How to find ith item in zigzag ordering? - algorithm

Question

A question last week defined the zig zag ordering on an n by m matrix and asked how to list the elements in that order.
My question is how to quickly find the ith item in the zigzag ordering? That is, without traversing the matrix (for large n and m that's much too slow).
For example with n=m=8 as in the picture and (x, y) describing (row, column)
f(0) = (0, 0)
f(1) = (0, 1)
f(2) = (1, 0)
f(3) = (2, 0)
f(4) = (1, 1)
...
f(63) = (7, 7)
Specific question: what is the ten billionth (1e10) item in the zigzag ordering of a million by million matrix?

Answer 1

Let's assume that the desired element is located in the upper half of the matrix. The length of the diagonals are 1, 2, 3 ..., n.
Let's find the desired diagonal. It satisfies the following property:
sum(1, 2 ..., k) >= pos but sum(1, 2, ..., k - 1) < pos. The sum of 1, 2, ..., k is k * (k + 1) / 2. So we just need to find the smallest integer k such that k * (k + 1) / 2 >= pos. We can either use a binary search or solve this quadratic inequality explicitly.
When we know the k, we just need to find the pos - (k - 1) * k / 2 element of this diagonal. We know where it starts and where we should move(up or down, depending on the parity of k), so we can find the desired cell using a simple formula.
This solution has an O(1) or an O(log n) time complexity(it depends on whether we use a binary search or solve the inequation explicitly in step 2).
If the desired element is located in the lower half of the matrix, we can solve this problem for a pos' = n * n - pos + 1 and then use symmetry to get the solution to the original problem.
I used 1-based indexing in this solution, using 0-based indexing might require adding +1 or -1 somewhere, but the idea of the solution is the same.
If the matrix is rectangular, not square, we need to consider the fact the length of diagonals look this way: 1, 2, 3, ..., m, m, m, .., m, m - 1, ..., 1(if m <= n) when we search for the k, so the sum becomes something like k * (k + 1) / 2 if k <= m and k * (k + 1) / 2 + m * (k - m) otherwise.

Answer 2

import math, random
def naive(n, m, ord, swap = False):
dx = 1
dy = -1
if swap:
dx, dy = dy, dx
cur = [0, 0]
for i in range(ord):
cur[0] += dy
cur[1] += dx
if cur[0] < 0 or cur[1] < 0 or cur[0] >= n or cur[1] >= m:
dx, dy = dy, dx
if cur[0] >= n:
cur[0] = n - 1
cur[1] += 2
if cur[1] >= m:
cur[1] = m - 1
cur[0] += 2
if cur[0] < 0: cur[0] = 0
if cur[1] < 0: cur[1] = 0
return cur
def fast(n, m, ord, swap = False):
if n < m:
x, y = fast(m, n, ord, not swap)
return [y, x]
alt = n * m - ord - 1
if alt < ord:
x, y = fast(n, m, alt, swap if (n + m) % 2 == 0 else not swap)
return [n - x - 1, m - y - 1]
if ord < (m * (m + 1) / 2):
diag = int((-1 + math.sqrt(1 + 8 * ord)) / 2)
parity = (diag + (0 if swap else 1)) % 2
within = ord - (diag * (diag + 1) / 2)
if parity: return [diag - within, within]
else: return [within, diag - within]
else:
ord -= (m * (m + 1) / 2)
diag = int(ord / m)
within = ord - diag * m
diag += m
parity = (diag + (0 if swap else 1)) % 2
if not parity:
within = m - within - 1
return [diag - within, within]
if __name__ == "__main__":
for i in range(1000):
n = random.randint(3, 100)
m = random.randint(3, 100)
ord = random.randint(0, n * m - 1)
swap = random.randint(0, 99) < 50
na = naive(n, m, ord, swap)
fa = fast(n, m, ord, swap)
assert na == fa, "(%d, %d, %d, %s) ==> (%s), (%s)" % (n, m, ord, swap, na, fa)
print fast(1000000, 1000000, 9999999999, False)
print fast(1000000, 1000000, 10000000000, False)
So the 10-billionth element (the one with ordinal 9999999999), and the 10-billion-first element (the one with ordinal 10^10) are:
[20331, 121089]
[20330, 121090]

Answer 3

An analytical solution
In the general case, your matrix will be divided in 3 areas:
an initial triangle t1
a skewed part mid where diagonals have a constant length
a final triangle t2
Let's call p the index of your diagonal run.
We want to define two functions x(p) and y(p) that give you the column and row of the pth cell.
Initial triangle
Let's look at the initial triangular part t1, where each new diagonal is one unit longer than the preceding.
Now let's call d the index of the diagonal that holds the cell, and
Sp = sum(di) for i in [0..p-1]
We have p = Sp + k, with 0 <=k <= d and
Sp = d(d+1)/2
if we solve for d, it brings
d²+d-2p = 0, a quadratic equation where we retain only the positive root:
d = (-1+sqrt(1+8*p))/2
Now we want the highest integer value closest to d, which is floor(d).
In the end, we have
p = d + k with d = floor((-1+sqrt(1+8*p))/2) and k = p - d(d+1)/2
Let's call
o(d) the function that equals 1 if d is odd and 0 otherwise, and
e(d) the function that equals 1 if d is even and 0 otherwise.
We can compute x(p) and y(p) like so:
d = floor((-1+sqrt(1+8*p))/2)
k = p - d(d+1)/2
o = d % 2
e = 1 - o
x = e*d + (o-e)*k
y = o*d + (e-o)*k
even and odd functions are used to try to salvage some clarity, but you can replace
e(p) with 1 - o(p) and have slightly more efficient but less symetric formulaes for x and y.
Middle part
let's consider the smallest matrix dimension s, i.e. s = min (m,n).
The previous formulaes hold until x or y (whichever comes first) reaches the value s.
The upper bound of p such as x(i) <= s and y(i) <= s for all i in [0..p]
(i.e. the cell indexed by p is inside the initial triangle t1) is given by
pt1 = s(s+1)/2.
For p >= pt1, diagonal length remains equal to s until we reach the second triangle t2.
when inside mid, we have:
p = s(s+1)/2 + ds + k with k in [0..s[.
which yields:
d = floor ((p - s(s+1)/2)/s)
k = p - ds
We can then use the same even/odd trick to compute x(p) and y(p):
p -= s(s+1)/2
d = floor (p / s)
k = p - d*s
o = (d+s) % 2
e = 1 - o
x = o*s + (e-o)*k
y = e*s + (o-e)*k
if (n > m)
x += d+e
y -= e
else
y += d+o
x -= o
Final triangle
Using symetry, we can calculate pt2 = m*n - s(s+1)/2
We now face nearly the same problem as for t1, except that the diagonal may run in the same direction as for t1 or in the reverse direction (if n+m is odd).
Using symetry tricks, we can compute x(p) and y(p) like so:
p = n*m -1 - p
d = floor((-1+sqrt(1+8*p))/2)
k = p - d*(d+1)/2
o = (d+m+n) % 2
e = 1 - $o;
x = n-1 - (o*d + (e-o)*k)
y = m-1 - (e*d + (o-e)*k)
Putting all together
Here is a sample c++ implementation.
I used 64 bits integers out of sheer lazyness. Most could be replaced by 32 bits values.
The computations could be made more effective by precomputing a few more coefficients.
A good part of the code could be factorized, but I doubt it is worth the effort.
Since this is just a quick and dirty proof of concept, I did not optimize it.
#include <cstdio> // printf
#include <algorithm> // min
