# How many Wazirs can be placed on an N×N Chessboard?

Suppose a new fairy chess piece named the Wazir is introduced to chess. Wazirs can move from a position (x, y) to:
(x+1, y)
(x, y+1)
(x-1, y)
(x, y-1)

That is, they move orthogonally like the rook, but only one step at a time like the king. How many such wazirs can be placed on an N×N chessboard so that no two wazirs can attack each other?

On a 1×1 board, there can be only 1 such piece.
On a 2×2 board, there can be 2 such pieces.
On a 3×3 board, there can be 5 such pieces.

Given N, return the number of wazirs that can be placed on an N×N chessboard.

This is OEIS sequence A000982.

### More test cases

725

832

1005000

• So the Camel is to the Rook what the King is to the Queen? I.e. can only move orthogonally, and only one step at a time.
Commented Oct 2, 2017 at 8:32
• @SashaR May I rewrite your question as a proper code golf challenge?
Commented Oct 2, 2017 at 9:09
• As a new user of this site, you've been very lucky this time. A lot of (off-topic) programming questions on this site has been permanently closed and downvoted, not edited as a challenge and upvoted like this one. As other people already explained, this site is for programming competitions only, not for asking homework. You can use the sandbox (at codegolf.meta.stackexchange.com/questions/2140/… ) before posting a challenge to avoid common mistakes next time; and note that most users on this site, as you have seen, use "unreadable" languages. Commented Oct 2, 2017 at 14:52
• This question is pretty confusing in that the Camel is already the standard fairy chess name for a piece like a knight that makes longer jumps and the piece you describe already has a fairy chess name: Wazir. Commented Oct 3, 2017 at 11:26
• OTOH editing the question to the the standard name after answers have already been written using the name used in the question is also confusing. Commented Oct 4, 2017 at 8:40

# Whitespace, 45 bytes





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By the way, here is a proof that the ⌈n²/2⌉ formula is correct.

• We can always place at least ⌈n²/2⌉ wazirs: just lay them out in a checkerboard pattern! Assuming the top-left tile is white, there are ⌈n²/2⌉ white tiles and ⌊n²/2⌋ black tiles on the n × n board. And if we place wazirs on the white tiles, no two of them are attacking each other, as every wazir only “sees” black tiles.

Here’s how we place 13 wazirs on a 5 × 5 board (each W is a wazir).

• We can’t do any better: let’s arbitrarily tile the checkerboard with 2 × 1 domino pieces, optionally using a 1 × 1 piece for the final corner of an odd-length chessboard, like so:

We need ⌈n²/2⌉ dominoes to cover the chessboard. Clearly, putting two wazirs on one domino makes it so that they can attack one another! So each domino can only contain at most one wazir, meaning we can’t possibly place more than ⌈n²/2⌉ wazirs on the board.

• You don't need the pigeonhole principle for the last part: you have exactly ⌈n²/2⌉ tiles, and at most camel per tile, so you have at most ⌈n²/2⌉ camels. Commented Oct 3, 2017 at 7:29
• @ShreevatsaR What ensures you that you can't put x > ⌈n²/2⌉ camels in ⌈n²/2⌉ tiles? It's the pigeonhole principle... Commented Oct 3, 2017 at 7:58
• I thought at first the code didn't load, so I refreshed the page, and it still didn't. Then I realized what language name was written at the top. Commented Oct 4, 2017 at 9:28
• I appreciate that you went and changed your C's to W's in your proof illustration. Commented Oct 4, 2017 at 13:39
• I also appreciate that the W's are all on the WHITE SPACES with an answer in WHITESPACE. Commented Oct 4, 2017 at 16:13

# Oasis, 3 bytes

k>v


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square - increment - integer halve

# Prolog (SWI), 22 19 bytes

Saved 3 bytes thanks to Kevin Cruijssen

X*Y:-Y is(X*X+1)/2.


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• X*Y:-Y is(X*X+1)/2. (19 bytes) Commented Oct 2, 2017 at 13:43
• @KevinCruijssen: Doh! I suggested the same change to another user, but somehow I didn't take my own advice. Thank you :) Commented Oct 2, 2017 at 14:06

# APL (Dyalog), 97 6 bytes

Now uses Mr. Xcoder's formula.

This is an anonymous prefix tacit function which takes N as argument.

