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• Nth largest element in a binary search tree

  - by Codenotguru

  How to find the Nth largest node in a BST? Do I keep a count variable while doing In Order Traversal of a BST? Return the element when the count = N???

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• Thread scheduling C

  - by MRP

  include <pthread.h> include <stdio.h> include <stdlib.h> #define NUM_THREADS 4 #define TCOUNT 5 #define COUNT_LIMIT 13 int done = 0; int count = 0; int thread_ids[4] = {0,1,2,3}; int thread_runtime[4] = {0,5,4,1}; pthread_mutex_t count_mutex; pthread_cond_t count_threshold_cv; void *inc_count(void *t) { int i; long my_id = (long)t; long run_time = thread_runtime[my_id]; if (my_id==2 && done ==0) { for(i=0; i< 5 ; i++) { if( i==4 ){done =1;} pthread_mutex_lock(&count_mutex); count++; if (count == COUNT_LIMIT) { pthread_cond_signal(&count_threshold_cv); printf("inc_count(): thread %ld, count = %d Threshold reached.\n", my_id, count); } printf("inc_count(): thread %ld, count = %d, unlocking mutex\n", my_id, count); pthread_mutex_unlock(&count_mutex); } } if (my_id==3 && done==1) { for(i=0; i< 4 ; i++) { if(i == 3 ){ done = 2;} pthread_mutex_lock(&count_mutex); count++; if (count == COUNT_LIMIT) { pthread_cond_signal(&count_threshold_cv); printf("inc_count(): thread %ld, count = %d Threshold reached.\n", my_id, count); } printf("inc_count(): thread %ld, count = %d, unlocking mutex\n", my_id, count); pthread_mutex_unlock(&count_mutex); } } if (my_id==4&& done == 2) { for(i=0; i< 8 ; i++) { pthread_mutex_lock(&count_mutex); count++; if (count == COUNT_LIMIT) { pthread_cond_signal(&count_threshold_cv); printf("inc_count(): thread %ld, count = %d Threshold reached.\n",my_id, count); } printf("inc_count(): thread %ld, count = %d, unlocking mutex\n", my_id, count); pthread_mutex_unlock(&count_mutex); } } pthread_exit(NULL); } void *watch_count(void *t) { long my_id = (long)t; printf("Starting watch_count(): thread %ld\n", my_id); pthread_mutex_lock(&count_mutex); if (count<COUNT_LIMIT) { pthread_cond_wait(&count_threshold_cv, &count_mutex); printf("watch_count(): thread %ld Condition signal received.\n", my_id); count += 125; printf("watch_count(): thread %ld count now = %d.\n", my_id, count); } pthread_mutex_unlock(&count_mutex); pthread_exit(NULL); } int main (int argc, char *argv[]) { int i, rc; long t1=1, t2=2, t3=3, t4=4; pthread_t threads[4]; pthread_attr_t attr; pthread_mutex_init(&count_mutex, NULL); pthread_cond_init (&count_threshold_cv, NULL); pthread_attr_init(&attr); pthread_attr_setdetachstate(&attr,PTHREAD_CREATE_JOINABLE); pthread_create(&threads[0], &attr, watch_count, (void *)t1); pthread_create(&threads[1], &attr, inc_count, (void *)t2); pthread_create(&threads[2], &attr, inc_count, (void *)t3); pthread_create(&threads[3], &attr, inc_count, (void *)t4); for (i=0; i<NUM_THREADS; i++) { pthread_join(threads[i], NULL); } printf ("Main(): Waited on %d threads. Done.\n", NUM_THREADS); pthread_attr_destroy(&attr); pthread_mutex_destroy(&count_mutex); pthread_cond_destroy(&count_threshold_cv); pthread_exit(NULL); } so this code creates 4 threads. thread 1 keeps track of the count value while the other 3 increment the count value. the run time is the number of times the thread will increment the count value. I have a done value that allows the first thread to increment the count value first until its run time is up.. so its like a First Come First Serve. my question is, is there a better way of implementing this? I have read about SCHED_FIFO or SCHED_RR.. I guess I dont know how to implement them into this code or if it can be.

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• Can someone help me with this Java Chess game please?

