I am currently learning the Mono alphabetic Cipher Encryption. For an example, for 26 letters substitution ciphers.
Suppose the letter identified with each other.
So the total possible key is 26!
And the can I know how to prove that: 26!>= 4*10^26 ????
Thanks!
log_10(26!) = log_10(26) + log_10(25) + ... + log_10(2) + log_10(1)
= 26.60562
Thus,
26! = 10 ^ (26.60562) = (10 ^ 0.60562) * 10^26
Since
log_10(4) = 0.6020
We could get 26! > 4 * 10^26.
Related
I'm new to LUA but figured out that gsub is a global substitution function and tonumber is a converter function. What I don't understand is how the two functions are used together to produce an encoded string.
I've already tried reading parts of PIL (Programming in Lua) and the reference manual but still, am a bit confused.
local L0_0, L1_1
function L0_0(A0_2)
return (A0_2:gsub("..", function(A0_3)
return string.char((tonumber(A0_3, 16) + 256 - 13 + 255999744) % 256)
end))
end
encodes = L0_0
L0_0 = gg
L0_0 = L0_0.toast
L1_1 = "__loading__\226\128\166"
L0_0(L1_1)
L0_0 = encodes
L1_1 = --"The Encoded String"
L0_0 = L0_0(L1_1)
L1_1 = load
L1_1 = L1_1(L0_0)
pcall(L1_1)
I removed the encoded string where I put the comment because of how long it was. If needed I can upload the encoded string as well.
gsub is being used to get 2 digit sections of A0_2. This means the string A0_3 is a 2 digit hexadecimal number but it is not in a number format so we cannot preform math on the value. A0_3 being a hex number can be inferred based on how tonubmer is used.
tonumber from Lua 5.1 Reference Manual:
Tries to convert its argument to a number. If the argument is already a number or a string convertible to a number, then tonumber returns this number; otherwise, it returns nil.
An optional argument specifies the base to interpret the numeral. The base may be any integer between 2 and 36, inclusive. In bases above 10, the letter 'A' (in either upper or lower case) represents 10, 'B' represents 11, and so forth, with 'Z' representing 35. In base 10 (the default), the number can have a decimal part, as well as an optional exponent part (see ยง2.1). In other bases, only unsigned integers are accepted.
So tonumber(A0_3, 16) means we are expecting for A0_3 to be a base 16 number (hexadecimal).
Once we have the number value of A0_3 we do some math and finally convert it to a character.
function L0_0(A0_2)
return (A0_2:gsub("..", function(A0_3)
return string.char((tonumber(A0_3, 16) + 256 - 13 + 255999744) % 256)
end))
end
This block of code takes a string of hex digits and converts them into chars. tonumber is being used to allow for the manipulation of the values.
Here is an example of how this works with Hello World:
local str = "Hello World"
local hex_str = ''
for i = 1, #str do
hex_string = hex_string .. string.format("%x", str:byte(i,i))
end
function L0_0(A0_2)
return (A0_2:gsub("..", function(A0_3)
return string.char((tonumber(A0_3, 16) + 256 - 13 + 255999744) % 256)
end))
end
local encoded = L0_0(hex_str)
print(encoded)
Output
;X__bJbe_W
And taking it back to the orginal string:
function decode(A0_2)
return (A0_2:gsub("..", function(A0_3)
return string.char((tonumber(A0_3, 16) + 13) % 256)
end))
end
hex_string = ''
for i = 1, #encoded do
hex_string = hex_string .. string.format("%x", encoded:byte(i,i))
end
print(decode(hex_string))
I am implementing a task that i can use to obtain checksum from modified ip hdr. This is what i got:
task checksum_calc;
input [159:0] IP_hdr_data;
output [15:0] IP_chksum;
reg [19:0] IP_chksum_temp;
reg [19:0] IP_chksum_temp1;
reg [19:0] IP_chksum_temp2;
begin
IP_chksum_temp = IP_hdr_data[15:0] + IP_hdr_data[31:16] + IP_hdr_data[47:32] + IP_hdr_data[63:48] + IP_hdr_data[79:64] + IP_hdr_data[111:96] + IP_hdr_data[127:112] + IP_hdr_data[143:128] + IP_hdr_data[159:144];
IP_chksum_temp1 = IP_chksum_temp[15:0] + IP_chksum_temp[19:16];
IP_chksum_temp2 = IP_chksum_temp1[15:0] + IP_chksum_temp1[19:16];
IP_chksum = ! IP_chksum_temp2[15:0];
end
endtask
It's that correct? Or it will be some timing problems due to using cominational logic?
Looks like all you are doing is some combination logic calculation. A functions is a better choice. The primary purpose of a function is to return a value that is to be used in an expression.
This is huge combo logic, which in most of the scenario's will cause trouble for timing.
Better to run it through synthesis and timings check to know the exact result.
One suggestion as
IP_chksum_temp1 = IP_chksum_temp[15:0] + IP_chksum_temp[19:16];
can only generate flip the 16th bit. Hence, there is no need of 20 bits in next addition.
IP_chksum_temp2 = IP_chksum_temp1[15:0] + IP_chksum_temp1[19:16];
This can be done :-
reg [16:0] IP_chksum_temp1;
reg [16:0] IP_chksum_temp2;
I am doing a school project in which I need to make a "sort of" vigenere cipher in which the user inputs both the keyword and plaintext. However the vigenere assumes a=0 whereas I am to assume a=1 and I have changed this accordingly for my program. However I am required to make my cipher work for both lower and upper case, How could I make this also work for lower case, it may be a stupid question but I'm very confused at this point and I'm new to programming, thanks.
