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  • How do you solve the 15-puzzle with A-Star or Dijkstra's Algorithm?

    - by Sean
    I've read in one of my AI books that popular algorithms (A-Star, Dijkstra) for path-finding in simulation or games is also used to solve the well-known "15-puzzle". Can anyone give me some pointers on how I would reduce the 15-puzzle to a graph of nodes and edges so that I could apply one of these algorithms? If I were to treat each node in the graph as a game state then wouldn't that tree become quite large? Or is that just the way to do it?

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  • programming logic and design pleas friends i need a flowcharts or pseudocode

    - by alex
    ***the midvile park maintains records containing info about players on it's soccer teams . each record contain a players first name,last name,and team number . the team are team number team name 1 goal getters 2 the force 3 top gun 4 shooting stars 5 midfield monsters design a proggram that accept player data and creates a report that lists each** player a long with his or her team number and team name**

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  • I'm working on Peano Axioms in Agda and I've hit a bit of a sticking point

    - by Schroedinger
    PA6 : ?{m n} -> m = n -> n = m is the axiom I am trying to solve and support, I've tried using a cong (from the core library) but am having troubles with the cong constructor PA6 = cong gets me nowhere, I know for cong I am required to supply a refl for equality and a type, but I'm, not sure what type I'm supposed to supply. Ideas? This is for a small assignment at University, so I'd rather someone demonstrate what I've missed rather than write the acutual answer, but I'd appreciate any degree of support.

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  • Beginner problems with references to arrays in python 3.1.1

    - by Protean
    As part of the last assignment in a beginner python programing class, I have been assigned a traveling sales man problem. I settled on a recursive function to find each permutation and the sum of the distances between the destinations, however, I am have a lot of problems with references. Arrays in different instances of the Permute and Main functions of TSP seem to be pointing to the same reference. from math import sqrt class TSP: def __init__(self): self.CartisianCoordinates = [['A',[1,1]], ['B',[2,2]], ['C',[2,1]], ['D',[1,2]], ['E',[3,3]]] self.Array = [] self.Max = 0 self.StoredList = ['',0] def Distance(self, i1, i2): x1 = self.CartisianCoordinates[i1][1][0] y1 = self.CartisianCoordinates[i1][1][1] x2 = self.CartisianCoordinates[i2][1][0] y2 = self.CartisianCoordinates[i2][1][1] return sqrt(pow((x2 - x1), 2) + pow((y2 - y1), 2)) def Evaluate(self): temparray = [] Data = [] for i in range(len(self.CartisianCoordinates)): Data.append([]) for i1 in range(len(self.CartisianCoordinates)): for i2 in range(len(self.CartisianCoordinates)): if i1 != i2: temparray.append(self.Distance(i1, i2)) else: temparray.append('X') Data[i1] = temparray temparray = [] self.Array = Data self.Max = len(Data) def Permute(self,varray,index,vcarry,mcarry): #Problem Class array = varray[:] carry = vcarry[:] for i in range(self.Max): print ('ARRAY:', array) print (index,i,carry,array[index][i]) if array[index][i] != 'X': carry[0] += self.CartisianCoordinates[i][0] carry[1] += array[index][i] if len(carry) != self.Max: temparray = array[:] for j in range(self.Max):temparray[j][i] = 'X' index = i mcarry += self.Permute(temparray,index,carry,mcarry) else: return mcarry print ('pass',mcarry) return mcarry def Main(self): out = [] self.Evaluate() for i in range(self.Max): array = self.Array[:] #array appears to maintain the same reference after each copy, resulting in an incorrect array being passed to Permute after the first iteration. print (self.Array[:]) for j in range(self.Max):array[j][i] = 'X' print('I:', i, array) out.append(self.Permute(array,i,[str(self.CartisianCoordinates[i][0]),0],[])) return out SalesPerson = TSP() print(SalesPerson.Main()) It would be greatly appreciated if you could provide me with help in solving the reference problems I am having. Thank you.

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  • Java: Using Command line arguments to process the names of files

    - by Kat
    I'm a writing a program that will determine the number of lines, characters, and average word length for a text file. For the program, the specifications say that the file or files will be entered as a command line argument and that we should make a TestStatistic object for each file entered. I don't understand how to write the code for making the TestStatistic objects if the user enters more than one file.

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  • How to exploit Diffie-hellman to perform a man in the middle attack

    - by jfisk
    Im doing a project where Alice and Bob send each other messages using the Diffie-Hellman key-exchange. What is throwing me for a loop is how to incorporate the certificate they are using in this so i can obtain their secret messages. From what I understand about MIM attakcs, the MIM acts as an imposter as seen on this diagram: Below are the details for my project. I understand that they both have g and p agreed upon before communicating, but how would I be able to implement this with they both having a certificate to verify their signatures? Alice prepares ?signA(NA, Bob), pkA, certA? where signA is the digital signature algorithm used by Alice, “Bob” is Bob’s name, pkA is the public-key of Alice which equals gx mod p encoded according to X.509 for a fixed g, p as specified in the Diffie-Hellman key- exchange and certA is the certificate of Alice that contains Alice’s public-key that verifies the signature; Finally, NA is a nonce (random string) that is 8 bytes long. Bob checks Alice's signature, and response with ?signB{NA,NB,Alice},pkB,certB?. Alice gets the message she checks her nonce NA and calculates the joint key based on pkA, pkB according to the Diffie-Hellman key exchange. Then Alice submits the message ?signA{NA,NB,Bob},EK(MA),certA? to Bob and Bobrespondswith?SignB{NA,NB,Alice},EK(MB),certB?. where MA and MB are their corresponding secret messages.

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  • Proving that a function f(n) belongs to a Big-Theta(g(n))

    - by PLS
    Its a exercise that ask to indicate the class Big-Theta(g(n)) the functions belongs to and to prove the assertion. In this case f(n) = (n^2+1)^10 By definition f(n) E Big-Theta(g(n)) <= c1*g(n) < f(n) < c2*g(n), where c1 and c2 are two constants. I know that for this specific f(n) the Big-Theta is g(n^20) but I don't know who to prove it properly. I guess I need to manipulate this inequality but I don't know how

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  • Creating GUI for Bantumi game

    - by owca
    I've written backend for simple Bantumi game. Now I'd like to create a simple GUI for it, so that it would look like this : How to start ? What layout should I use, and what type of component each element should be? Classes : Basket Player Game Main Shared

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  • Help to solve "Robbery Problem"

    - by peiska
    Hello, Can anybody help me with this problem in C or Java? The problem is taken from here: http://acm.pku.edu.cn/JudgeOnline/problem?id=1104 Inspector Robstop is very angry. Last night, a bank has been robbed and the robber has not been caught. And this happened already for the third time this year, even though he did everything in his power to stop the robber: as quickly as possible, all roads leading out of the city were blocked, making it impossible for the robber to escape. Then, the inspector asked all the people in the city to watch out for the robber, but the only messages he got were of the form "We don't see him." But this time, he has had enough! Inspector Robstop decides to analyze how the robber could have escaped. To do that, he asks you to write a program which takes all the information the inspector could get about the robber in order to find out where the robber has been at which time. Coincidentally, the city in which the bank was robbed has a rectangular shape. The roads leaving the city are blocked for a certain period of time t, and during that time, several observations of the form "The robber isn't in the rectangle Ri at time ti" are reported. Assuming that the robber can move at most one unit per time step, your program must try to find the exact position of the robber at each time step. Input The input contains the description of several robberies. The first line of each description consists of three numbers W, H, t (1 <= W,H,t <= 100) where W is the width, H the height of the city and t is the time during which the city is locked. The next contains a single integer n (0 <= n <= 100), the number of messages the inspector received. The next n lines (one for each of the messages) consist of five integers ti, Li, Ti, Ri, Bi each. The integer ti is the time at which the observation has been made (1 <= ti <= t), and Li, Ti, Ri, Bi are the left, top, right and bottom respectively of the (rectangular) area which has been observed. (1 <= Li <= Ri <= W, 1 <= Ti <= Bi <= H; the point (1, 1) is the upper left hand corner, and (W, H) is the lower right hand corner of the city.) The messages mean that the robber was not in the given rectangle at time ti. The input is terminated by a test case starting with W = H = t = 0. This case should not be processed. Output For each robbery, first output the line "Robbery #k:", where k is the number of the robbery. Then, there are three possibilities: If it is impossible that the robber is still in the city considering the messages, output the line "The robber has escaped." In all other cases, assume that the robber really is in the city. Output one line of the form "Time step : The robber has been at x,y." for each time step, in which the exact location can be deduced. (x and y are the column resp. row of the robber in time step .) Output these lines ordered by time . If nothing can be deduced, output the line "Nothing known." and hope that the inspector will not get even more angry. Output a blank line after each processed case.

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  • Array Vs. Linked List

    - by Onorio Catenacci
    I apologize--this question may be a bit open-ended but I think there are probably definite, quantifiable answers to it so I'll post it anyway. A person I know is trying to learn C++ and software development (+1 to him) and he asked me why someone would want to use a linked list in preference to an array. Coding a linked list is, no doubt, a bit more work than using an array and he wondered what would justify the additional effort. I gave him the answer I know: insertion of new elements is trivial in linked list but it's a major chore in an array. But then I got to thinking about it a bit more. Besides the ease of insertion of a new element into a linked list are there other advantages to using a linked list to store a set of data vs. storing it in an array? As I said, I'm not meaning to start a long and drawn-out discussion. I'm just looking for other reasons that a developer might prefer a linked list to an array.

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  • Converting Unicode strings to escaped ascii string

    - by Ali
    How can I convert this string: This string contains the unicode character Pi(p) into an escaped ascii string: This string contains the unicode character Pi(\u03a0) and vice versa ? The current Encoding available in C#, converts the p character into "?". I need to preserve that character.

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  • How can I write query to output this format in SQLite?

    - by GivenPie
    I would like to output in this format: e.EE_id e.FNAME e.LNAME SUPer_id s.FNAME s.LNAME --- --------- -------------- --- ------------- ------------------- 1 Ziqiao Li 2 Charlie Li 1 Ziqiao Li 3 George Pee 2 Charlie Li 4 Jason Dee 2 Charlie Li 5 Petey Wee 2 Charlie Li From this table created : I need to display the Primary key and foreign key in the same results while displaying the foreign key name values for the primary key names. Create table Employees( ee_id integer, fname varchar(20), lname varchar(20), super_id integer, Constraint emp_Pk Primary Key (ee_id), Constraint emp_Fk Foreign Key (super_id) references employees (ee_id) ); INSERT INTO Employees VALUES(1,'Charlie','Li',null); INSERT INTO Employees VALUES(2,'Ziqiao','Lee',1); INSERT INTO Employees VALUES(3,'George','Pee',2); INSERT INTO Employees VALUES(4,'Jason','Dee',2); INSERT INTO Employees VALUES(5,'Petey','Wee',2); Select ee_id, fname, lname, super_id from employees; ee_id fname lname super_id ---------- ---------- ---------- ---------- 1 Charlie Li 2 Ziqiao Lee 1 3 George Pee 2 4 Jason Dee 2 5 Petey Wee 2 Do I need to create a view?

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  • polynomial multiplication using fastfourier transform

    - by mawia
    i am going through the above topic from CLRS(CORMEN) (page 834) and I got stuck at this point. Can anybody please explain how the following expression, A(x)=A^{[0]}(x^2) +xA^{[1]}(x^2) follows from, n-1 ` S a_j x^j j=0 Where, A^{[0]} = a_0 + a_2x + a_4a^x ... a_{n-2}x^{\frac{n}{2-1}} A^{[1]} = a_1 + a_3x + a_5a^x ... a_{n-1}x^{\frac{n}{2-1}} WITH REGARDS THANKS

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  • Which knowledge base/rule-based inference engine to choose for real time Runway incursion prevention

    - by Piligrim
    Hello, we are designing a project that would listen to dialog between airport controllers and pilots to prevent runway incursions (eg. one airplane is taking off while other is crossing the runway). Our professor wants us to use Jena for knowledge base (or anything else but it should be some sort of rule-based engine). Inference is not the main thing in Jena and there's not much documentation and examples of this. So we need an engine that would get messages from pilots as input and output possible risks of incursion or any other error in message protocol. It should be easy to write rules, and should be easy to provide engine with real time data. I image it something like this: A pilot sends a message that he lands on some runway, the system remembers that the runway is busy and no one should cross it If someone is given an instruction to cross this runway, the engine should fire a rule that something is wrong When the pilot sends a message that he left the runway and goes to the gate, the system clears the runway and lets other planes to use it. So is Jena, or prolog or any other rules engine suitable for this? I mean it is suitable, but do we really need to use it? I asked the prof. if we could just keep state of the runway and use some simple checks based on messages we receive and he said that it is not scalable and we need the knowledge base. Can someone give me any advise on which approach to use for this system? If you recommend k.b., then which one should we use? The project is written in java. Thank you.

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  • C++ help with getline function with ifstream

    - by John
    So I am writing a program that deals with reading in and writing out to a file. I use the getline() function because some of the lines in the text file may contain multiple elements. I've never had a problem with getline until now. Here's what I got. The text file looks like this: John Smith // Client name 1234 Hollow Lane, Chicago, IL // Address 123-45-6789 // SSN Walmart // Employer 58000 // Income 2 // Number of accounts the client has 1111 // Account Number 2222 // Account Number ifstream inFile("ClientInfo.txt"); if(inFile.fail()) { cout << "Problem opening file."; } else { string name, address, ssn, employer; double income; int numOfAccount; getline(inFile, name); getline(inFile, address); // I'll stop here because I know this is where it fails. When I debugged this code, I found that name == "John", instead of name == "John Smith", and Address == "Smith" and so on. Am I doing something wrong. Any help would be much appreciated.

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  • adding nodes to a binary search tree randomly deletes nodes

    - by SDLFunTimes
    Hi, stack. I've got a binary tree of type TYPE (TYPE is a typedef of data*) that can add and remove elements. However for some reason certain values added will overwrite previous elements. Here's my code with examples of it inserting without overwriting elements and it not overwriting elements. the data I'm storing: struct data { int number; char *name; }; typedef struct data data; # ifndef TYPE # define TYPE data* # define TYPE_SIZE sizeof(data*) # endif The tree struct: struct Node { TYPE val; struct Node *left; struct Node *rght; }; struct BSTree { struct Node *root; int cnt; }; The comparator for the data. int compare(TYPE left, TYPE right) { int left_len; int right_len; int shortest_string; /* find longest string */ left_len = strlen(left->name); right_len = strlen(right->name); if(right_len < left_len) { shortest_string = right_len; } else { shortest_string = left_len; } /* compare strings */ if(strncmp(left->name, right->name, shortest_string) > 1) { return 1; } else if(strncmp(left->name, right->name, shortest_string) < 1) { return -1; } else { /* strings are equal */ if(left->number > right->number) { return 1; } else if(left->number < right->number) { return -1; } else { return 0; } } } And the add method struct Node* _addNode(struct Node* cur, TYPE val) { if(cur == NULL) { /* no root has been made */ cur = _createNode(val); return cur; } else { int cmp; cmp = compare(cur->val, val); if(cmp == -1) { /* go left */ if(cur->left == NULL) { printf("adding on left node val %d\n", cur->val->number); cur->left = _createNode(val); } else { return _addNode(cur->left, val); } } else if(cmp >= 0) { /* go right */ if(cur->rght == NULL) { printf("adding on right node val %d\n", cur->val->number); cur->rght = _createNode(val); } else { return _addNode(cur->rght, val); } } return cur; } } void addBSTree(struct BSTree *tree, TYPE val) { tree->root = _addNode(tree->root, val); tree->cnt++; } The function to print the tree: void printTree(struct Node *cur) { if (cur == 0) { printf("\n"); } else { printf("("); printTree(cur->left); printf(" %s, %d ", cur->val->name, cur->val->number); printTree(cur->rght); printf(")\n"); } } Here's an example of some data that will overwrite previous elements: struct BSTree myTree; struct data myData1, myData2, myData3; myData1.number = 5; myData1.name = "rooty"; myData2.number = 1; myData2.name = "lefty"; myData3.number = 10; myData3.name = "righty"; initBSTree(&myTree); addBSTree(&myTree, &myData1); addBSTree(&myTree, &myData2); addBSTree(&myTree, &myData3); printTree(myTree.root); Which will print: (( righty, 10 ) lefty, 1 ) Finally here's some test data that will go in the exact same spot as the previous data, but this time no data is overwritten: struct BSTree myTree; struct data myData1, myData2, myData3; myData1.number = 5; myData1.name = "i"; myData2.number = 5; myData2.name = "h"; myData3.number = 5; myData3.name = "j"; initBSTree(&myTree); addBSTree(&myTree, &myData1); addBSTree(&myTree, &myData2); addBSTree(&myTree, &myData3); printTree(myTree.root); Which prints: (( j, 5 ) i, 5 ( h, 5 ) ) Does anyone know what might be going wrong? Sorry if this post was kind of long.

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  • Need some help on how to replay the last game of a java maze game

