X-Y==Min

[SarK0Y]

Good Time to All, Amici.

i got following math problem: to split up number array into pairs of arrays where X[t] & Y[t] are multiplications of numbers of gotten arrays (in t'th pair) with condition: |X(0)-Y(0)|==MIN(0), |MIN(0)-(X(1)-Y(1))|==MIN(1), ..., |MIN(n-1)-(X(n)-Y(n))|==MIN(n). for example, let's take an array: 2, 5, 7 then: first pair is {2, 5}, {7} ==> X(0)==2*5==10, Y(0)==7, MIN(0)==3; second is {2, 7}, {5} ==> X(0)==14, Y(0)==5, MIN(1)==6; third is ..
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Thanks a lot in Advance.

[lfm]

Originally Posted by SarK0Y

Good Time to All, Amici.

i got following math problem: to split up number array into pairs of arrays where X[t] & Y[t] are multiplications of numbers of gotten arrays (in t'th pair) with condition: |X(0)-Y(0)|==MIN(0), |MIN(0)-(X(1)-Y(1))|==MIN(1), ..., |MIN(n-1)-(X(n)-Y(n))|==MIN(n). for example, let's take an array: 2, 5, 7 then: first pair is {2, 5}, {7} ==> X(0)==2*5==10, Y(0)==7, MIN(0)==3; second is {2, 7}, {5} ==> X(0)==14, Y(0)==5, MIN(1)==6; third is ..
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Thanks a lot in Advance.

Sorry, you haven't described what you mean by "t'th pair" yet. Without a more complete description of the problem there's not much we can do to help.
Your example doesn't match the description either, you seem to say X[t] and Y[t] are multiplications of pairs but your example the Y is not any multiplication of any pair.

You should call the MIN function something else. MIN is conventionally the MINIMUM function and will cause confusion.

Is this a school math problem? It should be in the homework help section. Not the main math section.

[SarK0Y]

lfm

you seem to say X[t] and Y[t] are multiplications of pairs but your example the Y is not any multiplication of any pair.

thanks for reply. "t" is index of array's pair, not number's pair, array with one item gives number itself. for example, another original array & possible array's pairs: {3, 11, 23, 31, 37} then possible array's pairs: {3, 11, 31}, {23, 37};... ; {3}, {11, 23, 31, 37}.

Is this a school math problem?

no

[cheesehead]

Originally Posted by SarK0Y

for example, another original array & possible array's pairs: {3, 11, 23, 31, 37} then possible array's pairs: {3, 11, 31}, {23, 37};... ; {3}, {11, 23, 31, 37}.

From this example, it seems to me that you may be talking about set partitions ( http://mathworld.wolfram.com/SetPartition.html ).

I'm not sure how to translate all your other notations into terms I'm familiar with, but let me ask about some possibilities:

Start with same "original array" as in your example: {3, 11, 23, 31, 37}

Is that intended to be a well ordered set ( as in http://mathworld.wolfram.com/WellOrderedSet.html ) ?

One partition of that set is: {3, 11, 31}, {23, 37}.

Another partition of that set is: {3}, {11, 23, 31, 37}.

From the Mathworld entry for Bell number ( http://mathworld.wolfram.com/BellNumber.html ): "The number of ways a set of n elements can be partitioned into nonempty subsets is called a Bell number and is denoted B_n (not to be confused with the Bernoulli number, which is also commonly denoted B_n).

The number of partitions of a 5-element set such as {3, 11, 23, 31, 37} is 52, so there are 50 partitions besides the two you give as examples, {3, 11, 31},{23, 37} and {3},{11, 23, 31, 37}. (... except that your term "pair" may mean you're considering only the partitions that have exactly two subsets of the original set. See below.)

Does all this make sense? Am I correctly interpreting what you mean, SarK0Y?

|X(0)-Y(0)|==MIN(0), |MIN(0)-(X(1)-Y(1))|==MIN(1), ..., |MIN(n-1)-(X(n)-Y(n))|==MIN(n). for example, let's take an array: 2, 5, 7 then: first pair is {2, 5}, {7} ==> X(0)==2*5==10, Y(0)==7, MIN(0)==3; second is {2, 7}, {5} ==> X(0)==14, Y(0)==5, MIN(1)==6; third is ..

_If_ my interpretation so far is correct, then for your 5-element example {3, 11, 23, 31, 37}, we have:

Suppose partition 0 is {{3, 11, 31}, {23, 37}}. Then

X(0) = 3*11*31 = 1023 and Y(0) = 23*37 = 851

and, using "M(n)" instead of "MIN(n)" to avoid confusion with the minimum function,

|X(0)-Y(0)|==M(0), so M(0) = |1023-851| = 172.

Okay so far?

In your 3-element example {2, 5, 7}, partition 0 ("first pair") is {{2, 5}, {7}} and partition 1 ("second pair") is {{2, 7}, {5}}.

Is the "third pair" (partition 2) = ({5, 7}, {2}} ?

Do you have any use for the partition {{2}, {7}, {5}}, or does your term "pair" mean you're considering only the partitions that have exactly two subsets of the original set?

How are you ordering the partitions? What determines which partition is "first pair" = partition 0?

[axn]

N=2*5*7;
s=0;fordiv(N,x,if(x*x>=N,d=x-N/x;print(N/x, ",", x, " -> ", d, ":", d-s);s=d))

result:
7,10 -> 3:3
5,14 -> 9:6
2,35 -> 33:24
1,70 -> 69:36

[SarK0Y]

Thanks for reply, Amici.
cheesehead
Bell numbers is more common task.

|X(0)-Y(0)|==M(0), so M(0) = |1023-851| = 172... Is the "third pair" (partition 2) = ({5, 7}, {2}}

perfectly correct.

Do you have any use for the partition {{2}, {7}, {5}}, or does your term "pair" mean you're considering only the partitions that have exactly two subsets of the original set?

you're right, original Set(array) must be splited up into two subsets on the each iteration.

How are you ordering the partitions? What determines which partition is "first pair" = partition 0?

|X(0)-Y(0)|==M(0), |M(0)-(X(1)-Y(1))|==M(1),... , |M(n-1)-(X(n)-Y(n))|==M(n). M(t) is min value for current iteration.

[SarK0Y]

i got M(0) with following algo:
//(given: set I with n integers)
sort(I); //I(0)<I(1)..<I(n-1)
int x=n-1, y=n-2;
for(int i=n-3; i>-1; i--)
{
if(x>y)y*=I(i);
else x*=I(i);
}
-------
suggestions about other iterations shall be very appreciated:-)

[SarK0Y]

small correction: int x=M(n-1), y=M(n-2);