Introduction

The perimeter density matrix is an infinite binary matrix M defined as follows. Consider a (1-based) index (x, y), and denote by M[x, y] the rectangular sub-matrix spanned by the corner (1, 1) and (x, y). Suppose that all values of M[x, y] except M_{x, y}, the value at index (x, y), have already been determined. Then the value M_{x, y} is whichever of 0 or 1 that puts the average value of M[x, y] closer to 1 / (x + y). In case of a tie, choose M_{x, y} = 1.

This is the sub-matrix M[20, 20] with zeros replaced by dots for clarity:

1 . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

For example, we have M_{1, 1} = 1 at the upper left corner, since 1 / (1 + 1) = ½, and the average of the 1 × 1 sub-matrix M[1, 1] is either 0 or 1; that's a tie, so we choose 1.

Consider then the position (3, 4). We have 1 / (3 + 4) = 1/7, and the average of the sub-matrix M[3, 4] is 1/6 if we choose 0, and 3/12 if we choose 1. The former is closer to 1/7, so we choose M_{3, 4} = 0.

Here is the sub-matrix M[800, 800] as an image, showing some of its intricate structure.

The task

Given a positive integer N < 1000, output the N × N sub-matrix M[N, N], in any reasonable format. The lowest byte count wins.