Bioinformatics Tutorial Questions 1 – 15th January 2003

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Oct 1, 2013 (3 years and 10 months ago)

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Artificial Intelligence Tutorial 4
-

Answers


1a) Recall these sentences from last week’s tutorial:


(i)

All dogs are mammals

(ii)

Fido is a dog

(iii)
Fido is a mammal

(iv)

All mammals produce milk

(v)

There exists a dog which doesn’t produce milk


Using sentences (i) and (iv) in

CNF (see your previous answers), deduce something new about dogs.


[Note that to show you have done the step, you should use the notation from the notes, i.e., draw a
line from your two sentences to the resolved sentences, and show any substitutions you
needed to
make in
-
between the two lines]


1b) Resolve your answer from part (a) with sentence (ii) to deduce something new about fido.


1a) By the term “resolve”, we mean use the resolution inference rule to deduce something new from
瑷漠歮潷渠獴慴s浥湴献m
周T 瑨敯ty be桩湤hre獯汵s楯渠瑨敯te洠灲潶楮g i猠晡楲iy ex瑥湳楶tⰠ扵琠t桥
a灰汩ca瑩潮o潦o瑨t 楮摩癩摵慬v牵汥 楳i 牥a汬y 煵楴e 獩s灬攺p橵獴 汯潫l 景f a l楴e牡氠楮i 潮o 獥湴敮ne 景爠
睨楣栠 y潵o ca渠 晩湤f 瑨t 湥条瑩潮o 楮i 瑨攠 潴桥o 獥湴敮neⰠ 瑨敮t 睲楴e 瑨t 摩獪畮u瑩
潮o 潦o 瑨攠 瑷漠
獥湴敮ne猬s牥浥浢m物湧r瑯t潭o琠any 浥湴楯渠潦 瑨攠汩瑥ta氠⡯爠楴猠湥ga瑩潮⤬ 扥ca畳e 楴 楳i瑨敳t 睨楣h
a牥 re獯汶敤saway⸠佮l c潭灬楣i瑩潮o潣c畲猠睨wn a 獵扳s楴畴u潮o楳i牥煵楲q搠楮i潲摥爠瑯t浡步 a 汩瑥ta氠
汯潫楫攠楴i ga瑩潮o


f渠獥湴e湣es

⡩⤠a湤n⡩瘩Ⱐ瑨攠潮oy 灲p摩da瑥tsy浢m氠睨楣栠a灰pa牳r楮i扯瑨b楳i浡浭慬Ⱐa湤n汵捫楬y 楴
湯n
-
negated in (i) and negated in (iv). For other contexts, it’s important to remember that the X
癡物r扬敳b a牥
different

variables. However, for this situation, we ca
n simply resolve the two
sentences to give the following:



(i) ¬dog(X)


浡浭慬m堩††††††††††††††††
i瘩€慭浡氨堩


浩汫⡘)



⡮(€⁤ g(堩


浩汫⡘)


S漬睥 獥e 瑨慴t瑨攠浡m浡氠灲p摩捡te 桡猠bee渠e牡摩捡瑥搬 a湤n瑨t 湥眠se湴敮ne ⡷桥n
瑲t湳污瑥d
扡c欠晲潭⁃NF⤠獴)瑥猠瑨a琠慬氠摯t猠灲潤sce 汫l


ㅢ⤠乥x琬ta猠獥湴敮ne ⡩(⤠潮y ha猠潮 汩瑥ta氬l瑨t猠楳ic汥l牬y 瑨t 潮 睨wc栠睩汬 扥 牥獯汶敤sa睡y.
F潲瑵湡瑥tyⰠ瑨攠湥条瑩潮潦瑨攠汩瑥ta氠潣c畲猠楮i潵爠湥眠獥湴敮ce ⡮(⸠H潷癥爠楮i⡩椩Ⱐwe

桡癥 a
g牯畮r 癡物r扬攠 楮獩摥 瑨攠 摯g 灲e摩捡teⰠ 湡mely 晩摯Ⱐ 睨erea猠 楮 ⡮⤬) 睥 桡ve a 畮楶u牳r汬y
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摯d(堩 睩瑨w摯d⡦楤i⤮)呯摯d瑨t猬 睥 睩汬 獵扳t楴畴攠堠
睩瑨w晩摯f 䡥湣e 瑨攠牥獯汵s楯渠獴数sca渠扥
睲楴瑥渠t猠景汬潷猺


⡮(€⁤ g(堩


浩汫⡘⤠††††††† 楩⤠摯)⡦楤漩



筘⽦楤潽


⡮㈩ 汫⡦楤漩


乯瑥 瑨慴t睥 car物r搠潵琠瑨攠獵扳s楴畴u潮睨w牥癥r 睥 獡眠堠楮i瑨攠晩f獴s獥湴敮neⰠ獯s瑨慴t睥 e湤nd
異u獵扳
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fido: he produces milk. It’s important to remember to put the substitution: {X/fido} on the diagram,
so that whoever reads the diagram can understand exactly what
has occurred.



1c) Take sentences (i),(ii),(iii) and (v) as axioms about the world and show they are inconsistent with
sentence (iv). Do this by drawing a resolution proof tree. You will have to search for the correct
resolution steps at each stage to
carry out.


1d) Which steps in your proof are unit resolution steps?



1c) Remember that the resolution method performs proof by contradiction, so it derives an empty set
of clauses which we take to be the false predicate. As it has derived false, the met
hod must have
shown that there was a contradiction in the knowledge base supplied to it. So, the first thing to note
here is that we are being asked to show that (iv) is
false
. Normally, we are asked to show that
statements are
true
, in which case, we have

to
negate them

in order to use the resolution theorem
proving method. But we don’t need to negate (iv) here, as we want to derive a contradiction using (i)
瑯
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oe獯汵s楯渠
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瑯t 牥獯汶攠 瑯te瑨e爠 楮i o牤r爠 瑯t 摥物癥 瑨攠 晡汳l 獴慴s浥湴⸠ f琠 楳i 桥re 瑨慴t 瑨攠 䕮b汩獨s 污lg畡来
摥獣物灴楯渠潦o睨w琠楳i 睲潮o 楮i 瑨攠歮潷汥dge ba獥 獨潵s搠桥汰l g畩摥uy
潵爠獥a牣栮h乯ke 瑨慴t睥
didn’t use fido in the English explanation, so it’s unlikely that he’ll feature in the proof. Always bear
楮i 浩湤n 睨w琠 睥 are try楮g 瑯t 摯㨠 eve湴畡汬y 睥 睡湴n 瑯t re獯汶攠 t睯w獥湴敮ne猠 獯s 瑨慴t 瑨ty
c潭灬e瑥ty ca湣e氠eac栠潴桥o 潵琠t
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睥⁷楬 ⁢ ⁵獩湧 a牥:


⡩E€摯g⡘⤠


浡浭慬m堩

⡩瘩€慭浡氨X⤠


浩汫⡘)

⡶ㄩ⁤ g⡳潭(彡湩na氩

⡶㈩€ 汫⡳潭敟l湩nal)


周牥 y⁢e瑨敲⁷ays映摥物癩rg⁴桥⁥浰my⁳ 琬⁢畴⁴桩猠潮s⁷
猠瑨攠s楲獴sI⁦潵 携


††††††††††††††††††††††††††††
⡤(€慭浡 ⡘⤠


浩汫⡘⤠†††††††(瘲⤠浩v欨獯浥彡湩na氩


†††††††††††††††††††††††††††††††††††††
筘⽳潭e彡湩na汽


†††††††††
†††††††††††††††
⡡⤠)摯木g⤠


浡m浡氨堩††††† 慭浡氨獯浥彡湩na氩


††††††††††††††††††††††††††
筘⽳潭e彡湩na汽


⡶ㄩ⁤ g⡳潭(彡湩na氩†l††††††€⁤潧 獯浥彡湩浡氩



Fa汳攠††††††††††


S漬楮i瑨t

first resolution step, we worked out that some_animal couldn’t be a mammal (it doesn’t
produce milk, after all). If it wasn’t a mammal, then it couldn’t be a dog was decided in the next
牥獯汵s楯渠獴e瀮⁈潷e癥rⰠ瑨I猠s潮瑲o摩捴d搠潮d映潵爠ox楯i猠⡶ㄩⰠI
漠睥⁨ 搠景d湤⁡⁣o湴牡摩捴楯渮


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灲潯pⰠ瑨Iy⁡re⁡汬⁵湩 ⁲ 獯汵s楯渠獴e灳p




2) Consider the following knowledge base:

i.

boss(colin) = boris

ii.

boss(boris) =

anna

iii.

paycut(anna)

iv.


