Related
Good evening, I want to do the following modeling in GAMS language:
PART 1 : Let N={n1,n2,n3,n4} be a set. I'm going to partitioning the set N in two subsets: B1={n1} and B2={n2,n3,n4}.
PART 2 : Then I want to form a set B whose elements are the sets B1 and B2, i.e, B={B1,B2}. That is, B will have 2 elements, which is the number of subsets that were created in part 1.
PART 3 : I want to know if an element of N is in any element of B, for example, i want to know if n1 is in B1 or not.
All these three steps is because I want to obtain equations of the form: Ec(n1,B1), Ec(n2,B2),Ec(n3,B2), Ec(n4,B2) using only one equation something like that : Ec(N,B)$[CONDITION]. The condition is precisely the PART 3.
My attempt is the following:
PART 1
SETS
N /n1*n4/
B1(N) /n1/
B2(N) /n2*n4/
;
PART 2 , I don't know if this is correct, that is, if a set whose elements are previously defined sets is written like this
SETS
B /B1,B2/
PART 3
PARAMETER test(N);
test(B1)=1;
test(B2)=2;
But this doesn't help me at all because if my partition had 1000 subsets, i.e. B1 ... B1000 then I would need to write it as some kind of recursion using PART 2.
Please I need your help. Thank you very much.
This is an interview question, and the problem description is as follows:
There are n couples sitting in a row with 2n seats. Find the minimum number of swaps to make everyone sit next on his/her partner. For example, 0 and 1 are couple, and 2 and 3 are couple. Originally they are sitting in a row in this order: [2, 0, 1, 3]. The minimum number of swaps is 1, for example swapping 2 with 1.
I know there is a greedy solution for this problem. You just need to scan the array from left to right. Every time you see an unmatched pair, you swap the first person of the pair to his/her correct position. For example, in the above example for pair [2, 0], you will directly swap 2 with 1. There is no need to try swapping 0 with 3.
But I don't really understand why this works. One of the proofs I saw was something like this:
Consider a simple example: 7 1 4 6 2 3 0 5. At first step we have two choices to match the first couple: swap 7 with 0, or swap 1 with 6. Then we get 0 1 4 6 2 3 7 5 or 7 6 4 1 2 3 0 5. Pay attention that the first couple doesn't count any more. For the later part it is composed of 4 X 2 3 Y 5 (X=6 Y=7 or X=1 Y=0). Since different couples are unrelated, we don't care X Y is 6 7 pair or 0 1 pair. They are equivalent! Thus it means our choice doesn't count.
I feel that this is very reasonable but not compelling enough. In my opinion we have to prove that X and Y are couple in all possible cases and don't know how. Can anyone give a hint? Thanks!
I've split the problem into 3 examples. A's are a pair and so are B's in all examples. Note that throughout the examples a match requires that elements are adjacent and the first element occupy an index that satisfies index%2 = 0. An array looking like this [X A1 A2 ...] does not satisfy this condition, however this does [X Y A1 A2 ...]. The examples also do not look to the left at all, because looking to the left of A2 below is the same as looking to the right of A1.
First example
There's an even number of elements between two unmatched pairs:
A1 B1 ..2k.. A2 B2 .. for any number k in {0, 1, 2, ..} meaning A1 B1 A2 B2 .. is just a another case.
Both can be matched in one swap:
A1 A2 ..2k.. B1 B2 .. or B2 B1 ..2k.. A2 A1 ..
Order is not important, so it doesn't matter which pair is first. Once the pairs are matched, there will be no more swapping involving either pair. Finding A2 based on A1 will result in the same amount of swaps as finding B2 based on B1.
Second example
There's an odd number of elements between two pairs (2k + the element C):
A1 B1 ..2k.. C A2 B2 D .. (A1 B1 ..2k.. C B2 A2 D .. is identical)
Both cannot be matched in one swap, but like before it doesn't matter which pair is first nor if the matched pair is in the beginning or in the middle part of the array, so all these possible swaps are equally valid, and none of them creates more swaps later on:
A1 A2 ..2k .. C B1 B2 D .. or B2 B1 ..2k.. C A2 A1 D .. Note that the last pair is not matched
C B1 ..2k.. A1 A2 B2 D .. or A1 D ..2k.. C A2 B2 B1 .. Here we're not matching the first pair.
The important thing about this is that in each case, only one pair is matched and none of the elements of that pair will need to be swapped again. The result of the remaining non-matched pair are either one of:
..2k.. C B1 B2 D ..
..2k.. C A2 A1 D ..
C B1 ..2k.. B2 D ..
A1 D ..2k.. C A2 ..
They are clearly equivalent in terms of swaps needed to match the remaining A's or B's.
Third example
This is logically identical to the second. Both B1/A2 and A2/B2 can have any number of elements between them. No matter how elements are swapped, only one pair can be matched. m1 and m2 are arbitrary number of elements. Note that elements X and Y are just the elements surrounding B2, and they're only used to illustrate the example:
A1 B1 ..m1.. A2 ..m2.. X B2 Y .. (A1 B1 ..m1.. B2 ..m2.. X A2 Y .. is identical)
Again both pairs cannot be matched in one swap, but it's not important which pair is matched, or where the matched pair position is:
A1 A2 ..m1.. B1 ..m2.. X B2 Y .. or B2 B1 ..m1.. A2 ..m2.. X A1 Y .. Note that the last pair is not matched
A1 X ..m1.. A2 ..m2-1.. B1 B2 Y .. or A1 Y ..m1.. A2 ..m2.. X B2 B1.. depending on position of B2. Here we're not matching the first pair.
Matching the pair around A2 is equivalent, but omitted.
As in the second example, one swap can also be matching a pair in the beginning or in the middle of the array, but either choice doesn't change that only one pair is matched. Nor does it change the remaining amount of unmatched pairs.
A little analysis
Keeping in mind that matched pairs drop out of the list of unmatched/problem pairs, the list of unmatched pairs are either one fewer or two fewer pairs for each swap. Since it's not important which pair drops out of the problem, it might as well be the first. In that case we can assume that pairs to the left of the cursor/current index are all matched. And that we only need to match the first pair, unless it's already matched by coincidence and the cursor is then rightfully moved.
It becomes even more clear if the above examples are looked at with the cursor being at the second unmatched pair, instead of the first. It still doesn't matter which pairs are swapped for the amount of total swaps needed. So there's no need to try to match pairs in the middle. The resulting amount of swaps are the same.
The only time two pairs can be matched with only one swap are those in the first example. There is no way to match two pairs in one swap in any other setup. Looking at the result of the swap in the second and third examples, it also becomes clear that none of the results have any advantage to any of the others and that each result becomes a new problem that can be described as one of the three cases (two cases really, because second and third are equivalent in terms of match-able pairs).
Optimal swapping
There is no way to modify the array to prepare it for more optimal swapping later on. Either a swap will match one or two pairs, or it will count as a swap with no matches:
Looking at this: A1 B1 ..2k.. C B2 ... A2 ...
Swap to prepare for optimal swap:
A1 B1 ..2k.. A2 B2 ... C ... no matches
A1 A2 ..2k.. B1 B2 ... C ... two in one
Greedy swap:
B2 B1 ..2k.. C A1 ... A2 ... one
B2 B1 ..2k.. A2 A1 ... C ... one
Un-matching pairs
Pairs already matched will not become unmatched because that would require that:
For A1 B1 ..2k.. C A2 B2 D ..
C is identical to A1 or
D is identical to B1
either of which is impossible.
Likewise with A1 B1 ..m1.. (Z) A2 (V) ..m2.. X B2 Y ..
Or it would require that matched pairs are shifted one (or any odd number of) index inside the array. That's also not possible, because we always ever swap, so the array elements aren't being shifted at all.
[Edited for clarity 4-Mar-2020.]
There is no point doing a swap which does not put (at least) one couple together. To do so would add 1 to the swap count and leave us with the same number of unpaired couples.
So, each time we do a swap, we put one couple together leaving at most n-1 couples. Repeating the process we end up with 1 pair, who must by then be a couple. So, the worst case must be n-1 swaps.
Clearly, we can ignore couples who are already together.
Clearly, where we have two pairs a:B b:A, one swap will create the two couples a:A b:B.
And if we have m pairs a:Q b:A c:B ... q:P -- where the m pairs are a "disjoint subset" (or cycle) of couples, m-1 swaps will put them into couples.
So: the minimum number of swaps is going to be n - s where s is the number of "disjoint subsets" (and s >= 1). [A subset may, of course, contain just one couple.]
Interestingly, there is nothing clever you can do to reduce the number of swaps. Provided every swap creates a couple you will do the minimum number.
If you wanted to arrange each couple in height order as well, things may or may not be more interesting.
FWIW: having shown that you cannot do better than n-1 swaps for each disjoint set of n couples, the trick then is to avoid the O(n^2) search for each swap. That can be done relatively straightforwardly by keeping a vector with one entry per person, giving where they are currently sat. Then in one scan you pick up each person and if you know where their partner is sat, swap down to make a pair, and update the location of the person swapped up.
I will swap every even positioned member,
if he/she doesn't sit besides his/her partner.
Even positioned means array indexed 1, 3, 5 and so on.
The couples are [even, odd] pair. For example [0, 1], [2, 3], [4, 5] and so on.
