*Wednesday, May 30th, 2018*

In this post, we will discuss all the concepts we will require while solving questions related to games and tournaments. This topic can sometimes involve some basic understanding of the games mentioned in the puzzle. One should know some common terminologies used by the examiners in the questions. Games and tournaments is generally a data interpretation topic and we need to analyze the data given before attempting any questions, which could make it a little time consuming if we are not familiar with the concepts. Broadly, we have two types of game structures that are used in making the puzzles. We can expect most of the puzzles based on these structures. Another very common puzzle type is Win-Loss balance table. We will discuss the table type puzzles after getting familiar with the two structures.

**ROUND ROBIN**

Round Robin is a tournament where each team plays with every other team. Suppose there are 8 teams in a tournament then every teams will play seven matches. This is known as single round robin. When every team plays with every other team exactly once, then it is called single round robin. We all know about IPL, right. In IPL, every team plays with every other team twice. This is double round robin, where each team plays every other team two times. That means, for a tournament of eight teams, each team gets to play 14 matches. The winner (or the rankings) is decided based on all the matches. A few things to remember regarding the matches:

- Total number of matches for a single round robin is
^{N}C_{2}, where N is the total number of teams participating in the tournament. - If there is no draws, the number of wins is equal to the number of matches. It applies for number of losses as well.
- In case there are draws, total number of wins + (total draws/2) = total number of matches.

We will discuss some scenarios related to round robin questions before discussing a real situation.

- MINIMUM WINS TO QUALIFY:

*Similar to IPL (single round robin), if 4 teams qualify for the play-offs or the qualifier matches for the final, what should be the minimum number of matches a team needs to win in order to qualify.*

For a tournament of 8 teams (say Team A, B, C, D, E, F, G, H), total number of matches is ^{8}C_{2}, that is 28 matches. If the question asks minimum number of wins (say M1), you need to consider no draw case (unless mentioned). Also, minimum number of wins __DOES NOT MEAN__ that a team winning M1 matches will always qualify for the play-offs. It simply means that there is one particular case when a team winning only M1 matches (M1 can be unexpectedly small) can qualify for the play-offs. For 28 wins and 4 teams to qualify, we will maximize wins for the top 3 teams.

Teams |
Wins |

A | 7 |

B | 6 |

C | 5 |

D | 2 |

E | 2 |

F | 2 |

G | 2 |

H | 2 |

To maximize the number of wins for the top three teams, the top team (A) wins all the matches, i.e. 7, then the next team (B) will win 6 matches (the only match the second team lost is to the top team) and the third team (C) will win 5 matches. Summing up the wins for top 3 teams, it comes out to be 18 with only 10 wins remaining for the rest of the 5 teams. So, we distribute each team with 2 wins. The team D will qualify with the help of net run rate or some other parameter. In case of a football tournament, the parameter can be the number of goals. Similarly, this parameter will vary with different games. The answer for M1 is 2 in this case.

- MAXIMUM NUMBER OF WINS EVEN AFTER WHICH A TEAM CAN GO OUT OF THE TOURNAMENT

Like the previous case, we will consider no match drawn case. The team that will not qualify will be the fifth team (E). All the top 5 teams will have equal wins but the fifth team will lose out on the net run rate (or some other parameter mentioned). Top three teams win against the bottom three teams. Total 15 wins out of 28 are distributed. We are remaining with 13 wins to distribute. Now, each of top 5 teams have matches with each other as well, four matches for each team. Total matches among top 5 teams is^{5}C_{2}, i.e. 10. 10 matches, 10 wins, we distribute all the teams with 2 wins each. Hence, all the top 5 teams win 5 matches each but due to net run rate the fifth team will lose a place in the qualifiers.

Another approach can be to distribute minimum wins from the bottom. Let us say H is the last team and it lost all the matches. It has 0 wins. G will win only one match against H and lose other matches. I will win 2 matches against G and H and lose other matches. Out of 28 wins, 3 wins distributed and 25 remaining. We can distribute 25 wins equally among the top 5 teams. In case the division in not equal, distribute extra wins to the top team, then to the second team and so on.

