We Keep Coding

How to write a program with unspecified number of if-else

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.

Primitive Calculator - Dynamic & Recursive approach

I have tried to solve the Primitive calculator problem with dynamic and recursive approach , works fine for smaller inputs but taking long time for larger inputs (eg: 96234) .
You are given a primitive calculator that can perform the following three operations with
the current number 𝑥: multiply 𝑥 by 2, multiply 𝑥 by 3, or add 1 to 𝑥. Your goal is given a
positive integer 𝑛, find the minimum number of operations needed to obtain the number 𝑛
starting from the number 1.
import sys
def optimal_sequence(n,memo={}):
if n in memo:
return memo[n]
if (n==1):
return 0
c1 = 1+optimal_sequence(n-1,memo)
c2 = float('inf')
if n % 2 == 0 :
c2 = 1+optimal_sequence(n // 2,memo)
c3 = float('inf')
if n % 3 == 0 :
c3 = 1+optimal_sequence(n // 3,memo)
c = min(c1,c2,c3)
memo[n] = c
return c
input = sys.stdin.read()
n = int(input)
sequence = optimal_sequence(n)
print(sequence) # Only printing optimal no. of operations
Can anyone point out what is wrong in recursive solution as it works fine by using for loop.

There are a few things to consider here. The first is that you always check if you can subtract 1 away from n. This is always going to be true until n is 1. therefor with a number like 12. You will end up taking 1 away first, then calling the function again with n=11, then n=10, then n=9 etc......only once you have resolved how many steps it will take to resolve using the -1 method (in this case c1 will be 11) you then try for c2.
So for c2 you then half 12, then call the function which will start with the -1 again so you end up with n=12, n=6, n=5, n=4...etc. Even though you have n in the memo you still spend a lot of wasted time on function calls.
Instead you probably want to just shrink the problem space as fast as possible. So start with the rule that will reduce n the most. I.E divide by 3, if that doesnt work then divide by 2, only if neither of the first two worked then subtract 1.
With this method you dont even need to track n as n will always be getting smaller so there is no need to have a memo dict tracking the results.
from time import time
def optimal_sequence(n):
if n == 1:
return 0
elif n % 3 == 0:
c = optimal_sequence(n // 3)
elif n % 2 == 0:
c = optimal_sequence(n // 2)
else:
c = optimal_sequence(n - 1)
return 1 + c
n = int(input("Enter a value for N: "))
start = time()
sequence = optimal_sequence(n)
end = time()
print(f"{sequence=} took {end - start} seconds")
also input is a python function to read from teh terminal you dont need to uses stdin
OUTPUT
Enter a value for N: 96234
sequence=15 took 0.0 seconds

Python Fewest Lines Challenge (HOTPO/Collatz)

The Collatz Conjecture, otherwise known as Half Or Triple Plus One (HOTPO), takes any positive integer n to start.
If n is even, divide it by two. If n is odd, multiply it by 3 and add 1. The conjecture is that no matter what number you start with, it will always reach 1.
The challenge is to write a program that outputs all the values of n as well as the number of steps it took. For example, with n = 3, the output should be:
3
10
5
16
8
4
2
1
Finished. 7 steps required.
My best attempt takes 6 lines:
x = [3]
while x[-1] != 1:
if x[-1] % 2 == 1: x.append(3*x[-1]+1)
x.append(x[-1]/2)
for num in x: print num
print "Finished. " + str(len(x)-1) + " steps needed."
What is the minimum number of lines required to generate this output for any reasonably-sized n? Can my code be further reduced?

Some improvements making use of the lambda function:
x=[3]
while x[-1] != 1:
x.append((lambda n: n%2==0 and n/2 or 3*n+1)(x[-1]))
print(x[-1])
print "Finished. " + str(len(x)-1) + " steps needed."
I just stick more code in one line, and there is now only one loop instead of two, printing can be done from the same loop as calculating.
Or since it is for least number of lines, you can use two one line loops and get it in 3 lines:
x=[3]
while x[-1]!=1: x=(lambda y: y+[(lambda n: n%2==0 and n/2 or 3*n+1)(x[-1])])(x)
print("\n".join([str(y) for y in x])+"\nFinished. {} steps required.".format(len(x)-1))
With thanks to Pythonista for that last line :)

Another way to write it in three lines without the use of lambda function because they are unnecessary for a variable that can be referenced.
x = [3]
while x[-1] != 1:x.append((3*x[-1]+1) if x[-1] % 2 == 1 else (x[-1]/2))
print("\n".join([str(num) for num in x])+"\nFinished. " + str(len(x)-1) + " steps needed.")

