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mathnumberspascal
Heretiiik
Heretiiik

Reputation: 59

Pascal - odd and even number

I want to make an algorithm which return whether is ginen number odd or even without using built-in functions like mod, div, odd(). Only solution that I made up is following, but it isn't much robust and only for numbers higher then 0

Var
  n: integer;

begin
  ReadLn(n);
  While (n not in [1, 0]) do
    n:=n-2;
  if n = 1 then
    WriteLn('odd')
  else
    WriteLn('even');
end.

Thank you for your help.

Upvotes: 2

Views: 7912

Answers (5)

Rattata 2me
Rattata 2me

Reputation: 37

You can use the number truncation:

var
  n: integer;
begin
  readln(n);
  if(n-(n/2)*2 = 1) then
    writeln('odd')
  else
    writeln('even');
end.

Upvotes: 0

Marco van de Voort
Marco van de Voort

Reputation: 26356

I never really used J&W Pascal but I know many bit manipulation operators are missing.

However I did use Pascal's successor Modula2, and in M2 one cast an integer to a set the same size. If that also works for classic Pascal then you could do

Type 
  TIntBitset = [0..31]; // or whatever your word size is.

if 0 in TIntBitSet(i) then 
  begin
  (* odd! *)
  end;

Upvotes: 1

lhf
lhf

Reputation: 72312

Can you use cos? If so, try

abs(cos(n*1.570796326794896619231321691639751442)) > 0.9

It should really be = 1 but we cannot use a perfect value of π/2 and there'll be a small error in cos...

Upvotes: 0

thanos
thanos

Reputation: 5858

well, to make it work with negative numbers you can just check if the number is negative and if it is, multiply with -1 (or just use abs).

The only issue with your algorithm is efficiency; runs in o(n). Some other ideas:

  • Checking the least significant bit (which is probably the last)
  • Integer division with 2 (could be done by dividing and trunc), the multiply the result with 2 and check if it's the same number

  • Check if the last digit is 0,2,4,6,8 (really good if the number is given as a string/array) In each step of your algorithm increase the subtracted number till you hit negative then reverse the process. For example, let's say we want to check 64:

    1. 64 - 2 = 62, it's >0 but not 0 or 1
    2. 63 - 4 = 58, >0, not 0/1
    3. 50 - 8 = 42, >0, not 0/1
    4. 42 -16 = 26, >0, not 0/1
    5. 26 -32 = -6, <0 => reverse
    6. 26 -16 = 10, >0, not 0/1
    7. 10 - 8 = 2, >0, not 0/1
    8. 2 - 4 = -2, <0 => reverse
    9. 2 - 2 = 0, =0 => even

So 10 steps instead of 63.
Note that after the first reverse, we decrease the subtracted number on each step because we know that it's less than 2x the number we subtract. Of course, you can create a ton of variations, some of them which are probably more efficient.

Upvotes: 2

Erkan Haspulat
Erkan Haspulat

Reputation: 12562

If a number is odd, its last bit is 1, and 0 otherwise. You could use a bitwise operator to test against integer(1), which is represented as 0..00001.

My Pascal skills are a little rusty, but it should be something similar:

var
  n: integer;
begin
  readln(n);
  if(n&1 = 1) then
    writeln('odd')
  else
    writeln('even');
end.

Upvotes: 0

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