P systems and unique-sum sets Pierluigi Frisco School of Mathematical and Computer Sciences...

P systems and unique-sum sets

Pierluigi Frisco

School of Mathematical and Computer Sciences

Heriot-Watt University

Edinburgh

Conference on Membrane Computing 24-27 August 2010, Jena (Germany)

P system with several symbols

12

34

ac b

c

b

a

b

a4

c4

a1

b1

b3

c2

b2

(c, out; b, in)

(c2 b1 c1 b2)

... this is a rewriting system

P system with only one symbol

12

34

aa a

a

a

a

a

(aa, out; a, in)

?

Work settings

system: P systems with symport/antiport

feature: only one symbol

operational mode: maximal strategy

In each configuration the maximal number of applicable rules is applied, but in a configuration

each rule can be applied at most once

Maximal strategy: example

aa

aa

(a, out)1

maximal parallelism: all a’s go out in one transition

maximal strategy: the a’s go out one by one.

With maximal strategy it makes sense to have repeated rules.

(a, out)

Results on P systems with s/a and only one symbol

degree 2n+3

max. parall.+

partially blind

register machines

with n registers

[11]

[11] O. H. Ibarra, S. Woodworth. On symport/antiport P systems with small number of objects. International Journal of Computer Mathematics, 83(7):613-629, 2006

degree n+3

max. parall.+

[6]

[6] P. Frisco. Computing with Cells. Advances in Membrane Computing. Oxford University Press, 2009

degree 2n+3

= max. parall.+ register

machines with n

registers

[11]

+priorities

degree 2n+3

max. strategy+=

... and using unique-sum sets

... and using unique-sum sets

Unique-sum sets

[5] P. Frisco. On s-sum vectors. Technical report, Heriot-Watt University, 2008, HW-MACS-TR-0058

Sets of integer numbers whose sum can be obtained in a unique way using

only the elements in a set.

{2, 5, 7}

2 + 5 + 7 = 14

{ 4, 6, 7}

4 + 6 + 7 = 17 ... and there is no other way to obtain 17 adding 4, 6 and 7

= 2 + 2 + 5 + 5 = 7 + 7 = ...

This means that no subset of a unique-sum set can be obtained as a linear combination of the remaining elements in that set.

[5]

Simulating register machines with n registers and S states

n+2n+2'

...

1

n+1n+1'

22'

simulating registers trap

a

This means that no subset of a unique-sum set can be obtained as a linear combination of the remaining elements in that set.

<occurrences of symbol ‘a’ equal to the sum of a subset of a specific unique-sum set>


n+2n+2'

...

1

n+1n+1'

22'

a

The proof requires the use of a unique-sum set U with at least n + 18|S| + 1 elements, where S is the number of states of the register machine.

Different multiplicities of ‘a’, where the multiplicities are elements in U, are associated with each of the registers and instructions of the register machine.

This is performed by the function b: {1, ..., n} U and the eighteen functions c, c(1), ..., c(17) all from S to U, injective and with disjoint values.

n+2n+2'

...

1

n+1n+1'

22'

aab(1) ab(1) ab(n) ab(n)

The presence of b(i) occurrences of ‘a’ in the “register-compartments” represents 0 as content of register i.

The addition of 1 to register i is simulated adding 2 occurrences of ‘a’ to compartment i+1. Conversely for the subtraction.

The presence of just c(p) occurrences of ‘a’ in the skin compartment indicates the simulation of the register machine being in state p.


ac(p)

Simulation of (p, +1, q)

n+2n+2'

...

1

n+1n+1'

22'


ac(p)

(ac(p), out; ac(1)(q)+b(1)+2, in) 1


n+2n+2'

...

1

n+1n+1'

22'


ac(1)(q)+b(1)+2

(ac(p), out; ac(1)(q)+b(1)+2, in) 1

(ac(1)(q), out; ac(2)(q), in) 1 (ab(1), out; ab(1)+2; in) 2'


n+2n+2'

...

1

n+1n+1'

22'

aab(1) ab(1)+2 ab(n) ab(n)

ac(2)(q)+b(1)

(ac(p), out; ac(1)(q)+b(1)+2, in) 1


(ac(2)(q)+b(1), out; ac(q), in) 1 (ab(1), out; ab(1)+2; in) 2


n+2n+2'

...

1

n+1n+1'

22'

aab(1)+2 ab(1) ab(n) ab(n)

ac(q)

(ac(p), out; ac(1)(q)+b(1)+2, in) 1


(ac(2)(q)+b(1), out; ac(q), in) 1 (ab(1), out; ab(1)+2; in) 2

Activation trap

n+2n+2'

...

