Binary Numbers Magic Trick

Question

The challenge is simply; output the following six 2D integer arrays:

[[ 1, 11, 21, 31, 41, 51],
 [ 3, 13, 23, 33, 43, 53],
 [ 5, 15, 25, 35, 45, 55],
 [ 7, 17, 27, 37, 47, 57],
 [ 9, 19, 29, 39, 49, 59]]
[[ 2, 11, 22, 31, 42, 51],
 [ 3, 14, 23, 34, 43, 54],
 [ 6, 15, 26, 35, 46, 55],
 [ 7, 18, 27, 38, 47, 58],
 [10, 19, 30, 39, 50, 59]]
[[ 4, 13, 22, 31, 44, 53],
 [ 5, 14, 23, 36, 45, 54],
 [ 6, 15, 28, 37, 46, 55],
 [ 7, 20, 29, 38, 47, 60],
 [12, 21, 30, 39, 52]]
[[ 8, 13, 26, 31, 44, 57],
 [ 9, 14, 27, 40, 45, 58],
 [10, 15, 28, 41, 46, 59],
 [11, 24, 29, 42, 47, 60],
 [12, 25, 30, 43, 56]]
[[16, 21, 26, 31, 52, 57],
 [17, 22, 27, 48, 53, 58],
 [18, 23, 28, 49, 54, 59],
 [19, 24, 29, 50, 55, 60],
 [20, 25, 30, 51, 56]]
[[32, 37, 42, 47, 52, 57],
 [33, 38, 43, 48, 53, 58],
 [34, 39, 44, 49, 54, 59],
 [35, 40, 45, 50, 55, 60],
 [36, 41, 46, 51, 56]]

What are these 2D integer arrays? These are the numbers used in a magic trick with cards containing these numbers:

enter image description here

The magic trick asks someone to think of a number in the range [1, 60], and give the one performing the magic trick all the cards which contain this number. The one performing the magic trick can then sum the top-left numbers (all a power of 2) of the given cards to get to the number the person was thinking of. Some additional explanation of why this works can be found here.

Challenge rules:

• You can output the six 2D integer arrays in any reasonable format. Can be printed with delimiters; can be a 3D integer array containing the six 2D integer arrays; can be a string-list of lines; etc.
• You are allowed to fill the bottom right position of the last four cards with a negative value in the range [-60, -1] or character '*' instead of leaving it out to make the 2D integer arrays rectangular matrices (no, you are not allowed to fill them with 0 or a non-integer like null/undefined as alternative, with the exception of * since a star is also used in the actual cards).
• The order of the numbers in the matrices is mandatory. Although it doesn't matter for the physical magic trick, I see this challenge mainly as a matrix-kolmogorov-complexity one, hence the restriction on order.
  The order of the matrices themselves in the output list can be in any order, since it's clear from the top-left card which matrix is which.

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code (i.e. TIO).
• Also, adding an explanation for your answer is highly recommended.

• Kevin Cruijssen: Related. (As in, it's referring to the same magic trick, but not really useful to get inspiration from for this challenge I think. That challenge asks to output a truthy/falsey value whether number n appears on the k'th card; where my challenge is a KC-challenge to output the six matrices.)
• Erik the Outgolfer: @DigitalTrauma Hm, I'm not so sure if this is really a duplicate, because your challenge is ascii-art (not tagged as such, but it is), while this one allows you to output the array in a way more lenient format (not just four essentially identical ways). I can't vote to reopen, though, because I have a hammer.
• Kevin Cruijssen: @EriktheOutgolfer Woops.. Forgot I have a hammer as well >.> Sometimes being able to close/open hammer is pretty annoying.. It had 2 votes already though, so with yours and mine in addition there were 4 open votes. But if someone wants to close it again I don't mind. They are indeed very similar, although his challenge is indeed an [ascii-art] challenge with strict (MD5) output rules, where mine are very flexible (and the rows/columns are swapped, and range is [1,60] instead of [1,63]; pretty minor differences, but still).
• Erik the Outgolfer: Looks like you didn't try to VTRO with an attitude of "this is my precious challenge!!!" at least... :P
• Digital Trauma: I too forgot about the hammer. I still think this is close enough to vote to dup, though I'll defer to the wisdom of the community if it is reopened.

