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Problematics | The perfect draw at a Grand Slam

Sep 02, 2024 07:54 AM IST

How do you make a draw such that the top four players head for the semifinals and the top two for the final?

Someone I know expressed disappointment recently that two of her favourite players, Karolina Muchova and Naomi Osaka, were facing off in the second round of the women’s singles at the 2024 US Open. One of her favourites, she noted with dismay, would be knocked out early in the tournament. That loser turned out to be Osaka who, although a former champion, has a rank of 88 at present, well below Muchova who is ranked 52.

For representational purposes only.
For representational purposes only.

At such low ranks, neither Muchova nor Osaka was seeded going into the tournament. The US Open has only 32 seeds in the women’s singles, and they are given matches in accordance with those seedings. At least, this is the way it is supposed to be. It works on assumptions, of course. If any seeded player faces an unseeded one, it is assumed that the seed will win. Or, if two seeded players face off, the higher seed is expected to win.

Based on such assumptions, the draw is made so that you would expect the top two seeds to meet in the final. Before that, the top four seeds would have played in the semifinals, in which the top two would have defeated numbers three and four. And so on.

The US Open women’s singles had 128 players with 32 seeds, just like imaginary tournament in the following puzzle. The tournament is at a stage when all 32 seeds have played well enough to reach the round of 32; just as expected, they have knocked out all unseeded players along the way. Let’s take off from that stage.

#Puzzle 106.1

Solution to Puzzle 106.1.
Solution to Puzzle 106.1.

The letters in the illustration stand for the 32 players, whom we shall keep anonymous. Since we need 32 letters and there are only 26 letters in the alphabet, I had to use uppercase and lowercase letters, with each standing for a different player.

At any stage, a higher seed will get an opponent who will be “easier” to defeat ((in theory) than the opponent of a lower seed. For example, if seeds #1, #2, #3 and #4 remain in the contest, seed #1’s opponent must be “easier” than seed #2’s opponent, and so seed #1 play against #4 while seed #2 will play against #3. The same reasoning will apply when there are 8, 16 or 32 players remaining.

In a theoretically perfect draw, you expect every higher seed to defeat whichever lower seed she plays against. From the given 32 players, the top 16 seeds are expected to go into the next round by defeating the lower 16 players. After the round of 16, only the top 8 must remain, and then the top 4, until you have the top two seeds playing in the final.

Reproducing the above illustration as neatly as you can manage, can you substitute the seedings 1-32 for the given letters so that you have such a perfect draw?

Puzzle #106.2

Ten volumes of an encyclopaedia set are placed upright in a row on a shelf, their spines facing you, with volumes 1-10 in order from left to right. Each volume contains 1000 pages written in simple English, with page 1 coming immediately after the front cover (no introduction, foreword etc) and page 1000 coming immediately before the back cover (no index, bibliography etc).

A worm starts boring a hole on page 1 of Volume 1 and continues digging through the set, page by page, volume by volume, until it emerges from page 1000 of Volume 10.

Excluding covers, how many pages in total has the worm eaten through?

 

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 105.1

 

Solution to Puzzle 105.1.
Solution to Puzzle 105.1.

Hi Kabir,

Say the smallest date inside the selected section is x. The sum of these nine numbers is = 9 (x + 8).

The arithmetic works as follows. Sum of the numbers in each row:

Row #1 = (x + 1) × 3 = 3x + 3

Row #2 = (x + 8) × 3 = 3x + 24

Row #3 = (x + 15) × 3 = 3x + 45

Sum of all the above = 9x + 72 = 9 (x + 8)

So the trick is to add 8 to the smallest number and multiply the sum by 9.

— Shishir Gupta, Indore

 

#Puzzle 105.2

 

Solution to Puzzle 105.2.
Solution to Puzzle 105.2.

Hi Kabir,

The photograph shows the arrangement of the cards. The 3 on the left will be the topmost card.

— Aditya Krishnan, Coimbatore

***

Hi Kabir,

The order of the cards is: 3-8-7-A-Q-6-4-2-J-K-10-9-5. I play the smart alec myself and demonstrate this card trick to younger members of the family who have learnt to spell the numbers. The “showmanship” I do is to lay out the cards while telling the following story: “387 years ago, there lived a (A) Queen who had 642 Jacks and a King, of whom 10 were blind and 95 were deaf.” And then I collect the cards in that order and start the spelling and revealing routine.

— Akshay Bakhai, Mumbai

Solved both puzzles: Shishir Gupta (Indore), Aditya Krishnan (Coimbatore), Akshay Bakhai (Mumbai), Anil Khanna (Ghaziabad), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Hiten Jindal (Delhi), Shri Ram Aggarwal (Delhi), Sanjay Gupta (Delhi), Partha Bhattacharya (Bengaluru), Sabornee Jana (Mumbai), Ajay Ashok (Delhi), Professor Anshul Kumar (Delhi), YK Munjal (Delhi), Harshit Arora (IIT Delhi), Kanwarjit Singh (Chief Commissioner of Income-Tax, retired), Sampath Kumar V (Coimbatore)

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.

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