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

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5

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 May 9 at 0:39

fibonaccipistaccifibonaccipistacci

294

add a comment | 

5

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 May 9 at 0:39

fibonaccipistaccifibonaccipistacci

294

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 May 9 at 0:39

fibonaccipistaccifibonaccipistacci

294

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 May 9 at 0:39

fibonaccipistaccifibonaccipistacci

294

share|improve this question

asked May 9 at 0:39

fibonaccipistaccifibonaccipistacci

294

asked May 9 at 0:39

fibonaccipistaccifibonaccipistacci

294

asked May 9 at 0:39

fibonaccipistaccifibonaccipistacci

294

2 Answers
2

active

oldest

votes

8

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 May 9 at 5:15

answered May 9 at 0:45

paxdiablopaxdiablo

650k18012721697

add a comment | 

5

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

edited May 9 at 5:50

cs95

147k25186260

answered May 9 at 0:44

Ned BatchelderNed Batchelder

266k53452575

add a comment | 

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8

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 May 9 at 5:15

answered May 9 at 0:45

paxdiablopaxdiablo

650k18012721697

add a comment | 

8

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 May 9 at 5:15

answered May 9 at 0:45

paxdiablopaxdiablo

650k18012721697

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 May 9 at 5:15

answered May 9 at 0:45

paxdiablopaxdiablo

650k18012721697

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 May 9 at 5:15

answered May 9 at 0:45

paxdiablopaxdiablo

650k18012721697

answered May 9 at 0:45

paxdiablopaxdiablo

650k18012721697

answered May 9 at 0:45

paxdiablopaxdiablo

650k18012721697

5

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

edited May 9 at 5:50

cs95

147k25186260

answered May 9 at 0:44

Ned BatchelderNed Batchelder

266k53452575

add a comment | 

5

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

edited May 9 at 5:50

cs95

147k25186260

answered May 9 at 0:44

Ned BatchelderNed Batchelder

266k53452575

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

edited May 9 at 5:50

cs95

147k25186260

answered May 9 at 0:44

Ned BatchelderNed Batchelder

266k53452575

i = i + n
n = i - n

new_i = i + n
n = i
i = new_i

i, n = i + n, i

share|improve this answer

edited May 9 at 5:50

cs95

147k25186260

answered May 9 at 0:44

Ned BatchelderNed Batchelder

266k53452575

edited May 9 at 5:50

cs95

147k25186260

edited May 9 at 5:50

cs95

147k25186260

answered May 9 at 0:44

Ned BatchelderNed Batchelder

266k53452575

answered May 9 at 0:44

Ned BatchelderNed Batchelder

266k53452575