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i would also not like to ask, but if not ask, what should i do?

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will the result or accuracy different between cpu and GPU when doing numerical calculation?


sorry , 

i discover missing equal operator 

it works now , please delete this question.


I find distribution function

but still not work

I removed picture.

@Joe Riel 

I removed picture

@Carl Love 

difficult to express in screenshot

it is just my sense to test my observation hypothesis

i do not know how to use too

i think guess actual use is mark it as a change


sequence faster or for loop faster?


is there artificial intelligence work in this way?


thanks, i forget hashtable value  can use list too

yesterday my status is not well, today i am refresh to use python do below

chosen1 = {}
chosen2 = {}
chosen3 = {}
chosen4 = {}
chosen5 = {}
chosen1[1] = [2]
chosen1[2] = [3]
chosen2[2] = [3]
chosen2[2] = chosen2[2]  + [5]
chosen3[3] = [4]
chosen3[5] = [9]
chosen4[4] = [5]
chosen4[9] = [8]
chosen5[5] = [6]
chosen5[8] = [12]

for i0 in chosen1.keys():
    for i1 in chosen1[i0]:
        if i1 in chosen2.keys():
            for i2 in chosen2[i1]:
                if i2 in chosen3.keys():
                    for i3 in chosen3[i2]:
                        if i3 in chosen4.keys():
                            for i4 in chosen4[i3]:
                                if i4 in chosen5.keys():
                                    for i5 in chosen5[i4]: 
                                        print([i, i1, i2, i3, i4, i5]);


very good function

it seems not need accumulate function


my notebook is faster to run web now.

i attached my code,

but the result different from table of the book photo

at orbit size 3 , only two number of each, but book photo show all are 3 numbers

FixPoints := (a,n) -> {seq(i,i=1..n)} minus {op(map(op,a))}:
a := [[seq(i,i=1..6)],[seq(i,i=7..9)]]:
b := a:
aa := []:
countlist := []:
for i to 6 do
tt := b;
if tt = [] then
 tt := permgroup(9, {[]}):
end if:
count := 1:
orbitlist := []:
continuerun := 0:
for j from 1 to 7 do
if continuerun = 0 then
if b = [] then
if orbit(tt,j) in orbitlist then
 continuerun := 1:
 orbitlist := [op(orbitlist), orbit(tt,j)]:
end if:
if orbit({tt},j) in orbitlist then
 continuerun := 1:
 orbitlist := [op(orbitlist), orbit({tt},j)]:
end if:
end if:
end if:
if nops(orbitlist) in countlist then
 rr := 0:
 countlist := [op(countlist), nops(orbitlist)]:
 lprint('a'^i=b, "orbit size=",nops(orbitlist)):
end if:
aa := [op(aa), b]:
b := mulperms(b,a);



I have network security concern to not to use mobile carrier network

how about the next two questions?

start from a^2 , result different from book photo?

how to calculate the orbit result at orbit size 3 at bottom table of the book photo?


sorry for missing one photo

my notebook load maple prime web very slow, I can only use iphone

start from a^2 , result different from book above photo

and aa0 can not be calculated with orbit , and how to calculate the orbit result at orbit size 3 at bottom table of the book photo



i understand now

a^4 repeated for a^2

so not choose a^4

and if calculate larger than orbit size 

it will repeated too

but I do not understand , start from a^2

the result is different from the book

and aa0 , can not calculate with orbit

return error

moreover when aa[3]= a^3

but how calculate the result having 3 numbers 036, 147 etc at orbit size 3


why orbit size 1 is 678

how to calculate this?

a^6 = a^0

how to calculate (018) .. etc at the bottom table

i use fixpoints and orbit function not work


and why it choose orbit size 1,2,3,6 only?


but i need a while to do a further combination 

from this result I need to choose 51 different rows

and search

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