superprismatic Posted December 19, 2009 Report Share Posted December 19, 2009 Take the integers from 1 to 13 and distribute them among 4 sets such that the sum of the products of the contents of the sets is minimal. Each set must be non-empty. If a set contains only one number, that number is considered to be the product for that set. For example, if I divided the numbers as follows: {1,2,3,4,5,7}, {6,11,12}, {8,9,10}, and {13}, their products would be 840, 792, 720, and 13 respectively. The sum of these is 2365. So, 2365 is the sum of products for this particular way of distributing the 13 numbers into 4 sets. Two questions: 1. What is the minimum value for the sum of products? 2. In how many different ways can you achieve the minimum? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 19, 2009 Report Share Posted December 19, 2009 1,5,9,13 585 5,9,13 585 5,9,13 585 5,9,13 585 2,8,10 160 1,2,8,10 160 2,8,10 160 2,8,10 160 3,7,11 231 3,7,11 231 1,3,7,11 231 3,7,11 231 4,6,12 288 4,6,12 288 4,6,12 288 1,4,6,12 288 1264 1264 1264 1264 Take the integers from 1 to 13 and distribute them among 4 sets such that the sum of the products of the contents of the sets is minimal. Each set must be non-empty. If a set contains only one number, that number is considered to be the product for that set. For example, if I divided the numbers as follows: {1,2,3,4,5,7}, {6,11,12}, {8,9,10}, and {13}, their products would be 840, 792, 720, and 13 respectively. The sum of these is 2365. So, 2365 is the sum of products for this particular way of distributing the 13 numbers into 4 sets. Two questions: 1. What is the minimum value for the sum of products? 2. In how many different ways can you achieve the minimum? Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 19, 2009 Author Report Share Posted December 19, 2009 (edited) 1,5,9,13 585 5,9,13 585 5,9,13 585 5,9,13 585 2,8,10 160 1,2,8,10 160 2,8,10 160 2,8,10 160 3,7,11 231 3,7,11 231 1,3,7,11 231 3,7,11 231 4,6,12 288 4,6,12 288 4,6,12 288 1,4,6,12 288 1264 1264 1264 1264 Good try, but you can get a bit lower! By the way, use spoilers so others can look at the posts without seeing answers which might spoil their experiences working on the problems. Another thing: You can use the code /code tags to make sure your columns align correctly. Edited December 19, 2009 by superprismatic Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 19, 2009 Report Share Posted December 19, 2009 1124 Don't know if it's a minimum though. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 20, 2009 Report Share Posted December 20, 2009 (edited) (1,2,11,12)(3,9,10)(4,5,13)(6,7,8) = 1130 1124 Don't know if it's a minimum though. Edited December 20, 2009 by jruel Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 20, 2009 Report Share Posted December 20, 2009 (1,2,11,12)(3,9,10)(4,5,13)(6,7,8) = 1130 switching your 5 and 6 ←1 (drops 4) and then switching your 12 and 13 ←1 (drops 2)) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 22, 2009 Report Share Posted December 22, 2009 I can't get less than 1124. Here is my sequence: {13,11,2,1} {12,8,3} {10,7,4} {9,6,5} Of course the number 1 can be put in any set so this makes 4 different sequences Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 22, 2009 Author Report Share Posted December 22, 2009 I can't get less than 1124. Here is my sequence: {13,11,2,1} {12,8,3} {10,7,4} {9,6,5} Of course the number 1 can be put in any set so this makes 4 different sequences 1124 is the lowest, but there's a little more than 4 of them. Of course, it will be a multiple of 4 because you can move the 1 around. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 23, 2009 Report Share Posted December 23, 2009 (edited) 1124 is the lowest, but there's a little more than 4 of them. Of course, it will be a multiple of 4 because you can move the 1 around. Well, I found 12 different groups that make the minimum, but didn't find a systematic way of making sure I have them all. Good puzzle superprismatic -- did you make this one up? Edited December 23, 2009 by xamdam Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 23, 2009 Author Report Share Posted December 23, 2009 Well, I found 12 different groups that make the minimum, but didn't find a systematic way of making sure I have them all. Good puzzle superprismatic -- did you make this one up? Yes, I made it up. Thanks for telling me you liked it. I only found 8 which is actually 2 different ones with the 1s moving around. I thought I had exhausted the possibilities with a program but maybe there's a small bug. Anyway, I'd like to see your 12 which I suppose is 3 with the 1s moving around. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 24, 2009 Report Share Posted December 24, 2009 Yes, I made it up. Thanks for telling me you liked it. I only found 8 which is actually 2 different ones with the 1s moving around. I thought I had exhausted the possibilities with a program but maybe there's a small bug. Anyway, I'd like to see your 12 which I suppose is 3 with the 1s moving around. I'd like to see that third one too! Seems I have two, not three. Two I had were basically the same -- didn't notice. Here are the two I got (without the 1s): #1 ---- (13,11,2) ; (12,6,4); (10,3,9); (8,7,5) #2 ---- (13,11,2); (3,8,12); (6,9,5); (4,10,7) I got them by trying to get the four sums close to each other -- I think they were between 270 and 288. But I didn't have a strategy to make sure I had 'em all. What was your program? Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 24, 2009 Author Report Share Posted December 24, 2009 I'd like to see that third one too! Seems I have two, not three. Two I had were basically the same -- didn't notice. Here are the two I got (without the 1s): #1 ---- (13,11,2) ; (12,6,4); (10,3,9); (8,7,5) #2 ---- (13,11,2); (3,8,12); (6,9,5); (4,10,7) I got them by trying to get the four sums close to each other -- I think they were between 270 and 288. But I didn't have a strategy to make sure I had 'em all. What was your program? My program? Well, I wrote it in FORTRAN. Basically, I wrote a subroutine to enumerate all ways of breaking a set up into n subsets. I simply used this subroutine to generate all ways to break up {1,2,3,4,5,6,7,8,9,10,11,12,13} into 4 subsets. I computed the sum of products of each one and printed it out if its sum did at least as well as the best so far. The program ran in about a second and the last eight things printed out had a sum of 1124. That's when I saw the roving 1 phenomenon. I also notice that there were some which had a sum of 1125 printed out before the 1124 ones. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 24, 2009 Report Share Posted December 24, 2009 My program? Well, I wrote it in FORTRAN. Basically, I wrote a subroutine to enumerate all ways of breaking a set up into n subsets. I simply used this subroutine to generate all ways to break up {1,2,3,4,5,6,7,8,9,10,11,12,13} into 4 subsets. I computed the sum of products of each one and printed it out if its sum did at least as well as the best so far. The program ran in about a second and the last eight things printed out had a sum of 1124. That's when I saw the roving 1 phenomenon. I also notice that there were some which had a sum of 1125 printed out before the 1124 ones. Very good. Ah yes, I remember writing a FORTRAN program to find Magic Squares. Those punch cards were fun. It took a day or so to have the programs returned. The printout would be wrapped around the card deck with a rubber band. Often there would be a message like "Syntax error in line 30." Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
Take the integers from 1 to 13 and distribute
them among 4 sets such that the sum of the
products of the contents of the sets is minimal.
Each set must be non-empty. If a set contains
only one number, that number is considered to
be the product for that set.
For example, if I divided the numbers as follows:
{1,2,3,4,5,7}, {6,11,12}, {8,9,10}, and {13},
their products would be 840, 792, 720, and 13
respectively. The sum of these is 2365. So,
2365 is the sum of products for this particular
way of distributing the 13 numbers into 4 sets.
Two questions:
1. What is the minimum value for the sum of products?
2. In how many different ways can you achieve the minimum?
Link to comment
Share on other sites
12 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.