What is the fastest integer factorization algorithm?
I've written a program that attempts to find Amicable Pairs. This requires finding the sums of the proper divisors of numbers.
Here is my current sumOfDivisors() method:
int sumOfDivisors(int n)
{
int sum = 1;
int bound = (int) sqrt(n);
for(int i = 2; i <= 1 + bound; i++)
{
if (n % i == 0)
sum = sum + i + n / i;
}
return sum;
}
So I need to do lots of factorization and that is starting to become the real bottleneck in my application. I typed开发者_如何学Go a huge number into MAPLE and it factored it insanely fast.
What is one of the faster factorization algorithms?
Pulled directly from my answer to this other question.
The method will work, but will be slow. "How big are your numbers?" determines the method to use:
- Less than 2^16 or so: Lookup table.
- Less than 2^70 or so: Richard Brent's modification of Pollard's rho algorithm.
- Less than 10^50: Lenstra elliptic curve factorization
- Less than 10^100: Quadratic Sieve
- More than 10^100: General Number Field Sieve
Shor's Algorithm: http://en.wikipedia.org/wiki/Shor%27s_algorithm
Of course you need a quantum computer though :D
The question in the title (and the last line) seems to have little to do with the actual body of the question. If you're trying to find amicable pairs, or computing the sum of divisors for many numbers, then separately factorising each number (even with the fastest possible algorithm) is absolutely an inefficient way to do it.
The sum-of-divisors function, σ(n) = (sum of divisors of n), is a multiplicative function: for relatively prime m and n, we have σ(mn) = σ(m)σ(n), so
σ(p1k1…prkr) = [(p1k1+1-1)/(p1-1)]…[(prkr+1-1)/(pr-1)].
So you would use any simple sieve (e.g. an augmented version of the Sieve of Eratosthenes) to find the primes up to n, and, in the process, the factorisation of all numbers up to n. (For example, as you do your sieve, store the smallest prime factor of each n. Then you can later factorize any number n by iterating.) This would be faster (overall) than using any separate factorization algorithm several times.
BTW: several known lists of amicable pairs already exist (see e.g. here and the links at MathWorld) – so are you trying to extend the record, or doing it just for fun?
I would suggest starting from the same algorithm used in Maple, the Quadratic Sieve.
- Choose your odd number n to factorize,
- Choose a natural number k,
- Search all p <= k so that k^2 is not congruent to (n mod p) to obtain a factor base B = p1, p2, ..., pt,
- Starting from r > floor(n) search at least t+1 values so that y^2 = r^2 - n all have just factors in B,
- For every y1, y2, ..., y(t+1) just calculated you generate a vector v(yi) = (e1, e2, ..., et) where ei is calculated by reducing over modulo 2 the exponent pi in yi,
- Use Gaussian Elimination to find some of the vectors that added together give a null vector
- Set x as the product of ri related to yi found in the previous step and set y as p1^a * p2^b * p3^c * .. * pt^z where exponents are the half of the exponents found in the factorization of yi
- Calculate d = mcd(x-y, n), if 1 < d < n then d is a non-trivial factor of n, otherwise start from step 2 choosing a bigger k.
The problem about these algorithms is that they really imply a lot of theory in numerical calculus..
This is a paper of the Integer Factorization in Maple.
"Starting from some very simple instructions—“make integer factorization faster in Maple” — we have implemented the Quadratic Sieve factoring algorithm in a combination of Maple and C..."
http://www.cecm.sfu.ca/~pborwein/MITACS/papers/percival.pdf
Depends how big your numbers are. If you're searching for amicable pairs you're doing a lot of factorisations, so the key may not be to factor as quickly as possible, but to share as much work as possible between different calls. To speed up trial division you could look at memoization, and/or precalculating primes up to the square root of the biggest number you care about. It's quicker to get the prime factorisation, then calculate the sum of all factors from that, than it is to loop all the way up to sqrt(n) for every number.
If you're looking for really big amicable pairs, say bigger than 2^64, then on a small number of machines you can't do it by factorising every single number no matter how fast your factorisation is. The short-cuts which you're using to find candidates might help you factor them.
