Rotate matrix rows according to the row above

Question

Specifications

For this challenge, you will be given a matrix of some sort in any reasonable format for a 2-D array. For each row except the last, in order from top (first) to bottom (last), rotate the row below it by x to either direction (you choose a consistent direction to use) where x is the first element in the current row.

Reference

A reference implementation in Python can be found here

Example

Let's walk through an example. Take the following matrix for example:

1 5 8
4 7 2
3 9 6

Let's say we're rotating right. First we rotate [4 7 2] to the right by 1 because the first element of [1 5 8] is 1. Thus, we get [2 4 7] as the second row. The matrix is now like this:

1 5 8
2 4 7
3 9 6

Then, we rotate [3 9 6] to the right by 2 because the first element of the second row is 2. Thus we get:

1 5 8
2 4 7
9 6 3

Challenge

Implement the above

Test Cases

Test cases are given as a list of lists in Python style.

Input -> Output
[[6, 7, 10, 1, 10, 7], [7, 3, 7, 8, 9, 2], [6, 8, 3, 9, 3, 1], [6, 3, 8, 6, 4, 1], [7, 5, 2, 9, 7, 2], [2, 10, 9, 9, 7, 9], [8, 8, 10, 10, 8, 4]] -> [[6, 7, 10, 1, 10, 7], [7, 3, 7, 8, 9, 2], [1, 6, 8, 3, 9, 3], [1, 6, 3, 8, 6, 4], [2, 7, 5, 2, 9, 7], [7, 9, 2, 10, 9, 9], [4, 8, 8, 10, 10, 8]]
[[8, 10, 4, 3, 9, 4, 6], [8, 8, 8, 8, 6, 7, 5], [6, 2, 2, 4, 8, 9, 6], [1, 7, 9, 9, 10, 7, 8]] -> [[8, 10, 4, 3, 9, 4, 6], [5, 8, 8, 8, 8, 6, 7], [2, 4, 8, 9, 6, 6, 2], [7, 8, 1, 7, 9, 9, 10]]
[[6, 2, 4, 4, 4], [5, 9, 10, 5, 4], [3, 5, 7, 2, 2], [1, 5, 5, 10, 10], [8, 2, 3, 2, 1], [3, 3, 9, 5, 10], [9, 5, 9, 7, 2]] -> [[6, 2, 4, 4, 4], [4, 5, 9, 10, 5], [5, 7, 2, 2, 3], [1, 5, 5, 10, 10], [1, 8, 2, 3, 2], [10, 3, 3, 9, 5], [9, 5, 9, 7, 2]]
[[3, 6, 3, 7, 4], [8, 8, 1, 5, 8], [7, 5, 1, 9, 4], [3, 9, 10, 8, 6]] -> [[3, 6, 3, 7, 4], [1, 5, 8, 8, 8], [4, 7, 5, 1, 9], [9, 10, 8, 6, 3]]
[[5, 6, 1, 2, 6], [1, 10, 10, 1, 4], [6, 2, 7, 1, 7], [9, 5, 8, 6, 3], [4, 5, 1, 2, 1], [8, 6, 1, 1, 3]] -> [[5, 6, 1, 2, 6], [1, 10, 10, 1, 4], [7, 6, 2, 7, 1], [6, 3, 9, 5, 8], [1, 4, 5, 1, 2], [3, 8, 6, 1, 1]]
[[10, 1, 2], [9, 9, 10], [2, 4, 7]] -> [[10, 1, 2], [10, 9, 9], [7, 2, 4]]
[[4, 5, 6], [6, 9, 4], [9, 3, 2], [3, 5, 9], [5, 5, 3], [9, 1, 4]] -> [[4, 5, 6], [4, 6, 9], [2, 9, 3], [5, 9, 3], [5, 3, 5], [1, 4, 9]]
[[8, 7, 6, 5], [9, 8, 6, 6], [7, 9, 7, 10]] -> [[8, 7, 6, 5], [9, 8, 6, 6], [10, 7, 9, 7]]
[[10, 2, 8, 8, 1], [2, 10, 1, 4, 10], [3, 2, 5, 3, 8], [5, 1, 8, 1, 8], [6, 1, 3, 1, 2], [7, 9, 1, 1, 2]] -> [[10, 2, 8, 8, 1], [2, 10, 1, 4, 10], [3, 8, 3, 2, 5], [8, 1, 8, 5, 1], [3, 1, 2, 6, 1], [1, 1, 2, 7, 9]]
[[9, 1, 3, 2, 7], [3, 8, 10, 3, 3], [8, 10, 7, 9, 5], [8, 1, 4, 9, 9], [6, 8, 4, 10, 10], [4, 7, 6, 2, 2], [8, 5, 3, 7, 6]] -> [[9, 1, 3, 2, 7], [8, 10, 3, 3, 3], [7, 9, 5, 8, 10], [9, 9, 8, 1, 4], [8, 4, 10, 10, 6], [6, 2, 2, 4, 7], [6, 8, 5, 3, 7]]

