I am trying to implement Distance Vector Routing algorithm, using http://www.cs.bu.edu/fac/byers/courses/791/F99/scribe_notes/cs791-notes-990923.html
(in C++).
Here is what I have done so
far:
i) Read no. of nodes.
ii) Implement the points 1 to 4 in the article as:
for(i = 0; i < nodes; i++) //nodes is the no. of nodes
{
for(j = 0; j < nodes; j++)
{
if(distanceVector[i][j] != 0) //distanceVector holds the cost value between every pair of links, 0 if no link exists
{
for(k = 0; k < nodes; k++)
{
if((distanceVector[i][j] + distanceVector[j][k]) < distanceVector[i][k])
{
distanceVector[i][k] = distanceVector[i][j] + distanceVector[j][k];
via[i][j] = i; // intermediate node, in case no link exists
via[j][i] = j;
}
}
}
}
}
I get the same array / matrix as it is. I have also tried juggling i, j, and k, but of no use.
Am I right in my implementation...???
Two things bother me about your code. First, you are using "0" to represent "no link". This can get you in trouble. The code basically read as follows: "if there is an intermediary point j, that makes that path from i to k shorter, change the path from i to k to pass via k". Therefore, using "0" to represent "no link" might make your code choose wrong "via"s. Try instead to use infinity (if you are using floating point) or a really big value (e.g., MAX_INT).
Second, these lines look wrong:
via[i][j] = i; // intermediate node, in case no link exists
via[j][i] = j;
Since you found a path from i to k via j that is shorter, it should be:
via[i][k] = j; // intermediate node, in case no link exists
via[k][i] = j;
Related
I want to write a program that randomly generates a tournament tree using only the number of challengers. I read into another such problem, but the answer described how ranks would take part and seeding the players, which went a little over head.
The problem I am facing is that my algorithm produces an infinite loop for values between 1 and 4 inclusively. For all values otherwise, the program runs as desired.
My approach was to take in an array of strings for the competitors' names. Then, I would iterate over each position and randomly select a competitor's name to take that spot. Because I am swapping the names, I have to check for duplicates in the array. I believe this is where my code is experiencing issues.
Here is the snippet that actually determines the tree
for(int i = 0; i < no_players;) {
int index = rand() % ((no_players - i) + i);
// randomly choose an element from the remainder
string temp = players[index];
bool unique = true;
// check all the elements before the current position
for(int j = 0; j < i; j++) {
// if the element is already there, it is not unique
if(players[j] == temp)
unique = false;
}
// only if the element is unique, perform the swap
if(unique) {
players[index] = players[i];
players[i] = temp;
i++;
}
}
Any help is much appreciated!
I'm trying to work out what the equivalent a[++j]=*pr++; in the following code (which comes from a MatLab mex file) is in Python. I've found out that 'pr' is a pointer to the first element of the input array, but I can't get my head around what is happening to j. Can someone explain what is happening there in simple terms without pointers etc?
rf3(mxArray *array_ext, mxArray *hs[]) {
double *pr, *po, a[16384], ampl, mean;
int tot_num, index, j, cNr;
mxArray *array_out;
tot_num = mxGetM(array_ext) * mxGetN(array_ext);
pr = (double *)mxGetPr(array_ext);
array_out = mxCreateDoubleMatrix(3, tot_num-1, mxREAL);
po = (double *)mxGetPr(array_out);
j = -1;
cNr = 1;
for (index=0; index<tot_num; index++) {
a[++j]=*pr++;
while ( (j >= 2) && (fabs(a[j-1]-a[j-2]) <= fabs(a[j]-a[j-1])) ) {
ampl=fabs( (a[j-1]-a[j-2])/2 );
switch(j)
{
case 0: { break; }
case 1: { break; }
case 2: {
mean=(a[0]+a[1])/2;
a[0]=a[1];
a[1]=a[2];
j=1;
if (ampl > 0) {
*po++=ampl;
*po++=mean;
*po++=0.50;
}
break;
}
default: {
mean=(a[j-1]+a[j-2])/2;
a[j-2]=a[j];
j=j-2;
if (ampl > 0) {
*po++=ampl;
*po++=mean;
*po++=1.00;
cNr++;
}
break;
}
}
}
}
for (index=0; index<j; index++) {
ampl=fabs(a[index]-a[index+1])/2;
mean=(a[index]+a[index+1])/2;
if (ampl > 0){
*po++=ampl;
*po++=mean;
*po++=0.50;
}
}
/* you can free the allocated memeory */
/* for array_out data */
mxSetN(array_out, tot_num - cNr);
hs[0]=array_out;
}
Here's what happens:
Increment j by 1
assign to a[j] value pointed at by pr
increment pr.
In this order.
You specifically asked about:
a[++j]=*pr++;
j is being incremented before the assignment. In python the left hand side would be:
a[j+1]
and you would also then need to increment j before you use it next:
j += 1
The right hand side simply accesses the current position and then increments the position in the array. In python you would probably just use an iterator for your array.
BTW, you might find it difficult to do a line-by-line 'translation' of the code. I would suggest writing down the steps of the algorithm and then tackling it fresh in python, if that is what you need.
In Python, there aren't pointers, so how you translate this will depend on how you decide to represent pr. If you think of the pointer as a copy of the list pr = array_ext[:], the line you've highlighted would be something like
j = j + 1
a[j] = pr.pop(0)
For greater efficiency (and a closer parallel to what the C code is doing), you could use pr as an index into the list array_ext, starting it at 0. Then, the line you highlighted does this:
j = j + 1
a[j] = array_ext[pr]
pr = pr + 1
Suppose I have bunch of numbers. I have to first put the least significant digit into the corresponding bucket. Ex: 530 , I have to first put into the bucket 0. For number 61, I have to put into bucket 1.
I planned to use a multidimensional array to do this. So I create a 2-dimenional array, which nrows is 10 ( for 0~ 9) and ncolumns is 999999 ( because I don't know how large will the list be):
int nrows = 10;
int ncolumns = 999999;
int **array_for_bucket = (int **)malloc(nrows * sizeof(int *));
for(i = 0; i < nrows; i++)
array_for_bucket[i] = (int *)malloc(ncolumns * sizeof(int));
left = (a->value)%10;
array_for_bucket[left][?? ] = a->value;
Then I created one node call a. In this node a, there is a value 50. To find out which bucket I want to put it in, I calculate "left" and I got 0. So I want to put this a-> value into bucket 0. But now I am stuck. How do I put this value into the bucket? I have to use a pointer array to do this.
I thought for a long time but still couldn't find a good way to do it. So please share some ideas with me. thank you!
There is a much easier way of doing this, and instead of radix*nkeys space you only need an nkeys-sized buffer.
Allocate a second buffer that can fit nkeys keys. Now do a first pass through your data and simply count how many keys end up in each bucket. You now can create a radix-sized array of pointers where each pointer is to the start of that bucket in the output buffer. Finally, the second pass though the data moves the keys. Every time you move a key, increment that bucket pointer.
Here's some C code to make into C++:
void radix_sort(int *keys, int nkeys)
{
int *shadow = malloc(nkeys * sizeof(*keys));
int bucket_count[10];
int *bucket_ptrs[10];
int i;
for (i = 0; i < 10; i++)
bucket_count[i] = 0;
for (i = 0; i < nkeys; i++)
bucket_count[keys[i] % 10]++;
bucket_ptrs[0] = shadow;
for (i = 1; i < 10; i++)
bucket_ptrs[i] = bucket_ptrs[i-1] + bucket_count[i-1];
for (i = 0; i < nkeys; i++)
*(bucket_ptrs[keys[i] % 10]++) = keys[i];
//shadow now has the sorted keys
free(shadow);
}
But I may have misunderstood the question. If you are doing something a little different than radix sort, pleas add some details.
Look the Boost Pointer containers library if you want to store pointers.
C++ isn't my forte but this code from wikipedia-Raidx Sort is very comprehensive and probably is more C++-ish than what you've implemented so far. Hope it helps
This is C++, we don't use malloc anymore. We use containers. A two-dimensional array is a vector of vectors.
vector<vector<int> > bucket(10);
left = (a->value)%10;
bucket[left].push_back(a->value);
How to find first non-repeating element in an array.
Provided that you can only use 1 bit for every element of the array and time complexity should be O(n) where n is length of array.
Please make sure that I somehow imposed constraint on memory requirements. It is also possible that it can not be done with just an extra bit per element of the string. Also please let me know if it is possible or not?
I would say there is no comparison based algorithm, that can do it in O(n). As you have to compare the the first element of the array with all others, the 2nd with all except the first, the 3rd with all except the first = Sum i = O(n^2).
(But that does not necessarily mean that there is no faster algorithm, see sorting: There is a proof that you cant sort fast than O(n log n) if you are comparison based - and there is indeed one faster: Bucket Sort, which can do it in O(n)).
EDIT: In one of the other comments I said something about hash functions. I checked some facts about it, and here are the hashmap approach thoughts:
Obvious approach is (in Pseudocode):
for (i = 0; i < maxsize; i++)
count[i] = 0;
for (i = 0; i < maxsize; i++) {
h = hash(A[i]);
count[h]++;
}
first = -1;
for (i = 0; i < maxsize; i++)
if (count[i] == 0) {
first = i;
break;
}
}
for (i = 0; hash(A[i]) != first; i++) ;
printf("first unique: " + A[i]);
There are some caveats:
How to get hash. I did some research on perfect hash functions. And indeed you can generate one in O(n). (Optimal algorithms for minimal perfect hashing by George Havas et al. - Not sure how good this paper is, as it claims as Time Limit O(n) but speaks from non linear space limit (which is plan an error, I hope I am not the only seeing the flaw in the this, but according to all theorical computer science I know off time is an upper border for space (as you dont have time to write in more space)). But I believe them when they say it is possible in O(n).
