There is one confusion I've about load factor. Some sources say that it is just the number of keys in hash table divided by total number of slots which is same as expected chain length for each slot. But that is only in simple uniform hashing right?
Suppose hash table T has n elements and we expand T into T1 by redistributing elements in slot T[0] by rehashing them using h'(k) = k mod 2m. The hash function of T1 is h(k) = k mod 2m if h(k) < 1 and k mod m if h(k) >= 1. Many sources say that we "Expand and rehash to maintain the load factor (does this imply expected chain length is still same?) Since this is not simple uniform hashing, I think the probability that any key k enters a slot is (1/4 + 1/2(m-1)).
For a key k (randomly selected), h(k) is first evaluated (there are 50-50 chances whether it is less than 1 or greater than or equal to 1) and then if it's less than 1, key k has JUST two ways - slot 0 or slot m. Hence, probability 1/4 (1/2 * 1/2) But if it is greater than or equal to 1, it has m-1 slots and could enter any and hence probability (1/2 * 1/m-1). So expected chain length would now be n/4 + n/2(m-1). Am I on right track?
The calculation for linear hashing should be the same as for "non-linear" hashing. With a certain initial number of buckets, uniform distribution of hash values would result in uniform placement. With enough expansions to double the size of the table, each of those values would be randomly split over the larger space via the incremental re-hashing, and new values would also have been distributed over the larger space. Incrementally, each point is equally likely to be at (initial bucket position) and (2x initial bucket position) as the table expands to that length.
There is a paper here which goes into detail about the chain length calculation under different circumstances (not just the average), specifically for linear hashing.
Related
we have a hash table with size 16, using double hashing method.
h1(x) = k mod 16
h2(x) = 2*(k mod 8)
I know that h2 hash function is bad, probably because mod 8 and times 2, but I don't know how to explain it. is there any explanation like "h2 hash function should mod prime or it will cause ____ problem "
It is bad because it increases the number of collisions.
The (mod 8) means that you are only looking for 8 pigeonholes in your 16-pigeonhole table.
Multiplying it by 2 just spreads those 8 pigeonholes out so that you don’t have to search too many slots past the hashed index to find an empty hole...
You should always compute modulo the size of your table.
h(x) ::= x (mod N) // where N is the table size
The purpose of making the table size a prime number just has to do with how powers of two are very common in computer science. If your data is random, then the size of the table doesn’t matter.
— As long as it is big enough for your expected load factor. A 16-element table is very small. You shouldn’t expect to store more than 6-12 random values in your table without a high-probability of collisions.
A very good linked thread is What is a good Hash Function?, which is totally worth a read just for the links to further reading alone.
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 want to determine the space complexity of the go to example of a simple streaming algorithm.
If you get a permutation of n-1 different numbers and have to detect the one missing number, you calculate the sum of all numbers 1 to n using the formula n (n + 1) / 2 and then you subtract each incoming number. The result is your missing number. I found a german wikipedia article stating that the space complexity of this algorithm is O(log n). (https://de.wikipedia.org/wiki/Datenstromalgorithmus)
What I do not understand is: The amount of bits needed to store a number n is log2(n). ok.. but I do have to calculate the sum, tough. So n (n + 1) / 2 is larger than n and therefore needs more space than just log (n) right?
Can someone help me with this? Thanks in advance!
If integer A in binary coding requires Na bits and integer B requires Nb bits then A*B requires no more than Na+Nb bits (not Na * Nb). So, expression n(n+1)/2 requires no more than log2(n) + log2(n+1) = O(2log2(n)) = O(log2(n)) bits.
Even more, you may raise n to any fixed power i and it still will use O(log2(n)) space. n itself, n10, n500, n10000000 all require O(log(n)) bits of storage.
I am reading about Universal hashing on integers. The prerequisite and mandatory precondition seems to be that we choose a prime number p greater than the set of all possible keys.
I am not clear on this point.
If our set of keys are of type int then this means that the prime number needs to be of the next bigger data type e.g. long.
But eventually whatever we get as the hash would need to be down-casted to an int to index the hash table. Doesn't this down-casting affect the quality of the Universal Hashing (I am referring to the distribution of the keys over the buckets) somehow?
If our set of keys are integers then this means that the prime number
needs to be of the next bigger data type e.g. long.
That is not a problem. Sometimes it is necessary otherwise the hash family cannot be universal. See below for more information.
But eventually whatever we get as the hash would need to be
down-casted to an int to index the hash table.
Doesn't this down-casting affect the quality of the Universal Hashing
(I am referring to the distribution of the keys over the buckets)
somehow?
The answer is no. I will try to explain.
Whether p has another data type or not is not important for the hash family to be universal. Important is that p is equal or larger than u (the maximum integer of the universe of integers). It is important that p is big enough (i.e. >= u).