Optimise the solution to Project Euler 12 (Python)

I have the following code for Project Euler Problem 12. However, it takes a very long time to execute. Does anyone have any suggestions for speeding it up?
n = input("Enter number: ")
def genfact(n):
t = []
for i in xrange(1, n+1):
if n%i == 0:
t.append(i)
return t
print "Numbers of divisors: ", len(genfact(n))
print
m = input("Enter the number of triangle numbers to check: ")
print
for i in xrange (2, m+2):
a = sum(xrange(i))
b = len(genfact(a))
if b > 500:
print a
For n, I enter an arbitrary number such as 6 just to check whether it indeed returns the length of the list of the number of factors.
For m, I enter entered 80 000 000
It works relatively quickly for small numbers. If I enter b > 50 ; it returns 28 for a, which is correct.

My answer here isn't pretty or elegant, it is still brute force. But, it simplifies the problem space a little and terminates successfully in less than 10 seconds.
Getting factors of n:
Like #usethedeathstar mentioned, it is possible to test for factors only up to n/2. However, we can do better by testing only up to the square root of n:
let n = 36
=> factors(n) : (1x36, 2x18, 3x12, 4x9, 6x6, 9x4, 12x3, 18x2, 36x1)
As you can see, it loops around after 6 (the square root of 36). We also don't need to explicitly return the factors, just find out how many there are... so just count them off with a generator inside of sum():
import math
def get_factors(n):
return sum(2 for i in range(1, round(math.sqrt(n)+1)) if not n % i)
Testing the triangular numbers
I have used a generator function to yield the triangular numbers:
def generate_triangles(limit):
l = 1
while l <= limit:
yield sum(range(l + 1))
l += 1
And finally, start testing:
def test_triangles():
triangles = generate_triangles(100000)
for i in triangles:
if get_factors(i) > 499:
return i
Running this with the profiler, it completes in less than 10 seconds:
$ python3 -m cProfile euler12.py
361986 function calls in 8.006 seconds
The BIGGEST time saving here is get_factors(n) testing only up to the square root of n - this makes it heeeaps quicker and you save heaps of memory overhead by not generating a list of factors.
As I said, it still isn't pretty - I am sure there are more elegant solutions. But, it fits the bill of being faster :)

