Deriving boolean expressions from hand drawn logic gate diagrams with python OpenCv - python-3.x

Using tensorflow I identified all the gates, letters and nodes.
When identifying all above components, it draws a rectangle box around each component.
Therefore, in the following array it contains list of all the components detected. Each list is in the following order.
Name of component
X, Y coordinates of top left corner of rectangle
X, Y coordinates of right bottom of rectangle
Nodes (Black Points) are used to indicate a bridge over a line without crossing.
Array of all above components.
labels =[['NODE',(1002.9702758789062, 896.4686675071716), (1220.212585389614, 1067.1142654418945)], ['NODE',(1032.444071739912, 635.7160077095032),(1211.6839590370655, 763.4382424354553)],['M', (57.093908578157425,607.6229677200317),(311.9765570014715,833.807623386383)],['NODE', (344.5295810997486, 806.3690414428711), (501.8982524871826, 930.6454839706421)], ['Z', (21.986433800309896, 1327.9791088104248), (266.36098374426365, 1565.158670425415)], ['OR', (476.0066536962986, 574.401759147644), (918.3125713765621, 1177.1423168182373)], ['NODE', (333.50814148783684, 1058.0092916488647), (497.6142471432686, 1202.9034795761108)], ['K', (37.06201596558094, 870.0414619445801), (311.77860628068447, 1105.8665227890015)], ['AND', (665.9987451732159, 1227.940999031067), (1062.7052736580372, 1594.6843948364258)],['AND', (1373.9987451732159, 204.940999031067), (1703.7052736580372, 612.6843948364258)], ['NOT', (694.2882044911385, 260.5083291530609), (1027.812717139721, 450.35294365882874)], ['XOR', (2027.6711627840996, 593.0362477302551), (2457.9011510014534, 1093.9836854934692)], ['J', (85.69029207900167, 253.8458535671234), (334.48535946011543, 456.5887498855591)], ['OUTPUT', (2657.3825285434723, 670.8418045043945), (2929.8974316120148, 975.4852895736694)]]
Then, using line detection algorithm I identified all the 17 lines connecting each component.
All above 17 lines take as a list of arrays, and those lines are not in a correct order and each line has 2 end points.
lines = [[(60, 1502), (787, 1467)], [(125, 1031), (691, 988)], [(128, 772), (685, 758)], [(131, 336),(709,347)], [(927,350),(1455, 348)], [(400, 1361), (792, 1369)], [(834, 843), (2343, 939)], [(915, 1430), (1119, 1424)], [(1125, 468), (1453, 470)], [(1587, 399), (1911, 405)], [(1884, 755), (2245, 814)],[(2372, 831), (2918, 859)], [(1891, 397), (1901, 767)], [(1138, 457), (1128, 738)], [(441, 738), (421, 903)], [(1125, 946), (1101, 1437)], [(420, 1098), (408, 1373)]]
When connecting those lines, it is must to consider following scenario.
It means nodes are used to indicate a bridge without crossing lines.
M is an input to the AND gate and (M.Z) is an input to the other AND gate.
So how can I generate the following Boolean Expression using above 2 arrays and above scenario? This should be a function that works for all logic gates.
It can be assumed that image is always going to read from left to right.

Related

detect highest peaks automatically from noisy data python

Is there any way to detect the highest peaks using a python library without setting any parameter?. I'm developing a user interface and I want the algorithm to be able to detect highest peaks automatically...
I want it to be able to detect these peaks in picture below:
graph here
Data looks like this:
8.60291e-07
-1.5491e-06
5.64568e-07
-9.51195e-07
1.07203e-06
4.6521e-07
6.43967e-07
-9.86092e-07
-9.82323e-07
6.38977e-07
-1.93884e-06
-2.98309e-08
1.33543e-06
1.05064e-06
1.17332e-06
-1.53549e-07
-8.9357e-07
1.59176e-06
-2.17331e-06
1.46756e-06
5.63301e-07
-8.77556e-07
7.47681e-09
-8.30101e-07
-3.6647e-07
5.27046e-07
-1.94983e-06
1.89018e-07
1.22533e-06
8.00735e-07
-8.51166e-07
1.13437e-06
-2.75787e-07
1.79601e-06
-1.67875e-06
1.13529e-06
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9.9688e-07
-9.34486e-07
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1.15124e-06
-4.23569e-07
-1.8029e-07
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4.10736e-07
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-3.62984e-07
2.97916e-06
-1.95828e-06
-1.07398e-06
2.422e-06
-6.33202e-07
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1.6694e-06
-4.71764e-07
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-1.0071e-06
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2.64193e-06
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-1.30419e-06
1.48369e-07
1.26033e-06
-2.59872e-07
4.28284e-07
-6.44356e-07
2.99934e-07
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3.53226e-07
-7.08252e-07
4.1243e-07
2.41525e-06
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4.31945e-06
3.75152e-06
1.091e-06
3.8204e-06
-1.21356e-06
3.35564e-06
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-5.99808e-07
2.18155e-06
5.90652e-07
-1.36728e-06
-4.97017e-07
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4.37645e-07
-1.26514e-06
2.26413e-06
-8.52966e-07
-7.35596e-07
4.11911e-07
1.7585e-06
-inf
1.10779e-08
-1.49507e-06
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-3.85296e-06
4.31265e-06
-9.89227e-07
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3.67615e-07
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1.39078e-06
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1.8172e-06
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4.00137e-06
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1.32196e-06
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2.8837e-06
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1.59641e-06
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2.53528e-06
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3.55881e-06
2.31015e-06
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4.45251e-08
-1.94791e-06
2.27081e-06
-3.34717e-06
3.05688e-06
4.57062e-07
3.87326e-06
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1.83841e-06
2.65482e-06
-3.10474e-06
-3.46741e-07
2.45557e-06
2.01328e-06
-3.92606e-06
inf
-8.11737e-07
5.72174e-07
1.57245e-06
8.02612e-09
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1.22079e-06
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-1.43695e-06
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3.47308e-06
3.78232e-06
4.01093e-07
-1.58435e-06
-3.47375e-06
1.34943e-06
1.11768e-06
1.95212e-06
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1.53705e-06
6.38031e-07
-1.84702e-06
1.34689e-06
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1.81653e-06
-2.42355e-06
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The task you are describing could be treated like anomaly/outlier detection.
One possible solution is to use a Z-score transformation and treat every value with a z score above a certain threshold as an outlier. Because there is no clear definition of an outlier it won't be able to detect such peaks without setting any parameters (threshold).
One possible solution could be:
import numpy as np
def detect_outliers(data):
outliers = []
d_mean = np.mean(data)
d_std = np.std(data)
threshold = 3 # this defines what you would consider a peak (outlier)
for point in data:
z_score = (point - d_mean)/d_std
if np.abs(z_score) > threshold:
outliers.append(point)
return outliers
# create normal data
data = np.random.normal(size=100)
# create outliers
outliers = np.random.normal(100, size=3)
# combine normal data and outliers
full_data = data.tolist() + outliers.tolist()
# print outliers
print(detect_outliers(full_data))
If you only want to detect peaks, remove the np.abs function call from the code.
This code snippet is based on a Medium Post, which also provides another way of detecting outliers.

How to draw a horizontal line at the large y-axis integer?

For the following data.dat file:
08:01:59 451206975237005878
08:04:07 451207335040839108
08:05:56 451207643872368805
08:07:49 451207961547842270
08:09:56 451208317883903787
08:10:12 451208364811411904
08:14:09 451209030026853864
08:16:19 451209395116787156
08:17:14 451209552481002386
08:20:22 451210080432357203
08:25:36 451210963309583903
08:30:23 451211772783766177
08:34:04 451212394854723707
08:35:53 451212702239472024
08:48:46 451214876715294857
08:49:56 451215072475511660
08:51:24 451215321890488523
08:52:39 451215533925588479
08:52:42 451215542324801784
08:54:30 451215845971562410
08:55:08 451215951262906948
08:58:30 451216519498960432
I'd like to draw a horizontal line at the specific level (e.g. 451211772783766177). However, the number is too large to process.
Here is my attempt (based on this post):
$ gnuplot -p -e 'set arrow from 451211772783766177 to 451211772783766177; plot "data.dat" using 2 with lines'
which gives the following error:
line 0: warning: integer overflow; changing to floating point
How I should proceed?

I would treat your large number as a constant function, and plot it right after your data. Also, writing it on a exponential notation X.XE+YY = X.X times 10 to the +YY power (more info on format specifiers here) also takes care of the error:
plot "data.dat" using 2 with lines, 4.51211772783766177E17 with lines
Let me know if this works!

linearK error in seq. default() cannot be NA, NaN

I am trying to learn linearK estimates on a small linnet object from the CRC spatstat book (chapter 17) and when I use the linearK function, spatstat throws an error. I have documented the process in the comments in the r code below. The error is as below.
Error in seq.default(from = 0, to = right, length.out = npos + 1L) : 'to' cannot be NA, NaN or infinite
I do not understand how to resolve this. I am following this process:
# I have data of points for each data of the week
# d1 is district 1 of the city.
# I did the step below otherwise it was giving me tbl class
d1_data=lapply(split(d1, d1$openDatefactor),as.data.frame)
# I previously create a linnet and divided it into districts of the city
d1_linnet = districts_linnet[["d1"]]
# I create point pattern for each day
d1_ppp = lapply(d1_data, function(x) as.ppp(x, W=Window(d1_linnet)))
plot(d1_ppp[[1]], which.marks="type")
# I am then converting the point pattern to a point pattern on linear network
d1_lpp <- as.lpp(d1_ppp[[1]], L=d1_linnet, W=Window(d1_linnet))
d1_lpp
Point pattern on linear network
3 points
15 columns of marks: ‘status’, ‘number_of_’, ‘zip’, ‘ward’,
‘police_dis’, ‘community_’, ‘type’, ‘days’, ‘NAME’,
‘DISTRICT’, ‘openDatefactor’, ‘OpenDate’, ‘coseDatefactor’,
‘closeDate’ and ‘instance’
Linear network with 4286 vertices and 6183 lines
Enclosing window: polygonal boundary
enclosing rectangle: [441140.9, 448217.7] x [4640080, 4652557] units
# the errors start from plotting this lpp object
plot(d1_lpp)
"show.all" is not a graphical parameter
Show Traceback
Error in plot.window(...) : need finite 'xlim' values
coords(d1_lpp)
x y seg tp
441649.2 4649853 5426 0.5774863
445716.9 4648692 5250 0.5435492
444724.6 4646320 677 0.9189631
3 rows
And then consequently, I also get error on linearK(d1_lpp)
Error in seq.default(from = 0, to = right, length.out = npos + 1L) : 'to' cannot be NA, NaN or infinite
I feel lpp object has the problem, but I find it hard to interpret the errors and how to resolve them. Could someone please guide me?
Thanks

I can confirm there is a bug in plot.lpp when trying to plot the marked point pattern on the linear network. That will hopefully be fixed soon. You can plot the unmarked point pattern using
plot(unmark(d1_lpp))
I cannot reproduce the problem with linearK. Which version of spatstat are you running? In the development version on my laptop spatstat_1.51-0.073 everything works. There has been changes to this code recently, so it is likely that this will be solved by updating to development version (see https://github.com/spatstat/spatstat).

How to estimate camera pose according to a projective transformation matrix of two consecutive frames?

I'm working on the kitti visual odometry dataset. I use projective transformation to register two 2D consecutive frames(see projective transformation example here
). I want to know how this 3*3 projective transformation matrix is related to the ground truth poses provided by the kitti dataset.
This dataset gives the ground truth poses (trajectory) for the sequences, which is described below:
Folder 'poses':
The folder 'poses' contains the ground truth poses (trajectory) for the
first 11 sequences. This information can be used for training/tuning your
method. Each file xx.txt contains a N x 12 table, where N is the number of
frames of this sequence. Row i represents the i'th pose of the left camera
coordinate system (i.e., z pointing forwards) via a 3x4 transformation
matrix. The matrices are stored in row aligned order (the first entries
correspond to the first row), and take a point in the i'th coordinate
system and project it into the first (=0th) coordinate system. Hence, the
translational part (3x1 vector of column 4) corresponds to the pose of the
left camera coordinate system in the i'th frame with respect to the first
(=0th) frame. Your submission results must be provided using the same data
format.
Some samples of the given groud-truth poses:
1.000000e+00 9.043680e-12 2.326809e-11 5.551115e-17 9.043683e-12 1.000000e+00 2.392370e-10 3.330669e-16 2.326810e-11 2.392370e-10 9.999999e-01 -4.440892e-16
9.999978e-01 5.272628e-04 -2.066935e-03 -4.690294e-02 -5.296506e-04 9.999992e-01 -1.154865e-03 -2.839928e-02 2.066324e-03 1.155958e-03 9.999971e-01 8.586941e-01
9.999910e-01 1.048972e-03 -4.131348e-03 -9.374345e-02 -1.058514e-03 9.999968e-01 -2.308104e-03 -5.676064e-02 4.128913e-03 2.312456e-03 9.999887e-01 1.716275e+00
9.999796e-01 1.566466e-03 -6.198571e-03 -1.406429e-01 -1.587952e-03 9.999927e-01 -3.462706e-03 -8.515762e-02 6.193102e-03 3.472479e-03 9.999747e-01 2.574964e+00
9.999637e-01 2.078471e-03 -8.263498e-03 -1.874858e-01 -2.116664e-03 9.999871e-01 -4.615826e-03 -1.135202e-01 8.253797e-03 4.633149e-03 9.999551e-01 3.432648e+00
9.999433e-01 2.586172e-03 -1.033094e-02 -2.343818e-01 -2.645881e-03 9.999798e-01 -5.770163e-03 -1.419150e-01 1.031581e-02 5.797170e-03 9.999299e-01 4.291335e+00
9.999184e-01 3.088363e-03 -1.239599e-02 -2.812195e-01 -3.174350e-03 9.999710e-01 -6.922975e-03 -1.702743e-01 1.237425e-02 6.961759e-03 9.998991e-01 5.148987e+00
9.998890e-01 3.586305e-03 -1.446384e-02 -3.281178e-01 -3.703403e-03 9.999605e-01 -8.077186e-03 -1.986703e-01 1.443430e-02 8.129853e-03 9.998627e-01 6.007777e+00
9.998551e-01 4.078705e-03 -1.652913e-02 -3.749547e-01 -4.231669e-03 9.999484e-01 -9.229794e-03 -2.270290e-01 1.649063e-02 9.298401e-03 9.998207e-01 6.865477e+00
9.998167e-01 4.566671e-03 -1.859652e-02 -4.218367e-01 -4.760342e-03 9.999347e-01 -1.038342e-02 -2.554151e-01 1.854788e-02 1.047004e-02 9.997731e-01 7.724036e+00
9.997738e-01 5.049868e-03 -2.066463e-02 -4.687329e-01 -5.289072e-03 9.999194e-01 -1.153730e-02 -2.838096e-01 2.060470e-02 1.164399e-02 9.997198e-01 8.582886e+00
9.997264e-01 5.527315e-03 -2.272922e-02 -5.155474e-01 -5.816781e-03 9.999025e-01 -1.268908e-02 -3.121547e-01 2.265686e-02 1.281782e-02 9.996611e-01 9.440275e+00
9.996745e-01 6.000540e-03 -2.479692e-02 -5.624310e-01 -6.345160e-03 9.998840e-01 -1.384246e-02 -3.405416e-01 2.471098e-02 1.399530e-02 9.995966e-01 1.029896e+01
9.996182e-01 6.468772e-03 -2.686440e-02 -6.093087e-01 -6.873365e-03 9.998639e-01 -1.499561e-02 -3.689250e-01 2.676374e-02 1.517453e-02 9.995266e-01 1.115757e+01
9.995562e-01 7.058450e-03 -2.894213e-02 -6.562052e-01 -7.530449e-03 9.998399e-01 -1.623192e-02 -3.973964e-01 2.882292e-02 1.644266e-02 9.994492e-01 1.201541e+01
9.995095e-01 5.595311e-03 -3.081450e-02 -7.018788e-01 -6.093682e-03 9.998517e-01 -1.610315e-02 -4.239119e-01 3.071983e-02 1.628303e-02 9.993953e-01 1.286965e+01

