Wednesday, January 27, 2016

Low Energy Nuclear Reactor Creates Gold and Platinum

Thursday March 06, 2014 13:03
The transmutation from lead to gold has been mankind’s dream for millennia. Lattice Energy LLC, a company from Chicago, IL, claims to have developed a process for energy production, utilizing a low-energy nuclear reactor (LENR) that, as a byproduct of neutron captures on tungsten, will create a mix of precious metals.
To learn more about the technology, Tech Metals Insider spoke with Lewis Larsen, president and CEO of Lattice.
Lattice was founded in 2001 upon the ruins of the “cold fusion” failures that had caused much hope and disappointment back in the late 1980’s. Larsen is part of a team that learned from cold fusion’s mistakes: “their heat production measurements were right”, said Larsen with respect to cold fusion, “but their conclusions about the heat being produced by a fusion process were completely wrong.”
What enabled Lattice’s new approach were recent advances in nanotechnology. “Nanotechnology and LENR are joined at the hip”, said Larsen. “It is one of the reasons why this could not be done back in 1989-90. Before our work, nobody had a grasp on the theory of neutron creation from protons and electrons in tabletop apparatus; nor on exactly how to apply advanced nanotechnology to build well-performing prototype devices.”
Combining the know-how of experts from a variety of disciplines including electro-dynamics, quantum electro-dynamics, nuclear physics and solid state physics, lead to the development of a theoretical foundation which is now ready to be prototyped, and put to the test.
The goal of Lattice is to build high performance thermal sources with outputs ranging from single watts to 100 kilowatts, the ultimate application being the use of LENRs in cars. Patents have been filed and some were issued. At this point, financing is provided by insiders and several angel investors, but larger amounts of capital are needed to take the technology to its next level.
Larsen is labeling the LENR as “green nuclear technology” – green because commercial systems could be operated very similar to aluminum production using an electric arc. The process would emit no energetic neutrons (LENR ultra low energy neutrons are all absorbed locally deep inside the reactor and are thus not a safety problem), and no gamma radiation.
When asked about differences compared to the deuterium-tritium fusion process presented by the Lawrence Livermore National Laboratory last week (please click here for Tech Metals Insider’s report) Larsen said: “Their dirty little secret they don’t talk about is that they produce deadly, very energetic neutrons and gamma radiation. Harvesting the energy from these neutrons produced by fusion is quite difficult. Furthermore, shielding requirements will make fusion unusable for mobile and portable power generation applications.”
Larsen’s theory that gold, platinum and several other metals can be created by his process is based on findings by Japanese physicist Prof. Hantaro Nagaoka who successfully transmuted tungsten into gold back in 1924. Nagaoka’s results have been verified by several institutions in recent independent experiments but so far there has been no effort to commercialize the process. “Now that the LENR transmutation process is well understood the use of nanotechnology may change all that”, believes Larsen.
“The neutron-catalyzed LENR process follows rows of the periodic table of elements”, he went on, meaning that heavier metals than the starting targets’ will be created. The work published by Larsen and his team suggests that a tungsten target, for instance, will absorb neutrons and gradually be transmuted to gold, platinum and other platinum group metals. “And because LENR products are not dangerously radioactive”, Larsen added, “conventional metal recovery processes can be utilized.”
“Can we scale this up to a commercial process that makes money?” – Larsen is convinced it may be possible.

http://www.kitco.com/ind/Albrecht/2014-02-25-Alchemy-2-0-Low-Energy-Nuclear-Reactor-Creates-Gold-and-Platinum.html

Monday, January 25, 2016

Rogue Access Point Toolkit : MANA


The MANA Toolkit evilAP attacks first presented at Defcon 22
More specifically, it contains the improvements to KARMA attacks with hostapd, as well as some useful configs for conducting MitM once you’ve managed to get a victim to connect.

