...What represents the root partition in Linux? / 4. The utility that can be used to copy a file, including text and executable program (binary) files is _cp_. 5. Which Linux utility is used to rename a file? mv 6. Which command displays the last 10 lines of the file abc? tail 7. Which Linux utility is used to view a file one screen at a time? Less 8. Which Linux utility is used to move a file from one directory to another? mv 9. Which Linux utility is used to search for a particular string within a file or a group of files? grap 10. Which Linux utility lists the names of files in a single directory? LS 11. Which Linux utility do you use to make your home directory the working directory? CD 12. An absolute path name always starts with a /, which is the home directory of the user. False 13. Which of the following commands do you use to ensure that all the users on a computer can execute a particular script? Chmod 14. Which Linux utility is used to display the pathname of a current working directory? PWD 15. Which symbol instructs the shell to redirect the output of a command to the specified file instead of the screen and overwrite the file? > Greater than 16. Which Linux utility do you use to move a process from the background to the foreground? FG 17. Which operator causes the output to be appended to an existing file? >> 18. Following the job number, the shell displays the _PID_ number—a larger number assigned...
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...UEFI Shell Specification September 25, 2008 Revision 2.0 The material contained herein is not a license, either expressly or impliedly, to any intellectual property owned or controlled by any of the authors or developers of this material or to any contribution thereto. The material contained herein is provided on an "AS IS" basis and, to the maximum extent permitted by applicable law, this information is provided AS IS AND WITH ALL FAULTS, and the authors and developers of this material hereby disclaim all other warranties and conditions, either express, implied or statutory, including, but not limited to, any (if any) implied warranties, duties or conditions of merchantability, of fitness for a particular purpose, of accuracy or completeness of responses, of results, of workmanlike effort, of lack of viruses and of lack of negligence, all with regard to this material and any contribution thereto. Designers must not rely on the absence or characteristics of any features or instructions marked "reserved" or "undefined." The Unified EFI Forum, Inc. reserves any features or instructions so marked for future definition and shall have no responsibility whatsoever for conflicts or incompatibilities arising from future changes to them. ALSO, THERE IS NO WARRANTY OR CONDITION OF TITLE, QUIET ENJOYMENT, QUIET POSSESSION, CORRESPONDENCE TO DESCRIPTION OR NON-INFRINGEMENT WITH REGARD TO THE SPECIFICATION AND ANY CONTRIBUTION THERETO. IN NO EVENT WILL ANY AUTHOR OR DEVELOPER OF THIS...
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...often our own worst enemies. The Chinese philosopher and military tactician Sun Tzu likened the successful soldier to water. "Water shapes its course according to the nature of the ground over which it flows; the soldier works out his victory in relation to the foe whom he is facing." The brilliance of this idea is not in the likening of life to war, rather in revealing the simple truth that we are a product of our experiences and how we adapt determines our success, along with what path we take. Each hill and valley we encounter provides us with an opportunity to adapt and change, gives us the chance to use our talents within the realms of our potential to evolve our life in new and sometimes unexpected ways. God's plan may be unknown and unfathomable to us as mere mortals, and there is no way we can predict where the road will lead us. We can, however, embrace our role in the game that is life, and welcome the changes that take place. We must do this with a serene state of mind, because life's paths and turns require a clear-headed approach to navigate. In the process, our own individual success is defined and revealed as in an artist who...
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...IOS110 Absolute and Relative Paths Worksheet Answers 1. Create the directory tree outlined below. 2. Use copy con or edit to create the files. 3. After you have created the directory tree, you may wish to record changes to it as you execute the commands. [pic] Step 1: 1. Create the directory tree above 2. Use tree /f to confirm the tree structure is correct. Step 2: Write out the command path statement using absolute and relative paths. Execute the command. Check your work and make corrections to your path as necessary. If you don't know a command or switch use HELP to find the correct syntax. 1. You are in the LIB directory move the file gbush.txt to the BRITN directory and rename it tblair.txt A: move \us\gbush.txt \britn\tblair.txt R: move ..\..\us\gbush.txt ..\..\britn\tblair.txt 2. You are in the PC directory; create a directory listing of the CA directory including subdirectories and files, organized by date and displayed one page at a time. A: dir /p /s /od \ca R: dir /p /s /od ..\.. 3. You are in the CONS directory; display a graphical view of the GOV directory including files A: tree /f \ca\gov R: tree /f ..\gov 4. You are in the GOV directory; delete the BRITN directory. A: rd /s \britn R: rd /s ..\..\britn 5. You are in the CONS directory; create a new file called harper3.txt. A: edit \ca\cons\harper3.txt R:...