Generate more testcases here

we rotate [4 7 2] to the right by 1 ... we get [2 7 4] Wait, what? Shouldn't it be [2 4 7]? Or am I misunderstanding "rotation"? — totallyhuman
– totallyhuman, Commented Sep 7, 2017 at 21:58
For each row except the last Shouldn't that be except the first? — totallyhuman
– totallyhuman, Commented Sep 7, 2017 at 22:00
@icrieverytim rotate the row below it technically this is inconsistent with my title but eh — hyperneutrino
– hyperneutrino ♦, Commented Sep 7, 2017 at 22:03
Are we guaranteed that the matrix is positive integers and that the matrix has at least two rows? — Giuseppe
– Giuseppe, Commented Sep 7, 2017 at 22:14
Since x can be larger than the length of the row below (according to the test cases), what exactly is [2, 5, 8] rotated by 4? The second test case says it's [2, 5, 8] which doesn't make a lot of sense to me... — totallyhuman
– totallyhuman, Commented Sep 7, 2017 at 22:20

Leo · Accepted Answer · 2017-09-08 00:24:50Z

6

Husk, 4 bytes

G(ṙ←

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Rotates rows to the left.

Explanation

G in Husk is scanl (or cumulative reduce from left, as some languages call it). When given a function of type x->x->x (like in this case) it can use as starting value the first element of the list.

( is needed here to combine the following two functions into a single one.

ṙ is "rotate to the left" and takes a number and a list as arguments.

← returns the first element of a list.

answered Sep 8, 2017 at 0:24

Leo

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miles · Accepted Answer · 2017-09-07 22:51:43Z

3

Jelly, 6 5 bytes

ṙ@Ḣð\

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Rotates to the left.

Saved a byte thanks to @Jonathan Allan.

Explanation

ṙ@Ḣð\  Input: array of arrays M
   ð\  Cumulative reduce over each array using
ṙ@       Rotate the RHS array using each in the LHS array
  Ḣ      Head

edited Sep 7, 2017 at 22:51

answered Sep 7, 2017 at 22:41

miles

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\$\begingroup\$ You can dump the € \$\endgroup\$

Jonathan Allan
– Jonathan Allan

2017-09-07 22:48:39 +00:00
Commented Sep 7, 2017 at 22:48
\$\begingroup\$ @JonathanAllan Thanks. Isn't ɓ supposed to work here, as in ṙḢɓ\. \$\endgroup\$

miles
– miles

2017-09-07 22:52:47 +00:00
Commented Sep 7, 2017 at 22:52
\$\begingroup\$ Ah, that actually may require a code change to reduce_cumulative; it would make sense to do so... \$\endgroup\$

Jonathan Allan
– Jonathan Allan

2017-09-07 23:00:20 +00:00
Commented Sep 7, 2017 at 23:00

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HatsuPointerKun · Accepted Answer · 2017-09-08 17:15:02Z

