week 9 slides state identification and experiments
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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
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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
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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.
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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
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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.
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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
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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”]
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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.
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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).
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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)]
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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) .
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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
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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
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(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
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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.
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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
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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.