Brocard's problem is n! + 1 = m^2. The solutions to this problems are pairs of integers called Brown numbers (4,5), etc, of which only three are known.
A very literal implementation to Brocard's problem:
import math
def brocard(n,m):
if math.factorial(n)+1 == m**2:
return (n,m)
else:
return
a=10000
for n in range(a):
for m in range(a):
b=brocard(n,m)
if b is not None:
print(b)
The time complexity of this should be O(n^2) because of the nested for loops with differing variables and the complexity of whatever math.factorial algorithm is (apparently divide-and-conquer). Is there any way to improve upon O(n^2)?
There are other interpretations on SO like this. How does the time complexity of this compare with my implementation?
Your algorithm is O(n^3).
You have two nested loops, and inside you use factorial(), having O(n) complexity itself.
Your algorithm tests all (n,m) combinations, even those where factorial(n) and m^2 are far apart, e.g. n=1 and m=10000.
You always recompute the factorial(n) deep inside the loop, although it's independent of the inner loop variable m. So, it could be moved outside of the inner loop.
And, instead of always computing factorial(n) from scratch, you could do that incrementally. Whenever you increment n by 1, you can multiply the previous factorial by n.
A different, better approach would be not to use nested loops, but to always keep n and m in a number range so that factorial(n) is close to m^2, to avoid checking number pairs that are vastly off. We can do this by deciding which variable to increment next. If the factorial is smaller, then the next brocard pair needs a bigger n. If the square is smaller, we need a bigger m.
In pseudo code, that would be
n = 1; m = 1; factorial = 1;
while n < 10000 and m < 10000
if factorial + 1 == m^2
found a brocard pair
// the next brocard pair will have different n and m,
// so we can increment both
n = n + 1
factorial = factorial * n
m = m + 1
else if factorial + 1 < m^2
// n is too small for the current m
n = n + 1
factorial = factorial * n
else
// m is too small for the given n
m = m + 1
In each loop iteration, we either increment n or m, so we can have at most 20000 iterations. There is no inner loop in the algorithm. We have O(n). So, this should be fast enough for n and m up to the millions range.
P.S. There are still some optimizations possible.
Factorials (after n=1, known to have no brocard pair) are always even numbers, so m^2 must be odd to satisfy the brocard condition, meaning that we can always increment m by 2, skipping the even number in between.
For larger n values, the factorial increases much faster than the square. So, instead of incrementing m until its square reaches the factorial+1 value, we could recompute the next plausible m as integer square root of factorial+1.
Or, using the square root approach, just compute the integer square root of factorial(n), and check if it matches, without any incremental steps for m.
. Is there any Direct formula or System to find out the Numbers of Zero's between a Distinct Range ... Let two Integer M & N are given . if I have to find out the total number of zero's between this Range then what should I have to do ?
Let M = 1234567890 & N = 2345678901
And answer is : 987654304
Thanks in advance .
Reexamining the Problem
Here is a simple solution in Ruby, which inspects each integer from the interval [m,n], determines the string of its digits in the standard base 10 positional system, and counts the occuring 0 digits:
def brute_force(m, n)
if m > n
return 0
end
z = 0
m.upto(n) do |k|
z += k.to_s.count('0')
end
z
end
If you run it in an interactive Ruby shell you will get
irb> brute_force(1,100)
=> 11
which is fine. However using the interval bounds from the example in the question
m = 1234567890
n = 2345678901
you will recognize that this will take considerable time. On my machine it does need more than a couple of seconds, I had to cancel it so far.
So the real question is not only to come up with the correct zero counts but to do it faster than the above brute force solution.
Complexity: Running Time
The brute force solution needs to perform n-m+1 times searching the base 10 string for the number k, which is of length floor(log_10(k))+1, so it will not use more than
O(n (log(n)+1))
string digit accesses. The slow example had an n of roughly n = 10^9.
Reducing Complexity
Yiming Rong's answer is a first attempt to reduce the complexity of the problem.
If the function for calculating the number of zeros regarding the interval [m,n] is F(m,n), then it has the property
F(m,n) = F(1,n) - F(1,m-1)
so that it suffices to look for a most likely simpler function G with the property
G(n) = F(1,n).
Divide and Conquer
Coming up with a closed formula for the function G is not that easy. E.g.
the interval [1,1000] contains 192 zeros, but the interval [1001,2000] contains 300 zeros, because a case like k = 99 in the first interval would correspond to k = 1099 in the second interval, which yields another zero digit to count. k=7 would show up as 1007, yielding two more zeros.
What one can try is to express the solution for some problem instance in terms of solutions to simpler problem instances. This strategy is called divide and conquer in computer science. It works if at some complexity level it is possible to solve the problem instance and if one can deduce the solution of a more complex problem from the solutions of the simpler ones. This naturally leads to a recursive formulation.
E.g. we can formulate a solution for a restricted version of G, which is only working for some of the arguments. We call it g and it is defined for 9, 99, 999, etc. and will be equal to G for these arguments.
It can be calculated using this recursive function:
# zeros for 1..n, where n = (10^k)-1: 0, 9, 99, 999, ..
def g(n)
if n <= 9
return 0
end
n2 = (n - 9) / 10
return 10 * g(n2) + n2
end
Note that this function is much faster than the brute force method: To count the zeros in the interval [1, 10^9-1], which is comparable to the m from the question, it just needs 9 calls, its complexity is
O(log(n))
Again note that this g is not defined for arbitrary n, only for n = (10^k)-1.
Derivation of g
It starts with finding the recursive definition of the function h(n),
which counts zeros in the numbers from 1 to n = (10^k) - 1, if the decimal representation has leading zeros.
Example: h(999) counts the zero digits for the number representations:
001..009
010..099
100..999
The result would be h(999) = 297.
Using k = floor(log10(n+1)), k2 = k - 1, n2 = (10^k2) - 1 = (n-9)/10 the function h turns out to be
h(n) = 9 [k2 + h(n2)] + h(n2) + n2 = 9 k2 + 10 h(n2) + n2
with the initial condition h(0) = 0. It allows to formulate g as
g(n) = 9 [k2 + h(n2)] + g(n2)
with the intital condition g(0) = 0.
From these two definitions we can define the difference d between h and g as well, again as a recursive function:
d(n) = h(n) - g(n) = h(n2) - g(n2) + n2 = d(n2) + n2
with the initial condition d(0) = 0. Trying some examples leads to a geometric series, e.g. d(9999) = d(999) + 999 = d(99) + 99 + 999 = d(9) + 9 + 99 + 999 = 0 + 9 + 99 + 999 = (10^0)-1 + (10^1)-1 + (10^2)-1 + (10^3)-1 = (10^4 - 1)/(10-1) - 4. This gives the closed form
d(n) = n/9 - k
This allows us to express g in terms of g only:
g(n) = 9 [k2 + h(n2)] + g(n2) = 9 [k2 + g(n2) + d(n2)] + g(n2) = 9 k2 + 9 d(n2) + 10 g(n2) = 9 k2 + n2 - 9 k2 + 10 g(n2) = 10 g(n2) + n2
Derivation of G
Using the above definitions and naming the k digits of the representation q_k, q_k2, .., q2, q1 we first extend h into H:
H(q_k q_k2..q_1) = q_k [k2 + h(n2)] + r (k2-kr) + H(q_kr..q_1) + n2
with initial condition H(q_1) = 0 for q_1 <= 9.
Note the additional definition r = q_kr..q_1. To understand why it is needed look at the example H(901), where the next level call to H is H(1), which means that the digit string length shrinks from k=3 to kr=1, needing an additional padding with r (k2-kr) zero digits.
Using this, we can extend g to G as well:
G(q_k q_k2..q_1) = (q_k-1) [k2 + h(n2)] + k2 + r (k2-kr) + H(q_kr..q_1) + g(n2)
with initial condition G(q_1) = 0 for q_1 <= 9.
