# A puzzle...

Bert Bates

author

Sheriff

Sheriff

Posts: 8919

11

posted 13 years ago

Actually three puzzles, rated medium, hard, really, really hard!

1) Given a balance scale and eight numbered balls, state the stategy to determine which one of the balls is either heavier or lighter than the others, given three chances to use the balance scale. Also determine whether the odd ball is heavier or lighter.

2) Same deal, with 10 balls.

3) Same deal, with 12 balls.

1) Given a balance scale and eight numbered balls, state the stategy to determine which one of the balls is either heavier or lighter than the others, given three chances to use the balance scale. Also determine whether the odd ball is heavier or lighter.

2) Same deal, with 10 balls.

3) Same deal, with 12 balls.

Spot false dilemmas now, ask me how!

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

Frank Carver

Sheriff

Posts: 6920

posted 13 years ago

Ah yes.

Strangely enough, the 12-ball version of this was used as a sort of "are you smart enough" test by my Father-in-Law. He set the puzzle both to me, and to the chap who eventually married his other daughter. We both solved it (although apparently I was quicker ), so we got the girls!

I wonder what happened to the poor guys who failed

Strangely enough, the 12-ball version of this was used as a sort of "are you smart enough" test by my Father-in-Law. He set the puzzle both to me, and to the chap who eventually married his other daughter. We both solved it (although apparently I was quicker ), so we got the girls!

I wonder what happened to the poor guys who failed

Melvin Menezes

Ranch Hand

Posts: 156

posted 13 years ago

Given three balls a,b,c if u know a is not the odd ball, you can weigh a against b and find out if

a > b, then b is the odd and is lighter

a < b, then b is the odd and is heavier

a == b, then c is the odd. to find out lighter of heavier, we have to weigh a and c once more. in this case it is a 2-step process.

Let's call this process as 3-by-1 rule or Rule31.

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

Let the balls be numbered as 1,2,3...8.

Divide them in three groups as 123 456 78

chance #1123 against 456 (keep 78 aside for now)

if they are equal, then one of the 7 or 8 is odd.

find out which by using balls 1,7,8 and Rule31

In this case Rule31 may use two chances.

chance #2weigh 1 against 7

chance #3weigh 1 against 8

(if 1 and 7 are equal, then we know 8 is the odd, but we still have to find out if 8 is lighter or heavier)

So atmost three chances.

or

if they are not equal then one of the 123456 is odd.

But we also know two more points

1. Which group 123 or 456 is ligher and which is heavier based on the shift of the balance.

2. 7 and 8 are ok.

Chance #212 against 34 (kkep 56 aside for now)

if they are equal, then one of 56 is odd.

chance#3 Use Rule31 with 1,5,6 to find out odd. We already know light or heavy

if the balance between 12 against 34 shifts the other way, then the culprit was 3. Because we moved 3 from one side to other. In this case we solved it in two chances only.

If the balance shifts the same way, then the culprit is one of 124, but 3 is ok

chance #31 against 2

if balance still shifts the same way, 1 is odd

if balance still shifts the other same way, 2 is odd

if equal, 4 is odd

Did I miss anything ?

Still trying with 10 and 12 version

a > b, then b is the odd and is lighter

a < b, then b is the odd and is heavier

a == b, then c is the odd. to find out lighter of heavier, we have to weigh a and c once more. in this case it is a 2-step process.

Let's call this process as 3-by-1 rule or Rule31.

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

Let the balls be numbered as 1,2,3...8.

Divide them in three groups as 123 456 78

chance #1123 against 456 (keep 78 aside for now)

if they are equal, then one of the 7 or 8 is odd.

find out which by using balls 1,7,8 and Rule31

In this case Rule31 may use two chances.

chance #2weigh 1 against 7

chance #3weigh 1 against 8

(if 1 and 7 are equal, then we know 8 is the odd, but we still have to find out if 8 is lighter or heavier)

So atmost three chances.

or

if they are not equal then one of the 123456 is odd.

