# Dice Question

Varun Khanna

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posted 13 years ago

I dropped a few (standard) dice on the floor. I did some quick addition and said, "I threw a 35."

My friend, who was sitting nearby, said, "I can arrange the dice in two groups such that both groups show the same total."

This surprised me because my friend is totally blind! I replied, "But you don't even know how many dice are there."

He said, "I don't need to."

How does he intend to arrange the dice?

My friend, who was sitting nearby, said, "I can arrange the dice in two groups such that both groups show the same total."

This surprised me because my friend is totally blind! I replied, "But you don't even know how many dice are there."

He said, "I don't need to."

How does he intend to arrange the dice?

- Varun

HS Thomas

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Varun Khanna

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posted 13 years ago

I havn't fully explored the solution here for correctness, but he takes 1/2 the dice and sets them aside. He then takes the other half and flips them over. Now both groups show the same total.

I think that this works for an even number of dice. I havn't figured out what to do if you had dropped an odd number of dice.

[ September 18, 2003: Message edited by: Joel McNary ]

I think that this works for an even number of dice. I havn't figured out what to do if you had dropped an odd number of dice.

[ September 18, 2003: Message edited by: Joel McNary ]

Piscis Babelis est parvus, flavus, et hiridicus, et est probabiliter insolitissima raritas in toto mundo.

posted 13 years ago
Piscis Babelis est parvus, flavus, et hiridicus, et est probabiliter insolitissima raritas in toto mundo.

No, never mind, it doesn't necessarily work for an even number, either:

Minimum # of dice: 6 (6, 6, 6, 6, 6, 5)

Maximum # of dice: 35 (1, 1, 1, 1,..., 1)

I'll have to think some more...maybe get out 35 dice and try it.

Minimum # of dice: 6 (6, 6, 6, 6, 6, 5)

Maximum # of dice: 35 (1, 1, 1, 1,..., 1)

I'll have to think some more...maybe get out 35 dice and try it.

HS Thomas

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Jim Yingst

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posted 13 years ago

This isn't making a lot of sense. Will our friend be arranging the dice via phone, too? Will he be able to receive any more info about the dice during this process? When we say "arange the dice", are you talking about moving them around, without changing the number they show? Or can they be turned around to show different numbers?

Let's forget about the phone, and go back to having our friend right there. Can we assume that he's allowed to touch the dics and move them around (which means that he'll also be able to eventually find out by touch how many dice there are) but that he's unable to tell by touch which number is on a side?

"Standard" dice have indented pips that would make it easy for a blind person to identify a number by touch. Shall we assume that these have perfectly smooth sides, and the number is indicated by paint and the blind man can't tell one numbe from the other?

Standard dice also have the potentially useful property that two opposite sides add up to 7. Is that true here?

When you say "arrange the dice in two groups" does that mean that (a) ever die is in one of the two groups, and (b) no die is in both groups?

When you say "show the same total", you mean you calculate the total for a group by adding up the values of the top face shown on each die in the group, right? You don't, say, arrange the dice in a line to look like "1", (orsomething similarly silly), right?

I know many of these questions may seem silly or obvious, but I styrongly suspect that somewhere, one of our "obvious" assumptions is incorrect, so I need to check them all (or, as many as possible).

Let's forget about the phone, and go back to having our friend right there. Can we assume that he's allowed to touch the dics and move them around (which means that he'll also be able to eventually find out by touch how many dice there are) but that he's unable to tell by touch which number is on a side?

"Standard" dice have indented pips that would make it easy for a blind person to identify a number by touch. Shall we assume that these have perfectly smooth sides, and the number is indicated by paint and the blind man can't tell one numbe from the other?

Standard dice also have the potentially useful property that two opposite sides add up to 7. Is that true here?

When you say "arrange the dice in two groups" does that mean that (a) ever die is in one of the two groups, and (b) no die is in both groups?

When you say "show the same total", you mean you calculate the total for a group by adding up the values of the top face shown on each die in the group, right? You don't, say, arrange the dice in a line to look like "1", (orsomething similarly silly), right?

I know many of these questions may seem silly or obvious, but I styrongly suspect that somewhere, one of our "obvious" assumptions is incorrect, so I need to check them all (or, as many as possible).

"I'm not back." - Bill Harding, *Twister*

Varun Khanna

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posted 13 years ago

Originally posted by Jim Yingst:

This isn't making a lot of sense.