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...Lab I – Probability Models When finished fill in the table at the end of this document and email it to your teacher Cards: Use the tab for Cards in the excel file and get the total probability for the various hands All of your inputs are on the yellow cells (a) The first task is to figure how many chances you have for each pick that will result in a successful hand INPUT: columns B/C/D/E/F (b) Then if any of the events (one of the five cards) has 100% chance then you need multiply the one path by five to account for the parallel paths to a successful hand INPUT: column H (c) Finally, if there are more than one way to get the result (example four of a kind can be done with 13 different cards) then you need to add that number too INPUT: column J Note: this step will be more complicated when dealing with royal hands, and multiple runs. (d) Your final probability will be the percentage in the last column in the “prob” row (I’ve formattesd it such that these all fall in rows 7,17,27, etc) Dice: Use the tab for Dice in the excel file and get the probabilities for all the outcomes when tossing three dice Note: this problem has been started with the first result for the outcomes of 3 and 4 The probability also has been simplified to two events (rather that 3) by using the results from lab H for the rolls for dice #2 and #3… see below (also on your excel file) Outcome | theoretical % | total | possible outcomes (die 1 / die 2) | 2 |...
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...that the following files are in the working directory: $ ls intro notesb ref2 section1 section3 section4b notesa ref1 ref3 section2 section4a sentrev Give commands for each of the following, using wildcards to express filenames with as few characters as possible. a. List all files that begin with section. $ ls section* b. List the section1, section2, and section3 files only. $ ls section[1-3] c. List the intro file only. $ ls i* d. List the section1, section3, ref1, and ref3 files. 1. $ ls *[13] 2. Give an example of a command that uses grep a. With both input and output redirected. $ grep \$Id < *.c > id_list Chapter 9 Number 1 Explain the following unexpected result: $ whereis date date: /bin/date ... $ echo $PATH .:/usr/local/bin:/usr/bin:/bin $ cat > date echo "This is my own version of date." $...
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...NT1430 Final Review 1. Know what a swap partition is used for. a. The swap partition is an independent section of the hard disk used solely for swapping; no other files can reside there. Swap partition is used to turn the hard drive into extra RAM. 2. Know what an architecture is, and which are supported by Fedora. a. 32 and 64 bit b. Fedora CDs and DVDs hold programs compiled to run on a specific processor architecture (class of processors, or CPUs) c. There are several that Fedora supports: i. EM64T/AMD64 ii. A 64 bit distribution allows each process to address more than 4 gigabytes of RAM. iii. I383/i686 (Intel x86) iv. X86_64 (AMD64 and Intel EM64T) v. Mac PowerPC (PPC) 3. Know how the root of the file system is represented. a. The root is represented by a / b. Fedora uses the Filesystem Hierarchy Standard (FHS) file system structure, which defines the names, locations, and permissions for many file types and directories. c. The two most important elements of FHS compliance are: i. Compatibility with other FHS-compliant systems ii. The ability to mount a /usr/ partition as read-only. This is especially crucial, since /usr/ contains common executables and should not be changed by users. In addition, since /usr/ is mounted as read-only, it should be mountable from the CD-ROM drive or from another machine via a read-only NFS mount. 4. Know what the cp command is used for. a. The Linux cp command is usually used to copy files and copy directory in Linux...
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...Week 4 Individual Assignment 11.1 (513-518) 3. For the graph in Fig. 11.7, how many paths are there from b to f ? There are 6 paths from b to f: (b, a, c, d, e, f), (b, c, d, e, f), (b, e, f), (b, a, c, d, e, g, f), (b, c, d, e, g, f), and (b, e, g, f) 11. Let G be a graph that satisfies the condition in Exercise 10. a) Must G be loop-free? b) Could G be a multigraph? c) If G has n vertices, can we determine how many edges it has? a) Yes, G must be loop-free because an edge is a bridge only if that edge is not contained in any cycles. A loop is a cycle. b) Yes, G can be a multi-graph. For example, the multi-graph in Figure 11.6 (pg. 518) becomes disconnected if we remove edge (c, e). This would leave two components of (a, b, c) and (d, e). c) Yes, G would have n – 1 edges for n vertices; the same vertex closes a graph and there is always a vertex at the start and end, which means there is one more vertex than an edge. 11.2 (520-528) 4. If G = (V, E) is an undirected graph, how many spanning subgraphs of G are also induced subgraphs? The undirected graph G = (V, E) has 2|E| spanning subgraphs, one for each subset of the edge set, and 2|V| induced subgraphs, one for each subset of the vertex set. 11.3 (530-537) 5. Let G1 = (V1, E1) and G2 = (V2, E2) be the loop-free undirected connected graphs in Fig. 11.42. a) Determine |V1|, |E1|, |V2|, and |E2|. Counting the vertices and edges in both graphs: |V1| = |V2| = 8 |E1| = |E2| = 14 11.4...
