Mathematical Script - Dynamical System Stoptime 3

I need a maths script developed. Make responsive PHP please.

Overview:

Start with a random INTEGER X.

If X is EVEN divide by 2

if X is ODD then Multiply by 3, then add 1 then divide by 2

so for X even X maps to X/2

and for X odd X maps to (3X+1) / 2

Part 2:

we now need to do 2 things

1. compute the STOPTIME for each number in the interval entered

2. organise them into classes modulo 2 to power S. (S is the stop time)

1. the stoptime is the LEAST number of computations (iterations) required to reach an INTEGER less than X.

Example:

suppose X = 3

then applying the map rules we get

3 > (3.3 +1)/2 = 5

5 > (3.5 +1)/2 = 8

8 > (8/2) = 4

4 > (4/2) = 2

2 > (2/2) = 1

so the sequence generated (the trajectory) is

3 > 5 > 8 > 4 > 2 > 1

so the STOPTIME for X = 3 is 4, since it takes 4 computations for the output to first reach a number less than 3.

Note: its not 5 since 5 is not the lowest number of iterations.

its the least number of iterations to reach the target.

Similarly for X = 7 we have a sequence 7,11,17,26,13,20,10,5,8,4,2,1

so the stop time for X=7 is 7

you can think of the stoptime S as a function of X i.e it depends on X.

S(3) = 4 and S(7) = 7.

Suppose our interval is 3 to 31 (increment of 4) e.g 3,7,11,15,19,23,27,31

then we can display the stop times for each

X= {3,7,11,15,19,23,27,31} S(X) = {4,7,5,7,4,5,59,56}

But this is the first part we now have to arrange them in classes modulo 2 to power of S.

so we construct the classes for each S

C[S(X)=4] = {3,19,35...}

C[S(X)=7] = {7,135,263...}

C[S(X)=5] = {11,43,75...}

C[S(X)=7] = {15,143,271....}

C[S(X)=5] = {23,55,87...}

C[S(X)=59] = {27,27+2^59,27+2.2^59...}

C[S(X)=56] = {31,31+2^56,27+2.2^56...}

Note: its quite technical, u need to make sure no classes get duplicated they should all be DISJOINT.

youll notice in the above example there is only 1 class for S = 4 but two for S=5 and S=7, this will happen but the important thing is they are DISJOINT (i.e contain no common integers) or their intersection is zero.

the easieat way to find all the disjoint classes is to check from 3 up to ( 2 to power of S) - 1

e.g for S=5, we check from 3 up to 2^5 - 1 = 31 i.e from 3 to 31

we find the values 11 and 23 have S = 5. this is the only values from 3 to 31.

so this means there is exactly 2 classes modulo 2^5 for S = 5.

11,43,75,107......

23,55,87,119.....

there are NO OTHERS as the first in each class will be between 3 and 2^S -1

for S = 7 we check from 3 up to 127 and again we find the following classes modulo 128 (2^7)

7,135,263....

15,143,271...

59,187,315....

There are no others as the next one after 59 is 135 but that actually lives in the first class 7,135,263, its the 2nd element of the first class.

OK so now we know how to arrange each X into different Stoptime classes modulo 2^S (S is stoptime)

Now we need some meaningful data output

all i need is 2 columns of output in a csv format for excel.

col 1 will be the stop time and col 2 will be the number of classes for that stoptime... ill type the first few rows below

S C

4 1

5 2

7 3

the first 3 rows are above... S is the stoptime and C is the number of disjoint classes modulo 2^S

REQUIREMENTS

-----------------------------

Enter Start End and Increment

e.g

Start = 3

End = 31

Increment = 4

this will calculate the stoptimes for 3 up to 31 inclusive (increment of 4)

then it will compute the number of disjoint classes for each of these stop times up to 31.

we output this to csv file with 2 columns as described above

Note: no time wasters, i will only pay on delivery of quality work, the script must be fast and be able to handle very large numbers. you must communicate with me when needed, no ignorance or arrogance, all i want is quality work, no moaning, complaining or bad attitudes.

Note: im open to other suggestions to make the script even faster so please be flexible also.

PLEASE NO TIME WASTERS

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ID projekta: #30099877

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