Jim, I added my p.s. comment just in order to avoid any potential questions in the same lines ... anyways !!!

Can we assume that he's allowed to touch the dics and move them around (which means that he'll also be able to eventually find out by touch how many dice there are) but that he's unable to tell by touch which number is on a side?

No (see thats why I added that p.s. comment)

"Standard" dice have indented pips that would make it easy for a blind person to identify a number by touch. Shall we assume that these have perfectly smooth sides, and the number is indicated by paint and the blind man can't tell one numbe from the other?

"touch" option is eliminated already

Standard dice also have the potentially useful property that two opposite sides add up to 7. Is that true here?

This is very much true ... and is th main key to the solution.

When you say "arrange the dice in two groups" does that mean that (a) ever die is in one of the two groups, and (b) no die is in both groups?

Sorry !!! your question not clear

When you say "show the same total", you mean you calculate the total for a group by adding up the values of the top face shown on each die in the group, right?

Right !!!

I know many of these questions may seem silly or obvious, but I styrongly suspect that somewhere, one of our "obvious" assumptions is incorrect

Right

- Varun

Bert Bates

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Jim Yingst

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posted 13 years ago

Yes, but then you didn't answer my question about how the blind guy was going to arrange the dice via phone. He has to give instructions to the other guy, who has vision, right? So my question is, what type of information can be used here? I had several other questions which you didn't answer.

How about this: say you [VK] are the one who dropped dice on the floor, and I'm the guy on the phone. I may or may not be blind; I guess it doesn't matter now. Here is are my instructions for how to arrange the dice into two groups with the same sum:

First, set every die to "1". Now let N be the number of dice.. If N is even, group the dice into two groups of N/2 dice each. Congratulations, you're done. If N is odd, then put (N+1)/2 dice in one group, and (N-1)/2 dice in the second group. Take one of the dice in the second group, and change it from "1" to "2". Both groups sum to (N+1)/s.

I'm guessing that this solution is not what you meant. But what, exactly, is wrong with it?

My other questions can wait until this is answered.

**[VK]: Jim, I added my p.s. comment just in order to avoid any potential questions in the same lines ... anyways !!!**

Yes, but then you didn't answer my question about how the blind guy was going to arrange the dice via phone. He has to give instructions to the other guy, who has vision, right? So my question is, what type of information can be used here? I had several other questions which you didn't answer.

How about this: say you [VK] are the one who dropped dice on the floor, and I'm the guy on the phone. I may or may not be blind; I guess it doesn't matter now. Here is are my instructions for how to arrange the dice into two groups with the same sum:

First, set every die to "1". Now let N be the number of dice.. If N is even, group the dice into two groups of N/2 dice each. Congratulations, you're done. If N is odd, then put (N+1)/2 dice in one group, and (N-1)/2 dice in the second group. Take one of the dice in the second group, and change it from "1" to "2". Both groups sum to (N+1)/s.

I'm guessing that this solution is not what you meant. But what, exactly, is wrong with it?

My other questions can wait until this is answered.

"I'm not back." - Bill Harding, *Twister*

Jignesh Malavia

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posted 13 years ago

I guess flipping is required because :

a) The current sum is 35, an odd number, and there is no way to divide them into two equal-valued groups.

b) Varun said that two opposite sides add up to 7 is the main key to the solution.

Also, if the current sum is odd and if you flip over "all" the dice, then you get an even number and vice versa. So extending Joel's idea, I guess the first step would be to flip "all" the dice to make the sum even. After that we can think of all the possibilities probably.

Originally posted by Bert Bates:

Can any of the die be flipped to have another side up?

I guess flipping is required because :

a) The current sum is 35, an odd number, and there is no way to divide them into two equal-valued groups.

b) Varun said that two opposite sides add up to 7 is the main key to the solution.

Also, if the current sum is odd and if you flip over "all" the dice, then you get an even number and vice versa. So extending Joel's idea, I guess the first step would be to flip "all" the dice to make the sum even. After that we can think of all the possibilities probably.

Jim Yingst

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posted 13 years ago

Say you have

5, 5, 5, 5, 5, 5, 3, 2

Flip them, and you get

2, 2, 2, 2, 2, 2, 4, 5

How is this better?

Also, in general, how would you group dice into groups with equal sums, unless you're able to look at the dice to see what the values are? If the blind guy can make use of the sighted guy's vision for this, cool. But if he's allowed to do that, and he's also allowed to "flip" each die to its opposite side - why wouldn't he be allowed to move the die to any side he wants? Like 1, for example, since he's evidently able to look at the dice. So, why not just use my solution?

**So extending Joel's idea, I guess the first step would be to flip "all" the dice to make the sum even.**

Say you have

5, 5, 5, 5, 5, 5, 3, 2

Flip them, and you get

2, 2, 2, 2, 2, 2, 4, 5

How is this better?

Also, in general, how would you group dice into groups with equal sums, unless you're able to look at the dice to see what the values are? If the blind guy can make use of the sighted guy's vision for this, cool. But if he's allowed to do that, and he's also allowed to "flip" each die to its opposite side - why wouldn't he be allowed to move the die to any side he wants? Like 1, for example, since he's evidently able to look at the dice. So, why not just use my solution?

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Jignesh Malavia

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Bert Bates

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posted 13 years ago

Would it be possible at this point to totally restate a revised version of this puzzle? You know, with all the necessary corrections, additions, and modifications, so that we can take a clean whack at it?

Spot false dilemmas now, ask me how!

(If you're not on the edge, you're taking up too much room.)

Jignesh Malavia

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Jim Yingst

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posted 13 years ago

Say you have p odd dice and q even dice. If the sum is 35, p must be odd, but q may be even or odd. After you flip them, you have p even dice and q odd dice. If q is even, the new sum is even; if q is odd, the new sum is odd.

There may be solutions in which you flip only some of the dice. But if you use vision to decide which ones to flip, then why not use vision to set each die to 1? (With a single 2 if N is odd.) That seems to make the problem too easy, but I don't see where the rules would prohibit it.

[ September 18, 2003: Message edited by: Jim Yingst ]

There may be solutions in which you flip only some of the dice. But if you use vision to decide which ones to flip, then why not use vision to set each die to 1? (With a single 2 if N is odd.) That seems to make the problem too easy, but I don't see where the rules would prohibit it.

[ September 18, 2003: Message edited by: Jim Yingst ]

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Jim Yingst

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posted 13 years ago

If we can figure out what the rules actually are. So far, all the rules variations I've thought of lead to a puzzle that is either too easy, or impossible. I'm not convinced there's any real puzzle here - just confusing rules. (Or lack of rules.) But I may be missing something...

**Would it be possible at this point to totally restate a revised version of this puzzle? You know, with all the necessary corrections, additions, and modifications, so that we can take a clean whack at it?**

If we can figure out what the rules actually are. So far, all the rules variations I've thought of lead to a puzzle that is either too easy, or impossible. I'm not convinced there's any real puzzle here - just confusing rules. (Or lack of rules.) But I may be missing something...

"I'm not back." - Bill Harding, *Twister*

HS Thomas

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posted 13 years ago

Take one die and place in one group.

Take another die and place it with the number that adds to 7 in the other group.

and carry on.

--------------------------

If you have one die left over add the smallest number ,1 to one group , then flip one die in the other group to reduce it's total by one.

, if this is wrong I shall eat my blog!

Credits to Jim Yingst for this little known fact (unless you are a serial gambler).

Hope " for every complex problem there's a solution that's simple,neat and wrong" doesn't apply here!

regards

[ September 19, 2003: Message edited by: HS Thomas ]

Take another die and place it with the number that adds to 7 in the other group.

and carry on.

--------------------------

If you have one die left over add the smallest number ,1 to one group , then flip one die in the other group to reduce it's total by one.

, if this is wrong I shall eat my blog!

Credits to Jim Yingst for this little known fact (unless you are a serial gambler).

Standard dice also have the potentially useful property that two opposite sides add up to 7. Is that true here?

Hope " for every complex problem there's a solution that's simple,neat and wrong" doesn't apply here!

regards

[ September 19, 2003: Message edited by: HS Thomas ]

Varun Khanna

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Varun Khanna

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Varun Khanna

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posted 13 years ago

This is fine.You can assume my friend will just instruct me to arrange the dice in two groups showing same total.

This is perfect.But as i mentioned, my friend is ready to do the arrangement without asking me whether the number of dice are odd or even or say, any other question

Believe me you can.If you know the current sum been displayed is 35 !!!

[ September 19, 2003: Message edited by: varun Khanna ]

**[JY]: Yes, but then you didn't answer my question about how the blind guy was going to arrange the dice via phone. He has to give instructions to the other guy, who has vision, right?**

This is fine.You can assume my friend will just instruct me to arrange the dice in two groups showing same total.

[JY]:

How about this: say you [VK] are the one who dropped dice on the floor, and I'm the guy on the phone. I may or may not be blind; I guess it doesn't matter now. Here is are my instructions for how to arrange the dice into two groups with the same sum:

First, set every die to "1". Now let N be the number of dice.. If N is even, group the dice into two groups of N/2 dice each. Congratulations, you're done. If N is odd, then put (N+1)/2 dice in one group, and (N-1)/2 dice in the second group. Take one of the dice in the second group, and change it from "1" to "2". Both groups sum to (N+1)/s.

I'm guessing that this solution is not what you meant. But what, exactly, is wrong with it?

[JY]:

How about this: say you [VK] are the one who dropped dice on the floor, and I'm the guy on the phone. I may or may not be blind; I guess it doesn't matter now. Here is are my instructions for how to arrange the dice into two groups with the same sum:

First, set every die to "1". Now let N be the number of dice.. If N is even, group the dice into two groups of N/2 dice each. Congratulations, you're done. If N is odd, then put (N+1)/2 dice in one group, and (N-1)/2 dice in the second group. Take one of the dice in the second group, and change it from "1" to "2". Both groups sum to (N+1)/s.

I'm guessing that this solution is not what you meant. But what, exactly, is wrong with it?

This is perfect.But as i mentioned, my friend is ready to do the arrangement without asking me whether the number of dice are odd or even or say, any other question

[JY]:how would you group dice into groups with equal sums, unless you're able to look at the dice to see what the values are?

[JY]:how would you group dice into groups with equal sums, unless you're able to look at the dice to see what the values are?

Believe me you can.If you know the current sum been displayed is 35 !!!

[ September 19, 2003: Message edited by: varun Khanna ]

- Varun

HS Thomas

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Varun Khanna

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posted 13 years ago

Smart try but NO, as in this scenario you need to know the number of dice indirectly, right?!!

You can arrange in a way if dice count = 6 and 've to re-arange again in case count is different say 35?

moreover there is a hidden "if" in your solution ... as u mentioned "If you have one die left over ......."

the blind man never asked for the number of dice directly or indirectly ...

he knew gave the solution moment he came to know tht the total count is 35

[ September 19, 2003: Message edited by: varun Khanna ]

Originally posted by HS Thomas:

Is my answer correct ?

Smart try but NO, as in this scenario you need to know the number of dice indirectly, right?!!

You can arrange in a way if dice count = 6 and 've to re-arange again in case count is different say 35?

moreover there is a hidden "if" in your solution ... as u mentioned "If you have one die left over ......."

the blind man never asked for the number of dice directly or indirectly ...

he knew gave the solution moment he came to know tht the total count is 35

[ September 19, 2003: Message edited by: varun Khanna ]

- Varun

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Jim Yingst

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posted 13 years ago

And yet you said my own was "perfect". Yet what you complain about in HS' solution is what I did - make use of whether the number of dice is even or odd. So, should "perfect" be translated as "wrong"?

Where "do the arrangement" means "tell you how to do the arrangement", since your friend is on the phone, right? I remain confused about what constitutes a "question" here. In my instructions, I did not ask you to relay any information back to me. I expected you to use the information yourself. Evidently the number of dice is a specific bit of information which you don't think is necessary, fine. But I'm pretty sure you're going to use some info, somewhere, which is based on your own visual ability, and I haven't yet understood any good rules for determining what's OK and what's not. I suspect your rules will turn out to be arbitrary in this respect. I may have to wait until I see your solution to know this, however.

[ September 19, 2003: Message edited by: Jim Yingst ]

**Smart try but NO**

And yet you said my own was "perfect". Yet what you complain about in HS' solution is what I did - make use of whether the number of dice is even or odd. So, should "perfect" be translated as "wrong"?

**But as i mentioned, my friend is ready to do the arrangement without asking me whether the number of dice are odd or even or say, any other question**

Where "do the arrangement" means "tell you how to do the arrangement", since your friend is on the phone, right? I remain confused about what constitutes a "question" here. In my instructions, I did not ask you to relay any information back to me. I expected you to use the information yourself. Evidently the number of dice is a specific bit of information which you don't think is necessary, fine. But I'm pretty sure you're going to use some info, somewhere, which is based on your own visual ability, and I haven't yet understood any good rules for determining what's OK and what's not. I suspect your rules will turn out to be arbitrary in this respect. I may have to wait until I see your solution to know this, however.

