I have an index set A and another set dim, and I need to define variables x[i,j] where i is in A and j starts from one and is <= dim[A] for each member of A:
set A := { 11 .. 20};
dim := {<11>1, <12>3, <13>1, <14>1, <15>5, <16>1, <17>2, <18>1, <19>1, <20>3};
For example, in the above, I will have:
x[11,1], x[12,1], x[12,2], x[12,3], x[13,1], x[14,1], ...
I can do something like:
set B := { 1 .. maxDim)};
var x[A *B];
where maxDim is the maximum number in dim. How can I do this without creating many extra variables?
I do not understand the question. What is dimlist indexed with?
What would be the size of the dimensions 3..5
Likely the best way to achieve this is to have a full dimensional table
and just exclude the ones you do not like, e.g.
set I := { <a,b,c,d,e,f> in A*B*C*D*E*F with a > b or whatever restriction you like };
var x[I];
or you can mess around with the bounds of the variables like in
var x[<a,b,c,d,e,f> in AxB*D3*D4*D5] real <= (if dimlist[a,b] > 0 ? dimlist[a,b] : 0);
After some research I found a solution, which I think should work:
var x[ <i, j> in A*B with j <= dim[i]];
Related
In java I got this construction
for (let i = 0; i < x.length-1; I++
And here to avoid outOfBoundsException we are using x.length-1 but how to do the same thing in Kotlin? I got this code so far
x.forEachIndexed { index, _ ->
output.add((x[index+1]-x[index])*10)
}
And it crashes on the last element when we call x[index+1] so I need to handle the last element somehow
Input list
var x = doubleArrayOf(0.0, 0.23, 0.46, 0.69, 0.92, 1.15, 1.38, 1.61)
For a classic Java for loop you got two options in Kotlin.
One would be something like this.
val x = listOf(1,2,3,4)
for (i in 0 .. x.lastIndex){
// ...
}
Using .. you basically go from 0 up to ( and including) the number coresponding to the second item, in this case the last index of the list.( so from 0 <= i <= x.lastIndex)
The second option is using until
val x = listOf(1,2,3,4)
for (i in 0 until x.size){
// ...
}
This is similar to the previous approach, except the fact that until is not inclusive with the last element.(so from 0 <= i < x.size ).
What you probably need is something like this
val x = listOf(1,2,3,4)
for (i in 0 .. x.lastIndex -1){
// ...
}
or alternative, using until, like this
val x = listOf(1,2,3,4)
for (i in 0 until x.size-1){
// ...
}
This should probably avoid the IndexOut of bounds error, since you go just until the second to last item index.
Feel free to ask more if something is not clear.
This is also a great read if you want to learn more about ranges. https://kotlinlang.org/docs/ranges.html#progression
You already have an answer, but this is another option. If you would use a normal list, you would have access to zipWithNext(), and then you don't need to worry about any index, and you can just do:
list.zipWithNext { current, next ->
output.add((next - current)*10)
}
As mentioned by k314159, we can also do asList() to have direct access to zipWithNext and other list methods, without many drawbacks.
array.asList().zipWithNext { current, next ->
output.add(next - current)
}
How can I create a fixed multidimensional array in Specman/e using varibles?
And then access individual elements or whole rows?
For example in SystemVerilog I would have:
module top;
function automatic my_func();
bit [7:0] arr [4][8]; // matrix: 4 rows, 8 columns of bytes
bit [7:0] row [8]; // array : 8 elements of bytes
row = '{1, 2, 3, 4, 5, 6, 7, 8};
$display("Array:");
foreach (arr[i]) begin
arr[i] = row;
$display("row[%0d] = %p", i, row);
end
$display("\narr[2][3] = %0d", arr[2][3]);
endfunction : my_func
initial begin
my_func();
end
endmodule : top
This will produce this output:
Array:
row[0] = '{'h1, 'h2, 'h3, 'h4, 'h5, 'h6, 'h7, 'h8}
row[1] = '{'h1, 'h2, 'h3, 'h4, 'h5, 'h6, 'h7, 'h8}
row[2] = '{'h1, 'h2, 'h3, 'h4, 'h5, 'h6, 'h7, 'h8}
row[3] = '{'h1, 'h2, 'h3, 'h4, 'h5, 'h6, 'h7, 'h8}
arr[2][3] = 4
Can someone rewrite my_func() in Specman/e?
There are no fixed arrays in e. But you can define a variable of a list type, including a multi-dimensional list, such as:
var my_md_list: list of list of my_type;
It is not the same as a multi-dimensional array in other languages, in the sense that in general each inner list (being an element of the outer list) may be of a different size. But you still can achieve your purpose using it. For example, your code might be rewritten in e more or less like this:
var arr: list of list of byte;
var row: list of byte = {1;2;3;4;5;6;7;8};
for i from 0 to 3 do {
arr.add(row.copy());
print arr[i];
};
print arr[2][3];
Notice the usage of row.copy() - it ensures that each outer list element will be a copy of the original list.
If we don't use copy(), we will get a list of many pointers to the same list. This may also be legitimate, depending on the purpose of your code.
In case of a field (as opposed to a local variable), it is also possible to declare it with a given size. This size is, again, not "fixed" and can be modified at run time (by adding or removing items), but it determines the original size of the list upon creation, for example:
struct foo {
my_list[4][8]: list of list of int;
};
I want to make a cave explorer game in game maker 8.0.
I've made a block object and an generator But I'm stuck. Here is my code for the generator
var r;
r = random_range(0, 1);
repeat(room_width/16) {
repeat(room_height/16) {
if (r == 1) {
instance_create(x, y, obj_block)
}
y += 16;
}
x += 16;
}
now i always get a blank frame
You need to use irandom(1) so you get an integer. You also should put it inside the loop so it generates a new value each time.
In the second statement, you are generating a random real value and storing it in r. What you actually require is choosing one of the two values. I recommend that you use the function choose(...) for this. Here goes the corrected statement:
r = choose(0,1); //Choose either 0 or 1 and store it in r
Also, move the above statement to the inner loop. (Because you want to decide whether you want to place a block at the said (x,y) location at every spot, right?)
Also, I recommend that you substitute sprite_width and sprite_height instead of using the value 16 directly, so that any changes you make to the sprite will adjust the resulting layout of the blocks accordingly.
Here is the code with corrections:
var r;
repeat(room_width/sprite_width) {
repeat(room_height/sprite_height) {
r = choose(0, 1);
if (r == 1)
instance_create(x, y, obj_block);
y += sprite_height;
}
x += sprite_width;
}
That should work. I hope that helps!
Looks like you are only creating a instance if r==1. Shouldn't you create a instance every time?
Variable assignment r = random_range(0, 1); is outside the loop. Therefore performed only once before starting the loop.
random_range(0, 1) returns a random real number between 0 and 1 (not integer!). But you have if (r == 1) - the probability of getting 1 is a very small.
as example:
repeat(room_width/16) {
repeat(room_height/16) {
if (irandom(1)) {
instance_create(x, y, obj_block)
}
y += 16;
}
x += 16;
}
Here's a possible, maybe even better solution:
length = room_width/16;
height = room_height/16;
for(xx = 0; xx < length; xx+=1)
{
for(yy = 0; yy < height; yy+=1)
{
if choose(0, 1) = 1 {
instance_create(xx*16, yy*16, obj_block); }
}
}
if you want random caves, you should probably delete random sections of those blocks,
not just single ones.
For bonus points, you could use a seed value for the random cave generation. You can also have a pathway random generation that will have a guaranteed path to the finish with random openings and fake paths that generate randomly from that path. Then you can fill in the extra spaces with other random pieces.
But in regards to your code, you must redefine the random number each time you are placing a block, which is why all of them are the same. It should be called inside of the loops, and should be an integer instead of a decimal value.
Problem is on the first line, you need to put r = something in the for cycle
I have a bunch of variables named index1, index2, ..., indexn. I want to calculate i = array[index1] + array[index2] + ... + array[indexn]. I heard that I can do that in a loop, getting the current variable name from the loop index. How can I do that?
Instead of having individual variables like this:
int index1, index2, index3, ...indexN:
you should consider using an array of indices:
int index[N];
and then you can sum in a loop, e.g.
sum = 0;
for (i = 0; i < N; ++i)
{
sum += array[index[i]];
}
Sorry, this is not possible in objective-c. It works for example in php.
There are ways to get objects by name if your data model allows for this, but in general variable names cannot be synthesized by name.
I'm looking for a way to generate combinations of objects ordered by a single attribute. I don't think lexicographical order is what I'm looking for... I'll try to give an example. Let's say I have a list of objects A,B,C,D with the attribute values I want to order by being 3,3,2,1. This gives A3, B3, C2, D1 objects. Now I want to generate combinations of 2 objects, but they need to be ordered in a descending way:
A3 B3
A3 C2
B3 C2
A3 D1
B3 D1
C2 D1
Generating all combinations and sorting them is not acceptable because the real world scenario involves large sets and millions of combinations. (set of 40, order of 8), and I need only combinations above the certain threshold.
