How to analyze time complexity here? - java

Assume you are playing the following Flip Game with your friend: Given a string that contains only these two characters: + and -, you and your friend take turns to flip two consecutive "++" into "--". The game ends when a person can no longer make a move and therefore the other person will be the winner.
Write a function to determine if the starting player can guarantee a win.
For example, given s = "++++", return true. The starting player can guarantee a win by flipping the middle "++" to become "+--+".
Here is my code:
public boolean canWin(String s) {
if(s==null || s.length()<2) return false;
char[] arr=s.toCharArray();
return canWinHelper(arr);
}
public boolean canWinHelper(char[] arr){
for(int i=0; i<arr.length-1; i++){
if(arr[i]=='+' && arr[i+1]=='+'){
arr[i]='-';
arr[i+1]='-';
boolean win=!canWinHelper(arr);
arr[i]='+';
arr[i+1]='+';
if(win) return true;
}
}
return false;
}
It works, but I'm not sure how to calculate the time complexity here since the function will keep calling itself until a false is returned. Anyone share some idea here?
Also during the search, we will encounter duplicate computation, so I think I can use a hashmap to avoid those duplicates. Key: String, Value: Boolean.
My updated code using a hashmap:
public boolean canWin(String s){
if(s==null || s.length()<2) return false;
HashMap<String,Boolean> map=new HashMap<String,Boolean>();
return helper(s,map);
}
public boolean helper(String s, HashMap<String,Boolean> map){
if(map.containsKey(s)) return map.get(s);
for(int i=0; i<s.length()-1; i++){
if(s.charAt(i)=='+' && s.charAt(i+1)=='+'){
String fliped=s.substring(0,i)+"--"+s.substring(i+2);
if(!helper(fliped,map)){
map.put(s,true);
return true;
}
}
}
map.put(s,false);
return false;
}
Still, I wanna know how to analyze the time and space complexity here?

Take that n = arr.length - 1
First pass you have n recursive calls. For each you have removed two +'s so each will have at most n-2 recursive calls, and so on.
So you have at most n+n(n-2)+n(n-2)(n-4)+... recursive calls.
In essence this is n!!(1+1/2+1/(2*4)+1/(2*4*8)+...) Since 1+1/2+1/(2*4)+1/(2*4*8)+... is convergent, ≤2, you have O(n!!)
Regarding memory, you have an array of length n for each recursive call, so you have n + nn + nnn + n ... (n/2 times) ... *n = n(n^(n/2)-1)/(n-1) and this is O(n^(n/2))
This is obviously pointing to not much better performance than with an exhaustive search.
For the hashed improvement, you are asking for all possible combinations that you have managed to create with your code. However, your code is not much different than the code that would actually create all combinations, apart from the fact that you are replacing two +'s with two -'s, which is reducing the complexity by some factor but not the level of it. Overall, the worst case scenario is the same as with the number of combinations of bits among n/2 locations which is 2^(n/2). Observe that hash function itself has probably some hidden log so the total complexity would be for search O(2^(n/2)*ln(n/2)) and memory O(2^(n/2)).
This is the worst case scenario. However, if there are arrangements where you cannot win, when there is no winning strategy, this above is really the complexity you need to count on.
The question of the average scenario is then the question of the number of cases where you can/cannot win and their distribution among all arrangements. This question has not much to do with your algorithm and requires a totally different set of tools in order to be solved.
After a few moments of checking whether the above reasoning is correct and to the point or not, I would be quite happy with the result, since it is telling me all that I need to know. You cannot expect that you will have an arrangement that will be favorable, and I really doubt that you have like only 0.01% of worst case arrangements, so you need to prepare the worst case scenario anyway and unless this is some special project the back-of-the-envelope calculation is your friend.
Anyway, these type of calculations are there to have test cases correctly prepared, not to have a correct and final implementation. Using the tests you can find what the hidden factors in O() really are, taking into account the compiler, memory consumption, pagination and so on.
Still not to leave this as it is, we can always improve the back-of-the-envelope reasoning, of course. For example, you actually do not have n-2 at each step, because it depends on the parity. For example for ++++++++... if you replace third +++--+++++... it is obvious that you are going to have n-3, not n-2 recursive calls, or even n-4. So the half number of calls may have n-3 recursive calls which would be n/2(n-3)+n/2(n-2)=n(n-5/2)
Observe that since n!=n!!(n-1)!! we can take n!!≈√n!, again n!=n!!!(n-1)!!!(n-2)!!! or n!!!≈∛n! This might lead to a conclusion that we should have something like O((n!)^(5/2)). The testing would tell me how much we can reduce x=3 in O((n!)^(x)).
(It is quite normal to look for the complexity in one particular form just like we have O((n!)^(x)), although it can be expressed differently. So I would continue with the complexity form O((n!)^(x)),1≤x≤3)

Related

Recursion - Identifying potential for memoization

I am trying to solve this problem I found on leetcode:
https://leetcode.com/problems/house-robber
We need to select a non adjacent combination with the largest sum and return it. Here is my attempt at solving this recursively
class Solution {
int[] arr;
int ans = 0;
private void helper(int i,boolean robprev,int curr){
/* i denotes index of array, robprev denotes whether previous house was selected,
curr denotes the current amount robbed */
ans = Math.max(ans,curr);
if(i>=arr.length)return;
if(!robprev){
helper(i+1,true,curr+arr[i]);
/* selected current house, move to next index, mark robprevious
as true and increment current amount */
}
helper(i+1,false,curr);
/* not selecting the current house, move next index,
current amount remains unchanged */
}
public int rob(int[] nums) {
arr = nums;
helper(0,false,0);
return ans;
}
}
As expected, I got a time limit exceeded error and my next thought was memoization, but I am unable to find where my algorithm is doing repeated word, as I am moving forward by one index each call.
Can someone suggest me how do I incorporate memoization in this approach ?
Yes, you're moving one index forward per call, but you're spawning two recursive calls, so that's a very high branching factor. Exponential O(2^n) to be precise.
The subproblems we're trying to memoize and avoid recomputation have to do with an index i. If we can compute the optimal robbery decisions for the subarray after i, then there's no need to revisit that.
The reason you may be struggling to memoize is that you're passing your result down the call stack in a parameter rather than up as a return value, making it difficult to determine which parameters constitute a key in the memoization lookup table. This table represents the solution to the subproblem starting at a particular index. An important rule of thumb for recursion, and especially for DP, is to pass data down and results up.
Additionally, robprev is unnecessary and frustrates memoization. You can simply skip an index when traversing when you're choosing not to rob a house. This makes your recursive calls look like i + 1 and i + 2 with respect to the indices.
Try this method header:
private int helper(int[] nums, int i, HashMap<Integer, Integer> memo);
First rewrite it naively expoentially with this header that returns an int, but ignore the memo parameter and make sure it works. Once you've ensured correctness, store each result before your return in the memo table and add a lookup to see if a parameter set is available before executing the recursive computation for that i.
If you're still stuck after that, see this answer which provides the solution in Python and should be easily translatable to Java. OP in that thread was also struggling to memoize House Robber because they were passing results down the call tree rather than up.
Using a bottom-up approach with a 2d array would be a good exercise once you have memoization. The array length on LeetCode is only 100 elements, but if it were 10k or so, a linear recursive algorithm will fail.

Correct runtime analysis on a Bubblesort-algorithm

Hey I did a runtime analysis on Bubblesort and I wanted to ask you if there were any mistakes since I was not sure at a certain point
heres an extract of the algorithm:
boolean sorted = false;
while(!sorted)
{
int a = 0;
int b = 1;
sorted = true;
while(a < sortArray.length && b < sortArray.length)
{
if(sortArray[a].getWertigkeit() < sortArray[b].getWertigkeit())
{
Karte tmp1 = sortArray[a];
Karte tmp2 = sortArray[b];
sortArray[a] = tmp2;
sortArray[b] = tmp1;
sorted = false;
}
a++;
b++;
}
}
So the problem I got is in the first while-loop, I solved it as following:
Best Case: In the best Case the sorted never gets set back to the false (through the if(..){..}) so it only goes once through the loop;
So the runtime is, if I am not wrong, 2*2n*1 = 4n => O(n) for the best Case;
Worst Case:In the worst Case the variable sorted gets set on false everytime the loop starts, as far as I know, so it needs another "n" comparisons so the runtime should be: n*2n*1 = 2n^2 => O(n^2)
I am really not sure if my thoughts about the while(!sorted) are correct or if the runtime makes any sense (since the big o notation seems fine, but im not sure about the precise runtime)
I really hope that my problem is relatable and I look forward to hear from you.
Thx already
Your analysis of the best-case runtime of O(n) is correct. Nice job!
For the worst-case, you need to be a little bit more precise. You're right that every time the flag gets reset you have to make another pass over the array to get it more sorted than before, so you know it's going to be O(n) times the number of times the loop runs. What you haven't yet done in your analysis is talk about how many times that loop can run before everything is going to end up sorted.
One observation you can make about bubble sort is that after the first pass over the array, the largest element is guaranteed to be in the last position - can you explain why? After the second pass, the second-largest element is guaranteed to be in the second-to-last position - again, can you explain why? Based on this pattern, can you argue why the number of times the outer loop runs is at most O(n)?

Behind the scenes of recursion? [duplicate]

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
One of the topics that seems to come up regularly on mailing lists and online discussions is the merits (or lack thereof) of doing a Computer Science Degree. An argument that seems to come up time and again for the negative party is that they have been coding for some number of years and they have never used recursion.
So the question is:
What is recursion?
