I have been investigating various collision resolution techniques for hashtables implemented via open addressing. However, all the collision resolution methods I have investigated so far (linear probing, quadratic probing, double hashing) have the pitfall that there exists a probing sequence that produces a cycle whose length is less than the size of the table. This becomes problematic when you're trying to insert an element with the open addressing scheme because there are free buckets to insert an entry but they might not be reachable if they're not part of the cycle.
For example, if we're using linear probing on a table of size 12 with the linear function: H(k, i) = (h(k) + 4*i) mod 12 then a cycle would occur if a particular key hashes to 8 and all the slots 0, 4, and 8 are already filled:
H(k, 0) = 8 + 0 mod 12 = 8
H(k, 1) = 8 + 4 mod 12 = 0
H(k, 2) = 8 + 8 mod 12 = 4
H(k, 3) = 8 + 12 mod 12 = 8
H(k, 4) = 8 + 16 mod 12 = 0
H(k, 5) = 8 + 20 mod 12 = 4
H(k, 6) = 8 + 24 mod 12 = 8
...
Similar cycles can also be found with quadratic and double hashing if the probing sequence is bad, so my question then is how are cycles handled? Or do we always pick hash functions/special table sizes which do not permit cycles which are too short?
Related
------ Context ------
i am completely new to swift and metal(kit?).
Problem is: i need to realize a parallel prefixSum(to be specific: the Blelloch scan) calculation in the next 2 weeks.
Due to me not having a NVIDIA GPU but an M1 Macbook, i chose metal, so please be patient with me..
After watching this video (the code is in the video-description), I now try to modify the code so that the prefixSum is calculated instead of just adding two arrays at index i.
------ Question ------
In the first step,I only want to pass an integer myVal and just add up the three values.
Unfortunately, the output for the first 3 values looks like this(note the last digit, which has the expected value but without the additional myVal):
3 + 5 + 10 = 42949672978
2 + 6 + 10 = 42949672978
8 + 3 + 10 = 42949672981
instead of this:
3 + 5 + 10 = 18
2 + 6 + 10 = 18
8 + 3 + 10 = 21
The main.swift code is in this pastebin. The compute.metal can be found here
If I have this matrix:
A:
X Y Z
1 1 2
0 3 4
0 5 6
2 7 8
7 9 10
8 11 12
3 13 14
12 14 16
15 17 18
How could I create new matrix B, C, D and E which contains:
B:
0 3 4
0 5 6
C:
X Y Z
1 1 2
2 7 8
3 13 14
D:
7 9 10
8 11 12
E:
12 14 16
15 17 18
The idea is to construct a loop asking if 0<A<1 else 1<A<5 else 6<A<10 else 11<A<15. and create new matrix from that condition. Any idea about how to store the results of the loop?
I suggest you an approach that uses the discretize function in order to group the matrix rows into different categories based on their range. Here is the full implementation:
A = [
1 1 2;
0 3 4;
0 5 6;
2 7 8;
7 9 10;
8 11 12;
3 13 14;
12 14 16;
15 17 18
];
A_range = [0 1 5 10 15];
bin_idx = discretize(A(:,1),A_range);
A_split = arrayfun(#(bin) A(bin_idx == bin,:),1:(numel(A_range) - 1),'UniformOutput',false);
celldisp(A_split);
Since you want to consider 5 different ranges based on the first column values, the arguments passed to discretize must be the first matrix column and a vector containing the group limits (first number inclusive left, second number exclusive right, second number inclusive left, third number exclusive right, and so on...). Since your ranges are a little bit messed up, feel free to adjust them to respect the correct output. The latter is returned in the form of a cell array of double matrices in which every element contains the rows belonging to a distinct group:
A_split{1} =
0 3 4
0 5 6
A_split{2} =
1 1 2
2 7 8
3 13 14
A_split{3} =
7 9 10
8 11 12
A_split{4} =
12 14 16
15 17 18
Instead of using a loop, use logical indexing to achieve what you want. Use the first column of A and check for the ranges that you want to look for, then use this to subset into the final matrix A to get what you want.
For example, to create the matrix C, find all locations in the first column of A that are between 1 and 5, then subset the matrix along the rows using these locations:
m = A(:,1) >= 1 & A(:,1) <= 5;
C = A(m,:);
You can repeat this in a similar way for the rest of the matrices you want to create.
I use factor(n) in matlab but it don't work for numbers like ((10^11-1)/11) - what can I do? My source is here.
m=input('Enter your number: ');
i=0;
while (i<m)
if(isprime(i))
% sum=factor((10^(i-1)-1));
sum=factor((10^(i-1)-1)/i);
disp(sum);
end
i =i+1;
end
but for large n it returns errors!!
>> FactorGen
Enter your number: 45
3 3
3 3 11
3 3 11 101
3 3 3 7 11 13 37
3 3 11 41 271 9091
3 3 3 7 11 13 37 101 9901
Error using factor (line 26) When n is single or double, its maximum
allowed value is FLINTMAX.
Error in FactorGen (line 7) sum=factor((10^(i-1)-1));
I want the function factor((10^(i-1)-1)) to work for same number. How can I solve my problem?
