Hello, could you post the right code?
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Write a function called max_sum that takes v, a row vector of numbers, and n, a positive integer as inputs. The function needs to find the n consecutive elements of v whose sum is the largest possible.
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In other words, if v is [1 2 3 4 5 4 3 2 1] and n is 3, it will find 4 5 and 4 because their sum of 13 is the largest of any 3 consecutive elements of v. If multiple such sequences exist in v, max_sum returns the first one. The function returns summa, the sum as the first output argument and index, the index of the first element of the n consecutive ones as the second output.
I tried the follwing code but when there are two occurences for the same number, they both get removed in the same iteration and thus causes trouble.
function [summa, index] = max_sum(v,n)
summa = 0;
i = 0;
j = v;
m = [];
while i < n
summa = summa + max(j);
b = find(v==max(j));
m = [m b];
j = j(j<max(j));
i = i + 1 ;
end
t = sort(m);
index = t(1,1);
end
12 个评论
采纳的回答
Guillaume
2019-2-13
Don't use the question title to post the content of your assignment as that get truncated.
Assuming that your assigment is to find the sequence of length n of consecutive numbers with the highest sum, then I don't see how your algorithm even attempts that. You don't care about the maximum of the vector, it's completely irrevelant to the sequence sum. You want to calculate a sliding sum (i.e. first sum(v(1:n)), then sum(v(2:n+1)), etc. and find the maximum of these.
If you're allowed to use movsum, it can be trivially done in just two lines.
19 个评论
Maximilian Schmidt
2019-5-1
@Guillaume: Thank you for your post. I was looking for hints on the same assignment and came across your comment. It seems to be doing the job fine:
h = [6 45 9 67 -36 -34 99 64 67 8]; n = 4;
>> [summa, index] = max(movsum(h, n, 'Endpoints', 'discard'))
summa =
238
index =
7
What I don't understand is how Matlab knows what to do with 'index'. In the documentation for movsum there was no hint about what the funciton returns if you ask for two output arguments. To check that 'index' wasn't some kind of keyword; I tried the above again but used the names 'a' and 'b' instead of summa and index. As I assumed, the output was the same.
So my question: What part of the code tells matlab to output the index of the starting point of the sum in the second output argument?
Stephen23
2019-5-1
编辑：Stephen23
2019-5-1
@Maximilian Schmidt: movesum is called inside the function max, and it is max is called with two output arguments. So you need to read the max documentation.
A function called inside another function, like movesum inside max in your example, returns exactly one output argument as an input argument to the wrapper function, so
your example:
[summa, index] = max(movsum(h, n, 'Endpoints', 'discard'))
is exactly equivalent to
tmp = movsum(h, n, 'Endpoints', 'discard');
[summa, index] = max(tmp)
更多回答（37 个）
lalit choudhary
2019-4-17
After hours of brainstorming, i finally figured it out
try this code-
function [summa, index]=max_sum(b,n)
y=length(b)
if n>y
summa=0;
index=-1;
else
[summa, index] = max(movsum(b, n, 'Endpoints', 'discard'));
end
end
6 个评论
Shandilya Kiran Bhatt
2020-5-11
Can you explain nx-n+1 term? i did not understand why are you making a vector containing zeros of 1x nx-n+1 dimension
Walter Roberson
2020-5-11
When you have a vector of length nx to be taken in full windows of length n, sliding along 1 at a time, then you have nx - n + 1 full windows. For example, data = [1 2 3 4 5 6 7 8 9 10], n = 7, then nx = 10, nx - n + 1 is 10 - 7 + 1 = 4, corresponding to the windows 1:7, 2:8, 3:9, 4:10
This codes is specifically for the case where you only want full windows -- which is what the 'discard' option of movsum is intended to indicate, that you want to discard any sum that was made with a window that was not full (such as 5:10 not being the full length of 7)
Shubham Pandey
2020-4-3
function [s,i] = max_sum(v,n)
s = 0; i = 0;
len = length(v);
l = len-n+1;
if n>len
s = 0;
i = -1;
else
for j=1:l
if j==1
s = sum(v(j:n));
i=1;
end
if s<sum(v(j:n))
s = sum(v(j:n));
i = j;
end
if n<len
n=n+1;
end
end
end
end
3 个评论
Vipin Parthan
2020-5-29
编辑：Vipin Parthan
2020-5-29
if j==1
s = sum(v(j:n));
i=1;
Could you explain how this works?
