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  • Algorithm to Render a Horizontal Binary-ish Tree in Text/ASCII form

    - by Justin L.
    It's a pretty normal binary tree, except for the fact that one of the nodes may be empty. I'd like to find a way to output it in a horizontal way (that is, the root node is on the left and expands to the right). I've had some experience expanding trees vertically (root node at the top, expanding downwards), but I'm not sure where to start, in this case. Preferably, it would follow these couple of rules: If a node has only one child, it can be skipped as redundant (an "end node", with no children, is always displayed) All nodes of the same depth must be aligned vertically; all nodes must be to the right of all less-deep nodes and to the left of all deeper nodes. Nodes have a string representation which includes their depth. Each "end node" has its own unique line; that is, the number of lines is the number of end nodes in the tree, and when an end node is on a line, there may be nothing else on that line after that end node. As a consequence of the last rule, the root node should be in either the top left or the bottom left corner; top left is preferred. For example, this is a valid tree, with six end nodes (node is represented by a name, and its depth): [a0]------------[b3]------[c5]------[d8] \ \ \----------[e9] \ \----[f5] \--[g1]--------[h4]------[i6] \ \--------------------[j10] \-[k3] Which represents the horizontal, explicit binary tree: 0 a / \ 1 g * / \ \ 2 * * * / \ \ 3 k * b / / \ 4 h * * / \ \ \ 5 * * f c / \ / \ 6 * i * * / / \ 7 * * * / / \ 8 * * d / / 9 * e / 10 j (branches folded for compactness; * representing redundant, one-child nodes; note that *'s are actual nodes, storing one child each, just with names omitted here for presentation sake) (also, to clarify, I'd like to generate the first, horizontal tree; not this vertical tree) I say language-agnostic because I'm just looking for an algorithm; I say ruby because I'm eventually going to have to implement it in ruby anyway. Assume that each Node data structure stores only its id, a left node, and a right node. A master Tree class keeps tracks of all nodes and has adequate algorithms to find: A node's nth ancestor A node's nth descendant The generation of a node The lowest common ancestor of two given nodes Anyone have any ideas of where I could start? Should I go for the recursive approach? Iterative?

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  • Data Modeling: Logical Modeling Exercise