Rogue Access Point Toolkit

It contains:
  • kali/ubuntu-install.sh – simple installers for Kali 1.0.9 and Ubuntu 14.04 (trusty)
  • slides – an explanation of what we’re doing here
  • run-mana – the controller scripts
  • hostapd-manna – modified hostapd that implements our new karma attacks
  • crackapd – a tool for offloading the cracking of EAP creds to an external tool and re-adding them to the hostapd EAP config (auto crack ‘n add)
  • sslstrip-hsts – our modifications to LeonardoNVE’s & moxie’s cool tools
  • apache – the apache vhosts for the noupstream hacks; deploy to /etc/apache2/ and /var/www/ respectivley

Installation

  • The simplest way to get up and running is it “apt-get install mana-toolkit” on Kali.
  • To get up and running setup a Kali 1.0.9 box (VM or otherwise), update it, then run kali-install.sh
  • To get up and running setup a Ubuntu 14.04 box (VM or otherwise), update it, then run ubuntu-install.sh
  • The ubuntu installer has much more dependency info than the kali one if you’re looking for a template.

Rogue Access Point Evil Rogue Access Point Rogue Access PointPrerequisites

Software
Check the ubuntu installer for more details on software pre-requisites.
Hardware 
You’ll need a wifi card that supports master mode. You can check whether it does by running: iw list You want to see “AP” in the output. Something like:
Supported interface modes:
         * IBSS
         * managed
         * AP
         * AP/VLAN
         * monitor
         * mesh point

Running

Mana has several components, these can be started using the example start scripts, or you can use these as templates to mix your own.
Mana will be installed to several directories:
  • The mana tools are installed to /usr/share/mana-toolkit
  • The start scripts are in /usr/share/mana-toolkit/run-mana
  • The captured traffic will be in /var/lib/mana-toolkit
The different start scripts are listed below and must be edited to point to the right wifi device (default is wlan0, this may not be right for your installation):
  • start-nat-full.sh – Will fire up MANA in NAT mode (you’ll need an upstream link) with all the MitM bells and whistles.
  • start-nat-simple.sh – Will fire up MANA in NAT mode, but without any of the firelamb, sslstrip, sslsplit etc.
  • start-noupstream.sh – Will start MANA in a “fake Internet” mode. Useful for places where people leave their wifi on, but there is no upstream Internet. Also contains the captive portal.
  • start-noupstream-eap.sh – Will start MANA with the EAP attack and noupstream mode.
While these should all work, it’s advisable that you craft your own based on your specific needs.



Source && Download

https://n0where.net/rogue-access-point-toolkit-mana/

CHECK IT OUT : TACACS+ authentication server

IT LOOKS PRETTY EASY, BUT IS IT?

welcome back to war!

"Information in passports is protected by the Basic Access Control protocol, the standard can be found here. In order to read from the passport you need certain pieces of information; the passport number, the D.O.B and the date of expiry.
NFC TagInfo is quite a nice app that can read passports amongst other things, make sure you enter the correct information before attempting the read though. It can take a few seconds to read due to the amount of information being transferred.
In the USA there are foil covers over the passport which prevent the NFC working without the passport being open however with my UK passport I can read it just fine when it's closed."

because this is not the way it looks EASY  : 

"Third party software usage for key distribution If you have OpenView, Nagios or similar monitoring system installed you can use their capabilities to distribute the key.
You can use xargs submitting to it list of servers and using scp command to distribute the files. Parallel is a Perl script written by Ole Tange that extends and improves capabilities of xargs that can optimize this operation creating multiple threads, one for each server. To transfer file to remote computer you can use option --transfer:
Especially convenient are parallel command execution packages. Among the latter we can mention pdsh -- a variant of the rsh(1) command. ROMs are available from Fedora EPEL (pdsh-2.26-4.el6.x86_64.rpm CentOS 6 Download), and SourceForge.net. Unlike rsh(1), which runs commands on a single remote host, pdsh can run multiple remote commands in parallel. pdsh uses a "sliding window" (or fanout) of threads to conserve resources on the initiating host while allowing some connections to time out. pdcp included in pdsh package

http://www.softpanorama.org/Net/Application_layer/SSH/passwordless_ssh_login.shtml#Third_party_software_usage_for_key_distribution_


 I Think I have an attack here Emoji smile 
To enable the TACACS+ password on the switch and specify authentication using the password provided in the ENABLE PASSWORD command is attempted if a TACACS+ server is not available, use the following commands:

awplus> enable
awplus# configure terminal
awplus(config)# aaa authentication enable default group tacacs local
To enable the TACACS+ password on the switch, use the following commands:
awplus> enable
awplus# configure terminal
awplus(config)# aaa authentication enable default group tacacs

Saturday, January 9, 2016

How to Hide Files in JPG Images Let's see this with real stuff for fake money "VOID" security scanning test ... PKzip Coomad Line



if you encrypt a message with OpenPGP and then "ACSII armor" the result, i.e. encode it in Base64, then this encoding enlarges the data by 34%: 3 bytes become 4 characters (plus the odd newline). Compression with DEFLATE will be effective at cancelling this enlargement (thanks to Huffman codes). That's a case of usefulness of compression after encryption -- but, really, that's more compression over Base64, rather than compression over encryption.

http://security.stackexchange.com/questions/19969/encryption-and-compression-of-data

1. https://users.cs.jmu.edu/buchhofp/forensics/formats/pkzip.html

2. http://ricardo.ecn.wfu.edu/~cottrell/pkzip.html

encode 
creates archive in the UUEncode format 
note: PKZIP will create two files when the encode option is invoked; a .ZIP archive (e.g. save.zip) as well as UUEncoded version of the .ZIP file (e.g. save.uue) are created

Friday, January 8, 2016

I am trying to teach myself to program using the interactivepython.org website. I have run into a problem that seems to be way over my head. I've been working on it for 3 hours straight and am racking my noggin. Not sure how to break this down at all.
Problem:
Decoding a secret message:
The description may seem daunting, but the solution is not that hard. You can use the built-in string datatype with the associated built-in functions and while loop (with ‘len’ function) or a for loop (with ‘in’ operator) to traverse the string. Also, use the ’chr’ and ’ord’ functions (which are based on ASCII code) discussed in course material. Make sure to look at the examples in the course material and do #18 and #19 in Exercises 2. Answer for #19 is provided and it can give valuable hints for solving this problem.

Your country is at war and your enemies are using a secret code to communicate with each other. You have managed to intercept a message that read as follows:
:mmZ\dxZmx]Zpgy
The message is obviously encrypted using the enemy’s secret code. You have just learned that their encryption method is based upon the ASCII code (you can find this set easily by searching online). Individual characters in a string are encoded using this system. For example, the letter ‘A’ is encoded using the number 65 and ‘B’ is encoded using the number 66.
Your enemy’s secret code takes each letter of the message and encrypts it as follows (using a secret key):
If (OriginalChar + Key > 126) then
    EncryptedChar = ((OriginalChar + Key) - 127) + 32
Else 
    EncryptedChar = (OriginalChar + Key)
For example, if the enemy uses Key = 10 then the message ”Hey” would be encrypted as:
Character   ASCII
H         72
e         101
y         121

Encrypted H = (72 + 10) = 82 = R in ASCII
Encrypted e = (101 + 10) = 111 = o in ASCII
Encrypted y = 32 + ((121 + 10) - 127) = 36 = $ in ASCII
Consequently, “Hey” would be transmitted as “Ro$”.
Write a program that decrypts the intercepted message. You only know that the key used is a number between 1 and 100. Your program should try to decode the message using all possible keys between 1 and 100. When you try the valid key, the message will make sense. For all other keys, the message will appear as gibberish.
HINT: You will need to implement a decrypt function that takes in an encrypted message as string and a key as integer and returns the decrypted message as string. You can decrypt each letter of the message as follows:
If (EncryptedChar - Key < 32) then
    DecryptedChar = ((EncryptedChar - Key) + 127) - 32
Else
    DecryptedChar = (EncryptedChar - Key)
NOTE: You should also implement an encrypt function that takes in a regular message as string and a key as integer and returns the corresponding encrypted message as string (the algorithm to encrypt a message is mentioned above in the problem description). This function would help you in encrypting any regular message, which then can be passed to your decrypt function to be decrypted.