    - by Marty
    Hello, I am working on creating a Java maze game for a project. The maze is displayed on the console as standard output not in an applet. I have created most of hte code I need, however I am stuck at one problem and that is I need a user to be able to replay the last game i.e redraw the maze with the users moves but without any input from the user. I am not sure on what course of action to take, i was thinking about copying each users move or the position of each move into another array, as you can see i have 2 variables which hold the position of the player, plyrX and plyrY do you think copying these values into a new array after each move would solve my problem and how would i go about this? I have updated my code, apologies about the textIO.java class not being present, not sure how to resolve that exept post a link to TextIO.java [TextIO.java][1] My code below is updated with a new array of type char to hold values from the original maze (read in from text file and displayed using unicode characters) and also to new variables c_plyrX and c_plyrY which I am thinking should hold the values of plyrX and plyrY and copy them into the new array. When I try to call the replayGame(); method from the menu the maze loads for a second then the console exits so im not sure what I am doing wrong Thanks public class MazeGame { //unicode characters that will define the maze walls, //pathways, and in game characters. final static char WALL = '\u2588'; //wall final static char PATH = '\u2591'; //pathway final static char PLAYER = '\u25EF'; //player final static char ENTRANCE = 'E'; //entrance final static char EXIT = '\u2716'; //exit //declaring member variables which will hold the maze co-ordinates //X = rows, Y = columns static int entX = 0; //entrance X co-ordinate static int entY = 1; //entrance y co-ordinate static int plyrX = 0; static int plyrY = 1; static int exitX = 24; //exit X co-ordinate static int exitY = 37; //exit Y co-ordinate //static member variables which hold maze values //used so values can be accessed from different methods static int rows; //rows variable static int cols; //columns variable static char[][] maze; //defines 2 dimensional array to hold the maze //variables that hold player movement values static char dir; //direction static int spaces; //amount of spaces user can travel //variable to hold amount of moves the user has taken; static int movesTaken = 0; //new array to hold player moves for replaying game static char[][] mazeCopy; static int c_plyrX; static int c_plyrY; /** userMenu method for displaying the user menu which will provide various options for * the user to choose such as play a maze game, get instructions, etc. */ public static void userMenu(){ TextIO.putln("Maze Game"); TextIO.putln("*********"); TextIO.putln("Choose an option."); TextIO.putln(""); TextIO.putln("1. Play the Maze Game."); TextIO.putln("2. View Instructions."); TextIO.putln("3. Replay the last game."); TextIO.putln("4. Exit the Maze Game."); TextIO.putln(""); int option; //variable for holding users option TextIO.put("Type your choice: "); option = TextIO.getlnInt(); //gets users option //switch statement for processing menu options switch(option){ case 1: playMazeGame(); case 2: instructions(); case 3: if (c_plyrX == plyrX && c_plyrY == plyrY)replayGame(); else { TextIO.putln("Option not available yet, you need to play a game first."); TextIO.putln(); userMenu(); } case 4: System.exit(0); //exits the user out of the console default: TextIO.put("Option must be 1, 2, 3 or 4"); } } //end of userMenu /**main method, will call the userMenu and get the users choice and call * the relevant method to execute the users choice. */ public static void main(String[]args){ userMenu(); //calls the userMenu method } //end of main method /**instructions method, displays instructions on how to play * the game to the user/ */ public static void instructions(){ TextIO.putln("To beat the Maze Game you have to move your character"); TextIO.putln("through the maze and reach the exit in as few moves as possible."); TextIO.putln(""); TextIO.putln("Your characer is displayed as a " + PLAYER); TextIO.putln("The maze exit is displayed as a " + EXIT); TextIO.putln("Reach the exit and you have won escaped the maze."); TextIO.putln("To control your character type the direction you want to go"); TextIO.putln("and how many spaces you want to move"); TextIO.putln("for example 'D3' will move your character"); TextIO.putln("down 3 spaces."); TextIO.putln("Remember you can't walk through walls!"); boolean insOption; //boolean variable TextIO.putln(""); TextIO.put("Do you want to play the Maze Game now? (Y or N) "); insOption = TextIO.getlnBoolean(); if (insOption == true)playMazeGame(); else userMenu(); } //end of instructions method /**playMazeGame method, calls the loadMaze method and the charMove method * to start playing the Maze Game. */ public static void playMazeGame(){ loadMaze(); plyrMoves(); } //end of playMazeGame method /**loadMaze method, loads the 39x25 maze from the MazeGame.txt text file * and inserts values from the text file into the maze array and * displays the maze on screen using the unicode block characters. * plyrX and plyrY variables are set at their staring co ordinates so that when * a game is completed and the user selects to play a new game * the player character will always be at position 01. */ public static void loadMaze(){ plyrX = 0; plyrY = 1; TextIO.readFile("MazeGame.txt"); //now reads from the external MazeGame.txt file rows = TextIO.getInt(); //gets the number of rows from text file to create X dimensions cols = TextIO.getlnInt(); //gets number of columns from text file to create Y dimensions maze = new char[rows][cols]; //creates maze array of base type char with specified dimnensions //loop to process the array and read in values from the text file. for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ maze[i][j] = TextIO.getChar(); } TextIO.getln(); } //end for loop TextIO.readStandardInput(); //closes MazeGame.txt file and reads from //standard input. //loop to process the array values and display as