Introduction to AI and the perceptron

8/13/2019 Introduction to AI and the perceptron

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CS344: Introduction to Artificial

Intelligence(associated lab: CS386)

Pushpak BhattacharyyaCSE Dept.,

IIT Bombay

From 29 Jan, 2014

(lecture 7 was by Vinita and Rahul on

Sentiment and Deep Learning)


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AI Perspective (post-web)

Planning

Computer

Vision

NLP

ExpertSystems

Robotics

Search,Reasoning,Learning

IR


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Search: Everywhere


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Planning (a) which block to pick, (b) which to stack, (c) which to unstack, (d)

whether to stack a block or (e) whether to unstackan already stackedblock. These options have to be searched in order to arrive at the rightsequence of actions.

A CB A

B

C

Table


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Vision A search needs to be carried out to find which point in the image of L

corresponds to which point in R. Naively carried out, this can becomean O(n2) process where n is the number of points in the retinal

images.

World

Two eyesystem

R L


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Robot Path Planning searching amongst the options of moving Left, Right, Up or Down.

Additionally, each movement has an associated cost representing therelative difficulty of each movement. The search then will have to findthe optimal, i.e., the least cost path.

O1

R

D

O2

Robot

Path


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Natural Language Processing search among many combinations of parts of speech on the way to

deciphering the meaning. This applies to every level of processing-syntax, semantics, pragmatics and discourse.

The man would like to play.

Noun Verb NounVerb VerbPreposition


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Expert SystemsSearch among rules, many of which can apply to a

situation:

If-conditionsthe infection is primary-bacteremiaAND the site of the culture is one of the sterile sitesAND the suspected portal of entry is the gastrointestinal tract

THENthere is suggestive evidence (0.7) that infection is bacteroid

(from MYCIN)


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Search building blocks

State Space : Graph of states (Express constraints

and parameters of the problem)

Operators : Transformations applied to the states.

Start state : S0(Search starts from here)Goal state : {G} - Search terminates here.

Cost : Effort involved in using an operator.

Optimal path : Least cost path


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Examples

Problem 1 : 8puzzle

8

4

6

5

1

7

2

1

4

7

63 3

5

8

S

2

G

Tile movement represented as the movement of the blank space.

Operators:L : Blank moves left

R : Blank moves right

U : Blank moves up

D : Blank moves downC(L) = C(R) = C(U) = C(D) = 1


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Problem 2: Missionaries and Cannibals

ConstraintsThe boat can carry at most 2 people

On no bank should the cannibals outnumber the missionaries

River

R

L

Missionaries Cannibals

boat

boat

Missionaries Cannibals


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State :

#M= Number of missionaries on bankL

#C= Number of cannibals on bankL

P= Position of the boat

S0 =

G = < 0, 0, R >

Operations

M2= Two missionaries take boat

M1= One missionary takes boat

C2= Two cannibals take boatC1= One cannibal takes boat

MC = One missionary and one cannibal takes boat


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C2 MC

Partial searchtree


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Problem 3

B B W W WB

G: States where no Bis to the left of any W

Operators:1) A tile jumps over another tile into a blank tile with cost

2

2) A tile translates into a blank space with cost 1


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Algorithmics of Search


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General Graph search Algorithm

SS

AA CB

F

ED

G

1 103

5 4 6

23

7

Graph G = (V,E)

A CB

D E

F G

S


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1) Open List : S (, 0)

Closed list :

2) OL : A(S,1)

, B(S,3)

, C(S,10)

CL : S

3) OL : B(S,3), C(S,10), D(A,6)

CL : S, A

4) OL : C(S,10), D(A,6), E(B,7)

CL: S, A, B

5) OL : D(A,6), E(B,7)

CL : S, A, B , C

6) OL : E(B,7), F(D,8), G(D, 9)

CL : S, A, B, C, D

7) OL : F(D,8)

, G(D,9)

CL : S, A, B, C, D, E

8) OL : G(D,9)

CL : S, A, B, C, D, E, F

9) OL :

CL : S, A, B, C, D, E,

F, G

St f GGS


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Steps of GGS(principles of AI, Nilsson,)

1. Create a search graph G, consisting solely of thestart node S; put Son a list called OPEN.

2. Create a list called CLOSEDthat is initially empty.

3. Loop: if OPENis empty, exit with failure.

4. Select the first node on OPEN, remove from OPENand put on CLOSED, call this node n.

5. if nis the goal node, exit with the solutionobtained by tracing a path along the pointers from n

to sin G. (ointers are established in step 7). 6. Expand node n, generating the set Mof its

successors that are not ancestors of n. Install thesememes of Mas successors of nin G.


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GGS steps (contd.)

7. Establish a pointer to nfrom those members of Mthat were not already in G(i.e., not already on eitherOPEN or CLOSED). Add these members of Mto

OPEN. For each member of M that was already onOPENor CLOSED, decide whether or not to redirectits pointer to n. For each member of M already onCLOSED, decide for each of its descendents in G

whether or not to redirect its pointer. 8. Reorder the list OPENusing some strategy.

9. Go LOOP.


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GGS is a general umbrella

S

n1

n2

g

C(n1,n2)

h(n2)

h(n1)

)(),()( 2211 nhnnCnh

OL is a

queue

(BFS)

OL is

stack

(DFS)

OL is accessed by

using a functions

f= g+h

(Algorithm A)


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Algorithm A

A functionfis maintained with each node

f(n) = g(n) + h(n), nis the node in the open list

Node chosen for expansion is the one with leastfvalue

For BFS: h= 0,g= number of edges in the

path to S

For DFS: h= 0,g=


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Algorithm A* One of the most important advances in AI

g(n)= least cost path to n from S found so far

h(n)


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A* AlgorithmDefinition andProperties

f(n) = g(n) + h(n) The node with the least

value of fis chosen from theOL.

f*(n) = g*(n) + h*(n),where,

g*(n)= actual cost ofthe optimal path (s, n)

h*(n)= actual cost ofoptimal path (n, g)

g(n) g*(n)

By definition, h(n) h*(n)

S s

n

goal

State space graph G

g(n)

h(n)


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8-puzzle: heuristics

2 1 4

7 8 3

5 6

1 6 7

4 3 2

5 8

1 2 3

4 5 6

7 8s n g

Example: 8 puzzle

h*(n)= actual no. of moves to transform nto g

1. h1(n)= no. of tiles displaced from their destined

position.2. h2(n)= sum of Manhattan distances of tiles from

their destined position.

h1(n) h*(n) andh1(n) h*(n)

h*

h2

h1

Comparison


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A* critical points

Goal

1. Do we know the goal?

2. Is the distance to the goal known?

3. Is there a path (known?) to the goal?


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A* critical points

About the pathAny time before A* terminates there

exists on the OL, a node from the optimalpath all whose ancestors in the optimalpath are in the CL.

This means,in the OL always a node n s.t.

g(n) = g*(n)


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Key point about A* search

S

Statement:

Let S -n1-n2-n3ni-nk-1-nk(=G) be an optimal path.

At any time during thesearch:

1. There is a node nifrom theoptimal path in the OL

2. For niall its ancestorsS,n1,n2,,ni-1are in CL

3. g(ni) = g*(ni)

S

|

n1|

n2|

.

.

ni.

.nk-1|

nk=g


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Proof of the statement

Proof by induction on iteration no. j

Basis : j = 0, S is on the OL, S satisfies

the statement

Hypothesis : Let the statement be true forj = p (pthiteration)

Let nibe the node satisfying thestatement


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Proof (continued)

Induction : Iteration no. j = p+1Case 1 : niis expanded and moved to

the closed list

Then, ni+1from the optimal pathcomes to the OL

Node ni+1satisfies the statement

(note: if ni+1is in CL, then ni+2satisfiesthe property)

Case 2 : Node x niis expanded

Here, nisatisfies the statement


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Admissibility: An algorithm is called admissible if italways terminates and terminates in optimal path

Theorem: A* is admissible.

Lemma: Any time before A* terminates there existson OLa node nsuch that f(n)


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f*(ni) = f*(s), ni sand ni g

Following set of equations show the above equality:

f*(ni) = g*(ni) + h*(ni)f*(ni+1) = g*(ni+1) + h*(ni+1)

g*(ni+1) = g*(ni) + c(ni , ni+1)

h*(ni+1) = h*(ni) - c(ni , ni+1)

Above equations hold since the path is optimal.

A* Properties (contd.)


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Lemma

Any time before A* terminates there exists in the open list a node n'

such thatf(n')


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If A* does not terminate

Let ebe the least cost of all arcs in the search graph.

