Trouble with my scheme function that tries to create production values with a given grammar
So I am trying to create a function where given the following input '((a (xz) (c)) (b (wy) (d)))
I am supposed to do something like this'((a (xz)) (a (c)) (b (wy)) (b (d)))
I tried to write this
(define productionValues
(lambda (input)
(let ((lhs (map (lambda (x) (car x)) input)))
(let ((rhs (map (lambda (y) (cdr y)) input)))
(map (lambda (l) (cons l (map (lambda (r) (car r)) rhs))) lhs)
)
)
))
This does not work and gets me ((a (xz) (c)) (b (xz) (c)))
My logic behind this is I have a variable lhs that stores (a b) and rhs that stores (((xz) (c)) ((wy) (d))) and I would try to use another set of map functions to grab from the between the two variables, but I feel like I'm on somewhat the right track but just don't understand how i could get my desired output.
scheme racket
asked Nov 21 at 4:40
onbu
358
add a comment |
So I am trying to create a function where given the following input '((a (xz) (c)) (b (wy) (d)))
I am supposed to do something like this'((a (xz)) (a (c)) (b (wy)) (b (d)))
I tried to write this
(define productionValues
(lambda (input)
(let ((lhs (map (lambda (x) (car x)) input)))
(let ((rhs (map (lambda (y) (cdr y)) input)))
(map (lambda (l) (cons l (map (lambda (r) (car r)) rhs))) lhs)
)
)
))
This does not work and gets me ((a (xz) (c)) (b (xz) (c)))
My logic behind this is I have a variable lhs that stores (a b) and rhs that stores (((xz) (c)) ((wy) (d))) and I would try to use another set of map functions to grab from the between the two variables, but I feel like I'm on somewhat the right track but just don't understand how i could get my desired output.
scheme racket
asked Nov 21 at 4:40
onbu
358
add a comment |
So I am trying to create a function where given the following input '((a (xz) (c)) (b (wy) (d)))
I am supposed to do something like this'((a (xz)) (a (c)) (b (wy)) (b (d)))
I tried to write this
(define productionValues
(lambda (input)
(let ((lhs (map (lambda (x) (car x)) input)))
(let ((rhs (map (lambda (y) (cdr y)) input)))
(map (lambda (l) (cons l (map (lambda (r) (car r)) rhs))) lhs)
)
)
))
This does not work and gets me ((a (xz) (c)) (b (xz) (c)))
My logic behind this is I have a variable lhs that stores (a b) and rhs that stores (((xz) (c)) ((wy) (d))) and I would try to use another set of map functions to grab from the between the two variables, but I feel like I'm on somewhat the right track but just don't understand how i could get my desired output.
scheme racket
asked Nov 21 at 4:40
onbu
358
So I am trying to create a function where given the following input '((a (xz) (c)) (b (wy) (d)))
I am supposed to do something like this'((a (xz)) (a (c)) (b (wy)) (b (d)))
I tried to write this
(define productionValues
(lambda (input)
(let ((lhs (map (lambda (x) (car x)) input)))
(let ((rhs (map (lambda (y) (cdr y)) input)))
(map (lambda (l) (cons l (map (lambda (r) (car r)) rhs))) lhs)
)
)
))
This does not work and gets me ((a (xz) (c)) (b (xz) (c)))
My logic behind this is I have a variable lhs that stores (a b) and rhs that stores (((xz) (c)) ((wy) (d))) and I would try to use another set of map functions to grab from the between the two variables, but I feel like I'm on somewhat the right track but just don't understand how i could get my desired output.
scheme racket
scheme racket
asked Nov 21 at 4:40
onbu
358
asked Nov 21 at 4:40
onbu
358
edited Nov 21 at 4:47
asked Nov 21 at 4:40
onbu
358
asked Nov 21 at 4:40
onbu
358
asked Nov 21 at 4:40
onbu
358
358
add a comment |
add a comment |
1 Answer
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I'm not entirely sure what you're trying to accomplish there - but this produced the desired output:
(define (productionValues input)
; flatten the sublists
(apply append
; create the lists as per the sample
(map (lambda (x) (list (list (first x) (second x))
(list (first x) (third x))))
input)))
For example:
(define input '((a (xz) (c)) (b (wy) (d))))
(productionValues input)
=> '((a (xz)) (a (c)) (b (wy)) (b (d)))
answered Nov 21 at 9:13
Óscar López
175k22225320
The overall goal of this project is to build an LL(1) parser and this is one of my functions that takes functions with the same non terminal and merge them onto one line. For example [a ::= (xz) | (c) ] b ::= (wy) | (d) ] is converted to this in scheme '((a (xz) (c)) (b (wy) (d))), and i have to take each left hand symbol and apply it to the appropriate rule. Hope this makes sense.
– onbu
Nov 21 at 16:50
@onbu ok, thanks for the context :) anyway, that doesn't change my answer, it's giving the correct output for your sample input. Unless you have a counter-example, I think we can mark is as correct now ;)
– Óscar López
Nov 21 at 16:57
yea thanks, took me a while to apply your logic into what i wanted.
– onbu
Nov 21 at 21:14
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
I'm not entirely sure what you're trying to accomplish there - but this produced the desired output:
(define (productionValues input)
; flatten the sublists
(apply append
; create the lists as per the sample
(map (lambda (x) (list (list (first x) (second x))
(list (first x) (third x))))
input)))
For example:
(define input '((a (xz) (c)) (b (wy) (d))))
(productionValues input)
=> '((a (xz)) (a (c)) (b (wy)) (b (d)))
answered Nov 21 at 9:13
Óscar López
175k22225320
The overall goal of this project is to build an LL(1) parser and this is one of my functions that takes functions with the same non terminal and merge them onto one line. For example [a ::= (xz) | (c) ] b ::= (wy) | (d) ] is converted to this in scheme '((a (xz) (c)) (b (wy) (d))), and i have to take each left hand symbol and apply it to the appropriate rule. Hope this makes sense.
– onbu
Nov 21 at 16:50
@onbu ok, thanks for the context :) anyway, that doesn't change my answer, it's giving the correct output for your sample input. Unless you have a counter-example, I think we can mark is as correct now ;)
– Óscar López
Nov 21 at 16:57
yea thanks, took me a while to apply your logic into what i wanted.
– onbu
Nov 21 at 21:14
add a comment |
I'm not entirely sure what you're trying to accomplish there - but this produced the desired output:
(define (productionValues input)
; flatten the sublists
(apply append
; create the lists as per the sample
(map (lambda (x) (list (list (first x) (second x))
(list (first x) (third x))))
input)))
For example:
(define input '((a (xz) (c)) (b (wy) (d))))
(productionValues input)
=> '((a (xz)) (a (c)) (b (wy)) (b (d)))
answered Nov 21 at 9:13
Óscar López
175k22225320
The overall goal of this project is to build an LL(1) parser and this is one of my functions that takes functions with the same non terminal and merge them onto one line. For example [a ::= (xz) | (c) ] b ::= (wy) | (d) ] is converted to this in scheme '((a (xz) (c)) (b (wy) (d))), and i have to take each left hand symbol and apply it to the appropriate rule. Hope this makes sense.
– onbu
Nov 21 at 16:50
@onbu ok, thanks for the context :) anyway, that doesn't change my answer, it's giving the correct output for your sample input. Unless you have a counter-example, I think we can mark is as correct now ;)
– Óscar López
Nov 21 at 16:57
yea thanks, took me a while to apply your logic into what i wanted.
