Link:
http://richardbowles.tripod.com/neural/source/source1.htm
//
// This is the source code for the Hopfield Net demonstration written in // C++. The simplest way to download this file is to go the // 'View Source Code' option of your browser and select the 'Save' option // from the file menu. It should run without any changes. // (Ignore the PRE and HTML above - they are just to ensure that your // browser doesn't mess around with the text formatting) // The line below may come out wrongly. The symbol before the 'iostream' // should be the greater than sign #include <iostream.h> const int MAX_CLASSES = 2; // Max. number of input classes const int PATTERN_LENGTH = 8; // Max. number of 'bits' per pattern const int MAX_TIME = 10; // Max. 'time steps' for convergence // These arrays have 1 added to the limits as I prefer to count from 1 // rather than from 0 // This is the vocabulary of patterns which the inputs are compared to const int X[MAX_CLASSES+1][PATTERN_LENGTH+1] = {{0,0,0,0,0,0,0,0,0}, // First row and column is dummy {0,1,1,1,1,-1,-1,-1,-1}, // Represents ****.... {0,-1,-1,-1,-1,1,1,1,1}}; // Represents ....**** // t[i][j] = Connection weight from node i to node j int t[PATTERN_LENGTH+1][PATTERN_LENGTH+1]; int mu[MAX_TIME][PATTERN_LENGTH+1]; // Contains output of net at time 't' // First index really does run from 0 void assign_connection_weights () { int sum; for (int i = 1; i <= PATTERN_LENGTH; i++) for (int j = 1; j <= PATTERN_LENGTH; j++) if (i == j) t[i][j] = 0; // Weight=0 for connection from a node to itself else { sum = 0; for (int s = 1; s <= MAX_CLASSES; s++) sum += X[s][i] * X[s][j]; t[i][j] = sum; } } // Get the user to type a test pattern using * to represent 1 and . to // represent -1. The pattern is read directly into the mu[0] slot of the // time array void get_test_pattern () { char s[PATTERN_LENGTH+2]; // 1 extra because of array indices starting // from 1 and 1 extra because of /0 character // at end int valid; // Some compilers don't have a bool type int i; do { cout << "Please enter a pattern using * to represent 1 and . to " << "represent -1" << endl; cout << "You should enter " << PATTERN_LENGTH << " symbols and press Enter at the end." << endl; cin >> s; // Go through the string to make sure each symbol is either * or . valid = 1; // Assume valid until proven otherwise // Translate string positions starting from 0 into mu // positions starting from 1 for (i = 0; i < PATTERN_LENGTH; i++) switch (s[i]) { case '*' : mu[0][i+1] = 1; break; case '.' : mu[0][i+1] = -1; break; default : valid = 0; } } while (valid == 0); } // Write the values of the mu array for time slot tt using * and . void write_output (int tt) { for (int i = 1; i <= PATTERN_LENGTH; i++) if (mu[tt][i] == 1) cout << "*"; else cout << "."; cout << endl; } // Iterate the net, moving from one time slot to another and printing out // the outputs at each stage void iterate () { int sum; for (int tt = 1; tt <= MAX_TIME; tt++) // Go through all time slots { for (int j= 1; j <= PATTERN_LENGTH; j++) { sum = 0; for (int i = 1; i <= PATTERN_LENGTH; i++) sum += t[i][j] * mu[tt-1][i]; // Now pass sum through hard-limiter so it is 1 or -1 if (sum > 0) mu[tt][j] = 1; else mu[tt][j] = -1; } write_output(tt); } } void main () { char k; assign_connection_weights(); do { get_test_pattern(); iterate(); cout << endl; cout << "Would you like another go? (Press Y or N, then Enter) : "; cin >> k; } while (k != 'Y' && k != 'N'); } //