Not sure why this function returns a reversed array
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1
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I'm working on a JavaScript exercise and having some trouble untangling the logic as to why it works. It basically has a function called "mystery" that uses a bunch of very simple functions, and returns an array you give it but in reversed order. I've been sitting in front of a whiteboard for an hour trying to figure out the logic behind it, and not getting it. Can a kind soul take a look at these functions and explain how the mystery function returns a reversed array? Thank you!
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
javascript arrays
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up vote
1
down vote
favorite
I'm working on a JavaScript exercise and having some trouble untangling the logic as to why it works. It basically has a function called "mystery" that uses a bunch of very simple functions, and returns an array you give it but in reversed order. I've been sitting in front of a whiteboard for an hour trying to figure out the logic behind it, and not getting it. Can a kind soul take a look at these functions and explain how the mystery function returns a reversed array? Thank you!
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
javascript arrays
Try to actually run it, use the debugger to trace through it, line by line...
– obe
Nov 10 at 21:14
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm working on a JavaScript exercise and having some trouble untangling the logic as to why it works. It basically has a function called "mystery" that uses a bunch of very simple functions, and returns an array you give it but in reversed order. I've been sitting in front of a whiteboard for an hour trying to figure out the logic behind it, and not getting it. Can a kind soul take a look at these functions and explain how the mystery function returns a reversed array? Thank you!
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
javascript arrays
I'm working on a JavaScript exercise and having some trouble untangling the logic as to why it works. It basically has a function called "mystery" that uses a bunch of very simple functions, and returns an array you give it but in reversed order. I've been sitting in front of a whiteboard for an hour trying to figure out the logic behind it, and not getting it. Can a kind soul take a look at these functions and explain how the mystery function returns a reversed array? Thank you!
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
javascript arrays
javascript arrays
edited Nov 10 at 21:09
asked Nov 10 at 21:06
amacdonald
529
529
Try to actually run it, use the debugger to trace through it, line by line...
– obe
Nov 10 at 21:14
add a comment |
Try to actually run it, use the debugger to trace through it, line by line...
– obe
Nov 10 at 21:14
Try to actually run it, use the debugger to trace through it, line by line...
– obe
Nov 10 at 21:14
Try to actually run it, use the debugger to trace through it, line by line...
– obe
Nov 10 at 21:14
add a comment |
8 Answers
8
active
oldest
votes
up vote
2
down vote
your first function "rest" removes the first element, as slice will return elements from 1 to the end of the array, then the "conj" function will take the first element that was removed (through the "first" function) and put it in the end, and doing so recursively it'll take elements from the beginning and put them to the end.
add a comment |
up vote
2
down vote
mystery is a recursive function.
It calls itself using the return value of the rest
function, which returns everything except the first element.
It uses the result of that + the result of first
, which returns the first character, and concatenates them again (using conj
), but with the first element at the end.
So, say you put in [H e l l o],
it will return conj(mystery([e l l o], H)
mystery([e l l o])
will return conj(mystery([l l o], e)
mystery([l l o])
will return conj(mystery([l o], l)
and so on, until the array that goes into mistery
is empty, in which case the recursion ends, and we bubble back up to the first call.
Side note, recursion is often used for exercises like this, but although it has some specific uses, in many cases it's way more efficient to not use recursion, because the overhead of making another function call is relatively hard work, compared to other solutions using a simple loop to move or swap items around.
You can see what is happening if you output some information:
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array, level)
if (array.length === 0)
console.log('mystery level '+level+' is called with an empty array. Recursion ends here.. Stay tuned for the answer.');
return ;
console.log('mystery level '+level+' is called with '+array+
'. I will move '+first(array)+' to the end.');
var result = conj(mystery(rest(array), level+1), first(array));
console.log('returning '+result+' for level '+level);
return result;
console.log(mystery(['H','e','l','l','o'], 0));
add a comment |
up vote
1
down vote
To understand a function that uses recursion it can help to just assume for a moment that the recursive (nested) call returns what it should and then see how it builds on that to produce a correct result.
Let's for example suppose that array
is [1, 2, 3, 4]
So this line:
conj(mystery(rest(array)), first(array));
... has a recursive call of mystery
. It gets as argument the array
, but with the first element removed from it (that is what rest
returns), so it gets [2, 3, 4]
Now we will just assume that this recursive call of mystery
does the right thing and reverses that array into [4, 3, 2]
. Then in the above quoted code we see this result is concatenated with first(array)
(which is the first value, i.e. 1). So we get [4, 3, 2, 1]
. Correct!
This teaches us that if we assume mystery
does the job right for an array with n-1 values, it also does it right for n values.
Now remains to see whether mystery
deals correctly with the smallest case, i.e. when the array is empty. It is easy to see it returns the correct result in that case, i.e. an empty array.
