In a two player game, each player removes 1 disc or two adjacent discs in alternate turns from 10 adjacent disks. Player removing last disc wins the game.

#### The Remove discs from 10 touching discs riddle

Place 10 small discs in series touching each other. In the two player game, each player is to remove *one disc or two adjacent discs* in alternate turns. *The player removing the last disc wins the game.* The first player to remove one disc in the first turn.

You have to decide whether you will take the first turn or the second to win the game.

Decide with reasons.

**Part 2:** How many ways can you win the game?

**Total time to solve:** 60 minutes.

None of the two parts is easy, but it is also not very difficult to solve if you apply systematic problem solving strategies and techniques.

### Solution to Remove discs from 10 touching discs riddle: Losing combinations, winning strategies

*The key to solving an arrangement puzzle of this nature is to identify an effective pattern that you can use for limiting the number of favorable arrangements at each turn.*

You will be at a loss to decide an arrangement as favorable for you and its corresponding move if you start from the first turn. Analysis from the beginning is further complicated as it is not yet decided who will take the first turn.

Would you try out possible (not knowing whether it will ultimately be favorable) moves turn after turn to find out how to win the game? *It will take a long time to evaluate all possible turns. The number of possible combinations is too large.*

Strategy:The most reasonable approach is to evaluate the End Conditions of winning the game with total certainty.

It is easy to figure out that in only two conditions you will win with certainty—*if at your last turn you are left with one disc or two adjacent discs.* Starting from this end point of certain win,** you will know how your opponent will certainly lose** when in the previous turn (last minus 1) he (or she) faces,

1. Two separate non-adjacent discs.

2. Or, two separate pairs of adjacent discs.

These are the **two certain losing configurations.**

A player loses facing either of the two arrangements in his turn. This certainty of loss gives you a breakthrough idea,

Strategy:You will adopt the strategy ofidentifying situations that are certain to lose. You will reject the losing combinations forthwith.

**You should be able to find all solutions,** *if you can identify losing situations in a short time and reject the combinations. The combination not rejected becomes a winning combination.*

Emphasis is on rejecting possible combinations rapidly.

So **you will first find more losing configurations** building upon the already known two losing configurations—*two separate groups of discs with one or two discs in each group.*

Next part involves using the losing patterns systematically on each of the **five unique starting choices.**

You will take up each starting choice and reject it as losing when facing a losing configuration for player 1. *If player 1 does not face any losing configuration for a starting choice, it becomes a winning choice for player 1.*

**This will be a comprehensive method for finding all possible solutions.**

*Instead of trying to find winning combinations, you will try to find a single losing combination in a chain of turns to identify the whole chain as a losing choice.* That will be the fastest way to find the solutions. This is

**Rough-edge rejection technique.**

*Find a single rough-edge and reject the possibility forthwith.*

Find a known weakness in a situation and eliminate the whole chain of possibilities swiftly. This is much faster than trying to evaluate all segments of a chain of possibilities to figure out whether it indeed is a winning chain.

#### Finding more losing disc arrangements for the current player: Remove discs from 10 touching discs riddle

The** first two losing configurations** are (1, 1) and (2, 2):

The **third easy to identify losing configuration** is four single non-adjacent discs (1, 1, 1, 1):

A black dot is an empty place between the two discs.

The **fourth losing configuration** is 1 separate single and 4 adjacent discs (1, 4):

The **fifth losing configuration** is (1, 2, 3):

Verify the losing configurations.

These 5 should be enough for the time being. If we need, we’ll form more losing patterns or weak points (or rough-edges) in a combination.

#### Identify unique starting points: Remove discs from 10 touching discs riddle

We’ll now take the bold approach of first identifying the unique starting points and moving forward from each starting point.

Let us show the ten discs with arrows showing the 5 unique starting points,

As disc 1 and 10 are equivalent as are, 2 and 9, 3 and 8, 4 and 7 and 5 and 6, you have five unique starting choices of removing disc 1, 2, 3, 4 or 5.

We’ll evaluate each of the choices

trying to prove it as losingas swiftly as possible.This is equivalent to playing the game for player 2.In the same way,

After all, both players must try to win and we will play for both taking up their roles in their turns.we’ll try to defeat the player 2 when playing as the player 1.

#### Choice of removing disc 5 in first turn as player 1

This splits the 10 disc series into two groups of 4 and 5 adjacent discs (4, 5).

Player 2 removes disc 6 in next turn to divide the discs into two groups of 4 and 4 discs (4, 4):

If Player 1 removes one disc 1, 2, 3, or 4 or any of the two discs of 1 to 4, *player 2 will remove 1 or 2 discs of the same group to present player 1 with the losing configuration (1, 4):*

If player 1 removes disc 1, player 2 will remove both inner discs 8 and 9. If player 1 removes disc 7 or disc 10, player 2 will again remove both 8 and 9. If player 1 removes disc 8 or disc 9, player 2 will remove discs 9, 10 or discs 7, 8.

So *starting choice of removing disc 5 is a losing move that results in sixth losing configuration (4, 4) for player 1.*

#### Choice of removing disc 4 in first turn as player 1

This divides the ten discs into two groups of 3 and 6 discs.

Player 2 evaluates and decides none of the six discs in the six disc group can be removed. Whichever be such a removal, player 1 will convert the combination to a (1, 2, 3) losing combination for player 2. For example, if player 2 removes disc 5 or 10, player 1 will remove discs 8,9 or 6,7.

In the same way, player 2 cannot remove disc 1 or disc 3. Player 1 will then remove discs 6,7 to present again losing configuration (1, 2, 3) to player 2.

On the other hand, if player 2 removes two discs from the three disc group, player 1 will remove two outer discs from the six disc group to present player 2 with the other losing configuration of (1, 4).

The only choice for player 2 is to remove disc 2 resulting in configuration (1, 1, 6).

But player 1 removes now discs 7,8 to convert to a combination (1, 1, 2, 2) for player 2.

For any move of player 2 now, player 1 wins from this position.

For example, if player 2 removes either disc 1 or 3, player 1 will remove disc 3 or 1 to present a losing (2, 2) configuration to player 2. If player 2 removes one of 5, 6 or 9, 10, player 1 will present player 2 with a losing (1, 1, 1, 1) configuration. If player 2 removes 5,6 or 9,10, player 1 presents a losing disc combination (1, 1) to player 2.

*Combination (1, 1, 2, 2) is the seventh losing configuration.*

For this first choice of removing disc 4 by player 1, for every move of player 2, the player 1 wins. This is the first winning choice for player 1.

#### Choice of removing disc 3 in first turn as player 1

This divides the ten discs into two groups of 2 and 7 discs.

If player 2 removes either disc 1 or disc 2, player 1 will remove disc 6,7 to hand defeat to player 2 with losing configuration of (1, 2, 3). If player 2 removes both 1,2, player 1 will remove 5,6 to defeat player 2 with losing configuration of (1, 4). **So player 2 must remove one or two discs from the group of 7 discs.**

If player 2 removes disc 4 or disc 10, player 1 will remove either discs 6,7 or discs 7,8 to defeat player 2 with losing configuration (1, 2, 3). Same fate awaits player 2 if he (or she) removes disc 5, disc 9, discs 4,5 or discs 9,10. Verify.

In case player 2 removes disc 6 or disc 8, a new configuration is created (2, 2, 4) for the turn of player 1.

But player 1 already knows that the combination (1, 1, 2, 2) will be a losing combination for player 2. So player 1 removes discs 8,9. If player 2 removes discs 5,6 (6,7 or 8,9) instead, player 1 removes disc 8 (disc 9 or disc 6) for the same winning result.

If player 2 removes disc 7 (or discs 7,8), player 1 will remove discs 5,6 (or disc 9 or 10) to defeat player 2 with losing combination (1, 2, 3).

Whichever action player 2 takes, player 1 will win. So this is the second win choice for player 1.

#### Choice of removing disc 2 in first turn as player 1

This divides the ten discs into two groups of 1 and 8 discs.

If player 2 removes disc 1, player 1 removes discs 6,7 to create a losing (3, 3) combination for player 2. This is the **eighth losing configuration for the current player.** Let us see how.

Player 1 will remove disc 8 to present losing (2, 2) combination for player 2 if player 2 removes either disc 3 or disc 5. If player 2 removes discs 3,4, player 1 will remove discs 8,9 to create losing combination (1, 1) for player 2. Instead, if player 2 removes disc 4, player 1 will remove disc 9. It will create losing combination (1, 1, 1, 1) for player 2.

**So player 2 won’t remove disc 1** and **instead will focus on 7 discs group.** If player 2 removes disc 3 (or 10), player 1 will remove discs 6,7 to create losing combination (1, 2, 3) for player 2. Same result will happen if discs 3,4 or 5,6 are removed.

*Player 1 will remove discs 7,8 (or disc 8) if player 2 removes disc 4 (or discs 4,5). This move of player 1 creates the losing combination (1, 1, 2, 2) for player 2.*

*Player 1 will remove discs 6,7 if disc 5 (or disc 8) is removed by player 2. This move creates the losing combination of (1, 2, 3) for player 2.*

*Player 1 will remove disc 1 and create losing combination (3, 3) for player 2, if player 2 removes discs 6,7.*

*Last, if disc 6 is removed by player 2, player 1 will remove discs 7,8 to create the losing combination of (1, 2, 3) for player 2.*

These are all possible moves for player 2 and in every case player 1 wins. So first choice of disc 2 by player 1 is the third winning choice by player 1.

#### Choice of removing disc 1 in first turn as player 1

This creates a single group of 9 contiguous discs.

Player 2 recognizes the opportunity and removes disc 6 **to present losing (4, 4) configuration to player 1.**

**Player 1 then wins in three cases,**

- If disc 2 is removed first,
- If disc 3 is removed first, and
- If disc 4 is removed first.

With so many opportunities, **you decide to take the first turn and win the game.**

#### Comment

The combinatorial riddle first looked to be beyond any method. It looked all solutions could only be found by long and arduous identification by trial and error of every chain from starting choice to end result.

*The strategic rough-edge rejection method with the help from analysis of end state, resolves this seemingly intractable problem.*

The method should be useful for solving similar combinatorial problems and real-life problems such as match-making.

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