permutation

问题 Peter Norvig's PAIP book contains this code as a solution to the permutation problem (some sections are removed for brevity) (defun permutations (bag) ;; If the input is nil, there is only one permutation: ;; nil itself (if (null bag) '(()) ;; Otherwise, take an element, e, out of the bag. ;; Generate all permutations of the remaining elements, ;; And add e to the front of each of these. ;; Do this for all possible e to generate all permutations. (mapcan #'(lambda (e) (mapcar #'(lambda (p)

问题 I've come across this post: How to generate all permutations of a list in Python But I require something more, namely all of the permutations of a string as well as all the permutations of all the substrings. I know it's a big number, but is it possible? 回答1: import itertools def all_permutations_substrings(a_str): return ( ''.join(item) for length in xrange(1, len(a_str)+1) for item in itertools.permutations(a_str, length)) Note, however, that this is true permutations - as in, hello will

问题 I have a calendar, typically a csv file containing a number of lines. Each line corresponds to an individual and is a sequence of consecutive values '0' and '1' where '0' refers to an empty time slot and '1' to an occupied slot. There cannot be two separated sequences in a line ( e.g. two sequences of '1' separated by a '0' such as '1,1,1,0,1,1,1,1'). The problem is to minimize the number of lines by combining the individuals and resolving the collisions between time slots. Note the time

问题 I want to rank and unrank permutations with one cycle in lexicographical order with a given len. A permutation with one cycles is where you can visit in this cycle each element. p:= (2,3,1) is a permutation with one cycle. Has rank 1. p:= (3,1,2) has 1 cycle too, but rank 2, because the permutation is lexicographical greater the frist so it becomes a greater rank. p:= (1,2,3) is a permutation with 3 cycles. (1),(2),(3) How can I efficently rank (permutation with one cycle to rank) and unrank

问题 I am trying to understand this recursion. I know how recursion works in factorial function but when it gets to this complex recursion like this I am confused. The most confusing part to me is this code str.split('').map( (char, i) => permutations( str.substr(0, i) + str.substr(i + 1) )map( p => char + p)) First, with "abc" , say, it will split into ["a","b","c"] and go through the map function, then go through the second map function to wrap each return with a , b , c , respectively. However,

问题 I am trying to understand this recursion. I know how recursion works in factorial function but when it gets to this complex recursion like this I am confused. The most confusing part to me is this code str.split('').map( (char, i) => permutations( str.substr(0, i) + str.substr(i + 1) )map( p => char + p)) First, with "abc" , say, it will split into ["a","b","c"] and go through the map function, then go through the second map function to wrap each return with a , b , c , respectively. However,

问题 I have this set of (Greek) strings: ἸἼΙἹἽ, ῇηἤήῃὴῆἡἠἢᾖἥἣῄἦᾗᾐἧᾔᾑ, σς, οὸόὀὄὅὂ, ὺὖυῦύὐὑὔΰϋὕὗὓὒῢ I'd like to find all possible permutations of the characters in these 5 strings. For example, Ἰῇσοὺ, Ἰῇσοὖ, Ἰῇσου, etc. I know it should involve recursion since the number of strings is not fixed but I'm a beginner and I'm completely dumbfounded by recursion. I did the following in Python and it does give me all combinations of the characters in each string. But I need the 'ἸἼΙἹἽ' to always come

问题 I have this set of (Greek) strings: ἸἼΙἹἽ, ῇηἤήῃὴῆἡἠἢᾖἥἣῄἦᾗᾐἧᾔᾑ, σς, οὸόὀὄὅὂ, ὺὖυῦύὐὑὔΰϋὕὗὓὒῢ I'd like to find all possible permutations of the characters in these 5 strings. For example, Ἰῇσοὺ, Ἰῇσοὖ, Ἰῇσου, etc. I know it should involve recursion since the number of strings is not fixed but I'm a beginner and I'm completely dumbfounded by recursion. I did the following in Python and it does give me all combinations of the characters in each string. But I need the 'ἸἼΙἹἽ' to always come

问题 I have a matrix, for example, 5x5. [,1] [,2] [,3] [,4] [,5] [1,] 22 -2 -2 -2 2 [2,] -2 22 2 2 2 [3,] -2 2 22 2 2 [4,] -2 2 2 22 2 [5,] 2 2 2 2 22. As you can see, the matrix is symmetric. Above the main diagonal, there are 4+3+2+1=10 positions, and I find via combn all the possible (permutation) matrices, which have (-2) 3 times in these 10 positions. That means 10!/3!*7!=120 matrices. But some of them are equivalent. So,my problem is how to find the non-equivalent matrices from the 120. I am

问题 I have a matrix, for example, 5x5. [,1] [,2] [,3] [,4] [,5] [1,] 22 -2 -2 -2 2 [2,] -2 22 2 2 2 [3,] -2 2 22 2 2 [4,] -2 2 2 22 2 [5,] 2 2 2 2 22. As you can see, the matrix is symmetric. Above the main diagonal, there are 4+3+2+1=10 positions, and I find via combn all the possible (permutation) matrices, which have (-2) 3 times in these 10 positions. That means 10!/3!*7!=120 matrices. But some of them are equivalent. So,my problem is how to find the non-equivalent matrices from the 120. I am