phil1882 Posted November 3, 2012 Report Share Posted November 3, 2012 can you generate the next 4 rows? at what point would the number of black overtake the number of red? Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 5, 2012 Report Share Posted November 5, 2012 Quote Link to comment Share on other sites More sharing options...
0 phil1882 Posted November 6, 2012 Author Report Share Posted November 6, 2012 (edited) congrats current. that is indeed the next four rows. any clue on my second question? I'm honestly not sure myself. there actually was an outer most layer to the picture that i deliberately removed to make the puzzle more challenging. the picture is based on the primes, or more specifically the prime factorization. Edited November 6, 2012 by phil1882 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted November 6, 2012 Report Share Posted November 6, 2012 Are you sure that yours and curr3ent's images are identical? Does your second question which row is mostly black, or at which row there are more cumulative blacks? Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 6, 2012 Report Share Posted November 6, 2012 Looks like I have a square that is black where yours is red on the fourth row. I had no idea what the tree was supposed to be. I was just extending patterns. Quote Link to comment Share on other sites More sharing options...
0 phil1882 Posted November 6, 2012 Author Report Share Posted November 6, 2012 (edited) your right, should be black, not red. (should alternate black red black red) for my second question, ignoring the outer most layer (my solution spoiler), which row is cumulative black more than cumulative red? Edited November 6, 2012 by phil1882 Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 6, 2012 Report Share Posted November 6, 2012 How are you determining red and black? I see the rows for the primes where red is flagged when the mod is equal to 0. So I was using red when n1 mod n2 = 0 and black when n1 and n2 share a factor but that gives me the reverse for the non-primes. I also had to use red if prime and black if not when n1 = n2. I get the following which matches the pattern of squares but not the coloring... Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 6, 2012 Report Share Posted November 6, 2012 wait... in your triangle the "12th" diagonal has a square on 2,3,4,6,8 and 10 but not 9? Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted November 6, 2012 Report Share Posted November 6, 2012 (edited) False alarm Edited November 6, 2012 by CaptainEd Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted November 6, 2012 Report Share Posted November 6, 2012 wait... in your triangle the "12th" diagonal has a square on 2,3,4,6,8 and 10 but not 9? Another apparent oddity: the "diagonal" for 4 has a BLACK square for 4, 8, 12, 16, 20, but RED for 24 ALso, the "diagonal" for 9 has Black for 9, but Red for 18 Of course, these wouldn't be oddities if we understood the pattern...but I don't Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 6, 2012 Report Share Posted November 6, 2012 He said the 24 on the 4th diagonal was supposed to be black. Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted November 6, 2012 Report Share Posted November 6, 2012 (edited) It appears that there's a white box (ie. no box at all) for n1, n2 exactly when n1 and n2 are relatively prime. EXCEPT for the white boxes on (9,12) as curr3nt pointed out, and (10,15).Yes, sorry curr3nt, I was slow catching up to your observation about (4,24), and his corroboration. Edited November 6, 2012 by CaptainEd Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 6, 2012 Report Share Posted November 6, 2012 (edited) So... What is special about the following that doesn't apply to anything else? 4: 4, 8, 12, 16, 20, 24 6: 12, 18 8: 4, 8, 12, 16, 20 9: 10: 12: 4, 6, 8, 16 14: 15: 16: 4, 8, 12 18: 6 20: 4, 8 The non primes down the middle are obvious. Why the multples of 4 and 6 for the rest? (Only 4 and 6 for the non-prime diagonals too...) Edited November 6, 2012 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted November 6, 2012 Report Share Posted November 6, 2012 Here's a rule, and a few exceptions. Since the OP is acknowledged to have an error, maybe the exceptions are errors as well. Rules, applied sequentially until a rule applies: cell (n1, n2) is--- * white if n1 and n2 are relatively prime, else * red if n1 is prime or n2 is prime, else * black if n1 divides n2 or n2 divides n1, else * red if GCD (n1, n2) is prime, else * black Exceptions: * (4,24) is red, should be black (acknowledged by poster) * (10,15) is white, should be red * (9,12) is white, should be red * (9,18) is red, should be black Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted November 6, 2012 Report Share Posted November 6, 2012 Rule: cell (n1,n2) is--- * white if n1 and n2 share no prime factor * red if n1 and n2 share exactly one prime factor * black if n1 and n2 share a composite factor Still assuming those exceptions are errors in the OP. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 6, 2012 Report Share Posted November 6, 2012 (edited) I updated to match his simplier rules. edit... gah I'm only counting a factor value once. So 8 and 12 are showing red when it should be black from 2^2 in both. Need to figure out how to update to fix that. Edited November 6, 2012 by curr3nt Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 6, 2012 Report Share Posted November 6, 2012 Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted November 6, 2012 Report Share Posted November 6, 2012 (edited) If we assume this coloration rule is correct, we are asked to consider what is now the collection of NW-SE diagonals, and answer where there's one with more blacks than reds. At a snail's pace, I point out that the diagonal that intersects (k,1) consists of all the n1, n2 where n1+n2 = k+1. And I find it fascinating that there are some that are all white (never mind black and red!) k = 4, 6, 10, 12, 16, for example. There must be an obvious mathematical reason why all pairs summing to those numbers are mutually prime... (blushing) yes, if K is prime, and n1 and n2 add to K, then they don't share any factor, otherwise, K would share that factor. Edited November 6, 2012 by CaptainEd Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted November 7, 2012 Report Share Posted November 7, 2012 34 13 Black vs 10 Red vs 10 White Black: (first 6 counted twice for reverse) 32,4 30,6 28,8 27,9 24,12 20,16 18,18 Red: (Counted twice for reverse) 34,2 33,3 26,10 22,16 21,15 White: 31,5 29,7 25,11 23,13 19,17 Other numbers with more black than red: 58: 23 vs 20 70: 27 vs 20 82: 31 vs 28 98: 31 vs 28 106: 41 vs 30 118: 47 vs 40 130: 47 vs 44 138: 47 vs 44 142: 55 vs 40 154: 55 vs 52 166: 63 vs 56 178: 79 vs 52 194: 57 vs 54 198: 63 vs 56 202: 71 vs 68 214: 83 vs 60 218: 71 vs 68 226: 79 vs 76 238: 95 vs 80 250: 107 vs 72 258: 83 vs 80 262: 95 vs 88 274: 95 vs 92 278: 95 vs 88 286: 111 vs 80 298: 123 vs 96 Quote Link to comment Share on other sites More sharing options...
0 phil1882 Posted November 7, 2012 Author Report Share Posted November 7, 2012 wow you guys really put a lot of time and effort into this. i appreciate that. the rules are pretty simple, or at least the ones i was trying to follow. take the horizontal pattern, and repeat it diagonally, starting at 2. note that the white space is cut in half. if a number is all white horizontally, then the outer most pixel should be red, else black. Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted November 8, 2012 Report Share Posted November 8, 2012 I've got to confess that I couldn't relate what you just said to the analysis curr3nt and I have been doing. I fear we are off target. Here's a question. Did you originate this problem? (For example, when curr3nt spotted the surprising color on 4x24, you agreed that it should have been black. That makes me think you created this.) Or did you find it elsewhere (for example, you said you had removed the outer shell. That made me think someone else originated it). Quote Link to comment Share on other sites More sharing options...
0 phil1882 Posted November 8, 2012 Author Report Share Posted November 8, 2012 (edited) as far as i can tell i originated it, but i was inspired by a similar pattern created by someone else. they used blue triangles all the way through, rather than red and black squares. Edited November 8, 2012 by phil1882 Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted November 9, 2012 Report Share Posted November 9, 2012 Thank you, phil1882, for finding and extending such an interesting puzzle! Now that I view the video a few times, I understand what you were saying about how the diagram is created. Overcoming my disbelief that such a simple mechanism could create a chart representing presence or absence of common factors was a pleasurable and enlightening experience. Thank you, curr3nt, for attacking this puzzle and sharing your progress with us. Quote Link to comment Share on other sites More sharing options...
0 TheChad08 Posted November 10, 2012 Report Share Posted November 10, 2012 Now I finally understand it. Thanks for the video, I was confused as all hell before. Quote Link to comment Share on other sites More sharing options...
Question
phil1882
can you generate the next 4 rows?
at what point would the number of black overtake the number of red?
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