Mixing
Assigning random values to the elements of a list
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From this question we can construct an interleaved list out of two lists with a different length (according to Sylwester answer making use of circular-list). My question is how can we do the same but taking the elements of the second list randomly.
That is, I want to assign to the elements of a list a random value out of a list. For example: given the assigning list '(0 1) and whatever list '(a b c d) I want as the output a list such as '((a 0) (b 0) (c 1) (d 0)) or '((a 1) (b 0) (c 0) (d 1)) for example.
My attempts are:
(map cons '(a b c d) (circular-list (random '(0 1))))
which gives '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)) and nothing different and
(map cons '(a b c d) (list (random '(0 1))))
which gives '((a . 0)) or '((a . 1)) and nothing different.
PS: I am making use of a function called random and defined as
(define (random lst) (list-ref lst (random (length lst))))
which in my example would take randomly 0 or 1 out of a list '(0 1).
list random racket
asked Jan 3 at 9:56
gibariangibarian
646
add a comment |
From this question we can construct an interleaved list out of two lists with a different length (according to Sylwester answer making use of circular-list). My question is how can we do the same but taking the elements of the second list randomly.
That is, I want to assign to the elements of a list a random value out of a list. For example: given the assigning list '(0 1) and whatever list '(a b c d) I want as the output a list such as '((a 0) (b 0) (c 1) (d 0)) or '((a 1) (b 0) (c 0) (d 1)) for example.
My attempts are:
(map cons '(a b c d) (circular-list (random '(0 1))))
which gives '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)) and nothing different and
(map cons '(a b c d) (list (random '(0 1))))
which gives '((a . 0)) or '((a . 1)) and nothing different.
PS: I am making use of a function called random and defined as
(define (random lst) (list-ref lst (random (length lst))))
which in my example would take randomly 0 or 1 out of a list '(0 1).
list random racket
asked Jan 3 at 9:56
gibariangibarian
646
What is the result of(random '(0) '(1))? (Which language are you using? (Racket, one of the teaching languages, other)
– soegaard
Jan 3 at 10:33
See answer below.
– soegaard
Jan 3 at 10:38
add a comment |
From this question we can construct an interleaved list out of two lists with a different length (according to Sylwester answer making use of circular-list). My question is how can we do the same but taking the elements of the second list randomly.
That is, I want to assign to the elements of a list a random value out of a list. For example: given the assigning list '(0 1) and whatever list '(a b c d) I want as the output a list such as '((a 0) (b 0) (c 1) (d 0)) or '((a 1) (b 0) (c 0) (d 1)) for example.
My attempts are:
(map cons '(a b c d) (circular-list (random '(0 1))))
which gives '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)) and nothing different and
(map cons '(a b c d) (list (random '(0 1))))
which gives '((a . 0)) or '((a . 1)) and nothing different.
PS: I am making use of a function called random and defined as
(define (random lst) (list-ref lst (random (length lst))))
which in my example would take randomly 0 or 1 out of a list '(0 1).
list random racket
asked Jan 3 at 9:56
gibariangibarian
646
From this question we can construct an interleaved list out of two lists with a different length (according to Sylwester answer making use of circular-list). My question is how can we do the same but taking the elements of the second list randomly.
That is, I want to assign to the elements of a list a random value out of a list. For example: given the assigning list '(0 1) and whatever list '(a b c d) I want as the output a list such as '((a 0) (b 0) (c 1) (d 0)) or '((a 1) (b 0) (c 0) (d 1)) for example.
My attempts are:
(map cons '(a b c d) (circular-list (random '(0 1))))
which gives '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)) and nothing different and
(map cons '(a b c d) (list (random '(0 1))))
which gives '((a . 0)) or '((a . 1)) and nothing different.
PS: I am making use of a function called random and defined as
(define (random lst) (list-ref lst (random (length lst))))
which in my example would take randomly 0 or 1 out of a list '(0 1).
list random racket
list random racket
asked Jan 3 at 9:56
gibariangibarian
646
asked Jan 3 at 9:56
gibariangibarian
646
edited Jan 3 at 13:46
gibarian
asked Jan 3 at 9:56
gibariangibarian
646
asked Jan 3 at 9:56
gibariangibarian
646
asked Jan 3 at 9:56
gibariangibarian
646
646
What is the result of(random '(0) '(1))? (Which language are you using? (Racket, one of the teaching languages, other)
– soegaard
Jan 3 at 10:33
See answer below.
– soegaard
Jan 3 at 10:38
add a comment |
What is the result of(random '(0) '(1))? (Which language are you using? (Racket, one of the teaching languages, other)
– soegaard
Jan 3 at 10:33
See answer below.
– soegaard
Jan 3 at 10:38
What is the result of
(random '(0) '(1)) ? (Which language are you using? (Racket, one of the teaching languages, other)– soegaard
Jan 3 at 10:33
What is the result of
(random '(0) '(1)) ? (Which language are you using? (Racket, one of the teaching languages, other)– soegaard
Jan 3 at 10:33
See answer below.
– soegaard
Jan 3 at 10:38
See answer below.
– soegaard
Jan 3 at 10:38
add a comment |
2 Answers
2
active
oldest
votes
To avoid confusion, you should rename your new "random" function on lists to random-element or something similar. That way both people reading your code, and Racket, will know the difference.
random : PositiveNaturalNumber -> NaturalNumber
random-element : [NonEmptyListof X] -> X
These two different functions need to have two different names so that when you want to refer to the first one random from Racket, you can do so. Otherwise Racket (as well as other people reading your code) will think you mean random-element when you really want random.
This confusion matters in the body of your definition of "random":
; /--------------<< This should be renamed to `random-element`
; /
(define (random lst)
(list-ref lst (random (length lst))))
; /
; ------<< This is meant to be the original Racket `random`
; but it ends up referring to "random-element" because
; of the naming conflict
Because of this name conflict, when this definition of random is put in a File, and I run a Repl for that File, I get an error like this:
> (random '(A B C D E F G))
length: contract violation
expected: list?
given: 7
When this is renamed, the definition should look like this:
;; random-element : [NonEmptyListof X] -> X
(define (random-element lst)
(list-ref lst (random (length lst))))
; /
; -------<< This is the `random` from Racket, not the "new" one
Using it:
> (random-element '(A B C D E F G))
'E
> (random-element '(A B C D E F G))
'B
> (random-element '(A B C D E F G))
'D
> (random-element '(A B C D E F G))
'F
> (random-element '(A B C D E F G))
'C
answered Jan 3 at 17:29
Alex KnauthAlex Knauth
5,2221823
Thank you very much for the clarifcation. It has been formative. However, my function still gives the same value for all (a b c d): '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)). @PetSerAl answer works fine but I don't understand why we need to convert the list into a vector yet (let alone the fact that it does not work in DrRacket, only in the Ubuntu terminal).
– gibarian
Jan 3 at 18:44
You shouldn't need to convert it to a vector. However, I suspect that my answer here combined with an answer to How to create a list with n applications of a procedure will solve that problem.
– Alex Knauth
Jan 3 at 19:47
add a comment |
The comments can't contain code fragments, so ...
What does this program give?
#lang racket
(require srfi/1)
(random '(0) '(1))
On my computer with Racket 7.0 I get:
random: contract violation
expected: exact-integer?
given: '(0)
argument position: 1st
other arguments...:
answered Jan 3 at 10:38
soegaardsoegaard
24.6k44080
'random' is a function I have written. It is actually more complex, I wanted to simplify the question since my interest is on the "interleaving part". For example: (define (random lista) (list-ref lista (random (length lista)))). Sorry for that.
– gibarian
Jan 3 at 11:15
In my Ubuntu terminal (random '(0 1)) returns 0 or 1.
– gibarian
Jan 3 at 11:21
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
To avoid confusion, you should rename your new "random" function on lists to random-element or something similar. That way both people reading your code, and Racket, will know the difference.
random : PositiveNaturalNumber -> NaturalNumber
random-element : [NonEmptyListof X] -> X
These two different functions need to have two different names so that when you want to refer to the first one random from Racket, you can do so. Otherwise Racket (as well as other people reading your code) will think you mean random-element when you really want random.
