Expected Token Density in Random Stream


A

Andrew Tomazos

Summary: We want to find out how often a given token appears in a
random stream formed by concatenating randomly chosen strings from a
given set of strings.

Details:

Given S, an array of n (non-empty) strings; and T, a string of length
m;

We create a random stream of characters by the following process:

1. assign i = a (uniformly) random integer between 1 and n
inclusive
2. write the characters of the ith element of S to the stream
3. goto 1

(The elements of S are not necessarily the same length)

For example:

if S = {"a", "bug", "ug"}

then the stream may start as follows:

augbugugugaabug...

Each time T appears in the stream we call that a hit. More precisely:

1. Initialize a queue Q with m null elements

2. If Q equals T record a hit
3. Pop front of Q
4. Push to back of Q next char from stream
5. Goto 2

For example:

If T = "ugu"...

then:

augbugugugaabug...
1 2

We get two hits so far.

(Note hits can overlap each other)

What is the expected frequency of hits in terms of S and T?

More precisely:

let y be the number of chars read from the stream so far
let x be the number of hits

What is the expected limit of x/y as y approaches infinity?

We can approximate this empirically as follows in C++:

$ cat > HitFinder.cpp
<paste below code>
$ g++ -o HitFinder HitFinder.cpp
$ ./HitFinder ugu a bug ug
T = ugu
A = {a, bug, ug}

(x, y, x/y) = (0, 1, 0)
(x, y, x/y) = (0, 2, 0)
(x, y, x/y) = (0, 4, 0)
(x, y, x/y) = (0, 8, 0)
(x, y, x/y) = (1, 16, 1/16)
(x, y, x/y) = (3, 32, 1/10.6667)
(x, y, x/y) = (8, 64, 1/8)
(x, y, x/y) = (16, 128, 1/8)
...
(x, y, x/y) = (29821708, 268435456, 1/9.00134)
(x, y, x/y) = (59648899, 536870912, 1/9.00052)
(x, y, x/y) = (119304725, 1073741824, 1/8.99999)
(x, y, x/y) = (238593314, 2147483648, 1/9.0006)
(x, y, x/y) = (477198099, 4294967296, 1/9.00039)

As you can see the answer for this converges on 1/9

How can we calculate this figure by direct computation?

ie How would you implement a function with signature...

double ExpectedHitLimit(string T, vector<string> A)

....that quickly calculates this limit for any given T and A?

Thanks and have fun,
Andrew.

--
Andrew Tomazos <[email protected]> <http://www.tomazos.com>


// ======================= HitFinder.cpp Cut Here
========================
// (C) 2011, Andrew Tomazos <[email protected]>. All Rights
Reserved.

#include <queue>
#include <string>
#include <iostream>
#include <limits>
#include <cstdlib>

using namespace std;

typedef long long int64;
bool IsPow2(int64 i)
{
return i != 0 && !(i & (i - 1));
}

int64 x = 0;

void Hit()
{
x++;
}

int Rand(int n)
{
return rand() % n;
}

class RandCharStream
{
public:
vector<string> S;
size_t n, istr, ichar;

RandCharStream(vector<string> S_) :
S(S_),
n(S.size()),
istr(Rand(n)),
ichar(0)
{
}

char nextChar()
{
if (ichar == S[istr].size())
{
istr = Rand(n);
ichar = 0;
}

return S[istr][ichar++];
}
};

class HitFinder
{
public:
RandCharStream& stream;
queue<char> T;
int m;
queue<char> Q;

HitFinder(RandCharStream& stream_, string T_) :
stream(stream_),
m(T_.size())
{
for (int i = 0; i < m; i++)
{
T.push(T_);
Q.push('\0');
}
}

void nextChar()
{
Q.pop();
Q.push(stream.nextChar());

if (Q == T)
Hit();
}
};

int main(int argc, char** argv)
{
if (argc < 3)
{
cerr << "Usage: " << argv[0] << " <T> <S1> <S2> ... <Sn>" << endl;
return -1;
}

string T = argv[1];
cout << "T = " << T << endl;

cout << "A = {";
vector<string> A;
for (int i = 2; i < argc; i++)
{
string s = argv;
cout << s << (i < argc - 1 ? ", " : "");
A.push_back(s);
if (s.empty())
{
cerr << "Element S" << i - 1 << " is empty" << endl;
return -1;
}
}
cout << "}" << endl;
cout << endl;

RandCharStream stream(A);
HitFinder finder(stream, T);

for (int64 y = 1; y < __LONG_LONG_MAX__; y++)
{
finder.nextChar();

if (IsPow2(y))
{
cout << "(x, y, x/y) = (" << x << ", " << y << ", ";
if (x == 0)
cout << "0";
else
cout << "1/" << double(y) / double(x);
cout << ")" << endl;
}
}

return 0;
}
//======================= HitFinder.cpp Cut Here
========================
 
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