xor

问题 Using php 7.1.0 I'm running this little test: <?php $a = true; $b = true; $value = $a xor $b; if ($value == true) { print "bad!\n"; } else { print "good\n"; } and it's coming back and saying bad. Why? An xor of two true values should be FALSE, not true. 回答1: The problem is operator precedence. The xor operator has lower precedence than = , so your statement is equivalent to: ($value = $a) xor $b; You need to write: $value = ($a xor $b); or $value = $a ^ $b; The ^ operator is bit-wise XOR, not

问题 I've always heard that the XOR problem can not be solved by a single layer perceptron (not using a hidden layer) since it is not linearly separable. I understand that there is no linear function that can separate the classes. However, what if we use a non-monotonic activation function like sin() or cos() is this still the case? I would imagine these types of functions might be able to separate them. 回答1: Yes , a single layer neural network with a non-monotonic activation function can solve

问题 I've always heard that the XOR problem can not be solved by a single layer perceptron (not using a hidden layer) since it is not linearly separable. I understand that there is no linear function that can separate the classes. However, what if we use a non-monotonic activation function like sin() or cos() is this still the case? I would imagine these types of functions might be able to separate them. 回答1: Yes , a single layer neural network with a non-monotonic activation function can solve

问题 I've always heard that the XOR problem can not be solved by a single layer perceptron (not using a hidden layer) since it is not linearly separable. I understand that there is no linear function that can separate the classes. However, what if we use a non-monotonic activation function like sin() or cos() is this still the case? I would imagine these types of functions might be able to separate them. 回答1: Yes , a single layer neural network with a non-monotonic activation function can solve

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 2 years ago . We are doing a school project where we need to explain code line by line. We need to explain the following 3 lines: // TODO: This is where the magic of XOR is happening. Are you able to explain what is going on? for (int i = 0; i < rawString.length(); i++) { thisIsEncrypted.append((char) (rawString

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 2 years ago . We are doing a school project where we need to explain code line by line. We need to explain the following 3 lines: // TODO: This is where the magic of XOR is happening. Are you able to explain what is going on? for (int i = 0; i < rawString.length(); i++) { thisIsEncrypted.append((char) (rawString

问题 I'm building the engine for a tiny game, and right now I have just got a red circle with two little eyes as a main character. I have keyPress functions to detect movement, and that works, but I wanted to use something I used a long time ago in QBASIC to remove the character and redraw at a new location: XOR Basically, on keypress this happens: if (code == 39) { mainChar.drawChar(); mainChar.x += 1; mainChar.leftEye.x += 1; mainChar.rightEye.x += 1; mainChar.drawChar(); } I thought drawing the

问题 I'm building the engine for a tiny game, and right now I have just got a red circle with two little eyes as a main character. I have keyPress functions to detect movement, and that works, but I wanted to use something I used a long time ago in QBASIC to remove the character and redraw at a new location: XOR Basically, on keypress this happens: if (code == 39) { mainChar.drawChar(); mainChar.x += 1; mainChar.leftEye.x += 1; mainChar.rightEye.x += 1; mainChar.drawChar(); } I thought drawing the

问题 I've created a program that allows me to create flexible Neural networks of any size/length, however I'm testing it using the simple structure of an XOR setup(Feed forward, Sigmoid activation, back propagation, no batching). EDIT: The following is a completely new approach to my original question which didn't supply enough information EDIT 2: I started my weight between -2.5 and 2.5, and fixed a problem in my code where I forgot some negatives. Now it either converges to 0 for all cases or 1

问题 Starting with two images im1 and im2 created with the PIL module, we have the corresponding black and white images, bw_im1 = im1.convert('1') and bw_im2 = im2.convert('1') Each pixel of bw_im2 and bw_im2 is either 0 or 256. Suppose both bw_im2 and bw_im2 have the same size. How do you XOR all the corresponding entries and then sum them up? My Work I wrote the following stub / proof of concept Python program, but worried that using the code (unpacking/translating over) would be complicated.