Sat Sep 19, 2020
In this post, we will talk about all the concepts we will require while solving questions identified with games and tournaments. This subject can now and again involve some essential understanding of the games mentioned in the riddle. One should realize some common terminologies utilized by the examiners in the questions. We have to dissect the information given before attempting any questions, which could make it a brief period consuming in the event that we are curious about the concepts. Comprehensively, we have two kinds of game structures that are utilized in making the riddles. We can expect the greater part of the riddles dependent on these structures. Another common riddle type is Win-Loss balance table. We will talk about the table kind riddles in the wake of getting acquainted with the two structures.
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:
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 8C2, 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.
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.
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 2k, where k is a natural number. There are cases when the number of players is not 2k. We will discuss them later. First, we will discuss the cases where we have 2k players. Few things to remember regarding such tournaments:
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 ith 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.
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 2k, 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), 2k-1< N < 2k, 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 26 < 70 < 27. The condition for walkover or bye could be either random or based on rank of the player; it will be mentioned in the question. Let us understand this with the help of an example:
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 quarterfinal and 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.
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.
This table looks difficult to fill and involves a lot of iterations.We know the following things:
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.
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.
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