X (paycut(boss(X))


paycut(X))


Draw a resolution proof tree to show that Colin is getting a paycut. You will need to use the
demodulation inference rule.


Recall that demodulation allows us to rewrite with equalities during resolutio
n proofs. The
demodulation inference rule is

X = Y A[S]


Unify(A, S) =


S畢獴(

Ⱐ䅛奝)


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䅰灬楣A瑩潮猠潦⁤浯摵ma瑩潮⁡牥 牫r搠䑅:


††
⡎吩€payc畴uc潬温††n†††††
楶⤠i灡yc畴⡢潳猨s⤩


payc畴⡘)


†††††††††††††††††
{
堯X潬湽


††††††††††
¬灡yc畴⡢潳猨s潬温n††††††† 椩⁢潳猨c潬温‽⁢潲楳


†††††††††††††††††††††††††††



††††††††††††††††††††††
¬灡yc畴⡢u物猩†r†††† †††† 楶⤠i灡yc畴⡢潳猨u⤩


灡yc畴⡘)


†††††††††††††††††††††††††††††
筘⽢潲楳i


†††††††††††††††††††††††††††††††††
¬灡yc畴u扯獳⡢潲楳⤩†i††††
楩⤠)潳猨扯楳⤠i⁡湮

††††††††††††††††††††††††††
D
E



⡩楩⤠灡yc畴⡡湮⤠†††††† † ⁰ yc畴ua湮n



Fa汳l


周牥⁩猠慮潴桥爠獯汵瑩潮⁷桥牥⁢潴栠摥浯摵污瑩潮⁳瑥灳⁡牥⁤潮e畳 ⁢ 景fe⁴桥⁦楮慬⁲e獯汵s楯渮


㌩3S異灯獥 y潵a牥 瑲y楮g 瑯t睯w欠潵琠睨y ce牴ri渠exa浳m睩汬 扥 摩晦楣畬i a湤n潴桥猠睩汬 扥
easy.
奯甠桡癥⁣潭灩汥搠瑨楳⁴慢汥⁢ 獥搠潮⁰a獴⁥s灥物r湣e:


䕸am

啳U⁃a汣畬l瑯t

䑵Da瑩潮
牳r

Lec瑵牥r

呥牭

䑩晦楣畬iy

1

yes

3

J潮s

獵浭er

easy

2

yes

3

J潮s

獰物湧

摩晦楣畬i

3



3

S浩瑨

獰物湧

摩晦楣畬i

4



2

䅲浳m牯湧

獵浭er

easy

5

yes

2

J潮s

獵浭er

easy


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-
S 浥瑨潤⸠
奯甠獰Yc楦y⁴桡琠瑨攠灯獩瑩癥猠s牥⁴桥 exa浳⁷桩c栠h牥
摩晦ic畬u
.


3a) Which would be the most specific hypothesis that the agent would
initially

construct?


The FIND
-
S method takes the first positive example and uses that with all it’s variables ground as in
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楳i


<ye猬″Ⱐs潮o猬⁳灲楮g>




㍢⤠䡯e 睯畬搠瑨攠a来n琠gene牡汩獥 瑨t猠hy灯瑨p獩s 楮i汩g桴h潦o瑨攠獥c潮o 灯獩瑩癥 exa浰me? t桡t
潴桥o⁰潳獩扬攠来湥ra汩獡瑩潮猠o牥⁴桥 e?



周T 湥x琠灯獩瑩癥 exam灬攠楳i exa洠湵浢敲 ㌬3睨楣栠桡猠瑨敳t a瑴物扵瑥猺s<湯Ⱐ㌬3獭
楴栬h獰物湧s⸠
oe浥浢m爠瑨慴t潵爠hy灯瑨敳楳t楳i <ye猬s㌬3橯湥猬j獰物湧>⸠周楳T摩晦e牳r晲o洠瑨攠a瑴物扵瑥猠景f exam
湵浢敲 ㌠楮i瑨攠晩f獴sa湤 瑨t牤ra瑴物扵瑥献s䡥湣eⰠ睥 浵獴 c桡nge 瑨e獥 瑯t癡物r扬敳b楮i潲摥爠瑯tge琠a
gene牡汩獥搠 hy灯瑨e獩s 睨楣栠 c潶o牳r 扯b
栠 灯獩瑩癥猠 ⡥xa洠 ㈠ a湤n exa洠 ㌩⸠ p漬o 瑨攠 来湥ra汩獥d
hy灯瑨敳楳⁢pc潭敳o


<堬″ⰠuⰠ獰物Ig>


睨w牥 堠u湤⁙na牥⁶a物r扬敳b


瑨ty ca渠瑡ne渠 ny⁶ 汵e.


周T 潴桥o 灯獳p扬攠来湥牡汩獡瑩潮猠a牥 <堬夬wⱳ灲楮朾 <堬㌬uI娾 a湤n<tⱘⱙⱚI 睩瑨wtⰠ堬uv
a湤nw 扥楮g v
a物r扬敳⸠k潴攠瑨慴tcf乄
-
p 摯d猠湯n 灲潤pce 瑨敳t gene牡汩獡瑩潮猬o扥ca畳u 楴 汯潫l 景f
瑨攠
least general

(i.e., most specific) generalisation


remember that the ‘S’ inf FIND
-
p 獴慮摳s景f
‘specific’. These three are not as specific as the first one, becau
獥⁴桥 晩f獴⁳灥c楦楥猠瑨e畭扥 ″ and
瑨攠獥a獯渠獰物sg.




㍣⤠䡯e 睯畬搠y潵o畳u 瑨攠瑷漠hy灯瑨敳e猠⡦牯洠灡牴猠㍡ a湤n㍢⤠瑯t灲e摩捴d睨w瑨敲 a渠exa洠w楬l
扥⁤楦 楣畬琿


㍤⤠t桡琠獣潲o 睯畬搠瑨e 瑷漠hy灯瑨敳p猠来琠楮i瑥t浳m潦o灲e摩捴楶攠acc畲acy 潶o爠
瑨攠瑲a楮i湧n獥琬t
a湤⁷桩n栠潮h⁷潵汤⁢e c桯獥渠n猠瑨攠汥a牮r搠dy灯瑨敳楳?




The first hypothesis states that an exam is difficult if you use a calculator, it’s 3 hours long, Dr.
Smith is the lecturer and it’s in the spring term. It also states that the e
xa洠楳ieasy 楦iany o映瑨敳e
c潮摩o楯湳ia牥 湯琠浥琮m乡瑵牡汬yⰠ瑨t猠co牲ec瑬y 灲p摩捴猠exa洠㈠⡴桥 晩牳琠灯獩瑩癥 exa浰me⤮Ff琠al獯s
c潲oec瑬y 灲e摩捴猠exa浳mㄬN㐠a湤n㔠a猠扥楮i easy. 䡯ee癥爬r楴 楮捯i牥c瑬y 灲p摩捴猠exa洠㌬3扥ca畳u
瑨t猠楳i 灲p摩捴d搠瑯t 扥
easy⸠p漬o楴 獣潲o猠㐯㔠= 〮㠠景f 灲p摩捴楶攠accu牡cy 潶o爠瑨攠瑲t楮ing 獥琠
⡥xa浳‱m瑯‵⤮


周T 獥c潮搠hy灯瑨敳楳p獴慴s猠瑨慴ta渠exa洠楳i摩晦ic畬u 楦i楴 楳 瑨牥e 桯畲猠汯湧 a湤n瑡步渠楮i瑨攠獰物湧s
瑥t洬m睩瑨w瑨攠exa洠扥楮g easy 潴桥牷楳攮i周楳Tc潲oec瑬y p
re摩捴猠扯瑨bexa洠㈠a湤n㌠beca畳e 楴 wa猠
gene牡汩獥d⁴漠摯⁳漮of琠al獯⁣潲oec瑬y⁰牥摩c瑳⁥ta洠ㄬ⁢散a畳e⁴桩猠睡猠sake渠楮⁴桥⁳畭浥m⸠c楮慬iyI
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