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  • How to obtain dependency metrics from Java source code?

    - by Bram Schoenmakers
    For an assignment we have to extract some software metrics from the Hibernate project. We have to extract the afferent coupling and efferent coupling metrics (dependency fan-in, fan-out) from each revision of each package in Hibernate. Some tools were provided which are able to extract these metrics, such as ckjm and JDepend. Other tools I have checked were Sonar, javancss and AOP. There is also the Metrics Eclipse plugin which I didn't get to work either. What these tools have in common, as far as I can see, is that they all operate on bytecode (*.class files). This is a problem, because I have to build every revision from source in order to run, say, JDepend on it. Older revisions won't build because my development stack is too recent. What I would like to do is to do this kind of analysis on source files so that I don't have to build each revision. Is this possible? Or is there a good reason why all these tools only operate on bytecode?

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  • How to change Matlab program for solving equation with finite element method?

    - by DSblizzard
    I don't know is this question more related to mathematics or programming and I'm absolute newbie in Matlab. Program FEM_50 applies the finite element method to Laplace's equation -Uxx(x, y) - Uyy(x, y) = F(x, y) in Omega. How to change it to apply FEM to equation -Uxx(x, y) - Uyy(x, y) + U(x, y) = F(x, y)? At this page: http://sc.fsu.edu/~burkardt/m_src/fem_50/fem_50.html additional code files in case you need them. function fem_50 ( ) %% FEM_50 applies the finite element method to Laplace's equation. % % Discussion: % % FEM_50 is a set of MATLAB routines to apply the finite % element method to solving Laplace's equation in an arbitrary % region, using about 50 lines of MATLAB code. % % FEM_50 is partly a demonstration, to show how little it % takes to implement the finite element method (at least using % every possible MATLAB shortcut.) The user supplies datafiles % that specify the geometry of the region and its arrangement % into triangular and quadrilateral elements, and the location % and type of the boundary conditions, which can be any mixture % of Neumann and Dirichlet. % % The unknown state variable U(x,y) is assumed to satisfy % Laplace's equation: % -Uxx(x,y) - Uyy(x,y) = F(x,y) in Omega % with Dirichlet boundary conditions % U(x,y) = U_D(x,y) on Gamma_D % and Neumann boundary conditions on the outward normal derivative: % Un(x,y) = G(x,y) on Gamma_N % If Gamma designates the boundary of the region Omega, % then we presume that % Gamma = Gamma_D + Gamma_N % but the user is free to determine which boundary conditions to % apply. Note, however, that the problem will generally be singular % unless at least one Dirichlet boundary condition is specified. % % The code uses piecewise linear basis functions for triangular elements, % and piecewise isoparametric bilinear basis functions for quadrilateral % elements. % % The user is required to supply a number of data files and MATLAB % functions that specify the location of nodes, the grouping of nodes % into elements, the location and value of boundary conditions, and % the right hand side function in Laplace's equation. Note that the % fact that the geometry is completely up to the user means that % just about any two dimensional region can be handled, with arbitrary % shape, including holes and islands. % clear % % Read the nodal coordinate data file. % load coordinates.dat; % % Read the triangular element data file. % load elements3.dat; % % Read the quadrilateral element data file. % load elements4.dat; % % Read the Neumann boundary condition data file. % I THINK the purpose of the EVAL command is to create an empty NEUMANN array % if no Neumann file is found. % eval ( 'load neumann.dat;', 'neumann=[];' ); % % Read the Dirichlet boundary condition data file. % load dirichlet.dat; A = sparse ( size(coordinates,1), size(coordinates,1) ); b = sparse ( size(coordinates,1), 1 ); % % Assembly. % for j = 1 : size(elements3,1) A(elements3(j,:),elements3(j,:)) = A(elements3(j,:),elements3(j,:)) ... + stima3(coordinates(elements3(j,:),:)); end for j = 1 : size(elements4,1) A(elements4(j,:),elements4(j,:)) = A(elements4(j,:),elements4(j,:)) ... + stima4(coordinates(elements4(j,:),:)); end % % Volume Forces. % for j = 1 : size(elements3,1) b(elements3(j,:)) = b(elements3(j,:)) ... + det( [1,1,1; coordinates(elements3(j,:),:)'] ) * ... f(sum(coordinates(elements3(j,:),:))/3)/6; end for j = 1 : size(elements4,1) b(elements4(j,:)) = b(elements4(j,:)) ... + det([1,1,1; coordinates(elements4(j,1:3),:)'] ) * ... f(sum(coordinates(elements4(j,:),:))/4)/4; end % % Neumann conditions. % if ( ~isempty(neumann) ) for j = 1 : size(neumann,1) b(neumann(j,:)) = b(neumann(j,:)) + ... norm(coordinates(neumann(j,1),:) - coordinates(neumann(j,2),:)) * ... g(sum(coordinates(neumann(j,:),:))/2)/2; end end % % Determine which nodes are associated with Dirichlet conditions. % Assign the corresponding entries of U, and adjust the right hand side. % u = sparse ( size(coordinates,1), 1 ); BoundNodes = unique ( dirichlet ); u(BoundNodes) = u_d ( coordinates(BoundNodes,:) ); b = b - A * u; % % Compute the solution by solving A * U = B for the remaining unknown values of U. % FreeNodes = setdiff ( 1:size(coordinates,1), BoundNodes ); u(FreeNodes) = A(FreeNodes,FreeNodes) \ b(FreeNodes); % % Graphic representation. % show ( elements3, elements4, coordinates, full ( u ) ); return end

