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A157639
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Least number of lattice points from which every point of a square n x n lattice is visible.
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5
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1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,4
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COMMENTS
| Adhikari and Granville give bounds on the size of a(n).
Contribution from Jon E. Schoenfield (jonscho(AT)hiwaay.net), Aug 03 2009: (Start)
Consider an arbitrarily large lattice. Define S1 as the square having both X and Y in the closed interval [1,3]. From a single viewpoint at (2,2), every lattice point on or inside S1 is visible. S1 (including its edges) covers 3 rows x 3 columns of lattice points, so a(3)=1. Also, since an n-row x n-column subset of S1 that includes the one viewpoint can be selected for n=1 and for n=2, this 1-viewpoint example is sufficient to show that a(n)=1 for n=1..3.
Define S2 as the square having both X and Y in [1,5]. Every lattice point in S2 is visible from at least one of the two viewpoints (3,1) and (3,4). S2 covers 5 rows x 5 columns of points, so a(5)<=2. Since a 4-row x 4-column subset of S2 that includes the two viewpoints can also be selected, a(4)<=2. Since no 1-viewpoint solution exists for n>3, we have a(n)=2 for n=4 and n=5.
Proceeding similarly, every lattice point where X and Y are both in [1,23] is visible from at least one of the three viewpoints (14,14), (15,14), and (14,15), so a(n)<=3 for n=2..23. Since no 2-viewpoint solution exists for n>5, we have a(n)=3 for n=6..23.
Together, the three 4-viewpoint solutions {(20,20), (21,20), (20,21), (21,21)}, {(43,51), (72,65), (58,80), (57,66)}, and {(28,34), (105,105), (34,99), (99,40)} are sufficient to show that a(n)<=4 for n=24..132: in these solutions, every lattice point where X and Y are both in [1,m], where m=40, 128, and 132, respectively, is visible from at least one viewpoint, and all four viewpoints would fit in a k-row by k-column subset of the m-row by m-column square, for k as small as 2, 30, and 78, respectively. Thus these three solutions demonstrate that a(n)<=4 for the overlapping ranges n=2..40, n=30..128, and n=78..132, respectively. Since (per an exhaustive search) no 3-viewpoint solution exists for n=24..132, we have a(n)=4 for n=24..132.
Per exhaustive search, no 4-viewpoint solution exists for n=133, so a(133)=5.
In summary: a(1..3)=1, a(4..5)=2, a(6..23)=3, a(24..132)=4, a(133)=5. (End)
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LINKS
| Sukumar Das Adhikari and Andrew Granville, Visibility in the Plane
Eric Weisstein, MathWorld: Visible Point
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EXAMPLE
| a(3) = 1 because all 9 points are visible from the central (2,2) point.
a(4) = 2 because all 16 points are visible from (1,2) or (2,1).
a(6) = 3 because all 36 points are visible from (1,1), (1,2), or (2,1).
a(24)= 4 because all 576 points are visible from (1,1), (1,2), (1,3), or (2,24).
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MATHEMATICA
| Table[sees=Table[{}, {n^2}]; Do[pt1=(c-1)*n+d; lst={}; Do[pt2=(a-1)*n+b; If[GCD[c-a, d-b]<2, AppendTo[lst, pt2]], {a, n}, {b, n}]; sees[[pt1]]=lst, {c, n}, {d, n}]; done=False; k=0; While[ !done, k++; len=Binomial[n^2, k]; i=0; While[i<len, i++; s=Subsets[Range[n^2], {k}, {i}][[1]]; If[Length[Union@@sees[[s]]]==n^2, done=True; Break[]]]]; k, {n, 10}]
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CROSSREFS
| A141224
Sequence in context: A122462 A115230 A165024 * A010096 A063510 A156878
Adjacent sequences: A157636 A157637 A157638 * A157640 A157641 A157642
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KEYWORD
| hard,nice,nonn
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AUTHOR
| T. D. Noe (noe(AT)sspectra.com), Mar 03 2009
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EXTENSIONS
| Terms after a(24) from Jon E. Schoenfield (jonscho(AT)hiwaay.net), Aug 03 2009
Corrected/updated comment to include 4-viewpoint solution showing a(n)<=4 for n=78..132 Jon E. Schoenfield (jonscho(AT)hiwaay.net), Aug 09 2009
Incorporated result that a(133)=5 Jon E. Schoenfield (jonscho(AT)hiwaay.net), Sep 02 2009
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