[ September 19, 2003: Message edited by: Jim Yingst ]

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Jim Yingst

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HS Thomas

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posted 13 years ago

Hi Jim,

You are right. I thought I had laid my groupings wrong ,but didn't stop to check enough. Does look correct on first pass, though!

In your example you have 7 dice.

The groupings would be obtained as follows.

Take a die and put it one group. Put the same number in the other group

In this case , 2. Match it in the next group.

And repeat till you are complete or have one remaining die.

I presume friend will ask Varun to pick up next die, and hopefully Varun won't piss around , and offer the information that there are no more dice.

Then blind friend will tell Varun that to take the last die, turn it to 1 and add to first group , then incrementdie n the second group by 1 .

In your example the groups would be 5,2,5,1 and 5,2,6.

If Varun picks a 1 first, he has to match it in the second group.

The next lots of die will be turned to 6.

Actually this is a really boring solution. You don't really have to go to diametrically opposite lengths. Just keep matching numbers.

VK might object to it's sheer simplicity.

regards

[ September 19, 2003: Message edited by: HS Thomas ]

You are right. I thought I had laid my groupings wrong ,but didn't stop to check enough. Does look correct on first pass, though!

In your example you have 7 dice.

The groupings would be obtained as follows.

Take a die and put it one group. Put the same number in the other group

**And flip the next die to be it's diametrically opposite number.**In this case , 2. Match it in the next group.

And repeat till you are complete or have one remaining die.

I presume friend will ask Varun to pick up next die, and hopefully Varun won't piss around , and offer the information that there are no more dice.

Then blind friend will tell Varun that to take the last die, turn it to 1 and add to first group , then incrementdie n the second group by 1 .

In your example the groups would be 5,2,5,1 and 5,2,6.

If Varun picks a 1 first, he has to match it in the second group.

The next lots of die will be turned to 6.

**The matching numbers was the key to a sccessful solution, not flipping opposites**Actually this is a really boring solution. You don't really have to go to diametrically opposite lengths. Just keep matching numbers.

VK might object to it's sheer simplicity.

regards

[ September 19, 2003: Message edited by: HS Thomas ]

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Varun Khanna

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posted 13 years ago

Okey here's the solution:

The blind man's statement was ---> "Pick any 5 dices from the set of the dices, lets assume they form group A and invert (flip) all of them !!!

remaining dices, assume them to be part of group B.

Now both the groups i.e. A & B, will be showing the same total."

EXAMPLE

In a standard dice,opposite two side will adds up to 7

since the total is 35,number of dices will be more thn 5,

lets take 7 dices as example and assume the displayed numbers are -->

5,4,6,5,6,6,3 (sum= 35)

now lets form group A by picking any 5 dices (say first 5 dices)

<<Group A>>

5,4,6,5,6

lets invert(flip) Group A--->

2,3,1,2,1 (sum= 9)

<<Group B>>

6,3 (sum= 9)

I got this puzzle frm the IIT's forum and found it worth sharing ... and to be honest i cudn't crack it & only after looking into the answer i understood the beauty of the question ... but offcourse thts only my opinion !!!

For every complex problem, there is a solution that is simple, neat, perfect and wrong.

[ September 19, 2003: Message edited by: varun Khanna ]

The blind man's statement was ---> "Pick any 5 dices from the set of the dices, lets assume they form group A and invert (flip) all of them !!!

remaining dices, assume them to be part of group B.

Now both the groups i.e. A & B, will be showing the same total."

EXAMPLE

In a standard dice,opposite two side will adds up to 7

since the total is 35,number of dices will be more thn 5,

lets take 7 dices as example and assume the displayed numbers are -->

5,4,6,5,6,6,3 (sum= 35)

now lets form group A by picking any 5 dices (say first 5 dices)

<<Group A>>

5,4,6,5,6

lets invert(flip) Group A--->

2,3,1,2,1 (sum= 9)

<<Group B>>

6,3 (sum= 9)

I got this puzzle frm the IIT's forum and found it worth sharing ... and to be honest i cudn't crack it & only after looking into the answer i understood the beauty of the question ... but offcourse thts only my opinion !!!

**JY:So, should "perfect" be translated as "wrong"?**For every complex problem, there is a solution that is simple, neat, perfect and wrong.