When would I use recursion?
Why don't people use recursion?
There are a number of good explanations of recursion in this thread, this answer is about why you shouldn't use it in most languages.* In the majority of major imperative language implementations (i.e. every major implementation of C, C++, Basic, Python, Ruby,Java, and C#) iteration is vastly preferable to recursion.
To see why, walk through the steps that the above languages use to call a function:
space is carved out on the stack for the function's arguments and local variables
the function's arguments are copied into this new space
control jumps to the function
the function's code runs
the function's result is copied into a return value
the stack is rewound to its previous position
control jumps back to where the function was called
Doing all of these steps takes time, usually a little bit more than it takes to iterate through a loop. However, the real problem is in step #1. When many programs start, they allocate a single chunk of memory for their stack, and when they run out of that memory (often, but not always due to recursion), the program crashes due to a stack overflow.
So in these languages recursion is slower and it makes you vulnerable to crashing. There are still some arguments for using it though. In general, code written recursively is shorter and a bit more elegant, once you know how to read it.
There is a technique that language implementers can use called tail call optimization which can eliminate some classes of stack overflow. Put succinctly: if a function's return expression is simply the result of a function call, then you don't need to add a new level onto the stack, you can reuse the current one for the function being called. Regrettably, few imperative language-implementations have tail-call optimization built in.
* I love recursion. My favorite static language doesn't use loops at all, recursion is the only way to do something repeatedly. I just don't think that recursion is generally a good idea in languages that aren't tuned for it.
** By the way Mario, the typical name for your ArrangeString function is "join", and I'd be surprised if your language of choice doesn't already have an implementation of it.
Simple english example of recursion.
A child couldn't sleep, so her mother told her a story about a little frog,
who couldn't sleep, so the frog's mother told her a story about a little bear,
who couldn't sleep, so the bear's mother told her a story about a little weasel...
who fell asleep.
...and the little bear fell asleep;
...and the little frog fell asleep;
...and the child fell asleep.
In the most basic computer science sense, recursion is a function that calls itself. Say you have a linked list structure:
struct Node {
Node* next;
};
And you want to find out how long a linked list is you can do this with recursion:
int length(const Node* list) {
if (!list->next) {
return 1;
} else {
return 1 + length(list->next);
}
}
(This could of course be done with a for loop as well, but is useful as an illustration of the concept)
Whenever a function calls itself, creating a loop, then that's recursion. As with anything there are good uses and bad uses for recursion.
The most simple example is tail recursion where the very last line of the function is a call to itself:
int FloorByTen(int num)
{
if (num % 10 == 0)
return num;
else
return FloorByTen(num-1);
}
However, this is a lame, almost pointless example because it can easily be replaced by more efficient iteration. After all, recursion suffers from function call overhead, which in the example above could be substantial compared to the operation inside the function itself.
So the whole reason to do recursion rather than iteration should be to take advantage of the call stack to do some clever stuff. For example, if you call a function multiple times with different parameters inside the same loop then that's a way to accomplish branching. A classic example is the Sierpinski triangle.
You can draw one of those very simply with recursion, where the call stack branches in 3 directions:
private void BuildVertices(double x, double y, double len)
{
if (len > 0.002)
{
mesh.Positions.Add(new Point3D(x, y + len, -len));
mesh.Positions.Add(new Point3D(x - len, y - len, -len));
mesh.Positions.Add(new Point3D(x + len, y - len, -len));
len *= 0.5;
BuildVertices(x, y + len, len);
BuildVertices(x - len, y - len, len);
BuildVertices(x + len, y - len, len);
}
}
If you attempt to do the same thing with iteration I think you'll find it takes a lot more code to accomplish.
Other common use cases might include traversing hierarchies, e.g. website crawlers, directory comparisons, etc.
Conclusion
In practical terms, recursion makes the most sense whenever you need iterative branching.
Recursion is a method of solving problems based on the divide and conquer mentality.
The basic idea is that you take the original problem and divide it into smaller (more easily solved) instances of itself, solve those smaller instances (usually by using the same algorithm again) and then reassemble them into the final solution.
The canonical example is a routine to generate the Factorial of n. The Factorial of n is calculated by multiplying all of the numbers between 1 and n. An iterative solution in C# looks like this:
public int Fact(int n)
{
int fact = 1;
for( int i = 2; i <= n; i++)
{
fact = fact * i;
}
return fact;
}
There's nothing surprising about the iterative solution and it should make sense to anyone familiar with C#.
The recursive solution is found by recognising that the nth Factorial is n * Fact(n-1). Or to put it another way, if you know what a particular Factorial number is you can calculate the next one. Here is the recursive solution in C#:
public int FactRec(int n)
{
if( n < 2 )
{
return 1;
}
return n * FactRec( n - 1 );
}
The first part of this function is known as a Base Case (or sometimes Guard Clause) and is what prevents the algorithm from running forever. It just returns the value 1 whenever the function is called with a value of 1 or less. The second part is more interesting and is known as the Recursive Step. Here we call the same method with a slightly modified parameter (we decrement it by 1) and then multiply the result with our copy of n.
When first encountered this can be kind of confusing so it's instructive to examine how it works when run. Imagine that we call FactRec(5). We enter the routine, are not picked up by the base case and so we end up like this:
// In FactRec(5)
return 5 * FactRec( 5 - 1 );
// which is
return 5 * FactRec(4);
If we re-enter the method with the parameter 4 we are again not stopped by the guard clause and so we end up at:
// In FactRec(4)
return 4 * FactRec(3);
If we substitute this return value into the return value above we get
// In FactRec(5)
return 5 * (4 * FactRec(3));
This should give you a clue as to how the final solution is arrived at so we'll fast track and show each step on the way down:
return 5 * (4 * FactRec(3));
return 5 * (4 * (3 * FactRec(2)));
return 5 * (4 * (3 * (2 * FactRec(1))));
return 5 * (4 * (3 * (2 * (1))));
That final substitution happens when the base case is triggered. At this point we have a simple algrebraic formula to solve which equates directly to the definition of Factorials in the first place.
It's instructive to note that every call into the method results in either a base case being triggered or a call to the same method where the parameters are closer to a base case (often called a recursive call). If this is not the case then the method will run forever.
Recursion is solving a problem with a function that calls itself. A good example of this is a factorial function. Factorial is a math problem where factorial of 5, for example, is 5 * 4 * 3 * 2 * 1. This function solves this in C# for positive integers (not tested - there may be a bug).
public int Factorial(int n)
{
if (n <= 1)
return 1;
return n * Factorial(n - 1);
}
Recursion refers to a method which solves a problem by solving a smaller version of the problem and then using that result plus some other computation to formulate the answer to the original problem. Often times, in the process of solving the smaller version, the method will solve a yet smaller version of the problem, and so on, until it reaches a "base case" which is trivial to solve.
For instance, to calculate a factorial for the number X, one can represent it as X times the factorial of X-1. Thus, the method "recurses" to find the factorial of X-1, and then multiplies whatever it got by X to give a final answer. Of course, to find the factorial of X-1, it'll first calculate the factorial of X-2, and so on. The base case would be when X is 0 or 1, in which case it knows to return 1 since 0! = 1! = 1.
Consider an old, well known problem:
In mathematics, the greatest common divisor (gcd) … of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.
The definition of gcd is surprisingly simple:
where mod is the modulo operator (that is, the remainder after integer division).
In English, this definition says the greatest common divisor of any number and zero is that number, and the greatest common divisor of two numbers m and n is the greatest common divisor of n and the remainder after dividing m by n.
If you'd like to know why this works, see the Wikipedia article on the Euclidean algorithm.
Let's compute gcd(10, 8) as an example. Each step is equal to the one just before it:
gcd(10, 8)
gcd(10, 10 mod 8)
gcd(8, 2)
gcd(8, 8 mod 2)
gcd(2, 0)
2
In the first step, 8 does not equal zero, so the second part of the definition applies. 10 mod 8 = 2 because 8 goes into 10 once with a remainder of 2. At step 3, the second part applies again, but this time 8 mod 2 = 0 because 2 divides 8 with no remainder. At step 5, the second argument is 0, so the answer is 2.
Did you notice that gcd appears on both the left and right sides of the equals sign? A mathematician would say this definition is recursive because the expression you're defining recurs inside its definition.
Recursive definitions tend to be elegant. For example, a recursive definition for the sum of a list is
sum l =
if empty(l)
return 0
else
return head(l) + sum(tail(l))
where head is the first element in a list and tail is the rest of the list. Note that sum recurs inside its definition at the end.
Maybe you'd prefer the maximum value in a list instead:
max l =
if empty(l)
error
elsif length(l) = 1
return head(l)
else
tailmax = max(tail(l))
if head(l) > tailmax
return head(l)
else
return tailmax
You might define multiplication of non-negative integers recursively to turn it into a series of additions:
a * b =
if b = 0
return 0
else
return a + (a * (b - 1))
If that bit about transforming multiplication into a series of additions doesn't make sense, try expanding a few simple examples to see how it works.
Merge sort has a lovely recursive definition:
sort(l) =
if empty(l) or length(l) = 1
return l
else
(left,right) = split l
return merge(sort(left), sort(right))
Recursive definitions are all around if you know what to look for. Notice how all of these definitions have very simple base cases, e.g., gcd(m, 0) = m. The recursive cases whittle away at the problem to get down to the easy answers.
With this understanding, you can now appreciate the other algorithms in Wikipedia's article on recursion!
A function that calls itself
When a function can be (easily) decomposed into a simple operation plus the same function on some smaller portion of the problem. I should say, rather, that this makes it a good candidate for recursion.
