A Tree-Based Definition of Business Process Conformance (Talk @ EDOC 2024)

A T�ee-Based
Deﬁn�ion o�
Busines� P�ocess
Confo�mance_
A T�ee-Based
Deﬁn�ion o�
Busines� P�ocess
Confo�mance_
Sylvain Hallé
Université du Québec à Chicoutimi, Canada
September
12th,
2024

Awards for players
Players are eligible for a trophy T if they
satisfy a specific condition φT

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
hit a minimum of 50 homeruns

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
HOMERUNS
3 70

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
HOMERUNS
51 70

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
won all games or threw a no-hitter

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
won all games or threw a no-hitter
ALL GAMES
NO-HITTER

Observations
It is easy to award the trophy when one player
satisfies the condition and the other does not

Observations
Although both players satisfy the condition,
one may satisfy it in a "stronger way"
numerical condition exceeded by a larger margin
"more components" of the condition are satisfied

Observations
numerical condition exceeded by a larger margin
Sometimes the two players cannot be ranked

Awards for players
Players are eligible for a dunce hat H if they
violate a specific condition φH
D
D
D
Players are eligible for a trophy T if they
satisfy a specific condition φT

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
never dropped the ball

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
DROPS
15 0

Awards for players
D
D
D
PLAYER A
PLAYER A PLAYER B
PLAYER B
DROPS
15 0

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
DROPS
15 1

Awards for players
D
D
D
PLAYER A
PLAYER A PLAYER B
PLAYER B
DROPS
15 1

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
stole at least one base and scored at least one run

Awards for players
PLAYER A
PLAYER A PLAYER B
PLAYER B
STOLE BASE
SCORED RUN

Awards for players
D
D
D
PLAYER A
PLAYER A PLAYER B
PLAYER B
STOLE BASE
SCORED RUN

Observations (again)
It is easy to award the dunce hat when one player
violates the condition and the other does not

Although both players violate the condition,
one may violate it in a "stronger way"
numerical condition not met by a larger margin
more components of the condition are violated

one may violate it in a "stronger way"
more components of the condition are violated
Sometimes the two players cannot be ranked

Ranking players
For a given condition φ, we can define a relation
that ranks players
D
D
D
"bad" "good"

One player satisfies the condition and the
other does not
Ranking players
that ranks players
D
D
D
"bad" "good"

One player satisfies the condition "in a
stronger way"
Ranking players
that ranks players
D
D
D
"bad" "good"

One player violates the condition "in a
stronger way"
Ranking players
that ranks players
D
D
D
"bad" "good"

Sometimes the players cannot be
ranked
Ranking players
that ranks players
D
D
D
"bad" "good"

Satisfaction
false true
⊤
⊤
⊥
⊥
For a given condition φ, satisfaction can be
generalized to a partial order

The execution of a business process generates a
sequence of events called a trace σ
Business processes
Obtain
Customer
Info
Identify
Customer
Info
Retrieve
Full
Customer
Dtl
Analyze
Customer
Relation
Select
Deposit
Service
Submit
Deposit
Prepare
Prop. Doc
Propose
Account
Opening
Schedule
Status
Review
Open Acc
Status
Review
Verify
Customer
ID
Open
Account
Record
Customer
Info
Receive
customer
request
Validate
Acc Info
Close Acc
Apply Acc
Policy
Activate
Acc
Record
Acc
Info
Evaluate
Deposit
Val
Do
Deposit
Report
Large
Deposit
Notify
Customer
Non-VIP
VIP

Business processes
a
b
e
c d
f
g h i
j
k
l
m
n
o
q r
p
s
t
u
v
w
¬α
α

Business processes
a
b
e
c d
f
g h i
j
k
l
m
n
o
q r
p
s
t
u
v
w
¬α
α
σ = a b e f i k j m l n o p w

Each event is modeled as a set of attribute-value
pairs.
Business processes
timestamp
action
location
agent
organization
duration
. . .
Possible attributes:
a v

A business process may be subject to a set of
conformance constraints, expressed as
conditions on a trace:
Conformance constraints
When n occurs, b and m occurred before
If α holds, the process duration when n occurs
is at most 7 days
At least one of i, n and o is performed by a
manager
Activity u is immediately preceded by t
etc.