Note: It is likely that one can simplify the above expressions like in case of g above. E.g. trying to express G just in terms of G and not using h and H. I might do this in the future. The above is already enough to implement a fast zero calculation.
Test Result
recursive(1234567890, 2345678901) =
987654304
expected:
987654304
success
See the source and log for details.
Update: I changed the source and log according to the more detailed problem description from that contest (allowing 0 as input, handling invalid inputs, 2nd larger example).
You can use a standard approach to find m = [1, M-1] and n = [1, N], then [M, N] = n - m.
Standard approaches are easily available: Counting zeroes.
I am trying to understand implementation of linear time suffix array creation algorithm by Karkkainen, P. Sanders. Details of algorithm can be found here.
I managed to understand overall concept but failing to match it with provided implementation and hence not able to grasp it clearly.
Here are initial code paths which are confusing me.
As per paper : n0, n1, n2 represent number of triplets starting at i mod 3 = (0,1,2)
As per code : n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3; => How these initialisations has been derived?
As per paper : We need to create T` which is concatenation of triplets at i mod 3 != 0
As per code : n02 = n0 + n2; s12 = [n02] ==> How came n02? It should be n12 i.e n1 + n2.
As per code : for (int i = 0, j = 0; i < n + (n0 - n1); i++) fill s12 with triplets such that i%3 != 0; => Why for loop runs for n + (n0 - n1) times ? It should be simply n1 + n2. Should't be ?
I am not able to proceed because of these :( Please to help.
Consider the following example where the length of the input is n=13:
STA | CKO | WER | FLO | W
As per code : n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3; => How these initialisations has been derived?
Note that the number of triplets i mod3 = 0 is n/3 if n mod3 = 0 and n/3+1 otherwise (if n mod3 = 1 or n mod3 = 2). In the current example n/3 = 4 but since the last triplet 'W' is not complete it is not counted in the integer division. A 'trick' to make this computation directly is to use (n+2)/3. Effectively, if n mod3 = 0 then the result of the integer divisions (n+2)/3 and n/3 will be the same. However, if n mod3 = 1 or 2 then the result of (n+2)/3 will be n/3+1. The same applies to n1 and n2.
As per code : n02 = n0 + n2; s12 = [n02] ==> How came n02? It should be n12 i.e n1 + n2.
As per code : for (int i = 0, j = 0; i < n + (n0 - n1); i++) fill s12 with triplets such that i%3 != 0; => Why for loop runs for n + (n0 - n1) times ? It should be simply n1 + n2. Should't be ?
Both questions have the same answer. In our example we'd have a B12 buffer like this:
B12 = B1 U B2 = {TA KO ER LO}
So you'd first sort the suffixes and end up with a suffix array of B12, which has 8 elements. To proceed to the merging step we first need to compute the suffix array of B0, which is obtained by sorting the tuples (B0(i),rank(i+1))... But this concrete case in which the last triplet has only one element (W) has a problem, because rank(i+1) is not defined for the last element of B0:
B0 = {0,3,6,9,12}
which sorted alphabetically results in
SA0 = {3, 9, 0, ?, ?}
Since the indices 6 and 12 contain a 'W', it is not enough to sort alphabetically, we need to check which goes first in the rank table, so let's check the rank of their suffixes.. oh, wait! rank(13) is not defined!
And that's why we add a dummy 0 to the last triplet of the input when the last triplet only contains one element (if n mod3 = 0). So then the size of B12 is n0+n2, no matter the size of n1, and one needs to add an extra element to B12 if B0 is larger than B1 (in which case n0-n1 = 1).
Hope it was clear.
Someone told me that the Frobenius pseudoprime algorithm take three times longer to run than the Miller–Rabin primality test but has seven times the resolution. So then if one where to run the former ten times and the later thirty times, both would take the same time to run, but the former would provide about 233% more analyse power. In trying to find out how to perform the test, the following paper was discovered with the algorithm at the end:
A Simple Derivation for the Frobenius Pseudoprime Test
There is an attempt at implementing the algorithm below, but the program never prints out a number. Could someone who is more familiar with the math notation or algorithm verify what is going on please?