  - by Chris Edwards

  Hey guys, Please can someone have a look at this code and let me know whether I am on the right track with the "check_somefigure_move"s and the "check_black/white_promotion"s please? And also any other help you can give would be greatly appreciated! Thanks! P.S. I know the code is not the best implementation, but its a template I have to follow :( Code: class Moves { private final Board B; private boolean regular; public Moves(final Board b) { B = b; regular = regular_position(); } public boolean get_regular_position() { return regular; } public void set_regular_position(final boolean new_reg) { regular = new_reg; } // checking whether B represents a "normal" position or not; // if not, then only simple checks regarding move-correctness should // be performed, only checking the direct characteristics of the figure // moved; // checks whether there is exactly one king of each colour, there are // no more figures than promotions allow, and there are no pawns on the // first or last rank; public boolean regular_position() { int[] counts = new int[256]; for (char file = 'a'; file <= 'h'; ++file) for (char rank = '1'; rank <= '8'; ++rank) ++counts[(int) B.get(file,rank)]; if (counts[Board.white_king] != 1 || counts[Board.black_king] != 1) return false; if (counts[Board.white_pawn] > 8 || counts[Board.black_pawn] > 8) return false; int count_w_promotions = 0; count_w_promotions += Math.max(counts[Board.white_queen]-1,0); count_w_promotions += Math.max(counts[Board.white_rook]-2,0); count_w_promotions += Math.max(counts[Board.white_bishop]-2,0); count_w_promotions += Math.max(counts[Board.white_knight]-2,0); if (count_w_promotions > 8 - counts[Board.white_pawn]) return false; int count_b_promotions = 0; count_b_promotions += Math.max(counts[Board.black_queen]-1,0); count_b_promotions += Math.max(counts[Board.black_rook]-2,0); count_b_promotions += Math.max(counts[Board.black_bishop]-2,0); count_b_promotions += Math.max(counts[Board.black_knight]-2,0); if (count_b_promotions > 8 - counts[Board.black_pawn]) return false; for (char file = 'a'; file <= 'h'; ++file) { final char fig1 = B.get(file,'1'); if (fig1 == Board.white_pawn || fig1 == Board.black_pawn) return false; final char fig8 = B.get(file,'8'); if (fig8 == Board.white_pawn || fig8 == Board.black_pawn) return false; } return true; } public boolean check_normal_white_move(final char file0, final char rank0, final char file1, final char rank1) { if (! Board.is_valid_white_figure(B.get(file0,rank0))) return false; if (! B.is_empty(file1,rank1) && ! Board.is_valid_black_figure(B.get(file1,rank1))) return false; if (B.get_active_colour() != 'w') return false; if (! check_move_simple(file0,rank0,file1,rank1)) return false; if (! regular) return true; final Board test_board = new Board(B); test_board.normal_white_move_0(file0,rank0,file1,rank1); final Moves test_move = new Moves(test_board); final char[] king_pos = test_move.white_king_position(); assert(king_pos.length == 2); return test_move.black_not_attacking(king_pos[0],king_pos[1]); } public boolean check_normal_black_move(final char file0, final char rank0, final char file1, final char rank1) { // ADDED THE CHECK NORMAL BLACK MOVE BASED ON THE CHECK NORMAL WHITE MOVE if (! Board.is_valid_black_figure(B.get(file0,rank0))) return false; if (! B.is_empty(file1,rank1) && ! Board.is_valid_white_figure(B.get(file1,rank1))) return false; if (B.get_active_colour() != 'b') return false; if (! check_move_simple(file0,rank0,file1,rank1)) return false; if (! regular) return true; final Board test_board = new Board(B); test_board.normal_black_move_0(file0,rank0,file1,rank1); final Moves test_move = new Moves(test_board); final char[] king_pos = test_move.black_king_position(); assert(king_pos.length == 2); return test_move.white_not_attacking(king_pos[0],king_pos[1]); } // for checking a normal move by just applying the move-rules private boolean check_move_simple(final char file0, final char rank0, final char file1, final char rank1) { final char fig = B.get(file0,rank0); if (fig == Board.white_king || fig == Board.black_king) return check_king_move(file0,rank0,file1,rank1); if (fig == Board.white_queen || fig == Board.black_queen) return check_queen_move(file0,rank0,file1,rank1); if (fig == Board.white_rook || fig == Board.black_rook) return check_rook_move(file0,rank0,file1,rank1); if (fig == Board.white_bishop || fig == Board.black_bishop) return check_bishop_move(file0,rank0,file1,rank1); if (fig == Board.white_knight || fig == Board.black_knight) return check_knight_move(file0,rank0,file1,rank1); if (fig == Board.white_pawn) return check_white_pawn_move(file0,rank0,file1,rank1); else return check_black_pawn_move(file0,rank0,file1,rank1); } private boolean check_king_move(final char file0, final char rank0, final char file1, final char rank1) { // ADDED KING MOVE int fileChange = file0 - file1; int rankChange = rank0 - rank1; return fileChange <= 1 && fileChange >= -1 && rankChange <= 1 && rankChange >= -1; } private boolean check_queen_move(final char file0, final char rank0, final char file1, final char rank1) { // ADDED QUEEN MOVE int fileChange = file0 - file1; int rankChange = rank0 - rank1; return fileChange <=8 && fileChange >= -8 && rankChange <= 8 && rankChange >= -8; } private boolean check_rook_move(final char file0, final char rank0, final char file1, final char rank1) { // ADDED ROOK MOVE int fileChange = file0 - file1; int rankChange = rank0 - rank1; return fileChange <=8 || fileChange >= -8 || rankChange <= 8 || rankChange >= -8; } private boolean check_bishop_move(final char file0, final char rank0, final char file1, final char rank1) { // ADDED BISHOP MOVE int fileChange = file0 - file1; int rankChange = rank0 - rank1; return fileChange <= 8 && rankChange <= 8 || fileChange <= 8 && rankChange >= -8 || fileChange >= -8 && rankChange >= -8 || fileChange >= -8 && rankChange <= 8; } private boolean check_knight_move(final char file0, final char rank0, final char file1, final char rank1) { // ADDED KNIGHT MOVE int fileChange = file0 - file1; int rankChange = rank0 - rank1; /* IS THIS THE CORRECT WAY? * return fileChange <= 1 && rankChange <= 2 || fileChange <= 1 && rankChange >= -2 || fileChange <= 2 && rankChange <= 1 || fileChange <= 2 && rankChange >= -1 || fileChange >= -1 && rankChange <= 2 || fileChange >= -1 && rankChange >= -2 || fileChange >= -2 && rankChange <= 1 || fileChange >= -2 && rankChange >= -1;*/ // OR IS THIS? return fileChange <= 1 || fileChange >= -1 || fileChange <= 2 || fileChange >= -2 && rankChange <= 1 || rankChange >= - 1 || rankChange <= 2 || rankChange >= -2; } private boolean check_white_pawn_move(final char file0, final char rank0, final char file1, final char rank1) { // ADDED PAWN MOVE int fileChange = file0 - file1; int rankChange = rank0 - rank1; return fileChange == 0 && rankChange <= 1; } private boolean check_black_pawn_move(final char file0, final char rank0, final char file1, final char rank1) { // ADDED PAWN MOVE int fileChange = file0 - file1; int rankChange = rank0 - rank1; return fileChange == 0 && rankChange >= -1; } public boolean check_white_kingside_castling() { // only demonstration code: final char c = B.get_white_castling(); if (c == '-' || c == 'q') return false; if (B.get_active_colour() == 'b') return false; if (B.get('e','1') != 'K') return false; if (! black_not_attacking('e','1')) return false; if (! free_white('f','1')) return false; // XXX return true; } public boolean check_white_queenside_castling() { // only demonstration code: final char c = B.get_white_castling(); if (c == '-' || c == 'k') return false; if (B.get_active_colour() == 'b') return false; // ADDED BASED ON KINGSIDE CASTLING if (B.get('e','1') != 'Q') return false; if (! black_not_attacking('e','1')) return false; if (! free_white('f','1')) return false; // XXX return true; } public boolean check_black_kingside_castling() { // only demonstration code: final char c = B.get_black_castling(); if (c == '-' || c == 'q') return false; if (B.get_active_colour() == 'w') return false; // ADDED BASED ON CHECK WHITE if (B.get('e','8') != 'K') return false; if (! black_not_attacking('e','8')) return false; if (! free_white('f','8')) return false; // XXX return true; } public boolean check_black_queenside_castling() { // only demonstration code: final char c = B.get_black_castling(); if (c == '-' || c == 'k') return false; if (B.get_active_colour() == 'w') return false; // ADDED BASED ON KINGSIDE CASTLING if (B.get('e','8') != 'Q') return false; if (! black_not_attacking('e','8')) return false; if (! free_white('f','8')) return false; // XXX return true; } public boolean check_white_promotion(final char pawn_file, final char figure) { // XXX // ADDED CHECKING FOR CORRECT FIGURE AND POSITION - ALTHOUGH IT SEEMS AS THOUGH // PAWN_FILE SHOULD BE PAWN_RANK, AS IT IS THE REACHING OF THE END RANK THAT // CAUSES PROMOTION OF A PAWN, NOT FILE if (figure == P && pawn_file == 8) { return true; } else return false; } public boolean check_black_promotion(final char pawn_file, final char figure) { // XXX // ADDED CHECKING FOR CORRECT FIGURE AND POSITION if (figure == p && pawn_file == 1) { return true; } else return false; } // checks whether black doesn't attack the field: public boolean black_not_attacking(final char file, final char rank) { // XXX return true; } public boolean free_white(final char file, final char rank) { // XXX return black_not_attacking(file,rank) && B.is_empty(file,rank); } // checks whether white doesn't attack the field: public boolean white_not_attacking(final char file, final char rank) { // XXX return true; } public boolean free_black(final char file, final char rank) { // XXX return white_not_attacking(file,rank) && B.is_empty(file,rank); } public char[] white_king_position() { for (char file = 'a'; file <= 'h'; ++file) for (char rank = '1'; rank <= '8'; ++rank) if (B.get(file,rank) == Board.white_king) { char[] result = new char[2]; result[0] = file; result[1] = rank; return result; } return new char[0]; } public char[] black_king_position() { for (char file = 'a'; file <= 'h'; ++file) for (char rank = '1'; rank <= '8'; ++rank) if (B.get(file,rank) == Board.black_king) { char[] result = new char[2]; result[0] = file; result[1] = rank; return result; } return new char[0]; } public static void main(final String[] args) { // checking regular_position { Moves m = new Moves(new Board()); assert(m.regular_position()); m = new Moves(new Board("8/8/8/8/8/8/8/8 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("KK6/8/8/8/8/8/8/8 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("kk6/8/8/8/8/8/8/8 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("Kk6/8/8/8/8/8/8/8 w - - 0 1")); assert(m.regular_position()); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/8 w - - 0 1")); assert(m.regular_position()); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/n7 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/N7 w - - 0 1")); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/b7 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/B7 w - - 0 1")); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/r7 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/R7 w - - 0 1")); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/q7 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("Kk6/qqqqqqqq/QQQQQQQQ/Q7/q7/rrbbnn2/RRBBNN2/Q7 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("Kkp5/8/8/8/8/8/8/8 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("KkP5/8/8/8/8/8/8/8 w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("Kk6/8/8/8/8/8/8/7p w - - 0 1")); assert(!m.regular_position()); m = new Moves(new Board("Kk6/8/8/8/8/8/8/7P w - - 0 1")); assert(!m.regular_position()); } // checking check_white/black_king/queenside_castling { Moves m = new Moves(new Board("4k2r/8/8/8/8/8/8/4K2R w Kk - 0 1")); assert(!m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); m = new Moves(new Board("4k2r/8/8/8/8/8/8/4K2R b Kk - 0 1")); assert(!m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); m = new Moves(new Board("4k2r/4pppp/8/8/8/8/4PPPP/4K2R w KQkq - 0 1")); assert(m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); m = new Moves(new Board("4k2r/4pppp/8/8/8/8/4PPPP/4K2R b KQkq - 0 1")); assert(!m.check_white_kingside_castling()); assert(m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); m = new Moves(new Board("r3k3/8/8/8/8/8/8/R3K3 w Qq - 0 1")); assert(!m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); m = new Moves(new Board("r3k3/8/8/8/8/8/8/R3K3 b Qq - 0 1")); assert(!m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); m = new Moves(new Board("r3k3/p7/8/8/8/8/8/R3K3 w Qq - 0 1")); assert(!m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); m = new Moves(new Board("r3k3/p7/8/8/8/8/8/R3K3 b Qq - 0 1")); assert(!m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(m.check_black_queenside_castling()); m = new Moves(new Board("r3k3/p7/8/8/8/n7/8/R3K3 w Qq - 0 1")); assert(!m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); m = new Moves(new Board("r3k3/p7/B7/8/8/8/8/R3K3 b Qq - 0 1")); assert(!m.check_white_kingside_castling()); assert(!m.check_black_kingside_castling()); assert(!m.check_white_queenside_castling()); assert(!m.check_black_queenside_castling()); // XXX } } }

