8 years, 83 days

## CONTINUE STATEMENT...

Maple

In C++, we use continue statement to continue with the loop. What is maple's equivalent statement of continue as used in C++?

Thanks:

## Using Linsolve for large systems of line...

Maple

Hi,
I am solving large systems of linear equations (mod 2) which are of this form ,
systems:= {0 = f[[1, 4]]-f[[1, 3]], 0 = f[[1, 5]]-f[[1, 3]], 0 = f[[1, 5]]-f[[1, 4]], 0 = f[[1, 6]]-f[[1, 3]], 0 = f[[1, 6]]-f[[1, 4]], 0 = f[[1, 6]]-f[[1, 5]], 0 = f[[1, 7]]-f[[1, 3]], 0 = f[[1, 7]]-f[[1, 4]], 0 = f[[1, 7]]-f[[1, 5]], 0 = f[[1, 7]]-f[[1, 6]], 0 = f[[2, 4]]-f[[1, 3]], 0 = f[[2, 5]]-f[[1, 3]], 0 = f[[2, 5]]-f[[1, 4]], 0 = f[[2, 6]]-f[[1, 3]], 0 = f[[2, 6]]-f[[1, 4]], 0 = f[[2, 6]]-f[[1, 5]], 0 = f[[2, 6]]-f[[2, 4]], 0 = f[[2, 6]]-f[[2, 5]], 0 = f[[2, 7]]-f[[1, 3]], 0 = f[[2, 7]]-f[[1, 4]], 0 = f[[2, 7]]-f[[1, 5]], 0 = f[[2, 7]]-f[[1, 6]], 0 = f[[2, 7]]-f[[2, 4]], 0 = f[[2, 7]]-f[[2, 5]], 0 = f[[3, 1]]-f[[1, 4]], 0 = f[[3, 1]]-f[[1, 5]], 0 = f[[3, 1]]-f[[1, 6]], 0 = f[[3, 1]]-f[[2, 7]], 0 = f[[3, 5]]-f[[1, 4]], 0 = f[[3, 5]]-f[[1, 5]], 0 = f[[3, 5]]-f[[2, 4]], 0 = f[[3, 5]]-f[[2, 5]], 0 = f[[3, 6]]-f[[1, 3]], 0 = f[[3, 6]]-f[[1, 4]], 0 = f[[3, 6]]-f[[1, 5]], 0 = f[[3, 6]]-f[[1, 6]], 0 = f[[3, 6]]-f[[2, 4]], 0 = f[[3, 6]]-f[[2, 5]], 0 = f[[3, 6]]-f[[2, 6]], 0 = f[[3, 7]]-f[[1, 4]], 0 = f[[3, 7]]-f[[1, 5]], 0 = f[[3, 7]]-f[[1, 6]], 0 = f[[3, 7]]-f[[1, 7]], 0 = f[[3, 7]]-f[[2, 4]], 0 = f[[3, 7]]-f[[2, 5]], 0 = f[[3, 7]]-f[[2, 6]], 0 = f[[3, 7]]-f[[2, 7]], 0 = f[[3, 7]]-f[[3, 6]], 0 = f[[4, 1]]-f[[1, 5]], 0 = f[[4, 1]]-f[[1, 7]], 0 = f[[4, 1]]-f[[2, 7]], 0 = f[[4, 1]]-f[[3, 5]], 0 = f[[4, 1]]-f[[3, 7]], 0 = f[[4, 2]]-f[[2, 6]], 0 = f[[4, 2]]-f[[2, 7]], 0 = f[[4, 6]]-f[[1, 5]], 0 = f[[4, 6]]-f[[1, 6]], 0 = f[[4, 6]]-f[[2, 5]], 0 = f[[4, 6]]-f[[2, 6]], 0 = f[[4, 6]]-f[[3, 5]], 0 = f[[4, 6]]-f[[3, 6]], 0 = f[[4, 7]]-f[[1, 5]], 0 = f[[4, 7]]-f[[1, 6]], 0 = f[[4, 7]]-f[[1, 7]], 0 = f[[4, 7]]-f[[2, 5]], 0 = f[[4, 7]]-f[[2, 6]], 0 = f[[4, 7]]-f[[2, 7]], 0 = f[[4, 7]]-f[[3, 5]], 0 = f[[4, 7]]-f[[3, 6]], 0 = f[[4, 7]]-f[[4, 6]], 0 = f[[5, 1]]-f[[2, 6]], 0 = f[[5, 1]]-f[[2, 7]], 0 = f[[5, 1]]-f[[3, 6]], 0 = f[[5, 1]]-f[[4, 6]], 0 = f[[5, 2]]-f[[2, 7]], 0 = f[[5, 2]]-f[[3, 5]], 0 = f[[5, 2]]-f[[3, 6]], 0 = f[[5, 2]]-f[[3, 7]], 0 = f[[5, 2]]-f[[4, 6]], 0 = f[[5, 2]]-f[[4, 7]], 0 = f[[5, 3]]-f[[4, 7]], 0 = f[[5, 7]]-f[[1, 6]], 0 = f[[5, 7]]-f[[1, 7]], 0 = f[[5, 7]]-f[[2, 6]], 0 = f[[5, 7]]-f[[2, 7]], 0 = f[[5, 7]]-f[[3, 6]], 0 = f[[5, 7]]-f[[3, 7]], 0 = f[[5, 7]]-f[[4, 6]], 0 = f[[5, 7]]-f[[4, 7]], 0 = f[[6, 1]]-f[[1, 7]], 0 = f[[6, 1]]-f[[3, 6]], 0 = f[[6, 1]]-f[[4, 7]], 0 = f[[6, 1]]-f[[5, 7]], 0 = f[[6, 2]]-f[[3, 7]], 0 = f[[6, 2]]-f[[4, 7]], 0 = f[[6, 3]]-f[[4, 6]], 0 = f[[6, 3]]-f[[5, 7]], 0 = f[[6, 4]]-f[[5, 7]], 0 = f[[7, 2]]-f[[3, 7]], 0 = f[[7, 2]]-f[[5, 7]], 1 = f[[3, 1]]-f[[1, 7]], 1 = f[[3, 1]]-f[[2, 4]], 1 = f[[3, 1]]-f[[2, 5]], 1 = f[[3, 1]]-f[[2, 6]], 1 = f[[3, 1]]-f[[3, 6]], 1 = f[[4, 1]]-f[[1, 6]], 1 = f[[4, 1]]-f[[2, 5]], 1 = f[[4, 1]]-f[[2, 6]], 1 = f[[4, 1]]-f[[3, 6]], 1 = f[[4, 2]]-f[[3, 5]], 1 = f[[4, 2]]-f[[3, 6]], 1 = f[[4, 2]]-f[[3, 7]], 1 = f[[5, 1]]-f[[1, 6]], 1 = f[[5, 1]]-f[[1, 7]], 1 = f[[5, 1]]-f[[3, 5]], 1 = f[[5, 1]]-f[[3, 7]], 1 = f[[5, 1]]-f[[4, 7]], 1 = f[[5, 2]]-f[[2, 6]], 1 = f[[5, 3]]-f[[4, 6]], 1 = f[[6, 1]]-f[[2, 7]], 1 = f[[6, 1]]-f[[3, 7]], 1 = f[[6, 1]]-f[[4, 6]], 1 = f[[6, 2]]-f[[3, 6]], 1 = f[[6, 2]]-f[[4, 6]], 1 = f[[6, 2]]-f[[5, 7]], 1 = f[[6, 3]]-f[[3, 7]], 1 = f[[6, 3]]-f[[4, 7]], 1 = f[[6, 4]]-f[[4, 7]], 1 = f[[7, 1]]-f[[3, 7]], 1 = f[[7, 1]]-f[[4, 7]], 1 = f[[7, 1]]-f[[5, 7]], 1 = f[[7, 2]]-f[[4, 7]], 1 = f[[7, 3]]-f[[5, 7]], 1 = f[[7, 4]]-f[[5, 7]]};

