Understanding the Problem
We are given two numbers: x
and n
. Our goal is to find the nth root of x
, which means finding a number r
such that rn = x
.
However, in many cases, the exact root may not be an integer or may not exist in the real number system. So, we want to find an approximation with high precision using an efficient algorithm that works for both small and large values of x
and n
.
To do this efficiently, we use binary search because the function f(r) = rn
is monotonic — it increases as r
increases.
Step-by-Step Solution with Example
step 1: Understand special cases
Before diving into the binary search logic, we must handle edge cases:
- If
x = 0
: The result is 0 (since 0 raised to any positive power is 0).
- If
x = 1
: The result is 1 (since 1 raised to any power is still 1).
- If
n = 1
: The root of degree 1 is the number itself, so return x
.
- If
n ≤ 0
: The root is not defined or non-standard — return -1
.
- If
x < 0
and n
is even: The root doesn't exist in real numbers — return -1
.
- If
x < 0
and n
is odd: The answer exists and is negative.
step 2: Setup binary search range
We define our search space:
- If
x ≥ 1
, the nth root lies between 1
and x
.
- If
0 < x < 1
, the nth root lies between 0
and 1
.
Initialize:
low = 0
high = max(1, x)
epsilon = 1e-6 # acceptable precision
step 3: Apply binary search
Repeat the following steps until the high and low pointers are close enough (within epsilon
):
while (high - low) > epsilon:
mid = (low + high) / 2
if mid**n < x:
low = mid
else:
high = mid
Once the loop ends, either low
or high
is an accurate enough approximation.
step 4: Return result
We return either low
or high
as the answer — both are close enough due to the epsilon
margin.
step 5: Example – Find cube root of 27
Let’s walk through the algorithm for x = 27
and n = 3
(i.e., find the cube root of 27).
- Initial range:
low = 0
, high = 27
- Mid = 13.5 → 13.53 = 2460.375 → Too high →
high = 13.5
- Mid = 6.75 → 6.753 = 307.546875 → Too high →
high = 6.75
- Mid = 3.375 → 3.3753 = 38.44 → Still too high →
high = 3.375
- Mid = 1.6875 → 1.68753 = 4.80 → Too low →
low = 1.6875
- Keep repeating...
- Eventually we’ll get
mid ≈ 3.0
with mid3 ≈ 27
Edge Cases
- Negative inputs with even roots: Example:
x = -16, n = 2
→ Invalid in real numbers.
- Fractional x: Example:
x = 0.008, n = 3
→ Root lies between 0 and 1.
- Large x and n: The method works well due to logarithmic time — just ensure high precision (
epsilon
).
Finally
This solution breaks down the problem clearly and solves it with binary search — a very efficient and beginner-friendly technique for continuous, monotonic functions like rn
. By handling special cases and edge cases explicitly, we ensure correctness and robustness.
Always remember to test your logic against different input types: small, large, negative, and fractional values. And tune your precision epsilon
based on how accurate the result needs to be.
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