A trace can either satisfy or violate a constraint

is at most 7 days

is at most 7 days
a b c d e f i j k l m n o p w

is at most 7 days
= 5 d.

is at most 7 days
= 5 d.
a b c d e f g h i k j m l n o p w
= 8 d.

Constraints can be formalized as expressions of
Linear Temporal Logic*
*including past modalities
When n occurs, b and m occurred before
is at most 7 days
manager

H ( = n → (O ( = b) ∧ O ( = m))
is at most 7 days
manager

H ( = n → (O ( = b) ∧ O ( = m))
α → G ( = n → < 7)
manager

H ( = n → (O ( = b) ∧ O ( = m))
α → G ( = n → < 7)
F ( = i ∧ = M) F ( = n ∧ = M)
∨
∨ F ( = o ∧ = M)

H ( = n → (O ( = b) ∧ O ( = m))
α → G ( = n → < 7)
F ( = i ∧ = M) F ( = n ∧ = M)
∨
∨ F ( = o ∧ = M)
H ( = u → (Y ( = t))

Goal
false true
⊤
⊤
⊥
⊥
Generalize the satisfaction relation of LTL to a
partial order

Goal
false true
⊤
⊤
⊥
⊥
Generalize the satisfaction relation of LTL to a
partial order
σ
σ
σ
σ
σ
σ
σ

Labeled tree
A labeled colored tree is a recursive structure of the
form:
T = ⟨ℓ, [T, ..., T]⟩ | ε
node label
children
ℓ
ℓ ℓ
ℓ ℓ ℓ ℓ ℓ
A tree node n is also
associated with a color
c(n) ∈ { , , }
ℓ ℓ ℓ ℓ ℓ ℓ
ℓ ℓ

Evaluation tree
From a trace σ and an LTL condition φ, function
τ(σ, φ) constructs a labeled tree as follows:
τ(σ, p(v1, ..., vn)) ⟨p, [v1, ..., vn]⟩
τ(σ, ¬φ) ⟨¬, [τ(σ, φ)]⟩
τ(σ, X φ) ⟨X, [τ(σ[1..], φ)]⟩
τ(σ, φ1 ∧ ... ∧ φn) ⟨∧, [τ(σ, φ1), ..., τ(σ, φn)]⟩
τ(σ, φ1 ∨ ... ∨ φn) ⟨∨, [τ(σ, φ1), ..., τ(σ, φn)]⟩
τ(σ, G φ) ⟨G, [τ(σ[0..], φ), τ(σ[1..], φ), ..., τ(σ[|σ| - 1], φ)]⟩
τ(σ, F φ) ⟨F, [τ(σ[0..], φ), τ(σ[1..], φ), ..., τ(σ[|σ| - 1], φ)]⟩
⟨O, [τ(σ[0..|σ| - 2], φ), τ(σ[|σ| - 3], φ), ..., τ(σ[0], φ)]⟩
τ(σ, O φ)
τ(σ, Y φ) ⟨Y, [τ(σ[0..|σ| - 2], φ)]⟩
τ(σ, H φ) ⟨H, [τ(σ[0..|σ| - 2], φ), τ(σ[|σ| - 3], φ), ..., τ(σ[0], φ)]⟩
=
=
=
=
=
=
=
=
=
=

Evaluation tree
τ(σ, p(v1, ..., vn)) ⟨p, [v1, ..., vn]⟩
τ(σ, ¬φ) ⟨¬, [τ(σ, φ)]⟩
τ(σ, X φ) ⟨X, [τ(σ[1..], φ)]⟩
τ(σ, φ1 ∧ ... ∧ φn) ⟨∧, [τ(σ, φ1), ..., τ(σ, φn)]⟩
τ(σ, φ1 ∨ ... ∨ φn) ⟨∨, [τ(σ, φ1), ..., τ(σ, φn)]⟩
⟨O, [τ(σ[0..|σ| - 2], φ), τ(σ[|σ| - 3], φ), ..., τ(σ[0], φ)]⟩
τ(σ, O φ)
τ(σ, Y φ) ⟨Y, [τ(σ[0..|σ| - 2], φ)]⟩
=
=
=
=
=
=
=
=
=
=
suffix of σ starting
at index 1

