Poggi analytics - concepts - 1a

Buenos Aires, abril de 2016Eduardo Poggi

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Concept Learning

Definitions Search Space and General-Specific Ordering Concept learning as search FIND-S The Candidate Elimination Algorithm Inductive Bias

First definition

The problem is to learn a function mapping examples into two classes: positive and negative.

We are given a database of examples already classified as positive or negative.

Concept learning: the process of inducing a function mapping input examples into a Boolean output.

Notation

Set of instances X Target concept c : X {+,-} Training examples E = {(x , c(x))} Data set D X Set of possible hypotheses H h H / h : X {+,-} Goal: Find h / h(x)=c(x)

Representation of Examples

Features:• color {red, brown, gray}• size {small, large}• shape {round,elongated}• land {humid,dry}• air humidity {low,high}• texture {smooth, rough}

The Input and Output Space

XOnly a small subset is contained

in our database.

Y = {+,-}

X : The space of all possible examples (input space). Y: The space of classes (output space).An example in X is a feature vector X.For instance: X = (red,small,elongated,humid,low,rough) X is the cross product of all feature values.

The Training Examples

D: The set of training examples. D is a set of pairs { (x,c(x)) }, where c is the target concept

Example of D:((red,small,round,humid,low,smooth), +)((red,small,elongated,humid,low,smooth),+)((gray,large,elongated,humid,low,rough), -)((red,small,elongated,humid,high,rough), +)

Instances from the input space Instances fromthe output space

Hypothesis Representation

Consider the following hypotheses: (*,*,*,*,*,*): all mushrooms are poisonous (0,0,0,0,0,0): no mushroom is poisonous

Special symbols: * Any value is acceptable 0 no value is acceptable

Any hypothesis h is a function from X to Y h: X Y

We will explore the space of conjunctions.

Hypothesis Space

The space of all hypotheses is represented by H

Let h be a hypothesis in H. Let X be an example of a mushroom. if h(X) = + then X is poisonous,

otherwise X is not-poisonous Our goal is to find the hypothesis, h*, that is very “close” to target concept c. A hypothesis is said to “cover” those examples it classifies as positive.

X

h

Assumption 1

We will explore the space of all conjunctions.We assume the target concept falls within this space.

Target concept c

H

Assumption 2

A hypothesis close to target concept c obtained afterseeing many training examples will result in high accuracy on the set of unobserved examples.

Training set D

Hypothesis h* is good

Complement set D’

Hypothesis h* is good

Concept Learning as Search

There is a general to specific ordering inherent to any hypothesis space.

Consider these two hypotheses:

h1 = (red,*,*,humid,*,*)h2 = (red,*,*,*,*,*)

We say h2 is more general than h1 because h2 classifies more instances than h1 and h1 is covered by h2.

General-Specific

For example, consider the following hypotheses:

h1

h2 h3

h1 is more general than h2 and h3. h2 and h3 are neither more specific nor more general than each other.

Let hj and hk be two hypotheses mapping examples into {+,-}.We say hj is more general than hk iff

For all examples X, hk(X) = + hj(X) = +

We represent this fact as hj >= hk

The >= relation imposes a partial ordering over the hypothesis space H (reflexive, antisymmetric, and transitive).

Definition

Lattice

Any input space X defines then a lattice of hypotheses orderedaccording to the general-specific relation:

h1

h3 h4

h2

h5 h6

h7 h8

Working Example: Mushrooms

Class of Tasks: Predicting poisonous mushroomsPerformance: Accuracy of ClassificationExperience: Database describing mushrooms with their class

Knowledge to learn: Function mapping mushrooms to {+,-}

where -:not-poisonous and +:poisonousRepresentation of target knowledge:

conjunction of attribute values.Learning mechanism:

Find-S

Finding a Maximally-Specific Hypothesis

Algorithm to search the space of conjunctions: Start with the most specific hypothesis Generalize the hypothesis when it fails to cover a positive

example

Algorithm:1. Initialize h to the most specific hypothesis2. For each positive training example X For each value a in h

If example X and h agree on a, do nothing else generalize a by the next more general constraint3. Output hypothesis h

Example

Let’s run the learning algorithm above with the following examples:((red,small,round,humid,low,smooth), +)((red,small,elongated,humid,low,smooth),+)((gray,large,elongated,humid,low,rough), -)((red,small,elongated,humid,high,rough), +)

We start with the most specific hypothesis:h = (0,0,0,0,0,0)

The first example comes and since the example is positive and h fails to cover it, we simply generalize h to cover exactly this example: h = (red,small,round,humid,low,smooth)

Example

Hypothesis h basically says that the first example is the onlypositive example, all other examples are negative.