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  • What's wrong with this conditional?

    - by David
    I am trying to make a method that tests to see if 3 lengths can make a triangle. I think i'm making some kind of syntax error but i can't figure out what it is. Here is the relevant bit of code: (its in java) public static void trya (int a, int b, int c) { if (c>(a+b)) { System.out.println ("yes") ; } else { if (b>(a+c)) { System.out.println ("yes") ; } } else { if (a>(b+c)) { System.out.println ("yes") ; } } else { System.out.println ("no") ; } } this is the error message i get: tryangle.java:17: 'else' without 'if' else ^

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  • How to write a C program using the fork() system call that generates the Fibonacci sequence in the

    - by Ellen
    The problem I am having is that when say for instance the user enters 7, then the display shows: 0 11 2 3 5 8 13 21 child ends. I cannot seem to figure out how to fix the 11 and why is it displaying that many numbers in the sequence! Can anyone help? The number of the sequence will be provided in the command line. For example, if 5 is provided, the first five numbers in the Fibonacci sequence will be output by the child process. Because the parent and child processes have their own copies of the data, it will be necessary for the child to output the sequence. Have the parent invoke the wait() call to wait for the child process to complete before exiting the program. Perform necessary error checking to ensure that a non-negative number is passed on the command line. #include <stdio.h> #include <sys/types.h> #include <unistd.h> int main() { int a=0, b=1, n=a+b,i,ii; pid_t pid; printf("Enter the number of a Fibonacci Sequence:\n"); scanf("%d", &ii); if (ii < 0) printf("Please enter a non-negative integer!\n"); else { pid = fork(); if (pid == 0) { printf("Child is producing the Fibonacci Sequence...\n"); printf("%d %d",a,b); for (i=0;i<ii;i++) { n=a+b; printf("%d ", n); a=b; b=n; } printf("Child ends\n"); } else { printf("Parent is waiting for child to complete...\n"); wait(NULL); printf("Parent ends\n"); } } return 0; }

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  • Data Flow Diagram

    - by Nilesh
    Can anyone help me to draw a data flow diagram for a travel request form for a company in which an employee can request for travel and request approval by his/her by project manager and HR department. Regards Nils

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  • Need help with basic ASM

    - by Malfist
    Hello, I'm trying to convert some c code to assmebly, and I need some help. char encode(char plain){ __asm{ mov eax, plain add eax, 2 ret } //C code /* char code; code = plain+2; return code;*/ } First problem is that visual studio complains that the register size doesn't match, i.e. eax is too small/large for char. I was under the impression that they were both DWORDs. Also, if I leave the variable in eax, and ret in assembly, it'll actually return that variable, right?

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  • Computer science final year project ideas

    - by roul
    I'm a Computer Science undergraduate student in UK and should be deciding the subject of my final year project soon. The school is pretty flexible with the subject... "The topic can be any area of the subject which is of mutual interest to both the student and supervisor. Topics can range from purely theoretical studies to practical work building a system for some third party, although most projects aim to provide a balance between the theoretical and practical aspects of the subject." ...so I'm a bit lost since I want to do something in software engineering but have no idea what (subject) or with what (languages)! :) a) Languages: I've had experience with Java, C# and ASP.NET mostly but I would definitely be interested in learning new languages/frameworks. I'm kind of drawn by the idea of dynamic languages at the moment so IronPython seems likely. b) Subject: Anything that will keep me interested through the year and will give me the opportunity to learn a lot of stuff. Maybe something that has to do with music, or a fancy website, or a website about music :P anything really. Open to any thoughts/ideas, geeky or cool! Edit: Professors do usually supervise projects in their research areas but I currently have the choice to approach any of them according to my interest - whatever that is.

