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  • bulls and cows game -- programming algorithm(python)

    - by IcyFlame
    This is a simulation of the game Cows and Bulls with three digit numbers I am trying to get the number of cows and bulls between two numbers. One of which is generated by the computer and the other is guessed by the user. I have parsed the two numbers I have so that now I have two lists with three elements each and each element is one of the digits in the number. So: 237 will give the list [2,3,7]. And I make sure that the relative indices are maintained.the general pattern is:(hundreds, tens, units). And these two lists are stored in the two lists: machine and person. ALGORITHM 1 So, I wrote the following code, The most intuitive algorithm: cows and bulls are initialized to 0 before the start of this loop. for x in person: if x in machine: if machine.index(x) == person.index(x): bulls += 1 print x,' in correct place' else: print x,' in wrong place' cows += 1 And I started testing this with different type of numbers guessed by the computer. Quite randomly, I decided on 277. And I guessed 447. Here, I got the first clue that this algorithm may not work. I got 1 cow and 0 bulls. Whereas I should have got 1 bull and 1 cow. This is a table of outputs with the first algorithm: Guess Output Expected Output 447 0 bull, 1 cow 1 bull, 1 cow 477 2 bulls, 0 cows 2 bulls, 0 cows 777 0 bulls, 3 cows 2 bulls, 0 cows So obviously this algorithm was not working when there are repeated digits in the number randomly selected by the computer. I tried to understand why these errors are taking place, But I could not. I have tried a lot but I just could not see any mistake in the algorithm(probably because I wrote it!) ALGORITHM 2 On thinking about this for a few days I tried this: cows and bulls are initialized to 0 before the start of this loop. for x in range(3): for y in range(3): if x == y and machine[x] == person[y]: bulls += 1 if not (x == y) and machine[x] == person[y]: cows += 1 I was more hopeful about this one. But when I tested this, this is what I got: Guess Output Expected Output 447 1 bull, 1 cow 1 bull, 1 cow 477 2 bulls, 2 cows 2 bulls, 0 cows 777 2 bulls, 4 cows 2 bulls, 0 cows The mistake I am making is quite clear here, I understood that the numbers were being counted again and again. i.e.: 277 versus 477 When you count for bulls then the 2 bulls come up and thats alright. But when you count for cows: the 7 in 277 at units place is matched with the 7 in 477 in tens place and thus a cow is generated. the 7 in 277 at tens place is matched with the 7 in 477 in units place and thus a cow is generated.' Here the matching is exactly right as I have written the code as per that. But this is not what I want. And I have no idea whatsoever on what to do after this. Furthermore... I would like to stress that both the algorithms work perfectly, if there are no repeated digits in the number selected by the computer. Please help me with this issue. P.S.: I have been thinking about this for over a week, But I could not post a question earlier as my account was blocked(from asking questions) because I asked a foolish question. And did not delete it even though I got 2 downvotes immediately after posting the question.

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  • definition of wait-free (referring to parallel programming)

    - by tecuhtli
    In Maurice Herlihy paper "Wait-free synchronization" he defines wait-free: "A wait-free implementation of a concurrent data object is one that guarantees that any process can complete any operation in a finite number of steps, regardless the execution speeds on the other processes." www.cs.brown.edu/~mph/Herlihy91/p124-herlihy.pdf Let's take one operation op from the universe. (1) Does the definition mean: "Every process completes a certain operation op in the same finite number n of steps."? (2) Or does it mean: "Every process completes a certain operation op in any finite number of steps. So that a process can complete op in k steps another process in j steps, where k != j."? Just by reading the definition i would understand meaning (2). However this makes no sense to me, since a process executing op in k steps and another time in k + m steps meets the definition, but m steps could be a waiting loop. If meaning (2) is right, can anybody explain to me, why this describes wait-free? In contrast to (2), meaning (1) would guarantee that op is executed in the same number of steps k. So there can't be any additional steps m that are necessary e.g. in a waiting loop. Which meaning is right and why? Thanks a lot, sebastian

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  • C++ Programming in Linux Platform

    - by viswanathan
    I am a software engineer and i work in VC++, C++ in WIndows OS. Are there any major differences when it comes to coding in C++ in Linux environment. Or is it just some adjustments that we have to make when we need to code in C++ in Linux.

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  • Representing a number in a byte array (java programming)

    - by Mark Roberts
    I'm trying to represent the port number 9876 (or 0x2694 in hex) in a two byte array: class foo { public static void main (String args[]) { byte[] sendData = new byte[1]; sendData[0] = 0x26; sendData[1] = 0x94; } } But I get a warning about possible loss of precision: foo.java:5: possible loss of precision found : int required: byte sendData[1] = 0x94; ^ 1 error How can I represent the number 9876 in a two byte array without losing precision?

