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How could I check if a number is a perfect square?
Speed is of no concern, for now, just working.
See also: Integer square root in python.
The problem with relying on any floating point computation (math.sqrt(x), or x**0.5) is that you can't really be sure it's exact (for sufficiently large integers x, it won't be, and might even overflow). Fortunately (if one's in no hurry;-) there are many pure integer approaches, such as the following...:
def is_square(apositiveint):
x = apositiveint // 2
seen = set([x])
while x * x != apositiveint:
x = (x + (apositiveint // x)) // 2
if x in seen: return False
seen.add(x)
return True
for i in range(110, 130):
print i, is_square(i)
Hint: it's based on the "Babylonian algorithm" for square root, see wikipedia. It does work for any positive number for which you have enough memory for the computation to proceed to completion;-).
Edit: let's see an example...
x = 12345678987654321234567 ** 2
for i in range(x, x+2):
print i, is_square(i)
this prints, as desired (and in a reasonable amount of time, too;-):
152415789666209426002111556165263283035677489 True
152415789666209426002111556165263283035677490 False
Please, before you propose solutions based on floating point intermediate results, make sure they work correctly on this simple example -- it's not that hard (you just need a few extra checks in case the sqrt computed is a little off), just takes a bit of care.
And then try with x**7 and find clever way to work around the problem you'll get,
OverflowError: long int too large to convert to float
you'll have to get more and more clever as the numbers keep growing, of course.
If I was in a hurry, of course, I'd use gmpy -- but then, I'm clearly biased;-).
>>> import gmpy
>>> gmpy.is_square(x**7)
1
>>> gmpy.is_square(x**7 + 1)
0
Yeah, I know, that's just so easy it feels like cheating (a bit the way I feel towards Python in general;-) -- no cleverness at all, just perfect directness and simplicity (and, in the case of gmpy, sheer speed;-)...
Use Newton's method to quickly zero in on the nearest integer square root, then square it and see if it's your number. See isqrt.
Python ≥ 3.8 has math.isqrt. If using an older version of Python, look for the "def isqrt(n)" implementation here.
import math
def is_square(i: int) -> bool:
return i == math.isqrt(i) ** 2
Since you can never depend on exact comparisons when dealing with floating point computations (such as these ways of calculating the square root), a less error-prone implementation would be
import math
def is_square(integer):
root = math.sqrt(integer)
return integer == int(root + 0.5) ** 2
Imagine integer is 9. math.sqrt(9) could be 3.0, but it could also be something like 2.99999 or 3.00001, so squaring the result right off isn't reliable. Knowing that int takes the floor value, increasing the float value by 0.5 first means we'll get the value we're looking for if we're in a range where float still has a fine enough resolution to represent numbers near the one for which we are looking.
If youre interested, I have a pure-math response to a similar question at math stackexchange, "Detecting perfect squares faster than by extracting square root".
My own implementation of isSquare(n) may not be the best, but I like it. Took me several months of study in math theory, digital computation and python programming, comparing myself to other contributors, etc., to really click with this method. I like its simplicity and efficiency though. I havent seen better. Tell me what you think.
def isSquare(n):
## Trivial checks
if type(n) != int: ## integer
return False
if n < 0: ## positivity
return False
if n == 0: ## 0 pass
return True
## Reduction by powers of 4 with bit-logic
while n&3 == 0:
n=n>>2
## Simple bit-logic test. All perfect squares, in binary,
## end in 001, when powers of 4 are factored out.
if n&7 != 1:
return False
if n==1:
return True ## is power of 4, or even power of 2
## Simple modulo equivalency test
c = n%10
if c in {3, 7}:
return False ## Not 1,4,5,6,9 in mod 10
if n % 7 in {3, 5, 6}:
return False ## Not 1,2,4 mod 7
if n % 9 in {2,3,5,6,8}:
return False
if n % 13 in {2,5,6,7,8,11}:
return False
## Other patterns
if c == 5: ## if it ends in a 5
if (n//10)%10 != 2:
return False ## then it must end in 25
if (n//100)%10 not in {0,2,6}:
return False ## and in 025, 225, or 625
if (n//100)%10 == 6:
if (n//1000)%10 not in {0,5}:
return False ## that is, 0625 or 5625
else:
if (n//10)%4 != 0:
return False ## (4k)*10 + (1,9)
## Babylonian Algorithm. Finding the integer square root.
## Root extraction.
s = (len(str(n))-1) // 2
x = (10**s) * 4
A = {x, n}
while x * x != n:
x = (x + (n // x)) >> 1
if x in A:
return False
A.add(x)
return True
Pretty straight forward. First it checks that we have an integer, and a positive one at that. Otherwise there is no point. It lets 0 slip through as True (necessary or else next block is infinite loop).
The next block of code systematically removes powers of 4 in a very fast sub-algorithm using bit shift and bit logic operations. We ultimately are not finding the isSquare of our original n but of a k<n that has been scaled down by powers of 4, if possible. This reduces the size of the number we are working with and really speeds up the Babylonian method, but also makes other checks faster too.
The third block of code performs a simple Boolean bit-logic test. The least significant three digits, in binary, of any perfect square are 001. Always. Save for leading zeros resulting from powers of 4, anyway, which has already been accounted for. If it fails the test, you immediately know it isnt a square. If it passes, you cant be sure.
Also, if we end up with a 1 for a test value then the test number was originally a power of 4, including perhaps 1 itself.
Like the third block, the fourth tests the ones-place value in decimal using simple modulus operator, and tends to catch values that slip through the previous test. Also a mod 7, mod 8, mod 9, and mod 13 test.
The fifth block of code checks for some of the well-known perfect square patterns. Numbers ending in 1 or 9 are preceded by a multiple of four. And numbers ending in 5 must end in 5625, 0625, 225, or 025. I had included others but realized they were redundant or never actually used.
Lastly, the sixth block of code resembles very much what the top answerer - Alex Martelli - answer is. Basically finds the square root using the ancient Babylonian algorithm, but restricting it to integer values while ignoring floating point. Done both for speed and extending the magnitudes of values that are testable. I used sets instead of lists because it takes far less time, I used bit shifts instead of division by two, and I smartly chose an initial start value much more efficiently.
By the way, I did test Alex Martelli's recommended test number, as well as a few numbers many orders magnitude larger, such as:
x=1000199838770766116385386300483414671297203029840113913153824086810909168246772838680374612768821282446322068401699727842499994541063844393713189701844134801239504543830737724442006577672181059194558045164589783791764790043104263404683317158624270845302200548606715007310112016456397357027095564872551184907513312382763025454118825703090010401842892088063527451562032322039937924274426211671442740679624285180817682659081248396873230975882215128049713559849427311798959652681930663843994067353808298002406164092996533923220683447265882968239141724624870704231013642255563984374257471112743917655991279898690480703935007493906644744151022265929975993911186879561257100479593516979735117799410600147341193819147290056586421994333004992422258618475766549646258761885662783430625 ** 2
for i in range(x, x+2):
print(i, isSquare(i))
printed the following results:
1000399717477066534083185452789672211951514938424998708930175541558932213310056978758103599452364409903384901149641614494249195605016959576235097480592396214296565598519295693079257885246632306201885850365687426564365813280963724310434494316592041592681626416195491751015907716210235352495422858432792668507052756279908951163972960239286719854867504108121432187033786444937064356645218196398775923710931242852937602515835035177768967470757847368349565128635934683294155947532322786360581473152034468071184081729335560769488880138928479829695277968766082973795720937033019047838250608170693879209655321034310764422462828792636246742456408134706264621790736361118589122797268261542115823201538743148116654378511916000714911467547209475246784887830649309238110794938892491396597873160778553131774466638923135932135417900066903068192088883207721545109720968467560224268563643820599665232314256575428214983451466488658896488012211237139254674708538347237589290497713613898546363590044902791724541048198769085430459186735166233549186115282574626012296888817453914112423361525305960060329430234696000121420787598967383958525670258016851764034555105019265380321048686563527396844220047826436035333266263375049097675787975100014823583097518824871586828195368306649956481108708929669583308777347960115138098217676704862934389659753628861667169905594181756523762369645897154232744410732552956489694024357481100742138381514396851789639339362228442689184910464071202445106084939268067445115601375050153663645294106475257440167535462278022649865332161044187890625 True
1000399717477066534083185452789672211951514938424998708930175541558932213310056978758103599452364409903384901149641614494249195605016959576235097480592396214296565598519295693079257885246632306201885850365687426564365813280963724310434494316592041592681626416195491751015907716210235352495422858432792668507052756279908951163972960239286719854867504108121432187033786444937064356645218196398775923710931242852937602515835035177768967470757847368349565128635934683294155947532322786360581473152034468071184081729335560769488880138928479829695277968766082973795720937033019047838250608170693879209655321034310764422462828792636246742456408134706264621790736361118589122797268261542115823201538743148116654378511916000714911467547209475246784887830649309238110794938892491396597873160778553131774466638923135932135417900066903068192088883207721545109720968467560224268563643820599665232314256575428214983451466488658896488012211237139254674708538347237589290497713613898546363590044902791724541048198769085430459186735166233549186115282574626012296888817453914112423361525305960060329430234696000121420787598967383958525670258016851764034555105019265380321048686563527396844220047826436035333266263375049097675787975100014823583097518824871586828195368306649956481108708929669583308777347960115138098217676704862934389659753628861667169905594181756523762369645897154232744410732552956489694024357481100742138381514396851789639339362228442689184910464071202445106084939268067445115601375050153663645294106475257440167535462278022649865332161044187890626 False
And it did this in 0.33 seconds.
In my opinion, my algorithm works the same as Alex Martelli's, with all the benefits thereof, but has the added benefit highly efficient simple-test rejections that save a lot of time, not to mention the reduction in size of test numbers by powers of 4, which improves speed, efficiency, accuracy and the size of numbers that are testable. Probably especially true in non-Python implementations.
Roughly 99% of all integers are rejected as non-Square before Babylonian root extraction is even implemented, and in 2/3 the time it would take the Babylonian to reject the integer. And though these tests dont speed up the process that significantly, the reduction in all test numbers to an odd by dividing out all powers of 4 really accelerates the Babylonian test.
I did a time comparison test. I tested all integers from 1 to 10 Million in succession. Using just the Babylonian method by itself (with my specially tailored initial guess) it took my Surface 3 an average of 165 seconds (with 100% accuracy). Using just the logical tests in my algorithm (excluding the Babylonian), it took 127 seconds, it rejected 99% of all integers as non-Square without mistakenly rejecting any perfect squares. Of those integers that passed, only 3% were perfect Squares (a much higher density). Using the full algorithm above that employs both the logical tests and the Babylonian root extraction, we have 100% accuracy, and test completion in only 14 seconds. The first 100 Million integers takes roughly 2 minutes 45 seconds to test.
EDIT: I have been able to bring down the time further. I can now test the integers 0 to 100 Million in 1 minute 40 seconds. A lot of time is wasted checking the data type and the positivity. Eliminate the very first two checks and I cut the experiment down by a minute. One must assume the user is smart enough to know that negatives and floats are not perfect squares.
import math
def is_square(n):
sqrt = math.sqrt(n)
return (sqrt - int(sqrt)) == 0
A perfect square is a number that can be expressed as the product of two equal integers. math.sqrt(number) return a float. int(math.sqrt(number)) casts the outcome to int.
If the square root is an integer, like 3, for example, then math.sqrt(number) - int(math.sqrt(number)) will be 0, and the if statement will be False. If the square root was a real number like 3.2, then it will be True and print "it's not a perfect square".
It fails for a large non-square such as 152415789666209426002111556165263283035677490.
My answer is:
def is_square(x):
return x**.5 % 1 == 0
It basically does a square root, then modulo by 1 to strip the integer part and if the result is 0 return True otherwise return False. In this case x can be any large number, just not as large as the max float number that python can handle: 1.7976931348623157e+308
It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.
