Can i get hits for pythagros triplet que

#1

https://online.codingblocks.com/player/7545/content/5167?s=1435

#2

Hey Ankush, what exact issue are you facing with this question. In this problem, you will be given a number n and you just have to print its Pythagoras pair in increasing order if they exist, otherwise you are supposed to print ā€œ-1ā€. If you are facing some problem with the code then share the link of your code we will help you in debugging the code.

#3

https://ide.codingblocks.com/s/51248

#4

my loop is running infinite times

#5

Hey Ankush, your logic is not correct, as you are iterating in the for loop till infinity
for(int i = m + 1; i <= INT_MAX; i++ ), if a no.'s Pythagoras triplets doesnā€™t exist then your codeā€™s loop will never break, and run infinite times.

#6

can i get more hints for logic or is my code is near the logic pls tell

#7

Hey Ankush, here I am sharing the approach for this problem

  • If m>n>0 are integers, then (m2-n2, 2mn, m2-n2) is a Pythagorean triple. This is easily seen with a bit of algebra. Thus, plugging in various (m,n) will give various triples.

  • Now, our goal is to find a triple of the form (m2-n2, 2mn, m2-n2) in which our input value is one of the catheti. Note that the catheti are m2-n2 and 2mn. Also, note that the second cathetus is always even. Thus, it makes sense to consider two cases:

  • If a is even, then letā€™s try to equate it with the second cathetus, 2mn, i.e., letā€™s try to find an (m,n) pair such that 2mn = a. In this case, is even, so we can divide by 2. We have mn = a/2. Thus, all we have to do is to find a factorization of a/2 into mn such that m>n. You can try various factoring methods ā€” even naĆÆve O(āˆša) time method will pass - but in this case , we donā€™t really even need to do that, since the trivial (m,n) = (a/2,1) works!. This gives us the solution triple ( (a^2)/4 - 1, a, (a^2)/4 + 1 ). Note that (a^2)/4 is an integer since a is even.

  • If a is odd, then we canā€™t equate it with the second cathetus, so letā€™s try instead m2-n2, i.e. we want to find an (m,n) such that m2-n2 = a. Node that m2-n2 = (m+n)(m-n), so all boils down again to factoring! Letā€™s try our trivial factorization again: (m+n, m-n) = (a,1) What we get is a system of two linear equations: m+n=a and m-n=1 which has solution m=(a+1)/2 and n=(a-1)/2. Substituting back, this gives us the triple (a, (a^2 - 1)/2, (a^2 + 1)/2). Note that (a^2 + 1)/2 is an integer as a is odd.

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