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  • Exchange not delivering the mail

    - by wolfvilleian
    I'm having an issue where my Exchange Edge Transport server receives mail (found in logs) and then it vanishes, never ending up in the users mailbox, I have a edge subscription setup between it and the main Exchange server, how can I go about tracing the message to figure out what is broken? I also have found records of the message in the logs on the main Exchange server. Thanks a ton for any help Edit: If I change port 25 on my main router to point to the main exchange server as opposed to the Edge Transport, email comes through fine form external domains and delivered in the correct mailbox

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  • Exchange Not Delivering Email

    - by wolfvilleian
    I'm having an issue where my Exchange Edge Transport server receives mail (found in logs) and then it vanishes, never ending up in the users mailbox, I have a edge subscription setup between it and the main Exchange server, how can I go about tracing the message to figure out what is broken? I also have found records of the message in the logs on the main Exchange server. Thanks a ton for any help Edit: If I change port 25 on my main router to point to the main exchange server as opposed to the Edge Transport, email comes through fine form external domains and delivered in the correct mailbox

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  • How can I move a polygon edge 1 unit away from the center?

    - by Stephen
    Let's say I have a polygon class that is represented by a list of vector classes as vertices, like so: var Vector = function(x, y) { this.x = x; this.y = y; }, Polygon = function(vectors) { this.vertices = vectors; }; Now I make a polygon (in this case, a square) like so: var poly = new Polygon([ new Vector(2, 2), new Vector(5, 2), new Vector(5, 5), new Vector(2, 5) ]); So, the top edge would be [poly.vertices[0], poly.vertices[1]]. I need to stretch this polygon by moving each edge away from the center of the polygon by one unit, along that edge's normal. The following example shows the first edge, the top, moved one unit up: The final polygon should look like this new one: var finalPoly = new Polygon([ new Vector(1, 1), new Vector(6, 1), new Vector(6, 6), new Vector(1, 6) ]); It is important that I iterate, moving one edge at a time, because I will be doing some collision tests after moving each edge. Here is what I tried so far (simplified for clarity), which fails triumphantly: for(var i = 0; i < vertices.length; i++) { var a = vertices[i], b = vertices[i + 1] || vertices[0]; // in case of final vertex var ax = a.x, ay = a.y, bx = b.x, by = b.y; // get some new perpendicular vectors var a2 = new Vector(-ay, ax), b2 = new Vector(-by, bx); // make into unit vectors a2.convertToUnitVector(); b2.convertToUnitVector(); // add the new vectors to the original ones a.add(a2); b.add(b2); // the rest of the code, collision tests, etc. } This makes my polygon start slowly rotating and sliding to the left, instead of what I need. Finally, the example shows a square, but the polygons in question could be anything. They will always be convex, and always with vertices in clockwise order.

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  • LYNC / OCS... problems getting edge server working.

    - by TomTom
    New setup Lync 2010 (i.e. OCS 2010). I have serious problems getting my edge system going. Internally things work fine. Externally I am stuck. I have used the tester at https://www.testocsconnectivity.com/ and it also fails. NOTE: I use the domain xample.com / xample.local here just as example. Here is the setup. I have 2 internal hosts (lync.xample.local, edge.xample.local). edge.xample.com is also correctly in dns. and points to the edge.xample.local external assigned ip address (external interface). Externally, I have the following dns entries: edge.xample.com _sip._tcp - edge.xample.com 443 _sipfederationtls._tcp - edge.xample.com 5061 _sipinternaltls._tcp - lync.xample.local 5061 _sip._tls - edge.xample.com 443 My problem is that the ocs connection test always ends up trying to contact lync.xample.local (i.e. the internal address) when connecting to [email protected]. The error is: Attempting to Resolve the host name lync.xample.local in DNS. This shows me it clearly manages to connect to SOMETHING, but it does either fall through to the _sipinternaltls._tcp entry, OR it does get that internal entry wrongly from the edge system. Am I missing some entries or have some wrong?

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  • IRQ problem with 2.6.32/2.6.39 kernel on Debian Squeeze x86_64