    - by swisscheese
    In trying to learn the art of data storage I have been trying to take in as much solid information as possible. PerformanceDBA posted some really helpful tutorials/examples in the following posts among others: is my data normalized? and Relational table naming convention. I already asked a subset question of this model here. So to make sure I understood the concepts he presented and I have seen elsewhere I wanted to take things a step or two further and see if I am grasping the concepts. Hence the purpose of this post, which hopefully others can also learn from. Everything I present is conceptual to me and for learning rather than applying it in some production system. It would be cool to get some input from PerformanceDBA also since I used his models to get started, but I appreciate all input given from anyone. As I am new to databases and especially modeling I will be the first to admit that I may not always ask the right questions, explain my thoughts clearly, or use the right verbage due to lack of expertise on the subject. So please keep that in mind and feel free to steer me in the right direction if I head off track. If there is enough interest in this I would like to take this from the logical to physical phases to show the evolution of the process and share it here on Stack. I will keep this thread for the Logical Diagram though and start new one for the additional steps. For my understanding I will be building a MySQL DB in the end to run some tests and see if what I came up with actually works. Here is the list of things that I want to capture in this conceptual model. Edit for V1.2 The purpose of this is to list Bands, their members, and the Events that they will be appearing at, as well as offer music and other merchandise for sale Members will be able to match up with friends Members can write reviews on the Bands, their music, and their events. There can only be one review per member on a given item, although they can edit their reviews and history will be maintained. BandMembers will have the chance to write a single Comment on Reviews about the Band they are associated with. Collectively as a Band only one Comment is allowed per Review. Members can then rate all Reviews and Comments but only once per given instance Members can select their favorite Bands, music, Merchandise, and Events Bands, Songs, and Events will be categorized into the type of Genre that they are and then further subcategorized into a SubGenre if necessary. It is ok for a Band or Event to fall into more then one Genre/SubGenre combination. Event date, time, and location will be posted for a given band and members can show that they will be attending the Event. An Event can be comprised of more than one Band, and multiple Events can take place at a single location on the same day Every party will be tied to at least one address and address history shall be maintained. Each party could also be tied to more then one address at a time (i.e. billing, shipping, physical) There will be stored profiles for Bands, BandMembers, and general members. So there it is, maybe a bit involved but could be a great learning tool for many hopefully as the process evolves and input is given by the community. Any input? EDIT v1.1 In response to PerformanceDBA U.3) That means no merchandise other than Band merchandise in the database. Correct ? That was my original thought but you got me thinking. Maybe the site would want to sell its own merchandise or even other merchandise from the bands. Not sure a mod to make for that. Would it require an entire rework of the Catalog section or just the identifying relationship that exists with the Band? Attempted a mod to sell both complete albums or song. Either way they would both be in electronic format only available for download. That is why I listed an Album as being comprised of Songs rather then 2 separate entities. U.5) I understand what you bring up about the circular relation with Favorite. I would like to get to this “It is either one Entity with some form of differentiation (FavoriteType) which identifies its treatment” but how to is not clear to me. What am I missing here? u.6) “Business Rules This is probably the only area you are weak in.” Thanks for the honest response. I will readdress these but I hope to clear up some confusion in my head first with the responses I have posted back to you. Q.1) Yes I would like to have Accepted, Rejected, and Blocked. I am not sure what you are referring to as to how this would change the logical model? Q.2) A person does not have to be a User. They can exist only as a BandMember. Is that what you are asking? Minor Issue Zero, One, or More…Oops I admit I forgot to give this attention when building the model. I am submitting this version as is and will address in a future version. I need to read up more on Constraint Checking to make sure I am understanding things. M.4) Depends if you envision OrderPurchase in the future. Can you expand as to what you mean here? EDIT V1.2 In response to PerformanceDBA input... Lessons learned. I was mixing the concept of Identifying / Non-Identifying and Cardinality (i.e. Genre / SubGenre), and doing so inconsistently to make things worse. Associative Tables are not required in Logical Diagrams as their many-to-many relationships can be depicted and then expanded in the Physical Model. I was overlooking the Cardinality in a lot of the relationships The importance of reading through relationships using effective Verb Phrases to reassure I am modeling what I want to accomplish. U.2) In the concept of this model it is only required to track a Venue as a location for an Event. No further data needs to be collected. With that being said Events will take place on a given EventDate and will be hosted at a Venue. Venues will host multiple events and possibly multiple events on a given date. In my new model my thinking was that EventDate is already tied to Event . Therefore, Venue will not need a relationship with EventDate. The 5th and 6th bullets you have listed under U.2) leave me questioning my thinking though. Am I missing something here? U.3) Is it time to move the link between Item and Band up to Item and Party instead? With the current design I don't see a possibility to sell merchandise not tied to the band as you have brought up. U.5) I left as per your input rather than making it a discrete Supertype/Subtype Relationship as I don’t see a benefit of having that type of roll up. Additional Revisions AR.1) After going through the exercise for FavoriteItem, I feel that Item to Review requires a many-to-many relationship so that is indicated. Necessary? Ok here we go for v1.3 I took a few days on this version, going back and forth with my design. Once the logical process is complete, as I want to see if I am on the right track, I will go through in depth what I had learned and the troubles I faced as a beginner going through this process. The big point for this version was it took throwing in some Keys to help see what I was missing in the past. Going through the process of doing a matrix proved to be of great help also. Regardless of anything, if it wasn't for the input given by PerformanceDBA I would still be a lost soul wondering in the dark. Who knows my current design might reaffirm that I still am, but I have learned a lot so I am know I at least have a flashlight in my hand. At this point in time I admit that I am still confused about identifying and non-identifying relationships. In my model I had to use non-identifying relationships with non nulls just to join the relationships I wanted to model. In reading a lot on the subject there seems to be a lot of disagreement and indecisiveness on the subject so I did what I thought represented the right things in my model. When to force (identifying) and when to be free (non-identifying)? Anyone have inputs? EDIT V1.4 Ok took the V1.3 inputs and cleaned things up for this V1.4 Currently working on a V1.5 to include attributes.