The loop will be like that:
for(i=1; i<n*2; i+=2) // when n = # of couples.
Now, we will check i-th and (i-1)-th index member. If they are not couple, then we will look for the partner of (i-1)-th member and once we have it, we should swap it with i-th member.
For an example, say at i=1, we got 6, now if (i-1)-th element is 7 then they form a couple (if (i-1)-th element is 5 then [5, 6] is not a couple.) and we don't need any swap, otherwise we should look for the partner of (i-1)-th element and will swap with i-th element. So, (i-1)-th and i-th will form a couple.
It ensure that we need to check only half of the total members, that means, n.
And, for any non-matched couple, we need a linear search from i-th position to the rest of the array. Which is O(2n), eventually O(n).
So, the overall technique complexity will be O(n^2).
In worst case, minimum swap will be n-1. (this is maximum as well).
Very straightforward. If you need help to code, let us know.
I am counting average of numbers using Hadoop/Mapreduce
with structure
guid banid countview
g1 b1 1
g1 b2 1
g1 b1 2
g1 b1 1
g2 b1 1
g2 b2 1
g2 b1 1
g2 b3 1
g3 b1 1
I want count average countview of each banid of guid?
(my mind is average=5/2 with guid g1 (2 is total numbers another banid: b1,b2))
So if i understand what you're asking, the answer you're looking for might look like:
g1 b1 1
g1 b2 1
g1 b1 2
g1 b1 1
Average for "g1" = 5/2 (total count / unique banid count)
First you need to break the problem down into your Map and Reduce stages. The objective is to group all the counts and banids for each "guid" in the reducer.
Mapper:
Output Key/Value types: Text / Text
The output key is probably going to be a Text Writable which will contain the guid. The Value will contain the banid and the count (ie b1:1). This will group all the banids and counts for each guid.
Reducer:
Output Key/Value types: Text / FloatWritable
You will now get a list of Text objects for each guid in the Key. Iterate through each Value object, spliting up the banid and the count. Create a set of the banids and sum the counts as you iterate. Once you've done this you should be able to calculate the average. Write out the average as a FloatWritable (or Text.. its up to you). The Key will be the same as the input key to the reduce.
This is a simple approach to dealing with a value that needs to contain multiple pieces of information. A more advanced approach would be to create your own Writable object that wraps a Text and VIntWritable object.
Have these two tables:
TableA
ID Opt1 Opt2 Type
1 A Z 10
2 B Y 20
3 C Z 30
4 C K 40
and
TableB
ID Opt1 Type
1 Z 57
2 Z 99
3 X 3000
4 Z 3000
What would be a good algorithm to find arbitrary relations between these two tables? In this example, I'd like it to find the apparent relation between records containing Op1 = C in TableA and Type = 3000 in TableB.
I could think of apriori in some way, but doesn't seems too practical. what you guys say?
thanks.
It sounds like a relational data mining problem. I would suggest trying Ross Quinlan's FOIL: http://www.rulequest.com/Personal/
In pseudocode, a naive implementation might look like:
1. for each column c1 in table1
2. for each column c2 in table2
3. if approximately_isomorphic(c1, c2) then
4. emit (c1, c2)
approximately_isomorphic(c1, c2)
1. hmap = hash()
2. for i = 1 to min(|c1|, |c2|) do
3. hmap[c1[i]] = c2[i]
4. if |hmap| - unique_count(c1) < error_margin then return true
5. else then return false
The idea is this: do a pairwise comparison of the elements of each column with each other column. For each pair of columns, construct a hash map linking corresponding elements of the two columns. If the hash map contains the same number of linkings as unique elements of the first column, then you have a perfect isomorphism; if you have a few more, you have a near isomorphism; if you have many more, up to the number of elements in the first column, you have what probably doesn't represent any correlation.
Example on your input:
ID & anything : perfect isomorphism since all of ID are unique
Opt1 & ID : 4 mappings and 3 unique values; not a perfect
isomorphism, but not too far away.
Opt1 & Opt1 : ditto above
Opt1 & Type : 3 mappings & 3 unique values, perfect isomorphism
Opt2 & ID : 4 mappings & 3 unique values, not a perfect
isomorphism, but not too far away
Opt2 & Opt2 : ditto above
Opt2 & Type : ditto above
Type & anything: perfect isomorphism since all of ID are unique
For best results, you might do this procedure both ways - that is, comparing table1 to table2 and then comparing table2 to table1 - to look for bijective mappings. Otherwise, you can be thrown off by trivial cases... all values in the first are different (perfect isomorphism) or all values in the second are the same (perfect isomorphism). Note also that this technique provides a way of ranking, or measuring, how similar or dissimilar columns are.
