wikipedia
standard manual
when calculating the SHA-1, we need a sequence of logical functions, f0, f1,…, f79,
I noticed that the function definitions in Wikipedia and the standard manual are different.
oddly, when I chose the ones in the standard manual, the SHA-1 result went wrong.
I used online sha-1 calculators and found that everyone uses the functions written in wikipedia.
Why?
Here are the truth tables for both versions of 'choose' (0..19) and 'majority' (40..59) (for 'parity' 20..39 and 60..79 both sources use xor). Please identify the rows for which the ior result is different from the xor result; those are the cases for which the two formulas produce different results.
x
y
z
x^y
¬x^z
ior
xor
0
0
0
0
0
0
0
0
0
1
0
1
1
1
0
1
0
0
0
0
0
0
1
1
0
1
1
1
1
0
0
0
0
0
0
1
0
1
0
0
0
0
1
1
0
1
0
1
1
1
1
1
1
0
1
1
x
y
z
x^y
x^z
y^z
ior
xor
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
1
0
0
1
1
1
1
0
0
0
0
0
0
0
1
0
1
0
1
0
1
1
1
1
0
1
0
0
1
1
1
1
1
1
1
1
1
1
Hint: there are no differences. The results are always the same, and it doesn't matter which formula you use, as long as you do it correctly you get the correct result.
In fact, on checking my copy of 180-4 this is even stated in section 4.1, immediately above the section you quoted:
... Each of the algorithms [for SHA-1, SHA-256 group, and SHA-512 group] include Ch(x, y, z)
and Maj(x, y, z) functions; the exclusive-OR operation (⊕ ) in these functions may be replaced
by a bitwise OR operation (∨) and produce identical results.
If something you did 'went wrong', it's because you did something wrong, but nobody here is psychic so we have absolutely no idea at all what you did wrong.
There is an incomplete graph (e.g. including 5 vertices). The adjacency matrix "a" is available. I want to define the set which includes all edges but exclude any other pair of vertices. That is, the pair of vertices belongs to the set of edges iff the element in matrix "a" is positive.
The last line of following code does not work!
sets i "Set of vertices" /1*5/ ;
alias(i,j);
set a(i,j) "Adjacency matrix" ;
Table a(i,j)
1 2 3 4 5
1 0 1 0 1 1
2 1 0 1 0 0
3 0 1 0 0 0
4 1 0 0 0 1
5 1 0 0 1 0;
Set edges(i,j);
edges(i,j) = a(i,j)$(a(i,j)>0);
If you want to have edge , you must define a set and parameter like this :
sets i "Set of vertices" /1*5/ ;
alias(i,j);
set a(i,j) "Adjacency matrix" ;
Table a(i,j)
1 2 3 4 5
1 0 1 0 1 1
2 1 0 1 0 0
3 0 1 0 0 0
4 1 0 0 0 1
5 1 0 0 1 0;
Set edges(i,j);
edges(i,j) $ a(i,j) =yes;
You can simplify your last line to
edges(i,j) = a(i,j);
This automatically acts as if you wrote something like $(a<>0). However, since you defined your symbol a as set already and not as parameter, I think you actually do not have to do anything. A just is what you are looking for. Just do
display a;
and look at the result in the lst file.
Why does a double of 65555 converted to byte produce a result of 19 in Kotlin?
That's because of a numerical conversion from a wider type to a type of a smaller size. The Double (IEEE 754 double precision number) has its integral part factored to the powers of two as 65555 = 217 + 24 + 22 + 20 = 65536 + 16 + 2 + 1, which is stored in binary form as (higher bits to lower):
... 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1
When this number is converted to Byte, only its lowest 8 bits are retained:
... _ _ _ _ _ _ _ _ _ _ 0 0 0 1 0 0 1 1
And that results into 24 + 22 + 20 = 16 + 2 + 1 = 19.
Because when you convert 65555 (or 65555.0) to a binary representation it takes more than one byte. So calling .toByte() takes the lowest one, which is 19.
65555 -> Binary == 1 00000000 00010011
^ ^^^^^^^^ ^^^^^^^^
1 0 19
Double 65555.0 gets converted to integer 65555 which is 0x10013. Conversion to Byte takes lower byte which is 0x13 (19 decimal).
Is somebody able to explain what the below query is actually doing?
SELECT (Convert(int, 33558529) & 4096),((Convert(int, 33558529) & 1048576))
FROM dbo.example
Why does the first part return 4096 and the second part returns 0?
& performs a bitwise logical AND operation between two integer values. See the doc.
Here are the integer values converted to binary :
33558529 = 10000000000001000000000001
4096 = 1000000000000 1 bit match hence 1000000000000 or 4096
1048576 = 100000000000000000000 0 bit match hence 0
The & sign in T-SQL is the bitwise AND. It is used for bitwise comparison on numbers.
Why does the first part return 4096 and the second part returns 0?
Because the big number (33558529) includes the 4096 bit, but does not contain the 1048576 bit.
