xor

Bitwise XOR java long

这一生的挚爱 提交于 2019-12-01 05:47:59
I am using Oracle Java 7.51 on Ubuntu 12.04, and trying to do this long a = 0x0000000080000001 ^ 0x4065DE839A6F89EEL; System.out.println("result "+ Long.toHexString(a)); Output: result bf9a217c1a6f89ef But I was expecting result to be 4065de831a6f89ef , since ^ operator is a bitwise XOR in Java. Which part of Java specification am I reading wrong? You need an L at the end of the first integer literal: long a = 0x0000000080000001L ^ 0x4065DE839A6F89EEL; Otherwise it is an int literal, not a long (the leading zeroes being ignored). The ^ operator then promotes the first operand value from

fast XORing bytes in python 3 [duplicate]

蓝咒 提交于 2019-12-01 04:12:45
问题 This question already has answers here : Simple Python Challenge: Fastest Bitwise XOR on Data Buffers (11 answers) Closed 5 years ago . I need to xor 2 bytes objects. I use this code: def bxor(b1, b2): # use xor for bytes result = b"" for b1, b2 in zip(b1, b2): result += bytes([b1 ^ b2]) return result It works fine when the bytes objects are small, but if I xor big objects (a few MB) it takes very long time (a few hours). How can I make it faster? 回答1: When XORing bytes objects with one

XOR of two short integers

心已入冬 提交于 2019-12-01 04:12:08
I am calculating XOR of two short integers using XOR ^ operator in a traditional fashion. Below is the method- short a=197; short b=341; short y = (short) (a ^ b); However the XOR always returned integer but in my case inputs are short integer, that is why i am casting short to the XOR output. The XOR can be calculated in different manners (example: using BigInteger etc.) but performance wise (less time) which is the best for short integers? While keeping performance in mind, should i first convert each short integer to binary number using Integer.toBinaryString(number) then apply bitwise XOR?

Effects of randomizing the order of inputs to a neural network

风格不统一 提交于 2019-12-01 03:55:39
问题 For my Advanced Algorithms and Data Structures class, my professor asked us to pick any topic that interested us. He also told us to research it and to try and implement a solution in it. I chose Neural Networks because it's something that I've wanted to learn for a long time. I've been able to implement an AND, OR, and XOR using a neural network whose neurons use a step function for the activator. After that I tried to implement a back-propagating neural network that learns to recognize the

How to solve systems of XOR equations?

折月煮酒 提交于 2019-12-01 03:17:46
问题 I have to solve a system that consists of 32 xor equations, each involving 15 of 32 variables. One would look like this: i[0] = p[0] ^ p[4] ^ p[5] ^ p[10] ^ p[11] ^ p[20] ^ p[21] ^ p[22] ^ p[23] ^ p[25] ^ p[26] ^ p[27] ^ p[28] ^ p[30] ^ p[31] i[n] and p[n] being 16 bit integers. So as I understand I would end up with a 32x32 matrix (containing only 1s and 0s) and a 32 result vector. Apparently Gaussian elimination is what I need but I can't wrap my mind around the problem, could someone give

Bitwise XOR java long

好久不见. 提交于 2019-12-01 03:14:59
问题 I am using Oracle Java 7.51 on Ubuntu 12.04, and trying to do this long a = 0x0000000080000001 ^ 0x4065DE839A6F89EEL; System.out.println("result "+ Long.toHexString(a)); Output: result bf9a217c1a6f89ef But I was expecting result to be 4065de831a6f89ef , since ^ operator is a bitwise XOR in Java. Which part of Java specification am I reading wrong? 回答1: You need an L at the end of the first integer literal: long a = 0x0000000080000001L ^ 0x4065DE839A6F89EEL; Otherwise it is an int literal, not

How can I use SIMD to accelerate XOR two blocks of memory?

删除回忆录丶 提交于 2019-12-01 02:13:41
问题 I want to XOR two blocks of memory as quickly as possible, How can I use SIMD to accelerate it? My original code is below: void region_xor_w64( unsigned char *r1, /* Region 1 */ unsigned char *r2, /* Region 2 */ int nbytes) /* Number of bytes in region */ { uint64_t *l1; uint64_t *l2; uint64_t *ltop; unsigned char *ctop; ctop = r1 + nbytes; ltop = (uint64_t *) ctop; l1 = (uint64_t *) r1; l2 = (uint64_t *) r2; while (l1 < ltop) { *l2 = ((*l1) ^ (*l2)); l1++; l2++; } } I wrote one myself, but

How to use keras for XOR

只愿长相守 提交于 2019-11-30 20:58:59
I want to practice keras by code a xor, but the result is not right, the followed is my code, thanks for everybody to help me. from keras.models import Sequential from keras.layers.core import Dense,Activation from keras.optimizers import SGD import numpy as np model = Sequential()# two layers model.add(Dense(input_dim=2,output_dim=4,init="glorot_uniform")) model.add(Activation("sigmoid")) model.add(Dense(input_dim=4,output_dim=1,init="glorot_uniform")) model.add(Activation("sigmoid")) sgd = SGD(l2=0.0,lr=0.05, decay=1e-6, momentum=0.11, nesterov=True) model.compile(loss='mean_absolute_error',

Why is a bias neuron necessary for a backpropagating neural network that recognizes the XOR operator?

此生再无相见时 提交于 2019-11-30 20:45:49
I posted a question yesterday regarding issues that I was having with my backpropagating neural network for the XOR operator. I did a little more work and realized that it may have to do with not having a bias neuron. My question is, what is the role of the bias neuron in general, and what is its role in a backpropagating neural network that recognizes the XOR operator? Is it possible to create one without a bias neuron? Kiril It's possible to create a neural network without a bias neuron... it would work just fine, but for more information I would recommend you see the answers to this

Python equivalent of sum() using xor()

喜欢而已 提交于 2019-11-30 18:43:01
I like the Python sum function : >>> z = [1] * 11 >>> zsum = sum(z) >>> zsum == 11 True I want the same functionality with using xor (^) not add (+). I want to use map. But I can not work out how to do this. Any hints? I am not satisfied with this : def xor(l): r = 0 for v in l: r ^= v return v I want a 1 liner using map. Hints? zxor = reduce(lambda a, b: a ^ b, z, 0) import operator zxor = reduce(operator.xor, z, 0) Note that starting Python 3.8 , and the introduction of assignment expressions (PEP 572) ( := operator), we can use and update a variable within a list comprehension and thus