For Encryption: You should ask the user for any regular message and a key and output the corresponding encrypted message.
Sample run:
Enter a regular message to encode:
Attack at dawn!
Enter a key value (between 0 and 100) for encoding:
88
The encoded message is: 
:mmZ\dxZmx]Zpgy

For Decryption: You should ask the user for an encrypted message and output 100 well-formatted, decrypted messages (using keys between 1 and 100) along with the corresponding key value.
Sample run (the gibberish messages below are not accurate):
Enter an encrypted message to decode:
:mmZ\dxZmx]Zpgy 
The following are the decoded messages for keys 1 to 100:
Key: 1 –> Decoded Message: whfuihwuiidh89
Key: 2 –> Decoded Message: 9ehkaOY3ewine
...
Key: 87 –> Decoded Message: Buubdl!bu!ebxo
Key: 88 –> Decoded Message: Attack at dawn!
...
Key: 100 –> Decoded Message: on3dwp389/wi8
This is the code I currently have:
def encrypt(message, key):
    result = ""
    for char in message:
        result += encryptedChar
    return result
shareimprove this question
  
The description of your problem seems daunting, but the solution is not that hard. You can use the built-instring datatype with the associated built-in functions and while loop (with 'len' function) or a forloop (with 'in' operator) to traverse the string. Also, use the 'chr' and 'ord' functions (which are based on ASCII code). – Brian Cain Nov 14 '13 at 17:43
  
You should try & use string.make_trans('plaintext', 'ciphertex') in the string library. – Ashish Nitin Patil Nov 14 '13 at 17:44
  
Yah i've read that about 500 times. Thanks! – TYPKRFT Nov 14 '13 at 17:45
  
the code you have probably raises an error? since you have no encryptedChar? it tells you exactly how to get the encrypted char ... – Joran Beasley Nov 14 '13 at 17:52 
  
Thanks Joran I have had no issues with most of the problems until now. I still dont understand how to do this, though. – TYPKRFT Nov 14 '13 at 17:54

2 Answers


Here is a simpler (but longer) answer as Joran Beasley's.
After you understand, that you can get the 'number' of a character with ord() and 'recover' the character with chr() it is quiet simple to 'translate' the code you got into correct python code.
Beginning with the following part:
If (OriginalChar + Key > 126) then
    EncryptedChar = ((OriginalChar + Key) - 127) + 32
Else 
    EncryptedChar = (OriginalChar + Key)
If you start out with the code you already wrote you can translate the above into:
def encrypt(message, key):
    result = ""
    for char in message:
        if (ord(char) + key > 126):
            result += chr(ord(char) + key - 127 + 32)
        else:
            result += chr(ord(char) + key)
    return result
You can do the same with the decryption part and write then a simple menu.
Here is the remaining code (you have to add the encrypt function at the top:
def decrypt(message):
    for key in range(1, 101):
        result = ""
        for char in message:
            if (ord(char) - key < 32):
                result += chr(ord(char) - key + 127 - 32)
            else:
                result += chr(ord(char) - key)
        print('key: {} -'.format(key), result)

if __name__ == '__main__':
    print('1 - Encrypt')
    print('2 - Decrypt')
    inp = input('select 1 or 2: ')
    if inp == '1':
        msg = input('Enter message: ')
        key = int(input('Enter key (1-100): '))
        print('Encrypted message:')
        print(encrypt(msg, key))
    else:
        msg = input('Enter message: ')
        decrypt(msg)
shareimprove this answer
  
I was trying to get him here :P (+1 all the same) – Joran Beasley Nov 14 '13 at 18:24 
  
I am going to try and comment every line of this.. could I private message either of you, maybe tom with the comment and maybe you can assess if I have grasped what I need to. I dont want to move on until I understand this like the back of my hand. – TYPKRFT Nov 14 '13 at 18:40 
1
I have created a so chatroom. You can post your questions/comments/... there. Now you, Joran Beasley, and I have write access, but everyone can see it. – TobiMarg Nov 14 '13 at 20:15
  
