This post looks at an extraordinary game of cricket that has just taken place in India. I also provide a solution to the mathematical conundrum from brilliant.org that I posed yesterday and of course a few photographs.
England were unchanged, India had two changes. Ishan Kishan had a minor injury and was replaced by Suryakumar Yadav. Yuzvendra Chahal was dropped and replaced by another leg spinner, Rahul Chahar. Eoin Morgan won the toss and chose to bowl.
THE INDIAN INNINGS
Neither of India’s openers were massively convincing, and Kohli at no 4 also failed with the bat. However, Suryakumar Yadav played a quite magnificent innings, at one stage threatening to record a century, and Rishabh Pant also played very nicely. India put up 185-8 in the end, a total that looked defensible but not unassailable. Jofra Archer took four wickets, Mark Wood was also impressive, but Adil Rashid had an off day for once, and Jordan, Stokes and Curran were all unimpressive as well.
THE ENGLAND RESPONSE
Buttler failed, Malan got a bit of a start but did not go on, Roy reached 40 for the third time of the series and for the third time of the series got out with a seriously big score apparently beckoning. Bairstow and Stokes batted well together before Bairstow was out, and then it looked like Stokes and Morgan were taking England close. However, both fell to Thakur in consecutive deliveries at the start of the 18th. Curran and Jordan played decently for the rest of that over, but then Curran fell in the 19th. A four off the last ball of the 19th by Archer reduced the requirement to 23 off the final over. Thakur, who had put India in command with his bowling at the start of the 18th now lost his bearings and at one point the ask was down to ten off three balls, but then he regathered his nerve, and India emerged victorious by eight runs, setting up a final game decider on Saturday. Although the standard of play was high an both sides it is not really acceptable for 40 overs of cricket to occupy four and a quarter hours of playing time as happened today.
SOLUTION TO A TEASER
Yesterday I set you the following:
In total there are 512 small cubes in the structure. Of these 216 (6x6x6) are purely internal and therefore unpainted, eight are corner cubes and painted on three faces, which leaves 288 cubes painted on either one or two faces. The cubes painted on one face are those in the centre of each visible face, which number 36 on each face = 216 in total. This leaves 72 cubes painted on two faces, and 216 – 72 = 144. For a cube with side length n, there will eight corner pieces, (n-2) ^ 3 centre pieces that are thus unpainted, 6 ((n-2)^2) pieces that are painted on one face only and 12 (n-2) pieces that are painted on exactly two faces. Though these equations only start to work once n is greater than 2 – a 2 x 2 x 2 cube has eight blocks each of which are painted on three faces.
My usual sign off…