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...Oracle Backup There are two type of back • Cold backup (Offline) • Hot Backup (Export) Cold backup Procedure Setup 1. Copy backup from Source Check the Oracle version & Path Open Comd Prompt and typecomd Tnsping [database_name] Then type Sqlplus Type username /password @ DB name as sysdba Check the How many DB install on system Click on Start Button and type services.msc and press enter this screen display Press O and search Oracleservice[DB name] Stop the service Now press the stop button on the left side Now go to your Oracle home directory and open it . Go to DB directory and Copy the Folder and past to the backup path After finish the copy then start service of “Orlaceservice[DB Name] Your cold/hard back Complete. Restore the Backup (On Same Version and installation Path) Step 2. Create New DB Click Start Button and type dbca and press next Select Create Database and Press Next Select General Purpose and Press Next Type Globaldatabase and SID and Press Next Uncheck Configure the database with Enterprise Manager Enter the Password for Sys and System and press Next Select File System and Press Next Select Use Common Location for all database and click Browser Select the Path and Press Ok and then Next Uncheck Specify Flash Recovery Area Press next on simple Schemas Press Next Press Finish Press Ok Press Finish Database Creation...
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...========================================================================= * Final Report * ========================================================================= Final Results RTL Top Level Output File Name : shift_register.ngr Top Level Output File Name : shift_register Output Format : NGC Optimization Goal : Speed Keep Hierarchy : NO Design Statistics # IOs : 8 Cell Usage : # BELS : 4 # INV : 1 # LUT3 : 3 # FlipFlops/Latches : 3 # FDC : 3 # Clock Buffers : 1 # BUFGP : 1 # IO Buffers : 7 # IBUF : 4 # OBUF : 3 ========================================================================= Device utilization summary: --------------------------- Selected Device : 2s15cs144-6 Number of Slices: 2 out of 192 1% Number of Slice Flip Flops: 3 out of 384 0% Number of 4 input LUTs: 4 out of 384 1% Number of IOs: 8 Number of bonded IOBs: 8 out of 86 9% Number of GCLKs: 1 out of 4 25% ---------------------------...
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...stored Minimal Guarantee No updates to data occur Trigger User accesses system Main Success Scenario 1. System provide option to locate a specific Product record by Item_Id 2. User enters Item_Id 3. System provides option to add a file 4. User selects option to add a file 5. System provides option to enter file path 6. User enters file path 7. System provides option to save file 8. User selects option to save file 9. File is stored and associated to product Use Case: Store Product Information Id: UC- 200 Description User accesses system to store product knowledge file Primary Actor User Pre-Conditions User must be logged in Product must be loaded Post Conditions Success end condition File is stored and associated to product Failure end condition: File is not stored Minimal Guarantee No updates to data occur Trigger User accesses system Main Success Scenario 1. System provide option to locate a specific Product record by Item_Id 2. User enters Item_Id 3. System provides option to add a file 4. User selects option to add a file 5. System provides option to enter file path 6. User enters file path 7. System provides option to save file 8. User selects option to save file 9. File is stored and associated to...
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...원천마디(source node) 중계되는 점 중계마디(intermediate node) 종료되는 점 종료마디(sink node) 서울대학교 경영대학 안상형 교수 7 가지: 유 · 무형의 재화 서비스 흐름의 통로 (1) 방향성의 유무 (a) 방향이 없음(bi-directed arc) (무 방향/양 방향) (i,j) = (j,i) (b) 방향이 있음(directed arc) (i,j) (j,i) 서울대학교 경영대학 안상형 교수 8 가지: 유 · 무형의 재화 서비스 흐름의 통로 (2) 가지 사용의 비용(cost): cij 가지의 길이(거리), 시간, 비용 등 (3) 가지의 용량 : aij 한번에 흐를 수 있는 용량 서울대학교 경영대학 안상형 교수 9 Bi-directed network (무방향/양방향 네트워크) 2 1 3 4 6 7 5 Directed network (유방향 네트워크) 2 1 3 4 6 7 5 서울대학교 경영대학 안상형 교수 11 bi-directed arc를 directed arc로 변환 i j bi-directed i j directed 서울대학교 경영대학 안상형 교수 12 1) bi-directed graph (1) 경로(path) 마디와 가지의 유한 순서 P = {s1,e1,s2,e2,s3…,sn-1,en-1,sn} 홀수 요소는 distinct 마디, 짝수 요소는 distinct 가지 여기서 ei = (si,sj) 서울대학교 경영대학 안상형 교수 13 (계속) 경로(path) 마디를 제외하고 가지만으로도 표시 P = {1, (1,3), 3, (3,6), 6, (6,7), 7} P = {e1, e2, …, en-1,en} 앞 가지의 꼬리가 뒷 가지의 머리 서울대학교 경영대학 안상형 교수 14 2 1 4 3 6 7 1-7 경로 • P = {1, (1,3), 3, (3,6), 6, (6,7), 7} {(1,3), (3,6), (6,7)} 서울대학교 경영대학 안상형 교수 5 15 2 1 4 3 6 7 1-7 경로: 여러 개 존재 • P ={(1,3), (3,6), (6,7)} • P = {(1,3), (3,4), (4,6), (6,7)} 서울대학교 경영대학 안상형 교수 5 16 (2) cycle s-t경로로서 s t인 경우 cycle이 없는 graph를 acyclic graph라 함 서울대학교 경영대학 안상형 교수 17 2 1...