This confusion matters in the body of your definition of "random":
; /--------------<< This should be renamed to `random-element`
; /
(define (random lst)
(list-ref lst (random (length lst))))
; /
; ------<< This is meant to be the original Racket `random`
; but it ends up referring to "random-element" because
; of the naming conflict
Because of this name conflict, when this definition of random is put in a File, and I run a Repl for that File, I get an error like this:
> (random '(A B C D E F G))
length: contract violation
expected: list?
given: 7
When this is renamed, the definition should look like this:
;; random-element : [NonEmptyListof X] -> X
(define (random-element lst)
(list-ref lst (random (length lst))))
; /
; -------<< This is the `random` from Racket, not the "new" one
Using it:
> (random-element '(A B C D E F G))
'E
> (random-element '(A B C D E F G))
'B
> (random-element '(A B C D E F G))
'D
> (random-element '(A B C D E F G))
'F
> (random-element '(A B C D E F G))
'C
answered Jan 3 at 17:29
Alex KnauthAlex Knauth
5,2221823
Thank you very much for the clarifcation. It has been formative. However, my function still gives the same value for all (a b c d): '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)). @PetSerAl answer works fine but I don't understand why we need to convert the list into a vector yet (let alone the fact that it does not work in DrRacket, only in the Ubuntu terminal).
– gibarian
Jan 3 at 18:44
You shouldn't need to convert it to a vector. However, I suspect that my answer here combined with an answer to How to create a list with n applications of a procedure will solve that problem.
– Alex Knauth
Jan 3 at 19:47
add a comment |
To avoid confusion, you should rename your new "random" function on lists to random-element or something similar. That way both people reading your code, and Racket, will know the difference.
random : PositiveNaturalNumber -> NaturalNumber
random-element : [NonEmptyListof X] -> X
These two different functions need to have two different names so that when you want to refer to the first one random from Racket, you can do so. Otherwise Racket (as well as other people reading your code) will think you mean random-element when you really want random.
This confusion matters in the body of your definition of "random":
; /--------------<< This should be renamed to `random-element`
; /
(define (random lst)
(list-ref lst (random (length lst))))
; /
; ------<< This is meant to be the original Racket `random`
; but it ends up referring to "random-element" because
; of the naming conflict
Because of this name conflict, when this definition of random is put in a File, and I run a Repl for that File, I get an error like this:
> (random '(A B C D E F G))
length: contract violation
expected: list?
given: 7
When this is renamed, the definition should look like this:
;; random-element : [NonEmptyListof X] -> X
(define (random-element lst)
(list-ref lst (random (length lst))))
; /
; -------<< This is the `random` from Racket, not the "new" one
Using it:
> (random-element '(A B C D E F G))
'E
> (random-element '(A B C D E F G))
'B
> (random-element '(A B C D E F G))
'D
> (random-element '(A B C D E F G))
'F
> (random-element '(A B C D E F G))
'C
answered Jan 3 at 17:29
Alex KnauthAlex Knauth
5,2221823
Thank you very much for the clarifcation. It has been formative. However, my function still gives the same value for all (a b c d): '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)). @PetSerAl answer works fine but I don't understand why we need to convert the list into a vector yet (let alone the fact that it does not work in DrRacket, only in the Ubuntu terminal).
– gibarian
Jan 3 at 18:44
You shouldn't need to convert it to a vector. However, I suspect that my answer here combined with an answer to How to create a list with n applications of a procedure will solve that problem.
– Alex Knauth
Jan 3 at 19:47
add a comment |
To avoid confusion, you should rename your new "random" function on lists to random-element or something similar. That way both people reading your code, and Racket, will know the difference.
random : PositiveNaturalNumber -> NaturalNumber
random-element : [NonEmptyListof X] -> X
These two different functions need to have two different names so that when you want to refer to the first one random from Racket, you can do so. Otherwise Racket (as well as other people reading your code) will think you mean random-element when you really want random.
