No. Semester-Examination of CS674 Name: 1. The sentence John drank a glass of milk can be paraphrased as John ingested the milk by getting the milk to his mouth from glass by transferring the glass containing milk from the table to his mouth by moving his hand to the glass and by grasping the glass of milk to his hand. Using CD primitive acts INGEST, PTRANS, MOVE and GRASP to write the CD representation of this sentence. (15%) [Answer] The CD representation of John drank a glass of milk. is as follows: Actor: John Action: INGEST Object: milk Direction TO: mouth of John FROM: glass Instrument: Actor: John Action: PTRANS Object: glass containing milk Direction TO: mouth of John FROM: table Instrument: Action: John Action: MOVE Object: hand of John Direction TO: glass FROM: unknown Instrument: Actor: John Action: GRASP Object: glass of milk Direction TO: hand of John Here we see a case of infinite expansion of instruments.
2. Design a context-sensitive grammar to generate the symbol string a k b k c k, write the derivation process of aaabbbccc step by step. (15%) [Answer] The context-sensitive grammar to generate a k b k c k is as follows: G = (V, Vt, S, P) V = {S, B, C, D 1, D 2, a, b, c} Vt = {a, b, c} S: S P = {S asbc, 1 S abc, 2 CB D 1 B, D 1 B D 1 D 2, D 1 D 2 BD 2, BD 2 BC, 3a 3b 3c bb bb, 4 bc bc, 5 cc cc} 6 The rules CB D 1 B D 1 B D 1 D 2 D 1 D 2 D 1 B D 1 B CB 3d are all context-sensitive rules (e.g. CB D 1 B means that in the context ~B, replace CB to D 1 B), they jointly have the same effect as the rule CB BC So this rule CB BC is also a context-sensitive rule. Rule 4, 5, 6 are all context-sensitive rules. The derivation process of aaabbbccc : Intermediate chains S rules asbc 1 aasbcbc 1 aaabcbcbc 2 aaabbccbc 3 aaabbcbcc 3 aaabbbccc 3 aaabbbccc 4 aaabbbccc 4 aaabbbccc 5 aaabbbccc 6 aaabbbccc 6
3. Using the category grammar to analyze the sentence Paul thought that John slept soundly, and explain the difference of meaning in the results. (15%) [Answer] We list the categories of every word in the bottom of corresponding word: Paul thought that John slept soundly n n n n n\s (n\s)\n\s n\s n\n (n\s/n)\n\s/n n\s/n n/s n\s/s We assume the categories of word thought are only 4 as above, the categories of word soundly are only 2 as above, then we can obtain 24 initiating category sequences: (1 4 3 1 1 2=24). Now we calculate these category sequences: In category sequence n n n n n\s (n\s)\n\s, n n n n n\s (n\s)\n\s S We use rule (α)(α\β) β to category 4 and 5, then we obtain n n n S (n\s)\n\s. For this sequence, any rule cannot be further used. We may also use rule (α)(α\β) β to category 5 and 6, then we obtain n n n n n\s. For this sequence, any rule cannot be further used. In the last, we can obtain two results: First result: n n\s/n n/s n n\s (n\s)\n\s n\s S n n\s S Second result: n n\s/s n/n n n\s (n\s)\n\s n n\s S n\s
S Here we obtain two results. Every result represents different meaning: First result means: Paul thought, John slept soundly. Here, that is the relative pronoun which introduces a clause John slept soundly. Second result means: Paul thought, that John sleepy soundly. Here, that is demonstrative pronoun which modifies John. The results show that this sentence is ambiguous. 4. What are three properties of garden path sentence? Compare the MV/RR probabilistic ratio in the garden path sentence The horse raced past the barn fell with the ratio in the sentence The horse found in the barn died and show that the dispreferred parse is the correct parse for the garden path sentence. (15%) [Answer] The garden-path sentence is a specific class of temporarily ambiguous sentences. For example, The horse raced past the barn fell is a garden path sentence with temporary ambiguity in the understanding process. (a) S (b) S NP VP NP VP Det N V PP NP VP V The horse raced P NP? Det N V PP fell past Det N V the horse raced P NP the barn fell past Det N the barn The garden-path sentences are the sentences that are cleverly constructed to have three properties that combine to make them very difficult for people to parse: They are temporarily ambiguous: the sentence is ambiguous, but its initial portion is ambiguous. One of the two or three parses in the initial portion is somehow preferable to the human parsing mechanism. But the dispreferred parse is the correct one for the sentence. The result of these three properties is that people are led down the garden path toward the incorrect parse, and then are confused when they realize it is the wrong way.
In the sentence The horse raced past the barn fell, verb raced is preferable to be used as Main Verb (MV), and it dispreferable to be used as Reduced-Relative (RR). But correct one just the dispreferable (RR). The probabilistic ration MV/RR = 387 In the sentence The horse found in the barn died, since the verb found is transitive, the reduced-relative (RR) interpretation becomes much more than it was for raced. Its MV/RR probabilistic ratio (lower than 5) is less than MV/RR probabilistic ratio of raced (387). So this sentence cannot become the garden-path sentence. The MV/RR probability ratio for raced much more than the MV/RR probabilistic ratio for found. It is the reasonable explanation for garden-path sentence. Log(MV/RR) MV/RR=387 raced WV/RR=5 (threshold) found The horse X-ed PP Fig. MV/RR probabilistic ratio The model assumes that people are unable to maintain very many interpretation at one time. Whether because of memory limitations, or just because they have a strong desire to come up with a simple interpretation, they prune away low-ranking interpretation. An interpretation is pruned if its probability is 5 times lower than the most-probable interpretation. The result is that they sometimes prune away the correct interpretation. Leaving a highest but incorrect interpretation. This is what happens with the probability (but correct ) reduced-relative (RR) interpretation in the sentence The horse raced past the barn fell. In above Figure, the WV/RR probabilistic ratio for raced falls above the threshold and the RR interpretation is pruned. For found its interpretation is active in the disambiguating region.
