All the single eights

25

2

Given a non-empty rectangular array of integers from 0 to 9, output the amount of cells that are 8 and do not have a neighbour that is 8. Neighbouring is here understood in the Moore sense, that is, including diagonals. So each cell has 8 neighbours, except for cells at the edges of the array.

For example, given the input

8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2

the output should be 3. The three qualifying cells would be the following, marked with an asterisk (but only the amount of such entries should be output):

* 4 5 6 5
9 3 8 4 *
0 8 6 1 5
6 7 9 * 2
8 8 7 4 2

Additional rules

Test cases

  1. Input:

    8 4 5 6 5
    9 3 8 4 8
    0 8 6 1 5
    6 7 9 8 2
    8 8 7 4 2
    

    Output: 3

  2. Input

    8 8
    2 3
    

    Output: 0

  3. Input:

    5 3 4
    2 5 2
    

    Output: 0

  4. Input:

    5 8 3 8
    

    Output: 2

  5. Input:

    8
    0
    8
    

    Output: 2.

  6. Input:

    4 2 8 5
    2 6 1 8
    8 5 5 8
    

    Output: 1

  7. Input:

    4 5 4 3 8 1 8 2
    8 2 7 7 8 3 9 3
    9 8 7 8 5 4 2 8
    4 5 0 2 1 8 6 9
    1 5 4 3 4 5 6 1
    

    Output 3.

  8. Input:

    8
    

    Output: 1

  9. Input:

    8 5 8 1 6 8 7 7
    9 9 2 8 2 7 8 3
    2 8 4 9 7 3 2 7
    9 2 9 7 1 9 5 6
    6 9 8 7 3 1 5 2
    1 9 9 7 1 8 8 2
    3 5 6 8 1 4 7 5
    

    Output: 4.

  10. Input:

    8 1 8
    2 5 7
    8 0 1
    

    Output: 3.

Inputs in MATLAB format:

[8 4 5 6 5; 9 3 8 4 8; 0 8 6 1 5; 6 7 9 8 2; 8 8 7 4 2]
[8 8; 2 3]
[5 3 4; 2 5 2]
[5 8 3 8]
[8; 0; 8]
[4 2 8 5; 2 6 1 8; 8 5 5 8]
[4 5 4 3 8 1 8 2; 8 2 7 7 8 3 9 3; 9 8 7 8 5 4 2 8; 4 5 0 2 1 8 6 9; 1 5 4 3 4 5 6 1]
[8]
[8 5 8 1 6 8 7 7; 9 9 2 8 2 7 8 3; 2 8 4 9 7 3 2 7; 9 2 9 7 1 9 5 6; 6 9 8 7 3 1 5 2; 1 9 9 7 1 8 8 2; 3 5 6 8 1 4 7 5]
[8 1 8; 2 5 7; 8 0 1]

Inputs in Python format:

[[8, 4, 5, 6, 5], [9, 3, 8, 4, 8], [0, 8, 6, 1, 5], [6, 7, 9, 8, 2], [8, 8, 7, 4, 2]]
[[8, 8], [2, 3]]
[[5, 3, 4], [2, 5, 2]]
[[5, 8, 3, 8]]
[[8], [0], [8]]
[[4, 2, 8, 5], [2, 6, 1, 8], [8, 5, 5, 8]]
[[4, 5, 4, 3, 8, 1, 8, 2], [8, 2, 7, 7, 8, 3, 9, 3], [9, 8, 7, 8, 5, 4, 2, 8], [4, 5, 0, 2, 1, 8, 6, 9], [1, 5, 4, 3, 4, 5, 6, 1]]
[[8]]
[[8, 5, 8, 1, 6, 8, 7, 7], [9, 9, 2, 8, 2, 7, 8, 3], [2, 8, 4, 9, 7, 3, 2, 7], [9, 2, 9, 7, 1, 9, 5, 6], [6, 9, 8, 7, 3, 1, 5, 2], [1, 9, 9, 7, 1, 8, 8, 2], [3, 5, 6, 8, 1, 4, 7, 5]]
[[8, 1, 8], [2, 5, 7], [8, 0, 1]]

Outputs:

3, 0, 0, 2, 2, 1, 3, 1, 4, 3

Luis Mendo

Posted 2018-11-11T19:52:34.853

Reputation: 87 464

18If you like it then you should have put a vote on it – Luis Mendo – 2018-11-11T19:52:42.933

When I read "cells that equal 8", for a moment I thought you meant that a cell could be larger than a 1x1 chuck (NxN) of the grid. Should probably rephrase that to "cells that are 8" to clarify no math needed. =P – Tezra – 2018-11-12T17:46:19.567

@Tezra Edited. I find the new wording a little less natural, but I’m not a native speaker so I’ll trust your criterion – Luis Mendo – 2018-11-12T18:57:34.840

Answers

2

MATL, 21 17 10 bytes

8=t3Y6Z+>z

Try it online!

Thanks to Luis Mendo for help in chat, and for suggesting 2D convolution.

Explanation:

	#implicit input, m
8=	#equal to 8? matrix of 1 where m is 8, 0 otherwise
t	#duplicate
3Y6	#push [1 1 1; 1 0 1; 1 1 1], "neighbor cells" for convolution
Z+	#2D convolution; each element is replaced by the number of neighbors that are 8
>	#elementwise greater than -- matrix of 1s where an 8 is single, 0s otherwise
z	#number of nonzero elements -- number of single eights
	#implicit output

Giuseppe

Posted 2018-11-11T19:52:34.853

Reputation: 21 077

You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept – Luis Mendo – 2018-11-12T17:31:01.010

1@LuisMendo 2D convolution is one of those things where I don't understand 1D convolution either so there's no hope for me there...sounds like an opportunity to learn both! – Giuseppe – 2018-11-12T17:33:30.363

1If you need help with that let me know in the chat room. Convolution is a very useful operation. If you want to learn convolution start with 1D. The generalization to 2D is immediate – Luis Mendo – 2018-11-12T17:39:42.297

9

R, 117 63 59 bytes

function(m)sum(colSums(as.matrix(dist(which(m==8,T)))<2)<2)

Try it online!

dist computes distances (default is Euclidean) among rows of a matrix. which with second argument TRUE returns the coordinates where the predicate is true.

Coordinates are neighbours if the distance between them is not more than the square root of 2, but the inner <2 is good enough because the possible distance jumps from sqrt(2) ro 2.

ngm

Posted 2018-11-11T19:52:34.853

Reputation: 3 974

it's a shame numerical imprecision doesn't allow colSums()^2<=2 to work. – Giuseppe – 2018-11-12T22:28:05.647

@Giuseppe of course there are only a few possible distances and sqrt(2) jumps to 2 (e.g. sort(c(dist(expand.grid(1:6,1:6))), decreasing = TRUE))) so we were being too clever there. – ngm – 2018-11-13T15:33:21.520

7

APL (Dyalog Classic), 29 28 25 bytes

≢∘⍸16=2⊥¨3,⌿3,/8=(⍉0,⌽)⍣4

Try it online!

ngn

Posted 2018-11-11T19:52:34.853

Reputation: 11 449

Note: 0 index origin is not even needed. – Zacharý – 2018-11-12T16:36:07.200

@Zacharý I always use it as a default, to avoid surprises. – ngn – 2018-11-12T20:03:15.597

Ah, so like others with 1 (except not explicitly set). That makes sense. – Zacharý – 2018-11-12T20:05:26.030

Surprised this doesn't use Stencil. Is there something that makes stencil inconvenient here? – lirtosiast – 2018-11-20T07:49:17.737

@lirtosiast it's just longer with it :) – ngn – 2018-11-20T09:39:51.860

5

Jelly, 18 15 bytes

8=µ+Ż+ḊZµ⁺ỊṖḋµS

Try it online!

How it works

8=µ+Ż+ḊZµ⁺ỊṖḋµS    Main link (monad). Input: digit matrix
8=              1) Convert elements by `x == 8`
  µ             2) New chain:
   +Ż+Ḋ              x + [0,*x] + x[1:] (missing elements are considered 0)
                     Effectively, vertical convolution with [1,1,1]
       Z             Transpose
        µ⁺      3) Start new chain, apply 2) again
          ỊṖ       Convert elements by `|x| <= 1` and remove last row
            ḋ      Row-wise dot product with result of 1)
             µS 4) Sum

Previous solution, 18 bytes

æc7B¤ZḊṖ
8=µÇÇỊḋµS

Try it online!