I find it easier to understand when you use smaller numbers, and write it out in binary. Suppose the big number you're checking is actually 9, written as binary 9 is
8 4 2 1
=======
1 0 0 1 <-- 9
If we were to perform bitwise AND logic on the above with the number 8 we would get
8 4 2 1
=======
1 0 0 1 <-- 9
1 0 0 0 <-- 8
-------
1 0 0 0 < -- result of ANDing 9 & 8 = 8
If we did the same exercise but with 2
8 4 2 1
=======
1 0 0 1 <-- 9
0 0 1 0 <-- 2
-------
0 0 0 0 <-- result of ANDing 9 & 2 = 0
& is the bitwise logical and operator - It performs the operation on
2 integer values.
Please refer :
Ampersand (&) operator in a SQL Server WHERE Clause
SELECT
(Convert(int, 33558529) & 4096),((Convert(int, 33558529) & 1048576))
FROM dbo.example
Its compare the bitwise data and gives the result
Check this link and compare the bit value
33558529 - 00000010 00000000 00010000 00000001
4096 - 00000000 00000000 00010000 00000000 means 1
1048576 - 00000000 00000000 00000000 00000000
Same you check the logic behind.
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Write code to determine if a number is divisible by 3. The input to the function is a single bit, 0 or 1, and the output should be 1 if the number received so far is the binary representation of a number divisible by 3, otherwise zero.
Examples:
input "0": (0) output 1
inputs "1,0,0": (4) output 0
inputs "1,1,0,0": (6) output 1
This is based on an interview question. I ask for a drawing of logic gates but since this is stackoverflow I'll accept any coding language. Bonus points for a hardware implementation (verilog etc).
Part a (easy): First input is the MSB.
Part b (a little harder): First input is the LSB.
Part c (difficult): Which one is faster and smaller, (a) or (b)? (Not theoretically in the Big-O sense, but practically faster/smaller.) Now take the slower/bigger one and make it as fast/small as the faster/smaller one.
There's a fairly well-known trick for determining whether a number is a multiple of 11, by alternately adding and subtracting its decimal digits. If the number you get at the end is a multiple of 11, then the number you started out with is also a multiple of 11:
47278 4 - 7 + 2 - 7 + 8 = 0, multiple of 11 (47278 = 11 * 4298)
52214 5 - 2 + 2 - 1 + 4 = 8, not multiple of 11 (52214 = 11 * 4746 + 8)
We can apply the same trick to binary numbers. A binary number is a multiple of 3 if and only if the alternating sum of its bits is also a multiple of 3:
4 = 100 1 - 0 + 0 = 1, not multiple of 3
6 = 110 1 - 1 + 0 = 0, multiple of 3
78 = 1001110 1 - 0 + 0 - 1 + 1 - 1 + 0 = 0, multiple of 3
109 = 1101101 1 - 1 + 0 - 1 + 1 - 0 + 1 = 1, not multiple of 3
It makes no difference whether you start with the MSB or the LSB, so the following Python function works equally well in both cases. It takes an iterator that returns the bits one at a time. multiplier alternates between 1 and 2 instead of 1 and -1 to avoid taking the modulo of a negative number.
def divisibleBy3(iterator):
multiplier = 1
accumulator = 0
for bit in iterator:
accumulator = (accumulator + bit * multiplier) % 3
multiplier = 3 - multiplier
return accumulator == 0
Here... something new... how to check if a binary number of any length (even thousands of digits) is divisible by 3.
-->((0))<---1--->()<---0--->(1) ASCII representation of graph
From the picture.
You start in the double circle.
When you get a one or a zero, if the digit is inside the circle, then you stay in that circle. However if the digit is on a line, then you travel across the line.
Repeat step two until all digits are comsumed.
If you finally end up in the double circle then the binary number is divisible by 3.
You can also use this for generating numbers divisible by 3. And I wouldn't image it would be hard to convert this into a circuit.
1 example using the graph...
11000000000001011111111111101 is divisible by 3 (ends up in the double circle again)
Try it for yourself.
You can also do similar tricks for performing MOD 10, for when converting binary numbers into base 10 numbers. (10 circles, each doubled circled and represent the values 0 to 9 resulting from the modulo)
EDIT: This is for digits running left to right, it's not hard to modify the finite state machine to accept the reverse language though.
NOTE: In the ASCII representation of the graph () denotes a single circle and (()) denotes a double circle. In finite state machines these are called states, and the double circle is the accept state (the state that means its eventually divisible by 3)
Heh
State table for LSB:
S I S' O
0 0 0 1
0 1 1 0
1 0 2 0
1 1 0 1
2 0 1 0
2 1 2 0
Explanation: 0 is divisible by three. 0 << 1 + 0 = 0. Repeat using S = (S << 1 + I) % 3 and O = 1 if S == 0.
State table for MSB:
S I S' O
0 0 0 1
0 1 2 0
1 0 1 0
1 1 0 1
2 0 2 0
2 1 1 0
Explanation: 0 is divisible by three. 0 >> 1 + 0 = 0. Repeat using S = (S >> 1 + I) % 3 and O = 1 if S == 0.
