Connected components of the graph on $[n]$ in which $i,j$ are connected if $\mathrm{gcd}(i,j) > g$

Question

I recently got asked the following question:

A set of $n$ cities are numbered from 1 to $n$. Given a positive integer $g$, two cities are connected if their greatest common divisor is greater than $g$. The number $n$ may be as large as $10^5$.

Alice has a list of $q$ cities that she'd like to visit. Given Alice's list of $q$ origin cities and $q$ destination cities, return a boolean vector with a 1 in the $i$th position indicating that it's possible to reach the $i$th destination city from the $i$th origin city, and 0 if not. The length of the list $q$ may be as large as $10^5$ as well.

I solved this problem with a quadratic time solution. However, the large upper bound for $n$ and $q$ clearly implies there is a better runtime. I haven't yet seen a way to speed up the creation of the graph, however.

My solution builds the graph in quadratic time, using a union find data structure to handle the component merging, and Euclid's algorithm for computing gcd's.

The fundamental question is whether there exists a faster than $O(n^2)$ algorithm to construct and query an undirected graph, when edges are defined by a pairwise rule, and it appears that every pair $(i, j)$ for $1 \leq i,j \leq n$ needs to be looked at.

Answer 1

Given parameters $n$ and $g$, you want to find the connected components of the graph on $[n] := \{1,\ldots,n\}$ whose edges are $\{(i,j) \in [n] : (i,j) > g\}$. Let us write these edges in a different way: $$ \bigcup_{d=g+1}^n \{(i,j) \in [n]: d \mid (i,j) \} = \bigcup_{d=g+1}^n \{(dI,dJ) : 1 \leq I,J \leq n/d \}. $$ For each $d$, we have a clique on $\lfloor n/d \rfloor$ vertices. We can preserve the same connectivity properties by replacing it with a path: $$ \bigcup_{d=g+1}^n \{(dI,d(I+1)) : 1 \leq I < \lfloor n/d \rfloor \}. $$ The number of edges in the new graph is $$ \sum_{d=g+1}^n \lfloor n/d \rfloor \leq n \sum_{d=g+1}^n \frac{1}{d} = O(n\log n). $$ You can find the connected components using BFS/DFS in time $O(n\log n)$, a large improvement over your quadratic time algorithm.

Answer 2

Pseudo code for finding the minimum connected edges.

    n=100
    lst = range(1,101)
    g=10
    ans = list(tuple())
    for d in range(g+1,n+1):
       for i in range(1,n/d):
         ans.append((d*i,d*(i+1)))
    print ans
    print "size="+str(len(ans))

Output:

   [(11, 22), (22, 33), (33, 44), (44, 55), (55, 66), (66, 77), (77, 88), (88, 99), (12, 24), (24, 36), (36, 48), (48, 60), (60, 72), (72, 84), (84, 96), (13, 26), (26, 39), (39, 52), (52, 65), (65, 78), (78, 91), (14, 28), (28, 42), (42, 56), (56, 70), (70, 84), (84, 98), (15, 30), (30, 45), (45, 60), (60, 75), (75, 90), (16, 32), (32, 48), (48, 64), (64, 80), (80, 96), (17, 34), (34, 51), (51, 68), (68, 85), (18, 36), (36, 54), (54, 72), (72, 90), (19, 38), (38, 57), (57, 76), (76, 95), (20, 40), (40, 60), (60, 80), (80, 100), (21, 42), (42, 63), (63, 84), (22, 44), (44, 66), (66, 88), (23, 46), (46, 69), (69, 92), (24, 48), (48, 72), (72, 96), (25, 50), (50, 75), (75, 100), (26, 52), (52, 78), (27, 54), (54, 81), (28, 56), (56, 84), (29, 58), (58, 87), (30, 60), (60, 90), (31, 62), (62, 93), (32, 64), (64, 96), (33, 66), (66, 99), (34, 68), (35, 70), (36, 72), (37, 74), (38, 76), (39, 78), (40, 80), (41, 82), (42, 84), (43, 86), (44, 88), (45, 90), (46, 92), (47, 94), (48, 96), (49, 98), (50, 100)]
   size=101

We can see that edges like (11,33), (11, 44) etc are not part of the set even though they are directly connected.

After we get the minimum number of connected edges we can apply disjoints sets to solve the above question.

Answer 3

Hint: to determine whether there is a path from source $s$ to destination $t$, factor both $s$ and $t$. What properties need to hold true about their factorization, for such a path to exist?