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Measuring the Execution Time in -Calculusand linear logic

Daniel de Carvalho (Roma)

(Partially joint workwith Michele Pagani (Torino)

and Lorenzo Tortora de Falco (Roma))

Concerto, Bologna

February 17, 2009

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The problem for nets

Let 1 and 2 be two cut free nets.

1. There exists 0 cut free such that

1 2

cut

0

?

2. Given 1 and 2, how many steps ?

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1 2

cut

0

?


Which nets ?

. p.2/3


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1 2

cut

0

?


Which nets ?

Which strategy ?

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1 2

cut

0

?


Which nets ?

Which strategy ?

Which semantics ?. p.2/3


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The problem for-terms

Let v1 and v2 be two normal -terms.

1. There exists a normal -term v0 such that

(v1)v2 v0 ?

2. Given v1 and v2, how many steps ?

-terms are type free.

Steps are steps in Krivines machine. Semantics is a -algebra induced by the

multiset based relational semantics of linear

logic.. p.3/3


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The problem



1 2

cut

0

?


Which nets ?

Which strategy ?



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Ground-structures

A ground-structure is an acyclic directed graphwhose nodes are among the following ones:

identities: ax cut

multiplicatives: ` 1

exponentials: !

?

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Untyped nets

An untyped -structure is a ground-structure with

a function:

!

n

an untyped -structure withn conclusions with and one conclusion without.

An untyped -net is an untyped -structure thatsatisfies the Danos-Rgnier criterion.

A net is an untyped -net without conclusion la-

beled with .. p.6/3


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Cut-elimination

cut

ax

b ca t

c

cut

`

ba

f g h i

t cut cut

f g h i

cut ba

1

t

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Cut-elimination

?

cutba

t

!

o

o

?

!

!

!

w

bk

c1

1

!

!

b1

vk

ck

k

!

v1

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Cut-elimination

!

cuto

?

!

c1

1

!

!

!

k

o

cut

ck

!

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The problem



1 2

cut

0

?


Which nets ?

Which strategy ?



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Krivines machine (states)

We denote by S the set of states.A state is a non-empty stack.

A stack is a finite sequence of closures.

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We denote by C the set of closures.

c C := (t, e)where t is a -term and e E.

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We denote by C the set of closures.

c C := (t, e)where t is a -term and e E.

We denote by E the set of environments.It is the set of partial functions V C

whose domain is finite.

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Krivines machine (execution)

current subterm environment stack

x V e with e(x) = (u, e)

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x V e with e(x) = (u, e) u e

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x V x / dom(e)

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x V x / dom(e)

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x.u e c.

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x.u e c.

u e {x c}

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x.u e

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x.u e

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(v)u e

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(v)u e v e (u, e).

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Krivines machine (extension)

We extend -terms with states.

k K ::= x1....xm.s | x1....xm.(x)t1...tp

where t1, . . . , tp K,

s S

and x, x1, . . . , xm V.

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Krivines machine (an example)

Computing (1)u

where 1 = f.x.(f)x and u = y.(y)yy.

outputcurrent

subtermenvironment stack arguments

(1)u

We denote the closure (u, ) by u.We denote the closure (x, f u) by x.

We denote the closure (y, y x) by y.

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Computing (1)u


outputcurrent


1 u



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Computing (1)u


outputcurrent


x.(f)x f u



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Computing (1)u


outputcurrent


x. (f)x f u



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Computing (1)u


outputcurrent


x. f f u x



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Computing (1)u


outputcurrent


x. u x



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Computing (1)u


outputcurrent


x. (y)yy y x



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Computing (1)u


outputcurrent


x. (y)y y x y



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Computing (1)u


outputcurrent


x. y y x y.y



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Computing (1)u


outputcurrent


x. x f u y.y



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Computing (1)u


outputcurrent


x.(x) y y x y



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Computing (1)u


outputcurrent


x.(x) x f u y



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Computing (1)u


outputcurrent


x.(x)x y y x



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Computing (1)u


outputcurrent


x.(x)x x f u



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K i i hi ( l )


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Computing (1)u


outputcurrent


x.(x)xx

We denote the closure (u, ) by u.

We denote the closure (x, f u) by x.We denote the closure (y, y x) by y.

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R d i


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Reduction strategy

We allow only stratified reduction steps.

For any reducible cut-link t of , a reduction step t() is said to be stratified when for everycut-link t of we have

depth(t) depth(t) .

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Th bl


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The problem



1 2

cut

0

?


Which nets ?

Which strategy ?


