2, X3, X4 are in the LAN zone. I tried to drag X3 into the WAN zone but I can't. Nor does clicking the "Configure" button allow me to move an unused interface from LAN to WAN. (3) I've read the post entitled Splitting up multiple WAN's on Sonicwall. This doesn't seem applicable to me. Any thoughts?

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  • how to send classes defined in .proto (protocol-buffers) over a socket

    - by make
    Hi, I am trying to send a proto over a socket, but i am getting segmentation error. Could someone please help and tell me what is wrong with this example? file.proto message data{ required string x1 = 1; required uint32 x2 = 2; required float x3 = 3; } client.cpp ... // class defined in proto data data_snd; data data_rec; char *y1 = "operation1"; uint32_t y2 = 123 ; float y3 = 3.14; // assigning data to send() data_snd.set_x1(y1); data_snd.set_x2(y2); data_snd.set_x3(y3); //sending data to the server if (send(socket, &data_snd, sizeof(data_snd), 0) < 0) { cerr << "send() failed" ; exit(1); } //receiving data from the client if (recv(socket, &data_rec, sizeof(data_rec), 0) < 0) { cerr << "recv() failed"; exit(1); } //printing received data cout << data_rec.x1() << "\n"; cout << data_rec.x2() << "\n"; cout << data_rec.x3() << "\n"; ... server.cpp ... //receiving data from the client if (recv(socket, &data_rec, sizeof(data_rec), 0) < 0) { cerr << "recv() failed"; exit(1); } //printing received data cout << data_rec.x1() << "\n"; cout << data_rec.x2() << "\n"; cout << data_rec.x3() << "\n"; // assigning data to send() data_snd.set_x1(data_rec.x1()); data_snd.set_x2(data_rec.x2()); data_snd.set_x3(data_rec.x3()); //sending data to the server if (send(socket, &data_snd, sizeof(data_snd), 0) < 0) { cerr << "send() failed" ; exit(1); } ... Thanks for help and replies-

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  • How to calculate a measure of a total error in this clustering

    - by Vafello
    I have the following points and clustering of data S1. Can anyone tell me how to calculate the total error associated with this clustering? I know it's not a strictly programming question, but I need it for my algorithm. I think the answer should be 4/3 but I have no idea how to calculate this. Can anyone help me? x1= (2.0,1.0) x2= (2.0,2.0) x3= (1.0,2.0) S1={ x1, x2, x3 }

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  • C++ STL make_heap and pop_heap not working.

    - by Henrique
    I need to use a Heap, so i've searched about the STL one, but it doesn't seem to work, i wrote some code to explain what i mean: #include <stdio.h> #include <stdlib.h> #include <vector> #include <algorithm> struct data { int indice; int tamanho; }; bool comparator2(const data* a, const data* b) { return (a->tamanho < b->tamanho); } int main() { std::vector<data*> mesas; data x1, x2, x3, x4, x5; x1.indice = 1; x1.tamanho = 3; x2.indice = 2; x2.tamanho = 5; x3.indice = 3; x3.tamanho = 2; x4.indice = 4; x4.tamanho = 6; x5.indice = 5; x5.tamanho = 4; mesas.push_back(&x1); mesas.push_back(&x2); mesas.push_back(&x3); mesas.push_back(&x4); mesas.push_back(&x5); make_heap(mesas.begin(), mesas.end(), comparator2); for(int i = 0 ; i < 5 ; i++) { data* mesa = mesas.front(); pop_heap(mesas.begin(),mesas.end()); mesas.pop_back(); printf("%d, %d\n", mesa->indice, mesa->tamanho); } return 0; }; and this is what i get: 4, 6 2, 5 1, 3 3, 2 5, 4 So it's not working as a heap, as the maximum element on the vector is not being returned right. Or am i doing something wrong?

