A Cone and a pile of disks.

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
You have 30 circular disks which when stacked from largest to smallest form a cone. The cone of disks are sitting over a wooden pole and sitting alongside are two more poles. You decide that you will attempt to move the disks from their current position to the pole at the end without at anytime allowing a larger disk to sit on top of a smaller disk. To achieve this you will use the middle pole and the current pole when needed.

If you move one disk every second,..how long will it take to move all the disks to the end pole,..from largest at the base to smallest at the top? Give your answer to the nearest whole number along with the units,...be they minutes, hours, weeks..etc.

At no time while doing the puzzle are you allowed to place a disk on the table or anywhere else, nor to have two disks on the move at the same time. Only one disk is to be held at anyone time, and that is the one being moved.

This problem requires no Calculus, rather a logical iterative approach is called for.

Ron.
 
I thought it seemed a long time

The first one took 1 second

The second 2

The third 4

The fourth 8

and the fifth 16

I assumed (bad move perhaps) that the time doubled on each disc.

Double 1 second 29 times and you get 17.012 years (allowing 365.25 days for a year)

Am I far out?

Richard
 
Hello Richard,

Calculate the number of moves first, and then determine the time.

You are correct in that your assumption of doubling the time (or number of moves) per disk could be wrong...

17 years is a long way from the correct answer.

Ron.
 
Assuming that Richard is on the right sort-of track with the "twice the last"

Would 68 years be far out? I've got a theory... not sure it works exactly though!
 
Hello Darth,

It is not 68 years. Start by drawing a diagram and determine the number of total moves required per disk in order to move the first, second, third and so on. Write the information down and look closely at what you see and you will realise that no guessing is required.

Ron.
 
SydneyRoverP6B said:
Hello Darth,

It is not 68 years. Start by drawing a diagram and determine the number of total moves required per disk in order to move the first, second, third and so on. Write the information down and look closely at what you see and you will realise that no guessing is required.

Ron.
Click to expand...

I give up! Richard, I think that the crown is more suited on your head than mine this time! :p

I'm quite interested to know if I was on the right track or not....
 
darth sidious wrote...
I give up! Richard, I think that the crown is more suited on your head than mine this time!
Click to expand...

Hello Richard,

It looks like for the moment Darth has stood aside, so how are you travelling?

Ron.
 
SydneyRoverP6B said:
Hello Richard,

Calculate the number of moves first, and then determine the time.

You are correct in that your assumption of doubling the time (or number of moves) per disk could be wrong...

17 years is a long way from the correct answer.

Ron.
Click to expand...

You'll have to give me a clue.

I have done the first 5 moves and get - 1,2,4,8,16

The 30th disc takes 17 years (light appears above head)

Total number of moves is 1,073,741,823

So that /60/60/24/365.25 = 34.02483 years

Richard

PS I'm not going to try it however :shock:
 
quattro wrote,..
I have done the first 5 moves and get - 1,2,4,8,16

The 30th disc takes 17 years (light appears above head)

Total number of moves is 1,073,741,823

So that /60/60/24/365.25 = 34.02483 years

Richard
Click to expand...

Hello Richard,

Hmmm.....you have the correct answer, but the number of moves you have for the first 5 disks,....1, 2, 4, 8, 16 is incorrect. If you based your answer on this sequence, you won't achieve the answer you obtained. How did obtain the total of moves?

Ron.
 
I may be measuring the number of moves in a slightly different way Ron

Lets call the poles 1 (currently with cones on), 2 and 3, and the discs a,b,c,d,etc from the top.

The first move is obviously 1 where a is put onto pole 2

The second move is an extra 2 where b is put onto pole 3 and a is put on top of it

The third move is an extra 4 where c is put onto pole 2, a goes back to pole 1, b is put on top of c on pole 2, then a is put onto b on pole 2

etc giving an extra 8 moves for disc d, and 16 for disc e

It could be said that disc one took 1 move, disc 2 took 3 (1+2), disc 3 took 7, i.e. 1,3,7,15,31

I made the mistake of only counting how many moves it would take to move the last disc onto pole 3, and then all of the other discs onto it, i.e. 17 years

So you add, 1 to 2 to 4 to 8 to 16 etc and after moving the 30 discs you get 1,073,741,823

You would have to be very bored to do that :)
 
Hello Richard,

Yes indeed,...1, 3, 7, 15, 31... is the correct sequence for the number of moves required to place each disk onto the third pole in correct order.