Edit 1: The code below has corrections added, but the implementation for compute_wm_wm1 is missing. Could someone explain the recursive definition from an algorithmic standpoint? It is not "clicking" for me.
Edit 2: The erroneous code has been removed, and an implementation of the compute_wm_wm1 function has been added below. It appears to work but may require further optimization to be practical.
from random import SystemRandom
from fractions import gcd
random = SystemRandom().randrange
def find_prime_number(bits, test):
number = random((1 << bits - 1) + 1, 1 << bits, 2)
while True:
for _ in range(test):
if not frobenius_pseudoprime(number):
break
else:
return number
number += 2
def frobenius_pseudoprime(integer):
assert integer & 1 and integer >= 3
a, b, d = choose_ab(integer)
w1 = (a ** 2 * extended_gcd(b, integer)[0] - 2) % integer
m = (integer - jacobi_symbol(d, integer)) >> 1
wm, wm1 = compute_wm_wm1(w1, m, integer)
if w1 * wm != 2 * wm1 % integer:
return False
b = pow(b, (integer - 1) >> 1, integer)
return b * wm % integer == 2
def choose_ab(integer):
a, b = random(1, integer), random(1, integer)
d = a ** 2 - 4 * b
while is_square(d) or gcd(2 * d * a * b, integer) != 1:
a, b = random(1, integer), random(1, integer)
d = a ** 2 - 4 * b
return a, b, d
def is_square(integer):
if integer < 0:
return False
if integer < 2:
return True
x = integer >> 1
seen = set([x])
while x * x != integer:
x = (x + integer // x) >> 1
if x in seen:
return False
seen.add(x)
return True
def extended_gcd(n, d):
x1, x2, y1, y2 = 0, 1, 1, 0
while d:
n, (q, d) = d, divmod(n, d)
x1, x2, y1, y2 = x2 - q * x1, x1, y2 - q * y1, y1
return x2, y2
def jacobi_symbol(n, d):
j = 1
while n:
while not n & 1:
n >>= 1
if d & 7 in {3, 5}:
j = -j
n, d = d, n
if n & 3 == 3 == d & 3:
j = -j
n %= d
return j if d == 1 else 0
def compute_wm_wm1(w1, m, n):
a, b = 2, w1
for shift in range(m.bit_length() - 1, -1, -1):
if m >> shift & 1:
a, b = (a * b - w1) % n, (b * b - 2) % n
else:
a, b = (a * a - 2) % n, (a * b - w1) % n
return a, b
print('Probably prime:\n', find_prime_number(300, 10))
You seem to have misunderstood the algorithm completely due to not being familiar with the notation.
def frobenius_pseudoprime(integer):
assert integer & 1 and integer >= 3
a, b, d = choose_ab(integer)
w1 = (a ** 2 // b - 2) % integer
That comes from the line
W0 ≡ 2 (mod n) and W1 ≡ a2b−1 − 2 (mod n)
But the b-1 doesn't mean 1/b here, but the modular inverse of b modulo n, i.e. an integer c with b·c ≡ 1 (mod n). You can most easily find such a c by continued fraction expansion of b/n or, equivalently, but with slightly more computation, by the extended Euclidean algorithm. Since you're probably not familiar with continued fractions, I recommend the latter.
m = (integer - d // integer) // 2
comes from
n − (∆/n) = 2m
and misunderstands the Jacobi symbol as a fraction/division (admittedly, I have displayed it here even more like a fraction, but since the site doesn't support LaTeX rendering, we'll have to make do).
The Jacobi symbol is a generalisation of the Legendre symbol - denoted identically - which indicates whether a number is a quadratic residue modulo an odd prime (if n is a quadratic residue modulo p, i.e. there is a k with k^2 ≡ n (mod p) and n is not a multiple of p, then (n/p) = 1, if n is a multiple of p, then (n/p) = 0, otherwise (n/p) = -1). The Jacobi symbol lifts the restriction that the 'denominator' be an odd prime and allows arbitrary odd numbers as 'denominators'. Its value is the product of the Legendre symbols with the same 'numerator' for all primes dividing n (according to multiplicity). More on that, and how to compute Jacobi symbols efficiently in the linked article.
The line should correctly read
m = (integer - jacobi_symbol(d,integer)) // 2
The following lines I completely fail to understand, logically, here should follow the calculation of
Wm and Wm+1 using the recursion
W2j ≡ Wj2 − 2 (mod n)
W2j+1 ≡ WjWj+1 − W1 (mod n)
An efficient method of using that recursion to compute the required values is given around formula (11) of the PDF.
w_m0 = w1 * 2 // m % integer
w_m1 = w1 * 2 // (m + 1) % integer
w_m2 = (w_m0 * w_m1 - w1) % integer
The remainder of the function is almost correct, except of course that it now gets the wrong data due to earlier misunderstandings.
if w1 * w_m0 != 2 * w_m2:
The (in)equality here should be modulo integer, namely if (w1*w_m0 - 2*w_m2) % integer != 0.
return False
b = pow(b, (integer - 1) // 2, integer)
return b * w_m0 % integer == 2
Note, however, that if n is a prime, then
b^((n-1)/2) ≡ (b/n) (mod n)
where (b/n) is the Legendre (or Jacobi) symbol (for prime 'denominators', the Jacobi symbol is the Legendre symbol), hence b^((n-1)/2) ≡ ±1 (mod n). So you could use that as an extra check, if Wm is not 2 or n-2, n can't be prime, nor can it be if b^((n-1)/2) (mod n) is not 1 or n-1.
Probably computing b^((n-1)/2) (mod n) first and checking whether that's 1 or n-1 is a good idea, since if that check fails (that is the Euler pseudoprime test, by the way) you don't need the other, no less expensive, computations anymore, and if it succeeds, it's very likely that you need to compute it anyway.
Regarding the corrections, they seem correct, except for one that made a glitch I previously overlooked possibly worse:
if w1 * wm != 2 * wm1 % integer:
That applies the modulus only to 2 * wm1.
Concerning the recursion for the Wj, I think it is best to explain with a working implementation, first in toto for easy copy and paste:
def compute_wm_wm1(w1,m,n):
a, b = 2, w1
bits = int(log(m,2)) - 2
if bits < 0:
bits = 0
mask = 1 << bits
while mask <= m:
mask <<= 1
mask >>= 1
while mask > 0:
if (mask & m) != 0:
a, b = (a*b-w1)%n, (b*b-2)%n
else:
a, b = (a*a-2)%n, (a*b-w1)%n
mask >>= 1
return a, b
Then with explanations in between:
def compute_wm_wm1(w1,m,n):
We need the value of W1, the index of the desired number, and the number by which to take the modulus as input. The value W0 is always 2, so we don't need that as a parameter.
Call it as
wm, wm1 = compute_wm_wm1(w1,m,integer)
in frobenius_pseudoprime (aside: not a good name, most of the numbers returning True are real primes).
a, b = 2, w1
We initialise a and b to W0 and W1 respectively. At each point, a holds the value of Wj and b the value of Wj+1, where j is the value of the bits of m so far consumed. For example, with m = 13, the values of j, a and b develop as follows:
consumed remaining j a b
1101 0 w_0 w_1
1 101 1 w_1 w_2
11 01 3 w_3 w_4
110 1 6 w_6 w_7
1101 13 w_13 w_14
The bits are consumed left-to-right, so we have to find the first set bit of m and place our 'pointer' right before it
bits = int(log(m,2)) - 2
if bits < 0:
bits = 0
mask = 1 << bits
I subtracted a bit from the computed logarithm just to be entirely sure that we don't get fooled by a floating point error (by the way, using log limits you to numbers of at most 1024 bits, about 308 decimal digits; if you want to treat larger numbers, you have to find the base-2 logarithm of m in a different way, using log was the simplest way, and it's just a proof of concept, so I used that here).
