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Computing Theory COSC 1107/1105
Creation date：2024-04-18 15:03:09
Belonging category：计算机computer science
Computing Theory

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Computing Theory

COSC 1107/1105
Final Exercise (Practice version Answers)
1 Assessment details
1. Consider the grammar derivations below.
S ⇒ aSb⇒ aaSbb⇒ aacSdbb⇒ aacdbb
S ⇒ A⇒ xAy ⇒ xxAyy ⇒ xxB@yy ⇒ xxxB@yy ⇒ xxxx@yy
S ⇒ A⇒ @C ⇒ @Cy ⇒ @Cyy ⇒ @yyy
(a) From the above derivations, construct rules that must exist in any context-free grammar G
for which these derivations are correct.
Answer: From the first derivation we can see that the rules must include S → aSb,
S → cSd and S → λ. From the second derivation we can add the rules S → A, A → xAy,
A → B@, B → xB and B → x. From the third derivation we can add the rules A → @C,
C → Cy and C → y.
As this covers all the derivation steps in all three derivations above, we get the rules below.
S → aSb | cSd | A | λ
A→ xAy | B@ | @C
B → xB | x
C → Cy | y
(b) Assuming that these are all the rules in G, give L(G) in set notation.
Answer: L(G) = {wxi@yj(bd(w))R | i 6= j, i, j ≥ 0, w ∈ {a, c}∗} ∪ {λ} where bd(w) is w
with all a’s replaced by b’s and all c’c replaced by d’s, and wR is the reverse of w.
This may seem a little complicated but consider a string like aabax@yycdcc. If we replaced
all d’s in the grammar with c’s and b’s with a’s, then the language would be {wxi@yjwR|i 6=
j, i, j ≥ 0, w1 ∈ {a, c}∗}. So all we need to do to get the language of the original grammar is
replace the a’s and c’s in wR with b’s and d’s respectively.
Another point to note is that this language is not the same as the one below.
{w1xi@yjw2 | i 6= j, i, j ≥ 0, w1 ∈ {a, c}∗, w2 ∈ {b, d}∗, |w1| = |w2|}
Note that this latter language contains strings such as aabxx@yddd, which cannot be derived
by the grammar.
where na(w) is the number of a’s in w, and similarly for b, c and d.
(c) Is there a regular grammar for L(G)? Explain your answer.
Answer: There is no regular grammar for L(G). This language is context-free but not
regular because there is a need to count the number of x’s and y’s to make sure that they
are different, as well as counting the number of a’s or c’s and making sure there is an equal
number of b’s or d’s.
(d) Construct a context-free grammar for the language below.
L = {xi w1@w2 yj | i 6= 2j, i, j ≥ 0, |w1| = |w2|, w1 ∈ {a, c}∗, w2 ∈ {b, d}∗}
Answer: It is best to split this up into two cases and then combine the two grammars. So
let L = L1 ∪ L2 where
L1 = {xi w1@w2 yj | i < 2j, i, j ≥ 0, |w1| = |w2|, w1 ∈ {a, c}∗, w2 ∈ {b, d}∗}
and L2 = {xi w1@w2 yj | i > 2j, i, j ≥ 0, |w1| = |w2|, w1 ∈ {a, c}∗, w2 ∈ {b, d}∗}
For a constraint like i < 2j it can be helpful to use a table like the one below.
i j 2j
0 1 2
1 1 2
2 2 4
3 2 4
4 3 6
5 3 6
So for L1 we get G1 below.
S → xxSy | XA
A→ Ay | By
X → x | λ
B → aBb | aBd | cBb | cBd | @
For L2 with the constraint i > 2j the corresponding table is below.
i j 2j
1 0 0
3 1 2
5 2 4
7 3 6
This leads us to the grammar G2 below.
S → xxSy | C
C → Cy | Dy
D → aDb| | aDd | cDb | cDd | @
We can then combine these together for a grammar G for L = L1 ∪ L2.
S → T | U
T → xxTy | XA
A→ Ay | By
X → x | λ
B → aBb| | aBd | cBb | cBd | @ U → xxUy | C
C → Cy | Dy
D → aDb | aDd | cDb | cDd | @
There are ways this could be simplified, but that is not required. Constructing the grammar
this way gives confidence in its correctness, which is less obvious otherwise.
2
2. Drogo the Dreary, a distant relative of Thorin Oakenshield, has written the following discussion
of intractability. There are 5 incorrect statements in the paragraph below. Identify all 5 incorrect
statements and justify each of your answers.
“There are a number of problems which can be solved in principle, but in practice can be very
difficult to solve. These problems are often referred to as NP-complete problems, and include
the Travelling Salesperson problem, 3-SAT, factorisation and vertex cover. These problems are
certainly intractable, i.e. all algorithms for these problems have exponential running times. This
means that they can be solved for small instances, but the rate of growth of their complexity is so
fast that they cannot be solved in practice for any reasonable size. For example, the best known
algorithm for the Travelling Salesperson problem can take up to 2n10 +7n2 operations for a graph
of size n. This means it is in the class O(n10) and is thus intractable. Fortunately it is possible to
use approximation and heuristic algorithms to find some kind of solutions to these problems, either
by removing the guarantee that an optimal solution will be found, or by removing the constraint
that the running time will be polynomial or less (or removing both). There are also some similar
problems, such as the Hamiltonian circuit problem, which are known to be simpler to solve than
the Travelling Salesperson problem and are tractable. The name NP-complete problems comes
from the property that such problems have run in at most polynomial-time on a nondeterministic
Turing machine.”
Answer:
(a) Factorisation is probably intractable, but it is not known to be NP-complete. So it is incorrect
to list it as an NP-complete problem.
(b) NP-complete problems are almost certainly intractable, but it is incorrect to say that these
are certainly intractable.
(c) The best known algorithm for the Travelling Salesperson problem is exponential, and so the
statement of the running time here is certainly incorrect.
(d) Algorithms with a running time of O(n10) are in the polynomial class, and hence are con-
sidered tractable.
(e) The Hamiltonian Circuit problem is NP-complete, as is the Travelling Salesperson problem.
So these are either both intractable or both tractable, i.e. in the same complexity class.
3. The generalised Platypus game with Gandalf the White is played as follows (we will abbreviate
the name of this to GPGGW). There are three machines, with two being the usual platypus
machines (as in the generalised Platypus game from Assignment 2), with the third machine being
Gandalf the White (which we will abbreviate to GW ), which has the transition table as below.
For simplicity we assume all three machines have the same alphabet Σ. The tape is infinite in
both directions, and is initially blank.
q0 blank blank R q1
q0 X X R q1
q1 blank blank L q0
q1 X X L q0
where X denotes any non-blank symbol in Σ.
(a) Show that the halting problem for the GPGGW is undecidable. You may use any reduction
you like. Note that you may assume that the generalised Platypus problem from Assignment
2 is undecidable if you would find that helpful.
3
Answer: The simplest proof of this will be to reduce the generalised Platypus problem
from Assignment 2 (which you can assume is undecidable) to this problem. The key obser-
vation is that the Gandalf the White machine never changes any cell on the tape, and never
terminates. This means that the generalised Platypus game with Gandalf the White halts
iff the generalised Platypus game (from Assignment 2) halts. In terms of machines, consider
the diagram below. Note that the machine the GPGGW only needs the machines M1 and
M2 as input.
It is also possible to use a reduction from the blank tape problem to the GPGGW problem
as follows. Let M be the machine we want to analyse for the blank tape problem. Then
M will halt on the blank tape iff the generalised Platypus problem with Gandalf the White
halts for M and GW .
In terms of machines, consider the diagram below.
(b) Suppose the GPGGW is played on a Turing machine with a finite tape (making the halting
problem decidable), and also that there is a decidable problem A for which there is a reduction
from A to the GPGGW. This information could be used as an argument that the GPGGW
is NP-complete, provided that some further information is known. What further information
is needed? Explain your answer.
4
Answer: To show that a problem is NP-complete, we need to show that the problem is
in NP, and that the problem is NP-hard, i.e. that there is a polynomial-time reduction to it
from every other problem in NP. The simplest way to show the latter property is to find a
polynomial-time reduction from a known NP-complete problem to it.
So given the information above, we also need to know the following.
i. That GPGGW is in NP.
ii. That the problem A is NP-complete.
iii. That the reduction from A to the GPGGW is polynomial-time.
(c) Freddo the optimistic Frog likes playing Platypus tournaments. He particularly likes the 3-
player version, for which a tournament of n machines will require n(n+1)(n+2)/6 matches.
He ran a tournament for 100 machines which took 42.42 seconds on the family desktop
computer. Encouraged with his success, he attempts to run a tournament with 10,000
machines, but when it was discovered the computer took well over a day without coming
close to finishing, he was given a strict limit of 8 hours for all such tournament play (so that
tournaments could be run at night when all the other frogs were asleep). What is the largest
tournament size that Freddo can play within this limit? Show your working. We will call
this number n1.
Answer: A tournament of 100 machines will require 100×101×102/6 = 171, 100 matches.