[ September 19, 2003: Message edited by: varun Khanna ]

- Varun

HS Thomas

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posted 13 years ago

Then there's no solution. (Other than the many loopholes in teh insturctions which we found.) The given solution is only possible if the sum is a multiple of 7. Here's why:

Let:

S = initial sum of dice

x = number of dice in group A

a = initial sum of group A

a' = sum of dice in group A after being flipped

b = sum of dice in group B

Now for the first x dice let the value of each die be val(i). Initially we have

a = sum(i=1,x) val(i)

when you flip them, you get

a' = sum(i=1,x) (flip(val(i)))

= sum(i=1,x) (7 - val(i))

= 7x - sum(i = 1,x) val(i)

= 7x - a

Meanwhile, for the remaining dice in group B we have

b = S - a

So to make the two groups match, we need

a' = b

7x - a = S - a

Now the a cancels, and we get x = S/7. This means that S needs to be a multiple of 7. If it is, then take S/7 dice for group A, flip them, and they have the same sum as the remaing dice in group B.

Varun: this is a cool problem; it's unfortunate that so much time was spent on trying to understand the specs. Invoking the telephone did not really help, because it just brought up the fact that the other person on the line

The truth is, being able to identify the number of dice isn't really important to the solution, and it was an unnecessary distraction to us. As long as the blind man is unable to identify the numbers on the dice, that will be sufficient to rule out any other solutions suggested here, I think. E.g. he can't use my solution, since he can't tell which side of a die is "1". I don't think there are any other solutions, but of course I may be overlooking something.

[ September 19, 2003: Message edited by: Jim Yingst ]

**Why did he say flip 5 dice and not 4 ? you need 6 dice, at least, to get 35, but 5 to form Group B with more than two dice.**

What if the sum was 27 with 5 dice?

What if the sum was 27 with 5 dice?

Then there's no solution. (Other than the many loopholes in teh insturctions which we found.) The given solution is only possible if the sum is a multiple of 7. Here's why:

Let:

S = initial sum of dice

x = number of dice in group A

a = initial sum of group A

a' = sum of dice in group A after being flipped

b = sum of dice in group B

Now for the first x dice let the value of each die be val(i). Initially we have

a = sum(i=1,x) val(i)

when you flip them, you get

a' = sum(i=1,x) (flip(val(i)))

= sum(i=1,x) (7 - val(i))

= 7x - sum(i = 1,x) val(i)

= 7x - a

Meanwhile, for the remaining dice in group B we have

b = S - a

So to make the two groups match, we need

a' = b

7x - a = S - a

Now the a cancels, and we get x = S/7. This means that S needs to be a multiple of 7. If it is, then take S/7 dice for group A, flip them, and they have the same sum as the remaing dice in group B.

Varun: this is a cool problem; it's unfortunate that so much time was spent on trying to understand the specs. Invoking the telephone did not really help, because it just brought up the fact that the other person on the line

*does*have vision, and can actually do a lot of other things which you did not initially prohibit. So here's my version of the same puzzle:

I dropped a few (standard) dice on the floor. I did some quick addition and said, "I threw a 35." My friend, who was sitting nearby, said, "I can arrange all the dice in two separate groups such that both groups show the same total."

This surprised me because my friend is totally blind! I replied, "But you don't even know how many dice are there."

He said, "I don't need to."

I thought about it some more and said "But you can read Braile! So you can probably feel the sides of the dice with your sensitive fingers, and read the numbers on the dice that way, and use that knowledge to help you arrange the dice. That's too easy!"

He said, "No, I don't need to feel the sides. To prove it, I will wear these gloves, which are too thick for me to identify the numbers on the dice by touch, but still allow me to feel the dice and move them around however I desire."

I said, "OK, prove it," and he did. I was impressed. How did he manage this feat?

The truth is, being able to identify the number of dice isn't really important to the solution, and it was an unnecessary distraction to us. As long as the blind man is unable to identify the numbers on the dice, that will be sufficient to rule out any other solutions suggested here, I think. E.g. he can't use my solution, since he can't tell which side of a die is "1". I don't think there are any other solutions, but of course I may be overlooking something.

[ September 19, 2003: Message edited by: Jim Yingst ]

"I'm not back." - Bill Harding, *Twister*

Varun Khanna

Ranch Hand

Posts: 1400

posted 13 years ago

JIM,

after reading the question and reviewing the answer, i too raised some question and thought of some possible alteration (for my sake ) in the question ... !!

but after noticing in the forum that couple of correct solution were provided, importantly frm the ppl who read the same question, same lines,

I prefered writing -->

But still, I fully agree with most of your post answer comments.

