[math-fun] Fwd: harder 7x7x7?
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf ---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill
The conjecture that the 2x3x1s make the puzzle strictly harder is still alive. (I.e., there are significantly fewer solutions, harder to find.) But the stronger conjecture, that you might as well glue the 2x3x1s back into a 2x2x3, is false. http://gosper.org/7x7x7.jpg shows a solution where they're not even touching. (Whitish, open-faced.) --rwg On Thu, Nov 1, 2012 at 9:37 PM, Bill Gosper <billgosper@gmail.com> wrote:
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf
---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com
Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute
a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill
A friend of mine had a nice magic cube puzzle. It was something like this: a 30x30x30 box, containing 27 almost-cubes, each of which was 9x10x11. Does this sound familiar to anyone? Where would I find one? Cris On Mar 24, 2013, at 4:45 PM, Bill Gosper wrote:
The conjecture that the 2x3x1s make the puzzle strictly harder is still alive. (I.e., there are significantly fewer solutions, harder to find.) But the stronger conjecture, that you might as well glue the 2x3x1s back into a 2x2x3, is false. http://gosper.org/7x7x7.jpg shows a solution where they're not even touching. (Whitish, open-faced.) --rwg
On Thu, Nov 1, 2012 at 9:37 PM, Bill Gosper <billgosper@gmail.com> wrote:
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf
---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com
Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute
a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill
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Cristopher Moore Professor, Santa Fe Institute
This sounds like Dean Hoffman's 1978 packing puzzle. Do a Google search to see if anyone out there is making/selling them. I got mine in the 1980s from Bill Cutler with the blocks measuring 15x18x22 deci-inches. On Mar 24, 2013, at 7:29 PM, Cris Moore <moore@santafe.edu> wrote:
A friend of mine had a nice magic cube puzzle. It was something like this: a 30x30x30 box, containing 27 almost-cubes, each of which was 9x10x11. Does this sound familiar to anyone? Where would I find one?
Here is some stuff on a Dean Hoffman puzzle from 1978 (it's from an essay David Singmaster wrote about David Klarner) http://plambeck.org/oldhtml/mathematics/klarner/singmaster/index.htm On Sun, Mar 24, 2013 at 4:51 PM, Hans Havermann <gladhobo@teksavvy.com>wrote:
This sounds like Dean Hoffman's 1978 packing puzzle. Do a Google search to see if anyone out there is making/selling them. I got mine in the 1980s from Bill Cutler with the blocks measuring 15x18x22 deci-inches.
On Mar 24, 2013, at 7:29 PM, Cris Moore <moore@santafe.edu> wrote:
A friend of mine had a nice magic cube puzzle. It was something like this: a 30x30x30 box, containing 27 almost-cubes, each of which was 9x10x11. Does this sound familiar to anyone? Where would I find one?
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-- Thane Plambeck tplambeck@gmail.com http://counterwave.com/
I also think it is the Hoffman packing puzzle. An article about it is in "The Mathematical Gardner", edited by David Klarner (reprinted now by Dover, I think the new version is called "Mathematical Recreations : A Collection in Honor of Martin Gardner. ") The article is "Packing Problems and Inequalities", by D.G. Hoffman. Hoffman's problem is: "Fit twenty-seven blocks, measuring A x B x C into a cubic box with sides of A + B + C. A, B and C must be different and the smallest dimension must be larger than (A + B + C) / 4." Also look at a puzzle called "Perfect Packing", designed by Donald Knuth and made by George Miller (http://www.puzzlepalace.com/#puzzle=200625) which looks at the case where the smallest dimension is exactly equal to (A=B=C)/4. Stan Isaacs 210 East Meadow Drive Palo Alto, CA 94306 stan@isaacs.com On Mar 24, 2013, at 4:51 PM, Hans Havermann wrote:
This sounds like Dean Hoffman's 1978 packing puzzle. Do a Google search to see if anyone out there is making/selling them. I got mine in the 1980s from Bill Cutler with the blocks measuring 15x18x22 deci-inches.
On Mar 24, 2013, at 7:29 PM, Cris Moore <moore@santafe.edu> wrote:
A friend of mine had a nice magic cube puzzle. It was something like this: a 30x30x30 box, containing 27 almost-cubes, each of which was 9x10x11. Does this sound familiar to anyone? Where would I find one?
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Cooked! By puzzle wizard Abel Garcia, who found a (so far) few solutions violating a mathematical requisite in the original, 30 piece version. Now I'm unsure which version has more solutions. Or is easier by some other metric. But even though it's less "mathematical", I think I prefer the 29 piece version. It can be converted to a 28 piece, "mathematical" version, slightly harder than the 30, by gluing the 2x3x1s face to face, as in the (failed) strong conjecture. Unfortunately, knowing this is possible makes the 29 easier, unless you forbid the glued configuration. --rwg On Sun, Mar 24, 2013 at 3:45 PM, Bill Gosper <billgosper@gmail.com> wrote:
The conjecture that the 2x3x1s make the puzzle strictly harder is still alive. (I.e., there are significantly fewer solutions, harder to find.) But the stronger conjecture, that you might as well glue the 2x3x1s back into a 2x2x3, is false. http://gosper.org/7x7x7.jpg shows a solution where they're not even touching. (Whitish, open-faced.) --rwg
On Thu, Nov 1, 2012 at 9:37 PM, Bill Gosper <billgosper@gmail.com> wrote:
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf
---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com
Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute
a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill
Omg, it's trivial! You can maneuver the face-to-face 2x3x1s<http://gosper.org/omg.jpg>to adjoin a 3x3x1, making a V corner that you can flip!<http://gosper.org/itstrivial.jpg> I'll bet Garcia knew this all along. --rwg On Tue, Mar 26, 2013 at 2:23 AM, Bill Gosper <billgosper@gmail.com> wrote:
Cooked! By puzzle wizard Abel Garcia, who found a (so far) few solutions violating a mathematical requisite in the original, 30 piece version. Now I'm unsure which version has more solutions. Or is easier by some other metric. But even though it's less "mathematical", I think I prefer the 29 piece version. It can be converted to a 28 piece, "mathematical" version, slightly harder than the 30, by gluing the 2x3x1s face to face, as in the (failed) strong conjecture. Unfortunately, knowing this is possible makes the 29 easier, unless you forbid the glued configuration. --rwg
On Sun, Mar 24, 2013 at 3:45 PM, Bill Gosper <billgosper@gmail.com> wrote:
The conjecture that the 2x3x1s make the puzzle strictly harder is still alive. (I.e., there are significantly fewer solutions, harder to find.) But the stronger conjecture, that you might as well glue the 2x3x1s back into a 2x2x3, is false. http://gosper.org/7x7x7.jpg shows a solution where they're not even touching. (Whitish, open-faced.) --rwg
On Thu, Nov 1, 2012 at 9:37 PM, Bill Gosper <billgosper@gmail.com> wrote:
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf
---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com
Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute
a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill
Very interesting! This guy Abel Garcia has done a great job, as have you in posing the question and following the developments. Amazing. -- Mike ----- Original Message ----- From: "Bill Gosper" <billgosper@gmail.com> To: <math-fun@mailman.xmission.com> Sent: Monday, April 01, 2013 12:24 AM Subject: Re: [math-fun] harder 7x7x7?
Omg, it's trivial! You can maneuver the face-to-face 2x3x1s<http://gosper.org/omg.jpg>to adjoin a 3x3x1, making a V corner that you can flip!<http://gosper.org/itstrivial.jpg> I'll bet Garcia knew this all along. --rwg
On Tue, Mar 26, 2013 at 2:23 AM, Bill Gosper <billgosper@gmail.com> wrote:
Cooked! By puzzle wizard Abel Garcia, who found a (so far) few solutions violating a mathematical requisite in the original, 30 piece version. Now I'm unsure which version has more solutions. Or is easier by some other metric. But even though it's less "mathematical", I think I prefer the 29 piece version. It can be converted to a 28 piece, "mathematical" version, slightly harder than the 30, by gluing the 2x3x1s face to face, as in the (failed) strong conjecture. Unfortunately, knowing this is possible makes the 29 easier, unless you forbid the glued configuration. --rwg
On Sun, Mar 24, 2013 at 3:45 PM, Bill Gosper <billgosper@gmail.com> wrote:
The conjecture that the 2x3x1s make the puzzle strictly harder is still alive. (I.e., there are significantly fewer solutions, harder to find.) But the stronger conjecture, that you might as well glue the 2x3x1s back into a 2x2x3, is false. http://gosper.org/7x7x7.jpg shows a solution where they're not even touching. (Whitish, open-faced.) --rwg
On Thu, Nov 1, 2012 at 9:37 PM, Bill Gosper <billgosper@gmail.com> wrote:
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf
---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com
Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute
a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill
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The translucency of the physob tricked Neil into seeing the lower left 4x2x1 as a 2x2x1, which he cranked into Burrtools and concluded I was April Fooling! When he sent his evidence, I overlooked his oversight, and we only now have sorted things out. For those with Mma, Graphics3D[{Yellow, Cuboid[{4, 6, 0}, {7, 7, 3}], Cuboid[{0, 0, 6}, {3, 3, 7}], Cuboid[{3, 3, 3}, {4, 6, 6}], Magenta, Table[Cuboid[{0, y, 4}, {3, y + 1, 6}], {y, 0, 1}], Red, Cuboid[{0, 6, 4}, {4, 7, 6}], Green, Cuboid[{0, 6, 0}, {4, 7, 4}], Table[Cuboid[{3, 2, z}, {7, 6, z + 1}], {z, 0, 2}], Table[Cuboid[{x, 3, 3}, {1 + x, 7, 7}], {x, 4, 6}], Opacity[.1], Cuboid[{0, 3, 6}, {4, 7, 7}], Red, Cuboid[{2, 0, 0}, {3, 2, 4}]}] Then greedily add all the 4x4x1s you can, and 2x4x1s should fill the rest. --rwg On Sun, Mar 31, 2013 at 9:24 PM, Bill Gosper <billgosper@gmail.com> wrote:
Omg, it's trivial! You can maneuver the face-to-face 2x3x1s<http://gosper.org/omg.jpg>to adjoin a 3x3x1, making a V corner that you can flip!<http://gosper.org/itstrivial.jpg> I'll bet Garcia knew this all along. --rwg
On Tue, Mar 26, 2013 at 2:23 AM, Bill Gosper <billgosper@gmail.com> wrote:
Cooked! By puzzle wizard Abel Garcia, who found a (so far) few solutions violating a mathematical requisite in the original, 30 piece version. Now I'm unsure which version has more solutions. Or is easier by some other metric. But even though it's less "mathematical", I think I prefer the 29 piece version. It can be converted to a 28 piece, "mathematical" version, slightly harder than the 30, by gluing the 2x3x1s face to face, as in the (failed) strong conjecture. Unfortunately, knowing this is possible makes the 29 easier, unless you forbid the glued configuration. --rwg
On Sun, Mar 24, 2013 at 3:45 PM, Bill Gosper <billgosper@gmail.com>wrote:
The conjecture that the 2x3x1s make the puzzle strictly harder is still alive. (I.e., there are significantly fewer solutions, harder to find.) But the stronger conjecture, that you might as well glue the 2x3x1s back into a 2x2x3, is false. http://gosper.org/7x7x7.jpg shows a solution where they're not even touching. (Whitish, open-faced.) --rwg
