0976HardMathEuclidean AlgorithmGreatest Common Divisor

Growing a Pair by Adding One to the Other

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A pair of whole numbers starts as (sx, sy). One step replaces the pair either by its first number plus its second alongside the second, or by the first alongside the first plus the second.

Return true when some sequence of steps turns the starting pair into (tx, ty).

Examples

Example 1

Input: sx = 1, sy = 1, tx = 3, ty = 2
Output: true

From 1 and 1, growing the second gives 1 and 2, then growing the first gives 3 and 2.

Example 2

Input: sx = 1, sy = 1, tx = 2, ty = 2
Output: false

Any step from 1 and 1 makes the two numbers different, and no later step can bring them level again, so an equal pair is out of reach.

Example 3

Input: sx = 2, sy = 2, tx = 4, ty = 2
Output: true

Growing the first number by the second takes 2 and 2 straight to 4 and 2.

Constraints

1 <= sx <= 10^9
1 <= sy <= 10^9
1 <= tx <= 10^9
1 <= ty <= 10^9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def reaching_points(sx: int, sy: int, tx: int, ty: int) -> bool:

Java

public boolean reachingPoints(int sx, int sy, int tx, int ty)

The intended approach

Time O(log(max(tx, ty))) for the chain of remainders.Space O(1).

Both numbers stay positive and every step makes one of them larger, so walking forwards branches endlessly. Walking backwards does not: from a pair, the step that produced it is forced, because the larger number must be the one that was just grown, and taking the smaller off it undoes the step. Only when the two are equal is there no way back. That gives a walk from the target back towards the start, and it can be hurried along: while the larger number is still above the start's matching number, taking the smaller off it repeatedly is the same as one remainder step, so the whole walk costs no more than a chain of remainders. When one of the pair reaches its target value, the other must be reachable by adding whole multiples of it, which is one divisibility test.

Pythonentry point: reaching_points

def reaching_points(sx, sy, tx, ty):
    while tx > sx and ty > sy:
        if tx > ty:
            tx %= ty
        else:
            ty %= tx
    if tx == sx and ty >= sy:
        return (ty - sy) % tx == 0
    if ty == sy and tx >= sx:
        return (tx - sx) % ty == 0
    return False

The obvious version

Walk forwards from the starting pair, taking both steps at every turn and dropping any pair whose numbers have passed the target, until the target turns up or nothing is left to try. Explores the steps rather than undoing them, so it only copes with small targets.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_reaching_points(sx, sy, tx, ty):
    from collections import deque
    start = (sx, sy)
    seen = set([start])
    queue = deque([start])
    while queue:
        a, b = queue.popleft()
        if a == tx and b == ty:
            return True
        for nxt in ((a + b, b), (a, a + b)):
            if nxt[0] > tx or nxt[1] > ty:
                continue
            if nxt not in seen:
                seen.add(nxt)
                queue.append(nxt)
    return False

How this problem was checked

Reference solution executed against 16 test cases — 16 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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