⌈2÷⍨×⍨


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×⍨ square N (lit. multiplication selfie, i.e. multiply by self)

2÷⍨ divide by 2

⌈ ceiling (round up)

• Wow!I have no idea how you did this !!Didn't get the logic thoughsigh Commented Oct 2, 2017 at 8:49
• Darn, someone already found out the pattern. Commented Oct 2, 2017 at 18:26
• Huh, just realized the formula is on the OEIS page. Probably shouldn't have linked that. Commented Oct 2, 2017 at 18:27

# JS (ES6) / C# polyglot, 11 bytes

n=>n*n+1>>1


### Test cases

let f =

n=>n*n+1>>1

console.log(f(7))
console.log(f(8))
console.log(f(100))

• C# polyglot Commented Oct 2, 2017 at 16:41
• @Shaggy Thanks for noticing! Commented Oct 2, 2017 at 16:49
• JS and C# polyglot seems extremely rare, given the differences between them. +1 Commented Oct 4, 2017 at 11:36

# 05AB1E, 3 bytes

Straight implementation of the formula given by A000982

n;î


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Explanation

n      # square
;     # divide by 2
î    # round up

• *;î works too Commented May 10, 2021 at 21:45

# dc, 6 bytes

2^2~+p


2^: square; 2~: divide by 2, pushing the quotient then the remainder; +p: add the remainder to the quotient & print.

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# Python 2, 17 bytes

1 byte thanks to @ovs.

lambda x:x*x+1>>1


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• Alternatively: lambda x:-~x**2/2 Commented Oct 2, 2017 at 23:46

# C (gcc), 2318 17 bytes

• Saved a byte thanks to Tahg; golfing n/2+n%2 to n+1>>1.
f(n){n=n*n+1>>1;}


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# C (gcc), 22 bytes (not using undefined behavior)

f(n){return n*n+1>>1;}


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Some people really do not like exploiting a certain compiler's undefined bahavior when using specific compiler flags. Doing so does save bytes, though.

• Odd way of providing an answer IMO, but: f(n){n=n*n+1>>1;} to save a byte.
– Tahg
Commented Oct 2, 2017 at 9:56
• @Tahg Thanks; though in what way do you find my way of providing an answer odd? Commented Oct 2, 2017 at 10:51
• I didn't think changing the input argument was a normal way of returning a value in C.
– Tahg
Commented Oct 2, 2017 at 13:41
• @YSC Yet in the compiler's opinion it is understandable and creates an executable which works. Commented Oct 3, 2017 at 15:10
• @YSC We believe here on PPCG that, if the program works on one interpreter, it is a valid submission. It works on an online interpreter, it is therefore valid without any further remarks. Commented Oct 3, 2017 at 22:37

# Brachylog, 6 bytes

^₂/₂⌉₁


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# Python 3, 19 bytes

lambda x:-(-x*x//2)


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lambda x:-(-x*x//2)  # Unnamed function
lambda x:            # Given input x:
x*x      # Square
-         # Negate
//2   # Halve and Floor (equivalent of Ceil)
-(       )  # Negate again (floor -> ceil)


-1 byte thanks to Mr. Xcoder

• x**2 -> x*x Commented Oct 2, 2017 at 13:00
• @Mr.Xcoder Facepalm thanks Commented Oct 2, 2017 at 13:06
• What about lambda x:x*x+1>>1 ? Commented Oct 2, 2017 at 15:47
• Or lambda x:x*x+1//2 Disclaimer: I don't know this language's syntax or order of operations, so I guessed; I'm saying add 1 before you //2 instead of negating twice. Commented Oct 2, 2017 at 15:57
• @DanHenderson You still need parentheses otherwise it's parsed as (x*x) + (1//2), so that isn't actually shorter. Commented Oct 2, 2017 at 17:09

# x86_64 machine language (Linux), 9 8 bytes

0:       97                      xchg   %eax,%edi
1:       f7 e8                   imul   %eax
3:       ff c0                   inc    %eax
5:       d1 f8                   sar    %eax
7:       c3                      retq


To Try it online!, compile and run the following C program.

#include<stdio.h>
const char *f="\x97\xf7\xe8\xff\xc0\xd1\xf8\xc3";
int main() {
for(int i=1; i<10; i++) {
printf("%d\n", ((int(*)())f)(i));
}
}

• Segmentation fault on TIO
– l4m2
Commented Apr 30, 2021 at 20:20
• @l4m2 Fixed, thanks! Commented May 9, 2021 at 20:36

# J, 8 bytes

Anonymous tacit prefix function.

