Problematics | Five-way tie in race to World Cup semifinals

The men’s ODI World Cup is now at a stage when you start to calculate how many victories in nine matches are likely to get you into the semifinals. It’s a stage that comes after the first few matches in most tournaments that feature a large number of teams, including the IPL.

When 10 teams are playing, it might be tempting to assume that 6 wins will help you finish fourth if the top three teams win 9, 8 and 7 matches respectively. This reasoning, however, can be flawed. What if, for example, A defeats B defeats C?

One possibility that seems to escape many of us at such a stage is that even 7 victories in 9 matches may not be enough. It is possible for 5 teams to win 7 matches each, in which case one of them would have to miss the semifinals on account of net run rate. Below is one such combination.

#Puzzle 60.1

Although it should be pretty obvious, here’s the disclaimer to prevent any misinterpretation: the results depicted in the table above are entirely fictional and any resemblance with an actual World Cup is purely coincidental.

As in the actual ongoing World Cup, every team plays every other team: 9 matches for each, 45 in total. The results of some matches will be immediately obvious from the table. For the remaining matches, here are some conditions:

• Afghanistan defeated two teams that both defeated Sri Lanka
• Bangladesh defeated two teams that both defeated England
• Afghanistan defeated England

Can you work out the results of all 45 matches?

#Puzzle 60.2

Suppose I offered you the following challenge. I bet you ₹100 that if you gave me ₹200, I would give you ₹300 in return.

If you accepted the bet, would either of us make a profit?

Mailbox: Last week’s solvers

#Puzzle 59.1

Hi Kabir,

Say, the batter played X innings, remained not out Y times, and scored N runs.

N/(X – Y) = 47

If he had scored 5 runs more in Y innings, then N + 5Y/(X – Y) = 55

This gives 8X = 13Y

Taking X = 13 and Y= 8 gives N = 235. Assuming that the batter’s scores (other than the three 0s) were in arithmetic progression with a common difference of 1, his highest score was 28*.

— Anil Khanna, Ghaziabad

While the above was the answer I had in mind, readers have pointed out that there could also be other solutions: if the batter’s scores were not in AP but otherwise all different (other than the 0s), the highest possible score would be 190*; if not, it increases to 226*.

#Puzzle 59.2

Hi Kabir,

If we assume that the diagonal is not already shaded gray then it took me 10 “shade” operations. But then, that's no better than just going row by row or column by column.

— Amit Khanna (Fremont, California)

Indeed, 10 operations are correct under the given constraints. Some readers have said 9, but they have missed the point that you also need to grey out the diagonal that is shown in red. As Amit notes, shading 10 times is the same as going row by row or column by column. One reader, meanwhile, has pointed out something that will make it easier for me to create crossword grids in the future:

Dear Kabir,

In a “one-handed” mouse operation, the “Shade” button will need to be pressed at least 10 times. However, MS Word in its current avatar allows us to select multiple cell ranges if we hold down the “Ctrl” key. Therefore, all 10 columns or all 10 rows can be selected, with lengths as required, and then the “Shade” button needs to be pressed only once to shade the entire selection.

— Group Captain RK Shrivastava (retired), Delhi

Solved both puzzles: Anil Khanna (Ghaziabad), Amit Khanna (Fremont, California), Prof Anshul Kumar (Delhi), Dr Nakul Makkar (Noida),

Solved #Puzzle 59.1: Dr Sunita Gupta (Delhi), Sumit Malhotra (Delhi), Amar Lal Miglani (Mohali), Siddharth Nagarajan (Krea University) & Nagarajan Subramanian (Gurgaon), Yadvendra Somra (Sonipat), Group Captain RK Shrivastava (retd, Delhi)

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