week 9 slides state identification and experiments

8/7/2019 Week 9 Slides State Identification and Experiments

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STATE IDENTIFICATION

AND

EXPERIMENTS

Experiment (applying an input sequence):

Simple, Multiple, Preset, Adaptive.

If we know that M can be in either state A,

or C, or E; then the uncertainty is (ACE).

Mealy M1 (we’ll use M1 in all our examples):

M1 NS/out

PS x=0 x=1

A C/0 D/1B C/0 A/1

C A/1 B/0

D B/0 C/1

Successor uncertainties:

Assume that the initial uncertainty is (ABCD)If now we apply a 1 input and observe a 0

we write: (ABCD) → (ACD)1/0



If we apply a 1 input and observe a 1 output,

we write: (ABCD) → (ACD)

Consider the case that we just applied a 1 input,

then the successor uncertainties will be (B) or

(ACD), depending whether the output is 0 or 1,

respectively. In this case

we write: (ABCD) → (B) (ACD) .

We shall refer a collection of uncertainties, such

as (B)(ACD) above, as an uncertainty vector

(U).

(B) and (ACD) in this example, are called the

components of the uncertainty vector.

Similarly, for M1, we may write:

(ABCD) → (BCC)(A); (ABC) → (CC)(A);

(ACD) → (B)(CD); (BCC)(A) → (C)(AA)(C).

1/1

1

00

01



Note: For a component, the order of statesis not important. Repetition of states areallowed and meaningful.

The successor of an uncertainty vector UV1 is found by constructing a UV whosecomponents are the successors of UV1’scomponents.

DEFINITONS (for components only) [Beware: not the same as in Kohavi]

Singleton: A component having a single state

without duplication. Ex: (A), or (B). Note:

(AA) is not a singleton.

Homogeneous: A component having only one

state, with or without duplications. Ex: (A), or

(AA), or (BBB).

Note: (AAB) is not homogeneous.

Nonhomogeneous: A component containing at

least two nonidentical states. Ex: (AB) or(AAB) or (AAAC) or (AABC).

Note: A component is either homogeneous

or nonhomogeneous.



SUCCESSOR TREE (M1):

M1 NS/outPS x=0 x=1

A C/0 D/1

B C/0 A/1

C A/1 B/0

D B/0 C/1

U0=(ABCD): (ABCD)

(BCC)(A) (B)(ACD)

{dead end for DS}

(C)(AA)(C) (BB)(A)(D) (C)(BC)(A) (A)(B)(CD){HS found} {HS found}

(A)(CC)(A) (B)(DD)(B) (A)(C)(A)(C) (B)(B)(A)(D)

{ DS found} { DS found}

(CC)(C)(B) (AA)(D)(C) (C)(C)(B)(A) (D)(A)(B)(C)

{ DS found} { DS found}

x=0 x=1



U0=(ACD): (ACD)

(BC)(A) (B)(CD)

(C)(A)(C) (B)(A)(D) (C)(B)(A) (A)(B)(C)

{DS found} {DS found} {DS found} {DS found}

Some Properties of the successor tree:

P1-Multiple component > multiple component[dead end for DS].

Ex:

(BCC)(A) → (C)(AA)(C); (BCC)(A) → (BB)(A)(D)

P2-Identical nonhomogeneous components >

similar subtrees [repeat].Ex: (ABB), (AB), (AABB) are considered identical

P3-Trivial vector: We have found a DS.

Each state in the initial uncertainty vector

responds to the input sequence JD leading to the

trivial vector with a distict output sequence.



Ex:

(A) → (B)

(C) → (B)

(D) → (A)

P4- Homogeneous vector: We have found a HS.

The output sequence of the input sequence JD

leading to the homogeneous vector allows us to

uniquely determine the final state.

(ABCD) → (BCC)(A) → (BB)(A)(D)

homogeneous and hence

(A) → (B)

(B) → (B)(C) → (D)

(D) → (A)

HOMING EXPERIMENTS:

An input sequence J is said to be a homing

sequence (HS), if the final state of M can be

determined uniquely from M’s response to J,

regardless of the initial state.

