Include missing exhaustivePermutations file.

+rsa/+stat/exhaustivePermutations.m

@@ -0,0 +1,115 @@
+% permutations = exhaustivePermutations(n[, cap])
+%
+% n 			number of items to be permuted
+% cap			(optional) stop after finding this many
+% permutations	each row is a permutation; they'll be in lexicographic order
+%
+% Cai Wingfield 3-2010
+
+function permutations = exhaustivePermutations(varargin)
+
+	% The following algorithm generates the next permutation lexicographically after
+	% a given permutation. It changes the given permutation in-place.
+	% 
+	%    1. Find the largest index j such that a[j] < a[j + 1]. If no such index
+	%       exists, the permutation is the last permutation.
+	%    2. Find the largest index l such that a[j] < a[l]. Such a l exists and
+	%       satisfies j < l, since j + 1 is such an index.
+	%    3. Swap a[j] with a[l].
+	%    4. Reverse the sequence from a[j + 1] up to an including the final element
+	%       a[n].
+	%
+	% After step 1, one knows that all of the elements strictly after position j
+	% form a weakly decreasing sequence, so no permutation of these elements will
+	% make it advance in lexicographic order; to advance one must increase a[j].
+	% Step 2 finds the smallest value a[l] to replace a[j] by, and swapping them in
+	% step 3 leaves the sequence after position j in weakly decreasing order.
+	% Reversing this sequence in step 4 then produces its lexicographically minimal
+	% permutation, and the lexicographic successor of the initial state for the
+	% whole sequence.
+
+	switch nargin
+		case 1
+			n = varargin{1};
+			cap = inf;
+		case 2
+			n = varargin{1};
+			cap = varargin{2};
+		otherwise
+			error('exhaustivePermutations:nargin', 'Only one or two arguments, please.');
+	end%switch:nargin
+
+	absoluteMaximum = 21; % Larger than this and n! is not accurately represented by a double. Does this matter?
+
+	initialLexOrder = 1:n;
+
+	%if n > absoluteMaximum, permutations = []; return; end%if
+	if n > absoluteMaximum, warning('exhaustivePermutations:tooMany', ['21! is the largest number accurately represented by a double.\nAttempting ' num2str(n) '!, therefore, may give\nunpredictable results.']); end%if
+
+	% The first in the list is just the initial lexical order
+	permutations = initialLexOrder;
+	i = 1;
+	allPermutationsFound = false;
+
+	while ~allPermutationsFound && i < cap
+
+		currentPermutation = permutations(i,:);
+		[nextPermutation, allPermutationsFound] = getNextPermutation(currentPermutation);
+		permutations = [permutations; nextPermutation];
+		i = i + 1;
+
+	end%while:~allPermutationsFound
+
+end%function
+
+%%%%%%%%%%%%%%%%%%
+%% Subfunctions %%
+%%%%%%%%%%%%%%%%%%
+
+function [nextPermutation, allPermutationsFound] = getNextPermutation(currentPermutation)
+
+	n = numel(currentPermutation);
+
+	allPermutationsFound = false;
+
+	% Search through the current permutation, beginning to end, to find the largest index j such that currentPermutation(j) < currentPermutation(j+1)
+	j = 0;
+	for jSearch = 1:n-1
+		if currentPermutation(jSearch) < currentPermutation(jSearch+1)
+			j = jSearch;
+		end%if
+	end%for:jSearch
+
+	% If no such j exists, the current permutation is the last permutation in lexicographical order.
+	% Otherwise, find the largest index k such that currentPermutation(k) > currentPermutation(j).
+	if j == 0
+
+		nextPermutation = [];
+		allPermutationsFound = true;
+
+	else
+
+		k = 0;
+		for kSearch = j:n
+			if currentPermutation(kSearch) > currentPermutation(j)
+				k = kSearch;
+			end%if
+		end%for:kSearch
+
+		nextPermutation = currentPermutation;
+
+		% Swap currentPermutation(j) and currentPermutation(k)
+
+		nextPermutation_j = nextPermutation(j);
+		nextPermutation_k = nextPermutation(k);
+
+		nextPermutation(j) = nextPermutation_k;
+		nextPermutation(k) = nextPermutation_j;
+
+		% Reverse order of currentPermutation(j+1:end)
+
+		nextPermutation(j+1:end) = nextPermutation(end:-1:j+1);
+
+	end%if:j == 0
+
+end%function