Answer 1

Perl 6, (削除) 63 (削除ここまで) 46 bytes

say grep(*+&2**$_,^61)[$_,*+5...*for ^5]for ^6

Try it online!

Outputs as 2D arrays on multiple lines, with the last array of each one cut off if necessary.

Answer 2

Python 2, 76 bytes

r=range;print[[[i for i in r(61)if i&2**k][j::5]for j in r(5)]for k in r(6)]

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The method here is to create a list of all possible numbers r(61) and then whittle that down to the list of numbers for a card i&2**k.

Then, by using list slicing, that 1D list of numbers is rearranged to the correct 6x5 card size [card nums][j::5]for j in r(5).

Then, this generator is just repeated for 6 cards for k in r(6).

---

While I could not find any solutions less than 76 bytes, here are two others that are also 76 bytes:

r=range;print[[[i for i in r(61)if i&1<<k][j::5]for j in r(5)]for k in r(6)]

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This next one is inspired by Jonathan Allan.

k=32
while k:print[[i for i in range(61)if i&k][j::5]for j in range(5)];k/=2

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Any comments are greatly appreciated.

Answer 3

MATL, (削除) 12 (削除ここまで) 11 bytes

-1 byte thanks to the master himself:)

60:B"@fQ6eq

Explanation:

60: % create a vector [1,2,3,...,60]
 B % convert to binary matrix (each row corresponds to one number)
 " % loop over the columns and execute following commands:
 @f % "find" all the nonzero entries and list their indices
 Q % increment everything
 6e % reshape and pad with a zero at the end
 q % decrement (reverts the increment and makes a -1 out of the zero
 % close loop (]) implicitly
 % display the entries implicitly

Try it online!

Answer 4

Charcoal, 26 bytes

E6E5⪫E6§+§⪪Φ61&πX2ι5ν⟦*⟧λ 

Try it online! Link is to verbose version of code. I tried calculating the entries directly but this was already 27 bytes before adjusting for the * in the bottom right. Outputs each row joined with spaces and a blank line between cards. Explanation:

E6 Loop over 6 cards
 E5 Loop over 5 rows
 E6 Loop over 6 columns
 Φ61 Filter over 0..60 where
 π Current value
 & Bitwise And
 2 Literal 2
 X Raised to power
 ι Card index
 ⪪ 5 Split into groups of 5
 § ν Indexed by column
 + Concatenated with
 * Literal string `*`
 ⟦ ⟧ Wrapped in an array
 § λ Indexed by row
 ⪫ Joined with spaces
 Implicitly print

• Kevin Cruijssen: I added that rule about * for fun after I saw the stars on the actual cards. Was wondering if there would be any languages using it, but I'm glad to see at least one did. :) Nice answer!
• Neil: @KevinCruijssen Charcoal doesn't have a transpose operator, and the golfiest transpose requires a rectangular array of known size, so I need to add something to make up the size, and * is at least as short as anything else would be.
• Tvde1: I don't think this is 26 bytes...
• Neil: @Tvde1 Charcoal, like many of the esolangs on this site, uses a custom code page. Characters from that page cost 1 byte, while other characters cost up to 4 bytes.

Answer 5

05AB1E, 16 bytes

60L2вíƶ0ζε0K5ô®ζ

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Explanation

60L # push [1 ... 60]
 2в # convert each to a list of binary digits
 í # reverse each
 ƶ # multiply each by its 1-based index
 0ζ # transpose with 0 as filler
 ε # apply to each list
 0K # remove zeroes
 5ô # split into groups of 5
 ®ζ # zip using -1 as filler

05AB1E, 17 bytes

6F60ÝNoôāÈÏ ̃5ô®ζ,

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Explanation

6F # for N in [0 ... 5] do
 60Ý # push [0 ... 60]
 Noô # split into groups of 2^N numbers
 āÈÏ # keep every other group
 ̃ # flatten
 5ô # split into groups of 5
 ®ζ # transpose with -1 as filler
 , # print

Answer 6

Husk, 13 bytes

ṠMöTC5Wnünḣ60

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Explanation

 ḣ60 Range [1..60]
 ü Uniquify using equality predicate
 n bitwise AND: [1,2,4,8,16,32]
 M For each number x in this list,
Ṡ W take the indices of elements of [1..60]
 n that have nonzero bitwise AND with x,
 C5 cut that list into chunks of length 5
 öT and transpose it.

