C++ Neural Networks and Fuzzy Logic by Valluru B. Rao M&T Books, IDG Books Worldwide, Inc. ISBN: 1558515526   Pub Date: 06/01/95

Comments on the C++ Program for Hopfield Network

Note the use of the output stream operator cout<< to output text strings or numerical output. C++ has istream and ostream classes from which the iostream class is derived. The standard input and output streams are cin and cout, respectively, used, correspondingly, with the operators >> and <<. Use of cout for the output stream is much simpler than the use of the C function printf. As you can see, there is no formatting suggested for output. However, there is a provision that allows you to format the output, while using cout.

Also note the way comments are introduced in the program. The line with comments should start with a double slash //. Unlike C, the comment does not have to end with a double slash. Of course, if the comments extend to subsequent lines, each such line should have a double slash at the start. You can still use the pair, /* at the beginning with */ at the end of lines of comments, as you do in C. If the comment continues through many lines, the C facility will be handier to delimit the comments.

The neurons in the network are members of the network class and are identified by the abbreviation nrn. The two patterns, 1010 and 0101, are presented to the network one at a time in the program.

Output from the C++ Program for Hopfield Network

The output from this program is as follows and is self-explanatory. When you run this program, you’re likely to see a lot of output whiz by, so in order to leisurely look at the output, use redirection. Type Hop > filename, and your output will be stored in a file, which you can edit with any text editor or list by using the type filename | more command.

THIS PROGRAM IS FOR A HOPFIELD NETWORK WITH A SINGLE LAYER OF 4 FULLY
INTERCONNECTED NEURONS. THE NETWORK SHOULD RECALL THE PATTERNS 1010 AND
0101 CORRECTLY.

 nrn[0].weightv[0] is  0
 nrn[0].weightv[1] is  -3
 nrn[0].weightv[2] is  3
 nrn[0].weightv[3] is  -3
activation is 3
output value is  1

 nrn[1].weightv[0] is  -3
 nrn[1].weightv[1] is  0
 nrn[1].weightv[2] is  -3
 nrn[1].weightv[3] is  3
activation is -6
output value is  0

 nrn[2].weightv[0] is  3
 nrn[2].weightv[1] is  -3
 nrn[2].weightv[2] is  0
 nrn[2].weightv[3] is  -3
activation is 3
output value is  1

 nrn[3].weightv[0] is  -3
 nrn[3].weightv[1] is  3
 nrn[3].weightv[2] is  -3
 nrn[3].weightv[3] is  0
activation is -6
output value is  0

 pattern= 1  output = 1  component matches
 pattern= 0  output = 0  component matches
 pattern= 1  output = 1  component matches
 pattern= 0  output = 0  component matches

 nrn[0].weightv[0] is  0
 nrn[0].weightv[1] is  -3
 nrn[0].weightv[2] is  3
 nrn[0].weightv[3] is  -3
activation is -6
output value is  0

 nrn[1].weightv[0] is  -3
 nrn[1].weightv[1] is  0
 nrn[1].weightv[2] is  -3
 nrn[1].weightv[3] is  3
activation is 3
output value is  1

 nrn[2].weightv[0] is  3
 nrn[2].weightv[1] is  -3
 nrn[2].weightv[2] is  0
 nrn[2].weightv[3] is  -3
activation is -6
output value is  0

 nrn[3].weightv[0] is  -3
 nrn[3].weightv[1] is  3
 nrn[3].weightv[2] is  -3
 nrn[3].weightv[3] is  0
activation is 3
output value is  1

 pattern= 0  output = 0  component matches
 pattern= 1  output = 1  component matches
 pattern= 0  output = 0  component matches
 pattern= 1  output = 1  component matches

Further Comments on the Program and Its Output

Let us recall our previous discussion of this example in Chapter 1. What does the network give as output if we present a pattern different from both A and B? If C = (0, 1, 0, 0) is the input pattern, the activation (dot products) would be –3, 0, –3, 3 making the outputs (next state) of the neurons 0,1,0,1, so that B would be recalled. This is quite interesting, because if we intended to input B, and we made a slight error and ended up presenting C instead, the network would recall B. You can run the program by changing the pattern to 0, 1, 0, 0 and compiling again, to see that the B pattern is recalled.

Another element about the example in Chapter 1 is that the weight matrix W is not the only weight matrix that would enable the network to recall the patterns A and B correctly. If we replace the 3 and –3 in the matrix with 2 and –2, respectively, the resulting matrix would facilitate the same performance from the network. One way for you to check this is to change the wt1, wt2, wt3, wt4 given in the program accordingly, and compile and run the program again. The reason why both of the weight matrices work is that they are closely related. In fact, one is a scalar (constant) multiple of the other, that is, if you multiply each element in the matrix by the same scalar, namely 2/3, you get the corresponding matrix in cases where 3 and –3 are replaced with 2 and –2, respectively.