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Parsing with CCGs - lambda part

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1

I am following this video tutorial, starting 6th minute

I would like to parse the following sentence

square blue or round yellow pillow. 

For now I am interested in only how square and blue are combined.

In particular with start with the following representation

square -> ADJ: lambda x. square(x)

blue -> ADJ: lambda x. blue(x)

Next step is we raise types:

square -> N/N: lambda f. lambda  x. f(x) / square (x)

blue ->  N/N: lambda f. lambda  x. f(x) / blue (x)

Now we create representation for square blue. I indicate substitution by brackets

lambda x. [ lambda f. lambda  x. f(x) / blue (x) ] (x) / square (x)

Next I simply substitute z for x everywhere outside of square brackets, so that we do not confuse different xs.

lambda z. [ lambda f. lambda  x. f(x) / blue (x) ] (z) / square (z)

Next I push z into square brackets

    lambda z.  lambda  x. z(x) / blue (x)  / square (z)

This is different from what stated in the lecture:

lambda z.  lambda  x. z(x) /   square (x)  /   blue (x) 

Where did I make a mistake?

semantics syntax-trees lambda-calculus ccg

share|improve this question

edited 3 hours ago

lemontree♦

4,85731030

asked 5 hours ago

user1700890user1700890

1678

add a comment | 

1

I am following this video tutorial, starting 6th minute

I would like to parse the following sentence

square blue or round yellow pillow. 

For now I am interested in only how square and blue are combined.

In particular with start with the following representation

square -> ADJ: lambda x. square(x)

blue -> ADJ: lambda x. blue(x)

Next step is we raise types:

square -> N/N: lambda f. lambda  x. f(x) / square (x)

blue ->  N/N: lambda f. lambda  x. f(x) / blue (x)

Now we create representation for square blue. I indicate substitution by brackets

lambda x. [ lambda f. lambda  x. f(x) / blue (x) ] (x) / square (x)

Next I simply substitute z for x everywhere outside of square brackets, so that we do not confuse different xs.

lambda z. [ lambda f. lambda  x. f(x) / blue (x) ] (z) / square (z)

Next I push z into square brackets

    lambda z.  lambda  x. z(x) / blue (x)  / square (z)

This is different from what stated in the lecture:

lambda z.  lambda  x. z(x) /   square (x)  /   blue (x) 

Where did I make a mistake?

semantics syntax-trees lambda-calculus ccg

share|improve this question

edited 3 hours ago

lemontree♦

4,85731030

asked 5 hours ago

user1700890user1700890

1678

add a comment | 

I am following this video tutorial, starting 6th minute

I would like to parse the following sentence

square blue or round yellow pillow. 

For now I am interested in only how square and blue are combined.

In particular with start with the following representation

square -> ADJ: lambda x. square(x)

blue -> ADJ: lambda x. blue(x)

Next step is we raise types:

square -> N/N: lambda f. lambda  x. f(x) / square (x)

blue ->  N/N: lambda f. lambda  x. f(x) / blue (x)

Now we create representation for square blue. I indicate substitution by brackets

lambda x. [ lambda f. lambda  x. f(x) / blue (x) ] (x) / square (x)

Next I simply substitute z for x everywhere outside of square brackets, so that we do not confuse different xs.

lambda z. [ lambda f. lambda  x. f(x) / blue (x) ] (z) / square (z)

Next I push z into square brackets

    lambda z.  lambda  x. z(x) / blue (x)  / square (z)

This is different from what stated in the lecture:

lambda z.  lambda  x. z(x) /   square (x)  /   blue (x) 

Where did I make a mistake?

semantics syntax-trees lambda-calculus ccg

share|improve this question

edited 3 hours ago

lemontree♦

4,85731030

asked 5 hours ago

user1700890user1700890

1678

square blue or round yellow pillow. 

square -> ADJ: lambda x. square(x)

blue -> ADJ: lambda x. blue(x)

square -> N/N: lambda f. lambda  x. f(x) / square (x)

blue ->  N/N: lambda f. lambda  x. f(x) / blue (x)

lambda x. [ lambda f. lambda  x. f(x) / blue (x) ] (x) / square (x)

lambda z. [ lambda f. lambda  x. f(x) / blue (x) ] (z) / square (z)

    lambda z.  lambda  x. z(x) / blue (x)  / square (z)

lambda z.  lambda  x. z(x) /   square (x)  /   blue (x) 

share|improve this question

edited 3 hours ago

lemontree♦

4,85731030

asked 5 hours ago

user1700890user1700890

1678

share|improve this question

edited 3 hours ago

lemontree♦

4,85731030

asked 5 hours ago

user1700890user1700890

1678

edited 3 hours ago

lemontree♦

4,85731030

edited 3 hours ago

lemontree♦

4,85731030

asked 5 hours ago

user1700890user1700890

1678

asked 5 hours ago

user1700890user1700890

1678

2 Answers
2

active

oldest

votes

2

I'll leave the lambdas to you, but you might like to know the syntactic structure. It is a RNR (right node-raising) construction, with "pillow" the raised node. The intonation I think makes that obvious. Thus:

[square blue GAP or round yellow GAP] (pillow), where the noun "pillow" fills the GAP.