A more 2015 C++ version 227 lookup table implementation for 1GB memory:
#include <iostream.h> // cerr, cout, and NULL
#include <string.h> // memcpy()
#define uint unsigned __int32
uint *factors;
const uint MAX_F=134217728; // 2^27
void buildFactors(){
factors=new (nothrow) uint [(MAX_F+1)*2]; // 4 * 2 * 2^27 = 2^30 = 1GB
if(factors==NULL)return; // not able to allocate enough free memory
int i;
for(i=0;i<(MAX_F+1)*2;i++)factors[i]=0;
//Sieve of Eratosthenese
factors[1*2]=1;
factors[1*2+1]=1;
for(i=2;i*i<=MAX_F;i++){
for(;factors[i*2] && i*i<=MAX_F;i++);
factors[i*2]=1;
factors[i*2+1]=i;
for(int j=2;i*j<=MAX_F;j++){
factors[i*j*2]=i;
factors[i*j*2+1]=j;
}
}
for(;i<=MAX_F;i++){
for(;i<=MAX_F && factors[i*2];i++);
if(i>MAX_F)return;
factors[i*2]=1;
factors[i*2+1]=i;
}
}
uint * factor(uint x, int &factorCount){
if(x > MAX_F){factorCount=-1;return NULL;}
uint tmp[70], at=x; int i=0;
while(factors[at*2]>1){
tmp[i++]=factors[at*2];
cout<<"at:"<<at<<" tmp:"<<tmp[i-1]<<endl;
at=factors[at*2+1];
}
if(i==0){
cout<<"at:"<<x<<" tmp:1"<<endl;
tmp[i++]=1;
tmp[i++]=x;
}else{
cout<<"at:"<<at<<" tmp:1"<<endl;
tmp[i++]=at;
}
factorCount=i;
uint *ret=new (nothrow) uint [factorCount];
if(ret!=NULL)
memcpy(ret, tmp, sizeof(uint)*factorCount);
return ret;
}
void main(){
cout<<"Loading factors lookup table"<<endl;
buildFactors(); if(factors==NULL){cerr<<"Need 1GB block of free memory"<<endl;return;}
int size;
uint x=30030;
cout<<"\nFactoring: "<<x<<endl;
uint *f=factor(x,size);
if(size<0){cerr<<x<<" is too big to factor. Choose a number between 1 and "<<MAX_F<<endl;return;}
else if(f==NULL){cerr<<"ran out of memory trying to factor "<<x<<endl;return;}
cout<<"\nThe factors of: "<<x<<" {"<<f[0];
for(int i=1;i<size;i++)
cout<<", "<<f[i];
cout<<"}"<<endl;
delete [] f;
x=30637;
cout<<"\nFactoring: "<<x<<endl;
f=factor(x,size);
cout<<"\nThe factors of: "<<x<<" {"<<f[0];
for(int i=1;i<size;i++)
cout<<", "<<f[i];
cout<<"}"<<endl;
delete [] f;
delete [] factors;
}
Update: or sacrificing some simplicity for a bit more range just past 228
#include <iostream.h> // cerr, cout, and NULL
#include <string.h> // memcpy(), memset()
//#define dbg(A) A
#ifndef dbg
#define dbg(A)
#endif
#define uint unsigned __int32
#define uint8 unsigned __int8
#define uint16 unsigned __int16
uint * factors;
uint8 *factors08;
uint16 *factors16;
uint *factors32;
const uint LIMIT_16 = 514; // First 16-bit factor, 514 = 2*257
const uint LIMIT_32 = 131074;// First 32-bit factor, 131074 = 2*65537
const uint MAX_FACTOR = 268501119;
//const uint64 LIMIT_64 = 8,589,934,594; // First 64-bit factor, 2^33+1
const uint TABLE_SIZE = 268435456; // 2^28 => 4 * 2^28 = 2^30 = 1GB 32-bit table
const uint o08=1, o16=257 ,o32=65665; //o64=4294934465
// TableSize = 2^37 => 8 * 2^37 = 2^40 1TB 64-bit table
// => MaxFactor = 141,733,953,600
/* Layout of factors[] array
* Indicies(32-bit) i Value Size AFactorOf(i)
* ---------------- ------ ---------- ----------------
* factors[0..128] [1..513] 8-bit factors08[i-o08]
* factors[129..65408] [514..131073] 16-bit factors16[i-o16]
* factors[65409..268435455] [131074..268501119] 32-bit factors32[i-o32]
*
* Note: stopping at i*i causes AFactorOf(i) to not always be LargestFactor(i)
*/
void buildFactors(){
dbg(cout<<"Allocating RAM"<<endl;)
factors=new (nothrow) uint [TABLE_SIZE]; // 4 * 2^28 = 2^30 = 1GB
if(factors==NULL)return; // not able to allocate enough free memory
uint i,j;
factors08 = (uint8 *)factors;
