Version Space 2001 16 Version Space 1 n

Version Space 2001. 16 윤지은

Version Space 1. n Attribute value – cs – ee – faculty – staff – four – five

n Positive ( attribute value ) ··· n Negative ¬ ( attribute value ··· attribute value )

The De morgan’s law n ¬ ( attribute value ··· attribute value ) = n ¬ attribute value ··· ¬ attribute value ) 각 attribute value들을 다른 attribute value 들과 독립적으로 디자인 할 것을 제안

n cs cs A’ n ee ee A’ n faculty A’ n staff A’ n four A’ n five A’

(1) ∧ cs cs ∧ faculty ee cs ∧ staff cs ∧ faculty ∧ four faculty ee ∧ faculty cs ∧ faculty ∧ five staff ee ∧ staff four five faculty ∧ four ee ∧ faculty ∧ four faculty ∧ five cs ∧ staff ∧ five

ee A cs A faculty A four A staff A five A

(2) first positive - cs faulty four ee A cs A faculty A cs A’ faculty A’ four A staff A four A’ five A

Double strand selection cs A’ faculty A’ four A’

(3) first negative ¬cs ¬staff ¬ five negative를 positive와 hybridization faculty A cs A’ four A staff A’ five A’

First postive와 first negative결과 Positive에서 만들어진 double strand는 cs, faculty, four이며 이것을 denature시켜서 negative와 hybridization시키면 positive와 negative에 공통으로 들어있는 것만 얻어 낼 수 있다. n 공통으로 들어있지 않은 negative attribute value는 영향을 주지 않으므로 고려대상 에서 제외됨 n

이 방법의 가정이 attribute를 찾아내는 것 이 아니라 attribute value들을 찾아내서 나 중에 conjunction을 시키는 것이므로, positive의 모든 subset을 element로 같는 powerset에서positive negative의 power set을 빼주면 찾고자 하는 해답이 된다. n Solution = (positive) - (S(positive) S(negative)) n

= { {cs}, {faculty}, {four}, {cs, faculty}, {cs, four}, {faculty, four}, {cs, faculty, four}, {} } { {cs}, {}} = {{faculty}, {four}, {cs, faculty}, {cs, four}, {faculty, four}, {cs, faculty, four}}

(4) second positive cs A’ five A’ faculty A’ four A cs A’ faculty A’

Seletion cs A’ faculty A’

(5) second negative ¬ee ¬faculty ¬ four positive 2와 hybridization cs A faculty A ee A’ four A’ faculty A’

(Positive 2) - ( ( (S(positive 1) S(negative 1)) (S(positive 2) S(negative 2))) n {{cs}, {faculty}, {cs, faculty}, {}} {{cs}, {faculty}, {}} = {{cs, faculty}} n

1. 마지막 Positive에 남은 attribute value들 을polymerase를 이용해, powerset을 만들 어낸다. (single strand로) n 2. negative들도 동일. n 1과 2에서 hybridization된 double strand를 제외한 single strand들의 conjunction이 solution. n hybridization되지 않고 남아 있는 single strand를 농도차이로 구분. n

Tip Example의 수를 예측할 수 있을때 병렬성 보장 n strand의 길이 : n – ( attribute value 길이 + special code ) * example 수 n strand의 개수 : – attibute value의 갯수

example 4개, attribute value 6개 cs A cs B cs C cs D ee A ee B ee C ee D faculty A faculty B faculty C faculty D staff A staff B staff C staff D four A four B four C four D five A five B five C five D

n Positive 1 cs n A’ faculty B’ four A’ five B’ four A’ negative 1 four A’ n four A’ positive 2 cs B’ n faculty A’ negative 2 four A’

cs A cs B cs C cs D cs A’ cs B’ ee A ee B ee C ee D faculty A faculty B faculty C faculty D faculty A’ faculty B’ staff A staff B staff C staff D five A five B five C five D four C four D five B’ four A’ four B