Then comes examples 2: ((red,small,elongated,humid,low,smooth), poisonous)

This example is positive. All attributes match hypothesis h except for attribute shape: it has the value elongated, not round.

We generalize this attribute using symbol * yielding:h: (red,small,*,humid,low,smooth)The third example is negative and so we just ignore it. Why is it we don’t need to be concerned with negative

examples?

Example

Upon observing the 4th example, hypothesis h is generalized to the following:

h = (red,small,*,humid,*,*)

h is interpreted as any mushroom that is red, small and found on humid land should be classified as poisonous.

Analyzing the Algorithm

• The algorithm is guaranteed to find the hypothesis that is most specific and consistent with the set of training examples.

• It takes advantage of the general-specific ordering to move on the corresponding lattice searching for the next most specific hypothesis.

h1

h3 h4

h2

h5 h6

h7 h8

X-H Relation

X-H Relation

Points to Consider

There are many hypotheses consistent with the training data D.

Why should we prefer the most specific hypothesis? What would happen if the examples are not consistent? What would happen if they have errors, noise? What if there is a hypothesis space H where one can find more

that one maximally specific hypothesis h? The search over the lattice must then be different to allow for

this possibility.

Summary FIND-S

The input space is the space of all examples; the output space is the space of all classes.

A hypothesis maps examples into classes. We want a hypothesis close to target concept c. The input space establishes a partial ordering over the

hypothesis space. One can exploit this ordering to move along the

corresponding lattice.

Working Example: Mushrooms

Class of Tasks: Predicting poisonous mushroomsPerformance: Accuracy of ClassificationExperience: Database describing mushrooms with their class

Knowledge to learn: Function mapping mushrooms to {+,-}

where -:not-poisonous and +:poisonousRepresentation of target knowledge:

conjunction of attribute values.Learning mechanism:

candidate-elimination

Candidate Elimination

The algorithm that finds the maximally specific hypothesis is limited in that it only finds one of many hypotheses consistent with the training data.

The Candidate Elimination Algorithm (CEA) finds ALL hypotheses consistent with the training data.

CEA does that without explicitly enumerating all consistent hypotheses.

Consistency vs Coverage

h1

h2

h1 covers a different set of examples than h2h2 is consistent with training set Dh1 is not consistent with training set D

Positive examplesNegative examples

Training set D

-

-

-

-++

+++ +

+

Version Space VS

Hypothesis space H

Version space: Subset of hypothesis from H consistent with training set D.

List-Then-Eliminate Algorithm

Algorithm:

1. Version Space VS: All hypotheses in H2. For each training example X Remove every hypothesis h in H inconsistent

with X: h(x) = c(x)

3. Output the version space VS

Comments: This is unfeasible. The size of H is unmanageable.

Previous Exercise: Mushrooms

Let’s remember our exercise in which we tried to classifymushrooms as poisonous (+) or not-poisonous (-).

Training set D:

((red,small,round,humid,low,smooth), +)((red,small,elongated,humid,low,smooth), +)((gray,large,elongated,humid,low,rough), -)((red,small,elongated,humid,high,rough), +)

Consistent Hypotheses

Our first algorithm found only one out of six consistent hypotheses:

(red,small,*,humid,*,*)

(*,small,*,humid,*,*)(red,*,*,humid,*,*) (red,small,*,*,*,*)

(red,*,*,*,*,*) (*,small,*,*,*,*)G:

S:S: Most specificG: Most general

Candidate-Elimination Algorithm

(red,small,*,humid,*,*)

(red,*,*,*,*,*)(*,small,*,*,*,*)G:

S:

The candidate elimination algorithm keeps two listsof hypotheses consistent with the training data:The list of most specific hypotheses S andThe list of most general hypotheses G

This is enough to derive the whole version space VS.

VS

Candidate-Elimination Algorithm

• Initialize G to the set of maximally general hypotheses in H• Initialize S to the set of maximally specific hypotheses in H• For each training example X do

• If X is positive: generalize S if necessary• If X is negative: specialize G if necessary

• Output {G,S}

Candidate-Elimination Algorithm

Initialize G to the set of maximally general hypotheses in H Initialize S to the set of maximally specific hypotheses in H For each training example d, do

If d+ Remove from G any hypothesis inconsistent with d For each hypothesis s in S that is not consistent with d

Remove s from S Add to S all minimal generalizations h of s such that h is

consistent with d and some member of G is more general than h Remove from S any hipothesis that is more general than

another hypothesis in S If d-

Remove from S any hypothesis inconsistent with d For each hypothesis g in G that is not consistent with d

Remove g from G Add to G all minimal specializations h of g such that h is

consistent with d and some member of S is more general than h Remove from G any hipothesis that is less general than another

hypothesis in G

Positive Examples

a) If X is positive: Remove from G any hypothesis inconsistent with X For each hypothesis h in S not consistent with X

