Issue with 3Sum problem!

Alright so I took some stuff out and now I'm down to this. For some reason though, it's just coming up with random numbers from the array instead of using them to find a 1D array which adds to zero. I think what the issue is, is that it's not using the integers to try and find where T would equal zero.

F = [-5,-4,-3,-2,-1,1,2,3,4,5];
for T = @(a,b,c) (a + b + c);
a = randsample (F,1);
b = randsample (F,1);
c = randsample (F,1);
if T(a,b,c) == 0
disp [a b c]
end
end

Answer 1

[Honglei Chen]

Your line

for T = @(a,b,c) (a + b + c)

does not really specify a loop. What you want is to find some a, b, and c whose sum is 0, so you need to loop through a, b, c. For example, if you want to do it 10 times, you can do

F = [-5,-4,-3,-2,-1,1,2,3,4,5];
T = @(a,b,c) (a + b + c);
for m = 1:10
    a = randsample (F,1);
    b = randsample (F,1);
    c = randsample (F,1);
    if T(a,b,c) == 0
        disp([a b c])
    end
end

[Walter Roberson]

Note that

disp [a b c]

means

disp('[a', 'b', 'c]')

Consider instead using

disp([a b c])

[Honglei Chen]

Thanks Walter, didn't pay attention to that. I updated the post.

[Nathaniel Ewing]

Okay, so I have all of that fixed. I tried using a while loop to get it to repeat until it hit the right numbers instead of having to repeatedly hit it, but that doesn't work. Last question I swear: how would I have that automatically repeat instead of requiring my input?

[Walter Roberson]

Please show us your attempt with a while loop, and tell us what did not work about it.

[Nathaniel Ewing]

F = [-5,-4,-3,-2,-1,1,2,3,4,5];

T = @(a,b,c) (a + b + c);

m=F;

while (0 > T(a,b,c) > 0)

a = randsample (F,1);

b = randsample (F,1);

c = randsample (F,1);

if T(a,b,c) == 0

disp ([a b c])

break

end

end

[Nathaniel Ewing]

For some reason it's just not looping like it should, I still have to click it constantly until it gets a solution.

[Walter Roberson]

0 > T(a,b,c) > 0

would be interpreted as

((0 > T(a,b,c) > 0)

The first subexpression is a logical comparison and so would return 0 (false) or 1 (true). That 0 or 1 would then (second subexpression) be compared to 0; 0 or 1 is greater than 0 only if the first subexpression is 1 (true). But then you can see that that is a redundant logical condition, the same as

0 > T(a,b,c)

I am not sure what you are trying to test there. Loop until you find a 0 value? But your existing "break" condition would already exit the loop when a match is found, so testing for non-zero would be unnecessary. You might as well just loop forever:

while true

Note with your existing code, you are going to have problems because the "while" is using T(a,b,c) before a, b, or c have been given values.

[Nathaniel Ewing]

Yes I'm trying to loop until I find a zero value. I removed the break and supressed the disp ([a b c]) which seemed to help, it's now displaying multiple answers which is partially what I was looking for. I just need the automation part now.

[Walter Roberson]

If you commented out the disp() then there should be nothing left there that can display multiple values.

If it is already displaying multiple answers, then how does that differ from the automation you are looking for?

If you want multiple answers displayed, but you also want to stop at the first answer, that seems to be a contradiction.

When I look at another question (by someone else) on the same topic, I find that _they_ were required to produce the entire list of solutions. Is that a requirement for you as well? If it is then you are going about it the wrong way.

[Nathaniel Ewing]

Yes it's supposed to be a list of solutions, which I can't seem to get.

[Nathaniel Ewing]

What will happen is it will display two sets of numbers once and awhile if it finds them, but otherwise I have to keep clicking until it does so.

Answer 2

[Nathaniel Ewing]

You guys are terrific, thank you for helping me learn!