This can be solved using the decimal module to get arbitrary precision square roots and easy checks for "exactness":
import math
from decimal import localcontext, Context, Inexact
def is_perfect_square(x):
# If you want to allow negative squares, then set x = abs(x) instead
if x < 0:
return False
# Create localized, default context so flags and traps unset
with localcontext(Context()) as ctx:
# Set a precision sufficient to represent x exactly; `x or 1` avoids
# math domain error for log10 when x is 0
ctx.prec = math.ceil(math.log10(x or 1)) + 1 # Wrap ceil call in int() on Py2
# Compute integer square root; don't even store result, just setting flags
ctx.sqrt(x).to_integral_exact()
# If previous line couldn't represent square root as exact int, sets Inexact flag
return not ctx.flags[Inexact]
For demonstration with truly huge values:
# I just kept mashing the numpad for awhile :-)
>>> base = 100009991439393999999393939398348438492389402490289028439083249803434098349083490340934903498034098390834980349083490384903843908309390282930823940230932490340983098349032098324908324098339779438974879480379380439748093874970843479280329708324970832497804329783429874329873429870234987234978034297804329782349783249873249870234987034298703249780349783497832497823497823497803429780324
>>> sqr = base ** 2
>>> sqr ** 0.5 # Too large to use floating point math
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: int too large to convert to float
>>> is_perfect_power(sqr)
True
>>> is_perfect_power(sqr-1)
False
>>> is_perfect_power(sqr+1)
False
If you increase the size of the value being tested, this eventually gets rather slow (takes close to a second for a 200,000 bit square), but for more moderate numbers (say, 20,000 bits), it's still faster than a human would notice for individual values (~33 ms on my machine). But since speed wasn't your primary concern, this is a good way to do it with Python's standard libraries.
Of course, it would be much faster to use gmpy2 and just test gmpy2.mpz(x).is_square(), but if third party packages aren't your thing, the above works quite well.
I just posted a slight variation on some of the examples above on another thread (Finding perfect squares) and thought I'd include a slight variation of what I posted there here (using nsqrt as a temporary variable), in case it's of interest / use:
import math
def is_square(n):
if not (isinstance(n, int) and (n >= 0)):
return False
else:
nsqrt = math.sqrt(n)
return nsqrt == math.trunc(nsqrt)
It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.
A variant of #Alex Martelli's solution without set
When x in seen is True:
In most cases, it is the last one added, e.g. 1022 produces the x's sequence 511, 256, 129, 68, 41, 32, 31, 31;
In some cases (i.e., for the predecessors of perfect squares), it is the second-to-last one added, e.g. 1023 produces 511, 256, 129, 68, 41, 32, 31, 32.
Hence, it suffices to stop as soon as the current x is greater than or equal to the previous one:
def is_square(n):
assert n > 1
previous = n
x = n // 2
while x * x != n:
x = (x + (n // x)) // 2
if x >= previous:
return False
previous = x
return True
x = 12345678987654321234567 ** 2
assert not is_square(x-1)
assert is_square(x)
assert not is_square(x+1)
Equivalence with the original algorithm tested for 1 < n < 10**7. On the same interval, this slightly simpler variant is about 1.4 times faster.
This is my method:
def is_square(n) -> bool:
return int(n**0.5)**2 == int(n)
Take square root of number. Convert to integer. Take the square. If the numbers are equal, then it is a perfect square otherwise not.
It is incorrect for a large square such as 152415789666209426002111556165263283035677489.
If the modulus (remainder) leftover from dividing by the square root is 0, then it is a perfect square.
def is_square(num: int) -> bool:
return num % math.sqrt(num) == 0
I checked this against a list of perfect squares going up to 1000.
It is possible to improve the Babylonian method by observing that the successive terms form a decreasing sequence if one starts above the square root of n.
def is_square(n):
assert n > 1
a = n
b = (a + n // a) // 2
while b < a:
a = b
b = (a + n // a) // 2
return a * a == n
If it's a perfect square, its square root will be an integer, the fractional part will be 0, we can use modulus operator to check fractional part, and check if it's 0, it does fail for some numbers, so, for safety, we will also check if it's square of the square root even if the fractional part is 0.
import math
def isSquare(n):
root = math.sqrt(n)
if root % 1 == 0:
if int(root) * int(root) == n:
return True
return False
isSquare(4761)
You could binary-search for the rounded square root. Square the result to see if it matches the original value.
You're probably better off with FogleBirds answer - though beware, as floating point arithmetic is approximate, which can throw this approach off. You could in principle get a false positive from a large integer which is one more than a perfect square, for instance, due to lost precision.
A simple way to do it (faster than the second one) :
def is_square(n):
return str(n**(1/2)).split(".")[1] == '0'
Another way:
def is_square(n):
if n == 0:
return True
else:
if n % 2 == 0 :
for i in range(2,n,2):
if i*i == n:
return True
else :
for i in range(1,n,2):
if i*i == n:
return True
return False
This response doesn't pertain to your stated question, but to an implicit question I see in the code you posted, ie, "how to check if something is an integer?"
The first answer you'll generally get to that question is "Don't!" And it's true that in Python, typechecking is usually not the right thing to do.
For those rare exceptions, though, instead of looking for a decimal point in the string representation of the number, the thing to do is use the isinstance function:
>>> isinstance(5,int)
True
>>> isinstance(5.0,int)
False
Of course this applies to the variable rather than a value. If I wanted to determine whether the value was an integer, I'd do this:
>>> x=5.0
>>> round(x) == x
True
But as everyone else has covered in detail, there are floating-point issues to be considered in most non-toy examples of this kind of thing.
If you want to loop over a range and do something for every number that is NOT a perfect square, you could do something like this:
def non_squares(upper):
next_square = 0
diff = 1
for i in range(0, upper):
if i == next_square:
next_square += diff
diff += 2
continue
yield i
If you want to do something for every number that IS a perfect square, the generator is even easier:
(n * n for n in range(upper))
I think that this works and is very simple:
import math
def is_square(num):
sqrt = math.sqrt(num)
return sqrt == int(sqrt)
It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.
a=int(input('enter any number'))
flag=0
for i in range(1,a):
if a==i*i:
print(a,'is perfect square number')
flag=1
break
if flag==1:
pass
else:
print(a,'is not perfect square number')
In kotlin :
It's quite easy and it passed all test cases as well.
really thanks to >> https://www.quora.com/What-is-the-quickest-way-to-determine-if-a-number-is-a-perfect-square
fun isPerfectSquare(num: Int): Boolean {
var result = false
var sum=0L
var oddNumber=1L
while(sum<num){
sum = sum + oddNumber
oddNumber = oddNumber+2
}
result = sum == num.toLong()
return result
}
def isPerfectSquare(self, num: int) -> bool:
left, right = 0, num
while left <= right:
mid = (left + right) // 2
if mid**2 < num:
left = mid + 1
elif mid**2 > num:
right = mid - 1
else:
return True
return False
Decide how long the number will be.
take a delta 0.000000000000.......000001
see if the (sqrt(x))^2 - x is greater / equal /smaller than delta and decide based on the delta error.
import math
def is_square(n):
sqrt = math.sqrt(n)
return sqrt == int(sqrt)
It fails for a large non-square such as 152415789666209426002111556165263283035677490.
The idea is to run a loop from i = 1 to floor(sqrt(n)) then check if squaring it makes n.
bool isPerfectSquare(int n)
{
for (int i = 1; i * i <= n; i++) {
// If (i * i = n)
if ((n % i == 0) && (n / i == i)) {
return true;
}
}
return false;
}
I have written a code based on the two pointer algorithm to find the sum of two squares. My problem is that I run into a memory error when running this code for an input n=55555**2 + 66666**2. I am wondering how to correct this memory error.
def sum_of_two_squares(n):
look=tuple(range(n))
i=0
j = len(look)-1
while i < j:
x = (look[i])**2 + (look[j])**2
if x == n:
return (j,i)
elif x < n:
i += 1
else:
j -= 1
return None
n=55555**2 + 66666**2
print(sum_of_two_squares(n))
The problem Im trying to solve using two pointer algorithm is:
return a tuple of two positive integers whose squares add up to n, or return None if the integer n cannot be so expressed as a sum of two squares. The returned tuple must present the larger of its two numbers first. Furthermore, if some integer can be expressed as a sum of two squares in several ways, return the breakdown that maximizes the larger number. For example, the integer 85 allows two such representations 7*7 + 6*6 and 9*9 + 2*2, of which this function must therefore return (9, 2).
You're creating a tuple of size 55555^2 + 66666^2 = 7530713581
So if each element of the tuple takes one byte, the tuple will take up 7.01 GiB.
You'll need to either reduce the size of the tuple, or possibly make each element take up less space by specifying the type of each element: I would suggest looking into Numpy for the latter.
Specifically for this problem:
Why use a tuple at all?
You create the variable look which is just a list of integers:
look=tuple(range(n)) # = (0, 1, 2, ..., n-1)
Then you reference it, but never modify it. So: look[i] == i and look[j] == j.
So you're looking up numbers in a list of numbers. Why look them up? Why not just use i in place of look[i] and remove look altogether?
As others have pointed out, there's no need to use tuples at all.
One reasonably efficient way of solving this problem is to generate a series of integer square values (0, 1, 4, 9, etc...) and test whether or not subtracting these values from n leaves you with a value that is a perfect square.
You can generate a series of perfect squares efficiently by adding successive odd numbers together: 0 (+1) → 1 (+3) → 4 (+5) → 9 (etc.)
There are also various tricks you can use to test whether or not a number is a perfect square (for example, see the answers to this question), but — in Python, at least — it seems that simply testing the value of int(n**0.5) is faster than iterative methods such as a binary search.
def integer_sqrt(n):
# If n is a perfect square, return its (integer) square
# root. Otherwise return -1
r = int(n**0.5)
if r * r == n:
return r
return -1
def sum_of_two_squares(n):
# If n can be expressed as the sum of two squared integers,
# return these integers as a tuple. Otherwise return <None>
# i: iterator variable
# x: value of i**2
# y: value we need to add to x to obtain (i+1)**2
i, x, y = 0, 0, 1
# If i**2 > n / 2, then we can stop searching
max_x = n >> 1
while x <= max_x:
r = integer_sqrt(n-x)
if r >= 0:
return (i, r)
i, x, y = i+1, x+y, y+2
return None
This returns a solution to sum_of_two_squares(55555**2 + 66666**2) in a fraction of a second.
You do not need the ranges at all, and certainly do not need to convert them into tuples. They take a ridiculous amount of space, but you only need their current elements, numbers i and j. Also, as the friendly commenter suggested, you can start with sqrt(n) to improve the performance further.
def sum_of_two_squares(n):
i = 1
j = int(n ** (1/2))
while i < j:
x = i * i + j * j
if x == n:
return j, i
if x < n:
i += 1
else:
j -= 1
Bear in mind that the problem takes a very long time to be solved. Be patient. And no, NumPy won't help. There is nothing here to vectorize.
The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
a = 1
b = 1
i = 2
while(1):
c = a + b
i += 1
length = len(str(c))
if length == 1000:
print(i)
break
a = b
b = c
I got the answer(works fast enough). Just looking if there's a better way for this question
If you've answered the question, you'll find plenty of explanations on answers in the problem thread. The solution you posted is pretty much okay. You may get a slight speedup by simply checking that your c>=10^999 at every step instead of first converting it to a string.
The better method is to use the fact that when the Fibonacci numbers become large enough, the Fibonacci numbers converge to round(phi**n/(5**.5)) where phi=1.6180... is the golden ratio and round(x) rounds x to the nearest integer. Let's consider the general case of finding the first Fibonacci number with more than m digits. We are then looking for n such that round(phi**n/(5**.5)) >= 10**(m-1)
We can easily solve that by just taking the log of both sides and observe that
log(phi)*n - log(5)/2 >= m-1 and then solve for n.
If you're wondering "well how do I know that it has converged by the nth number?" Well, you can check for yourself, or you can look online.
Also, I think questions like these either belong on the Code Review SE or the Computer Science SE. Even Math Overflow might be a good place for Project Euler questions, since many are rooted in number theory.
Your solution is completely fine for #25 on project euler. However, if you really want to optimize for speed here you can try to calculate fibonacci using the identities I have written about in this blog post: https://sloperium.github.io/calculating-the-last-digits-of-large-fibonacci-numbers.html
from functools import lru_cache
#lru_cache(maxsize=None)
def fib4(n):
if n <= 1:
return n
if n % 2:
m = (n + 1) // 2
return fib4(m) ** 2 + fib4(m - 1) ** 2
else:
m = n // 2
return (2 * fib4(m - 1) + fib4(m)) * fib4(m)
def binarySearch( length):
first = 0
last = 10**5
found = False
while first <= last and not found:
midpoint = (first + last) // 2
length_string = len(str(fib4(midpoint)))
if length_string == length:
return midpoint -1
else:
if length < length_string:
last = midpoint - 1
else:
first = midpoint + 1
print(binarySearch(1000))
This code tests about 12 times faster than your solution. (it does require an initial guess about max size though)
I'm trying to write a program to look for a number, n, between 0 and 100 such that n! + 1 is a perfect square. I'm trying to do this because I know there are only three so it was meant as a test of my Python ability.
Refer to Brocard's problem.
math.sqrt always returns a float, even if that float happens to be, say, 4.0. As the docs say, "Except when explicitly noted otherwise, all return values are floats."
So, your test for type(math.sqrt(x)) == int will never be true.
You could try to work around that by checking whether the float represents an integer, like this:
sx = math.sqrt(x)
if round(sx) == sx:
There's even a built-in method that does this as well as possible:
if sx.is_integer():
But keep in mind that float values are not a perfect representation of real numbers, and there are always rounding issues. For example, for a too-large number, the sqrt might round to an integer, even though it really wasn't a perfect square. For example, if math.sqrt(10000000000**2 + 1).is_integer() is True, even though obviously the number is not a perfect square.
I could tell you whether this is safe within your range of values, but can you convince yourself? If not, you shouldn't just assume that it is.
So, is there a way we can check that isn't affected by float roading issues? Sure, we can use integer arithmetic to check:
sx = int(round(math.sqrt(x)))
if sx*sx == x:
But, as Stefan Pochmann points out, even if this check is safe, does that mean the whole algorithm is? No; sqrt itself could have already been rounded to the point where you've lost integer precision.
So, you need an exact sqrt. You could do this by using decimal.Decimal with a huge configured precision. This will take a bit of work, and a lot of memory, but it's doable. Like this:
decimal.getcontext().prec = ENOUGH_DIGITS
sx = decimal.Decimal(x).sqrt()
But how many digits is ENOUGH_DIGITS? Well, how many digits do you need to represent 100!+1 exactly?
So:
decimal.getcontext().prec = 156
while n <= 100:
x = math.factorial(n) + 1
sx = decimal.Decimal(x).sqrt()
if int(sx) ** 2 == x:
print(sx)
n = n + 1
If you think about it, there's a way to reduce the needed precision to 79 digits, but I'll leave that as an exercise for the reader.
The way you're presumably supposed to solve this is by using purely integer math. For example, you can find out whether an integer is a square in logarithmic time just by using Newton's method until your approximation error is small enough to just check the two bordering integers.
For very large numbers it's better to avoid using floating point square roots altogether because you will run into too many precision issues and you can't even guarantee that you will be within 1 integer value of the correct answer. Fortunately Python natively supports integers of arbitrary size, so you can write an integer square root checking function, like this:
def isSquare(x):
if x == 1:
return True
low = 0
high = x // 2
root = high
while root * root != x:
root = (low + high) // 2
if low + 1 >= high:
return False
if root * root > x:
high = root
else:
low = root
return True
Then you can run through the integers from 0 to 100 like this:
n = 0
while n <= 100:
x = math.factorial(n) + 1
if isSquare(x):
print n
n = n + 1
Here's another version working only with integers, computing the square root by adding decreasing powers of 2, for example intsqrt(24680) will be computed as 128+16+8+4+1.
def intsqrt(n):
pow2 = 1
while pow2 < n:
pow2 *= 2
sqrt = 0
while pow2:
if (sqrt + pow2) ** 2 <= n:
sqrt += pow2
pow2 //= 2
return sqrt
factorial = 1
for n in range(1, 101):
factorial *= n
if intsqrt(factorial + 1) ** 2 == factorial + 1:
print(n)
The number math.sqrt returns is never an int, even if it's an integer.How to check if a float value is a whole number
How could I check if a number is a perfect square?
Speed is of no concern, for now, just working.
See also: Integer square root in python.
The problem with relying on any floating point computation (math.sqrt(x), or x**0.5) is that you can't really be sure it's exact (for sufficiently large integers x, it won't be, and might even overflow). Fortunately (if one's in no hurry;-) there are many pure integer approaches, such as the following...:
def is_square(apositiveint):
x = apositiveint // 2
seen = set([x])
while x * x != apositiveint:
x = (x + (apositiveint // x)) // 2
if x in seen: return False
seen.add(x)
return True
for i in range(110, 130):
print i, is_square(i)
Hint: it's based on the "Babylonian algorithm" for square root, see wikipedia. It does work for any positive number for which you have enough memory for the computation to proceed to completion;-).
Edit: let's see an example...
x = 12345678987654321234567 ** 2
for i in range(x, x+2):
print i, is_square(i)
this prints, as desired (and in a reasonable amount of time, too;-):
152415789666209426002111556165263283035677489 True
152415789666209426002111556165263283035677490 False
Please, before you propose solutions based on floating point intermediate results, make sure they work correctly on this simple example -- it's not that hard (you just need a few extra checks in case the sqrt computed is a little off), just takes a bit of care.
And then try with x**7 and find clever way to work around the problem you'll get,
OverflowError: long int too large to convert to float
you'll have to get more and more clever as the numbers keep growing, of course.
If I was in a hurry, of course, I'd use gmpy -- but then, I'm clearly biased;-).
>>> import gmpy
>>> gmpy.is_square(x**7)
1
>>> gmpy.is_square(x**7 + 1)
0
Yeah, I know, that's just so easy it feels like cheating (a bit the way I feel towards Python in general;-) -- no cleverness at all, just perfect directness and simplicity (and, in the case of gmpy, sheer speed;-)...
Use Newton's method to quickly zero in on the nearest integer square root, then square it and see if it's your number. See isqrt.
Python ≥ 3.8 has math.isqrt. If using an older version of Python, look for the "def isqrt(n)" implementation here.
import math
def is_square(i: int) -> bool:
return i == math.isqrt(i) ** 2
Since you can never depend on exact comparisons when dealing with floating point computations (such as these ways of calculating the square root), a less error-prone implementation would be
import math
def is_square(integer):
root = math.sqrt(integer)
return integer == int(root + 0.5) ** 2
Imagine integer is 9. math.sqrt(9) could be 3.0, but it could also be something like 2.99999 or 3.00001, so squaring the result right off isn't reliable. Knowing that int takes the floor value, increasing the float value by 0.5 first means we'll get the value we're looking for if we're in a range where float still has a fine enough resolution to represent numbers near the one for which we are looking.
If youre interested, I have a pure-math response to a similar question at math stackexchange, "Detecting perfect squares faster than by extracting square root".
My own implementation of isSquare(n) may not be the best, but I like it. Took me several months of study in math theory, digital computation and python programming, comparing myself to other contributors, etc., to really click with this method. I like its simplicity and efficiency though. I havent seen better. Tell me what you think.
def isSquare(n):
## Trivial checks
if type(n) != int: ## integer
return False
if n < 0: ## positivity
return False
if n == 0: ## 0 pass
return True
## Reduction by powers of 4 with bit-logic
while n&3 == 0:
n=n>>2
## Simple bit-logic test. All perfect squares, in binary,
## end in 001, when powers of 4 are factored out.
if n&7 != 1:
return False
if n==1:
return True ## is power of 4, or even power of 2
## Simple modulo equivalency test
c = n%10
if c in {3, 7}:
return False ## Not 1,4,5,6,9 in mod 10
if n % 7 in {3, 5, 6}:
return False ## Not 1,2,4 mod 7
if n % 9 in {2,3,5,6,8}:
return False
if n % 13 in {2,5,6,7,8,11}:
return False
## Other patterns
if c == 5: ## if it ends in a 5
if (n//10)%10 != 2:
return False ## then it must end in 25
if (n//100)%10 not in {0,2,6}:
return False ## and in 025, 225, or 625
if (n//100)%10 == 6:
if (n//1000)%10 not in {0,5}:
return False ## that is, 0625 or 5625
else:
if (n//10)%4 != 0:
return False ## (4k)*10 + (1,9)
## Babylonian Algorithm. Finding the integer square root.
## Root extraction.
s = (len(str(n))-1) // 2
x = (10**s) * 4
A = {x, n}
while x * x != n:
x = (x + (n // x)) >> 1
if x in A:
return False
A.add(x)
return True
Pretty straight forward. First it checks that we have an integer, and a positive one at that. Otherwise there is no point. It lets 0 slip through as True (necessary or else next block is infinite loop).
The next block of code systematically removes powers of 4 in a very fast sub-algorithm using bit shift and bit logic operations. We ultimately are not finding the isSquare of our original n but of a k<n that has been scaled down by powers of 4, if possible. This reduces the size of the number we are working with and really speeds up the Babylonian method, but also makes other checks faster too.
The third block of code performs a simple Boolean bit-logic test. The least significant three digits, in binary, of any perfect square are 001. Always. Save for leading zeros resulting from powers of 4, anyway, which has already been accounted for. If it fails the test, you immediately know it isnt a square. If it passes, you cant be sure.
Also, if we end up with a 1 for a test value then the test number was originally a power of 4, including perhaps 1 itself.
Like the third block, the fourth tests the ones-place value in decimal using simple modulus operator, and tends to catch values that slip through the previous test. Also a mod 7, mod 8, mod 9, and mod 13 test.
The fifth block of code checks for some of the well-known perfect square patterns. Numbers ending in 1 or 9 are preceded by a multiple of four. And numbers ending in 5 must end in 5625, 0625, 225, or 025. I had included others but realized they were redundant or never actually used.
Lastly, the sixth block of code resembles very much what the top answerer - Alex Martelli - answer is. Basically finds the square root using the ancient Babylonian algorithm, but restricting it to integer values while ignoring floating point. Done both for speed and extending the magnitudes of values that are testable. I used sets instead of lists because it takes far less time, I used bit shifts instead of division by two, and I smartly chose an initial start value much more efficiently.