The additional space - here I dont see a solution. Above papers cites some research that says that you need 2.7 bits for the perfect hash function. With the additional count array (which you can shorten to the states: Empty + 1 Element + More than 1 Element) you need 2 additional bits per element (1.58 if you assume you can it somehow combine with the above 2.7), which sums up to additional 5 bits.
Here I'm just taking one assumption that the string is Character String, just containing small alphabets, so that I can use one Integer (32 bit) so that with 26 alphabets it will be sufficient to take one bit per alphabet. Earlier I thought to take an array of 256 elements but then it will have 256*32 bits in total. 32 bits per element. But finally I found that I will be unable to do it without one more variable. So the solution is like this with just one integer (32 bits) for 26 alphabets:
int print_non_repeating(char* str)
{
int bitmap = 0, bitmap_check = 0;
int length = strlen(str);
for(int i=0;i<len;i++)
{
if(bitmap & 1<<(str[i] - 'a'))
{
bitmap_check = bitmap_check | ( 1 << (str[i] - 'a');
}
else
bitmap = bitmap | (1 << str[i] - 'a');
}
bitmap = bitmap ^ bitmap_check;
i = 0;
if(bitmap != 0)
{
while(!bitmap & (1<< (str[i])))
i++;
cout<<*(str+i);
return 1;
}
else
return 0;
}
You can try doing a modified bucketsort as exemplified below. However, you need to know the max value in the array passed into the firstNonRepeat method. So this runs at O(n).
For comparison based methods, the theoretical fastest (at least in terms of sorting) is O(n log n). Alternatively, you can even use modified versions of radix sort to accomplish this.
public class BucketSort{
//maxVal is the max value in the array
public int firstNonRepeat(int[] a, int maxVal){
int [] bucket=new int[maxVal+1];
for (int i=0; i<bucket.length; i++){
bucket[i]=0;
}
for (int i=0; i<a.length; i++){
if(bucket[a[i]] == 0) {
bucket[a[i]]++;
} else {
return bucket[a[i]];
}
}
}
}
This code finds the first repeating element. havent figured out yet if in the same for loop if it is possible to find the non-repeating element without introducing another for (to keep the code O(n)). Other answers suggest bubble sort which is O(n^2)
#include <iostream>
using namespace std;
#define max_size 10
int main()
{
int numbers[max_size] = { 1, 2, 3, 4, 5, 1, 3, 4 ,2, 7};
int table[max_size] = {0,0,0,0,0,0,0,0,0,0};
int answer = 0, j=0;
for (int i = 0; i < max_size; i++)
{
j = numbers[i] %max_size;
table[j]++;
if(table[j] >1)
{
answer = 1;
break;
}
}
std::cout << "answer = " << answer ;
}
Hey all, I'm trying to write a sort function but am having trouble figuring out how to initialize a value, and making this function work as a generic template. The sort works by:
Find a pair =(ii,jj)= with a minimum value = ii+jj = such at A[ii]>A[jj]
If such a pair exists, then
swap A[ii] and A[jj] else
break;
The function I have written is as follows:
template <typename T>
void sort(T *A, int size)
{
T min =453;
T temp=0;
bool swapper = false;
int index1 = 0, index2 = 0;
for (int ii = 0; ii < size-1; ii++){
for (int jj = ii + 1; jj < size; jj++){
if((min >= (A[ii]+A[jj])) && (A[ii] > A[jj])){
min = (A[ii]+A[jj]);
index1 = ii;
index2 = jj;
swapper = true;
}
}
}
if (!swapper)
return;
else
{
temp = A[index1];
A[index1] = A[index2];
A[index2] = temp;
sort(A,size);
}
}
This function will successfully sort an array of integers, but not an array of chars. I do not know how to properly initialize the min value for the start of the comparison. I tried initializing the value by simply adding the first two elements of the array together (min = A[0] + A[1]), but it looks to me like for this algorithm it will fail. I know this is sort of a strange type of sort, but it is practice for a test, so thanks for any input.
most likely reason it fails, is because char = 453 does not produce 453 but rather different number, depending what char is (signed versus unsigned). your immediate solution would be to use numerical_limits, http://www.cplusplus.com/reference/std/limits/numeric_limits/
you may also need to think about design, because char has small range, you are likely to overflow often when adding two chars.
The maximum value of any type is std::numeric_limits<T>::max(). It's defined in <limits>.
Also, consider a redesign. This is not a good algorithm. And I would make sure I knew what I was doing before calling my sort function recursively.
I haven't put too much time reading your algorithm, but as an alternative to std::numeric_limits, you can use the initial element in your array as the initial minimum value. Then you don't have to worry about what happens if you call the function with a class that doesn't specialize std::numeric_limits, and thus can't report a maximum value.