I have been using msolve as follows; msolve(systems,2);  which returns NULL if there is no solution. This however takes a lot of my computer memory and also take a lot of time since the equations are very many.  I decided to try LinearSolve, but it doesn't give me any solutions even for the ones that I know that the solutions exist. My code was writen as follows:

systems:=as above..

vars := [f[[1, 3]], f[[1, 4]], f[[1, 5]], f[[1, 6]], f[[1, 7]], f[[2, 4]], f[[2, 5]], f[[2, 6]], f[[2, 7]], f[[3, 1]], f[[3, 5]], f[[3, 6]], f[[3, 7]], f[[4, 1]], f[[4, 2]], f[[4, 6]], f[[4, 7]], f[[5, 1]], f[[5, 2]], f[[5, 3]], f[[5, 7]], f[[6, 1]], f[[6, 2]], f[[6, 3]], f[[6, 4]], f[[7, 1]], f[[7, 2]], f[[7, 3]], f[[7, 4]]];

A,b:=GenerateMatrix(systems,var):
LinearSolve(A,b) mod 2;
what I'm I doing wrong? How can I make this fast and in such a way that the solution is computed without taking too much memory?

I also understand that LinearSolve returns the following when there is no solution, "Error, (in LinearAlgebra:-LinearSolve) inconsistent system
", in my case, if there is no solution, I don't want that error to be printed, but I want some variable (call it x) to be printed if a solution is found.

Any help is appreciated.

Vic.

## Is it possible to use option cache when ...

Maple

I am getting the error that maple was unable to allocate enough memmory for the computation. I have seen that some of the values could have been stored in a remember table so that they are accessed whenever needed. I am however using arrays and maple cannot tell when the entries in an array change, how about option cache? can that be useful? if not, how can I optimize my storage space?

## Count number of lists...

Hi,
I have this lists and I have written a program to process these lists and produce only the lists that certisfy a certain condition. Instead of printing these lists, I want just the total number of the lists produced. How can I do these? The only thing I can think of now is to append the lists in an Array and get the number of elements in the array, but this will be very inefficient since I am looking at a very big number here.

Any help is appreciated,

Thanks.

Vic.

## Using break to stop the entire computati...

Maple

Hi,

I would like to thank everyone who takes their time to respond to posts on this page. I have another question.

I have about 11 trillion lists that are of the form A shown below. The lists ofcourse have more elements than A (about 50 elements). What I want is to compute this value I call f, and the moment this value is found to be 1, we stop, and if all the values of f are 0, then we print(A). Here is my code, which has a problem and doesn't give me the result I want. How do I modify this so that if any value of f=1, then we stop, elif no 1 is found, we print just one value for A and not for every 0 produced?

with(ListTools);
A := [[1, 2], [1, 7], [5, 6], [1, 6], [1, 9], [6, 5], [9, 1], [2, 1], [7, 1], [6, 1], [5, 6]]; n := 9;
for i to n do for j to n do for k to nops(A) do if [i, j] = A[k] then a := Search([i, j], A); b := Search([j, i], A); if a < b then f := `mod`(b-a, 2); if f = 1 then break else print(A) end if end if end if end do end do end do;

Presently what I am doing is to Append all values of f into an Array and check that 1 is not an element of that Array. This is however very inconvenient since my lists are huge, and they are many, and we do not have to cpompute all the values of f once we come across a 1. Other than this, is there a way I can make my program run faster? Any suggestions are welcomed.

Thanks,

VIC.

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