ex-6: EMD section completed
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@ -38,8 +38,8 @@ int show_help(char **argv) {
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/* Performs an experiment consisting in
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*
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* 1. Measuring the distribution I(θ) by reverse
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* sampling from an RNG;
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* 1. Measuring the distribution I(θ) sampling from
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* an RNG;
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* 2. Convolving the I(θ) sample with a kernel
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* to simulate the instrumentation response;
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* 3. Applying a gaussian noise with σ=opts.noise
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@ -423,36 +423,22 @@ deconvolved outcome with the original signal was quantified using the earth
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mover's distance.
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In statistics, the earth mover's distance (EMD) is the measure of distance
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between two probability distributions [@cock41]. Informally, the distributions
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are interpreted as two different ways of piling up a certain amount of dirt over
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a region and the EMD is the minimum cost of turning one pile into the other,
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where the cost is the amount of dirt moved times the distance by which it is
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moved. It is valid only if the two distributions have the same integral, that
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is if the two piles have the same amount of dirt.
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between two distributions [@cock41]. Informally, if one imagines the two
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distributions as two piles of different amount of dirt in their respective
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regions, the EMD is the minimum cost of turning one pile into the other,
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making the first one the most possible similar to the second one, where the
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cost is the amount of dirt moved times the distance by which it is moved.
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Computing the EMD is based on a solution to the transportation problem, which
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can be formalized as follows.
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Consider two vectors $P$ and $Q$ which represent the two probability
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distributions whose EMD has to be measured:
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Consider two vectors $P$ and $Q$ which represent the two distributions whose
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EMD has to be measured:
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$$
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P = \{ (p_1, w_{p1}) \dots (p_m, w_{pm}) \} \et
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Q = \{ (q_1, w_{q1}) \dots (q_n, w_{qn}) \}
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$$
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L'istogramma P deve essere distrutto in modo tale da ottenere l'istogramma Q,
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che in partenza è vuoto ma so che vorrò avere w_qj in ogni bin che sta alla
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posizione qj.
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- sposto solo da P a Q
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- sposto non più di ogni ingresso di P
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- ottengo non più di ogni ingreddo di Q
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- sposto tutto quello che posso: o ottengo tutto Q o ho finito P
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e non devono venire uguali, quindi!
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where $p_i$ and $q_i$ are the 'values' (that is, the location of the dirt) and
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$w_{pi}$ and $w_{qi}$ are the 'weights' (that is, the quantity of dirt). A
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ground distance matrix $D_{ij}$ is defined such as its entries $d_{ij}$ are the
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@ -464,28 +450,40 @@ $$
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W (P, Q, F) = \sum_{i = 1}^m \sum_{j = 1}^n f_{ij} d_{ij}
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$$
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with the constraints:
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The fact is that the $Q$ region is to be considerd empty at the beginning: the
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'dirt' present in $P$ must be moved to $Q$ in order to reach the same
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distribution as close as possible. Namely, the following constraints must be
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satisfied:
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\begin{align*}
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&f_{ij} \ge 0 \hspace{15pt} &1 \le i \le m \wedge 1 \le j \le n \\
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&\sum_{j = 1}^n f_{ij} \le w_{pi} &1 \le i \le m \\
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&\sum_{i = 1}^m f_{ij} \le w_{qj} &1 \le j \le n
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\end{align*}
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$$
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\sum_{j = 1}^n f_{ij} \sum_{j = 1}^m f_{ij} \le w_{qj}
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&\text{1.} \hspace{20pt} f_{ij} \ge 0 \hspace{15pt}
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&1 \le i \le m \wedge 1 \le j \le n
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\\
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&\text{2.} \hspace{20pt} \sum_{j = 1}^n f_{ij} \le w_{pi}
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&1 \le i \le m
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\\
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&\text{3.} \hspace{20pt} \sum_{i = 1}^m f_{ij} \le w_{qj}
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&1 \le j \le n
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\\
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&\text{4.} \hspace{20pt} \sum_{j = 1}^n f_{ij} \sum_{j = 1}^m f_{ij} \le w_{qj}
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= \text{min} \left( \sum_{i = 1}^m w_{pi}, \sum_{j = 1}^n w_{qj} \right)
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$$
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\end{align*}
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The first constraint allows moving 'dirt' from $P$ to $Q$ and not vice versa.
