Donate to Remove ads

Got a credit card? use our Credit Card & Finance Calculators

Thanks to Rhyd6,eyeball08,Wondergirly,bofh,johnstevens77, for Donating to support the site

Cube2

cinelli
Lemon Slice
Posts: 553
Joined: November 9th, 2016, 11:33 am
Has thanked: 234 times
Been thanked: 161 times

Cube2

#61578

Postby cinelli » June 21st, 2017, 11:44 am

A carpenter has a wooden cube 3 inches on each side, which he wishes to cut into 27 one-inch cubes. He can do it by making six cuts through the cube, keeping the pieces together in the cube shape. Can he reduce the number of necessary cuts by rearranging the pieces after each cut? If so how, if not why not?

Cinelli

kiloran
Lemon Quarter
Posts: 4112
Joined: November 4th, 2016, 9:24 am
Has thanked: 3250 times
Been thanked: 2855 times

Re: Cube2

#61580

Postby kiloran » June 21st, 2017, 11:49 am

No, can't be done, unless he has a saw which is infinitesimally thin.

--kiloran (smart alec since he was born)

Gengulphus
Lemon Quarter
Posts: 4255
Joined: November 4th, 2016, 1:17 am
Been thanked: 2628 times

Re: Cube2

#61587

Postby Gengulphus » June 21st, 2017, 12:20 pm

cinelli wrote:A carpenter has a wooden cube 3 inches on each side, which he wishes to cut into 27 one-inch cubes. He can do it by making six cuts through the cube, keeping the pieces together in the cube shape. Can he reduce the number of necessary cuts by rearranging the pieces after each cut? If so how, if not why not?

Spoiler, assuming an infinitesimally thin saw...

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

No, he cannot. To show that's the case, look at the three dimensions of the largest piece remaining after each saw cut. Each of them is in the range 1-3, only one of them can be reduced by the next saw cut, and it can only be reduced by 1. So the sum of those three dimensions starts at 3+3+3 = 9 after no saw cuts, can be reduced by at most 1 by each saw cut, and must end at 1+1+1 = 3. Therefore at least 9-3 = 6 saw cuts are needed.

If it were a 4x4x4 cube, he would be able to reduce the obvious 9 saw cuts to 6 by first sawing the cube in half, then stacking the two resulting half-cubes to saw them both in half with the same cut, and repeating for the other two dimensions, and something similar goes for all bigger cubes. But not for the 3x3x3 cube.

Gengulphus

argoal
2 Lemon pips
Posts: 131
Joined: November 4th, 2016, 10:51 am
Has thanked: 19 times
Been thanked: 68 times

Re: Cube2

#61604

Postby argoal » June 21st, 2017, 1:16 pm

I came to the answer by a different line of thought.

One of the 27 cubed that will result is the one in the very center of the 3 x 3 cube and it needs all 6 of its faces to be cut to expose them as none are currently exposed.

There is physically no possible way for a single cut to expose more than one of this cube's faces.

Therefore 6 cuts is the minimum required.

Gengulphus
Lemon Quarter
Posts: 4255
Joined: November 4th, 2016, 1:17 am
Been thanked: 2628 times

Re: Cube2

#61607

Postby Gengulphus » June 21st, 2017, 1:22 pm

Yes, a more elegant solution than mine!

Gengulphus

cinelli
Lemon Slice
Posts: 553
Joined: November 9th, 2016, 11:33 am
Has thanked: 234 times
Been thanked: 161 times

Re: Cube2

#62466

Postby cinelli » June 24th, 2017, 11:12 am

There is some interesting analysis from Gengulphus but I agree with him that argoal's explanation is more elegant.

Cinelli


Return to “Games, Puzzles and Riddles”

Who is online

Users browsing this forum: No registered users and 9 guests