I have a problem understanding this specific code and managing to convert it to Python from Javascript. The problem lies in the Buffer method used by Javascript which creates a different hash output than in Python. The main goal is to get the merkleRoot of the transactions ["a","b"].
Javascript: (The hashes of "a" and "b" individually are the same as with a python SHA256 implementation. However, the method (Buffer.concat([hashA, hashB])) makes the difference apparently, however I cannot figure out how to implement it in Python. In python I get a merkleRoot of "ca978112ca1bbdcafac231b39a23dc4da786eff8147c4e72b9807785afee48bb3e23e8160039594a33894f6564e1b1348bbd7a0088d42c4acb73eeaed59c009d", which is not correct. I posted the correct merkleRoot below.
const sha256 = (tx) => crypto.createHash("sha256").update(tx).digest();
const hashPair = (hashA, hashB, hashFunction = sha256) =>
hashFunction(Buffer.concat([hashA, hashB]));
const a = sha256("a");
const b = sha256("b");
hashPair(a, b).toString("hex");
e5a01fee14e0ed5c48714f22180f25ad8365b53f9779f79dc4a3d7e93963f94a
├─ ca978112ca1bbdcafac231b39a23dc4da786eff8147c4e72b9807785afee48bb
└─ 3e23e8160039594a33894f6564e1b1348bbd7a0088d42c4acb73eeaed59c009d
I have tried some approaches like with base64 and encodings, however due to my limitation in cryptography knowledge I can't seem to figure out the right approach. My approach in python was:
Get SHA256 of the string "a"
Get SHA256 of the string "b"
Get SHA256 of the concatenated hashes of "a"+"b":
ca978112ca1bbdcafac231b39a23dc4da786eff8147c4e72b9807785afee48bb3e23e8160039594a33894f6564e1b1348bbd7a0088d42c4acb73eeaed59c009d
Here is the Python Implementation from: https://www.geeksforgeeks.org/introduction-to-merkle-tree/
Python:
# Python code for implemementing Merkle Tree
from typing import List
import hashlib
class Node:
def __init__(self, left, right, value: str, content, is_copied=False) -> None:
self.left: Node = left
self.right: Node = right
self.value = value
self.content = content
self.is_copied = is_copied
#staticmethod
def hash(val: str) -> str:
return hashlib.sha256(val.encode('utf-8')).hexdigest()
def __str__(self):
return (str(self.value))
def copy(self):
"""
class copy function
"""
return Node(self.left, self.right, self.value, self.content, True)
class MerkleTree:
def __init__(self, values: List[str]) -> None:
self.__buildTree(values)
def __buildTree(self, values: List[str]) -> None:
leaves: List[Node] = [Node(None, None, Node.hash(e), e) for e in values]
if len(leaves) % 2 == 1:
leaves.append(leaves[-1].copy()) # duplicate last elem if odd number of elements
self.root: Node = self.__buildTreeRec(leaves)
def __buildTreeRec(self, nodes: List[Node]) -> Node:
if len(nodes) % 2 == 1:
nodes.append(nodes[-1].copy()) # duplicate last elem if odd number of elements
half: int = len(nodes) // 2
if len(nodes) == 2:
return Node(nodes[0], nodes[1], Node.hash(nodes[0].value + nodes[1].value), nodes[0].content+"+"+nodes[1].content)
left: Node = self.__buildTreeRec(nodes[:half])
right: Node = self.__buildTreeRec(nodes[half:])
value: str = Node.hash(left.value + right.value)
content: str = f'{left.content}+{right.content}'
return Node(left, right, value, content)
def printTree(self) -> None:
self.__printTreeRec(self.root)
def __printTreeRec(self, node: Node) -> None:
if node != None:
if node.left != None:
print("Left: "+str(node.left))
print("Right: "+str(node.right))
else:
print("Input")
if node.is_copied:
print('(Padding)')
print("Value: "+str(node.value))
print("Content: "+str(node.content))
print("")
self.__printTreeRec(node.left)
self.__printTreeRec(node.right)
def getRootHash(self) -> str:
return self.root.value
def mixmerkletree() -> None:
elems = ["a", "b"]
#as there are odd number of inputs, the last input is repeated
print("Inputs: ")
print(*elems, sep=" | ")
print("")
mtree = MerkleTree(elems)
print("Root Hash: "+mtree.getRootHash()+"\n")
mtree.printTree()
mixmerkletree()
#This code was contributed by Pranay Arora (TSEC-2023).
