Very Computer

Here's one that's confusing me...  While solving a problem in
a programing book I came up with this, but I don't understand
why it works.  (I just kept trying things until I got the right
answer)

i=1
while [ $(($i%3)) -ne 2 ] || [ $(($i%5)) -ne 3 ] || [ $(($i%7)) -ne 4 ]
do
        let i=$i+1
done
echo The Number is ${i}.

The part that is confusing to me is the while condition.
I am "OR"ing 3 test conditions.  When they are all false,
the while loop is exited.  

It seems to work like,
when [ false OR false OR false ]
then
        exit loop
done

I'd think it would work like this,
when [ true AND true AND true ]
then
        exit loop
done

Here is the last 10 lines of a trace on the loop:

+ let i=50+1
+ [ 0 -ne 2 ]
+ let i=51+1
+ [ 1 -ne 2 ]
+ let i=52+1
+ [ 2 -ne 2 ]                   # false condition
+ [ 3 -ne 3 ]                   # false condition
+ [ 4 -ne 4 ]                   # false condition
+ echo The Number is 53.
The Number is 53.

Can anyone explain why OR'ing false statements works this way?

p.s. this isn't for a class or anything, so I don't have a
teacher to ask.

p.p.s.
Here is the original problem:
Write a program to find the smallest positive integer n
which corresponds to the following conditions:
n / 3 = integer x and remainder 2
n / 5 = integer y and remainder 3
n / 3 = integer z and remainder 4

--
Randall W. Hron
Roswell, GA

uunet!gatech!kd4nc!vdbsan!willard!hrandoz!hronr

Quote:>Here's one that's confusing me...  While solving a problem in a programing
>book I came up with this, but I don't understand why it works.  (I just kept
>trying things until I got the right answer)
>i=1
>while [ $(($i%3)) -ne 2 ] || [ $(($i%5)) -ne 3 ] || [ $(($i%7)) -ne 4 ]
>do
>    let i=$i+1
>done
>echo The Number is ${i}.
>The part that is confusing to me is the while condition.  I am "OR"ing 3 test
>conditions.  When they are all false, the while loop is exited.
>It seems to work like,
>when [ false OR false OR false ]
>then
>    exit loop
>done

    when [ false AND false AND false ]
    then
        exit loop
    done

That's because of simple Boolean algebra:

    NOT(a OR b) = (NOT a) AND (NOT b).

Your `when' statement uses this incorrect rule:

    NOT(a OR b) = (NOT a) OR (NOT b)

Michael.

Quote:>Can anyone explain why OR'ing false statements works this way?
>Here is the original problem:
>Write a program to find the smallest positive integer n
>which corresponds to the following conditions:
>n / 3 = integer x and remainder 2
>n / 5 = integer y and remainder 3
>n / 3 = integer z and remainder 4

if this is the original problem, then your program will run forever
because there is no solution!

the way I see it we can rephrase your problem like
so

n = 3x+2
n = 5y+4
n = 3z+4

which means 3x+2 = 3z+4

rearranging gives 3x = 3z + 2
dividing by 3 gives x = z +2/3

You tell me which integer x equals another integer z + 2/3

Besides this is nothing more than a simultaneous equation problem and does
not require a loop to begin with!  Solve the equations in terms of one
variable and choose 1 as you lowest integer for that variable and figure
out what the other variable would be.

Joe Carl

You must have misquoted the problem
--
ShadowMaster "If you ever drop your watch in a pool of molten lava, let it go,
              'cause man it's gone"          Saturday Nigh Live "Deep Thoughts"

Quote:

>Here's one that's confusing me...  While solving a problem in
>a programing book I came up with this, but I don't understand
>why it works.  (I just kept trying things until I got the right
>answer)

>i=1
>while [ $(($i%3)) -ne 2 ] || [ $(($i%5)) -ne 3 ] || [ $(($i%7)) -ne 4 ]
>do
>    let i=$i+1
>done
>echo The Number is ${i}.

>The part that is confusing to me is the while condition.
>I am "OR"ing 3 test conditions.  When they are all false,
>the while loop is exited.  

|| isn't a boolean arithmetic operator, it is a conditional execution
operator.  It executes the antecedent only when the precedent returns a
false exit code.

-r

Quote:>false exit code.

-r