- MINIMUM WINS TO ENSURE A PLACE IN PLAY-OFFS

From the previous case, we can say even at 5 wins a team can lose a place in the play-offs. In order to qualify, a team must win at least 6 matches (with regards to the case we are discussing) to ensure a place in the play-offs. This case is different from the first case as in the first case, it is just one possibility whereas third case is always valid, it provide a guaranteed place in the playoffs.

**KNOCKOUT TOURNAMENTS**

Knockout tournament is another very commonly asked structure in CAT.Every game is a knockout, which means the winner must have won all the matches. A knockout tournament has typically a number of rounds and in each round, a certain number of players are knocked out of the tournament. Tennis is one particular game, which is frequently asked in CAT where knockout tournament is concerned.

The total number of players are generally 2^{k}, where k is a natural number. There are cases when the number of players is not 2^{k}. We will discuss them later. First, we will discuss the cases where we have 2^{k} players. Few things to remember regarding such tournaments:

- Total number of matches = Total number of players – 1.
- Knockout tournaments will generally have seeded (ranked) players.
- Knockout tournaments will not have a tie. There will always be a tiebreaker. It is knockout after all.
- Upset means a lower seeded player defeat upper seeded player.
- The total number of rounds is k (from 2
^{k}) and k^{th}round is the final round. - First seeded player (rank 1) will play with last seeded player, rank 2 with second last seeded and so on.

**Question: **Let us take a case of 64 players. We know 2^{6}is 64; therefore, there are 6 rounds. If player seeded 2 reached semi-final, then who could be his likely opponent.

**Approach:** First, we should know the number of possibilities. Semi-final is the fifth round that means any player in fifth round has won four matches. Let us say i^{th} seeded player is second seeded player’s opponent. Every time we will expand the value for i by subtracting 2 (initially for rank 2) from total players in round + 1. Let us form a chart to make the things easier. The chart shows the player second seeded player might have defeated(or played against) in different rounds.

ROUND |
1 (64 players) | 2 (32 Players) | 3 (16 P) | 4 (8P) | 5 ( SF- 4P) |

PLAYER DEFEATED BY RANK 2 |
(Total player + 1)- 2
(64+1) – 2 = |
Winner of match 31 of R1 33 – 2 = 3165 – 31 = 34 |
17 -2 =1533-15 = 1865-15 = 5065-18 = 47 |
9-2=717-7= 1033-7=2633-10= 2365-7= 5865-10= 5565-26= 3965-23= 42 |
5-2=39-3= 617-3= 1417-6= 1133-3= 3033-6= 2733-14= 1933-11= 2265-3= 6265-6=5965-14=5165-11= 5465-30= 3565-27= 3865-19= 4665-22= 43 |

The bold number in the tables indicate the possibilities of opponents for the second seeded player in that particular round. Semi-final has 16 possibilities or 2^{(round number -1)} possibilities. The method is simple as we can see from the table. We subtract the rank from total number of players in that round +1. Next, the value obtained is minus from total players + 1 in previous round (say in R3, the previous round is R2). All the rank values obtained after this step is subtracted from the total players + 1 of the round before R2 (i.e. R1) and so on.

In case of no upsets, we will always take the lowest rank from the set of values, or just subtract from total players + 1 for that particular round.

In case of upset, you should try to find out the rank of the player, the higher seeded player would have played against.

If the number of player is not 2^{k}, where k is a natural number, then there will be walkovers. Remember, at the end we need a final match and it will require just two players. So, at some point in the tournament we will encounter an odd number which will make it difficult to have matches for all the players. Suppose, at some point there are 25 players, then we can have only 12 matches and player will get a walkover to next round or he/she might be knocked out at random. In any case, we can only have even number of players at every round. In such cases to find out the number of rounds (say k rounds), **2 ^{k-1}< N < 2^{k}, **where N is the number of players at the start of the tournament.For example, if we have 70 players, then the number of rounds will be 7 as 2

**Question: **75 players participated in a tennis tournament, where every match is a knockout. If there are odd number of players in any round, then the top seeded player in that round gets a walkover to next round.