unicode characters for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ if (i == plyrX && j == plyrY){ plyrX = i; plyrY = j; TextIO.put(PLAYER); //puts the player character at player co-ords } else{ if (maze[i][j] == '0') TextIO.putf("%c",WALL); //puts wall block if (maze[i][j] == '1') TextIO.putf("%c",PATH); //puts path block if (maze[i][j] == '2') { entX = i; entY = j; TextIO.putf("%c",ENTRANCE); //puts entrance character } if (maze[i][j] == '3') { exitX = i; //holds value of exit exitY = j; //co-ordinates TextIO.putf("%c",EXIT); //puts exit character } } } TextIO.putln(); } //end for loop } //end of loadMaze method /**redrawMaze method, method for redrawing the maze after each move. * */ public static void redrawMaze(){ TextIO.readFile("MazeGame.txt"); //now reads from the external MazeGame.txt file rows = TextIO.getInt(); //gets the number of rows from text file to create X dimensions cols = TextIO.getlnInt(); //gets number of columns from text file to create Y dimensions maze = new char[rows][cols]; //creates maze array of base type char with specified dimnensions //loop to process the array and read in values from the text file. for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ maze[i][j] = TextIO.getChar(); } TextIO.getln(); } //end for loop TextIO.readStandardInput(); //closes MazeGame.txt file and reads from //standard input. //loop to process the array values and display as unicode characters for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ if (i == plyrX && j == plyrY){ plyrX = i; plyrY = j; TextIO.put(PLAYER); //puts the player character at player co-ords } else{ if (maze[i][j] == '0') TextIO.putf("%c",WALL); //puts wall block if (maze[i][j] == '1') TextIO.putf("%c",PATH); //puts path block if (maze[i][j] == '2') { entX = i; entY = j; TextIO.putf("%c",ENTRANCE); //puts entrance character } if (maze[i][j] == '3') { exitX = i; //holds value of exit exitY = j; //co-ordinates TextIO.putf("%c",EXIT); //puts exit character } } } TextIO.putln(); } //end for loop } //end redrawMaze method /**replay game method * */ public static void replayGame(){ c_plyrX = plyrX; c_plyrY = plyrY; TextIO.readFile("MazeGame.txt"); //now reads from the external MazeGame.txt file rows = TextIO.getInt(); //gets the number of rows from text file to create X dimensions cols = TextIO.getlnInt(); //gets number of columns from text file to create Y dimensions mazeCopy = new char[rows][cols]; //creates maze array of base type char with specified dimnensions //loop to process the array and read in values from the text file. for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ mazeCopy[i][j] = TextIO.getChar(); } TextIO.getln(); } //end for loop TextIO.readStandardInput(); //closes MazeGame.txt file and reads from //standard input. //loop to process the array values and display as unicode characters for (int i = 0; i<rows; i++){ for (int j = 0; j<cols; j++){ if (i == c_plyrX && j == c_plyrY){ c_plyrX = i; c_plyrY = j; TextIO.put(PLAYER); //puts the player character at player co-ords } else{ if (mazeCopy[i][j] == '0') TextIO.putf("%c",WALL); //puts wall block if (mazeCopy[i][j] == '1') TextIO.putf("%c",PATH); //puts path block if (mazeCopy[i][j] == '2') { entX = i; entY = j; TextIO.putf("%c",ENTRANCE); //puts entrance character } if (mazeCopy[i][j] == '3') { exitX = i; //holds value of exit exitY = j; //co-ordinates TextIO.putf("%c",EXIT); //puts exit character } } } TextIO.putln(); } //end for loop } //end replayGame method /**plyrMoves method, method for moving the players character * around the maze. */ public static void plyrMoves(){ int nplyrX = plyrX; int nplyrY = plyrY; int pMoves; direction(); //UP if (dir == 'U' || dir == 'u'){ nplyrX = plyrX; nplyrY = plyrY; for(pMoves = 0; pMoves <= spaces; pMoves++){ if (maze[nplyrX][nplyrY] == '0'){ TextIO.putln("Invalid move, try again."); } else if (pMoves != spaces){ nplyrX =plyrX + 1; } else { plyrX = plyrX-spaces; c_plyrX = plyrX; movesTaken++; } } }//end UP if //DOWN if (dir == 'D' || dir == 'd'){ nplyrX = plyrX; nplyrY = plyrY; for (pMoves = 0; pMoves <= spaces; pMoves ++){ if (maze[nplyrX][nplyrY] == '0'){ TextIO.putln("Invalid move, try again"); } else if (pMoves != spaces){ nplyrX = plyrX+1; } else{ plyrX = plyrX+spaces; c_plyrX = plyrX; movesTaken++; } } } //end DOWN if //LEFT if (dir == 'L' || dir =='l'){ nplyrX = plyrX; nplyrY = plyrY; for (pMoves = 0; pMoves <= spaces; pMoves++){ if (maze[nplyrX][nplyrY] == '0'){ TextIO.putln("Invalid move, try again"); } else if (pMoves != spaces){ nplyrY = plyrY + 1; } else{ plyrY = plyrY-spaces; c_plyrY = plyrY; movesTaken++; } } } //end LEFT if //RIGHT if (dir == 'R' || dir == 'r'){ nplyrX = plyrX; nplyrY = plyrY; for (pMoves = 0; pMoves <= spaces; pMoves++){ if (maze[nplyrX][nplyrY] == '0'){ TextIO.putln("Invalid move, try again."); } else if (pMoves != spaces){ nplyrY += 1; } else{ plyrY = plyrY+spaces; c_plyrY = plyrY; movesTaken++; } } } //end RIGHT if //prints message if player escapes from the maze. if (maze[plyrX][plyrY] == '3'){ TextIO.putln("****Congratulations****"); TextIO.putln(); TextIO.putln("You have escaped from the maze."); TextIO.putln(); userMenu(); } else{ movesTaken++; redrawMaze(); plyrMoves(); } } //end of plyrMoves method /**direction, method * */ public static char direction(){ TextIO.putln("Enter the direction you wish to move in and the distance"); TextIO.putln("i.e D3 = move down 3 spaces"); TextIO.putln("U - Up, D - Down, L - Left, R - Right: "); dir = TextIO.getChar(); if (dir =='U' || dir == 'D' || dir == 'L' || dir == 'R' || dir == 'u' || dir == 'd' || dir == 'l' || dir == 'r'){ spacesMoved(); } else{ loadMaze(); TextIO.putln("Invalid direction!"); TextIO.put("Direction must be one of U, D, L or R"); direction(); } return dir; //returns the value of dir (direction) } //end direction method /**spaces method, gets the amount of spaces the user wants to move * */ public static int spacesMoved(){ TextIO.putln(" "); spaces = TextIO.getlnInt(); if (spaces <= 0){ loadMaze(); TextIO.put("Invalid amount of spaces, try again"); spacesMoved(); } return spaces; } //end spacesMoved method } //end of MazeGame class

