The Brain Teaser Thread

[jonks]

4
14
11-14
31-14
13-21-14


What's the next in the progression?

 

[hypn0tik]

Originally posted by: sirjonk
4
14
11-14
31-14
13-21-14


What's the next in the progression?

I've seen this one before. Just count out the number of numbers in the previous line.

i.e.

4
14 (One four)
11-14 (One one, one four)
31-14
13-21-14
31-13-12-14

 

[jonks]

A man has a wolf, a sheep, and some cabbages and needs to ferry them across a river...

 

[Kyteland]

You have 12 golf balls, 11 of the same weight and one of a deviating weight. You are allowed to use a balance scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether its heavier or lighter than the rest.

You are a prisoner in a foreign land. your fate will be determined by a little game. there are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)?

5 robbers hold up a diamond store and make out with 100 diamonds, all of equal value. Instead of splitting the diamonds equally among them, they decide that they will have a bidding contest with these rules:
- Each robber draws a number out of a hat (1, 2, 3, 4 or 5)
- Starting with the smallest number, each robber will have the chance to make a proposal on how to best split the diamonds. Eg, 20-20-20-20-20 or 0-10-20-30-40 are possibile bids.
- Each robber votes on whether to accept the proposal and if there is a majority (>50% vote yes) it passes.
- If is does not pass then the robber who proposed it is killed and the process is repeated with the remaining, smaller group.
- Each robber is assumed to be equally smart and will veto any proposal when they know they can do better.

What should the first robber bid to keep themselves alive and to maximize their diamonds?

 

[Atomic Playboy]

Originally posted by: brikis98

Originally posted by: sirjonk
Some of the versions of this riddle I found limit the sun to <100, if that helps. Having read the answer, it wouldn't help me. Unless phrases like "of course, all even numbers can be expressed as the sum of two primes" roll off your tongue with ease, I suggest skipping.

It's an interesting riddle, because you definitely CAN see patterns and eliminate lots of possibilities... but it is WAY too much paper work to track it down to the final answer. I could prob write a computer program to simplify it... but, while I'm bored at work, I don't think i'm THAT bored

any1 else got interesting puzzles to post?

No, seriously, why can it not be any combination of numbers? Suppose the numbers are 10 and 27. The conversation goes as follows:

Ann says "I know their product [270, but she doesn't tell Bob], but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum [37, again, not telling Ann]."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

Ann guesses 9 and 30, Bob guesses 13 and 14, they're both wrong. Where's the riddle?

 

[jonks]

http://www.puzzle.dse.nl/tease...ex_us.html#know_or_not

From this version it appears the info in the statements is slightly different, i.e. the first line is Product guy: "I don't know what YOUR sum is, Sum guy"

 

[QED]

Ok... another oldie:

Given a standard 8-by-8 chessboard with two opposing corner squares removed and a stack
of 2-by-1 domino pieces, is it possible to cover the entire chessboard using the domino pieces
without any overlaps? Each half (a square) of each domino must cover exactly one square on the chessboard,
and each square on the chessboard must be covered. If so, how?

 

[Atomic Playboy]

Originally posted by: sirjonk
http://www.puzzle.dse.nl/tease...ex_us.html#know_or_not

From this version it appears the info in the statements is slightly different, i.e. the first line is Product guy: "I don't know what YOUR sum is, Sum guy"

The people in this riddle are very quick, unless their conversation is taking place over a period of a couple hours or so. The fact that anyone thought to come up with this in the first place makes me want to destroy the world.

 

[QED]

Originally posted by: Atomic Playboy

Originally posted by: brikis98

Originally posted by: sirjonk
Some of the versions of this riddle I found limit the sun to <100, if that helps. Having read the answer, it wouldn't help me. Unless phrases like "of course, all even numbers can be expressed as the sum of two primes" roll off your tongue with ease, I suggest skipping.

It's an interesting riddle, because you definitely CAN see patterns and eliminate lots of possibilities... but it is WAY too much paper work to track it down to the final answer. I could prob write a computer program to simplify it... but, while I'm bored at work, I don't think i'm THAT bored

any1 else got interesting puzzles to post?

No, seriously, why can it not be any combination of numbers? Suppose the numbers are 10 and 27. The conversation goes as follows:

Ann says "I know their product [270, but she doesn't tell Bob], but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum [37, again, not telling Ann]."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

Ann guesses 9 and 30, Bob guesses 13 and 14, they're both wrong. Where's the riddle?

10 and 27 does not work because Bob wouldn't have necessarily known BEFORE HAND that Ann wouldn't be able to divine the two numbers from knowin the product.

