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
Related
A homework assignment for an introductory C++ course.
The task is to search a two dimensional array for two numbers adding up to a given number 'search_sum.' Enter all found addends as {startRow,startColumn,endRow,endColumn} in the array 'summations,' then return the array 'summations'.
The line with summations[*sums_found] = new size_t[kIndices_size]; in the if statement was written by the instructor, as well as the comment. I am under the impression that this allocates a new space in memory where data can be assigned to and stored, so, in the next line of code, I attempted to take the column and row variables from the for loop and place them in the newly allocated memory as such.
summations[*sums_found]={startRow, column, endRow, column};
This threw a 'too many assignments' and 'segmentation fault' error.
I assumed you just could not do it this way, and data must be added to a 2D array in another fashion. I removed the faulty assignment code and ran the code as shown below through a debugger, just out of curiosity. The error thrown was once again a Segmentation Fault.
// Parameters:
// - matrix: a two-dimension integer array
// - matrix_size: a two-element size_t array storing sizes of matrix
// - search_sum: the integer value for which the function seeks sums
// - sums_found: an OUTPUT PARAMETER pointing to a size_t
//
const size_t** FindSum(const int** matrix,
const size_t* matrix_size,
int search_sum,
size_t* sums_found) {
*sums_found = 0; // init sums found to 0
size_t summations_size = 2; // start assuming no more than 2 summations
// build structure to hold all found summations
size_t** summations = new size_t*[summations_size];
switch (search_sum)
{
case -92:{ //column search
for(size_t column = 0; column < matrix_size[1]; column++){
for(size_t startRow = 0; startRow < matrix_size[0]; startRow++){
for(size_t endRow = 0; endRow < matrix_size[0]; endRow++){
int j = matrix[startRow][column];
int k = matrix[endRow][column];
int sum = j + k;
if(sum = search_sum){
summations[*sums_found] = new size_t[kIndices_size]; // only done when summation is found
*sums_found++;
}
}
}
}
}
break;
case 60:{ //row search
for(size_t row = 0; row < matrix_size[0]; row++){
for(size_t startColumn = 0; startColumn < matrix_size[1]; startColumn++){
for(size_t endColumn = 0; endColumn < matrix_size[1]; endColumn++){
int j = matrix[row][startColumn];
int k = matrix[row][endColumn];
int sum = j + k;
if(sum = search_sum){
summations[*sums_found] = new size_t[kIndices_size]; // only done when summation is found
*sums_found++;
}
}
}
}
}
break;
case 1203:{ //desc/ascending diagonal
}
break;
case 412:{ //single entry
}
break;
default:{ //large array
}
break;
}
return const_cast<const size_t**>(summations);
}
I did not know what this was, so I researched the error and found that you are not allowed to perform a read/write action on read-only code, which makes sense. What I do not understand is what exactly makes this code read-only, when it seems like its function is to assign a new space for data to be assigned to, which (to me), sounds like a 'write-like' action? I more than likely am misunderstanding the full scope of the codes function, and I am further confused with as to how I should go about assigning the data to the summations array.
Also, this is a university course taught by a grad student who is (seemingly) less than well versed in c++. It is a very 'teach yourself' type class. I understand that this assignment is an exercise on pointers and references, but it feels like I am very poorly equipped to solve a problem like this, and I am unsure what exactly to research and study independently to improve my knowledge of this particular topic. Please, if you can tell what it is I am struggling with just by looking at the code I've written, let me know what I should prioritize my studying on.
need your help and better if you can help me fast. It is very trivial problem but still can't understand what exactly i need to put in one line.
The following code i have
for (busRequest = apointCollection.begin(); busRequest != apointCollection.end(); busRequest++)
{
double Min = DBL_MAX;
int station = 0;
for (int i = 0; i < newStations; i++)
{
distance = sqrt(pow((apointCollection2[i].x - busRequest->x1), 2) + pow((apointCollection2[i].y - busRequest->y1), 2));
if (distance < Min)
{
Min = distance;
station = i;
}
}
if (people.find(station) == people.end())
{
people.insert(pair<int, int>(station, i));
}
else
{
how can i increment "i" if the key of my statation is already in the map.
}
}
Just briefly what i do , i take the first busrequest go to the second loop take the first station and find the minimum distance. After i go over the second loop , i add that station with minimum distance to my map . After i proceed with all my loops and if there is the same station , i need to increment it , so it means that that station is using two times and etc.
I need the help just give me hint or provide the line that i need to add.
I thank you in advance and waiting for your help.
And I think you meant Min Distance instead of i? Check and let me know.
for (busRequest = apointCollection.begin(); busRequest != apointCollection.end(); busRequest++)
{
double Min = DBL_MAX;
int station = 0;
for (int i = 0; i < newStations; i++)
{
distance = sqrt(pow((apointCollection2[i].x - busRequest->x1), 2) + pow((apointCollection2[i].y - busRequest->y1), 2));
if (distance < Min)
{
Min = distance;
station = i;
}
}
if (people.find(station) == people.end())
{
people.insert(pair<int, int>(station, i)); // here???
}
else
{
// This routine will increment the value if the key already exists. If it doesn't exist it will create it for you
YourMap[YourKey]++;
}
}
In C++ you can directly access a map key without inserting it. C++ will automatically create it with default value.
In your case, if a station is not present in people map and you will access people[station] then people[station] will automatically be set to 0 ( default value of int is 0 ).
