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ramsey2879
guest
Sep-09-06, 08:14 AM (EST)

"Paired sums using two sets of consecutive integers"

 Let A(n) = {1,2,3,4, … 8} and B(n) = {1,2,3,4, …. 16} be two sets ofconsecutive integers with no repetition.Divide the set of elements B(n) into two subsets C(n) and D(n) eachcomprising 8 different integers and such that every element of B(n)is used with the condition that 8 equations of the form A(i) + C(i) =D(i) can be simultaneously formed without using any element of A(n), C(n) and D(n) more than once. For bookkeeping purposes i goes from 1to 8 and matches the value of A(i) but the C(i) and D(i) are not soordered.There are 48 unique solutions each involving the additionallimitation that C(1) = 1 and D(1) = 2. Thus the equation for A(1) is1+1=2 with each of these 48 solutions. But what I am looking forfrom you goes still further and there is only one such solution. Thesolution which I am looking for also has C(1) = 1 and D(1) = 2 butfurthermore has the unique quality that a simple manipulationconverts the solution to one for A(n) equal {1,2,3,4 ... 9} and B(n)= {1,2,3,4, ... 18}! This is done as follows: A(1) to A(8) and thefirst half of C(n) (C(1) to C(4)) are not changed. A(9) = 9. Old C(5)to C(8) (The second half) are each respectively promoted to becomenew C(6) to C(9). New C(5)= a number between 1 and 16 that wasorigionally move from B(n) to set D(n). The new values for D(n) aredetermined by these conditions and comprise the remaining numbersfrom the set {1,2,3 ... 18} (that are not used as part of new C(n)).I ask this for A(n) having the starting numbers 1-8 but I canillustrate the problem with A(n) = {1-4} since A(n) can be anymultiple of 4 even with this added complexity. It is possible toprove that solutions to the general problem are possible only for n =4m or for n = 4m +1. After noticing a similarity between a solutionfor A(n) = 1 to 4 and another solution for A(n) = 1 to 5. I checkedto see if the pattern could be repeated for higher values of m and Itproved true. Using the pattern as a guide, I solved this problem for A(n) = {1,2,3, ... 4m} where m = 2-4 with only paper and a pencil.Before noting this pattern, I needed a computer to find solutions form > 2, and it took my program many hours to find a solution for m =4. Now it is possible to find solutions easily for still highervalues of m. Look at my example for m = 1. See if you can spot a pattern. If not try other ways to solve the stated problem. If you solve the problem above you should be able to see the pattern and solve for m = 3 and 4 also. I have already solve it for m = 5 and 6 without having to write any computer algorithm.Solution for m = 11+1 = 22+4 = 63+5 = 84+3 = 7Here the number 5 is added to A(n) and 9,10 added to D(n). Inthis case the new number C(3) = old D(4) and the second half of old C(n) are promoted. The new sets are.1 plus 1 = 22 plus 4 = 63 plus 7 = 104 plus 5 = 95 plus 3 = 8

Ramsey2879
guest
Sep-17-06, 06:48 AM (EST)

1. "RE: Paired sums using two sets of consecutive integers"
In response to message #0

 Here is the solution set for m = 2 to 6, only I substituted lettersfor like even numbers. Can anyone solve this matrix for the complete solution sets? Note that the odd number pattern gives 3/4 of the solution. solution matrix for m = 2 to 6: 16. 1 A 3 B 11 9 7 5 18. 1 A 3 B 13 11 9 7 5 24. 1 C 3 B 5 D 17 15 13 11 9 7 26. 1 C 3 B 5 D 19 17 15 13 11 9 7 32. 1 E 3 B 5 D 7 F 23 21 19 17 15 13 11 9 34. 1 E 3 B 5 D 7 F 25 23 21 19 17 15 13 11 9 40. 1 G 3 B 5 E 7 F 9 D 29 27 25 23 21 19 17 15 13 11 42. 1 G 3 B 5 E 7 F 9 D 31 29 27 25 23 21 19 17 15 13 11 48. 1 H 3 B 5 G 7 D 9 F 11 J 35 33 31 29 27 25 23 21 19 17 15 13 50. 1 H 3 B 5 G 7 D 9 F 11 J 37 35 33 31 29 27 25 23 21 19 17 15 13