They do!
The canonical example is the factorial which looks like:
int fact(int a)
{
if(a==1)
return 1;
return a*fact(a-1);
}
In general, recursion isn't necessarily fast (function call overhead tends to be high because recursive functions tend to be small, see above) and can suffer from some problems (stack overflow anyone?). Some say they tend to be hard to get 'right' in non-trivial cases but I don't really buy into that. In some situations, recursion makes the most sense and is the most elegant and clear way to write a particular function. It should be noted that some languages favor recursive solutions and optimize them much more (LISP comes to mind).
A recursive function is one which calls itself. The most common reason I've found to use it is traversing a tree structure. For example, if I have a TreeView with checkboxes (think installation of a new program, "choose features to install" page), I might want a "check all" button which would be something like this (pseudocode):
function cmdCheckAllClick {
checkRecursively(TreeView1.RootNode);
}
function checkRecursively(Node n) {
n.Checked = True;
foreach ( n.Children as child ) {
checkRecursively(child);
}
}
So you can see that the checkRecursively first checks the node which it is passed, then calls itself for each of that node's children.
You do need to be a bit careful with recursion. If you get into an infinite recursive loop, you will get a Stack Overflow exception :)
I can't think of a reason why people shouldn't use it, when appropriate. It is useful in some circumstances, and not in others.
I think that because it's an interesting technique, some coders perhaps end up using it more often than they should, without real justification. This has given recursion a bad name in some circles.
Recursion is an expression directly or indirectly referencing itself.
Consider recursive acronyms as a simple example:
GNU stands for GNU's Not Unix
PHP stands for PHP: Hypertext Preprocessor
YAML stands for YAML Ain't Markup Language
WINE stands for Wine Is Not an Emulator
VISA stands for Visa International Service Association
More examples on Wikipedia
Recursion works best with what I like to call "fractal problems", where you're dealing with a big thing that's made of smaller versions of that big thing, each of which is an even smaller version of the big thing, and so on. If you ever have to traverse or search through something like a tree or nested identical structures, you've got a problem that might be a good candidate for recursion.
People avoid recursion for a number of reasons:
Most people (myself included) cut their programming teeth on procedural or object-oriented programming as opposed to functional programming. To such people, the iterative approach (typically using loops) feels more natural.
Those of us who cut our programming teeth on procedural or object-oriented programming have often been told to avoid recursion because it's error prone.
We're often told that recursion is slow. Calling and returning from a routine repeatedly involves a lot of stack pushing and popping, which is slower than looping. I think some languages handle this better than others, and those languages are most likely not those where the dominant paradigm is procedural or object-oriented.
For at least a couple of programming languages I've used, I remember hearing recommendations not to use recursion if it gets beyond a certain depth because its stack isn't that deep.
A recursive statement is one in which you define the process of what to do next as a combination of the inputs and what you have already done.
For example, take factorial:
factorial(6) = 6*5*4*3*2*1
But it's easy to see factorial(6) also is:
6 * factorial(5) = 6*(5*4*3*2*1).
So generally:
factorial(n) = n*factorial(n-1)
Of course, the tricky thing about recursion is that if you want to define things in terms of what you have already done, there needs to be some place to start.
In this example, we just make a special case by defining factorial(1) = 1.
Now we see it from the bottom up:
factorial(6) = 6*factorial(5)
= 6*5*factorial(4)
= 6*5*4*factorial(3) = 6*5*4*3*factorial(2) = 6*5*4*3*2*factorial(1) = 6*5*4*3*2*1
Since we defined factorial(1) = 1, we reach the "bottom".
Generally speaking, recursive procedures have two parts:
1) The recursive part, which defines some procedure in terms of new inputs combined with what you've "already done" via the same procedure. (i.e. factorial(n) = n*factorial(n-1))
2) A base part, which makes sure that the process doesn't repeat forever by giving it some place to start (i.e. factorial(1) = 1)
It can be a bit confusing to get your head around at first, but just look at a bunch of examples and it should all come together. If you want a much deeper understanding of the concept, study mathematical induction. Also, be aware that some languages optimize for recursive calls while others do not. It's pretty easy to make insanely slow recursive functions if you're not careful, but there are also techniques to make them performant in most cases.
Hope this helps...
I like this definition:
In recursion, a routine solves a small part of a problem itself, divides the problem into smaller pieces, and then calls itself to solve each of the smaller pieces.
I also like Steve McConnells discussion of recursion in Code Complete where he criticises the examples used in Computer Science books on Recursion.
Don't use recursion for factorials or Fibonacci numbers
One problem with
computer-science textbooks is that
they present silly examples of
recursion. The typical examples are
computing a factorial or computing a
Fibonacci sequence. Recursion is a
powerful tool, and it's really dumb to
use it in either of those cases. If a
programmer who worked for me used
recursion to compute a factorial, I'd
hire someone else.
