Warm up for some IPL mathematics with 11 batsmen called Rohit, Dhawan, Kohli, Rahane, Raina, Dhoni, Jadeja, Ashwin, Shami, Mohit and Umesh. Send them out in that order. Starc clean bowls six batsmen in his first over, then Hazlewood does the same to four batsmen with four deliveries. Team all out 0 in 1.4 overs. Who’s the not out batsman?
Yes, it’s Ashwin, who escaped without facing a ball when the bowling ends changed from the first over (odd) to the second (even). What if Starc had bowled only five batsmen and conceded a single off delivery no. 6, followed by Hazlewood clean bowling five in five? The not out batsman then would have been Dhawan, having migrated from the even batting end to odd when the single was run (by Jadeja).
For more fun with odd and even parity, give the same team 53 for 1 in 11.4 overs, with Dhawan the batsman out. If the extras are 2 wides, 1 no-ball and 2 leg-byes, who’s facing delivery no. 5 of this over?
If you remove the wides and the no-balls, you have 50 runs scored between the wickets or through boundaries. For odd and even parity, hitting a four or a six works the same as running an even number between the wickets. Since 50 is even, the original parity remains, with Rohit facing odd overs and Dhawan (or his replacement) facing even overs. And since the 12th is an even over, the batsman facing is Dhawan-replaced-by-Kohli. Had the score minus wides and no-balls been odd, the original parity would have been reversed, with Rohit facing the even over.
At times, however, odds and evens can lead you astray. Take 20 for no loss in the fifth over, an even score with no extras. Rohit, facing the odd over, calls for a single and runs himself out at the non-striker’s end. Now it’s 20 for 1, still even, but the parity equations have swung the other way, for it’s now Dhawan who’s at the striker’s end, facing an odd over, while Rohit’s replacement has switched to even.
Puzzle#10A (i): Still in the fifth over, Dhawan is caught and it’s 20 for 2. Can that dismissal restore the original parity by bringing Rohit’s replacement to the striker’s end (odd) and Dhawan’s replacement to the even end? You should have no trouble with that one.
Enough of fictional situations, though. The following scorecard is fact; it’s from actual play in the current edition of the IPL.
Batsman 1 c Point Fielder b Bowler 1 1 (2 balls)
Batsman 2 batting 0 (0 balls)
Total 1 for 1 (0.2 overs)
Fall of wickets: 1-1 (0.2)
Puzzle#10A (ii): In the true story above, the first batsman faced both deliveries bowled till then. The second ball got him. What happened on the first ball? If you think he took a single, why wasn’t his partner facing the second ball?
Puzzle#10B: A calculating captain wants his strike bowler to close an ODI innings, so that the 50th over is the bowler’s 10th and last. If the bowler has bowled x overs in earlier spells, the captain should bring him back at over no. ____?
What you wrote
A couple of readers wrote to say Puzzle#9A of last week was too easy. On the other hand, it drew in a couple of new readers.
Dear Mr Firaque, here are my answers to Puzzle#9A. How did I do?
(i) Adult males past expiry cannot provide a narrative = Dead men tells no tales.
(ii) The person releasing the terminal cachinnation achieves the most audible cachinnation = He who laughs last laughs the loudest.
(iii) Let inspection precede your escalation = Look before you leap.
(iv) It is futile to attempt inculcating in a senile canine the ability to perform modern gimmicks = You can’t teach an old dog new tricks.
(v) Selective tactics are beyond the means of mendicants = Beggars can’t be choosers.
(vi) Success is prejudiced towards the intrepid = Luck favours the brave.
(vii) The smallest prime number of inaccuracies does not construct an angle of 90 degrees = Two wrongs don’t make a right.
(viii) A visual representation is equal in value to a kilo of etymological units = A picture is worth a thousand words.
(ix) Not a single masculine human is a landmass surrounded by a potable fluid = No man is an island.
(x) When progress acquires hardness, the hard achieve progress = When the going gets tough, the tough get going.
Thank you for your column. I’ll be following it now.
Anindita Basu (editor with IBM India, Allahabad)
It’s “fortune” that favours the brave, Anindita, but I don’t see a problem with “luck” either if other readers don’t. Solved also by Biren Parmar (Ph D student, Texas A&M University), Harsha T R (NIT Karnataka, batch of 2012), Rishi Iyengar (reporter, Time, Hong Kong), Sampath Kumar V (IIM Kozhikode alumnus), Bindia George (IT engineer, Kochi) and Sanjay Gupta (New Delhi). Sanjay misquotes one proverb as “Dead men don’t talk”, though, while more than one reader writes that “He who laughs last laughs best”, although the correct proverb I think is “laughs the loudest”.
I found Bindia’s solution too complex to draw with my tools. “Each triangle ends of the star is made of 10 dots. The inside of the star will be a mirror image of the triangles plus a dot at centre. Hence total will be 10×6×2 + 1 =121,” Bindia writes.
The simplest way is by breaking up the board into one large triangle of side 13 (slots = 13×14/2 = 91) and three smaller triangles of 10 slots each. Biren, Harsha (whose initial solution matched Sanjay’s) and Sampath got there on their second attempts.
Please mail your replies to: