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
>>
>>
>