I thought this was a very interesting point to raise and may be a reason why recursion is often misunderstood.
EDIT:
This was not a dig at Dav's answer - I had not seen that reply when I posted this
1.)
A method is recursive if it can call itself; either directly:
void f() {
... f() ...
}
or indirectly:
void f() {
... g() ...
}
void g() {
... f() ...
}
2.) When to use recursion
Q: Does using recursion usually make your code faster?
A: No.
Q: Does using recursion usually use less memory?
A: No.
Q: Then why use recursion?
A: It sometimes makes your code much simpler!
3.) People use recursion only when it is very complex to write iterative code. For example, tree traversal techniques like preorder, postorder can be made both iterative and recursive. But usually we use recursive because of its simplicity.
Here's a simple example: how many elements in a set. (there are better ways to count things, but this is a nice simple recursive example.)
First, we need two rules:
if the set is empty, the count of items in the set is zero (duh!).
if the set is not empty, the count is one plus the number of items in the set after one item is removed.
Suppose you have a set like this: [x x x]. let's count how many items there are.
the set is [x x x] which is not empty, so we apply rule 2. the number of items is one plus the number of items in [x x] (i.e. we removed an item).
the set is [x x], so we apply rule 2 again: one + number of items in [x].
the set is [x], which still matches rule 2: one + number of items in [].
Now the set is [], which matches rule 1: the count is zero!
Now that we know the answer in step 4 (0), we can solve step 3 (1 + 0)
Likewise, now that we know the answer in step 3 (1), we can solve step 2 (1 + 1)
And finally now that we know the answer in step 2 (2), we can solve step 1 (1 + 2) and get the count of items in [x x x], which is 3. Hooray!
We can represent this as:
count of [x x x] = 1 + count of [x x]
= 1 + (1 + count of [x])
= 1 + (1 + (1 + count of []))
= 1 + (1 + (1 + 0)))
= 1 + (1 + (1))
= 1 + (2)
= 3
When applying a recursive solution, you usually have at least 2 rules:
the basis, the simple case which states what happens when you have "used up" all of your data. This is usually some variation of "if you are out of data to process, your answer is X"
the recursive rule, which states what happens if you still have data. This is usually some kind of rule that says "do something to make your data set smaller, and reapply your rules to the smaller data set."
If we translate the above to pseudocode, we get:
numberOfItems(set)
if set is empty
return 0
else
remove 1 item from set
return 1 + numberOfItems(set)
There's a lot more useful examples (traversing a tree, for example) which I'm sure other people will cover.
Well, that's a pretty decent definition you have. And wikipedia has a good definition too. So I'll add another (probably worse) definition for you.
When people refer to "recursion", they're usually talking about a function they've written which calls itself repeatedly until it is done with its work. Recursion can be helpful when traversing hierarchies in data structures.
An example: A recursive definition of a staircase is:
A staircase consists of:
- a single step and a staircase (recursion)
- or only a single step (termination)
To recurse on a solved problem: do nothing, you're done.
To recurse on an open problem: do the next step, then recurse on the rest.
In plain English:
Assume you can do 3 things:
Take one apple
Write down tally marks
Count tally marks
You have a lot of apples in front of you on a table and you want to know how many apples there are.
start
Is the table empty?
yes: Count the tally marks and cheer like it's your birthday!
no: Take 1 apple and put it aside
Write down a tally mark
goto start
The process of repeating the same thing till you are done is called recursion.
I hope this is the "plain english" answer you are looking for!
A recursive function is a function that contains a call to itself. A recursive struct is a struct that contains an instance of itself. You can combine the two as a recursive class. The key part of a recursive item is that it contains an instance/call of itself.
Consider two mirrors facing each other. We've seen the neat infinity effect they make. Each reflection is an instance of a mirror, which is contained within another instance of a mirror, etc. The mirror containing a reflection of itself is recursion.
A binary search tree is a good programming example of recursion. The structure is recursive with each Node containing 2 instances of a Node. Functions to work on a binary search tree are also recursive.
This is an old question, but I want to add an answer from logistical point of view (i.e not from algorithm correctness point of view or performance point of view).
I use Java for work, and Java doesn't support nested function. As such, if I want to do recursion, I might have to define an external function (which exists only because my code bumps against Java's bureaucratic rule), or I might have to refactor the code altogether (which I really hate to do).
Thus, I often avoid recursion, and use stack operation instead, because recursion itself is essentially a stack operation.
You want to use it anytime you have a tree structure. It is very useful in reading XML.
Recursion as it applies to programming is basically calling a function from inside its own definition (inside itself), with different parameters so as to accomplish a task.
"If I have a hammer, make everything look like a nail."
Recursion is a problem-solving strategy for huge problems, where at every step just, "turn 2 small things into one bigger thing," each time with the same hammer.
Example
Suppose your desk is covered with a disorganized mess of 1024 papers. How do you make one neat, clean stack of papers from the mess, using recursion?
Divide: Spread all the sheets out, so you have just one sheet in each "stack".
Conquer:
Go around, putting each sheet on top of one other sheet. You now have stacks of 2.
Go around, putting each 2-stack on top of another 2-stack. You now have stacks of 4.
Go around, putting each 4-stack on top of another 4-stack. You now have stacks of 8.
... on and on ...
You now have one huge stack of 1024 sheets!
Notice that this is pretty intuitive, aside from counting everything (which isn't strictly necessary). You might not go all the way down to 1-sheet stacks, in reality, but you could and it would still work. The important part is the hammer: With your arms, you can always put one stack on top of the other to make a bigger stack, and it doesn't matter (within reason) how big either stack is.
Recursion is the process where a method call iself to be able to perform a certain task. It reduces redundency of code. Most recurssive functions or methods must have a condifiton to break the recussive call i.e. stop it from calling itself if a condition is met - this prevents the creating of an infinite loop. Not all functions are suited to be used recursively.
hey, sorry if my opinion agrees with someone, I'm just trying to explain recursion in plain english.
suppose you have three managers - Jack, John and Morgan.
Jack manages 2 programmers, John - 3, and Morgan - 5.
you are going to give every manager 300$ and want to know what would it cost.
The answer is obvious - but what if 2 of Morgan-s employees are also managers?
HERE comes the recursion.
you start from the top of the hierarchy. the summery cost is 0$.
you start with Jack,
Then check if he has any managers as employees. if you find any of them are, check if they have any managers as employees and so on. Add 300$ to the summery cost every time you find a manager.
when you are finished with Jack, go to John, his employees and then to Morgan.
You'll never know, how much cycles will you go before getting an answer, though you know how many managers you have and how many Budget can you spend.
Recursion is a tree, with branches and leaves, called parents and children respectively.
When you use a recursion algorithm, you more or less consciously are building a tree from the data.
In plain English, recursion means to repeat someting again and again.
In programming one example is of calling the function within itself .
Look on the following example of calculating factorial of a number:
public int fact(int n)
{
if (n==0) return 1;
else return n*fact(n-1)
}
Any algorithm exhibits structural recursion on a datatype if basically consists of a switch-statement with a case for each case of the datatype.
for example, when you are working on a type
tree = null
| leaf(value:integer)
| node(left: tree, right:tree)
a structural recursive algorithm would have the form
function computeSomething(x : tree) =
if x is null: base case
if x is leaf: do something with x.value
if x is node: do something with x.left,
do something with x.right,
combine the results
this is really the most obvious way to write any algorith that works on a data structure.
now, when you look at the integers (well, the natural numbers) as defined using the Peano axioms
integer = 0 | succ(integer)
you see that a structural recursive algorithm on integers looks like this
function computeSomething(x : integer) =
if x is 0 : base case
if x is succ(prev) : do something with prev
the too-well-known factorial function is about the most trivial example of
this form.
function call itself or use its own definition.

More efficient alternative to these "for" loops?

I'm taking an introductory course to Java and one of my latest projects involve making sure an array doesn't contain any duplicate elements (has distinct elements). I used a for loop with an inner for loop, and it works, but I've heard that you should try to avoid using many iterations in a program (and other methods in my classes have a fair number of iterations as well). Is there any efficient alternative to this code? I'm not asking for code of course, just "concepts." Would there potentially be a recursive way to do this? Thanks!
The array sizes are generally <= 10.
/** Iterates through a String array ARRAY to see if each element in ARRAY is
* distinct. Returns false if ARRAY contains duplicates. */
boolean distinctElements(String[] array) { //Efficient?
for (int i = 0; i < array.length; i += 1) {
for (int j = i + 1; j < array.length; j += 1) {
if (array[i] == array[j]) {
return false;
}
}
} return true;
}
"Efficiency" is almost always a trade-off. Occasionally, there are algorithms that are simply better than others, but often they are only better in certain circumstances.
For example, this code above: it's got time complexity O(n^2).
One improvement might be to sort the strings: you can then compare the strings by comparing if an element is equal to its neighbours. The time complexity here is reduced to O(n log n), because of the sorting, which dominates the linear comparison of elements.
However - what if you don't want to change the elements of the array - for instance, some other bit of your code relies on them being in their original order - now you also have to copy the array and then sort it, and then look for duplicates. This doesn't increase the overall time or storage complexity, but it does increase the overall time and storage, since more work is being done and more memory is required.
Big-oh notation only gives you a bound on the time ignoring multiplicative factors. Maybe you only have access to a really slow sorting algorithm: actually, it turns out to be faster just to use your O(n^2) loops, because then you don't have to invoke the very slow sort.