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• forks in C - exercise

  - by Zka

  I try to repeat and learn more advanced uses and options when cutting trees with forks in the jungle of C. But foolishly I find an example which should be very easy as I have worked with forks before and even written some code, but i can't understand it fully. Here comes : main() { if (fork() == 0) { if (fork() == 0) { printf("3"); } else if ((wait(NULL)) > 0) { printf("2"); } } else { if (fork() == 0) { printf("1"); exit(0); } if (fork() == 0) { printf("4"); } } printf("0"); return 0; } Possible solutions are : 3201040 3104200 1040302 4321000 4030201 1403020 where 2, 5 and 6 are correct answers. First of all, shouldn't there be four zeroes in the output? Second... How does one come to the solution at all? Been doing this on paper for almost an hour and I'm not even close to understanding why the given solution are more correct than the false ones (except for nr3 as it can't end with 2 since a 0 must follow). Anyone with his forks in check who can offer some good explanation?

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• Regular expression for a phone number

  - by Zerobu

  Hello, I would like a regular expression in this format. It Must match one of the following formats: * (###)###-#### * ###-###-#### * ###.###.#### * ########## Strip all whitespace. Make sure it's a valid phone number, then (if necessary) translate it to the first format listed above.

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• Fread binary file dynamic size string [C]

  - by Blackbinary

  I've been working on this assignment, where I need to read in "records" and write them to a file, and then have the ability to read/find them later. On each run of the program, the user can decide to write a new record, or read an old record (either by Name or #) The file is binary, here is its definition: typedef struct{ char * name; char * address; short addressLength, nameLength; int phoneNumber; }employeeRecord; employeeRecord record; The way the program works, it will store the structure, then the name, then the address. Name and address are dynamically allocated, which is why it is necessary to read the structure first to find the size of the name and address, allocate memory for them, then read them into that memory. For debugging purposes I have two programs at the moment. I have my file writing program, and file reading. My actual problem is this, when I read a file I have written, i read in the structure, print out the phone # to make sure it works (which works fine), and then fread the name (now being able to use record.nameLength which reports the proper value too). Fread however, does not return a usable name, it returns blank. I see two problems, either I haven't written the name to the file correctly, or I haven't read it in correctly. Here is how i write to the file: where fp is the file pointer. record.name is a proper value, so is record.nameLength. Also i am writing the name including the null terminator. (e.g. 'Jack\0') fwrite(&record,sizeof record,1,fp); fwrite(record.name,sizeof(char),record.nameLength,fp); fwrite(record.address,sizeof(char),record.addressLength,fp); And i then close the file. here is how i read the file: fp = fopen("employeeRecord","r"); fread(&record,sizeof record,1,fp); printf("Number: %d\n",record.phoneNumber); char *nameString = malloc(sizeof(char)*record.nameLength); printf("\nName Length: %d",record.nameLength); fread(nameString,sizeof(char),record.nameLength,fp); printf("\nName: %s",nameString); Notice there is some debug stuff in there (name length and number, both of which are correct). So i know the file opened properly, and I can use the name length fine. Why then is my output blank, or a newline, or something like that? (The output is just Name: with nothing after it, and program finishes just fine) Thanks for the help.