Evaluation tree
τ(σ, p(v1, ..., vn)) ⟨p, [v1, ..., vn]⟩
τ(σ, ¬φ) ⟨¬, [τ(σ, φ)]⟩
τ(σ, X φ) ⟨X, [τ(σ[1..], φ)]⟩
τ(σ, φ1 ∧ ... ∧ φn) ⟨∧, [τ(σ, φ1), ..., τ(σ, φn)]⟩
τ(σ, φ1 ∨ ... ∨ φn) ⟨∨, [τ(σ, φ1), ..., τ(σ, φn)]⟩
⟨O, [τ(σ[0..|σ| - 2], φ), τ(σ[|σ| - 3], φ), ..., τ(σ[0], φ)]⟩
τ(σ, O φ)
τ(σ, Y φ) ⟨Y, [τ(σ[0..|σ| - 2], φ)]⟩
=
=
=
=
=
=
=
=
=
=
length of σ

Evaluation tree: example
Consider the trace of 4 events:
σ = {a ↦ 0}, {a ↦ 1}, {a ↦ 1}, {a ↦ 0}
and the LTL condition:
φ = G (a = 0)
Then τ(σ, φ) is:

σ = {a ↦ 0}, {a ↦ 1}, {a ↦ 1}, {a ↦ 0}
φ = G (a = 0)
Then τ(σ, φ) is:
G

σ = {a ↦ 0}, {a ↦ 1}, {a ↦ 1}, {a ↦ 0}
φ = G (a = 0)
Then τ(σ, φ) is:
=
0
a
G

σ = {a ↦ 0}, {a ↦ 1}, {a ↦ 1}, {a ↦ 0}
φ = G (a = 0)
Then τ(σ, φ) is:
=
0
a
=
0
a
G

σ = {a ↦ 0}, {a ↦ 1}, {a ↦ 1}, {a ↦ 0}
φ = G (a = 0)
Then τ(σ, φ) is:
=
0
a
=
0
a
=
0
a
G

σ = {a ↦ 0}, {a ↦ 1}, {a ↦ 1}, {a ↦ 0}
φ = G (a = 0)
Then τ(σ, φ) is:
=
0
a
=
0
a
=
0
a
=
0
a
G

The color of the root of τ(σ, φ) is defined as:
c(τ(σ, φ)) =
{ if σ satisfies φ
if σ violates φ
otherwise
The resulting colored tree is called an evaluation tree.

For a given condition φ, ranking two traces σ and
σ' will be done by comparing the evaluation trees they
induce.
Subsumption relation

induce.
σ
a b e f g h i k a b c d e f h
σ'

induce.
σ
σ'
τ

induce.
σ
σ'
τ
τ(σ,φ)

induce.
σ
σ'
τ
τ(σ,φ)
τ

induce.
σ
σ'
τ
τ(σ,φ)
τ
τ(σ',φ)

induce.
σ
σ'
τ
τ(σ,φ)
τ
τ(σ',φ)
vs.

. . .
T1
. . .
T2

. . .
T1
. . .
T2
We say that T1 is subsumed by T2 (noted T1 T2) if
their roots have the same label, and...

. . .
T1
. . .
T2
CASE 1 The root of T1 is green
T1

. . .
T1
. . .
T2
T1
the root of T2 is also green
T2

. . .
T1
. . .
T2
T1
T2
for every green child T of T1, there exists a
distinct child T' of T2 such that T T'

. . .
T1
. . .
T2
T1
T2
T

. . .
T1
. . .
T2
T1
T2
T T'

. . .
T1
. . .
T2
T1
T2
T T'
. . .
T1
. . .
T2
T1 T2
T T'

. . .
T1
. . .
T2
CASE 2 The root of T1 is red
T1

. . .
T1
. . .
T2
T1
the root of T2 is red or green
T2

. . .
T1
. . .
T2
T1
T2
for every red child T' of T2, there exists a
distinct child T of T1 such that T T'