Remove h from S Add all minimal generalizations of h consistent with X

such that some member of G is more general than h Remove from S any hypothesis more general than any other hypothesis in S

G:

S:

hinconsistent

add minimal generalizations

Negative Examples

b) If X is negative:Remove from S any hypothesis inconsistent with XFor each hypothesis h in G not consistent with X

Remove g from GAdd all minimal generalizations of h consistent with X

such that some member of S is more specific than hRemove from G any hypothesis less general than any other

hypothesis in G

G:

S: h inconsistent

add minimal specializations

An Exercise

Initialize the S and G sets:

S: (0,0,0,0,0,0)G: (*,*,*,*,*,*)

Let’s look at the first two examples:

((red,small,round,humid,low,smooth), +)((red,small,elongated,humid,low,smooth), +)

An Exercise: two positives

The first two examples are positive:((red,small,round,humid,low,smooth), +)((red,small,elongated,humid,low,smooth), +)

S: (0,0,0,0,0,0)

(red,small,round,humid,low,smooth)

(red,small,*,humid,low,smooth)

G: (*,*,*,*,*,*)

generalize

specialize

An Exercise: first negative

The third example is a negative example:((gray,large,elongated,humid,low,rough), -)

S:(red,small,*,humid,low,smooth)

G: (*,*,*,*,*,*)

generalize

specialize

(red,*,*,*,*,*,*) (*,small,*,*,*,*) (*,*,*,*,*,smooth)

Why is (*,*,round,*,*,*) not a valid specialization of G

An Exercise: another positive

The fourth example is a positive example:((red,small,elongated,humid,high,rough), +)

S:(red,small,*,humid,low,smooth)generalize

specializeG: (red,*,*,*,*,*,*) (*,small,*,*,*,*) (*,*,*,*,*,smooth)

(red,small,*,humid,*,*)

The Learned Version Space VS

G: (red,*,*,*,*,*,*) (*,small,*,*,*,*)

S: (red,small,*,humid,*,*)

(red,*,*,humid,*,*) (red,small,*,*,*,*) (*,small,*,humid,*,*)

Points to Consider

Will the algorithm converge to the right hypothesis? The algorithm is guaranteed to converge to the right

hypothesis provided the following: No errors exist in the examples The target concept is included in the hypothesis space H

What happens if there exists errors in the examples? The right hypothesis would be inconsistent and thus eliminated. If the S and G sets converge to an empty space we have

evidence that the true concept lies outside space H.

Query Learning

Remember the version space VS after seeing our 4 exampleson the mushroom database:

G: (red,*,*,*,*,*,*) (*,small,*,*,*,*)

S: (red,small,*,humid,*,*)

(red,*,*,humid,*,*) (red,small,*,*,*,*) (*,small,*,humid,*,*)

What would be a good question to pose to the algorithm? What example is best next?

Query Learning

Remember there are three settings for learning: Tasks are generated by a random process outside the learner The learner can pose queries to a teacher The learner explores its surroundings autonomously

Here we focus on the second setting; posing queries to an expert.

Version space strategy: Ask about the class of an example that would prune half of the space.

Example: (red,small,round,dry,low,smooth)

Query Learning

In general if we are able to prune the version space by half on each new query then we can find an optimal hypothesis in the following

Number of steps: log2 |VS|

Can you explain why?

Classifying Examples

What if the version space VS has not collapsed into a single hypothesis and we are asked to classify a new instance?

Suppose all hypotheses in set S agree that the instance is positive.

Then we are sure that all hypotheses in VS agree the instance is positive. Why?

The same can be said if the instance is negative by all members of set G. Why?

In general one can vote over all hypotheses in VS if there is no unanimous agreement.

Inductive Bias

Inductive bias is the preference for a hypothesis space H and a search mechanism over H.

What would happen if we choose an H that contains all possible hypotheses?

What would the size of H be? |H| = Size of the power set of the input space X.

Example: You have n Boolean features. |X| = 2n And the size of H is 2^2^n

Inductive Bias

In this case, the candidate elimination algorithm would simply classify as positive the training examples it has seen. This is because H is so large, every possible hypothesis is contained within it.

A Property of any Inductive Algorithm:It must have some embedded assumptions about the nature of H.

Without assumptions learning is impossible.

Summary

The candidate elimination algorithm exploits the general-specific ordering of hypotheses to find all hypotheses consistent with the training data.

The version space contains all consistent hypotheses and is simply represented by two lists: S and G.

Candidate elimination algorithm is not robust to noise and assumes the target concept is included in the hypothesis space.

Any inductive algorithm needs some assumptions about the hypothesis space, otherwise it would be impossible to perform predictions.

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Bibliografía

Capítulo 2 de Mitchell