Thanks that is really too nice of you! – TYPKRFT Nov 15 '13 at 3:06

heres a fun solution
import string,codecs
class RotEncoder:
    def __init__(self,rot):
        self._rot = rot
    def _encChar(self,ch):
        return chr((ord(ch) + self._rot) if ord(ch) + self.rot =< 126 else  (((ord(ch) + self._rot) - 127) + 32))
    def _decChar(self,ch):
        return chr((ord(ch) - self._rot) if ord(ch) - self._rot >= 32 else (((ord(ch) - self._rot) + 127) - 32))
    def encode(self,txt,errors=[]):
        return "".join(map(self._encChar,txt)),1
    def decode(self,txt,errors=[]):
        return "".join(map(self._decChar,txt)),1
import re
def find_rot(search):
    t = re.match("rot\s?([0-9]+)",search.lower())
    if t.groups():
        val = int(t.groups()[0])
        return codecs.CodecInfo(
            name='rotcipher',
            encode=RotEncoder(val).encode,
            decode=RotEncoder(val).decode
            )

codecs.register(find_rot)

print ":mmZ\dxZmx]Zpgy".decode('rot88')
shareimprove this answer
  
Sorry that is way way above my head. I appreciate your input though. – TYPKRFT Nov 14 '13 at 18:03
  
ok what part of the original assignment do you not understand? maybe that would be a better question ... because from where I sit it looks like you posted your assignment and just wanted an answer ... so why do you think your existing solution does not work? – Joran Beasley Nov 14 '13 at 18:10
  
Well I am new and I dont understand how to break it down. I feel like I got through the lessons fine...then bam no clue. I know that I need two functions one for the encrypt and one for the decrypt. I know the encrypt needs to take in a message and a key 1-100 i think. I think that I need to grab the ascii value for each character but Im stuck after that. – TYPKRFT Nov 14 '13 at 18:16
  
so do you know how to get the ascii value for a character? if so why is that not part of your "attempt" above?– Joran Beasley Nov 14 '13 at 18:17
  
An answer would be helpful. I have answered most of my own questions on here but am just baffled by this. The instructions are becoming lost in translation to me and I don't know anyone else that knows this kind of stuff..That is why I am on here. I fear that I have gotten to a stuck point and will not understand python past this. – TYPKRFT Nov 14 '13 at 18:19
  
I know what needs to be done I think but I dont know how to write it. I know that I need to do ord() for each character right? – TYPKRFT Nov 14 '13 at 18:20
1
ok thats a good start ... so you have originalChar as char in your attempt above... what do you need to do to get encryptedChar? as far as an answer... if thats all you want I gave it above ... if you want to understand both the problem and the solution I will do my best to walk you to that... once you understand what you need to do here you can apply it to many other problems – Joran Beasley Nov 14 '13 at 18:21 
1
This answer is short, very nice and useful if one has a bit more experience with python, but hard to understand for a beginner. So I added a simpler answer (but not with such an elegant code). – TobiMarg Nov 14 '13 at 18:26
  
Joran I am still lost I appreciate your trying to hold my hand through this. I fear that maybe going back to school for something other than computer science will be in my future. Do you have any recommendations on maybe something I have missed in my studies and should have a better understanding of? – TYPKRFTNov 14 '13 at 18:31
1
it just takes practice ... and being able to recognize the class of problem and know what potential solutions are to that class of problem ... there is no innate programmer gene ... you just need to work at it (I posted this answer specifically because I knew it would make no sense to a beginner and I wanted you to make the logic leap on your own(with some help), believe me it happens you will be totally lost and all of a sudden some small thing will click and you will fully understand both the problem and the solution) ... 

http://stackoverflow.com/questions/19984548/how-can-i-tackle-this-cryptography

Portugal Ukraine Russia (mercenaires killed)

  https://sicnoticias.pt/especiais/guerra-russia-ucrania/2025-04-03-video-russia-diz-ter-abatido-mercenarios-portugueses-na-ucrania-755f2fec