How to generate sequences with distinct subsums? - sum

I'm looking for a way to generate some (6 for default) equations where all subsums are unique.
For example,
a+b+c=50
d+e+f=50
g+h+i=50
a, d and g have to be distinct.
a+b and d+e have to be distinct.
e+f and h+i have to be distinct.
a+c and d+f have to be distinct.
But, a+b and e+f can be the same. So I only care about the subsums of aligned parameters..
I could only found ways to check whether some sequence is subsum-distinct, but I found nothing on how to generate such a sequence..

You didn't state whether you need it to be a random sequence, so suppose that this is not required.
One simple approach is this:
1 + 2 + 47 = 50
3 + 4 + 43 = 50
5 + 6 + 39 = 50
7 + 8 + 35 = 50
9 + 10 + 31 = 50
11 + 12 + 27 = 50
First two numbers are 2 smallest available numbers, the third number is final sum - those numbers.
a and b are always increasing, c is always decreasing
a + b is always increasing, b + c and a + c are always decreasing
You can generate it this way in a loop.
EDIT after comment that it has to be a random sequence:
Possibly you could create several sets (some sort of hashset/hashmap would be the most appropriate)
set of first summands
set of sums of first and second summands
set of sums of second and third summands
set of sums of first and third summands
set of previously generated triples
You would generate random triples this way:
If total number of demanded triples was not achieved generate a random triple, otherwise finish.
Check if the triple was not previously generated, if not proceed with step 3.
Conduct checks for first four sets. If no sums are contained within those sets, add triple and proceed with step 1.
However, I am not sure if this approach guarantees that you will get results (especially in small final sums).
So, I would add an counter, if too many consecutive attempts are not successful, then I would switch to brute force approach (which should not be problem if final sums are small and on other hand is very unlikely to happen if a final sum is large).
Overall, performance should be good.

Related

Sorting/Optimization problem with rows in a pandas dataframe [duplicate]

So if I was given a sorted list/array i.e. [1,6,8,15,40], the size of the array, and the requested number..
How would you find the minimum number of values required from that list to sum to the requested number?
For example given the array [1,6,8,15,40], I requested the number 23, it would take 2 values from the list (8 and 15) to equal 23. The function would then return 2 (# of values). Furthermore, there are an unlimited number of 1s in the array (so you the function will always return a value)
Any help is appreciated
The NP-complete subset-sum problem trivially reduces to your problem: given a set S of integers and a target value s, we construct set S' having values (n+1) xk for each xk in S and set the target equal to (n+1) s. If there's a subset of the original set S summing to s, then there will be a subset of size at most n in the new set summing to (n+1) s, and such a set cannot involve extra 1s. If there is no such subset, then the subset produced as an answer must contain at least n+1 elements since it needs enough 1s to get to a multiple of n+1.
So, the problem will not admit any polynomial-time solution without a revolution in computing. With that disclaimer out of the way, you can consider some pseudopolynomial-time solutions to the problem which work well in practice if the maximum size of the set is small.
Here's a Python algorithm that will do this:
import functools
S = [1, 6, 8, 15, 40] # must contain only positive integers
#functools.lru_cache(maxsize=None) # memoizing decorator
def min_subset(k, s):
# returns the minimum size of a subset of S[:k] summing to s, including any extra 1s needed to get there
best = s # use all ones
for i, j in enumerate(S[:k]):
if j <= s:
sz = min_subset(i, s-j)+1
if sz < best: best = sz
return best
print min_subset(len(S), 23) # prints 2
This is tractable even for fairly large lists (I tested a random list of n=50 elements), provided their values are bounded. With S = [random.randint(1, 500) for _ in xrange(50)], min_subset(len(S), 8489) takes less than 10 seconds to run.
There may be a simpler solution, but if your lists are sufficiently short, you can just try every set of values, i.e.:
1 --> Not 23
6 --> Not 23
...
1 + 6 = 7 --> Not 23
1 + 8 = 9 --> Not 23
...
1 + 40 = 41 --> Not 23
6 + 8 = 14 --> Not 23
...
8 + 15 = 23 --> Oh look, it's 23, and we added 2 values
If you know your list is sorted, you can skip some tests, since if 6 + 20 > 23, then there's no need to test 6 + 40.

SQL Max Consecutive Values in a number set using recursion

The following SQL query is supposed to return the max consecutive numbers in a set.
WITH RECURSIVE Mystery(X,Y) AS (SELECT A AS X, A AS Y FROM R)
UNION (SELECT m1.X, m2.Y
FROM Mystery m1, Mystery m2
WHERE m2.X = m1.Y + 1)
SELECT MAX(Y-X) + 1 FROM Mystery;
This query on the set {7, 9, 10, 14, 15, 16, 18} returns 3, because {14 15 16} is the longest chain of consecutive numbers and there are three numbers in that chain. But when I try to work through this manually I don't see how it arrives at that result.
For example, given the number set above I could create two columns:
m1.x
m2.y
7
7
9
9
10
10
14
14
15
15
16
16
18
18
If we are working on rows and columns, not the actual data, as I understand it WHERE m2.X = m1.Y + 1 takes the value from the next row in Y and puts it in the current row of X, like so
m1.X
m2.Y
9
7
10
9
14
10
15
14
16
15
18
16
18
Null?
The main part on which I am uncertain is where in the SQL recursion actually happens. According to Denis Lukichev recursion is the R part - or in this case the RECURSIVE Mystery(X,Y) - and stops when the table is empty. But if the above is true, how would the table ever empty?
Since I don't know how to proceed with the above, let me try a different direction. If WHERE m2.X = m1.Y + 1 is actually a comparison, the result should be:
m1.X
m2.Y
14
14
15
15
16
16
But at this point, it seems that it should continue recursively on this until only two rows are left (nothing else to compare). If it stops here to get the correct count of 3 rows (2 + 1), what is actually stopping the recursion?
I understand that for the above example the MAX(Y-X) + 1 effectively returns the actual number of recursion steps and adds 1.
But if I have 7 consecutive numbers and the recursion flows down to 2 rows, should this not end up with an incorrect 3 as the result? I understand recursion in C++ and other languages, but this is confusing to me.
Full disclosure, yes it appears this is a common university question, but I am retired, discovered this while researching recursion for my use, and need to understand how it works to use similar recursion in my projects.
Based on this db<>fiddle shared previously, you may find it instructive to alter the CTE to include an iteration number as follows, and then to show the content of the CTE rather than the output of final SELECT. Here's an amended CTE and its content after the recursion is complete:
Amended CTE
WITH RECURSIVE Mystery(X,Y) AS ((SELECT A AS X, A AS Y, 1 as Z FROM R)
UNION (SELECT m1.X, m2.A, Z+1
FROM Mystery m1
JOIN R m2 ON m2.A = m1.Y + 1))
CTE Content
x
y
z
7
7
1
9
9
1
10
10
1
14
14
1
15
15
1
16
16
1
18
18
1
9
10
2
14
15
2
15
16
2
14
16
3
The Z field holds the iteration count. Where Z = 1 we've simply got the rows from the table R. The, values X and Y are both from the field A. In terms of what we are attempting to achieve these represent sequences consecutive numbers, which start at X and continue to (at least) Y.
Where Z = 2, the second iteration, we find all the rows first iteration where there is a value in R which is one higher than our Y value, or one higher than the last member of our sequence of consecutive numbers. That becomes the new highest number, and we add one to the number of iterations. As only three numbers in our original data set have successors within the set, there are only three rows output in the second iteration.
Where Z = 3, the third iteration, we find all the rows of the second iteration (note we are not considering all the rows of the first iteration again), where there is, again, a value in R which is one higher than our Y value, or one higher than the last member of our sequence of consecutive numbers. That, again, becomes the new highest number, and we add one to the number of iterations.
The process will attempt a fourth iteration, but as there are no rows in R where the value is one more than the Y values from our third iteration, no extra data gets added to the CTE and recursion ends.
Going back to the original db<>fiddle, the process then searches our CTE content to output MAX(Y-X) + 1, which is the maximum difference between the first and last values in any consecutive sequence, plus one. This finds it's value from the record produced in the third iteration, using ((16-14) + 1) which has a value of 3.
For this specific piece of code, the output is always equivalent to the value in the Z field as every addition of a row through the recursion adds one to Z and adds one to Y.

repeated sum of digits big o complexity

Lets say for example we have the number 12345.
This sums to 15 when you add 1 + 2 + 3 + 4 + 5, which sums to 6 when you add 1 + 5.
My question is, what would the time complexity be for a repetitive adding algorithm like this be? This process is happens until there is only a single digit left.
I know that for any given number, the # of digits is approximately ln(n). Im thinking that this means that the big o would look something like (ln(n))^k, for some k. However, I am not confident because each time you sum, the number of digits gets smaller (first summed 5 digits, then only 2).
How would I go about figuring this out?