But we also know two more points

1. Which group 123 or 456 is ligher and which is heavier based on the shift of the balance.

2. 7 and 8 are ok.

Chance #212 against 34 (kkep 56 aside for now)

if they are equal, then one of 56 is odd.

chance#3 Use Rule31 with 1,5,6 to find out odd. We already know light or heavy

if the balance between 12 against 34 shifts the other way, then the culprit was 3. Because we moved 3 from one side to other. In this case we solved it in two chances only.

If the balance shifts the same way, then the culprit is one of 124, but 3 is ok

chance #31 against 2

if balance still shifts the same way, 1 is odd

if balance still shifts the other same way, 2 is odd

if equal, 4 is odd

Did I miss anything ?

Still trying with 10 and 12 version

Melvin Menezes

Ranch Hand

Posts: 156

posted 13 years ago

For 10 balls, divide in three groups as 123 456 789A

Chance#1 as above. 123 against 456

if they are not equal, then one of 123456 is odd. Proceed as Chance #2 that we did in the 8-ball example.

If they are equal, that is if 123 == 456, then one of 789A is odd. In that case, proceed as follows

Chance #2, 123 against 789

If equal, A was odd.

Use chance #3, 1 against A to find out of A is light or heavy

If not equal, then A is ok. One of 789 is odd. And based on the shift in balance, we know whether it is heavy or light.

Chance #3, 7 aganst 8.

If the balance shifts as before (the way in 123 and 789), 8 is odd.

If the balance shifts in oppsoite direction 7 is odd.

I equal, 9 is odd.

Is that right Bert?

Chance#1 as above. 123 against 456

if they are not equal, then one of 123456 is odd. Proceed as Chance #2 that we did in the 8-ball example.

If they are equal, that is if 123 == 456, then one of 789A is odd. In that case, proceed as follows

Chance #2, 123 against 789

If equal, A was odd.

Use chance #3, 1 against A to find out of A is light or heavy

If not equal, then A is ok. One of 789 is odd. And based on the shift in balance, we know whether it is heavy or light.

Chance #3, 7 aganst 8.

If the balance shifts as before (the way in 123 and 789), 8 is odd.

If the balance shifts in oppsoite direction 7 is odd.

I equal, 9 is odd.

Is that right Bert?

Melvin Menezes

Ranch Hand

Posts: 156

posted 13 years ago

For 12 balls, divide them in three groups as 1234 5678 9abc

chance #1,1234 against 5678 (keep 9abc aside)

if 1234 == 5678, then proceed with 9abc as in 10-balls from chance #2

if 1234 != 5678, then one of 12345678 is odd and 9abc are ok.

Note the balance(B) shift.

Chance #2129 against 345 (keep 678 unknowns aside, we know abc are ok)

if the balance shifts as before, then one of 125 is odd. proceed as 3.1 below

if the balance shifts other way, 3 or 4 is odd. proceed as 3.2

if the balance is equal, then one of 678 is odd. Proceed as 3.3

Chance #3.1, weigh 1 against 2 (kkep 5 aside)

if shifts as before (that we noted when we did 1234 against 5678), then 1 is odd.

if shifts opposite (that we noted when we did 1234 against 5678), then 2 is odd.

if balanced, then 5 is odd.

We already know whether the odd ball is light or heavy based on the previously noted shift in balance(B).

Chance #3.2, use Rule31, with balls 3,4, and 9

Chance #3.3, weigh 6 against 7 (keep 8 aside)

if shifts as before (that we noted when we did 1234 against 5678), then 7 is odd.

if shifts opposite (that we noted when we did 1234 against 5678), then 6 is odd.

if balanced, then 8 is odd

We already know whether the odd ball is light or heavy based on the previously noted shift in balance(B).

You there Bert, Frank?

[ March 24, 2003: Message edited by: Melvin Menezes ]

chance #1,1234 against 5678 (keep 9abc aside)

if 1234 == 5678, then proceed with 9abc as in 10-balls from chance #2

if 1234 != 5678, then one of 12345678 is odd and 9abc are ok.

