## how to simplify this logic?...

would like to return K map of P1

Summation expression for logic only consider 1 but how about wildcard x ?

if consider wildcard x as 1 too, then will use below

source = [[0,0,1,0],[0,0,1,1],[0,1,0,1],[0,1,1,0],[0,1,1,1],[1,0,0,0],[1,0,0,1],[1,0,1,0],[1,0,1,1],[1,1,0,0],[1,1,0,1],[1,1,1,0],[1,1,1,1]];
i use Quine Mccluskey algorithm

got result below

wildcard is 5 or x
[[0, 5, 1, 5], [5, 0, 1, 5], [5, 5, 1, 0], [1, 0, 5, 5], [1, 5, 0, 5], [1, 5, 5, 0], [5, 5, 1, 1], [5, 1, 5, 1], [5, 1, 1, 5], [1, 5, 5, 1], [1, 5, 1, 5], [1, 1, 5, 5]]
A'C + B'C + CD' + AB' + AC' + AD' + CD + BD + BC + AD + AC + AB

table1 = [[0,0,0,0],
[0,0,0,1],
[0,0,1,0],
[0,0,1,1],
[0,1,0,0],
[0,1,0,1],
[0,1,1,0],
[0,1,1,1],
[1,0,0,0],
[1,0,0,1],
[1,0,1,0],
[1,0,1,1],
[1,1,0,0],
[1,1,0,1],
[1,1,1,0],
[1,1,1,1]];

loand(lonot(tt[0]),tt[2])
loand(lonot(tt[1]),tt[2])
loand(lonot(tt[3]),tt[2])
loand(lonot(tt[1]),tt[0])
loand(lonot(tt[2]),tt[0])
loand(lonot(tt[3]),tt[0])
loand(tt[2],tt[3])
loand(tt[1],tt[3])
loand(tt[1],tt[2])
loand(tt[0],tt[3])
loand(tt[0],tt[2])
loand(tt[0],tt[1])

loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loand(lonot(tt[0]),tt[2]),
loand(lonot(tt[1]),tt[2])),
loand(lonot(tt[3]),tt[2])),
loand(lonot(tt[1]),tt[0])),
loand(lonot(tt[2]),tt[0])),
loand(lonot(tt[3]),tt[0])),
loand(tt[2],tt[3])),
loand(tt[1],tt[3])),
loand(tt[1],tt[2])),
loand(tt[0],tt[3])),
loand(tt[0],tt[2])),
loand(tt[0],tt[1]));
def lonot(z):
if z == 1:
return 0
else:
return 1
def loand(a, b):
if a == 1 and b == 1:
return 1
else:
return 0
def loor(a, b):
if a == 0 and b == 0:
return 0
else:
return 1
#A'C + B'C + AB' + CD + A'BD
for tt in table1:
print loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loand(lonot(tt[0]),tt[2]),
loand(lonot(tt[1]),tt[2])),
loand(lonot(tt[3]),tt[2])),
loand(lonot(tt[1]),tt[0])),
loand(lonot(tt[2]),tt[0])),
loand(lonot(tt[3]),tt[0])),
loand(tt[2],tt[3])),
loand(tt[1],tt[3])),
loand(tt[1],tt[2])),
loand(tt[0],tt[3])),
loand(tt[0],tt[2])),
loand(tt[0],tt[1]));

finally i use python to verify
return
0
0
1
1
0
1
1
1
1
1
1
1
1
1
1
1

seems correct if wildcard is 1 too, but
can boolean simplify function simplify this
A'C + B'C + CD' + AB' + AC' + AD' + CD + BD + BC + AD + AC + AB

to

C + A + B.D   which is
P1 = D + Q0 + Q1.N in png file ?

## how to compute the most simplified result with sol...

solve(diff(-1/x,x) = (-1/x)^(b), b);

originally is 2, but it use ln(....) to express

if start from substitute, it seems need to replace manually.

solve(subs(a(x)=-1/x,diff(a(x),x) = (a(x))^(b)), b);

goal is to find b in equation below
solve(diff((x^2+x+1)/(-1+x)^2,x) = ((x^2+x+1)/(-1+x)^2)^(b), b);
(2*x+1)/(-1+x)^2-(2*(x^2+x+1))/(-1+x)^3 = ((x^2+x+1)/(-1+x)^2)^(b)

solve(diff((x^2+x+1)/(-1+x)^2,x) = ((x^2+x+1)/(-1+x)^2)*(b), b);

## is it possible to evaluate or how to evaluate this...

updated

after refer from

https://en.wikipedia.org/wiki/List_of_representations_of_e

exponential1 := sum((1/n!), n=0..infinity);
exponential1 is not a decimal number, it is exp(1)

hoyeung1:= sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x)), x=0..infinity);

hoyeung2:= sum((Int(exp(LambertW(1/(-1+x!))*(-1+x!)), x)), x=0..infinity);

how to evalute hoyeung1 or hoyeung2 as a decimal number?

how to evalute hoyeung^x as a decimal number function is func1 := proc(x) return hoyeung^x end proc:

but i do not know whether sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x))*m^x, x=0..infinity) = hoyeung^x

can limit(1+(Int(exp(LambertW(1/(-1+x))*(-1+x)), x)))^x, x=infinity) = hoyeung^1 ?

## how to dsolve this case and evaluate the solution?...

sol := dsolve(diff(ln(y(x)),x) = y(x)^(1/(1-y(x))), y(x));
x-Intat(_a^(-(-2+_a)/(-1+_a)), _a = y(x))+_C1 = 0

the solution is not y(x) = , but y(x) at the right hand side

## how to find back the term in summation?...

Lee := (-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
sum(unknown, n=1..infinity) = Lee

how to find unknown?

## how to complex plot this function?...

complexpoint run a long time
there is no option numpoints in complexplot, how to fasten it?

Lee := (-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
complexplot(Lee, x = 0 .. 1);
Lee := Re(-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
plot(Lee, x = 0 .. 2, numpoints = 5);
Lee := Im(-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
plot(Lee, x = 0 .. 2, numpoints = 5);

## Is there geometric or statistical meaning for ln(d...

Is there geometric or statistical meaning for ln(dy/dx) = 0?

is there any feature in vector field plot when ln(dy/dx) = 0?

## how to evaluate or how to use these functions?...

Int(exp(LambertW(1/(-1+x))*(-1+x)), x)+1

x-Intat(1/exp((-1+_a)*LambertW(1/(-1+_a))), _a = y(x))-_C1 = 0

i use dsolve two equations, get two possible results,
how to evaluate these functions or how to use these functions?

## error when using variable...

mas := proc(f)
return ln(diff(rhs(subs(_C1=0,dsolve(diff(y(x),x) = f))), x\$2));
end proc:
mas(exp(x));
mas(mas(exp(x)));
mas(x^2);
mas(x^2+x^3);

when i hard code x, there is no problem in above code.
but when i op to get variable x and run below, it do not have problem when run line by line, but it has problem when run in
procedure
Error, (in mas) invalid input: diff received exp(x), which is not valid for its 2nd argument

mas := proc(f)
local martin:
martin := op(f):
return ln(diff(rhs(subs(_C1=0,dsolve(diff(y(martin),martin) = f))), martin\$2));
end proc:

mas(exp(x));
mas(mas(exp(x)));
mas(x^2);
mas(x^2+x^3);

## How to find this function?...

F(exp(t)) = t

F(F(exp(t))) = 0

what is F ?

is it diff(ln(x),t) ?

## How to draw this graph?...

A system of algebraic equation

in terms of x, y, z

how draw 3 different circles to show the range of possible values for x, y and z respectively?

it may not be a circle

It may be 3 bounded area graph to show the range of x , y , z respectively

updated

like the graph in many examples in

algebraic and geometric ideas in the theory of discrete optimization

bound area have color

## is it possible to change ODE to PDE?...

is it possible to change ODE to PDE?

the ODE has diff(a(t),t) and diff(b(t),t)

how to convert to diff(t, a), diff(t, b) ?