Python Output:
Inputs:
a | b
Root Hash: 62af5c3cb8da3e4f25061e829ebeea5c7513c54949115b1acc225930a90154da
Left: ca978112ca1bbdcafac231b39a23dc4da786eff8147c4e72b9807785afee48bb
Right: 3e23e8160039594a33894f6564e1b1348bbd7a0088d42c4acb73eeaed59c009d
Value: 62af5c3cb8da3e4f25061e829ebeea5c7513c54949115b1acc225930a90154da
Content: a+b
So my main question is, how can I correctly implement the Buffer method from javascript into Python to get the same hash of when combining the hashes of "a" and "b". The correct merkleRoot as shown above should be: e5a01fee14e0ed5c48714f22180f25ad8365b53f9779f79dc4a3d7e93963f94a
SOLVED, thanks to the great explanation of Michael Butscher above.
The JavaScript "Buffer" objects translate to Python "bytes" (or sometimes "bytearray") objects. Ask the hash object for its "digest" instead of the hexadecimal representation "hexdigest". The digest is a "bytes" object you can concatenate with another with a simple plus sign. A "bytes" object can be fed into a hash function (as the code does already) and has a "hex" method to return a hexadecimal string representation for printing. –
Michael Butscher
Here is a simplified python solution to get the same MerkleRoot as with the Javascript Buffer method:
import hashlib
def hashab(string):
x = string.encode()
return hashlib.sha256(x).digest()
a = hashab("a")
b = hashab("b")
ab = a+b
print(ab)
print(hashlib.sha256(ab).hexdigest())
Output:
b'\xca\x97\x81\x12\xca\x1b\xbd\xca\xfa\xc21\xb3\x9a#\xdcM\xa7\x86\xef\xf8\x14|Nr\xb9\x80w\x85\xaf\xeeH\xbb>#\xe8\x16\x009YJ3\x89Oed\xe1\xb14\x8b\xbdz\x00\x88\xd4,J\xcbs\xee\xae\xd5\x9c\x00\x9d'
e5a01fee14e0ed5c48714f22180f25ad8365b53f9779f79dc4a3d7e93963f94a
Related
What's the difference between String[] and [String] in typescript ?
What is the best option to choose between the two ?
They are not the same!
Update: Since TypeScript 2.7 this is not entirely correct anymore (see below for additions):
string[] says the variable is an array with values of type string (it can be of any size, even empty).
[string] says the variable is an array of size >= 1 and the first entry is a string
since 2.7: size === 1
the [type] syntax can be extended, like [type1,type2,type3,typeN] and then requires the array to have a size of at least N and the first N types are as specified, while the following types are the union of those types.
Some different examples that illustrate this issue:
const testA: string[] = []; // good
const testB: [string] = []; // Error, array must have a string at position 0
const testC: [string, number] = ["a", 0]; // good
const testC1 = testC[0]; // testC1 is detected to be string
const testC2 = testC[1]; // testC2 is detected to be number
const testD: [string, number] = ["a", 0, "1"]; // good before 2.7, bad afterwards
const testD1 = testD[2]; // testD1 is detected to be string | number
const testE: [string, number] = ["a", 0, 1]; // good before 2.7, bad afterwards
const testE1 = testE[2]; // testE1 is detected to be string | number
const testF: [string, number] = ["a", 0, null]; // Error, null is not of type string|number
Since TypeScript 2.7
The size is by default fixed. So if you want [string] to allow more than one entry, you need to specify in a very ugly way:
interface MinimumStrTuple extends Array<number | string> {
0: string;
}
Returning to the basics , what is TypeScript ?
TypeScript is a typed superset of JavaScript that compiles to javascript in the end
Javascript doesn't support types if you write code like this
// its OK to write code like
let str = 'sfafsa' ;
str = true ;
but TypeScript helps you write a typed code and you'll get error before compiling if you're trying to assign a value that is not the type of the variable
if you try the differnet syntax for writing array
They are the same why ? because typescript will compile them the same
for example
let str: String[] = ['ge', 'gege', 'egeg'];
let str2: [String] = ['ge', 'gege', 'egeg'];
will compilied to
var str = ['ge', 'gege', 'egeg'];
var str2 = ['ge', 'gege', 'egeg'];
so they compiled to the same code so use whatever you'll find better for you and I recommend to use the same convention for your code base
BTW :- I use this syntax let str: String[]=[];
Note :- If you use this syntax [String] that means you should have an array that has at least one String >> so in this case empty array will give you error
thanks #Lusito for paying attention to complete the answer .
I have been handed a project at work where I need to find duplicate pairings from multiple rows within a dataset. While the data set is much larger, the main portion revolves around the date of a training, the location of a training, and the names of the trainers. So every row of data has a date, a location, and then a comma separated list of names:
Date Location Names
1/13/2014 Seattle A, B, D
1/16/2014 Dallas C, D, E
1/20/2014 New York A, D
1/23/2014 Dallas C, E
1/27/2014 Seattle B, D
1/30/2014 Houston C, A, F
2/3/2014 Washington DC D, A, F
2/6/2014 Phoenix B, E
2/10/2014 Seattle C, B
2/13/2014 Miami A, B, E
2/17/2014 Miami C, D
2/20/2014 New York B, E, F
2/24/2014 Houston A, B, F
My goal is to be able to find rows with similar pairings of names. One example would be to know that A & B were in paired in Seattle on 1/13, Miami on 2/13, and Houston on 2/24, even though the third name is different in each occurrence. So instead of just simply finding duplicates among the entire string of names, I would also like to find pairings among partial segments of the “Names” column.
Is this possible to do within Excel or would I need to use a programming language to accomplish the task?
While I can manually do this, it represents a lot of time that could be used towards other things. If there was a way that I could automate this it would make this portion of my task a lot simpler.
Thank you in advance for any assistance or advice on a way forward.
You can do it with VBA. The solution below assumes
Your data is on the active sheet in columns A:C
You results will be output in columns E:G
The output will be a list sorted by pairs, and then by dates, so you can easily see where pairs repeated.
The routine assumes no more than three trainers at a time, but could be modified add more possible combinations.
Cities with just a single trainer will be ignored.
The routine uses a Class module to gather the information, and two Collections to process the data. It also makes use of the feature that collections will not allow addition of two items with the same key.
Class Module
Rename the Class Module: cPairs
Option Explicit
Private pTrainer1 As String
Private pTrainer2 As String
Private pCity As String
Private pDT As Date
Public Property Get Trainer1() As String
Trainer1 = pTrainer1
End Property
Public Property Let Trainer1(Value As String)
pTrainer1 = Value
End Property
Public Property Get Trainer2() As String
Trainer2 = pTrainer2
End Property
Public Property Let Trainer2(Value As String)
pTrainer2 = Value
End Property
Public Property Get City() As String
City = pCity
End Property
Public Property Let City(Value As String)
pCity = Value
End Property
Public Property Get DT() As Date
DT = pDT
End Property
Public Property Let DT(Value As Date)