Wanted to share another approach, though this is 1 byte longer than Jonathan Allan's solution.

How it works

æc7B¤ZḊṖ    Auxiliary link (monad). Input: integer matrix
æc7B¤       Convolution with [1,1,1] on each row
     ZḊṖ    Zip (transpose), remove first and last elements

8=µÇÇỊḋµS    Main link (monad). Input: digit matrix
8=           Convert 8 to 1, anything else to 0 (*A)
  怀        Apply aux.link twice (effective convolution with [[1,1,1]]*3)
     Ịḋ      Convert to |x|<=1, then row-wise dot product with A
       µS    Sum the result

Bubbler

Posted 2018-11-11T19:52:34.853

Reputation: 16 616

4

JavaScript (Node.js), 88 85 bytes

a=>g=(x,y=0,c=0)=>a.map(p=>p.map((q,j)=>c+=q-8?0:1/x?(x-j)**2+y*y<3:g(j,-y)<2)&--y)|c

Try it online!

Thank Arnauld for 2 bytes

l4m2

Posted 2018-11-11T19:52:34.853

Reputation: 5 985

4

J,43, 40 37 bytes

-3 bytes thanks to Bubbler

1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4

Try it online!

Explanation:

The first part of the algorithm assures that we can apply a 3x3 sliding window to he input. This is achieved by prepending a row of zeroes and 90 degrees rotation, repeated 4 times.

1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
                         (       )^:4 - repeat 4 times
                          0|:@,|.     - reverse, prepend wit a row of 0 and transpose
                     ;._3             - cut the input (already outlined with zeroes)
      3 3                             - into matrices with size 3x3
         (          )                 - and for each matrix do
                   ,                  - ravel (flatten)
             8    =                   - check if each item equals 8
              #.@:                    - and convert the list of 1s and 0s to a decimal
          16=                         - is equal to 16?
   1#.                                - add (the result has the shape of the input)
1#.                                   - add again

Galen Ivanov

Posted 2018-11-11T19:52:34.853

Reputation: 13 815

137 bytes using @: and moving |.. Note that @ in place of @: doesn't work. – Bubbler – 2018-11-12T10:26:03.493

@Bubbler Thank you! – Galen Ivanov – 2018-11-12T11:07:12.453

This is nice. Probably worth adding at least a high level explanation of how it works, if not a code breakdown. It took me 10m or so to figure it out. Also, it's interesting how much shorter the APL version (which uses the same approach) is. Looks like that's mostly the result of digraphs instead of single char symbols... – Jonah – 2018-11-12T15:00:01.603

@Jonah I'll add an explanation. For comparison with APL you can look at the revisions of ngn's solution, especially the 28 byte version

– Galen Ivanov – 2018-11-12T15:16:47.813

1@Jonah Explanation added – Galen Ivanov – 2018-11-12T18:26:02.533

3

Jelly, 17 bytes

=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL

Try it online! Or see the test-suite.

How?

=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL - Link: list of lists of integers (digits)
=8                - equals 8?
  ŒṪ              - multidimensional truthy indices (pairs of row & column indices of 8s)
    µ             - start a new monadic chain
     Œc           - all pairs (of the index-pairs) 
            Ƈ     - filter keep if:  (keep those that represent adjacent positions)
           Ʋ      -   last four links as a monad:
       Z          -     transpose
        I         -     incremental differences
         Ị        -     insignificant? (abs(x) <= 1)
          Ȧ       -     all?
             Ẏ    - tighten (to a list of all adjacent 8's index-pairs, at least once each)
              ⁸   - chain's left argument (all the index-pairs again)
               ḟ  - filter discard (remove those found to be adjacent to another)
                L - length (of the remaining pairs of indices of single 8s)

Jonathan Allan

Posted 2018-11-11T19:52:34.853

Reputation: 67 804

3

Retina 0.8.2, 84 bytes

.+
_$&_
m`(?<!(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.|8)8(?!8|.(.)*¶.*8.?.?(?<-2>.)*$(?(2).))