This could be the case when you have very small inputs. An oft-cited example of an algorithm that has poor time complexity but actually is useful in practice is Bubble Sort: it's O(n^2) in the worst case, but if you have a small and/or nearly-sorted array, it can actually be pretty darn fast, and pretty darn simple to implement - never forget the inefficiency of you having to write and debug the code, and to have to ask questions on SO when it doesn't work as you expect.
What if you know that the elements are already sorted, because you know something about their source. Now you can simply iterate through the array, comparing neighbours, and the time complexity is now O(n). I can't remember where I read it, but I once saw a blog post saying (I paraphrase):
A given computer can never be made to go quicker; it can only ever do less work.
If you can exploit some property to do less work, that improves your efficiency.
So, efficiency is a subjective criterion:
Whenever you ask "is this efficient", you have to be able to answer the question: "efficient with respect to what?". It might be space; it might be time; it might be how long it takes you to write the code.
You have to know the constraints of the hardware that you're going to run it on - memory, disk, network requirements etc may influence your choices.
You need to know the requirements of the user on whose behalf you are running it. One user might want the results as soon as possible; another user might want the results tomorrow. There is never a need to find a solution better than "good enough" (although that can be a moving goal once the user sees what is possible).
You also have to know what inputs you want it to be efficient for, and what properties of that input you can exploit to avoid unnecessary work.
First, array[i] == array[j] tests reference equality. That's not how you test String(s) for value equality.
I would add each element to a Set. If any element isn't successfully added (because it's a duplicate), Set.add(E) returns false. Something like,
static boolean distinctElements(String[] array) {
Set<String> set = new HashSet<>();
for (String str : array) {
if (!set.add(str)) {
return false;
}
}
return true;
}
You could render the above without a short-circuit like
static boolean distinctElements(String[] array) {
Set<String> set = new HashSet<>(Arrays.asList(array));
return set.size() == array.length;
}

KMP prefix table running time

I wrote a code for filling the prefix table for KMP. It is small variation of this algorithm. I'm unable to convince myself that this algorithm/implementation runs in O(n) time. I have hard time figuring out the second recursive call affect on the total run time. Any help?
public void fillFailTable(int[] failTable,String p){
failTable[failTable.length-1] = preLength(failTable,p);
}
private int preLength(int[] failTable,String s){
if(s.length() == 1){
return 0;
}
int n = s.length();
int k = preLength(failTable,s.substring(0,n-1));
failTable[n-2] = k;
if(s.charAt(k) == s.charAt(n-1)){
return k+1;
}else{
return preLength(failTable,s.substring(n-1-k));
}
}
It's actually pretty interesting (I'm still wondering why no one smarter than me answered this yet). Please take this explanation with a grain of salt as I'm not 100% sure this is even close to being correct (although I can tell you for 100% that this method runs in O(n) since that's what they told me at the University years ago but they didn't bother explaining it though, d'uh, so I had to come up with it on my own).
Ok so let's start with a very basic example of s.length = 2. Two things to mention beforehand:
during each example lets only worry about the worst case scenarion, since we're interested in Big Oh, meaning we enter the second preLength() method.
we can observe, when looking for the Big Oh, that "k" (and the values returned by preLength()) in this code will always be 0, which you will notice in the images below and which is really important.
s.length == 2
We first enter the first preLength() method (lets call it *), which is now invoked with s.length = 1 and return immediately with a 0. Now since we're considering only the worst case scenario (meaning s.charAt(k) != s.charAt(n-1)) we enter the second preLength() with also a string of length = 1 (since n=2 and k=0). This one also returns a 0 immediately to our *. This ends our method invocation. In total we had 3 method invocations. Our * and two preLength(). Here's an image:
s.length == 3
Now lets look at an example with a starting s.length = 3. As you can notice we immediately invoke a preLength() with s.length = 2 and, from our previous example, we know that this one need 3 method invocations. Now we need to remember that when the method preLength(2) returns this time it returns to our native preLength(3) which will now invoke again preLength(2) (the one in the else) which will again need 3 method invocations. So in total we need 2*3+1 method invocations. This gives us 7. Again, here's an image (a circle is an invocation of preLength with a string of the length shown in the circle):
Conclusions
Now as you can see all those method invocations are symmetrical - and that is because our k is always equal to 0 which means that the second preLengt() will be invoked with a string of the same size as the first one - and we can see how many of them we will need for s.length = m when we know how many of them we need for m-1 since f(m) = 2*f(m-1)+1 where f(m) is the function telling us how many method invocations we need to compute the table for a string of size m. This works since as I said before the method invocations are symmetrical (that's because in worst case k=0 always and preLenght() always return 0, hence the 2* and we need to add 1 method invocation, the first one we ever invoke).
So basically with each incrementation of our input (size of m) the computational time grows 2 times plus one (2*m+1) which to my understanding means that this method is, in worst case, O(n).
As I said please do take this with a grain of salt but I hope this makes some sense :)

Categories