So putting those two things together you can see that mystery
does the job correctly for all possible array sizes.
add a comment |
up vote
1
down vote
The .push
method places the item to end of array..slice(1)
means “except the first item”
Pseudocode
- Get array A (arg of
mystery
). If it is empty, return it - Take
rest
(everything except the first). We will call the rest B - Run this program on B (recursively)
- Append the first item of A to end of B
conj
= append value
to arr
first
= get first item of arr
rest
= return everything except the first item
mystery
when array
is empty = return empty array
mystery
when array
is not empty = Take rest(array)
, run mystery
on it, then append first
of array
add a comment |
up vote
1
down vote
Yeah, the magic of recursion. To understand think about what it does if you call mystery
with a 2-element array [1,2]
.
rest(array)
will then be [2]
mystery(rest(array))
will also be [2]
first(array)
will be 1
.
Then you return conj([2], 1)
which locically results in [2,1]
.
Now the trick is the recursion. If you have 3 elements [0,1,2]
and call mystery with it this will happen:
- it will call
mystery(rest(array))
with essentially ismystery([1,2])
. That this returns[2,1]
have we already seen. first(array)
will be0
- so it returns
conj([2,1],0)
which is logically[2,1,0]
.
this now recusivly works for as many elements as you wish. Essentially mystery
will be called for every element to place it after all elements.
add a comment |
up vote
1
down vote
One thing to note here is that mystery()
is called recursively. I added some comments to illustrate what's going on on each step.
function rest(arr)
console.log('take first n-1 elements of', arr);
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
console.log('combine', arr, 'with', value)
return arr;
function mystery(array)
console.log('call mystery on ', array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
mystery([0,1,2])
add a comment |
up vote
1
down vote
It's pretty simple. The reason for you not being able to see it is RECURSION. Couple things, that I would like you to notice:
- mystery(array) is recursive function because it call itself until the array passed in is emplty
- all of your work is happening here: return conj(mystery(rest(array)), first(array));
I'm not going to talk more about recursive function here rather I will show how you can track each recursive call using console.log(). Check out my code below, I've added console.log() to make things more clear for you. Try running mystery with some array and see results. This will make sense to you.
function rest(arr)
return arr.slice(1);
function first(arr)
console.log("Returning ",arr[0], "from first(arr).")
return arr[0];
function conj(arr, value)
console.log("Pushing ",value, " to ",arr, " in conj(arr,value)");
arr.push(value);
console.log("Returning ",arr, " from Conj(arr,value)");
return arr;
function mystery(array)
if (array.length === 0)
console.log("array is emplty. So, returning empty array from mystery");
return ;
console.log("array is not emplty. So, calling mystery(array) again.");
return conj(mystery(rest(array)), first(array));
var reverse =mystery([1,2,3,4]);
console.log("The final result", reverse);
add a comment |
up vote
0
down vote
Not gonna lie, this is a weird way to reverse the order of the array. It's basically using recursion to slice the first element of the array and conjoin the elements with first element at the end.
So here's a walk though:
mystery(['h','e','l','l','o'])
->
first check if array is empty then method rest
is called ->
rest(['h','e','l','l','o'])
->
rest slices the array at index 1 returning a new array of ['e','l','l','o']
->
then mystery
is called which repeats the previous steps with the return value of rest
->
mystery(['e','l','l','o'])
->
this means the mystery will be keep getting called on until rest
returns an empty array ->
mystery(['e','l','l','o'])
->mystery(['l','l','o'])
->mystery(['l','o'])
->mystery(['o'])
->mystery()
->
when mystery has an empty array it returns the empty array and then first
is called ->
first returns just the first element of the array and happens after mystery is returned it would look something like this before mystery is returned ->
mystery(['e','l','l','o']) , first(['h','e','l','l','o'])
->
so at the lowest level when mystery returns an empty array and first return the first element in ['o']
it would look like this ->
, ['o']
->conj
would be called with these values ->conj( , ['o'])
->conj
returns the combination of the two values ->conj( , ['o'])
->['o']
->
this is then returned and repeated ->conj( , ['o'])
->conj(['o'] , ['l'])
->conj(['o','l'] , ['l'])
->conj(['o', 'l','l'] , ['e'])
->conj(['o', 'l','l', 'e'] , ['h'])
->['o', 'l','l', 'e','h']
add a comment |
8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
your first function "rest" removes the first element, as slice will return elements from 1 to the end of the array, then the "conj" function will take the first element that was removed (through the "first" function) and put it in the end, and doing so recursively it'll take elements from the beginning and put them to the end.
add a comment |
up vote
2
down vote
your first function "rest" removes the first element, as slice will return elements from 1 to the end of the array, then the "conj" function will take the first element that was removed (through the "first" function) and put it in the end, and doing so recursively it'll take elements from the beginning and put them to the end.
add a comment |
up vote
2
down vote
up vote
2
down vote
your first function "rest" removes the first element, as slice will return elements from 1 to the end of the array, then the "conj" function will take the first element that was removed (through the "first" function) and put it in the end, and doing so recursively it'll take elements from the beginning and put them to the end.
your first function "rest" removes the first element, as slice will return elements from 1 to the end of the array, then the "conj" function will take the first element that was removed (through the "first" function) and put it in the end, and doing so recursively it'll take elements from the beginning and put them to the end.
answered Nov 10 at 21:16
Reda Drissi
169111
169111
add a comment |
add a comment |
up vote
2
down vote
mystery is a recursive function.
