Skip to main content

Faster way to test possible points-to-plane-fitting identity?


Summary: I want to confirm the three sum-defined quantities in https://github.com/barrycarter/bcapps/blob/master/STACK/planetest.m (also below) are identically zero for all values of n>=3.


While attempting to solve https://stats.stackexchange.com/questions/196655 (fitting points to a plane), I came up with these (probably either wrong or previously derived by someone else) formulas for a,b,c such that z=a*x+b*y+c is a best fit for n points x[i], y[i], and z[i]:



a =
-((Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[y[i], {i, 1, n}]^2*Sum[x[i]*z[i], {i, 1, n}] +

n*Sum[y[i]^2, {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] +
Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}] -
n*Sum[x[i]*y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]))

b =
-((-(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}]) +

Sum[x[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}] +
Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] -
n*Sum[x[i]*y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]^2*Sum[y[i]*z[i], {i, 1, n}] +
n*Sum[x[i]^2, {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]))


c =
(Sum[x[i]*y[i], {i, 1, n}]^2*Sum[z[i], {i, 1, n}] -
Sum[x[i]^2, {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] +
Sum[x[i], {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] +
Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*

Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*Sum[y[i]^2, {i, 1, n}])

To confirm these values, I'd compute the sum of the differences squared. Each term would look like this:


diff[i_] = (z[i]-(a*x[i]+b*y[i]+c))^2


Treating the sum as a function of a,b,c, I would take partials with respect to these three variables and set equal to 0.


Since derivatives add, I would be adding the sum of the derivatives of each term:



derva[i_] = -2*x[i]*(-c - a*x[i] - b*y[i] + z[i])
dervb[i_] = -2*y[i]*(-c - a*x[i] - b*y[i] + z[i])
dervc[i_] = -2*(-c - a*x[i] - b*y[i] + z[i])


and setting each sum equal to 0.


Mathematica won't solve that for arbitrary n (which I sort of expected):



Solve[{
Sum[derva[i],{i,1,n}] == 0,
Sum[dervb[i],{i,1,n}] == 0,
Sum[dervc[i],{i,1,n}] == 0
}, {a,b,c}]


Out[74] = {}

and Reduce doesn't help either. Keeping the derivative outside the sum doesn't work either, albeit with a different error message (the standard Solve::nsmet: This system cannot be solved with the methods available to Solve.).


Mathematica will solve for a,b,c for specific values of n, which led me to the guess above.


To test my guess, I need to confirm that each partial derivative is zero. In other words, I need to confirm:



Sum[-2*x[i]*(-c - a*x[i] - b*y[i] + z[i]),{i,1,n}] == 0
Sum[-2*y[i]*(-c - a*x[i] - b*y[i] + z[i]),{i,1,n}] == 0
Sum[-2*(-c - a*x[i] - b*y[i] + z[i]),{i,1,n}] == 0


are identically zero for all values of x[i], y[i], z[i], and n, when I put in my guesses above.


In other words (these are the big ugly equations which I'm intentionally giving in "full" form for those who don't want to read the rest of this question)...:



atest[n_] :=

Sum[-2*x[i]*(-((Sum[x[i]*y[i], {i, 1, n}]^2*Sum[z[i], {i, 1, n}] -
Sum[x[i]^2, {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*
Sum[x[i]*z[i], {i, 1, n}] + Sum[x[i], {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*
Sum[x[i]*z[i], {i, 1, n}] + Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]*

Sum[y[i]*z[i], {i, 1, n}] - Sum[x[i], {i, 1, n}]*
Sum[x[i]*y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 - 2*Sum[x[i], {i, 1, n}]*
Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*
Sum[y[i]^2, {i, 1, n}])) +
((Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[y[i], {i, 1, n}]^2*Sum[x[i]*z[i], {i, 1, n}] +

n*Sum[y[i]^2, {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] +
Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}] -
n*Sum[x[i]*y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])*x[i])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*
Sum[y[i]^2, {i, 1, n}]) +
((-(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}]) +
Sum[x[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}] +

Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] -
n*Sum[x[i]*y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]^2*Sum[y[i]*z[i], {i, 1, n}] +
n*Sum[x[i]^2, {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])*y[i])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*
Sum[y[i]^2, {i, 1, n}]) + z[i]), {i, 1, n}];


btest[n_] :=

Sum[-2*y[i]*(-((Sum[x[i]*y[i], {i, 1, n}]^2*Sum[z[i], {i, 1, n}] -
Sum[x[i]^2, {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*
Sum[x[i]*z[i], {i, 1, n}] + Sum[x[i], {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*
Sum[x[i]*z[i], {i, 1, n}] + Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]*
Sum[y[i]*z[i], {i, 1, n}] - Sum[x[i], {i, 1, n}]*
Sum[x[i]*y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 - 2*Sum[x[i], {i, 1, n}]*

Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*
Sum[y[i]^2, {i, 1, n}])) +
((Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[y[i], {i, 1, n}]^2*Sum[x[i]*z[i], {i, 1, n}] +
n*Sum[y[i]^2, {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] +
Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}] -
n*Sum[x[i]*y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])*x[i])/