01/00

01/00

01/11

01

01/01

10/10

10/00

10/11



Homing tree is the successor tree with the

following termination rules:

Rule H1 [Repeat] : A node becomes terminal if

the nonhomogeneous components are asso-

ciated with some node in a preceding level (P2).

[(AB)(C)(D) is “identical” with (AAB)(B)(E)]

Rule H2 [HS found] : All k th

level nodes

become terminal if any kth level node is

associated with a homogeneous vector (P4)

[only if we are seeking the shortest, or

one of the shortest, HS].

DISTINGUISHING EXPERIMENTS:

An input sequence J is said to be a

distinguishing sequence (DS), if the output

sequence produced by the machine in response

to J is different for each initial state.

[Do not confuse: “DS for pair of states (A,B)”

and “DS”]



Distinguishing tree is the successor tree with

the following termination rules:

Rule D1 [Repeat] : Identical to Rule H1.

Rule D2 [Dead end] : A node becomes terminal

if it is associated with a vector having a

multiple component (P1).

Rule D3 [DS found] : All k th level nodesbecome terminal if any k

thlevel node is

associated with a trivial vector (P3)

[only if we are seeking the shortest, or

one of the shortest, DS].

Property: Every distinguishing sequence is

also a homing sequence (the converse is not true)

Note that every singleton component is

homogeneous, but not vice versa, and compare

rules H2 [HS found] and D3 [DS found].

Example: For machine M1, 111 is a DS, henceit is also an HS; 01 is an HS, but it is not a DS.



SYNCHRONIZING EXPERIMENTS

A synchronizing sequence (SS) of a machine

M is a sequence that takes M to a specified

state, regardless of the output or the initial state

of M.

Since the output of M is not observable (or is

not used), a simpler form for uncertainty

vectors and successor tree can be used.

A synch uncertainty vector (SUV) is a UV

with the following modifications:

(a) All the components are combined into a

single component,(b) If duplicate states exist, all but one is

deleted.

For example, (A)(BCC) becomes (ABC),

(ACD)(B) becomes (ABCD),

(AA)(C)(C) becomes (AC).



Synch tree is a successor tree whose nodes are

associated with synch UV’s, with the following

termination rules:

Rule S1 [Repeat] : A jth

level node becomes

terminal whenever the synch UV associated

with the node is identical to a synch UV

associated with some previous-level node [only

if we are seeking the shortest SS].

Rule S2 [SS found] : All jth

level nodes nodes

become terminal if any jth level node is

associated with a synch UV which is a singleton.

Ex: SS tree for M1, with U0 = (ABCD)

[underlined: terminating by rule S1 (repeat)]



EXISTENCE THEOREMS AND BOUNDS

ON LENGTHS

Theorem. A preset homing sequence,whose length is at most (n - 1)

2, exists for

every reduced machine (Proof: Kohavi).

Theorem. A preset homing sequence,

whose length is at most n(n - 1)/2, exists

for every reduced machine.

While every machine has at least one homing

sequence, not every machine has a distinguishing

sequence. If a machine has at least one DS, it is

called a diagnosable machine.

Example: M1 is diagnosable, since it possessesat least one DS. For example, 100, 101, 110,

and 111 are DS’s (note that any string of length

≥ 3 and starting with 1 is a DS for M1) .



Example: Consider M2. ABCD

M2 NS/out

PS x=0 x=1

A B/0 D/0 (AB)(DD) (ABCD) B A/0 B/0 [multiple [repeat]

C D/1 A/0 component::

D D/1 C/0 dead end]

There is no DS for this machine. The shortest

HS for this machine is 010 [find].

ADAPTIVE SYNCH EXPERIMENTS

Theorem: If a synch sequence exists for

machine M, then its length is at most n (n-1)2/2.

Example: Consider M5. This machine has nosychronizing sequence (construct the SS tree.

You’ll see that every node becomes terminal by

rule S1 [repeat].