It calls itself using the return value of the rest
function, which returns everything except the first element.
It uses the result of that + the result of first
, which returns the first character, and concatenates them again (using conj
), but with the first element at the end.
So, say you put in [H e l l o],
it will return conj(mystery([e l l o], H)
mystery([e l l o])
will return conj(mystery([l l o], e)
mystery([l l o])
will return conj(mystery([l o], l)
and so on, until the array that goes into mistery
is empty, in which case the recursion ends, and we bubble back up to the first call.
Side note, recursion is often used for exercises like this, but although it has some specific uses, in many cases it's way more efficient to not use recursion, because the overhead of making another function call is relatively hard work, compared to other solutions using a simple loop to move or swap items around.
You can see what is happening if you output some information:
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array, level)
if (array.length === 0)
console.log('mystery level '+level+' is called with an empty array. Recursion ends here.. Stay tuned for the answer.');
return ;
console.log('mystery level '+level+' is called with '+array+
'. I will move '+first(array)+' to the end.');
var result = conj(mystery(rest(array), level+1), first(array));
console.log('returning '+result+' for level '+level);
return result;
console.log(mystery(['H','e','l','l','o'], 0));
add a comment |
up vote
2
down vote
mystery is a recursive function.
It calls itself using the return value of the rest
function, which returns everything except the first element.
It uses the result of that + the result of first
, which returns the first character, and concatenates them again (using conj
), but with the first element at the end.
So, say you put in [H e l l o],
it will return conj(mystery([e l l o], H)
mystery([e l l o])
will return conj(mystery([l l o], e)
mystery([l l o])
will return conj(mystery([l o], l)
and so on, until the array that goes into mistery
is empty, in which case the recursion ends, and we bubble back up to the first call.
Side note, recursion is often used for exercises like this, but although it has some specific uses, in many cases it's way more efficient to not use recursion, because the overhead of making another function call is relatively hard work, compared to other solutions using a simple loop to move or swap items around.
You can see what is happening if you output some information:
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array, level)
if (array.length === 0)
console.log('mystery level '+level+' is called with an empty array. Recursion ends here.. Stay tuned for the answer.');
return ;
console.log('mystery level '+level+' is called with '+array+
'. I will move '+first(array)+' to the end.');
var result = conj(mystery(rest(array), level+1), first(array));
console.log('returning '+result+' for level '+level);
return result;
console.log(mystery(['H','e','l','l','o'], 0));
add a comment |
up vote
2
down vote
up vote
2
down vote
mystery is a recursive function.
It calls itself using the return value of the rest
function, which returns everything except the first element.
It uses the result of that + the result of first
, which returns the first character, and concatenates them again (using conj
), but with the first element at the end.
So, say you put in [H e l l o],
it will return conj(mystery([e l l o], H)
mystery([e l l o])
will return conj(mystery([l l o], e)
mystery([l l o])
will return conj(mystery([l o], l)
and so on, until the array that goes into mistery
is empty, in which case the recursion ends, and we bubble back up to the first call.
Side note, recursion is often used for exercises like this, but although it has some specific uses, in many cases it's way more efficient to not use recursion, because the overhead of making another function call is relatively hard work, compared to other solutions using a simple loop to move or swap items around.
You can see what is happening if you output some information:
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array, level)
if (array.length === 0)
console.log('mystery level '+level+' is called with an empty array. Recursion ends here.. Stay tuned for the answer.');
return ;
console.log('mystery level '+level+' is called with '+array+
'. I will move '+first(array)+' to the end.');
var result = conj(mystery(rest(array), level+1), first(array));
console.log('returning '+result+' for level '+level);
return result;
console.log(mystery(['H','e','l','l','o'], 0));
mystery is a recursive function.
It calls itself using the return value of the rest
function, which returns everything except the first element.
It uses the result of that + the result of first
, which returns the first character, and concatenates them again (using conj
), but with the first element at the end.
So, say you put in [H e l l o],
it will return conj(mystery([e l l o], H)
mystery([e l l o])
will return conj(mystery([l l o], e)
mystery([l l o])
will return conj(mystery([l o], l)
and so on, until the array that goes into mistery
is empty, in which case the recursion ends, and we bubble back up to the first call.
Side note, recursion is often used for exercises like this, but although it has some specific uses, in many cases it's way more efficient to not use recursion, because the overhead of making another function call is relatively hard work, compared to other solutions using a simple loop to move or swap items around.
You can see what is happening if you output some information:
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array, level)
if (array.length === 0)
console.log('mystery level '+level+' is called with an empty array. Recursion ends here.. Stay tuned for the answer.');
return ;
console.log('mystery level '+level+' is called with '+array+
'. I will move '+first(array)+' to the end.');
var result = conj(mystery(rest(array), level+1), first(array));
console.log('returning '+result+' for level '+level);
return result;
console.log(mystery(['H','e','l','l','o'], 0));
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array, level)
if (array.length === 0)
console.log('mystery level '+level+' is called with an empty array. Recursion ends here.. Stay tuned for the answer.');
return ;
console.log('mystery level '+level+' is called with '+array+
'. I will move '+first(array)+' to the end.');
var result = conj(mystery(rest(array), level+1), first(array));
console.log('returning '+result+' for level '+level);
return result;
console.log(mystery(['H','e','l','l','o'], 0));
function rest(arr)
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
return arr;
function mystery(array, level)
if (array.length === 0)
console.log('mystery level '+level+' is called with an empty array. Recursion ends here.. Stay tuned for the answer.');
return ;
console.log('mystery level '+level+' is called with '+array+
'. I will move '+first(array)+' to the end.');
var result = conj(mystery(rest(array), level+1), first(array));
console.log('returning '+result+' for level '+level);
return result;
console.log(mystery(['H','e','l','l','o'], 0));
edited Nov 11 at 0:50
answered Nov 10 at 21:18
GolezTrol
96.9k9129171
96.9k9129171
add a comment |
add a comment |
up vote
1
down vote
To understand a function that uses recursion it can help to just assume for a moment that the recursive (nested) call returns what it should and then see how it builds on that to produce a correct result.
Let's for example suppose that array
is [1, 2, 3, 4]
So this line:
conj(mystery(rest(array)), first(array));
... has a recursive call of mystery
. It gets as argument the array
, but with the first element removed from it (that is what rest
returns), so it gets [2, 3, 4]
Now we will just assume that this recursive call of mystery
does the right thing and reverses that array into [4, 3, 2]
. Then in the above quoted code we see this result is concatenated with first(array)
(which is the first value, i.e. 1). So we get [4, 3, 2, 1]
. Correct!
This teaches us that if we assume mystery
does the job right for an array with n-1 values, it also does it right for n values.
Now remains to see whether mystery
deals correctly with the smallest case, i.e. when the array is empty. It is easy to see it returns the correct result in that case, i.e. an empty array.
So putting those two things together you can see that mystery
does the job correctly for all possible array sizes.
add a comment |
up vote
1
down vote
To understand a function that uses recursion it can help to just assume for a moment that the recursive (nested) call returns what it should and then see how it builds on that to produce a correct result.
Let's for example suppose that array
is [1, 2, 3, 4]
So this line:
conj(mystery(rest(array)), first(array));
... has a recursive call of mystery
. It gets as argument the array
, but with the first element removed from it (that is what rest
returns), so it gets [2, 3, 4]
Now we will just assume that this recursive call of mystery
does the right thing and reverses that array into [4, 3, 2]
. Then in the above quoted code we see this result is concatenated with first(array)
(which is the first value, i.e. 1). So we get [4, 3, 2, 1]
. Correct!
This teaches us that if we assume mystery
does the job right for an array with n-1 values, it also does it right for n values.
Now remains to see whether mystery
deals correctly with the smallest case, i.e. when the array is empty. It is easy to see it returns the correct result in that case, i.e. an empty array.
So putting those two things together you can see that mystery
does the job correctly for all possible array sizes.
add a comment |
up vote
1
down vote
up vote
1
down vote
To understand a function that uses recursion it can help to just assume for a moment that the recursive (nested) call returns what it should and then see how it builds on that to produce a correct result.
Let's for example suppose that array
is [1, 2, 3, 4]
So this line:
conj(mystery(rest(array)), first(array));
... has a recursive call of mystery
. It gets as argument the array
, but with the first element removed from it (that is what rest
returns), so it gets [2, 3, 4]
Now we will just assume that this recursive call of mystery
does the right thing and reverses that array into [4, 3, 2]
. Then in the above quoted code we see this result is concatenated with first(array)
(which is the first value, i.e. 1). So we get [4, 3, 2, 1]
. Correct!
This teaches us that if we assume mystery
does the job right for an array with n-1 values, it also does it right for n values.
Now remains to see whether mystery
deals correctly with the smallest case, i.e. when the array is empty. It is easy to see it returns the correct result in that case, i.e. an empty array.
So putting those two things together you can see that mystery
does the job correctly for all possible array sizes.
To understand a function that uses recursion it can help to just assume for a moment that the recursive (nested) call returns what it should and then see how it builds on that to produce a correct result.
Let's for example suppose that array
is [1, 2, 3, 4]
So this line:
conj(mystery(rest(array)), first(array));
... has a recursive call of mystery
. It gets as argument the array
, but with the first element removed from it (that is what rest
returns), so it gets [2, 3, 4]
Now we will just assume that this recursive call of mystery
does the right thing and reverses that array into [4, 3, 2]
. Then in the above quoted code we see this result is concatenated with first(array)
(which is the first value, i.e. 1). So we get [4, 3, 2, 1]
. Correct!
