[bonanova]

This puzzle not mine, but I believe I have found the minimum answer.

If goes like this:

You are given a square pane of glass 1 meter on a side.

You are required to cut the pane into 100 pieces, all of the same area.

There is no requirement what shape the pieces are, nor do they have to all be the same shape.

Just so all 100 pieces have the same area.

The instrument you use is a cutting laser, and you may assume it will cut through glass of arbitrary thickness.

However, the laser consumes vast amounts of energy as it cuts.

For both cost and "green" reasons, you want to accomplish the cutting using as little energy as possible.

If you make multiple cuts - i.e. when the cut reaches the edge of the glass - you may turn the laser off immediately.

So the problem is: minimize the total length of your cuts.

Enjoy

you can stack pieces for cuts after the first.

[Guest]

still thinking of more. I can get it done in 3.1m. Is there a better answer?

[Izzy]

Is this like a solid 1m by 1m by 1m block of glass?

Let's just drop the glass and have it shatter.

My initial thought was cut it into 10 rows, stack them, and then cut them into little blocks by making another 10 rows. That's a total of 20 and took about 15 seconds to come up with.

I'm going to think about this for at least another half hour until I decide whether or not this is my final answer.

[Guest]

Not necessarily minimized but...

Cut square in half = 1m of laser cutting

Stack

cut stack again in half = 1m + 0.5m =1.5m so far

(now have 4 equal pieces)

etc. until 100 pieces

such that you reach the limit of 2m for infinite number of pieces

I'm too lazy to get an exact number for 100

Unless I'm missing something

[preflop]

Not necessarily minimized but...

Cut square in half = 1m of laser cutting

Stack

cut stack again in half = 1m + 0.5m =1.5m so far

(now have 4 equal pieces)

etc. until 100 pieces

such that you reach the limit of 2m for infinite number of pieces

I'm too lazy to get an exact number for 100

Unless I'm missing something

I agree with lupetu. I think this is the smallest and would would look like this.

1 --------- 2

0.5 -------- 4

0.25 ------- 8

0.125 ------ 16

0.0625 ---- 32

0.03125 --- 64

0.015625 -- 100

total length

1.984375

Edited October 3, 2008 by preflop

[Guest]

I agree with lupetu. I think this is the smallest and would would look like this.

1 --------- 2

0.5 -------- 4

0.25 ------- 8

0.125 ------ 16

0.0625 ---- 32

0.03125 --- 64

0.015625 -- 100

total length

1.984375

Wouldn't you end up with 128 pieces instead of 100 doing it that way?

Edited October 3, 2008 by Sinistral

[Guest]

I agree with lupetu. I think this is the smallest and would would look like this.

1 2

0.5 -------- 4

0.25 ------- 8

0.125 ------ 16

0.0625 ---- 32

0.03125 --- 64

0.015625 -- 100

total length

1.984375

I think the problem with this is that you don't actually get 100 equal pieces. This process can only give you equal sized pieces for powers of 2. You would get 128 equal pieces. Or you could not cut all of them in half the last time and get 100 pieces, but you would get 28 of one size and 72 of another. Unless I am misunderstanding either the OP or your solution, I don't feel like this works.

[Guest]

So we must get exactly 100 pieces with the same area??? Not 1 more or 1 less???

[Guest]

I think the problem with this is that you don't actually get 100 equal pieces. This process can only give you equal sized pieces for powers of 2. You would get 128 equal pieces. Or you could not cut all of them in half the last time and get 100 pieces, but you would get 28 of one size and 72 of another. Unless I am misunderstanding either the OP or your solution, I don't feel like this works.

The other problem is that your third cut needs to be as long as your second cut. See picture:

[Guest]

This puzzle not mine, but I believe I have found the minimum answer.

If goes like this:

You are given a square pane of glass 1 meter on a side.

You are required to cut the pane into 100 pieces, all of the same area.

There is no requirement what shape the pieces are, nor do they have to all be the same shape.

Just so all 100 pieces have the same area.

The instrument you use is a cutting laser, and you may assume it will cut through glass of arbitrary thickness.

However, the laser consumes vast amounts of energy as it cuts.

For both cost and "green" reasons, you want to accomplish the cutting using as little energy as possible.

If you make multiple cuts - i.e. when the cut reaches the edge of the glass - you may turn the laser off immediately.

So the problem is: minimize the total length of your cuts.

Enjoy

you can stack pieces for cuts after the first.

Best I can come up with is 2.995m with rectangular pieces. I think a different shape might be more efficient, but I haven't worked it out yet.