Theng(n) >= e.l(n)where l(n)= # of arcs in the path from Sto

nfound so far. If A* does not terminate,g(n)and hence

f(n) = g(n) + h(n) [h(n) >= 0]will become unbounded.

This is not consistent with the lemma. So A* has to terminate.


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2ndpart of admissibility of A*

The path formed by A* is optimal when it has terminated

Proof

Suppose the path formed is not optimal

Let Gbe expanded in a non-optimal path.

At the point of expansion of G,

f(G) = g(G) + h(G)

= g(G) + 0

> g*(G) = g*(S) + h*(S)

= f*(S)[f*(S)= cost of optimal path]

This is a contradiction

So path should be optimal


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Summary on Admissibility

1. A* algorithm halts

2.A* algorithm finds optimal path

3. If f(n) < f*(S) then node nhas to be expandedbefore termination

4. If A* does not expand a node nbefore termination

then f(n) >= f*(S)


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Exercise-1

Prove that if the distance of every node from the goalnode is known, then no search: is necessary

Ans:

For every node n, h(n)=h*(n). The algo is A*.

Lemma proved: any time before A* terminates, there is a nodemin the OL that has f(m)


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Exercise-2If the hvalue for every node over-estimates the h*value of the

corresponding node by a constant, then the path found need

not be costlier than the optimal path by that constant. Provethis.

Ans:

Under the condition of the problem, h(n)


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Better Heuristic PerformsBetter

h


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Theorem

A version A2* of A* that has a better heuristic than another version

A1* of A* performs at least as well as A1*Meaning of better

h2(n) > h1(n)for all n

Meaning of as well asA

1* expands at least all the nodes of A

2*

h*(n)

h2(n)

h1(n) For all nodes n,

except the goal

node

P f b i d i h h f A *


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Proof by induction on the search tree of A2*.

A* on termination carves out a tree out of G

Induction

on the depth k of the search tree of A2*. A

1* before termination

expands all the nodes of depth kin the search tree of A2*.

k=0. True since start node Sis expanded by both

Suppose A1* terminates without expanding a node nat depth (k+1)of

A2* search tree.

Since A1* has seen all the parents of nseen by A

2*

g1(n)


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Proof for g1(nk+1)


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Proof for g1(nk+1)


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Proof for g1(nk+1)


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k+1

S

G

Since A1* has terminated without

expanding n,

f1(n) >= f*(S) (2)

Any node whosef value is strictly less

thanf*(S)has to be expanded.

Since A2* has expanded n

f2(n) = h2(n) which is a contradiction. Therefore, A1* has to expand

all nodes that A2* has expanded.

Exercise

If better meansh2(n) > h1(n)for some nand h2(n) = h1(n)for others,

then Can you prove the result ?


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Monotonicity

Steps of GGS


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Steps of GGS(principles of AI, Nilsson,)

1. Create a search graph G, consisting solely of thestart node S; put Son a list called OPEN.

2. Create a list called CLOSEDthat is initially empty.

3. Loop: if OPENis empty, exit with failure.

4. Select the first node on OPEN, remove from OPENand put on CLOSED, call this node n.

5. if nis the goal node, exit with the solutionobtained by tracing a path along the pointers from n

to sin G. (ointers are established in step 7). 6. Expand node n, generating the set Mof its

successors that are not ancestors of n. Install thesememes of Mas successors of nin G.


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GGS steps (contd.)

7. Establish a pointer to nfrom those members of Mthat were not already in G(i.e., not already on eitherOPEN or CLOSED). Add these members of Mto

OPEN. For each member of M that was already onOPENor CLOSED, decide whether or not to redirectits pointer to n. For each member of M already onCLOSED, decide for each of its descendents in G

whether or not to redirect its pointer. 8. Reorder the list OPENusing some strategy.

9. Go LOOP.

I ustration or CL parent pointer


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I ustration or CL parent pointerredirection recursively

S

1

23

54

6

Node in CL

Node in OL

Parent Pointer


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Definition of monotonicity

A heuristic h(p)is said to satisfy themonotone restriction, if for all p,

h(p)


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Grounding the Monotone Restriction7 3

1 2 4

8 5 6

1 2 3

4 5 6

7 8

nG

7 3 4

1 2

8 5 6

h(n) -: number of displaced tiles

Is h(n) monotone ?h(n) = 8h(n) = 8

C(n,n) = 1

Hence monotone

n

M i i f # f Di l d


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Monotonicity of # of DisplacedTile Heuristic

h(n) < = h(n) + c(n, n)

Any move changes h(n) by at most 1

c = 1

Hence, h(parent) < = h(child) + 1

If the empty cell is also included in thecost, then hneed not be monotone(try!)