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  • Can't find how to import as one object or how to merge

    - by Aaron
    I need write a script in blender that creates some birds which fly around some obstacles. The problem is that I need to import a pretty large Collada model (a building) which consists of multiple objects. The import works fine, but the the building is not seen as 1 object. I need to resize and move this building, but I can only get the last object in the building (which is a camera)... Does anyone know how to merge this building in 1 object, group, variable... so I can resize and move it correctly? Part of the code I used: bpy.ops.wm.collada_import(filepath="C:\\Users\\me\\building.dae") building= bpy.context.object building.scale = (100, 100, 100) building.name = "building"

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  • Array: mathematical sequence

    - by VaioIsBorn
    An array of integers A[i] (i 1) is defined in the following way: an element A[k] ( k 1) is the smallest number greater than A[k-1] such that the sum of its digits is equal to the sum of the digits of the number 4* A[k-1] . You need to write a program that calculates the N th number in this array based on the given first element A[1] . INPUT: In one line of standard input there are two numbers seperated with a single space: A[1] (1 <= A[1] <= 100) and N (1 <= N <= 10000). OUTPUT: The standard output should only contain a single integer A[N] , the Nth number of the defined sequence. Input: 7 4 Output: 79 Explanation: Elements of the array are as follows: 7, 19, 49, 79... and the 4th element is solution. I tried solving this by coding a separate function that for a given number A[k] calculates the sum of it's digits and finds the smallest number greater than A[k-1] as it says in the problem, but with no success. The first testing failed because of a memory limit, the second testing failed because of a time limit, and now i don't have any possible idea how to solve this. One friend suggested recursion, but i don't know how to set that. Anyone who can help me in any way please write, also suggest some ideas about using recursion/DP for solving this problem. Thanks.

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  • Taking User Input in Java

    - by jdbeverly87
    I'm creating a program that checks if a word or phrase is a palindrome. I have the actual "palindrome tester" figured out. What I'm stuck with is where and what to place in my code to have the console read out "Enter palindrome..." and then text. I've tried with IO but it doesnt work out right. Also, how do I create a loop to keep going? This code only allows one at a time `public class Palindrome { public static void main(String args[]) { String s=""; int i; int n=s.length(); String str=""; for(i=n-1;i>=0;i--) str=str+s.charAt(i); if(str.equals(s)) System.out.println(s+ " is a palindrome"); else System.out.println(s+ " is not a palindrome"); } }

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  • Why is my Pre to Postfix code not working?

    - by Anthony Glyadchenko
    For a class assignment, I have to use two stacks in C++ to make an equation to be converted to its left to right equivalent: 2+4*(3+4*8) -- 35*4+2 -- 142 Here is the main code: #include <iostream> #include <cstring> #include "ctStack.h" using namespace std; int main (int argc, char * const argv[]) { string expression = "2+4*2"; ctstack *output = new ctstack(expression.length()); ctstack *stack = new ctstack(expression.length()); bool previousIsANum = false; for(int i = 0; i < expression.length(); i++){ switch (expression[i]){ case '(': previousIsANum = false; stack->cmstackPush(expression[i]); break; case ')': previousIsANum = false; char x; while (x != '('){ stack->cmstackPop(x); output->cmstackPush(x); } break; case '0': case '1': case '2': case '3': case '4': case '5': case '6': case '7': case '8': case '9': cout << "A number" << endl; previousIsANum = true; output->cmstackPush(expression[i]); break; case '+': previousIsANum = false; cout << "+" << endl; break; case '-': previousIsANum = false; cout << "-" << endl; break; case '*': previousIsANum = false; cout << "*" << endl; break; case '/': previousIsANum = false; cout << "/" << endl; break; default: break; } } char i = ' '; while (stack->ltopOfStack > 0){ stack->cmstackPop(i); output->cmstackPush(i); cout << i << endl; } return 0; } Here is the stack code (watch out!): #include <cstdio> #include <assert.h> #include <new.h> #include <stdlib.h> #include <iostream> class ctstack { private: long* lpstack ; // the stack itself long ltrue ; // constructor sets to 1 long lfalse ; // constructor sets to 0 // offset to top of the stack long lmaxEleInStack ; // maximum possible elements of stack public: long ltopOfStack ; ctstack ( long lnbrOfEleToAllocInStack ) { // Constructor lfalse = 0 ; // set to zero ltrue = 1 ; // set to one assert ( lnbrOfEleToAllocInStack > 0 ) ; // assure positive argument ltopOfStack = -1 ; // ltopOfStack is really an index lmaxEleInStack = lnbrOfEleToAllocInStack ; // set lmaxEleInStack to max ele lpstack = new long [ lmaxEleInStack ] ; // allocate stack assert ( lpstack ) ; // assure new succeeded } ~ctstack ( ) { // Destructor delete [ ] lpstack ; // Delete the stack itself } ctstack& operator= ( const ctstack& ctoriginStack) { // Assignment if ( this == &ctoriginStack ) // verify x not assigned to x return *this ; if ( this -> lmaxEleInStack < ctoriginStack . lmaxEleInStack ) { // if destination stack is smaller than delete [ ] this -> lpstack ; // original stack, delete dest and alloc this -> lpstack = // sufficient memory new long [ ctoriginStack . lmaxEleInStack ] ; assert ( this -> lpstack ) ; // assure new succeeded // reset stack size attribute this -> lmaxEleInStack = ctoriginStack . lmaxEleInStack ; } // copy original to destination stack for ( long i = 0 ; i < ctoriginStack . lmaxEleInStack ; i ++ ) *( this -> lpstack + i ) = *( ctoriginStack . lpstack + i ) ; this -> ltopOfStack = ctoriginStack . ltopOfStack ; // reset stack position attribute return *this ; } long cmstackPush (char lplaceInStack ) { // Push Method if ( ltopOfStack == lmaxEleInStack - 1 ) // stack is full can't add element return lfalse ; ltopOfStack ++ ; // acquire free slot *(lpstack + ltopOfStack ) = lplaceInStack ; // add element return ltrue ; // any number other than zero is true } long cmstackPop (char& lretrievedStackEle ) { // Pop Method if ( ltopOfStack < 0 ) { // stack has no elements lretrievedStackEle = -1 ; // dummy element return lfalse ; } lretrievedStackEle = *( lpstack + ltopOfStack ) ; // stack has element -- return it ltopOfStack -- ; // stack is pop'd return ltrue ; // any number other than zero is true } long cmstackLookAtTop (char& lretrievedStackEle ) { // Pop Method if ( ltopOfStack < 0 ) { // stack has no elements lretrievedStackEle = -1 ; // dummy element return lfalse ; } lretrievedStackEle = *( lpstack + ltopOfStack ) ; // stack has element -- return it return ltrue ; // any number other than zero is true } long cmstackHasAnEle (char& lretrievedTopOfStack ) { // Has element method lretrievedTopOfStack = ltopOfStack ; return ltopOfStack < 0 ? lfalse : ltrue ; // 0 - false stack does not have any ele } // 1 - true stack has at least one element long cmstackMaxNbrOfEle (char& lretrievedMaxStackEle ) { // Maximum element method lretrievedMaxStackEle = lmaxEleInStack ; // return stack size in reference var return ltrue ; // Return Maximum Size of Stack } } ; Thanks, Anthony.

