Given an integer z<=10^100, find the smallest row of Pascal's triangle that contains z

Question

How can I find an algorithm to solve this problem using C++: given an integer z<=10^100, find the smallest row of Pascal's triangle that contains the number z.

    1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1

For example if z=6 => result is on the 4th row.

Another way to describe the problem: given integer z<=10^100, find the smallest integer n: exist integer k so that C(k,n) = z.

C(k,n) is combination of n things taken k at a time without repetition

Answer 1

EDIT This solution needs Logarithmic time, it's O(Log z). Or maybe O( (Log z)^2 ).

Say you are looking for n,k where Binomial(n,k)==z for a given z.

1. Each row has its largest value in the middle, so starting from n=0 you increase the row number, n, as long as the middle value is smaller than the given number. Actually, 10^100 isn't that big, so before row 340 you find a position n0,k0=n0/2 where the value from the triangle is larger than or equal to z: Binomial(n0,k0)>=z

2. You walk to the left, i.e. you decrease the column number k, until eventually you find a value smaller than z. If there was a matching value in that row you would have hit it by now. k is not very large, less than 170, so this step won't be executed more often than that and does not present a performance problem.

3. From here you walk down, increasing n. Here you will find a steadily increasing value of Binomial[n,k]. Continue with 3 until the value gets bigger than or equal to z, then goto 2.

EDIT: This step 3 can loop for a very long time when the row number n is large, so instead of checking each n linearly you can do a binary search for n with Binomial(n,k) >= z > Binomial(n-1,k), then it only needs Log(n) time.

A Python implementation looks like this, C++ is similar but somewhat more cumbersome because you need to use an additional library for arbitrary precision integers:

# Calculate (n-k+1)* ... *n
def getnk( n, k ):
    a = n
    for u in range( n-k+1, n ):
        a = a * u
    return a

# Find n such that Binomial(n,k) >= z and Binomial(n-1,k) < z
def find_n( z, k, n0 ):
    kfactorial = k
    for u in range(2, k):
        kfactorial *= u

    xk = z * kfactorial            

    nk0 = getnk( n0, k )
    n1=n0*2
    nk1 = getnk( n1, k )

    # duplicate n while the value is too small
    while nk1 < xk:
        nk0=nk1
        n0=n1
        n1*=2
        nk1 = getnk( n1, k )
    # do a binary search
    while n1 > n0 + 1:
        n2 = (n0+n1) // 2
        nk2 = getnk( n2, k )
        if nk2 < xk:
            n0 = n2
            nk0 = nk2
        else:
            n1 = n2
            nk1 = nk2

    return n1, nk1 // kfactorial


def find_pos( z ):
    n=0
    k=0
    nk=1

    # start by finding a row where the middle value is bigger than z
    while nk < z:
        # increase n
        n = n + 1
        nk = nk * n // (n-k)
        if nk >= z:
            break
        # increase both n and k
        n = n + 1
        k = k + 1
        nk = nk * n // k

    # check all subsequent rows for a matching value
    while nk != z:
        if nk > z:
            # decrease k
            k = k - 1
            nk = nk * (k+1) // (n-k)
        else:
            # increase n
            # either linearly
            #  n = n + 1
            #  nk = nk * n // (n-k)
            # or using binary search:
            n, nk = find_n( z, k, n )
    return n, k

z = 56476362530291763837811509925185051642180136064700011445902684545741089307844616509330834616
print( find_pos(z) )

It should print

(5864079763474581, 6)

Answer 2

Stirling estimation for n! can be used to find first row in triangle with binomial coefficient bigger or equal to a given x. Using this estimation we can derive lower and upper bound for

and then by observation that this is the maximum coefficient in row that expands 2n:

P( 2n, 0), P( 2n, 1), P( 2n, 2), ..., P( 2n, 2n -1), P( 2n, 2n)

we can find first row with maximum binomial coefficient bigger or equal to a given x. This is the first row in which x can be looking for, this is not possible that x can be found in the row smaller than this. Note: this may be right hint and give an answer immediately in some cases. At the moment I cannot see other way than to start a brute force search from this row.

template <class T>
T binomial_coefficient(unsigned long n, unsigned long k) {
    unsigned long i;
    T b;
    if (0 == k || n == k) {
        return 1;
    }
    if (k > n) {
        return 0;
    }
    if (k > (n - k)) {
        k = n - k;
    }
    if (1 == k) {
        return n;
    }
    b = 1;
    for (i = 1; i <= k; ++i) {
        b *= (n - (k - i));
        if (b < 0) return -1; /* Overflow */
        b /= i;
    }
    return b;
}

Stirling:

double stirling_lower_bound( int n) {
    double n_ = n / 2.0;
    double res = pow( 2.0, 2 * n_);
    res /= sqrt( n_ * M_PI);
    return res * exp( ( -1.0) / ( 6 * n_));
}

double stirling_upper_bound( int n) {
    double n_ = n / 2.0;
    double res = pow( 2.0, 2 * n_) ;
    res /= sqrt( n_ * M_PI);
    return res * exp( 1.0 / ( 24 * n_));
}

int stirling_estimate( double x) {
    int n = 1;
    while ( stirling_lower_bound( n) <= x) {
        if ( stirling_upper_bound( n) > x) return n;
        ++n;
    }
    return n;
}

usage:

long int search_coefficient( unsigned long int &n, unsigned long int x) {
    unsigned long int k = n / 2;
    long long middle_coefficient = binomial_coefficient<long long>( n, k);
    if( middle_coefficient == x) return k;

    unsigned long int right = binomial_coefficient<unsigned long>( n, ++k);
    while ( x != right) {

        while( x < right ||  x < ( right * ( n + 1) / ( k + 1))) {
            right = right * ( n + 1) / ( ++k) - right;
        }
        if ( right == x) return k;
        right = right * ( ++n) / ( ++k);
        if( right > x) return -1;
    }
    return k;
}


/*
 * 
 */
int main(int argc, char** argv) {

    long long x2 = 1365;
    unsigned long int n = stirling_estimate( x2);
    long int k = search_coefficient( n, x2);
    std::cout << "row:" << n <<", column: " << k; 
    return 0;
}

output:

row:15, column: 11