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  • Java Programming Help

    - by user215049
    hi, im a bit confused on this and i dont know how to solve this question that i have been asked, id be grateful if you could assist me on this question, maybe try to tell me what needs to be done, and how. the question is: Write a method called countChars which takes an InputStream as a parameter, reads the stream and returns the number of characters it contains as an int. Any IOExceptions which might occur in the method should be passed back to the method's caller. Note that the method header should show that it is possible that an exception might occur. i attempted this question with the following code: public class countChars { public int countChars(int InputStream) { return InputStream; } } and i get an error message saying : Main.java:26: cannot find symbol symbol : method countChars(java.io.InputStream) location: class Main s1 = "" + countChars(f1); ^ 1 error

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  • help for hindi programming

    - by yogeshbablu
    #include<wchar.h> #include<iostream> using namespace std; int main(int argc,char* argv[]) { fputws(L"?? ?? ?????? ????",stdout); return 0; } ?? ?? ?????? ??? is displayed when i run it on ubuntu can anybody help me out?

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  • Android programming help

    - by Tuffy G
    I want to start making a program for a local charity on the Android platform. Where can go I for resources and tutorials to learn? I'm very new at this, so would like something simple that can be followed by someone with minimal technical knowledge.

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  • Dual Boot issues with Windows 7 and Ubuntu

