# Subset problem

by SeventhSigma
Tags: subset
 P: 688 Here is some rationale as to why the above program appears to work. Consider, for example, the case N=5, where you want all subsets of {1,2,3,4} whose sum is larger than 5. I have listed all subsets below.  Sum 1 { 1} 1 2 { 2 } 2 3 { 2 1} 3 4 { 3 } 3 5 { 3 1} 4 6 { 3 2 } 5 7 { 3 2 1} 6 -------------------- 8 {4 } 4 -------------------- 9 {4 1} 5 10 {4 2 } 6 11 {4 2 1} 7 12 {4 3 } 7 13 {4 3 1} 8 14 {4 3 2 } 9 15 (4 3 2 1} 10 The middle row (#8) contains a subset with only one element, N-1 = 4. And notice that, by construction, all sums *after* the middle row are equal to 4 plus the corresponding sums *before* the middle row. Therefore, the sums larger than 5 are: - the sums after the middle (rows 9-15) larger than 5, which are as many as the sums before the middle (rows 1-7) that are larger than 5-4 = 1; - plus 1 if the middle row were larger than 5, which it isn't; so nothing added here; - plus the sums before the middle (rows 1-7) which are larger than 5. So we are referring now *only* to the sums in the first rows (1-7). These sums before the middle can be figured out recursively, as they follow the same pattern:  Sum 1 { 1} 1 2 { 2 } 2 3 { 2 1} 3 ----------------- 4 {3 } 3 ------------------ 5 {3 1} 4 6 {3 2 } 5 7 {3 2 1} 6 where the three sums after this new middle (rows 5-7) add to 3 plus the corresponding sums in rows 1-3. The final line in the 'compare' function return compare(n - p, p - 1) + (p > n) + compare(n, p - 1) is the one responsible for subdividing the problem in the three parts I mentioned before (after the middle, middle, and before the middle). The variable 'p' is meant to represent a power of 2.
 P: 250 However those results aren't quite right... should be 2, 7, 19, 47, 108, 236, 501, onward
P: 688
 Quote by SeventhSigma However those results aren't quite right... should be 2, 7, 19, 47, 108, 236, 501, onward
Hmm, curious... I tried generating the subsets by brute force and I get the same numbers as I listed above, not yours. Let me re-check.
 P: 250 some raw output: http://justpaste.it/yc5
 P: 688 Ahhh... I was always using {1,2,3,...,N} (with all integers from 1 to N) as my primary set of length N. If you choose an arbitrary set of length N, then the count for N will obviously depend on the particular set, not just on N or on the largest integer in the set. Edit: Oh, wait, you said in post #6 that the list comes from a specific recurrence sequence. Also, I see now that I have the comparison criterion wrong: a subset is valid if the largest element in the subset (not N, nor the largest in the original set) is smaller than the sum of the others; this renders my subdivision strategy useless.
 P: 250 right, the subset,sorry for the confusion
 P: 688 Hmm, not all is lost. Here is a variation of the program; it produces the output below in under a second, but it will get terribly slower for larger N, it is not linear. Output: 4 2 5 7 6 19 7 47 8 108 9 236 10 501 11 1045 12 2149 13 4378 14 8869 15 17892 16 35988 17 72254 18 144886 19 290271 20 581207 Program: #!/usr/bin/python def main(): sequence = [1,2,3] for n in range(4, 21): addOneMore(sequence) print n, count(sequence, n) def addOneMore(sequence): sequence.append(sequence[-1] + sequence[-3]) def count(sequence, n): sum = 0 for largest in range(3, n): sum += compare(sequence, sequence[largest], largest) return sum def compare(sequence, top, p): if top <= 0: return 2**p - 1 if p <= 0: return 0 return compare(sequence, top - sequence[p - 1], p - 1) \ + (sequence[p - 1] > top) \ + compare(sequence, top, p - 1) main() Perhaps the only Python-specifics needed to understand this are that sequence indexes are 0-based (from 0 to length-1), and that negative indexes pick items from the end: sequence[-1] is the last, sequence[-2] the next to last, and so on. The iterator range(a,b) will iterate starting from 'a' and until less than 'b'; that is, up to b-1.
 P: 250 sum-k's fall into an arithmetic progression across N's... but it's not clear how to derive them otherwise
P: 688
 Quote by SeventhSigma sum-k's fall into an arithmetic progression across N's... but it's not clear how to derive them otherwise
Sorry, I don't know which sum do you mean.
 P: 250 say N=5 or [1, 2, 3, 4, 6] and we count invalid solutions instead: there are X ways to have an inner sum of k: there are 4 ways to have an inner sum of 1 (1, 2) (1, 3) (1, 4) (1, 6) there are 3 ways to have an inner sum of 2 (2, 3) (2, 4) (2, 6) there are 5 ways to have an inner sum of 3 (1, 2, 3) (1, 2, 4) (1, 2, 6) (3, 4) (3, 6) there are 3 ways to have an inner sum of 4 (1, 3, 4) (1, 3, 6) (4, 6) and so forth
 P: 250 lots of patterns but they all seem random
 P: 250 Does anyone have any further thoughts on this problem?
 Homework Sci Advisor HW Helper Thanks P: 9,928 This is connected with partition numbers, in particular, restricted partitions: http://en.wikipedia.org/wiki/Partiti...ted_partitions. Let PD(n) be the number of partitions of n into distinct positive integers. An 'invalid' set in the OP is one in which the largest in the set, Y, exceeds or equals the sum, X, of the rest. For a given X and any N >= Y >= X, there are PD(X) of these. (Except, when Y=X we have to exclude the partition of X as just X, otherwise the invalid set would consist of the same number twice.) So in total there are N.PD(1) - 1 + (N-1).PD(2) - 1 + ... + 1.PD(N) - 1 that are subsets of {1..N}. Check: N = 5: 5.PD(1) + 4.PD(2) + 3.PD(3) + 2.PD(4) + 1.PD(5) - 5 = 5.1 + 4.1 + 3.2 + 2.2 + 1.3 - 5 = 17 But these are only the invalid sets of 2 or more numbers. Any single member set is necessarily invalid, for a total of 22. The number of valid sets = 31 - 22 = 9, which I believe is correct.
P: 36
 Quote by SeventhSigma I've tried most of these techniques already; in all actuality my set S is more like {1, 2, 3, 4, 6, 9, 13, 19, 28, 41} where the first three terms are hardcoded and the rest: a(n) = a(n-1) + a(n-3) Problem: When I try to even list out all possible sets by hand, I don't see any patterns, nor do I know how to take advantage of combinatorics. Every time I think I've found a pattern, it diverges. I am aware of the points stated so far in this thread though. I think my issue is understanding the combinatorics behind the counting.
This is a very different and much more tractable problem. It is easier than finding the solutions for sets, S'={1,2,3,4,5,6,7,...,n} because S={a(1),a(2),a(3),...,a(n)} has much more structure.

For starts try to find a formula that will tell you exactly how many subsets of S have totals that are exactly equal to a(n). For any of these subsets, if you add an extra a(i) to it, it will form the basis for one of the subsets you're counting. So for each of these subsets check how many ways you can add to them without creating duplicates.

The case of S'=(1,2,3,...,n) is interesting though.
 P: 250 Can you provide an example ? Not sure I understand
 P: 36 Let's take the general case of count(n) so that S={a(1),a(2),a(3),...a(n-3),a(n-2),a(n-1),a(n)} Now we know the a(i) are positive and increasing, and you should be able to prove it if needed. We also know that a(n)=a(n-1)+a(n-3). But they are increasing so a(n-2)>a(n-3). So it follows that after we add a(n-1) to both we get a(n-1)+a(n-2)>a(n-1)+a(n-3)=a(n). In other words any subset of S that contains both a(n-1) and a(n-2) is one we should count. The other smaller elements can either be present or not present (two possibilities) and there are exactly n-3 of these other elements so there are 2^(n-3) different subsets that go into our count so far (where 2^(n-3) should be read as 2 to the power of (n-3)). Of course that isn't all by a long shot. So far we just have the subsets counted that contain both a(n-1) and a(n-2). Let's stick with a(n-1) and drop a(n-2). Now as we started out noticing a(n)=a(n-1)+a(n-3). So if we take any subset that contains a(n-1) and a(n-3) and at least one more element then we have a subset of S we should count. If the subset contained a(n-2) we would have already counted it in our previous steps, so there are only n-4 other elements that might be present or not. The only possibility we don't want to count is where they all aren't present and just leave us with {a(n-3),a(n-1)} which isn't quite enough and so we have (2^(n-4))-1 additional subsets to count. So far we have count(n)=(2^(n-3))+(2^(n-4))-1+??? Have we counted everything that contains a(n-1)? We've counted everything that contains a(n-1) and either a(n-2) or a(n-3) or both. Are there other combinations that add up greater than a(n-3) (so that when combined with a(n-1) will be greater than a(n))? Why yes there are. You should be able to see there are count(n-3) of them because of the way count is defined. So we now have count(n)=(2^(n-3))+(2^(n-4))-1+count(n-3)+??? with just having counted the Subsets containing a(n-1). Can you take it from there???
 P: 250 not so much. I am trying to use a smaller set of S to see if I can count them properly. I am not sure that approach works because it's easier to count the extremities than it is to count the nitty gritty stuff towards the center of S where stuff gets messier
 Homework Sci Advisor HW Helper Thanks P: 9,928 I'm puzzled there's been no response to my post (#31). Isn't it the answer to the originally posed question? Do I need to explain more?

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