Introduction
I am taking an online Introduction to Computer Science course for which I was given the task to create a function that takes in a coordinate and returns its corresponding "spiral index". I have a working function, but the online learning platform tells me that my code takes too long to execute (I am currently learning about code complexity and Big O notation).
The Problem Statement
All numbers on an infinite grid that extends in all four directions can be identified with a single number in the following manner:
Where 0 corresponds with the coordinate (0, 0), 5 corresponds with the coordinate (-1, 0), and 29 with (3, 2).
Create a function which returns the spiral index of any pair of coordinates that are input by the user.
Examples:
spiral_index(10, 10) returns 380.
spiral_index(10, -10) returns 440.
spiral_index(3, 15) returns 882.
spiral_index(1000, 1000) returns 3998000.
My Approach
def spiral_index(x, y):
if x == 0 and y == 0:
return 0
pos = [1, 0]
num = 1
ring_up = 0
ring_left = 0
ring_down = 0
ring_right = 0
base = 3
while pos != [x, y]:
if ring_up < base - 2:
pos[1] += 1
ring_up += 1
num += 1
elif ring_left < base - 1:
pos[0] -= 1
ring_left += 1
num += 1
elif ring_down < base - 1:
pos[1] -= 1
ring_down += 1
num += 1
elif ring_right < base:
pos[0] += 1
ring_right += 1
num += 1
else:
base = base + 2
ring_up = 0
ring_left = 0
ring_down = 0
ring_right = 0
return num
The above code is able to find the correct index for every coordinate, and (on my laptop) computes spiral_index(1000, 1000) in just over 2 seconds (2.06).
Question
I have seen some solutions people have posted to a similar problem. However, I am wondering what is making my code so slow? To my knowledge, I believe that it is executing the function in linear time (is that right?). How can I improve the speed of my function? Can you post a faster function?
The course told me that a for loop is generally faster than a while loop, and I am guessing that the conditional statements are slowing the function down as well.
Any help on the matter is greatly appreciated!
Thanks in advance.
First of all, your solution takes two coordinates, A and B, and is O(AB), which can be considered quadratic. Second of all, the similar problem involves constructing the entire spiral, which means there is no better solution. You only have to index it, which means there's no reason to traverse the entire spiral. You can simply find which ring of the spiral you're on, then do some math to figure out the number. The center has 1 element, the next ring has 8, the next 16, the next 24, and it always increases by 8 from there. This solution is constant time and can almost instantly calculate spiral_index(1000, 1000).
def spiral_index(x, y):
ax = abs(x)
ay = abs(y)
# find loop number in spiral
loop = max(ax, ay)
# one less than the edge length of the current loop
edgelen = 2 * loop
# the numbers in the inner loops
prev = (edgelen - 1) ** 2
if x == loop and y > -loop:
# right edge
return prev + y - (-loop + 1)
if y == loop:
# top edge
return prev + loop - x + edgelen - 1
if x == -loop:
# left edge
return prev + loop - y + 2 * edgelen - 1
if y == -loop:
# bottom edge
return prev + x + loop + 3 * edgelen - 1
raise Exception("this should never happen")
Related
I have a little problem which looks like a “combinatorics” problem.
We know that 1+2+3+…+k = (k^2 + k)/2; so, let’s take the set of numbers S = {1,2,3,4,…,(k^2 + k)/2} and divide this collection into k parts:
The 1st part has 1 element 1; the 2nd has 2 elements 2,3; the 3rd has 3 elements 4,5,6; …and so on…; the kth having k elements (k^2 – k + 2)/2,…,(k^2 + k)/2.
Then I have to draw a random integer in S, say i = random.randint(1, (k^2 + k)/2) and I have to do some operations according to the element that was drawn:
if i == 1:
`something`
else if 2 <= i <= 3:
`something else`
else if 4 <= i <= 6:
`something else`
…
else: # last line when `i` is in the last `kth` part
`something else`
The number k I have to use is variable, so I can't actually write the above program, because I don't know a priori where it should stop...
It seems to me that the best would be to define a function:
def cases(k):
i = random.randint(1, (k^2 + k)/2)
if i == 1:
`something`
else if 2 <= i <= 3:
… and so on…
But the problem remains: how could I write such a function without a specific k? There may be a trick in Python to do this, but I don't see how.