This confusion matters in the body of your definition of "random":
; /--------------<< This should be renamed to `random-element`
; /
(define (random lst)
(list-ref lst (random (length lst))))
; /
; ------<< This is meant to be the original Racket `random`
; but it ends up referring to "random-element" because
; of the naming conflict
Because of this name conflict, when this definition of random is put in a File, and I run a Repl for that File, I get an error like this:
> (random '(A B C D E F G))
length: contract violation
expected: list?
given: 7
When this is renamed, the definition should look like this:
;; random-element : [NonEmptyListof X] -> X
(define (random-element lst)
(list-ref lst (random (length lst))))
; /
; -------<< This is the `random` from Racket, not the "new" one
Using it:
> (random-element '(A B C D E F G))
'E
> (random-element '(A B C D E F G))
'B
> (random-element '(A B C D E F G))
'D
> (random-element '(A B C D E F G))
'F
> (random-element '(A B C D E F G))
'C
answered Jan 3 at 17:29
Alex KnauthAlex Knauth
5,2221823
To avoid confusion, you should rename your new "random" function on lists to random-element or something similar. That way both people reading your code, and Racket, will know the difference.
random : PositiveNaturalNumber -> NaturalNumber
random-element : [NonEmptyListof X] -> X
These two different functions need to have two different names so that when you want to refer to the first one random from Racket, you can do so. Otherwise Racket (as well as other people reading your code) will think you mean random-element when you really want random.
This confusion matters in the body of your definition of "random":
; /--------------<< This should be renamed to `random-element`
; /
(define (random lst)
(list-ref lst (random (length lst))))
; /
; ------<< This is meant to be the original Racket `random`
; but it ends up referring to "random-element" because
; of the naming conflict
Because of this name conflict, when this definition of random is put in a File, and I run a Repl for that File, I get an error like this:
> (random '(A B C D E F G))
length: contract violation
expected: list?
given: 7
When this is renamed, the definition should look like this:
;; random-element : [NonEmptyListof X] -> X
(define (random-element lst)
(list-ref lst (random (length lst))))
; /
; -------<< This is the `random` from Racket, not the "new" one
Using it:
> (random-element '(A B C D E F G))
'E
> (random-element '(A B C D E F G))
'B
> (random-element '(A B C D E F G))
'D
> (random-element '(A B C D E F G))
'F
> (random-element '(A B C D E F G))
'C
answered Jan 3 at 17:29
Alex KnauthAlex Knauth
5,2221823
answered Jan 3 at 17:29
Alex KnauthAlex Knauth
5,2221823
answered Jan 3 at 17:29
Alex KnauthAlex Knauth
5,2221823
answered Jan 3 at 17:29
Alex KnauthAlex Knauth
5,2221823
5,2221823
Thank you very much for the clarifcation. It has been formative. However, my function still gives the same value for all (a b c d): '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)). @PetSerAl answer works fine but I don't understand why we need to convert the list into a vector yet (let alone the fact that it does not work in DrRacket, only in the Ubuntu terminal).
– gibarian
Jan 3 at 18:44
You shouldn't need to convert it to a vector. However, I suspect that my answer here combined with an answer to How to create a list with n applications of a procedure will solve that problem.
– Alex Knauth
Jan 3 at 19:47
add a comment |
Thank you very much for the clarifcation. It has been formative. However, my function still gives the same value for all (a b c d): '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)). @PetSerAl answer works fine but I don't understand why we need to convert the list into a vector yet (let alone the fact that it does not work in DrRacket, only in the Ubuntu terminal).
– gibarian
Jan 3 at 18:44
You shouldn't need to convert it to a vector. However, I suspect that my answer here combined with an answer to How to create a list with n applications of a procedure will solve that problem.
– Alex Knauth
Jan 3 at 19:47
Thank you very much for the clarifcation. It has been formative. However, my function still gives the same value for all (a b c d): '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)). @PetSerAl answer works fine but I don't understand why we need to convert the list into a vector yet (let alone the fact that it does not work in DrRacket, only in the Ubuntu terminal).
– gibarian
Jan 3 at 18:44
Thank you very much for the clarifcation. It has been formative. However, my function still gives the same value for all (a b c d): '((a . 0) (b . 0) (c . 0) (d . 0)) or '((a . 1) (b . 1) (c . 1) (d . 1)). @PetSerAl answer works fine but I don't understand why we need to convert the list into a vector yet (let alone the fact that it does not work in DrRacket, only in the Ubuntu terminal).
– gibarian
Jan 3 at 18:44
You shouldn't need to convert it to a vector. However, I suspect that my answer here combined with an answer to How to create a list with n applications of a procedure will solve that problem.