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...SPICE Netlist Simulation Using NI LabVIEW and NI Multisim Estimated time to complete: 45 minutes Introduction This tutorial demonstrates the step-by-step process to run text-based SPICE netlist simulations from NI LabVIEW. NI Multisim is used as the simulation engine and it is run in a background process connected to LabVIEW using the LabVIEW Multisim API Toolkit. The LabVIEW Multisim API Toolkit is installed by Multisim and it provides over 80 VIs to control a Multisim application instance from LabVIEW. In this tutorial, you will explore the functionality of the SPICE command line interface this toolkit offers. Tip Consult the toolkit Help to find programming guides, flowcharts and the toolkit VI reference. From LabVIEW, select Help»NI Multisim API Toolkit Help. From Multisim, select Help»Multisim help, and select the Automation API topic. Objectives Create and modify a netlist file from a Multisim circuit file Create and modify a SPICE nutmeg command file Use LabVIEW to build a VI that will execute the SPICE nutmeg command file Display results from the simulation There are multiple use-cases for this type of simulation, especially tailored for advanced SPICE users. Some of them may range from old library of SPICE files that you wish to simulate, make a batch process to run analyses on a directory of files, dynamically modify netlist files or analysis parameters to see differences, and so on. This demo script assumes...
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...* Reference: ** Basics: *** Getting help: # View the manual for target command man command # Get help with a target command (probably the same as above, but not always): command -h # In case you forget the name of a command, print possible commands relating to any given word: apropos word # View index of help pages: info *** Command Line Utilities: **** Basic File and Directory Operations: # Print current working directory: pwd # Show files in current directory: ls # Show maximum information about all files, including hidden: ls -a # Recurse into subdirectories and list those as well: ls -R # List files by modification time, most recent first. ls -lt # Move/rename a file or directory (be careful that you don't move the source over a destination with the same name): mv source destination # Delete target forever (be very careful), use -r recursive flag for directories: rm target # Copy file or directory: cp source destination # Mount filesytem: mount /dev/device_name /media/device_name # Unmount: umount /media/device_name # Forensically clone filesystems and do other low-level operations on files. Be careful with this one. Can be destructive: dd # Work with disk partitions: parted # Filesystem creation tool: mkfs **** System Administration: # Execute command as an administrator (can be destructive/insecure. Use only for system administration tasks): sudo command # Become system administrator: sudo...
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...Q1. The above represents a random looking undirected graph where the circles represent nodes and the boundary circumscribes all the nodes of the graph. The length of the segment joining two nodes represents the length of the shortest path from one node to another. Let’s assume nodes A and B are the farthest pair of nodes according to the definition in the question. Thus A and B have to be on the circumscribing boundary. Thus AB represents the diameter of the graph. Let C be any random node in the graph. Let D be a point which is farthest away from C. So if a BFS is launched on node C, the depth(in terms of the summation of the edges) of the BFS tree rooted at C will be equal to CD. As a result CD >= CA and CD>=CB. Now let AB = d. So from triangle inequality CA+CB>=d. Thus d/2 <= max(CA,CB)<=d. Now since CD is greater than or equals to both CA and CB, it has to be greater than or equals to d/2 as well. CD has to be less than d as well because if not then CD would have been the diameter. But that is not the case. Hence d/2<=CD<=d thus if we can find the length of CD we are done with approximating the value of the diameter within a factor of 2 and that we can do by creating BFS tree from C and calculating depth of the tree which is nothing but...
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