S' is different from above, but O works the same, since S' is 0 for the same cases (00 and 11). Since O is the same in both cases, O_LSB = O_MSB, so to make MSB as short as LSB, or vice-versa, just use the shortest of both.
Here is an simple way to do it by hand.
Since 1 = 22 mod 3, we get 1 = 22n mod 3 for every positive integer.
Furthermore 2 = 22n+1 mod 3. Hence one can determine if an integer is divisible by 3 by counting the 1 bits at odd bit positions, multiply this number by 2, add the number of 1-bits at even bit posistions add them to the result and check if the result is divisible by 3.
Example: 5710=1110012.
There are 2 bits at odd positions, and 2 bits at even positions. 2*2 + 2 = 6 is divisible by 3. Hence 57 is divisible by 3.
Here is also a thought towards solving question c). If one inverts the bit order of a binary integer then all the bits remain at even/odd positions or all bits change. Hence inverting the order of the bits of an integer n results is an integer that is divisible by 3 if and only if n is divisible by 3. Hence any solution for question a) works without changes for question b) and vice versa. Hmm, maybe this could help to figure out which approach is faster...
You need to do all calculations using arithmetic modulo 3. This is the way
MSB:
number=0
while(!eof)
n=input()
number=(number *2 + n) mod 3
if(number == 0)
print divisible
LSB:
number = 0;
multiplier = 1;
while(!eof)
n=input()
number = (number + multiplier * n) mod 3
multiplier = (multiplier * 2) mod 3
if(number == 0)
print divisible
This is general idea...
Now, your part is to understand why this is correct.
And yes, do homework yourself ;)
The idea is that the number can grow arbitrarily long, which means you can't use mod 3 here, since your number will grow beyond the capacity of your integer class.
The idea is to notice what happens to the number. If you're adding bits to the right, what you're actually doing is shifting left one bit and adding the new bit.
Shift-left is the same as multiplying by 2, and adding the new bit is either adding 0 or 1. Assuming we started from 0, we can do this recursively based on the modulo-3 of the last number.
last | input || next | example
------------------------------------
0 | 0 || 0 | 0 * 2 + 0 = 0
0 | 1 || 1 | 0 * 2 + 1 = 1
1 | 0 || 2 | 1 * 2 + 0 = 2
1 | 1 || 0 | 1 * 2 + 1 = 0 (= 3 mod 3)
2 | 0 || 1 | 2 * 2 + 0 = 1 (= 4 mod 3)
2 | 1 || 2 | 2 * 2 + 1 = 2 (= 5 mod 3)
Now let's see what happens when you add a bit to the left. First, notice that:
22n mod 3 = 1
and
22n+1 mod 3 = 2
So now we have to either add 1 or 2 to the mod based on if the current iteration is odd or even.
last | n is even? | input || next | example
-------------------------------------------
d/c | don't care | 0 || last | last + 0*2^n = last
0 | yes | 1 || 0 | 0 + 1*2^n = 1 (= 2^n mod 3)
0 | no | 1 || 0 | 0 + 1*2^n = 2 (= 2^n mod 3)
1 | yes | 1 || 0 | 1 + 1*2^n = 2
1 | no | 1 || 0 | 1 + 1*2^n = 0 (= 3 mod 3)
1 | yes | 1 || 0 | 2 + 1*2^n = 0
1 | no | 1 || 0 | 2 + 1*2^n = 1
input "0": (0) output 1
inputs "1,0,0": (4) output 0
inputs "1,1,0,0": (6) output 1
shouldn't this last input be 12, or am i misunderstanding the question?
Actually the LSB method would actually make this easier. In C:
MSB method:
/*
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_msb(char *input) {
unsigned value = 0;
char *p = input;
if (*p == '1') {
value &= 1;
}
p++;
while (*p) {
if (*p != ',') {
value <<= 1;
if (*p == '1') {
ret &= 1;
}
}
p++;
}
return (value % 3 == 0) ? 1 : 0;
}
LSB method:
/*
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_lsb(char *input) {
unsigned value = 0;
unsigned mask = 1;
char *p = input;
while (*p) {
if (*p != ',') {
if (*p == '1') {
value &= mask;
}
mask <<= 1;
}
p++;
}
return (value % 3 == 0) ? 1 : 0;
}
Personally I have a hard time believing one of these will be significantly different to the other.
I think Nathan Fellman is on the right track for part a and b (except b needs an extra piece of state: you need to keep track of if your digit position is odd or even).
I think the trick for part C is negating the last value at each step. I.e. 0 goes to 0, 1 goes to 2 and 2 goes to 1.
A number is divisible by 3 if the sum of it's digits is divisible by 3.
So you can add the digits and get the sum:
if the sum is greater or equal to 10 use the same method
if it's 3, 6, 9 then it is divisible
if the sum is different than 3, 6, 9 then it's not divisible