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• Scheme sorting a list

  - by John

  Okay so I am trying to take in a list and sort it from greatest to smallest. Example: > (maxheap (list 5 6 2 1 18 7)) ;output: > (18 7 6 5 2 1) So here's what I got so far: (define (mkmaxheap heaplist) (let ((max (mymax(heaplist)))) ;mymax is a func that returns max number, it works (let (( head (car heaplist)) (tail (cdr heaplist))) (if (null? tail) newlist)))) Thats all I could get to compile, all the other code I wrote failed. Any help on solving this would be much appreciated.

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• How to sort 100GB worth of strings.

  - by Christian

  Given a harddrive with 120GB, 100 of which are filled with the strings of length 256 and 2 GB Ram how do I sort those strings in Java most efficiently? How long will it take?

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• Sorting the order of growth of the functions?

  - by rachel7660

  Please order the function belows by growth rate from fastest to slowest: n^10 2^n nlog(n) 10^6 And my answer is: 2^n n^10 nlog(n) 10^6 Is my answer correct?

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• pointers to functions

  - by DevAno1

  I have two basic Cpp tasks, but still I have problems with them. First is to write functions mul1,div1,sub1,sum1, taking ints as arguments and returning ints. Then I need to create pointers ptrFun1 and ptrFun2 to functions mul1 and sum1, and print results of using them. Problem starts with defining those pointers. I thought I was doing it right, but devcpp gives me errors in compilation. #include <iostream> using namespace std; int mul1(int a,int b) { return a * b; } int div1(int a,int b) { return a / b; } int sum1(int a,int b) { return a + b; } int sub1(int a,int b) { return a - b; } int main() { int a=1; int b=5; cout << mul1(a,b) << endl; cout << div1(a,b) << endl; cout << sum1(a,b) << endl; cout << sub1(a,b) << endl; int *funPtr1(int, int); int *funPtr2(int, int); funPtr1 = sum1; funPtr2 = mul1; cout << funPtr1(a,b) << endl; cout << funPtr2(a,b) << endl; system("PAUSE"); return 0; } 38 assignment of function int* funPtr1(int, int)' 38 cannot convertint ()(int, int)' to `int*()(int, int)' in assignment Task 2 is to create array of pointers to those functions named tabFunPtr. How to do that ?

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• Anagram of a word

  - by Codenotguru

  How to print anagrams of a word from a dictionary? how to optimize it if we have to do this repeatedly?? is there any java code explanation ??

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• Relational Clausal Logic question: what is a Herbrand interpretation

  - by anotherstat

  I'm having a hard time coming to grips with relational clausal logic, and I'm not sure if this is the place to ask but it would be help me so much with revision if anyone could provide guidance with the following questions. Let P be the program: academic(X); student(X); other_staff(X):- works_in(X, university). :-student(john). :-other_staff(john). works_in(john, university) Question: Which are the Herbrand interpreations of P? AS

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• How do you display a binary search tree?

  - by fakeit

  I'm being asked to display a binary search tree in sorted order. The nodes of the tree contain strings. I'm not exactly sure what the best way is to attack this problem. Should I be traversing the tree and displaying as I go? Should I flatten the tree into an array and then use a sorting algorithm before I display? I'm not looking for the actual code, just a guide where to go next.

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• Dynamic programming solution to the subset-sum decision problem

  - by Gail

  How can a dynamic programming solution for the unbounded knapsack decision problem be used to come up with a dynamic programming solution to the subset-sum decision problem? This limitation seems to render the unbounded knapsack problem useless. In the unbounded knapsack, we simply store true or false for if some subset of integers sum up to our target value. However, if we have a limit on the frequency of the use of these integers, the optimal substructure at least appears to fail. How can this be done?

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• Pthread Queue System

  - by Wallace

  Hi. I'm working on my assignment on pthreads. I'm new and never touched on pthreads before. Is there any sample codes or resources out there that anyone of you have, that might aid me in my assignment? Here are my assignment details. A pthread program about queue system: Write a C/C++ Pthread program for a Dental clinic’s queuing system that declares an array of integers of size N, where N is the maximum number of queue for the day. The pthread program uses two threads. Whenever there is a new dental appointment, the first thread (the creator) puts the queue numbers in the array, one after the other. The second thread (the remover) removes the queue numbers from the array whenever the dentist has seen the patient. This is done in a FIFO fashion (First In First Out). The algorithm of the creator is as follows: • If the array is not full then put a new number in it (the numbers start at 1 and are incremented by one each time, so the creator create queue number 1, 2, 3 etc.) • sleep for 1 to 10 seconds, randomly • repeat The algorithm of the remover is as follows: • If the array is not empty then remove its smallest queue number • sleep for 1 to 10 seconds, randomly • repeat You should use mutex locks to protect things that must be protected. Each thread should print on the screen what it is doing (eg: "number 13 is added into the queue", "number 7 is removed from the queue", etc.). The program should run forever. Any help will be appreciated. Thanks.