– onbu
Nov 21 at 21:14
add a comment |
I'm not entirely sure what you're trying to accomplish there - but this produced the desired output:
(define (productionValues input)
; flatten the sublists
(apply append
; create the lists as per the sample
(map (lambda (x) (list (list (first x) (second x))
(list (first x) (third x))))
input)))
For example:
(define input '((a (xz) (c)) (b (wy) (d))))
(productionValues input)
=> '((a (xz)) (a (c)) (b (wy)) (b (d)))
answered Nov 21 at 9:13
Óscar López
175k22225320
I'm not entirely sure what you're trying to accomplish there - but this produced the desired output:
(define (productionValues input)
; flatten the sublists
(apply append
; create the lists as per the sample
(map (lambda (x) (list (list (first x) (second x))
(list (first x) (third x))))
input)))
For example:
(define input '((a (xz) (c)) (b (wy) (d))))
(productionValues input)
=> '((a (xz)) (a (c)) (b (wy)) (b (d)))
answered Nov 21 at 9:13
Óscar López
175k22225320
answered Nov 21 at 9:13
Óscar López
175k22225320
answered Nov 21 at 9:13
Óscar López
175k22225320
answered Nov 21 at 9:13
Óscar López
175k22225320
175k22225320
The overall goal of this project is to build an LL(1) parser and this is one of my functions that takes functions with the same non terminal and merge them onto one line. For example [a ::= (xz) | (c) ] b ::= (wy) | (d) ] is converted to this in scheme '((a (xz) (c)) (b (wy) (d))), and i have to take each left hand symbol and apply it to the appropriate rule. Hope this makes sense.
– onbu
Nov 21 at 16:50
@onbu ok, thanks for the context :) anyway, that doesn't change my answer, it's giving the correct output for your sample input. Unless you have a counter-example, I think we can mark is as correct now ;)
– Óscar López
Nov 21 at 16:57
yea thanks, took me a while to apply your logic into what i wanted.
– onbu
Nov 21 at 21:14
add a comment |
The overall goal of this project is to build an LL(1) parser and this is one of my functions that takes functions with the same non terminal and merge them onto one line. For example [a ::= (xz) | (c) ] b ::= (wy) | (d) ] is converted to this in scheme '((a (xz) (c)) (b (wy) (d))), and i have to take each left hand symbol and apply it to the appropriate rule. Hope this makes sense.
– onbu
Nov 21 at 16:50
@onbu ok, thanks for the context :) anyway, that doesn't change my answer, it's giving the correct output for your sample input. Unless you have a counter-example, I think we can mark is as correct now ;)
– Óscar López
Nov 21 at 16:57
yea thanks, took me a while to apply your logic into what i wanted.
– onbu
Nov 21 at 21:14
The overall goal of this project is to build an LL(1) parser and this is one of my functions that takes functions with the same non terminal and merge them onto one line. For example [a ::= (xz) | (c) ] b ::= (wy) | (d) ] is converted to this in scheme '((a (xz) (c)) (b (wy) (d))), and i have to take each left hand symbol and apply it to the appropriate rule. Hope this makes sense.
– onbu
Nov 21 at 16:50
The overall goal of this project is to build an LL(1) parser and this is one of my functions that takes functions with the same non terminal and merge them onto one line. For example [a ::= (xz) | (c) ] b ::= (wy) | (d) ] is converted to this in scheme '((a (xz) (c)) (b (wy) (d))), and i have to take each left hand symbol and apply it to the appropriate rule. Hope this makes sense.
– onbu
Nov 21 at 16:50
@onbu ok, thanks for the context :) anyway, that doesn't change my answer, it's giving the correct output for your sample input. Unless you have a counter-example, I think we can mark is as correct now ;)
– Óscar López
Nov 21 at 16:57
@onbu ok, thanks for the context :) anyway, that doesn't change my answer, it's giving the correct output for your sample input. Unless you have a counter-example, I think we can mark is as correct now ;)
– Óscar López
Nov 21 at 16:57
yea thanks, took me a while to apply your logic into what i wanted.
– onbu
Nov 21 at 21:14
yea thanks, took me a while to apply your logic into what i wanted.
– onbu
Nov 21 at 21:14
add a comment |
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