    - by Michael
    I'm finding myself in a rather unique situation. I've read through just about every resource I can find about this and while things have helped me understand some background, I haven't yet been able to find a solution. So I'm asking here. I originally had just a Windows 7 64-bit OS installation on my desktop. Learning that I couldn't do anything with Apache, PHP and MySql from within a 64-bit system, I did some research and found out that I could use Ubuntu. I've installed the latest version: 11.04. I created a CD to install Ubuntu from and the install went just fine. I installed it side-by-side with Windows 7. I can boot into Ubuntu just fine through the dual-boot option. When I reboot to load Windows though, the Grub2 list shows Windows 7 (loader) and when I select this option the Windows System Recovery loads instead of the actual OS. I haven't made it past there because I didn't know what to do. I just shut the computer down and rebooted into Ubuntu. I've been working for the last hour and a half to try to figure out how to boot into the Windows 7 OS and I haven't got a clue. While I'm somewhat proficient with Windows 7, I'm totally new to Ubuntu, so if you do know what needs to happen, please keep it simple enough that I'll be able to understand. Thanks for all your help in advance. Here's the results after using the Boot Info Script: Boot Info Script 0.55 dated February 15th, 2010 ============================= Boot Info Summary: ============================== => Grub 2 is installed in the MBR of /dev/sda and looks on the same drive in partition #5 for cbh. => Windows is installed in the MBR of /dev/sdb => Grub 2 is installed in the MBR of /dev/mapper/pdc_bdadcfbdif and looks on the same drive in partition #5 for cbh. sda1: _________________________________________________________________________ File system: ntfs Boot sector type: Windows Vista/7 Boot sector info: No errors found in the Boot Parameter Block. Mounting failed: fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy sda2: _________________________________________________________________________ File system: ntfs Boot sector type: Windows Vista/7 Boot sector info: No errors found in the Boot Parameter Block. Mounting failed: fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy sda3: _________________________________________________________________________ File system: ntfs Boot sector type: Windows Vista/7 Boot sector info: No errors found in the Boot Parameter Block. Mounting failed: fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy sdb1: _________________________________________________________________________ File system: ntfs Boot sector type: Windows Vista/7 Boot sector info: No errors found in the Boot Parameter Block. Operating System: Boot files/dirs: /bootmgr /Boot/BCD sdb2: _________________________________________________________________________ File system: ntfs Boot sector type: Windows Vista/7 Boot sector info: No errors found in the Boot Parameter Block. Operating System: Boot files/dirs: sdb3: _________________________________________________________________________ File system: ntfs Boot sector type: Windows Vista/7 Boot sector info: No errors found in the Boot Parameter Block. Operating System: Boot files/dirs: /bootmgr /boot/BCD sdb4: _________________________________________________________________________ File system: Extended Partition Boot sector type: - Boot sector info: sdb5: _________________________________________________________________________ File system: ext4 Boot sector type: - Boot sector info: Operating System: Ubuntu 11.04 Boot files/dirs: /boot/grub/grub.cfg /etc/fstab /boot/grub/core.img sdb6: _________________________________________________________________________ File system: swap Boot sector type: - Boot sector info: pdc_bdadcfbdif1: _________________________________________________________________________ File system: ntfs Boot sector type: Windows Vista/7 Boot sector info: No errors found in the Boot Parameter Block. Operating System: Boot files/dirs: /bootmgr /Boot/BCD pdc_bdadcfbdif2: _________________________________________________________________________ File system: ntfs Boot sector type: Windows Vista/7 Boot sector info: No errors found in the Boot Parameter Block. Operating System: Windows 7 Boot files/dirs: /bootmgr /Boot/BCD /Windows/System32/winload.exe pdc_bdadcfbdif3: _________________________________________________________________________ File system: Boot sector type: Unknown Boot sector info: Mounting failed: fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy fuse: mount failed: Device or resource busy mount: unknown filesystem type '' =========================== Drive/Partition Info: ============================= Drive: sda ___________________ _____________________________________________________ Disk /dev/sda: 750.2 GB, 750156374016 bytes 255 heads, 63 sectors/track, 91201 cylinders, total 1465149168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes Partition Boot Start End Size Id System /dev/sda1 * 2,048 206,847 204,800 7 HPFS/NTFS /dev/sda2 206,911 1,440,372,735 1,440,165,825 7 HPFS/NTFS /dev/sda3 1,440,372,736 1,464,856,575 24,483,840 7 HPFS/NTFS Drive: sdb ___________________ _____________________________________________________ Disk /dev/sdb: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes Partition Boot Start End Size Id System /dev/sdb1 * 2,048 206,847 204,800 7 HPFS/NTFS /dev/sdb2 206,911 1,342,554,688 1,342,347,778 7 HPFS/NTFS /dev/sdb3 1,930,344,448 1,953,521,663 23,177,216 7 HPFS/NTFS /dev/sdb4 1,342,556,158 1,930,344,447 587,788,290 5 Extended /dev/sdb5 1,342,556,160 1,896,806,399 554,250,240 83 Linux /dev/sdb6 1,896,808,448 1,930,344,447 33,536,000 82 Linux swap / Solaris Drive: pdc_bdadcfbdif ___________________ _____________________________________________________ Disk /dev/mapper/pdc_bdadcfbdif: 750.0 GB, 749999947776 bytes 255 heads, 63 sectors/track, 91182 cylinders, total 1464843648 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes Partition Boot Start End Size Id System /dev/mapper/pdc_bdadcfbdif1 * 2,048 206,847 204,800 7 HPFS/NTFS /dev/mapper/pdc_bdadcfbdif2 206,911 1,440,372,735 1,440,165,825 7 HPFS/NTFS /dev/mapper/pdc_bdadcfbdif3 1,440,372,736 1,464,856,575 24,483,840 7 HPFS/NTFS /dev/mapper/pdc_bdadcfbdif3 ends after the last sector of /dev/mapper/pdc_bdadcfbdif blkid -c /dev/null: ____________________________________________________________ Device UUID TYPE LABEL /dev/mapper/pdc_bdadcfbdif1 888E54CC8E54B482 ntfs SYSTEM /dev/mapper/pdc_bdadcfbdif2 C2766BF6766BEA1D ntfs OS /dev/mapper/pdc_bdadcfbdif: PTTYPE="dos" /dev/sda1 888E54CC8E54B482 ntfs SYSTEM /dev/sda2 C2766BF6766BEA1D ntfs OS /dev/sda3 BE6CA31D6CA2CF87 