Problem 44481. How many Fibonacci numbers?
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8 Comments
Hello,
As soon as I do test 3, the 499th fibonacci term with 104 digits is converted to a double  should the test be char or strings for the inputs?
if I use int2str() it loses some of the accuracy...
Hello, Bob Tivnan. I agree with Ahmos Sansom that there is a flaw in the current Test Suite. The problem is demonstrated in "Solution 1418487". Please address this, either by changing the input data type (not my preference) or by using numbers with precision compatible with double or uint64 data type (my preference). . . . If you are concerned about hardcoded 'solutions' (e.g. "Solution 1418491"), then you should add more tests to your Test Suite, and can also implement random permuting of the order of elements within each of your specified input vectors.
—DIV
By the way, this might be avoided if you develop your own reliable Reference Solution beforehand.
David Verrelli, my reference test suite passes without the problem you mention.
what happens if you change the last digit  does it still pass?
i.e:
x = [2 2 3 3 3 3 3 3 5 5 6 6 6 7 86168291600238450732788312165664788095941068326060883324529903470149056115823592713458328176574447204501];
to
x = [2 2 3 3 3 3 3 3 5 5 6 6 6 7 86168291600238450732788312165664788095941068326060883324529903470149056115823592713458328176574447204500];
Hi Bob, in the last testsuite problem the last element in your x vector cannot be properly represented using double precision (t will be indistinguishable from similar numbers differing by anything within the last 88 digits). Please either switch to using other suitable representations (e.g. java big integers) or modify the testsuite to keep the values there a bit smaller (double precision can reasonably work with integers below ~1e15 and int64 could go up to ~1e20)
I went ahead and changed the test suite so the last number can be safely represented as an integer in double precision floating point.
having issue with that large number e^15 even using format long g...precision is reduced.. method of perfect square identfying doesnt work here
Solution Comments

2 Comments
Any ideas why this keeps failing Test 3 while it works perfectly on my PC?
@RAHUL DEY: Are you sure that your solution gives you the correct answer for test 3 on your PC? When I run your solution on tests 3 and 4, they both give me the same result (correct for test 4, but not for test 3).
Note what Alfonso NietoCastanon mentioned in a comment on this problem: "double precision can reasonably work with integers below ~1e15." The method you are using squares the values in x, which makes that large number (about 8.9e15) equal to about 8e31, which is beyond the allowable precision in double values, causing what should be different values to appear the same. That is why you solution returns the same result for tests 3 and 4. You'll have to find a different method that does not rely upon squaring (or otherwise increasing) the values of x.

1 Comment
Your solution was the first one to work! However, I changed the test suite and your code did not pass the new test. I basically added a really really big fibonacci number. Can you modify your code to make it work?
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