factors16 = (uint16 *)factors;
factors32 = factors;
dbg(cout<<"Zeroing RAM"<<endl;)
memset(factors,0,sizeof(uint)*TABLE_SIZE);
//for(i=0;i<TABLE_SIZE;i++)factors[i]=0;
//Sieve of Eratosthenese
//8-bit values
dbg(cout<<"Setting: 8-Bit Values"<<endl;)
factors08[1-o08]=1;
for(i=2;i*i<LIMIT_16;i++){
for(;factors08[i-o08] && i*i<LIMIT_16;i++);
dbg(cout<<"Filtering: "<<i<<endl;)
factors08[i-o08]=1;
for(j=2;i*j<LIMIT_16;j++)factors08[i*j-o08]=i;
for(;i*j<LIMIT_32;j++)factors16[i*j-o16]=i;
for(;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
for(;i<LIMIT_16;i++){
for(;i<LIMIT_16 && factors08[i-o08];i++);
dbg(cout<<"Filtering: "<<i<<endl;)
if(i<LIMIT_16){
factors08[i-o08]=1;
j=LIMIT_16/i+(LIMIT_16%i>0);
for(;i*j<LIMIT_32;j++)factors16[i*j-o16]=i;
for(;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
}i--;
dbg(cout<<"Setting: 16-Bit Values"<<endl;)
//16-bit values
for(;i*i<LIMIT_32;i++){
for(;factors16[i-o16] && i*i<LIMIT_32;i++);
factors16[i-o16]=1;
for(j=2;i*j<LIMIT_32;j++)factors16[i*j-o16]=i;
for(;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
for(;i<LIMIT_32;i++){
for(;i<LIMIT_32 && factors16[i-o16];i++);
if(i<LIMIT_32){
factors16[i-o16]=1;
j=LIMIT_32/i+(LIMIT_32%i>0);
for(;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
}i--;
dbg(cout<<"Setting: 32-Bit Values"<<endl;)
//32-bit values
for(;i*i<=MAX_FACTOR;i++){
for(;factors32[i-o32] && i*i<=MAX_FACTOR;i++);
factors32[i-o32]=1;
for(j=2;i*j<=MAX_FACTOR;j++)factors32[i*j-o32]=i;
}
for(;i<=MAX_FACTOR;i++){
for(;i<=MAX_FACTOR && factors32[i-o32];i++);
if(i>MAX_FACTOR)return;
factors32[i-o32]=1;
}
}
uint * factor(uint x, int &factorCount){
if(x > MAX_FACTOR){factorCount=-1;return NULL;}
uint tmp[70], at=x; int i=0;
while(at>=LIMIT_32 && factors32[at-o32]>1){
tmp[i++]=factors32[at-o32];
dbg(cout<<"at32:"<<at<<" tmp:"<<tmp[i-1]<<endl;)
at/=tmp[i-1];
}
if(at<LIMIT_32){
while(at>=LIMIT_16 && factors16[at-o16]>1){
tmp[i++]=factors16[at-o16];
dbg(cout<<"at16:"<<at<<" tmp:"<<tmp[i-1]<<endl;)
at/=tmp[i-1];
}
if(at<LIMIT_16){
while(factors08[at-o08]>1){
tmp[i++]=factors08[at-o08];
dbg(cout<<"at08:"<<at<<" tmp:"<<tmp[i-1]<<endl;)
at/=tmp[i-1];
}
}
}
if(i==0){
dbg(cout<<"at:"<<x<<" tmp:1"<<endl;)
tmp[i++]=1;
tmp[i++]=x;
}else{
dbg(cout<<"at:"<<at<<" tmp:1"<<endl;)
tmp[i++]=at;
}
factorCount=i;
uint *ret=new (nothrow) uint [factorCount];
if(ret!=NULL)
memcpy(ret, tmp, sizeof(uint)*factorCount);
return ret;
}
uint AFactorOf(uint x){
if(x > MAX_FACTOR)return -1;
if(x < LIMIT_16) return factors08[x-o08];
if(x < LIMIT_32) return factors16[x-o16];
return factors32[x-o32];
}
void main(){
cout<<"Loading factors lookup table"<<endl;
buildFactors(); if(factors==NULL){cerr<<"Need 1GB block of free memory"<<endl;return;}
int size;
uint x=13855127;//25255230;//30030;
cout<<"\nFactoring: "<<x<<endl;
uint *f=factor(x,size);
if(size<0){cerr<<x<<" is too big to factor. Choose a number between 1 and "<<MAX_FACTOR<<endl;return;}
else if(f==NULL){cerr<<"ran out of memory trying to factor "<<x<<endl;return;}
cout<<"\nThe factors of: "<<x<<" {"<<f[0];
for(int i=1;i<size;i++)
cout<<", "<<f[i];
cout<<"}"<<endl;
delete [] f;
x=30637;
cout<<"\nFactoring: "<<x<<endl;
f=factor(x,size);
cout<<"\nThe factors of: "<<x<<" {"<<f[0];
for(int i=1;i<size;i++)
cout<<", "<<f[i];
cout<<"}"<<endl;
delete [] f;
delete [] factors;
}
This is an important open mathematical problem as of 2020