pDT = Value
End Property
Regular Module
Option Explicit
Option Compare Text
Public cP As cPairs, colP As Collection
Public colCityPairs As Collection
Public vSrc As Variant
Public vRes() As Variant
Public rRes As Range
Public I As Long, J As Long
Public V As Variant
Public sKey As String
Sub FindPairs()
vSrc = Range("A1", Cells(Rows.Count, "C").End(xlUp))
Set colP = New Collection
Set colCityPairs = New Collection
'Collect Pairs
For I = 2 To UBound(vSrc)
V = Split(Replace(vSrc(I, 3), " ", ""), ",")
If UBound(V) >= 1 Then
'sort the pairs
SingleBubbleSort V
Select Case UBound(V)
Case 1
AddPairs V(0), V(1)
Case 2
AddPairs V(0), V(1)
AddPairs V(0), V(2)
AddPairs V(1), V(2)
End Select
End If
Next I
ReDim vRes(0 To colCityPairs.Count, 1 To 3)
vRes(0, 1) = "Date"
vRes(0, 2) = "Location"
vRes(0, 3) = "Pairs"
For I = 1 To colCityPairs.Count
With colCityPairs(I)
vRes(I, 1) = .DT
vRes(I, 2) = .City
vRes(I, 3) = .Trainer1 & ", " & .Trainer2
End With
Next I
Set rRes = Range("E1").Resize(UBound(vRes, 1) + 1, UBound(vRes, 2))
With rRes
.EntireColumn.Clear
.Value = vRes
With .Rows(1)
.HorizontalAlignment = xlCenter
.Font.Bold = True
End With
.Sort key1:=.Columns(3), order1:=xlAscending, key2:=.Columns(1), order2:=xlAscending, _
Header:=xlYes
.EntireColumn.AutoFit
V = VBA.Array(vbYellow, vbGreen)
J = 0
For I = 2 To rRes.Rows.Count
If rRes(I, 3) = rRes(I - 1, 3) Then
.Rows(I).Interior.Color = .Rows(I - 1).Interior.Color
Else
J = J + 1
.Rows(I).Interior.Color = V(J Mod 2)
End If
Next I
End With
End Sub
Sub AddPairs(T1, T2)
Set cP = New cPairs
With cP
.Trainer1 = T1
.Trainer2 = T2
.City = vSrc(I, 2)
.DT = vSrc(I, 1)
sKey = .Trainer1 & "|" & .Trainer2
On Error Resume Next
colP.Add cP, sKey
If Err.Number = 457 Then
Err.Clear
colCityPairs.Add colP(sKey), sKey & "|" & colP(sKey).DT & "|" & colP(sKey).City
colCityPairs.Add cP, sKey & "|" & .DT & "|" & .City
Else
If Err.Number <> 0 Then Stop
End If
On Error GoTo 0
End With
End Sub
Sub SingleBubbleSort(TempArray As Variant)
'copied directly from support.microsoft.com
Dim Temp As Variant
Dim I As Integer
Dim NoExchanges As Integer
' Loop until no more "exchanges" are made.
Do
NoExchanges = True
' Loop through each element in the array.
For I = LBound(TempArray) To UBound(TempArray) - 1
' If the element is greater than the element
' following it, exchange the two elements.
If TempArray(I) > TempArray(I + 1) Then
NoExchanges = False
Temp = TempArray(I)
TempArray(I) = TempArray(I + 1)
TempArray(I + 1) = Temp
End If
Next I
Loop While Not (NoExchanges)
End Sub
Ok. I got bored and did this whole thing in Python code. I assume you are familiar with the language; however, you should be able to get the following piece of code to work on any computer with Python installed.
I have made a few assumptions. For instance, I have used your example input as definite input.
A few things which will mess up the program:
Not entering with case sensitivity. Beware of capital letters etc.
Having a inputfile which has the following row: "Date Location Names". Just remove and keep straight facts in the file. I got lazy and do not bother adjusting this.
A ton of other small stuff. Just do what the program asks you to do and dont enter funky input.
About program:
Revolves around using a dictionary with person names as keys. The values in the dictionary is a set with tuples containing the places they've been during what date. By then comparing these sets and getting the intersection, we can find the answer.
Kinda messy since I took this as Python practice. Have not coded in Python for a while and I got a thrill out of doing it all without utilizing objects. Just follow the "instructions" and keep the inputfile, which stores all information, in the same folder as the piece of code are running.
As a side note, you might want to check that the program yields correct output.