All ideas will be welcome.
Sorry for the indenting; I acknowledge it's wrong. Following the comment of Zach Munro, I allow myself an answer to better clarify the idea that I had in mind and which was not very clear.
Below is the kind of program I was thinking of; the something and something else are actually similar, for they use the same function, but with a different domain each time:
def cases(start, end, k):
delta = (end - start) / k
i = random.randint(1, (k**2 + k)/2)
if i == 1:
# random in the 1st part
x = random.uniform(start, start + 1*delta)
elif 2 <= i <= 3:
# random in the 2nd part
x = random.uniform(start + 1*delta, start + 2*delta)
elif 4 <= i <= 6:
# random in the 3rd part
x = random.uniform(start + 2*delta, start + 3*delta)
...
elif (k**2 – k + 2)/2 <= i <= (k**2 + k)/2:
# random in the kth part
x = random.uniform(start + (k-1)*delta, start + k*delta)
The question is always how to stop using ... in the program, but to make something that runs when the parameters are provided (start is the beginning of an interval, end is the end of the interval and k is in fact the number of parts into which we separate this interval).
Basically, we sample more and more as we move away from the origin of the interval.
I am a new programmer and this is my first question here so excuse me if its pretty easy.
M doing a multiplication table but each time it stops after the 1 and it doesnt increment the number
I am trying to do it with 2 while loops
nb = 1
i = 0
while nb<10 :
while i<=10 :
print(nb * i)
i+=1
nb+=1
with that code I only have the 1 multiplication table then the program stops
Reset variable inside loop. Variable i reaches its max value after the first iteration of inner loop, which we need to set back to 0 for the next iteration to work:
nb = 1
while nb < 10 :
i = 0
while i <= 10 :
print(nb * i)
i += 1
nb += 1
You can do the same using for, which in my opinion is more readable and you don't need to worry about incrementing/resetting variables:
for x in range(1, 10):
for y in range(11):
print(x * y)
I was curious if any of you could come up with a more streamline version of code to calculate Brown numbers. as of the moment, this code can do ~650! before it moves to a crawl. Brown Numbers are calculated thought the equation n! + 1 = m**(2) Where M is an integer
brownNum = 8
import math
def squareNum(n):
x = n // 2
seen = set([x])
while x * x != n:
x = (x + (n // x)) // 2
if x in seen: return False
seen.add(x)
return True
while True:
for i in range(math.factorial(brownNum)+1,math.factorial(brownNum)+2):
if squareNum(i) is True:
print("pass")
print(brownNum)
print(math.factorial(brownNum)+1)
break
else:
print(brownNum)
print(math.factorial(brownNum)+1)
brownNum = brownNum + 1
continue
break
print(input(" "))
Sorry, I don't understand the logic behind your code.
I don't understand why you calculate math.factorial(brownNum) 4 times with the same value of brownNum each time through the while True loop. And in the for loop:
for i in range(math.factorial(brownNum)+1,math.factorial(brownNum)+2):
i will only take on the value of math.factorial(brownNum)+1
Anyway, here's my Python 3 code for a brute force search of Brown numbers. It quickly finds the only 3 known pairs, and then proceeds to test all the other numbers under 1000 in around 1.8 seconds on this 2GHz 32 bit machine. After that point you can see it slowing down (it hits 2000 around the 20 second mark) but it will chug along happily until the factorials get too large for your machine to hold.
I print progress information to stderr so that it can be separated from the Brown_number pair output. Also, stderr doesn't require flushing when you don't print a newline, unlike stdout (at least, it doesn't on Linux).