## how to buildsym in this case?...

with(DEtools, buildsym, equinv, symtest):
ans := dsolve([eq2,eq3,eq4], Lie);
Error, (in dsolve) too many arguments; some or all of the following are wrong: [{a(t), b(t), c(t)}, Lie]

ans := dsolve([eq2+eq3+eq4 = exp(t)], Lie);
Error, (in PDEtools/sdsolve) too many arguments; some or all of the following are wrong: [{a(t), b(t), c(t)}, Lie]

ans := dsolve([eq2,eq3,eq4]);
sym2 := buildsym(ans);
Error, (in buildsym) invalid input: `ODEtools/buildsym` expects its 1st argument, sol, to be of type {algebraic, algebraic = algebraic}, but received [{c(t) = ...}, {b(t) = ...}, {a(t) = ...)}]

PDEtools[declare](a(t), b(t), c(t), prime = t):
symgen(eq2+eq3+eq4=0);
a(t) will now be displayed as a
b(t) will now be displayed as b
c(t) will now be displayed as c
derivatives with respect to t of functions of one variable will now be
displayed with 'symgen(....)'

update
if it can not do for 3 function a(t),b(t),c(t) system of differential equations
then

i change to use
eq2 := subs(b(t)=a(t),subs(c(t)=a(t),eq2));
eq3 := subs(b(t)=a(t),subs(c(t)=a(t),eq3));
eq4 := subs(b(t)=a(t),subs(c(t)=a(t),eq4));

with(DEtools, buildsym, equinv, symtest):
ans := dsolve(eq2 = 0, Lie);
buildsym(ans[1], a(t));
buildsym(ans[2], a(t));
buildsym(ans[3], a(t));

there are 3 answers, can i use one of it to recover the equation eq2 or  eq3 or eq4?

ans := dsolve(eq3=0, Lie);
buildsym(ans[1], a(t));
sym2 := buildsym(ans[2], a(t));
buildsym(ans[3], a(t));

sym := [_xi=rhs(sym2[2]),_eta=rhs(sym2[1])];
ODE := equinv(sym, a(t));
eq3 - ODE;
sym := [_xi=rhs(sym2[1]),_eta=rhs(sym2[2])];
ODE := equinv(sym, a(t));
eq3 - ODE;
but ODE is not equal to original eq3
ans := dsolve(eq4=0, Lie);
buildsym(ans[1], a(t));
buildsym(ans[2], a(t));

ans := dsolve(eq2+eq3+eq4=0, Lie);
sym := buildsym(ans[1], a(t));
ODE := equinv(sym, a(t));
eq2+eq3+eq4 - ODE;
sym := buildsym(ans[2], a(t));
ODE := equinv(sym, a(t));
eq2+eq3+eq4 - ODE;
sym := buildsym(ans[3], a(t));
ODE := equinv(sym, a(t));
simplify(eq2+eq3+eq4 - - ODE);

can not recover the original result

## how to calculate probability of matrix in this cas...

i count the number among group
but when the list a large such as over 1000 records, the count will be over 30,000
use which denominator to find probability?
is there any functions in maple for this case?

with(LinearAlgebra):
correlationlist1 := [[1,2,3],[1,3,5]....]:
PAB := Matrix(50):
for ii from 1 to nops(correlationlist) do
for jj from 1 to nops(correlationlist[ii]) do
for kk from 1 to nops(correlationlist) do
for qq from 1 to nops(correlationlist[kk]) do
if ii <> kk then
#print("scan=",correlationlist2[kk],"kk=",kk,"qq=",qq,"row=",correlationlist[ii][jj],"column=",correlationlist[kk][qq]):
PAB[correlationlist[ii][jj],correlationlist[kk][qq]] := PAB[correlationlist[ii][jj],correlationlist[kk][qq]] + 1: # group to group relations
end if:
od:
od:
od:
od:

## How to use statistics correlate function with a li...

If there is a list

[[1,2],[2,2],[3,3]...

how to use correlate function?

assume [1,2] and [2,1] count as 2

when find correlation between 1and 2