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  • How to download a file from a server using Java Socket?

    - by Ada
    Hi, I have an assignment about uploading and downloading a file to a server. I managed to do the uploading part using Java Sockets however I am having a hard time doing the downloading part. I should use Range: for downloading parellel. In my request, I should have the Range: header. But I don't understand how I will receive the file with that HTTP GET request. All the examples I have seen was about uploading a file. I already did it. I can upload .exe, image, .pdf, anything and when I download them back (by my browser), there are no errors. Can you help me with the downloading part? Can you give me an example beacuse I really didn't get it.

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  • HttpURLConnection timeout question

    - by Malachi
    I want to return false if the URL takes more then 5 seconds to connect - how is this possible using java? Here is the code I am using to check if the URL is valid HttpURLConnection.setFollowRedirects(false); HttpURLConnection con = (HttpURLConnection) new URL(url).openConnection(); con.setRequestMethod("HEAD"); return (con.getResponseCode() == HttpURLConnection.HTTP_OK);

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  • Need help with dynamic programming problem

    - by John Retallack
    I have the following problem : I am given a tree with N apples, for each apple I am given it's weight and height,I can pick apples up to a given height H,each time I pick an apple the height of every apple is increased with U(also given).I have to find out the maximum weight of apples I can pick. e.g: N=4 H=100 U=10 (height-eight) apple1: 91 10 apple2: 82 30 apple3: 93 5 apple4: 94 15 The answer is 45 : I first pick the apple with the weight of 15 then the one with the weight of 30. I would like to know if someone here could help me with giving me an hint on how I should approach this problem. Thank you.

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  • Why isn't my operator overloading working properly?

    - by Mithrax
    I have the following Polynomial class I'm working on: #include <iostream> using namespace std; class Polynomial { //define private member functions private: int coef[100]; // array of coefficients // coef[0] would hold all coefficients of x^0 // coef[1] would hold all x^1 // coef[n] = x^n ... int deg; // degree of polynomial (0 for the zero polynomial) //define public member functions public: Polynomial::Polynomial() //default constructor { for ( int i = 0; i < 100; i++ ) { coef[i] = 0; } } void set ( int a , int b ) //setter function { //coef = new Polynomial[b+1]; coef[b] = a; deg = degree(); } int degree() { int d = 0; for ( int i = 0; i < 100; i++ ) if ( coef[i] != 0 ) d = i; return d; } void print() { for ( int i = 99; i >= 0; i-- ) { if ( coef[i] != 0 ) { cout << coef[i] << "x^" << i << " "; } } } // use Horner's method to compute and return the polynomial evaluated at x int evaluate ( int x ) { int p = 0; for ( int i = deg; i >= 0; i-- ) p = coef[i] + ( x * p ); return p; } // differentiate this polynomial and return it Polynomial differentiate() { if ( deg == 0 ) { Polynomial t; t.set ( 0, 0 ); return t; } Polynomial deriv;// = new Polynomial ( 0, deg - 1 ); deriv.deg = deg - 1; for ( int i = 0; i < deg; i++ ) deriv.coef[i] = ( i + 1 ) * coef[i + 1]; return deriv; } Polynomial Polynomial::operator + ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] += b.coef[i]; c.deg = c.degree(); return c; } Polynomial Polynomial::operator += ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] += b.coef[i]; c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator -= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator *= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) for ( int j = 0; j <= b.deg; j++ ) c.coef[i+j] += ( a.coef[i] * b.coef[j] ); c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator - ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); return c; } Polynomial Polynomial::operator * ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) for ( int j = 0; j <= b.deg; j++ ) c.coef[i+j] += ( a.coef[i] * b.coef[j] ); c.deg = c.degree(); return c; } }; int main() { Polynomial a, b, c, d; a.set ( 7, 4 ); //7x^4 a.set ( 1, 2 ); //x^2 b.set ( 6, 3 ); //6x^3 b.set ( -3, 2 ); //-3x^2 c = a - b; // (7x^4 + x^2) - (6x^3 - 3x^2) a -= b; c.print(); cout << "\n"; a.print(); cout << "\n"; c = a * b; // (7x^4 + x^2) * (6x^3 - 3x^2) c.print(); cout << "\n"; d = c.differentiate().differentiate(); d.print(); cout << "\n"; cout << c.evaluate ( 2 ); //substitue x with 2 cin.get(); } Now, I have the "-" operator overloaded and it works fine: Polynomial Polynomial::operator - ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); return c; } However, I'm having difficulty with my "-=" operator: Polynomial Polynomial::operator -= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); // overwrite value of 'a' with the newly computed 'c' before returning 'a' for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } I just slightly modified my "-" operator method to overwrite the value in 'a' and return 'a', and just use the 'c' polynomial as a temp. I've put in some debug print statement and I confirm that at the time of computation, both: c = a - b; and a -= b; are computed to the same value. However, when I go to print them, their results are different: Polynomial a, b; a.set ( 7, 4 ); //7x^4 a.set ( 1, 2 ); //x^2 b.set ( 6, 3 ); //6x^3 b.set ( -3, 2 ); //-3x^2 c = a - b; // (7x^4 + x^2) - (6x^3 - 3x^2) a -= b; c.print(); cout << "\n"; a.print(); cout << "\n"; Result: 7x^4 -6x^3 4x^2 7x^4 1x^2 Why is my c = a - b and a -= b giving me different results when I go to print them?

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  • describe class versions of primitive type data

    - by paula
    please some explain to me this.... Explain why java has class versios of primitive data types? describe the class versions of primitive java types and expain pros and cons? expalin why class versions of primitive java types are available, but why the primitive types are still frequently used? expalin how primitive and non-primitive data types are passed as parameters to methods in java and how that affects altering the value of data passed? primitve data types are: int, float, char, long, short, byte, boolean and double.

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  • C - array count, strtok, etc

    - by Pedro
    Hi... i have a little problem on my code... HI open a txt that have this: LEI;7671;Maria Albertina da silva;[email protected]; 9;8;12;9;12;11;6;15;7;11; LTCGM;6567;Artur Pereira Ribeiro;[email protected]; 6;13;14;12;11;16;14; LEI;7701;Ana Maria Carvalho;[email protected]; 8;13;11;7;14;12;11;16;14; LEI, LTCGM are the college; 7671, 6567, 7701 is student number; Maria, Artur e Ana are the students name; [email protected], ...@gmail are emails from students; the first number of every line is the total of classes that students have; after that is students school notes; example: College: LEI Number: 7671 Name: Maria Albertina da Silva email: [email protected] total of classes: 9 Classe Notes: 8 12 9 12 11 6 15 7 11. My code: typedef struct aluno{ char sigla[5];//college char numero[80];//number char nome[80];//student name char email[20];//email int total_notas;// total of classes char tot_not[40]; // total classes char notas[20];// classe notes int nota; //class notes char situacao[80]; //situation (aproved or disaproved) }ALUNO; void ordena(ALUNO*alunos, int tam)//bubble sort { int i=0; int j=0; char temp[100]; for( i=0;i<tam;i++) for(j=0;j<tam-1;j++) if(strcmp( alunos[i].sigla[j], alunos[i].sigla[j+1])>0){ strcpy(temp, alunos[i].sigla[j]); strcpy(alunos[i].sigla[j],alunos[i].sigla[j+1]); strcpy(alunos[i].sigla[j+1], temp); } } void xml(ALUNO*alunos, int tam){ FILE *fp; char linha[60];//line int soma, max, min, count;//biggest note and lowest note and students per course count float media; //media of notes fp=fopen("example.txt","r"); if(fp==NULL){ exit(1); } else{ while(!(feof(fp))){ soma=0; media=0; max=0; min=0; count=0; fgets(linha,60,fp); if(linha[0]=='L'){ if(ap_dados=strtok(linha,";")){ strcpy(alunos[i].sigla,ap_dados);//copy to struct // i need to call bubble sort here, but i don't know how printf("College: %s\n",alunos[i].sigla); if(ap_dados=strtok(NULL,";")){ strcpy(alunos[i].numero,ap_dados);//copy to struct printf("number: %s\n",alunos[i].numero); if(ap_dados=strtok(NULL,";")){ strcpy(alunos[i].nome, ap_dados);//copy to struct printf("name: %s\n",alunos[i].nome); if(ap_dados=strtok(NULL,";")){ strcpy(alunos[i].email, ap_dados);//copy to struct printf("email: %s\n",alunos[i].email); } } } }i++; } if(isdigit(linha[0])){ if(info_notas=strtok(linha,";")){ strcpy(alunos[i].tot_not,info_notas); alunos[i].total_notas=atoi(alunos[i].tot_not);//total classes for(z=0;z<=alunos[i].total_notas;z++){ if(info_notas=strtok(NULL,";")){ strcpy(alunos[i].notas,info_notas); alunos[i].nota=atoi(alunos[i].notas); // student class notes } soma=soma + alunos[i].nota; media=soma/alunos[i].total_notas;//doesn't work if(alunos[i].nota>max){ max=alunos[i].nota;;//doesn't work } else{ if(min<alunos[i].nota){ min=alunos[i].nota;;//doesn't work } } //now i need to count the numbers of students in the same college, but doesn't work /*If(strcmp(alunos[i].sigla, alunos[i+1].sigla)=0){ count ++; printf("%d\n", count); here for LEI should appear 2 students and for LTCGM appear 1, don't work }*/ //Now i need to see if student is aproved or disaproved // Student is disaproved if he gets 3 notes under 10, how can i do that? } printf("media %d\n",media); //media printf("Nota maxima %d\n",max);// biggest note printf("Nota minima %d\n",min); //lowest note }i++; } } } fclose(fp); } int main(int argc, char *argv[]){ ALUNO alunos; FILE *fp; int tam; fp=fopen(nomeFicheiro,"r"); alunos = (ALUNO*) calloc (tam, sizeof(ALUNO)); xml(alunos,nomeFicheiro, tam); system("PAUSE"); return 0; }

