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These are questions asked by ComputerUser

I find some book named automatic construction of graph for CAD

Then searched some paper related with it about geometric constraint solver

then find YouTube has AutoCAD can automatic design

Do maple has these things?

I have no idea in design, which input should be for these automatic design from nothing?

is there any mechanical knowledge as constraint by default in this automatic design?

F(n,n+1),2) = F(Product(x, k=1..n), 2)

how to solve function F?


Above equation is wrong, below is newly updated 

Expect F is product operator

F(2,3) + 2 = F(2,4)

F(3,4) + 3 = F(3,5)

solve([ F(1,2)+1=F(1,3), F(n,n+1) + n = F(n,n+2), F(n,n+1)= F(summation(k,k=1..n),2)],F) assuming n > 0

Expect F is new operator
solve([ F(1,2)*1=F(1,3), n*F(n,n+1) = F(n,n+2), F(n,n+1)= F(product(k,k=1..n),2)],F) assuming n > 0

Would like to export a list into a text file

when I tried , it show ... 1234 items... , can not export a whole list into text file.

[[[0,0,0,0]], [[0,0,0,0]], .....]
thousands of [[...]] elements


[[1,1,["ab","xy"]], [0,1,["xy"]]]

how to sort by first number , second number , and list of string ?

how to count and group by third list of strings having first two number 0,0 or 0,1 or 1,0 or 1,1 

for example 

["xy"] at most count 4  , at least count 1

would like to find which is [1,1,... ] and only count 1


second has rsolve solution

compare original sequence

accsum(ListTools[Reverse]([seq(patterngroup[k], k=(mm)..(mm+lengthofsequence))])),

with rsolve solution , it start from third number wrong, some number are wrong but correct at next number, 

but i just want to guess next number is 1 or 2

[seq(subs(k=kk, sol), kk=0..(lengthofsequence+1))],

accsum := proc(xx)
local result,mm:
result := []:
for mm from 1 to nops(xx) do
temp := Summation(xx[k], k = 1 .. mm):
result := [op(result), temp]:
return result:
end proc:

patterngroup := [2, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1]: # left is latest, 2,1,1..
lengthofsequence := 5:
for mm from 1 to 12 do
if guessgf(accsum(Reverse([seq(patterngroup[k], k=(mm)..(mm+lengthofsequence))])), x) = FAIL then
sol := rsolve(listtorec(accsum(ListTools[Reverse]([seq(patterngroup[k], k=(mm)..(mm+lengthofsequence))])), u(n))[1], u(k)):
acxxc := accsum(ListTools[Reverse]([seq(patterngroup[k], k=(mm)..(mm+lengthofsequence))])):
olist := [seq(patterngroup[k], k=(mm)..(mm+5))]:
#start from 0
print(patterngroup[mm], olist, accsum(ListTools[Reverse]([seq(patterngroup[k], k=(mm)..(mm+lengthofsequence))])), [seq(subs(k=kk, sol), kk=0..(lengthofsequence+1))], sol, subs(k=(lengthofsequence+1), sol), subs(k=(lengthofsequence+1), sol)-acxxc[nops(acxxc)]):
end if:

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