But i won't agree with ur modified question, it should rather be -->

For every complex problem, there is a hint involving a telephone that is simple, neat, and wrong.

[ September 20, 2003: Message edited by: varun Khanna ]

after reading the question and reviewing the answer, i too raised some question and thought of some possible alteration (for my sake ) in the question ... !!

but after noticing in the forum that couple of correct solution were provided, importantly frm the ppl who read the same question, same lines,

I prefered writing -->

**and to be honest i cudn't crack it & only after looking into the answer i understood the beauty of the question**But still, I fully agree with most of your post answer comments.

But i won't agree with ur modified question, it should rather be -->

I dropped .........

............................

.......<<same as earlier>>.............

.........

He said, "I don't need to."

I thought about it some more and said "But you can read Braile! So you can probably feel the sides of the dice with your sensitive fingers, and read the numbers on the dice that way, and use that knowledge to help you arrange the dice. That's too easy!"

He said, "No, I don't need to feel the sides. To prove it, I will wear these gloves, which are too thick for me to identify the numbers on the dice by touch, but still allow me to feel the dice and move them around however I desire."

I asked whether the two groups will show the same total of 35 or any other number .. he said "any other number".

I asked can a die be member of both group and can it be flipped?

He replied "No and Yes", respectively.

I asked him to give me some hint, he said he knows opposite side of dices adds up to 7 ....

I said "So what?"

He added "He will seperate out five dices into one group and rest into other" !!!

I said, "OK, prove it," and he did. I was impressed. How did he manage this feat?

For every complex problem, there is a hint involving a telephone that is simple, neat, and wrong.

[ September 20, 2003: Message edited by: varun Khanna ]

- Varun

HS Thomas

Ranch Hand

Posts: 3404

posted 13 years ago

Varun,

As Jim pointed out this is an excellent problem.

The statement could have been put together a bit better.

The blind person and telephone

In your revised statement , "He" and "I" are a bit mixed up which adds to the confusion. A bit more thought and you'd have a killer of a problem statement ( that's not just you, I've been there and goofed around too till I realised that some value was lost in the badly stated problem ).

But thanks for giving us this problem , we are richer for that.

It's just as well Jim is here to explain why a solution for a sum of 27 doesn't work.

Thanks Jim.

regards

As Jim pointed out this is an excellent problem.

The statement could have been put together a bit better.

The blind person and telephone

**were**irrelevant to the problem. (In fact when I try it out on my friends I shall definitely leave these out.)In your revised statement , "He" and "I" are a bit mixed up which adds to the confusion. A bit more thought and you'd have a killer of a problem statement ( that's not just you, I've been there and goofed around too till I realised that some value was lost in the badly stated problem ).

But thanks for giving us this problem , we are richer for that.

It's just as well Jim is here to explain why a solution for a sum of 27 doesn't work.

Thanks Jim.

regards

Jim Yingst

Wanderer

Sheriff

Sheriff

Posts: 18671

posted 13 years ago

Well, that could be because they happened to interpret the puzzle the intended way. But it could also be because they had heard the puzzle before somewhere else.

Note that my version had some differences from the original you posted. E.g. I replaced

"I can arrange the dice in two groups"

with

"I can arrange all the dice in two separate groups"

to make it a little more certain that (a) no dice were left out, and (b) no dice were in both groups simultaneously.

Later, when I added

"...but still allow me to feel the dice and move them around however I desire"

that implies that flipping dice is a allowed, doesn't it? Heck, it implies the dice could be turned on any side, but without knowledge of the numbers, only the opposite side can be used effectively.

My goal was to avoid hints as much as possible, but still rule out the various "incorrect" solutions we had come up with. Although the hint about adding up to 7 is probably something that should be in the problem, since many people don't know that fact and the problem is pretty much impossible without it. And that's enough to get people thinking about the possibility of flipping dice, since the fact that the sum is 7 is irrelevant unless you do something involving the other side.

Really? I thought the blindness was a good way of ensuring that the person arranging the dice did not have any knowledge of the numbers on the dice. If you leave that out, how to you prevent people from offering the same sorts of solutions that we did? The

This is probably a lot easier to tell people in person, since any questions and misunerstandings can be cleared up immediately. Online is more difficult, because while we wait for clarification of things, we come up with other solutions which satisfy our current understanding of the problem, but later turn out to be incorrect.