"square", "blue", "round", "yellow" are adjectives which modify nouns or modified nouns.

share|improve this answer

answered 2 hours ago

Greg LeeGreg Lee

9,52511023

add a comment | 

1

If I am not mistaken, your computation is correct, and the video simply shows the wrong result.

The order square(x) ^ blue(x) comes from the fact that squared is applied to blue where blue will be substituted for f in the term f(x), which comes before square(x) in the conjunction.

To change the order in which the terms will appear in the conjunction without changing the order in which the terms are applied to each other (of course you could also do the backward composition blue square so square goes in for f in blue, but that wouldn't be in line with the intended forward composition), one could simply change the order of the expressions in the definition of square to

square -> N/N: lambda f. lambda  x. square (x) ^ f(x)

which changes the rule for type raising to

lambda x. g(x) -> lambda f. lambda  x. g(x) ^ f(x)

---

Edit:

I might at a total loss, but it actually looks like the computation in the video doesn't work out at all:

(λf.[λx.[f(x) ^ square(x)]])(λf.[λy.[f(y) ^ blue(y)]]) reduces to
(λx.[(λf.[λy.[f(y) ^ blue(y)]])(x) ^ square(x)]) which reduces to
(λx.[(λy.[x(y) ^ blue(y)]) ^ square(x)]),

not (λf.[λx.[f(x) ^ square (x) ^ blue (x)]]).

I have no idea how they arrive at this result, it must be a mistake.

share|improve this answer

edited 2 hours ago

answered 3 hours ago

lemontree♦lemontree

4,85731030

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for reply. Also I believe type raising is not very accurate. It should go from ADJ to NP/N. Even in simplest case blue pillow we apply blue to noun pillow and end up with noun phrase.
  
  – user1700890
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @user1700890 No, I the type N/N is okay. blue pillow should only be an N because the corredponding NP would be the blue pillow. Otherweise, you wouldn't be able to combine the result with more adjectives (in theory, you can append an unlimited number of adjectives) or determiners like the, both of which require an input of type N. In general, it is the inherent property of modifiers (which adjectives belong to) that they don't change the syntactic status of the element they combine with, and are therefore always of type X/X: A noun goes in, a noun comes out.
  
  – lemontree♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  But see the edit to my post on the lambda computation.
  
  – lemontree♦
  3 hours ago
  

  
  
  
  

add a comment | 

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2

I'll leave the lambdas to you, but you might like to know the syntactic structure. It is a RNR (right node-raising) construction, with "pillow" the raised node. The intonation I think makes that obvious. Thus:

[square blue GAP or round yellow GAP] (pillow), where the noun "pillow" fills the GAP.

"square", "blue", "round", "yellow" are adjectives which modify nouns or modified nouns.

share|improve this answer

answered 2 hours ago

Greg LeeGreg Lee

9,52511023

add a comment | 

2

I'll leave the lambdas to you, but you might like to know the syntactic structure. It is a RNR (right node-raising) construction, with "pillow" the raised node. The intonation I think makes that obvious. Thus:

[square blue GAP or round yellow GAP] (pillow), where the noun "pillow" fills the GAP.

"square", "blue", "round", "yellow" are adjectives which modify nouns or modified nouns.

share|improve this answer

answered 2 hours ago

Greg LeeGreg Lee

9,52511023

add a comment | 

I'll leave the lambdas to you, but you might like to know the syntactic structure. It is a RNR (right node-raising) construction, with "pillow" the raised node. The intonation I think makes that obvious. Thus:

[square blue GAP or round yellow GAP] (pillow), where the noun "pillow" fills the GAP.

"square", "blue", "round", "yellow" are adjectives which modify nouns or modified nouns.

share|improve this answer

answered 2 hours ago

Greg LeeGreg Lee

9,52511023

share|improve this answer

answered 2 hours ago

Greg LeeGreg Lee

9,52511023

answered 2 hours ago

Greg LeeGreg Lee

9,52511023

answered 2 hours ago

Greg LeeGreg Lee

9,52511023

1

If I am not mistaken, your computation is correct, and the video simply shows the wrong result.

The order square(x) ^ blue(x) comes from the fact that squared is applied to blue where blue will be substituted for f in the term f(x), which comes before square(x) in the conjunction.

To change the order in which the terms will appear in the conjunction without changing the order in which the terms are applied to each other (of course you could also do the backward composition blue square so square goes in for f in blue, but that wouldn't be in line with the intended forward composition), one could simply change the order of the expressions in the definition of square to

square -> N/N: lambda f. lambda  x. square (x) ^ f(x)

which changes the rule for type raising to

lambda x. g(x) -> lambda f. lambda  x. g(x) ^ f(x)

---

Edit:

I might at a total loss, but it actually looks like the computation in the video doesn't work out at all:

(λf.[λx.[f(x) ^ square(x)]])(λf.[λy.[f(y) ^ blue(y)]]) reduces to
(λx.[(λf.[λy.[f(y) ^ blue(y)]])(x) ^ square(x)]) which reduces to
(λx.[(λy.[x(y) ^ blue(y)]) ^ square(x)]),

not (λf.[λx.[f(x) ^ square (x) ^ blue (x)]]).