By the way, I did test Alex Martelli's recommended test number, as well as a few numbers many orders magnitude larger, such as:
x=1000199838770766116385386300483414671297203029840113913153824086810909168246772838680374612768821282446322068401699727842499994541063844393713189701844134801239504543830737724442006577672181059194558045164589783791764790043104263404683317158624270845302200548606715007310112016456397357027095564872551184907513312382763025454118825703090010401842892088063527451562032322039937924274426211671442740679624285180817682659081248396873230975882215128049713559849427311798959652681930663843994067353808298002406164092996533923220683447265882968239141724624870704231013642255563984374257471112743917655991279898690480703935007493906644744151022265929975993911186879561257100479593516979735117799410600147341193819147290056586421994333004992422258618475766549646258761885662783430625 ** 2
for i in range(x, x+2):
print(i, isSquare(i))
printed the following results:
1000399717477066534083185452789672211951514938424998708930175541558932213310056978758103599452364409903384901149641614494249195605016959576235097480592396214296565598519295693079257885246632306201885850365687426564365813280963724310434494316592041592681626416195491751015907716210235352495422858432792668507052756279908951163972960239286719854867504108121432187033786444937064356645218196398775923710931242852937602515835035177768967470757847368349565128635934683294155947532322786360581473152034468071184081729335560769488880138928479829695277968766082973795720937033019047838250608170693879209655321034310764422462828792636246742456408134706264621790736361118589122797268261542115823201538743148116654378511916000714911467547209475246784887830649309238110794938892491396597873160778553131774466638923135932135417900066903068192088883207721545109720968467560224268563643820599665232314256575428214983451466488658896488012211237139254674708538347237589290497713613898546363590044902791724541048198769085430459186735166233549186115282574626012296888817453914112423361525305960060329430234696000121420787598967383958525670258016851764034555105019265380321048686563527396844220047826436035333266263375049097675787975100014823583097518824871586828195368306649956481108708929669583308777347960115138098217676704862934389659753628861667169905594181756523762369645897154232744410732552956489694024357481100742138381514396851789639339362228442689184910464071202445106084939268067445115601375050153663645294106475257440167535462278022649865332161044187890625 True
1000399717477066534083185452789672211951514938424998708930175541558932213310056978758103599452364409903384901149641614494249195605016959576235097480592396214296565598519295693079257885246632306201885850365687426564365813280963724310434494316592041592681626416195491751015907716210235352495422858432792668507052756279908951163972960239286719854867504108121432187033786444937064356645218196398775923710931242852937602515835035177768967470757847368349565128635934683294155947532322786360581473152034468071184081729335560769488880138928479829695277968766082973795720937033019047838250608170693879209655321034310764422462828792636246742456408134706264621790736361118589122797268261542115823201538743148116654378511916000714911467547209475246784887830649309238110794938892491396597873160778553131774466638923135932135417900066903068192088883207721545109720968467560224268563643820599665232314256575428214983451466488658896488012211237139254674708538347237589290497713613898546363590044902791724541048198769085430459186735166233549186115282574626012296888817453914112423361525305960060329430234696000121420787598967383958525670258016851764034555105019265380321048686563527396844220047826436035333266263375049097675787975100014823583097518824871586828195368306649956481108708929669583308777347960115138098217676704862934389659753628861667169905594181756523762369645897154232744410732552956489694024357481100742138381514396851789639339362228442689184910464071202445106084939268067445115601375050153663645294106475257440167535462278022649865332161044187890626 False
And it did this in 0.33 seconds.
In my opinion, my algorithm works the same as Alex Martelli's, with all the benefits thereof, but has the added benefit highly efficient simple-test rejections that save a lot of time, not to mention the reduction in size of test numbers by powers of 4, which improves speed, efficiency, accuracy and the size of numbers that are testable. Probably especially true in non-Python implementations.
Roughly 99% of all integers are rejected as non-Square before Babylonian root extraction is even implemented, and in 2/3 the time it would take the Babylonian to reject the integer. And though these tests dont speed up the process that significantly, the reduction in all test numbers to an odd by dividing out all powers of 4 really accelerates the Babylonian test.
I did a time comparison test. I tested all integers from 1 to 10 Million in succession. Using just the Babylonian method by itself (with my specially tailored initial guess) it took my Surface 3 an average of 165 seconds (with 100% accuracy). Using just the logical tests in my algorithm (excluding the Babylonian), it took 127 seconds, it rejected 99% of all integers as non-Square without mistakenly rejecting any perfect squares. Of those integers that passed, only 3% were perfect Squares (a much higher density). Using the full algorithm above that employs both the logical tests and the Babylonian root extraction, we have 100% accuracy, and test completion in only 14 seconds. The first 100 Million integers takes roughly 2 minutes 45 seconds to test.
EDIT: I have been able to bring down the time further. I can now test the integers 0 to 100 Million in 1 minute 40 seconds. A lot of time is wasted checking the data type and the positivity. Eliminate the very first two checks and I cut the experiment down by a minute. One must assume the user is smart enough to know that negatives and floats are not perfect squares.
import math
def is_square(n):
sqrt = math.sqrt(n)
return (sqrt - int(sqrt)) == 0
A perfect square is a number that can be expressed as the product of two equal integers. math.sqrt(number) return a float. int(math.sqrt(number)) casts the outcome to int.
If the square root is an integer, like 3, for example, then math.sqrt(number) - int(math.sqrt(number)) will be 0, and the if statement will be False. If the square root was a real number like 3.2, then it will be True and print "it's not a perfect square".
It fails for a large non-square such as 152415789666209426002111556165263283035677490.
My answer is:
def is_square(x):
return x**.5 % 1 == 0
It basically does a square root, then modulo by 1 to strip the integer part and if the result is 0 return True otherwise return False. In this case x can be any large number, just not as large as the max float number that python can handle: 1.7976931348623157e+308
It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.
This can be solved using the decimal module to get arbitrary precision square roots and easy checks for "exactness":
import math
from decimal import localcontext, Context, Inexact
def is_perfect_square(x):
# If you want to allow negative squares, then set x = abs(x) instead
if x < 0:
return False
# Create localized, default context so flags and traps unset
with localcontext(Context()) as ctx:
# Set a precision sufficient to represent x exactly; `x or 1` avoids
# math domain error for log10 when x is 0
ctx.prec = math.ceil(math.log10(x or 1)) + 1 # Wrap ceil call in int() on Py2
# Compute integer square root; don't even store result, just setting flags
ctx.sqrt(x).to_integral_exact()
# If previous line couldn't represent square root as exact int, sets Inexact flag
return not ctx.flags[Inexact]
For demonstration with truly huge values:
# I just kept mashing the numpad for awhile :-)
>>> base = 100009991439393999999393939398348438492389402490289028439083249803434098349083490340934903498034098390834980349083490384903843908309390282930823940230932490340983098349032098324908324098339779438974879480379380439748093874970843479280329708324970832497804329783429874329873429870234987234978034297804329782349783249873249870234987034298703249780349783497832497823497823497803429780324
>>> sqr = base ** 2
>>> sqr ** 0.5 # Too large to use floating point math
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: int too large to convert to float
>>> is_perfect_power(sqr)
True
>>> is_perfect_power(sqr-1)
False
>>> is_perfect_power(sqr+1)
False
If you increase the size of the value being tested, this eventually gets rather slow (takes close to a second for a 200,000 bit square), but for more moderate numbers (say, 20,000 bits), it's still faster than a human would notice for individual values (~33 ms on my machine). But since speed wasn't your primary concern, this is a good way to do it with Python's standard libraries.