    - by MasterM
    I recently assembled a new computer so that all hardware is pretty new. Since then I've been experiencing some problem with IRQs when running Debian 6.0. On random occasions, usually after an hour or so of running I hear a beep and this shows up in dmesg: [ 3537.762795] irq 16: nobody cared (try booting with the "irqpoll" option) [ 3537.762797] Pid: 0, comm: swapper Tainted: P W O 2.6.39-2-amd64 #1 [ 3537.762798] Call Trace: [ 3537.762799] <IRQ> [<ffffffff810924d4>] ? __report_bad_irq+0x3a/0xa2 [ 3537.762803] [<ffffffff810926a4>] ? note_interrupt+0x168/0x1da [ 3537.762805] [<ffffffff81090dd4>] ? handle_irq_event_percpu+0x171/0x18f [ 3537.762807] [<ffffffff8100e0e2>] ? read_tsc+0x5/0x16 [ 3537.762809] [<ffffffff8106b8a2>] ? update_ts_time_stats+0x32/0x6b [ 3537.762810] [<ffffffff81090e26>] ? handle_irq_event+0x34/0x52 [ 3537.762812] [<ffffffff81063fb7>] ? sched_clock_idle_wakeup_event+0x12/0x1c [ 3537.762813] [<ffffffff81092df2>] ? handle_fasteoi_irq+0x82/0xa4 [ 3537.762815] [<ffffffff8100aadb>] ? handle_irq+0x1a/0x23 [ 3537.762816] [<ffffffff8100a384>] ? do_IRQ+0x45/0xaa [ 3537.762818] [<ffffffff81332c93>] ? common_interrupt+0x13/0x13 [ 3537.762818] <EOI> [<ffffffff81332c8e>] ? common_interrupt+0xe/0x13 [ 3537.762821] [<ffffffff81026800>] ? native_safe_halt+0x2/0x3 [ 3537.762829] [<ffffffffa016ed58>] ? acpi_idle_do_entry+0x39/0x62 [processor] [ 3537.762831] [<ffffffffa016edde>] ? acpi_idle_enter_c1+0x5d/0xad [processor] [ 3537.762834] [<ffffffff81261033>] ? cpuidle_idle_call+0x11f/0x1cc [ 3537.762835] [<ffffffff81008dd2>] ? cpu_idle+0xab/0xe1 [ 3537.762837] [<ffffffff8169fc60>] ? start_kernel+0x3e0/0x3eb [ 3537.762838] [<ffffffff8169f3c8>] ? x86_64_start_kernel+0x102/0x10f [ 3537.762839] handlers: [ 3537.762840] [<ffffffffa0358d5a>] (rtl8169_interrupt+0x0/0x2d7 [r8169]) [ 3537.762842] [<ffffffffa08ff2ca>] (nv_kern_isr+0x0/0x54 [nvidia]) [ 3537.762902] Disabling IRQ #16 After that Xorg either hogs on CPU or is unstable (up to hanging the system completely). When I restart Xorg everything is fine again and the problem doesn't occur until next reboot. I tried to upgrade the kernel from stock 2.6.32 to 2.6.39 from unstable repository but that didn't help. Booting with irqpoll option only seems to prolong the initial time period after which the problem occurs. I'm using latest NVIDIA drivers and Realtek firmware from firmware-realtek package. I have two GTX 560Ti that run in SLI. Disabling SLI or taking out one card completely doesn't solve the problem either. Output of uname -a is: Linux whitestar 2.6.39-2-amd64 #1 SMP Wed Jun 8 11:01:04 UTC 2011 x86_64 GNU/Linux Output of lspci is: 00:00.0 Host bridge: Intel Corporation Sandy Bridge DRAM Controller (rev 09) 00:01.0 PCI bridge: Intel Corporation Sandy Bridge PCI Express Root Port (rev 09) 00:01.1 PCI bridge: Intel Corporation Sandy Bridge PCI Express Root Port (rev 09) 00:16.0 Communication controller: Intel Corporation Cougar Point HECI Controller #1 (rev 04) 00:19.0 Ethernet controller: Intel Corporation 82579V Gigabit Network Connection (rev 05) 00:1a.0 USB Controller: Intel Corporation Cougar Point USB Enhanced Host Controller #2 (rev 05) 00:1b.0 Audio device: Intel Corporation Cougar Point High Definition Audio Controller (rev 05) 00:1c.0 PCI bridge: Intel Corporation Cougar Point PCI Express Root Port 1 (rev b5) 00:1c.1 PCI bridge: Intel Corporation Cougar Point PCI Express Root Port 2 (rev b5) 00:1c.2 PCI bridge: Intel Corporation Cougar Point PCI Express Root Port 3 (rev b5) 00:1c.4 PCI bridge: Intel Corporation Cougar Point PCI Express Root Port 5 (rev b5) 00:1c.6 PCI bridge: Intel Corporation 82801 PCI Bridge (rev b5) 00:1d.0 USB Controller: Intel Corporation Cougar Point USB Enhanced Host Controller #1 (rev 05) 00:1f.0 ISA bridge: Intel Corporation Cougar Point LPC Controller (rev 05) 00:1f.2 SATA controller: Intel Corporation Cougar Point 6 port SATA AHCI Controller (rev 05) 00:1f.3 SMBus: Intel Corporation Cougar Point SMBus Controller (rev 05) 01:00.0 VGA compatible controller: nVidia Corporation Device 1200 (rev a1) 01:00.1 Audio device: nVidia Corporation Device 0e0c (rev a1) 02:00.0 VGA compatible controller: nVidia Corporation Device 1200 (rev a1) 02:00.1 Audio device: nVidia Corporation Device 0e0c (rev a1) 04:00.0 USB Controller: NEC Corporation uPD720200 USB 3.0 Host Controller (rev 04) 06:00.0 USB Controller: NEC Corporation uPD720200 USB 3.0 Host Controller (rev 04) 07:00.0 PCI bridge: Device 1b21:1080 (rev 01) 08:02.0 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL-8110SC/8169SC Gigabit Ethernet (rev 10) 08:03.0 FireWire (IEEE 1394): VIA Technologies, Inc. VT6306/7/8 [Fire II(M)] IEEE 1394 OHCI Controller (rev c0) Contents of /proc/interrupts: CPU0 CPU1 CPU2 CPU3 CPU4 CPU5 CPU6 CPU7 0: 77 0 0 0 0 0 0 0 IO-APIC-edge timer 1: 2 0 0 0 0 0 0 0 IO-APIC-edge i8042 8: 1 0 0 0 0 0 0 0 IO-APIC-edge rtc0 9: 0 0 0 0 0 0 0 0 IO-APIC-fasteoi acpi 12: 4 0 0 0 0 0 0 0 IO-APIC-edge i8042 16: 699083 0 0 0 0 0 0 0 IO-APIC-fasteoi nvidia, eth0 17: 87810 0 0 0 0 0 0 0 IO-APIC-fasteoi firewire_ohci, hda_intel, nvidia 18: 242 0 0 0 0 0 0 0 IO-APIC-fasteoi hda_intel 23: 85925 0 0 0 0 0 0 0 IO-APIC-fasteoi ehci_hcd:usb5, ehci_hcd:usb6 40: 0 0 0 0 0 0 0 0 PCI-MSI-edge PCIe PME 41: 0 0 0 0 0 0 0 0 PCI-MSI-edge PCIe PME 42: 0 0 0 0 0 0 0 0 PCI-MSI-edge PCIe PME 43: 0 0 0 0 0 0 0 0 PCI-MSI-edge PCIe PME 44: 0 0 0 0 0 0 0 0 PCI-MSI-edge PCIe PME 45: 0 0 0 0 0 0 0 0 PCI-MSI-edge PCIe PME 46: 79853 0 0 0 0 0 0 0 PCI-MSI-edge ahci 48: 1 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 49: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 50: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 51: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 52: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 53: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 54: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 55: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 56: 1 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 57: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 58: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 59: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 60: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 61: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 62: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 63: 0 0 0 0 0 0 0 0 PCI-MSI-edge xhci_hcd 64: 173506 0 0 0 0 0 0 0 PCI-MSI-edge hda_intel NMI: 482 89 25 13 277 24 11 10 Non-maskable interrupts LOC: 783857 194752 114133 70577 372438 179065 117179 162016 Local timer interrupts SPU: 0 0 0 0 0 0 0 0 Spurious interrupts PMI: 482 89 25 13 277 24 11 10 Performance monitoring interrupts IWI: 0 0 0 0 0 0 0 0 IRQ work interrupts RES: 131917 46750 7432 3291 150003 9576 3435 3067 Rescheduling interrupts CAL: 2759 6563 7150 6997 5387 7140 7269 6678 Function call interrupts TLB: 4396 2038 1336 492 5434 1896 1121 606 TLB shootdowns TRM: 0 0 0 0 0 0 0 0 Thermal event interrupts THR: 0 0 0 0 0 0 0 0 Threshold APIC interrupts MCE: 0 0 0 0 0 0 0 0 Machine check exceptions MCP: 37 37 37 37 37 37 37 37 Machine check polls ERR: 0 MIS: 0 Last but not least, right after boot-up those lines are usually present in dmesg: [ 18.367094] hda-intel: IRQ timing workaround is activated for card #1. Suggest a bigger bdl_pos_adj. [ 18.458859] hda-intel: IRQ timing workaround is activated for card #2. Suggest a bigger bdl_pos_adj. I'm not sure if it's related or a symptom of a bigger problem so I'm posting it just in case. I don't really know what other information might be of relevance here. Don't hesitate to ask for more in the comments.