M t i it f M h tt


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Monotonicity of ManhattanDistance Heuristic (1/2)

Manhattan distance= X-dist+Y-distfromthe target position

Refer to the diagram in the first slide: hmn(n) = 1 + 1 + 1 + 2 + 1 + 1 + 2 + 1 =

10 hmn(n) = 1 + 1 + 1 + 3 + 1 + 1 + 2 + 1

= 11 Cost = 1 Again, h(n) < = h(n) + c(n, n)

M t i it f M h tt


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Monotonicity of ManhattanDistance Heuristic (2/2)

Any move can either increase the h valueor decrease it by at most 1.

Cost again is 1. Hence, this heuristic also satisfies

Monotone Restriction If empty cell is also included in the cost

then manhattan distance does not satisfymonotone restriction (try!)

Apply this heuristic for Missionaries andCannibals problem

R l ti hi b t


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Relationship betweenMonotonicity and Admissibility

Observation:

Monotone Restriction Admissibility

but not vice-versa

Statement: If h(ni)


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Proof of MR leading to optimal path


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Let S-N1- N2- N3- N4... NmNk be an optimal path from S to Nk (all ofwhich might or might not have been explored). Let Nm be the lastnode on this path which is on the open list, i.e., allthe ancestors from Sup to Nm-1are in the closed list.

For every node Np on the optimal path,

g*(Np)+h(Np)


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Now if Nk is chosen in preference to Nm,f(Nk)


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Monotonicity of f(.) values

Statement:fvalues of nodes expanded by A*

increase monotonically, if his

monotone.Proof:

Suppose niand njare expanded with

temporal sequentiality, i.e., njisexpanded after ni


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Proof (1/3)

niexpanded before nj

niand njco-existing njcomes to open list as aresult of expandingni and isexpanded immediately

njsparent pointerchanges to niandexpanded

njexpandedafter ni


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Proof (2/3)

All the previous cases are forms of thefollowing two cases (think!)

CASE 1:njwas on open list when niwas expanded

Hence, f(ni)


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Proof (3/3)

ni

nj

Case 2:

f(ni) = g(ni) + h(ni) (Defn of f)f(nj) = g(nj) + h(nj) (Defn off)

f(ni) = g(ni) + h(ni) = g*(ni) + h(ni) ---Eq 1

(since niis picked for expansion niis on optimal path)

With the similar argument for njwe can write the following:f(nj) = g(nj) + h(nj) = g*(nj) + h(nj) ---Eq 2

Also,h(ni) < = h(nj) + c(ni, nj) ---Eq 3 (Parent- child

relation)g*(nj) = g*(ni) + c(ni, nj) ---Eq 4 (both nodes on

optimal path)From Eq 1, 2, 3 and 4

f(ni)


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Better way to understandmonotonicity of f()

Let f(n1), f(n2), f(n3), f(n4) f(nk-1), f(nk) be the fvalues of k expanded nodes.

The relationship between two consecutive expansionsf(ni) and f(ni+1) nodes always remains the same, i.e.,

f(ni)


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Monotonicity of f()

f(n1), f(n2), f(n3), ,f(ni), f(ni+1), ,f(nk)Sequence of expansion ofn1, n2,n3 ni nk

f values increase monotonically

f(n) = g(n) + h(n)

Consider two successive expansions - > ni, ni+1

Case 1:ni & ni+1 Co-existing in OL

ni precedes ni+1

By definition of A*f(ni)


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Monotonicity of f()

Case 2:ni+1 came to OL because of expanding ni and ni+1

is expanded

f(ni) = g(ni) + h(ni)


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A list of AI Search Algorithms

A* AO* IDA* (Iterative Deepening)

Minimax Search on Game Trees

Viterbi Search on Probabilistic FSA Hill Climbing Simulated Annealing Gradient Descent

Stack Based Search Genetic Algorithms Memetic Algorithms

Introduction to AI and the perceptron

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