[Guest]

The first puzzle i've posted... so here goes...

U have 12 mice, one of them being special. You also have an infinite supply of cakes. What is the minimum number of cakes u would use to identify the special mouse, given that-

Scenario 1: The special mouse eats slower than the rest of 'em

Scenario 2: the special one eats faster...

Scenario 3: it's not known whether the special mouse is faster or slower than the others

For all scenarios u do not have any knife or timer... just the mice and the cakes...

Have fun solving...

[Guest]

Seem a bit to me like a variation of the weighing puzzles. Intuitively I'd like to say 6 cakes in all cases.

[Guest]

The first 2 only take 5 cakes.

[Guest]

You are assuming all eat at the same speed except the special mouse. So simple. Give them all one cake each at the same time. 11 will finish together and 1 will finish either sooner or later.

If you have a problem in administering 12 cakes to 12 mice at the same time, you can do it in parts. Give 3 cakes to 3 mice and you know whether the special mouse is in this group or not. Repeat till you find your special mouse. The maximum number of cakes is still 12 but the expected value will come down to 6. Time is not optimized though and you have to do max 4 (expected value 2) iterations.

[Guest]

The first puzzle i've posted... so here goes...

U have 12 mice, one of them being special. You also have an infinite supply of cakes. What is the minimum number of cakes u would use to identify the special mouse, given that-

Scenario 1: The special mouse eats slower than the rest of 'em

Scenario 2: the special one eats faster...

Scenario 3: it's not known whether the special mouse is faster or slower than the others

For all scenarios u do not have any knife or timer... just the mice and the cakes...

Have fun solving...

Scenario 1

Separate the mice into 2 groups. Then I identify the slower of the two. Then separate it further to 2 groups, 3-3. Then give each a piece. Identify the slower group. Give each one, except one, a piece each.. If they are the same speed, it is the last one. If one is slower, it's that one. So the answer is 6.

Scenario 2

Separate the mice into 2 groups. Then I identify the faster of the two. Then separate it further to 2 groups, 3-3. Then give each a piece. Identify the faster group. Give each one, except one, a piece each.. If they are the same speed, it is the last one. If one is faster, it's that one. So the answer is 6.

Scenario 3

Separate the mice into 3 groups. Give each group, one piece to eat. The one that is different will be the one the mice is in. Then separate that group into 4 and give each, except for 1, a piece to eat each. From the first, you know that the special mouse eats faster/slower. So look out for the one that is different - faster/slower depending. If all the same, it is the last one. So the answer is 6.

Edited May 5, 2010 by Kragusification

[Guest]

You are assuming all eat at the same speed except the special mouse. So simple. Give them all one cake each at the same time. 11 will finish together and 1 will finish either sooner or later.

If you have a problem in administering 12 cakes to 12 mice at the same time, you can do it in parts. Give 3 cakes to 3 mice and you know whether the special mouse is in this group or not. Repeat till you find your special mouse. The maximum number of cakes is still 12 but the expected value will come down to 6. Time is not optimized though and you have to do max 4 (expected value 2) iterations.

Twelve is not the minimum number of cakes.

[Guest]

The first puzzle i've posted... so here goes...

U have 12 mice, one of them being special. You also have an infinite supply of cakes. What is the minimum number of cakes u would use to identify the special mouse, given that-

Scenario 1: The special mouse eats slower than the rest of 'em

Scenario 2: the special one eats faster...

Scenario 3: it's not known whether the special mouse is faster or slower than the others

For all scenarios u do not have any knife or timer... just the mice and the cakes...

Have fun solving...

It will be 11 cakes in all the above 3 scenarios

Edited May 5, 2010 by Sumit Modi

[Guest]

I'll post an explanation for my answer above in about an hour when I get a longer moment.

Edited May 5, 2010 by Tuckleton

[Guest]

The first 2 only take 5 cakes.

How do you do it with 5 cakes?

[Guest]

How do you do it with 5 cakes?

For all three:

Separate the mice into 4 groups. Have 3 groups eat a cake each while one group sits out. If one takes a different amount of time than the others, our mouse is in that group (and if it's scenario 3 we now know if he's faster or slower). If all three take the same time the mouse is in the group that sat out (and in the case of Scenario 3, if this happens go to the **'d portion).

Take the group we know the mouse is in and have 2 of the mice eat a cake each while the third sits out. If they are consumed at different rates then we know which mouse it is (since we know if the mouse is faster or slower). If they take the same time then it's the mouse who sat out. (5 cakes used)

**In scenario 3, if the first 3 cakes are eaten at the same rate then take the group of 3 that sat out and give them a cake each. Odd one out is the mouse in question. (6 cakes used)**

[Guest]

The perfect answer I believe has been provided by Tuckleton. But I have an interesting solution with five cakes for scenario 2 where the special mouse eats faster.

