Code Crosswords - Solutions

Question

This is the companion question for Code Crosswords. Robber answers go here.

See Where should we put robbers? for meta-info.

Answer 1

Solution to es1024's C puzzle

I had to fudge a bit on 7-Down by invoking undefined behavior. es1024 has indicated that this use of UB was the intended solution. It will work on most people's computers though. I came up with various expressions achieving the desired result such as -1 << 30, 3 << 30, 6 << 29, and ~(~0U/4) but all of them made it impossible for me to get 5-Across. So I used the Intel architecture-specific property that only the least significant 5 bits are used to determine the size of a left shift. Note that it needs to be ~_ and not ~1 so that the amount to shift by is not a compile-time constant. I tested the expressions with the following code:

#define T(X) _ = c; _ = X; printf("%d\n", _);
z[4] = {9};
int main(c)
{
    int _;

    T("01"[0])
    T(-8)
    T(-2)
    T(11<-328111)
    T(+2+71)
    T(9+0)
    T(0**z)
    T(5;)
    T(0<-000)
    T(2+~3)
    T(!91)
    T(!_)
    T(2**z)
    T('_T;')

    T("11"?5*9:2)
    T(15<<9)
    T(0+22)
    T(-211*0-97;)
    T(-17*0)
    T(3+0<<~_)
    T(8+000)
    T(+0)
    T(42)
    T(+!z)
    T(~_)
}

Answer 2

professorfish, CJam, 41 darks

(This solution actually requires several spaces, so it's not the one professorfish was looking for.)

#K###I#'32
#HDJ*s\ ##
2##2#`#`#-
4Zm*`##L#3
m##6##]`' 
f####e#`#'
`#0#'d1+eu
## #!####9
## '{;'"'m
C5(#}####q

That was great fun. Note that the # beneath the 6 is actually code and not a dark cell. Let's go through this:

Across

Hint 2: [[4 3]]. This was one of the trickier ones, because I really got stuck on trying 4Z]]` or similar. Turns out you can use the Cartesian product m* on stuff that isn't an array, and it will make an array for you. So here it is:

4Zm*`

Hint 4: 24717. By the time I got to this, the H, J, s and trailing space were already in place. The H gave away that I could probably just reuse the 17 and do \ at the end. The J pushes a 19 and 247 == 13 * 19, so:

HDJ*s\ 

Hint 7: 32. There are a bunch of ways to do this: Y5#, 3 2, ZY, YZ\, 4(2, 2)2, '32. I went with the last one, because starting with a character seemed promising for 7-down, and that turned out to be right.

Hint 8: E. I already had the 'd when I got there, so it was between choosing 'd1+eu, 'deu1+ or a variant where I used ) and a space instead of 1+ (for the non-CJam people, this taking the d character and increment and upper-casing it, in either order). However, the u in the final column looked useful for A-down. So I picked the first of those. In the end, 'd) eu would have worked, too.

Hint 9: "". Well, this had to be "empty string, get string representation, push a space". But also needed the ` for string representation in 7-down, and a space in A-down seemed useful, too, so I chose

]`' 

Note that ] might also have been one of LMOQR.

Hint B: "m. I just really had to fit this one in with the rest, but there were few characters that mattered. I already had the { and the m. So instead of using a block, I turned { into a character, discarded it, and then pushed the two required characters:

 '{;'"'m

Hint D: 124. I solved this one together with C-down, which was easiest with a decrement at the end. So I push a 12, a 5, and decrement the latter:

C5(

Down

Hint 1: [2 2 2 3]. That looked too suspiciously like a prime factorisation for it not to be one. :)

24mf`

Hint 3: 3010936384. Factoring this showed that it's actually 38⁶. The only question was how to get the 38 in compliance with 2-across. In the end I needed a * in the third place, so doubling 19 it was:

J2*6#

Hint 5: 2017. Two characters for such a large number? Just use built-in two-digit variables:

KH

Hint 6: "18". I think there's only a single way to do this in 3 characters. Use the built-in 18, convert it to a string, and then to its string representation.

Is`

Hint 7: ' "\"\"". Probably the hardest part of the puzzle. In particular, I needed to get the "\"\"" in only three characters. The trick was to get the string representation of the empty string twice. That led to:

' `L``+

THe + isn't necessary, but was needed by 8-across.

