Great question! Optimizing that square root step is such a smart way to speed up Hart's one-pass factorization—since you’ve already noted that’s where most of the time is spent, reusing the previous result instead of computing each square root from scratch is going to save you a ton of cycles. Let’s break down how to do this efficiently:

Core Approach: Incremental Adjustment #

The key insight is that ni = n(i-1) + n, so the square root of ni is just slightly larger than the square root of n(i-1) (especially as i grows). Instead of starting from scratch, we can use the previous ceiling square root s_prev = ⌈√(n(i-1))⌉ as our starting point and adjust it upward (or rarely, check if it’s already sufficient) to get s_curr = ⌈√(ni)⌉.

Step-by-Step Method #

Let’s formalize this with integer-only operations (no floating-point math needed, which avoids precision errors):

• Define your values:
  • Let prev_val = n*(i-1) and curr_val = ni = prev_val + n
  • You already know s_prev satisfies (s_prev - 1)² < prev_val ≤ s_prev²
• Initialize the candidate:
  Start with s_curr = s_prev—since curr_val > prev_val, s_curr will never be smaller than s_prev (the only edge case is when ni is a perfect square equal to s_prev², but that’s rare and easy to handle)
• Adjust to the ceiling square root:
  Loop to refine s_curr until it’s the smallest integer where s_curr² ≥ curr_val:
  1. If s_curr² < curr_val, calculate a smart delta to jump closer to the target:
     Use delta = max(1, (curr_val - s_curr*s_curr) // (2*s_curr))—this comes from the expansion (s + delta)² = s² + 2s*delta + delta², so the delta approximates how much we need to add to s_curr to cover the gap between s_curr² and curr_val
  2. Add this delta to s_curr
  3. Once s_curr² ≥ curr_val, double-check if we can reduce it by 1 (in case the delta overshot) to ensure it’s the minimal ceiling value

Critical Optimizations #

• Avoid integer overflow: For large n or i, multiplying s_curr * s_curr might overflow your integer type. Instead, compare s_curr > curr_val // s_curr—this is equivalent to checking s_curr² > curr_val without direct multiplication. For perfect squares, you’ll also need to verify curr_val % s_curr == 0
• Large i shortcut: As i grows, the difference between ni and n(i-1) becomes tiny relative to their size. In these cases, s_curr will usually be either s_prev or s_prev + 1—you can skip the delta calculation entirely and just check those two values

Example Pseudocode #

Here’s how this might look in practice (adjust for your language’s integer handling):

def get_next_ceil_sqrt(n, i, s_prev):
    curr_val = n * i
    s_curr = s_prev

    while True:
        # Check if s_curr is too small, using division to avoid overflow
        if s_curr <= curr_val // s_curr:
            # s_curr² <= curr_val; calculate how much to add
            if s_curr == 0:
                delta = 1
            else:
                delta = max(1, (curr_val - s_curr * s_curr) // (2 * s_curr))
            s_curr += delta
        else:
            # s_curr² > curr_val; check if we can go smaller
            if (s_curr - 1) > curr_val // (s_curr - 1):
                s_curr -= 1
            else:
                break

    # Final sanity check to ensure we have the ceiling
    if (s_curr - 1) ** 2 >= curr_val:
        s_curr -= 1
    elif s_curr ** 2 < curr_val:
        s_curr += 1

    return s_curr

Why This Beats Recomputing the Square Root #

Traditional square root algorithms (like Newton-Raphson) start from a rough initial guess and iterate to converge. Here, your initial guess (s_prev) is already extremely close to the target—especially as i increases. For small i, the delta calculation lets you jump most of the way to the target in one step, and for large i, you’ll often get the correct value with zero or one adjustments.

内容的提问来源于stack exchange，提问作者user448810