Answer 7

Python 2, (削除) 82 (削除ここまで) (削除) 80 (削除ここまで) (削除) 78 (削除ここまで) 74 bytes

i=1 
exec"print zip(*zip(*[(n for n in range(61)+[-1]if n&i)]*5));i*=2;"*6

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_{-4 bytes, thanks to Jonathan Allan}

• Jonathan Allan: Nice. The iter keyword is redundant here since a generator will do the job just as well.
• TFeld: @JonathanAllan Cool, thanks :D

Answer 8

Japt, 14 bytes

6Æ60õ f&2pX)ó5

Try it

6Æ Create a range from 0 to 5 (inclusive) and map each X into
 60õ Elements in the range [1..60]
 f Where
 &2pX) The number bitwise AND with X is not 0
 ó5 Split into 5 arrays, where each array contains every 5th element
-Q flag is just for formatting purposes

Answer 9

JavaScript (ES6), (削除) 90 (削除ここまで) 88 bytes

_=>[1,2,4,8,16,32].map(n=>(g=i=>i<60?g(++i,i&n?m[y%5]=[...m[y++%5]||[],i]:0):m)(y=m=[]))

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Commented

_ => // anonymous function taking no argument
 [1, 2, 4, 8, 16, 32] // list of powers of 2, from 2**0 to 2**5
 .map(n => // for each entry n in this list:
 ( g = i => // g = recursive function taking a counter i
 i < 60 ? // if i is less than 60:
 g( // recursive call:
 ++i, // increment i
 i & n ? // if a bitwise AND between i and n is non-zero:
 m[y % 5] = // update m[y % 5]:
 [ ...m[y++ % 5] // prepend all previous values; increment y
 || [], // or prepend nothing if it was undefined so far
 i // append i
 ] // end of update
 : // else:
 0 // do nothing
 ) // end of recursive call
 : // else:
 m // return m[]
 )(y = m = []) // initial call to g with i = y = m = []
 // (i and y being coerced to 0)
 ) // end of map()

Answer 10

Python 2, 73 bytes

Inspiration taken from both TFeld's and The Matt's.

k=32
while k:print zip(*zip(*[(i for i in range(61)+[-1]if i&k)]*5));k/=2

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Answer 11

C (gcc), 95 bytes

i,j,k;f(int o[][5][6]){for(i=6;i;)for(o[--i][4][5]=j=k=-1;j<60;)++j&1<<i?o[i][++k%5][k/5]=j:0;}

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Returns the matrices as a 3D int array in o.

The last 4 matrices have -1 as their last value.

Saved 2 bytes thanks to Kevin Cruijssen.

Saved (削除) 7 (削除ここまで) 8 bytes thanks to Arnauld.

• Kevin Cruijssen: You can save 2 bytes by changing o[i][4][5]=-1;for(j=k=0; to for(o[i][4][5]=-1,j=k=0; so the brackets can be removed. Nice answer btw, +1 from me.
• Arnauld: 95 bytes
• Arnauld: (Note that I'm not 100% sure if passing a 3D array already allocated with the correct dimensions is allowed. But I'll let regular C golfers provide a better insight about that.)
• Matej Mulej: @Arnauld I was thinking about that, but decided against it.
• ASCII-only: better to leave out #include to show that it works without it

Answer 12

CJam (18 bytes)

6{61{2A#&},5/zp}fA

Online demo. This is a full program which outputs to stdout.

Dissection

6{ }fA # for A = 0 to 5
 61{2A#&}, # filter [0,61) by whether bit 2^A is set
 5/z # break into chunks of 5 and transpose to get 5 lists
 p # print

Answer 13

Jelly, 13 bytes

60&ƇⱮs5ドルLÐṂZ€

A niladic Link which yields a list of (6) lists of lists of integers. (It outputs the using the default option of having no * or negative filler.)

Try it online!

How?

Each matrix contains, in column-major order, the numbers up to \60ドル\$ which share the single set-bit with the top-left (minimal) number.

This program first makes all \60ドル\$ possible ordered lists of numbers in \$[1,60]\$ which share any set-bit(s) with their index number. It then splits each into chunks of \5ドル\$ and keeps only those with minimal length - which will be the ones where the index has only a single set-bit (and hence also being its minimal value). Finally it transposes each to put them into column-major order.

60&ƇⱮs5ドルLÐṂZ€ - Link: no arguments
60 - set the left argument to 60
 Ɱ - map across ([1..60]) with: (i.e. [f(60,x) for x in [1..60]])
 Ƈ - filter keep if: (N.B. 0 is falsey, while non-zeros are truthy)
 & - bitwise AND
 € - for each:
 s 5 - split into chunks of five
 ÐṂ - keep those with minimal:
 L - length
 Z€ - transpose each

---

Lots of 15s without realising the "minimal by length when split into fives" trick:

5Ż2*Ɱ60&ƇⱮs5ドルZ€
6μ’2*60&Ƈ)s5ドルZ€
60&ƇⱮ`LÞḣ6s5ドルZ€

...and, while attempting to find shorter, I got another 13 without needing the trick at all:

60B€Uz0Ts5ZƊ€

Answer 14

Wolfram Language (Mathematica), 88 bytes

Transpose@Partition[#~Append~-1,5]&/@Last@Reap[Sow[,NumberExpand[,2]]~Do~{,60},Except@0]

• Kevin Cruijssen: I took the liberty of adding a TIO link (based on @J42161217's answer). +1 from me.
• Mr. Xcoder: 91 bytes?: Transpose@Partition[#~Append~-1,5]&/@Last@Reap[Do[Sow[k,k~NumberExpand~2],{k,60}],Except@0]
• Bruno Le Floch: @Mr.Xcoder Thanks. I've used this ~ trick in one more place and replaced the variable k by Null. Sorry, no time to add a tio link.

Answer 15

Wolfram Language (Mathematica), 99 bytes

Transpose@Partition[#~FromDigits~2&/@Last@GatherBy[{0,1}~Tuples~6,#[[-k]]&],5]~Table~{k,6}/. 61->-1

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• Bruno Le Floch: You can save a few char by: doing Transpose@ instead of Transpose[...]; padding to 30 entries before partitioning; using Table[...,{k,6}] to avoid needing k=#.
• ZaMoC: @Bruno Le Floch Table may save one byte. Did you try transpose@? Because it doesn't work if you watch carefully. I'm afk but will golf later
• Bruno Le Floch: Sorry, Transpose@ works after you move PadRight inside Partition. Another comment is that the question does not seem to allow "" for the placeholder; you can replace it by -1 without losing any byte.

Answer 16

Jelly, 13 bytes

60B€Uz0μTs5Z)

Try it online!

Loosely based on flawr’s MATL answer. A niladic link which outputs a list of lists as required.

Answer 17

R, 73 bytes

`!`=as.raw;lapply(0:5,function(i)matrix(c((a=1:60)[(!a&!2^i)>0],-1),5,6))

I am not entirely sure if I have met the requirement for order, since R by default fills matrices by column, so the order such that it appears physically on the cards is the same as the way matrices are allocated in R.

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• Kevin Cruijssen: The output looks good. And if R fills matrices by columns before row instead of row before column like almost all other languages, it just means it's a good programming language to use for this challenge I guess. :)

Answer 18

T-SQL, ((削除) 1,168 (削除ここまで) 1,139 bytes)

I just wanted to know I could do it.

