Find the perfect square in the string - c++

Question

A perfect square is taken in binary and some bits are replaced with "?" for example 1??, the number would be 4.(or 1????000???0000)
I need to find that perfect square.(there will be only such possible number)
number of '?'s in the string be n
To find that number what I am doing is iterating through 2**n numbers(111,110,101,100) and checking if it is a perfect square. I am using following function to check if it is a perfect square.
bool issqr(int n){
int d=(int)(sqrt(n));
if(d*d==n) return true;
else return false;
}
Even though in python I did it, it is taking a lot of time, so I shifted to C++ using only bit operations for populating 2**n numbers(which was much faster than the python version)
but this fails if the number has more than 64 bits
How to avoid this problem? How can I do the same thing if a number has say 120 bits.
(10100110???1?1?01?1?011000?1100?00101000?1?11001101100110001010111?0?1??0110?110?01?1100?1?0110?1?10111?01?0111000?10??101?01)

Answer 1

Rather than re-writing in C++ you should first have looked at improving your algorithm. The lowest possible answer is the square root from the original value with all '?' replace by 0 rounded up, the highest possible answer is the square root of the pattern with the '?'s replaced by 1 rounded down. Find those two values, iterate through them, square and check against the pattern.
This is faster both because you are iterating through many fewer numbers and because you aren't calculating any square roots in the loop: squaring is much easier.
You don't need to compare string to check for a match:
mask = int(pattern.replace('0', '1').replace('?', '0'), 2)
test = int(pattern.replace('?', '0'), 2)
def is_match(n):
return (n&mask)==test
So putting it all together:
def int_sqrt(x):
if x < 0:
raise ValueError('square root not defined for negative numbers')
n = int(x)
if n == 0:
return 0
a, b = divmod(n.bit_length(), 2)
x = 2**(a+b)
while True:
y = (x + n//x)//2
if y >= x:
return x
x = y
def find_match(pattern):
lowest = int(pattern.replace('?', '0'), 2)
highest = int(pattern.replace('?', '1'), 2)
mask = int(pattern.replace('0', '1').replace('?', '0'), 2)
lowsqrt = int_sqrt(lowest)
if lowsqrt*lowsqrt != lowest:
lowsqrt += 1
highsqrt = int_sqrt(highest)
for n in range(lowsqrt, highsqrt+1):
if (n*n & mask)==lowest:
return n*n
print(find_match('1??1??1'))
print(find_match('1??0??1'))
print(find_match('1??????????????????????????????????????????????????????????????????????1??0??1'))
Output:
121
81
151115727461209345152081
N.B. This only works in Python 3.x, the last test will overflow range in Python 2.x

Answer 2

From my understanding, given an integer n you are trying to find a square number sq that matches :
2n - 1 < sq < 2n+1 - 1
This condition is the mathematic translation of "my number must have the form 1????" where there are n "?".
First, you can notice that if n is even, the number 2n is a perfect square and matches your condition (in binary, it is the number 1000...000 - n zeroes -).
If n is uneven (say n = 2.p + 1), then 2n+1 is a perfect square ((2p+1)2). Computing the following number will give you a perfect square :
(2p+1 - 1)2
To satisfy the first inequality, p must satisfy :
2n - 1 < (2p+1 - 1)2
Then
0 < 2n+1 - 2p+2 + 1 - 2n + 1,
Finally,
2n + 2 - 2p+2 > 0
Or
22p - 2p+1 + 1 > 0
If we consider the function that matches p with f(p) such that :
f(p) = 22p - 2p+1 + 1
This function is defined for each positive real number, and is strictly increasing. Moreover, f(0) = 0. Finally, the initial condition is satisfied when p > 0 !
For p = 0 - or n = 1 -, the problem does not have a valid solution.

Answer 3

You don't need to iterate trough all the 2**n numbers to find the perfect square, in fact you only need one fractional square operation:
Say you have integer n and you want to find the largest perfect square smaller or equal than n, let's call it m.
Then:
d = (int)sqrt(n);
m = d*d;
Explanation:
Assume there is a perfect square m' larger than m , this implies there is an integer d' so that: d' > d and d'*d' = m'.
But d' >=d+1 and (d+1)*(d+1) > n so m' > n in contradiction to our requirement m' <= n.
Now to your question:
in order to find the perfect squares just change to "1" all of the "?" and find the perfect square if it conforms to your string you got the number you're looking for, if not change just enough "?" from the msb to "0" so that the resulting number is smaller or equal to the perfect square you just found, and keep going until you find the perfect square or run out of options.

Answer 4

Your operations may be returning something too large for an integer...
http://www.cplusplus.com/doc/tutorial/variables/