– Alex Knauth
Jan 3 at 19:47
You shouldn't need to convert it to a vector. However, I suspect that my answer here combined with an answer to How to create a list with n applications of a procedure will solve that problem.
– Alex Knauth
Jan 3 at 19:47
add a comment |
The comments can't contain code fragments, so ...
What does this program give?
#lang racket
(require srfi/1)
(random '(0) '(1))
On my computer with Racket 7.0 I get:
random: contract violation
expected: exact-integer?
given: '(0)
argument position: 1st
other arguments...:
answered Jan 3 at 10:38
soegaardsoegaard
24.6k44080
'random' is a function I have written. It is actually more complex, I wanted to simplify the question since my interest is on the "interleaving part". For example: (define (random lista) (list-ref lista (random (length lista)))). Sorry for that.
– gibarian
Jan 3 at 11:15
In my Ubuntu terminal (random '(0 1)) returns 0 or 1.
– gibarian
Jan 3 at 11:21
add a comment |
The comments can't contain code fragments, so ...
What does this program give?
#lang racket
(require srfi/1)
(random '(0) '(1))
On my computer with Racket 7.0 I get:
random: contract violation
expected: exact-integer?
given: '(0)
argument position: 1st
other arguments...:
answered Jan 3 at 10:38
soegaardsoegaard
24.6k44080
'random' is a function I have written. It is actually more complex, I wanted to simplify the question since my interest is on the "interleaving part". For example: (define (random lista) (list-ref lista (random (length lista)))). Sorry for that.
– gibarian
Jan 3 at 11:15
In my Ubuntu terminal (random '(0 1)) returns 0 or 1.
– gibarian
Jan 3 at 11:21
add a comment |
The comments can't contain code fragments, so ...
What does this program give?
#lang racket
(require srfi/1)
(random '(0) '(1))
On my computer with Racket 7.0 I get:
random: contract violation
expected: exact-integer?
given: '(0)
argument position: 1st
other arguments...:
answered Jan 3 at 10:38
soegaardsoegaard
24.6k44080
The comments can't contain code fragments, so ...
What does this program give?
#lang racket
(require srfi/1)
(random '(0) '(1))
On my computer with Racket 7.0 I get:
random: contract violation
expected: exact-integer?
given: '(0)
argument position: 1st
other arguments...:
answered Jan 3 at 10:38
soegaardsoegaard
24.6k44080
answered Jan 3 at 10:38
soegaardsoegaard
24.6k44080
answered Jan 3 at 10:38
soegaardsoegaard
24.6k44080
answered Jan 3 at 10:38
soegaardsoegaard
24.6k44080
24.6k44080
'random' is a function I have written. It is actually more complex, I wanted to simplify the question since my interest is on the "interleaving part". For example: (define (random lista) (list-ref lista (random (length lista)))). Sorry for that.
– gibarian
Jan 3 at 11:15
In my Ubuntu terminal (random '(0 1)) returns 0 or 1.
– gibarian
Jan 3 at 11:21
add a comment |
'random' is a function I have written. It is actually more complex, I wanted to simplify the question since my interest is on the "interleaving part". For example: (define (random lista) (list-ref lista (random (length lista)))). Sorry for that.
– gibarian
Jan 3 at 11:15
In my Ubuntu terminal (random '(0 1)) returns 0 or 1.
– gibarian
Jan 3 at 11:21
'random' is a function I have written. It is actually more complex, I wanted to simplify the question since my interest is on the "interleaving part". For example: (define (random lista) (list-ref lista (random (length lista)))). Sorry for that.
– gibarian
Jan 3 at 11:15
'random' is a function I have written. It is actually more complex, I wanted to simplify the question since my interest is on the "interleaving part". For example: (define (random lista) (list-ref lista (random (length lista)))). Sorry for that.
– gibarian
Jan 3 at 11:15
In my Ubuntu terminal (random '(0 1)) returns 0 or 1.
– gibarian
Jan 3 at 11:21
In my Ubuntu terminal (random '(0 1)) returns 0 or 1.
– gibarian
Jan 3 at 11:21
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What is the result of
(random '(0) '(1))? (Which language are you using? (Racket, one of the teaching languages, other)– soegaard
Jan 3 at 10:33
See answer below.
– soegaard
Jan 3 at 10:38