I have no idea how they arrive at this result, it must be a mistake.

share|improve this answer

edited 2 hours ago

answered 3 hours ago

lemontree♦lemontree

4,85731030

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for reply. Also I believe type raising is not very accurate. It should go from ADJ to NP/N. Even in simplest case blue pillow we apply blue to noun pillow and end up with noun phrase.
  
  – user1700890
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @user1700890 No, I the type N/N is okay. blue pillow should only be an N because the corredponding NP would be the blue pillow. Otherweise, you wouldn't be able to combine the result with more adjectives (in theory, you can append an unlimited number of adjectives) or determiners like the, both of which require an input of type N. In general, it is the inherent property of modifiers (which adjectives belong to) that they don't change the syntactic status of the element they combine with, and are therefore always of type X/X: A noun goes in, a noun comes out.
  
  – lemontree♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  But see the edit to my post on the lambda computation.
  
  – lemontree♦
  3 hours ago
  

  
  
  
  

add a comment | 

1

If I am not mistaken, your computation is correct, and the video simply shows the wrong result.

The order square(x) ^ blue(x) comes from the fact that squared is applied to blue where blue will be substituted for f in the term f(x), which comes before square(x) in the conjunction.

To change the order in which the terms will appear in the conjunction without changing the order in which the terms are applied to each other (of course you could also do the backward composition blue square so square goes in for f in blue, but that wouldn't be in line with the intended forward composition), one could simply change the order of the expressions in the definition of square to

square -> N/N: lambda f. lambda  x. square (x) ^ f(x)

which changes the rule for type raising to

lambda x. g(x) -> lambda f. lambda  x. g(x) ^ f(x)

---

Edit:

I might at a total loss, but it actually looks like the computation in the video doesn't work out at all:

(λf.[λx.[f(x) ^ square(x)]])(λf.[λy.[f(y) ^ blue(y)]]) reduces to
(λx.[(λf.[λy.[f(y) ^ blue(y)]])(x) ^ square(x)]) which reduces to
(λx.[(λy.[x(y) ^ blue(y)]) ^ square(x)]),

not (λf.[λx.[f(x) ^ square (x) ^ blue (x)]]).

I have no idea how they arrive at this result, it must be a mistake.

share|improve this answer

edited 2 hours ago

answered 3 hours ago

lemontree♦lemontree

4,85731030

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for reply. Also I believe type raising is not very accurate. It should go from ADJ to NP/N. Even in simplest case blue pillow we apply blue to noun pillow and end up with noun phrase.
  
  – user1700890
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @user1700890 No, I the type N/N is okay. blue pillow should only be an N because the corredponding NP would be the blue pillow. Otherweise, you wouldn't be able to combine the result with more adjectives (in theory, you can append an unlimited number of adjectives) or determiners like the, both of which require an input of type N. In general, it is the inherent property of modifiers (which adjectives belong to) that they don't change the syntactic status of the element they combine with, and are therefore always of type X/X: A noun goes in, a noun comes out.
  
  – lemontree♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  But see the edit to my post on the lambda computation.
  
  – lemontree♦
  3 hours ago
  

  
  
  
  

add a comment | 

If I am not mistaken, your computation is correct, and the video simply shows the wrong result.

The order square(x) ^ blue(x) comes from the fact that squared is applied to blue where blue will be substituted for f in the term f(x), which comes before square(x) in the conjunction.

To change the order in which the terms will appear in the conjunction without changing the order in which the terms are applied to each other (of course you could also do the backward composition blue square so square goes in for f in blue, but that wouldn't be in line with the intended forward composition), one could simply change the order of the expressions in the definition of square to

square -> N/N: lambda f. lambda  x. square (x) ^ f(x)

which changes the rule for type raising to

lambda x. g(x) -> lambda f. lambda  x. g(x) ^ f(x)

---

Edit:

I might at a total loss, but it actually looks like the computation in the video doesn't work out at all:

(λf.[λx.[f(x) ^ square(x)]])(λf.[λy.[f(y) ^ blue(y)]]) reduces to
(λx.[(λf.[λy.[f(y) ^ blue(y)]])(x) ^ square(x)]) which reduces to
(λx.[(λy.[x(y) ^ blue(y)]) ^ square(x)]),

not (λf.[λx.[f(x) ^ square (x) ^ blue (x)]]).

I have no idea how they arrive at this result, it must be a mistake.

share|improve this answer

edited 2 hours ago

answered 3 hours ago

lemontree♦lemontree

4,85731030

square -> N/N: lambda f. lambda  x. square (x) ^ f(x)

lambda x. g(x) -> lambda f. lambda  x. g(x) ^ f(x)

share|improve this answer

edited 2 hours ago

answered 3 hours ago

lemontree♦lemontree

4,85731030

answered 3 hours ago

lemontree♦lemontree

4,85731030

answered 3 hours ago

lemontree♦lemontree

4,85731030