5. If the CFG grammar rules are as follows: S NP VP NP PrN NP DET N NP DET N WH VP VP V NP DET the a N table leg PrN Jack V lacks hits WH that Use CYK approach to analyze sentence the table that lacks a leg hits Jack. To transform the rewriting rules to Chomsky Normal Form (CNF); To calculate the b ij of every non-terminal symbols: To draw a pyramid table to show the result of CYK parsing. (20%) [Answer] Transformation of rewriting rules to Chomsky Normal Form: S NP VP NP PrN It is not CNF and must be transformed to NP Jack NP DET N NP DET N WH VP It must be transformed to: NP NP CL NP DET N CL WH VP Here CL is WH clause, it = (that + VP) VP V NP DET the a N table leg PrN Jack V lacks hits WH that Calculation of the b ij of non-terminal symbols: --To arrange POS non-terminal symbols and calculate their b ij The table that lacks a leg hits Jack DET N WH V DET N V PrN (NP) b 11 b 21 b 31 b 41 b 51 b 61 b 71 b 81 --To calculate the b ij of phrase non-terminal symbols b ij (NP1): i=1. j=1+1=2 b ij (NP2): i=5, j=1+1=2
b ij (VP1): i=7, j=1+1=2 b ij (VP2): i=4, j=1+2=3 b ij (CL): i=3, j=1+3=4 b ij (NP3): i=1, j=2+4=6 b ij (S): i=1, j=2+6=8 The length of this sentence is 8, and we get box line number of S is also 8, so the sentence was recognized. To draw a pyramid table to show the structure of this sentence: S 18 (S NP VP) NP3 (NP NP CL) b 16 CL (CL WH VP) b 34 VP2 (VP V NP) b 43 NP1 (NP DET N) NP2 (NP DET N) VP1 (VP V NP) b 12 b 52 b 72 DET N WH V DET N V NP b 11 b 21 b 31 b 41 b 51 b 61 b 71 b 81 The table that lacks a leg hits Jack By the CYK approach, we can create the pyramid in above figure. This pyramid is also a tree graph.
6. If we have following small CFG for English: S NP VP S AUX NP VP S VP NP Det Nom NP Ord Nom NP PrN Nom N Nom N Nom VP V VP V NP Det that this a Ord first second N section class V have include prefer Aux does PrN ASIANA KA-852 CAAC Using Earley algorithm to parse sentence Does KA-852 have a first class section? Describe the state sequence in chart step by step; Describe the parsing process step by step; Draw the chart to represent the parsing result. (20%) [Answer] The states are as follows: Does KA-852 have first class section 0 1 2 3 4 5 6 The state sequence in chart: Chart [0] γ.s [0,0] Dummy start state S.NP VP [0,0] Predictor NP.Ord Nom [0,0] Predictor NP.PrN [0,0] Predictor S.Aux NP VP [0,0] Predictor S.VP [0,0] Predictor VP.V [0,0] Predictor VP.V NP [0,0] Predictor Chart [1] Aux does. [0,1] Scanner S Aux. NP VP [0,1] Completer
NP.Ord Nom [1,1] Predictor NP.PrN [1,1] Predictor Chart [2] PrN KA-852. [1,2] Scanner NP PrN. [1,2] Completer S Aux NP. VP [0,2] Completer VP.V [2,2] Predictor VP.V NP [2,2] Predictor Chart [3] V have. [2,3] Scanner VP V. [2,3] Completer VP V. NP [2,3] Completer NP..Ord Nom [3,3] Predictor Chart [4] Ord first. [3,4] Scanner NP Ord. Nom [3,4] Completer Nom.N Nom [4,4] Predictor Nom.N. [4,4] Predictor Chart [5] N class. [4,5] Scanner Nom N. [4,5] Completer NP Ord Nom. [3,5] Completer VP V NP. [2,5] Completer S Aux NP VP. [0,5] Completer (S span is 5, 5 < 6) Nom N. Nom [4,5] Completer Nom.N [5,5] Predictor Chart [6] N section. [5,6] Scanner Nom N. [5,6] Completer Nom N Nom. [4,6] Completer NP Ord Nom. [3,6] Completer VP V NP. [2,6] Completer S Aux NP VP. [0,6] Completer [Success!] The parsing process: Aux does. [0,1] Scanner S Aux. NP VP [0,1] Completer
PrN KA-852. [1,2] Scanner NP PrN. [1,2] Completer S Aux NP. VP [0,2] Completer V have. [2,3] Scanner VP V. [2,3] Completer VP V. NP [2,3] Completer Ord first. [3,4] Scanner NP Ord. Nom [3,4] Completer N class. [4,5] Scanner N section. [5,6] Scanner Nom N. [5,6] Completer Nom N Nom. [4,6] Completer NP Ord Nom. [3,6] Completer VP V NP. [2,6] Completer S Aux NP VP. [0,6] Completer [Success!] The chart representing the parsing result: S Aux NP VP. VP V NP. NP Ord Nom. NP PrN. Nom N Nom. Nom N. Aux does. PrN KA-852. V have. Ord first. N class. N section. 0 1 2 3 4 5 6