Time microsecondsToRun: [ |n| n := (2 raisedToInteger: 8 * 8) - 1. Transcript nextPutAll: 'The number of grains on an 8x8 chessboard is '; print: n; cr; endEntry]. <PRINT-IT> On my laptop, this reports 194 microseconds. Why would you use recursion, anyway? Time microsecondsToRun: [ |n| n := (1 to: 8 * 8) inject: 0 into: [:acc :each | acc+acc+1]. Transcript nextPutAll: 'The number of grains on an 8x8 chessboard is '; print: n; cr; endEntry]. <PRINT-IT> On the same laptop, this reports 118 microseconds. One of the lessons of 'functional' languages, promptly adopted by Smalltalk, is to encapsulate control structures into reusable methods, such as #inject:into:, more commonly known as `foldl` in functional languages. It's then none of my business whether such a method works by recursion, iteration, or gangs of otherwise seasonally unemployed Christmas elves. In my own Smalltalk library, (GeometricSeries new: 64 from: 1 byFactor: 2) sum only takes 15 microseconds. I do note that you are calling validateInput repeatedly. Why? On Sun, 5 Jan 2020 at 07:41, Roelof Wobben via Pharo-users <pharo-users@lists.pharo.org> wrote:
Oke,
So I can better not use recursion for this problem if I understand you well, Richard.
Roelof
Op 4-1-2020 om 19:02 schreef Richard Sargent:
On Sat, Jan 4, 2020 at 9:47 AM Roelof Wobben via Pharo-users <pharo-users@lists.pharo.org> wrote:
Hello,
For a exercism challenge I need to calculate the total grains on a chessboard. So I did :
atSquare: anInteger self validateInput: anInteger. ^ anInteger = 1 ifTrue: [ 1 ] ifFalse: [ 2 * (self atSquare: anInteger - 1) ]
but when I run the tests , the vm seems to be not responsive for some 4 - 5 seconds.
Is there a way I can use this code and take care that the vm stays responsive.
What do you want the VM to do in addition to calculating that sum while it is calculating that sum?
The best way to keep the VM responsive is to take a page from Gauss' notebook and pay attention to the numbers[1]. Let's consider the first four squares and extrapolate from there.
In binary, the squares hold the following grains: 2r1, 2r10, r2100, and 2r1000. When we add them up, we get 2r1111. If we use Gauss' tricks, we can notice that 2r1111 is equal to 2r10000 - 1. So, the sum of the grains on the first 4 squares is 2^5 - 1. You can easily generalize that pattern, of course. Then your program can calculate the answer quickly.
[1] The story goes that Gauss' teacher was frustrated with his student's abilities and set him a challenge to occupy him for some time: sum the numbers from 1 through 100. Gauss immediately answered 5,050, much to his teacher's chagrin.
Regards,
Roelof