As far as I know, counting every way to give change to a set sum and a starting till configuration is a classic Dynamic Programming problem.
I was wondering if there was a way to also display (or store) the actual change structures that could possibly amount to the given sum while preserving the DP complexity.
I have never saw this issue being discussed and I would like some pointers or a brief explanation of how this can be done or why this cannot be done.
DP for change problem has time complexity O(Sum * ValuesCount) and storage complexity O(Sum).
You can prepare extra data for this problem in the same time as DP for change, but you need more storage O(O(Sum*ValuesCount), and a lot of time for output of all variants O(ChangeWaysCount).
To prepare data for way recovery, make the second array B of arrays (or lists). When you incrementing count array A element from some previous element, add used value to corresponding element of B. At the end, unwind all the ways from the last element.
Example: values 1,2,3, sum 4
index 0 1 2 3 4
A 0 1 2 3 4
B - 1 1 2 1 2 3 1 2 3
We start unwinding from B[4] elements:
1-1-1-1 (B[4]-B[3]-B[2]-B[1])
2-1-1 (B[4]-B[2]-B[1])
2-2 (B[4]-B[2])
3-1 (B[4]-B[1])
Note that I have used only ways with non-increasing values to avoid permutation variants (i.e. 1-3 and 3-1)
Related
In the card game cribbage, counting the runs for a hand during the show (one of the stages of a turn in the game) is reporting the longest increasing subsequence which consists of only values that increase by 1. If duplicate values exist are apart of this subsequence than a double run (or triple, quadruple, et cetera) is reported.
Some examples:
("A","2","3","4","5") => (1,5) Single run for 5
("A","2","3","4","4") => (2,4) Double run for 4
("A","2","3","3","3") => (3,3) Triple run for 3
("A","2","3","4","6") => (1,4) Single run for 4
("A","2","3","5","6") => (1,3) Single run for 3
("A","2","4","5","7") => (0,0) No runs
To address cases that arise with hands larger than the cribbage hand size of 5. A run will be selected if it has the maximum product of the number duplicates of a subsequence and that subsequences length.
Some relevant examples:
("A","2","2","3","5","6","7","8","9","T","J") => (1,7) Single run for 7
("A","2","2","3","5","6","7","8") => (2,3) Double run for 3
My method for finding the maximum scoring run is as follows:
Create a list of ranks and sort it. O(N*log(N))
Create a list to store the length of the maximum run length and how many duplicates of it exist. Initialize it to [1 duplicate, 1 long].
Create an identical list as above to store the current run.
Create a flag that indicates whether the duplicate you've encountered is not the initial duplicate of this value. Initialize it to False.
Create a variable to store the increase in duplicate subsequences if additional duplicates values are found after the initial duplicate. Initialize it to 1.
Iterate over the differences between adjacent elements. O(N)
If the difference is greater than one, the run has ended. Check if the product of the elements of the max run is less than the current run and the current run has length 3 or greater. If this is true, the current run becomes the maximum run and the current run list is reset to [1,1]. The flag is reset to False. The increment for duplicate subsequences is reset to 1. Iterate to next value.
If the difference is 1, increment the length of the current run by 1 and set the flag to False. Iterate to next value.
If the difference is 0 and the flag is False, set the increment for duplicate subsequences equal to the current number of duplicates for the run. Then, double the number of duplicates for the run and set the flag to True. Iterate to the next value
If the difference is 0 and the flag is True, increase the number of the runs by the increment for duplicate subsequences value.
After the iteration, check the current run list as in step 7 against the max run and set max run accordingly.
I believe this has O(N*(1+log(N)). I believe this is the best time complexity, but I am not sure how to prove this or what a better algorithm would look like. Is there a way to do this without sorting the list first that achieves a better time complexity? If not, how does one go about proving this is the best time complexity?
iterate over the differences between
Time complexity of an algorithm is a well-traveled path. Proving the complexity of an algorithm varies slightly among mathematician clusters; rather, the complexity community usually works with modular pseudo-code and standard reductions. For instance, a for loop based on the input length is O(N) (surprise); sorting a list is known to be O(log N) at best (in the general case). For an good treatment, see Big O, how do you calculate/approximate it?.
Note: O(N x (1+log(N)) is slightly sloppy notation. Only the greatest complexity factor -- the one that dominates as N approaches infinity -- is used. Drop the 1+: it's simply O(N log N).
As I suggested in a comment, you can simply count elements. Keep a list of counts, indexed by your card values. For discussing the algorithm, don't use the "dirty" data of character representations: "A23456789TJQK"; simply use their values, either 0-12 or 1-13.
for rank in hand:
count[rank] += 1
This is a linear pass through the data, O(N).
Now, traverse your array of counts, finding the longest sequence of non-zero values. This is a fixed-length list of 13 elements, touching each element only once: O(1). If you accumulate a list of multiples (card counts, then you'll also have your combinatoric factors at the end.
The resulting algorithm and code are, therefore, O(N).
For instance, let's shorten this to 7 card values, 0-6. Given the input integers
1 2 1 3 6 1 3 5 0
You make the first pass to count items:
[1 3 1 2 0 1 1]
A second pass gives you a max run length of 4, with counts [1 3 1 2].
You report a run of 4, a triple and a double, or the point count
4 * (1 * 3 * 1 * 2)
You can also count the pair values:
2 * 3! + 2 * 2!
I need to iterate over every possible combination of elements (with repetitions) up to n elements long.
I've found multiple solutions for this problem. But all of these are recursively generating collection of every possible combination, then iterating over it. while this works, for large element collections and combinations size it results in heavy memory use, so I'm looking for a solution that would allow me to calculate next combination from previous one, knowing number of elements, and maximum size of combination.
Is this even possible and is there any particular algorith that would work here?
Generate the combinations so that each combination is sorted. (This assumes the elements themselves can easily be placed in order. The precise ordering relationship is not important as long as it is a total order.)
Start with the combination consisting of n repetitions of the smallest element. To produce the next combination from any given combination:
Scan backwards until you find an element which is not the largest element. If you can't find one, you are done.
Replace that element and all following elements with the next larger element of that element.
If you want combinations of all lengths up to n, run that algorithm for each length up to n. Or start with a vector which contains empty slots and use the above algorithm with the understanding that the "next larger element" after an empty slot is the smallest element.
Example: length 3 of 3 values:
1 1 1
1 1 2
1 1 3
1 2 2
1 2 3
1 3 3
2 2 2
2 2 3
2 3 3
3 3 3
I am solving Segment tree and Quad Tree related problems; while I noticed that in segment tree we split the 1D array into 2 (2^1) segments and recursively do this until base case arrives. Similarly, in Quad tree We subdivide the 2D grid into 4 (2^2) segments in every step. All these divide-and-Conquer mechanism is for achieving logarithmic time complexity. No offense!
But why don't we subdivide the array into 4 (4^1) parts or more instead of 2 parts in segment tree? And why we don't split the grid into 16 (4^2) parts instead of 4? By doing these, We can achieve O(log(N)) performance, but it would be a better log as log(N)(base 4) is better than log(N)(base 2).
I know in this case, the implementation would be little bit difficult. Is there a memory overhead problem? Or anything?
Please correct me if I am wrong anywhere. Thanks!
It wouldn't actually work faster. Let's assume that we divided it into 4 parts. Then we would have to merge 4 values instead of 2 in each node to answer the query. Assuming that merging 4 values takes 3 times longer(for example, to get the maximum of 2 numbers we need 1 call to max function, but to get the maximum of 4 values 3 calls are required), we have log4(n) * 3 > log2(n) * 1. Moreover, it would be harder to implement(more cases to be considered and so on).
log 4 (N) = log 2 (N) / log 2 (4) = log 2 (N) / 2
in general,the time complexity are both O(logn) , while four segments is much harder than two segments to maintain. In fact ，（In acm/icpc） two segments are much easy to code and it is suffcient to work.
I asked whether this problem was NP-complete on the Computer Science forum, but asking for programming heuristics seems better suited for this site. So here it goes.
You are given an NxN grid of unit squares and 2N binary strings of length N. The goal is to fill the grid with 0's and 1's so that each string appears once and only once in the grid, either horizontally (left to right) or vertically (top down). Or determine that no such solution exists. If N is not fixed I suspect this is an NP-complete problem. However are there any heuristics that can hopefully speed up the search to faster than brute force trying all ways to fill in the grid with N vertical strings?
I remember programming this for my friend that had the 5x5 physical version of this game, but I used brute force back then. I can only think of this heuristic:
Consider a 4x4 map with these 8 strings (read each from left to right):
1 1 0 1
1 0 0 1
1 0 1 1
1 0 1 0
1 1 1 1
1 0 0 0
0 0 1 1
1 1 1 0
(Note that this is already solved, since the second 4 is the first 4 transposed)
First attempt:
We will choose columns from left to right. Since 7 of 8 strings start with 1, we will try to put the one with most 1s to the first column (so that we can lay rows more easily when columns are done).
In the second column, most string have 0, so you can also try putting a string with most zeros to the second row, and so on.
This i would call a wide-1 prediction, since it only looks at one column at a time
(Possible) Improvement:
You can look at 2 columns at a time (a wide-2 prediction, if i may call it like that). In this case, from the 8 strings, the most common combination of first two bits is 10 (5/8), so you would like to choose first two columns so the the combination 10 occurring as much as possible (in this case, 1111 followed by 1000 has 3 of 4 10 at start).
(Of course you don't have to stop at 2)
Weaknesses:
I don't know if this would work. I just made it up and thought it might work.
If you choose to he wide-X prediction, the number of possibilities is exponential with X
This can absolutely fail if the distribution of combinations if even.
What you can do:
As i said, this game has physical 5x5 adaptation, only there you can also lay the string from right-to-left and bottom-to-top, if you found that name, you could google further. I unfortunately don't remember it.
Sounds like you want the crossword grid filling algorithm:
First, build 2N subsets of your 2N strings -- each subset has all the strings with a particular bit at a particular postion. So subset(0,3) is all the strings that have a 0 in the 3rd position and subset(1,5) is all the strings that have a 1 in the 5th position.
The algorithm is a basic brute-force depth fist search trying all possible mappings of strings to slots in the grid, with severe pruning of impossible branches
Your search state is a set of assignments of strings to slots and a set of sets of possible assignments to the remaining slots. The initial state has 0 assignments and 2N sets, all of which contain all 2N strings.
At each step of the search, pick the most constrained set (the set with the fewest elements) from the set of possible sets. Try each element of the set in turn in that slot (adding it to the assigments and removing it from the set of sets), and constrain all the remaining sets of sets by removing the chosen string and intersecting the crossing sets with subset(X,N) (computed in step 1) where X is the bit from the chosen string and N is the row/column number of the chosen string
If you find an empty set when picking above, there is no solution with the choices so far, so backtrack up the tree to a different choice
This is still EXPTIME, but it is about as fast as you can get it. Since the main time consuming step is the set intersections, using 2N bit binary strings for your set representation is very fast -- for N=32, the sets fit in a 64-bit word and can be intersected with a single AND instruction. It also helps to have a POPCOUNT instruction, since you also need set sizes.
This can be solved as a 0/1 integer linear program with O(N^2) variables and constraints. First there are variables Xij which are 1 if string i is assigned to line j (where j=1 to N are rows and j = (N+1) to 2N are columns). Then there is a variable for each square in the grid, which indicates if the entry is 0 or 1. If the position of the square is (i,j) with variable Yij then the sum of all X variables for line j that correspond to strings that have a 1 in position i is equal to Yij, and the sum of all X variables for line j that correspond to strings that have a 0 in position i is equal to (1 - Yij). And similarly for line i and position j. Finally, the sum of all X variables Xij for each string i (summed over all lines j) is equal to 1.
There has been a lot of research in speeding up solvers for 0/1 integer programming so this may be able to often handle fairly large N (like N=100) for many examples. Also, in some cases, solving the relaxed non-integer linear program and rounding the solution off to 0/1 may produce a valid solution, in polynomial time.
We could choose the first lg 2N rows out of the 2N strings, and then since 2^(lg 2N) = 2N, in a lot of cases there shouldn't be very many ways to assign the N columns so that the prefixes of length lg 2N are respected. Then all the rows are filled in so they can be checked to see if a solution has been found. We can also try assigning more rows in the beginning, and fill in different combinations of rows besides the initial rows. (e.g. we can try filling in contiguous rows starting anywhere in the grid).
Running time for assigning lg 2N rows out of 2N strings is O((2N)^(lg 2N)) = O(2^((lg 2N)^2)), which grows slower than 2^N. Assigning columns to match the prefixes is the part that's the hardest to predict run time. If a prefix occurs K times among the assigned rows, and there are M remaining strings that have the prefix, then the number of assignments for this prefix is M*(M-1)...(M-K+1). The total number of possible column assignments is the product of these terms over all prefixes that occur among the rows. If this gets to be too large, the number of rows initially assigned can be increased. But it's hard to predict the worst-case run time unless an assumption is made like the NxN grid is filled in randomly.
I have a language-agnostic question about an algorithm.
This comes from a (probably simple) programming challenge I read. The problem is, I'm too stupid to figure it out, and curious enough that it is bugging me.
The goal is to sort a list of integers to ascending order by swapping the positions of numbers in the list. Each time you swap two numbers, you have to add their sum to a running total. The challenge is to produce the sorted list with the smallest possible running total.
