Commit 98e18fc2 by Hrishee Shastri

### Implementation of a Reversal Class and helpful methods, as well as an

`implementation of odd-even sort`
parent dcf8876e
SBR.py 0 → 100644
 """ Interface for a Reversal Class. """ # Returns the cost of the given reversal def rev_cost(reversal): return abs(reversal.beg - reversal.end) + 2 # Currently omits the 1/3 factor class Reversal: def __init__(self, i, j, time=0): if i > j: raise ValueError('Reversal\'s left index greater than right index') self.beg = i self.end = j if time != 0: self.time_remaining = time else: self.time_remaining = rev_cost(self) # Decrease the time remaining for the reversal by the given time (does not fall below zero) # Returns the subsequent time remaining def dec_time(self, t): self.time_remaining = max(0, self.time_remaining - t) return self.time_remaining def get_time_remaining(self): return self.time_remaining def get_length(self): return abs(self.beg - self.end) + 1 def __lt__(self, other): return self.beg < other.beg def __str__(self): return "(" + str(self.beg) + ", " + str(self.end) + ")" # Perform the given reversal on the given permutation def reverse(permutation, reversal): i = reversal.beg j = reversal.end permutation[i:j+1] = permutation[i:j+1][::-1] # Determines if the reversals in the given list are independent, where perm_length is the length of the permutation def independent(rev_list, perm_length): busy = [0]*perm_length for rev in rev_list: for i in range(rev.beg, rev.end+1): if busy[i]: return False busy[i] = 1 return True # Simpler function that checks if rev is independent to all reversals in the # active rev list, when we know that the latter are already all indepdendent def check_independent(rev, active_rev_list): for a_rev in active_rev_list: if not (rev.beg > a_rev.end or rev.end < a_rev.beg): return False return True # Checks whether the item at index i is busy; namely, if i is participating in # a reversal in the rev_list def is_busy(rev_list, i): for rev in rev_list: if rev.beg <= i and rev.end >= i: return True return False # Determines if the permutation is sorted (i.e. if it is the identity permutation from start to end, inclusive) def is_sorted(permutation, start, end): return permutation == list(range(start, end+1)) # Determines if the permutation is a valid permutation on the elements from start to end, inclusive def is_valid_permutation(permutation, start, end): return sorted(permutation) == list(range(start,end+1)) # Determines the time (in time steps) taken to perform the reversals in the given list in the given order in parallel def compress_reversals(revlist, permsize): timesteps = [] for rev in revlist: # print(rev, ' cost =', rev_cost(rev)) init_step = len(timesteps) # if rev.beg == 5 and rev.end == 7: # print('Here') for step in timesteps[::-1]: busy = False for i in range(rev.beg, rev.end+1): if step[i]: busy = True break if busy: break init_step -= 1 while init_step + rev_cost(rev) - 1 > len(timesteps) - 1: timesteps.append([0]*permsize) # if init_step + rev_cost(rev) - 1 > len(timesteps) - 1: # timesteps += [[0] * permsize] * (init_step + rev_cost(rev) - 1 - (len(timesteps) - 1)) for i in range(init_step, init_step + rev_cost(rev)): # print('Affecting row', i) for j in range(rev.beg, rev.end+1): timesteps[i][j] = 1 # for t in range(len(timesteps)): # print(t, '\t-->', timesteps[t]) # print() return len(timesteps) class ReversalCompresser: """ Alternate implementation of the compress_reversals function. """ def __init__(self): """ List to be sorted """ self.model = [] self.active_revs = [] self.counter = 0 def reset(self): self.__init__() def is_busy(self, start_index, end_index): """ checks if chain is busy between start index and end index, inclusive """ return 1 in self.model[start_index:end_index+1] def make_busy(self, start_index, end_index): """inclusive""" for i in range(start_index, end_index + 1): self.model[i] = 1 def make_free(self, start_index, end_index): for i in range(start_index, end_index + 1): self.model[i] = 0 def independent(self, rev, revlist): """ checks if rev is independent of all reversals in revlist, when we allow None in revlist """ for a_rev in revlist: if a_rev is None: continue if not (rev.beg > a_rev.end or rev.end < a_rev.beg): return False return True def update(self): for rev in self.active_revs: rev.step -= 1 if rev.step == 0: self.make_free(rev.beg, rev.end) self.active_revs = [rev for rev in self.active_revs if rev.step > 0] self.counter += 1 def compress(self, revlist, permsize): revlist = revlist[:] self.reset() self.model = [0]*permsize self.active_revs = [] while len(revlist) > 0: for r in range(len(revlist)): rev = revlist[r] if not self.is_busy(rev.beg, rev.end) and self.independent(rev, revlist[:r]): self.make_busy(rev.beg, rev.end) self.active_revs.append(rev) rev.step = rev.get_length() + 1 # to account for the + 1 in the cost fucntion revlist[r] = None revlist = list(filter(lambda i : i is not None, revlist)) self.update() # wait for final reversals to finish, if there are any while self.active_revs != []: self.update() return self.counter def odd_even_sort(L): """ L: List to be sorted. Since we can reduce the problem of sorting according to a permutation pi to just sorting according to the identity permutation, we just sort normally. Odd even sort is performed in parallel. Each pass is an odd pass or even pass, where we look at every odd/even pair and swap if they are out of order. Each pass takes time 1, since there are n/2 swaps per pass (each taking time 1) done in parallel. sorts in place. Returns the cost """ cost = 0 sortflag = False addflag = False while not sortflag: sortflag = True addflag = False for i in range(1, len(L) - 1, 2): if L[i] > L[i+1]: L[i], L[i+1] = L[i+1], L[i] sortflag = False addflag = True cost += int(addflag) addflag = False for i in range(0, len(L) - 1, 2): if L[i] > L[i+1]: L[i], L[i+1] = L[i+1], L[i] sortflag = False addflag = True cost += int(addflag) return cost def apply_revs(revs, perm): """ Applies reversals in list revs in order to permutation perm, in place """ for rev in revs: reverse(perm, rev)
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