ntfs HP_RECOVERY /dev/sda promise_fasttrack_raid_member /dev/sdb1 20B65685B6565B7C ntfs SYSTEM /dev/sdb2 B4467A314679F508 ntfs HP /dev/sdb3 6E10B7A410B77227 ntfs FACTORY_IMAGE /dev/sdb4: PTTYPE="dos" /dev/sdb5 266f9801-cf4f-4acc-affa-2092be035f0c ext4 /dev/sdb6 1df35749-a887-45ff-a3de-edd52239847d swap /dev/sdb: PTTYPE="dos" error: /dev/mapper/pdc_bdadcfbdif3: No such file or directory error: /dev/sdc: No medium found error: /dev/sdd: No medium found error: /dev/sde: No medium found error: /dev/sdf: No medium found error: /dev/sdg: No medium found ============================ "mount | grep ^/dev output: =========================== Device Mount_Point Type Options /dev/sdb5 / ext4 (rw,errors=remount-ro,commit=0) =========================== sdb5/boot/grub/grub.cfg: =========================== # # DO NOT EDIT THIS FILE # # It is automatically generated by grub-mkconfig using templates # from /etc/grub.d and settings from /etc/default/grub # ### BEGIN /etc/grub.d/00_header ### if [ -s $prefix/grubenv ]; then set have_grubenv=true load_env fi set default="0" if [ "${prev_saved_entry}" ]; then set saved_entry="${prev_saved_entry}" save_env saved_entry set prev_saved_entry= save_env prev_saved_entry set boot_once=true fi function savedefault { if [ -z "${boot_once}" ]; then saved_entry="${chosen}" save_env saved_entry fi } function recordfail { set recordfail=1 if [ -n "${have_grubenv}" ]; then if [ -z "${boot_once}" ]; then save_env recordfail; fi; fi } function load_video { insmod vbe insmod vga insmod video_bochs insmod video_cirrus } insmod part_msdos insmod ext2 set root='(/dev/sdb,msdos5)' search --no-floppy --fs-uuid --set=root 266f9801-cf4f-4acc-affa-2092be035f0c if loadfont /usr/share/grub/unicode.pf2 ; then set gfxmode=auto load_video insmod gfxterm fi terminal_output gfxterm insmod part_msdos insmod ext2 set root='(/dev/sdb,msdos5)' search --no-floppy --fs-uuid --set=root 266f9801-cf4f-4acc-affa-2092be035f0c set locale_dir=($root)/boot/grub/locale set lang=en_US insmod gettext if [ "${recordfail}" = 1 ]; then set timeout=-1 else set timeout=10 fi ### END /etc/grub.d/00_header ### ### BEGIN /etc/grub.d/05_debian_theme ### set menu_color_normal=white/black set menu_color_highlight=black/light-gray if background_color 44,0,30; then clear fi ### END /etc/grub.d/05_debian_theme ### ### BEGIN /etc/grub.d/10_linux ### if [ ${recordfail} != 1 ]; then if [ -e ${prefix}/gfxblacklist.txt ]; then if hwmatch ${prefix}/gfxblacklist.txt 3; then if [ ${match} = 0 ]; then set linux_gfx_mode=keep else set linux_gfx_mode=text fi else set linux_gfx_mode=text fi else set linux_gfx_mode=keep fi else set linux_gfx_mode=text fi export linux_gfx_mode if [ "$linux_gfx_mode" != "text" ]; then load_video; fi menuentry 'Ubuntu, with Linux 2.6.38-8-generic-pae' --class ubuntu --class gnu-linux --class gnu --class os { recordfail set gfxpayload=$linux_gfx_mode insmod part_msdos insmod ext2 set root='(/dev/sdb,msdos5)' search --no-floppy --fs-uuid --set=root 266f9801-cf4f-4acc-affa-2092be035f0c linux /boot/vmlinuz-2.6.38-8-generic-pae root=UUID=266f9801-cf4f-4acc- affa-2092be035f0c ro quiet splash vt.handoff=7 initrd /boot/initrd.img-2.6.38-8-generic-pae } menuentry 'Ubuntu, with Linux 2.6.38-8-generic-pae (recovery mode)' --class ubuntu --class gnu-linux --class gnu --class os { recordfail set gfxpayload=$linux_gfx_mode insmod part_msdos insmod ext2 set root='(/dev/sdb,msdos5)' search --no-floppy --fs-uuid --set=root 266f9801-cf4f-4acc-affa-2092be035f0c echo 'Loading Linux 2.6.38-8-generic-pae ...' linux /boot/vmlinuz-2.6.38-8-generic-pae root=UUID=266f9801-cf4f-4acc-affa-2092be035f0c ro single echo 'Loading initial ramdisk ...' initrd /boot/initrd.img-2.6.38-8-generic-pae } ### END /etc/grub.d/10_linux ### ### BEGIN /etc/grub.d/20_linux_xen ### ### END /etc/grub.d/20_linux_xen ### ### BEGIN /etc/grub.d/20_memtest86+ ### menuentry "Memory test (memtest86+)" { insmod part_msdos insmod ext2 set root='(/dev/sdb,msdos5)' search --no-floppy --fs-uuid --set=root 266f9801-cf4f-4acc-affa-2092be035f0c linux16 /boot/memtest86+.bin } menuentry "Memory test (memtest86+, serial console 115200)" { insmod part_msdos insmod ext2 set root='(/dev/sdb,msdos5)' search --no-floppy --fs-uuid --set=root 266f9801-cf4f-4acc-affa-2092be035f0c linux16 /boot/memtest86+.bin console=ttyS0,115200n8 } ### END /etc/grub.d/20_memtest86+ ### ### BEGIN /etc/grub.d/30_os-prober ### menuentry "Windows 7 (loader) (on /dev/sdb1)" --class windows --class os { insmod part_msdos insmod ntfs set root='(/dev/sdb,msdos1)' search --no-floppy --fs-uuid --set=root 20B65685B6565B7C chainloader +1 } menuentry "Windows Recovery Environment (loader) (on /dev/sdb3)" --class windows --class os { insmod part_msdos insmod ntfs set root='(/dev/sdb,msdos3)' search --no-floppy --fs-uuid --set=root 6E10B7A410B77227 drivemap -s (hd0) ${root} chainloader +1 } ### END /etc/grub.d/30_os-prober ### ### BEGIN /etc/grub.d/40_custom ### # This file provides an easy way to add custom menu entries. Simply type the # menu entries you want to add after this comment. Be careful not to change # the 'exec tail' line above. ### END /etc/grub.d/40_custom ### ### BEGIN /etc/grub.d/41_custom ### if [ -f $prefix/custom.cfg ]; then source $prefix/custom.cfg; fi ### END /etc/grub.d/41_custom ### =============================== sdb5/etc/fstab: =============================== # /etc/fstab: static file system information. # # Use 'blkid -o value -s UUID' to print the universally unique identifier # for a device; this may be used with UUID= as a more robust way to name # devices that works even if disks are added and removed. See fstab(5). # # <file system> <mount point> <type> <options> <dump> <pass> proc /proc proc nodev,noexec,nosuid 0 0 # / was on /dev/sdb5 during installation UUID=266f9801-cf4f-4acc-affa-2092be035f0c / ext4 errors=remount-ro 0 1 # swap was on /dev/sdb6 during installation UUID=1df35749-a887-45ff-a3de-edd52239847d none swap sw 0 0 =================== sdb5: Location of files loaded by Grub: =================== 900.1GB: boot/grub/core.img 825.0GB: boot/grub/grub.cfg 688.7GB: boot/initrd.img-2.6.38-8-generic-pae 688.0GB: boot/vmlinuz-2.6.38-8-generic-pae 688.7GB: initrd.img 688.0GB: vmlinuz =========================== Unknown MBRs/Boot Sectors/etc ======================= Unknown BootLoader on pdc_bdadcfbdif3 =======Devices which don't seem to have a corresponding hard drive============== sdc sdd sde sdf sdg =============================== StdErr Messages: =============================== ERROR: dos: partition address past end of RAID device hexdump: /dev/mapper/pdc_bdadcfbdif3: No such file or directory hexdump: /dev/mapper/pdc_bdadcfbdif3: No such file or directory ERROR: dos: partition address past end of RAID device

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  • BizTalk: History of one project architecture

    - by Leonid Ganeline
    "In the beginning God made heaven and earth. Then he started to integrate." At the very start was the requirement: integrate two working systems. Small digging up: It was one system. It was good but IT guys want to change it to the new one, much better, chipper, more flexible, and more progressive in technologies, more suitable for the future, for the faster world and hungry competitors. One thing. One small, little thing. We cannot turn off the old system (call it A, because it was the first), turn on the new one (call it B, because it is second but not the last one). The A has a hundreds users all across a country, they must study B. A still has a lot nice custom features, home-made features that cannot disappear. These features have to be moved to the B and it is a long process, months and months of redevelopment. So, the decision was simple. Let’s move not jump, let’s both systems working side-by-side several months. In this time we could teach the users and move all custom A’s special functionality to B. That automatically means both systems should work side-by-side all these months and use the same data. Data in A and B must be in sync. That’s how the integration projects get birth. Moreover, the specific of the user tasks requires the both systems must be in sync in real-time. Nightly synchronization is not working, absolutely.   First draft The first draft seems simple. Both systems keep data in SQL databases. When data changes, the Create, Update, Delete operations performed on the data, and the sync process could be started. The obvious decision is to use triggers on tables. When we are talking about data, we are talking about several entities. For example, Orders and Items [in Orders]. We decided to use the BizTalk Server to synchronize systems. Why it was chosen is another story. Second draft   Let’s take an example how it works in more details. 1.       User creates a new entity in the A system. This fires an insert trigger on the entity table. Trigger has to pass the message “Entity created”. This message includes all attributes of the new entity, but I focused on the Id of this entity in the A system. Notation for this message is id.A. System A sends id.A to the BizTalk Server. 2.       BizTalk transforms id.A to the format of the system B. This is easiest part and I will not focus on this kind of transformations in the following text. The message on the picture is still id.A but it is in slightly different format, that’s why it is changing in color. BizTalk sends id.A to the system B. 3.       The system B creates the entity on its side. But it uses different id-s for entities, these id-s are id.B. System B saves id.A+id.B. System B sends the message id.A+id.B back to the BizTalk. 4.       BizTalk sends the message id.A+id.B to the system A. 5.       System A saves id.A+id.B. Why both id-s should be saved on both systems? It was one of the next requirements. Users of both systems have to know the systems are in sync or not in sync. Users working with the entity on the system A can see the id.B and use it to switch to the system B and work there with the copy of the same entity. The decision was to store the pairs of entity id-s on both sides. If there is only one id, the entities are not in sync yet (for the Create operation). Third draft Next problem was the reliability of the synchronization. The synchronizing process can be interrupted on each step, when message goes through the wires. It can be communication problem, timeout, temporary shutdown one of the systems, the second system cannot be synchronized by some internal reason. There were several potential problems that prevented from enclosing the whole synchronization process in one transaction. Decision was to restart the whole sync process if it was not finished (in case of the error). For this purpose was created an additional service. Let’s call it the Resync service. We still keep the id pairs in both systems, but only for the fast access not for the synchronization process. For the synchronizing these id-s now are kept in one main place, in the Resync service database. The Resync service keeps record as: ·       Id.A ·       Id.B ·       Entity.Type ·       Operation (Create, Update, Delete) ·       IsSyncStarted (true/false) ·       IsSyncFinished (true/false0 The example now looks like: 1.       System A creates id.A. id.A is saved on the A. Id.A is sent to the BizTalk. 2.       BizTalk sends id.A to the Resync and to the B. id.A is saved on the Resync. 3.       System B creates id.B. id.A+id.B are saved on the B. id.A+id.B are sent to the BizTalk. 4.       BizTalk sends id.A+id.B to the Resync and to the A. id.A+id.B are saved on the Resync. 5.       id.A+id.B are saved on the B. Resync changes the IsSyncStarted and IsSyncFinished flags accordingly. The Resync service implements three main methods: ·       Save (id.A, Entity.Type, Operation) ·       Save (id.A, id.B, Entity.Type, Operation) ·       Resync () Two Save() are used to save id-s to the service storage. See in the above example, in 2 and 4 steps. What about the Resync()? It is the method that finishes the interrupted synchronization processes. If Save() is started by the trigger event, the Resync() is working as an independent process. It periodically scans the Resync storage to find out “unfinished” records. Then it restarts the synchronization processes. It tries to synchronize them several times then gives up.     One more thing, both systems A and B must tolerate duplicates of one synchronizing process. Say on the step 3 the system B was not able to send id.A+id.B back. The Resync service must restart the synchronization process that will send the id.A to B second time. In this case system B must just send back again also created id.A+id.B pair without errors. That means “tolerate duplicates”. Fourth draft Next draft was created only because of the aesthetics. As it always happens, aesthetics gave significant performance gain to the whole system. First was the stupid question. Why do we need this additional service with special database? Can we just master the BizTalk to do something like this Resync() does? So the Resync orchestration is doing the same thing as the Resync service. It is started by the Id.A and finished by the id.A+id.B message. The first works as a Start message, the second works as a Finish message.     Here is a diagram the whole process without errors. It is pretty straightforward. The Resync orchestration is waiting for the Finish message specific period of time then resubmits the Id.A message. It resubmits the Id.A message specific number of times then gives up and gets suspended. It can be resubmitted then it starts the whole process again: waiting [, resubmitting [, get suspended]], finishing. Tuning up The Resync orchestration resubmits the id.A message with special “Resubmitted” flag. The subscription filter on the Resync orchestration includes predicate as (Resubmit_Flag != “Resubmitted”). That means only the first Sync orchestration starts the Resync orchestration. Other Sync orchestration instantiated by the resubmitting can finish this Resync orchestration but cannot start another instance of the Resync   Here is a diagram where system B was inaccessible for some period of time. The Resync orchestration resubmitted the id.A two times. Then system B got the response the id.A+id.B and this finished the Resync service execution. What is interesting about this, there were submitted several identical id.A messages and only one id.A+id.B message. Because of this, the system B and the Resync must tolerate the duplicate messages. We also told about this requirement for the system B. Now the same requirement is for the Resunc. Let’s assume the system B was very slow in the first response and the Resync service had time to resubmit two id.A messages. System B responded not, as it was in previous case, with one id.A+id.B but with two id.A+id.B messages. First of them finished the Resync execution for the id.A. What about the second id.A+id.B? Where it goes? So, we have to add one more internal requirement. The whole solution must tolerate many identical id.A+id.B messages. It is easy task with the BizTalk. I added the “SinkExtraMessages” subscriber (orchestration with one receive shape), that just get these messages and do nothing. Real design Real architecture is much more complex and interesting. In reality each system can submit several id.A almost simultaneously and completely unordered. There are not only the “Create entity” operation but the Update and Delete operations. And these operations relate each other. Say the Update operation after Delete means not the same as Update after Create. In reality there are entities related each other. Say the Order and Order Items. Change on one of it could start the series of the operations on another. Moreover, the system internals are the “black boxes” and we cannot predict the exact content and order of the operation series. It worth to say, I had to spend a time to manage the zombie message problems. The zombies are still here, but this is not a problem now. And this is another story. What is interesting in the last design? One orchestration works to help another to be more reliable. Why two orchestration design is more reliable, isn’t it something strange? The Synch orchestration takes all the message exchange between systems, here is the area where most of the errors could happen. The Resync orchestration sends and receives messages only within the BizTalk server. Is there another design? Sure. All Resync functionality could be implemented inside the Sync orchestration. Hey guys, some other ideas?

