# 核心函數 def findSquareRoot(x):if x < 0:print ('Sorry, imaginary numbers are out of scope!')returnans = 0while ans**2 < x:ans = ans + 1if ans**2 != x:print (x, 'is not a perfect square')print ('Square root of ' + str(x) + ' is close to ' + str(ans - 1))else:print ('Square root of ' + str(x) + ' is ' + str(ans))return# 基于迭代查找在某個誤差范圍內找到平方根 def findSquareRootWithinError(x, epsilon, increment):if x < 0:print ('Sorry, imaginary numbers are out of scope!')returnnumGuesses = 0ans = 0.0while x - ans**2 > epsilon:ans += incrementnumGuesses += 1print ('numGuesses =', numGuesses)if abs(x - ans**2) > epsilon:print ('Failed on square root of', x)else:print (ans, 'is close to square root of', x)return# 基于二分查找在某個誤差范圍內找到平方根 def bisectionSearchForSquareRoot(x, epsilon):if x < 0:print ('Sorry, imaginary numbers are out of scope!')returnnumGuesses = 0low = 0.0high = xans = (high + low)/2.0while abs(ans**2 - x) >= epsilon:if ans**2 < x:low = anselse:high = ansans = (high + low)/2.0numGuesses += 1# print('low = ', low, 'high = ', high, 'guess = ', ans)print ('numGuesses =', numGuesses)print (ans, 'is close to square root of', x)return

測試代碼：

findSquareRoot(65536) findSquareRootWithinError(65535, .01, .001) findSquareRootWithinError(65535, .01, .0001) findSquareRootWithinError(65535, .01, .00001) bisectionSearchForSquareRoot(65535, .01)

運行結果：

Square root of 65536 is 256 numGuesses = 255999 Failed on square root of 65535 numGuesses = 2559981 Failed on square root of 65535 numGuesses = 25599803 255.99803007005775 is close to square root of 65535 numGuesses = 24 255.99804684519768 is close to square root of 65535

首先第一個函數，是迭代查找，猜想并檢查，得到的應該是相當于向下取整的結果了。
我們的測試數據是65536，得到256。
如果測試數據是65535，將得到255，然而結果顯然更接近256，這就無奈了。

第二個函數修改了while循環，遞增一個遠小于1的數字，由用戶決定誤差和每次遞增的數。
這樣可能跳過結果或者導致運行過慢。
這個函數努力規避浮點數嚴格相等判定時精度的影響。

print(0.1+0.2)

輸出結果：

0.30000000000000004

最后一個運用了二分查找（折半查找）的思想，done

總結

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