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  • Updating a Minimum spanning tree when a new edge is inserted

    - by Lynette
    Hello, I've been presented the following problem in University: Let G = (V, E) be an (undirected) graph with costs ce = 0 on the edges e € E. Assume you are given a minimum-cost spanning tree T in G. Now assume that a new edge is added to G, connecting two nodes v, tv € V with cost c. a) Give an efficient algorithm to test if T remains the minimum-cost spanning tree with the new edge added to G (but not to the tree T). Make your algorithm run in time O(|E|). Can you do it in O(|V|) time? Please note any assumptions you make about what data structure is used to represent the tree T and the graph G. b)Suppose T is no longer the minimum-cost spanning tree. Give a linear-time algorithm (time O(|E|)) to update the tree T to the new minimum-cost spanning tree. This is the solution I found: Let e1=(a,b) the new edge added Find in T the shortest path from a to b (BFS) if e1 is the most expensive edge in the cycle then T remains the MST else T is not the MST It seems to work but i can easily make this run in O(|V|) time, while the problem asks O(|E|) time. Am i missing something? By the way we are authorized to ask for help from anyone so I'm not cheating :D Thanks in advance

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  • Ubuntu snap-to-edge feature for XP

    - by Wesley
    Hi all. I am running Windows XP SP3 dual booting with Ubuntu 9.10. I really like the snap to edge feature of Ubuntu and wondered if you could get that feature in XP. This would basically prevent any windows from straying from the workspace and would allow windows to snap to the edges of the screen and to the edges of other windows. Thanks in advance!