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  • Code structure for multiple applications with a common core

    - by Azrael Seraphin
    I want to create two applications that will have a lot of common functionality. Basically, one system is a more advanced version of the other system. Let's call them Simple and Advanced. The Advanced system will add to, extend, alter and sometimes replace the functionality of the Simple system. For instance, the Advanced system will add new classes, add properties and methods to existing Simple classes, change the behavior of classes, etc. Initially I was thinking that the Advanced classes simply inherited from the Simple classes but I can see the functionality diverging quite significantly as development progresses, even while maintaining a core base functionality. For instance, the Simple system might have a Project class with a Sponsor property whereas the Advanced system has a list of Project.Sponsors. It seems poor practice to inherit from a class and then hide, alter or throw away significant parts of its features. An alternative is just to run two separate code bases and copy the common code between them but that seems inefficient, archaic and fraught with peril. Surely we have moved beyond the days of "copy-and-paste inheritance". Another way to structure it would be to use partial classes and have three projects: Core which has the common functionality, Simple which extends the Core partial classes for the simple system, and Advanced which also extends the Core partial classes for the advanced system. Plus having three test projects as well for each system. This seems like a cleaner approach. What would be the best way to structure the solution/projects/code to create two versions of a similar system? Let's say I later want to create a third system called Extreme, largely based on the Advanced system. Do I then create an AdvancedCore project which both Advanced and Extreme extend using partial classes? Is there a better way to do this? If it matters, this is likely to be a C#/MVC system but I'd be happy to do this in any language/framework that is suitable.

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  • Is there a common programming term for the problems of adding features to an already-featureful program?

    - by Jeremy Friesner
    I'm looking for a commonly used programming term to describe a software-engineering phenomenon, which (for lack of a better way to describe it) I'll illustrate first with a couple of examples-by-analogy: Scenario 1: We want to build/extend a subway system on the outskirts of a small town in Wyoming. There are the usual subway-problems to solve, of course (hiring the right construction company, choosing the best route, buying the subway cars), but other than that it's pretty straightforward to implement the system because there aren't a huge number of constraints to satisfy. Scenario 2: Same as above, except now we need to build/extend the subway system in downtown Los Angeles. Here we face all of the problems we did in case (1), but also additional problems -- most of the applicable space is already in use, and has a vocal constituency which will protest loudly if we inconvenience them by repurposing, redesigning, or otherwise modifying the infrastructure that they rely on. Because of this, extensions to the system happen either very slowly and expensively, or they don't happen at all. I sometimes see a similar pattern with software development -- adding a new feature to a small/simple program is straightforward, but as the program grows, adding further new features becomes more and more difficult, if only because it is difficult to integrate the new feature without adversely affecting any of the large number of existing use-cases or user-constituencies. (even with a robust, adaptable program design, you run into the problem of the user interface becoming so elaborate that the program becomes difficult to learn or use) Is there a term for this phenomenon?