. . .
T1
. . .
T2
T1
T2
T'

. . .
T1
. . .
T2
T1
T2
T'
T

. . .
T1
. . .
T2
T1
T2
T'
T
. . .
T1
. . .
T2
T1
T T'
T2

Some examples
a = 0 ∨ b = 0
∨
=
a 0
=
b 0
∨
=
a 0
=
b 0
t1 t2
{a ↦ 0, b ↦ 1} {a ↦ 0, b ↦ 0}
⊑

Some examples
a = 0 ∨ b = 0
∨
=
a 0
=
b 0
∨
=
a 0
=
b 0
t1 t2
{a ↦ 0, b ↦ 1} {a ↦ 0, b ↦ 0}
⊑
t1 t2
⊑

Some examples
a = 0 ∧ b = 0
∧
=
a 0
=
b 0
∧
=
a 0
=
b 0
t1 t2
{a ↦ 1, b ↦ 1} {a ↦ 0, b ↦ 1}
⊑

t1 t2
⊑
Some examples
a = 0 ∧ b = 0
∧
=
a 0
=
b 0
∧
=
a 0
=
b 0
t1 t2
{a ↦ 1, b ↦ 1} {a ↦ 0, b ↦ 1}
⊑

t1 t2
⊑
Some examples
a = 0 ∧ b = 0
∧
=
a 0
=
b 0
∧
=
a 0
=
b 0
t1 t2
{a ↦ 1, b ↦ 1} {a ↦ 0, b ↦ 1}
⊑
"more components" of the condition are violated
one may violate it in a "stronger way" D
D
D

Some examples
a = 0 ∨ (b = 0 ∧ c = 0)
t1
∨
=
a 0
b 0 c 0
∧
∨
=
a 0 =
b 0 c 0
∧
t2
{a ↦ 1, b ↦ 0, c ↦ 1} {a ↦ 1, b ↦ 1, c ↦ 0}
=
=
=

Some examples
a = 0 ∨ (b = 0 ∧ c = 0)
t1
∨
=
a 0
b 0 c 0
∧
∨
=
a 0 =
b 0 c 0
∧
t2
{a ↦ 1, b ↦ 0, c ↦ 1} {a ↦ 1, b ↦ 1, c ↦ 0}
=
=
=
t1 t2
⊑

Some examples
a = 0 ∨ (b = 0 ∧ c = 0)
t1
∨
=
a 0
b 0 c 0
∧
∨
=
a 0 =
b 0 c 0
∧
t2
{a ↦ 1, b ↦ 0, c ↦ 1} {a ↦ 1, b ↦ 1, c ↦ 0}
=
=
=
t1 t2
⊑ t2 t1
⊑

Some examples
a = 0 ∨ (b = 0 ∧ c = 0)
t1
∨
=
a 0
b 0 c 0
∧
∨
=
a 0 =
b 0 c 0
∧
t2
{a ↦ 1, b ↦ 0, c ↦ 1} {a ↦ 1, b ↦ 1, c ↦ 0}
=
=
=
t1 t2
⊑ t2 t1
⊑
Sometimes the players cannot be ranked

F (a = 0)
Some examples
⊑ F
=
0
=
0
t2
{a ↦ 0}, {a ↦ 1}, {a ↦ 1}, {a ↦ 0}
=
0
=
0
{a ↦ 1}, {a ↦ 0}
F
=
0
=
a a
a a
a a 0
t1

t1 t2
⊑
F (a = 0)
Some examples
⊑ F
=
0
=
0
t2
{a ↦ 0}, {a ↦ 1}, {a ↦ 1}, {a ↦ 0}
=
0
=
0
{a ↦ 1}, {a ↦ 0}
F
=
0
=
a a
a a
a a 0
t1

a = 4
Some examples
t2
{a ↦ 5}
=
4
{a ↦ 2}
=
4 a
a
t1

a = 4
Some examples
t2
{a ↦ 5}
=
4
{a ↦ 2}
=
4 a
a
t1
t1 t2
⊑

a = 4
Some examples
t2
{a ↦ 5}
=
4
{a ↦ 2}
=
4 a
a
t1
t1 t2
⊑ t2 t1
⊑

a = 4
Some examples
t2
{a ↦ 5}
=
4
{a ↦ 2}
=
4 a
a
t1
t1 t2
⊑ t2 t1
⊑
0 1 2 3 4 5 6 7 8

