Body Fitness Gym Website Design Free Download – Responsive Website Design Tutorials | HTML – CSS – JavaScript

Hello everyone, In this tutorial, we are building a responsive fitness or gym website by using HTML, CSS, and JavaScript. By following this video tutorial you can learn website design and development with HTML CSS and javascript. We provide a boilerplate of this gym or fitness website so that you can download and start coding based on the video tutorials.

Here you can learn how to create a responsive header, hero section, footer section, service section, and so on.

After learning with this video tutorial, you can develop any website by yourself.

## Body Fitness Gym Website Design Free Download – Responsive Website Design Tutorials | HTML – CSS – JavaScript

To get a complete gym or fitness website design, You have to code by following this video tutorial. All demo images are also included on this link, You Can Download them.

**Dynamic Placing-3 (Longest Increasing Subsequence, Path Printing)**

Now that we have learned the basics of Dynamic Placing, we can find the shortest path in Fibonacci and DAG. We will learn to find the Longest Increasing Sequence or LIS. So far we have only learned to find out the optimal solution, we have not learned to find the path to reach the solution. We’ll also learn how to use the next-pointer.

My name is A

.

An array subsequence means deleting some elements from the array and the rest is that sequence. Element matching cannot be changed. 2n of an array of size n

T can be subsequence advantages (2 choices for each element, video or deletion). An increasing subsequence is a subsequence whose position value is greater than its own position value.

For example {0,9},{5,9},{0,2,7},{0,2,3,4}

Some increasing subsequences. Longest increasing subsequence or list {0,2,3,4}

. The way we try to figure out the endurance of LIS, learn to track the original subsequence as well.

Parameters or State of the Problem Right:

We can think of the problem inductively: we are currently at index i

To be there and we need to find the longest subsequence i starting at the th index. Let the function f(i) calculate exactly that. Now how to find f(0)? We don’t know exactly which index LIS starts from. We need to perform the convenience result – max(f(0),f(1).f(n–1)).

State Transition and Recursion:

Now we know the index i

At which index will we find the longest subsequence? Now we have to guess. i

Moving on from that to the index we’ll look at the length of the LIS and discuss the most.

Index i

From which index can go? j

The index must satisfy the condition:

Same main to main, so new indj

must be greater than (i<j<n)

.

Also the value of A[j] must be greater than A[i] (A[j]>A[i]

.

terms (i < j < n) and (A[j] > A[i])

J from only us

can go to Then the recursion will happen

f(i)=1+max(f(j)) where i<j<n & A[j]>A[i]

Confused? Think of the Date Shortest Path problem as if each index is a city. Answer the answer by dragging the arrow from which city the group advances.

Array Lines We are modeling with graphics for our visualization. Our problem now becomes harder to find the longest path in a directed acyclic graph!

Note that we define no base case in recursion. The reason for not being Rika during the time of hope in this love is that the normal status will go out from one stat (we are thinking about the arc, one can’t move forward anymore). There is also no reading in the recursion loop.

Here’s the power:

Complexity: Like the shirt path problem, it also has O(n∗n) complexity. There are n different states and state is n for u

Running the size loop.

Iterative version:

The problem is that it is not very difficult to write iterative versions of passwords like short paths. It takes a long time to sort the states into a topological table and we move from small to large. n−1, n−2….,1,0

This medal can be built up deductively DP table too.

Note that we no longer have to fill the mem table with −1. I’m inside its loop and the rest is like my recursion. i

Don’t work by letting the loop start from 0.

Printing the solution path

Both iterative and recursive methods have the same rules for printing the same path. We need to maintain a table named next_pointer. Whenever subproblem any better reza you have to save in which table from which state. Finally the path can be found from the first state to the next point.

So in this episode you learned:

Visualization with a graph of the strength of the dirp can be useful for moving from state to state.

Sends to Nextpoints for every start.

**Body Fitness Gym Website Design Free Download – Responsive Website Design Tutorials | HTML – CSS – JavaScript**

**Dynamic Programming-II (Shortest Path)**

We discussed Fibonacci numbers. We have seen how to generate Fibonacci numbers recursively and iteratively using dynamic programming. In this section we will talk about the shortest path problem and learn some new properties of DP while discussing it.

Although the shortest path is a problem in graph theory, this text can be understood without knowledge of graphs.

Suppose we have to find the shortest path from one city to another. There are total n cities

T. 0 is the first city, and n–1 is the last city. From which city you can go directly to which city and the distance between the cities is shown by arrows as shown in the image below:

Now our problem is 0 to n−1

What is the length of the shortest path? Now we will formulate the problem in a few steps.

Determining the parameters or state of the problem:

First we need to figure out what parameters or states can express the problem. Remember the word state, it will be used many times in the future. Our state will be in which city you are currently in. Assuming you are currently u

is in the th city and we define our problem as f(u)

I will express it with function. We now need to derive a recursive relation, similar to Fibonacci.

State Transition and Recursion:

We don’t know the city u

From which city I can reach the destination in the fastest way. We will assume that in this step. Each hypothesis will be a sub-problem. Cities 0 to 1 and 3 can be visited in the picture. then f(0)

From we get the following shortest path with the formula:

f(0)=min(f(1)+2,f(3)+1)

The idea is that from each city we will go to all the other cities and try to find the shortest path from there. The smallest of all guesses will be the answer. Generalized to write:

f(n–1)=0f(u)=min(f(u,v)+w(u,v)) where (u,v)∈E for u≠n–1

where w(u,v)

If u to v

distance to go to

Subproblem/State Ordering:

Now think about how the function call will happen if we continue this recursion.

Now we’re in a bit of trouble, a cycle is created between the subproblems. To know the value of f(0) we need to know f(2) but f(2) is again f(0).

Depends on it. Implementing this solution will cause our code to get stuck in an infinite loop.

The purpose of going to great lengths to find the wrong solution is to introduce you to the DAG or Directed Acyclic Graph concept. A DAG stands for a directed graph with no cycles. Dynamic programming solutions will work only when the subproblems form a DAG. If there is a cycle, the subproblems will get stuck in an infinite loop.

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