Order complexity of a pairing function
I have a code that involves a 2 to 1 pairing of numbers. The code for this is
function [ A ] = CantorPairing( B )
[~, b] =(size(B));
%b=sqrt(b);
k=1;
A=zeros(1,b/2);
for i=1:2:(b)
if( B(i)< B(i+1))
A(k)= B(i)+(B(i+1))^2;
else
A(k)= (B(i))^2+B(i)+B(i+1);
end
k=k+1;
end
This is a matlab code that i am implementing. What this code does is, it pairs the adjacent elements of an array B depending on which is greater. So for an n size array the number of elements it returns is n/2. I have a few questions for this code.
Since I am new to computer science, I don't think this is a very efficient code in MATLAB, can I optimize this code.
What is the order complexity of this code and it's optimized version?
What if I again run this function on the new set A to get an array C of size n/4.
complexity matlab
asked 6 mins ago
user311790
1
New contributor
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I have a code that involves a 2 to 1 pairing of numbers. The code for this is
function [ A ] = CantorPairing( B )
[~, b] =(size(B));
%b=sqrt(b);
k=1;
A=zeros(1,b/2);
for i=1:2:(b)
if( B(i)< B(i+1))
A(k)= B(i)+(B(i+1))^2;
else
A(k)= (B(i))^2+B(i)+B(i+1);
end
k=k+1;
end
This is a matlab code that i am implementing. What this code does is, it pairs the adjacent elements of an array B depending on which is greater. So for an n size array the number of elements it returns is n/2. I have a few questions for this code.
Since I am new to computer science, I don't think this is a very efficient code in MATLAB, can I optimize this code.
What is the order complexity of this code and it's optimized version?
What if I again run this function on the new set A to get an array C of size n/4.
complexity matlab
asked 6 mins ago
user311790
1
New contributor
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I have a code that involves a 2 to 1 pairing of numbers. The code for this is
function [ A ] = CantorPairing( B )
[~, b] =(size(B));
%b=sqrt(b);
k=1;
A=zeros(1,b/2);
for i=1:2:(b)
if( B(i)< B(i+1))
A(k)= B(i)+(B(i+1))^2;
else
A(k)= (B(i))^2+B(i)+B(i+1);
end
k=k+1;
end
This is a matlab code that i am implementing. What this code does is, it pairs the adjacent elements of an array B depending on which is greater. So for an n size array the number of elements it returns is n/2. I have a few questions for this code.
Since I am new to computer science, I don't think this is a very efficient code in MATLAB, can I optimize this code.
What is the order complexity of this code and it's optimized version?
What if I again run this function on the new set A to get an array C of size n/4.
complexity matlab
asked 6 mins ago
user311790
1
New contributor
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a code that involves a 2 to 1 pairing of numbers. The code for this is
function [ A ] = CantorPairing( B )
[~, b] =(size(B));
%b=sqrt(b);
k=1;
A=zeros(1,b/2);
for i=1:2:(b)
if( B(i)< B(i+1))
A(k)= B(i)+(B(i+1))^2;
else
A(k)= (B(i))^2+B(i)+B(i+1);
end
k=k+1;
end
This is a matlab code that i am implementing. What this code does is, it pairs the adjacent elements of an array B depending on which is greater. So for an n size array the number of elements it returns is n/2. I have a few questions for this code.
Since I am new to computer science, I don't think this is a very efficient code in MATLAB, can I optimize this code.
What is the order complexity of this code and it's optimized version?
What if I again run this function on the new set A to get an array C of size n/4.
complexity matlab
complexity matlab
asked 6 mins ago
user311790
1
New contributor
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
user311790
1
New contributor
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
user311790
1
New contributor
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
user311790
1
asked 6 mins ago
user311790
1
1
New contributor
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user311790 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "196"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
user311790 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f210468%2forder-complexity-of-a-pairing-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
user311790 is a new contributor. Be nice, and check out our Code of Conduct.
user311790 is a new contributor. Be nice, and check out our Code of Conduct.
user311790 is a new contributor. Be nice, and check out our Code of Conduct.
user311790 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Code Review Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f210468%2forder-complexity-of-a-pairing-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
