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Fibonacci sequence calculator seems correct but can't find similar code online. Is there something wrong?

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6

I made a simple Fibonacci sequence calculator for the first 22 terms:

i=1
n=0
while i<=20000:
 i = i + n
 n = i - n
 print(i)

Looks like the result is correct

1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657

but I can't seem to find similar code anywhere online. I think that's a big red flag. Can someone tell me what is wrong here? Is this inefficient code?

python fibonacci

share|improve this question

asked 5 hours ago

fibonaccipistaccifibonaccipistacci

311

New contributor

fibonaccipistacci is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

6

I made a simple Fibonacci sequence calculator for the first 22 terms:

i=1
n=0
while i<=20000:
 i = i + n
 n = i - n
 print(i)

Looks like the result is correct

1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657

but I can't seem to find similar code anywhere online. I think that's a big red flag. Can someone tell me what is wrong here? Is this inefficient code?

python fibonacci

share|improve this question

asked 5 hours ago

fibonaccipistaccifibonaccipistacci

311

New contributor

fibonaccipistacci is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

I made a simple Fibonacci sequence calculator for the first 22 terms:

i=1
n=0
while i<=20000:
 i = i + n
 n = i - n
 print(i)

Looks like the result is correct

1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657

but I can't seem to find similar code anywhere online. I think that's a big red flag. Can someone tell me what is wrong here? Is this inefficient code?

python fibonacci

share|improve this question

asked 5 hours ago

fibonaccipistaccifibonaccipistacci

311

New contributor

fibonaccipistacci is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

i=1
n=0
while i<=20000:
 i = i + n
 n = i - n
 print(i)

1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657

share|improve this question

asked 5 hours ago

fibonaccipistaccifibonaccipistacci

311

New contributor

fibonaccipistacci is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 5 hours ago

fibonaccipistaccifibonaccipistacci

311

New contributor

fibonaccipistacci is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

fibonaccipistaccifibonaccipistacci

311

New contributor

fibonaccipistacci is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

fibonaccipistaccifibonaccipistacci

311

2 Answers
2

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7

No, that code is fine. The probable reason you can't find similar code online is that it's unusual to use the subtraction operator in Fibonacci, which is a purely additive function, tn = tn-2 + tn-1.

It works, of course, since addition/subtraction is both commutative and associative, meaning that order and grouping of terms is unimportant:

i = i + n # iNew = iOld + nOld

n = i - n # nNew = (iNew) - nOld
 # = (iOld + nOld) - nOld
 # = iOld + (nOld - nOld)
 # = iOld + (0)
 # = iOld

Use of subtraction allows you to bypass needing a third variable, which would be something like this in a lesser language than Python:

nextN = i + n
i = n
n = nextN

In Python, you don't actually need that since you can use tuple assignment such as:

(n, i) = (i, n + i)

With that, everything on the right of the = is evaluated before any assignments to the left.

share|improve this answer

edited 27 mins ago

answered 4 hours ago

paxdiablopaxdiablo

648k17912691695

add a comment | 

3

It's an unusual way to do it, but it's correct. Your lines:

i = i + n
n = i - n

are the same as doing:

new_i = i + n
n = i
i = new_i

or,

i, n = i + n, i

which would be the usual way in Python

share|improve this answer

answered 4 hours ago

Ned BatchelderNed Batchelder

265k53450572

add a comment | 

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7

No, that code is fine. The probable reason you can't find similar code online is that it's unusual to use the subtraction operator in Fibonacci, which is a purely additive function, tn = tn-2 + tn-1.

It works, of course, since addition/subtraction is both commutative and associative, meaning that order and grouping of terms is unimportant:

i = i + n # iNew = iOld + nOld

n = i - n # nNew = (iNew) - nOld
 # = (iOld + nOld) - nOld
 # = iOld + (nOld - nOld)
 # = iOld + (0)
 # = iOld

Use of subtraction allows you to bypass needing a third variable, which would be something like this in a lesser language than Python:

nextN = i + n
i = n
n = nextN

In Python, you don't actually need that since you can use tuple assignment such as:

(n, i) = (i, n + i)

With that, everything on the right of the = is evaluated before any assignments to the left.

share|improve this answer

edited 27 mins ago

answered 4 hours ago

paxdiablopaxdiablo

648k17912691695

add a comment | 

7

No, that code is fine. The probable reason you can't find similar code online is that it's unusual to use the subtraction operator in Fibonacci, which is a purely additive function, tn = tn-2 + tn-1.