Prasad Reddy
2020-4-14
function [summa,index] = max_sum(v,n) % defining a function with name "max_sum", input augements are (v,n) and out put augemnts are [summa,index]
[r,e]=size(v); % reading the size of vector "v" to "r,e"
summa=0; % asaining 0 to summa
index=-1; % assaining -1 to index, as they asked in the question
sums=zeros(1,e-(n-1)); % creating a zro vector of requirted size. you can try itby taling two or th trial cases
if n>e % checking if n is greatr than th length of the vector so that summa and index can be rturnd as
summa=summa; % 0 and -1 respectively as askeed in the question.
index=index;
else % if n is less than the length of vctor then
for i=1:e-(n-1) % for loop to run across the length of the vector
for j=0:n-1 % for loop to run across the required number of elements to be summed
sums(i)=sums(i)+v(i+j); % elments of sums which are initially zeros are being updated step by step with th sum of 'n' numbers in vector
end
end
[s,i]=max(sums); % reading th maximum valu and its indx from sums to 's' and 'i'
summa=s;
index=i;
end
end
0 个评论
asad jaffar
2019-4-4
jan and walter take a look at this code this is giving me correct answers except for negative elements array
function [summa, index]=max_sum(b,n)
y=length(b)
q=movsum(b,n)
[maxsum,maxidx]=max(q)
summa=maxsum
index=(maxidx-floor(n/2))
if n==y
j=1
end
end
command window
max_sum([ 79 15 -18 -28 -30 52 -81 31 -74 4 57 -96],4)
y =
12
maxsum=
94 76 48 -61 -24 -87 -28 -72 -120 18 -109 -35
summa =
94
index =
1
index =
1
but i think my code is giving the right value plz take a look and tell me what to do
6 个评论
Jan
2019-4-5
编辑：Jan
2019-4-5
@asad: Again: Read the documentation of movsum:
doc movsum
Then try it by your own in the command windows, e.g.:
x = rand(1, 6)
movsum(x, 2)
movsum(x, 2, 'Endpoints', 'discard')
You asked: "are guys even do coursera" - no, of course we do not solve a course for beginners. we have done this about 20 or 30 years ago (a bold but maybe matching guess). We are not going to do exactly what you are doing only to answer very easy programming questions.
"i am waiting" - this is definitely the wrong approach. Exhaustive hints and working code have been posted some hours earlier already.
Some comments to your code:
function [summa, index]=max_sum(b,n)
y=length(b) % This is not useful
q=movsum(b,n)
[maxsum,maxidx]=max(q)
summa=maxsum % Why not directly: [summa,maxidx]=max(q)
index=(maxidx-floor(n/2))
if n==y % While j is not used anywhere, omit this
j=1
end
end
A nicer version:
function [summa, index] = max_sum(b, n)
q = movsum(b, n);
[summa, maxidx] = max(q);
index = maxidx - floor(n / 2);
end
But the result is not correct: The first and last two elements of the sum (or in general: floor(n / 2) elements) are not the sum of n elements of the input vector. See: doc movsum. A solution:
function [summa, index] = max_sum(b, n)
q = movsum(b, n);
m = floor(n / 2);
[summa, index] = max(q(m+1:end-m+1));
end
Try to understand, why q is cropped here. Instead of cropping it is smarter to let movsum exclude the marginal elements already with setting 'Endpoints' to 'discard'.
