# Compute the square root of a positive integer using binary search

The requirement is to find the square root of a positive integer using binary search and the math property that square root of a number n is between 0 and n/2, and the required answer is "floored", meaning mySqrt(8) is to return 2.

Please comment on the efficiency, and if possible, the loop invariants in terms of correctness:

class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int

Loop invariant:

The answer is always in the range [low, high] inclusive,
except possibly:

1) low == high == mid, and mid * mid == x and
any of low, high, or mid can be returned as the answer.

2) if there is no exact answer and the floor is to be
returned, then low > high by 1. Since sq != x,
so either low or high is set inside the loop.

If low gets set and gets pushed up, it is pushed up too much.
So when low > high by 1, low - 1 is the answer and it is the same
as high, because low > high by 1.

If high gets set and gets pushed down, high can be
the correct answer. When low > high, it is by 1,
and high is the correct floor value to be returned.
(since there is no perfect square root and the floor is required)

0 <= low <= answer <= high <= n//2 + 1

where answer is floor(sqrt(x)) to be found,
except if low > high and the loop will exit.

Each loop iteration always makes the range smaller.

If the range is empty at the end, low is > high by 1, and high is
the correct floored value, and low is the ceiling value, so high is returned.
"""

low = 0;
high = x//2 + 1;

while (low <= high):
mid = low + (high - low) // 2;
sq = mid * mid;
if (sq == x):
return mid;
elif (sq > x):
high = mid - 1;  # sq exceeds target, so mid cannot be the answer floored, but when high is set to mid - 1, then it can be the answer
else:
low = mid + 1;   # (here sq < x, and mid might be the answer floored, so when low is set to mid + 1, then low might be too big, while high is correct)

return high;

• Please do not update the code in your question after receiving feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers. – Mast Aug 18 at 7:47

• Your comments on the elif / else part are too long to be just after the statements

• Don't use semicolons (;) in Python.

• This is a refactored version of the code

import math

class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int

Returns floor(sqrt(x))
"""
low = 0
high = x//2 + 1

"""
It is proved that 0 <= sqrt(x) <= x/2, so
we run a dichotomic in [0, x/2] to find floor(sqrt(x))

Loop analysis:
* Initialization: low = 0 and high = x/2 + 1
* Termination: |high-low| is reduced each iteration,
as shown in lines high = mid - 1 and low = mid + 1.
* Invariant: low <= floor(sqrt(x)) <= high.
Let mid be (low + high)/2.
- If mid^2 <= x < (mid+1)^2,
then mid is floor(sqrt(x)) and just return it.
- If mid^2 > x, search for values smaller than mid.
- Otherwise, if mid^2 < x, search within higher values.
"""
while (low <= high):
mid = (low + high) // 2
sq = mid * mid
sq_next = (mid+1)*(mid+1)
if (sq <= x < sq_next):
return mid
elif (sq > x):
high = mid - 1
else:
low = mid + 1

for i in range(1, 26):
assert(math.floor(math.sqrt(i)) == Solution().mySqrt(i))

• I suppose if it is Python, it can be mid = (low + high) // 2 because there is infinite precision arithmetics... but it just depends whether you want it first to overflow (either 32 bit or 64 bit int) to become bignum, and then divided by 2 to make it back to a 32 bit or 64 bit int – nopole Aug 17 at 20:28
• @太極者無極而生 Take a look to the new answer; now there is no need for returning high or low after the loop, so it is easier to reason about. – JnxF Aug 17 at 21:04
• you are using $$mid^2$$ and $$(mid+1)^2$$ to check for the answer and return and no need to consider how low or high gets set... interesting... it looks like it can make it simpler loop invariants – nopole Aug 17 at 21:19
• how come you assert from 1 to 26 instead of from 1 to some larger number like... 5000 or a million – nopole Aug 17 at 21:26
• @太極者無極而生 Code is proved to be correct mathematically, so you can check up to any number you want. Checked up to $1000000$ and it works (surely, you can try up to any number you want). – JnxF Aug 17 at 21:32