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  • What are basic programs like, recursion, Fibonacci, small trick programs?

    - by Mike
    This question may seem daft (I'm a new to 'programming' and should probably stop if this is the type of question I'm required to ask)... What are: "basic programs like, recursion, fibonacci, factorial, string manipulation, small trick programs"? I've recently read Coding Horror - the non programmer and followed the links to Kegel and How to get hired. Then I delved through some similar questions here (hence the block quote) and I realised that as a fully fledged non-programmer I probably wouldn't know if I knew recursion (or any of the others) because I wouldn't know what it looked like, or why it was used, and what the results would look like after it was used. I suppose I'm trying to get a picture of "the basics". What the principles are and why we learn them - where they'll be used and what result/s your looking for. If they'll be used as an interview question during my first interview sometime in 2020 I would like to look less ignorant than those 199 out of 200 who just don't know the how, or the why, of programming. As always...I'll get my coat. Thanks Mike

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  • perl multiple tasks problem

    - by Alice Wozownik
    I have finished my earlier multithreaded program that uses perl threads and it works on my system. The problem is that on some systems that it needs to run on, thread support is not compiled into perl and I cannot install additional packages. I therefore need to use something other than threads, and I am moving my code to using fork(). This works on my windows system in starting the subtasks. A few problems: How to determine when the child process exits? I created new threads when the thread count was below a certain value, I need to keep track of how many threads are running. For processes, how do I know when one exits so I can keep track of how many exist at the time, incrementing a counter when one is created and decrementing when one exits? Is file I/O using handles obtained with OPEN when opened by the parent process safe in the child process? I need to append to a file for each of the child processes, is this safe on unix as well. Is there any alternative to fork and threads? I tried use Parallel::ForkManager, but that isn't installed on my system (use Parallel::ForkManager; gave an error) and I absolutely require that my perl script work on all unix/windows systems without installing any additional modules.

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  • Why is the Clojure Hello World program so slow compared to Java and Python?

    - by viksit
    Hi all, I'm reading "Programming Clojure" and I was comparing some languages I use for some simple code. I noticed that the clojure implementations were the slowest in each case. For instance, Python - hello.py def hello_world(name): print "Hello, %s" % name hello_world("world") and result, $ time python hello.py Hello, world real 0m0.027s user 0m0.013s sys 0m0.014s Java - hello.java import java.io.*; public class hello { public static void hello_world(String name) { System.out.println("Hello, " + name); } public static void main(String[] args) { hello_world("world"); } } and result, $ time java hello Hello, world real 0m0.324s user 0m0.296s sys 0m0.065s and finally, Clojure - hellofun.clj (defn hello-world [username] (println (format "Hello, %s" username))) (hello-world "world") and results, $ time clj hellofun.clj Hello, world real 0m1.418s user 0m1.649s sys 0m0.154s Thats a whole, garangutan 1.4 seconds! Does anyone have pointers on what the cause of this could be? Is Clojure really that slow, or are there JVM tricks et al that need to be used in order to speed up execution? More importantly - isn't this huge difference in performance going to be an issue at some point? (I mean, lets say I was using Clojure for a production system - the gain I get in using lisp seems completely offset by the performance issues I can see here). The machine used here is a 2007 Macbook Pro running Snow Leopard, a 2.16Ghz Intel C2D and 2G DDR2 SDRAM. BTW, the clj script I'm using is from here and looks like, #!/bin/bash JAVA=/System/Library/Frameworks/JavaVM.framework/Versions/1.6/Home/bin/java CLJ_DIR=/opt/jars CLOJURE=$CLJ_DIR/clojure.jar CONTRIB=$CLJ_DIR/clojure-contrib.jar JLINE=$CLJ_DIR/jline-0.9.94.jar CP=$PWD:$CLOJURE:$JLINE:$CONTRIB # Add extra jars as specified by `.clojure` file if [ -f .clojure ] then CP=$CP:`cat .clojure` fi if [ -z "$1" ]; then $JAVA -server -cp $CP \ jline.ConsoleRunner clojure.lang.Repl else scriptname=$1 $JAVA -server -cp $CP clojure.main $scriptname -- $* fi

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  • What does it mean that "Lisp can be written in itself?"

    - by Mason Wheeler
    Paul Graham wrote that "The unusual thing about Lisp-- in fact, the defining quality of Lisp-- is that it can be written in itself." But that doesn't seem the least bit unusual or definitive to me. ISTM that a programming language is defined by two things: Its compiler or interpreter, which defines the syntax and the semantics for the language by fiat, and its standard library, which defines to a large degree the idioms and techniques that skilled users will use when writing code in the language. With a few specific exceptions, (the non-C# members of the .NET family, for example,) most languages' standard libraries are written in that language for two very good reasons: because it will share the same set of syntactical definitions, function calling conventions, and the general "look and feel" of the language, and because the people who are likely to write a standard library for a programming language are its users, and particularly its designer(s). So there's nothing unique there; that's pretty standard. And again, there's nothing unique or unusual about a language's compiler being written in itself. C compilers are written in C. Pascal compilers are written in Pascal. Mono's C# compiler is written in C#. Heck, even some scripting languages have implementations "written in itself". So what does it mean that Lisp is unusual in being written in itself?

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  • NSPredicate as a constraint solver?

    - by Felixyz
    I'm working on a project which includes some slightly more complex dynamic layout of interface elements than what I'm used to. I always feel stupid writing complex code that checks if so-and-so is close to such-and-such and in that case move it x% in some direction, etc. That's just not how programming should be done. Programming should be as declarative as possible! Precisely because what I'm going to do is fairly simple, I thought it would be a good opportunity to try something new, and I thought of using NSPredicate as a simple constraints solver. I've only used NSPredicate for very simple tasks so far, but I know that it capable of much more. Are there any ideas, experiences, examples, warnings, insights that could be useful here? I'll give a very simple example so there will be something concrete to answer. How could I use NSPredicate to solve the following constraints: viewB.xmid = (viewB.leftEdge + viewB.width) / 2 viewB.xmid = max(300, viewA.rightEdge + 20 + viewB.width/2) ("viewB should be horizontally centered on coordinate 300, unless its left edge gets within 20 pixels of viewB's right edge, in which case viewA's left edge should stay fixed at 20 pixels to the right of viewB's right edge and viewA's horizontal center get pushed to the right.") viewA.rightEdge and viewB.width can vary, and those are the 'input variables'. EDIT: Any solution would probably have to use the NSExpression method -(id)expressionValueWithObject:(id)object context:(NSMutableDictionary *)context. This answer is relevant.

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  • Need some ignition for learning Embedded Systems

    - by Rahul
    I'm very much interested in building applications for Embedded Devices. I'm in my 3rd year Electrical Engineering and I'm passionate about coding, algorithms, Linux OS, etc. And also by Googling I found out that Linux OS is one of the best OSes for Embedded devices(may be/may not be). I want to work for companies which work on mobile applications. I'm a newbie/naive to this domain & my skills include C/C++ & MySQL. I need help to get started in the domain of Embedded Systems; like how/where to start off, Hardware prerequisites, necessary programming skills, also what kind of Embedded Applications etc. I've heard of ARM, firmware, PIC Micorcontrollers; but I don't know anything & just need proper introduction about them. Thanx. P.S: I'm currently reading Bjarne Struotsup's lecture in C++ at Texas A&M University, and one chapter in it describes about Embedded Systems Programming.

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  • Need some ignition in Embedded Systems

    - by Rahul
    I'm very much interested in building applications for Embedded Devices. I'm in my 3rd year Electrical Engineering and I'm passionate about coding, algorithms, Linux OS, etc. And also by Googling I found out that Linux OS is one of the best OSes for Embedded devices(may be/may not be). I want to work for companies which work on mobile applications. I'm a newbie/naive to this domain & my skills include C/C++ & MySQL. I need help to get started in the domain of Embedded Systems; like how/where to start off, Hardware prerequisites, necessary programming skills, also what kind of Embedded Applications etc. I've heard of ARM, firmware, PIC Micorcontrollers; but I don't know anything & just need proper introduction about them. Thanx. P.S: I'm currently reading Bjarne Struotsup's lecture in C++ at Texas A&M University, and one chapter in it describes about Embedded Systems Programming.

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  • minimum L sum in a mxn matrix - 2

    - by hilal
    Here is my first question about maximum L sum and here is different and hard version of it. Problem : Given a mxn *positive* integer matrix find the minimum L sum from 0th row to the m'th row . L(4 item) likes chess horse move Example : M = 3x3 0 1 2 1 3 2 4 2 1 Possible L moves are : (0 1 2 2), (0 1 3 2) (0 1 4 2) We should go from 0th row to the 3th row with minimum sum I solved this with dynamic-programming and here is my algorithm : 1. Take a mxn another Minimum L Moves Sum array and copy the first row of main matrix. I call it (MLMS) 2. start from first cell and look the up L moves and calculate it 3. insert it in MLMS if it is less than exists value 4. Do step 2. until m'th row 5. Choose the minimum sum in the m'th row Let me explain on my example step by step: M[ 0 ][ 0 ] sum(L1 = (0, 1, 2, 2)) = 5 ; sum(L2 = (0,1,3,2)) = 6; so MLMS[ 0 ][ 1 ] = 6 sum(L3 = (0, 1, 3, 2)) = 6 ; sum(L4 = (0,1,4,2)) = 7; so MLMS[ 2 ][ 1 ] = 6 M[ 0 ][ 1 ] sum(L5 = (1, 0, 1, 4)) = 6; sum(L6 = (1,3,2,4)) = 10; so MLMS[ 2 ][ 2 ] = 6 ... the last MSLS is : 0 1 2 4 3 6 6 6 6 Which means 6 is the minimum L sum that can be reach from 0 to the m. I think it is O(8*(m-1)*n) = O(m*n). Is there any optimal solution or dynamic-programming algorithms fit this problem? Thanks, sorry for long question

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