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

조회 수: 14 (최근 30일)

편집: John D'Errico 2024년 7월 15일
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.

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답변 (2개)

Milan Bansal 2024년 7월 15일
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));
Hope this helps!
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John D'Errico 2024년 7월 15일
편집: John D'Errico 2024년 7월 15일
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.

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