In this case, Ann wouldn't have been able to say "Based on what you told me, I now know what the two numbers are", which is why the solution couldn't be 10 and 27.

I think you might be under the impression that Bob would always be able to say to Ann, "I knew that you would't know what the two numbers are", but that isn't always the case. There are some number combinations that Ann would be able to deduce from their product alone. For instance, if the product is 85 then Ann would know the two numbers are 5 and 17. So if Bob only knows the sum is 22, he cannot be sure whether Ann knows the numbers or not (their product might be 85, in which case she would-- or their product might be 4*18=72, in which case, she wouldn't).

In short, in order for Bob to make his first statement, he has to know, given the sum, that every possible two numbers that add to that sum are not both prime.

 

[Mo0o]

Originally posted by: Kyteland

You have 12 golf balls, 11 of the same weight and one of a deviating weight. You are allowed to use a balance scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether its heavier or lighter than the rest.

Measure 5 vs 5. If they're even, measure the two not used. If one set of 5 is heavier, measure 2 vs 2. If they're even , the one left out is the different one. Otherwise, measure among the 2 that's heavier.

Edit: hmm just realized you dont know whether it's heavier or lighter... That stumps me

You are a prisoner in a foreign land. your fate will be determined by a little game. there are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)?

One white marble into one jar. 49 White and 50 black into the other. Chance of survival should be around 74.9 ish

5 robbers hold up a diamond store and make out with 100 diamonds, all of equal value. Instead of splitting the diamonds equally among them, they decide that they will have a bidding contest with these rules:
- Each robber draws a number out of a hat (1, 2, 3, 4 or 5)
- Starting with the smallest number, each robber will have the chance to make a proposal on how to best split the diamonds. Eg, 20-20-20-20-20 or 0-10-20-30-40 are possibile bids.
- Each robber votes on whether to accept the proposal and if there is a majority (>50% vote yes) it passes.
- If is does not pass then the robber who proposed it is killed and the process is repeated with the remaining, smaller group.
- Each robber is assumed to be equally smart and will veto any proposal when they know they can do better.

What should the first robber bid to keep themselves alive and to maximize their diamonds?

98 0 1 1 0 or 98 0 1 0 1 works as well

 

[brikis98]

Originally posted by: Atomic Playboy

Originally posted by: brikis98

Originally posted by: sirjonk
Some of the versions of this riddle I found limit the sun to <100, if that helps. Having read the answer, it wouldn't help me. Unless phrases like "of course, all even numbers can be expressed as the sum of two primes" roll off your tongue with ease, I suggest skipping.

It's an interesting riddle, because you definitely CAN see patterns and eliminate lots of possibilities... but it is WAY too much paper work to track it down to the final answer. I could prob write a computer program to simplify it... but, while I'm bored at work, I don't think i'm THAT bored

any1 else got interesting puzzles to post?

No, seriously, why can it not be any combination of numbers? Suppose the numbers are 10 and 27. The conversation goes as follows:

Ann says "I know their product [270, but she doesn't tell Bob], but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum [37, again, not telling Ann]."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

Ann guesses 9 and 30, Bob guesses 13 and 14, they're both wrong. Where's the riddle?

The point of the puzzle is that both Ann and Bob do a LOT of thinking before they get together to talk and in between sentences (this prob is more realistic as an email exchange). Once Ann tells Bob she doesn't know the numbers and Bob reveals to Ann that he doesn't know what the two numbers are and that he knew in advance Ann wouldn't know, we get a lot of information. For example, the number Ann knows (AxB) must not be a product of primes (e.g. 3x5 = 15). If it was, Ann would have known what the two original numbers were. You can also deduce that AxB is not a cube of a prime, or again, there would only be one choice from A and B.

I imagine you can keep extending this logic to narrow the list of possibilities until you're down to one, but it seems awfully tedious.

 

[jonks]

Originally posted by: Mo0o

5 robbers hold up a diamond store and make out with 100 diamonds, all of equal value. Instead of splitting the diamonds equally among them, they decide that they will have a bidding contest with these rules:
- Each robber draws a number out of a hat (1, 2, 3, 4 or 5)
- Starting with the smallest number, each robber will have the chance to make a proposal on how to best split the diamonds. Eg, 20-20-20-20-20 or 0-10-20-30-40 are possibile bids.
- Each robber votes on whether to accept the proposal and if there is a majority (>50% vote yes) it passes.
- If is does not pass then the robber who proposed it is killed and the process is repeated with the remaining, smaller group.
- Each robber is assumed to be equally smart and will veto any proposal when they know they can do better.

What should the first robber bid to keep themselves alive and to maximize their diamonds?