Actually I need count of combinations above a threshold grouped by a sum of a given attribute, but I think it is far more difficult to do - so I'd settle for developing all combinations above a threshold and counting them. If that's possible at all.
EDIT - My original question wasn't very precise... I don't actually need these combinations ordered, just thought it would help to isolate combinations above a threshold. To be more precise, in the above example, giving a threshold of 5, I'm looking for an information that the given set produces 1 combination with a sum of 6 ( A3 B3 ) and 2 with a sum of 5 ( A3 C2, B3 C2). I don't actually need the combinations themselves.
I was looking into subset-sum problem, but if I understood correctly given dynamic solution it will only give you information is there a given sum or no, not count of the sums.
Thanks
Actually, I think you do want lexicographic order, but descending rather than ascending. In addition:
It's not clear to me from your description that A, B, ... D play any role in your answer (except possibly as the container for the values).
I think your question example is simply "For each integer at least 5, up to the maximum possible total of two values, how many distinct pairs from the set {3, 3, 2, 1} have sums of that integer?"
The interesting part is the early bailout, once no possible solution can be reached (remaining achievable sums are too small).
I'll post sample code later.
Here's the sample code I promised, with a few remarks following:
public class Combos {
/* permanent state for instance */
private int values[];
private int length;
/* transient state during single "count" computation */
private int n;
private int limit;
private Tally<Integer> tally;
private int best[][]; // used for early-bail-out
private void initializeForCount(int n, int limit) {
this.n = n;
this.limit = limit;
best = new int[n+1][length+1];
for (int i = 1; i <= n; ++i) {
for (int j = 0; j <= length - i; ++j) {
best[i][j] = values[j] + best[i-1][j+1];
}
}
}
private void countAt(int left, int start, int sum) {
if (left == 0) {
tally.inc(sum);
} else {
for (
int i = start;
i <= length - left
&& limit <= sum + best[left][i]; // bail-out-check
++i
) {
countAt(left - 1, i + 1, sum + values[i]);
}
}
}
public Tally<Integer> count(int n, int limit) {
tally = new Tally<Integer>();
if (n <= length) {
initializeForCount(n, limit);
countAt(n, 0, 0);
}
return tally;
}
public Combos(int[] values) {
this.values = values;
this.length = values.length;
}
}
Preface remarks:
This uses a little helper class called Tally, that just isolates the tabulation (including initialization for never-before-seen keys). I'll put it at the end.
To keep this concise, I've taken some shortcuts that aren't good practice for "real" code:
This doesn't check for a null value array, etc.
I assume that the value array is already sorted into descending order, required for the early-bail-out technique. (Good production code would include the sorting.)
I put transient data into instance variables instead of passing them as arguments among the private methods that support count. That makes this class non-thread-safe.
Explanation:
An instance of Combos is created with the (descending ordered) array of integers to combine. The value array is set up once per instance, but multiple calls to count can be made with varying population sizes and limits.
The count method triggers a (mostly) standard recursive traversal of unique combinations of n integers from values. The limit argument gives the lower bound on sums of interest.
The countAt method examines combinations of integers from values. The left argument is how many integers remain to make up n integers in a sum, start is the position in values from which to search, and sum is the partial sum.
The early-bail-out mechanism is based on computing best, a two-dimensional array that specifies the "best" sum reachable from a given state. The value in best[n][p] is the largest sum of n values beginning in position p of the original values.
The recursion of countAt bottoms out when the correct population has been accumulated; this adds the current sum (of n values) to the tally. If countAt has not bottomed out, it sweeps the values from the start-ing position to increase the current partial sum, as long as:
enough positions remain in values to achieve the specified population, and
the best (largest) subtotal remaining is big enough to make the limit.
