McCaw won the steps competition for part 2 and placed 2nd for part 3. Part 2:
Oh, some cool ideas here.
Generated code to handle all the possible cases.
Coming from both ends.
Being able to handle all the necessary information with states + 5 heads.
Reading the string, rewriting it 1 row above.
And part 3:
This one has a sort of exploding wave front. The 1st time a “=” is encountered, replace it with “~”, but also send some heads 1 row down. As long as a new “=” is encountered, keep doing that. This of course has to stop when all 10 heads are on different rows, so there’s a different mechanism to finish the deep ponds in that case. V’s and B’s to keep track of how much is left to do. Oh, and a funny “with the very deep ponds, keep 1 head on the top row, and when the deep pond is finished, let that head travel again, alone for a while”.
I don’t quite know what to say. It’s elegant. And a lot of people used similar techniques to get good scores.
Use a step to spread the heads. Then use 2×2 heads to read new and previous information. With this string, ABBAAB, the first time this is AB-AB. Delete the outer characters and replace the inner characters with lower case copies. Next time it’s bB-Aa. As bB and Aa represent pairs, 2 @’s are written. Again, the inner characters are changed to lower case. In the next step we get ba-ba, actually the same ba seen twice. Write nothing, delete the characters, stop.
Apparently there’s a set of rules for each step. Yeah, from what I can tell, the MOVE1 group is almost exactly a duplicate of the MOVE0 group etc. So, this program was quick and easy to write, but could easily have been shorter. However! Right now we’re counting steps, and then it doesn’t matter.
For part 2, more than 1 string of characters is possible.
With 1 string, the program seems to be the same as before. With 2 strings, there are 2×4 heads, and the strings are still processed from both ends at the same time. But with 3-5 strings…
The 4 heads at the end of the strings are simply left there, useless.
The 1st rule spreads the heads a little. A and B are deleted, but remembered in the state name. The 2nd rule encounters a blank space, and this must just have been created by the 1st rule. So we move into 2 string mode.
PREPAB does not see a blank space in front of head 2, so we go a different path.
Finally, for part 3, pairs are counted both horizontally and vertically.
Apart from a few things in the beginning to detect the number of strings, this feels very familiar. Once we’re set, 4 heads look at the situation and react accordingly. Extra @’s can be written in front of the strings.
The golf for the whole program is 1829. 1 rule to spread the heads a little. S000 and S001 each spreads them a little more. And then S01A goes through the rest of the line. At 1st glance, there are 21 rules with S000 and S001 each. S000 writes 2 P’s, probably with the idea, that they might be deleted later.
From what I can tell, the purpose of a rule like S032 is to say “remember to write 32 more P’s”. (Maybe that 32 is in hex?) Let’s look at the extreme example. Note that the program keeps running for a while after the D’s end.
The first rule spreads the heads. The next rule writes 4 P’s, as the heads have only spread to 4 positions. Go to S007, meaning 8 P’s are owed? S007 spreads the heads even more, and now the 10 heads are at 10 different positions. Meanwhile 8 P’s are written and we go to S026. From there on, each rule writes 10 P’s, but as the input is DD, that means the debt rises with 2 P’s, so we go from S026 to S028 to S02A etc. At S05A we finally run out of D’s, and the rest of the rules write P’s as fast as possible. I guess some of the optimization here is, that instead of using the 10 heads to write 12 P’s in 2 steps, all the debt is written at the end. I’m not sure about the numbering of those rules. 26hex = 38dec doesn’t quite mean 38 P’s are owed. So maybe there’s an overlap, where S01C means S018 + S004, 18hex to signify some state and 4 to signify writing 4 more P’s. The 1st example stays in S01A, and nothing is ever owed.
I looked a tiny bit at Peter’s solution, coming in at no. 2. He operates with MOVE6, MOVE8 and MOVE10, each meaning 6, 8 or 10 P’s are owed. But then at MOVE10 and when encountering DD, let the reading heads stay at these DD, write 8 P’s and go to MOVE2. A nice way for the debt not to grow too large.
This week the #puzzle is: Can You Fit the Stars on the Flag? #coding #bisection #desmos
When designing her new nation’s flag, Retsy Boss wanted to compactly arrange some stars. These stars were positioned along a square grid, but she only wanted to include stars whose centers were at most two units away from some point on the plane.
