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im writing a boolean function that calculate if two binary trees are identical.
Lets see the program:
boolean func(Node head1 , Node head2){
if(head1 == null || head2 == null) return head1 == null && head2 == null;
return func(head1.left , head2.left) && (head1.right,head2.right);
}
i know that in the worst case the program check n element so its O(n).
and i want to describe this recurrence function in T(n).
i dont know were to start becuse i dont know what is the value of the stopping point ,
i think that the function is T(n,m) = 2*t(n-1,m-1) + n + m.
Let n be the number of nodes in the binary trees. If the binary trees have different number of nodes, then n is smaller of the two sizes (since you'll stop comparing once you reach the end of one tree).
On each call you are doing some comparisons between the nodes of the tree. This will be in constant time, so you can call this time spent d which represents some arbitrary constant.
Lastly, you're making 2 recursive calls for the current node's 2 children. Note that in the worst case, each of the children are the root of a subtree that holds half the total number of nodes in the tree. In other words if you have a tree with n nodes, and you are looking at the root node, then each child of the root has about n/2 nodes below it (and including it).
So your recurrence is as follows:
T(n) = 2*T((n-1)/2) + d
You can simplify this to:
T(n) = 2*T(n/2) + d
Consider all Binary Search Trees of height ≤H that can be created using the first N natural numbers. Find the sum of the roots of those Binary Search Trees.
For example, for N = 3, H = 3: There are 2 trees with 1 as root, 1 tree with 2 as root and 2 trees with 3 as root.
Hence, Sum = 2∗(1)+1∗(2)+2∗(3)=10
I am trying to solve this problem by writing a function f(n,h) which is related to f(n−1,h−1) and f(n−1,h) in some way, but unable to find the solution.
Note: All numbers [1,N] must be present in the tree and the height should be ≤H
Ok let us start with basics.
The number of BST that can be created using first N natural numbers can be very easily calculated using the following algorithm.
natural_number compute_no_of_BST(N)
{
if(N<=1)
return 1;
else
{
left=0,right=0,sum=0;
for(root = 1 to N)
{
left = compute_no_of_BST(root-1);
right = compute_no_of_BST(N-root);
sum = sum + (left*right);
}
return sum;
}
}
Explanation:
The key to understand this algorithm is this:
No matter what the distinct keys are, the number of BST only depends on number of distinct keys
So, this is what we use in recursion.For the left subtree number of distinct values are root-1 and for the right subtree the number of distinct values are N-root.Also we give every key the chance of being the root using the for loop.
Now, let us handle the constraint of height H.I am assuming the height to be the number of edges from root to leaf path. This can also be handled by focusing on the above algorithm and the trick is:
We will not call the recursive function calls for height > H and for this we must keep track of the number of edges traversed from root, which initially is 0.
So that kind of narrows it down to what are new function call will look like.
natural_number compute_no_of_BST(N,H,0);
And every time we make a recursive call, we increment the third variable to indicate an edge traversal.
We will also use an extra data structure, which is an array of length N where
arr[i] = number of BST with root i+1.
Here goes the algorithm for this
natural_number compute_no_of_BST(N,H,l)
{
if(N<=1)
return 1;
else
{
left=0,right=0,sum=0;
for(root = 1 to N)
{
if(l+1<=H)
{
left = compute_no_of_BST(root-1,H,l+1);
right = compute_no_of_BST(N-root,H,l+1);
if(l==0)
arr[root-1] = (left*right);
sum = sum + (left*right);
}
}
return sum;
}
}
Now sum can be easily computed as
arr[0]*1 + arr[1]*2 + ..... arr[N-1]*N.
Here is just a DP conversion of the above recursive algorithm.
int bottom_up_specific_height(int n,int h){
int i,j,l;
for(l=0;l<=h;l++){
dp[0][l]=1;
dp[1][l]=1;
}
int s=0;
for(i=2;i<=n;i++){
for(j=1;j<=i;j++){
for(l=h;l>=0;l--){
if(l==h)
dp[i][l]=0;
else
dp[i][l]+=(dp[j-1][l+1]*dp[i-j][l+1]);
if(l==0 && i==n)
s+=(j)*(dp[j-1][l+1]*dp[i-j][l+1]);
}
}
}
return s;
}
Here complexity reduces to O(h*n^2).
Is it possible to optimize it further!!
I'm trying to understand the solution provided in a book to the following question:
"A child is running up a staircase with n steps and can hop either 1 step, 2 steps or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs."
The book's solution is as follows, stemming from the fact that "the last move may be a single step hop from n - 1, a double step hop from step n - 2 or a triple step hop from step n - 3"
public static int countWaysDP(int n, int[] map) {
if (n < 0)
return 0;
else if (n == 0)
return 1;
else if (map[n] > -1)
return map[n];
else {
map[n] = countWaysDP(n - 1, map) + countWaysDP(n - 2, map) + countWaysDP(n - 3, map);
return map[n]; }
}
My confusion is:
Why should the program return 1 if the number of steps is zero? The way I think about it, if the number of steps is zero, then there are zero ways to traverse the staircase. Wouldn't a better solution be something like "if (n <= 0) return 0; else if (n == 1) return 1"?
I'm not sure I understand the rationale behind making this a static method? Google says that a static method is one that is called by the entire class, and not by an object of the class. So the book's intention seems to be something like:
.
class Staircase {
static int n;
public:
static int countWaysDP(int n, int[] map); }
instead of:
class Staircase {
int n;
public:
int countWaysDP(int n, int[] map); }
Why? What's the problem with there being multiple staircases instantiated by the class?
Thanks.
(Note: Book is Cracking the Coding Interview)
To answer your first question, it turns it is the beauty of mathematics: if there is 1 step for the staircase, there is 1 way to solve it. If there is 0 steps, there is also 1 way to solve it, which is to do nothing.
It is like, for an n-step staircase, for m times, you can either walk 1, 2, or 3 steps to finish it. So if n is 1, then m is 1, and there is 1 way. If n is 0, m is 0, and there is also 1 way -- the way of not taking any step at all.
If you write out all the ways for a 2-step staircase, it is [[1, 1], [2]], and for 1-step staircase, it is [[1]], and for 0-staircase, it is [[]], not []. The number of elements inside of the array [[]] is 1, not 0.
This will become the fibonacci series if the problem is that you can walk 1 step or 2 steps. Note that fib(0) = 1 and fib(1) = 1, and it corresponds to the same thing: when staircase is 1 step, there is 1 way to solve it. When there is 0 steps, there is 1 way to solve it, and it is by doing nothing. It turns out the number of ways to walk a 2-step staircase is fib(2) is 2 and it equals fib(1) + fib(0) = 1 + 1 = 2, and it wouldn't have worked if fib(0) were equal to 0.
Answer 2:
A Static method means the function doesn't need any information from the object.
The function just takes an input (in the parameters), processes it and returns something.
When you don't see any "this" in a function, you can set it as static.
Non-static methods usually read some properties (this-variables) and/or store values in some properties.
Answer 1:
I converted this to javascript, just to show what happens.
http://jsbin.com/linake/1/edit?html,js,output
I guess this is the point. Recursion often works opposite to what you could expect. It often returns values in the opposite order.
For 5 staircases:
First it returns n=1; then n=2, ... up to n=5;
n=5 has to wait until n=4 is ready, n=4 has to wait until n=3 is ready, ...
So here is your n=0 and n<0:
The first return of the function has n=1; that calls this
map[n] = countWaysDP(n - 1, map) + countWaysDP(n - 2, map) + countWaysDP(n - 3, map)
So that is
map[n] = countWaysDP(0, map) + countWaysDP(-1, map) + countWaysDP(-2, map)
There countWaysDP(0, map) returns 1; the other terms are meaningless, so they return 0. That's why there are these clauses for n==0 and n<0
notice, you can add
+ countWaysDP(n - 4, map)
if you want to see what happens when the child can also jump 4 cases
Also notice:
As I said in answer 2, you see this function doesn't require any object. It just processes data and returns something.
So, in your case, having this function in your class is useful because your functions are grouped (they 're not just loose functions scattered around your script), but making it static means the compiler doesn't have to carry around the memory of the object (especially the properties of the object).
I hope this makes sense. There are surely people that can give more accurate answers; I'm a bit out of my element answering this (I mostly do javascript).
To try and answer your first question, why it returns 1 instead of 0, say you're looking at a stair with 2 steps in total, the recursive call then becomes:
countWaysDP(2 - 1, map) + countWaysDP(2 - 2, map) + countWaysDP(2 - 3, map);
The second recursive call is the one where n becomes zero, that's when we have found a successful path, because from 2 steps, there's obviously a path of taking 2 steps. Now, if you write as you suggested:
n == 1: return 1
you would not accept taking two steps from the two stepped stair! What the statement means is that you only count the path if it ends with a single step!
You need to think about it has a tree with 3 possible options on each node.
If the size of the staircase is 4
we will have something like this:
(4)--1-->(3)--..(Choose a step and keep branching)...
|__2-->(2)--..(Until you get size of zero)............
|__3-->(1)--1-->(0) # <--Like this <--
At the end if you count all the leafs with size of zero you will get all the possible ways.
So you can think it like this, what if you take a step and then consider update the size of the stair like this size-step, where your steps can be (1,2,3)
Doing that you can code something like this:
choices = (1, 2, 3)
counter = 0
def test(size):
global counter
if size == 0:
counter += 1
for choice in choices:
if size - choice >= 0:
test(size - choice)
return counter
The example code here solves a project Euler problem:
Starting with the number 1 and moving to the right in a clockwise
direction a 5 by 5 spiral is formed as follows:
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is
101.
What is the sum of the numbers on the diagonals in a 1001 by 1001
spiral formed in the same way?
but my question is a matter of functional programming style rather than about how to get the answer (I already have it). I am trying to teach myself a bit about functional programming by avoiding imperative loops in my solutions, and so came up with the following recursive function to solve problem 28:
let answer =
let dimensions = 1001
let max_number = dimensions * dimensions
let rec loop total increment increment_count current =
if current > max_number then total
else
let new_inc, new_inc_count =
if increment_count = 4 then increment + 2, 0
else increment, increment_count + 1
loop (total + current) new_inc new_inc_count (current + increment)
loop 0 2 1 1
However, it seems to me my function is a bit of a mess. The following imperative version is shorter and clearer, even after taking into account the fact that F# forces you to explicitly declare variables as mutable and doesn't include a += operator:
let answer =
let dimensions = 1001
let mutable total = 1
let mutable increment = 2
let mutable current = 1
for spiral_layer_index in {1..(dimensions- 1) / 2} do
for increment_index in {1..4} do
current <- current + increment
total <- total + current
increment <- increment + 2
total
Disregarding the fact that people with more maths ability have solved the problem analytically, is there a better way to do this in a functional style? I also tried using Seq.unfold to create a sequence of values and then piping the resulting sequence into Seq.sum, but this ended up being even messier than my recursive version.
Since you didn't describe the problem you're trying to solve, this answer is based only on the F# code you posted. I agree that the functional version is a bit messy, but I believe it could be clearer. I don't really understand the nested for loop in your imperative solution:
for increment_index in {1..4} do
current <- current + increment
total <- total + current
You're not using the increment_index for anything, so you could just multiply increment and current by four and get the same result:
total <- total + 4*current + 10*increment
current <- current + 4*increment
Then your imperative solution becomes:
let mutable total = 0
let mutable increment = 2
let mutable current = 1
for spiral_layer_index in {1..(dimensions- 1) / 2} do
total <- total + 4*current + 10*increment
current <- current + 4*increment
increment <- increment + 2
total
If you rewrite this to a recursive function, it becomes just:
let rec loop index (total, current, increment) =
if index > (dimensions - 1) / 2 then total
else loop (index + 1) ( total + 4*current + 10*increment,
current + 4*increment, increment + 2 )
let total = loop 1 (0, 2, 1)
The same thing could be also written using Seq.fold like this (this is even more "functional", because in functional programming, you use recursion only to implement basic functions, like fold that can then be re-used):
let total, _, _=
{1 .. (dimensions - 1) / 2} |> Seq.fold (fun (total, current, increment) _ ->
(total + 4*current + 10*increment, current + 4 * increment, increment + 2)) (0, 1, 2)
NOTE: I'm not sure if this actually implements what you want. It is just a simplification of your imperative solution and then rewrite of that using a recursive function...
In fact, this is Project Euler Problem 28 and my F# solution circa November 21, 2011 is quite similar to one suggested in Tomas' answer:
let problem028 () =
[1..500]
|> List.fold (fun (accum, last) n ->
(accum + 4*last + 20*n, last + 8*n)) (1,1)
|> fst
Indeed, solution of the original problem takes just one-liner simple fold over the list of all involved squares with corners at diagonal nodes while threading through the accumulated sum and value of current diagonal element. Folding is one of the major idioms of functional programming; there is a great classic paper A tutorial on the universality and expressiveness of fold that covers many important facets of this core pattern.
Recently I have been studying recursion; how to write it, analyze it, etc. I have thought for a while that recurrence and recursion were the same thing, but some problems on recent homework assignments and quizzes have me thinking there are slight differences, that 'recurrence' is the way to describe a recursive program or function.
This has all been very Greek to me until recently, when I realized that there is something called the 'master theorem' used to write the 'recurrence' for problems or programs. I've been reading through the wikipedia page, but, as usual, things are worded in such a way that I don't really understand what it's talking about. I learn much better with examples.
So, a few questions:
Lets say you are given this recurrence:
r(n) = 2*r(n-2) + r(n-1);
r(1) = r(2)
= 1
Is this, in fact, in the form of the master theorem? If so, in words, what is it saying? If you were to be trying to write a small program or a tree of recursion based on this recurrence, what would that look like? Should I just try substituting numbers in, seeing a pattern, then writing pseudocode that could recursively create that pattern, or, since this may be in the form of the master theorem, is there a more straightforward, mathematical approach?
Now, lets say you were asked to find the recurrence, T(n), for the number of additions performed by the program created from the previous recurrence. I can see that the base case would probably be T(1) = T(2) = 0, but I'm not sure where to go from there.
Basically, I am asking how to go from a given recurrence to code, and the opposite. Since this looks like the master theorem, I'm wondering if there is a straightforward and mathematical way of going about it.
EDIT: Okay, I've looked through some of my past assignments to find another example of where I'm asked, 'to find the recurrence', which is the part of this question I'm having the post trouble with.
Recurrence that describes in the best
way the number of addition operations
in the following program fragment
(when called with l == 1 and r == n)
int example(A, int l, int r) {
if (l == r)
return 2;
return (A[l] + example(A, l+1, r);
}
A few years ago, Mohamad Akra and Louay Bazzi proved a result that generalizes the Master method -- it's almost always better. You really shouldn't be using the Master Theorem anymore...
See, for example, this writeup: http://courses.csail.mit.edu/6.046/spring04/handouts/akrabazzi.pdf
Basically, get your recurrence to look like equation 1 in the paper, pick off the coefficients, and integrate the expression in Theorem 1.
Zachary:
Lets say you are given this
recurrence:
r(n) = 2*r(n-2) + r(n-1); r(1) = r(2)
= 1
Is this, in fact, in the form of the
master theorem? If so, in words, what
is it saying?
I think that what your recurrence relation is saying is that for function of "r" with "n" as its parameter (representing the total number of data sets you're inputting), whatever you get at the nth position of the data-set is the output of the n-1 th position plus twice whatever is the result of the n-2 th position, with no non-recursive work being done. When you try to solve a recurrence relation, you're trying to go about expressing it in a way that doesn't involve recursion.
However, I don't think that that is in the correct form for the Master Theorem Method. Your statement is a "second order linear recurrence relation with constant coefficients". Apparently, according to my old Discrete Math textbook, that's the form you need to have in order to solve the recurrence relation.
Here's the form that they give:
r(n) = a*r(n-1) + b*r(n-2) + f(n)
For 'a' and 'b' are some constants and f(n) is some function of n. In your statement, a = 1, b = 2, and f(n) = 0. Whenever, f(n) is equal to zero the recurrence relation is known as "homogenous". So, your expression is homogenous.
I don't think that you can solve a homogenous recurrence relation using the Master Method Theoerm because f(n) = 0. None of the cases for Master Method Theorem allow for that because n-to-the-power-of-anything can't equal zero. I could be wrong, because I'm not really an expert at this but I don't that it's possible to solve a homogenous recurrence relation using the Master Method.