The common name for your "projective transformation" is homography. In a calibrated setup (i.e. if you know your camera's field of view or, equivalently, its focal length) a homography can be decomposed into 3D rotation and translation, the latter only up to scale. The decomposition algorithm additionally produces the normal to the 3D plane inducting the homography. The algorithm has up to 4 solutions, of which only one is feasible when you apply additional constraints, such as that the matched image points triangulate in front of the camera, and that the general direction of the translation match a known prior.
More information about the method is in a well-known paper by Malis and Vargas. There is an implementation in OpenCV, under the name decomposeHomographyMat.

R simplify heatmap to pdf

I want to plot a simplified heatmap that is not so difficult to edit with the scalar vector graphics program I am using (inkscape). The original heatmap as produced below contains lots of rectangles, and I wonder if they could be merged together in the different sectors to simplify the output pdf file:
nentries=100000
ci=rainbow(nentries)
set.seed=1
mean=10
## Generate some data (4 factors)
i = data.frame(
a=round(abs(rnorm(nentries,mean-2))),
b=round(abs(rnorm(nentries,mean-1))),
c=round(abs(rnorm(nentries,mean+1))),
d=round(abs(rnorm(nentries,mean+2)))
)
minvalue = 10
# Discretise values to 1 or 0
m0 = matrix(as.numeric(i>minvalue),nrow=nrow(i))
# Remove rows with all zeros
m = m0[rowSums(m0)>0,]
# Reorder with 1,1,1,1 on top
ms =m[order(as.vector(m %*% matrix(2^((ncol(m)-1):0),ncol=1)), decreasing=TRUE),]
rowci = rainbow(nrow(ms))
colci = rainbow(ncol(ms))
colnames(ms)=LETTERS[1:4]
limits=c(which(!duplicated(ms)),nrow(ms))
l=length(limits)
toname=round((limits[-l]+ limits[-1])/2)
freq=(limits[-1]-limits[-l])/nrow(ms)
rn=rep("", nrow(ms))
for(i in toname) rn[i]=paste(colnames(ms)[which(ms[i,]==1)],collapse="")
rn[toname]=paste(rn[toname], ": ", sprintf( "%.5f", freq ), "%")
heatmap(ms,
Rowv=NA,
labRow=rn,
keep.dendro = FALSE,
col=c("black","red"),
RowSideColors=rowci,
ColSideColors=colci,
)
dev.copy2pdf(file="/tmp/file.pdf")

Why don't you try RSvgDevice? Using it you could save your image as svg file, which is much convenient to Inkscape than pdf

I use the Cairo package for producing svg. It's incredibly easy. Here is a much simpler plot than the one you have in your example:
require(Cairo)
CairoSVG(file = "tmp.svg", width = 6, height = 6)
plot(1:10)
dev.off()
Upon opening in Inkscape, you can ungroup the elements and edit as you like.
Example (point moved, swirl added):

I don't think we (the internet) are being clear enough on this one.
Let me just start off with a successful export example
png("heatmap.png") #Ruby dev's think of this as kind of like opening a `File.open("asdfsd") do |f|` block
heatmap(sample_matrix, Rowv=NA, Colv=NA, col=terrain.colors(256), scale="column", margins=c(5,10))
dev.off()
The dev.off() bit, in my mind, reminds me of an end call to a ruby block or method, in that, the last line of the "nested" or enclosed (between png() and dev.off()) code's output is what gets dumped into the png file.
For example, if you ran this code:
png("heatmap4.png")
heatmap(sample_matrix, Rowv=NA, Colv=NA, col=terrain.colors(32), scale="column", margins=c(5,15))
heatmap(sample_matrix, Rowv=NA, Colv=NA, col=greenred(32), scale="column", margins=c(5,15))
dev.off()
it would output the 2nd (greenred color scheme, I just tested it) heatmap to the heatmap4.png file, just like how a ruby method returns its last line by default

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