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• about Quick Sort

  - by matin1234

  Hi I have written this code but it will print these stack traces in the console please help me thanks! (Aslo "p" and "q" are the first and last index of our array ,respectively) public class JavaQuickSort { public static void QuickSort(int A[], int p, int q) { int i, last = 0; Random rand = new Random(); if (q < 1) { return; } **swap(A, p, rand.nextInt() % (q+1));** for (i = p + 1; i <= q; i++) { if (A[i] < A[p]) { swap(A, ++last, i); } } swap(A, p, last); QuickSort(A, p, last - 1); QuickSort(A, last + 1, q); } private static void swap(int[] A, int i, int j) { int temp; temp = A[i]; **A[i] = A[j];** A[j] = temp; } public static void main(String[] args){ int[] A = {2,5,7,3,9,0,1,6,8}; **QuickSort(A, 0,8 );** System.out.println(Arrays.toString(A)); } } the Stack traces : run: Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: -3 at JavaQuickSort.swap(JavaQuickSort.java:38) at JavaQuickSort.QuickSort(JavaQuickSort.java:22) at JavaQuickSort.main(JavaQuickSort.java:45) Java Result: 1 BUILD SUCCESSFUL (total time: 2 seconds) I also bold those statements that cause these stack traces. like == ** ...**

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• Bash and regex problem : check for tokens entered into a Coke vending machine

  - by Michael Mao

  Hi all: Here is a "challenge question" I've got from Linux system programming lecture. Any of the following strings will give you a Coke if you kick: L = { aaaa, aab, aba, baa, bb, aaaa"a", aaaa"b", aab"a", … ab"b"a, ba"b"a, ab"bbbbbb"a, ... } The letters shown in wrapped double quotes indicate coins that would have fallen through (but those strings are still part of the language in this example). Exercise (a bit hard) show this is the language of a regular expression And this is what I've got so far : #!/usr/bin/bash echo "A bottle of Coke costs you 40 cents" echo -e "Please enter tokens (a = 10 cents, b = 20 cents) in a sequence like 'abba' :\c" read tokens #if [ $tokens = aaaa ]||[ $tokens = aab ]||[ $tokens = bb ] #then # echo "Good! now a coke is yours!" #else echo "Thanks for your money, byebye!" if [[ $token =~ 'aaaa|aab|bb' ]] then echo "Good! now a coke is yours!" else echo "Thanks for your money, byebye!" fi Sadly it doesn't work... always outputs "Thanks for your money, byebye!" I believe something is wrong with syntax... We didn't provided with any good reference book and the only instruction from the professor was to consult "anything you find useful online" and "research the problem yourself" :( I know how could I do it in any programming language such as Java, but get it done with bash script + regex seems not "a bit hard" but in fact "too hard" for anyone with little knowledge on something advanced as "lookahead"(is this the terminology ?) I don't know if there is a way to express the following concept in the language of regex: Valid entry would consist of exactly one of the three components : aaaa, aab and bb, regardless of order, followed by an arbitrary sequence of a or b's So this is what is should be like : (a{4}Ua{2}bUb{2})(aUb)* where the content in first braces is order irrelevant. Thanks a lot in advance for any hints and/or tips :)

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• Neural Network Always Produces Same/Similar Outputs for Any Input

  - by l33tnerd

  I have a problem where I am trying to create a neural network for Tic-Tac-Toe. However, for some reason, training the neural network causes it to produce nearly the same output for any given input. I did take a look at Artificial neural networks benchmark, but my network implementation is built for neurons with the same activation function for each neuron, i.e. no constant neurons. To make sure the problem wasn't just due to my choice of training set (1218 board states and moves generated by a genetic algorithm), I tried to train the network to reproduce XOR. The logistic activation function was used. Instead of using the derivative, I multiplied the error by output*(1-output) as some sources suggested that this was equivalent to using the derivative. I can put the Haskell source on HPaste, but it's a little embarrassing to look at. The network has 3 layers: the first layer has 2 inputs and 4 outputs, the second has 4 inputs and 1 output, and the third has 1 output. Increasing to 4 neurons in the second layer didn't help, and neither did increasing to 8 outputs in the first layer. I then calculated errors, network output, bias updates, and the weight updates by hand based on http://hebb.mit.edu/courses/9.641/2002/lectures/lecture04.pdf to make sure there wasn't an error in those parts of the code (there wasn't, but I will probably do it again just to make sure). Because I am using batch training, I did not multiply by x in equation (4) there. I am adding the weight change, though http://www.faqs.org/faqs/ai-faq/neural-nets/part2/section-2.html suggests to subtract it instead. The problem persisted, even in this simplified network. For example, these are the results after 500 epochs of batch training and of incremental training. Input |Target|Output (Batch) |Output(Incremental) [1.0,1.0]|[0.0] |[0.5003781562785173]|[0.5009731800870864] [1.0,0.0]|[1.0] |[0.5003740346965251]|[0.5006347214672715] [0.0,1.0]|[1.0] |[0.5003734471544522]|[0.500589332376345] [0.0,0.0]|[0.0] |[0.5003674110937019]|[0.500095157458231] Subtracting instead of adding produces the same problem, except everything is 0.99 something instead of 0.50 something. 5000 epochs produces the same result, except the batch-trained network returns exactly 0.5 for each case. (Heck, even 10,000 epochs didn't work for batch training.) Is there anything in general that could produce this behavior? Also, I looked at the intermediate errors for incremental training, and the although the inputs of the hidden/input layers varied, the error for the output neuron was always +/-0.12. For batch training, the errors were increasing, but extremely slowly and the errors were all extremely small (x10^-7). Different initial random weights and biases made no difference, either. Note that this is a school project, so hints/guides would be more helpful. Although reinventing the wheel and making my own network (in a language I don't know well!) was a horrible idea, I felt it would be more appropriate for a school project (so I know what's going on...in theory, at least. There doesn't seem to be a computer science teacher at my school). EDIT: Two layers, an input layer of 2 inputs to 8 outputs, and an output layer of 8 inputs to 1 output, produces much the same results: 0.5+/-0.2 (or so) for each training case. I'm also playing around with pyBrain, seeing if any network structure there will work. Edit 2: I am using a learning rate of 0.1. Sorry for forgetting about that. Edit 3: Pybrain's "trainUntilConvergence" doesn't get me a fully trained network, either, but 20000 epochs does, with 16 neurons in the hidden layer. 10000 epochs and 4 neurons, not so much, but close. So, in Haskell, with the input layer having 2 inputs & 2 outputs, hidden layer with 2 inputs and 8 outputs, and output layer with 8 inputs and 1 output...I get the same problem with 10000 epochs. And with 20000 epochs. Edit 4: I ran the network by hand again based on the MIT PDF above, and the values match, so the code should be correct unless I am misunderstanding those equations. Some of my source code is at http://hpaste.org/42453/neural_network__not_working; I'm working on cleaning my code somewhat and putting it in a Github (rather than a private Bitbucket) repository. All of the relevant source code is now at https://github.com/l33tnerd/hsann.

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• Matlab: Analysis of signal

  - by Mateusz

  Hi, I have a problem with this task: For free route perform frequency analysis and give parametrs of each signal component: time of beginning and ending of each component beginning and ending frequency amplitude (in time domain) in the beginning and end of each signal's component level of noise in dB Assume, that, the parametrs of each component like amplitude, frequency is changing lineary in time. Frequency of sampling is 1000Hz For example I have signal like this: Nx=64; fs=1000; t=1/fs*(0:Nx-1); %========================== A1=1; A2=4; f1=500; f2=1000; x1=A1*cos(2*pi*f1*t); x2=A2*sin(2*pi*f2*t); %========================== x=x1+x2;

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• loops and array help

  - by dalton

  public void arrayCalculation(int[][]scores,float[]averages, int[]temp) { int total; for(int a=0; a<5; a++) { for (int b=0; b<5; b++) { scores[a][b] = temp[a+b*5]; } } for(int a = 0; a <5; a++) { total = total + scores[a]; } scores[5][0] = total; } i need to add up the values stored in the first row and store it in the 6th positon in the row

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• Problem building a complete binary tree of height 'h' in Python

  - by Jack

  Here is my code. The complete binary tree has 2^k nodes at depth k. class Node: def __init__(self, data): # initializes the data members self.left = None self.right = None self.data = data root = Node(data_root) def create_complete_tree(): row = [root] for i in range(h): newrow = [] for node in row: left = Node(data1) right = Node(data2) node.left = left node.right = right newrow.append(left) newrow.append(right) row = copy.deepcopy(newrow) def traverse_tree(node): if node == None: return else: traverse_tree(node.left) print node.data traverse_tree(node.right) create_complete_tree() print 'Node traversal' traverse_tree(root) The tree traversal only gives the data of root and its children. What am I doing wrong?

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• how to run/compile java code from JTextArea at Runtime? ----urgent!!! college project