Optimised version

 WITH g AS(SELECT 1 AS n UNION ALL SELECT n+1 FROM g WHERE n+1<61),B as(SELECT cast(cast(n&32 as bit)as CHAR(1))+cast(cast(n&16 as bit)as CHAR(1))+cast(cast(n&8 as bit)as CHAR(1))+cast(cast(n&4 as bit)as CHAR(1))+cast(cast(n&2 as bit)as CHAR(1))+cast(cast(n&1 as bit)as CHAR(1))as b FROM g),P as(SELECT * from (values(1), (2), (4), (8), (16), (32)) as Q(p)),S as(select distinct p,p+(substring(b,6,1)*1)*(case when p=1 then 0 else 1 end)+(substring(b,5,1)*2)*(case when p=2 then 0 else 1 end)+(substring(b,4,1)*4)*(case when p=4 then 0 else 1 end)+(substring(b,3,1)*8)*(case when p=8 then 0 else 1 end)+(substring(b,2,1)*16)*(case when p=16 then 0 else 1 end)+(substring(b,1,1)*32)*(case when p=32 then 0 else 1 end)as e from P cross apply B),D as(select * from S where e>=p and e<61),R as(select p,(row_number()over(partition by p order by cast(e as int)))%5 as r,e from D),H as(select k.p,'['+stuff((select','+cast(l.e as varchar)from R l where l.p=k.p and l.r=k.r for xml path('')),1,1,'')+']'as s from R k group by k.p,k.r)select stuff((select','+cast(x.s as varchar)from H x where x.p=z.p for xml path('')),1,1,'')from H z group by z.p

Online demo

Try it online!

Verbose version - with notes as SQL comments

WITH gen -- numbers 1 to 60
AS (
 SELECT 1 AS num
 UNION ALL
 SELECT num+1 FROM gen WHERE num+1<=60
),
BINARIES -- string representations of binaries 000001 through 111111
as (
SELECT 
 +cast( cast(num & 32 as bit) as CHAR(1))
 +cast( cast(num & 16 as bit) as CHAR(1))
 +cast( cast(num & 8 as bit) as CHAR(1))
 +cast( cast(num & 4 as bit) as CHAR(1))
 +cast( cast(num & 2 as bit) as CHAR(1))
 +cast(cast(num & 1 as bit) as CHAR(1)) as binry FROM gen
),
POWERS -- first 6 powers of 2
as (
SELECT * from (values(1), (2), (4), (8), (16), (32)) as Q(powr)
),
SETELEMENTS -- cross apply the six powers of 2 against the binaries
-- returns 2 cols. col 1 = the power of 2 in question.
-- col 2 is calculated as that power of 2 plus the sum of each power of 2 other than the current row's power value, 
-- but only where a given power of 2 is switched "on" in the binary string, 
-- ie. where the first digit in the string represents 32, the second represents 16 and so on. 
-- That is, if the binary is 100100 then the number will be 
-- the sum of (32 x 1) + (16 x 0) + (8 x 0) + (4 x 1) + (2 x 0) + (1 x 0) 
-- but if the current row's power is 32 or 4, then just that number (32 or 4) is excluded from the sum.
-- rows are distinct.
as (
select distinct powr,
powr+
 (substring(binry,6,1) * 1) * (case when powr = 1 then 0 else 1 end)
 +(substring(binry,5,1) * 2) * (case when powr = 2 then 0 else 1 end)
 +(substring(binry,4,1) * 4) * (case when powr = 4 then 0 else 1 end)
 +(substring(binry,3,1) * 8) * (case when powr = 8 then 0 else 1 end)
 +(substring(binry,2,1) * 16) * (case when powr = 16 then 0 else 1 end)
 +(substring(binry,1,1) * 32) * (case when powr = 32 then 0 else 1 end) as elt
from POWERS cross apply BINARIES
),
DISTINCTELEMENTS -- purge calculated numbers smaller than the power of 2 or greater than 60
as (
select * from SETELEMENTS where elt >= powr and elt < 61
)--,
,
ROWNUMBERED -- for each power, number the rows repeatedly from 0 through 5, then back to 0 through 5 again, etc
as (
select powr, (row_number() over (partition by powr order by cast(elt as int)))%5 as r, elt from DISTINCTELEMENTS
),
GROUPEDSETS -- for each row number, within each power, aggregate the numbers as a comma-delimited list and wrap in square brackets - the inner arrays
as (
select r1.powr, '['+stuff((select ',' + cast(r2.elt as varchar) from ROWNUMBERED r2 where r2.powr = r1.powr and r2.r = r1.r for xml path('')),1,1,'')+']' as s
from ROWNUMBERED r1
group by r1.powr,r1.r
)
select -- now aggregate all the inner arrays per power
stuff((select ',' + cast(g2.s as varchar) from GROUPEDSETS g2 where g2.powr = g1.powr for xml path('')),1,1,'')
from GROUPEDSETS g1
group by g1.powr