It works, of course, since addition/subtraction is both commutative and associative, meaning that order and grouping of terms is unimportant:

i = i + n # iNew = iOld + nOld

n = i - n # nNew = (iNew) - nOld
 # = (iOld + nOld) - nOld
 # = iOld + (nOld - nOld)
 # = iOld + (0)
 # = iOld

Use of subtraction allows you to bypass needing a third variable, which would be something like this in a lesser language than Python:

nextN = i + n
i = n
n = nextN

In Python, you don't actually need that since you can use tuple assignment such as:

(n, i) = (i, n + i)

With that, everything on the right of the = is evaluated before any assignments to the left.

share|improve this answer

edited 27 mins ago

answered 4 hours ago

paxdiablopaxdiablo

648k17912691695

add a comment | 

No, that code is fine. The probable reason you can't find similar code online is that it's unusual to use the subtraction operator in Fibonacci, which is a purely additive function, tn = tn-2 + tn-1.

It works, of course, since addition/subtraction is both commutative and associative, meaning that order and grouping of terms is unimportant:

i = i + n # iNew = iOld + nOld

n = i - n # nNew = (iNew) - nOld
 # = (iOld + nOld) - nOld
 # = iOld + (nOld - nOld)
 # = iOld + (0)
 # = iOld

Use of subtraction allows you to bypass needing a third variable, which would be something like this in a lesser language than Python:

nextN = i + n
i = n
n = nextN

In Python, you don't actually need that since you can use tuple assignment such as:

(n, i) = (i, n + i)

With that, everything on the right of the = is evaluated before any assignments to the left.

share|improve this answer

edited 27 mins ago

answered 4 hours ago

paxdiablopaxdiablo

648k17912691695

i = i + n # iNew = iOld + nOld

n = i - n # nNew = (iNew) - nOld
 # = (iOld + nOld) - nOld
 # = iOld + (nOld - nOld)
 # = iOld + (0)
 # = iOld

nextN = i + n
i = n
n = nextN

(n, i) = (i, n + i)

share|improve this answer

edited 27 mins ago

answered 4 hours ago

paxdiablopaxdiablo

648k17912691695

answered 4 hours ago

paxdiablopaxdiablo

648k17912691695

answered 4 hours ago

paxdiablopaxdiablo

648k17912691695

3

It's an unusual way to do it, but it's correct. Your lines:

i = i + n
n = i - n

are the same as doing:

new_i = i + n
n = i
i = new_i

or,

i, n = i + n, i

which would be the usual way in Python

share|improve this answer

answered 4 hours ago

Ned BatchelderNed Batchelder

265k53450572

add a comment | 

3

It's an unusual way to do it, but it's correct. Your lines:

i = i + n
n = i - n

are the same as doing:

new_i = i + n
n = i
i = new_i

or,

i, n = i + n, i

which would be the usual way in Python

share|improve this answer

answered 4 hours ago

Ned BatchelderNed Batchelder

265k53450572

add a comment | 

It's an unusual way to do it, but it's correct. Your lines:

i = i + n
n = i - n

are the same as doing:

new_i = i + n
n = i
i = new_i

or,

i, n = i + n, i

which would be the usual way in Python

share|improve this answer

answered 4 hours ago

Ned BatchelderNed Batchelder

265k53450572

i = i + n
n = i - n

new_i = i + n
n = i
i = new_i

i, n = i + n, i

share|improve this answer

answered 4 hours ago

Ned BatchelderNed Batchelder

265k53450572

answered 4 hours ago

Ned BatchelderNed Batchelder

265k53450572

answered 4 hours ago

Ned BatchelderNed Batchelder

265k53450572