98 0 1 1 0 or 98 0 1 0 1 works as well

Why would 3 of 5 robbers vote to approve that breakdown?

 

[ColdFusion718]

Originally posted by: merlocka
The last guy has a 50-50 to live.

But he counts the remaining hats. If it's even, he says red. If it's odd, he says black.

The second guy counts the hats. He knows what color he has. He then repeats the above.

98% survival rate.

This was in CS101, it's a parity check.

Sucks to be the other 2% though.

 

[Mo0o]

Originally posted by: sirjonk

Originally posted by: Mo0o

5 robbers hold up a diamond store and make out with 100 diamonds, all of equal value. Instead of splitting the diamonds equally among them, they decide that they will have a bidding contest with these rules:
- Each robber draws a number out of a hat (1, 2, 3, 4 or 5)
- Starting with the smallest number, each robber will have the chance to make a proposal on how to best split the diamonds. Eg, 20-20-20-20-20 or 0-10-20-30-40 are possibile bids.
- Each robber votes on whether to accept the proposal and if there is a majority (>50% vote yes) it passes.
- If is does not pass then the robber who proposed it is killed and the process is repeated with the remaining, smaller group.
- Each robber is assumed to be equally smart and will veto any proposal when they know they can do better.

What should the first robber bid to keep themselves alive and to maximize their diamonds?

98 0 1 1 0 or 98 0 1 0 1 works as well

Why would 3 of 5 robbers vote to approve that breakdown?

I worked backwards
Pirates A B C D E

Scenario 1: D E
D knows he's fucked because theres nothing he can propose that E would approve since E can veto everything then keep all 100 for himself. So the best D can expect is 1 coin from one of the earlier rounds. What D does NOT want is for it to get to Scenario 1

Scenario 2: C D E
Here, E will Veto no matter what because he wants the situation to turn into scenario 1. So instead, C will distribute 99 to himself and 1 to D, claiming a majority while ignoring E

Scenario 3: B C D E
While in this situation, everyone knows the voting can (and would) end at Scenario 2 so B will distribute the diamond accordingly. C would expect to make 99 diamonds next round, so clearly his vote can not be bought. D and E would expect 1 and 0 respectively. From the parameters, I assume B would give 1 to D again and give 1 to E to buy his vote. (This is where it gets fuzzy because I'm not sure under the rules whether D would vote for a situation where he gets 1 in an earlier round or is vindictive and will opt to Veto then get 1 anyways, in the next round)

Scenario 4: A B C D E
So A, knowing B could end the game at Scenario 3, will alter his distributions accordingly as well. B's vote can not be bought since he stand to make 98 diamonds next round. C's vote can be bought w/ 1 because he would be making 0 next round. D and E, you just need one of their votes and that can be bought for 1.

 

[ColdFusion718]

Originally posted by: JTsyo

Originally posted by: brikis98
ok, one more puzzle from me. after this, someone else please post something interesting for us to figure out

In your laboratory, you are mixing up a huge batch of very dangerous chemicals. The reaction is so dangerous, that if you don't do your job properly, it will explode and take out a 50 mile radius. Fortunately, your job is simple. You have two chemicals - chemical A and chemical B - which come in the form of small sticks. All you have to do is toss in exactly 1 stick of chemical A and 1 stick of chemical B every ten minutes. The reaction lasts 100 minutes, so you have 10 sticks of each chemical. Unfortunately, your foolish lab assistant dropped all 20 sticks on the floor, and they all got mixed up! Worse yet, chemical A and B are virtually indistinguishable: they look, taste, smell, feel, and weigh the same!

The question: what strategy can you use to be sure to toss in exactly 1 stick of each chemical every 10 minutes for 100 minutes? Remember, if you put in too little during a 10 minute interval, you die. If you put in too much, you won't have enough alter, and you die.

Split them into two piles and call black when you toss them in?

Since it sounds like the process hasn't started yet, I'd quit the job and drive 51 miles away from the lab.

EDIT: oh someone else has already said it.

 

[Mo0o]

Here's an easy one:

You have two ropes. Each takes exactly 60 minutes to burn. They are made of different material so even though they take the same amount of time to burn, they burn at separate rates. In addition, each rope burns inconsistently, so half way mark on each rope does not represent 30 minutes. How do you measure out exactly 45 minutes?

 

[brikis98]

You have 12 golf balls, 11 of the same weight and one of a deviating weight. You are allowed to use a balance scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether its heavier or lighter than the rest.

Ugh. This involves careful numbering of the 12 balls and weighing them in various permutations to narrow down the possibilities. I remember I worked it out a while ago, but it's not coming to me now.