(i) If there are no walkovers from round 2 onwards, then how many matches are played in round 1.

(ii) If top seeded players of quarterfinaland semi-final are defeated in the upsets, what is the sum of the ranks of the players reaching the finals?Assume only one walkover per round.

Solution:

- Since round 2 onwards there are no walkovers, we need to make number of players 2
^{k}in round 2 itself. The 2^{k}nearest to 75 and smaller than 75 is 64. Hence, we need to have 64 players in round 2. 75-64 = 11 players are knocked out in round 1, which means 11 matches have been played in round 1. This also means that 75-22 (11 matches between 22 players) = 53 players were granted a walkover to round 2. - Since, the only two upsets mentioned are in quarterfinal and semi-final, top seeded player (rank 1) must have reached quarterfinal. Top seeded player is defeated in quarterfinal. Only 1 walkover allowed in every round to the top seeded player of that round, that top seeded player will always be rank 1 till quarterfinal. We need to know the number of player in every round.

Round | 1 | 2 | 3 | 4 | 5 (Q-F) | 6 (S-F) | 7 (final) |

No. of Players | 75 | 38 | 19 | 10 | 5 | 3 | 2 |

Walkover granted | 1 | 0 | 1 | 0 | 1 | 1 | 0 |

We see that in quarterfinal, a walkover is granted. Since rank 1 is top-most seeded and is the one eligible for a walkover, the next top seeded player will definitely be rank 2. Rank 2 is defeated in Q-F. Again, in semifinal a player is granted a walkover, which will again be rank 1. Hence, rank 1 reached the final. Rank 2 is defeated in Q-F and since no other upsets in previous rounds are mentioned, rank 2 must have played against rank 5 in Q-F and rank 3 against rank 4 (rank 4 defeated rank 3 as top seeded players lose in QF and SF). In semifinal, rank 5 defeated rank 3. Final match is rank 1 vs rank 5. Sum of the ranks is 6.

We see that the question asked is tricky and can be confusing at times. One needs to be clear with the terms used and have a understanding of the structure in the question. Otherwise, you might spend unnecessary time figuring out the structure itself.

- We talked about the two structures. Now, we will quickly discuss the win-loss table. The table is based mostly on either of the two structures, sometimes with little more information. Consider the following table:

** **

The following table shows the points scored by the teams in a cricket tournament. A win is rewarded with 2 points. In case of a tie, each team gets 1 point. Each team plays with every other team exactly once in the tournament. Only two teams have same points.Two pairs have same number of wins and one pair has same number of losses. Some of the entries in the table are missing:

STANDING |
TEAM |
WIN |
LOSS |
DRAW |
POINTS |

1 | DELHI | 1 | |||

2 | KOLKATA | 0 | 6 | ||

3 | CHENNAI | 2 | |||

4 | PUNE | 2 | |||

5 | MUMBAI | 3 | |||

6 | PUNJAB | 4 | 2 |

This table looks difficult to fill and involves a lot of iterations.We know the following things:

- 6 teams, total matches are
^{6}C_{2}= 15. - Total wins + (total draw/2) = 15
- Each team played 5 matches

From the table, we see that Punjab has 2 points. 2 points are possible either by 1 win or by 2 ties and Punjab had already lost 4 matches, therefore, only possibility is a win. Punjab 1 W, 4 L, 0 D. Kolkata has 6 points with 0 draw. Hence, 3 W for Kolkata and remaining 2 matches were lost. We know that two teams have same score. The three possibilities are Kolkata and Chennai or Chennai and Pune or Pune and Mumbai. WHY? Delhi has one draw and hence will have odd points. Mumbai and Punjab is not possible.

Let us start with lower one. If Pune and Mumbai have same points, Mumbai should have 2 wins and 4 points. Pune will be 2 W, 3 L, 0 D, 4 points. Chennai would therefore be with 5 points. As Chennai already has 2 W, for 5 points it needs a draw. So Chennai is finally 2 W, 2 L, 1 D, 5 points. Total wins till now = 3 (Kolkata) + 2 (Chennai) + 2 (Pune) + 2(Mumbai) + 1(Punjab) = 10. We have 1 draw match and 4 wins remaining. All 4 are won by Delhi. We missed out the condition that two pairs have same number of wins. Hence, our assumption that Pune and Mumbai have same points is wrong.

If Chennai and Pune have same points then, Chennai – 2 W, 3 L, 0 D; Pune – 2 W, 3 L, 0 D; Mumbai – 1 W, 3 L, 1 D; Delhi will have 5 wins and 1 draw, which is impossible. Another possibility is, , Chennai – 2 W, 2 L, 1 D; Pune – 2 W, 2 L, 1 D; Mumbai – 1 W, 3 L, 1 D. Here, we have three teams with same number of losses (Kolkata, Chennai, and Pune) but as per the question only one pair (two teams) has same number of losses.

Finally, we try to balance Kolkata and Chennai for same points. We get, Kolkata – 3 W, 2 L, 0 D; Chennai – 2 W, 1 L, 2 D; Pune – 2 W, 3 L, 0 D; Mumbai – 1 W, 3 L, 1 D; Delhi 4 W, 0 L, 1 D. We got two teams with same points, two pairs with same number of wins and one pair with same number of losses. The procedure involved much iteration which makes it time consuming.

STANDING |
TEAM |
WIN |
LOSS |
DRAW |
POINTS |

1 | DELHI | 4 | 0 | 1 | 9 |

2 | KOLKATA | 3 | 2 | 0 | 6 |

3 | CHENNAI | 2 | 1 | 2 | 6 |

4 | PUNE | 2 | 3 | 0 | 4 |

5 | MUMBAI | 1 | 3 | 1 | 3 |

6 | PUNJAB | 1 | 4 | 0 | 2 |

Win- loss table is sometimes time consuming and should be practiced thoroughly. Games like race events can be solved with linear arrangement concepts. Try solving more and more tables to get acquainted with the basic details used in the table.

Games and Tournaments for CAT Exam Logical Reasoning – Part 1

Games and Tournaments for CAT Exam Logical Reasoning – Part 2

How to prepare for LRDI

How to prepare for Logical Reasoning

How to Prepare for Data Interpretation

How to solve Family Tree, Relationships problems based on coded symbols

All questions from CAT Exam Logical Reasoning and Data Interpretation

Logical Reasoning – Set 1: A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands.

Logical Reasoning – Set 2: There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular skilled employees (RE).

Logical Reasoning – Set 3: Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are going to Delhi from Kolkata by a flight operated by Cheap Air.

Logical Reasoning – Set 4: In a square layout of size 5m × 5m, 25 equal sized square platforms of different heights are built.

Logical Reasoning – Set 5: Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N).

Data Interpretation – Set 1: Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET).

Data Interpretation – Set 2: An old woman had the following assets: (a) Rs. 70 lakh in bank deposits

Data Interpretation – Set 3: There were seven elective courses – El to E7 – running in a specific term in a college.

Data Interpretation – Set 4: Funky Pizzaria was required to supply pizzas to three different parties.

Data Interpretation – Set 5: At a management school, the oldest 10 dorms, numbered 1 to 10, need to be repaired urgently, The following diagram represents the estimated repair costs (in Rs. Crores) for the 10 dorms.

How to prepare for LRDI (Logical Reasoning and Data Interpretation) for the CAT 2018 Exam?

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Why Can’t we have odd points for Chennai , Pune , Mumbai.So that Delhi can form pair with them.

‘C’stands for??

If u keep chennai points to be 5 Pune won’t have total of 5 matches.

And, if u keep Chennai points to be 7 conditions won’t match.