Others have answered from a practical point of view, and there is high probability that for the problem sizes encountered in practice, that those algorithms are close to the optimal.
However, I would also like to highlight, that the more general mathematical problem (in asymptotic computation complexity, i.e. as the number of bits tends to infinity) is completely unsolved.
No one has ever been able to prove what is the minimal optimal time for of what is the fastest possible algorithm.
This is shown on the Wikipedia page: https://en.wikipedia.org/wiki/Integer_factorization The algorithm also figures on Wiki's "List of unsolved problems in computer science" page: https://en.wikipedia.org/wiki/List_of_unsolved_problems_in_computer_science
This is important, because until we have such an optimality proof, we won't be able to be sure that RSA encryption is actually secure. Actually, we don't have any such proofs for any existing public key system, all we know is that they seem safe because no one managed to break them publicly yet. See also this related post.
All that we know is that the best we currently have is the general number sieve. And until 2018, we didn't even have a non-heuristic proof for its complexity. The complexity of that algorithm in terms of the number of digits of the integer to be factored is something like:
e^{(k + o(1))(n^(1/3) * ln(n)^(2/3)}
which as mentioned at: Polynomial time and exponential time is not truly exponential, but it is superpolynomial.
As of 2020, we haven't even proved if the problem is NP-complete or not (although it is obviously NP since all you have to do to verify a solution is to multiply the numbers)! Although it is widely expected for it to be NP-complete. We can't be that bad at finding algorithms, can we?
There is of course the HAL Algorithm by Professor Hal Mahutan, (Feb 2021), which is on the edge of factorisation research.
PLEASE VIEW LATEST UPDATE HERE
https://docs.google.com/document/d/1BlDMHCzqpWNFblq7e1F-rItCf7erFMezT7bOalvhcXA/edit?usp=sharing
Solving for two large primes for the public key is as follows...
Any AxB = Public Key can be drawn on the postive X and Y axis that forms a continuous curve where all non-integer factors solve for the public key. Of course, this is not useful, it's just an observation at this point.
Hal's insight is this: If we insist that we are only interested in the points that A is a whole number, particularly the points that B present when A is whole.
Previous attempts with this approach failed when mathematicians struggled with the apparent randomness of the remainder of B, or at least lack of predictability.
What Hal is saying is that predictability is universal for any public key providing the ratio of a/b is the same. Basically, when a series of different public keys are presented for analysis, all of them can be processed identically, provided they share the same point during processing where a/b is constant, ie they share the same tangent.
Take a look at this sketch I drew to try explain what Prof Mahutan has got going on here.
So, here's Hal's genius. Hal makes use of powerful super-computers to generate a series of hashtables (in diagram, Q, R, S & T). What you can see in the 3 A x B = Key curves on the left is that they all share tangent T and S (the only ones highlighted there) but what the diagram is showing is that given any Public Key, over a region of the curve where the tangent is the same, then you can share hashtables that preside over that region.
Just a technical note, obviously in the curve of AxB= Key, things are shifting continuously as one changes the values of AxB, so in theory, the shared tangent that maps to a hashtable is going to go out of date, but the interesting thing is that with really large keys (ironically, this makes them easier to crack because they share longer runs where a hashtable is useful.). So that's great news as key sizes are expected to get much bigger as advances in factorisation and computation accelerate. What actually happens is that the predictability of the hashtable is going to literally 'go out of focus' as the tangents to which they apply begin to diverge. Luckily, this is not a problem, because you jump to the next hashtable that is appropriately mapped to the new Tangent.
Just to make it clear, all Public Keys ever generated will always make use of the same set of hashtables, so it's a kind of one off investment that can be stored online, literally millions of terabytes of lookup data, since all keys obey the same tangential ratios.
So, what do the hashtables do to accelerate the finding of primes. The hashtables are indexed with the remainder when the Public Key is divided by B. Basically, Hal is saying that for all Keys, any ratio of A x B can be looked up. The only difference between the different curves that share the same Tangent is that they require a different offset as determined by the 'Control Curve'. The 'Control Curve' is any arbitrary curve for which you generate appropriate factors for. Let's say for the 'Control Curve', the Key is 15, and the tangent being mapped is when B = 5, so A is 3 and the remainder is zero. In another mapping with the same Tangent, say the Key is now 16. We need to find the same Tangent which lies at 5.33 for B and 3.2 for A. So the remainder for A is .2, so a public key of 16 can use the same lookup table as 15 provided it is suitably offset by .2.
So what's in the hashtables? Indexed by the offset and the value always returns the distance along the AxB curve for which you do not find another whole number of B. What Hal is saying is that it is safe to jump ahead and not check those numbers for being factors. And that's basically it. Hal punches holes into the curve that need never be checked and that just speeds up the whole game.
Thanks Professor Mahutan!
For those of you still following, here are some of our working notes:
Bullet Points for Fast Factorisation Attack Algorithm
- All public key's can be represented along a curve A x B = 'Key'
- This is an observation that maps all real numbers (is this the right term for non-integers?) that all multiply together to equal the key... So far not useful
- We are only interested in the points where A is whole and B are both whole numbers.
- We can step through the entire line where A is whole. This is half way there but has problems. Firstly, we don't know where B is whole and also, it would take too much processing power to calculate all the points.
- What we are interested in is predicting where B is also whole, so we want a mechanism to be able to 'jump' along the curve where we know that B is definitely still a real number (non-whole). If we can make big enough jumps, then we reduce the processing required.
Now follows the strategy of the algorithm to predict B
Another observation is that for sufficiently large values of the 'Key', as we step through changing the value of A in whole number increments, we observe that the ratio of A/B or the tangential angle will remain mostly the same.
An important side point to this observation is that as the Key size increases, the tangent remains more constant with each iteration. Fundamentally, the means that any algorithm using this property is going to become more efficient as the Key increases in size, which is opposite of traditional approaches where increasing the Key size makes it exponentially harder to guess the factors. This is a very important point... (Elaborate on this please Nick)
The algorithm itself is as follows
- Purchase sufficient storage and processing power on a cloud
- Break the problem into pieces that can be run on different processes in parallel. To do this, we step through different values of A and assign the search to different processors in the cloud.
- For any value of A that is being checked use a universal lookup table to predict a safe distance along the curve that we can move without having to calculate if B is going to be a whole number
- Check only those positions along the curve for which the lookup table shows that the probability of it being a whole number is high enough to warrant checking.
The important concept here is that the lookup tables can be shared for any 'Key' for which the ratio of A/B (the tangent) is sufficiently close enough before the lookup becomes inaccurate (and goes out of focus).
(Also to note is that as the key size shifts, then you either need a new set of tables or you should make an appropriate mapping to the existing ratio tables in order to reuse them.)
Another really important point Nick is that all Keys can share the same set of lookup tables.
Fundamentally, the lookup table maps the remainder from any calculation of Key / A. We are interested in the remainder because when the remainder is = zero, then we have done it because A already is a whole number.
I suggest that we have enough hash tables to ensure that we can scan ahead without having to calculate the actual remainder. Let's say we start off with 1 trillion, but that can obviously be changed depending on how much computational power we have.
The hashtable for any suitably close A/b ratio is as follows. The table is indexed (keyed) with the appropriate remainder and the value at any given remainder is the 'safe' distance that can be traversed along the A * B curve without the remainder touching Zero.
You can visualise the remainder oscillating (pseudo-randomly) between 0 and 1.
The algorithm picks on a number A along the curve, then looks up the safe distance and jumps to the next hashtable, or at least the algorithm does those factor checks until the next hashtable becomes available. Given enough hashtables, I'm thinking we can pretty much avoid most of the checking.
Notes on lookup tables.
For any key, you can generate a table that curves appropriately for A/B ratio Reuse of tables is imperative. Suggested approach Generate a series of control curves for A/B from say Square Root of N (The appropriate key) and do so midway by squaring. Let’s say each is 0.0001% larger than the previous key Let’s also make the size of the table say 1 % of the A/B ratio When calculating the co-prime factors, pick the closest lookup table that matches the key. Pick your entry point into the hashtable. This means remembering the offset that the entry point in the table has with your actual entry point. The lookup table should provide a series of points for the entry point for which the corresponding co-prime entry could be very close to zero and needs to be checked manually. For each point in the series, calculate the actual offset using the appropriate mathematical formula. (This will be an integral calculation, we need to have a mathematician take a look at it) Why? Because our control table was calculated when A/B was the square root of Key. As we shift along the curve, we need to space out appropriately. As a bonus, can the mathematician collapse the Key space into a point on the A/B curve. If so, we only need to generate a single table.
Key Concepts
The number A x B = Key plots the following:
X-Axis maps A and Y Axis maps B.
The closeness of the curve to the A and B axis is dependent on the size of the public key (where A x B = Key). Basically, the curve is going to shift to the right as the Key becomes larger.
Now the idea I would like you to digest or have questions about is
- Given any point on the curve, there exists a tangent (ratio of A/B)
- We are interested in the value of B as A increases in whole number increments and is a whole number itself. In particular, we really are only interested in the Remainder of B when Key / A is something remainder ZERO. We will have found the factors for this public key. Specifically, it will be the last value of A (also and always a whole number) that we tried and the value B for which the remainder is zero (so also is a whole number).
- The basic algorithm is simple enough. -1- Pick a point on the curve where A is a whole number -2- Find the remainder of B where Key/A is B -3- Check if remainder of B is ZERO, (if it is ZERO then we are done.) -4- Go back to step 1, (next you will pick the next whole number for A)
This will work, but is way too slow. What we are doing with the HAL algorithm is improving the basic algorithm above so that we can jump chunks of the curve where we know the remainder doesn't get too close to zero.
The more we can jump, the more efficient the algorithm.
Before we step into the improved HAL algorithm, let's review a key concept.
For very large values of Key, (remember A x B = Key), the ratio of A/B will be pretty much constant, the RSA key is 2 power 4096, and that is big.
Let us assume we have created a set of tables already preloaded into the cloud that are optimised for a particular (roughly) sized key.
- Let's say we have 1 million tables to be used only against this particularly narrow range of key sizes
- Each table maps for a slightly different tangent or A/B ratio, but remember that all of these tables can only be used for the appropriate key size
- These tables are spread evenly along the curve so that for any point I pick, there is a table that can tell me how many whole numbers of A that I can safely jump before the possibility arises that I will hit a remainder of zero for B in Key/A = B. Remember, we don't want to miss the point where B is zero or HAL won't work.
- These tables are indexed with the current remainder. (Programmers or developers note these lookup tables can be implemented with a Hashtable, say in C# or Java) Let's assume that we have to check all the points where A moves along the curve, each time producing a remainder B. As soon as B is close enough to any of the indexes, then you can use the table to work out how many whole numbers you can jump safely without missing when B has a remainder of zero.
This piece is a critical concept.
- Each set of lookup tables that is indexed with some reasonable offset only really works for a particular Key size.
- As the key grows or shrinks for the series of tables, the results of the lookup 'go out of focus' or become more inaccurate.
- Therefore, we also need to create a new series of tables as the Key size grows. Perhaps we need to create the tables every 0.001 % growth of the key.
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