The sequence forms a recurrence relation where each element of the sequence less the first is composed of twice the value of that element that preceeded it plus one. In other words,...if m is the number of moves, then the first requires one move,...ie m(subscipt1), the second 2m(subscript1) + 1, the third 2m(subscript 2) + 1,..., the 30th..2m(subscipt 29) + 1. In order to calculate the solution for this recurrence relation derive an equation for the sum of the k elements within the sequence. It can be seen that 2 with an exponent of k where k is the mth move and then deduct 1 will be such a solution.

So the number of moves required to move the first disk is 2^k - 1 which is 2^1 - 1 = 1. The second 2^2 - 1 = 3, the third 2^3 - 1 = 7 and it follows that the number of moves required for the 30th disk is 2^30 - 1 = 1.073741823 X 10^9.

Well done Richard.. :D

Ron.
 
SydneyRoverP6B said:
Hello Richard,

Yes indeed,...1, 3, 7, 15, 31... is the correct sequence for the number of moves required to place each disk onto the third pole in correct order.

The sequence forms a recurrence relation where each element of the sequence less the first is composed of twice the value of that element that preceeded it plus one. In other words,...if m is the number of moves, then the first requires one move,...ie m(subscipt1), the second 2m(subscript1) + 1, the third 2m(subscript 2) + 1,..., the 30th..2m(subscipt 29) + 1. In order to calculate the solution for this recurrence relation derive an equation for the sum of the k elements within the sequence. It can be seen that 2 with an exponent of k where k is the mth move and then deduct 1 will be such a solution.

So the number of moves required to move the first disk is 2^k - 1 which is 2^1 - 1 = 1. The second 2^2 - 1 = 3, the third 2^3 - 1 = 7 and it follows that the number of moves required for the 30th disk is 2^30 - 1 = 1.073741823 X 10^9.

Well done Richard.. :D

Ron.
Click to expand...

I KNEW Richard and I were on the right track! He estimated 17 years, I estimated 68. Both a factor of 2 out!

Well done though Richard! :)
 
SydneyRoverP6B said:
Hello Richard,

Yes indeed,...1, 3, 7, 15, 31... is the correct sequence for the number of moves required to place each disk onto the third pole in correct order.
Click to expand...

Yep, I understand that.

SydneyRoverP6B said:
The sequence forms a recurrence relation where each element of the sequence less the first is composed of twice the value of that element that preceeded it plus one. In other words,...if m is the number of moves, then the first requires one move,...ie m(subscipt1), the second 2m(subscript1) + 1, the third 2m(subscript 2) + 1,..., the 30th..2m(subscipt 29) + 1. In order to calculate the solution for this recurrence relation derive an equation for the sum of the k elements within the sequence. It can be seen that 2 with an exponent of k where k is the mth move and then deduct 1 will be such a solution.

So the number of moves required to move the first disk is 2^k - 1 which is 2^1 - 1 = 1. The second 2^2 - 1 = 3, the third 2^3 - 1 = 7 and it follows that the number of moves required for the 30th disk is 2^30 - 1 = 1.073741823 X 10^9.
Click to expand...

:shock: nope - not the foggiest about that bit. :?

I got a spread sheet to double '1', 29 times then add it all up.
 
darth sidious said:
I KNEW Richard and I were on the right track! He estimated 17 years, I estimated 68. Both a factor of 2 out!

Well done though Richard! :)
Click to expand...

Well done darth, it seems I doubled it 29 times and forgot to add it all up, where it looks like you did add it all up, but doubled it 30 times.

Richard
 
SydneyRoverP6B said:
Hello Richard and Darth,

As young Mr Grace would say,..."you've all done very well" :D

Ron.
Click to expand...

Thanks to both of you!

I now realise what I did wrong... I noticed the 2^n - 1.. but instead of realising that to move 30 discs would take 2^30 - 1 seconds, I for some reason thought it necessary to sum {2^n -1} from 1 to 30, which boils down to 2^31 - 32, which is roughly twice what it should have been!
 
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