How to solve this problem combining sequences and reasoning using code?

An arithmetic sequence:range(4m + 2) , has two elements removed, leaving ( 4m ) elements. These remaining elements are evenly divided into ( m ) groups, with each group containing 4 numbers. After removing the two elements, each of the ( m ) groups must form an arithmetic sequence.how to automatically generates all possible cases of arithmetic sequences obtained by removing two elements from range(4m + 2) when a specific positive integer ( m ) is input, and lists them one by one.

function generateArithmeticSequences(m)

% Generate the original arithmetic sequence

originalSequence = 1:(4*m + 2);

% Generate all combinations of two elements to remove

removeCombinations = nchoosek(originalSequence, 2);

% Preallocate the cell array to store valid cases

validCases = {};

% Loop through all combinations of removals

for i = 1:size(removeCombinations, 1)

% Create a copy of the original sequence without the removed elements

remainingSequence = setdiff(originalSequence, removeCombinations(i, :));

% Check if the remaining sequence can be divided into m groups of 4

% such that each group forms an arithmetic sequence

if isDivisibleIntoArithmeticSequences(remainingSequence, m)

validCases{end+1} = remainingSequence;

end

end

% Display the valid cases

displayValidCases(validCases);

end

% Check function to see if the sequence can be divided into m groups of 4

% where each group is an arithmetic sequence

function result = isDivisibleIntoArithmeticSequences(sequence, m)

result = false;

% Divide the sequence into m groups of 4

groups = mat2cell(sequence, 4*ones(1, m), 4);

% Check if each group forms an arithmetic sequence

for i = 1:m

if ~isArithmeticSequence(groups{i})

return;

end

end

result = true;

end

% Check function to determine if a sequence is an arithmetic sequence

function result = isArithmeticSequence(sequence)

result = false;

if length(sequence) < 4

return;

end

differences = diff(sequence);

if all(differences(1) == differences)

result = true;

end

end

% Function to display the valid cases

function displayValidCases(cases)

for i = 1:length(cases)

disp(['Case ', num2str(i), ': ', mat2str(cases{i})]);

end

end

The code provided above may not produce the correct answers.

Answer 1

[Milan Bansal]

Hi 代丝

The code attached in the question produces the following error:

"Error using mat2cell

Input arguments, D1 through D2, must sum to each dimension of the input matrix size, [1 12]."

The error is due to the incorrect dimensions specified while using mat2Cell function in the following line of code.

groups = mat2cell(sequence, 4*ones(1, m), 4);

To resolve the error please replace the above line with the below line of code.

groups = mat2cell(sequence, 1, 4*ones(1, m));

Please refer to following documentation link to learn more about mat2cell function:

• https://www.mathworks.com/help/matlab/ref/mat2cell.html

Hope this helps!

Answer 2

[John D'Errico]

I would probably think first, and only then write code. I gather the global sequence will always be 1:(4*m+2), so the first 4*m+2 natural numbers.

What properties must these subsequences enjoy?

First, assume the stride is always positive. So the subsequence [1 2 3 4] is equivalent to the subsequence [4 3 2 1]. Thus there is no need to check for decreasing subsequences.

Second, consider the stride of any such subsequence. What is the stride of a sequence? It is the difference between consecutuve values in that sequence. You can think of the stride as the length of your steps if you were walking, thus the name stride.

Is there a maximum possible stride, assuming all subsequences you wil consider are of length 4? YES. For any value of m, suppose the largest possible stride you need consider is S. What is the largest possible value for S? We would then have one possible arithmetic subsequence of length 4 as:

1, 1+S, 1+2*S, 1+3*S

As such, we would have the requirement

1+3*S <= 4*m+2

and therefore,

S <= (4*m+1)/3

Since S must be an integer, we would have

S = floor((4*m+1)/3)

Next, what are the possible starting elements of the 4 disjoint subsequences? How many possible starting elements are there? This is easy enough, as since the stride must always be at least 1 (zero is not an option for the stride), then the first element of a subsequence may be no larger than 4*m-1. Then the possible starting elements must come from the set of numbers given by nchoosek, as follows.

startElements = @(m) nchoosek(1:4*m-1,4)

startElements = function_handle with value:
    @(m)nchoosek(1:4*m-1,4)

For example, when m==2, the global sequence is formed from the natural numbers 1:10.

startElements(2)

ans = 35x4

1 2 3 4 1 2 3 5 1 2 3 6 1 2 3 7 1 2 4 5 1 2 4 6 1 2 4 7 1 2 5 6 1 2 5 7 1 2 6 7

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Then there would be 35 possible starting points for those subsequences. Combine each of those sets of start points, with all possible strides, and we see there are at most

possibleSeqs = @(m) nchoosek(4*m-1,4)*floor((4*m+1)/3)^m

possibleSeqs = function_handle with value:
    @(m)nchoosek(4*m-1,4)*floor((4*m+1)/3)^m

possible subsequences. You can now use brute force to test every possible sets of subsequences to see if they are valid. I would postulate there are better schemes, less CPU intensive, but this would be certainly viable. Note that as m would itself grow large, the number of subsequence sets will grow to be very large.

for example, with m==2,

possibleSeqs(2)

ans = 315

possibleSeqs(3)

ans = 21120

possibleSeqs(4)

ans = 853125

possibleSeqs(5)

ans = 65143932

That is a simple upper bound on the number of possible sequences you would need to check, using any reasonably designed brute force scheme. You should see it grows at a rate that is more than exponentially fast. And that suggests perhaps you need to use more intelligence than brute force to perform this assignment, at least if m is anything larger than 5.

Anyway, surely we can do better than brute force. For example, list all subsequences of length 4, that start with 1, for any given value of m. This is easy to do. Pick some value of m, arbitrarily perhaps 3.

m = 3;
S = floor((4*m+1)/3) % the largest possible stride for m = 4

S = 4

Next, we can prove trivially that at least one subsequence MUST start with either 1, 2, or 3. (This happens because we know we are dropping 2 elements always from the list.)

subSeq1 = [1 + (1:S)'.*(0:3);2 + (1:S)'.*(0:3);3 + (1:S)'.*(0:3)]

subSeq1 = 12x4

1 2 3 4 1 3 5 7 1 4 7 10 1 5 9 13 2 3 4 5 2 4 6 8 2 5 8 11 2 6 10 14 3 4 5 6 3 5 7 9

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So when m == 3, we KNOW that one of the subsequences MUST be one of those 12 sub-sequences listed. Next, each of those sub-sequences reduces the set of possible sub-sequences, since the sub-sequences must be disjoint. With a little effort, I think this would provide a viable start at a solution.