Vikas Kumar
2019-6-11
function [summa, index] = max_sum(A,n)
if length(A)<n
summa = 0;
index = -1;
else
B = maxk(A,n)
summa = sum(B);
z = zeros(1,length(B));
m = [];
for i = 1:length(B)
z = find(A==B(i));
m = [m z]
end
T = sort(m)
index = T(1,1)
end
end
Jaimin Motavar
2019-6-29
function [summa,index]=max_sum(a,b)
n=length(a);
summa=0;
total=0;
if b>n
summa=0;
index=-1;
return
end
for i=1:(n-b+1)
for j=i:(b-1+i)
total=total+a(1,j);
end
if total>summa
summa=total;
index=i;
end
total=0;
end
end
2 个评论
Jan
2019-6-30
Some simplifications:
function [summa, index] = max_sum(a, b)
n = length(a);
summa = 0;
index = -1;
if b <= n
for i = 1:(n - b + 1)
total = sum(a(i:b - 1 + i));
if total > summa
summa = total;
index = i;
end
end
end
end
Roshan Singh
2019-9-15
function [summa, index]=max_sum(v,n)
L=length(v);
p=L-(n-1);
k=0;
t=0;
for i=1:p
z=n+k;
k=k+1;
s=0;
for d=i:z
s=s+v(d);
end
t(i)=s;
end
if (n<=L)
[summa index]=max(t);
else
summa=0;
index=-1;
end
end
0 个评论
Hussain Bhavnagarwala
2020-3-13
function [summa,index] = max_sum(v,n)
if n>length(v)
summa = 0;
index = -1;
return
end
x = (length(v) - n) + 1;
tot =[];
t=0;
total = 0;
for k = 1:x
for i =1:n
total = total + v(t+i);
end
tot(k) = total;
t=t+1;
total = 0;
end
summa = max(tot);
ind= find(tot==max(tot));
index = ind(1);
end
0 个评论
Shashidhar Rai
2020-3-15
编辑：Walter Roberson
2020-3-15
I have written the code but it's showing error. Can anyone tell me what's wrong in it.
function [summa, index] = max_sum(v, m)
ii=1;
if length(v)
index=-1;
summa=0;
elseif length(v) ==n
index=1;
summa=sum(v);
else
summa=0;
start=1;
for ii =start : start +n-1
if summa < sum (v [start : start+n-1] )
summa=sum(v[ start : start +n-1] ) ;
index = start;
end
start =start +1;
if start
length(v) - n+1
break;
end
end
end
2 个评论
Walter Roberson
2020-3-15
What is your expectation of what this code will do:
sum (v [start : start+n-1] )
?
Guillaume
2020-3-15
To add to the problem Walter pointed out:
- the function has an input m that is never used, but use the undefined variable n instead.
- there's a ii loop that never uses ii.
- that loops will only ever do one iteration because the if start is always going to be true and break out of it.
Irfan Hussain
2020-3-31
function [summa, index] = max_sum(v,n)
leng = length(v);
if leng < n
summa = 0;
index = -1;
else
w = [];
summa = 0;
for i = 1:leng
if summa < sum(v(i:n));
summa = sum(v(i:n));
index = i;
% x = sum(v(i:n));
%w = [w ,x];
if n < leng
n = n + 1;
%else
% summa = max(w);
% index = find(x == );
% end
end
end
end
end
1 个评论
Shubham Pandey
2020-4-3
this code has some error, wont work correctly for this case:
[summa index]=max_sum([ 45 -32 -25 27 78 -9 32 -1 -85 2 -32 -81 -32 -14 -31 46 -70 87 94 ], 15)
Jaimin Patel
2020-4-7
function [X,Y] = max_sum(v,n) % X and Y are output arguments
X = -inf; %take -inf value of X for getting minus sum
p = n;
Y = 1;
if n > length(v)
X = 0;
Y = -1;
end
for i = 1:(length(v)-(n-1))
a = sum(v(i:p));
p = p + 1;
if a > X %condition for taking maximum sum of consecutive elements of vector
X = a;
Y = i;
end
end
simple code for solution...
0 个评论
Emine Ertugrul
2020-4-13
function [summa,index]=max_sum(v,n)
m=length(v);
if n>m
summa=0;
index=-1;
else
M = movsum(v,n,'Endpoints', 'discard')
[a,b]=max(M)
summa = a
index = b
end
end
1 个评论
Walter Roberson
2020-4-13
I think the point of the assignment was not to use movsum() or other similar library functions, and to write the algorithm in more basic MATLAB.
Garvit Kukreja
2020-4-14
I have written this code. Can anyone tell me what's wrong in it.