If you have any questions, feel free to contact me.
def readWord(line, stringIndex):
word = ""
while(line[stringIndex] != " "):
word += line[stringIndex]
stringIndex += 1
return word, stringIndex
def removeSpacing(line, stringIndex):
while(line[stringIndex] == " "):
stringIndex += 1
return stringIndex
def readPeople(line, stringIndex):
lineSize = len(line)
people = []
while(stringIndex < lineSize):
people.append(line[stringIndex])
stringIndex += 3
return people
def readLine(travels, line):
stringIndex = 0
date, stringIndex = readWord(line, stringIndex)
stringIndex = removeSpacing(line, stringIndex)
location, stringIndex = readWord(line, stringIndex)
stringIndex = removeSpacing(line, stringIndex)
people = readPeople(line, stringIndex)
for person in people:
if(person not in travels.keys()):
travels[person] = set()
travels[person].add((date, location))
return travels
def main():
f = open(input("Enter filename (must be in same folder as this program code. For instance, name could be: testDocument.txt\n\n"))
travels = dict()
for line in f:
travels = readLine(travels, line)
print("\n\n\n\n PROGRAM RUNNING \n \n")
while(True):
persons = []
userInput = "empty"
while(userInput):
userInput = input("Enter person name (Type Enter to finish typing names): ")
if(userInput):
persons.append(userInput)
output = travels[persons[0]]
for person in persons[1:]:
output = output.intersection(travels[person])
print("")
for hit in output:
print(hit)
print("\nFINISHED WITH ONE RUN. STARTING NEW ONE\n")
Basically, I have a cgi script that prints out valid json, I have checked and I have a similar functions that work the same way but this one doesn't some reason and I can't find it.
Javascript:
function updateChat(){
$.ajax({
type: "get",
url: "cgi-bin/main.py",
data: {'ajax':'1', 'chat':'1'},
datatype:"html",
async: false,
success: function(response) {
alert(response); //Returns an empty string
},
error:function(xhr,err)
{
alert("Error connecting to server, please contact system administator.");
}
});
Here is the JSON that python prints out:
[
"jon: Hi.",
"bob: Hello."
]
I used json.dumps to create the JSON it worked in previous functions that have pretty much the same JSON layout only different content.
There is a whole bunch more of server code, I tried to copy out the relevant parts. Basically I'm just trying to filter an ugly chat log for learning purposes. I filter it with regex and then create a json out of it.
#!/usr/bin/env python
# -*- coding: UTF-8 -*-
print "Content-type: text/html\n\n"
print
import cgi, sys, cgitb, datetime, re, time, random, json
cgitb.enable()
formdata = cgi.FieldStorage()
def tail( f, window=20 ):
BUFSIZ = 1024
f.seek(0, 2)
bytes = f.tell()
size = window
block = -1
data = []
while size > 0 and bytes > 0:
if (bytes - BUFSIZ > 0):
# Seek back one whole BUFSIZ
f.seek(block*BUFSIZ, 2)
# read BUFFER
data.append(f.read(BUFSIZ))
else:
# file too small, start from begining
f.seek(0,0)
# only read what was not read
data.append(f.read(bytes))
linesFound = data[-1].count('\n')
size -= linesFound
bytes -= BUFSIZ
block -= 1
return '\n'.join(''.join(data).splitlines()[-window:])
def updateChatBox():
try:
f = open('test.txt', 'r')
lines = tail(f, window = 20)
chat_array = lines.split("\n")
f.close()
except:
print "Failed to access data"
sys.exit(4)
i = 0
while i < len(chat_array):
#remove timer
time = re.search("(\[).*(\])", chat_array[i])
result_time = time.group()
chat_array[i] = chat_array[i].replace(result_time, "")
#Removes braces around user
user = re.search("(\\().*?(_)", chat_array[i])
result_user = user.group()
chat_array[i] = chat_array[i].replace("(", "")
chat_array[i] = chat_array[i].replace(")", "")
#Removes underscore and message end marker
message = re.search("(_).*?(\|)", chat_array[i])
result_message = message.group()
chat_array[i] = chat_array[i].replace("_", ":")
chat_array[i] = chat_array[i].replace("|", "")
data += chat_array[i] + "\n"
i = i + 1
data_array = data.split("\n")
json_string = json.dumps(data_array)
print json_string
if formdata.has_key("ajax"):
ajax = formdata["ajax"].value
if ajax == "1": #ajax happens
if formdata.has_key("chat"):
chat = formdata["chat"].value
if chat == 1:
updateChatBox()
else:
print "ERROR"
elif formdata.has_key("get_all_stats"):
get_all_stats = formdata["get_all_stats"].value
if get_all_stats == "1":
getTopScores()
else:
print "ERROR"
Here is also a function that works perfectly and is in the same python file
def getTopScores():
try:
f = open('test_stats.txt', 'r')
stats = f.read()
stats_list = stats.split("\n")
f.close()
except:
print "Failed reading file"
sys.exit(4)
json_string = json.dumps(stats_list)
print json_string
The only difference is using the tail function and regex, the end result JSON actually looks identical.