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The next two constraints limits the amount of supplies that can be sent by the
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values in $P$ to their weights, and the values in $Q$ to receive no more
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supplies than their weights; the last constraint forces to move the maximum
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amount of supplies possible. The total moved amount is the total flow. Once the
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transportation problem is solved, and the optimal flow is found, the earth
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mover's distance $D$ is defined as the work normalized by the total flow:
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The first constraint allows moving dirt from $P$ to $Q$ and not vice versa; the
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second limits the amount of dirt moved by each position in $P$ in order to not
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exceed the available quantity; the third sets a limit to the dirt moved to each
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position in $Q$ in order to not exceed the required quantity and the last one
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forces to move the maximum amount of supplies possible: either all the dirt
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present in $P$ has be moved, or the $Q$ distibution is obtained.
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The total moved amount is the total flow. If the two distributions have the
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same amount of dirt, hence all the dirt present in $P$ is necessarily moved to
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$Q$ and the flow equals the total amount of available dirt.
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Once the transportation problem is solved and the optimal flow is found, the
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EMD is defined as the work normalized by the total flow:
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$$
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D (P, Q) = \frac{\sum_{i = 1}^m \sum_{j = 1}^n f_{ij} d_{ij}}
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\text{EMD} (P, Q) = \frac{\sum_{i = 1}^m \sum_{j = 1}^n f_{ij} d_{ij}}
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{\sum_{i = 1}^m \sum_{j=1}^n f_{ij}}
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$$
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@ -494,28 +492,29 @@ procedure simplifies a lot. By representing both histograms with two vectors $u$
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and $v$, the equation above boils down to [@ramdas17]:
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$$
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D (u, v) = \sum_i |U_i - V_i|
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\text{EMD} (u, v) = \sum_i |U_i - V_i|
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$$
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where the sum runs over the entries of the vectors $U$ and $V$, which are the
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cumulative vectors of the histograms.
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In the code, the following equivalent recursive routine was implemented.
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cumulative vectors of the histograms. In the code, the following equivalent
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recursive routine was implemented.
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$$
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D (u, v) = \sum_i |D_i| \with
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\text{EMD} (u, v) = \sum_i |\text{EMD}_i| \with
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\begin{cases}
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D_i = v_i - u_i + D_{i-1} \\
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D_0 = 0
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\text{EMD}_i = v_i - u_i + \text{EMD}_{i-1} \\
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\text{EMD}_0 = 0
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\end{cases}
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$$
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In fact:
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\begin{align*}
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D (u, v) &= \sum_i |D_i| = |D_0| + |D_1| + |D_2| + |D_3| + \dots \\
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&= 0 + |v_1 - u_1 + D_0| +
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|v_2 - u_2 + D_1| +
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|v_3 - u_3 + D_2| + \dots \\
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\text{EMD} (u, v) &= \sum_i |\text{EMD}_i| = |\text{EMD}_0| + |\text{EMD}_1|
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+ |\text{EMD}_2| + |\text{EMD}_3| + \dots \\
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&= 0 + |v_1 - u_1 + \text{EMD}_0| +
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|v_2 - u_2 + \text{EMD}_1| +
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|v_3 - u_3 + \text{EMD}_2| + \dots \\
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&= |v_1 - u_1| +
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|v_1 - u_1 + v_2 - u_2| +
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|v_1 - u_1 + v_2 - u_2 + v_3 - u_3| + \dots \\
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@ -526,19 +525,8 @@ In fact:
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&= \sum_i |U_i - V_i|
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\end{align*}
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\textcolor{red}{EMD}
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These distances were used to build their empirical cumulative distribution.
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\textcolor{red}{empirical distribution}
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At 95% confidence level, the compatibility of the deconvolved signal with
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the original one cannot be disporoved if its distance from the original signal
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is grater than \textcolor{red}{value}.
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\textcolor{red}{counts}
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This simple formula enabled comparisons to be made between a great number of
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results.
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## Results comparison {#sec:conv_results}
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