I got my answer to run in 1.8 seconds with Python.
import time
from math import sqrt
def count_divisors(n):
d = {}
count = 1
while n % 2 == 0:
n = n / 2
try:
d[2] += 1
except KeyError:
d[2] = 1
for i in range(3, int(sqrt(n+1)), 2):
while n % i == 0 and i != n:
n = n / i
try:
d[i] += 1
except KeyError:
d[i] = 1
d[n] = 1
for _,v in d.items():
count = count * (v + 1)
return count
def tri_number(num):
next = 1 + int(sqrt(1+(8 * num)))
return num + (next/2)
def main():
i = 1
while count_divisors(i) < 500:
i = tri_number(i)
return i
start = time.time()
answer = main()
elapsed = (time.time() - start)
print("result %s returned in %s seconds." % (answer, elapsed))
Here is the output showing the timedelta and correct answer:
$ python ./project012.py
result 76576500 returned in 1.82238006592 seconds.
Factoring
For counting the divisors, I start by initializing an empty dictionary and a counter. For each factor found, I create key of d[factor] with value of 1 if it does not exist, otherwise, I increment the value d[factor].
For example, if we counted the factors 100, we would see d = {25: 1, 2: 2}
The first while loop, I factor out all 2's, dividing n by 2 each time. Next, I begin factoring at 3, skipping two each time (since we factored all even numbers already), and stopping once I get to the square root of n+1.
We stop at the square_root of n because if there's a pair of factors with one of the numbers bigger than square_root of n, the other of the pair has to be less than 10. If the smaller one doesn't exist, there is no matching larger factor.
https://math.stackexchange.com/questions/1343171/why-only-square-root-approach-to-check-number-is-prime
while n % 2 == 0:
n = n / 2
try:
d[2] += 1
except KeyError:
d[2] = 1
for i in range(3, int(sqrt(n+1)), 2):
while n % i == 0 and i != n:
n = n / i
try:
d[i] += 1
except KeyError:
d[i] = 1
d[n] = 1
Now that I have gotten each factor, and added it to the dictionary, we have to add the last factor (which is just n).
Counting Divisors
Now that the dictionary is complete, we loop through each of the items, and apply the following formula: d(n)=(a+1)(b+1)(c+1)...
https://www.wikihow.com/Determine-the-Number-of-Divisors-of-an-Integer
All this formula means is taking all of the counts of each factor, adding 1, then multiplying them together. Take 100 for example, which has factors 25, 2, and 2. We would calculate d(n)=(a+1)(b+1) = (1+1)(2+1) = (2)(3) = 6 total divisors
for _,v in d.items():
count = count * (v + 1)
return count
Calculate Triangle Numbers
Now, taking a look at tri_number(), you can see that I opted to calculate the next triangle number in a sequence without manually adding each whole number together (saving me millions of operations). Instead I used T(n) = n (n+1) / 2
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html
We are providing a whole number to the function as an argument, so we need to solve for n, which is going to be the whole number to add next. Once we have the next number (n), we simply add that single number to num and return
S=n(n+1)2
S=n2+n2
2S=n2+n
n2+n−2S=0
At this point, we use the quadratic formula for : ax2+bx+c=0.
n=−b±√b2−4ac / 2a
n=−1±√1−4(1)(−2S) / 2
n=−1±√1+8S / 2
https://socratic.org/questions/how-do-you-solve-for-n-in-s-n-n-1-2
So all tri_number() does is evaluate n=1+√1+8S / 2 (we ignore the negative equation here). The answer that is returned is the next triangle number in the sequence.
def tri_number(num):
next = 1 + int(sqrt(1+(8 * num)))
return num + (next/2)
Main Loop
Finally, we can look at main(). We start at whole number 1. We count the divisor of 1. If it is less than 500, we get the next triangle number, then try again and again until we get a number with > 500 divisors.
def main():
i = 1
while count_divisors(i) < 500:
i = tri_number(i)
return i
I am sure there are additional ways to optimize but I am not smart enough to understand those ways. If you find any better ways to optimize python, let me know! I originally solved project 12 in Golang, and that run in 25 milliseconds!
$ go run project012.go
76576500
2018/07/12 01:56:31 TIME: main() took 23.581558ms

one of the hints i can give is
def genfact(n):
t = []
for i in xrange(1, n+1):
if n%i == 0:
t.append(i)
return t
change that to
def genfact(n):
t=[]
for i in xrange(1,numpy.sqrt(n)+1):
if(n%i==0):
t.append(i)
t.apend(n/i)
since if a is a divisor than so is b=n/a, since a*b=a*n/b=n, That should help a part already (not sure if in your case a square is possible, but if so, add another case to exclude adding the same number twice)
You could devise a recursive thing too, (like if it is something like for 28, you get 1,28,2,14 and at the moment you are at knowing 14, you put in something to actually remember the divisors of 14 (memoize), than check if they are alraedy in the list, and if not, add them to the list, together with 28/d for each of the divisors of 14, and at the end just take out the duplicates
If you think my first answer is still not fast enough, ask for more, and i will check how it would be done to solve it faster with some more tricks (could probably make use of erastothenes sieve or so too, and some other tricks could be thought up as well if you would wish to really blow up the problem to huge proportions, like to check the first one with over 10k divisors or so)

while True:
c=0
n=1
m=1
for i in range(1,n+1):
if n%i==0:
c=c+1
m=m+1
n=m*(m+1)/2
if c>500:
break
print n

this is not my code but it is so optimized.
source: http://code.jasonbhill.com/sage/project-euler-problem-12/
import time
def num_divisors(n):
if n % 2 == 0: n = n / 2
divisors = 1
count = 0
while n % 2 == 0:
count += 1
n = n / 2
divisors = divisors * (count + 1)
p = 3
while n != 1:
count = 0
while n % p == 0:
count += 1
n = n / p
divisors = divisors * (count + 1)
p += 2
return divisors
def find_triangular_index(factor_limit):
n = 1
lnum, rnum = num_divisors(n), num_divisors(n + 1)
while lnum * rnum < 500:
n += 1
lnum, rnum = rnum, num_divisors(n + 1)
return n
start = time.time()
index = find_triangular_index(500)
triangle = (index * (index + 1)) / 2
elapsed = (time.time() - start)
print("result %s returned in %s seconds." % (triangle, elapsed))

Optimizing python code

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