Personal portfolio website design free download – HTML CSS JavaScript Free Download
Personal portfolio website design is a project that every artist needs. It can also be a big undertaking. So to make the process easier, you will need a professional website designer to help you through the process. But what if you are not ready to shell out $5000+ for a website design? There is a solution. This tutorial will walk you through the process of creating your own personal portfolio website. This tutorial will teach you a basic website design, how to use HTML and CSS, and how to get your website up and running in a short amount of time.
The Internet is a creative and collaborative medium for people all over the world. It is also a place where businesses and people can communicate, share ideas, and connect. One of the best ways to do this is through personal websites. These can be simple sites that serve as marketing tools or elaborate sites that showcase a creative vision. If you are considering building a personal website for your business or personal brand, you may want to consider using a pre-made template. To learn how to create a personal website, use these six templates as a starting point.
Personal portfolio website design free download – HTML CSS JavaScript Free Download
Then
(A1∗A2)∗A3
Total number to be multiplied in this bracketing (10×100×5) + (10×5×50) = 7500 times
A1∗(A2∗A3)
In this bracketing, the total number to be multiplied is (100×5×50) + (10×100×50) = 75,000 times.
(Here multiplication refers to the number of times the scalar is multiplied inside the matrix multiplication, see condition number 3)
The amount of calculation increased by 10 times in the 2nd way! Using the matrix chain multiplication method, we will find a “bracketing” that minimizes number multiplication. You A1,A2……..An
Many such matrices will have only given dimensions, say the minimum number of times A1∗A2∗………..An
can be extracted We assume the dimensions are valid, i.e. satisfy the first condition.
We will solve it by divide and conquer method. Dynamic programming will be needed because the same subproblem will come up again and again. Let n be
value is 5, then there are 5 matrices A1,A2,A3,A4,A5, you A1∗A2∗A3∗A4∗A5
Find out how many numbers it takes to multiply. Now see that we can bisect matrices in different ways, one way is as follows:
(A1∗A2)∗(A3∗A4∗A5)
That means we have a bracketing Aleft=A1∗A2
I will extract and extract Aright=A3∗A4∗A5 in any bracketing. We don’t know how to figure them out, but if someone tells us Aleft and Aright
To find out how many numbers need to be multiplied then we can tell how many numbers need to be multiplied in total. The total number multiplied by the number will be:
Total number of times = Aleft
Number of product to determine + Aright Number of product to determine + Aleft∗Aright
Number of multiplications to be determined
Or we can write the last term from the 3rd term: Aleft
Number of Rows×Aleft Number of Columns×Aright
Its column number
Total number of times = Aleft
Number of multiplications to determine + Aright Number of multiplications to determine + Number of rows of Aleft×Number of columns of Aleft×Aright
Its column number
That is, if someone is magically Aleft
Aright of
Only then we can find out the total number.
But we could have divided it in many other ways, such as:
(A1)∗(A2∗A3∗A4∗A5)
(A1∗A2∗A3)∗(A4∗A5)
(A1∗A2∗A3∗A4)∗(A5)
We will divide each way and magically Aleft after dividing
and Aright
To find out how many numbers need to be multiplied, I will find out the total number. My answer is that the total number is minimum if you divide it!
If you have a good understanding of recursion then by now you know how to magically do the work. We will solve the left and right sides recursively in the same way, that is, we will divide them again in many ways and bring out the optimal answer! So our algorithm is very simple:
Share as many ways as possible
Recursively solve for smaller parts
Find the total number of times by merging the left and right sides
Take the best of all means
So think what will be the parameter or state of our recursion? What information can I give to solve the problem?
I just need to know which part of the input matrices I’m working with now. That means we will keep the start point and end point as states.
int f(int beg, int end)
Personal portfolio website design free download – HTML CSS JavaScript Free Download
Such will be the parameter or state of the function. f
The function will tell you the total number of multiplications needed to multiply the matrices from beg to end optimally! When does recursion stop? That is, what is the base case? When we see that there is one or fewer matrices (b>=e)
Then we know we don’t need to do any multiplication, we will return zero.
So now we write a code. It’s best if you try it yourself first, the joy of solving it without help is unmatched! Even if you can’t, there’s nothing to be upset about. As input you take the number of rows and columns of each matrix. The output will be the number of operations.
After getting the state, our task will be to find out how to go from one state to another state, that is, the recurrence relation. In this problem we will recursively solve the problem for the left and right sides by selecting one midpoint and merging them. The recurrence will then be: MCM
Now we convert it to code:
#define MAX 100
int row[MAX], col[MAX];
int dp[MAX][MAX];
bool visited[MAX][MAX];
int f(int beg,int end)
{
if(beg>=end) return 0;
if(visited[beg][end])return dp[beg][end];
int ans=1<<30; // Taking 2^30 to infinity
for(int mid=beg; mid<end;mid++) //Dividing in two
{
int opr_left = f(beg, mid); //opr = multiplication operation
int opr_right = f(mid+1, end);
int opr_to_multiply_left_and_right = row[beg]*col[mid]*col[end];
int total = opr_left + opr_right + opr_to_multiply_left_and_right;
ans = min(ans, total);
}
visited[beg][end] = 1;
dp[beg][end] = ans;
return dp[beg][end];
}
int main()
{
int n;
cin>>n;
rep(i,n)cin>>row[i]>>col[i];
cout<<f(0,n-1)<<endl;
}
#define MAX 100
int row[MAX], col[MAX];
int dp[MAX][MAX];
bool visited[MAX][MAX];
int f(int beg,int end)
{
if(beg>=end) return 0;
if(visited[beg][end])return dp[beg][end];
int ans=1<<30; // Taking 2^30 to infinity
for(int mid=beg; mid<end;mid++) //Dividing in two
{
int opr_left = f(beg, mid); //opr = multiplication operation
int opr_right = f(mid+1, end);
int opr_to_multiply_left_and_right = row[beg]*col[mid]*col[end];
int total = opr_left + opr_right + opr_to_multiply_left_and_right;
ans = min(ans, total);
}
visited[beg][end] = 1;
dp[beg][end] = ans;
return dp[beg][end];
}
int main()
{
int n;
cin>>n;
rep(i,n)cin>>row[i]>>col[i];
cout<<f(0,n-1)<<endl;
}
A very simple code, dividing it into two parts and solving the smaller part. I don’t want to solve for the same part again and again, so I am saving it in DP array, if I see that any state has been visited before, then I am returning the old result!
Complexity:
beg and end can have values of 1
from to n. Then there are about n2 different states. The loop can run up to n times in each state. Then the time complexity is O(n^3). Memory required is O(n2).
.
Related problems:
You may not get direct matrix chain multiplication problems in the contest, but you can solve many problems with the idea of dividing both sides by states in different ways, which is why it’s so important to learn.
1. http://www.spoj.com/problems/MIXTURES/
2. Cutting Sticks
Happy Coding
Before Download
You must Join our Facebook Group and Subscribe YouTube Channel
All Links in Below:
Join Our FreeWebsiteCreate Facebook Group to get an instant update for projects, templates, design resources, and solutions.
Join Our YouTube Channel & Subscribe with Bell Icon for New Video:
Join Our Official Facebook Page For the Latest updates All Code Projects are Free:
Visit our service page to get premium services.
Free Website Create – HTML CSS, PHP, JavaScript Programming Projects For Free
Follow Us
Thank You,
Before Download
You must Join our Facebook Group and Subscribe YouTube Channel
FreeWebsiteCreate.net tries to provide HTML, CSS, SCSS, JavaScript, React, Android Studio, Java, PHP, Laravel, Python, Django, C#(C Sharp), and ASP.net-related projects 100% free. We try to make learning easier. Free Website Create always tries to give free projects to new learners. Free projects and source code will help to learn quickly.
They can save time and learn more. In this post, we share a free portfolio project website code with HTML and CSS. This free code portfolio contains a single landing page with a responsive design. In this post, we get a free best carpenter and craftsman service website designed by FreeWebsiteCreate with HTML, CSS, Bootstrap, and JavaScript.