# reversible integer transform

Let be x, y two natural integers. Define the following transform: X=x+(x-y). Y=y-(x-y). If the difference d=x-y tends to 0, we have X=x and Y=y. The above transform has the property that is reversible in integer domain. I have $(x,y) \in \{0,1..63\}$ . We know that pairs $(x_i,y_i)$, where i=$\overline{0,63}$, have a Laplace distribution of the histogram differences. What part of mathematics dealing with this transform? I want to know more reversible integer transforms.

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## 1 Answer

Your transform is the linear transform $$T:\quad \left[\matrix{x \cr y\cr}\right]\quad \mapsto\quad \left[\matrix{x' \cr y'\cr}\right]\ :=\ \left[\matrix{2 &-1 \cr -1 &2\cr}\right]\ \left[\matrix{x \cr y\cr}\right]\ .$$ Its inverse is given by $$T^{-1}:\quad \left[\matrix{x' \cr y'\cr}\right]\quad \mapsto\quad \left[\matrix{x \cr y\cr}\right]\ :=\ {\textstyle\left[\matrix{{2\over3} &{1\over3} \cr {1\over3} &{2\over3}\cr}\right]}\ \left[\matrix{x' \cr y'\cr}\right]\ .$$ The inverse does not map integer pairs to integer pairs. Consider the example $(x',y'):=(2,1)\in{\mathbb Z}^2$. Here $T^{-1}(x',y')=\bigl({5\over3},{4\over3}\bigr)\notin{\mathbb Z}^2$.

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2*x'+y'=2*(x+(x-y))+y-(x-y)=2*x+x-y+y=3*x and 2*y'+x'=2*(y-(x-y))+x+(x-y)=2y-2x+2y+2x-y=3y. So the transform map integer pairs to integer pairs. –  Ion Caciula Jan 12 '13 at 13:00
Thanks for clarifications. –  Ion Caciula Jan 12 '13 at 19:10