Why This Vectorization of my Code Does Not Work?
Afficher commentaires plus anciens
Consider the following two codes:
n = 4;
% Code 1:
c1 = zeros(floor(n/2)+1,n+1); c1(1,1) = 1; c1(1,2:n+1) = 2.^(0:n-1);
for k = 2:n
for l = 1:floor(k/2), c1(l+1,k+1) = -((k - 2*l + 2)/(4*l*(k - l)))*(k - 2*l + 1)*c1(l,k+1); end
end
c1
% Code 2:
c2 = zeros(floor(n/2)+1,n+1); c2(1,1) = 1; c2(1,2:n+1) = 2.^(0:n-1);
for k = 2:n
lend = floor(k/2); c2(2:lend+1,k+1) = -((k-2*(1:lend)+2)./(4*(1:lend).*(k-(1:lend)))).*(k-2*(1:lend)+ 1).*c2(1:lend,k+1)';
end
c2
The first code gives
while the second gives

Réponses (1)
KSSV
le 14 Juil 2020
Might be some problem with indexing. Try this:
clc; clear all ;
n = 4;
% Code 1:
c1 = zeros(floor(n/2)+1,n+1);
c1(1,1) = 1;
c1(1,2:n+1) = 2.^(0:n-1);
for k = 2:n
for l = 1:floor(k/2)
c1(l+1,k+1) = -((k - 2*l + 2)/(4*l*(k - l)))*(k - 2*l + 1)*c1(l,k+1);
end
end
c1
% Code 2:
c2 = zeros(floor(n/2)+1,n+1);
c2(1,1) = 1;
c2(1,2:n+1) = 2.^(0:n-1);
for k = 2:n
lend = floor(k/2);
% c2(2:lend+1,k+1) = -((k-2*(1:lend)+2)./(4*(1:lend).*(k-(1:lend)))).*(k-2*(1:lend)+ 1).*c2(1:lend,k+1)';
l = 1:lend ;
c2(2:lend+1,k+1) = -((k - 2*l + 2)./(4*l.*(k - l))).*(k - 2*l + 1).*c1(l,k+1)';
end
c2
1 commentaire
Kareem Elgindy
le 14 Juil 2020
Modifié(e) : Kareem Elgindy
le 14 Juil 2020
Catégories
En savoir plus sur String Parsing dans Centre d'aide et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!