3

C++, 187 149 bytes

38 bytes thanks to Karl Napf

Output is done by the reference type parameter The parameter's type have to be a std::vector<std::vector<int>> type

auto r=[](auto&a){for(int i=1,j,v;i<a.size();++i){v=a[i-1][0]%a[i].size();for(j=0;j<v;++j){a[i].insert(a[i].begin(),a[i].back());a[i].pop_back();}}};

Trying to do as #define A a[i] and replacing all occurences of a[i] by A will result with same byte count code

Code to help for the test :

//Shift operator overload
std::ostream& operator<<(std::ostream& os, std::vector<std::vector<int>>& v) {
    os << "{\n";
    for (auto&a : v) {
        os << "\t{ ";
        for (auto&b : a) {
            os << b << ' ';
        }
        os << "},\n";
    }
    return os << "}\n";
}

And in the main function

std::vector<std::vector<int>> t{
    {1,5,8},
    {4,7,2},
    {3,9,6}
};

std::cout << t;

r(t);

std::cout << t;

edited Sep 8, 2017 at 17:15

answered Sep 8, 2017 at 14:30

HatsuPointerKun

2,2111 gold badge9 silver badges13 bronze badges

1

\$\begingroup\$ You can drop the #include and std::vector<... declarations by using an unnamed generic lambda [](auto&a){...} and demanding the input to be like vector<vector<int>> as I did in my challenges, e.g. codegolf.stackexchange.com/questions/100205/… \$\endgroup\$

Karl Napf
– Karl Napf

2017-09-08 16:08:26 +00:00
Commented Sep 8, 2017 at 16:08
\$\begingroup\$ 141 bytes \$\endgroup\$

ceilingcat
– ceilingcat

2021-01-07 07:23:52 +00:00
Commented Jan 7, 2021 at 7:23

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ბიმო · Accepted Answer · 2018-05-05 00:29:58Z

2

Haskell, 84 59 45 bytes

Shifts to the left (now using Leo's approach):

scanl1(\a r->(drop<>take$a!!0`mod`length r)r)

Try it online!

edited May 5, 2018 at 0:29

answered May 4, 2018 at 22:20

ბიმო

17k3 gold badges43 silver badges106 bronze badges

\$\begingroup\$ 59 bytes: Try it online! \$\endgroup\$

Laikoni
– Laikoni

2018-05-04 23:36:12 +00:00
Commented May 4, 2018 at 23:36
\$\begingroup\$ @Laikoni: Damn that's so much nicer and saves a whole lot of bytes, thanks a lot! \$\endgroup\$

ბიმო
– ბიმო

2018-05-05 00:23:03 +00:00
Commented May 5, 2018 at 0:23

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Mr. Xcoder · Accepted Answer · 2017-09-07 22:21:55Z

1

Python 2, 65 bytes

l=input();K=0
for i in l:g=-K%len(i);e=i[g:]+i[:g];print e;K=e[0]

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Python 2, 55 bytes (possibly broken)

If this version is found to fail for some test cases, I can assure you the above works. I am currently half-asleep, so feel free to remove this version in case it is invalid by editing.

l=input();K=0
for i in l:e=i[-K:]+i[:-K];print e;K=e[0]

Try it online!

edited Sep 7, 2017 at 22:21

answered Sep 7, 2017 at 22:10

Mr. Xcoder

42.9k9 gold badges87 silver badges221 bronze badges

\$\begingroup\$ You need to add modulo somewhere in there. The rotation numbers are sometimes longer than the lengths. \$\endgroup\$

totallyhuman
– totallyhuman

2017-09-07 22:15:26 +00:00
Commented Sep 7, 2017 at 22:15
\$\begingroup\$ @icrieverytim I think it's fixed \$\endgroup\$

Mr. Xcoder
– Mr. Xcoder

2017-09-07 22:17:32 +00:00
Commented Sep 7, 2017 at 22:17
\$\begingroup\$ ...Actually, the test cases don't reflect that, which is in turn because the reference implementation doesn't use modulo. I'll ask OP whether it's needed. \$\endgroup\$

totallyhuman
– totallyhuman

2017-09-07 22:19:11 +00:00
Commented Sep 7, 2017 at 22:19

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Riley · Accepted Answer · 2017-09-08 13:26:42Z