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  • Fastest pathfinding for static node matrix

    - by Sean Martin
    I'm programming a route finding routine in VB.NET for an online game I play, and I'm searching for the fastest route finding algorithm for my map type. The game takes place in space, with thousands of solar systems connected by jump gates. The game devs have provided a DB dump containing a list of every system and the systems it can jump to. The map isn't quite a node tree, since some branches can jump to other branches - more of a matrix. What I need is a fast pathfinding algorithm. I have already implemented an A* routine and a Dijkstra's, both find the best path but are too slow for my purposes - a search that considers about 5000 nodes takes over 20 seconds to compute. A similar program on a website can do the same search in less than a second. This website claims to use D*, which I have looked into. That algorithm seems more appropriate for dynamic maps rather than one that does not change - unless I misunderstand it's premise. So is there something faster I can use for a map that is not your typical tile/polygon base? GBFS? Perhaps a DFS? Or have I likely got some problem with my A* - maybe poorly chosen heuristics or movement cost? Currently my movement cost is the length of the jump (the DB dump has solar system coordinates as well), and the heuristic is a quick euclidean calculation from the node to the goal. In case anyone has some optimizations for my A*, here is the routine that consumes about 60% of my processing time, according to my profiler. The coordinateData table contains a list of every system's coordinates, and neighborNode.distance is the distance of the jump. Private Function findDistance(ByVal startSystem As Integer, ByVal endSystem As Integer) As Integer 'hCount += 1 'If hCount Mod 0 = 0 Then 'Return hCache 'End If 'Initialize variables to be filled Dim x1, x2, y1, y2, z1, z2 As Integer 'LINQ queries for solar system data Dim systemFromData = From result In jumpDataDB.coordinateDatas Where result.systemId = startSystem Select result.x, result.y, result.z Dim systemToData = From result In jumpDataDB.coordinateDatas Where result.systemId = endSystem Select result.x, result.y, result.z 'LINQ execute 'Fill variables with solar system data for from and to system For Each solarSystem In systemFromData x1 = (solarSystem.x) y1 = (solarSystem.y) z1 = (solarSystem.z) Next For Each solarSystem In systemToData x2 = (solarSystem.x) y2 = (solarSystem.y) z2 = (solarSystem.z) Next Dim x3 = Math.Abs(x1 - x2) Dim y3 = Math.Abs(y1 - y2) Dim z3 = Math.Abs(z1 - z2) 'Calculate distance and round 'Dim distance = Math.Round(Math.Sqrt(Math.Abs((x1 - x2) ^ 2) + Math.Abs((y1 - y2) ^ 2) + Math.Abs((z1 - z2) ^ 2))) Dim distance = firstConstant * Math.Min(secondConstant * (x3 + y3 + z3), Math.Max(x3, Math.Max(y3, z3))) 'Dim distance = Math.Abs(x1 - x2) + Math.Abs(z1 - z2) + Math.Abs(y1 - y2) 'hCache = distance Return distance End Function And the main loop, the other 30% 'Begin search While openList.Count() != 0 'Set current system and move node to closed currentNode = lowestF() move(currentNode.id) For Each neighborNode In neighborNodes If Not onList(neighborNode.toSystem, 0) Then If Not onList(neighborNode.toSystem, 1) Then Dim newNode As New nodeData() newNode.id = neighborNode.toSystem newNode.parent = currentNode.id newNode.g = currentNode.g + neighborNode.distance newNode.h = findDistance(newNode.id, endSystem) newNode.f = newNode.g + newNode.h newNode.security = neighborNode.security openList.Add(newNode) shortOpenList(OLindex) = newNode.id OLindex += 1 Else Dim proposedG As Integer = currentNode.g + neighborNode.distance If proposedG < gValue(neighborNode.toSystem) Then changeParent(neighborNode.toSystem, currentNode.id, proposedG) End If End If End If Next 'Check to see if done If currentNode.id = endSystem Then Exit While End If End While If clarification is needed on my spaghetti code, I'll try to explain.

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  • Multiple column sorting (SQL SERVER 2005)

    - by Newbie
    I have a table which looks like Col1 col2 col3 col4 col5 1 5 1 4 6 1 4 0 3 7 0 1 5 6 3 1 8 2 1 5 4 3 2 1 4 The script is declare @t table(col1 int, col2 int, col3 int,col4 int,col5 int) insert into @t select 1,5,1,4,6 union all select 1,4,0,3,7 union all select 0,1,5,6,3 union all select 1,8,2,1,5 union all select 4,3,2,1,4 I want the output to be every column being sorted in ascending order i.e. Col1 col2 col3 col4 col5 0 1 0 1 3 1 3 1 1 4 1 4 2 3 5 1 5 2 4 6 4 8 5 6 7 I already solved tye problem by the folowing program Select x1.col1 ,x2.col2 ,x3.col3 ,x4.col4 ,x5.col5 From (Select Row_Number() Over(Order By col1) rn1, col1 From @t)x1 Join(Select Row_Number() Over(Order By col2) rn2, col2 From @t)x2 On x1.rn1=x2.rn2 Join(Select Row_Number() Over(Order By col3) rn3, col3 From @t)x3 On x1.rn1=x3.rn3 Join(Select Row_Number() Over(Order By col4) rn4, col4 From @t)x4 On x1.rn1=x4.rn4 Join(Select Row_Number() Over(Order By col5) rn5, col5 From @t)x5 On x1.rn1=x5.rn5 But I am not happy with this solution. Is there any better way to achieve the same? (Using set based approach) If so, could any one please show an example. Thanks