(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*
Sum[y[i]^2, {i, 1, n}]) +
((-(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}]) +
Sum[x[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}] +
Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] -
n*Sum[x[i]*y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]^2*Sum[y[i]*z[i], {i, 1, n}] +

n*Sum[x[i]^2, {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])*y[i])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*
Sum[y[i]^2, {i, 1, n}]) + z[i]), {i, 1, n}];

ctest[n_] :=

Sum[-2*(-((Sum[x[i]*y[i], {i, 1, n}]^2*Sum[z[i], {i, 1, n}] -

Sum[x[i]^2, {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*
Sum[x[i]*z[i], {i, 1, n}] + Sum[x[i], {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*
Sum[x[i]*z[i], {i, 1, n}] + Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]*
Sum[y[i]*z[i], {i, 1, n}] - Sum[x[i], {i, 1, n}]*
Sum[x[i]*y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 - 2*Sum[x[i], {i, 1, n}]*
Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*

Sum[y[i]^2, {i, 1, n}])) +
((Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]*Sum[y[i]^2, {i, 1, n}]*Sum[z[i], {i, 1, n}] -
Sum[y[i], {i, 1, n}]^2*Sum[x[i]*z[i], {i, 1, n}] +
n*Sum[y[i]^2, {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] +
Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}] -
n*Sum[x[i]*y[i], {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])*x[i])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +
n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*

Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*
Sum[y[i]^2, {i, 1, n}]) +
((-(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}]) +
Sum[x[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}]*Sum[z[i], {i, 1, n}] +
Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] -
n*Sum[x[i]*y[i], {i, 1, n}]*Sum[x[i]*z[i], {i, 1, n}] -
Sum[x[i], {i, 1, n}]^2*Sum[y[i]*z[i], {i, 1, n}] +
n*Sum[x[i]^2, {i, 1, n}]*Sum[y[i]*z[i], {i, 1, n}])*y[i])/
(Sum[x[i]^2, {i, 1, n}]*Sum[y[i], {i, 1, n}]^2 -
2*Sum[x[i], {i, 1, n}]*Sum[y[i], {i, 1, n}]*Sum[x[i]*y[i], {i, 1, n}] +

n*Sum[x[i]*y[i], {i, 1, n}]^2 + Sum[x[i], {i, 1, n}]^2*
Sum[y[i]^2, {i, 1, n}] - n*Sum[x[i]^2, {i, 1, n}]*
Sum[y[i]^2, {i, 1, n}]) + z[i]), {i, 1, n}];

I'm claiming atest[n], btest[n], ctest[n] are identially 0 for all values of n>=3.


How far I've gotten on my own:




t = Table[Simplify[{atest[i], btest[i], ctest[i]}], {i,3,6}]


Out[1]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}, {0, 0, 0}}


However, Simplify[atest[7]] just hangs. Even if I waited for it, I suspect Simplifying atest, btest, and ctest for larger numbers would take even longer.


I realize I'm asking Mathematica to do a lot, especially since I'm using symbols instead of actual numbers, but is there a good way to verify this identity for all n>=3 (or prove its false, of course).


For anyone interested, https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solve-stats-196655.m has some notes re how I 'derived' this.



Answer



Here is my take at generating the identities you are trying to test; I will confess that I am not completely sure of what testing you need to carry out, so I hope that you will be able to compare the expressions obtained below with those you already have.


To set up the 3D linear regression problem I will use generic 3D points {x[i], y[i], z[i]}, and parameters $(a,b,c)$.


(* Generate the sum of squares to be minimized *)

Sum[Expand[(a x[i] + b y[i] + c - z[i])^2], {i, 1, n}]

sum of squares


(* Calculate the derivatives *)
Grad[%, {a, b, c}] // TableForm

gradient


(* Use linearity of sums *)
Distribute /@ % // TableForm


distributed form


At this point, you will want to pull out the numerical and parameter constants from the summations to simplify the problem. Below is a set of very handy replacement rules that will help you do that, courtesy of Peltio who originally used them with Integrate, but they work with Sum right out of the box after swapping the right function name in:


outrules = {
Sum[f_ + g_, it : {x_Symbol, __}] :> Sum[f, it] + Sum[g, it],
Sum[c_ f_, it : {x_Symbol, __}] :> c Sum[f, it] /; FreeQ[c, x],
Sum[c_, it : {x_Symbol, __}] :> c Sum[1, it] /; FreeQ[c, x]
};

These must be applied repeatedly using ReplaceRepeated (//.) until the result doesn't change anymore:


(* Pull out any constants from summations *)

% //. outrules // TableForm

sum of squares in separated form


Now it's just a matter of setting the derivatives to zero to generate a system of three linear equations:


(* Set the derivatives equal to zero to generate the system of equations *)
Simplify[Thread[% == 0]] // TableForm

simplified equations


Now is a good time to simplify the notation a bit by introducing names for the summation results that act as constants in this system:


% //. {

Sum[x[i], {i, 1, n}] -> sumX, Sum[y[i], {i, 1, n}] -> sumY, Sum[z[i], {i, 1, n}] -> sumZ,

Sum[x[i]*y[i], {i, 1, n}] -> sumXY,
Sum[y[i]*z[i], {i, 1, n}] -> sumYZ,
Sum[x[i]*z[i], {i, 1, n}] -> sumXZ,

Sum[x[i]^2, {i, 1, n}] -> sumX2, Sum[y[i]^2, {i, 1, n}] -> sumY2
} // TableForm

simplified equations with named constants



Finally we can Solve this linear equation for $(a,b,c)$:


Solve[%, {a, b, c}]

explicit solutions


Comments

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

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[ ...