S t R


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System R

We are given a non-empty set A.We define a set D in such a way that

D = A (D D) .

The contexts are partial functions

: V Mf(D) s.t. {x V / (x) = []} is finite.x : [] R x :

, x : a R v :

R x.v : (a, )

0 R v : ([1, . . . , n], ) 1 R u : 1, . . . , n R u : n0 + 1 + . . . + n R (v)u :

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A ti f t


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A semantics for pure nets

We consider the multiset based relationalsemantics of MELL+MIX.

We are given a set A.We define a set D in such a way that

D = AA 1(D D)(D`

D)!D?D .

Example. We have (, [(+, ), (+, )]) D .

. p.18/3

E i t


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Experiments

An experiment e is a function edge x D with

a function!

[e1, . . . , em] such that

ax

xx

cut

x x

1

(+, )

(, )

(+, x , y)

yx`

x y

(, x , y)

x

(,[x])

?(, m)(, 1)

,jm

j

(, j)

,jm

j!

xi

[eo1, . . . , eom]

o

(+,[x1, . . . , x

m])

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Th bl


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The problem



1 2

cut

0

?


Which nets ? Which strategy ?


N li ti i l l


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Normalization in -calculus

Definition. For any D, is said to beexhaustive if [] has only negative occurrences in

.We denote by Dex the set of D s.t. isexhaustive.

Theorem. A -term t is normalizable if, and only

if, there exists Dex s.t. R t : .

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N li ti i MELL


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Normalization in MELL

We set ex = {x ; (+, []) doesnt appear in x}.

Theorem. Let be a net.There exists a cut free net 0 s.t.

0 if, and

only if, ex = .

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The problem


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The problem



1 2

cut

0

?


Which nets ? Which strategy ?


Size of points for calculus


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Size of points for-calculus

Definition. For D, we define the size of ,denoted by s():

if A, s() = 1 and s() = 0; if = ([1, . . . , n], 0), then

s

() = ni=1

s

(i) + s

(

0) + 1;

s() =n

i=1 s(i) + s(0) + 1.

For any a = [1, . . . , n], we set

s(a) =n

i=1s(i) .

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A bound for execution-time in


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-calculusTheorem [Carvalho 2006]. Let v1 and v2 twonormal closed -terms. For

a Mfin(v2),

Dex,

(a, ) v1,

we have l(((v1)v2, ).) 2s(a) + s() + 2 .

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Indeed we can be very precise


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Indeed, we can be very precise

Definition. Let X, Y D. We denote byUex(X, Y) the following set:

((a, ), a) (X \ A) Mfin(Y)/( S)((a) = (a) and () Dex)

,

where S is the set of substitutions on D.Theorem [Carvalho 2006]. Assume A is infinite.We have

l(((v)u, ).)

= inf s(a, ) + s(a) + 1/

((a, ), a

) Uex

(v, u) .

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An example


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An example

We have

R 1 : ([([0, 1, 4], 2)], ([0, 1, 4], 2));

R y.(y)yy : ([0, 1, ([0], ([1], 2))], 2).

We set

= ([0, 1, 4], 2); a = [];

a

= [([0, 1, ([0], ([1], 2))], 2)].

We have s(a, ) = 7 and s(a) = 6.

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Size of points for MELL


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Size of points for MELL

Definition. For x D, we define the size of x,denoted by s(x) :

if x A or x = (p, ), then s(x) = 1; s(p,y,z) = 1 + s(y) + s(z);

s(p, [x1, . . . , xm]) = 1 +m

j=1 s(xj).For every (x1, . . . , xn)

ni=1 D, we set

s(x1, . . . , xn) = ni=1

s(xi).

Example. We have s(, [(+, ), (+, )]) = 3 .

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A bound for execution-time in


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MELL

=

1 2

cutd c c

d

0

1 and 2 cut free R 0 cut free

Theorem [Carvalho-Pagani-Tortora de Falco2008]. For

y = (x, x) 1,

y = (x, x) 2,

x,x exhaustive,

we havelength(R) s(y)+s(y

)2 .

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Indeed we can be very precise


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Indeed, we can be very precise

Theorem [Carvalho-Pagani-Tortora de Falco2008]. Assume A is infinite.

length(R) = i n f

s(z)+s(z)sinf(ex)2 /

z 1, z 2 and

( S)((zc) = (z

c)

and (zd), (zd) exhaustive)

where sinf(ex) = inf{s(x) / x ex}

and S is the set of substitutions.

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But we are not able to compute 0...

END

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