Answer:

You divide them to 3 groups and give each group a cake. As soon as one group finishes it's cake, grab the cake remains of the other two groups. Divide the faster group into two and give each the remains of the other groups (theoretically the same in quantity). Find out who is faster and give both mice a cake each to find out the faster thus the special one. (five cakes used)

Thinking this way lead me to a better solution where 4 cakes are only used in scenario two.

Answer:

Divide them into 4 groups and give each group one cake. As soon as one group finishes it's cake, grab the cake remains of the other three groups and give them to each individual in the faster group to find out the faster one. (only four cakes are used)

Wow!! Since I broke the record, I thought why can't I break my own record, and I did. For scenario two, ONLY THREE CAKES ARE NEEDED to find the faster mouse. I'll wait and hear from you regarding this solution and I will post it later if no one answers correctly. But I'm sure you will.

Thank you

[Guest]

For all three:

Separate the mice into 4 groups. Have 3 groups eat a cake each while one group sits out. If one takes a different amount of time than the others, our mouse is in that group (and if it's scenario 3 we now know if he's faster or slower). If all three take the same time the mouse is in the group that sat out (and in the case of Scenario 3, if this happens go to the **'d portion).

Take the group we know the mouse is in and have 2 of the mice eat a cake each while the third sits out. If they are consumed at different rates then we know which mouse it is (since we know if the mouse is faster or slower). If they take the same time then it's the mouse who sat out. (5 cakes used)

**In scenario 3, if the first 3 cakes are eaten at the same rate then take the group of 3 that sat out and give them a cake each. Odd one out is the mouse in question. (6 cakes used)**

i'm sorry Tuckleton... nice thinking, though its too much on the lines of the weighing puzzle

i'd say its not that similar to the weighing puzzle...

[Guest]

The perfect answer I believe has been provided by Tuckleton. But I have an interesting solution with five cakes for scenario 2 where the special mouse eats faster.

Answer:

You divide them to 3 groups and give each group a cake. As soon as one group finishes it's cake, grab the cake remains of the other two groups. Divide the faster group into two and give each the remains of the other groups (theoretically the same in quantity). Find out who is faster and give both mice a cake each to find out the faster thus the special one. (five cakes used)

Thinking this way lead me to a better solution where 4 cakes are only used in scenario two.

Answer:

Divide them into 4 groups and give each group one cake. As soon as one group finishes it's cake, grab the cake remains of the other three groups and give them to each individual in the faster group to find out the faster one. (only four cakes are used)

Wow!! Since I broke the record, I thought why can't I break my own record, and I did. For scenario two, ONLY THREE CAKES ARE NEEDED to find the faster mouse. I'll wait and hear from you regarding this solution and I will post it later if no one answers correctly. But I'm sure you will.

Thank you

Moze,u've got it... 4 is the ans for Scenario 2

Waiting for more replies... ^_^

i'd like to know your explanation of 3 cakes though...

Edited May 5, 2010 by Tralala!

[Guest]

Whoops! Thanks for busting open the box I was thinking in guys!

[Guest]

2 groups of 6 on a cake each. The slow group leaves a fragment behind. Now split the slow group into 3 and 3 and fill the groups to 6 with the other mice. a cake per group. The slow group leaves a fragment the same size as the other fragment. Now we have three posible mice and 2 equal size fragments. pick 2 mice and give them a fragment each. From the results we should be able to find the slow one. Anyways, I'll think about 3 some more on the way home.

[Guest]

2 groups of 6 on a cake each. The slow group leaves a fragment behind. Now split the slow group into 3 and 3 and fill the groups to 6 with the other mice. a cake per group. The slow group leaves a fragment the same size as the other fragment. Now we have three posible mice and 2 equal size fragments. pick 2 mice and give them a fragment each. From the results we should be able to find the slow one. Anyways, I'll think about 3 some more on the way home.

perfect!!!

[Guest]

I don't know how I saw the light for only 3 cakes. I guess I just got carried away while I was typing. Sorry

[Guest]

I don't know how I saw the light for only 3 cakes. I guess I just got carried away while I was typing. Sorry

:lol: :lol:

[Guest]

I think the best I can do is 5. Split them into 3 groups of 4. Give 2 groups a cake. If they eat at the same rate, our mouse is in the group that didn't get a cake. Give three of those mice a cake. Odd one out is the winner. A tie goes to the mouse that hasn't had any cake yet.

If the first 2 groups don't eat at the same rate, then take one mouse from the group that ate slower and add it to the fast group, take one mouse from the group that didn't eat and add it to the slow group and take one mouse from the fast group and add it to the group that didn't eat. We should now have 3 slightly mixed up groups. Give each group a cake.

