What is Heron’s algorithm for finding the square root of a number?
Question
What is Heron’s algorithm for finding the square root of a number?
Solution
Heron's algorithm, also known as Hero's method, is a way to find the square root of a number. Here are the steps:
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Start with an initial guess. This can be any number, but the closer it is to the actual square root, the fewer steps the algorithm will take.
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Divide the number you want to find the square root of by your current guess. This gives you a second number.
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Take the average of your current guess and the second number. This gives you a new guess.
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Repeat steps 2 and 3 until your guess is as close to the actual square root as you want it to be. The more times you repeat the steps, the closer your guess will get to the actual square root.
This method is based on the fact that if x is an overestimate to the square root of a number N then N/x will be an underestimate, or vice versa. So, the average of the two numbers should be closer to the actual square root than either of the two starting numbers.
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Mary is working to find the square root of a number in C++ language using a binary search algorithm. Which of the following options can she perform to get an optimized solution?Constraints0<=x<=2^31-1Optionsint fun(int x) { if(x==1) return x; int low=0,high=x; int res=0; while(low<high){ int mid=(low+high)/2; if(mid>x/mid) high=mid; else{ res=mid; low=mid+1; } } return res;}int fun(int x) { if(x==1) return x; int low=0,high=x; int res=0; while(low<=high){ int mid=low+(high-low)/2; if(mid>x/(mid*mid) high=mid; else{ res=mid; low=mid+1; } } return res;}int fun(int x) { if(x==1) return x; int low=0,high=x; int res=0; while(low<high){ int mid=low+(high-low)/2; else if((mid*mid*mid)>x) high=mid; else{ res=mid; low=mid+1; } } return res;}int fun(int x) { int res=0; for(int i=0;i<=x;i++){ if((i*i)<=x) res=i; } return res;}
Heron's Formula is used to find the area of SAS triangles.Group of answer choicesTrueFalse PreviousNext
Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.You must not use any built-in exponent function or operator.For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python. Example 1:Input: x = 4Output: 2Explanation: The square root of 4 is 2, so we return 2.Example 2:Input: x = 8Output: 2Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest integer, 2 is returned.
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