Please find the error

1 Ansicht (letzte 30 Tage)
vidushi Chaudhary
vidushi Chaudhary am 19 Mai 2020
Beantwortet: Geoff Hayes am 19 Mai 2020
%Write a function called max_sum that takes v as a row vector of numbers, & n,a postive integer as inputs.The function needs to find n consecutive
%elements of v whose sum is the largest possible.If multiple such sequence exists in v,max_sum returns first one.The function returns summa & index
%,the index of first element of n consecutive ones.If n>v,then the function returns summa as 0 & index=-1.
%Example-[summa,index]=max_sum([1 2 3 4 5 4 3 2 1],3])
%summa=13
%index=4
function [summa,index]=max_sum(v,n)
total=0;
if n>v
summa=0;
index=-1;
else
for ii=1:length(v)
jj=ii+(n-1);
if jj<=length(v)
total=[total,sum(v(ii):v(jj))]; %Here I'm trying to create a row vector with sum of consecutive n integers
end
end
[summa,index]=max(total);
end
end
  2 Kommentare
KSSV
KSSV am 19 Mai 2020
Your jj gets equals to 9. Where as your v is of size 1*8. Code tries to extract v(9) so the error.
vidushi Chaudhary
vidushi Chaudhary am 19 Mai 2020
Bearbeitet: vidushi Chaudhary am 19 Mai 2020
The code is working now,but it's giving incorrect value of summa(It is giving 12 for the above example..).What is wrong?

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Geoff Hayes
Geoff Hayes am 19 Mai 2020
vidushi - the problem is with this line
total=[total,sum(v(ii):v(jj))];
and in particular the inputs to the sum function. Here you are providing two integers (based on the input array) and using the : colon to produce an array of elements between these two. So if your two integers are 1 and 8, then
>> 1:8
ans =
1 2 3 4 5 6 7 8
which is an array of eight integers which isn't what you want. Instead, you want to extract from v a subset or sub-array of consecurtive elements of length 3. So you need to do
total=[total,sum(v(ii:jj))];

Kategorien

Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by