Error : Variable summa has an incorrect value. max_sum([ -91 -57 -45 -8 -21 8 75 -80 89 -36 59 -71 -57 14 3 88 -79 -68 -6 ], 5) returned sum = 325 and index = 7 which is incorrect...
function [s,i]=max_sum(a,b)
B = maxk(a,b)
[n m]=size(B);
[y z] = size(a);
f=[]; %empty matrix
kk=0; %counter
if z>=b; %no. of element in 'a' more than or equal to'b'.
s=sum(B)
for ii=1:z;
for jj=1:m;
if B(jj)==a(ii);
kk=kk+1;
f(kk)=ii ;
end
end
end
i=min(f)
else
s=0
i= -1
end
2 个评论
Walter Roberson
2020-4-14
The sum that has to be returned is the deduced sum of the subset, not the sum of the entire vector.
Garvit Kukreja
2020-4-14
编辑：Garvit Kukreja
2020-4-14
Thanks for your response.
it is sum of deduced matrix 'B',
where B=Maxk(a,b) ( If a is a vector, then maxk returns a vector containing the k largest elements of a)
program worked for this command : [summa, index] = max_sum([1 2 3 4 5 4 3 2 1],3)
Ahmed J. Abougarair
2020-4-21
function [summa, index]=max_sum(v,n)
L=length(v);
p=L-(n-1);
k=0;
t=0;
for i=1:p
z=n+k;
k=k+1;
s=0;
for d=i:z
s=s+v(d);
end
t(i)=s;
end
if (n<=L)
[summa index]=max(t);
else
summa=0;
index=-1;
end
end
0 个评论
Ahmed J. Abougarair
2020-4-21
function [summa, index] = max_sum(v,n)
k =length(v);
if k==n
summa = sum(sum(v));
index =1;
return
elseif k<n
summa = 0;
index =-1;
return
else
z = length(v);
summa = 0;
index = -1;
if n <= z
for i = 1:(z - n + 1)
total = sum(v(i:n - 1 + i));
if total > summa
summa = total;
index = i;
end
end
end
end
0 个评论
Roshan Barnwal
2020-4-27
function [summa, index] = maxsum(v,n)
summa = sum(v(1:n-1));
for j = 2:length(v) -2;
r = sum(v(j:j+n-1));
if summa<r
summa=r;
index = j;
end
end
%% this works well but in few cases it asks for some error. can any one help me to short out the problem?
0 个评论
Salman P H
2020-4-30
function [summa,index] = max_sum(v,n)
if n>length(v)
summa = 0;
index = -1;
return;
end
x=1;
y=n;
great = -inf;
while y<=(length(v))
z = sum(v(1,x:y));
if z>great
great=z;
a=x;
x=x+1;
y=y+1;
elseif great==z
a=x-(x-a);
x=x+1;
y=y+1;
else
x=x+1;
y=y+1;
end
end
summa = great;
index = a;
end
0 个评论
Olel Arem
2020-4-30
function [summa,index]=max_sum(v,n)
len_v=length(v);summa=0;poss=[];ii=1;jj=n;
if n>len_v
summa=0;
index=-1;
return
end
while jj<=len_v
poss(ii)= sum(v(ii:jj));
[summa,index]=max(poss);
ii=ii+1;
jj=jj+1;
end
0 个评论
Prithvi Shams
2020-5-5
Here's my version of the code.
While the problem can be solved with movsum() in a single line, it's recommended to utilize if and for loops as a learning exercise.
function [summa, index] = max_sum(v, n)
if n > numel(v)
summa = 0;
index = -1;
else
j = n;
for i = 1:(numel(v) - n + 1)
s(i) = [sum(v(i:j))];
j = j + 1;
end
if numel(max(s)) > 1
m = max(s);
[summa index] = m(1);
else
[summa index] = max(s);
end
end
0 个评论
anuj chaudhari
2020-5-5
编辑：anuj chaudhari
2020-5-5
function [a, b] = max_sum(v,n)
if ~isscalar(n)||n<0||n~=fix(n)
error('n must be a positive integer')
end
[~,c]=size(v);
d=length(v);
r = zeros(1,d-n+1);
if n>c
b=-1;
a=0;
else
for k=1:(d-n+1) %from here, this is a means of movsum
r(k)=sum(v(k:k+n-1));
[a, b]=max(r);
end
end
end
0 个评论
Shandilya Kiran Bhatt
2020-5-11
You can also solve this question by using while loop if you don't know movsum function like me.Look at following code:-
function [summa , index ] = max_sum(A,n)
b = length(A);
c = zeros(1,b-n+1);
i = 1;
if b < n
summa = 0;
index = -1;
else
while n<=b && i<=b
c(1,i) = sum(A(i:n));
n = n+1;
i = i+1;
end
[summa , index ] = max(c);
end
0 个评论
Tahsin Oishee
2020-5-17
function [summa,index]=max_sum(v,n)
summa=0;
index=0;
w=1;
[b a]=size(v);
if n>a
index=-1
summa=0;
else
if n<=a
for i=1:(a+1-n)
sum=0;
for j=1:n
sum=sum+v(w);
w=w+1;
end
w=w-n+1;
if sum>summa;
summa=sum;
index=v(i);
end
summa
index
end
end
end
end
2 个评论
Walter Roberson
2020-5-17
You already have two loop control variables, i and j: use 2D indexing instead of doing strange things with w.
We recommend against naming a variable sum: it is very common to want to use the sum() function after having used sum as a variable.
khyathi beeram
2020-5-18
编辑：khyathi beeram
2020-5-18
function [summa index]=max_sum(v,n)
a=[ ];
if n<=length(v)
for i=1:length(v)-(n-1)
sum=0;
for j=i:i+n-1
sum=sum+v(j);
end
a(i)=sum;
end
summa=max(a);
g=find(a==summa);
index=g(1,1);
else
summa=0;
index=-1;
end
end
0 个评论
vighnesh rana
2020-5-19
function [summa,index] = max_sum(A,n)
if length(A)< n
summa = 0;
index = -1;
return;
end
summa = -inf;
index = -1;
for i = 1:(length(A)-n+1)
total = sum(A(i:(i+n-1)));
if total > summa
summa = total;
index = i;
end
end
end
0 个评论
Taif Ahmed BIpul
2020-5-20
function[summa,index]=max_sum(v,n)
if size(v,2)<n
summa=0;
index=-1;
return
end
p=size(v,2)-(n-1);
k=1;
A=zeros(1,n);
summ1=zeros(1,p);
for i=1:p
jj=i;
while jj<=(i+n-1)
while k<=n
A(k)=v(jj);
jj=jj+1;
k=k+1;
end
end
summ1(i)=sum(A);
k=1;
end
summa=max(summ1);
index1=find(summ1==max(summ1));
if size(index1,2)>1
index=index1(1);
return
else
index=index1;
end
0 个评论
utkarsh singh
2020-5-21
function [summa, index]=max_sum(m,n)
if numel(m)<n
summa=0;
index=-1;
else
summa=-inf;
for i=1:(numel(m)-n+1)
j=m(i:i+n-1);
maxsumma=sum(j);
if(maxsumma>summa)
summa=maxsumma;
index=i;
end
end
end
end
0 个评论
sai teja
2020-5-28
function [summa, ind] = max_sum(v,n)
% If n is greater than v return the specified values
% Using return keyword exits the function so no further code is
% evaluated if n > length(v) summa = 0; ind = -1;
return;
end
% Initialize summa to -inf.
% Then work through the vector, checking if each sum is larger than the
% current value of summa summa = -inf; ind = -1;
% Once we get to length(v)-n+1
we stop moving through the vector for ii = 1:length(v)-n+1
currentV = v(ii:(ii+n-1));
currentSumma = sum(currentV);
% If currentSumma greater than summa, update summa and ind
if currentSumma > summa
summa = currentSumma;
ind = ii;
end
end
end
0 个评论
Kumar Shubham
2020-6-2
编辑：Kumar Shubham
2020-6-2
function [summa, index]= max_sum(v,n)
if n > length(v)
summa = 0;
index = -1;
return;
elseif n<=length(v)
a=movsum(v,[0,n-1]);
b=a(1:length(v)-n+1);
[summa,index]= max(b);
end
end
0 个评论
Sumit Kumar Sharma
2020-6-2
function [summa, index]=max_sum(v,n)
l=length(v);
if n>l
summa=0;
index=-1;
else
p=l-n+1;
ii=0;
total=-inf;
for j=1:p
s=sum(v(j:n+ii));
ii=ii+1;
if s>total
total=s;
index=j;
else
continue;
end
end
summa=total;
end
end
0 个评论
KAVITI BHARGAV RAM NAIDU
2020-6-3
编辑：KAVITI BHARGAV RAM NAIDU
2020-6-5
function [summa,index] = max_sum(v,n)
b=zeros(1,length(v)-n+1); % initializing the b vector(only zeros) of length = length(v)-n+1
% this will decrease time taken for getting answer
if n > (length(v)) % if n is larger than length of v sum = 0 and index = -1 as per question (last line)
summa = 0 ;
index = -1;
else
for p = 1:[length(v)-n+1]
b(1,p) = sum (v(p:p+n-1)); % b(1,1)= v(1,1)+v(1,2) (if n = 2)
end % giving values to vector b using sum function
[summa index] = max(b); % if n = 2 sum should be between 2 consecutive elements.