Are you certain that updateChatBox is even getting called? Note that you compare ajax to the string "1" but you compare chat to the integer 1. I bet one of those doesn't match (in particular the chat one). If that doesn't match, your script will fall through without ever returning a value.
Also, though it isn't the root cause, you should clean up your content types for correctness. Your Javascript AJAX call is declared as expecting html in response, and your cgi script is also set to return content-type:text/html. These should be changed to json and content-type:application/json, respectively.
I'm trying to decrypt a ciphertext created by CryptoJS using PyCrypto. I am using AES-256-CTR, with a 12-byte random prefix and 4-byte counter. So far, I've had limited success. Please read this previous post where I made a first attempt.
This works in Javascript:
Install the CryptoCat extension
Run CryptoCat
Fire up the developer console (F12 in Chrome/Firefox)
Run these lines of code
key = 'b1df40bc2e4a1d4e31c50574735e1c909aa3c8fda58eca09bf2681ce4d117e11';
msg = 'LwFUZbKzuarvPR6pmXM2AiYVD2iL0/Ww2gs/9OpcMy+MWasvvzA2UEmRM8dq4loB\ndfPaYOe65JqGQMWoLOTWo1TreBd9vmPUZt72nFs=';
iv = 'gpG388l8rT02vBH4';
opts = {mode: CryptoJS.mode.CTR, iv: CryptoJS.enc.Base64.parse(iv), padding: CryptoJS.pad.NoPadding};
CryptoJS.AES.decrypt(msg, CryptoJS.enc.Hex.parse(key), opts).toString(CryptoJS.enc.Utf8);
Expected output: "Hello, world!ImiAq7aVLlmZDM9RfhDQgPp0CrAyZE0lyzJ6HDq4VoUmIiKUg7i2xpTSPs28USU8".
Here is a script I wrote in Python that partially(!) decrypts the ciphertext:
import struct
import base64
import Crypto.Cipher.AES
import Crypto.Util.Counter
def bytestring_to_int(s):
r = 0
for b in s:
r = r * 256 + ord(b)
return r
class IVCounter(object):
def __init__(self, prefix="", start_val=0):
self.prefix = prefix
self.first = True
self.current_val = start_val
def __call__(self):
if self.first:
self.first = False
else:
self.current_val += 1
postfix = struct.pack(">I", self.current_val)
n = base64.b64decode(self.prefix) + postfix
return n
def decrypt_msg(key, msg, iv):
k = base64.b16decode(key.upper())
ctr = IVCounter(prefix=iv)
#ctr = Crypto.Util.Counter.new(32, prefix=base64.b64decode(iv), initial_value=0, little_endian=False)
aes = Crypto.Cipher.AES.new(k, mode=Crypto.Cipher.AES.MODE_CTR, counter=ctr)
plaintext = aes.decrypt(base64.b64decode(msg))
return plaintext
if __name__ == "__main__":
#original:
key = 'b1df40bc2e4a1d4e31c50574735e1c909aa3c8fda58eca09bf2681ce4d117e11'
msg = 'LwFUZbKzuarvPR6pmXM2AiYVD2iL0/Ww2gs/9OpcMy+MWasvvzA2UEmRM8dq4loB\ndfPaYOe65JqGQMWoLOTWo1TreBd9vmPUZt72nFs='
iv = 'gpG388l8rT02vBH4'
decrypted = decrypt_msg(key, msg, iv)
print "Decrypted message:", repr(decrypt_msg(key, msg, iv))
print decrypted
The output is:
'Hello, world!Imi\xfb+\xf47\x04\xa0\xb1\xa1\xea\xc0I\x03\xec\xc7\x13d\xcf\xe25>l\xdc\xbd\x9f\xa2\x98\x9f$\x13a\xbb\xcb\x13\xd2#\xc9T\xf4|\xd8\xcbaO)\x94\x9aq<\xa7\x7f\x14\x11\xb5\xb0\xb6\xb5GQ\x92'
The problem is, only the first 16 bytes of the output match the first 16 bytes of the expected output!