This teaches us that if we assume mystery
does the job right for an array with n-1 values, it also does it right for n values.
Now remains to see whether mystery
deals correctly with the smallest case, i.e. when the array is empty. It is easy to see it returns the correct result in that case, i.e. an empty array.
So putting those two things together you can see that mystery
does the job correctly for all possible array sizes.
answered Nov 10 at 21:15
trincot
113k1477109
113k1477109
add a comment |
add a comment |
up vote
1
down vote
The .push
method places the item to end of array..slice(1)
means “except the first item”
Pseudocode
- Get array A (arg of
mystery
). If it is empty, return it - Take
rest
(everything except the first). We will call the rest B - Run this program on B (recursively)
- Append the first item of A to end of B
conj
= append value
to arr
first
= get first item of arr
rest
= return everything except the first item
mystery
when array
is empty = return empty array
mystery
when array
is not empty = Take rest(array)
, run mystery
on it, then append first
of array
add a comment |
up vote
1
down vote
The .push
method places the item to end of array..slice(1)
means “except the first item”
Pseudocode
- Get array A (arg of
mystery
). If it is empty, return it - Take
rest
(everything except the first). We will call the rest B - Run this program on B (recursively)
- Append the first item of A to end of B
conj
= append value
to arr
first
= get first item of arr
rest
= return everything except the first item
mystery
when array
is empty = return empty array
mystery
when array
is not empty = Take rest(array)
, run mystery
on it, then append first
of array
add a comment |
up vote
1
down vote
up vote
1
down vote
The .push
method places the item to end of array..slice(1)
means “except the first item”
Pseudocode
- Get array A (arg of
mystery
). If it is empty, return it - Take
rest
(everything except the first). We will call the rest B - Run this program on B (recursively)
- Append the first item of A to end of B
conj
= append value
to arr
first
= get first item of arr
rest
= return everything except the first item
mystery
when array
is empty = return empty array
mystery
when array
is not empty = Take rest(array)
, run mystery
on it, then append first
of array
The .push
method places the item to end of array..slice(1)
means “except the first item”
Pseudocode
- Get array A (arg of
mystery
). If it is empty, return it - Take
rest
(everything except the first). We will call the rest B - Run this program on B (recursively)
- Append the first item of A to end of B
conj
= append value
to arr
first
= get first item of arr
rest
= return everything except the first item
mystery
when array
is empty = return empty array
mystery
when array
is not empty = Take rest(array)
, run mystery
on it, then append first
of array
answered Nov 10 at 21:16
jiwopene
1688
1688
add a comment |
add a comment |
up vote
1
down vote
Yeah, the magic of recursion. To understand think about what it does if you call mystery
with a 2-element array [1,2]
.
rest(array)
will then be [2]
mystery(rest(array))
will also be [2]
first(array)
will be 1
.
Then you return conj([2], 1)
which locically results in [2,1]
.
Now the trick is the recursion. If you have 3 elements [0,1,2]
and call mystery with it this will happen:
- it will call
mystery(rest(array))
with essentially ismystery([1,2])
. That this returns[2,1]
have we already seen. first(array)
will be0
- so it returns
conj([2,1],0)
which is logically[2,1,0]
.
this now recusivly works for as many elements as you wish. Essentially mystery
will be called for every element to place it after all elements.
add a comment |
up vote
1
down vote
Yeah, the magic of recursion. To understand think about what it does if you call mystery
with a 2-element array [1,2]
.
rest(array)
will then be [2]
mystery(rest(array))
will also be [2]
first(array)
will be 1
.
Then you return conj([2], 1)
which locically results in [2,1]
.
Now the trick is the recursion. If you have 3 elements [0,1,2]
and call mystery with it this will happen:
- it will call
mystery(rest(array))
with essentially ismystery([1,2])
. That this returns[2,1]
have we already seen. first(array)
will be0
- so it returns
conj([2,1],0)
which is logically[2,1,0]
.
this now recusivly works for as many elements as you wish. Essentially mystery
will be called for every element to place it after all elements.
add a comment |
up vote
1
down vote
up vote
1
down vote
Yeah, the magic of recursion. To understand think about what it does if you call mystery
with a 2-element array [1,2]
.
rest(array)
will then be [2]
mystery(rest(array))
will also be [2]
first(array)
will be 1
.
Then you return conj([2], 1)
which locically results in [2,1]
.
Now the trick is the recursion. If you have 3 elements [0,1,2]
and call mystery with it this will happen:
- it will call
mystery(rest(array))
with essentially ismystery([1,2])
. That this returns[2,1]
have we already seen. first(array)
will be0
- so it returns
conj([2,1],0)
which is logically[2,1,0]
.
this now recusivly works for as many elements as you wish. Essentially mystery
will be called for every element to place it after all elements.
Yeah, the magic of recursion. To understand think about what it does if you call mystery
with a 2-element array [1,2]
.
rest(array)
will then be [2]
mystery(rest(array))
will also be [2]
first(array)
will be 1
.
Then you return conj([2], 1)
which locically results in [2,1]
.
Now the trick is the recursion. If you have 3 elements [0,1,2]
and call mystery with it this will happen:
- it will call
mystery(rest(array))
with essentially ismystery([1,2])
. That this returns[2,1]
have we already seen. first(array)
will be0
- so it returns
conj([2,1],0)
which is logically[2,1,0]
.
this now recusivly works for as many elements as you wish. Essentially mystery
will be called for every element to place it after all elements.
answered Nov 10 at 21:16
Lux
10.5k32658
10.5k32658
add a comment |
add a comment |
up vote
1
down vote
One thing to note here is that mystery()
is called recursively. I added some comments to illustrate what's going on on each step.
function rest(arr)
console.log('take first n-1 elements of', arr);
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
console.log('combine', arr, 'with', value)
return arr;
function mystery(array)
console.log('call mystery on ', array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
mystery([0,1,2])
add a comment |
up vote
1
down vote
One thing to note here is that mystery()
is called recursively. I added some comments to illustrate what's going on on each step.
function rest(arr)
console.log('take first n-1 elements of', arr);
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
console.log('combine', arr, 'with', value)
return arr;
function mystery(array)
console.log('call mystery on ', array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
mystery([0,1,2])
add a comment |
up vote
1
down vote
up vote
1
down vote
One thing to note here is that mystery()
is called recursively. I added some comments to illustrate what's going on on each step.
function rest(arr)
console.log('take first n-1 elements of', arr);
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
console.log('combine', arr, 'with', value)
return arr;
function mystery(array)
console.log('call mystery on ', array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
mystery([0,1,2])
One thing to note here is that mystery()
is called recursively. I added some comments to illustrate what's going on on each step.
function rest(arr)
console.log('take first n-1 elements of', arr);
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
console.log('combine', arr, 'with', value)
return arr;
function mystery(array)
console.log('call mystery on ', array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
mystery([0,1,2])
function rest(arr)
console.log('take first n-1 elements of', arr);
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
console.log('combine', arr, 'with', value)
return arr;
function mystery(array)
console.log('call mystery on ', array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
mystery([0,1,2])
function rest(arr)
console.log('take first n-1 elements of', arr);
return arr.slice(1);
function first(arr)
return arr[0];
function conj(arr, value)
arr.push(value);
console.log('combine', arr, 'with', value)
return arr;
function mystery(array)
console.log('call mystery on ', array)
if (array.length === 0)
return ;
return conj(mystery(rest(array)), first(array));
mystery([0,1,2])
answered Nov 10 at 21:22
shkaper
603311
603311
add a comment |
add a comment |
up vote
1
down vote
It's pretty simple. The reason for you not being able to see it is RECURSION. Couple things, that I would like you to notice:
- mystery(array) is recursive function because it call itself until the array passed in is emplty
- all of your work is happening here: return conj(mystery(rest(array)), first(array));
I'm not going to talk more about recursive function here rather I will show how you can track each recursive call using console.log(). Check out my code below, I've added console.log() to make things more clear for you. Try running mystery with some array and see results. This will make sense to you.
function rest(arr)
return arr.slice(1);
function first(arr)
console.log("Returning ",arr[0], "from first(arr).")
return arr[0];
function conj(arr, value)
console.log("Pushing ",value, " to ",arr, " in conj(arr,value)");
arr.push(value);
console.log("Returning ",arr, " from Conj(arr,value)");
return arr;
function mystery(array)
if (array.length === 0)
console.log("array is emplty. So, returning empty array from mystery");
return ;
console.log("array is not emplty. So, calling mystery(array) again.");
return conj(mystery(rest(array)), first(array));
var reverse =mystery([1,2,3,4]);
console.log("The final result", reverse);
add a comment |
up vote
1
down vote
It's pretty simple. The reason for you not being able to see it is RECURSION. Couple things, that I would like you to notice:
- mystery(array) is recursive function because it call itself until the array passed in is emplty
- all of your work is happening here: return conj(mystery(rest(array)), first(array));
I'm not going to talk more about recursive function here rather I will show how you can track each recursive call using console.log(). Check out my code below, I've added console.log() to make things more clear for you. Try running mystery with some array and see results. This will make sense to you.
function rest(arr)
return arr.slice(1);
function first(arr)
console.log("Returning ",arr[0], "from first(arr).")
return arr[0];
function conj(arr, value)
console.log("Pushing ",value, " to ",arr, " in conj(arr,value)");
arr.push(value);
console.log("Returning ",arr, " from Conj(arr,value)");
return arr;
function mystery(array)
if (array.length === 0)
console.log("array is emplty. So, returning empty array from mystery");
return ;
console.log("array is not emplty. So, calling mystery(array) again.");
return conj(mystery(rest(array)), first(array));
var reverse =mystery([1,2,3,4]);
console.log("The final result", reverse);
add a comment |
up vote
1
down vote
up vote
1
down vote
It's pretty simple. The reason for you not being able to see it is RECURSION. Couple things, that I would like you to notice:
- mystery(array) is recursive function because it call itself until the array passed in is emplty
- all of your work is happening here: return conj(mystery(rest(array)), first(array));
I'm not going to talk more about recursive function here rather I will show how you can track each recursive call using console.log(). Check out my code below, I've added console.log() to make things more clear for you. Try running mystery with some array and see results. This will make sense to you.
function rest(arr)
return arr.slice(1);
function first(arr)
console.log("Returning ",arr[0], "from first(arr).")
return arr[0];
function conj(arr, value)
console.log("Pushing ",value, " to ",arr, " in conj(arr,value)");
arr.push(value);
console.log("Returning ",arr, " from Conj(arr,value)");
return arr;
function mystery(array)
if (array.length === 0)
console.log("array is emplty. So, returning empty array from mystery");
return ;
console.log("array is not emplty. So, calling mystery(array) again.");
return conj(mystery(rest(array)), first(array));
var reverse =mystery([1,2,3,4]);
console.log("The final result", reverse);
It's pretty simple. The reason for you not being able to see it is RECURSION. Couple things, that I would like you to notice:
- mystery(array) is recursive function because it call itself until the array passed in is emplty
- all of your work is happening here: return conj(mystery(rest(array)), first(array));
I'm not going to talk more about recursive function here rather I will show how you can track each recursive call using console.log(). Check out my code below, I've added console.log() to make things more clear for you. Try running mystery with some array and see results. This will make sense to you.
function rest(arr)
return arr.slice(1);
function first(arr)
console.log("Returning ",arr[0], "from first(arr).")
return arr[0];
function conj(arr, value)
console.log("Pushing ",value, " to ",arr, " in conj(arr,value)");
arr.push(value);
console.log("Returning ",arr, " from Conj(arr,value)");
return arr;
function mystery(array)
if (array.length === 0)
console.log("array is emplty. So, returning empty array from mystery");
return ;
console.log("array is not emplty. So, calling mystery(array) again.");
return conj(mystery(rest(array)), first(array));
var reverse =mystery([1,2,3,4]);
console.log("The final result", reverse);
function rest(arr)
return arr.slice(1);
function first(arr)
console.log("Returning ",arr[0], "from first(arr).")
return arr[0];
function conj(arr, value)
console.log("Pushing ",value, " to ",arr, " in conj(arr,value)");
arr.push(value);
console.log("Returning ",arr, " from Conj(arr,value)");
return arr;
function mystery(array)
if (array.length === 0)
console.log("array is emplty. So, returning empty array from mystery");
return ;
console.log("array is not emplty. So, calling mystery(array) again.");
return conj(mystery(rest(array)), first(array));
var reverse =mystery([1,2,3,4]);
console.log("The final result", reverse);
function rest(arr)
return arr.slice(1);
function first(arr)
console.log("Returning ",arr[0], "from first(arr).")
return arr[0];
function conj(arr, value)
console.log("Pushing ",value, " to ",arr, " in conj(arr,value)");
arr.push(value);
console.log("Returning ",arr, " from Conj(arr,value)");
return arr;
function mystery(array)
if (array.length === 0)
console.log("array is emplty. So, returning empty array from mystery");
return ;
console.log("array is not emplty. So, calling mystery(array) again.");
return conj(mystery(rest(array)), first(array));
var reverse =mystery([1,2,3,4]);
console.log("The final result", reverse);
answered Nov 10 at 22:11
b1k
737
737
add a comment |
add a comment |
up vote
0
down vote
Not gonna lie, this is a weird way to reverse the order of the array. It's basically using recursion to slice the first element of the array and conjoin the elements with first element at the end.
So here's a walk though:
mystery(['h','e','l','l','o'])
->
first check if array is empty then method rest
is called ->
rest(['h','e','l','l','o'])
->
rest slices the array at index 1 returning a new array of ['e','l','l','o']
->
then mystery
is called which repeats the previous steps with the return value of rest
->
mystery(['e','l','l','o'])
->
this means the mystery will be keep getting called on until rest
returns an empty array ->
mystery(['e','l','l','o'])
->mystery(['l','l','o'])
->mystery(['l','o'])
->mystery(['o'])
->mystery()
->
when mystery has an empty array it returns the empty array and then first
is called ->
first returns just the first element of the array and happens after mystery is returned it would look something like this before mystery is returned ->
mystery(['e','l','l','o']) , first(['h','e','l','l','o'])
->
so at the lowest level when mystery returns an empty array and first return the first element in ['o']
it would look like this ->
, ['o']
->conj
would be called with these values ->conj( , ['o'])
->conj
returns the combination of the two values ->conj( , ['o'])
->['o']
->
this is then returned and repeated ->conj( , ['o'])
->conj(['o'] , ['l'])
->conj(['o','l'] , ['l'])
->conj(['o', 'l','l'] , ['e'])
->conj(['o', 'l','l', 'e'] , ['h'])
->['o', 'l','l', 'e','h']
add a comment |
up vote
0
down vote
Not gonna lie, this is a weird way to reverse the order of the array. It's basically using recursion to slice the first element of the array and conjoin the elements with first element at the end.
So here's a walk though:
mystery(['h','e','l','l','o'])
->
first check if array is empty then method rest
is called ->
rest(['h','e','l','l','o'])
->
rest slices the array at index 1 returning a new array of ['e','l','l','o']
->
then mystery
is called which repeats the previous steps with the return value of rest
->
mystery(['e','l','l','o'])
->
this means the mystery will be keep getting called on until rest
returns an empty array ->
mystery(['e','l','l','o'])
->mystery(['l','l','o'])
->mystery(['l','o'])
->mystery(['o'])
->mystery()
->
when mystery has an empty array it returns the empty array and then first
is called ->
first returns just the first element of the array and happens after mystery is returned it would look something like this before mystery is returned ->
mystery(['e','l','l','o']) , first(['h','e','l','l','o'])
->
so at the lowest level when mystery returns an empty array and first return the first element in ['o']
it would look like this ->
, ['o']
->conj
would be called with these values ->conj( , ['o'])
->conj
returns the combination of the two values ->conj( , ['o'])
->['o']
->
this is then returned and repeated ->conj( , ['o'])
->conj(['o'] , ['l'])
->conj(['o','l'] , ['l'])
->conj(['o', 'l','l'] , ['e'])
->conj(['o', 'l','l', 'e'] , ['h'])
->['o', 'l','l', 'e','h']
add a comment |
up vote
0
down vote
up vote
0
down vote
Not gonna lie, this is a weird way to reverse the order of the array. It's basically using recursion to slice the first element of the array and conjoin the elements with first element at the end.
So here's a walk though:
mystery(['h','e','l','l','o'])
->
first check if array is empty then method rest
is called ->
rest(['h','e','l','l','o'])
->
rest slices the array at index 1 returning a new array of ['e','l','l','o']
->
then mystery
is called which repeats the previous steps with the return value of rest
->
mystery(['e','l','l','o'])
->
this means the mystery will be keep getting called on until rest
returns an empty array ->
mystery(['e','l','l','o'])
->mystery(['l','l','o'])
->mystery(['l','o'])
->mystery(['o'])
->mystery()
->
when mystery has an empty array it returns the empty array and then first
is called ->
first returns just the first element of the array and happens after mystery is returned it would look something like this before mystery is returned ->
mystery(['e','l','l','o']) , first(['h','e','l','l','o'])
->
so at the lowest level when mystery returns an empty array and first return the first element in ['o']
it would look like this ->
, ['o']
->conj
would be called with these values ->conj( , ['o'])
->conj
returns the combination of the two values ->conj( , ['o'])
->['o']
->
this is then returned and repeated ->conj( , ['o'])
->conj(['o'] , ['l'])
->conj(['o','l'] , ['l'])
->conj(['o', 'l','l'] , ['e'])
->conj(['o', 'l','l', 'e'] , ['h'])
->['o', 'l','l', 'e','h']
Not gonna lie, this is a weird way to reverse the order of the array. It's basically using recursion to slice the first element of the array and conjoin the elements with first element at the end.
So here's a walk though:
mystery(['h','e','l','l','o'])
->
first check if array is empty then method rest
is called ->
rest(['h','e','l','l','o'])
->
rest slices the array at index 1 returning a new array of ['e','l','l','o']
->
then mystery
is called which repeats the previous steps with the return value of rest
->
mystery(['e','l','l','o'])
->
this means the mystery will be keep getting called on until rest
returns an empty array ->
mystery(['e','l','l','o'])
->mystery(['l','l','o'])
->mystery(['l','o'])
->mystery(['o'])
->mystery()
->
when mystery has an empty array it returns the empty array and then first
is called ->
first returns just the first element of the array and happens after mystery is returned it would look something like this before mystery is returned ->
mystery(['e','l','l','o']) , first(['h','e','l','l','o'])
->
so at the lowest level when mystery returns an empty array and first return the first element in ['o']
it would look like this ->
, ['o']
->conj
would be called with these values ->conj( , ['o'])
->conj
returns the combination of the two values ->conj( , ['o'])
->['o']
->
this is then returned and repeated ->conj( , ['o'])
->conj(['o'] , ['l'])
->conj(['o','l'] , ['l'])
->conj(['o', 'l','l'] , ['e'])
->conj(['o', 'l','l', 'e'] , ['h'])
->['o', 'l','l', 'e','h']
answered Nov 10 at 21:26
Thatalent
3268
3268
add a comment |
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Try to actually run it, use the debugger to trace through it, line by line...
– obe
Nov 10 at 21:14