A hash family is universal when the collision probability is equal or
smaller than 1/m.
So the idea is to hold that constraint.
The value of p, in theory, can be as big as a long or more. It just needs to be an integer and prime.
u is the size of the domain/universe (or the number of keys). Given the universe U = {0, ..., u-1}, u denotes the size |U|.
m is the number of bins or buckets
p is a prime which must be equal or greater than n
the hash family is defined as H = {h(a,b)(x)} with h(a,b)(x) = ((a * x + b) mod p) mod m. Note that a and b are randomly chosen integers (from all possible integers, so theoretically can be larger than p) modulo a prime p (which can make them either smaller or larger than m, the number of bins/buckets); but here too the data type (domain of values does not matter). See Hashing integers on Wikipedia for notation.
Follow the proof on Wikipedia and you conclude that the collision probability is _p/m_ * 1/(p-1) (the underscores mean to truncate the decimals). For p >> m (p considerably bigger than m) the probability tends to 1/m (but this does not mean that the probability would be better the larger p is).
In other terms answering your question: p being a bigger data type is not a problem here and can be even required. p has to be equal or greater than u and a and b have to be randomly chosen integers modulo p, no matter the number of buckets m. With these constraints you can construct a universal hash family.
Maybe a mathematical example could help
Let U be the universe of integers that correspond to unsigned char (in C for example). Then U = {0, ..., 255}
Let p be (next possible) prime equal or greater than 256. Note that p can be any of these types (short, int, long be it signed or unsigned). The point is that the data type does not play a role (In programming the type mainly denotes a domain of values.). Whether 257 is short, int or long doesn't really matter here for the sake of correctness of the mathematical proof. Also we could have chosen a larger p (i.e. a bigger data type); this does not change the proof's correctness.
The next possible prime number would be 257.
We say we have 25 buckets, i.e. m = 25. This means a hash family would be universal if the collision probability is equal or less than 1/25, i.e. approximately 0.04.
Put in the values for _p/m_ * 1/(p-1): _257/25_ * 1/256 = 10/256 = 0.0390625 which is smaller than 0.04. It is a universal hash family with the chosen parameters.
We could have chosen m = u = 256 buckets. Then we would have a collision probability of 0.003891050584, which is smaller than 1/256 = 0,00390625. Hash family is still universal.
Let's try with m being bigger than p, e.g. m = 300. Collision probability is 0, which is smaller than 1/300 ~= 0.003333333333. Trivial, we had more buckets than keys. Still universal, no collisions.
Implementation detail example
We have the following:
x of type int (an element of |U|)
a, b, p of type long
m we'll see later in the example
Choose p so that it is bigger than the max u (element of |U|), p is of type long.
Choose a and b (modulo p) randomly. They are of type long, but always < p.
For an x (of type int from U) calculate ((a*x+b) mod p). a*x is of type long, (a*x+b) is also of type long and so ((a*x+b) mod p is also of type long. Note that ((a*x+b) mod p)'s result is < p. Let's denote that result h_a_b(x).
h_a_b(x) is now taken modulo m, which means that at this step it depends on the data type of m whether there will be downcasting or not. However, it does not really matter. h_a_b(x) is < m, because we take it modulo m. Hence the value of h_a_b(x) modulo m fits into m's data type. In case it has to be downcasted there won't be a loss of value. And so you have mapped a key to a bin/bucket.
Given a set of 100 different strings of equal length, how can you quantify the probability that a SHA1 digest collision for the strings is unlikely... ?
Are the 160 bit hash values generated
by SHA-1 large enough to ensure the
fingerprint of every block is unique?
Assuming random hash values with a
uniform distribution, a collection of
n different data blocks and a hash
function that generates b bits, the
probability p that there will be one
or more collisions is bounded by the
number of pairs of blocks multiplied
by the probability that a given pair
will collide.
(source : http://bitcache.org/faq/hash-collision-probabilities)
Well, the probability of a collision would be:
1 - ((2^160 - 1) / 2^160) * ((2^160 - 2) / 2^160) * ... * ((2^160 - 99) / 2^160)
Think of the probability of a collision of 2 items in a space of 10. The first item is unique with probability 100%. The second is unique with probability 9/10. So the probability of both being unique is 100% * 90%, and the probability of a collision is:
1 - (100% * 90%), or 1 - ((10 - 0) / 10) * ((10 - 1) / 10), or 1 - ((10 - 1) / 10)
It's pretty unlikely. You'd have to have many more strings for it to be a remote possibility.
Take a look at the table on this page on Wikipedia; just interpolate between the rows for 128 bits and 256 bits.
That's Birthday Problem - the article provides nice approximations that make it quite easy to estimate the probability. Actual probability will be very very very low - see this question for an example.