For example, if she had centered the circle on a star itself, then she could have placed a total of 13 stars on the flag, as shown below:
What is the greatest number of stars Retsy could have placed on the flag?
And for extra credit:
After 250 years, the nation has commissioned Retsy Boss VIII to design a new flag with one star for each of the nation’s current 58 states. As an homage to the original flag design, Retsy wants to select 58 stars from the square grid that are all at most some distance R from a point on the plane.
What is the minimum distance R that Retsy can use?
Method 1: Recreate the situation in Desmos and mess around. First define a function to count the number of stars (the stars have integer coordinates). The center of the circle:
The distance to the center from another point:
Use rounded values for the other point:
This function will be 1, if the middle of a unit square is within the circle with radius r1:
Compute the area of all the unit squares:
Method 2: Create a program to go through a lot of options.
Result 1: 14 (x1: 0.063600, y1: 0.499700)
Result: 14 stars.
And for extra credit:
Same methods.
My program is surprisingly sensitive to , the distance between different (x,y) combinations to try. I employ bisection to fine tune the solution.
Max stars for r = 4.155180, d = 0.000100: 57 (x1: 0.000000, y1: 0.000000) Max stars for r = 4.155190, d = 0.000100: 57 (x1: 0.000000, y1: 0.000000) Max stars for r = 4.155200, d = 0.000100: 58 (x1: 0.125000, y1: 0.500000) Max stars for r = 4.155210, d = 0.000100: 58 (x1: 0.125000, y1: 0.500000)
4.1550000 < r < 4.1560000 4.1550000 < r < 4.1555000 4.1550000 < r < 4.1552500 4.1551250 < r < 4.1552500 4.1551875 < r < 4.1552500 4.1551875 < r < 4.1552188 4.1551875 < r < 4.1552031
Result 2: 4.1551875 - 4.1551953 (x1: 0.125000, y1: 0.500000)
The 2 methods largely agree. Result: A radius of 4.1552.
Head 1 and 6 are reading. In the 1st step, there are 2 A’s, so they’re just deleted. Meanwhile the heads are spreading. In the 2nd step, there’s DA. Write 2 P’s and remember, that we need 3 more / go to state MOVE3.
Other cases employ PREP1, still preparing everything, and we need 1 more P. This goes up to PREP8. Similarly there’s a MOVE10. And whole families of MOVEC and MOVED. There’s nothing very complicated here, looking at each group of rules. There’s just a lot. 454 rules in all.
This one is in a perpetual debt, owing 8 P’s in the preparation rule and 4 in the move rules.
I’m sure this program was generated, it has 367 lines. It has a lot of lines like this:
S_AB B*****B*** S_ABBB _PP_**____ RUDLUDLUDR
2 heads begin at the ends and work their way in. The states keep track of what was just seen and what is here now. Also there are 10 heads in all, so there’s a lot of places to write the P’s. In the example above, there’s actually some backtracking. In the final step, 1 P is deleted. Let me just find which rules were actually used in this example (I added a little formatting):
For the 2 reading heads, ABBAC is seen as AC-BA-BB. (ABCDE would be seen as AE-BD-CC.) The 1st rule writes 2 P’s — that’s all it can do, as the heads haven’t spread out yet. So remember, we need to write 1 more P at some point. The 2nd rule writes 2 P’s, and we’re square. The 3rd rule writes 2 P’s, because it’s seeing 2 B’s. But then the final rule sees, that it was only 1 B seen by 2 heads and deletes a P. And done!
I’m actually not quite sure how this program detects, that BB is 1 B seen by 2 heads and therefore a stopping condition… Or how it detects, that there’s only 1 character left? Maybe this whole program is very specifically tailored to the input and knows, that CC-AA means, there’s only 1 character left? With larger strings, we run into stuff like:
This time there are a lot of heads. In the gif above, it’s especially visible, that the left and right vertical sides are done simultaneously. Something similar is true about the top and bottom horizontal sides. And I think it’s a feature in this part, that the 2 vertical sides are actually equal, and the same for the horizontal sides, because that means fewer rules. (Though it actually leads to fewer rules needed to be written, the number of used rules isn’t affected.)
For et par dage siden anbefalede Zach Wissner-Gross fra Fiddler on the Proof langnovellen “Understudies” af #GregEgan, og nu har jeg fået den læst. Wow! For fans af fiddler-opgaver og den slags, så er det her en gave. Og lad mig med det samme indrømme, jeg løste altså ikke alle opgaverne i teksten, mens jeg læste. De er svære!
