Forgot to add... in a more complex program where this calculation is just a small-part you might use something like a Semaphore to signal between different threads. cheers -ben On Mon, 10 Dec 2018 at 06:10, Ben Coman <btc@openinworld.com> wrote:

This is running the calculation in a new Process (in general terms, our "Process"es are "green-threads"). You can't "return" values between threads. You need to pass it some other way through a variable. If you put that code into an instance method, then you could store the result in a class-variable. Change the Transcript to ```self inform: 'FInished' ``` and then have a different method you call after you see that to return the answer.

cheers -ben

On Mon, 10 Dec 2018 at 04:16, Roelof Wobben <r.wobben@home.nl> wrote:

...

Hello,

I got this code from the discord channel

secret := 'abcdef'. zeros := '000'. zerosSize := zeros size. count := 0. [ [ candidate := secret , count printString. ((MD5 hashMessage: candidate) hex first: zerosSize) = zeros ] whileFalse: [ (count \\ 10000) = 0 ifTrue: [ Transcript crShow: count ]. count := count + 1. ]. Transcript crShow: 'ANSWER = ' , count printString. ] forkAt: 35 named: 'Mining'

but I like that there is no transcipt in the code

so my question is how can I change this part( Transcript crShow: 'ANSWER = ' , count printString. )

so it still returns the count.

I tried ^count printstring but that is not allowed

Regards,

Roelof

Op 9-12-2018 om 12:29 schreef Roelof Wobben:

Helllo,

a question about day4 of 2015 is it right that I can only solve this bruteforece so begin at for example 1 , encode it, check for 5 zeros if found then I have found it otherwise increase the counter and do it again https://adventofcode.com/2015/day/4

Regards,

Roelof

On Mon, 10 Dec 2018 at 14:28, Roelof Wobben <r.wobben@home.nl> wrote:

Hmm,

I thought I understand this idea but I do not.

I did this in the playground :

| answer | answer := Key5Solver new solver. ^ answer Result

You can't access the Result variable inside like that. You need a accessor method to do it for you like this... Key5Solver>>result ^ Result But also, you never set the value of Result, see below...

class :

Object subclass: #Key5Solver instanceVariableNames: '' classVariableNames: 'Result' package: 'AdventOfCode2015-Day4 - Part1'

Thats good. Actually I amend my suggestion from yesterday. An instance variable would do. I've renamed your method here... Key5Solver>>startCalculation

zeros := '000'. zerosSize := zeros size. count := 0. [ [ candidate := secret , count printString. ((MD5 hashMessage: candidate) hex first: zerosSize) = zeros ] whileFalse: [ count := count + 1. ].

Result := count "<<<<< need to store the result of the calculation" self inform: 'Mining calculation is complete. You can now ask me for the result'.

] forkAt: 35 named: 'Mining'.

but still I do not see the answer.

With the above, in Playground you can do... solver := Key5Solver new startCalculation . "<<<evaluate this line only" "wait for information message to appear" solver result inspect "<<<<now you can evaluate this line to get the answer" cheers -ben

Maybe I have to go for the blocking answer.

Roelof

Op 9-12-2018 om 23:10 schreef Ben Coman:

This is running the calculation in a new Process (in general terms, our "Process"es are "green-threads"). You can't "return" values between threads. You need to pass it some other way through a variable. If you put that code into an instance method, then you could store the result in a class-variable. Change the Transcript to ```self inform: 'FInished' ``` and then have a different method you call after you see that to return the answer.

cheers -ben

On Mon, 10 Dec 2018 at 04:16, Roelof Wobben <r.wobben@home.nl> wrote:

...

Hello,

I got this code from the discord channel

secret := 'abcdef'. zeros := '000'. zerosSize := zeros size. count := 0. [ [ candidate := secret , count printString. ((MD5 hashMessage: candidate) hex first: zerosSize) = zeros ] whileFalse: [ (count \\ 10000) = 0 ifTrue: [ Transcript crShow: count ]. count := count + 1. ]. Transcript crShow: 'ANSWER = ' , count printString. ] forkAt: 35 named: 'Mining'

but I like that there is no transcipt in the code

so my question is how can I change this part( Transcript crShow: 'ANSWER = ' , count printString. )

so it still returns the count.

I tried ^count printstring but that is not allowed

Regards,

Roelof

Op 9-12-2018 om 12:29 schreef Roelof Wobben:

Helllo,

a question about day4 of 2015 is it right that I can only solve this bruteforece so begin at for example 1 , encode it, check for 5 zeros if found then I have found it otherwise increase the counter and do it again https://adventofcode.com/2015/day/4

Regards,

Roelof

Op 10-12-2018 om 08:29 schreef Ben Coman:

solver := Key5Solver new startCalculation  .

Got it working There was a dot missing Roelof

PS: converting the result of MD5 hashMessage: to hex is a total waste of time. Just ask: - is the first byte zero? - is the second byte zero? - is the third byte less than 16? On Thu, 13 Dec 2018 at 00:07, Richard O'Keefe <raoknz@gmail.com> wrote:

This is the puzzle where you have to find the smallest integer such that '<some salt>' , (that integer) printString has an MD5 hash beginning with 5 hexadecimal zeros?

Brute force is just fine. Calling libnettle's MD5 functions from a C program, it found the right answer in under 0.2 seconds on a cheap laptop. If using the MD5 class in Pharo doesn't take a reasonable time, you may find that (MD5 isPluginAvailable) is false. On my cheap laptop, the obvious code took 30 seconds to find the right answer, which surely means that the plugin is *not* available. But there is absolutely no point in looking for anything faster.

So, (0 to: 1000000) findFirst: [:n | |digest| digest := MD5 hashMessage: 'YOUR SALT GOES HERE' , n printString. ... bytes begins with 00000 ...]

Linear search. Compute string. Compute its MD5 hash. Check the result.

The only time-consuming thing was browsing the MD5 class (and its superclass HashFunction) to figure out the interface.

Now it *could* have turned out to be too slow to be useful, in which case another strategy would have been needed. But it didn't. (I started with a rather smaller bound to get an idea of how fast it was, and it was fast enough.)

On Mon, 10 Dec 2018 at 00:30, Roelof Wobben <r.wobben@home.nl> wrote:

...

Helllo,

a question about day4 of 2015 is it right that I can only solve this bruteforece so begin at for example 1 , encode it, check for 5 zeros if found then I have found it otherwise increase the counter and do it again https://adventofcode.com/2015/day/4

Regards,

Roelof