Some examples
{a ↦ 5}
{a ↦ 2}
a = 4 0 1 2 3 4 5 6 7 8

Some examples
{a ↦ 5}
{a ↦ 2}
a = 4
δ(x,y) = |x-y| 0 1 2 3 4 5 6 7 8

Some examples
{a ↦ 5}
{a ↦ 2}
δ(a,4) ≤ 0 ∧
δ(a,4) ≤ 1 ∧
δ(a,4) ≤ 2
δ(x,y) = |x-y| 0 1 2 3 4 5 6 7 8

Some examples
{a ↦ 5}
{a ↦ 2}
δ(a,4) ≤ 0 ∧
δ(a,4) ≤ 1 ∧
δ(a,4) ≤ 2
δ(x,y) = |x-y|
t2
t1
∧
0
≤
δ
a 4
1
≤
δ
a 4
2
≤
δ
a 4
∧
0
≤
δ
a 4
1
≤
δ
a 4
2
≤
δ
a 4
0 1 2 3 4 5 6 7 8

t1 t2
⊑
Some examples
{a ↦ 5}
{a ↦ 2}
δ(a,4) ≤ 0 ∧
δ(a,4) ≤ 1 ∧
δ(a,4) ≤ 2
δ(x,y) = |x-y|
t2
t1
∧
0
≤
δ
a 4
1
≤
δ
a 4
2
≤
δ
a 4
∧
0
≤
δ
a 4
1
≤
δ
a 4
2
≤
δ
a 4
0 1 2 3 4 5 6 7 8

t1 t2
⊑
Some examples
{a ↦ 5}
{a ↦ 2}
δ(a,4) ≤ 0 ∧
δ(a,4) ≤ 1 ∧
δ(a,4) ≤ 2
δ(x,y) = |x-y|
t2
t1
∧
0
≤
δ
a 4
1
≤
δ
a 4
2
≤
δ
a 4
∧
0
≤
δ
a 4
1
≤
δ
a 4
2
≤
δ
a 4
0 1 2 3 4 5 6 7 8
one may violate it in a "stronger way" D
D
D

Positive points
The subsumption relation can compare two
executions of a process with respect to an
arbitrary* LTL conformance condition
*In Negated Normal Form (NNF)

Positive points
It does not require any additional information from
the user (weights, coefficients, etc.)

Positive points
It does not involve any explicit counting or
arithmetic calculations

Positive points
It does not output a numerical value

Positive points
It produces a ranking mostly matching our intuition

Positive points
It produces a ranking mostly matching our intuition
It is concretely implemented:
https://github.com/liflab/shaded-compliance

Some experimental results
Implementation tested on a sample of real-world and
synthetic logs and conformance properties

Challenges
Finding a mapping between sub-trees is
computationally expensive
more compact representation of evaluation trees as DAGs

Challenges
Subsumption may be excessively fine-grained
compare "simplified" versions of evaluation trees instead

Challenges
It does not directly accommodate user-defined
weights
accept an order relation between sub-formulas

Challenges
It does not directly accommodate user-defined
weights
accept an order relation between sub-formulas
How to evaluate incrementally on a stream?
incorporate into the library

Possible uses
All executions of a process are arranged into a
(finite) lattice
1
1
3 3
3 3
9
9 1
1
3
9
3
3
9
3
number of
equivalent traces

Possible uses
The traces of a specific log can be placed on this
lattice (a form of visualization)
0
0
0 1
0 0
1
0 6
12
0
0
0
0
5
1

Possible uses
"Layers" of the lattice can formally model service
level agreements (SLAs)
1
1
3 3
3 3
9
9 1
1
3
9
3
3
9
3

A Tree-Based Definition of Business Process Conformance (Talk @ EDOC 2024)

More Related Content

More from Sylvain Hallé (20)

Recently uploaded (20)

A Tree-Based Definition of Business Process Conformance (Talk @ EDOC 2024)