# Tag Info

12

Answer to Part 1 Answer to Part 2 This is an updated solution to part 2. It produces all the dragons through level 19 without needing to adjust the MathJax parameters, though the later levels take quite a long time to produce, and are beginning to push the limits of the browser itself to display the result (for levels past 15 or so, I have to wait a while ...

8

New Answer Here's a shorter version that improves on my earlier answer (below). One of the places where there is a lot of redundancy is in the initial definitions that had to be repeated in the \D...\D block for printing. I wondered if it would be possible to have the code for printing also be executed so that it would not have to be duplicated. This ...

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$$\require{begingroup}\begingroup \def\mirror#1#2 { #1\mirror#2 % #1} \mirror|look| \endgroup$$ Trace, with $\, \raise-.4ex{\unicode{8629}} \:$ for line breaks: \require{begingroup}\begingroup \def \Type #1#2#3{ \def \Typf % ##1#1##2#1%\Typf##3{\texttt{##1}##3\Typf % ##2#1%\Typf{##3}} \Typf % #3#1%\Typf{#2}#1%\Typf% } \def \RETURN #1 #2\... 5 The lines can be rearranged to form a The coordinates required for the lines (in the order provided in the MathJax) are: Which generates the following (sorry, can't spoilerize it):  \rlap{\kern 1 pc\raise 2 pc\Rule{0.1pc}{3.1pc}{0pc}} \rlap{\kern 2 pc\raise 3 pc\Rule{0.1pc}{3.1pc}{0pc}} \rlap{\kern 3 pc\raise 3 pc\Rule{0.1pc}{3.1pc}{0pc}} % ... 5 The Framework One approach to this problem is to consider the binary number as follows: \begin{align} 10011 &= 1\times 2^4 + 0\times 2^3 + 0\times 2^2 + 1\times 2^1 + 1\times 2^0\\ &= 2\cdot2\cdot2\cdot2\cdot1+2\cdot2\cdot2\cdot0+2\cdot2\cdot0+2\cdot1+1\\ &= 2\cdot(2\cdot2\cdot2\cdot1+2\cdot2\cdot0+2\cdot0+1)+1\\ &= 2\cdot(2\... 4 Last solution\require{begingroup}\begingroup \def \G |#1|{ | \R #1 |\R |{\let\ } } \def \R #1#2|#3{ #3 #2 |#3 #1 } \G |Medusa| \endgroup$$This has 54 non-space replacement characters and exploits two economies: • \small\texttt{\\}\color{violet}{\LARGE\raise-3mu\square} control space (... 3 Newest Solution Based on @humn's comment, it seems that he has produced a single-\def solution of 34 characters that he has not yet posted. Because a definition requires at least 8 characters (four for the \def, two for the control sequence being defined, and two for the braces surrounding the replacement text), I was not surprised that one could save 7 by ... 3 A \texttt{\binary} Klein bottle These 72 replacement code tokens form 2 static macros and 1 dynamic macro, while blurring the distinction between inside and outside, reminiscent of a Klein bottle, with data that are not just produced by recursive calls but are explicitly those calls themselves.$$\require{begingroup}\begingroup \def\safe{\text{\endgroup ...

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This appears to work: (I confess I got there by largely-undirected trial and error.) Here: $$\require{begingroup}\begingroup \def\1#1 {#1 #11 } \1\1 \endgroup$$ Trace: $$\require{begingroup}\begingroup \def\Rest#1{{\small\texttt{#1}}} \def\This#1{{ \small\, \rlap{ \texttt {#1} } \raise-.2ex{ \underline{\hphantom{ \... 2 Overhauled solution — just \, \texttt{\stars{}} \: and \, \texttt{\N} \: while borrowing \, \texttt{\endgroup} \: as a delimiter \small \texttt{\stars{57}} — solution with 47 code tokens, as \small\texttt{\stars{}} and one dynamic macro — base 10:$$\require{begingroup}\begingroup \def \stars #1#2 #3\endgroup{ \def ...

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A Hint (My Solution is Below) One of the key components of my solution (and one that others have asked about) is a macro that, given a single digit, produces that many stars. There are several approaches to this. One of them is to use one macro to produce another macro, and use the data given to the first macro as part of the template for the second. ...

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Plagiarized addition Nothing original here beyond what was initially in Davide Cervone's solution. This only contorts it for the sake of ever fewer lines (10) by doing the obvious, in tucking the comment into another line, and by doing the deranged, in using the terminal $\small\texttt{\endgroup}$ as a delimiter. \require{begingroup}\begingroup \def ...

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