If the fast group is still fastest then the mouse is fast and we have 3 equal fragments left and 3 possible mice to choose from. Give each a fragment and the fastest mouse wins.

If the slow group is still slowest then the mouse is slow and we have 2 equal fragments left and 3 possible mice to choose from. Give 2 of them a fragment. Slowest one wins. Tie goes to the one who didn't eat.

If the fast group is suddenly the slowest then the one we moved to that group wins. If the neutral group is fastest then the one we moved to that group wins.

[Guest]

I think the best I can do is 5. Split them into 3 groups of 4. Give 2 groups a cake. If they eat at the same rate, our mouse is in the group that didn't get a cake. Give three of those mice a cake. Odd one out is the winner. A tie goes to the mouse that hasn't had any cake yet.

If the first 2 groups don't eat at the same rate, then take one mouse from the group that ate slower and add it to the fast group, take one mouse from the group that didn't eat and add it to the slow group and take one mouse from the fast group and add it to the group that didn't eat. We should now have 3 slightly mixed up groups. Give each group a cake.

If the fast group is still fastest then the mouse is fast and we have 3 equal fragments left and 3 possible mice to choose from. Give each a fragment and the fastest mouse wins.

If the slow group is still slowest then the mouse is slow and we have 2 equal fragments left and 3 possible mice to choose from. Give 2 of them a fragment. Slowest one wins. Tie goes to the one who didn't eat.

If the fast group is suddenly the slowest then the one we moved to that group wins. If the neutral group is fastest then the one we moved to that group wins.

i have to say i myself don know the solution to Scenario 3... its been eatin my head...

but i got the answer 5 cakes through many more methods (2 more precisely)... and well the answer i know it to be 4 cakes for all scenarios.... so still thinkin....

[Guest]

Can we let the mice share the same cake and be able to measure which one is slower or faster? There are many factors gulp size, frequency etc. As posted before let 12 mice eating 12 cakes of finite size is the way to go.

Edited May 5, 2010 by sid2

[Guest]

U have 12 mice, one of them being special. You also have an infinite supply of cakes. What is the minimum number of cakes u would use to identify the special mouse, given that-

Scenario 1: The special mouse eats slower than the rest of 'em

Scenario 2: the special one eats faster...

Scenario 3: it's not known whether the special mouse is faster or slower than the others

For all scenarios u do not have any knife or timer... just the mice and the cakes...

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

Edited May 6, 2010 by Rational

[Guest]

i have to say i myself don know the solution to Scenario 3... its been eatin my head...

but i got the answer 5 cakes through many more methods (2 more precisely)... and well the answer i know it to be 4 cakes for all scenarios.... so still thinkin....

I think I'm stumped. But I'll keep thinking.

[Guest]

U have 12 mice, one of them being special. You also have an infinite supply of cakes. What is the minimum number of cakes u would use to identify the special mouse, given that-

Scenario 1: The special mouse eats slower than the rest of 'em

Scenario 2: the special one eats faster...

Scenario 3: it's not known whether the special mouse is faster or slower than the others

For all scenarios u do not have any knife or timer... just the mice and the cakes...

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

Well if u put it that way ure probably rite... but in puzzles such as these it is understood that 'minimum' implies 'the minimum to make sure' eg- even in the puzzle u mentioned u still wouldnt be sure of ur ans if u pull out 2 stones.. so the minimum would be 3... i hope what i wrote makes sense....

[Guest]

I think I'm stumped. But I'll keep thinking.

the guy i got this puzzle frm gave me some hints for Scenario 3- (tho i really don know what sum of em mean...)

1: Permutations

2: Mouse can eat from more than one cake

3: Time is the factor that helps determine the required mouse.

:blink:

hope that helps...

[plainglazed]

Put four mice on each of two cakes - 1,2,3,4 vs 5,6,7,8

If equal then the special mouse (fast or slow) is among the remaining four - 9,10,11,12

- Compare 9,10,11 vs X,X,X; if equal then 12 is fast or slow so compare it with one of the others; if fast or slow, compare 9 vs 10 like before

If 1,2,3,4 slower (or faster) than 5,6,7,8 then save the remaining morsel and compare the potentially slower group 1,2,3,4 vs 5,X,X,X

- if equal then one of 6,7,8 is fast (in this case) so compare 6 vs 7 as before with the remaining bit from the first trial

- if 1,2,3,4 is slower than 5,X,X,X then either 1,2,3,4 is slow or 5 is fast and the left over morsel should be the same as the first leftover bit.

- compare 1,2,3 vs X,X,X using one of the morsels; if same then either 4 is slow or 5 is fast; compare one of those to an X to see which

- if not the same then one of 1,2,3 is slow so compare 1 vs 2 with the last ort as before to see which is slow.

Four cakes total