Specielt en af opgaverne fangede mig, fordi jeg ikke helt kunne forstå løsningen. Så i det her indlæg vil jeg prøve at læse opgave og løsning, indtil jeg forstår det.
“A few years before the fall of the Berlin Wall,” she said, “an English journalist was imprisoned in East Germany. She was able to exchange letters with her husband, but they knew their mail was being read by authorities. Among the first things her husband sent her was a computer printout of a poem, written by their twelve-year-old daughter, which the journalist kept pinned to the wall of her prison cell. It read:
“Evening light fades / Quiet descends / The joking boy relaxed / Sweetly amazed / Night promises peaceful dreams / Until morning.
“Historians studying the correspondence between the couple found that the words of this poem showed up in their letters far more often than would have been expected, and so might have played a part in a code they used.
“Once, in response to a letter where her husband used ‘joking,’ ‘descends,’ ‘quiet,’ ‘fades,’ and ‘light,’ in that order, the journalist responded by using ‘evening,’ ‘amazed,’ ‘boy’ three times, ‘amazed,’ ‘evening,’ ‘joking’ twice, ‘relaxed,’ ‘sweetly,’ ‘amazed,’ ‘sweetly,’ then ‘relaxed.’
“What was she communicating?”
Og den korrekte løsning:
“The husband wasn’t saying anything in his letter,” she explained. “He was providing a key that the journalist could use to help encode her own message. She took his numbers—five, four, three, two, one, by their position in the poem, with ‘evening’ as zero and ‘the’ omitted—then used the dot matrix printout of the poem, with each letter of the alphabet forming a grid five dots wide, to transform them into a number for each row, adding up those with a matching dot, leaving out the rest. The lowercase letters ‘o’ and ‘k’ provided her response. She was saying ‘okay.’ ”
Lad os tage den igen, langsomt.
Digtet kan også skrives således:
0
Evening
4
descends
8
Sweetly
12
peaceful
1
light
5
joking
9
amazed
13
dreams
2
fades
6
boy
10
Night
14
Until
3
Quiet
7
relaxed
11
promises
15
morning
Det første brev:
“a letter where her husband used ‘joking,’ ‘descends,’ ‘quiet,’ ‘fades,’ and ‘light,’ in that order, …’”
Det bliver altså til 5, 4, 3, 2 og 1.
Og det andet brev:
“… the journalist responded by using ‘evening,’ ‘amazed,’ ‘boy’ three times, ‘amazed,’ ‘evening,’ ‘joking’ twice, ‘relaxed,’ ‘sweetly,’ ‘amazed,’ ‘sweetly,’ then ‘relaxed.’”
Dette svarer til 0, 9, 6, 6, 6, 9, 0, 5, 5, 7, 8, 9, 8, 7.
Jeg kan se, at der er noget struktur her, der passer med o og k. I første omgang med et tilfældigt bud på en font.
x xx x x x x x x x x x x x xx xx x x x x
Eller:
0 5 x 9 xx 5 x 6 x x 7 x x 6 x x 8 x x 6 x x 9 xx 9 xx 8 x x 0 7 x x
Ja, noget her fungerer. Samme linje af pixler i det samme bogstav giver det samme tal.
Men jeg har ikke brugt 5, 4, 3, 2, 1 til det her? Hov, jeg prøver lige igen, med en bedre font.
0 5 x 9 xxx 5 x 6 x x 7 x x 6 x x 8 x x 6 x x 9 xx 9 xxx 8 x x 0 7 x x
Ja, nu virker det.
“_xxx_” bliver parret sammen med “54321”. Det giver “04320”. Og tværsummen af det er 9. Og således bliver toppen af o’et til 9. “x___x” bliver til “50001”, bliver til 6. Osv.
Zach won the rules competitions for part 2 and 3. For part 2 there are 4 heads. In phase 1 the red head looks at the nails, and the yellow head looks for the top and bottom nails. When a nail head is detected and we go into phase 2, the green and yellow heads come running and fix the wrong nails.
A little extra work for part 3. When a nail head has been detected, there’s a marker written ($) for that nail, to keep track of it.
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Stuff from ommadawn.dk (Lise Andreasen) 