CODEWITHSUNDEEP
python
matin saber
matin saber

Reputation: 52

How to solve this equation with a loop?

This is the sequence:

1, 2, 3, -1, 7, -10, 3, -73, …

and actually it's like this:

t(n) = (t(n-3) * t(n-2)) - t(n-1)

for example: -10 = (3 * -1) - 7.

I used this code but it's not acting like this equation that I have provided.

n1 = 1
n2 = 2
n3 = 3
m = eval(input("number: "))
if m < 4:
    if m == n1:
        print(n1)
    elif m == n2:
        print(n2)
    elif m == n3:
        print(n3)
elif m >= 4:
    n4 = (n1 * n2 - n3)
    print(n4)

Upvotes: 1

Views: 215

Answers (5)

srishtigarg
srishtigarg

Reputation: 1204

This is how you can do this. This is similar to Fibonacci sequence.

n1 = 1
n2 = 2
n3 = 3
m = int(input("number: "))
if m < 4:
    if m == n1:
        print(n1)
    elif m == n2:
        print(n2)
    elif m == n3:
        print(n3)
    
else:
    for i in range(3, m):
        temp1, temp2 = n2, n3
        n3 = n1 * n2 - n3
        n1, n2 = temp1, temp2
    print(n3)

Upvotes: 2

wuerfelfreak
wuerfelfreak

Reputation: 2439

This is my solution to finding t(m)

n = [1,2,3]
m = int(input("number: ")) #eval is a big security risk!!!
while len(n) < m:
    n.append((n[-3] * n[-2] - n[-1]))
print(n[m-1])

where n[negative number] is a number from the end of the list.

Results in:

1, 2, 3, -1, 7, -10, 3, -73, 43, -262, -2877, -8389, 762163, 23372990, etc.

Upvotes: 2

Bigya Pradhan
Bigya Pradhan

Reputation: 206

sequence=[1,2,3]

def find_value(n):
    if n < 4:
        return sequence[n-1]
    else:
        count=3
        while count<n:
            sequence.append(sequence[0]*sequence[1]-sequence[2])
            sequence.pop(0)
            count=count+1
        return sequence[-1]

if __name__=="__main__":
    n=int(input("Enter digit"))
    print(find_value(n))

Upvotes: 1

dariofac
dariofac

Reputation: 267

This can be handled using a recursive approach, like the following:

m = eval(input("number: "))


def recursive_function(n):
    if 1 <= n < 4:
        return n
    else:
        return (recursive_function(n-3) * recursive_function(n-2)) - recursive_function(n-1)

res = recursive_function(m)
print(m)

You have to perform checks on your input (e.g. if you input a string, the function will crash).

Upvotes: -1

Mark Feng
Mark Feng

Reputation: 1009

A O(n) time and space complexity solution:

def t(n):
    arr = [1, 2, 3]
    if n < 3: return arr[n]
    for i in range(n - 3):
        # calculate the next number with your given equation,
        # and push the result into the temp array
        arr.append(arr[-3] * arr[-2] - arr[-1])

    print(arr)  # show arr content, can remove this line
    # return the last item of the array, which is the result
    return arr[-1]


print(t(10))

We won't need recursion here and it would be more efficient in both space and time cost.

Upvotes: 0

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