a ~= b*c;
It can't be boolean logic, so what's that?
~ gives the bitwise complement of a single integer....
So it could be that here b and c multiply and then bits of the number get reversed.
Related
I have the following question
Implement a circuit which shifts a 32-bit vector by two to the left and fills empty spots with zeros. Use only concatenation operator.
I'm just learning SystemVerilog, and I can't understand how I can use a concatenation operator for this. I would have just done 'assign y = a << 2' without concatenation ( with a declared as 32bit vector and y a 32 bit output), but I don't understand how can concatenation have anything to do with it.
Concatenation allows you to concatenate multiple parts of a vector with constants. As such, the left shift operator for a << 2 can be implemented as the following:
logic [7:0] x, y;
always_comb
y[7:0] = {a[5:0], 2'b0};
Just a concatenation of the lower 6 bits of a with 2 bits of 0 on the right hand side. Same what the left shift would do, but without any overflow.
For the hash function : h(k) = k mod m;
I understand that m=2^n will always give the last n LSB digits. I also understand that m=2^p-1 when K is a string converted to integers using radix 2^p will give same hash value for every permutation of characters in K. But why exactly "a prime not too close to an exact power of 2" is a good choice? What if I choose 2^p - 2 or 2^p-3? Why are these choices considered bad?
Following is the text from CLRS:
"A prime not too close to an exact power of 2 is often a good choice for m. For
example, suppose we wish to allocate a hash table, with collisions resolved by
chaining, to hold roughly n D 2000 character strings, where a character has 8 bits.
We don’t mind examining an average of 3 elements in an unsuccessful search, and
so we allocate a hash table of size m D 701. We could choose m D 701 because
it is a prime near 2000=3 but not near any power of 2."
Suppose we work with radix 2p.
2p-1 case:
Why that is a bad idea to use 2p-1? Let us see,
k = ∑ai2ip
and if we divide by 2p-1 we just get
k = ∑ai2ip = ∑ai mod 2p-1
so, as addition is commutative, we can permute digits and get the same result.
2p-b case:
Quote from CLRS:
A prime not too close to an exact power of 2 is often a good choice for m.
k = ∑ai2ip = ∑aibi mod 2p-b
So changing least significant digit by one will change hash by one. Changing second least significant bit by one will change hash by two. To really change hash we would need to change digits with bigger significance. So, in case of small b we face problem similar to the case then m is power of 2, namely we depend on distribution of least significant digits.
I am implementing cooley-tuckey fft(raddix - 2 DIF / DIT) algorithm in matlab.In that for the bit reversing i want to have reverse of an binary number. so can anyone suggest how can I get the reverse of a binary number(like 100111 -> 111001). One who have worked on fft implementation can help me with the algorithm also.
Topic: How to do bit reversal in Matlab? .
If you're using double precision floating point ('double') numbers
which are integers, you can do this:
dr = bin2dec(fliplr(dec2bin(d,n))); % Bits in dr are in reverse order
where n is the number of bits to be reversed and where 0 <= d < 2^n.
You will experience no precision problems at all as long as the
integers are no more than 52 bits long.
And
Re: How to do bit reversal in Matlab?
How large will the numbers be that you need to reverse? May I ask what
is the purpose of it? Maybe there is a more efficient way to solve the
whole problem. If the numbers are large you can just store the bits as
a string. To reverse it just read the string backwards! Or use
fliplr().
(There may be better places to ask).
If it were VHDL I'd suggest an alias with 'REVERSE'RANGE.
Taken from the help section;
Y = swapbytes(X) reverses the byte ordering of each element in array X, converting little-endian values to big-endian (and vice versa). The input array must contain all full, noncomplex, numeric elements.
I want a column vector that contains only the character A, that is,
A
A
A
A
like this. So i have tried
'A'*ones(4,1)
But in place of A, it takes on value 65. How can i get A ?
You can do it this way:
repmat('A',4,1)
Or use your approach but include char to convert back to string after the multiplication:
char('A'*ones(4,1))
Multiplication of chars with doubles gives doubles; the charis cast to double, using the corresponding ASCII value.
Just cast back to char:
char('A'*ones(4,1))
but probably Luis' answer is faster ;)
So I'm completely new to MATLAB and I'm trying to understand colon notation within mathematical operations. So, in this book I found this statement:
w(1:5)=j(1:5) + k(1:5);
I do not understand what it really does. I know that w(1:5) is pretty much iterating through the w array from index 1 through 5, but in the statement above, shouldn't all indexes of w be equal to j(5) + k(5) in the end? Or am I completely wrong on how this works? It'd be awesome if someone posted the equivalent in Java to that up there. Thanks in advance :-)
I am pretty sure this means
"The first 5 elements of w shall be the first 5 elements of j + the first 5 elements of k" (I am not sure if matlab arrays start with 0 or 1 though)
So:
w1 = j1+k1
w2 = j2+k2
w3 = j3+k3
w4 = j4+k4
w5 = j5+k5
Think "Vector addition" here.
w(1:5)=j(1:5) + k(1:5);
is the same that:
for i=1:5
w(i)=j(i)+k(i);
end
MATLAB uses vectors and matrices, and is heavily optimized to handle operations on them efficiently.
The expression w(1:5) means a vector consisting of the first 5 elements of w; the expression you posted adds two 5 element vectors (the first 5 elements of j and k) and assigns the result to the first five elements of w.
I think your problem comes from the way how do you call this statement. It is not an iteration, but rather simple assignment. Now we only need to understand what was assigned to what.
I will assume j,k, w are all vectors 1 by N.
j(1:5) - means elements from 1 to 5 of the vector j
j(1:5) + k(1:5) - will result in elementwise sum of both operands
w(1:5) = ... - will assign the result again elementwise to w
Writing your code using colon notation makes it less verbose and more efficient. So it is highly recommended to do so. Also, colon notation is the basic and very powerful feature of MATLAB. Make sure you understand it well, before you move on. MATLAB is very well documented so you can read on this topic here.