# How to get perfect square from the list element..

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 Okkes Dulgerci 1 Vote Hi All, I need some help to solve this problem.. Assume you have n=1,2,3,...,15 number. Put them in an order so that when you add two successive number you get perfect square. As seen one example below. 9+7=16, 7+2=9, 2+14=16, and so on. I did this using pen and paper work.. I tried to solve this using Mathematica. First I got a set of number that make each number perfect square.. But I don't know what should be next step. I tried to use graph but it takes time and is easy to make error (has not been completed yet). I also want to extend it to say n=25.. Any help appreciated. Thanks.. when n=15---->9-7-2-14-11-5-4-12-13-3-6-10-15-1-8when n=16----> 16-9-7-2-14-11-5-4-12-13-3-6-10-15-1-8when n=17----> 16-9-7-2-14-11-5-4-12-13-3-6-10-15-1-8-17when n=18----> no sequenceswhen n=19----> no sequenceswhen n=20----> I claimed no sequencesIt Must start with 18 since there is only one number make it perfect square namely 7. 18+7 but after this we have two options. 18+7+2 or 18+7+9Please see attachment.. Attachments:
2 years ago
9 Replies
 Paul Cleary 2 Votes Hi OkkesThis works for n >=32, they are circular paths that are squares i.e. the first and last is also square. Do[a = Range[q]; b = Subsets[a, {2}]; b = Select[b, IntegerQ[Sqrt[#[[1]] + #[[2]]]] &]; b = GatherBy[b, #[[1]] &]; b = Flatten[b, 1]; b = DirectedEdge @@@ b; g = Graph[b]; g1 = UndirectedGraph[g]; s = FindHamiltonianCycle[g1]; s = Flatten[s]; s = ToString /@ s; s = StringReplace[s, "\[UndirectedEdge]" .. -> ","]; p = DeleteDuplicates[ Flatten[StringCases[s, RegularExpression["\\d+"]]]]; Print[{q, p}], {q, 32, 40}] // Timing It will slow down if you go above 1000 or so.P.
2 years ago
 The sequences produced by Paul's procedure In[5]:= Do[a = Range[q]; b = Subsets[a, {2}]; b = Select[b, IntegerQ[Sqrt[#[[1]] + #[[2]]]] &]; b = GatherBy[b, #[[1]] &]; b = Flatten[b, 1]; b = DirectedEdge @@@ b; g = Graph[b]; g1 = UndirectedGraph[g]; s = FindHamiltonianCycle[g1]; s = Flatten[s]; s = ToString /@ s; s = StringReplace[s, "\[UndirectedEdge]" .. -> ","]; p = DeleteDuplicates[ Flatten[StringCases[s, RegularExpression["\\d+"]]]]; Print[{q, p}], {q, 32, 40}] // Timing During evaluation of In[5]:= {32,{1,15,10,26,23,2,14,22,27,9,16,20,29,7,18,31,5,11,25,24,12,13,3,6,30,19,17,32,4,21,28,8}} During evaluation of In[5]:= {33,{1,15,10,26,23,2,14,22,27,9,16,33,31,18,7,29,20,5,11,25,24,12,13,3,6,30,19,17,32,4,21,28,8}} During evaluation of In[5]:= {34,{1,8,28,21,15,10,26,23,13,12,24,25,11,5,4,32,17,19,6,30,34,2,14,22,27,9,16,20,29,7,18,31,33,3}} During evaluation of In[5]:= {35,{1,8,28,21,15,34,2,14,11,25,24,12,13,23,26,10,6,30,19,17,32,4,5,31,18,7,9,27,22,3,33,16,20,29,35}} During evaluation of In[5]:= {36,{1,8,17,32,4,12,13,36,28,21,15,34,30,19,6,10,26,23,2,14,35,29,20,16,33,3,22,27,9,7,18,31,5,11,25,24}} During evaluation of In[5]:= {37,{1,24,25,11,14,22,3,33,16,9,27,37,12,13,36,28,8,17,32,4,21,15,34,30,19,6,10,26,23,2,7,18,31,5,20,29,35}} During evaluation of In[5]:= {38,{1,8,17,32,4,5,31,18,7,9,27,37,12,24,25,11,38,26,23,13,36,28,21,15,10,6,19,30,34,2,14,22,3,33,16,20,29,35}} During evaluation of In[5]:= {39,{1,24,25,39,10,6,19,30,34,15,21,4,32,17,8,28,36,13,12,37,27,9,7,18,31,5,11,38,26,23,2,14,22,3,33,16,20,29,35}} During evaluation of In[5]:= {40,{1,24,40,9,27,37,12,13,36,28,8,17,32,4,21,15,34,30,19,6,10,39,25,11,38,26,23,2,14,22,3,33,16,20,5,31,18,7,29,35}} Out[5]= {0.156001, Null} have the property, that also the first and the last entry of them sum up to a perfect square. That comes from the hamiltonian cycle searched for in the graph containing only vertices whose node values sum up to a perfect square. To get rid of that over-doing, one should accept one non-square vertex, then search again for a hamiltonian cycle and cut the cycle by throwing the non-square vertex away.
2 years ago
 Udo Krause 1 Vote Doing so, a little rebuild of Paul's code gives Clear[dulgerciSquareSequence] dulgerciSquareSequence[n_Integer] := Module[{b = Subsets[Range[n], {2}], s, ns, o, g, g1, c, p, bFound = False}, s = Select[b, IntegerQ[Sqrt[Plus @@ #]] &]; ns = Complement[b, s]; (* try every from the ns non-square vertices *) For[o = 1, o <= Length[ns], ++o, g = Graph[Join[s, {ns[[o]]}]]; g1 = UndirectedGraph[g]; c = FindHamiltonianCycle[g1]; If[Length[c] > 0, bFound = True; c = ToString /@ Flatten[c]; c = StringReplace[c, "\[UndirectedEdge]" .. -> ","]; p = DeleteDuplicates[Flatten[StringCases[c, RegularExpression["\\d+"]]]]; (* should cut the non-square vertex *) Print[p] ] ]; If[! bFound, Print["No sequence found for n = ", n]] ] /; n > 5 lets test this dulgerciSquareSequence[15] {1,15,10,6,3,13,12,4,5,11,14,2,7,9 | 8} this is your solution for n=15 in reverse order: 9+8=17, here is the split. dulgerciSquareSequence[16] {1,15,10,6,3,13,12,4,5,11,14,2,7,9,16 | 8} dulgerciSquareSequence[17] {1,15,10,6,3,13,12,4,5,11,14,2,7,9,16 | 17,8} dulgerciSquareSequence[18] No sequence found for n = 18 dulgerciSquareSequence[19] No sequence found for n = 19 dulgerciSquareSequence[20] No sequence found for n = 20 dulgerciSquareSequence[21] No sequence found for n = 21 dulgerciSquareSequence[22] No sequence found for n = 22 dulgerciSquareSequence[23] {1,3,22,14,11,5,20,16,9,7,18 | 2,23,13,12,4,21,15,10,6,19,17,8} {1,3,22,14,11,5,20,16,9 | 18,7,2,23,13,12,4,21,15,10,6,19,17,8} {1,3,22 | 18,7,9,16,20,5,11,14,2,23,13,12,4,21,15,10,6,19,17,8} dulgerciSquareSequence[24] No sequence found for n = 24 dulgerciSquareSequence[25] {1,3,22,14,11,25,24,12,13,23,2 | 18,7,9,16,20,5,4,21,15,10,6,19,17,8} {1,24,25,11,5,20,16,9,7,18 | 3,22,14,2,23,13,12,4,21,15,10,6,19,17,8} {1,3,22,14,2,23,13,12,24,25,11,5,20,16,9,7,18 | 4,21,15,10,6,19,17,8} {1,3,22,14,2,23,13,12,4,21,15,10,6,19,17,8 | 18,7,9,16,20,5,11,25,24} {1,3,22,14,11,25,24,12,13,23,2,7,18 | 9,16,20,5,4,21,15,10,6,19,17,8} {1,8,17,19,6,3,22,14,2,23,13,12,4,21,15,10 | 18,7,9,16,20,5,11,25,24} {1,3,22,14,2,23,13,12,24,25,11 | 18,7,9,16,20,5,4,21,15,10,6,19,17,8} {1,3,22,14,2,23,13 | 18,7,9,16,20,5,11,25,24,12,4,21,15,10,6,19,17,8} {1,3,13,23,2,14,22 | 18,7,9,16,20,5,11,25,24,12,4,21,15,10,6,19,17,8} {1,8,17,19,6,10,15,21,4,12,13,3,22,14,2,23 | 18,7,9,16,20,5,11,25,24} the 18 is the at one end of a sequence.There are many solutions for 25 < n < 32. So experimentally the situation is clear (mod errors).
2 years ago
 Hi Paul,Thanks for your reply. Someone conjectured that you can always find a sequence for n>=25. It seems it is true for n>=32. How can we check it for 20<=n<31
2 years ago
 Anonymous User Might this work For n=17----> 16-9-7- 2- 14-11- 5- 4-12- 13- 3-6-10- 15- 1- 8-8......... Then you Get n=18----->16-9-7- 2- 14-11- 5- 4-12- 13- 3-6-10- 15- 1- 8-8-17 Please see attached pic of a analysis I used sorting this out. In this analysis is seen a pattern and how extending that pattern can become a calculator generating next a term iteratively. depicted in the pic and shown are several elements generated past 8-8-17 utilizing this method Attachments:
2 years ago
 Anonymous User Sorry wrong pic See better pic attached Attachments: