Cutting A Gold Bar 3

[bam720]

If I have 1 gold bar, equally divided into 7 pieces as such:
|=|=|=|=|=|=|=|
I need to pay an employee each day for one week. His wages are 1 piece of the gold bar per day. How can I cut the bar to allow for fair payment using only 2 cuts. You can not fold the bar because it is gold.

 

[Skie]

Alternate 2 cut solution:
Provided the top and bottom of the prior cut leaves 1/7th of the bar on the top and 1/7th of the bar on the bottom. Do this wave cut:
[tt]
|[red] [/red]|
|[red]/\/\/[/red]|
| |[/tt]

Place the two pieces on top of each other and cut the 1/7th pieces off like so:
[tt]|[red]_____[/red]|
|/\/\/|[/tt]

Giving you 6 pieces that are 1/7th pieces and 2 pieces that are 1/14th.

 

[mmerlinn]

[ ]

I can't believe this thread is still going since the first poster answered the question.

I can't either. LOL.

(BTW, I think the first reply by mmerlinn was the correct one.)

I won't claim my answer is the correct one, but I do think that it is the answer bam720 was looking for. Would be nice if he would comment on all of these comments.

I also notice that SkipVought has the same answer I had, just spelled out a bit more.


mmerlinn

"Political correctness is the BADGE of a COWARD!"

 

[BobRodes]

Well, I stated the position that providing a gold piece that one isn't able to spend doesn't constitute payment, and if one has to give it back one can't spend it. So it may be the answer bam720 was looking for, but MY answer was an improvement.

 

[skyline666]

Well my fiances surname is spelt rhodes, theres is an island called rhodes, and as im not interested in the exact spelling of ur name and just merely glanced at what ur name was i spelt it that way.

I also dont no whether to agree or disagree with the cutting. As I said previously, if you are cutting and you get to the edge, that first small piece will break off. If you are then still cutting (counting this as one cut), then you must be shaving some of the gold off to have it counting as a single cut must you not, and if this is then true, there wont be equal cuts at all as there will be bits of gold shavings on whatever your cutting the gold bar on. Unless what you are stating is that picture the sine wave, with a tangent at its amplitude. When you are cutting and you get to the tangent, you are infact not cutting the first small piece off, but leaving a hairline piece of gold on, which connects it to the rest of the gold bar. This will count as a single cut.

 

[akalinowski]

option 1: assuming the gold bar is divided into 7 pieces like a chocolate bar, break gold bits off like Hershey's, mmmmm... chocolate... cut it once for shits

if that's not the case...

option 2: cut the gold twice longways to expose more surface area, melt the gold, poor into 7 molds each able to hold 1/7th the volume of the original bar, let cool, remove from mold and use for payment.


akalinowski

http://www.nofear.com

MCSE 2000, A+, N+, LCP, CNE

 

[BobRodes]

<im not interested in the exact spelling of ur name and just merely glanced
Some of us are more meticulous thinkers than others...

<leaving a hairline piece of gold [etc]
This is hardly a refutation of my point of view. I again point out that a tangent is the point at which two lines, at least one of them curved, intersect. If there is a "hairline piece of gold left" it isn't a tangent. Furthermore, if you leave said "hairline piece" there, you aren't in fact cutting off a piece of gold, either.

 

[bam720]

My apologies for not responding. It has in fact been so long that I do not recall looking for a specific answer. If I recall correctly, I heard the puzzle and felt like sharing. I admit I do not read the forums as often as I should (I drop in unannounced time to time), I never responded because the multiple solutions continued to intrigue me. Without saying a "correct" solution has been found I like reading other "alternate and equally valid" solutions

mmerlinn I think that your answer solves the problem at hand in the most direct and straight forward manner. I really like the sine wave solution as well, BobRodes

 

[BobRodes]

<mmerlinn I think that your answer solves the problem at hand in the most direct and straight forward manner.

Glad you're not MY paymaster.

 

[mmerlinn]

[&nbsp;]
BobRodes:

I really like the sine wave solution as well, BobRodes

Glad I didn't need to twist my brain to come up with your solution.

mmerlinn

http://mmerlinn.com

"Political correctness is the BADGE of a COWARD!"

 

[BobRodes]

<Glad I didn't need to twist my brain to come up with your solution.

Yes, I'm glad too. I didn't have to twist my brain either, and the fact that you found it as simple to envision as I did gives it a certain validity it wouldn't otherwise have.

 

[COBOLScott]

Hi all, I have pondered this for about 5 years. I was asked this once by a student who was stuck on this word problem in some class or another, and I have come up with more than one solution, and although he didn't like any of them, it was a good way to kill a couple of beers.

I found this board and thought I would share, and probably stick around as there is a lot of info here.

BobRodes, I originally envisioned the solution you proposed, but I had to discard it, simply because it creates too many pieces. I'm pretty sure that the original question was designed to elicit a similar answer regarding sine wave measurement though. You could stretch the semantics to say you would melt the two ends together, because that is not specifically prohibited in the problem, but then you could melt it into 7 smaller ingots and not make any cuts, so I'm going to push the "fail" buzzer on that one too.

I guess I should start by making an assumption on the wording of the problem. "Equal Pieces" has to be quantifiable. I'm going to assume that volume, or mass, or for our purposes weight, is keeping within the guidelines. This is very important. We'll assume the bar is 1 ounce for reference below.