using namespace std;
typedef long long tCoord;
void panic(const char * msg)
{
printf("PANIC: %s\n", msg);
exit(-1);
}
struct tPoint {
tCoord x, y;
tPoint(tCoord x = 0, tCoord y = 0) : x(x), y(y) {}
tPoint operator+(const tPoint & p) const { return{ x + p.x, y + p.y }; }
bool operator!=(const tPoint & p) const { return x != p.x || y != p.y; }
};
class tMatrix {
tCoord n, m; // dimensions
tCoord s; // smallest dimension
tCoord pt1, pt2; // t1 / mid / t2 limits for p
public:
tMatrix(tCoord n, tCoord m) : n(n), m(m)
{
s = min(n, m);
pt1 = (s*(s + 1)) / 2;
pt2 = n*m - pt1;
}
tPoint diagonal_cell(tCoord p)
{
tCoord x, y;
if (p < pt1) // inside t1
{
tCoord d = (tCoord)floor((-1 + sqrt(1 + 8 * p)) / 2);
tCoord k = p - (d*(d + 1)) / 2;
tCoord o = d % 2;
tCoord e = 1 - o;
x = o*d + (e - o)*k;
y = e*d + (o - e)*k;
}
else if (p < pt2) // inside mid
{
p -= pt1;
tCoord d = (tCoord)floor(p / s);
tCoord k = p - d*s;
tCoord o = (d + s) % 2;
tCoord e = 1 - o;
x = o*s + (e - o)*k;
y = e*s + (o - e)*k;
if (m > n) // vertical matrix
{
x -= o;
y += d + o;
}
else // horizontal matrix
{
x += d + e;
y -= e;
}
}
else // inside t2
{
p = n * m - 1 - p;
tCoord d = (tCoord)floor((-1 + sqrt(1 + 8 * p)) / 2);
tCoord k = p - (d*(d + 1)) / 2;
tCoord o = (d + m + n) % 2;
tCoord e = 1 - o;
x = n - 1 - (o*d + (e - o)*k);
y = m - 1 - (e*d + (o - e)*k);
}
return{ x, y };
}
void check(void)
{
tPoint move[4] = { { 1, 0 }, { -1, 1 }, { 1, -1 }, { 0, 1 } };
tPoint pos;
tCoord dir = 0;
for (tCoord p = 0; p != n * m ; p++)
{
tPoint dc = diagonal_cell(p);
if (pos != dc) panic("zot!");
pos = pos + move[dir];
if (dir == 0)
{
if (pos.y == m - 1) dir = 2;
else dir = 1;
}
else if (dir == 3)
{
if (pos.x == n - 1) dir = 1;
else dir = 2;
}
else if (dir == 1)
{
if (pos.y == m - 1) dir = 0;
else if (pos.x == 0) dir = 3;
}
else
{
if (pos.x == n - 1) dir = 3;
else if (pos.y == 0) dir = 0;
}
}
}
};
void main(void)
{
const tPoint dim[] = { { 10, 10 }, { 11, 11 }, { 10, 30 }, { 30, 10 }, { 10, 31 }, { 31, 10 }, { 11, 31 }, { 31, 11 } };
for (tPoint d : dim)
{
printf("Checking a %lldx%lld matrix...", d.x, d.y);
tMatrix(d.x, d.y).check();
printf("done\n");
}
tCoord p = 10000000000;
tMatrix matrix(1000000, 1000000);
tPoint cell = matrix.diagonal_cell(p);
printf("Coordinates of %lldth cell: (%lld,%lld)\n", p, cell.x, cell.y);
}
Results are checked against "manual" sweep of the matrix.
This "manual" sweep is a ugly hack that won't work for a one-row or one-column matrix, though diagonal_cell() does work on any matrix (the "diagonal" sweep becomes linear in that case).
The coordinates found for the 10.000.000.000th cell of a 1.000.000x1.000.000 matrix seem consistent, since the diagonal d on which the cell stands is about sqrt(2*1e10), approx. 141421, and the sum of cell coordinates is about equal to d (121090+20330 = 141420). Besides, it is also what the two other posters report.
I would say there is a good chance this lump of obfuscated code actually produces an O(1) solution to your problem.