Introduction to Knowledge Space Theory:Part II

Christina Steiner,University of Graz, Austria

April 4, 2005

Surmise Function

for mastering a problem p there is a minimal set of problems that must have been

mastered before- prerequisites for problem p

- example: c

b

a

d

e

Surmise Function

there may be more than one set of prerequisites to a problem

- these different sets of prerequisites may represent alternative ways of solving a problem

- example:for the mastery of problem d, the problems (a and b) or e must have been mastered before

to capture the fact, that a problem may have more than one set of prerequisites, the notion of a surmise function has been introduced

- generalisation of the concept of a surmise relation

- allows for assigning multiple sets of prerequisites to a problem

Surmise Function

assigns to each problem p a family of subsets of problems, called ‚clauses‘

- denoted by σ(p)- they represent all possible ways of acquiring the mastery of problem p- minimal states containing problem p

can be depicted by an And/Or-Graph

example:σ(a) = {{a}} σ(b) = {{a, b}}σ(c) = {{a, b, c}}σ(d) = {{a, b, d}, {d, e}}σ(e) = {{e}}

if a person is found to have mastered a given problem,then at least one ot the clauses for the problem must be included in the person‘s knowledge state

c

b

a

d

e

v

Surmise Function

clauses satisfy the following conditions- for each problem p, there is at least one clause for p

- every clause for a problem p contains p

- if a problem q is in some clause C for p, then there must be some clause D for q included in C

- example:σ(a) = {{a}} σ(b) = {{a, b}}σ(c) = {{a, b, c}}σ(d) = {{a, b, d}, {d, e}}σ(e) = {{e}}

- any two clauses C, C‘ for the same problem are incomparable, i.e. neither C C‘ nor C‘ C

Surmise Function

a knowledge structure conforming to a surmise function is closed under union but not necessarily under intersection

example:

c

b

a

d

e

v K = { Ø, {a}, {e}, {a, b}, {a, e}, {d, e},{a, b, c}, {a, b, d}, {a, b, e}, {a, d, e},{a, b, c, d}, {a, b, c, e}, {a, b, d, e},{a, b, c, d, e}}

Exercise

Let us assume the followingsurmise function for thedomain Q = {a, b, c, d, e}

a) What are the clauses for the problems?

b) Find the collection of possible knowledge states corresponding to the surmise function!

K = { Ø, {d}, {e}, {b,d}, {d,e}, {a,b,d}, {b,d,e}, {c,d,e},{a,b,d,e}, {a,c,d,e}, {b,c,d,e}, {a,b,c,d,e}}

σ(a) = {{a,b,d}, {a,c,d,e}}σ(b) = {{b,d}}σ(c) = {{c,d,e}}σ(d) = {{d}}σ(e) = {{e}}

Base of a Knowledge Space

in practical application knowledge spaces can grow very large

the base B of a knowledge space provides a way of describing such a structure economically exploiting the property of being closed under union smallest subcollection of a knowledge space from which

the complete knowledge space can be reconstructed by closure under union

Base of a Knowledge Space

example:

K = { Ø, {a}, {e}, {a, b}, {a, e}, {a, b, e}, {a, b, c},{a, b, c, e}, {a, b, d, e}, {a, b, c, d, e}}

all states of the given knowledge space can be obtained by taking all arbitrary unions of the states included in the subcollection:

B = {{a}, {e}, {a, b}, {a, b, c}, {a, b, d, e}}

Base of a Knowledge Space

the base of a knowledge space is formed by the family of all knowledge states that are minimal for at least one problem

atoms of a knowledge space- for any problem p, an atom at p is a minimal knowledge state

containing p- a knowledge state K is minimal for an item p if for any other

knowledge state K‘ the condition K‘ K holds

Base of a Knowledge Space

example: K = { Ø, {a}, {e}, {a, b}, {a, e}, {a, b, e}, {a, b, c}, {a, b, c, e}, {a, b, d, e}, {a, b, c, d, e}}

atom at a: {a} atom at b: {a, b} atom at c: {a, b, c} atom at d: {a, b, d, e}atom at e: {e}

B = {{a}, {e}, {a, b}, {a, b, c}, {a, b, d, e}}

in case of a knowledge space induced by a surmise function

- each of the clauses is an element of the base

- each element of the base is a clause for some problem

c

b

a

d

e

Exercise

Let us assume the following base of a knowledge space

for the domain Q = {a, b, c, d, e}

B = {{b}, {c}, {c, d}, {a, b, c}, {c, d, e}}

- Find the collection of all possible knowledge states!

K = { Ø, {b}, {c}, {b, c}, {c, d}, {a, b, c}, {b, c, d}, {c, d, e},

{a, b, c, d}, {b, c, d, e}, {a, b, c, d, e}}

Exercise

Let us assume the following surmise relation and the corresponding knowledge space for the domainQ = {a, b, c, d, e}

- Determine the base!

B = {{a}, {b}, {a, c}, {a, b, c, d}, {a, b, c, e}, {a, b, c, d, e, f}}

a

c b

d e

f

K ={ Ø, {a}, {b}, {a, b}, {a, c}, {a, b, c}, {a, b, c, d}, {a, b, c, e}, {a, b, c, d, e}, {a, b, c, d, e, f}}

Learning Paths

a knowledge structure allows several learning paths starting from the naive knowledge state leading to the knowledge state of full mastery

Ø

{a} {e}

{a,e}{a,b}

{a,b,c} {a,b,e}

{a,b,c,e} {a,b,d,e}

{a,b,c,d,e}

Ø a e b d c

Ø a b c e d

Exercise

How many learning paths are possible for the given knowledge structure?

- Which sequences of problems do they suggest for learning?

Ø

{a} {e}

{a,e}{a,b}

{a,b,c} {a,b,e}

{a,b,c,e} {a,b,d,e}

{a,b,c,d,e}

c d

e d c

b c e d

a e b c d

Ø d c

e a b c d

d c

Well-Graded Knowledge Structure

a knowledge structure where learning can take place step by step is called well-graded

each knowledge state has at least one immediate successor state containing all the same problems,

plus exactly one

each knowledge state has at least one predecessor state containing exactly the same

problems, except one Ø

{a} {e}

{a,e}{a,b}

{a,b,c} {a,b,e}

{a,b,c,e} {a,b,d,e}

{a,b,c,d,e}

Fringes of a Knowledge State

outer fringe set of all problems p such that

adding p to K forms another knowledge state

- learning proceeds by mastering a new problem in the outer fringe

inner fringe set of all problems p such that

removing p from K forms another knowledge state

- reviewing previous material should take place in the inner fringe of the current knowledge state Ø

{a} {e}

{a,e}{a,b}

{a,b,c} {a,b,e}

{a,b,c,e} {a,b,d,e}

{a,b,c,d,e}

Fringes of a Knowledge State

for a well-graded knowledge structure the two fringes suffice to completely specify the knowledge state summarising the results of assessment the knowledge state of a learner can be characterized by

two lists- the inner fringe specifies what the student can do (the most

sophisticated problems in the knowledge state)

- the outer fringe specifies what the student is ready to learn

Exercise

Let us assume the following knowledge structure for the domainQ = {a, b, c, d, e}

- Determine the fringes of the encircled knowledge states!

Ø

{d} {e}

{d,e}{b,d}

{c,d,e}{b,d,e}{a,b,d}

{b,c,d,e} {a,c,d,e}{a,b,d,e}

{a,b,c,d,e}

knowledge state

outer fringe

inner fringe

{b,d} {a,e} {b}

{b,d,e} {a,c} {b,e}

{a,b,d,e} {c} {a,e}

Thank you for your attention!