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  • Firebox Edge 11 and SBS 2008 VPN produces error 721

    - by Charlie Bear
    My VPN has stopped working. I have sbs 2008 and have run the VPN wizard. I have opened the port 1723 on my firebox edge as it instructed me to do. It was working but I think that an upgrade of the firebox software to version 11 has affected it. The port is still open. When connecting I get to verifying username and password then I get Error 721. Not sure whats wrong here. can anyone help?

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  • Removing RAID on a DELL power-edge server

    - by Simon Callan
    We have a Dell Power-edge T110 server, with either a PERC S100 or S300 raid controller (I suspect S100), and 2 x 500GB SATA hard discs. These discs are running in RAID-1 configuration, (either 1 or 2 virtual discs), and on top of this, we have the C and D drives. As we are running low on disc space, we are looking at dropping the RAID to get some disc space back. Is it possible to split the discs back into 2 seperate hards discs that the OS can see, without losing all the data on the drives? Even better, would it be possible to split the D: drive, leaving the OS C: drive in RAID1 configuration?

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  • How can I force a USB modem to only connect via EDGE and not 3G?

    - by Anders Wallenquist
    How do I get Network-admin to restrict connection to GSM (Edge) instead of flipping between 3G and Edge and lose connection. Usually it works out-of-the box, but at my current location there are a lot of radio shadows, so the recommendation from my ISP is that I should lock to Edge - which can be done using their own driver in Windows. How can I do this in Ubuntu Ubuntu 11.04 Modem: Huawei E220 ISP: Telia mobilt bredband kontant

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  • Connect Lenovo Thinkpad Edge e120 to WQHD display (such as Dell U2711) through HDMI

    - by Fulvio
    On paper, the HDMI version of the Lenovo Thinkpad Edge e120 is 1.3, which supports WQHD. Does it actually work? I want to connect the e120 to a WQHD display, such as the Dell U2711 or an Apple Cinema Display. I run Windows 7 Ultimate and latest drivers. The e120 I think has the Intel HD Graphics 3000 onboard chipset. I don't have the chance to test out the e120 with neither of these displays, so I'm seeking for an opinion from those who have tried. Thank you

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  • Windows 8 trackpad edge swipe zones

    - by askvictor
    I'm running Windows 8 on a Lenovo x220 laptop; and have just inadvertently discovered the edge-swipe feature or that brings up the charms or switches between desktop and RT. Only problem is that the landing zones are a little too wide for my liking - I'd like to keep this feature, but to narrow the zone where it can start. I'd rather not disable it completely as per: Modify or disable Windows 8 swipe gestures on touchpad / laptop The Synaptic driver (latest available) doesn't seem to provide for changing this (though it does for other zones). Any ideas?

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  • Change Target of Edge in BGL

    - by Sunny
    If my BGL graph contain edge from node x to node y, and I want to change the target of this edge, so that now it's pointing from x to z, how it can be done? Are there any functions in BGL for that?

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  • Can't set correct desktop resolution on Windows 7 after playing Mirror's Edge

    - by DeadMG
    Ever since I launched Mirror's Edge, my entire system seems to think that my monitor is 1600x1200, when actually it's 1920x1080. The game would only go up to 1600x1200, Windows will only go up to 1600x1200, and even my video drivers will only go up that high. The monitor isn't even that aspect ratio. I've rebooted the system to no effect, and installed the latest graphics drivers for my Radeon 5770, unplugged it and re-plugged it, etc. My secondary screen is still working perfectly correctly. Any suggestions?

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  • How do i use GraphMLReader2 in Jung?