[preflop]

I think the problem with this is that you don't actually get 100 equal pieces. This process can only give you equal sized pieces for powers of 2. You would get 128 equal pieces. Or you could not cut all of them in half the last time and get 100 pieces, but you would get 28 of one size and 72 of another. Unless I am misunderstanding either the OP or your solution, I don't feel like this works.

yeah - i wasn't thinking properly before. I can get it down to 3.1125 (i think) - I am working on a shorter length.

Edited October 3, 2008 by preflop

[bonanova]

Kudos to anyone who has a better answer than 3.2 meters, which is my best result.

Details would be lovely to hear: what are the cuts, and what shape are the final pieces?

[Guest]

Kudos to anyone who has a better answer than 3.2 meters, which is my best result.

Details would be lovely to hear: what are the cuts, and what shape are the final pieces?

first to avoid decimal points I assume the length to be 10m.

It goes like

cut No---length of cut---available sizes

1---10m---2pcs(10x5)

2---5m---4pcs(5x5)

3---5m---4pcs(5x2),4pcs(5x3)

4---3m---4pcs(2x2),8pcs(3x2)4pcs(3x3)

5---3m---8pcs(2x1),20pcs(3x1)4pcs(3x2)

6---3m---36pcs(1x1),20pcs(2x1)8pcs(3x1)

7---1m---84pcs(1x1),8pcs(2x1)

8---1m---100pcs(1x1)

This totals up to 31m and scaling back to 3.1m

[Guest]

Cut in half so you have two .5 by 1m rectangles, stack(1m)

Cut in half so you have 4 .5 by .5 squares, stack(1.5)

Cut 5 times so you have 10 .5 by .1 rectangles, stack(1.5+.5*5 = 4)

Cut 5 times so you have 100 .1 by .1 rectanges, stack (4 + .1*5 = 4.5)

I saw some people had lower answers, but did not look at their work yet so this is just my thought process so far.

[Guest]

Details would be lovely to hear: what are the cuts, and what shape are the final pieces?

Drawing isn't my forte, so I'll try to explain in words the best I can.

The way I did it was to make the first cut 64 cm along one side. You then cut the larger piece into 64 pieces by stacking and cutting in half along the shortest side 6 times. This will give you 64 pieces that are 0.125X0.08 m.

While you are making the second cut described above, cut 0.111 m off the end of the smaller piece (1/9th). On subsequent cuts, you will cut and stack the smaller piece in half twice more (into 4), and the larger piece 5 more times (into 32), again, by always cutting along the shortest side. Because the shortest side of these are always shorter than those of the pieces coming from the 1X0.64 piece, they do not add to the length of any cuts. So, the total length is 1+0.64+0.5+0.32+0.25+0.16+0.125 = 2.995 m.

This gives you 64 pieces that are 12.5 cm X 8 cm, and 36 pieces that are 11.1 cm X 0.09 cm.

Originally, I thought that, because the ratio of area to perimeter of a polygon increases with the number of sides, that maybe the best option was to cut the original pane into polygons of as many sides as you could with your 7 cuts. However, you also want the cuts to be as short as possible (i.e. straight lines), and there is no close-packing scheme you can make with straight line cuts for polygons of more than 4 sides (none that I can think of anyway). Obviously, you have some problems around the edges of the original square as well. So, I don't think you do better by using a different shape.

D.

[bonanova]

Cut in half so you have two .5 by 1m rectangles, stack(1m)

Cut in half so you have 4 .5 by .5 squares, stack(1.5)

Cut 5 times so you have 10 .5 by .1 rectangles, stack(1.5+.5*5 = 4)

Cut 5 times so you have 100 .1 by .1 rectanges, stack (4 + .1*5 = 4.5)

I saw some people had lower answers, but did not look at their work yet so this is just my thought process so far.

Check your last two steps. You may have overcounted.

[Guest]

Because the shortest side of these are always shorter than those of the pieces coming from the 1X0.64 piece, they do not add to the length of any cuts.

That bit isn't true. I forgot you need to make three cuts of 0.36, not just two. So, it becomes 1+0.64+0.5+0.36+0.25+0.18+0.125 = 3.055m.

Oops.

[Guest]

Check your last two steps. You may have overcounted.

I don't think I overcounted, but looking at how imran did it, it is basically the same thing, he just went about it in a much more efficient way.

edit: I definitely messed up but it still isn't right... i'm going to edit my answer because right now it doesn't even produce the right number of squares...

Edited October 3, 2008 by telethar

[Guest]

Cut in half so you have two .5 by 1m rectangles, stack(1m)

Cut in half so you have 4 .5 by .5 squares, stack(1.5)

Cut 4 times so you have 10 .5 by .1 rectangles, stack(1.5+.5*4 = 3.5)

Cut 4 times so you have 100 .1 by .1 rectanges, stack (4 + .1*4 = 3.9)

[Prime]

That bit isn't true. I forgot you need to make three cuts of 0.36, not just two. So, it becomes 1+0.64+0.5+0.36+0.25+0.18+0.125 = 3.055m.