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  • What are logical and path queries

    - by NomeN
    I'm reading a paper which mentions that a language for refactoring has three specific requirements. functional features (like ML) logical queries (like Datalog) path queries (like Datalog) I know what they mean by functional features, but I'm not totally clear on the latter two and can't find a clear explanation either. Although I have a good idea after what I could find on the subjects, I need to be sure so here goes: Could the SO-community please clearly explain to me what logical queries and path queries are? Or at the very least what the people from the paper meant?

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  • How can I convert this PHP script to Ruby? (build tree from tabbed string)

    - by Jon Sunrays
    I found this script below online, and I'm wondering how I can do the same thing with a Ruby on Rails setup. So, first off, I ran this command: rails g model Node node_id:integer title:string Given this set up, how can I make a tree from a tabbed string like the following? <?php // Make sure to have "Academia" be root node with nodeID of 1 $data = " Social sciences Anthropology Biological anthropology Forensic anthropology Gene-culture coevolution Human behavioral ecology Human evolution Medical anthropology Paleoanthropology Population genetics Primatology Anthropological linguistics Synchronic linguistics (or Descriptive linguistics) Diachronic linguistics (or Historical linguistics) Ethnolinguistics Sociolinguistics Cultural anthropology Anthropology of religion Economic anthropology Ethnography Ethnohistory Ethnology Ethnomusicology Folklore Mythology Political anthropology Psychological anthropology Archaeology ...(goes on for a long time) "; //echo "Checkpoint 2\n"; $lines = preg_split("/\n/", $data); $parentids = array(0 => null); $db = new PDO("host", 'username', 'pass'); $sql = 'INSERT INTO `TreeNode` SET ParentID = ?, Title = ?'; $stmt = $db->prepare($sql); foreach ($lines as $line) { if (!preg_match('/^([\s]*)(.*)$/', $line, $m)) { continue; } $spaces = strlen($m[1]); //$level = intval($spaces / 4); //assumes four spaces per indent $level = strlen($m[1]); // if data is tab indented $title = $m[2]; $parentid = ($level > 0 ? $parentids[$level - 1] : 1); //All "roots" are children of "Academia" which has an ID of "1"; $rv = $stmt->execute(array($parentid, $title)); $parentids[$level] = $db->lastInsertId(); echo "inserted $parentid - " . $parentid . " title: " . $title . "\n"; } ?>

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  • How do I force all Tree itemrenderers to refresh?

    - by Richard Haven
    I have item renderers in an mx.controls.Tree that I need to refresh on demand. I have code in the updateDisplayList that fires for only some of the visible nodes no matter what I do. I've tried triggering a change that they should all be listening for; I have tried clearing and resetting the dataProvider and the itemRenderer properties. private function forceCategoryTreeRefresh(event : Event = null) : void { trace("forceCategoryTreeRefresh"); var prevDataProvider : Object = CategoryTree.dataProvider; CategoryTree.dataProvider = null; CategoryTree.validateNow(); CategoryTree.dataProvider = prevDataProvider; var prevItemRenderer : IFactory = CategoryTree.itemRenderer; CategoryTree.itemRenderer = null; CategoryTree.itemRenderer = prevItemRenderer as IFactory; _categoriesChangeDispatcher.dispatchEvent(new Event(Event.CHANGE)); } The nodes refresh properly when I scroll them into view (e.g. the .data gets set), but I cannot force the ones that already exist to refresh or reset themselves. Any ideas?

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  • How to find sum of node's value for given depth in binary tree?

    - by masato-san
    I've been scratching my head for several hours for this... problem: Binary Tree (0) depth 0 / \ 10 20 depth 1 / \ / \ 30 40 50 60 depth 2 I am trying to write a function that takes depth as argument and return the sum of values of nodes of the given depth. For instance, if I pass 2, it should return 180 (i.e. 30+40+50+60) I decided to use breath first search and when I find the node with desired depth, sum up the value, but I just can't figure out how to find out the way which node is in what depth. But with this approach I feel like going to totally wrong direction. function level_order($root, $targetDepth) { $q = new Queue(); $q->enqueue($root); while(!$q->isEmpty) { //how to determin the depth of the node??? $node = $q->dequeue(); if($currentDepth == $targetDepth) { $sum = $node->value; } if($node->left != null) { $q->enqueue($node->left); } if($node->right != null) { $q->enqueue($node->right); } //need to reset this somehow $currentDepth ++; } }

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  • SQL SERVER – Simple Demo of New Cardinality Estimation Features of SQL Server 2014

    - by Pinal Dave
    SQL Server 2014 has new cardinality estimation logic/algorithm. The cardinality estimation logic is responsible for quality of query plans and majorly responsible for improving performance for any query. This logic was not updated for quite a while, but in the latest version of SQL Server 2104 this logic is re-designed. The new logic now incorporates various assumptions and algorithms of OLTP and warehousing workload. Cardinality estimates are a prediction of the number of rows in the query result. The query optimizer uses these estimates to choose a plan for executing the query. The quality of the query plan has a direct impact on improving query performance. ~ Souce MSDN Let us see a quick example of how cardinality improves performance for a query. I will be using the AdventureWorks database for my example. Before we start with this demonstration, remember that even though you have SQL Server 2014 to see the effect of new cardinality estimates, you will need your database compatibility mode set to 120 which is for SQL Server 2014. If your server instance of SQL Server 2014 but you have set up your database compatibility mode to 110 or any other earlier version, you will get performance from your query like older version of SQL Server. Now we will execute following query in two different compatibility mode and see its performance. (Note that my SQL Server instance is of version 2014). USE AdventureWorks2014 GO -- ------------------------------- -- NEW Cardinality Estimation ALTER DATABASE AdventureWorks2014 SET COMPATIBILITY_LEVEL = 120 GO EXEC [dbo].[uspGetManagerEmployees] 44 GO -- ------------------------------- -- Old Cardinality Estimation ALTER DATABASE AdventureWorks2014 SET COMPATIBILITY_LEVEL = 110 GO EXEC [dbo].[uspGetManagerEmployees] 44 GO Result of Statistics IO Compatibility level 120 Table ‘Person’. Scan count 0, logical reads 6, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table ‘Employee’. Scan count 2, logical reads 7, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table ‘Worktable’. Scan count 0, logical reads 0, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table ‘Worktable’. Scan count 2, logical reads 7, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Compatibility level 110 Table ‘Worktable’. Scan count 2, logical reads 7, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table ‘Person’. Scan count 0, logical reads 137, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table ‘Employee’. Scan count 2, logical reads 7, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. Table ‘Worktable’. Scan count 0, logical reads 0, physical reads 0, read-ahead reads 0, lob logical reads 0, lob physical reads 0, lob read-ahead reads 0. You will notice in the case of compatibility level 110 there 137 logical read from table person where as in the case of compatibility level 120 there are only 6 physical reads from table person. This drastically improves the performance of the query. If we enable execution plan, we can see the same as well. I hope you will find this quick example helpful. You can read more about this in my latest Pluralsight Course. Reference: Pinal Dave (http://blog.SQLAuthority.com)Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL

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  • Approaches to create a nested tree structure of NSDictionaries?

    - by d11wtq
    I'm parsing some input which produces a tree structure containing NSDictionary instances on the branches and NSString instance at the nodes. After parsing, the whole structure should be immutable. I feel like I'm jumping through hoops to create the structure and then make sure it's immutable when it's returned from my method. We can probably all relate to the input I'm parsing, since it's a query string from a URL. In a string like this: a=foo&b=bar&a=zip We expect a structure like this: NSDictionary { "a" => NSDictionary { 0 => "foo", 1 => "zip" }, "b" => "bar" } I'm keeping it just two-dimensional in this example for brevity, though in the real-world we sometimes see var[key1][key2]=value&var[key1][key3]=value2 type structures. The code hasn't evolved that far just yet. Currently I do this: - (NSDictionary *)parseQuery:(NSString *)queryString { NSMutableDictionary *params = [NSMutableDictionary dictionary]; NSArray *pairs = [queryString componentsSeparatedByString:@"&"]; for (NSString *pair in pairs) { NSRange eqRange = [pair rangeOfString:@"="]; NSString *key; id value; // If the parameter is a key without a specified value if (eqRange.location == NSNotFound) { key = [pair stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; value = @""; } else { // Else determine both key and value key = [[pair substringToIndex:eqRange.location] stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; if ([pair length] > eqRange.location + 1) { value = [[pair substringFromIndex:eqRange.location + 1] stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; } else { value = @""; } } // Parameter already exists, it must be a dictionary if (nil != [params objectForKey:key]) { id existingValue = [params objectForKey:key]; if (![existingValue isKindOfClass:[NSDictionary class]]) { value = [NSDictionary dictionaryWithObjectsAndKeys:existingValue, [NSNumber numberWithInt:0], value, [NSNumber numberWithInt:1], nil]; } else { // FIXME: There must be a more elegant way to build a nested dictionary where the end result is immutable? NSMutableDictionary *newValue = [NSMutableDictionary dictionaryWithDictionary:existingValue]; [newValue setObject:value forKey:[NSNumber numberWithInt:[newValue count]]]; value = [NSDictionary dictionaryWithDictionary:newValue]; } } [params setObject:value forKey:key]; } return [NSDictionary dictionaryWithDictionary:params]; } If you look at the bit where I've added FIXME it feels awfully clumsy, pulling out the existing dictionary, creating an immutable version of it, adding the new value, then creating an immutable dictionary from that to set back in place. Expensive and unnecessary? I'm not sure if there are any Cocoa-specific design patterns I can follow here?

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  • How to structure classes in the filesystem?

    - by da_b0uncer
    I have a few (view) classes. Table, Tree, PagingColumn, SelectionColumn, SparkLineColumn, TimeColumn. currently they're flat under app/view like this: app/view/Table app/view/Tree app/view/PagingColumn ... I thought about restructuring it, because the Trees and Tables use the columns, but there are some columns, which only work in a tree, some who work in trees and tables and in the future there are probably some who only work in tables, I don't know. My first idea was like this: app/view/Table app/view/Tree app/view/column/PagingColumn app/view/column/SelectionColumn app/view/column/SparkLineColumn app/view/column/TimeColumn But since the SelectionColumn is explicitly for trees, I have the fear that future developers could get the idea of missuse them. But how to restructure it probably? Like this: app/view/table/panel/Table app/view/tree/panel/Tree app/view/tree/column/PagingColumn app/view/tree/column/SelectionColumn app/view/column/SparkLineColumn app/view/column/TimeColumn Or like this: app/view/Table app/view/Tree app/view/column/SparkLineColumn app/view/column/TimeColumn app/view/column/tree/PagingColumn app/view/column/tree/SelectionColumn

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  • Logical move of a server to UK, what do I do with the SSL certificates

    - by flyfishr64
    I have been asked to move a rails application from the US to the UK. This involves bringing up the rails stack on Ubuntu 8.04.4; that's completed. I'm stumped with the SSL configuration though. The plan was to bring this server up with the same domain name but temporarily use a subdomain (app2.xxx.com instead of app.xxx.com) during the move and for testing, then rename it to app.xxx.com when we're ready for the cutover (does that make sense?). In the meantime, we need a new cert for the app2 subdomain. So to generate a CSR, I need a server key but do I need a new one, or should I copy the one from the existing production server?

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  • How to Extending a logical volume in WMWare

    - by Mercer
    down vote favorite i have a CentOS 6.3 into my Virtual Machine. I have 2 Disk: Disk#1 = 18G Disk#2 = 20G [root@vm ~]# df -h Filesystem Filesystem Size Used Avail Use% Mounted on /dev/mapper/vg_system-lv_root 1008M 250M 708M 27% / tmpfs 1.9G 0 1.9G 0% /dev/shm /dev/sda1 194M 31M 154M 17% /boot /dev/mapper/vg_system-lv_home 504M 17M 462M 4% /home /dev/mapper/vg_system-lv_opt 2.0G 68M 1.9G 4% /opt /dev/mapper/vg_produits-lv_grid 6.9G 2.5G 4.1G 38% /opt/grid /dev/mapper/vg_produits-lv_oracle 6.9G 144M 6.4G 3% /opt/oracle /dev/mapper/vg_system-lv_tmp 2.8G 71M 2.6G 3% /tmp /dev/mapper/vg_system-lv_usr 2.5G 1.6G 799M 67% /usr /dev/mapper/vg_system-lv_var 2.0G 278M 1.6G 15% /var So i want to extend my /tmp and my /opt/oracle like this: 10Go in/tmp 13Go in /opt/oracle Thx.