    - by askus
    I want to use class GraphMLReader to read a Undirected Graph from graphML with JUNG2.0. The code is as follow: import edu.uci.ics.jung.io.*; import edu.uci.ics.jung.io.graphml.*; import java.io.*; import java.util.*; import org.apache.commons.collections15.Transformer; import edu.uci.ics.jung.graph.*; class Vertex{ int id; String type; String value; } class Edge{ int id ; String type; String value; } public class Loader{ static String src = "test.xsl"; public static void Main( String[] args){ Reader reader = new FileReader(src ); Transformer<NodeMetadata, Vertex> vtrans = new Transformer<NodeMetadata,Vertex>(){ public Vertex transform(NodeMetadata nmd ){ Vertex v = new Vertex() ; v.type = nmd.getProperty("type"); v.value = nmd.getProperty("value"); v.id = Integer.valueOf( nmd.getId() ); return v; } }; Transformer<EdgeMetadata, Edge> etrans = new Transformer<EdgeMetadata,Edge>(){ public Edge transform( EdgeMetadata emd ){ Edge e = new Edge() ; e.type = emd.getProperty("type"); e.value = emd.getProperty("value"); e.id = Integer.valueOf( emd.getId() ); return e; } }; Transformer<HyperEdgeMetadata, Edge> hetrans = new Transformer<HyperEdgeMetadata,Edge>(){ public Edge transform( HyperEdgeMetadata emd ){ Edge e = new Edge() ; e.type = emd.getProperty("type"); e.value = emd.getProperty("value"); e.id = Integer.valueOf( emd.getId() ); return e; } }; Transformer< GraphMetadata , UndirectedSparseGraph> gtrans = new Transformer<GraphMetadata,UndirectedSparseGraph>(){ public UndirectedSparseGraph<Vertex,Edge> transform( GraphMetadata gmd ){ return new UndirectedSparseGraph<Vertex,Edge>(); } }; GraphMLReader2< UndirectedSparseGraph<Vertex,Edge> , Vertex , Edge> gmlr = new GraphMLReader2< UndirectedSparseGraph<Vertex,Edge> ,Vertex, Edge>( reader, gtrans, vtrans, etrans, hetrans); UndirectedSparseGraph<Vertex,Edge> g = gmlr.readGraph(); return ; } } However, compiler alert that: Loader.java:60: cannot find symbol symbol : constructor GraphMLReader2(java.io.Reader,org.apache.commons.collections15.Transformer<edu.uci.ics.jung.io.graphml.GraphMetadata,edu.uci.ics.jung.graph.UndirectedSparseGraph>,org.apache.commons.collections15.Transformer<edu.uci.ics.jung.io.graphml.NodeMetadata,Vertex>,org.apache.commons.collections15.Transformer<edu.uci.ics.jung.io.graphml.EdgeMetadata,Edge>) location: class edu.uci.ics.jung.io.graphml.GraphMLReader2<edu.uci.ics.jung.graph.UndirectedSparseGraph<Vertex,Edge>,Vertex,Edge> new GraphMLReader2< UndirectedSparseGraph<Vertex,Edge> ,Vertex, Edge>( ^ 1 error How can i solve this problem? Thanks.

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  • Java: Object Array assignment in for loop

    - by Hackster
    I am trying to use Dijkstra's algorithm to find the shortest path from a specific vertex (v0) to the rest of them. That is solved and works well with this code from this link below: http://en.literateprograms.org/index.php?title=Special:DownloadCode/Dijkstra%27s_algorithm_(Java)&oldid=15444 I am having trouble with assigning the Edge array in a for loop from the user input, as opposed to hard-coding it like it is here. Any help assigning a new edge to Edge[] adjacencies from each vertex? Keeping in mind it could be 1 or multiple edges. class Vertex implements Comparable<Vertex> { public final String name; public Edge[] adjacencies; public double minDistance = Double.POSITIVE_INFINITY; public Vertex previous; public Vertex(String argName) { name = argName; } public String toString() { return name; } public int compareTo(Vertex other){ return Double.compare(minDistance, other.minDistance); } } class Edge{ public final Vertex target; public final double weight; public Edge(Vertex argTarget, double argWeight){ target = argTarget; weight = argWeight; } } public static void main(String[] args) { Vertex v[] = new Vertex[3]; Vertex v[0] = new Vertex("Harrisburg"); Vertex v[1] = new Vertex("Baltimore"); Vertex v[2] = new Vertex("Washington"); v0.adjacencies = new Edge[]{ new Edge(v[1], 1), new Edge(v[2], 3) }; v1.adjacencies = new Edge[]{ new Edge(v[0], 1), new Edge(v[2], 1),}; v2.adjacencies = new Edge[]{ new Edge(v[0], 3), new Edge(v[1], 1) }; Vertex[] vertices = { v0, v1, v2}; /*Three vertices with weight: V0 connects (V1,1),(V2,3) V1 connects (V0,1),(V2,1) V2 connects (V1,1),(V2,3) */ computePaths(v0); for (Vertex v : vertices){ System.out.println("Distance to " + v + ": " + v.minDistance); List<Vertex> path = getShortestPathTo(v); System.out.println("Path: " + path); } } } The above code works well in finding the shortest path from v0 to all the other vertices. The problem occurs when assigning the new edge[] to edge[] adjacencies. For example this does not produce the correct output: for (int i = 0; i < total_vertices; i++){ s = br.readLine(); char[] line = s.toCharArray(); for (int j = 0; j < line.length; j++){ if(j % 4 == 0 ){ //Input: vertex weight vertex weight: 1 1 2 3 int vert = Integer.parseInt(String.valueOf(line[j])); int w = Integer.parseInt(String.valueOf(line[j+2])); v[i].adjacencies = new Edge[] {new Edge(v[vert], w)}; } } } As opposed to this: v0.adjacencies = new Edge[]{ new Edge(v[1], 1), new Edge(v[2], 3) }; How can I take the user input and make an Edge[], to pass it to adjacencies? The problem is it could be 0 edges or many. Any help would be much appreciated Thanks!

