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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.

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Figure 6: Smart cheating...

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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:

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nk :≤ n+(h−1)∗(n−1)nk−nn−1

=h−1h=nk−nn−1

+1h=nk−1n−1

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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]

3D (Guessing the Plane Bound)

1 1 1 0 /// 12 7 7 0 6 73 13 14 1 7 144 21 26 5 12 265 31 43 12 17 406 43 64 21 21 597 57 89 32 25 828 73 117 44 28 1099 91 148 57 31 139

10 111 183 72 35 17311 133 222 89 39 21112 157 265 108 43 25313 183 311 128 46 29814 211 361 150 50 34715 241 415 174 54 40016 273 473 200 58 45717 307 535 228 62 51818 343 601 258 66 58319 381 670 289 69 65120 421 743 322 73 723

Table 2: Upper/Lower bounds in 2 and 3 dimensions.

N.B.

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Table 2 indicates the lower bounds for the nxnxn dots problem if we are searching plane-by-plane solutions only.

No intersections additional constraint:

Figure 9: Connecting dots without crossing a line.

h = (2*n-1)*2+(2*n-2)*2+(2*n-3)*4+(2*n-4)*4+(2*n-5)*6+(2*n-6)*6+(2*n-7)*8+…+(n-1) =

¿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⌉ ])(2)

Where i(max) is the maximum (integer) value of “i” inside the summation (the maximum value

such that n ≥∑i

[2∗⌈ i2⌉ ]).

A general method to reduce the gap between the 3D lower/upper bounds is to combine the pattern below with the square spiral one:

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Figure 10: 5x5 dots 8 lines basic pattern.

This is not the best upper bound we can define for a plane-by-plane problem as well. In fact:

Figure 11: 6x6x6 dots, 62 lines

[this pattern has been inspired by http://www.mathpuzzle.com/joe6x6.GIF].

The best upper bound I have currently proved for any arbitrary large value of n (following the pattern by figure 10) is as follows:

nUpper Bound

(nxnxn)n

Upper Bound (nxnxn)

1 1 31 17992 7 32 19193 14 33 20434 26 34 21715 43 35 2302

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6 63 36 24377 87 37 25768 115 38 27199 146 39 286610 181 40 301711 220 41 317212 263 42 333113 309 43 349314 359 44 365915 413 45 382916 471 46 400317 532 47 418118 597 48 436319 666 49 454920 739 50 473921 816 51 493222 897 52 512923 982 53 533024 1071 54 553525 1163 55 574426 1259 56 595727 1359 57 617428 1463 58 639529 1571 59 662030 1683 60 6849

Table 10: nxnxn dots puzzle upper bounds following the pattern by figure 10.

nnxnxn Solution by the Square Spiral

Best nxnxn Upper Bound Currently Discovered

Gap (square spiral-pattern by Fig 10)

1 1 1 02 7 7 03 16 14 24 29 26 35 45 43 26 65 63→62 2→37 89 87→85 2→48 117 115→112 2→59 148 146 2

10 183 181→178 2→5

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11 222 220 212 265 263→260 2→513 311 309 214 361 359→358 2→315 415 413 216 473 471 217 535 532 318 601 597 419 670 666 420 743 739 421 820 816 422 901 897 423 986 982 424 1075 1071 425 1167 1163 426 1263 1259 427 1363 1359 428 1467 1463 429 1575 1571 430 1687 1683 431 1803 1799 432 1923 1919 433 2046 2043 334 2173 2171 235 2304 2302 236 2439 2437 237 2578 2576 238 2721 2719 239 2868 2866 240 3019 3017 241 3173 3172 142 3331 3331 043 3493 3493 044 3659 3659 045 3829 3829 046 4003 4003 047 4181 4181 048 4363 4363 049 4549 4549 050 4739 4739 051 4932 4932 0

Table 11: nxnxn dots puzzle upper bounds following the Square Spiral pattern and the one by figure 10: if n≥42, 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−1 )∗n+n−1−(1+2∗(n−5 ) )=2∗n2−3∗n+8, following Roger Phillips’ 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:

nk−1n−1

≤~l ≤ (t +1 )∗n

k−3−1 (6)