MATLAB Answers

## Perms output as trials to solve an equation

Asked by B

### B (view profile)

on 30 Apr 2014
Latest activity Commented on by Image Analyst

### Image Analyst (view profile)

on 3 May 2014
Is there a way to use the perms output as the variable values of an equation and identify a unique solution?
For example, perms(1:2) gives [2 1; 1 2]
If I have an equation 4x + 2y, can I use the different values (x=2, y=1 and x=1, y=2) in an equation to determine a desired solution (i.e. 10)?
If I wanted to have a final value of 10, the identified values would x=2 and y=1.

#### 0 Comments

Sign in to comment.

## 2 Answers ### Image Analyst (view profile)

Answer by Image Analyst

### Image Analyst (view profile)

on 30 Apr 2014

I would not think so. What makes you think that you would happen to have the solution already in your list of numbers that you're going to make permutations of? If you want to do it numerically, you could use linspace() to make a list of values for x and y, then use meshgrid() to get every (x,y) coordinate in the space, then plug those x,y into your equation, then search for the coordinate that has a value closest to your desired value (10) with min() and ind2sub.
% Make x and y axes.
x=linspace(1,10,100);
y = linspace(1, 20, 100);
% Get x and y over the entire grid.
[xg, yg] = meshgrid(x, y);
% Make the equation for every x and y.
z = 4*xg + 2* yg
% Display it.
image(z);
colormap(gray);
colorbar
% Find where target value lives.
targetValue = 10;
diffMatrix = abs(z - targetValue);
[minDiff, linearIndex] = min(diffMatrix(:))
[row, col] = ind2sub(size(z), linearIndex)

Image Analyst

### Image Analyst (view profile)

on 1 May 2014
Do you fully understand this section of the FAQ: http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F and how you need to test floating point numbers for "matching" values?
B

### B (view profile)

on 1 May 2014
That FAQ section makes perfect sense. I'm great with the conceptualization of what needs to be done, I just don't know how to implement it. Also, all of the numbers I'm using (both variables and coefficients) are integers, so I'm not worried about issues with floating point numbers.
My equation is -402u^6 + 67v^5 + 48w^4 + 295x^3 - 720y^2 + 1145z - 14648, and the values of u, v, w, x, y, z can only be 1-6.
Image Analyst

### Image Analyst (view profile)

on 3 May 2014
At the very inside of the 6 for loops, do this:
value = -402u^6 + 67v^5 + 48w^4 + 295x^3 - 720y^2 + 1145z - 14648;
if value == 10
foundAnswer = true;
message = sprintf('It is 10 for u=%d, v=%d, w=%d, x=%d, y=%d, z=%d',...
u,v,w,x,y,z);
uiwait(helpdlg(message));
break;
end
Before the 6 for loops have
foundAnswer = false;
and then at the end of each for loop, just before its end statement, you'll have:
if foundAnswer = true;
break;
end

Sign in to comment.

Answer by B

### B (view profile)

on 3 May 2014

Any help would be appreciated.

#### 0 Comments

Sign in to comment.