# PE 1 - Board Games | LR - Board Games

**Answer the next 3 questions based on the information given below.**

Minesweeper is a game played on a 8 × 8 matrix. The columns are numbered A to H from left to right and rows are numbered 1 to 8 from top to bottom.

A cell in B^{th} column and 5^{th} row is termed as B5.

There is a mine in one of cells. If a player clicks on any of the cells in the same column, row or any diagonal as the cell with mine the player will lose.

The player has to click on all the safe cells (Safe cells are those cells which are neither in the same column, row or any diagonal as the cell with mine).

The game is designed such that first two clicks by a player are always safe and mine is placed accordingly after that.

**PE 1 - Board Games | LR - Board Games**

What is the difference between the maximum and minimum number of safe cells?

- A.
8

- B.
6

- C.
5

- D.
4

- E.
None of these

Answer: Option B

**Explanation** :

Maximum number of safe cells will be when the mine is placed at one of the corners.

If mine is placed at A8, cells marked with × denote unsafe cells.

∴ Number of unsafe cells = 22

∴ Number of safe cells = 64 – 22 = 42.

Minimum number of safe cells will be when the mine is placed in the middle.

If mine is placed at D4, cells marked with × denote unsafe cells.

∴ Number of unsafe cells = 28

∴ Number of safe cells = 64 – 28 = 36.

⇒ Difference between maximum and minimum safe cells = 42 – 36 = 6.

Hence, option (b).

Workspace:

**PE 1 - Board Games | LR - Board Games**

If the first 2 clicks are on A5 and G2, while the third click C3 is also safe. In how many cells mine can be placed now?

Answer: 14

**Explanation** :

Here × denotes safe cells.

Remaining cells = 14.

Mine can be placed in any of these 14 cells.

Hence, 14.

Workspace:

**PE 1 - Board Games | LR - Board Games**

If the first 2 clicks are on D4 and H7, in how many ways can 2 different mines be placed each in a different cell?

Answer: 420

**Explanation** :

Here × denotes safe cells.

∴ 2 mines can be placed on any of the remaining 21 cells.

⇒ Number of ways = ^{21}C_{2} × 2! = 420.

Hence, 210.

Workspace:

**PE 1 - Board Games | LR - Board Games**

**Answer the next 3 questions based on the information given below.**

The following questions relate to a game to be played by you and your friend. The game consists of a 4 × 4 board (see below) where each cell contains a positive integer. You and your friend make moves alternately. A move by any of the players consists of splitting the current board configuration into two equal halves and retaining one of them. In your moves you are allowed only to split the board vertically and decide to retain either the left half or the right half. Your friend, in his/her moves, can only split the board horizontally and can retain either the lower half or the upper half. After two moves by each player a single cell will remain which can no longer be split and the number in that cell will be treated as the gain (in rupees) of the person who has started the game. A sample game is shown below.

So, your gain is Re.1. With the same initial board configuration as above and assuming that you have to make the first move, answer the following questions.

- A.
(retain upper) (retain lower)

- B.
(retain lower) (retain upper)

- C.
(retain upper) (retain upper)

- D.
(retain lower) (retain lower)

Answer: Option D

**Explanation** :

The initial board is:

Since your first move is (retain right), the board configuration after the move is as shown below

Since your next move is (retain left), your final profit will be one from (2, 9, 4 and 1).

Since your friend wants to minimize your profit, he/she would like you to get Rs. 1.

Hence, he/she would retain lower in first attempt and again retain lower in second attempt.

Your gain now will be Rs. 1.

Hence, the moves are (retain lower), (retain lower).

Hence, option (d).

Workspace:

**PE 1 - Board Games | LR - Board Games**

With the initial board configuration as shown in the question, if both of you select your moves intelligently then at the end of the game, your gain will be?

- A.
Rs. 4

- B.
Rs. 3

- C.
Rs. 2

- D.
None of these

Answer: Option A

**Explanation** :

The initial board is:

Since both of you play intelligently, your objective is to get the maximum possible gain while your friend's objective is to give you the minimum possible gain.

**Case I:** You select (retain right).

Since your friend also plays intelligently, he will select (retain lower) so that your gain does not cross Rs. 6. If he selects (retain upper), your gain can possibly go up to Rs. 9

Now, you will select (retain left) to get a maximum possible gain of Rs. 6.

Finally, in order to minimise your gain, your friend will select (retain upper).

This will limit your gain to Rs. 4.

Thus, with intelligent moves, your maximum gain in this case is Rs. 4.

**Case II**: You select (retain left).

Now, if your friend selects (retain upper), the profit you can earn (both playing intelligently) is Rs. 4 when you again select (retain right).

On the other hand, if your friend selects (retain lower), the profit you can earn (both playing intelligently) is Rs. 2 when you again select (retain right).

Since your friend also plays intelligently, he will select (retain lower) so that your profit is only Rs. 2.

Since you want to maximize your gain, your first intelligent move will be (retain right) i.e., case I.

As per Case I, your gain will be Rs. 4.

Hence, option (a).

Workspace:

**PE 1 - Board Games | LR - Board Games**

In the previous question, if your first move is (retain left), then whatever moves your friend may select, you can always force a gain of no less than?

Answer: 2

**Explanation** :

Consider the solution to previous question of this set.

if your first move is (retain left), then whatever moves your friend may select, you can always force a gain of no less than Rs. 2.

Hence, 2.

Workspace:

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