Solution 1, we can modify your solution slightly, by cutting it in a sine wave, but in the middle, limiting the frequency to 1.5 wavelengths. That gives us 3 equal pieces of 1/7 oz and two ends of 2/7 oz each, if we've dome it correctly. Then we can stack the two ends and make one cut through them to divide them into 4 pieces of 1/7 oz each. 7 * 1/7 oz = 1 oz.

That was discarded because the stack-and-cut could be considered two cuts by itself depending on how you read it. No problem.

Solution 2, we don't cut all the way through, but more like we etch through the ingot or bar almost all the way. We still use our sine wave pattern, but this time 3 wavelengths so that we have 6 full "sine-pieces" with extra on the end. Now here is the tricky part, we turn the ingot in its edge, and slice down only in the middle, freeing the 6 pieces from the ingot, 1/7 oz each, and leaving a flat sheet with two partial pieces attached. If we calculated our cut depth and our sine frequency correctly by making it slightly longer, the two ends will be connected by a flattish sheet of leftover gold, still technically one piece, and weighing 1/7 oz total.

I liked that better but couldn't come up with enough math to convince my bud to use it, so I came up with yet another way.

Solution 3, the gold is in a cylindrical bar shape. Shape is critical here as I'll get to in a moment. We take our magic gold knife and start a cut only halfway through the bar to the centerline, and rotate the bar as we cut so that we get a spiral cut. The bar will still be in one piece at this point. Then we cut lengthwise, again halfway through to centerline, which causes the pieces to separate. It is not too different from cutting a spring across the coils; you end up with a bunch of rings at the end. I said shape was critical here, but it can be done. I spent awhile later on playing with modeling clay and a sharp knife. Try it.

Solution 4 came from solution 3, and assumes we start with a spring, or a thickish wire we can coil into a spring. Simply cut down the edge, one cut does it.

Those are what I came up with at the spur of the moment in June of 2000. It has haunted me ever since, LOL!

 

[COBOLScott]

Sorry I meant 2003 above. But if it compiles it goes, pre-editing is not a requirement for programmers, right?

 

[Moissenjeff]

I think I have the solution:

you first cut the bar into 2 piece slice + 5 piece slice

|=|=| + |=|=|=|=|=|
Then you put them on top of each other like so:

|=|=|
|=|=|=|=|=|

Then you cut your second cut between first & second and third and fourth, wich gives you:

|=| + |=| + |=| + |=|=|=|=|

day 1 |=|
day 2 |=| + |=|
day 3 |=| + |=| + |=|
day 4 |=|=|=|=| - 3 *(|=|)
day 5 |=|=|=|=| + |=|
day 6 |=|=|=|=| + |=| + |=|
day 7 |=|=|=|=| + |=| + |=| + |=|

It is solved

 

[Skie]

Solution 3, the gold is in a cylindrical bar shape. Shape is critical here as I'll get to in a moment. We take our magic gold knife and start a cut only halfway through the bar to the centerline, and rotate the bar as we cut so that we get a spiral cut. The bar will still be in one piece at this point. Then we cut lengthwise, again halfway through to centerline, which causes the pieces to separate. It is not too different from cutting a spring across the coils; you end up with a bunch of rings at the end. I said shape was critical here, but it can be done. I spent awhile later on playing with modeling clay and a sharp knife. Try it.

Why does it have to be cylinder? Provided it's a rectangular prism, this same method should work. It may work for other shapes, but I'd hate to figure the math to confirm it.

For ease, we'll imagine the bar has 24 sections to make cutting easier. Starting from the bottom right of the second section, cut diagonally from there to the top left of the same section. Turn, and cut from the bottom right of section 3 to the top left of section 3. So on, until the last cut going to the top left of section 22 (ending at the bottom right of 23) where you stop.

Then cutting from top to bottom, start in the corner you started from (should be same you ended in) and cut to the middle of the cube slicing through the entire bar.

If the size of the bar is 1" x 1" x 24", then you need only cut the middle sections in half to see you can make a cube with a volume of 3 cubic inches (1/7 total volume). With the ends you can do similar (you just cut it at 2/3s) to make a cube with a volume of 3 cubic inches (1/7 total volume).

I also did the actual math (incase my cube visualizations were incorrect) for volume (which was a pain considering all the triangles), but it does equal out into 7 equal pieces.

 

[BobRodes]

<it creates too many pieces

Your question was

How can I cut the bar to allow for fair payment using only 2 cuts.

I don't believe my solution fails to accomplish this, given that two halves have the same amount of gold as one whole and would therefore constitute fair payment.

However, if you wish to change the requirements document after seeing the solution (which never really happens, of course, but we're just having fun here), amending it to say "divide into 7 equally sized pieces" I concede that my solution fails, unless as some have said the two ends of the initial ingot are joined into a cylinder. I do NOT concede that my solution fails to solve the problem as stated above.

 

[strongm]

The way I have heard this problem expressed (and it helps clarify the origins of the original post's comment about "equally divided into 7 pieces", which in the version in the post doesn't make a lot of sense or add information to the problem) is:

Microsoft Interview Question
You've got someone working for you for seven days and a gold bar to pay them. The gold bar is segmented into seven connected pieces. You must give them a piece of gold at the end of every day. If you are only allowed to make two breaks in the gold bar, how do you pay your worker?

In this version it is pretty clear that the only way you can divide the bar is by physically breaking the connections between the segments. At which point the binary cut solution becomes the only viable one (ignoring the real world issue, rasied by several, that this would prevent the worker spending any of his money until the end of the week)

 

[traingamer]