Note the balance(B) shift.

Chance #2129 against 345 (keep 678 unknowns aside, we know abc are ok)

if the balance shifts as before, then one of 125 is odd. proceed as 3.1 below

if the balance shifts other way, 3 or 4 is odd. proceed as 3.2

if the balance is equal, then one of 678 is odd. Proceed as 3.3

Chance #3.1, weigh 1 against 2 (kkep 5 aside)

if shifts as before (that we noted when we did 1234 against 5678), then 1 is odd.

if shifts opposite (that we noted when we did 1234 against 5678), then 2 is odd.

if balanced, then 5 is odd.

We already know whether the odd ball is light or heavy based on the previously noted shift in balance(B).

Chance #3.2, use Rule31, with balls 3,4, and 9

Chance #3.3, weigh 6 against 7 (keep 8 aside)

if shifts as before (that we noted when we did 1234 against 5678), then 7 is odd.

if shifts opposite (that we noted when we did 1234 against 5678), then 6 is odd.

if balanced, then 8 is odd

We already know whether the odd ball is light or heavy based on the previously noted shift in balance(B).

You there Bert, Frank?

[ March 24, 2003: Message edited by: Melvin Menezes ]

Johannes de Jong

tumbleweed

Bartender

Bartender

Posts: 5089

posted 13 years ago

Does that mean you got the # 1 daugther Frank

[ March 24, 2003: Message edited by: Johannes de Jong ]

Originally posted by Frank Carver:

Ah yes.

Strangely enough, the 12-ball version of this was used as a sort of "are you smart enough" test by my Father-in-Law. He set the puzzle both to me, and to the chap who eventually married his other daughter. We both solved it (although apparently I was quicker ), so we got the girls!

I wonder what happened to the poor guys who failed

Does that mean you got the # 1 daugther Frank

[ March 24, 2003: Message edited by: Johannes de Jong ]

Melvin Menezes

Ranch Hand

Posts: 156

Frank Carver

Sheriff

Posts: 6920

Melvin Menezes

Ranch Hand

Posts: 156

Kathy Sierra

Cowgirl and Author

Rancher

Rancher

Posts: 1589

5

posted 13 years ago

Sorry -

Just out of town - in an internet cafe... will review the answer on Saturday!!

-Bert - yes it's really Bert

Just out of town - in an internet cafe... will review the answer on Saturday!!

-Bert - yes it's really Bert

Co-Author of Head First Design Patterns

Just a Jini girl living in a J2EE world.

Bert Bates

author

Sheriff

Sheriff

Posts: 8919

11

posted 13 years ago

Melvin -

Sorry for the delay! We had to go to the Software Developer's convention - to pick up JAVARANCH's PRODUCTIVITY AWARD !!!

YEAH!

Anyway, we're back now - I looked over your answer and it seems to match with mine...

To me, the key to cracking this is the following step from your post, on weighing #2...

===================

Chance #2129 against 345 (keep 678 unknowns aside, we know abc are ok)

===================

Once you've made that breakthrough the rest falls into place pretty easily.

I'm not even going to tell you how long it took me to solve that puzzle - Melvin my hat's off to you!

-Bert

[ March 30, 2003: Message edited by: Bert Bates ]

Sorry for the delay! We had to go to the Software Developer's convention - to pick up JAVARANCH's PRODUCTIVITY AWARD !!!

YEAH!

Anyway, we're back now - I looked over your answer and it seems to match with mine...

To me, the key to cracking this is the following step from your post, on weighing #2...

===================

Chance #2129 against 345 (keep 678 unknowns aside, we know abc are ok)

===================

Once you've made that breakthrough the rest falls into place pretty easily.

I'm not even going to tell you how long it took me to solve that puzzle - Melvin my hat's off to you!

-Bert

[ March 30, 2003: Message edited by: Bert Bates ]

Spot false dilemmas now, ask me how!

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