Hello, world!ImiAq7aVLlmZDM9RfhDQgPp0CrAyZE0lyzJ6HDq4VoUmIiKUg7i2xpTSPs28USU8
When I modify the script to do this:
def __init__(self, prefix="", start_val=1):
and
self.current_val += 0 #do not increment
which makes the counter output the same value (\x00\x00\x00\x01) every time it is called, the plaintext is:
\xf2?\xaf:=\xc0\xfd\xbb\xdf\xf6h^\x9f\xe8\x16I\xfb+\xf47\x04\xa0\xb1\xa1\xea\xc0I\x03\xec\xc7\x13dQgPp0CrAyZE0lyzJ\xa8\xcd!?h\xc9\xa0\x8b\xb6\x8b\xb3_*\x7f\xf6\xe8\x89\xd5\x83H\xf2\xcd'\xc5V\x15\x80k]
where the 2nd block of 16 bytes (dQgPp0CrAyZE0lyzJ) matches the expected output.
When I set the counter to emit \x00\x00\x00\x02 and \x00\x00\x00\x03, I get similar results- subsequent 16-byte blocks are revealed. The only exception is that with 0s, the first 32 bytes are revealed.
All 0s: reveals first 32 bytes.
'Hello, world!ImiAq7aVLlmZDM9RfhD\xeb=\x93&b\xaf\xaf\x8d\xc9\xdeA\n\xd2\xd8\x01j\x12\x97\xe2i:%}G\x06\x0f\xb7e\x94\xde\x8d\xc83\x8f#\x1e\xa0!\xfa\t\xe6\x91\x84Q\xe3'
All 1s: reveals next 16 bytes.
"\xf2?\xaf:=\xc0\xfd\xbb\xdf\xf6h^\x9f\xe8\x16I\xfb+\xf47\x04\xa0\xb1\xa1\xea\xc0I\x03\xec\xc7\x13dQgPp0CrAyZE0lyzJ\xa8\xcd!?h\xc9\xa0\x8b\xb6\x8b\xb3_*\x7f\xf6\xe8\x89\xd5\x83H\xf2\xcd'\xc5V\x15\x80k]"
All 2s: reveals next 16 bytes.
'l\xba\xcata_2e\x044\xb2J\xe0\xf0\xd7\xc8e\xae\x91yX?~\x7f1\x02\x93\x17\x93\xdf\xd2\xe5\xcf\xe25>l\xdc\xbd\x9f\xa2\x98\x9f$\x13a\xbb\xcb6HDq4VoUmIiKUg7i\x17P\xe6\x06\xaeR\xe8\x1b\x8d\xd7Z\x7f"'
All 3s: reveals next 13 bytes.
'I\x92\\&\x9c]\xa9L\xb1\xb6\xbb`\xfa\xbet;#\x86\x07+\xa5=\xe5V\x84\x80\x9a=\x89\x91q\x16\xea\xca\xa3l\x91\xde&\xb6\x17\x1a\x96\x0e\t/\x188\x13`\xd2#\xc9T\xf4|\xd8\xcb`aO)\x94\x9a2xpTSPs28USU8'
If you concat the "correct" blocks, you'll get the expected plaintext:
Hello, world!ImiAq7aVLlmZDM9RfhDQgPp0CrAyZE0lyzJ6HDq4VoUmIiKUg7i2xpTSPs28USU8
This is really strange. I am definitely doing something wrong on the Python end as things can be decrypted, but not all in one go. If anyone can help, I would be really grateful. Thank you.