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  • How do I work out IEEE 754 64-bit Floating Point Double Precision?

    - by yousef gassar
    enter code herehello i have done it in 32 but i could dont do it in 62bits please i need help I am stuck on this question and need help. I don't know how to work it out. This is the question. Below are two numbers represented in IEEE 754 64-bit Floating Point Double Precision, the bias of the signed exponent is -1023. Any particular real number ‘N’ represented in 64-bit form (i.e. with the following bit fields; 1-bit Sign, 11-bit Exponent, 52-bit Fraction) can be expressed in the form ±1.F2 × 2X by substituting the bit-field values using formula (IV.I): N = (-1) S × 1.F2 × 2(E – 1023) for 0 < E < 2047.........................….(IV.I) Where N= the number represented, S=Sign bit-value, E=Exponent=X +1023, F=Fraction or Mantissa are the values in the 1, 11 and 52-bit fields respectively in the IEEE 754 64-bit FP representation. Using formula (IV.I), express the 64-bit FP representation of each number as: (i) A binary number of the form:- ±1.F2 × 2X (ii) A decimal number of the form:- ±0.F10 × 10Y {limit F10 to 10 decimal places} Sign 0 1 Exponent 1000 0001 001 11 Fraction 1111 0111 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 52 Sign 1 1 Exponent 1000 0000 000 11 Fraction 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 52 I know I have to use the formula for each of the these but how do I work it out? Is it like this? N = (-1) S × 1.F2 × 2(E – 1023) = 1 x 1.1111 0111 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 x 1000 0001 00111 (-1023)?

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  • Update table.column with another table.column with common joined column

    - by Matt
    Hit a speed bump, trying to update some column values in my table from another table. This is what is supposed to happen when everything works Correct all the city, state entries in tblWADonations by creating an update statement that moves the zip city from the joined city/state zip field to the tblWADonations city state TBL NAME | COLUMN NAMES tblZipcodes with zip,city,State tblWADonations with zip,oldcity,oldstate This is what I have so far: UPDATE tblWADonations SET oldCity = tblZipCodes.city, oldState = tblZipCodes.state FROM tblWADonations INNER JOIN tblZipCodes ON tblWADonations.zip = tblZipCodes.zip Where oldCity <> tblZipcodes.city; There seems to be easy ways to do this online but I am overlooking something. Tried this by hand and in editor this is what it kicks back. Msg 8152, Level 16, State 2, Line 1 String or binary data would be truncated. The statement has been terminated. Please include a sql statement or where I need to make the edit so I can mark this post as a reference in my favorites. Thanks!