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  • Read a buffer of unknown size (Console input)

    - by Sanarothe
    Hi. I'm a little behind in my X86 Asm class, and the book is making me want to shoot myself in the face. The examples in the book are insufficient and, honestly, very frustrating because of their massive dependencies upon the author's link library, which I hate. I wanted to learn ASM, not how to call his freaking library, which calls more of his library. Anyway, I'm stuck on a lab that requires console input and output. So far, I've got this for my input: input PROC INVOKE ReadConsole, inputHandle, ADDR buffer, Buf - 2, ADDR bytesRead, 0 mov eax,OFFSET buffer Ret input EndP I need to use the input and output procedures multiple times, so I'm trying to make it abstract. I'm just not sure how to use the data that is set to eax here. My initial idea was to take that string array and manually crawl through it by adding 8 to the offset for each possible digit (Input is integer, and there's a little bit of processing) but this doesn't work out because I don't know how big the input actually is. So, how would you swap the string array into an integer that could be used? Full code: (Haven't done the integer logic or the instruction string output because I'm stuck here.) include c:/irvine/irvine32.inc .data inputHandle HANDLE ? outputHandle HANDLE ? buffer BYTE BufSize DUP(?),0,0 bytesRead DWORD ? str1 BYTE "Enter an integer:",0Dh, 0Ah str2 BYTE "Enter another integer:",0Dh, 0Ah str3 BYTE "The higher of the two integers is: " int1 WORD ? int2 WORD ? int3 WORD ? Buf = 80 .code main PROC call handle push str1 call output call input push str2 call output call input push str3 call output call input main EndP larger PROC Ret larger EndP output PROC INVOKE WriteConsole Ret output EndP handle PROC USES eax INVOKE GetStdHandle, STD_INPUT_HANDLE mov inputHandle,eax INVOKE GetStdHandle, STD_INPUT_HANDLE mov outputHandle,eax Ret handle EndP input PROC INVOKE ReadConsole, inputHandle, ADDR buffer, Buf - 2, ADDR bytesRead, 0 mov eax,OFFSET buffer Ret input EndP END main

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  • Why isnt this returning the new string?

    - by Evan Kimia
    I have a recursive method that reversed a string (HW assignment, has to be recursive). I did it....but its only returning the value of the string after the first pass. By analyzing the output after each pass i can see it does do its job correctly. heres my code, and the output i get below it: String s = "Hello, I love you wont you tell me your name?"; int k=0; public String reverseThisString(String s) { if(k!=s.length()) { String first =s.substring(0,k)+s.charAt(s.length()-1); String end = ""+s.substring(k, s.length()-1); k++; s=first+end; System.out.println(s); this.reverseThisString(s); } return s; } output: ?Hello, I love you wont you tell me your name

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  • How to spot empty parking spaces?

    - by mithila
    I want to do a final year B.Sc project on parking space detection. Can anybody give me some link related to it? Any text-book, tutorial or anything? What would be prerequisite for this project? What type of skills(programming/math) are needed? What are the initial steps to do? What type of readings(algorithms of image processing) are needed? Detail added in comments: i'm going to use camera, not infrared. i would like to use still images, or one camera which captures images from a parking lot. i think reak-time processing will be tough, at this moment i just need to start the project. so still images will work fine. but later it may b a real-time project

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