There are a couple problems here. First, the message is not a multiple of the block size, and you're not using padding. And second -- and the most crucial to this issue -- is that the IV is also not the correct size. It should be 16 bytes, but you only have 12. Probably both implementations should fail with an exception, and in the next major revision of CryptoJS, this will be the case.
Here's what happens due to this mistake: When the counter increments for the first time, it tries to increment the undefined value, because the last byte of the IV is missing. Undefined + 1 is NaN, and NaN | 0 is 0. That's how you end up getting 0 twice.
When using crypto mode CryptoJS.mode.CTR (where CTR stands for counter) the Initailization vector together with a counter is encrypted and then applied to the data to encrypt. This is done for each block of data you encrypt.
You explain that different parts of the message are decrypted correctly, when you apply different values to start_val, so I suspect that the counter is simply not correctly increased with each decrypt of a block.
Take a look at Block Cipher Mode: CTR at wikipedia
Caution: Please note that when using the CTR mode, the combination of the initialization vector + the counter should never be repeated.
Fixed. I simply made the counter start with 0 twice. Does anyone know if this is a vulnerability?
import struct
import base64
import Crypto.Cipher.AES
import Crypto.Util.Counter
import pdb
def bytestring_to_int(s):
r = 0
for b in s:
r = r * 256 + ord(b)
return r
class IVCounter(object):
def __init__(self, prefix="", start_val=0):
self.prefix = prefix
self.zeroth = True
self.first = False
self.current_val = start_val
def __call__(self):
if self.zeroth:
self.zeroth = False
self.first = True
elif self.first:
self.first = False
else:
self.current_val += 1
postfix = struct.pack(">I", self.current_val)
n = base64.b64decode(self.prefix) + postfix
return n
def decrypt_msg(key, msg, iv):
k = base64.b16decode(key.upper())
ctr = IVCounter(prefix=iv)
#ctr = Crypto.Util.Counter.new(32, prefix=base64.b64decode(iv), initial_value=0, little_endian=False)
aes = Crypto.Cipher.AES.new(k, mode=Crypto.Cipher.AES.MODE_CTR, counter=ctr)
plaintext = aes.decrypt(base64.b64decode(msg))
return plaintext
if __name__ == "__main__":
#original:
key = 'b1df40bc2e4a1d4e31c50574735e1c909aa3c8fda58eca09bf2681ce4d117e11'
msg = 'LwFUZbKzuarvPR6pmXM2AiYVD2iL0/Ww2gs/9OpcMy+MWasvvzA2UEmRM8dq4loB\ndfPaYOe65JqGQMWoLOTWo1TreBd9vmPUZt72nFs='
iv = 'gpG388l8rT02vBH4'
decrypted = decrypt_msg(key, msg, iv)
print "Decrypted message:", repr(decrypt_msg(key, msg, iv))
print decrypted
I have a java.util.ArrayList which contains the red marked value:
if I render it with render relations as JSON then in JavaScript the value is the following:
The milliseconds are gone, if I understand it right the value of receiveDate is a string so there is no way to parse it to another format.
Is there a way to render the value that the milliseconds are not gone ?
Method 1
import grails.converters.JSON
def r = [now: new Date(), name: 'Roong']
JSON.registerObjectMarshaller(Date) {o -> o.getTime()}
println(r as JSON)
Result 1
{
"now": 1356595418027,
"name": "Roong"
}
or you can set in Config.groovy
Method 2
grails.converters.json.date = "javascript"
Result 2
{
"now": new Date(1356595418027),
"name": "Roong"
}
You can save the missing part to another key-value pair, using method
java.sql.Timestamp.getNanos()
and restore the complete value in javascript by consolidating the two parts. For example:
String receiveDate = grails.converters.JSON.parse(jsonData).data[0].receiveDate
def formatDate = receiveDate.replace('Z','.'+ dt.getNanos().toString()).replace('T', ' ')
def tsDate = java.sql.Timestamp.valueOf(formatDate)
println 'The timestamp is : ' + tsDate