c++,sorting,radix-sort,floating-point-exceptions

Your radixSort function violates the Compare requirements, namely irreflexivity (that is, radixOrder(x, x) must return false but it returns true because the execution goes to the first if branch). So you get a classic example of undefined behavior here. I believe that piece of code should be rewritten somehow like...

c++,algorithm,radix-sort,radix

Answer The number of iterations in MSD radix depends on the input size, whereas the number of iterations in LSD radix sort depends on the key length. This often leads to MSD radix sort requiring significantly fewer iterations than LSD radix sort and is therefore faster. Memory allocations are not...

Here is a radix sort coded so someone reasonable can understand it. I stole most of it from here. def radixsort( aList ): RADIX = 10 maxLength = False tmp , placement = -1, 1 while not maxLength: maxLength = True # declare and initialize buckets buckets = [list() for...

algorithm,sorting,caching,radix-sort

I took at look at the paper you linked. You haven't made any mistakes, none that I can see. In fact, in my estimation, you corrected a mistake in the algorithm. I wrote out the algorithm and ended up with the exact same problem as you did. After reviewing Algorithm...

c++,arrays,string,sorting,radix-sort

After the second loop, count[0] should be zero, and the third loop is missing a -97. This example fixes the problem using count of size 27 instead of 26. The first loop in this example uses -96, so count[0] = 0, count[1] = # instances of 'a's, count[2] = #...

string,algorithm,sorting,radix-sort

I'm not quite sure what you mean by "variable-length strings" but you can perform a binary MSB radix sort in-place so the length of the string doesn't matter since there are no intermediate buckets. #include <stdio.h> #include <algorithm> static void display(char *str, int *data, int size) { printf("%s: ", str);...

Change this: def new_output(three_list): output = "" for queue_list in three_list: for numqueue in queue_list: while len(numqueue) > 0: output += numqueue.after() output += " " output += "\n" return output To this (mind the indentation in last but one line!): def new_output(three_list): output = "" for queue_list in three_list:...

java,sorting,time-complexity,radix-sort

for(int z = 0; z<bucket.length; z++){ for (int k=0; k<bucket[z].size(); k++){ The crucial thing here is that the sum total of all the bucket sizes will equal n, so these loops combined only take O(n). The loop over j takes O(n), and the while loop on i will run...