Lambda counting

Introduction

The question is: among programs, what is the probability of having a fixed property.

what kind of program : turing machines, cellular automata, combinatory logic, lambda calculus

what kind of properties : structural (for functional programs), behaviour (SN, weakly normalizable, ...

references to known results on : turing machines, cellular automata

we concentrate on combinatory logic, lambda calculus

Lambert function, Catalan and Motzkin numbers

Catalan numbers

• $C(n)$ : Catalan numbers

Usual equivalent: $C(n)\sim {\frac {4^{n}}{n^{3/2}{\sqrt {\pi }}}}$ which is obtained using Strirling formula. However, using stirling series: $n!={\sqrt {2\pi n}}\left({n \over e}\right)^{n}\left(1+{1 \over 12n}+{1 \over 288n^{2}}-{139 \over 51840n^{3}}-{571 \over 2488320n^{4}}+\cdots \right)$, we get that for $n\geq 1$ we have ${\sqrt {2\pi n}}\left({n \over e}\right)^{n}\leq n!\leq {\frac {7}{6}}{\sqrt {2\pi n}}\left({n \over e}\right)^{n}$

Thus, using this and $\left({\frac {n}{n+1}}\right)^{n}>e^{-1}$, we have:

$C(n)={\frac {(2n)!}{(n+1)!n!}}\geq {\frac {36}{49{\sqrt {\pi }}}}{\frac {4^{n}}{(n+1)^{\frac {3}{2}}}}$ for all $n\geq 1$

Motzkin numbers

• $M(n,k)$

Lambert W function

• $W(\alpha )$: $X=Ye^{Y}\;\Longleftrightarrow \;Y=W(X)$.

combinatory logic

results on combinatory logic

Generality on lambda calculus

what kind of distribution ?

we look only for densities,

for that we need size.

different size for variables: zero, one, binary with optimal size, binary with fixed size, debruijn indices in unary...

we concentrate on the simple one : variable of size zero (probably similar for size one ) more later for other size

generating functions

this does not work (by now) because radius of convergence 0

no known results for the number of terms of size n (denoted $L_{n}$)

our results

(the proof of result of section k needs the result of section (k-1))

Upper and lower bounds for $L_{n}$

For the lower bound, we will first count the number $LB(n,k)$ of lambda-terms of size $n$ starting with $k$ lambdas and having no other lambda below. This means that the lower part of the term is a binary tree of size $n-k$ with $k$ possibility for each leaf. Therefore we have:

$LB(n,k)=C(n-k)k^{n-k}$

And therefore, for $n>k$, using our lower bound for $C(n)$ and $n+1\geq n-k+1$, we get:

$LB(n,k)\geq K{\frac {(4k)^{n-k}}{(n+1)^{\frac {3}{2}}}}$ with $K={\frac {36}{49{\sqrt {\pi }}}}$

Now, for $n$ fixed, we define $f(\alpha )=\left(4n\alpha \right)^{n(1-\alpha )}$ (so $LB(n,k)\geq {\frac {K}{(n+1)^{\frac {3}{2}}}}f\left({\frac {k}{n}}\right)$) and look for the maximum of this function. We have $f'(\alpha )=nf(\alpha )\left(-\ln(4n\alpha )+{\frac {1-\alpha }{\alpha }}\right)$. Thus, $f'(\alpha )\geq 0$ is equivalent to Échec de l’analyse (fonction inconnue « \alphe »): {\displaystyle \frac{1}{\alphe}e^{\frac{1}{\alpha}}\geq 4en} . The Lambert function begin increasing this means that $f'(\alpha )\geq 0$ is equivalent to $\alpha \leq {\frac {1}{4en}}$.

at least $o({\sqrt {n/\ln(n)}})$ lambdas in head position and number of lambdas in one path

Remark: (may be 4) can be done directly without 3))

every fixed closed term (including the identity !) does not appear in a random term (in fact we have much more than that)

comment : so different situation in combinatory logic and lambda calculus ; the coding uses a big size so need to count variables in a different way

Experiments

results of the experiments we have done

some experiments that have to be done : e.g. density of terms having $\lambda x.y$ or big Omega pattern ...

to be done

Upper and lower bounds for $L_{n}$ with other size for variables especially one, binary with fixed size

Open questions and Future work

.....