2>.@%~*:


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*: square

>. ceiling (round up)
@ after
2%~ dividing by two

• Alternate solutions: <.@-:@*: and *:<.@%2: Commented Oct 3, 2017 at 22:39
• @ConorO'Brien 2>.@%~*:‽ Where did I get that from? I can't read that – looks like line noise to me…
Commented Oct 3, 2017 at 22:47
• >.@-:@*: gets my vote. Commented Oct 4, 2017 at 2:56
• @Jonah If you squint, you can see a camel.
Commented Oct 4, 2017 at 8:30

# Actually, 3 bytes

²½K


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# Mathematica, 12 bytes

⌈#^2/2⌉&

• 11 bytes: ⌈#.5#⌉& Commented Oct 5, 2017 at 0:48

# R, 22 21 bytes

cat((scan()^2+1)%/%2)


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Square, increment, integer divide. Easy peasy.

Input from stdin; it can take space or newline separated input and it will compute the max wazirs for each input boardsize. Output to stdout.

-1 byte thanks to plannapus

• @plannapus fixed, thank you. Commented Dec 11, 2017 at 20:48
• Is the cat necessary? if you run it as an Rscript, it will auto output to stdout.
– qwr
Commented Apr 30, 2021 at 21:34
• @qwr there was an older consensus that you had to explicitly write output out; we've mostly gone away from that at this point, but this answer was following that particular rule. Commented Apr 30, 2021 at 21:44

# Pyth, 6 bytes

.E**.5


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# Pyke, 3 bytes

Xhe


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## How?

X   - Square.
h  - Increment.
e - Floor halve.


# C# (.NET Core), 14 12 bytes

-2 bytes thanks to Emigna

n=>(n*n+1)/2


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• Save 2 bytes Commented Oct 2, 2017 at 12:46
• n=>n*n+1>>1 (11 bytes) Commented Oct 2, 2017 at 13:40
• I have noticed the javascript answer to do the same, and this change would make our answers byte to byte equivalent. Commented Oct 2, 2017 at 13:41

# Perl 5, 16 bytes

$_=0|$_*$_/2+.5  Try it online! • $_=$_*$_+1>>1 14 bytes Commented Oct 4, 2017 at 8:11

# MATL, 5 bytes

UH/Xk


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• or UQ2/k at the same byte count. Commented Oct 2, 2017 at 13:31

# Cubix, 11 bytes

Iu*:^\)2,O@


Heheh, :^\)

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Expands to the following cube:

    I u
* :
^ \ ) 2 , O @ .
. . . . . . . .
. .
. .


Which is the same algorithm that many use.

• ^Iu : read in input as int and change directions
• :* : dup top of stack, multiply
• \) : change direction, increment
• 2, : push 2, integer divide
• O@ : print output as int, end program.

# Pyke, 3 bytes

Xhe


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X   -   input ** 2
h  -  ^ + 1
e - floor_half(^)

• I ninja'd you by about 12 hours :-/ Commented Oct 3, 2017 at 11:56

# Ohm v2, 3 bytes

²½ı


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² squares, ½ halves, ı ceils.

# Jelly, 3 bytes

-1 thanks to Mr. Xcoder.

Based on my APL solution.

²HĊ


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² Square

HHalve

ĊCeiling (round up)

# Japt, 4 bytes

Been sitting on these since the challenge was closed.

²Ä z


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Explanation: Square, add 1, floor divide by 2

## Alternative

²ÄÁ1


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Explanation: Square, add 1, bit-shift right by 1.

# ,,,, 7 bytes

2*1+1»


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Yay, commata works for something!

Fudge, I didn't implement ceiling...

# Gaia, 3 bytes

sḥ⌉


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s squares, ḥ halves, ⌉ ceils.

# Commentator, 19 bytes

//{-//-}! {-#  -}<!


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Who needs golfing languages? I've got confusing languages!

Ungolfed version:

5//{-8}//{5-}
print(10!= 5)
x={-1,3,4} # Smiley :-}
print(5<!=10)*/ # Weird comparision.


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How does it work? I'll explain, with input 5

//                         - Take input.                           Tape: [5 0 0]
{-//-}!                  - Square the input.                     Tape: [25 0 0]
{-                         - Move one along the tape
//                       - Copy the input to the tape.         Tape: [5 5 0]
-}                     - Move one back along the tape
!                    - Take the product of the tape.       Tape: [25 5 0]
<space>           - Increment the tape head.              Tape: [26 5 0]
{-#  -}<! - Halve the tape head (floor division). Tape: [13 2 0]
{-          - Move one along the tape
#         - Set the tape head to 2.             Tape: [26 2 0]
-}     - Move one back along the tape
<!   - Reduce the tape by floor division.  Tape: [13 2 0]


# OCaml, 19 bytes

let f n=(n*n+1)/2;;


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I'm a bit bummed the name got changed from "camels" to "wazirs" before I managed to write this, but I figured I'd post it anyway.