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  • is it a good idea to change a recovery partition from primary to logical? [HP laptop]

    - by DiegoDD
    I have a new HP laptop, model dv6-6c85la, with 1TB hard drive, and it has 4 primary partitions, like this: |<- system [199 MB] -|<- c: [899.8 GB] -|<- d:(recovery) [27.5 GB] -|<- e:(hp_tools) [4 GB] -| I wanted to make another partition, splitting "C" which is the main partition, into TWO partitions, and leave the rest as it is. but it doesn't let me because they are already 4 primary partitions (the ones in the diagram). I read somewhere, that i could in fact split C into 2 partitions, but only if the adjacent partition (in this case d:(recovery) is converted into a "logical" partition. That way, the new unallocated part taken from C, and the recovery partition, would each be logical, "inside" an extended partition (right???) As i understand, the resulting partitions would be: primary (system, no letter), primary (c:), extended [ logical (x:) | logical(d:recovery) ], primary (e: hp_tools) "x" being the new one. am i correct? My question is, if i do convert the recovery partition to logical (and thus, it is inside an extended partition adjacent to the new "x:" one), would i have any problems when in case of a disaster i would like to restore the system using the now logical instead of primary RECOVERY partition? Or it is completely safe to change it to logical? My main concern is because i think i may need to be primary so the recovery can proceed in boot time? Or i am completely wrong? how does the recovery process happens? I also understand that i can simply create recovery media, in DVDs, and then even i would be able to delete that recovery partition completely, but as of now, i don't want to do that. I may create the disks, but i don't want to delete the partition, simply because it would be a lot faster and easier to recover from a hard drive than disks. Wrapping up: if i change a recovery partition from primary to logical, will the system still be capable of using it to recover? or it NEEDS to be primary to work? The whole point is that i want to split C:, but as things are, i cant directly, i'd need to change the recovery partition to logical. Or is there another way? thanks.

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  • Discs in DVD Drive not being read

    - by I Heart Ubuntu
    Does anyone have an experience with Ubuntu that is no longer reading discs in the DVD drive? This is my first time Ubuntu does not see the disc :| The disc is fine and works on my other Ubuntu computers. The drive is there and powered. I can even type in eject in a terminal and the drive will open. Using the command: sudo lshw -C disk I am able to see info about my drive too. Actually BOTH of my internal DVD drives cannot read discs anymore. If the output is not readable below, here is the info in pastebin. http://pastebin.com/GqqSCTPw *-cdrom:0 description: DVD writer product: DVD_RW ND-3500AG vendor: _NEC physical id: 0.0.0 bus info: scsi@0:0.0.0 logical name: /dev/cdrom1 logical name: /dev/cdrw1 logical name: /dev/dvd1 logical name: /dev/dvdrw1 logical name: /dev/scd0 logical name: /dev/sr0 version: 2.1B serial: [_NEC DVD_RW ND-3500AG2.1B06022300BT-LIGGY capabilities: removable audio cd-r cd-rw dvd dvd-r configuration: ansiversion=5 status=nodisc *-cdrom:1 description: DVD-RAM writer product: CDDVDW SH-S222A vendor: TSSTcorp physical id: 0.1.0 bus info: scsi@0:0.1.0 logical name: /dev/cdrom logical name: /dev/cdrw logical name: /dev/dvd logical name: /dev/dvdrw logical name: /dev/scd1 logical name: /dev/sr1 version: SB01 capabilities: removable audio cd-r cd-rw dvd dvd-r dvd-ram configuration: ansiversion=5 status=nodisc

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  • Logical file not working for SUBFILE/SETLL?

    - by user1516536
    I'm using three logical file with different Record Format where on the first subfile I'm using LF1 and LF2 where on the first subfile I cannot use *LOVAL SETLL it will give me Run Time Error. not sure why? then the program will lead me to second subfile and I'm using LF3 it seems fine. but then If I back to first subfile the subfile turn to blanks.???? why? this my subroutine for building my subfile: C CLRSR BEGSR C EVAL *IN55='0' C WRITE USQLSCTL C EVAL *IN55='1' C ENDSR C* C*BUILDING SUBFILE C BLDSR BEGSR C *LOVAL SETLL USRLGX C EVAL RECNO=0 C EXSR TMISR1 C EXSR REDSR1 C DOW NOT %EOF C IF USRID<>IDD C EXSR MVESR C EXSR DIMSR C MOVE USRID IDD C EVAL RECNO=RECNO+1 C WRITE USQLS C ENDIF C EXSR TMISR1 C EXSR REDSR1 C ENDDO C ENDSR and the related subroutine C TMISR1 BEGSR C READ USRLGX C MOVE USRTI MINTI C ENDSR C REDSR1 BEGSR C READ USRLG C MOVE USRTO MAXTO C ENDSR 6 n the LF I used is USRLG and USRLGX. where both LF refer to the same record format. but each LF has different sorted order. *record format has been RENAME on F-Spec I have this two problem which is: I only can use *LOVAL setll logical-file once only. n the result for coding above sometime it will give result for UserTimeIn some time it equals to blanks.(000000)