while mask <= m:
mask <<= 1
Shift the mask until it's greater than m,so the set bit points just before m's first set bit. Then shift one position back, so we point at the bit.
mask >>= 1
while mask > 0:
if (mask & m) != 0:
a, b = (a*b-w1)%n, (b*b-2)%n
If the next bit is set, the value of the initial portion of consumed bits of m goes from j to 2*j+1, so the next values of the W sequence we need are W2j+1 for a and W2j+2 for b. By the above recursion formula,
W_{2j+1} = W_j * W_{j+1} - W_1 (mod n)
W_{2j+2} = W_{j+1}^2 - 2 (mod n)
Since a was Wj and b was Wj+1, a becomes (a*b - W_1) % n and b becomes (b * b - 2) % n.
else:
a, b = (a*a-2)%n, (a*b-w1)%n
If the next bit is not set, the value of the initial portion of consumed bits of m goes from j to 2*j, so a becomes W2j = (Wj2 - 2) (mod n), and b becomes
W2j+1 = (Wj * Wj+1 - W1) (mod n).
mask >>= 1
Move the pointer to the next bit. When we have moved past the final bit, mask becomes 0 and the loop ends. The initial portion of consumed bits of m is now all of m's bits, so the value is of course m.
Then we can
return a, b
Some additional remarks:
def find_prime_number(bits, test):
while True:
number = random(3, 1 << bits, 2)
for _ in range(test):
if not frobenius_pseudoprime(number):
break
else:
return number
Primes are not too frequent among the larger numbers, so just picking random numbers is likely to take a lot of attempts to hit one. You will probably find a prime (or probable prime) faster if you pick one random number and check candidates in order.
Another point is that such a test as the Frobenius test is disproportionally expensive to find that e.g. a multiple of 3 is composite. Before using such a test (or a Miller-Rabin test, or a Lucas test, or an Euler test, ...), you should definitely do a bit of trial division to weed out most of the composites and do the work only where it has a fighting chance of being worth it.
Oh, and the is_square function isn't prepared to deal with arguments less than 2, divide-by-zero errors lurk there,
def is_square(integer):
if integer < 0:
return False
if integer < 2:
return True
x = integer // 2
should help.
I need to find the time complexity (in terms of theta) of this function:
int x = 0;
for (int i=1; i < n ; i++) {
for (double j=i; j <= n ; j+=sqrt(i)) {
++x;
}
}
I know that the outer loop does n-1 iterations and the inner loop does (n-i)/sqrt(i) iterations so I need to calculate sigma of i=1 to n-1 of (n-i)/sqrt(i). Any idea how to do that?
EDIT:
Assume sqrt() runs in O(1).
I don't know what sigma and theta mean, but sqrt is a constant time operation so it basically doesn't matter in big O notation, ie j+=sqrt(i); is the same as j+=i; is the same as j+=1;. Also (n-k) ~= n for k much less than n. This means as n gets large n-i is just n. So (n-i) * sqrt() = n * 1 = n. And you do this n times for the outer loop so n^2.
Addition:
As to your complicated series, I'm sure this is accurate, but it is not what we care about, we care about the order of the operation. So we need show your series is O(n^2) or K*n^2. So you have i + 2*i + ... (n-1)*i + n*i. Where i is constant so we can factor it out and wrap it up in K and are left with 1 + ... + n. This statement is dominated by n ie as n gets large n ~= (n-1), and (n-1) ~= (n-2) which implies that (n-2) ~= n. Now this doesn't hold as we approach zero, but n is so large we can drop the first say million terms. so we are left with some function that looks like
C*(n-k)*n + c. where C, k, and c are all constant. Since we don't care about constants we just care about growth as n grows we can drop all these constants and just save the n^2. Alternatively, you could show that your series is bounded by n^k*n where k goes to one as n approaches infinity, but a good logic argument is usually better. ~Ben