Doing this in 42.42 seconds means that this takes 0.000247059 seconds per match. An 8-hour
limit gives Freddo 8 × 60 × 60 = 28, 800 seconds, and hence 8 × 60 × 60/(0.000247059) =
116, 571, 345 matches. When n = 886, the numebr of matches is 116, 310, 536, and for
n = 887 it is 116,704,364. So n1 = 886, i.e. Freddo can play a tournament of up to 886
machines.
(d) Having despaired of realising his dream of a complete 3-player tournament, Freddo hears
of a similar tournament game, known as Krazy Koalas. His friend Choco tells him that he
can also run a 100-machine tournament in 42.42 seconds, but the Koala tournament “only”
requires n6/(1000000) matches. Given Freddo’s time limit of 8 hours, what is the largest
Koala tournament he can run? Show your working. We will call this number n2.
Answer: From the previous answer we know that Freddo can run up to 116, 571, 345
matches. When n = 221, we have n6/(1000000) = (221)6/(1000000) = 116, 507, 435, and
when n = 222, it is 119, 706, 531. So n2 = 221, i.e. Freddo can play a tournament of up to
221 machines.
(e) Freddo, being an optimist, decides he wants to investigate the two types of tournament a
little further. Given he knows it takes just under 8 hours to run a Platypus tournament with
n1 machines and a Koala tournament of n2 machines, how long will it take to run a Platypus
tournament with n2 machines? And how long will it take to run a Koala tournament with
n1 machines? Show your working in each case.
Answer: A Platypus tournament with n2 = 221 machines will take 1,823,471 matches,
which will take 450.5 seconds, i.e. 7.5 minutes. A Koala tournament with n1 = 886 machines
will take 483,729,230,338 matches which will take 119, 509, 574.55 seconds, i.e. 3.78 years.
(f) Freddo’s activities attract the attention of a spambot (secretly installed on the family desk-
top), and gets an unsolicited offer from Hammy Spam Solutions to provide a host server
for running his tournaments, at a cost of $(1.01)n for a Platypus tournament of n machines,
where n could be as high as 10,000. After a small amount of thought, Freddo deletes the offer
5
and tells all his family and friends to avoid Hammy Spam Solutions at all times. Explain
why Freddo did this, with particular reference to the cost for a tournaments involving 100,
1,000 and 10,000 machines.
Answer: The cost is exponential, and sooner or later will become far too expensive. Freddo
knows he can run a Platypus tournament of 886 machines at no cost (apart from 8 hours on
the familiy desktop). Consider the table below.
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COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 MAS316/414 ACCT2522 MSIN0181 ATMS 201 COMP3425 MATH7861 SOLA2060 MA 325 7CCEMCTH ECON 432 INFT2051 SOLA 2060 ECOS2002 Stats141XP BIOA02H3 S2022 MA4601 K5 COMP5329 EG 284 IB3A70 WECO1020 COMP3027 Energy ACS61011 MAST20026 STAT 533 MATA31 ENVS257 BEM2031 EC 486 COMP0172 CS3214 EECE5644 IB2B20 COSC2536 COSC1147 COMP6214 5CCM242A COMP0130 AMME2000 MTMG50: EG238 ACF6006 QBUS2810 COSC2391 ECON7560 MATH377 STA305 CS 486 COMP1117B EL648 ELEC3909 ECON 241 COMP 4901W BUSI4428 ENEN20002 ENGN4524 ELEC3115 CSC 413 PSYC 100B ECON1002 MATH256 ANSC20005 MTHS2008 MATH40082 COSC 2406 FIT3174 STA 238 CMT202 Differential Equations MTHM017 GARCH 101 MATH2069 CSC336S ACSC12 ETF2121 CEME 2001 DB B474F ARHT1001 EECS 3311 3220 STATS 786 CS 484 COMP20008 MATH4091 ECE 455 CSCI 204 BUS B200F BA 875 BU5562 COSC 1147 X11 GEOG 101 IMM250F CSE 584A MATH38172 LAB 4 ELEC5882M FINS5523 EE 