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  • Printer Properties not sending the correct page size to program

    - by Jeff
    We have a Zebra 2844 Label Printer with 4"x3" labels in it. When we check the page settings in Windows XP, it is set to the correct size. When we go to print out of any program, it prints 4"x11". I have checked the page settings in one of the programs that we are using and it does show correctly the 4"x3" settings but prints at the incorrect 4"x11". I have followed the steps here Paper size: Printer Properties vs Page Setup and here Is there a way to programatically set the printer properties in windows? but have not been able to glean enough information to figure it out. If anyone can help, I would appreciate it!

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  • Big Data Appliance X4-2 Release Announcement

    - by Jean-Pierre Dijcks
    Today we are announcing the release of the 3rd generation Big Data Appliance. Read the Press Release here. Software Focus The focus for this 3rd generation of Big Data Appliance is: Comprehensive and Open - Big Data Appliance now includes all Cloudera Software, including Back-up and Disaster Recovery (BDR), Search, Impala, Navigator as well as the previously included components (like CDH, HBase and Cloudera Manager) and Oracle NoSQL Database (CE or EE). Lower TCO then DIY Hadoop Systems Simplified Operations while providing an open platform for the organization Comprehensive security including the new Audit Vault and Database Firewall software, Apache Sentry and Kerberos configured out-of-the-box Hardware Update A good place to start is to quickly review the hardware differences (no price changes!). On a per node basis the following is a comparison between old and new (X3-2) hardware: Big Data Appliance X3-2 Big Data Appliance X4-2 CPU 2 x 8-Core Intel® Xeon® E5-2660 (2.2 GHz) 2 x 8-Core Intel® Xeon® E5-2650 V2 (2.6 GHz) Memory 64GB 64GB Disk 12 x 3TB High Capacity SAS 12 x 4TB High Capacity SAS InfiniBand 40Gb/sec 40Gb/sec Ethernet 10Gb/sec 10Gb/sec For all the details on the environmentals and other useful information, review the data sheet for Big Data Appliance X4-2. The larger disks give BDA X4-2 33% more capacity over the previous generation while adding faster CPUs. Memory for BDA is expandable to 512 GB per node and can be done on a per-node basis, for example for NameNodes or for HBase region servers, or for NoSQL Database nodes. Software Details More details in terms of software and the current versions (note BDA follows a three monthly update cycle for Cloudera and other software): Big Data Appliance 2.2 Software Stack Big Data Appliance 2.3 Software Stack Linux Oracle Linux 5.8 with UEK 1 Oracle Linux 6.4 with UEK 2 JDK JDK 6 JDK 7 Cloudera CDH CDH 4.3 CDH 4.4 Cloudera Manager CM 4.6 CM 4.7 And like we said at the beginning it is important to understand that all other Cloudera components are now included in the price of Oracle Big Data Appliance. They are fully supported by Oracle and available for all BDA customers. For more information: Big Data Appliance Data Sheet Big Data Connectors Data Sheet Oracle NoSQL Database Data Sheet (CE | EE) Oracle Advanced Analytics Data Sheet

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  • Java - Using Linear Coordinates to Check Against AI [closed]

    - by Oliver Jones
    I'm working on some artificial intelligence, and I want my AI not to run into given coordinates as these are references of a wall/boundary. To begin with, every time my AI hits a wall, it makes a reference to that position (x,y). When it hits the same wall three times, it uses linear check points to 'imagine' there is a wall going through these coordinates. I want to now prevent my AI from going into that wall again. To detect if my coordinates make a straight line, i use: private boolean collinear(double x1, double y1, double x2, double y2, double x3, double y3) { return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2); } This returns true is the given points are linear to one another. So my problems are: How do I determine whether my robot is approaching the wall from its current trajectory? Instead of Java 'imagining' theres a line from 1, to 3. But to 'imagine' a line all the way through these linear coordinantes, until infinity (or close). I have a feeling this is going to require some confusing trigonometry? (REPOST: http://stackoverflow.com/questions/13542592/java-using-linear-coordinates-to-check-against-ai)

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  • Oracle Day 2012

    - by Mark Hesse
    Normal.dotm 0 0 1 133 760 Sun Microsystems 6 1 933 12.0 0 false 18 pt 18 pt 0 0 false false false /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-ascii-font-family:Cambria; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Cambria; mso-hansi-theme-font:minor-latin;} As a keynote speaker at this year’s Oracle Day 2012, “Your Vision, Engineered” I had the honor and pleasure of speaking to a crowd of about 150 attendees about our recently released, fourth generation Exadata X3 In-Memory Machine in a presentation entitled “Oracle Exadata X3 - Transforming Data Management”. The general theme of the thirty-minute talk was how to improve performance, lower costs, and build the foundation for your cloud service platform using Exadata. Since its introduction in 2008, I’ve watched first-hand as Exadata has evolved from a data warehouse-only system to an OLTP and DW in-memory database machine capable of storing hundreds of terabytes of compressed user data in flash and main memory.  Many of my Exadata customers are now purchasing additional systems as they continue to standardize Oracle 11g deployments on the best database platform available.

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  • Database Insider - December 2012 issue

    - by Javier Puerta
    The December issue of the Database Insider newsletter is now available. (Full newsletter here) Big Data: From Acquisition to Analysis 2012 will likely be remembered as the year of big data, as a new generation of technologies enables organizations to acquire, organize, and analyze the exponentially growing and typically less-structured data generated from a variety of new sources. Oracle has produced a series of five short videos that offer a quick and compelling high-level introduction to big data. Read More Total Cost of Ownership Comparison: Oracle Exadata vs. IBM P-Series Read the research that found that over three years, the IBM hardware running Oracle Database cost 31 percent more in total cost of ownership than Oracle Exadata. Webcast - Oracle Exadata Database Machine X3 Learn about Oracle’s next-generation database machine, Oracle Exadata X3, that combines massive memory and low-cost disks to deliver the highest performance at the lowest cost. Available in an eight-rack configuration, it allows you to start small and grow.    Maximum Availability with Oracle GoldenGate Discover how to eliminate not only unplanned downtime but also planned downtime resulting from database upgrades, migrations, and consolidation.Thursday, December 1319:00 CET / 6 pm. UK   

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  • PRUEBAS DE ESPECIALIZACION 2013/2014

    - by agallego
    Consigue  tu Certificado de Especialista Oracle  de forma GRATUITA , 27 y 28 de Noviembre de 2013  Ahora puedes realizar los exámenes de implementación de las especializaciones de Oracle y convertirte en especialista. Podrás realizar cualquiera de los exámenes de implementación de la siguiente lista: Oracle Fusion Customer Relationship Management 11g Sales Certified Implementation Specialist (1Z0-456) Oracle Fusion Customer Relationship Management 11g Incentive Compensation Certified Implementation Specialist (1Z0-472) Oracle ATG Web Commerce 10 Implementation Developer Certified Implementation Specialist (1Z0-510) Oracle RightNow CX Cloud Service 2012 Certified Implementation Specialist (1Z0-465) Oracle RightNow CX Cloud Service 2012 Developer Certified Implementation Specialist (1Z0-480) Oracle Fusion Human Capital Management 11g Human Resources Certified Implementation Specialist (1Z0-584) Oracle Fusion Human Capital Management 11g Talent Management Certified Implementation Specialist (1Z0-585) Oracle Taleo Recruiting Cloud Service 2013 Certified Implementation Specialist  (1Z0-474) Oracle Fusion Financials 11g Accounts Payable Certified Implementation Specialist(1Z0-507) Oracle Fusion Financials 11g Accounts Receivable Certified Implementation Specialist(1Z0-506) Oracle Fusion Financials 11g General Ledger Certified Implementation Specialist (1Z0-508) Oracle Fusion Distributed Order Orchestration 11g Essentials (1Z0-469) Oracle Documaker Standard Edition 12 Implementation Essentials (1Z0-570) Oracle Hyperion Planning 11 Essentials (1Z0-533) Oracle Hyperion Financial Management 11 Essentials (1Z0-532) Oracle Business Intelligence Foundation Suite 11g Essentials (1Z0-591) Oracle Essbase 11 Essentials (1Z0-531) Oracle GoldenGate 10 Essentials (1Z0-539) Oracle GoldenGate 11g Certified Implementation Exam Essentials Oracle Business Intelligence Applications 7.9.6 for CRM Essentials (1Z0-524) Oracle Business Intelligence Applications 7.9.6 for ERP Essentials (1Z0-525) Oracle Oracle Endeca Information Discovery 2.3 Certified Implementation Specialist (1Z0-461) Oracle SOA Suite 11g Essentials (1Z0-478) Oracle Service-Oriented Architecture Certified Implementation Specialist (1Z0-451) 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Implementation Specialist (1Z0-455) Oracle Flexcube Universal Banking 11 Technical Implementation Essentials (1Z0-579) Oracle FlexCube Universal Banking 11 Basic Implementation Essentials (1Z0-561) Oracle Flexcube Universal Banking 11 Technical Implementation Essentials (1Z0-579) Oracle FLEXCUBE Direct Banking 6 Implementation Essentials (1Z0-594)   Puedes consultar la información acerca de los examenes en cada uno de los enlaces. Para prepararte los examenes sigue la Guia de estudio que encontrarás en la página de cada examen. Requisitos: ser  Partner Gold, Platinum o Diamond de Oracle y tener un usuario de Oracle Pearson Vue.  ¿Cuándo?: 27 y 28 de noviembre  a las (9:00, 12:00, 16:00)  ¿Dónde?: Core Networks, C.E.Parque Norte, Edificio Olmo, Planta 1 Serrano Galvache 56 | 28033, Madrid Para inscribirte: Create una cuenta en Pearson Vue (www.pearsonvue.com/oracle). Para Registrarte aquí. Para más información sobre el programa de especializaciones, haz clic aquí. No pierdas esta oportunidad e inscríbete hoy.  Para cualquier duda contactar con [email protected]. Ana María Gallego Partner Enablement Manager Spain and Portugal        

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  • Simplex Noise Help

    - by Alex Larsen
    Im Making A Minecraft Like Gae In XNA C# And I Need To Generate Land With Caves This Is The Code For Simplex I Have /// <summary> /// 1D simplex noise /// </summary> /// <param name="x"></param> /// <returns></returns> public static float Generate(float x) { int i0 = FastFloor(x); int i1 = i0 + 1; float x0 = x - i0; float x1 = x0 - 1.0f; float n0, n1; float t0 = 1.0f - x0 * x0; t0 *= t0; n0 = t0 * t0 * grad(perm[i0 & 0xff], x0); float t1 = 1.0f - x1 * x1; t1 *= t1; n1 = t1 * t1 * grad(perm[i1 & 0xff], x1); // The maximum value of this noise is 8*(3/4)^4 = 2.53125 // A factor of 0.395 scales to fit exactly within [-1,1] return 0.395f * (n0 + n1); } /// <summary> /// 2D simplex noise /// </summary> /// <param name="x"></param> /// <param name="y"></param> /// <returns></returns> public static float Generate(float x, float y) { const float F2 = 0.366025403f; // F2 = 0.5*(sqrt(3.0)-1.0) const float G2 = 0.211324865f; // G2 = (3.0-Math.sqrt(3.0))/6.0 float n0, n1, n2; // Noise contributions from the three corners // Skew the input space to determine which simplex cell we're in float s = (x + y) * F2; // Hairy factor for 2D float xs = x + s; float ys = y + s; int i = FastFloor(xs); int j = FastFloor(ys); float t = (float)(i + j) * G2; float X0 = i - t; // Unskew the cell origin back to (x,y) space float Y0 = j - t; float x0 = x - X0; // The x,y distances from the cell origin float y0 = y - Y0; // For the 2D case, the simplex shape is an equilateral triangle. // Determine which simplex we are in. int i1, j1; // Offsets for second (middle) corner of simplex in (i,j) coords if (x0 > y0) { i1 = 1; j1 = 0; } // lower triangle, XY order: (0,0)->(1,0)->(1,1) else { i1 = 0; j1 = 1; } // upper triangle, YX order: (0,0)->(0,1)->(1,1) // A step of (1,0) in (i,j) means a step of (1-c,-c) in (x,y), and // a step of (0,1) in (i,j) means a step of (-c,1-c) in (x,y), where // c = (3-sqrt(3))/6 float x1 = x0 - i1 + G2; // Offsets for middle corner in (x,y) unskewed coords float y1 = y0 - j1 + G2; float x2 = x0 - 1.0f + 2.0f * G2; // Offsets for last corner in (x,y) unskewed coords float y2 = y0 - 1.0f + 2.0f * G2; // Wrap the integer indices at 256, to avoid indexing perm[] out of bounds int ii = i % 256; int jj = j % 256; // Calculate the contribution from the three corners float t0 = 0.5f - x0 * x0 - y0 * y0; if (t0 < 0.0f) n0 = 0.0f; else { t0 *= t0; n0 = t0 * t0 * grad(perm[ii + perm[jj]], x0, y0); } float t1 = 0.5f - x1 * x1 - y1 * y1; if (t1 < 0.0f) n1 = 0.0f; else { t1 *= t1; n1 = t1 * t1 * grad(perm[ii + i1 + perm[jj + j1]], x1, y1); } float t2 = 0.5f - x2 * x2 - y2 * y2; if (t2 < 0.0f) n2 = 0.0f; else { t2 *= t2; n2 = t2 * t2 * grad(perm[ii + 1 + perm[jj + 1]], x2, y2); } // Add contributions from each corner to get the final noise value. // The result is scaled to return values in the interval [-1,1]. return 40.0f * (n0 + n1 + n2); // TODO: The scale factor is preliminary! } public static float Generate(float x, float y, float z) { // Simple skewing factors for the 3D case const float F3 = 0.333333333f; const float G3 = 0.166666667f; float n0, n1, n2, n3; // Noise contributions from the four corners // Skew the input space to determine which simplex cell we're in float s = (x + y + z) * F3; // Very nice and simple skew factor for 3D float xs = x + s; float ys = y + s; float zs = z + s; int i = FastFloor(xs); int j = FastFloor(ys); int k = FastFloor(zs); float t = (float)(i + j + k) * G3; float X0 = i - t; // Unskew the cell origin back to (x,y,z) space float Y0 = j - t; float Z0 = k - t; float x0 = x - X0; // The x,y,z distances from the cell origin float y0 = y - Y0; float z0 = z - Z0; // For the 3D case, the simplex shape is a slightly irregular tetrahedron. // Determine which simplex we are in. int i1, j1, k1; // Offsets for second corner of simplex in (i,j,k) coords int i2, j2, k2; // Offsets for third corner of simplex in (i,j,k) coords /* This code would benefit from a backport from the GLSL version! */ if (x0 >= y0) { if (y0 >= z0) { i1 = 1; j1 = 0; k1 = 0; i2 = 1; j2 = 1; k2 = 0; } // X Y Z order else if (x0 >= z0) { i1 = 1; j1 = 0; k1 = 0; i2 = 1; j2 = 0; k2 = 1; } // X Z Y order else { i1 = 0; j1 = 0; k1 = 1; i2 = 1; j2 = 0; k2 = 1; } // Z X Y order } else { // x0<y0 if (y0 < z0) { i1 = 0; j1 = 0; k1 = 1; i2 = 0; j2 = 1; k2 = 1; } // Z Y X order else if (x0 < z0) { i1 = 0; j1 = 1; k1 = 0; i2 = 0; j2 = 1; k2 = 1; } // Y Z X order else { i1 = 0; j1 = 1; k1 = 0; i2 = 1; j2 = 1; k2 = 0; } // Y X Z order } // A step of (1,0,0) in (i,j,k) means a step of (1-c,-c,-c) in (x,y,z), // a step of (0,1,0) in (i,j,k) means a step of (-c,1-c,-c) in (x,y,z), and // a step of (0,0,1) in (i,j,k) means a step of (-c,-c,1-c) in (x,y,z), where // c = 1/6. float x1 = x0 - i1 + G3; // Offsets for second corner in (x,y,z) coords float y1 = y0 - j1 + G3; float z1 = z0 - k1 + G3; float x2 = x0 - i2 + 2.0f * G3; // Offsets for third corner in (x,y,z) coords float y2 = y0 - j2 + 2.0f * G3; float z2 = z0 - k2 + 2.0f * G3; float x3 = x0 - 1.0f + 3.0f * G3; // Offsets for last corner in (x,y,z) coords float y3 = y0 - 1.0f + 3.0f * G3; float z3 = z0 - 1.0f + 3.0f * G3; // Wrap the integer indices at 256, to avoid indexing perm[] out of bounds int ii = i % 256; int jj = j % 256; int kk = k % 256; // Calculate the contribution from the four corners float t0 = 0.6f - x0 * x0 - y0 * y0 - z0 * z0; if (t0 < 0.0f) n0 = 0.0f; else { t0 *= t0; n0 = t0 * t0 * grad(perm[ii + perm[jj + perm[kk]]], x0, y0, z0); } float t1 = 0.6f - x1 * x1 - y1 * y1 - z1 * z1; if (t1 < 0.0f) n1 = 0.0f; else { t1 *= t1; n1 = t1 * t1 * grad(perm[ii + i1 + perm[jj + j1 + perm[kk + k1]]], x1, y1, z1); } float t2 = 0.6f - x2 * x2 - y2 * y2 - z2 * z2; if (t2 < 0.0f) n2 = 0.0f; else { t2 *= t2; n2 = t2 * t2 * grad(perm[ii + i2 + perm[jj + j2 + perm[kk + k2]]], x2, y2, z2); } float t3 = 0.6f - x3 * x3 - y3 * y3 - z3 * z3; if (t3 < 0.0f) n3 = 0.0f; else { t3 *= t3; n3 = t3 * t3 * grad(perm[ii + 1 + perm[jj + 1 + perm[kk + 1]]], x3, y3, z3); } // Add contributions from each corner to get the final noise value. // The result is scaled to stay just inside [-1,1] return 32.0f * (n0 + n1 + n2 + n3); // TODO: The scale factor is preliminary! } private static byte[] perm = new byte[512] { 151,160,137,91,90,15, 131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23, 190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33, 88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166, 77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244, 102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196, 135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123, 5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42, 223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9, 129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228, 251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107, 49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254, 138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180, 151,160,137,91,90,15, 131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23, 190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33, 88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166, 77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244, 102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196, 135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123, 5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42, 223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9, 129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228, 251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107, 49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254, 138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180 }; private static int FastFloor(float x) { return (x > 0) ? ((int)x) : (((int)x) - 1); } private static float grad(int hash, float x) { int h = hash & 15; float grad = 1.0f + (h & 7); // Gradient value 1.0, 2.0, ..., 8.0 if ((h & 8) != 0) grad = -grad; // Set a random sign for the gradient return (grad * x); // Multiply the gradient with the distance } private static float grad(int hash, float x, float y) { int h = hash & 7; // Convert low 3 bits of hash code float u = h < 4 ? x : y; // into 8 simple gradient directions, float v = h < 4 ? y : x; // and compute the dot product with (x,y). return ((h & 1) != 0 ? -u : u) + ((h & 2) != 0 ? -2.0f * v : 2.0f * v); } private static float grad(int hash, float x, float y, float z) { int h = hash & 15; // Convert low 4 bits of hash code into 12 simple float u = h < 8 ? x : y; // gradient directions, and compute dot product. float v = h < 4 ? y : h == 12 || h == 14 ? x : z; // Fix repeats at h = 12 to 15 return ((h & 1) != 0 ? -u : u) + ((h & 2) != 0 ? -v : v); } private static float grad(int hash, float x, float y, float z, float t) { int h = hash & 31; // Convert low 5 bits of hash code into 32 simple float u = h < 24 ? x : y; // gradient directions, and compute dot product. float v = h < 16 ? y : z; float w = h < 8 ? z : t; return ((h & 1) != 0 ? -u : u) + ((h & 2) != 0 ? -v : v) + ((h & 4) != 0 ? -w : w); } This Is My World Generation Code Block[,] BlocksInMap = new Block[1024, 256]; public bool IsWorldGenerated = false; Random r = new Random(); private void RunThread() { for (int BH = 0; BH <= 256; BH++) { for (int BW = 0; BW <= 1024; BW++) { Block b = new Block(); if (BH >= 192) { } BlocksInMap[BW, BH] = b; } } IsWorldGenerated = true; } public void GenWorld() { new Thread(new ThreadStart(RunThread)).Start(); } And This Is A Example Of How I Set Blocks Block b = new Block(); b.BlockType = = Block.BlockTypes.Air; This Is A Example Of How I Set Models foreach (Block b in MyWorld) { switch(b.BlockType) { case Block.BlockTypes.Dirt: b.Model = DirtModel; break; ect. } } How Would I Use These To Generate To World (The Block Array) And If Possible Thread It More? btw It's 1024 Wide And 256 Tall

