Find next multiple of 32 that follows given number without using *, / and loops

Question

Imagine simple math problem. You have a number, say it's 1000. And you want to find the next multiple of 32 that follows 1000. You automatically make some simple calculations and get the result. Typical algorithm looks like this:

int number = 1000, result;
result = (1 + (number/32)) * 32; // integer division

It's kind of easy and obvious, right? And now the challenge. Make this calculation without using *, / and loops.

Rules:

• Don't use *, / and loops.
• Assume that input number is already stored in variable n, store result in the variable r (already declared).
• The shortest code wins.

Sample input: (already stored in n)

1000

Sample output: (value of r)

1024

---

Sample input:

31

Sample output:

32

---

Sample input:

32

Sample output:

64

It is assumed that n is a positive number.

Answer 1

JavaScript (26 13 chars)

alert((+prompt()>>5)+1<<5)

Re-read the problem statement. Here's my amended answer:

r=(n>>5)+1<<5

And this seems to be the best javascript answer at 10 chars:

r=(n|31)+1

Answer 2

GolfScript (10 chars)

32n+31~&:r

Problems this trivial aren't interesting.

Answer 3

C (11 chars)

r=(n|31)+1;

Python (10 chars)

r=(n|31)+1

Work for negative numbers also

Answer 4

Mathematica - 15 13

⌈##⌉&[n+1,32]

It's quite simple if Ceiling (⌈...⌉) isn't forbidden

Answer 5

EXCEL (14 characters):

=n-MOD(n;32)+32

n is declared name for a an input cell. put the formula in a cell declared as 'r'. No use of / and * ...

Answer 6

Python (11 chars)

r=n+32-n%32

Using modulo

Alternative (10 chars), using bitwise OR like everbody else:

r=(n|31)+1

Answer 7

R, 12

r=n+32-n%%32

This uses modulo %%, subtraction -, and addition +.

Answer 8

Julia, 8

r=n|31+1

(This is based on the C solution.)

Some parentheses can be saved in Julia due to the order in which the functions are called:

:(r=n|31+1) # : returns the ATK
:(r = +(|(n,31),1))

Answer 9

R (10 16)

r=-n%%32+n

r=-(n=n+1)%%32+n

R (recursive, long)

r<-(function (n) if(n%%32) Recall(n+1) else n)(n+1)

Python (9 14)

r=-n%32+n

n+=1
r=-n%32+n

Answer 10

PowerShell: 14

$r=$n-$n%32+32

 

Answer 11

C (16 characters):

r=((n>>5)+1)<<5;

Answer 12

Kona (21)

 r:{_(1+(x%32))%(%32)}

No *, /, nor a loop in the above.

Answer 13

J - 11 characters

r=:2^>.2^.n

Answer 14

C - 19 characters

not the shortest, but a different method

j=i%32?i+32-i%32:i;

Answer 15

J (10)

This returns n if n is a multiple of 32. So it doesn't satisfy everything I think you want.

r=:n-_32|n

If we need to be stringent on that rule, the following works.

(14)

r=:>:n-_32|>:n

Answer 16

Perl 6 (9 bytes)

r=n+|31+1

r and n can be declared as sigilless variables (my \variable). In Perl 6, bitwise operators (like +|)'s precedence is better, so no parens are needed.

Answer 17

R, 40

Just for kicks, a solution that doesn't use the modulo operator either:

r=tail(which(!cbind(1:n,31:0)[,2]),1)+32

cbind(1:n,31:0) creates an array with one column going from 1 to n and another from 31 to 0. That column is recycled to fit the size of the first one. which(!cbind(1:n,31:0)[,2]) returns the indices of the elements for which the second column is 0 (0 being FALSE, !0 is TRUE), i.e. the multiples of 32. Finally we take the last one using tail and add 32.

Solution at 34 characters with rep:

r=tail(which(!rep(31:0,l=n)),1)+32

Answer 18

Perl5 14

$r=($n+32)&~31

In Perl5 variables needs sigilia. $n gets added 32 and we strip the last 5 bits by anding by the inverse of 31.

Answer 19

x86 Machine Code - 6 bytes

Assuming that the register 'ax' is 'n', and 'bx' is 'r' then

83C81F 40 89C3

Assembled from:

or ax, 0x1F
inc ax
mov bx, ax

Answer 20

Game Maker Language, 8

r=n|31+1

Yes, | is evaluated before + (see here)

Answer 21

GolfScript, 7 characters

31n|):r

Takes input from variable n and puts the result into r. Note that for testing it is advisable to use another variable name since n is also part of the p and puts built-ins.

Answer 22

TI-BASIC, 12 

32(1+N/32)→R

Answer 23

C 45

int a=32,b,c;b=31;c=b%a;c=a-c;b=b+c;return c;