430 COMP 465 6CCS3PRE 7CCSMNSE FINANCE ACMB02H3 ENGG2112 ELEC5616 SCM B371 CHEM3120 CS22 ECMT2130 MAST30011 ADSPBF706 CENG0013 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COMSM0093 COMP281 AFM 333 IBUS1001 CP2501 Math 105A VE320 TEC101 SOCY2166 COM240 EFIM20033 ENGI4467 EFIM20034 EC 979 BE432 7EJ502 CSSE6400 COMP90025 Econ 312 ENG2031 ECO0006 EXERSCI 207 CAN209 ECO 137W FT312 ECO 2199 MATH32052 BDW4213 APS 360 SAS 6 CHME0017 PO92Q Economics 01AV MATH39512 MATH 5486 COMP8620 ISE102 ECON235 ECON 206 FI 345 PHAY0020 MATH0099 MATH0085 EFB106 PHY 134 MAE115 circuits Chemistry CSE12 MATH-UA 233 PHAS0038 COMP2140 CONMGNT 7049 COMP151101 ECON 485 ACX5903 ENGN6536 ARCH1162 Law PHAS0042 CCF1C03 MATH 351 Phys 139 AG942 MGT001371 APH106 ECON1051 MA4AM MA3Z7 NUMBER THEORY REST0006 SESS0026 STAT0023 TABL 2741 CMP4008Y EFIM10015 Acct 560 PSYC735 GEOG 5 ACCT3104 CSEC5616 PHIL1012 ITLS6111 ECON8022 5CCM226A compX123 MA4XJ R001 EWB LLP120 COMP643 ENG 103 ECON 23950 ACCT 207 FINC 616 Bio99 Math 137A Psychology BCPM0054 MATH20212 EC3450 CENV6141 PSY 205 CE335 CS365 Calculus ACSD INFT6800 MCIT INFO6030 PHYS 111LV H63EEJ ECON2151 ENGI 2181 MTRX1701 PLIT08009 CS240 Math 270 CMP7000A CMP-7038B Math 263 LAW3016 CS371 STAT 334 FIN0008 ACC3004 STAT 371 ​COMP1921 MTH201 ACCT2000 UN3412 CIV6000 MATHS 3021 MATE7014 ECON1 CIBC6035 LAW121G COMM 321 CSC8204R TTM2033 CSCI 2600 MAE 115 LNG302 CENV6175W1 ECON30019 CPS2015 BST969 DDES9015 AMATH242 CS 1151 N1592 ACCT600 EG503G SMM519 EC318 Ecstatistics EEE 471 MARK2051 Topology COM00037H ACCT3000 MTHM501 EG501V COM00055H BUSN9085 GGR273 MATH 2B ITD102 MA214 ECON0060 ECE3114 ECMM447 Mechanics CCT361H5S BSAN3210 WELF2001 Fin3001 RESP5001 STA 137 Linguistics 101 CSCI368 CMPSC497 LUBS3655 OMBA 5355 BUSM2562 ​BFF5902 CYBR7001 CULT2005 ECON 7720 MTRX3760 FSE598 ​LUE1001 MGSC 7100 CS3240 COSC7500 ACCT3091 INFOGR TRAN 2080 DESIGN CIVE215001 CIVE2360 CPSC 8430 FNCE2003 Stoichiometry INFS6023 Quantitative Method Math 21D ECE5179 Design ​ECON8022 Chemical DMS 213 ADM1370 BFF5902 BFF2701 AGRI10051 AMN400 ECON7322 ​FINC5090 ​MATH1052 ​SciComp Elec4622 ​FINS5516 ENG 101 MMM143 CS431CA2 BSAN2201 EBSC7070 ES9v9-10 PLIN0009 DATA 8 MATH 101 BCPM0097 ​MAT 237 BISM2201 EEE8099 Physics 11 ECE2238B Biostat 234 ​ECON30009 INF10025 CHEM252 CCMA4008 CIVE 3007 STAT0031 ELEC ENG 3108 MGMT90140 CHEM 252 STAT3016 MMM095 ​MMGT6016 Legal STAT 311 ELEC9764 COMP 370 TCS 3053 EECS 1021 FIT1051 C/C++ EE3C COMP2003J COM398 COMP51915 Simulator FIN42330 8PRO102 PGD ELEC5160 XJCO1921 COMP5123M INFS 2042 CHE1148H ECA5313 IN3329 Microsoft Modeling B15 2TT COMP3687 COMP2432 DMS2030 KAN4TSF CFKG CSE 5322 CSE 445 STATS 3DA3 CHC5223 COMP1048 EL3105 EECS 3221 CS231 ME 4434 ITP4510 ITP4501 CSCI 5708 Controller EECE 5699 CPEN 231L CPN programs CSC8016 MATU9D2 ENGN4213 2. SVM. (10 points) Consider an SVM classifier using RBF kernel, and finish the following python code to search for the optimum kernel parameter γ. You are required to calculate the RBF function and search yourself, instead of using functions in Scikit-Learn (sklearn). def parameter search(gamma list ,Xtr,ytr,Xts,yts ,...): """ Parameters gamma list: A list of RBF kernel parameter gamma to search for Xtr: Training data, shape (ntr,nrow,ncol) ytr: Training labels, shape (ntr,) Xts: Test data, shape (nts,nrow,ncol) yts: Test labels, shape (nts,) ... Add other parameters as needed Returns: gamma optimum: The gamma with minimum error on Xts, yts """ ... return gamma optimum
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