Skip to main content

graphs and networks - Binary decision diagrams (BDD)


I would like to use Mathematica to help me simplify a given Binary Decision Diagram (i.e. pick the optimal variable ordering). I don't seem to find anything at all on the topic, although it is something everyone with a little CS background knows very well.



Given MMA's ability of dealing with graphs and boolean functions the task seems feasible, but I wouldn't want to waste a long time to solve a problem that is already solved. Is anybody aware of MMA resources on the matter?



Answer



This answer is a modification of my answer given in the discussion Creating Identification/Classification trees.


With this solution I am trying to achieve the simplification by using the impurity function applied to the data (the truth table in this case).


Make the truth table from the linked Wikipedia article (Binary Decision Diagram):


truthTable = {{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {0, 1, 1, 
1}, {1, 0, 0, 0}, {1, 0, 1, 0}, {1, 1, 0, 1}, {1, 1, 1, 1}};
truthTable = Map[ToString, truthTable, {-1}];
colNames = {"x1", "x2", "x3", "f"};
TableForm[truthTable, TableHeadings -> {None, colNames}]


enter image description here


Import the package:


Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/AVCDecisionTreeForest.m"]

Build a decision tree:


dtree = BuildDecisionTree[truthTable, "ImpurityFunction" -> "Gini"]

(* {{0.125, "0", 2, Symbol,
8}, {{0.125, "0", 1, Symbol,

4}, {{0.5, "0", 3, Symbol,
2}, {{{1, "1"}}}, {{{1, "0"}}}}, {{{2, "0"}}}}, {{0.125, "0", 1, Symbol,
4}, {{0.5, "0", 3, Symbol, 2}, {{{1, "0"}}}, {{{1, "1"}}}}, {{{2, "1"}}}}} *)

The impurity function "Gini" is used by default. Another possible value is "Entropy". Note that to make the decision tree I converted the 0-1 values in the truth table into strings.


Visualize the tree:


trules = dtree // DecisionTreeToRules;
trules = trules /. ({m_, v_, cInd_Integer, s_, n_, id_} :> {m, v, colNames[[cInd]], s, n, id});
LayeredGraphPlot[trules, VertexLabeling -> True]


The last commands produce the plot:


enter image description here


The second row of a leaf shows the corresponding number of records and label.


The tree rules can be manipulated to produce a graph closer to the one in the Wikipedia article, if that is desired.


See these blog posts for application examples, parameters explanations, and tweaking discussions:


Decision trees posts at MathematicaForPrediction at WordPress.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...