  - by Lokesh Kumar

  I have a JInternalFrame painted with a BufferedImage and contained in the JDesktopPane of a JFrame.I also have a JTextArea where i want to write some java code (function) that takes the current JInternalFrame's painted BufferedImage as an input and after doing some manipulation on this input it returns another manipulated BufferedImage that paints the JInternalFrame with new manipulated Image again!!. Manipulation java code of JTextArea:- public BufferedImage customOperation(BufferedImage CurrentInputImg) { Color colOld; Color colNew; BufferedImage manipulated=new BufferedImage(CurrentInputImg.getWidth(),CurrentInputImg.getHeight(),BufferedImage.TYPE_INT_ARGB); //make all Red pixels of current image black for(int i=0;i< CurrentInputImg.getWidth();i++) { for(int j=0;j< CurrentInputImg.getHeight(),j++) { colOld=new Color(CurrentInputImg.getRGB(i,j)); colNew=new Color(0,colOld.getGreen(),colOld.getBlue(),colOld.getAlpha()); manipulated.setRGB(i,j,colNew.getRGB()); } } return manipulated; } so,how can i run/compile this JTextArea java code at runtime and get a new manipulated image for painting on JInternalFrame???????   Here is my Main class: (This class is not actual one but i have created it for u for basic interfacing containing JTextArea,JInternalFrame,Apply Button) import java.awt.*; import java.awt.event.*; import javax.swing.*; import javax.swing.event.*; import javax.swing.JInternalFrame; import javax.swing.JDesktopPane; import java.awt.image.*; import javax.imageio.*; import java.io.*; import java.io.File; import java.util.*; class MyCustomOperationSystem extends JFrame **{** public JInternalFrame ImageFrame; public BufferedImage CurrenFrameImage; public MyCustomOperationSystem() **{** setTitle("My Custom Image Operations"); setSize((int)Toolkit.getDefaultToolkit().getScreenSize().getWidth(),(int)Toolkit.getDefaultToolkit().getScreenSize().getHeight()); JDesktopPane desktop=new JDesktopPane(); desktop.setPreferredSize(new Dimension((int)Toolkit.getDefaultToolkit().getScreenSize().getWidth(),(int)Toolkit.getDefaultToolkit().getScreenSize().getHeight())); try{ CurrenFrameImage=ImageIO.read(new File("c:/Lokesh.png")); }catch(Exception exp) { System.out.println("Error in Loading Image"); } ImageFrame=new JInternalFrame("Image Frame",true,true,false,true); ImageFrame.setMinimumSize(new Dimension(CurrenFrameImage.getWidth()+10,CurrenFrameImage.getHeight()+10)); ImageFrame.getContentPane().add(CreateImagePanel()); ImageFrame.setLayer(1); ImageFrame.setLocation(100,100); ImageFrame.setVisible(true); desktop.setOpaque(true); desktop.setBackground(Color.darkGray); desktop.add(ImageFrame); this.getContentPane().setLayout(new BorderLayout()); this.getContentPane().add("Center",desktop); this.getContentPane().add("South",ControlPanel()); pack(); setVisible(true); **}** public JPanel CreateImagePanel(){ JPanel tempPanel=new JPanel(){ public void paintComponent(Graphics g) { g.drawImage(CurrenFrameImage,0,0,this); } }; tempPanel.setPreferredSize(new Dimension(CurrenFrameImage.getWidth(),CurrenFrameImage.getHeight())); return tempPanel; } public JPanel ControlPanel(){ JPanel controlPan=new JPanel(new FlowLayout(FlowLayout.LEFT)); JButton customOP=new JButton("Custom Operation"); customOP.addActionListener(new ActionListener(){ public void actionPerformed(ActionEvent evnt){ JFrame CodeFrame=new JFrame("Write your Code Here"); JTextArea codeArea=new JTextArea("Your Java Code Here",100,70); JScrollPane codeScrollPan=new JScrollPane(codeArea,ScrollPaneConstants.VERTICAL_SCROLLBAR_ALWAYS, ScrollPaneConstants.HORIZONTAL_SCROLLBAR_ALWAYS); CodeFrame.add(codeScrollPan); CodeFrame.setVisible(true); } }); JButton Apply=new JButton("Apply Code"); Apply.addActionListener(new ActionListener(){ public void actionPerformed(ActionEvent event){ // What should I do!!! Here!!!!!!!!!!!!!!! } }); controlPan.add(customOP); controlPan.add(Apply); return controlPan; } public static void main(String s[]) { new MyCustomOperationSystem(); } } Note: in above class JInternalFrame (ImageFrame) is not visible even i have declared it visible. so, ImageFrame is not visible while compiling and running above class. U have to identify this problem before running it.

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• F# exercise help - Beginner

  - by bobjink

  I hope you can help me with these exercises I have been stuck with. I am not in particular looking for the answers. Tips that help me solve them myself are just as good :) 1. Write a function implode : char list - string so that implode s returns the characters concatenated into a string: let implode (cArray:char list) = List.foldBack (+) 0 cArray;; Error. I am thinking that i need to cast every char to strings before I do the above. I bet however, that there is a much simpler and better solution. 2. Write a function palindrome : string - bool, so that palindrome s returns true if the string s is a palindrome let isIdentical cArray1 cArray2 = (cArray1 = cArray2);; let palinDrome (word:string) = isIdentical(word.ToCharArray(), Array.rev(word.ToCharArray()));; I get the value: val palinDrome : string - (char [] * char [] - bool), which is not what I want. 3. Write a function combinePair : int list - int list so that combinePair xs returns the list with elements from xs combined into pairs. If xs contains an odd number of elements, then the last element is thrown away: I have no idea for this one.

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• Proving that the distance values extracted in Dijkstra's algorithm is non-decreasing?

  - by Gail

  I'm reviewing my old algorithms notes and have come across this proof. It was from an assignment I had and I got it correct, but I feel that the proof certainly lacks. The question is to prove that the distance values taken from the priority queue in Dijkstra's algorithm is a non-decreasing sequence. My proof goes as follows: Proof by contradiction. Fist, assume that we pull a vertex from Q with d-value 'i'. Next time, we pull a vertex with d-value 'j'. When we pulled i, we have finalised our d-value and computed the shortest-path from the start vertex, s, to i. Since we have positive edge weights, it is impossible for our d-values to shrink as we add vertices to our path. If after pulling i from Q, we pull j with a smaller d-value, we may not have a shortest path to i, since we may be able to reach i through j. However, we have already computed the shortest path to i. We did not check a possible path. We no longer have a guaranteed path. Contradiction.

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• Virtual Memory and Paging

  - by Kenshin

  Hello, I am doing some exercices to understand how the virtual memory and paging works, my question is as follows : Suppose we use a paged memory with pages of 1024 bytes, the virtual address space is of 8 pages but the physical memory can only contain 4 frames of pages. Replacement policy is LRU. What is the physical address in main memory that corresponds to virtual address 4096? and how do you get to that result? Same thing as question 1 but with virtual address 1024 A page fault occurs when accessing a word in page 0, which page frame will be used to receive the virtual page 0? Page Image

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