A first post on the Bob Willis Trophy, a bit of mathematics and some photographs.
This post looks at the start of the competition that has been devised to replace the county championship in this pandemic hit season, play in which got underway at 11AM this morning. The second ODI between England and Ireland gets underway at the Ageas Bowl shortly.
HONOURING A LEGEND
Bob Willis, the former fast bowler who was only the second England bowler to take as many as 300 test wickets, following in the footsteps of Fred Trueman, died on December 4th 2019. When it became apparent a normal county championship would be impossible to stage it was only natural that his name should be attached to the replacement competition. Willis’ finest hour came at Headingley in 1981. He he taken no wickets in either innings, when with Australia 56-1 in their second innings needing only a further 74 for victory he was given one last chance to save his test career. Just under an hour later Australia were 75-8 and Willis had taken six wickets in as many overs (the other, the adhesive Border, had fallen to Chris Old). Dennis Lillee and Ray Bright launched a counter attack that yielded 35 runs in four overs before Lillee mistimed a drive and Gatting ran in, dived and held the catch. Alderman was dropped twice of Botham, but the first ball of Willis’ 16th over of the innings and tenth off the reel uprooted Bright’s middle stump to give England victory by 18 runs, with Willis having figures of 8-43. Willis would play on another three years, captaining the side for a period.
AN EXPERIMENT BORN OF NECESSITY
The 18 First Class counties have been split into three regions, South, Central and North. Each region will be play five rounds of matches, so that each side plays each other side in their region twice. At the end of this the two teams with the most points will play a final at Lord’s, which will be contested over five days instead of the regular four for a county fixture. Certain other changes have been made to the normal format of county games: the number of points for a draw has been increased from five to eight so that teams who suffer a lot of adverse weather will not too badly affected, a new ball will only be available at 90 overs rather than 80, the first innings for each county cannot last beyond 120 overs, and the minimum lead to be able to enforce the follow-on will be 200 rather than 150 runs. One beneficial side effect of these arrangements should be that spinners come into the game more than at present (Surrey and Middlesex, whose game I have listened to some of, are each playing two spinners, in Surrey’s case first class debutant Daniel Moriarty and England hopeful Amar Virdi, who would be the most obvious replacement for Dom Bess in the off spinner;s role). England is somewhat overburdened with bowlers who move the ball around a bit at medium pace or fractionally above and short of both genuine pace and spin. Surrey and Warwickshire were going to be experimenting with letting in spectators, but that has been prevented by the fact that Covid-19 cases are spiking upward making caution once more the order of the day.
A MEASURE OF MATHEMATICS
I have solutions to provide to the two problems I posed in my previous post, and I also have a new problem to set. My first was this one:
This was a bit of trick question. The answer is the both final shapes have the same number of faces (14 as it happens). Here is a published solution from Mahdi Raza:
The second problem I posed was this one:
The fact that the result is not allowed to be negative at any stage means that only five square numbers need be considered as possible plays for Mei – 1, 4, 9, 16 and 25. 25 + 9 = 34, which means that if either of these numbers is chosen Yuri is left with a square number and reaches 0 at the first attempt, which leaves 1, 4 and 16 as options.
Case 1: Mei plays 16. This reduces the number to 18. Yuri’s choices are now 1, 4, 9 or 16, of which 9 is instantly ruled out since it gives the game to Mei. However a choice of 16 by Yuri reduces the number to 2. Mei’s next move is forced 0 she subtracts 1, leaving 1 remaining and a win for Yuri as he also subtracts 1.
Case 2: Mei plays 4. This reduces the number to 30. All Yuri now has to do is play 25, reducing the number to 5, and whether Mei subtracts 1 or 4 she leaves a square number which Yuri thus reduces to 0 winning the game.
Case 3: Mei plays 1 which reduces the number to 33. If Yuri plays 25 that reduces the number to 8. Mei has a choice between 1 and 4, and 4 reduces the total to 4 an a win for Yuri, so she has to play 1. If Yuri now plays 4 then Mei plays 1 and Yuri has to do likewise, giving the game to Mei. Thus Yuri plays 1 reducing the number to 6, and Mei can then win the game by playing 4 and making the number 2 with Yuri to play. Thus Yuri cannot play 25 as his first response. If he plays 16 that reduces the number to 17, from which Mei cannot play 1 as that gives Yuri the game. If she plays four that reduces the number to 13, and Yuri’s forced moved of 1 reduces the number to 12, from which Mei cannot play one or nine as they immediately allow winning moves for Yuri. So she plays four, making the number now eight, and Yuri counters with a one which makes the number seven, and whether Mei plays one or four Yuri is in control because his own next move makes the number two. If she plays a nine instantly from 17 that reduces the number to eight and again Yuri is in control. Similarly 16 hands the game straight to Yuri. Thus whatever number Mei starts with Yuri has a winning response. THus, if both players play optimally Mei cannot win.
The new problem involves fractal geometry:
There are five answers for you to choose from:
Solution in my next post.
My usual sign off: