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Tuesday, July 27, 2021

Problematics: The Beti Badhao formula

A village has a scheme to improve its child sex ratio, but will it work?

Written by Kabir Firaque | New Delhi |
Updated: June 3, 2015 3:51:16 pm
Problematics, math puzzles, beti bachao, beti badhao The structure of families of various sizes, as foreseen by the village headman. (Source: Express Illustration)

The first of this week’s puzzles is my adaptation of an old chestnut, which should still be new to most readers because the original has never been in wide circulation as far as I can make out. So come visit a village that is looking at ways to improve its child sex ratio (girls per 1,000 boys in age group 0-6).

“I have a family planning scheme,” the village headman tells the schoolmaster. “If all healthy couples follow it, which they have promised to, we should be able to populate our village with more new girls than new boys.”


Under his scheme, no couple will have more than one son in future. If a couple get a son, they stop having any more children. If their first child is a daughter, they look for a second child; if the second too is a daughter, they look for a third. And so on, as long as all the children are daughters. The moment a son arrives, he’s the last child.

“That way, some couples will have a son as their only child,” the headman admits. “But there will also be families with many daughters and just one son. That should improve our poor sex ratio, shouldn’t it?” he asks the schoolmaster, who teaches mathematics.

The teacher is too smart to commit himself. But he has a brainwave.

“Let’s offer it to Problematics.” he suggests.

Puzzle#13A: And so they did. Will it work?

What you wrote

Last week, we discussed the game of Hex, co-invented by John Nash and Piet Hein, the latter also the creator of the puzzle you solved.

Problematics 2

Dear Kabir — If White picks 5 in the next move, whatever remaining picks Black has, Black will fail everywhere. Thus 5 is a sure win for White.

Option 1: Black takes 8. White should take 3. White could now win by picking either 4 or 7 in the next move. Black can only block one.

Option 2: Black takes 7. White should take 4. White could now win by picking either 3 or 8 in the next move. Black can only block one.

Option 3: Black takes 4. White should take 7. White could now win by picking either 3 or 8 in the next move. Black can only block one.

Option 4: Black takes 3. White should take 8. White could now win by picking either 4 or 7 in the next move. Black can only block one.

Option 5: Black takes 1. Here there is no impending disaster for White to be tacked in the next move. However, if White moves 4, White could now win by picking either 3 or 8 in its next move. Black can only block one.
Jaysun Antony Alumkal (Coordinator, Academic Committee, IIM Raipur)

Brothers Sampath Kumar V (IIM Kozhikode alumnus) and Guhanesh Kishore (Viscom grad) have an alternative to one of Jaysun’s options. “If Black plays 1,” they suggest, “White should play 7.” That would open up two routes, 3-5-7 and 7-8-9. The only others to send a detailed analysis are Biren Parmar (PhD student, Texas A&M University) and S Sreenvas (IIM Raipur, PGP 2014-16), who notes, “Interestingly, White will have won just by trying to block black’s winning chains.”

Right, Sreenvas, it was something noted by Nash himself. “When the board is filled one or the other of the players will have connected but not both,” he wrote to author-columnist Martin Gardner in 1957.

Sanjay Gupta (New Delhi) sends a broad strategy without going step by step. Anwesa Banerjee (MSc student, IIT Kharagpur) sends an incomplete strategy. And Harsha T R (NIT Karnataka, batch of 2012) writes, “I went through all the possible moves. It was just too much type out everything, so didn’t want to.”

For the easy Scrabble problem (Puzzle#12B), the answer, as everyone found out, was TWELVE = 1 + 4+ 1+ 1+ 4 + 1 = 12. All those named above solved it. So did Tushar Menon (masters student, NTU Singapore), Neha Jain and Sathya Prakash (software developer, New Jersey).

The schoolmaster’s turn

The teacher’s scheme is to admit more girls than boys without compromising on Education or All. He will admit children in sequence, one by one, according to a pattern. No boy will be admitted exactly two places after another boy, no girl will be admitted exactly three places after another girl. That way, he figures, he can have a sequence of three girls as compared to a sequence of only two boys.

Puzzle#13B: Assume that the school takes in the maximum number of pupils possible with that pattern, what will the sequence be? And will it fulfil the objective of admitting more girls than boys?

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