1

05AB1E, 9 bytes

vyNFÁ}=нƒ

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vy        # For each row:
  NFÁ}    #   Rotate this row N times (initially 0)
      =   #   Print without popping
       н  #   Get the head
        ƒ #   Assign that to N

answered Sep 8, 2017 at 13:26

Riley

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Xcali · Accepted Answer · 2017-09-07 22:18:35Z

0

Perl 5, 41 + 1 (`-a`) = 42 bytes

push@F,shift@F for 1..$r;say"@F";$r=$F[0]

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Rotates to the left. Takes the input as lines of space separated integers.

How?

Remove an element from the left end of the list, put it on the right end as many times as we should rotate ($r). Since $r starts at undef (equivaltent to 0 in this usage), nothing happens to the first line. Output the list. Save the first element of the list in $r so we can rotate that many times on the next line.

edited Sep 7, 2017 at 22:18

answered Sep 7, 2017 at 22:13

Xcali

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Pyth, 11 bytes

VQ=ZhJ.>NZJ

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Explanation

VQ - For each list in the matrix input:
- =Z - Assign Z to the value of the below statement. Its initial value is 0, so that helps us with the first row.
- h - The first element of ...
- J.>NZ - ... The current list, cyclically rotated by Z places, and assigned to a variable J.
- J - Print J.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 7, 2017 at 22:27

Mr. Xcoder

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Giuseppe · Accepted Answer · 2017-09-08 17:47:07Z

0

R, 77 bytes

function(B){for(i in 2:nrow(B))B[i,]=rep(B[i,],sum(B))[1:ncol(B)+B[i-1,1]]
B}

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Rotates rows to the right.

Replicates each row of B sum(B) times (creating a huge vector), then grabs the appropriate values from that replicated vector to put into the correct row.

I also wrote a function in the header that converts from the python matrix format to R matrices.

answered Sep 8, 2017 at 17:47

Giuseppe

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Adám · Accepted Answer · 2017-10-16 06:09:08Z

APL (Dyalog), 14 bytes

Anonymous tacit prefix function. Takes list of lists as argument and prints rows on separate lines with the numbers in each row separated by single spaces. Rotates left (although the explanation mentions how rotate right).

{⎕←⍺⌽⍨⊃⍵}/0,⍨⌽

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⌽ reverse the argument (because reduction is right-to-left)

0,⍨ append a zero (for initial rotation of zero steps)

{…}/ reduce by the following anonymous lambda, where ⍺ is the element on the left and ⍵ on the right (the reduction goes right-to-left

⊃ pick the first of the right argument

⍺⌽⍨ rotate the left argument to the left* that many steps

⎕← output to STDOUT with a trailing line-break

* To rotate right, insert a - to the right of ⍨. This will negate the rotation amount.

Stack Exchange Network

Rotate matrix rows according to the row above

Specifications

Reference

Example

Challenge

Test Cases

10 Answers 10

Husk, 4 bytes

Explanation

Jelly, 6 5 bytes

Explanation

C++, 187 149 bytes

Haskell, 84 59 45 bytes

Python 2, 65 bytes

Python 2, 55 bytes (possibly broken)

05AB1E, 9 bytes

Perl 5, 41 + 1 (`-a`) = 42 bytes

Pyth, 11 bytes

Explanation

R, 77 bytes

APL (Dyalog), 14 bytes

Your Answer

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Rotate matrix rows according to the row above

Specifications

Reference

Example

Challenge

Test Cases

10 Answers 10

Husk, 4 bytes

Explanation

Jelly, 6 5 bytes

Explanation

C++, 187 149 bytes

Haskell, 84 59 45 bytes

Python 2, 65 bytes

Python 2, 55 bytes (possibly broken)

05AB1E, 9 bytes

Perl 5, 41 + 1 (-a) = 42 bytes

Pyth, 11 bytes

Explanation

R, 77 bytes

APL (Dyalog), 14 bytes

Your Answer

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Perl 5, 41 + 1 (`-a`) = 42 bytes