import sys
# Calculate the integer square root of `m` using Newton's method.
# Returns r: r**2 <= m < (r+1)**2
def int_sqrt(m):
if m <= 0:
return 0
n = m << 2
r = n >> (n.bit_length() // 2)
while True:
d = (n // r - r) >> 1
r += d
if -1 <= d <= 1:
break
return r >> 1
# Search for Browns numbers
fac = i = 1
while True:
if i % 100 == 0:
print('\r', i, file=sys.stderr, end='')
fac *= i
n = fac + 1
r = int_sqrt(n)
if r*r == n:
print('\nFound', i, r)
i += 1
You might want to:
pre calculate your square numbers, instead of testing for them on the fly
pre calculate your factorial for each loop iteration num_fac = math.factorial(brownNum) instead of multiple calls
implement your own, memoized, factorial
that should let you run to the hard limits of your machine
one optimization i would make would be to implement a 'wrapper' function around math.factorial that caches previous values of factorial so that as your brownNum increases, factorial doesn't have as much work to do. this is known as 'memoization' in computer science.
edit: found another SO answer with similar intention: Python: Is math.factorial memoized?
You should also initialize the square root more closely to the root.
e = int(math.log(n,4))
x = n//2**e
Because of 4**e <= n <= 4**(e+1) the square root will be between x/2 and x which should yield quadratic convergence of the Heron formula from the first iteration on.
This question already has an answer here:
Random walk's weird outcome in python 3?
(1 answer)
Closed 6 years ago.
I am having unexpected outputs with the following code:
import random
N = 30 # number of steps
n = random.random() # generate a random number
x = 0
y = 0
z = 0
count = 0
while count <= N:
if n < 1/3:
x = x + 1 # move east
n = random.random() # generate a new random number
if n >= 1/3 and n < 2/3:
y = y + 1 # move north
n = random.random() # generate a new random number
if n >= 2/3:
z = z + 1 # move up
n = random.random() # generate a new random number
print("(%d,%d,%d)" % (x,y,z))
count = count + 1
When I run the code, the problem is:
Code output displays 31 coordinates, 1 more than the number of steps (N) variable.
Each iteration for 1 step should take only 1 step but it sometimes take multiple steps.
When I tested the code, the problem is ensured. To test the code, I assigned N = 1, and saw the following output:
(-1,0,1) This should be the initial step, but it took multiple steps (both x-1 and z+1), how could this happen?
(-2,0,1) Number of step variable (N) = 1 but this is the second output, why was it displayed?
Thanks for helping
N is 30, so count goes from 0 to 30. Since 30 <= 30 you will run the loop for count=0, 1, ..., 29 AND 30 which is 31 times
When you take a step, you don't ensure that another step is NOT taken. If random happens, you could enter the second or third if after already being in a previous one in the same loop iteration
You are dividing two ints which will only result in another int. So basically your code is do the following:
if n < 0:
x = x + 1 # move east
n = random.random() # generate a new random number
if n >= 0 and n < 1:
y = y + 1 # move north
n = random.random() # generate a new random number
if n >= 1:
z = z + 1 # move up
n = random.random()
fix by changing each if line to include a floating point number
if n < 1.0/3
Any tips on optimizing this python code for finding next palindrome:
Input number can be of 1000000 digits
COMMENTS ADDED
#! /usr/bin/python
def inc(lst,lng):#this function first extract the left half of the string then
#convert it to int then increment it then reconvert it to string
#then reverse it and finally append it to the left half.
#lst is input number and lng is its length
if(lng%2==0):
olst=lst[:lng/2]
l=int(lng/2)
olst=int(olst)
olst+=1
olst=str(olst)
p=len(olst)
if l<p:
olst2=olst[p-2::-1]
else:
olst2=olst[::-1]
lst=olst+olst2
return lst
else:
olst=lst[:lng/2+1]
l=int(lng/2+1)
olst=int(olst)
olst+=1
olst=str(olst)
p=len(olst)
if l<p:
olst2=olst[p-3::-1]
else:
olst2=olst[p-2::-1]
lst=olst+olst2
return lst
t=raw_input()
t=int(t)
while True:
if t>0:
t-=1
else:
break
num=raw_input()#this is input number
lng=len(num)
lst=num[:]
if(lng%2==0):#this if find next palindrome to num variable
#without incrementing the middle digit and store it in lst.
olst=lst[:lng/2]
olst2=olst[::-1]
lst=olst+olst2
else:
olst=lst[:lng/2+1]
olst2=olst[len(olst)-2::-1]
lst=olst+olst2
if int(num)>=int(lst):#chk if lst satisfies criteria for next palindrome
num=inc(num,lng)#otherwise call inc function
print num
else:
print lst
I think most of the time in this code is spent converting strings to integers and back. The rest is slicing strings and bouncing around in the Python interpreter. What can be done about these three things? There are a few unnecessary conversions in the code, which we can remove. I see no way to avoid the string slicing. To minimize your time in the interpreter you just have to write as little code as possible :-) and it also helps to put all your code inside functions.
The code at the bottom of your program, which takes a quick guess to try and avoid calling inc(), has a bug or two. Here's how I might write that part:
def nextPal(num):
lng = len(num)
guess = num[:lng//2] + num[(lng-1)//2::-1] # works whether lng is even or odd
if guess > num: # don't bother converting to int
return guess
else:
return inc(numstr, n)
This simple change makes your code about 100x faster for numbers where inc doesn't need to be called, and about 3x faster for numbers where it does need to be called.
To do better than that, I think you need to avoid converting to int entirely. That means incrementing the left half of the number without using ordinary Python integer addition. You can use an array and carry out the addition algorithm "by hand":
import array
def nextPal(numstr):
# If we don't need to increment, just reflect the left half and return.
n = len(numstr)
h = n//2
guess = numstr[:n-h] + numstr[h-1::-1]
if guess > numstr:
return guess
# Increment the left half of the number without converting to int.
a = array.array('b', numstr)
zero = ord('0')
ten = ord('9') + 1
for i in range(n - h - 1, -1, -1):
d = a[i] + 1
if d == ten:
a[i] = zero
else:
a[i] = d
break
else:
# The left half was all nines. Carry the 1.
# Update n and h since the length changed.
a.insert(0, ord('1'))
n += 1
h = n//2
# Reflect the left half onto the right half.
a[n-h:] = a[h-1::-1]
return a.tostring()
This is another 9x faster or so for numbers that require incrementing.
You can make this a touch faster by using a while loop instead of for i in range(n - h - 1, -1, -1), and about twice as fast again by having the loop update both halves of the array rather than just updating the left-hand half and then reflecting it at the end.
You don't have to find the palindrome, you can just generate it.
Split the input number, and reflect it. If the generated number is too small, then increment the left hand side and reflect it again:
def nextPal(n):
ns = str(n)
oddoffset = 0
if len(ns) % 2 != 0:
oddoffset = 1
leftlen = len(ns) / 2 + oddoffset
lefts = ns[0:leftlen]
right = lefts[::-1][oddoffset:]
p = int(lefts + right)
if p < n:
## Need to increment middle digit
left = int(lefts)
left += 1
lefts = str(left)
right = lefts[::-1][oddoffset:]
p = int(lefts + right)
return p
def test(n):
print n
p = nextPal(n)
assert p >= n
print p
test(1234567890)
test(123456789)
test(999999)
test(999998)
test(888889)
test(8999999)
EDIT
NVM, just look at this page: http://thetaoishere.blogspot.com/2009/04/finding-next-palindrome-given-number.html
Using strings. n >= 0
from math import floor, ceil, log10
def next_pal(n):
# returns next palindrome, param is an int
n10 = str(n)
m = len(n10) / 2.0
s, e = int(floor(m - 0.5)), int(ceil(m + 0.5))
start, middle, end = n10[:s], n10[s:e], n10[e:]
assert (start, middle[0]) == (end[-1::-1], middle[-1]) #check that n is actually a palindrome
r = int(start + middle[0]) + 1 #where the actual increment occurs (i.e. add 1)
r10 = str(r)
i = 3 - len(middle)
if len(r10) > len(start) + 1:
i += 1
return int(r10 + r10[-i::-1])
Using log, more optized. n > 9
def next_pal2(n):
k = log10(n + 1)
l = ceil(k)
s, e = int(floor(l/2.0 - 0.5)), int(ceil(l/2.0 + 0.5))
mmod, emod = 10**(e - s), int(10**(l - e))
start, end = divmod(n, emod)
start, middle = divmod(start, mmod)
r1 = 10*start + middle%10 + 1
i = middle > 9 and 1 or 2
j = s - i + 2
if k == l:
i += 1
r2 = int(str(r1)[-i::-1])
return r1*10**j + r2