M5 NS/out

PS x=0 x=1

A B/0 C/1B C/0 D/0

C D/1 C/1

D A/1 B/0



However, since every machine has at least one

HS, M5 has an HS [shortest HS’s for this

machine are 00, 01 and 10].

Using an HS, one can design an adaptive

synchronizing experiment, and since every

machine has an HS, an adaptive synch

experiment can be designed for any machine.

Let us design an adaptive synch experiment for

M5.

(a) Find a (shortest) HS. Let’s choose 10 as our

shortest HS. Prepare a table for responses

and final states for U0 = (ABCD), when an

input 10 is applied.

M5 NS/out

PS x=0 x=1 init’l response to 10 final

A B/0 C/1 A 11 D

B C/0 D/0

B 01 A

C D/1 C/1 C 11 D

D A/1 B/0 D 00 C



(b) The adaptive sync experiment (to synch M5

to, say, state B) is as follows:

Apply 10 to M5, if the response to 10 is

00, M5 is now in C, apply T(C,B) = 01

01, M5 is now in A, apply T(A,B) = 0

11, M5 is now in D, apply T(D,B) = 1

[T(Si,Sj) is the transfer sequence]

The average length of this adaptive experiment,

if initial states are equally likely, is 13/4 = 3.25.

Note that for a machine which already

possesses a preset synchronizing sequence, it is

always possible to design a (possibly shorter)adaptive synchronizing experiment.

ADAPTIVE DISTINGUISHING

EXPERIMENTS

For a machine not possessing a DS, we may

want to design an adaptive distinguishingexperiment, this may be possible for some

machines and may not be possible for some

others. Clearly, the length of the minimal



adaptive experiment will never exceed the

length of the minimal [preset] distinguishing

sequence, if it exists. Hence, we may want to

design an adaptive distinguishing experiment

even for a machine having a [preset] DS, since

this may be shorter.

In order to design an adaptive successor tree,

noting that in the next step we now have the

opportunity of selecting an input symbolaccording to the response we’ve had up to that

time, we group the successors according to not

only the inputs, but also according to the

outputs. Just think that at every application of

an input symbol, a new experiment is being

started. Each nontrivial component of everysuccessor uncertainty vector is considered as an

initial uncertainty for an independent

experiment. A node is terminated whenever an

uncertainty containing either repeated entries or

just a single element is encountered. A minimal

adaptive experiment is derived from the

smallest set of paths emanating from the initialuncertainty and terminating on trivial

uncertainties.



Example: Adaptive distinguishing tree for M1

M1 NS/out

PS x=0 x=1

A C/0 D/1

B C/0 A/1

C A/1 B/0

D B/0 C/1

ABCD

BCC A B ACD

BC A B CD

C A B A B A B C

Four adaptive experiments come out.

Stop

(multiple)

x=0

x=0

x=0

x=0

x=1

x=1

x=1x=1

z=1 z=1

z=1z=1

z=1 z=1z=1 z=1z=0z=0z=0 z=0

z=0 z=0

z=0z=0



One of these experiments is as follows:

1. (ABCD) Apply x = 1

if z = 0, we’re in (B), the initial state was C,

if z = 1, we’re in (ACD), go to step 2

2. (ACD) Apply x = 0

if z = 0, we’re in (BC), go to step 3

if z = 1, we’re in (A), the initial state was D

3. (BC) Apply x = 1

if z = 0, we’re in (B), the initial state was Bif z = 1, we’re in (A), the initial state was A

Another experiment:

1. (ABCD) Apply x = 1

if z = 0, we’re in (B), the initial state was C

if z = 1, we’re in (ACD), go to step 2

2. (ACD) Apply x = 1if z = 0, we’re in (B), the initial state was D

if z = 1, we’re in (CD), go to step 3

3. (CD) Apply x = 0

if z = 0, we’re in (B), the initial state was B

if z = 1,we’re in (A), the initial state was A.

If all initial states are equally likely, then theaverage length of this adaptive experiment is

9/4 = 2.25. Compare this with the length of the

shortest preset DS, which is 3.

week 9 slides state identification and experiments

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