Voila!

Note 1: Some of the logic pertains to rendering square brackets and commas.

Note 2: Newer versions of SQLServer have more compact approaches to creating comma-delimited lists. (This was created on SQL Server 2016.)

Note 3: Arrays for a given card are not sorted (which is ok per the spec). Numbers within an array are correctly sorted. In this case, each "card" of the question, renders its arrays on a separate row in the results.

Shorter to hard-code arrays?

Yes.

Byte me.

• Jo King: Jeez, wouldn't it be shorter to just hardcode the result?
• youcantryreachingme: Haha. Neither as fun, nor extensible.
• Jonathan Frech: I do not fully understand -- are you saying that your solution only works by chance or are you convinced you followed the specifications correctly?
• youcantryreachingme: @JonathanFrech - I did not explicitly code for the ordering of the numbers, although there may be an implicit condition in the language resulting in a guaranteed order. They render in correct ascending order. Separately, after posting, I realised I had misunderstood how the data were to be presented (in striped arrays per card, rather than one single set per card) - so have yet to solve that problem. As such, the result currently renders the correct numbers, in ascending order, within each of the 6 expected sets - see the linked sql fiddle. Still to do: break the sets into 5 subsets each.
• Jonathan Frech: I appreciate your effort but if your solution is not correct, please fix it or delete your post. We generally do not allow for invalid answers to linger.

Answer 19

C# (Visual C# Interactive Compiler), 112 bytes

_=>"
 ".Select(x=>Enumerable.Range(1,60).Where(l=>(l&x)>0).Select((a,b)=>new{a,b}).GroupBy(i=>i.b%5,i=>i.a))

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Answer 20

Red, (削除) 108 (削除ここまで) 107 bytes

n: 32 until[b: collect[repeat k 60[if n and k = n[keep k]]]loop 5[print
extract b 5 b: next b]1 > n: n / 2]

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Answer 21

APL+WIN, (削除) 65 (削除ここまで) 62 bytes

v←∊+\ ̈n, ̈29⍴ ̈1↓ ̈(n⍴ ̈1), ̈1+n←2*0,⍳5⋄((v=61)/v)← ̄1⋄1 3 2⍉6 6 5⍴v

Try it online! Courtesy of Dyalog Classic

Answer 22

MATLAB, 155 bytes

cellfun(@disp,cellfun(@(x)x-repmat(62,5,6).*(x>60),cellfun(@(x)reshape(find(x,30),[5 6]),mat2cell(dec2bin(1:62)-48,62,ones(1,6)),'Uniform',0),'Uniform',0))

This could be shorter as multiple lines but I wanted to do it in one line of code.

• Kevin Cruijssen: Could you perhaps add a TIO link with test code, so I can verify the output?

Answer 23

05AB1E, 14 bytes

žOε60LDNo&ĀÏ5ι

Try it online!

• Kevin Cruijssen: Why the žO instead of just 6L? I know you're not using them in your map, but I'm curious why you've used aeiouy to create a list with 6 values. xD Nice answer, btw!
• Grimmy: @KevinCruijssen No particular reason, I just thought it was funnier than 6L, 6и, 5Ý, 5°, or 9!.
• Kevin Cruijssen: It certainly caught my eye, that's for sure. ;)
• Grimmy: @KevinCruijssen I just realized тœ, 5œ, 1œ, also work, those are pretty cool too (:
• Kevin Cruijssen: 6b would work as well ;)