Distribute numbers as close to possible

This seems to be a 2 step problem I'm trying to solve.
Let's say we have N records, and we are trying to distribute as evenly as possible into K groups.
The second problem - each group in K can only accept an M amount of records.
For example, if we have 5 records, and 3 groups, then we would distribute 2 into Group K1, 2 into Group K2 and 1 record into Group K3. However, if say in group 1, it only accepts at most 1 record. Then the arrangement would need to be 1 into Group K1, 2 into Group K2, and 2 into Group K3.
I'm not necessary after the solution but what algorithm I might need to use to solve this? Apparently for the distribution, I need to use the Greedy algorithm? But for the second step, this seems to be a bit more complicated
Edit:
The example I'm looking at is:
Number of records: 23
Groups: 10
Max records for each group
G1 = 4
G2 = 1
G3 = 0
G4 = 5
G5 = 0
G6 = 0
G7 = 2
G8 = 4
G9 = 2
G10 = 2
if N=12 and K=3 then in normal situation,you just split it V=12/3=4 for each group. but since you have M limitation, and for example K3 can only accept 1 then the distribution can be 6-5-1 which is not evenly distributed.
So i guess you need to sort K based on the M limitation, so for the example above the groups order become K3-K1-K2.
then if the distributed value V is bigger than the accepted amount M for that group, you need to take the remainder and distribute it again to the remaining group (K3=1, then 4-1=3 must be distributed to K1 and K2).
the implementation might be complicated, i hope you can find more simple solution for this
From what I understood, you need to separate all groups which allows a fixed number of values first and then equally distribute records among remaining groups. Let's take an example, let's say we have 15 records which needs to be distributed among 5 groups (G1, G2, G3, G4 and G5). Also let's assume that G2 and G4 allows max records of 2 and 4 respectively. Now algorithm should go like this:
Get average(ceiling integer) of records based on number of groups (In this example we'll get 3).
Add all max allowed records which are smaller than our average (In this example it's G2 only who's max limit(i.e. 2) is less than our average hence the number comes as 2).
Now subtract our number from step 2 from total records and also subtract the number of groups involved in step 2 from total groups. (remaining total records: 13, remaining total groups 4).
Get the new average(ceiling integer) using remaining records and groups. (New average 4).
Get average (Integer) (i.e. 3) and allot equal number of records to remaining groups - 1.
Get Mod (i.e. 1) and allot that number to the last group.
Now what we finally will have here:
G1(No limit): 4
G2(Limit 2): 2
G3(No limit): 4
G4(Limit 4): 4
G5(No limit): 1
Let me know if you think that this algo might fail for some scenarios.
Formula to get ceiling integer average
floor((#total_records + #total_groups-1) / #total_groups)

SPSS Compute Variable

Below is some data:
Test Day1 Day2 Score
A 1 2 100
B 1 3 62
C 3 4 90
D 2 4 20
E 4 5 80
I am trying to take the values from column 'day' and 'day2' and use them to select the row number for the column score. For example for Test A I would like to find the sum of 100 and 62 because that is the values of the first and second rows of score. Test B I would like to find the sum of 100, 62 and 90.
Is their anyway to do this in the Compute Variable window? Found in the menu Transform-Compute Variable?
I tried the following:
Score(MEAN(VALUE(Day1), VALUE(DAY2)))
This is not the proper way to call the cell location of Score and I received an error.
Can anyone help?
Thank you!
You really have two different datasets here. One is a dataset of scores numbered 1 through 5.
The other is a dataset that includes indexes into the score dataset. So the steps would be something like this.
First take the scores dataset and transpose it so that it has one row and 5 columns (Data>Transpose)
Then match that dataset to each case in the main dataset (Data>Merge Files>Add Variables).
Next you have to resort to using syntax directly.
You would declare a vector for the scores (VECTOR)
Finally, you use COMPUTE to index into the scores.
For your real problem, I suppose that you might have batches of scores and maybe there are some gaps. The Restructure Data Wizard can help you generalize this - convert cases into variables, but let's not go there yet.
HTH,
Jon Peck