A sample run with your question's data:
int[] values = {3, 3, 2, 1};
Combos mine = new Combos(values);
Tally<Integer> tally = mine.count(2, 5);
for (int i = 5; i < 9; ++i) {
int n = tally.get(i);
if (0 < n) {
System.out.println("found " + tally.get(i) + " sums of " + i);
}
}
produces the results you specified:
found 2 sums of 5
found 1 sums of 6
Here's the Tally code:
public static class Tally<T> {
private Map<T,Integer> tally = new HashMap<T,Integer>();
public Tally() {/* nothing */}
public void inc(T key) {
Integer value = tally.get(key);
if (value == null) {
value = Integer.valueOf(0);
}
tally.put(key, (value + 1));
}
public int get(T key) {
Integer result = tally.get(key);
return result == null ? 0 : result;
}
public Collection<T> keys() {
return tally.keySet();
}
}
I have written a class to handle common functions for working with the binomial coefficient, which is the type of problem that your problem falls under. It performs the following tasks:
Outputs all the K-indexes in a nice format for any N choose K to a file. The K-indexes can be substituted with more descriptive strings or letters. This method makes solving this type of problem quite trivial.
Converts the K-indexes to the proper index of an entry in the sorted binomial coefficient table. This technique is much faster than older published techniques that rely on iteration. It does this by using a mathematical property inherent in Pascal's Triangle. My paper talks about this. I believe I am the first to discover and publish this technique, but I could be wrong.
Converts the index in a sorted binomial coefficient table to the corresponding K-indexes.
Uses Mark Dominus method to calculate the binomial coefficient, which is much less likely to overflow and works with larger numbers.
The class is written in .NET C# and provides a way to manage the objects related to the problem (if any) by using a generic list. The constructor of this class takes a bool value called InitTable that when true will create a generic list to hold the objects to be managed. If this value is false, then it will not create the table. The table does not need to be created in order to perform the 4 above methods. Accessor methods are provided to access the table.
There is an associated test class which shows how to use the class and its methods. It has been extensively tested with 2 cases and there are no known bugs.
To read about this class and download the code, see Tablizing The Binomial Coeffieicent.
Check out this question in stackoverflow: Algorithm to return all combinations
I also just used a the java code below to generate all permutations, but it could easily be used to generate unique combination's given an index.
public static <E> E[] permutation(E[] s, int num) {//s is the input elements array and num is the number which represents the permutation
int factorial = 1;
for(int i = 2; i < s.length; i++)
factorial *= i;//calculates the factorial of (s.length - 1)
if (num/s.length >= factorial)// Optional. if the number is not in the range of [0, s.length! - 1]
return null;
for(int i = 0; i < s.length - 1; i++){//go over the array
int tempi = (num / factorial) % (s.length - i);//calculates the next cell from the cells left (the cells in the range [i, s.length - 1])
E temp = s[i + tempi];//Temporarily saves the value of the cell needed to add to the permutation this time
for(int j = i + tempi; j > i; j--)//shift all elements to "cover" the "missing" cell
s[j] = s[j-1];
s[i] = temp;//put the chosen cell in the correct spot
factorial /= (s.length - (i + 1));//updates the factorial
}
return s;
}
I am extremely sorry (after all those clarifications in the comments) to say that I could not find an efficient solution to this problem. I tried for the past hour with no results.
The reason (I think) is that this problem is very similar to problems like the traveling salesman problem. Until unless you try all the combinations, there is no way to know which attributes will add upto the threshold.
There seems to be no clever trick that can solve this class of problems.
Still there are many optimizations that you can do to the actual code.
Try sorting the data according to the attributes. You may be able to avoid processing some values from the list when you find that a higher value cannot satisfy the threshold (so all lower values can be eliminated).
If you're using C# there is a fairly good generics library here. Note though that the generation of some permutations is not in lexicographic order
Here's a recursive approach to count the number of these subsets: We define a function count(minIndex,numElements,minSum) that returns the number of subsets of size numElements whose sum is at least minSum, containing elements with indices minIndex or greater.
As in the problem statement, we sort our elements in descending order, e.g. [3,3,2,1], and call the first index zero, and the total number of elements N. We assume all elements are nonnegative. To find all 2-subsets whose sum is at least 5, we call count(0,2,5).
Sample Code (Java):
int count(int minIndex, int numElements, int minSum)
{
int total = 0;
if (numElements == 1)
{
// just count number of elements >= minSum
for (int i = minIndex; i <= N-1; i++)
if (a[i] >= minSum) total++; else break;
}
else
{
if (minSum <= 0)
{
// any subset will do (n-choose-k of them)
if (numElements <= (N-minIndex))
total = nchoosek(N-minIndex, numElements);
}
else
{
// add element a[i] to the set, and then consider the count
// for all elements to its right
for (int i = minIndex; i <= (N-numElements); i++)
total += count(i+1, numElements-1, minSum-a[i]);
}
}
return total;
}
Btw, I've run the above with an array of 40 elements, and size-8 subsets and consistently got back results in less than a second.