Of course, it would be much faster to use gmpy2 and just test gmpy2.mpz(x).is_square(), but if third party packages aren't your thing, the above works quite well.
I just posted a slight variation on some of the examples above on another thread (Finding perfect squares) and thought I'd include a slight variation of what I posted there here (using nsqrt as a temporary variable), in case it's of interest / use:
import math
def is_square(n):
if not (isinstance(n, int) and (n >= 0)):
return False
else:
nsqrt = math.sqrt(n)
return nsqrt == math.trunc(nsqrt)
It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.
A variant of #Alex Martelli's solution without set
When x in seen is True:
In most cases, it is the last one added, e.g. 1022 produces the x's sequence 511, 256, 129, 68, 41, 32, 31, 31;
In some cases (i.e., for the predecessors of perfect squares), it is the second-to-last one added, e.g. 1023 produces 511, 256, 129, 68, 41, 32, 31, 32.
Hence, it suffices to stop as soon as the current x is greater than or equal to the previous one:
def is_square(n):
assert n > 1
previous = n
x = n // 2
while x * x != n:
x = (x + (n // x)) // 2
if x >= previous:
return False
previous = x
return True
x = 12345678987654321234567 ** 2
assert not is_square(x-1)
assert is_square(x)
assert not is_square(x+1)
Equivalence with the original algorithm tested for 1 < n < 10**7. On the same interval, this slightly simpler variant is about 1.4 times faster.
This is my method:
def is_square(n) -> bool:
return int(n**0.5)**2 == int(n)
Take square root of number. Convert to integer. Take the square. If the numbers are equal, then it is a perfect square otherwise not.
It is incorrect for a large square such as 152415789666209426002111556165263283035677489.
If the modulus (remainder) leftover from dividing by the square root is 0, then it is a perfect square.
def is_square(num: int) -> bool:
return num % math.sqrt(num) == 0
I checked this against a list of perfect squares going up to 1000.
It is possible to improve the Babylonian method by observing that the successive terms form a decreasing sequence if one starts above the square root of n.
def is_square(n):
assert n > 1
a = n
b = (a + n // a) // 2
while b < a:
a = b
b = (a + n // a) // 2
return a * a == n
If it's a perfect square, its square root will be an integer, the fractional part will be 0, we can use modulus operator to check fractional part, and check if it's 0, it does fail for some numbers, so, for safety, we will also check if it's square of the square root even if the fractional part is 0.
import math
def isSquare(n):
root = math.sqrt(n)
if root % 1 == 0:
if int(root) * int(root) == n:
return True
return False
isSquare(4761)
You could binary-search for the rounded square root. Square the result to see if it matches the original value.
You're probably better off with FogleBirds answer - though beware, as floating point arithmetic is approximate, which can throw this approach off. You could in principle get a false positive from a large integer which is one more than a perfect square, for instance, due to lost precision.
A simple way to do it (faster than the second one) :
def is_square(n):
return str(n**(1/2)).split(".")[1] == '0'
Another way:
def is_square(n):
if n == 0:
return True
else:
if n % 2 == 0 :
for i in range(2,n,2):
if i*i == n:
return True
else :
for i in range(1,n,2):
if i*i == n:
return True
return False
This response doesn't pertain to your stated question, but to an implicit question I see in the code you posted, ie, "how to check if something is an integer?"
The first answer you'll generally get to that question is "Don't!" And it's true that in Python, typechecking is usually not the right thing to do.
For those rare exceptions, though, instead of looking for a decimal point in the string representation of the number, the thing to do is use the isinstance function:
>>> isinstance(5,int)
True
>>> isinstance(5.0,int)
False
Of course this applies to the variable rather than a value. If I wanted to determine whether the value was an integer, I'd do this:
>>> x=5.0
>>> round(x) == x
True
But as everyone else has covered in detail, there are floating-point issues to be considered in most non-toy examples of this kind of thing.
If you want to loop over a range and do something for every number that is NOT a perfect square, you could do something like this:
def non_squares(upper):
next_square = 0
diff = 1
for i in range(0, upper):
if i == next_square:
next_square += diff
diff += 2
continue
yield i
If you want to do something for every number that IS a perfect square, the generator is even easier:
(n * n for n in range(upper))
I think that this works and is very simple:
import math
def is_square(num):
sqrt = math.sqrt(num)
return sqrt == int(sqrt)
It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.
a=int(input('enter any number'))
flag=0
for i in range(1,a):
if a==i*i:
print(a,'is perfect square number')
flag=1
break
if flag==1:
pass
else:
print(a,'is not perfect square number')
In kotlin :
It's quite easy and it passed all test cases as well.
really thanks to >> https://www.quora.com/What-is-the-quickest-way-to-determine-if-a-number-is-a-perfect-square
fun isPerfectSquare(num: Int): Boolean {
var result = false
var sum=0L
var oddNumber=1L
while(sum<num){
sum = sum + oddNumber
oddNumber = oddNumber+2
}
result = sum == num.toLong()
return result
}
def isPerfectSquare(self, num: int) -> bool:
left, right = 0, num
while left <= right:
mid = (left + right) // 2
if mid**2 < num:
left = mid + 1
elif mid**2 > num:
right = mid - 1
else:
return True
return False
This is an elegant, simple, fast and arbitrary solution that works for Python version >= 3.8:
from math import isqrt
def is_square(number):
if number >= 0:
return isqrt(number) ** 2 == number
return False
Decide how long the number will be.
take a delta 0.000000000000.......000001
see if the (sqrt(x))^2 - x is greater / equal /smaller than delta and decide based on the delta error.
import math
def is_square(n):
sqrt = math.sqrt(n)
return sqrt == int(sqrt)
It fails for a large non-square such as 152415789666209426002111556165263283035677490.
The idea is to run a loop from i = 1 to floor(sqrt(n)) then check if squaring it makes n.
bool isPerfectSquare(int n)
{
for (int i = 1; i * i <= n; i++) {
// If (i * i = n)
if ((n % i == 0) && (n / i == i)) {
return true;
}
}
return false;
}