On Thu, Nov 1, 2012 at 9:37 PM, Bill Gosper <billgosper@gmail.com>wrote:
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf
---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com
Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute
a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill
On 2013-04-02 16:07, Bill Gosper wrote:
The translucency of the physob tricked Neil into seeing the lower left 4x2x1 as a 2x2x1, which he cranked into Burrtools and concluded I was April Fooling! When he sent his evidence, I overlooked his oversight, and we only now have sorted things out. For those with Mma,
Graphics3D[{Yellow, Cuboid[{4, 6, 0}, {7, 7, 3}], Cuboid[{0, 0, 6}, {3, 3, 7}], Cuboid[{3, 3, 3}, {4, 6, 6}], Magenta, Table[Cuboid[{0, y, 4}, {3, y + 1, 6}], {y, 0, 1}], Red, Cuboid[{0, 6, 4}, {4, 7, 6}], Green, Cuboid[{0, 6, 0}, {4, 7, 4}], Table[Cuboid[{3, 2, z}, {7, 6, z + 1}], {z, 0, 2}], Table[Cuboid[{x, 3, 3}, {1 + x, 7, 7}], {x, 4, 6}], Opacity[.1], Cuboid[{0, 3, 6}, {4, 7, 7}], Red, Cuboid[{2, 0, 0}, {3, 2, 4}]}]
gosper.org/7x7x7hoohah.png
Then greedily add all the 4x4x1s you can, and 2x4x1s should fill the rest. --rwg
Note: yellow+purple makes a reflectable V , proving that the 3x3x1s are no longer confined to disjoint planes. Mike Beeler has at last completed his analysis (http://gosper.org/new7x7x7.txt). Bottom line: Despite relaxing the constraints on the key three 3x3x1s, he finds only 460926 solutions, about 8/69 his original count of 3970180 for the "parity cube". (I have recovered that (mid 2006) analysis, if anyone wants it.) For reasons that now elude us, I doubled Mike's count and announced 7.9 million. ?? Whichever, I'm grateful to Mike for his huge efforts, and gratified by his result that the new puzzle is more evil. --rwg
On Sun, Mar 31, 2013 at 9:24 PM, Bill Gosper <billgosper@gmail.com> wrote:
Omg, it's trivial! You can maneuver the face-to-face 2x3x1s<http://gosper.org/omg.jpg>to adjoin a 3x3x1, making a V corner that you can flip!<http://gosper.org/itstrivial.jpg> I'll bet Garcia knew this all along. --rwg
On Tue, Mar 26, 2013 at 2:23 AM, Bill Gosper <billgosper@gmail.com> wrote:
Cooked! By puzzle wizard Abel Garcia, who found a (so far) few solutions violating a mathematical requisite in the original, 30 piece version. Now I'm unsure which version has more solutions. Or is easier by some other metric. But even though it's less "mathematical", I think I prefer the 29 piece version. It can be converted to a 28 piece, "mathematical" version, slightly harder than the 30, by gluing the 2x3x1s face to face, as in the (failed) strong conjecture. Unfortunately, knowing this is possible makes the 29 easier, unless you forbid the glued configuration. --rwg
On Sun, Mar 24, 2013 at 3:45 PM, Bill Gosper <billgosper@gmail.com>wrote:
The conjecture that the 2x3x1s make the puzzle strictly harder is still alive. (I.e., there are significantly fewer solutions, harder to find.) But the stronger conjecture, that you might as well glue the 2x3x1s back into a 2x2x3, is false. http://gosper.org/7x7x7.jpg shows a solution where they're not even touching. (Whitish, open-faced.) --rwg
On Thu, Nov 1, 2012 at 9:37 PM, Bill Gosper <billgosper@gmail.com>wrote:
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf
---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com
Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute
a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill
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participants (7)
-
Bill Gosper -
Cris Moore -
Hans Havermann -
Michael Beeler -
rwg -
Stan Isaacs -
Thane Plambeck