Try it online! Explanation:

.+
_$&_

Wrap each line in non-8 characters so that all 8s have at least one character on each side.

m`

This is the last stage, so counting matches is implied. The m modifier makes the ^ and $ characters match at the start or end of any line.

(?<!...|8)

Don't match a character directly after an 8, or...

(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.

... a character below an 8; the (?(1).)^(?<-1>.)* matches the same column as the ¶(.)* on the next line, but the .?.? allows the 8 to be 1 left or right of the character after the . on the next line.

8

Match 8s.

(?!8|...)

Don't match an 8 immediately before an 8, or...

.(.)*¶.*8.?.?(?<-2>.)*$(?(2).)

... a character with an 8 in the line below; again, the (?<-2>.)*$(?(2).) matches the same column as the (.)*¶ on the previous line, but the .?.? allows the 8 to be 1 left or right of the 8 before the . on the previous line.

Neil

Posted 2018-11-11T19:52:34.853

Reputation: 95 035

3

J, 42 bytes

[:+/@,8=]+[:+/(<:3 3#:4-.~i.9)|.!.0(_*8&=)

Try it online!

explanation

The high-level approach here is similar to the one used in the classic APL solution to the game of life: https://www.youtube.com/watch?v=a9xAKttWgP4.

In that solution, we shift our matrix in the 8 possible neighbor directions, creating 8 duplicates of the input, stack them up, and then add the "planes" together to get our neighbor counts.

Here, we use a "multiply by infinity" trick to adapt the solution for this problem.

[: +/@, 8 = ] + [: +/ (neighbor deltas) (|.!.0) _ * 8&= NB. 
                                                        NB.
[: +/@,                                                 NB. the sum after flattening
        8 =                                             NB. a 0 1 matrix created by
                                                        NB. elmwise testing if 8
                                                        NB. equals the matrix
            (the matrix to test for equality with 8   ) NB. defined by...
            ] +                                         NB. the original input plus
                [: +/                                   NB. the elmwise sum of 8
                                                        NB. matrices defined by
                                                _ *     NB. the elmwise product of 
                                                        NB. infinity and
                                                    8&= NB. the matrix which is 1
                                                        NB. where the input is 8
                                                        NB. and 0 elsewhere, thus
                                                        NB. creating an infinity-0
                                                        NB. matrix
                                        (|.!.0)         NB. then 2d shifting that 
                                                        NB. matrix in the 8 possible
                                                        NB. "neighbor" directions
                      (neighbor deltas)                 NB. defined by the "neighbor
                                                        NB. deltas" (see below)
                                                        NB. QED.
                                                        NB. ***********************
                                                        NB. The rest of the
                                                        NB. explanation merely
                                                        NB. breaks down the neighbor
                                                        NB. delta construction.


                      (neighbor deltas  )               NB. the neighbor deltas are
                                                        NB. merely the cross product
                                                        NB. of _1 0 1 with itself,
                                                        NB. minus "0 0"
                      (<: 3 3 #: 4 -.~ i.9)             NB. to create that...
                       <:                               NB. subtract one from
                          3 3 #:                        NB. the base 3 rep of
                                       i.9              NB. the numbers 0 - 8
                                 4 -.~                  NB. minus the number 4
                                                        NB.
                                                        NB. All of which produces
                                                        NB. the eight "neighbor"
                                                        NB. deltas:
                                                        NB. 
                                                        NB.       _1 _1
                                                        NB.       _1  0
                                                        NB.       _1  1
                                                        NB.        0 _1
                                                        NB.        0  1
                                                        NB.        1 _1
                                                        NB.        1  0
                                                        NB.        1  1

Jonah

Posted 2018-11-11T19:52:34.853

Reputation: 8 729

1You have forgotten to remove a space between ~ and > – Galen Ivanov – 2018-11-12T11:12:47.617

@GalenIvanov Fixed now. Thank you. – Jonah – 2018-11-12T13:58:22.070

3

Java 8, 181 157 156 bytes

(M,R,C)->{int z=0,c,f,t;for(;R-->0;)for(c=C;c-->0;z+=f>1?0:f)for(f=0,t=9;M[R][c]==8&t-->0;)try{f+=M[R+t/3-1][c+t%3-1]==8?1:0;}catch(Exception e){}return z;}

-24 bytes thanks to @OlivierGrégoire.