REM Variables
plaintext$=""
PRINT "Enter the text you would like to encrypt"
INPUT plaintext$
keyword$=""
PRINT "Enter the keyword you wish to use"
INPUT keyword$
encrypted$= FNencrypt(plaintext$, keyword$)
REM PRINTING OUTPUTS
PRINT "Key = " keyword$
PRINT "Plaintext = " plaintext$
PRINT "Encrypted = " encrypted$
PRINT "Decrypted = " FNdecrypt(encrypted$, keyword$)
END
DEF FNencrypt(plain$, keyword$)
LOCAL i%, offset%, Ascii%, output$
FOR i% = 1 TO LEN(plain$)
Ascii% = ASCMID$(plain$, i%)
IF Ascii% >= 65 IF Ascii% <= 90 THEN
output$ += CHR$((66 + (Ascii% + ASCMID$(keyword$, offset%+1)) MOD 26))
ENDIF
offset% = (offset% + 1) MOD LEN(keyword$)
NEXT
= output$
DEF FNdecrypt(encrypted$, keyword$)
LOCAL i%, offset%, n%, o$
FOR i% = 1 TO LEN(encrypted$)
n% = ASCMID$(encrypted$, i%)
o$ += CHR$(64 + (n% + 26 - ASCMID$(keyword$, offset%+1)) MOD 26)
offset% = (offset% + 1) MOD LEN(keyword$)
NEXT
= output$
You can always convert from upper to lowercase and the Stringlib library contains a function for doing this.
First import stringlib at the top of your program:
import #lib$+"stringlib"
then convert strings using:
plaintext$ = fn_lower(plaintext$)
I am reading a manual on sending commands via serial to a device as shown:
Assume that my equipment address is 000. I would send a command like:
">000P**cr".
what would **, my checksum be? According to the manual, I need the last two digits of the total char code of "000P".
Isn't that just the hex value of "P"? I can't seem to understand this.
Yes it's a bit confused, but my guess would be:
total = ascii('0') + ascii('0') + ascii('0') + ascii('P')
total = 48 + 48 + 48 + 80
total = 224
cksum = last2digits(total) = 24
If it does not work as is, maybe try in hexadecimal notation:
hex(total) = E0
hex_cksum = last2digits(hex(total)) = E0
Edit: This was originally on programmers.stackexchange.com, but since I already had some code I thought it might be better here.
I am trying to generate a random math problem/equation. After researching the best (and only) thing I found was this question: Generating random math expression. Because I needed all 4 operations, and the app that I will be using this in is targeted at kids, I wanted to be able to insure that all numbers would be positive, division would come out even, etc, I decided to use a tree.
I have the tree working, and it can generate an equation as well as evaluate it to a number. The problem is that I am having trouble getting it to only use parentheses when needed. I have tried several solutions, that primaraly involve:
Seeing if this node is on the right of the parent node
Seeing if the node is more/less important then it's parent node (* > +, etc).
Seeing if the node & it's parent are of the same type, and if so if the order matters for that operation.
Not that it matters, I am using Swift, and here is what I have so far:
func generateString(parent_node:TreeNode) -> String {
if(self.is_num){
return self.equation!
}
var equation = self.equation!
var left_equation : String = self.left_node!.generateString(self)
var right_equation : String = self.right_node!.generateString(self)
// Conditions for ()s
var needs_parentheses = false
needs_parentheses = parent_node.importance > self.importance
needs_parentheses = (
needs_parentheses
||
(
parent_node.right_node?.isEqualToNode(self)
&&
parent_node.importance <= self.importance
&&
(
parent_node.type != self.type
&&
( parent_node.order_matters != true || self.order_matters != true )
)
)
)
needs_parentheses = (
needs_parentheses
&&
(
!(
self.importance > parent_node.importance
)
)
)
if (needs_parentheses) {
equation = "(\(equation))"
}
return equation.stringByReplacingOccurrencesOfString("a", withString: left_equation).stringByReplacingOccurrencesOfString("b", withString: right_equation)
}
I have not been able to get this to work, and have been banging my head against the wall about it.
The only thing I could find on the subject of removing parentheses is this: How to get rid of unnecessary parentheses in mathematical expression, and I could not figure out how to apply it to my use case. Also, I found this, and I might try to build a parser (using PEGKit), but I wanted to know if anybody had a good idea to either determine where parentheses need to go, or put them everywhere and remove the useless ones.
EDIT: Just to be clear, I don't need someone to write this for me, I am just looking for what it needs to do.
EDIT 2: Since this app will be targeted at kids, I would like to use the least amount of parentheses possible, while still having the equation come out right. Also, the above algorithm does not put enough parentheses.
I have coded a python 3 pgrm that does NOT use a tree, but does successfully generate valid equations with integer solutions and it uses all four operations plus parenthetical expressions. My target audience is 5th graders practicing order of operations (PEMDAS).
872 = 26 * 20 + (3 + 8) * 16 * 2
251 = 256 - 3 - 6 + 8 / 2
367 = 38 * 2 + (20 + 15) + 16 ** 2
260 = 28 * 10 - 18 - 4 / 2
5000 = 40 * 20 / 4 * 5 ** 2
211 = 192 - 10 / 2 / 1 + 24
1519 = 92 * 16 + 6 + 25 + 16
If the python of any interest to you ... I am working on translating the python into JavaScript and make a web page available for students and teachers.