# pairing function

A pairing function  is a function $P\colon\mathbb{Z}_{+}^{2}\to\mathbb{Z}_{+}$ which establishes a one-to-one correspondence between $\mathbb{Z}_{+}^{2}$ and $\mathbb{Z}_{+}$. Such functions are useful in the theory of recursive functions  because they allow one to express recursive functions of $m$ variables in terms of recursive functions of $n$ variables with $m\neq n$.

Two examples of pairing functions are the following;

 $P_{1}(x,y)=(x+y)(x+y+1)/2+y$
 $P_{2}(x,y)=2^{x}(2y+1)-1$

It is not hard to see that these functions are recursive (actually, primitive recursive). For instance, one could use the recursion relations   and initial conditions

 $P_{1}(x+1,y)=P_{1}(x,y)+x+y+1$
 $P_{1}(0,y)=T(y)+y$
 $T(y+1)=T(y)+y+1$
 $T(0)=0$

where $T(n)$ is the $n$-th triangular number  to show that $P_{1}$ is recursive. Likewise, one could use the recursions

 $P_{2}(x+1,y)=P_{2}(x,y)+P_{2}(x,y)$
 $P_{2}(0,y)=y+y$

to show that $P_{2}$ is recursive.

An easy way to see that $P_{1}$ effects a one-to-one correspondence between $\mathbb{Z}_{+}^{2}$ and $\mathbb{Z}_{+}$ is as follows: Define the “successor    ” of a pair $(x,y)\in\mathbb{Z}_{+}^{2}$ to be the pair $(x-1,y+1)$ when $x\neq 0$; otherwise, when $x=0$, the successor is $(y+1,0)$. It is easy to see that every pair has a successor and that every pair except $(0,0)$ is the successor of exactly one other pair. With this definition of successor, the set of pairs of positive integers satisfies the Peano axioms and, hence, is isomorphic to the integers. From the definition of $P_{1}$ it follows that, if $(x^{\prime},y^{\prime})$ is the successor of $(x,y)$, then $P_{1}(x^{\prime},y^{\prime})=P(x,y)+1$ and that $P_{1}(0,0)=0$. This means that $P_{1}$ is the isomorphism     described two sentences  ago.

That $P_{2}$ effects a one-to-one correspondence between positive integers and pairs of positive integers follows readily from uniqueness of factorization of integers. On the one hand, for any number $z$, one can find numbers $x$ and $y$ such that $z=P_{2}(x,y)$ by factoring $z+1$ and letting $x$ be the power of $2$ which appears in the factorization. On the other hand, this is the only solution of $z=P_{2}(x,y)$ because prime factorization  is unique.

Since a pairing function $P$ sets up a 1-1 correspondence between $\mathbb{Z}_{+}$ and $\mathbb{Z}_{+}^{n}$, there exist uniquely defined unpairing functions $R$ and $L$ such that

 $P(L(x),R(x))=x$

It is not hard to show that, if $P$ is recursive, $R$ and $L$ will also be recursive.

Once one has a pairing function $P^{(2)}$, one can use it to set up 1-1 correspondences between $\mathbb{Z}_{+}$ and $\mathbb{Z}_{+}^{n}$ for any $n$. For instance, one could define

 $P^{(3)}(x,y,z)=P^{(2)}(x,P^{(2)}(y,z))$
 $P^{(4)}(x,y,z,w)=P^{(2)}(x,P^{(3)}(y,z,w))=P^{(2)}(x,P^{(2)}(y,P^{(2)}(z,w)))$

In general,

 $P^{(n+1)}(x_{1},\ldots,x_{n+1})=P^{(2)}(x_{1},P^{(n)}(x_{2},\ldots,x_{n}))$

(This manner of encoding a list one pair at a time will be familiar to anyone who has programmed a computer in LISP. In fact, LISP was designed to be serve as a mathematical definition of computability equivalent     to Turing machines  or recursive functions. A fun exercise is to write a compiler which translates LISP programs into recursive functions using the representation of lists by single integers defined above.)

An important consequence of the fact noted above is that there is a 1-1 correspondence between recursive functions of $n$ variables and recursive functions of a single variable. If we have a function $F\colon\mathbb{Z}_{+}^{n}\to\mathbb{Z}_{+}$, we can associate to it the function $G\colon\mathbb{Z}_{+}\to\mathbb{Z}_{+}$ by the formula   $F(x_{1},\ldots,x_{n})=G(P^{(n)}(x_{1},\ldots,x_{n}))$

Doing this can often save work by allowing one to draw conclusions  about recursive functions of several variables from the special case of functions of one variable.

Title pairing function PairingFunction 2013-03-22 14:34:46 2013-03-22 14:34:46 rspuzio (6075) rspuzio (6075) 18 rspuzio (6075) Definition msc 03D20 ExampleOfBijection