I have always been interested in algorithms, sort, crypto, binary trees, data compression, memory operations, etc.
I read Mark Nelson's article about permutations in C++ with the STL function next_perm(), very interesting and useful, after that I wrote one class method to get the next permutation in Delphi, since that is the tool I presently use most. This function works on lexographic order, I got the algo idea from a answer in another topic here on stackoverflow, but now I have a big problem. I'm working with permutations with repeated elements in a vector and there are lot of permutations that I don't need. For example, I have this first permutation for 7 elements in lexographic order:
6667778 (6 = 3 times consecutively, 7 = 3 times consecutively)
For my work I consider valid perm only those with at most 2 elements repeated consecutively, like this:
6676778 (6 = 2 times consecutively, 7 = 2 times consecutively)
In short, I need a function that returns only permutations that have at most N consecutive repetitions, according to the parameter received.
Does anyone know if there is some algorithm that already does this?
Sorry for any mistakes in the text, I still don't speak English very well.
Thank you so much,
Carlos
My approach is a recursive generator that doesn't follow branches that contain illegal sequences.
Here's the python 3 code:
def perm_maxlen(elements, prefix = "", maxlen = 2):
if not elements:
yield prefix + elements
return
used = set()
for i in range(len(elements)):
element = elements[i]
if element in used:
#already searched this path
continue
used.add(element)
suffix = prefix[-maxlen:] + element
if len(suffix) > maxlen and len(set(suffix)) == 1:
#would exceed maximum run length
continue
sub_elements = elements[:i] + elements[i+1:]
for perm in perm_maxlen(sub_elements, prefix + element, maxlen):
yield perm
for perm in perm_maxlen("6667778"):
print(perm)
The implentation is written for readability, not speed, but the algorithm should be much faster than naively filtering all permutations.
print(len(perm_maxlen("a"*100 + "b"*100, "", 1)))
For example, it runs this in milliseconds, where the naive filtering solution would take millenia or something.
So, in the homework-assistance kind of way, I can think of two approaches.
Work out all permutations that contain 3 or more consecutive repetitions (which you can do by treating the three-in-a-row as just one psuedo-digit and feeding it to a normal permutation generation algorithm). Make a lookup table of all of these. Now generate all permutations of your original string, and look them up in lookup table before adding them to the result.
Use a recursive permutation generating algorthm (select each possibility for the first digit in turn, recurse to generate permutations of the remaining digits), but in each recursion pass along the last two digits generated so far. Then in the recursively called function, if the two values passed in are the same, don't allow the first digit to be the same as those.
Why not just make a wrapper around the normal permutation function that skips values that have N consecutive repetitions? something like:
(pseudocode)
funciton custom_perm(int max_rep)
do
p := next_perm()
while count_max_rerps(p) < max_rep
return p
Krusty, I'm already doing that at the end of function, but not solves the problem, because is need to generate all permutations and check them each one.
consecutive := 1;
IsValid := True;
for n := 0 to len - 2 do
begin
if anyVector[n] = anyVector[n + 1] then
consecutive := consecutive + 1
else
consecutive := 1;
if consecutive > MaxConsecutiveRepeats then
begin
IsValid := False;
Break;
end;
end;
Since I do get started with the first in lexographic order, ends up being necessary by this way generate a lot of unnecessary perms.
This is easy to make, but rather hard to make efficient.
If you need to build a single piece of code that only considers valid outputs, and thus doesn't bother walking over the entire combination space, then you're going to have some thinking to do.
On the other hand, if you can live with the code internally producing all combinations, valid or not, then it should be simple.
Make a new enumerator, one which you can call that next_perm method on, and have this internally use the other enumerator, the one that produces every combination.
Then simply make the outer enumerator run in a while loop asking the inner one for more permutations until you find one that is valid, then produce that.
Pseudo-code for this:
generator1:
when called, yield the next combination
generator2:
internally keep a generator1 object
when called, keep asking generator1 for a new combination
check the combination
if valid, then yield it
Related
I have an array of some particular ints, and I want to find all the unique combinations (by addition) of these ints. I'm sure there's a way to do this functionally; I'm trying to avoid the iterative way of shoving for loops inside of for loops. I'm using Rust in this case, but this question is more generally a, "functional programming, how do?" take.
My first thought is that I should just zip every entry in v with every other entry, reduce them by addition to a single element, and filter duplicates. This is O(|v|^2), which feels bad, meaning I'm probably missing something fairly obvious to the functional whiz kids. Also I'm not even sure how to do it, I'd probably use a for loop to construct my new massive array.
My first pass: note that v holds all the numbers I care about.
let mut massive_arr = Vec::new();
for &elem in v.iter(){
for &elem2 in v.iter(){
massive_arr.push((elem,elem2));
}
}
let mut single_massive = Vec::new();
for &tuple in massive_arr.iter(){
single_massive.push(tuple.0 + tuple.1);
}
single_massive.dedup();
let summand: usize = single_massive.iter().sum(); println!("The sum of all that junk is {:?}", summand);```
Help me baptize my depraved iterations in the pure light of functional programming.
edited: I threw up an example before as I was still figuring out the implementation that actually worked, and the question was more of a, how do I do this better question. The thing above now actually works (but is still ugly!).
You can possibly use itertools (I do not have a compiler at hand, but you probably get the idea):
use itertools::Itertools;
iproduct!(v.iter(), v.iter()) // construct all pairs
.map(|tuple| tuple.0+tuple.1) // sum each pair
.unique() // de-duplicate (uses a HashMap internally)
.sum() // sum up
All this is still O(n^2) which is -- as far as I see -- asymptotically optimal, because all pairs of numbers might be needed.
To avoid the obvious duplicates, you can use tuple_combinations:
v.iter()
.tuple_combinations()
.map(|(a, b)| a+b)
.unique()
.sum()
Improving on #phimuemue's answer, you can avoid the obvious duplicates like this:
v.iter()
.enumerate()
.flat_map (|(i, a)| v[i+1..].iter().map (move |b| a+b))
.unique() // May not be needed or what you really want, see note below
.sum()
Playground
Note however that this may not give you the answer you really want if multiple pairs of numbers have the same sum. For example, given vec![1, 2, 3, 4] as input, which do you expect:
(1+2) + (1+3) + (1+4) + (2+3) + (2+4) + (3+4) = 30
or 3 + 4 + 5 + 6 + 7 = 25 because 1+4 == 2+3 == 5 is only counted once?
My mind is blown since I began, last week, trying to sort an array of N elements by condition: the difference between 2 elements being always less or equal to the next 2 elements. For example:
Α[4] = { 10, 2, 7, 4}
It is possible to rearrange that array this way:
{2, 7, 10, 4} because (2 - 7 = -5) < (7 - 10 = -3) < (10 - 4 = 6)
{4, 10, 7, 2} because (4 - 10 = -6) < (10 - 7 = 3) < (7 - 2 = 5)
One solution I considered was just shuffling the array and checking each time if it agreed with the conditions, an efficient method for a small number of elements, but time consuming or even impossible for a larger number of elements.
Another was trying to move elements around the array with loops, hoping again to meet the requirements, but again this method is very time consuming and also sometimes not possible.
Trying to find an algorithm doesn't seem to have any result but there must be something.
Thank you very much in advance.
I normally don't just provide code, but this question intrigued me, so here's a brute-force solution, that might get you started.
The concept will always be slow because the individual elements in the list to be sorted are not independent of each other, so they cannot be sorted using traditional O(N log N) algorithms. However, the differences can be sorted that way, which simplifies checking for a solution, and permutations could be checked in parallel to speed up the processing.
import os,sys
import itertools
def is_diff_sorted(qa):
diffs = [qa[i] - qa[i+1] for i in range(len(qa)-1)]
for i in range(len(diffs)-1):
if diffs[i] > diffs[i+1]:
return False
return True
a = [2,4,7,10]
#a = [1,4,6,7,20]
a.sort()
for perm in itertools.permutations(a):
if is_diff_sorted(perm):
print "Solution:",str(a)
break
This condition is related to differentiation. The (negative) difference between neighbouring elements has to be steady or increasing with increasing index. Multiply the condition by -1 and you get
a[i+1]-a[i] => a[i+2]-a[i+1]
or
0 => (a[i+2]-a[i+1])- (a[i+1]-a[i])
So the 2nd derivative has to be 0 or negative, which is the same as having the first derivative stay the same or changing downwards, like e.g. portions of the upper half of a circle. That does not means that the first derivative itself has to start out positive or negative, just that it never change upward.
The problem algorithmically is that it can't be a simple sort, since you never compare just 2 elements of the list, you'll have to compare three at a time (i,i+1,i+2).
So the only thing you know apart from random permutations is given in Klas` answer (values first rising if at all, then falling if at all), but his is not a sufficient condition since you can have a positive 2nd derivative in his two sets (rising/falling).
So is there a solution much faster than the random shuffle? I can only think of the following argument (similar to Klas' answer). For a given vector the solution is more likely if you separate the data into a 1st segment that is rising or steady (not falling) and a 2nd that is falling or steady (not rising) and neither is empty. Likely an argument could be made that the two segments should have approximately equal size. The rising segment should have the data that are closer together and the falling segment should contain data that are further apart. So one could start with the mean, and look for data that are close to it, move them to the first set,then look for more widely spaced data and move them to the 2nd set. So a histogram might help.
[4 7 10 2] --> diff [ 3 3 -8] --> 2diff [ 0 -11]
Here is a solution based on backtracking algorithm.
Sort input array in non-increasing order.
Start dividing the array's values into two subsets: put the largest element to both subsets (this would be the "middle" element), then place second largest one into arbitrary subset.
Sequentially put the remaining elements to either subset. If this cannot be done without violating the "difference" condition, use other subset. If both subsets are not acceptable, rollback and change preceding decisions.
Reverse one of the arrays produced on step 3 and concatenate it with other array.
Below is Python implementation (it is not perfect, the worst defect is recursive implementation: while recursion is quite common for backtracking algorithms, this particular algorithm seems to work in linear time, and recursion is not good for very large input arrays).
def is_concave_end(a, x):
return a[-2] - a[-1] <= a[-1] - x
def append_element(sa, halves, labels, which, x):
labels.append(which)
halves[which].append(x)
if len(labels) == len(sa) or split_to_halves(sa, halves, labels):
return True
if which == 1 or not is_concave_end(halves[1], halves[0][-1]):
halves[which].pop()
labels.pop()
return False
labels[-1] = 1
halves[1].append(halves[0][-1])
halves[0].pop()
if split_to_halves(sa, halves, labels):
return True
halves[1].pop()
labels.pop()
def split_to_halves(sa, halves, labels):
x = sa[len(labels)]
if len(halves[0]) < 2 or is_concave_end(halves[0], x):
return append_element(sa, halves, labels, 0, x)
if is_concave_end(halves[1], x):
return append_element(sa, halves, labels, 1, x)
def make_concave(a):
sa = sorted(a, reverse = True)
halves = [[sa[0]], [sa[0], sa[1]]]
labels = [0, 1]
if split_to_halves(sa, halves, labels):
return list(reversed(halves[1][1:])) + halves[0]
print make_concave([10, 2, 7, 4])
It is not easy to produce a good data set to test this algorithm: plain set of random numbers either is too simple for this algorithm or does not have any solutions. Here I tried to generate a set that is "difficult enough" by mixing together two sorted lists, each satisfying the "difference" condition. Still this data set is processed in linear time. And I have no idea how to prepare any data set that would demonstrate more-than-linear time complexity of this algorithm...
Not that since the diffence should be ever-rising, any solution will have element first in rising order and then in falling order. The length of either of the two "suborders" may be 0, so a solution could consist of a strictly rising or strictly falling sequence.
The following algorithm will find any solutions:
Divide the set into two sets, A and B. Empty sets are allowed.
Sort A in rising order and B in falling order.
Concatenate the two sorted sets: AB
Check if you have a solution.
Do this for all possible divisions into A and B.
Expanding on the #roadrunner66 analysis, the solution is to take two smallest elements of the original array, and make them first and last in the target array; take two next smallest elements and make them second and next-to-last; keep going until all the elements are placed into the target. Notice that which one goes to the left, and which one to the right doesn't matter.
Sorting the original array facilitates the process (finding smallest elements becomes trivial), so the time complexity is O(n log n). The space complexity is O(n), because it requires a target array. I don't know off-hand if it is possible to do it in-place.
For one of my homework problems, we had to write a function that creates an array containing n random numbers between 1 and 365. (Done). Then, check if any of these n birthdays are identical. Is there a shorter way to do this than doing several loops or several logical expressions?
Thank you!
CODE SO FAR, NOT DONE YET!!
function = [prob] bdayprob(N,n)
N = input('Please enter the number of experiments performed: N = ');
n = input('Please enter the sample size: n = ');
count = 0;
for(i=1:n)
x(i) = randi(365);
if(x(i)== x)
count = count + 1
end
return
If I'm interpreting your question properly, you want to check to see if generating n integers or days results in n unique numbers. Given your current knowledge in MATLAB, it's as simple as doing:
n = 30; %// Define sample size
N = 10; %// Define number of trials
%// Define logical array where each location tells you whether
%// birthdays were repeated for a trial
check = false(1, N);
%// For each trial...
for idx = 1 : N
%// Generate sample size random numbers
days = randi(365, n, 1);
%// Check to see if the total number of unique birthdays
%// are equal to the sample size
check(idx) = numel(unique(days)) == n;
end
Woah! Let's go through the code slowly shall we? We first define the sample size and the number of trials. We then specify a logical array where each location tells you whether or not there were repeated birthdays generated for that trial. Now, we start with a loop where for each trial, we generate random numbers from 1 to 365 that is of n or sample size long. We then use unique and figure out all unique integers that were generated from this random generation. If all of the birthdays are unique, then the total number of unique birthdays generated should equal the sample size. If we don't, then we have repeats. For example, if we generated a sample of [1 1 1 2 2], the output of unique would be [1 2], and the total number of unique elements is 2. Since this doesn't equal 5 or the sample size, then we know that the birthdays generated weren't unique. However, if we had [1 3 4 6 7], unique would give the same output, and since the output length is the same as the sample size, we know that all of the days are unique.
So, we check to see if this number is equal to the sample size for each iteration. If it is, then we output true. If not, we output false. When I run this code on my end, this is what I get for check. I set the sample size to 30 and the number of trials to be 10.
check =
0 0 1 1 0 0 0 0 1 0
Take note that if you increase the sample size, there is a higher probability that you will get duplicates, because randi can be considered as sampling with replacement. Therefore, the larger the sample size, the higher the chance of getting duplicate values. I made the sample size small on purpose so that we can see that it's possible to get unique days. However, if you set it to something like 100, or 200, you will most likely get check to be all false as there will most likely be duplicates per trial.
Here are some more approaches that avoid loops. Let
n = 20; %// define sample size
x = randi(365,n,1); %// generate n values between 1 and 365
Any of the following code snippets returns true (or 1) if there are two identical values in x, and false (or 0) otherwise:
Sort and then check if any two consecutive elements are the same:
result = any(diff(sort(x))==0);
Do all pairwise comparisons manually; remove self-pairs and duplicate pairs; and check if any of the remaining comparisons is true:
result = nnz(tril(bsxfun(#eq, x, x.'),-1))>0;
Compute the distance between distinct values, considering each pair just once, and then check if any distance is 0:
result = any(pdist(x(:))==0);
Find the number of occurrences of the most common value (mode):
[~, occurs] = mode(x);
result = occurs>1;
I don't know if I'm supposed to solve the problem for you, but perhaps a few hints may lead you in the right direction (besides I'm not a matlab expert so it will be in general terms):
Maybe not, but you have to ask yourself what they expect of you. The solution you propose requires you to loop through the array in two nested loops which will mean n*(n-1)/2 times through the loop (ie quadratic time complexity).
There are a number of ways you can improve the time complexity of the problem. The most straightforward would be to have a 365 element table where you can keep track if a particular number has been seen yet - which would require only a single loop (ie linear time complexity), but perhaps that's not what they're looking for either. But maybe that solution is a little bit ad-hoc? What we're basically looking for is a fast lookup if a particular number has been seen before - there exists more memory efficient structures that allows look up in O(1) time and O(log n) time (if you know these you have an arsenal of tools to use).
Then of course you could use the pidgeonhole principle to provide the answer much faster in some special cases (remember that you only asked to determine whether two or more numbers are equal or not).
There is an integer array, for eg.
{3,1,2,7,5,6}
One can move forward through the array either each element at a time or can jump a few elements based on the value at that index. For e.g., one can go from 3 to 1 or 3 to 7, then one can go from 1 to 2 or 1 to 2(no jumping possible here), then one can go 2 to 7 or 2 to 5, then one can go 7 to 5 only coz index of 7 is 3 and adding 7 to 3 = 10 and there is no tenth element.
I have to only count the number of possible paths to reach the end of the array from start index.
