Transfer a pile of 15 numbered discs from square A to square F on a board with 6 squares in minimum moves. Ensure: A disc can't be on a lower numbered disc.
The riddle: Transfer a pile of 15 numbered discs
First, make 15 small circular cardboard discs and write the digits 1, 2, 3, 4, 5.... to 15 on the center of the discs.
Then divide a cardboard or paper sheet into six square areas and mark the squares A, B, C, D, E and F.
In top row there will be squares A, B and C and in bottom row D, E and F.
Now place the disc number 15 in square A and place the other 14 numbered discs on top of it one by one to make a pile.
Make the pile in such a way that the top disc will be numbered 1 with consecutively increasing numbered discs to the bottom-most disc numbered 15.
Figure with the pile on the board is shown below.
Your job is to transfer the pile to square F with the numbered discs placed on top of each other in the same way of increasing number towards the bottom.
Transfer should be in least number of moves and in no move, you can place a disc on another with a number lower than its own number. At a time, you can move only one disc.
Time to solve 45 minutes.
Frankly, you may not need so much time if you imagine the likely moves strategically. Strategy and reasoning should help you solve the riddle quick.
Solution to Transfer a pile of 15 numbered discs riddle: Knowing a goal, understanding how to expose an intermediate disc and forming a strategy
I realize first what I must do.
Conclusion 1: At one stage, I must expose the bottom-most disc 15 as a single disc in square A and transfer it to the vacant square F.
This is like going straight to the heart of the problem. Without doing this, the riddle cannot be solved at all.
To achieve the goal, I MUST UNDERSTAND MORE ABOUT HOW TO EXPOSE AN INTERMEDIATE DISC.
First I take a simple example: how do I expose the disc number 3?
Conclusion 2: To expose a disc, I must remove all the discs above it and place them somehow in empty squares. To expose disc 3, I must remove first disc 1 and then disc 2 and place them in two empty squares as I cannot place disc 2 on disc 1.
This leads me to the important realization that,
Conclusion 3: At any stage the number of empty squares are precious for placing either single or multiple discs. I can manipulate the discs more easily with more empty squares.
This raises the question,
Question: Can I use square F in the beginning to spread out and manipulate the discs?
When I mull over this important question, I realize that nothing stops me to use all empty squares, including square F, for my disc placements TILL THE POINT I NEED TO PLACE DISC 15 IN SQUARE F.
Conclusion 4: I won't keep the square F reserved for the final pile at every stage. I will vacate it when needed.
Conclusion 5: I will use at the start all 5 empty squares for placing discs if I need.
The number of vacant squares is a precious asset. It is only 5 to create barriers for moving discs. The more empty squares I use, more flexibility I will have.
I decide my first moves with the knowledge gained till now.
I should take out first disc 1 and disc 2 and place the two say, in square F and square E. Why disc 1 in square F?
The reason forms part of a small strategy:
Strategy point: As the smallest numbered disc 1 will be removed again soon to place it on disc 2, if I place it in square F, I will make sure that square F is vacated soon enough. If I act like this at every stage, automatically I will get square F vacant when I need it.
I continue to remove the disc 3, disc 4 and disc 5 to place them in squares D, C and B.
Now I face a hurdle—I cannot move disc 6.
Every empty square is occupied. I cannot take the top disc 6 in the original pile and place it anywhere. The other single discs are numbered less than 6.
That forces my next resolve,
Conclusion 6: I should make the squares vacant one by one up to the largest number possible. So, I make a pile of 5 discs numbered 1 to 5 with increasing numbers at the bottom.
Placing 4 on 5, 3 on 4, 2 on 3 and lastly 1 on 2, I make the 5 disc pile in square B without breaking the basic rule of no disc on another disc with a lower number.
Status: A pile of 10 discs with disc 6 at the top in square A, a pile of 5 discs with disc 1 at the top in square B and squares C, D, E and F vacant.
The following figure shows the status.
What to do next? How to use the four vacant squares.
Answer is easy. I have to use the vacant squares to remove the next group of discs from the main pile to reduce its number of discs further. And I will do this way till I expose the disc 15.
I have formed the first part of my main strategy.
I will make small piles of discs in the empty squares one by one thus exposing disc 15 for transferring it to square F.
Solution to Transfer a pile of 15 numbered discs riddle: Acting on the first part of strategy to make piles of 4, 3 and 2 discs
A 5 disc pile already formed in square B.
Following the same method, I make a 4 disc pile with discs 6, 7, 8, and 9 in square C ensuring that the 3 squares D, E and F are vacant.
Now I see the steps to expose disc 15.
I make the third pile with 3 discs 10, 11 and 12 in square D.
I feel I am nearing the end of the first part of my solution, that is exposing disc 15, and stop for a moment to take stock.
In square A, I have a pile of discs 13, 14, 15.
In square B, I have a pile of discs 1, 2, 3, 4, 5.
In square C, I have a pile of discs 6, 7, 8, 9.
In square D, I have a pile of discs 10, 11, 12.
Two squares E and F are vacant. Figure below shows this.
Now place 13 in square F and 14 in square E. Transfer 13 on 14 to make a two disc pile in square E and also vacate square F.
Disc 15 is waiting exposed in square A.
Transfer disc 15 from square A to square F. Square A vacated.
This single vacant square provides breathing room for the next moves.
No unnecessary step taken yet.
Now time to make the new pile by breaking up each pile one by one and adding the discs thus spread out to the new pile in square F.
Second part of strategy: Break up smaller disc piles and add discs to new pile in square F
Transfer 13 to empty square A and move exposed 14 on 15. Move 13 on 14. A 3 disc pile formed in square F and two squares A and E fall vacant.
Following figure shows the status.
We'll now break up the 3 disc pile of 10, 11 and 12 to add the three in correct order to the new growing pile in square F.
To do it, first place 10 and 11 in the two empty squares exposing 12. The discs 10, 11 and 12 are exposed as single discs with the flexibility of moving any of the three. Status is shown.
Add 12 on 13, 11 on 12 and 10 on 11 to make a six disc pile in square F. Three squares, A, D, E are vacated. And this is what we need.
Last few steps are similar.
Break up the 4 disc pile in square C by placing 6, 7, and 8 in the three empty squares. Add these to the new pile in square F. Four vacant squares to be used for breaking up the last pile of 5 discs in square B.
To break up the 5 disc pile, place 1, 2, 3 and 4 in the 4 empty squares. Discs 1, 2, 3, 4 and 5 are all exposed.
First add disc 5 on top of 6, next 4 on top of 5, 3 on top of 4, 2 on top of 3 and 1 on top of 2 to complete the single pile of 15 discs in square F. Solution shown.
The riddle is solved, but are you sure it is solved in a minimum number of steps?
Can you solve it in fewer steps?
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