反转Arnold的Cat地图

反转Arnold的Cat地图 - 负数组索引(Invert Arnold's Cat map - negative array indexes)

我正在尝试使用以下公式为N * N图像实现Arnold的Cat图

for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { desMatrix[(i + j) % N][(i + 2 * j) % N] = srcMatrix[i][j]; } }

要反转我做的过程：

for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { srcMatrix[(j-i) % N][(2*i-j) % N] = destMatrix[i][j]; } }

实施是否正确？

在我看来，对于j和i的某些值，我可能从（ji）和（2 * ij）获得负指数; 我应该如何处理这些情况，因为矩阵指数只是正数？

I'm trying to implement Arnold's Cat map for N*N images using the following formula

for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { desMatrix[(i + j) % N][(i + 2 * j) % N] = srcMatrix[i][j]; } }

To invert the process I do:

for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { srcMatrix[(j-i) % N][(2*i-j) % N] = destMatrix[i][j]; } }

Is the implementation correct?

It seems to me that for certain values of j and i I might get negative indexes from (j-i) and (2*i-j); how should I handle those cases, since matrix indexes are only positive?

最满意答案

通常，当模数（％）操作需要处理负索引时，您可以根据需要简单地添加模数参数。 以来

x % N == ( x + a*N ) % N

对于所有自然a，并且在这种情况下你有i和j约束在[0，N]，那么你可以写（N + i - j）并确保即使我是0而j是N-1（甚至是N就此问题而言，结果总是非负的。 出于同样的原因，（2 * N + i - 2 * j）或等效地（i + 2 *（Nj））总是非负的。

但是，在这种情况下，这不是必需的。 要反转地图，您将重复前进步骤以反转分配 。 由于矩阵具有一元行列式并且是面积保持的，因此您可以确保最终获得所有积分（即覆盖M（i + 1）将产生M（i）的覆盖）。

for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { newMatrix[i][j] = desMatrix[(i + j) % N][(i + 2 * j) % N]; } }

在这一点上，newMatrix和srcMatrix应该完全相同。

^{（实际上，你已经在运行逆向转换作为前向转换了。我设置的反向转换是前向转换的一种常用形式）。}

In general, when a modulo (%) operation needs to work on negative indexes, you can simply add the modulo argument as many times as it's needed. Since

x % N == ( x + a*N ) % N

for all natural a's, and in this case you have i and j constrained in [0, N), then you can write (N + i - j) and ensure that even if i is 0 and j is N-1 (or even N for that matter), the result will always be non-negative. By the same token, (2*N + i - 2*j) or equivalently (i + 2*(N-j)) is always non-negative.

In this case, though, this is not necessary. To invert your map, you would repeat the forward step reversing the assignments. Since the matrix has unary determinant and is area-preserving, you're assured that you'll get all your points eventually (i.e. covering M(i+1) will yield a covering of M(i)).

for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { newMatrix[i][j] = desMatrix[(i + j) % N][(i + 2 * j) % N]; } }

At this point newMatrix and srcMatrix ought to be identical.

^{(Actually, you're already running the reverse transformation as your forward one. The one I set up to reverse yours is the one commonly used form for the forward transformation).}