My revision, or VK's revision? (Those pesky pronouns, always causing trouble...)

**[VK]: couple of correct solution were provided, importantly frm the ppl who read the same question, same lines**

Well, that could be because they happened to interpret the puzzle the intended way. But it could also be because they had heard the puzzle before somewhere else.

**[VK]: .......<<same as earlier>>.............**

Note that my version had some differences from the original you posted. E.g. I replaced

"I can arrange the dice in two groups"

with

"I can arrange all the dice in two separate groups"

to make it a little more certain that (a) no dice were left out, and (b) no dice were in both groups simultaneously.

Later, when I added

"...but still allow me to feel the dice and move them around however I desire"

that implies that flipping dice is a allowed, doesn't it? Heck, it implies the dice could be turned on any side, but without knowledge of the numbers, only the opposite side can be used effectively.

My goal was to avoid hints as much as possible, but still rule out the various "incorrect" solutions we had come up with. Although the hint about adding up to 7 is probably something that should be in the problem, since many people don't know that fact and the problem is pretty much impossible without it. And that's enough to get people thinking about the possibility of flipping dice, since the fact that the sum is 7 is irrelevant unless you do something involving the other side.

**[HST]: The blind person and telephone were irrelevant to the problem. (In fact when I try it out on my friends I shall definitely leave these out.)**

Really? I thought the blindness was a good way of ensuring that the person arranging the dice did not have any knowledge of the numbers on the dice. If you leave that out, how to you prevent people from offering the same sorts of solutions that we did? The

*telephone*caused confusion, I think, not the blindness. IMO of course.

This is probably a lot easier to tell people in person, since any questions and misunerstandings can be cleared up immediately. Online is more difficult, because while we wait for clarification of things, we come up with other solutions which satisfy our current understanding of the problem, but later turn out to be incorrect.

**[HST]: In your revised statement, "He" and "I" are a bit mixed up which adds to the confusion.**

My revision, or VK's revision? (Those pesky pronouns, always causing trouble...)

"I'm not back." - Bill Harding, *Twister*

HS Thomas

Ranch Hand

Posts: 3404

posted 13 years ago

[B] My version or VK's ? [\B]

Well, VK's. (But now I shall check your's. ).

Re: the blindness, I felt if it was omitted the problem-solvers would fully concentrate on the actual ham-dinger of a puzzle instead of being led astray. A sighted person who could see the numbers adding up to 35 has no advantage at all in solving it.

Well , we wouldn't want the problem-solvers to feel short-changed now , do we ? Just imagine being around a log fire Christmas time trying this problem out on festive punters.With my luck the boozy lot would probably solve it!

And Jim, you did say :

Now doesn't that translate to *blindness* not adding to the solution ?

regards

[ September 20, 2003: Message edited by: HS Thomas ]

Well, VK's. (But now I shall check your's. ).

He added "He will seperate out five dices into one group and rest into other

Re: the blindness, I felt if it was omitted the problem-solvers would fully concentrate on the actual ham-dinger of a puzzle instead of being led astray. A sighted person who could see the numbers adding up to 35 has no advantage at all in solving it.

Well , we wouldn't want the problem-solvers to feel short-changed now , do we ? Just imagine being around a log fire Christmas time trying this problem out on festive punters.With my luck the boozy lot would probably solve it!

And Jim, you did say :

The truth is, being able to identify the number of dice isn't really important to the solution, and it was an unnecessary distraction to us.

Now doesn't that translate to *blindness* not adding to the solution ?

regards

[ September 20, 2003: Message edited by: HS Thomas ]

Varun Khanna

Ranch Hand

Posts: 1400

posted 13 years ago

HS, I will def. try to convey your message to the person behind this puzzle!!!

HS, apologies as I was not sure whether you *really wanted to ask these 2 questions !!!

JY: I agree, I should have provided the hint of 7(sum of opposite sides), along with the puzzle !!!

HS: But thanks for giving us this problem , we are richer for that.

HS, I will def. try to convey your message to the person behind this puzzle!!!

HS: Why did he say flip 5 dice and not 4 ?

HS: What if the sum was 27 with 5 dice?

---------------------------------------------------------------------------

HS: It's just as well Jim is here to explain why a solution for a sum of 27 doesn't work.

Thanks Jim.

HS, apologies as I was not sure whether you *really wanted to ask these 2 questions !!!

JY: I agree, I should have provided the hint of 7(sum of opposite sides), along with the puzzle !!!

- Varun