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  • how to serialize / deserialize classes defined in .proto (protobuf)

    - by make
    Hi, Could someone please help me with serialization/deserialization classes defined in .proto (protobuf). here is an exp that I am trying to build: file.proto message Data{ required string x1 = 1; required uint32 x2 = 2; required float x3 = 3; } message DataExge { repeated Data data = 1; } client.cpp ... void serialize(const DataExge &data_snd){ try { ofstream ofs("DataExge"); data_snd.SerializeToOstream(&ofs); } catch(exception &e) { cerr << "serialize/exception: " << e.what() << endl; exit(1); } } void deserialize(DataExge &data_rec){ try { ifstream ifs("DataExge"); data_rec.ParseFromIstream(&ifs); } catch(exception& e) { cerr << "deserialize/exception: " << e.what() << endl; exit(1); } } int main(){ ... DataExge dataexge; Data *dat = dataexge.add_data(); char *y1 = "operation1"; uint32_t y2 = 123 ; float y3 = 3.14; // assigning data to send() dat->set_set_x1(y1); dat->set_set_x2(y2); dat->set_set_x3(y3); //sending data to the client serialize(dataexge); if (send(socket, &dataexge, sizeof(dataexge), 0) < 0) { cerr << "send() failed" ; exit(1); } //receiving data from the server deserialize(dataexge); if (recv(socket, &dataexge, sizeof(dataexge), 0) < 0) { cerr << "recv() failed"; exit(1); } //printing received data cout << dat->x1() << "\n"; cout << dat->x2() << "\n"; cout << dat->x3() << "\n"; ... } server.cpp ... void serialize(const DataExge &data_snd){ try { ofstream ofs("DataExge"); data_snd.SerializeToOstream(&ofs); } catch(exception &e) { cerr << "serialize/exception: " << e.what() << endl; exit(1); } } void deserialize(DataExge &data_rec){ try { ifstream ifs("DataExge"); data_rec.ParseFromIstream(&ifs); } catch(exception& e) { cerr << "deserialize/exception: " << e.what() << endl; exit(1); } } int main(){ ... DataExge dataexge; Data *dat = dataexge.add_data(); //receiving data from the client deserialize(dataexge); if (recv(socket, &dataexge, sizeof(dataexge), 0) < 0) { cerr << "recv() failed"; exit(1); } //printing received data cout << dat->x1() << "\n"; cout << dat->x2() << "\n"; cout << dat->x3() << "\n"; // assigning data to send() dat->set_set_x1("operation2"); dat->set_set_x2(dat->x2() + 1); dat->set_set_x3(dat->x3() + 1.1); //sending data to the client serialize(dataexge); //error// I am getting error at this line ... if (send(socket, &dataexge, sizeof(dataexge), 0) < 0) { cerr << "send() failed" ; exit(1); } ... } Thanks for your help and replies -

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  • Suggestion to reverse string in c#

    - by HasanGursoy
    Is this the right method to reverse a string? I'm planning to use it to reverse a string like: Products » X1 » X3 to X3 « X1 « Products I want it to be a global function which can be used elsewhere. public static string ReverseString(string input, string separator, string outSeparator) { string result = String.Empty; string[] temp = Regex.Split(input, separator, RegexOptions.IgnoreCase); Array.Reverse(temp); for (int i = 0; i < temp.Length; i++) { result += temp[i] + " " + outSeparator + " "; } return result; }

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  • One blank line in list item

    - by rima
    Hi there. I try to write a code to fill my list item in oracle form builder. I do it by write a function to handle this. list_index number(10) := 1; clear_list(list_item1); FOR I IN (Select id,desc FORM table1) LOOP ADD_LIST_ELEMENT('list_item1',list_index,desc,id); list_index := list_index + 1; END LOOP list_item1 := get_list_element_value('list_item1',1); my result in output is like this: x1 x2 x3 x4 <a blank field> but in my database table I just have x1 x2 x3 x4 would you help me please to how find what's my problem that I have one more space in my list item.

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  • Choose item from a list with bias?

    - by ooboo
    Given a list of items x1 ... xn and associated probabilties p1 ... pn that sum up to 1 there's a well known procedure to select a random item with its associated proabability by sorting the list according to weight, choosing a random value between 1 and 0, and adding up to a culmination sum until it exceeds the value selected and return the item at this point. So if we have x1 - 0.5, x2 - 0.3, x3 - 0.2, if the randomly chosen value is less than 0.5 x1 will be chosen, if between 0.5 and 0.8, x2, and else x3. This requires sorting so it needs O(nlogn) time. Is there anything more efficient than that?

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