You are a prisoner in a foreign land. your fate will be determined by a little game. there are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)?

1 white ball in one jar and the other 99 marbles go in the other jar. Half the time you'll pick the jar with one marble and live. The other half of the time, you'll pick the jar with 99 marbles and your odds of living are 49/99 = ~.495. Putting the two together, your odds of living are about .5 + .247 = ~.747.

5 robbers hold up a diamond store and make out with 100 diamonds, all of equal value. Instead of splitting the diamonds equally among them, they decide that they will have a bidding contest with these rules:
- Each robber draws a number out of a hat (1, 2, 3, 4 or 5)
- Starting with the smallest number, each robber will have the chance to make a proposal on how to best split the diamonds. Eg, 20-20-20-20-20 or 0-10-20-30-40 are possibile bids.
- Each robber votes on whether to accept the proposal and if there is a majority (>50% vote yes) it passes.
- If is does not pass then the robber who proposed it is killed and the process is repeated with the remaining, smaller group.
- Each robber is assumed to be equally smart and will veto any proposal when they know they can do better.

What should the first robber bid to keep themselves alive and to maximize their diamonds?

I believe you have an error in your problem. I've solved an almost identical one before with 5 pirates splitting 100 gold coins. The only difference there was that each pirate needed greater than or equal to 50% to get his vote to pass. If you only have greater than, you run into ambiguous scenarios.

For example, take the same problem, but with only two robbers, D and E, with D proposing first. If D needs greater than 50% of the vote, he knows he can't make a proposal where he gets ANY diamonds, or E will vote it down. What isn't clear, however, is if E would vote down the distribution where D gets 0 and E gets 100. There's no reason to assume E would want D's death since his desire is only to earn diamonds. However, this is very important as it can impact the final solution. That is, if D knows he will ALWAYS die in a 2 person scenario, when we discuss a 3 person scenario (C, D, E), he would vote for ANY proposal of C's as it would let him live. Otherwise, if D always lives in a 2 person scenario, but with no diamonds, C must offer D 1 diamond to secure his vote in a 3 person scenario. These same ambiguities go all the way up to the 5 person scenario and you end up with multiple possible answers.

If we solve the *proper* problem (with greater than or equal), we get:

D, E

D would propose a distribution of 100, 0 and since he needs only needs 50% of the vote to pass, his own vote pushes it through.

C, D, E

C knows that in a 2 person scenario, E gets nothing. So, he offers E one diamond for his vote, or a 99, 0, 1 distribution.

B, C, D, E

B knows that in the 3 person scenario, D gets nothing. So he offers D one diamond for his vote, or a 99, 0, 1, 0 distribution. Again, they need just 50% to pass, and this will work.

A, B, C, D, E

A knows in that in the 4 person scenario, C and E get nothing. He therefore offers them just one diamond, or a 98, 0, 1, 0, 1 distribution.

 

[brikis98]

Originally posted by: Mo0o
Here's an easy one:

You have two ropes. Each takes exactly 60 minutes to burn. They are made of different material so even though they take the same amount of time to burn, they burn at separate rates. In addition, each rope burns inconsistently, so half way mark on each rope does not represent 30 minutes. How do you measure out exactly 45 minutes?

At the same time, light both ends of the first rope and one end of the second rope. Exactly when the two ends of the first rope come together, light the remaining end of the second rope. When the two ends of the second rope touch, 45 minutes has elapsed.

 

[Kyteland]

Say you have a sheet of paper, what is the maximum number of pieces you can cut it into given n cuts. You cannot fold the sheet or rearrange already cut pieces (i.e., even after the nth cut, the pieces remain in the same position as they were before any cuts were made.)

And if you really want to geek out generalize the problem to any dimension. You have a X dimensional shape. You cut it N times in the X-1 dimension without moving or rearranging the pieces. How many pieces do you have?

 

[QED]

Originally posted by: Kyteland
Say you have a sheet of paper, what is the maximum number of pieces you can cut it into given n cuts. You cannot fold the sheet or rearrange already cut pieces (i.e., even after the nth cut, the pieces remain in the same position as they were before any cuts were made.)

And if you really want to geek out generalize the problem to any dimension. You have a X dimensional shape. You cut it N times in the X-1 dimension without moving or rearranging the pieces. How many pieces do you have?

I'm guessing you mean straight line cuts, right? Otherwise, I can cut a sheet of paper into as many little pieces as I want with 1 or two cuts.