I could only do it recursively and naively which runs in exponential time.
Somebody plz help.
My recommendation: use dynamic programming.
If this key word is sufficient and you want the challenge to find a possible solution on your own, dont read any further!
Here a possible DP-algorithm on the example input {3,1,2,7,5,6}. It will be your job to adjust on the general problem.
create array sol length 6 with just zeros in it. the array will hold the number of ways.
sol[5] = 1;
for (i = 4; i>=0;i--) {
sol[i] = sol[i+1];
if (i+input[i] < 6 && input[i] != 1)
sol[i] += sol[i+input[i]];
}
return sol[0];
runtime O(n)
As for the directed graph solution hinted in the comments :
Each cell in the array represents a node. Make an directed edge from each node to the node accessable. Basically you can then count more easily the number of ways by just looking at the outdegrees on the nodes (since there is no directed cycle) however it is a lot of boiler plate to actual program it.
Adjusting the recursive solution
another solution would be to pruning. This is basically equivalent to the DP-algorithm. The exponentiel time comes from the fact, that you calculate values several times. Eg function is recfunc(index). The initial call recFunc(0) calls recFunc(1) and recFunc(3) and so on. However recFunc(3) is bound to be called somewhen again, which leads to a repeated recursive calculation. To prune this you add a Map to hold all already calculated values. If you make a call recFunc(x) you lookup in the map if x was already calculated. If yes, return the stored value. If not, calculate, store and return it. This way you get a O(n) too.
Can anyone tell me if there is a way (in MATLAB) to check whether a certain value is equal to any of the values stored within another array?
The way I intend to use it is to check whether an element index in one matrix is equal to the values stored in another array (where the stored values are the indices of the elements which meet a certain criteria).
So, if the indices of the elements which meet the criteria are stored in the matrix below:
criteriacheck = [3 5 6 8 20];
Going through the main array (called array) and checking if the index matches:
for i = 1:numel(array)
if i == 'Any value stored in criteriacheck'
%# "Do this"
end
end
Does anyone have an idea of how I might go about this?
The excellent answer previously given by #woodchips applies here as well:
Many ways to do this. ismember is the first that comes to mind, since it is a set membership action you wish to take. Thus
X = primes(20);
ismember([15 17],X)
ans =
0 1
Since 15 is not prime, but 17 is, ismember has done its job well here.
Of course, find (or any) will also work. But these are not vectorized in the sense that ismember was. We can test to see if 15 is in the set represented by X, but to test both of those numbers will take a loop, or successive tests.
~isempty(find(X == 15))
~isempty(find(X == 17))
or,
any(X == 15)
any(X == 17)
Finally, I would point out that tests for exact values are dangerous if the numbers may be true floats. Tests against integer values as I have shown are easy. But tests against floating point numbers should usually employ a tolerance.
tol = 10*eps;
any(abs(X - 3.1415926535897932384) <= tol)
you could use the find command
if (~isempty(find(criteriacheck == i)))
% do something
end
Note: Although this answer doesn't address the question in the title, it does address a more fundamental issue with how you are designing your for loop (the solution of which negates having to do what you are asking in the title). ;)
Based on the for loop you've written, your array criteriacheck appears to be a set of indices into array, and for each of these indexed elements you want to do some computation. If this is so, here's an alternative way for you to design your for loop:
for i = criteriacheck
%# Do something with array(i)
end
This will loop over all the values in criteriacheck, setting i to each subsequent value (i.e. 3, 5, 6, 8, and 20 in your example). This is more compact and efficient than looping over each element of array and checking if the index is in criteriacheck.
NOTE: As Jonas points out, you want to make sure criteriacheck is a row vector for the for loop to function properly. You can form any matrix into a row vector by following it with the (:)' syntax, which reshapes it into a column vector and then transposes it into a row vector:
for i = criteriacheck(:)'
...
The original question "Can anyone tell me if there is a way (in MATLAB) to check whether a certain value is equal to any of the values stored within another array?" can be solved without any loop.
Just use the setdiff function.
I think the INTERSECT function is what you are looking for.
C = intersect(A,B) returns the values common to both A and B. The
values of C are in sorted order.
http://www.mathworks.de/de/help/matlab/ref/intersect.html
The question if i == 'Any value stored in criteriacheck can also be answered this way if you consider i a trivial matrix. However, you are proably better off with any(i==criteriacheck)