mutiplying a role of number and seach or max and min value

let say a=[1:1:3] and b=[3:1:5]

and c = a*b

do i need to do anyting to make sure it multiply so it goes like 1*3 1*4 1*5 , 2*3 2*4 etc instead of 1*3 2*4 3*5?

and how do you find a and b when c is max or min?

Answer 1

[Honglei Chen]

You can use kron to do what you want.

kron(a,b)

doc kron

As to finding out what a and b is, given the regular structure in the product, once you know the index for max and min in c, it should be fairly easy to find out what the corresponding index in a and b. Using your example, let's say the index for max in c is Nc, then the index in a (Na) and b (Nb) can be calculated like

Na = fix(Nc-1,3)+1
Nb = Nc-(Na-1)*3

HTH

Answer 2

[Thomas]

Kron is a good way to do it,

However you could code multiple for loops and work it.. I hope this was not a homework problem..

c=[];
for a=
for b=
    c1=a*b;
c=[c,c1];
if c1==max(c)
    maxab=[a,b];
end
end
end
max(c)
maxab

[Jan]

"close all; clear all" is useless here. See: http://www.mathworks.com/matlabcentral/answers/16484-good-programming-practice#answer_22301

Answer 3

[Andrei Bobrov]

a=1:3;
b=3:5;
c = a(:)*b(:).';
[~, a2] = cellfun(@(x)x(c(:)),{@min,@max});
[iout,jout] = ind2sub(size(c),a2);
c_min = [a(iout(1)) b(jout(1))]
c_max = [a(iout(2)) b(jout(2))]

ADD 19:01MDT 27.11.2011

out = bsxfun(@rdivide,a.',b)

or (20:14MDT 27 Nov 2011)

out = a.'*(1./b) 

[~, a3] = cellfun(@(x)x(out(:)),{@min,@max});
idx_a = rem(a3-1,numel(a))+1; % for min and max of a/b
idx_b = ceil(a3/numel(a));

Answer 4

[Image Analyst]

Here, run this code and I think you'll see what you're trying to figure out:

a=[1:1:3] % Row vector using your strange syntax.
b=[3:1:5] % Another row vector.

c1 = a' * b % Note transpose operator '
c2 = a .* b % Note dot

From your description, it sounds like you're wanting to make sure you get c1 and not c2. Results:

a =
     1     2     3
b =
     3     4     5
c1 =
     3     4     5
     6     8    10
     9    12    15
c2 =
     3     8    15

Now, for the max part:

% Find max of c1
cMax = c1 == max(c1(:))
% Find row(s) and columns(s) where c1 = cMax.
[rowsOfMax colsOfMax] = find(cMax)
% Extract out a and b where c == cMax.
aAtCMax = a(rowsOfMax)
bAtCMax = b(colsOfMax)

Do similar for the min of c1.

[kevin]

how would you do the same if it is a/b? also i dont find max part

[Andrei Bobrov]

see my ADD

[Image Analyst]

kevin, the max part was the code following where it says "Now, for the max part." Use / instead of * for division or matrix inverting.

andrei: How did that ADD and the time stamp get in there? Sometimes I see things like that or that say EDIT. Did you put that there or does it add it automatically? I never see anything with my code when I modify it.

[Andrei Bobrov]

I am put this is

[kevin]

if i do

a=[1:1:3]

b=[3:1:5]

c=a\b

i get

0 0 0

0 0 0

1.0000 1.3333 1.6667

if i do a/b i get c=0.52, both are not the answer i want, i want answer for 1/3,1/4,1/5,2/3... etc, how can i do that? thx

[Image Analyst]

Like this:

bReciprocal = 1 ./ b

c= (a * bReciprocal)'

c =

0.3333 0.6667 1.0000

0.2500 0.5000 0.7500

0.2000 0.4000 0.6000

[kevin]

it didnt work, here is the code i used

a=[1:1:3]

b=[3:1:5]

bReciprocal =1./b

c=(a*bReciprocal)'

and i get

bReciprocal = 0.3333 0.2500 0.2000

??? Error using ==> mtimes

Inner matrix dimensions must agree.

is there any way to treat a and b as a range of number instead as a matrix?

[kevin]

my actual question is more like a b and c is a number range from 1 to 10, d is range from 10 to 34)

and i try to f= (a/(b*c))*tan(d)