Oops.

I don't see why you are backing off from 299.5 cm. Here is a set of simple straightforward cuts adding up to it.

As shown on the picture:

The total cut length: 100+64+50+32+25+16+12.5=299.5cm.

And the resulting pieces are 4 4x25 and 96 8x12.5

The interesting problem is to prove what's the absolute minimum...

Edited October 4, 2008 by Prime

[bonanova]

Results so far, I believe, are:

4.5 - telethar

3.9 - telethar

3.2 - bonanova [details: cut into halves [1 cut], fifths [3 cuts], fifths [3], halves [1] -> 100 1x1 squares]

3.1 - Imran

2.995 - d3k3

3.055 - d3k3 oops

2.995 - Prime - modifying d3k3

Prime asks, can a minimum cut length be calculated?

I gave that some thought, and it seems 7 cuts and 2.75m are lower bounds.

Note that when a cut is made, the total perimeter is increased by twice the length of the cut.

We start with 4m perimeter and must end up with at least 40m perimeter [100 .1x.1 squares].

So there is an obvious upper bound of 18m of cuts to achieve squares.

Nine vertical and nine horizontal cuts with no stacking would do that.

The worst upper bound is a horrendous 99m of cuts: 99 vertical cuts yielding 100 1 x .01 rectangles!

You can't do worse than 99m of cuts.

But lower bounds are more interesting.

That would occur when the stack thickness doubles with each cut,

and each cut is made so as to minimize the length of the succeeding cut.

Let l_i be the length of the i^th cut.

Then P = 2 * sum [i=1,n] 2^[i-1] * l_i is the perimeter added after n cuts,

And L = sum n] l_i = the total length of the n cuts.

If we cut and stack halves, the values of l_i are 1, .5, .5, .25, .25, .125, .125, ...

Compute the results after i cuts:

...i....P....L

...1....2....1

...2....4....1.5

...3...12....2

...4...16....2.25

...5...24....2.5

...6...32....2.625

...7...48....2.75

And we can stop after 7 cuts, because the added perimeter has achieved the required 36m.

Thus, under the best of circumstances, we must make at least 7 total cuts with a total cut length of at least 2.75m

The lower bound of 2.75m cannot be achieved because the optimal scenario doubles the number of pieces for each cut, and we go from 64 to 128 equal-size pieces, not 100.

Still, 7 cuts and 2.75m are lower bounds. 2.75 could be higher, but 7 cannot.

[Guest]

The worst upper bound is a horrendous 99m of cuts: 99 vertical cuts yielding 100 1 x .01 rectangles!

You can't do worse than 99m of cuts.

I would disagree with this. I think it is also interesting to look at upper bounds. I have two ideas for that.

consider cutting 98 parallelograms and 2 triangles. then we require 99 cuts of length sqrt(1+0.4)=1.183m and the total amount of cutting required is 117.14m

Make cuts that are inscribed squares

[Guest]

Guessing ...

1st cut - 1m,

Pieces 2x 1m x .5m , Total length cut = 1m

2nd cut .5m

Pieces 4x .5m x .5m, Total length cut = 1m + .5m = 1.5m

3rd ... 6th cuts (4 cuts), Cuts .5 into .1 width, length of each cut = .5m

Pieces 20 x .5m x .1m, Total length cut = 1.5m + 2m = 3.5m

7th ... 10th cuts (4 cuts), Cuts .5 into .1 width, length of each cut = .1m

Pieces 100 x .1m x .1m, Total length cut = 3.5m + .4m = 3.9m

[Guest]

Guessing ...

1st cut - 1m,

Pieces 2x 1m x .5m , Total length cut = 1m

2nd cut .5m

Pieces 4x .5m x .5m, Total length cut = 1m + .5m = 1.5m

3rd ... 6th cuts (4 cuts), Cuts .5 into .1 width, length of each cut = .5m

Pieces 20 x .5m x .1m, Total length cut = 1.5m + 2m = 3.5m

7th ... 10th cuts (4 cuts), Cuts .5 into .1 width, length of each cut = .1m

Pieces 100 x .1m x .1m, Total length cut = 3.5m + .4m = 3.9m

3rd ... 6th cuts can be achieved in 3 cuts so saving .5m

7th ... 10th cuts can also be achieved in 3 cuts so saving ..1m

New Total = 3.3m

[Guest]

Make cuts that are inscribed squares

by inscribed I meant

cutting_glass.bmp

[Guest]

by inscribed I meant

cutting_glass.bmp

But of the same size ? How do you do it ? I do not think it is possible