Hint 8: !{}. The block needed to go in the code, so that left only two characters for !, which meant using another literal:

'!{}

Hint A: -3u3. With the u in place from 8-across, I started to put -3 and 3 in the corners where no other hint cared about them. But then I needed an m at the bottom. I think there are multiple ways to get a 3 with _m_, but the simplest was taking the square root of 9:

-3 'u9mq

Hint C: -1. I already had the decrement there, so I just put a 0 where no one else cared about:

0  (

Hint E: Stack: "". Well that was the simplest one. Just call the debugger:

ed

Answer 3

COTO, Javascript ES4, 37 Darks

 __________
|{}=51###6#|
|a##.#I-9<4|
|:##-##"#<#|
|5+Math.PI#|
|}##+##.#+#|
|["9"+0][0]|
|'##p##"###|
|a+-a#a=-10|
|'##c##=###|
|]##"\x48I"|
 ¯¯¯¯¯¯¯¯¯¯

• 6 across is 5+Math.PI or Math.PI+5; but the latter would leave 'M' and 'h' crossing over into other expressions, seemed unlikely.
• A across had to be a string; with 4 in the middle it looked like an escape, the only way you have room for that is with "\x48I".
• 2 down now ends in a string, so probably is "pac" appended to a number; we'll come back to that.
• 8 ac is now xxxa, evaluating to zero, so -a+a? a+-a? 3+-a? a+-a looked best since it gave me what looked like a char in a string in 1dn.
• 1 down contained :5, which could only be legal grammar in {x:5}xxa'x. Since the value returned is 5, it must be pulling it out of the object, so {x:5}['a'], which means the final missing character is also a: {a:5}['a']
• 1 across now has { at the start. I guessed this was a red herring assignment, tried t={}=51, and it worked. Did not know that!
• 2 down now has 5xxax"pac"=>"2pac". This has to be 5-a+"pac", somehow, so the second char has to be '.' for the floating point literal.
• 7 ac is now [xx"xxxx], returning "90". So this has to be an array literal with the value pulled out; there's only room for one value so we have [xx"xx][0]. There's not room for two strings in there, but either "9"+0 or 9+"0" fit.
• 3 dn; 3072 is 3*1024; the power of 2 is suspicious, and I already had 'I' in there blocking other ways of getting large numbers (like 1eX). So, guessed bitshift; 6<<I turned out to be the answer, leaving +0 to add on to the end.
• 4 ac now contained a comparison operator; and the second char had to be a valid unary operator (to fit in after both 't=' and 'I'). I guessed '-', and there's several possible solutions left (I-I<4, I-6<4, etc)
• 5 dn now contained -x..]xxx4. The ".." threw me - there's only a couple of ways that can be legal syntax and is why I asked if ES4 is what was intended - was this some odd feature of that abandoned spec? But then I saw that this was a red herring; -"" is NaN, so -"..]"xx4 must be what's there - a comparison to NaN, returning false; '==' will do, but need to look at the last answer for confirmation...
• 9ac had several possible solutions, but the restriction on characters that would be allowed for 5dn made me think this was yet another red herring assignment; something =-10. To be honest I also looked at the earlier version of 9dn, and realised that must be something =top (to get a Window back). The variable assigned to could be a or I, it doesn't matter.

Tricky puzzle!

Answer 4

grc's Python puzzle

For all the lengthy floating-point expressions, I made a C++ program to generate Python mathematical expressions by force and evaluates them. It assumes all numbers are floating-point and only supports operators +, -, *, /, //, **, and ~. I used it to get every clue longer than 5 characters except a**9*27%b and the hash. With 6 or less blanks it finishes within a couple of seconds, while there is a bit of a wait for 7.

Answer 5

Solution to COTO's MATLAB puzzle

I guess I golfed this one pretty well, as there are 14 spaces.

This test script:

g=4;
o=magic(3);
D=@disp;

D(max([  2]));
D( i^3);
D(o^0);
D(6 -7+eye  );
D((i));
D(.1  ^5* g );
D(~2);
D(diag(~o)  );

D(asin (1)*i);
D((93+7) +~g);
D(  10e15);
D(2*ones (2));
D(02 ^  9 );
D(-i );
D(~o);

produces the following output:

     2

        0 - 1.0000i

     1     0     0
     0     1     0
     0     0     1

     0

        0 + 1.0000i

   4.0000e-05

     0

     0
     0
     0

        0 + 1.5708i

   100

   1.0000e+16

     2     2
     2     2

   512

        0 - 1.0000i

     0     0     0
     0     0     0
     0     0     0