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fahana1710
Group Title
let A and B be two symmetric matrices of the same order. Develop an algorithm to compute C=A+B,taking advantage of symmetry for each matrix.Your algorithm should overwrite B with C. What is the flop count?
 11 months ago
 11 months ago
fahana1710 Group Title
let A and B be two symmetric matrices of the same order. Develop an algorithm to compute C=A+B,taking advantage of symmetry for each matrix.Your algorithm should overwrite B with C. What is the flop count?
 11 months ago
 11 months ago

This Question is Closed

myininaya Group TitleBest ResponseYou've already chosen the best response.1
So we know that this means A[i,j]=A[j,i] , B[i,j]=B[j,i], and C[i,j]=C[j,i]=A[i,j]+B[i,j] A and B are symmetrical. This means A+B is symmetrical.
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
okay, but how to write the pseudo code for the algorithm?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
You need to make a loop I think the loops usually use something like for...do
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
okay..it is correct if i do like this : Input: Dim n for \[a _{ij}\], j>=i and \[b _{ij}\],j>=i where i,j=1,...,n Output : B=C Step 1 : for j=1,...,n for i=1,...,j \[b _{ij}^{new}=a _{ij}+b _{ij}\]
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
but, i don't have any idea to make this as programming.
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I'm thinking... Does your algorithm use the fact that A and B are symmetrical? We should be able to do something like Step 1 : for j=1,...,n for i=1,...,j if i does not equal j, then output cij=cji=aij+bij if i equals j, then cij=aij+bij What do you think of this part? Do you think you can use it? Just trying to help.
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
yes. it's also correct. but do you know how to make it into pseudo code form?such as using mathematica or MATLAB?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
err...nope. It has been way to long. I think I could do it with maple after trail and error. But I don't have maple on this computer.
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
Owh, it's okay. One more question, do you know how to do calculate flopcount? My friend get \[\frac{ n(n+1) }{ 2 }\] is it correct?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
actually that is what I got. You want to know how I figured it out... Ok we want to basically know how many operations we did. And I had to look up the definition of flops. So we want to count the number of operations we did. I counted my diagonal elements which were n. Then for the first row we had (n1) entries left to calculate second row we had (n2) entries left to calculate third row we had (n3) entries left to calculate ....... nth row we had (nn) entries left to calculate So we had n diagonal entries + (n1)+(n2)+...+(nn) other entries to calculate. So we have n+sum(ni, i=1..n) Do you understand how I got this so far? We can simplify this to what your friend got.
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[n+\sum_{i=1}^{n}(ni)\] \[n+\sum_{i=1}^{n}n  \sum_{i=1}^{n}i\] \[n+n(n)\frac{n(n+1)}{2}\] \[n(n+1)\frac{n(n+1)}{2}=\frac{2n(n+1)n(n+1)}{2}=\frac{(n+1)(2nn)}{2}=\frac{1n(n+1)}{2}\]
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I counted each entry we had to calculate. I know the diagonal entries will be repeated once but all other entries will be counted twice. So you know if we had a 3 by 3 symmetric matrix. You find the sum of the operations we did by counting the diagonal entries and the entries above the diagonal. So for a 3 b3 symmetric matrix, we have 3 diagonal entries then the number of entries above that diagonal is (31)+(32)+(33) (31)=2 that is how many are above the diagonal in the first row (32)=1 that is how many are above the diagonal in the second row (33)=0 that is how many are above the diagonal in the 3rd row Now I ignored everything below the diagonal because we already performed those operations above the diagonal. We were suppose to use the fact that the matrix was symmetric after all.
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
So using this we have 3+(31)+(32)+(33) or you can use the simplified formula 3(3+1)/2
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Makes sense how I found the flops?
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
yes. i'm still trying to understand what you explain. =)
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
[ a11 a12 a13] [ a22 a23] [ a33] Or you could have looked at is in much simpler way lol 3+2+1 =3(3+1)/2
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
for a 4by 4, we could have said (4+3+2+1)
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I made it more complicated than it should have been Do you see what I mean so for an n by n, we do 1+2+...+n
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
But either way the formula can be simplified to that n(n+1)/2 one
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
ohhh..now i understand already how to get n(n+1)/2. it's simple but quite tricky. =)
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
dw:1381169885319:dw 1+2+...(n1)+n \[\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\]
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
dw:1381169994491:dw
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
woww..you tried so hard to make me understand. thank you so much. now, honestly, i understand already. =) are you a lecturer?
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
oh, i see. until right now, i'm still dont know how to develop an algorithm to compute c=a+b.
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Is it because you aren't sure what is acceptable to write exactly?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
And also because you want to input in the matlab and see if it works?
 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Yeah. I will have to do some review on that part. I'm going to class soon. I can't promise I will look at it tonight but I think @zarkon might know matlab. I could be wrong. No promises. lol.
 11 months ago

fahana1710 Group TitleBest ResponseYou've already chosen the best response.0
hehe. It's okay. You helped me a lot today. Actually, i'm going to sleep already because in Malaysia, right now is midnight. Thank you so much for your helped. May God bless you. =)
 11 months ago
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