I that that the way to solve a homogeneous recurrence relation is to go by 5 steps:
1) Form the characteristic equation, which is something of the form of:
x^k - c[1]*x^k-1 - c[2]*x^k-2 - ... - c[k-1]*x - c[k] = 0
If you've only got 2 recursive instances in your homogeneous recurrence relation then you only need to change your equation into the Quadratic Equation where
x^2 - a*x - b = 0
This is because a recurrence relation of the form of
r(n) = a*r(n-1) + b*r(n-2)
Can be re-written as
r(n) - a*r(n-1) - b*r(n-2) = 0
2) After your recurrence relation is rewritten as a characteristic equation, next find the roots (x[1] and x[2]) of the characteristic equation.
3) With your roots, your solution will now be one of the two forms:
if x[1]!=x[2]
c[1]*x[1]^n + c[2]*x[2]^n
else
c[1]*x[1]^n + n*c[2]*x[2]^n
for when n>2.
4) With the new form of your recursive solution, you use the initial conditions (r(1) and r(2)) to find c[1] and c[2]
Going with your example here's what we get:
1)
r(n) = 1*r(n-1) + 2*r(n-2)
=> x^2 - x - 2 = 0
2) Solving for x
x = (-1 +- sqrt(-1^2 - 4(1)(-2)))/2(1)
x[1] = ((-1 + 3)/2) = 1
x[2] = ((-1 - 3)/2) = -2
3) Since x[1] != x[2], your solution has the form:
c[1](x[1])^n + c[2](x[2])^n
4) Now, use your initial conditions to find the two constants c[1] and c[2]:
c[1](1)^1 + c[2](-2)^1 = 1
c[1](1)^2 + c[2](-2)^2 = 1
Honestly, I'm not sure what your constants are in this situation, I stopped at this point. I guess you'd have to plug in numbers until you'd somehow got a value for both c[1] and c[2] which would both satisfy those two expressions. Either that or perform row reduction on a matrix C where C equals:
[ 1 1 | 1 ]
[ 1 2 | 1 ]
Zachary:
Recurrence that describes in the best
way the number of addition operations
in the following program fragment
(when called with l == 1 and r == n)
int example(A, int l, int r) {
if (l == r)
return 2;
return (A[l] + example(A, l+1, r);
}
Here's the time complexity values for your given code for when r>l:
int example(A, int l, int r) { => T(r) = 0
if (l == r) => T(r) = 1
return 2; => T(r) = 1
return (A[l] + example(A, l+1, r); => T(r) = 1 + T(r-(l+1))
}
Total: T(r) = 3 + T(r-(l+1))
Else, when r==l then T(r) = 2, because the if-statement and the return both require 1 step per execution.
Your method, written in code using a recursive function, would look like this:
function r(int n)
{
if (n == 2) return 1;
if (n == 1) return 1;
return 2 * r(n-2) + r(n-1); // I guess we're assuming n > 2
}
I'm not sure what "recurrence" is, but a recursive function is simply one that calls itself.
Recursive functions need an escape clause (some non-recursive case - for example, "if n==1 return 1") to prevent a Stack Overflow error (i.e., the function gets called so much that the interpreter runs out of memory or other resources)
A simple program that would implement that would look like:
public int r(int input) {
if (input == 1 || input == 2) {
return 1;
} else {
return 2 * r(input - 2) + r(input -1)
}
}
You would also need to make sure that the input is not going to cause an infinite recursion, for example, if the input at the beginning was less than 1. If this is not a valid case, then return an error, if it is valid, then return the appropriate value.
"I'm not exactly sure what 'recurrence' is either"
The definition of a "recurrence relation" is a sequence of numbers "whose domain is some infinite set of integers and whose range is a set of real numbers." With the additional condition that that the function describing this sequence "defines one member of the sequence in terms of a previous one."
And, the objective behind solving them, I think, is to go from a recursive definition to one that isn't. Say if you had T(0) = 2 and T(n) = 2 + T(n-1) for all n>0, you'd have to go from the expression "T(n) = 2 + T(n-1)" to one like "2n+2".
sources:
1) "Discrete Mathematics with Graph Theory - Second Edition", by Edgar G. Goodair and Michael M. Parmenter
2) "Computer Algorithms C++," by Ellis Horowitz, Sartaj Sahni, and Sanguthevar Rajasekaran.