So you can just do this:
if (people[station] == 0)
{
// Do something
people[station] = station; // NOTE: i is not accessible here! check ur logic
}
else
{
people[station]++;
}
Also: In your code i cannot be accessed inside IF condition to insert into people map.
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;
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 ;
}
I'm trying to write a method that takes an array of integers (0-51, in that order), cuts it into two separate arrays (A and B in the below function by using the cut method, which I know for sure works) and then re-fuses the two arrays together by randomly selecting 0, 1 or 2 cards from the BOTTOM of either A or B and then adding them to the deck.
(ps- by "array" I mean linked list, I just said array because I thought it would be conceptually easier)
This is my code so far, it works, but there's a definite bias when it comes to where the cards land. Can anybody spot my logic error?
[code]
void Deck::shuffle(){
IntList *A = new IntList();
IntList *B = new IntList();
cut(A, B);
IntListNode *aMarker = new IntListNode;
aMarker = A->getSentinel()->next;
//cout<< A->getSentinel()->prev->prev->data <<'\n'<<'\n';
IntListNode *bMarker = new IntListNode;
bMarker = B->getSentinel()->next;
//cout<< B->getSentinel()->prev->data;
deckList.clear();
srand(time(NULL));
int randNum = 0, numCards = 0, totalNumCards = 0;
bool selector = true, aisDone = false, bisDone = false;
while(totalNumCards < 52){
randNum = rand() % 3;
if(randNum == 0){
selector = !selector;
continue;
}
numCards = randNum;
if(!aisDone && !bisDone){
if(selector){
for(int i = 0; i < numCards; i++){
deckList.push_back(aMarker->data);
aMarker = (aMarker->next);
if(aMarker == A->getSentinel()){
aisDone = true;
break;
}
}
selector = false;
}else{
for(int i = 0; i < numCards; i++){
deckList.push_back(bMarker->data);
bMarker = (bMarker->next);
if(bMarker == B->getSentinel()){
bisDone = true;
break;
}
}
selector = true;
}
}
if(aisDone && !bisDone){
for(int i = 0; i < (52 - totalNumCards); i++){
deckList.push_back(bMarker->data);
bMarker = (bMarker->next);
if(bMarker == B->getSentinel()){
bisDone = true;
break;
}
}
//return;
}
if(bisDone && !aisDone){
for(int i = 0; i < (52 - totalNumCards); i++){
deckList.push_back(aMarker->data);
aMarker = (aMarker->next);
if(aMarker == A->getSentinel()){
aisDone = true;
break;
}
}
//return;
}
totalNumCards += numCards;
}
int tempSum = 0;
IntListNode *tempNode = deckList.head();
for(int j = 0; j < 52; j++){
//cout<< (tempNode->data) << '\n';
tempSum += (tempNode->data);
tempNode = (tempNode ->next);
}
if(tempSum != 1326)
system("PAUSE");
return;
}
[/code]
What about just using std::random_shuffle? Yeah, it won't work for linked list, but you can change it to vector :)
If your instructor would have the moral to teach you programming the way it should be done then they'd encourage you to solve the problem like so, with four lines of code:
#include<algorithm>
#include<vector>
// ...
std::vector<int> cards; // fill it in ...
std::random_shuffle(cards.begin(), cards.end());
Using the standard library is the right way of doing things. Writing code on your own when you can solve the problem with the standard library is the wrong way of doing things. Your instructor doesn't teach you right. If they want to get a point across (say, have you practice using pointers) then they should be more attentive in selecting the exercise they give you.
That speech given, here is a solution worse than the above but better than your instructor's:
52 times do the following:
Choose two random none-equal integers in the range [0,52).
Swap the values in the array corresponding to these positions.
For most random number generators, the low bits are the least random ones. So your line
randNum = rand() % 3;
should be modified to get its value more from the high- to middle-order bits from rand.
Your expectations may be off. I notice that you swap the selector if your random value is 0. Coupled with the relative non-randomness of randNum, this may be your problem. Perhaps you need to make things less random to make them appear more random, such as swapping the selector every time through the loop, and always taking 1 or more cards from the selected deck.
Comments:
srand(time(NULL));
This should only be called once during an applications run. This it is usally best to call it in main() as you start.
int randNum = 0, numCards = 0, totalNumCards = 0;
bool selector = true, aisDone = false, bisDone = false;
One identifier per line. Every coding standard written has this rule. It also prevents some subtle errors that can creep in when using pointers. Get used to it.
randNum = rand() % 3;
The bottom bits of rand are the lest random.
rand Num = rand() / (MAX_RAND / 3.0);
Question:
if(!aisDone && !bisDone)
{
This can execute
and set one of the above to isDone
Example:
Exit state aisDone == false bsiDone == false // OK
Exit state aisDone == true bsiDone == false // Will run below
Exit state aisDone == false bsiDone == ture // Will run below
}
if(aisDone && !bisDone)
{
Is this allowed to run if the first block above is run?
}
if(bisDone && !aisDone)
{
Is this allowed to run if the first block above is run?
}
The rest is too complicated and I don't understand.
I can think of simpler techniques to get a good shuffle of a deck of cards:
for(loop = 0 .. 51)
{
rand = rand(51 - loop);
swap(loop, loop+rand);
}
The above simulates picking a card at random from the deck A and putting it on the top of deck B (deck B initially being empty). When the loop completes B is now A (as it was done in place).
Thus each card (from A) has the same probability of being placed at any position in B.