Sum to determine the largest multiple of 5 under 1,000 - java

I am currently trying to get a feature working in a Java application I am making, however I'm uncertain how to implement this in a single line.
I know that I could do something along the lines of (not exactly, but roughly):
while(i<995){
i=i+5
}
However I am eager to implement this all into one line, such as in a single
static int highestMult = *the equation*
I would not be using this specifically for the highest multiple of 5 in 1,000, however upon my own research I could not find a desired solution for this specific case, therefore this is an example.
The examples I have previously found all, generally, relate to finding only a highest multiple, not putting together the highest multiple, and a limit.
If this is not knowledge from the back of your head, it'd also be a great help just to understand the logic behind how you came up with the solution, it could save me being stuck on similar issues in the future.
Thanks,

If c is the under number (1000 in your case), and m the multiple (5 in your case), then
((c - 1) / m) * m
is one way. (Note to purists: you don't actually need the outer parentheses but I include them for clarity).
Here I'm exploiting integer arithmetic to force the truncation of ((c - 1) / m) to the flooring integer. Multiplication of this result by m means the final value is a multiple of m. Make sure that c and m are integral types or this will not work (unless you cast explicitly which is not as elegant).
This is undefined for c < 1 and m < 1

try:
int number=5;
int limit=999;
int i=limit-(limit%number);
where 999 is limit - 1
% is reminder
(999%5)=4
if we remove the reminder from limit, we will got it:
999-4=995
we could use limit=1000, but result could be 1000 too
reminder is very useful thing for programming :D
defined for number > 0 and limit >=0

Related

Using the definition of O, prove each of the following by finding explicit integer constants C and k so that k is minimal for the given C

a.) 9n + 77 = O(n) (find a C < 50)
d.) 11n^2 + 75n + 84 = O(n^2) (find a C < 40)
NOTE: I am second year Java student and am currently in a Data Structures and Algorithms class. I have no idea how to do these problems (there are more but I figured two should give me a basis). What is O(n) and how does it differ from O(n^2)? My professor just said we should look at this stuff in the book because these questions will be similar to what are on our exams. I have no idea how to attack these problems. Please explain. I'm not looking for just a solution to copy down, but rather understanding.
f(n) = O(g(n)) means there are positive constants c and k, such that 0 ≤ f(n) ≤ cg(n) for all n ≥ k. https://xlinux.nist.gov/dads/HTML/bigOnotation.html
is a definition.
Here is a sketch of where to go with that and what it means without doing these problems for you. [I am deliberately leaving parts for you to figure out.]
So if we are saying 9n+77=O(n) that mean there is a constant c such that for n big enough 9n+77<=Cn. The intuition here is that in the grand scheme of things 77 does not matter and 9n is basically what matters, and that is a multiple of n. But choosing the constant = 9 will not do because 9n+77 is actually > 9n always. I feel like anything bigger than 9 would do, but let us keep life simple and choose C=10. So we want 9n+77<=10n. Solve that and you will have your k that n must be >= to.
11n^2 + 75n + 84 = O(n^2) is intuitively true because in the long run the 11n^2 dwarfs everything else. Again, you would be tempted to pick C=11, but that will not quite do (why?). So you choose a bigger C and use it to say we want 11n^2+75n+84<=Cn^2. To find k, solve the inequality (choosing the part that looks like n>=the larger root of the corresponding quadratic equation). You may need to use the quadratic formula.

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.

Check if only one single bit is set within an integer (whatever its position)

I store flags using bits within a 64-bits integer.
I want to know if there is a single bit set whatever the position within the 64-bits integer (e.i. I do not care about the position of any specific bit).
boolean isOneSingleBitSet (long integer64)
{
return ....;
}
I could count number of bits using the Bit Twiddling Hacks (by Sean Eron Anderson), but I am wondering what is the most efficient way to just detect whether one single bit is set...
I found some other related questions:
(8051) Check if a single bit is set
Detecting single one-bit streams within an integer
and also some Wikipedia pages:
Find first one
Bit manipulation
Hamming weight
NB: my application is in java, but I am curious about optimizations using other languages...
EDIT: Lưu Vĩnh Phúc pointed out that my first link within my question already got the answer: see section Determining if an integer is a power of 2 in the Bit Twiddling Hacks (by Sean Eron Anderson). I did not realized that one single bit was the same as power of two.
If you just literally want to check if one single bit is set, then you are essentially checking if the number is a power of 2. To do this you can do:
if ((number & (number-1)) == 0) ...
This will also count 0 as a power of 2, so you should check for the number not being 0 if that is important. So then:
if (number != 0 && (number & (number-1)) == 0) ...
(using x as the argument)
Detecting if at least one bit is set is easy:
return x!=0;
Likewise detecting if bit one (second lowest bit) is set is easy:
return (x&2)!=0;
Exactly one bit is set iff it is a power of two. This works:
return x!=0 && (x & (x-1))==0;
The wrapper class java.lang.Long has a static function bitCount() that returns the number of bits in a long (64-bit int):
boolean isSingleBitSet(long l)
{
return Long.bitCount(l) == 1;
}
Note that ints are 32-bit in java.
Assuming you have already an efficient - or hardware - implementation of ffs() - find first set - you may act as follows:
bool isOneSingleBitSet (long integer64)
{
return (integer64 >> ffs(integer64)) == 0;
}
The ffs() function may be already available, or you may like to see your own link above
lets assume X is a 64bit inter full of 0s exept the one you are looking for;
return ((64bitinteger&X)==X)
Seems like you can do a bitwise AND with a long representation of the single bit you want to check. For example, to check the LSB
return( (integer64 & 1L)!=0 );
Or to check the 4th bit from the right
return( (integer64 & 8L)!=0 );

library for integer factorization in java or scala

There are a lot of questions about how to implement factorization, however for production use, I would rather use an open source library to get something efficient and well tested right away.
The method I am looking for looks like this:
static int[] getPrimeFactors(int n)
it would return {2,2,3} for n=12
A library may also have an overload for handling long or even BigInteger types
The question is not about a particular application, it is about having a library which handles well this problem. Many people argue that different implementations are needed depending on the range of the numbers, in this regard, I would expect that the library select the most reasonable method at runtime.
By efficient I don't mean "world fastest" (I would not work on the JVM for that...), I just mean dealing with int and long range within a second rather than a hour.
It depends what you want to do. If your needs are modest (say, you want to solve Project Euler problems), a simple implementation of Pollard's rho algorithm will find factors up to ten or twelve digits instantly; if that's what you want, let me know, and I can post some code. If you want a more powerful factoring program that's written in Java, you can look at the source code behind Dario Alpern's applet; I don't know about a test suite, and it's really not designed with an open api, but it does have lots of users and is well tested. Most of the heavy-duty open-source factoring programs are written in C or C++ and use the GMP big-integer library, but you may be able to access them via your language's foreign function interface; look for names like gmp-ecm, msieve, pari or yafu. If those don't satisfy you, a good place to ask for more help is the Mersenne Forum.
If you want to solve your problem, rather than get what you are asking for, you want a table. You can precompute it using silly slow methods, store it, and then look up the factors for any number in microseconds. In particular, you want a table where the smallest factor is listed in an index corresponding to the number--much more memory efficient if you use trial division to remove a few of the smallest primes--and then walk your way down the table until you hit a 1 (meaning no more divisors; what you have left is prime). This will take only two bytes per table entry, which means you can store everything on any modern machine more hefty than a smartphone.
I can demonstrate how to create this if you're interested, and show how to check that it is correct with greater reliability than you could hope to achieve with an active community and unit tests of a complex algorithm (unless you ran the algorithm to generate this table and verified that it was all ok).
I need them for testing if a polynomial is primitive or not.
This is faster than trying to find the factors of all the numbers.
public static boolean gcdIsOne(int[] nums) {
int smallest = Integer.MAX_VALUE;
for (int num : nums) {
if (num > 0 && smallest < num)
smallest = num;
}
OUTER:
for (int i = 2; i * i <= smallest; i = (i == 2 ? 3 : i + 2)) {
for (int num : nums) {
if (num % i != 0)
continue OUTER;
}
return false;
}
return true;
}
I tried this function in scala. Here is my result:
def getPrimeFactores(i: Int) = {
def loop(i: Int, mod: Int, primes: List[Int]): List[Int] = {
if (i < 2) primes // might be i == 1 as well and means we are done
else {
if (i % mod == 0) loop(i / mod, mod, mod :: primes)
else loop(i, mod + 1, primes)
}
}
loop(i, 2, Nil).reverse
}
I tried it to be as much functional as possible.
if (i % mod == 0) loop(i / mod, mod, mod :: primes) checks if we found a divisor. If we did we add it to primes and divide i by mod.
If we did not find a new divisor, we just increase the divisor.
loop(i, 2, Nil).reverse initializes the function and orders the result increasingly.

Exclusive or between N bit sets

I am implementing a program in Java using BitSets and I am stuck in the following operation:
Given N BitSets return a BitSet with 0 if there is more than 1 one in all the BitSets, and 1 otherwise
As an example, suppose we have this 3 sets:
10010
01011
00111
11100 expected result
For the following sets :
10010
01011
00111
10100
00101
01000 expected result
I am trying to do this exclusive with bit wise operations, and I have realized that what I need is literally the exclusive or between all the sets, but not in an iterative fashion,
so I am quite stumped with what to do. Is this even possible?
I wanted to avoid the costly solution of having to check each bit in each set, and keep a counter for each position...
Thanks for any help
Edit : as some people asked, this is part of a project I'm working on. I am building a time table generator and basically one of the soft constraints is that no student should have only 1 class in 1 day, so those Sets represent the attending students in each hour, and I want to filter the ones who have only 1 class.
You can do what you want with two values. One has the bits set at least once, the second has those set more than once. The combination can be used to determine those set once and no more.
int[] ints = {0b10010, 0b01011, 0b00111, 0b10100, 0b00101};
int setOnce = 0, setMore = 0;
for (int i : ints) {
setMore |= setOnce & i;
setOnce |= i;
}
int result = setOnce & ~setMore;
System.out.println(String.format("%5s", Integer.toBinaryString(result)).replace(' ', '0'));
prints
01000
Well first of all, you can't do this without checking every bit in each set. If you could solve this question without checking some arbitrary bit, then that would imply that there exist two solutions (i.e. two different ones for each of the two values that bit can be).
If you want a more efficient way of computing the XOR of multiple bit sets, I'd consider representing your sets as integers rather than with sets of individual bits. Then simply XOR the integers together to arrive at your answer. Otherwise, it seems to me that you would have to iterate through each bit, check its value, and compute the solution on your own (as you described in your question).

Categories