end % if p = 2 and n = 2 sum between 2 nd and 3rd element should be considered
% p+n-1 = 2+2-1 = 3 // sum (v(2:3)) means sum of 2nd and 3rd element of vector b.
% so sum(v(p:p+n-1))
0 个评论
zineb britel
2020-6-4
%what's wrong with my code
function [summa index]=max_sum(v,n)
index=0; summa=0;
if n > length(v)
summa= 0;
index=-1;
else
a=sort(v,'descend');
summa= sum(a(1:n));
index_row= find(v>=a(n));
index= min(index_row);
end
end
2 个评论
KAVITI BHARGAV RAM NAIDU
2020-6-5
编辑：KAVITI BHARGAV RAM NAIDU
2020-6-5
- By using sort function u will get all elements in descending order
- But according to question
- we should keep the order of elements same.
- case 1 : if v = [1 2 3 4 5]; n = 2;
- summa is maximum of (1+2),(2+3),(3+4),(4+5)
- index is position where we are getting maximum sum here maximum sum is 9 so index = 4
- case 2 : if v = [1 2 3 4 5 ];n = 3;
- summa is maximum of (1+2+3),(2+3+4),(3+4+5)
- here maximum sum is 12 so index = 3
- case 3 : if v = [1 2 3 4 5 5 4 3 ];n=2;
- summa is maximum of (1+2),(2+3),(3+4),(4+5),(5+5),(5+4),(4+3)
- here maximum sum is 10 so index = 5
Ujjawal Barnwal
2020-6-8
function [summa , index]=max_sum(v,n)
summa=sum(v(v<0));
for ii=1:(length(v)-(n-1))
t=0;
for jj=ii:(ii+(n-1))
t=t+v(jj);
end
if (t>summa)
summa=t;
index=ii;
end
end
end
0 个评论
AYUSH MISHRA
2020-6-13
编辑：AYUSH MISHRA
2020-6-13
function [summa, index] = max_sum(v,n)
if n > length(v)
summa = 0;
index = -1;
return;
end
summa = -inf;
index = -1;
for ii = 1:length(v)-n+1
currentV = v(ii:(ii+n-1));
currentSumma = sum(currentV);
if currentSumma > summa
summa = currentSumma;
index = ii;
end
end
end
0 个评论
Maddireddy Harshavardhan Reddy
2020-6-18
Create a row vector named x that starts at 1, ends at 10, and contains 5 elements.
1 个评论
Sneha Paul
2020-6-19
function [summa,index]=max_sum(v,n)
z=[];
if n>length(v)
summa=0;
index=-1;
return
elseif (n<=length(v))
for i=1:length(v)-(n-1)
z(i)=sum(v(i:i+(n-1)));
end
end
[summa,index]=max(z)
0 个评论
MICHAEL
2020-6-19
function [s,i] = max_sum(v,n)
s = 0; i = 0;
len = length(v);
l = len-n+1;
if n>len
s = 0;
i = -1;
else
for j=1:l
if j==1
s = sum(v(j:n));
i=1;
end
if s<sum(v(j:n))
s = sum(v(j:n));
i = j;
end
if n<len
n=n+1;
end
end
end
end
0 个评论
Sakib Javed
2020-6-22
编辑：Sakib Javed
2020-6-22
ffunction [summa index] = max_sum(A,n)
m=length(A);
if n > m
summa = 0;
index = -1;
return;
end
B = [];
q=m-(n-1);
for ii = 1:q
B=[B sum(A(ii:ii+n-1))];
end
[summa index] = max(B);
end
0 个评论
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