KaJaSim feature idea: suggest diving by symmetry number observables

[hmedina]

When an observable has two symmetries, KaSim reports twice what's there. If I define with initial condition one single lone symmetric dimer, KaSim reports in the output there are two dimers, even if the snapshot shows only one dimer (documented in chapter 8 of the manual).

Merritt of maintaining symmetry agnosticism aside, KaJaSim would benefit from issuing a notification to the user when dealing with symmetries, in rules as in observables.

[pirbo]

In the language of algebraic variable, | pattern | does not means "how many often pattern appears in the mixture?" but "How many embeddings of patternthere is in the mixture?"
Of course, the user cares about the first one but KaSim efficiently computes only the second one. The question (I discussed with @jkrivine months ago and I should ask @vd1 and @russharmer) is "How to compute statically the formula that correspond to the first question in term of requests of the second kind?"

When dealing with patterns with 1 connected component, it is indeed just a matter of deciding by the number of automorphisms but, for example, the number of A(y), A(x!1),B(x!1) in the mixture is
| A(y) | * | A(x!1), B(x!1) | - |A(x!1,y), B(x!1) |. What now when there are more connected components or both several connected components and symmetries ...

[feret]

Dear Pierre,

When you count instances of patterns, you discard clashes as well, don't
you ?

If so, I do not understand why counting instances is easier than counting
number of embeddings.

Anyway, I would bet 3 cents for a formula of this kind:

Consider a pattern P (P is a concrete pattern).
Consider the set of the foldings of P (agents in a folding are now
identified by a set of agent identifiers).
Define the polarity of a folding as 0 if the number of fold operations
(one fold operation consists in merging two agents, and propagating by rigidity)
is even, and as -1 otherwise.

Then decompose each such folding in connected components.

Bet:

number of embeddings of P in the mixture M =

\sum_{for each folding P'=C'1,...,C'{k_p'}}
(-1)^polarity * \prod number of embeddings of C'_i in the mixture M

The coefficient (-1)^polarity may sound surprising, but it comes from the
formal expansion of (1 + (-1))^n .
Ferdinanda is using a similar formula in ODE fragmentation (and this is
proved correct in her context). The bet is that the problem is similar,
but some care has to be made to check how each folding is counted.

Cheers.
Jerome.

---

Jerome Feret
DI - Ecole Normale Superieure
45, rue d'Ulm
75230 Paris Cedex 5

phone: +33 1 44 32 37 66
fax : +33 1 44 32 21 51
mail : feret@ens.fr

On Sun, 1 May 2016, Pierre Boutillier wrote:

In the language of algebraic variable, | pattern | does not means "how many often pattern appears in the mixture?" but "How many embeddings of patternthere is in the mixture?"
Of course, the user cares about the first one but KaSim efficiently computes only the second one. The question (I discussed with @jkrivine months ago and I should ask @vd1 and
@russharmer) is "How to compute statically the formula that correspond to the first question in term of requests of the second kind?"

When dealing with patterns with 1 connected component, it is indeed just a matter of deciding by the number of automorphisms but, for example, the number of A(y), A(x!1),B(x!1) in the
mixture is
| A(y) | * | A(x!1), B(x!1) | - |A(x!1,y), B(x!1) |. What now when there are more connected components or both several connected components and symmetries ...

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[russharmer]

hi Pierre,

Do you want to resolve all clashes statically?
My recollection of how simplx/KaSim3 works is that: each connected
component has its matchings stored independently and clashes are detected
dynamically.
This works even in cases like your example here: the 'correction' term [ -
|A(x!1,y), B(x!1) | ] is those cases where you've chosen twice the same A.
If you want to detect and resolve this statically, you will need to do

• either something like Jerome suggests to correct counting [by removing
  all instances of over-counting] without knowing the actual formula;
• or compute the actual formula statically by calculating minimal gluings.
  For your example, the minimal gluing of A(y) and A(x!1),B(x!1) has two
  members:
  A(y),A(x!1),B(x!1) or A(y,x!1),B(x!1)
  for which the latter explicitly uses the same A twice and so violates
  injectivity [and cannot be a matching]. From this, it is immediate to write
  the formula
  | A(y) | * | A(x!1), B(x!1) | - |A(x!1,y), B(x!1) |
  which you can then use throughout the whole simulation.

The issue of preserved automorphisms is completely orthogonal to this. All
you need to do there is to compute that number and divide the given rate
constant by it; IMHO, there is no point messing around trying to quotient
the space of matchings -- the end result will be exactly the same for a lot
more work done.

Cheers, russ

On Mon, May 2, 2016 at 5:26 AM, Pierre Boutillier notifications@github.com
wrote:

In the language of algebraic variable, | pattern | does not means "how
many often pattern appears in the mixture?" but "How many embeddings of
patternthere is in the mixture?"
Of course, the user cares about the first one but KaSim efficiently
computes only the second one. The question (I discussed with @jkrivine
https://github.com/jkrivine months ago and I should ask @vd1
https://github.com/vd1 and @russharmer https://github.com/russharmer)
is "How to compute statically the formula that correspond to the first
question in term of requests of the second kind?"

When dealing with patterns with 1 connected component, it is indeed just a
matter of deciding by the number of automorphisms but, for example, the
number of A(y), A(x!1),B(x!1) in the mixture is
| A(y) | * | A(x!1), B(x!1) | - |A(x!1,y), B(x!1) |. What now when there
are more connected components or both several connected components and
symmetries ...

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#217 (comment)

Russ Harmer http://perso.ens-lyon.fr/russell.harmer/
CNRS & ENS Lyon