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I have $2$ known grayscale images ($256 \times 256$ matrices) $A$ and $B$ and want to find the unknown variable $x$ so that

$$\text{minimize} \quad \| xA-B\|_F^2$$

The goal is to find the best multiplier $x$ that is able to scale the brightness of $A$ to $B$.

I don't know how I should approach a problem like this and how to solve it. Is it an optimization problem? If so, then of what kind and how can I solve it? I'd also appreciate mentioning books or references in this field of work.

Thanks.

Rodrigo de Azevedo
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Cypher
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  • minimizing $xA-B$ in which sense ? You know that $xA-B$ is a vector. So are you minimizing the sum of components, or are you minimizing the max of components ? The question lacking information – Ahmad Bazzi Sep 28 '19 at 09:08
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    Would be easier if you’d like to minimize sum of components squared. – mathcounterexamples.net Sep 28 '19 at 09:15
  • @mathcounterexamples.net Thanks. I changed the question to that. Could you please also mention why $xA-B$ would be harder than $(xA-B)^2$ ? – Cypher Sep 28 '19 at 09:16
  • See my answer below. – mathcounterexamples.net Sep 28 '19 at 09:24
  • @RodrigodeAzevedo You're right but Would looking at the images (matrices) as reshaped vectors help? for example reshaping the $[256*256]$ matrix to a $[65536*1]$ vector? That way only a single index is enough. Does reshaping the images to vectors and minimizing their squared difference result is a different answer than when we don't do this? – Cypher Sep 28 '19 at 09:42
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    @Cypher Vectorizing makes no difference whatsoever. You have a least-squares problems with matrices. The solution has the same form as if they were vector. Just use the Frobenius inner product instead of the standard inner product for vectors and use the Frobenius norm instead of the Euclidean norm. – Rodrigo de Azevedo Sep 28 '19 at 09:53

3 Answers3

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Let $f : \mathbb R \to \mathbb R_0^+$ be defined as follows

$$f (x) := \| x \mathrm A - \mathrm B \|_{\text{F}}^2 = x^2 \| \mathrm A \|_{\text{F}}^2 - 2 x \langle \mathrm A , \mathrm B \rangle + \| \mathrm B \|_{\text{F}}^2$$

Taking the 1st derivative and finding where it vanishes, we obtain the least-squares solution

$$\hat{x} := \frac{\langle \mathrm A , \mathrm B \rangle}{\| \mathrm A \|_{\text{F}}^2} = \frac{\langle \mathrm A , \mathrm B \rangle}{\langle \mathrm A , \mathrm A \rangle}$$

Rodrigo de Azevedo
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To find the best fit for $B=xA$, you might formulate one of three seemingly equivalent cost functions: $$\eqalign{ &(1)\;\quad &\min_x \;\|xA-B\|_F^2 \\ &(2)\;\quad &\min_x \;\|A-\tfrac{1}{x}B\|_F^2 \\ &(3)\;\quad &\min_x \;\|{\sqrt x}A-\tfrac{1}{{\sqrt x}}B\|_F^2 \\ }$$ each of which can be solved using the method that Rodrigo outlines for problem $(1)$ $$\eqalign{ &x_1 = \frac{\langle A,B\rangle}{\langle A,A\rangle},\quad &x_2 = \frac{\langle B,B\rangle}{\langle A,B\rangle},\quad &x_3 = {\rm sign}\Big(\langle A,B\rangle\Big){\sqrt\frac{\langle B,B\rangle}{\langle A,A\rangle}} }$$ The advantage of the third formula is that it provides a usable value when the matrices are nearly perpendicular, i.e. $\langle A,B\rangle\sim 0.\,$ When the matrices are parallel, all three formula are identical. In any other situation, it's not obvious which formula is "better".

You could also multiply each side of the fit equation by a third matrix $M$, and solve for $x$. $$\langle B,M\rangle=x\langle A,M\rangle \quad \implies x=\frac{\langle B,M\rangle}{\langle A,M\rangle}$$ Note that setting $M\!=\!A$ yields $x_1$, while $M\!=\!B$ yields $x_2.\,$

Another sensible choice might be a matrix, each of whose elements equals the mean grayscale value. This is equivalent to swapping the Frobenius norm for the Manhattan norm.

Or set $M=\tfrac{1}{2}(A+B)$, which yields the mediant of $(x_1,x_2)$.

greg
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Denote $(A_i)$ and $(B_i)$ the components of both images.

Minimizing

$$f_2(x)=\sum(xA_i-B_i)^2$$ is quite simple and leads to the solution

$$X= \frac{\sum A_iB_i}{\sum A_i^2}$$

Trying to minimize

$$f_1(x)=\sum \vert xA_i-B_i \vert$$ is harder as this map is not always differentiable.

mathcounterexamples.net
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  • Thanks for the quick response, Could I also ask for the full explanation of how we end up with the final solution. If it's time consuming, even giving a reference to an article or book would be enough. I want to start learning this field and a reference would really help. – Cypher Sep 28 '19 at 09:25
  • This may help. – mathcounterexamples.net Sep 28 '19 at 09:30