 

[jonks]

Originally posted by: brikis98

5 robbers hold up a diamond store and make out with 100 diamonds, all of equal value. Instead of splitting the diamonds equally among them, they decide that they will have a bidding contest with these rules:
- Each robber draws a number out of a hat (1, 2, 3, 4 or 5)
- Starting with the smallest number, each robber will have the chance to make a proposal on how to best split the diamonds. Eg, 20-20-20-20-20 or 0-10-20-30-40 are possibile bids.
- Each robber votes on whether to accept the proposal and if there is a majority (>50% vote yes) it passes.
- If is does not pass then the robber who proposed it is killed and the process is repeated with the remaining, smaller group.
- Each robber is assumed to be equally smart and will veto any proposal when they know they can do better.

What should the first robber bid to keep themselves alive and to maximize their diamonds?

I believe you have an error in your problem. I've solved an almost identical one before with 5 pirates splitting 100 gold coins. The only difference there was that each pirate needed greater than or equal to 50% to get his vote to pass. If you only have greater than, you run into ambiguous scenarios.

For example, take the same problem, but with only two robbers, D and E, with D proposing first. If D needs greater than 50% of the vote, he knows he can't make a proposal where he gets ANY diamonds, or E will vote it down. What isn't clear, however, is if E would vote down the distribution where D gets 0 and E gets 100. There's no reason to assume E would want D's death since his desire is only to earn diamonds. However, this is very important as it can impact the final solution. That is, if D knows he will ALWAYS die in a 2 person scenario, when we discuss a 3 person scenario (C, D, E), he would vote for ANY proposal of C's as it would let him live. Otherwise, if D always lives in a 2 person scenario, but with no diamonds, C must offer D 1 diamond to secure his vote in a 3 person scenario. These same ambiguities go all the way up to the 5 person scenario and you end up with multiple possible answers.

If we solve the *proper* problem (with greater than or equal), we get:

D, E

D would propose a distribution of 100, 0 and since he needs only needs 50% of the vote to pass, his own vote pushes it through.

C, D, E

C knows that in a 2 person scenario, E gets nothing. So, he offers E one diamond for his vote, or a 99, 0, 1 distribution.

B, C, D, E

B knows that in the 3 person scenario, D gets nothing. So he offers D one diamond for his vote, or a 99, 0, 1, 0 distribution. Again, they need just 50% to pass, and this will work.

A, B, C, D, E

A knows in that in the 4 person scenario, C and E get nothing. He therefore offers them just one diamond, or a 98, 0, 1, 0, 1 distribution.

If all the robbers or pirates were really this smart they'd never agree to this distribution system since it results in a 20% of getting everything, and an 80% of getting nothing. Whoever gets lucky enough to pick the #1 out of the hat walks away with 98% of everything. What intelligent person would bother with a life of crime if that were the payoff? Smart crooks would figure out a distribution system beforehand. And the smartest one would Kaiser Soze the rest of them during the robbery walking away with 100%.

 

[Kyteland]

Right, the cut is a straight line. If the problem was expressed in the third dimension: You have a cube. You can cut this cube with a plane. If you cut it N times, what is the maximum number of pieces you can have.

 

[brikis98]

Originally posted by: QED
Ok... another oldie:

Given a standard 8-by-8 chessboard with two opposing corner squares removed and a stack
of 2-by-1 domino pieces, is it possible to cover the entire chessboard using the domino pieces
without any overlaps? Each half (a square) of each domino must cover exactly one square on the chessboard,
and each square on the chessboard must be covered. If so, how?

I don't think it can be done.

On an 8x8 chessboard, squares are (or can be) colored black and white, so no two adjacent squares are of the same color. If you do this, you'll notice that the squares on opposite corners must be the same color. So, if you removed two corners, your board has either 32 black squares and 30 white squares or 32 white squares and 30 black squares.

The problem states that each half of a domino piece covers exactly one square on the chessboard. Since domino pieces cover two squares total, and these squares are adjacent, the two squares covered MUST be of different color. Since domino pieces can't overlap or hang off the board, it then follows that there is no way to get them to cover the entire board if the number of black and white squares is not equal.

 

[CPA]

I'm not good at these, but I have one. Probably too simple, though.

Picture a chocolate cake and a cake knife. You need to cut the cake into eight pieces with only three slices with the knife. How do you do it?

 

[jonks]

Originally posted by: CPA
I'm not good at these, but I have one. Probably too simple, though.

Picture a chocolate cake and a cake knife. You need to cut the cake into eight pieces with only three slices with the knife. How do you do it?

I've got this one.

You bring 3 chefs blindfolded into the room. You tell each one that you're going to put a marble in their left hand and a coin in their right. They have to flip the coin behind their back so that it lands in the hand with the marble, freeing their right hand. The first chef to do this walks up to the cake and cuts it horizontally and vertically and then once sideways through the middle.