I think this can be partially alleviated by converting your large number into uint64 format. For R2014b maximum integer that can be handled is:
n = intmax('uint64')
n = 1.8447e+19
While the maximum double that can be handled is:
n = flintmax('double')
n = 9.0072e+15
This can be verified by simple example. Let's use factor on the number larger than flintmax. First, try double:
factor(10^16)
Error using factor (line 26)
When n is single or double, its maximum allowed value is FLINTMAX.
Now, we try uint64:
factor(uint64(10^16))
ans = 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
So, using factor(uint64(x)) increases your limit by several orders. Indeed, there's no sense in using double format for factor function.
By the way, since using factor on large numbers may freeze your system, I suggest using this function:
function f = Factorize(n)
i = 0;
while mod(n,2) == 0
i = i + 1;
n = n/2;
f(i) = 2;
disp([2,n])
end
q = round(sqrt(double(n)));
q = q + 1 - mod(q,2);
for j = 3:2:q
while mod(n,j) == 0
i = i + 1;
f(i) = j;
n = n/j;
end
end
if n > 2;
i = i + 1;
f(i) = n;
end
It is much faster for large numbers and does not overload the system at large n
Hope that helps
I have got a nx3 adjacency matrix that contains nodes in the first two dimension and the correspondant weight in the third dimension. I want to filter the matrix for specific thresholds (for nodes indexing). For example, I want to keep the adjacency matrix for nodes smaller than 10.000, 20.000, etc. Which is the most efficient way to do so in matlab? I tried to do the following, find the index which correspond to nodes:
counter = 1;
for i=1: size(graph4, 1)
if (graph4(i,1) >30000) | (graph4(i,2) >30000)
bucket(counter) = i;
counter=counter+1;
end
end
Suppose the adjacency matrix is A as given below:
A =
8 1 6
3 5 7
4 9 2
11 4 9
6 8 10
7 12 5
17 10 15
12 14 16
13 18 11
If you want both column 1 and column 2 to be less than a value, you can do:
value = 10;
T = A(A(:,1) < value & A(:,2) < value, :)
T =
8 1 6
3 5 7
4 9 2
6 8 10
The following line seems to give the same results as your sample code (but it doesn't seem like it fits your description.
value = 10000;
bucket = find((A(:,1)>value) | A(:,2)>value)
I guess you made a mistake and want to increment the counter above the bucket-line and initialize it as counter = 0 before the loop? As it is now, it will be one more than the number of elements in the bucket-list.
I'm working on a program that converts between number bases. For example Octal is 8, decimal is 10. Letters A to Z could be considered as base 26.
I want to convert a number like "A" into 0, Z into 25, "AA" into 27 and "BA" into 53.
Before I start coding I'm doing it on paper so I understand the process. To start out I'm trying to convert 533 to base 26.
What algorithm is best for doing this?
You need to assign a "digit" to each letter, like:
A = 0 N = 13
B = 1 O = 14
C = 2 P = 15
D = 3 Q = 16
E = 4 R = 17
F = 5 S = 18
G = 6 T = 19
H = 7 U = 20
I = 8 V = 21
J = 9 W = 22
K = 10 X = 23
L = 11 Y = 24
M = 12 Z = 25
Then, your {20,13} becomes UN.
Converting back is UN -> {20,13} -> (20 * 26 + 13) -> 52.
By way of further example, let's try the number 10163, just plucked out of the air at random.
Divide that by 26 until you get a number less than 26 (i.e., twice), and you get 15 with a fractional part of 0.03402366.
Multiply that by 26 and you get 0 with a fractional part of 0.88461516.
Multiply that by 26 and you get 23 (actually 22.99999416 on my calculator but, since the initial division was only two steps, we stop here - the very slight inaccuracy is due to the fact that the floating point numbers are being rounded).
So the "digits" are {15,0,23} which is the "number" PAX. Wow, what a coincidence?
To convert PAX back into decimal, its
P * 262 + A * 261 + X * 260
or
(15 * 676) + (0 * 26) + 23
= 10140 + 0 + 23
= 10163
Let's take a step back for a second, and look at decimal.
What does a number like "147" mean? Or rather, what do the characters '1', '4' and '7', when arranged like that, indicate?
There are ten digits in decimal, and after that, we add another digit to the left of the first, and so on as our number increases. So after "9" = 9*1, we get "10" = 1*10 + 0*1. So "147" is 1*10^2 + 4*10 + 7*1 = 147. Similarly, we can go backwards - 147/10^2 = 1, which maps to the character '1'. (147 % 10^2) / 10 = 4, which maps to the character '4'. And 147 % 10 = 7, which maps to the character '7'.
This works works for any base N - if we get the number 0, that maps to the first character in our set. The number 1 maps to the second character, and so on until the number N-1 maps to the last character in our set of digits.
You convert 20 and 13 to the symbols that represent 20 and 13 in your base 26 notation. It sounds like you are using the letters of the alphabet so, that would be UN (where A is 0 and Z is 25).
What language are you writing this in? If you're doing this in Perl you can use the CPAN module Math::Fleximal that I wrote many years ago while I was bored. If you're using a language with infinite precision integers, then life becomes much easier. All you have to do is take characters, convert them into an array of integers, then do the calculation to turn that into a number.