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  • Pseudo LRU tree algorithm.

    - by patros
    A lot of descriptions of Pseudo LRU algorithms involve using a binary search tree, and setting flags to "point away" from the node you're searching for every time you access the tree. This leads to a reasonable approximation of LRU. However, it seems from the descriptions that all of the nodes deemed LRU would be leaf nodes. Is there a pseudo-LRU algorithm that deals with a static tree that will still perform reasonably well, while determining that non-leaf nodes are suitable LRU candidates?

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  • gunit syntax for tree walker with a flat list of nodes

    - by Kaleb Pederson
    Here's a simple gunit test for a portion of my tree grammar which generates a flat list of nodes: objectOption walks objectOption: <<one:"value">> -> (one "value") Although you define a tree in ANTLR's rewrite syntax using a caret (i.e. ^(ROOT child...)), gunit matches trees without the caret, so the above represents a tree and it's not surprising that it fails: it's a flat list of nodes and not a tree. This results in a test failure: 1 failures found: test2 (objectOption walks objectOption, line17) - expected: (one \"value\") actual: one \"value\" Another option which seems intuitive is to leave off the parenthesis, like this: objectOption walks objectOption: <<one:"value">> -> one "value" But gunit doesn't like this syntax. It seems to result in a parse failure in the gunit grammar: line 17:20 no viable alternative at input 'one' line 17:24 missing ':' at 'value' line 0:-1 no viable alternative at input '<EOF>' java.lang.NullPointerException at org.antlr.gunit.OutputTest.getExpected(OutputTest.java:65) at org.antlr.gunit.gUnitExecutor.executeTests(gUnitExecutor.java:245) ... What is the correct way to match a flat tree?

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  • Parsing a given binary tree using python?

    - by kaushik
    Parse a binary tree,referring to given set of features,answering decision tree question at each node to decide left child or right child and find the path to leaf node according to answer given to the decision tree.. input wil be a set of feature which wil help in answering the question at each level to choose the left or right half and the output will be the leaf node.. i need help in implementing this can anyone suggest methods?? Please answer... thanks in advance..

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  • Psuedo LRU tree algorithm.

    - by patros
    A lot of descriptions of Pseudo LRU algorithms involve using a binary search tree, and setting flags to "point away" from the node you're searching for every time you access the tree. This leads to a reasonable approximation of LRU. However, it seems from the descriptions that all of the nodes deemed LRU would be leaf nodes. Is there a pseudo-LRU algorithm that deals with a static tree that will still perform reasonably well, while determining that non-leaf nodes are suitable LRU candidates?

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  • Output of gcc -fdump-tree-original

    - by Job
    If I dump the code generated by GCC for a virtual destructor (with -fdump-tree-original), I get something like this: ;; Function virtual Foo::~Foo() (null) ;; enabled by -tree-original { <<cleanup_point <<< Unknown tree: expr_stmt (void) (((struct Foo *) this)->_vptr.Foo = &_ZTV3Foo + 8) >>> >>; } <D.20148>:; if ((bool) (__in_chrg & 1)) { <<cleanup_point <<< Unknown tree: expr_stmt operator delete ((void *) this) >>> >>; } My question is: where is the code after "<D.20148>:;" located? It is outside of the destructor so when is this code executed?

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  • What is validating a binary search tree?

    - by dotnetdev
    I read on here of an exercise in interviews known as validating a binary search tree. How exactly does this work? What would one be looking for in validating a binary search tree? I have written a basic search tree, but never heard of this concept. Thanks

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