1 The nine dots puzzle extended to n^2 dots Marco Ripà, sPIqr Society, Apr 01 2013 Figure 1: The 9 dots puzzle/solution Problem 1: Considering the nine dots puzzle (see http://en.wikipedia.org/wiki/Thinking_outside_the_box) simplest extension, I show that you can solve it (4x4 dots grid fitting all dots into the center with straight, thin, lines) staring from any dots of the grid. Just take a look at the picture below:
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Extended 9 Dots Puzzle to nxnx...xn Dots: General Solving Method
How to solve any nxnx...xn dots problem: 2D, 3D or k-dimensions, it doesn't matter. I can provide a strict upper bound and a lower bound for any dots&lines problem belonging to this family.
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The nine dots puzzle extended to n^2 dots
Marco Ripà, sPIqr Society, Apr 01 2013
Figure 1: The 9 dots puzzle/solution
Problem 1:
Considering the nine dots puzzle (see http://en.wikipedia.org/wiki/Thinking_outside_the_box) simplest extension, I show that you can solve it (4x4 dots grid fitting all dots into the center with straight, thin, lines) staring from any dots of the grid. Just take a look at the picture below:
Figure 2: The 16 dots // 6 lines puzzle solutions (starting from any dots).
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Problem 2:
Considering a generic nxn grid, what is the minimum number of straight lines that you need to fit all the dots (without lifting the pencil from your paper)?
Answer: for n=1, you need (at least) 1 line; for n=2, you need (at least) 3 lines…and, for n≥3, you need (at least) 2*n-2 lines.
To understand this result, you can look at the following solving pattern, using the same strategy to solve any nxn puzzle (for n>3). You can easily see that you need two more lines for any n’:=n+1 square.
Figure 3: To fit nxn dots (into the center and without lifting your pencil from the paper), for n≥3, you need (at least) 2n-2 straight lines.
N.B.A consequence of this proof is that, for n→∞, the size of the paper that you need to draw your
solution with your pencil, tends to be the same one of the area encompassed by the most external dots.
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Figure 4: The square spiral proof (2*n-2 straight lines proof).If n’:=n+1, you have to draw one horizontal and one vertical line more... a square spiral, indeed.
Figure 5: Solving the nxn problem inside the box for n=5. For n>5, it is possible to solve any nxn puzzle inside the box just using the square spiral method.
nxnxn Dots Puzzle in 3D and its Generalization to K-Dimensions!
Marco Ripà, sPIqr Society, May 01 2013
I am glad to announce that I have just solved the nxnx...xn dots puzzle, in any dimensions amount you like (n≥2)... the inconvenience is that, for three dimensions and more, I can provide only a lower plus an upper bound (considering the straight lines you need to connect nxnx...xn dots).
In 3D, the upper bound of the number of consecutive straight lines you need is given by the formula SL:=2(n^2)-n-1, for n>2 (if n=2, SL=7... it is trivial).
The solving pattern is the same one I have just shown in 2D, connecting every plane using one more line (so, there are n-1 lines more).
2D2(n-1) SL.
We have to reproduce the “square spiral” n-times, connecting every plane we have considered... so there are “n-1” lines more. Thus, 2(n-1)*n+(n-1)=2n*n-2n+n-1 (Q.E.D.).
Square spiral method in k-dimensions (general solving method):
The total straight lines amount, in k-dimensions (nxnx...xn where “n” appears k-times), would be:2(n-1)*(n^(k-2))+(n^(k-2)-1)=2(n^(k-1))-n^(k-2)+(n^(k-2)-1).This is a general result: it is an upper bound for any nxnx...xn dots problem!
Now, let’s focus ourselves on the 3D generalization.
My result/proof is as follows:
Lower Bound (Proof - by definition): k(l)>=[(n^3-n)/(n-1)]+1=n*(n+1)+1.Upper Bound (Square Spiral Proof): k(m)=2(n^2)-n-1.Corrected Upper Bound: for any n>3, k(m)=2(n^2)-n-2.
Thus, the GAP (by proofs) is given by 2(n^2)-n-2-(n^2+n+1)=n^2-2n-3.
For example, considering n=4, I can solve the puzzle using only 26 straight lines, instead of 2(4^2)-4-1=27.
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Figure 7: 4x4x4 dots, 26 straight lines.
The table below refers to the Upper Bound (by the Square Spiral) above, while the Lower Bound is based on the consideration that you cannot connect more than “n” dots using the first line and then the maximum number is “n-1” for any additional line:
Table 1: Upper/Lower bounds in 2 and 3 dimensions.
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The 3D approximated upper-bound pattern, implies that, for n>2, SL(n+1)=SL(n)+5+4(n-1).
Here is the Corrected Upper Bound solution for n=5 (43 straight lines only):
Figure 8: 5x5x5 dots, 43 straight lines.
Upper Bound in k dimensions (k≥2). Let “h” denote the straight lines number you need to fit every dot:
h=2∗( n−1 )∗nk−2+nk−2−1h=(2 n−1 )∗nk−2−1.
Lower Bound in k dimensions (k≥2). Let “h” denote the straight lines number you need to fit every dot:
My lower bound for the “best plane by plane” 3D solution (solving each nxn grid before skipping to a new grid using a “linking” straight line) is as follows:
h = (2*n-2)*2+(2*n-3)*2+(2*n-4)*4+(2*n-5)*4+(2*n-6)*6+(2*n-7)*6+(2*n-8)*8+…+(n-1) =
¿n−1+ ∑i=1
i (max )∨n≥∑i
2∗⌈ i2⌉
[2∗(2n−i−1 )∗⌈ i2⌉ ]+(2 n−i(max)−2 )∗(n− ∑
i=1
i(max)
[2∗⌈ i2⌉ ]) (1)
n3D (Lower Bound)
3D (Upper Bound)
Gap (Upper-Lower B.) Upper B. Increments [n-->n+1]
Table 11: nxnxn dots puzzle upper bounds following the Square Spiral pattern and the one by figure 10: if n>41, we have the same result.
If n=6, 8, 10, 12 or 14, the best plane-by-plane upper bound is given by
h=2∗( n−2 )∗n+n−1−(1+2∗(n−5 ) )=2∗n2−3∗n−8, following Roger Philips’ pattern
(http://www.mathpuzzle.com/dots10x10.gif).
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For any n≥42,
h=n−1+ ∑i=1
i ( max )∨n≥∑i
2∗⌈ i2⌉
[2∗(2n−i )∗⌈ i2⌉ ]+ (2n−i (max )−1 )∗(n− ∑
i=1
i ( max )
[2∗⌈ i2⌉ ])
So, we have proved that, if n≥42, the Square Spiral pattern is the best plane-by-plane pattern available: it is as good as the one by figure 10 (for any n>41 – n>50 considering a generic 5x5 pattern -, the last/external pieces of the two patterns overlap – it is a square spiral frame, indeed).
It is also the best pattern available with zero crossing lines (from 1 to k-dimensions).
Thus “h”, the upper bound for the k-dimensions dots problem, can be further lowered as:
∀n∈ℕ\{0 }, let “t” be defined as the lowest upper bound we have previously proved for the standard
nxnxn dots problem (see Table 11), e.g. n=6t=62, while, ∀n>41,
t=n−1+ ∑i=1
i (max )∨n≥∑i
2∗⌈ i2⌉
[2∗(2 n−i )∗⌈ i2⌉ ]+(2 n−i (max )−1 )∗(n−∑
i=1
i (max )
[2∗⌈ i2⌉ ]) (4),
h :=t∗nk −3+nk−3−1h :=( t+1 )∗nk −3−1.
Let ~l be the minimum amount of straight lines you need to solve the nk dots problem (k,n∈ℕ\
{0,1,2 }), we have just proved that:
nk−1n−1
≤~l ≤ (2 n−1 )∗nk−2−1 (5)
The (5) can be further improved, by the ( (4) + Table 10 ), as: