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#19
May212, 03:13 PM

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.
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 915) larger than 5, which are as many as the sums before the middle (rows 17) that are larger than 54 = 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 17) which are larger than 5. So we are referring now *only* to the sums in the first rows (17). These sums before the middle can be figured out recursively, as they follow the same pattern:
The final line in the 'compare' function



#20
May212, 03:17 PM

P: 250

However those results aren't quite right... should be 2, 7, 19, 47, 108, 236, 501, onward



#21
May212, 03:20 PM

P: 688




#22
May212, 03:26 PM

P: 250

some raw output: http://justpaste.it/yc5



#23
May212, 03:34 PM

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. 


#24
May212, 05:01 PM

P: 250

right, the subset,sorry for the confusion



#25
May212, 05:20 PM

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:



#26
May212, 05:39 PM

P: 250

sumk's fall into an arithmetic progression across N's... but it's not clear how to derive them otherwise



#27
May212, 05:47 PM

P: 688




#28
May212, 05:53 PM

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 


#29
May412, 11:01 AM

P: 250

lots of patterns but they all seem random



#30
May712, 11:37 AM

P: 250

Does anyone have any further thoughts on this problem?



#31
May1812, 02:34 AM

Homework
Sci Advisor
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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 + (N1).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. 


#32
May1812, 09:36 AM

P: 36

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. 


#33
May1812, 10:51 AM

P: 250

Can you provide an example ? Not sure I understand



#34
May1812, 03:38 PM

P: 36

Let's take the general case of count(n) so that S={a(1),a(2),a(3),...a(n3),a(n2),a(n1),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(n1)+a(n3). But they are increasing so a(n2)>a(n3). So it follows that after we add a(n1) to both we get a(n1)+a(n2)>a(n1)+a(n3)=a(n). In other words any subset of S that contains both a(n1) and a(n2) is one we should count. The other smaller elements can either be present or not present (two possibilities) and there are exactly n3 of these other elements so there are 2^(n3) different subsets that go into our count so far (where 2^(n3) should be read as 2 to the power of (n3)). Of course that isn't all by a long shot. So far we just have the subsets counted that contain both a(n1) and a(n2). Let's stick with a(n1) and drop a(n2). Now as we started out noticing a(n)=a(n1)+a(n3). So if we take any subset that contains a(n1) and a(n3) and at least one more element then we have a subset of S we should count. If the subset contained a(n2) we would have already counted it in our previous steps, so there are only n4 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(n3),a(n1)} which isn't quite enough and so we have (2^(n4))1 additional subsets to count. So far we have count(n)=(2^(n3))+(2^(n4))1+??? Have we counted everything that contains a(n1)? We've counted everything that contains a(n1) and either a(n2) or a(n3) or both. Are there other combinations that add up greater than a(n3) (so that when combined with a(n1) will be greater than a(n))? Why yes there are. You should be able to see there are count(n3) of them because of the way count is defined. So we now have count(n)=(2^(n3))+(2^(n4))1+count(n3)+??? with just having counted the Subsets containing a(n1). Can you take it from there??? 


#35
May1912, 12:12 AM

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



#36
May1912, 12:37 AM

Homework
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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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