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  • Avoiding RTTI In Java

    - by destructo_gold
    Hi, If I have a superclass, say Animal, and two subclasses: Zebra and Giraffe, If I decide to define a Vector of Animals: Vector <Animal> animals = new Vector(); and I want to say: You can add Giraffes, but you must own at least one Zebra first. What is the best way to do this without using RTTI? (instanceof)

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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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  • Python sudoku programming

    - by trevor
    I need your help on this. I have this program and I must finish it. It's missing 3 parts. Here is the program I'm working with: import copy def display(A): if A: for i in range(9): for j in range(9): if type(A[i][j]) == type([]): print A[i][j][0], else: print A[i][j], print print else: print A def has_conflict(A): for i in range(9): for j in range(9): for (x,y) in get_neighbors(i,j): if len(A[i][j])==1 and A[i][j]==A[x][y]: return True return False # HERE ARE THE PARTS THAT REQUIRE HELP!!!! def get_neighbors(x,y): return [] def update(A, i, j, value): return [] def solve(A): return [] # ENDS PARTS THAT REQUIRE HELP!!!! A = [] infile = open('puzzle1.txt', 'r') for i in range(9): A += [[]] for j in range(9): num = int(infile.read(2)) if num: A[i] += [[num]] else: A[i] += [[1,2,3,4,5,6,7,8,9]] for i in range(9): for j in range(9): if len(A[i][j])==1: A = update(A, i,j, A[i][j][0]) if A==[]: break if A==[]: break if A<>[]: A = solve(A) display(A) I need to solve the stuff formerly in bold letters, now explicitly marked in the code, specifically - get_neighbors(): - update(): - solve(): Thank you for your time and help.

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  • using java Calendar

    - by owca
    I have a simple task. There are two classes : Ticket and Date. Ticket contains event, event place and event date which is a Date object. I also need to provide a move() method for Date object, so I used Calendar and Calendar's add(). Everything looks fine apart of the output. I constantly get 5,2,1 as the date's day,month,year. Lines with asterix return proper date. The code : Ticket class : public class Ticket { private String what; private String where; private Date when; public Ticket(String s1, String s2, Data d){ this.what = s1; this.where = s2; this.when = d; } *public Date giveDate(){ System.out.println("when in giveDate() "+this.when); return this.when; } public String toString(){ return "what: "+this.what+"\n"+"where: "+this.where+"\n"+"when: "+this.when; } } Date class: import java.util.Calendar; import java.util.GregorianCalendar; public class Date { public int day; public int month; public int year; public Date(int x, int y, int z){ *System.out.println("x: "+x); *System.out.println("y: "+y); *System.out.println("z: "+z); this.day = x; this.month = y; this.year = z; *System.out.println("this.day: "+this.day); *System.out.println("this.month: "+this.month); *System.out.println("this.year: "+this.year); } public Date move(int p){ *System.out.println("before_change: "+this.day+","+this.month+","+this.year); Calendar gc = new GregorianCalendar(this.year, this.month, this.day); System.out.println("before_adding: "+gc.DAY_OF_MONTH+","+gc.MONTH+","+gc.YEAR); gc.add(Calendar.DAY_OF_YEAR, p); System.out.println("after_adding: "+gc.DAY_OF_MONTH+","+gc.MONTH+","+gc.YEAR); this.year = gc.YEAR; this.day = gc.DAY_OF_MONTH; this.month = gc.MONTH; return this; } @Override public String toString(){ return this.day+","+this.month+","+this.year; } } Main for testing : public class Main { public static void main(String[] args) { Date date1=new Date(30,4,2002); Ticket event1=new Ticket("Peter Gabriel's gig", "London",date1 ); Ticket event2=new Ticket("Diana Kroll's concert", "Glasgow",date1 ); Date date2=event2.giveDate(); date2.move(30); Ticket event3=new Ticket("X's B-day", "some place",date2 ); System.out.println(date1); System.out.println(event1); System.out.println(event2); System.out.println(event3); } } And here's my output. I just can't get it where 5,2,1 come from :/ x: 30 y: 4 z: 2002 this.day: 30 this.month: 4 this.year: 2002 when in giveDate() 6,12,2004 before_change: 6,12,2004 before_adding: 5,2,1 after_adding: 5,2,1 5,2,1 what: Peter Gabriel's gig where: London when: 5,2,1 (...)

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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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