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Introduction to Knowledge Space Theory: Part II. Christina Steiner, University of Graz, Austria April 4, 2005. c. d. b. e. a. 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 :.
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Introduction to Knowledge Space Theory:Part II Christina Steiner, University of Graz, Austria April 4, 2005
c d b e a 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:
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
c d v b e a 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
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
c d v b e a Surmise Function • a knowledge structure conforming to a surmise function • is closed under union but • not necessarily under intersection • example: 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} • What are the clauses for the problems? • 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 Bof 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
c d b e a 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
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}}
f d e c b a 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}} 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}}
{a,b,c,d,e} {a,b,c,e} {a,b,d,e} {a,b,c} {a,b,e} {a,e} {a,b} {a} {e} Ø 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 b d c Ø a b c e d
{a,b,c,d,e} {a,b,c,e} {a,b,d,e} {a,b,c} {a,b,e} {a,e} {a,b} {a} {e} Ø Exercise • How many learning paths are possible for the given knowledge structure? • Which sequences of problems do they suggest for learning?
{a,b,c,d,e} {a,b,c,e} {a,b,d,e} {a,b,c} {a,b,e} {a,e} {a,b} {a} {e} Ø 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
{a,b,c,d,e} {a,b,c,e} {a,b,d,e} {a,b,c} {a,b,e} {a,e} {a,b} {a} {e} Ø 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
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
{a,b,c,d,e} {a,b,d,e} {b,c,d,e} {a,c,d,e} {c,d,e} {a,b,d} {b,d,e} {b,d} {d,e} {d} {e} Ø Exercise • Let us assume the following knowledge structure for the domainQ = {a, b, c, d, e} • Determine the fringes of the encircled knowledge states!