1

n+1n+1'

22'


ac(p)

(a, in) n+2'

(a, out; a, in) n+2

Simulation of (p, =01, q) with empty counter

n+2n+2'

...

1

n+1n+1'

22'

aab(1) ab(1)

ab(n) ab(n)

ac(p)

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1

n+2n+2'

...

1

n+1n+1'

22'

aab(1) ab(1)

ab(n) ab(n)

ac(10)(q)+5b(1)+2

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1

(ac(10)(q), out; ac(11)(q), in) 1 (ab(1), out; a5b(1)+2; in) 2'


n+2n+2'

...

1

n+1n+1'

22'

aab(1) a5b(1)+2

ab(n) ab(n)

ac(11)(q)+b(1)

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1


(ac(11)(q)+b(1), out; ac(12)(q)+b(1)+2, in) 1 (ab(1), out; ab(1)+2; in) 2


n+2n+2'

...

1

n+1n+1'

22'

aab(1)+2 a5b(1)

ab(n) ab(n)

ac(12)(q)+b(1)+2

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1



(ac(12)(q), out; ac(13)(q), in) 1 (a5b(1), out; ab(1)+2; in) 2'


n+2n+2'

...

1

n+1n+1'

22'

aab(1)+2 ab(1)+2

ab(n) ab(n)

ac(13)(q)+5b(1)

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1




(ac(13)(q)+5b(1), out; ac(14)(q)+5b(1)+1, in) 1 (ab(1), out; ab(1)+2; in) 2


n+2n+2'

...

1

n+1n+1'

22'

aab(1)+4 ab(1)

ab(n) ab(n)

ac(14)(q)+b(1)+1

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1







n+2n+2'

...

1

n+1n+1'

22'

aab(1)+4 ab(1)+1

ab(n) ab(n)

ac(15)(q)+b(1)

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1






(ac(15)(q)+b(1), out; ac(16)(q)+b(1), in) 1 (ab(1)+3, out; ab(1)-1; in) 2


n+2n+2'

...

1

n+1n+1'

22'

aab(1) ab(1)+5

ab(n) ab(n)

ac(16)(q)+b(1)

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1







(ac(16)(q), out; ac(17)(q), in) 1 (ab(1)+5, out; ab(1); in) 2'


n+2n+2'

...

1

n+1n+1'

22'

aab(1) ab(1)

ab(n) ab(n)

ac(17)(q)+b(1)+5

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1








(ac(17)(q)+b(t)+5, out; ac(q), in) 1


n+2n+2'

...

1

n+1n+1'

22'

aab(1) ab(1)

ab(n) ab(n)

ac(q)

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1








(ac(17)(q)+b(t)+5, out; ac(q), in) 1


Simulation of (p, =01, q) with something in the counter

n+2n+2'

...

1

n+1n+1'

22'

aab(1)+2 ab(1)

ab(n) ab(n)

ac(p)

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1

n+2n+2'

...

1

n+1n+1'

22'

aab(1)+2 ab(1)

ab(n) ab(n)

ac(10)(q)+5b(1)+2

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1



n+2n+2'

...

1

n+1n+1'

22'

aab(1)+2 a5b(1)+2

ab(n) ab(n)

ac(11)(q)+b(1)

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1




(a, out; a2b(1); in) 2

(a, out; a2b(1); in) 2

(a5b(1), out; ab(1)+2; in) 2'

cannot be applied

n+2n+2'

...

1

n+1n+1'

22'

aa5b(1)+2 ab(1)+2

ab(n) ab(n)

ac(12)(q)+b(1)+2

(ac(p), out; ac(10)(q)+5b(1)+2, in) 1




(a, out; a2b(1); in) 2

(a, out; a2b(1); in) 2

(a5b(1), out; ab(1)+2; in) 2'

cannot be applied

Other results

Corollary 2: P systems with symport/antiport operating under maximal strategy, using a constant number of occurrences of only one symbol induce an infinite hierarchy on the number of compartments.

Accepting purely multi-catalytic P systems

cax c(ay, tar)

Theorem 5: Any accepting register machine with n registers can be simulated by an accepting purely multi-catalytic P system operating under maximal strategy, using only one symbol and with degree 2n+3.

... more in the paper

Thanks for

your

attention

P systems and unique-sum sets Pierluigi Frisco School of Mathematical and Computer Sciences...

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