Is this going in the right direction? By the way, this is O(ijk) where table1 has i columns, table 2 has j columns and each column has k elements. In theory, the best you could do for a method would be O(ik + jk), if you can find correlations without doing pairwise comparisons.
Let's say I have the three following lists
A1
A2
A3
B1
B2
C1
C2
C3
C4
C5
I'd like to combine them into a single list, with the items from each list as evenly distributed as possible sorta like this:
C1
A1
C2
B1
C3
A2
C4
B2
A3
C5
I'm using .NET 3.5/C# but I'm looking more for how to approach it then specific code.
EDIT: I need to keep the order of elements from the original lists.
Take a copy of the list with the most members. This will be the destination list.
Then take the list with the next largest number of members.
divide the destination list length by the smaller length to give a fractional value of greater than one.
For each item in the second list, maintain a float counter. Add the value calculated in the previous step, and mathematically round it to the nearest integer (keep the original float counter intact). Insert it at this position in the destination list and increment the counter by 1 to account for it. Repeat for all list members in the second list.
Repeat steps 2-5 for all lists.
EDIT: This has the advantage of being O(n) as well, which is always nice :)
Implementation of
Andrew Rollings' answer:
public List<String> equimix(List<List<String>> input) {
// sort biggest list to smallest list
Collections.sort(input, new Comparator<List<String>>() {
public int compare(List<String> a1, List<String> a2) {
return a2.size() - a1.size();
}
});
List<String> output = input.get(0);
for (int i = 1; i < input.size(); i++) {
output = equimix(output, input.get(i));
}
return output;
}
public List<String> equimix(List<String> listA, List<String> listB) {
if (listB.size() > listA.size()) {
List<String> temp;
temp = listB;
listB = listA;
listA = temp;
}
List<String> output = listA;
double shiftCoeff = (double) listA.size() / listB.size();
double floatCounter = shiftCoeff;
for (String item : listB) {
int insertionIndex = (int) Math.round(floatCounter);
output.add(insertionIndex, item);
floatCounter += (1+shiftCoeff);
}
return output;
}
First, this answer is more of a train of thought than a concete solution.
OK, so you have a list of 3 items (A1, A2, A3), where you want A1 to be somewhere in the first 1/3 of the target list, A2 in the second 1/3 of the target list, and A3 in the third 1/3. Likewise you want B1 to be in the first 1/2, etc...
So you allocate your list of 10 as an array, then start with the list with the most items, in this case C. Calculate the spot where C1 should fall (1.5) Drop C1 in the closest spot, (in this case, either 1 or 2), then calculate where C2 should fall (3.5) and continue the process until there are no more Cs.
Then go with the list with the second-to-most number of items. In this case, A. Calculate where A1 goes (1.66), so try 2 first. If you already put C1 there, try 1. Do the same for A2 (4.66) and A3 (7.66). Finally, we do list B. B1 should go at 2.5, so try 2 or 3. If both are taken, try 1 and 4 and keep moving radially out until you find an empty spot. Do the same for B2.
You'll end up with something like this if you pick the lower number:
C1 A1 C2 A2 C3 B1 C4 A3 C5 B2
or this if you pick the upper number:
A1 C1 B1 C2 A2 C3 A3 C4 B2 C5
This seems to work pretty well for your sample lists, but I don't know how well it will scale to many lists with many items. Try it and let me know how it goes.
Make a hash table of lists.
For each list, store the nth element in the list under the key (/ n (+ (length list) 1))
Optionally, shuffle the lists under each key in the hash table, or sort them in some way
Concatenate the lists in the hash by sorted key
I'm thinking of a divide and conquer approach. Each iteration of which you split all the lists with elements > 1 in half and recurse. When you get to a point where all the lists except one are of one element you can randomly combine them, pop up a level and randomly combine the lists removed from that frame where the length was one... et cetera.
Something like the following is what I'm thinking:
- filter lists into three categories
- lists of length 1
- first half of the elements of lists with > 1 elements
- second half of the elements of lists with > 1 elements
- recurse on the first and second half of the lists if they have > 1 element
- combine results of above computation in order
- randomly combine the list of singletons into returned list
You could simply combine the three lists into a single list and then UNSORT that list. An unsorted list should achieve your requirement of 'evenly-distributed' without too much effort.
Here's an implementation of unsort: http://www.vanheusden.com/unsort/.
A quick suggestion, in python-ish pseudocode:
merge = list()
lists = list(list_a, list_b, list_c)
lists.sort_by(length, descending)
while lists is not empty:
l = lists.remove_first()
merge.append(l.remove_first())
if l is not empty:
next = lists.remove_first()
lists.append(l)
lists.sort_by(length, descending)
lists.prepend(next)
This should distribute elements from shorter lists more evenly than the other suggestions here.