Examples:
3 2 1 - 4
1 8 9 7 6 - 41
8 4 5 3 2 7 - 34
Though you are free to just give the answer if you want, if you'd rather offer a "hint" in the right direction (if such a thing is possible), I would prefer that.
Only read the first two paragraph is you just want a hint. There is a an efficient solution to this (unless I made a mistake of course). First sort the list. Now we can write the original list as a list of products of disjoint cycles.
For example 5,3,4,2,1 has two cycles, (5,1) and (3,4,2). The cycle can be thought of as starting at 3, 4 is in 3's spot, 2 is in 4's spot, and 4 is in 3's. spot. The end goal is 1,2,3,4,5 or (1)(2)(3)(4)(5), five disjoint cycles.
If we switch two elements from different cycles, say 1 and 3 then we get: 5,1,4,2,3 and in cycle notation (1,5,3,4,2). The two cycles are joined into one cycle, this is the opposite of what we want to do.
If we switch two elements from the same cycle, say 3 and 4 then we get: 5,4,3,2,1 in cycle notation (5,1)(2,4)(3). The one cycle is split into two smaller cycles. This gets us closer to the goal of all cycles of length 1. Notice that any switch of two elements in the same cycle splits the cycle into two cycles.
If we can figure out the optimal algorithm for switching one cycle we can apply that for all cycles and get an optimal algorithm for the entire sort. One algorithm is to take the minimum element in the cycle and switch it with the the whose position it is in. So for (3,4,2) we would switch 2 with 4. This leaves us with a cycle of length 1 (the element just switched into the correct position) and a cycle of size one smaller than before. We can then apply the rule again. This algorithm switches the smallest element cycle length -1 times and every other element once.
To transform a cycle of length n into cycles of length 1 takes n - 1 operations. Each element must be operated on at least once (think about each element to be sorted, it has to be moved to its correct position). The algorithm I proposed operates on each element once, which all algorithms must do, then every other operation was done on the minimal element. No algorithm can do better.
This algorithm takes O(n log n) to sort then O(n) to mess with cycles. Solving one cycle takes O(cycle length), the total length of all cycles is n so cost of the cycle operations is O(n). The final run time is O(n log n).
I'm assuming memory is free and you can simulate the sort before performing it on the real objects.
One approach (that is likely not the fastest) is to maintain a priority queue. Each node in the queue is keyed by the swap cost to get there and it contains the current item ordering and the sequence of steps to achieve that ordering. For example, initially it would contain a 0-cost node with the original data ordering and no steps.
Run a loop that dequeues the lowest-cost queue item, and enqueues all possible single-swap steps starting at that point. Keep running the loop until the head of the queue has a sorted list.
I did a few attempts at solving one of the examples by hand:
1 8 9 7 6
6 8 9 7 1 (+6+1=7)
6 8 1 7 9 (7+1+9=17)
6 8 7 1 9 (17+1+7=25)
6 1 7 8 9 (25+1+8=34)
1 6 7 8 9 (34+1+6=41)
Since you needed to displace the 1, it seems that you may have to do an exhaustive search to complete the problem - the details of which were already posted by another user. Note that you will encounter problems if the dataset is large when doing this method.
If the problem allows for "close" answers, you can simply make a greedy algorithm that puts the largest item into position - either doing so directly, or by swapping the smallest element into that slot first.
Comparisons and traversals apparently come for free, you can pre-calculate the "distance" a number must travel (and effectively the final sort order). The puzzle is the swap algorithm.
Minimizing overall swaps is obviously important.
Minimizing swaps of larger numbers is also important.
I'm pretty sure an optimal swap process cannot be guaranteed by evaluating each ordering in a stateless fashion, although you might frequently come close (not the challenge).
I think there is no trivial solution to this problem, and my approach is likely no better than the priority queue approach.
Find the smallest number, N.
Any pairs of numbers that occupy each others' desired locations should be swapped, except for N.
Assemble (by brute force) a collection of every set of numbers that can be mutually swapped into their desired locations, such that the cost of sorting the set amongst itself is less than the cost of swapping every element of the set with N.
These sets will comprise a number of cycles. Swap within those cycles in such a way that the smallest number is swapped twice.
Swap all remaining numbers, which comprise a cycle including N, using N as a placeholder.
As a hint, this reeks of dynamic programming; that might not be precise enough a hint to help, but I'd rather start with too little!
You are charged by the number of swaps, not by the number of comparisons. Nor did you mention being charged for keeping other records.