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  • Finding edge and corner values of an image in matlab

    - by James
    Hi, this problem links to two other questions i've asked on here. I am tracing the outline of an image and plotting this to a dxf file. I would like to use the bwboundaries function to find the coordinates of the edges of the image, find the corner coordinates using the cornermetric function and then remove any edge coordinates that are not a corner. The important thing I need to be able to do is keep the order of the corner elements obtained from bwboundaries, so that the section traces properly. The dxf function I have that draws from the coordinates draws lines between coordinates that are next to each other, so the line has to be drawn "around" the section rather than straight between the corner points. The reason I am doing this is because there are less coordinates obtained this way, so it is easier to amend the dxf file (as there are less points to manipulate). The code I have so far is: %# Shape to be traced bw = zeros(200); bw(20:40,20:180) = 1; bw(20:180,90:110) = 1; bw(140:180,20:185) = 1; %# Boundary Finding Section [Boundary] = bwboundaries(bw); %Traces the boundary of each section figure, imshow(bw); hold on; colors=['b' 'g' 'r' 'c' 'm' 'y']; for k=1:length(Boundary) perim = Boundary{k}; %Obtains perimeter coordinates (as a 2D matrix) from the cell array cidx = mod(k,length(colors))+1;% Obtains colours for the plot plot(perim(:,2), perim(:,1),... colors(cidx),'LineWidth',2); end Coordmat = cell2mat(Boundary) %Converts the traced regions to a matrix X = Coordmat(:,1) Y = Coordmat(:,2) % This gives the edge coordinates in matrix form %% Corner Finding Section (from Jonas' answer to a previous question %# get corners cornerProbability = cornermetric(bw); cornerIdx = find(cornerProbability==max(cornerProbability(:))); %# Label the image. bwlabel puts 1 for the first feature, 2 for the second, etc. %# Since concave corners are placed just outside the feature, grow the features %# a little before labeling bw2 = imdilate(bw,ones(3)); labeledImage = bwlabel(bw2); %# read the feature number associated with the corner cornerLabels = labeledImage(cornerIdx); %# find all corners that are associated with feature 1 corners_1 = cornerIdx(cornerLabels==1) [Xcorners, Ycorners] = ind2sub(200,corners_1) % Convert subscripts The code I have is, to give a matrix Xfin for the final x coordinates (which are on the edge AND at a corner. Xfin = zeros(length(X),1) for i = Xcorners XFin(i) = Xcorners if i~= Xcorners XFin(i) = [] end end However, this does not work correctly, because the values in the solution are sorted into order, and only one of each value remains. As I said, I would like the corner elements to be in the same order as obtained from bwboundaries, to allow the image to trace properly. Thanks

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  • Drawing Directed Acyclic Graphs: Minimizing edge crossing?

    - by Robert Fraser
    Laying out the verticies in a DAG in a tree form (i.e. verticies with no in-edges on top, verticies dependent only on those on the next level, etc.) is rather simple without graph drawing algorithms such as Efficient Sugimiya. However, is there a simple algorithm to do this that minimizes edge crossing? (For some graphs, it may be impossible to completely eliminate edge crossing.) A picture says a thousand words, so is there an algorithm that would suggest: instead of: EDIT: As the picture suggests, a vertex's inputs are always on top and outputs are always below, which is another barrier to just pasting in an existing layout algorithm.

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  • Matlab - Propagate unit vectors on to the edge of shape boundaries

    - by Graham
    Hi I have a set of unit vectors which I want to propagate on to the edge of shape boundary defined by a binary image. The shape boundary is defined by a 1px wide white edge. I also have the coordinates of these points stored in a 2 row by n column matrix. The shape forms a concave boundary with no holes within itself made of around 2500 points. What would be the best method to do this? Are there some sort of ray tracing algorithms that could be used? Or would it be a case of taking the unit vector and multiplying it by a scalar and testing after multiplication if the end point of the vector is outside the shape boundary. When the end point of the unit vector is outside the shape, just find the point of intersection? Thank you very much in advance for any help!

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  • Twin edges - Half edge data structure

    - by Pradeep Kumar
    I have implemented a Half-edge data structure for loading 3d objects. I find that the part of assigning twin/pair edges takes the longest computation time (especially for objects which have hundreds of thousands half edges). The reason is that I use nested loops to accomplish this. Is there a simpler and efficient way of doing this? Below is the code which I've written. HE is the half-edge data structure. hearr is a vector containing all the half edges. vert is the starting vertex and end is the ending vertex. Thanks!! HE *e1,*e2; for(size_t i=0;i<hearr.size();i++){ e1=hearr[i]; for(size_t j=1;j<hearr.size();j++){ e2=hearr[j]; if((e1->vert==e2->end)&&(e2->vert==e1->end)){ e1->twin=e2; e2->twin=e1; } } }

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  • creating objects from trivial graph format text file. java. dijkstra algorithm.

    - by user560084
    i want to create objects, vertex and edge, from trivial graph format txt file. one of programmers here suggested that i use trivial graph format to store data for dijkstra algorithm. the problem is that at the moment all the information, e.g., weight, links, is in the sourcecode. i want to have a separate text file for that and read it into the program. i thought about using a code for scanning through the text file by using scanner. but i am not quite sure how to create different objects from the same file. could i have some help please? the file is v0 Harrisburg v1 Baltimore v2 Washington v3 Philadelphia v4 Binghamton v5 Allentown v6 New York # v0 v1 79.83 v0 v5 81.15 v1 v0 79.75 v1 v2 39.42 v1 v3 103.00 v2 v1 38.65 v3 v1 102.53 v3 v5 61.44 v3 v6 96.79 v4 v5 133.04 v5 v0 81.77 v5 v3 62.05 v5 v4 134.47 v5 v6 91.63 v6 v3 97.24 v6 v5 87.94 and the dijkstra algorithm code is Downloaded from: http://en.literateprograms.org/Special:Downloadcode/Dijkstra%27s_algorithm_%28Java%29 */ import java.util.PriorityQueue; import java.util.List; import java.util.ArrayList; import java.util.Collections; class Vertex implements Comparable<Vertex> { public final String name; public Edge[] adjacencies; public double minDistance = Double.POSITIVE_INFINITY; public Vertex previous; public Vertex(String argName) { name = argName; } public String toString() { return name; } public int compareTo(Vertex other) { return Double.compare(minDistance, other.minDistance); } } class Edge { public final Vertex target; public final double weight; public Edge(Vertex argTarget, double argWeight) { target = argTarget; weight = argWeight; } } public class Dijkstra { public static void computePaths(Vertex source) { source.minDistance = 0.; PriorityQueue<Vertex> vertexQueue = new PriorityQueue<Vertex>(); vertexQueue.add(source); while (!vertexQueue.isEmpty()) { Vertex u = vertexQueue.poll(); // Visit each edge exiting u for (Edge e : u.adjacencies) { Vertex v = e.target; double weight = e.weight; double distanceThroughU = u.minDistance + weight; if (distanceThroughU < v.minDistance) { vertexQueue.remove(v); v.minDistance = distanceThroughU ; v.previous = u; vertexQueue.add(v); } } } } public static List<Vertex> getShortestPathTo(Vertex target) { List<Vertex> path = new ArrayList<Vertex>(); for (Vertex vertex = target; vertex != null; vertex = vertex.previous) path.add(vertex); Collections.reverse(path); return path; } public static void main(String[] args) { Vertex v0 = new Vertex("Nottinghill_Gate"); Vertex v1 = new Vertex("High_Street_kensignton"); Vertex v2 = new Vertex("Glouchester_Road"); Vertex v3 = new Vertex("South_Kensignton"); Vertex v4 = new Vertex("Sloane_Square"); Vertex v5 = new Vertex("Victoria"); Vertex v6 = new Vertex("Westminster"); v0.adjacencies = new Edge[]{new Edge(v1, 79.83), new Edge(v6, 97.24)}; v1.adjacencies = new Edge[]{new Edge(v2, 39.42), new Edge(v0, 79.83)}; v2.adjacencies = new Edge[]{new Edge(v3, 38.65), new Edge(v1, 39.42)}; v3.adjacencies = new Edge[]{new Edge(v4, 102.53), new Edge(v2, 38.65)}; v4.adjacencies = new Edge[]{new Edge(v5, 133.04), new Edge(v3, 102.53)}; v5.adjacencies = new Edge[]{new Edge(v6, 81.77), new Edge(v4, 133.04)}; v6.adjacencies = new Edge[]{new Edge(v0, 97.24), new Edge(v5, 81.77)}; Vertex[] vertices = { v0, v1, v2, v3, v4, v5, v6 }; computePaths(v0); for (Vertex v : vertices) { System.out.println("Distance to " + v + ": " + v.minDistance); List<Vertex> path = getShortestPathTo(v); System.out.println("Path: " + path); } } } and the code for scanning file is import java.util.Scanner; import java.io.File; import java.io.FileNotFoundException; public class DataScanner1 { //private int total = 0; //private int distance = 0; private String vector; private String stations; private double [] Edge = new double []; /*public int getTotal(){ return total; } */ /* public void getMenuInput(){ KeyboardInput in = new KeyboardInput; System.out.println("Enter the destination? "); String val = in.readString(); return val; } */ public void readFile(String fileName) { try { Scanner scanner = new Scanner(new File(fileName)); scanner.useDelimiter (System.getProperty("line.separator")); while (scanner.hasNext()) { parseLine(scanner.next()); } scanner.close(); } catch (FileNotFoundException e) { e.printStackTrace(); } } public void parseLine(String line) { Scanner lineScanner = new Scanner(line); lineScanner.useDelimiter("\\s*,\\s*"); vector = lineScanner.next(); stations = lineScanner.next(); System.out.println("The current station is " + vector + " and the destination to the next station is " + stations + "."); //total += distance; //System.out.println("The total distance is " + total); } public static void main(String[] args) { /* if (args.length != 1) { System.err.println("usage: java TextScanner2" + "file location"); System.exit(0); } */ DataScanner1 scanner = new DataScanner1(); scanner.readFile(args[0]); //int total =+ distance; //System.out.println(""); //System.out.println("The total distance is " + scanner.getTotal()); } }

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  • BFS Shortest Path: Edge weight either 1 or 2

    - by Hackster
    I am trying to implement a shortest path algorithm using BFS. That is I am trying to find the shortest path from a specified vertex to every other vertex. However, its a special case where all edge weights are either 1 or 2. I know it could be done with Dijkstra's algorithm but I must use Breadth First Search. So far I have a working version of BFS that searches first for a vertex connected with an edge of weight 1. If it cannot find it, then returns a vertex connected with an edge of weight 2. After thinking about it, this is not the correct way to find the shortest path. The problem is I cannot think of any reasoning why BFS would work with weights 1 or 2, as opposed to any weight. Here is the code: public void addEdge(int start, int end, int weight) { adjMat[start][end] = 1; adjMat[end][start] = 1; edge_weight[start][end] = weight; edge_weight[end][start] = weight; } // ------------------------------------------------------------- public void bfs() // breadth-first search { // begin at vertex 0 vertexList[0].wasVisited = true; // mark it displayVertex(0); // display it theQueue.insert(0); // insert at tail int v2; while( !theQueue.isEmpty() ) // until queue empty, { int v1 = theQueue.remove(); // remove vertex at head // until it has no unvisited neighbors while( (v2=getAdjUnvisitedVertex(v1)) != -1 ){// get one, vertexList[v2].wasVisited = true; // mark it displayVertex(v2); // display it theQueue.insert(v2); // insert it } } // end while(queue not empty) // queue is empty, so we're done for(int j=0; j<nVerts; j++) // reset flags vertexList[j].wasVisited = false; } // end bfs() // ------------------------------------------------------------- // returns an unvisited vertex adj to v -- ****WITH WEIGHT 1**** public int getAdjUnvisitedVertex(int v) { for (int j = 0; j < nVerts; j++) if (adjMat[v][j] == 1 && vertexList[j].wasVisited == false && edge_weight[v][j] == 1){ //System.out.println("Vertex found with 1:"+ vertexList[j].label); return j; } for (int k = 0; k < nVerts; k++) if (adjMat[v][k] == 1 && vertexList[k].wasVisited == false && edge_weight[v][k] == 2){ //System.out.println("Vertex found with 2:"+vertexList[k].label); return k; } return -1; } // end getAdjUnvisitedVertex() // ------------------------------------------------------------- } //////////////////////////////////////////////////////////////// public class BFS{ public static void main(String[] args) { Graph theGraph = new Graph(); theGraph.addVertex('A'); // 0 (start for bfs) theGraph.addVertex('B'); // 1 theGraph.addVertex('C'); // 2 theGraph.addEdge(0, 1,2); // AB theGraph.addEdge(1, 2,1); // BC theGraph.addEdge(2, 0,1); // AD System.out.print("Visits: "); theGraph.bfs(); // breadth-first search System.out.println(); } // end main() } The problem then is, that I don't know why BFS can work for the shortest path problem with edges of weight 1 or 2 as opposed to any edges of any weight. Any help is appreciated. Thanks!

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  • Matlab - Propagate points orthogonally on to the edge of shape boundaries

    - by Graham
    Hi I have a set of points which I want to propagate on to the edge of shape boundary defined by a binary image. The shape boundary is defined by a 1px wide white edge. I also have the coordinates of these points stored in a 2 row by n column matrix. The shape forms a concave boundary with no holes within itself made of around 2500 points. I want to cast a ray from each point from the set of points in an orthogonal direction and detect at which point it intersects the shape boundary at. What would be the best method to do this? Are there some sort of ray tracing algorithms that could be used? Or would it be a case of taking orthogonal unit vector and multiplying it by a scalar and testing after multiplication if the end point of the vector is outside the shape boundary. When the end point of the unit vector is outside the shape, just find the point of intersection? Thank you very much in advance for any help!

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  • Compatibility of ESX 4 and LTO 3 with LSI2032 Dell Power Edge

    - by dbaur
    Hello, does anybody know something about the compatibility of an LTO 3 Tapedrive (PV HH LTO 3 400GB) attached to an LSI2032 SCSI-Controller? These devices are added to an Dell PowerEdge T610 Server which is listed in the certified Compatibility Guide. I can`t find any information for this controller and this tapdevice. Or are they automaticly compatible because they are options for an compatible Server? greetings Dennis

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