Takes the dimensions as additional parameters R (amount of rows) and C (amount of columns).

The cells are checked pretty similar as I did in my Fryer simulator answer.

Try it online.

Explanation:

(M,R,C)->{                    // Method with integer-matrix as parameter & integer return
  int z=0,                    //  Result-counter, starting at 0
      c,f,t;                  //  Temp-integers, starting uninitialized
  for(;R-->0;)                //  Loop over the rows:
    for(c=C;c-->0             //   Inner loop over the columns:
           ;                  //     After every iteration:
            z+=f==1?          //      If the flag-integer is larger than 1:
                0             //       Leave the result-counter the same by adding 0
               :              //      Else:
                f)            //       Add the flag-integer (either 0 or 1)
      for(f=0,                //    Reset the flag to 0
          t=9;M[R][c]==8&     //    If the current cell contains an 8:
              t-->0;)         //     Inner loop `t` in the range (9, 0]:
        try{f+=               //      Increase the flag by:
               M[R+t/3-1]     //       If `t` is 0, 1, or 2: Look at the previous row
                              //       Else-if `t` is 6, 7, or 8: Look at the next row
                              //       Else (`t` is 3, 4, or 5): Look at the current row
                [c+t%3-1]     //       If `t` is 0, 3, or 6: Look at the previous column
                              //       Else-if `t` is 2, 5, or 8: Look at the next column
                              //       Else (`t` is 1, 4, or 7): Look at the current column
                ==8?          //       And if the digit in this cell is 8:
                 1            //        Increase the flag-integer by 1
                :0;           //       Else: leave it the same
        }catch(Exception e){} //      Catch and ignore ArrayIndexOutOfBoundsExceptions
                              //      (try-catch saves bytes in comparison to if-checks)
  return z;}                  //  And finally return the counter

Kevin Cruijssen

Posted 2018-11-11T19:52:34.853

Reputation: 67 575

2

Python 2, 130 bytes

lambda a:sum(sum(u[~-c*(c>0):c+2].count(8)for u in a[~-r*(r>0):r+2])*8==8==a[r][c]for r in range(len(a))for c in range(len(a[0])))

Try it online!

Chas Brown

Posted 2018-11-11T19:52:34.853

Reputation: 8 959

Seems shorter if take length from args – l4m2 – 2018-11-12T02:54:01.557

2

Powershell, 121 bytes

param($a)(($b='='*4*($l=($a|% Le*)[0]))+($a|%{"!$_!"})+$b|sls "(?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3})" -a|% m*).Count

Less golfed test script:

$f = {

param($a)

$length=($a|% Length)[0]
$border='='*4*$length
$pattern="(?<=[^8]{3}.{$length}[^8])8(?=[^8].{$length}[^8]{3})"
$matches=$border+($a|%{"!$_!"})+$border |sls $pattern -a|% Matches
$matches.count

}

@(

,(3,"84565","93848","08615","67982","88742")
,(0,"88","23")
,(0,"534","252")
,(2,"5838")
,(2,"8","0","8")
,(1,"4285","2618","8558")
,(3,"45438182","82778393","98785428","45021869","15434561")
,(1,"8")
,(4,"85816877","99282783","28497327","92971956","69873152","19971882","35681475")
,(3,"818","257","801")
,(0,"")

) | % {
    $expected,$a = $_
    $result = &$f $a
    "$($result-eq$expected): $result : $a"
}

Output:

True: 3 : 84565 93848 08615 67982 88742
True: 0 : 88 23
True: 0 : 534 252
True: 2 : 5838
True: 2 : 8 0 8
True: 1 : 4285 2618 8558
True: 3 : 45438182 82778393 98785428 45021869 15434561
True: 1 : 8
True: 4 : 85816877 99282783 28497327 92971956 69873152 19971882 35681475
True: 3 : 818 257 801
True: 0 : 

Explanation:

First, the script calculates a length of the first string.

Second, it adds extra border to strings. Augmended reality string likes:

....=========!84565! !93848! !08615! !67982! !88742!===========....

represents the multiline string:

...=====
=======
!84565!
!93848!
!08615!
!67982!
!88742!
=======
========...

Note 1: the number of = is sufficient for a string of any length.

Note 2: a large number of = does not affect the search for eights.

Next, the regular expression (?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3}) looks for the digit 8 with the preceding non-eights (?<=[^8]{3}.{$l}[^8]) and the following non-eights (?=[^8].{$l}[^8]{3}):

.......
<<<....
<8>....
>>>....
.......

Finally, the number of matches is returned as a result.

mazzy

Posted 2018-11-11T19:52:34.853

Reputation: 4 832

2

Jelly, 12 bytes

œẹ8ạṀ¥þ`’Ạ€S

Try it online!

How it works

œẹ8ạṀ¥þ`’Ạ€S  Main link. Argument: M (matrix)

œẹ8           Find all multidimensional indices of 8, yielding an array A of pairs.
      þ`      Table self; for all pairs [i, j] and [k, l] in A, call the link to the
              left. Return the results as a matrix.
   ạ              Absolute difference; yield [|i - k|, |j - l|].
    Ṁ             Take the maximum.
        ’     Decrement all the maxmima, mapping 1 to 0.
         Ạ€   All each; yield 1 for each row that contains no zeroes.
           S  Take the sum.

Dennis

Posted 2018-11-11T19:52:34.853

Reputation: 196 637

1

JavaScript (ES6), 106 bytes

a=>a.map((r,y)=>r.map((v,x)=>k+=v==8&[...'12221000'].every((d,i,v)=>(a[y+~-d]||0)[x+~-v[i+2&7]]^8)),k=0)|k

Try it online!


Bitwise approach, 110 bytes

a=>a.map(r=>r.map(v=>x=x*2|v==8,x=k=0)|x).map((n,y,b)=>a[0].map((_,x)=>(n^1<<x|b[y-1]|b[y+1])*2>>x&7||k++))&&k

Try it online!

Arnauld

Posted 2018-11-11T19:52:34.853

Reputation: 111 334

Bitwise approach fail on [[7]] – l4m2 – 2018-11-12T02:52:34.467

@lm42 Oh, thanks. Now fixed. – Arnauld – 2018-11-12T09:29:29.940

1

Clojure, 227 198 bytes

(fn[t w h](let[c #(for[y(range %3 %4)x(range % %2)][x y])e #(= 8(get-in t(reverse %)0))m(fn[[o p]](count(filter e(c(dec o)(+ o 2)(dec p)(+ p 2)))))](count(filter #(= 1(m %))(filter e(c 0 w 0 h))))))

Ouch. Definitely not the shortest here by any means. 54 bytes of parenthesis is killer. I'm still relatively happy with it though.

-29 bytes by creating a helper function that generates a range since I was doing that twice, changing the reduce to a (count (filter setup, and getting rid of the threading macro after golfing.

(defn count-single-eights [td-array width height]
  ; Define three helper functions. One generates a list of coords for a given range of dimensions, another checks if an eight is
  ; at the given coord, and the other counts how many neighbors around a coord are an eight
  (letfn [(coords [x-min x-max y-min y-max]
            (for [y (range y-min y-max)
                  x (range x-min x-max)]
              [x y]))
          (eight? [[x y]] (= 8 (get-in td-array [y x] 0)))
          (n-eights-around [[cx cy]]
            (count (filter eight?
                           (coords (dec cx) (+ cx 2), (dec cy) (+ cy 2)))))]

    ; Gen a list of each coord of the matrix
    (->> (coords 0 width, 0 height)

         ; Remove any coords that don't contain an eight
         (filter eight?)

         ; Then count how many "neighborhoods" only contain 1 eight
         (filter #(= 1 (n-eights-around %)))
         (count))))

(mapv #(count-single-eights % (count (% 0)) (count %))
      test-cases)
=> [3 0 0 2 2 1 3 1 4 3]

Where test-cases is an array holding all the "Python test cases"

Try it online!

Carcigenicate

Posted 2018-11-11T19:52:34.853

Reputation: 3 295