# Big OOoooooo

Francis Siu

Ranch Hand

Posts: 867

posted 11 years ago

- 0

I have a question that I have used about a week to think but still can not get an answer , so hope ranchers could answer.

I require to write a program then estimate the time using Big Ooooo notation.

Thanks for your time

[ September 24, 2004: Message edited by: siu chung man ]

I require to write a program then estimate the time using Big Ooooo notation.

Thanks for your time

[ September 24, 2004: Message edited by: siu chung man ]

Francis Siu

SCJP, MCDBA

posted 11 years ago

- 0

Don't know anything about "Big OOOooooo" notation, but writing something like O(N) is often called "Big O" notation, so I'll assume that's what we're talking about.

When you do an estimate like that, you just take the worst case. The worst case is that you compute a random number every time through the inner loop, so you compute N^2/2 or O(N^2) random numbers. Am I missing something?

When you do an estimate like that, you just take the worst case. The worst case is that you compute a random number every time through the inner loop, so you compute N^2/2 or O(N^2) random numbers. Am I missing something?

Ilja Preuss

author

Sheriff

Sheriff

Posts: 14112

posted 11 years ago

Yes, the

which means that in the worst case scenario, it will run endlessly: O(infinity)

- 0

Originally posted by Ernest Friedman-Hill:

Am I missing something?

Yes, the

which means that in the worst case scenario, it will run endlessly: O(infinity)

The soul is dyed the color of its thoughts. Think only on those things that are in line with your principles and can bear the light of day. The content of your character is your choice. Day by day, what you do is who you become. Your integrity is your destiny - it is the light that guides your way. - Heraclitus

Francis Siu

Ranch Hand

Posts: 867

posted 11 years ago

- 0

Ernest Friedman-Hill and Ilja Preuss

thanks you very much

I agree that in the worst case scenario, it will run endlessly.

It is probability question that when n-->infinity, the probability for generating the last unique random number n that assigns to an array a[infinity] is 1/infinity. I do not know how to write it in Big O notation

In the first (for loop) it is used in N times

In the second (for loop) it is used about N times

Within the second for loop, (if condition) will be infinity

So I do not know what is the Big O notation which will be O( )

Ummm......Can I write it in O(infinity)?

thanks you very much

**it will run endlessly: O(infinity)**I agree that in the worst case scenario, it will run endlessly.

It is probability question that when n-->infinity, the probability for generating the last unique random number n that assigns to an array a[infinity] is 1/infinity. I do not know how to write it in Big O notation

In the first (for loop) it is used in N times

In the second (for loop) it is used about N times

Within the second for loop, (if condition) will be infinity

So I do not know what is the Big O notation which will be O( )

Ummm......Can I write it in O(infinity)?

Francis Siu

SCJP, MCDBA

posted 11 years ago

- 0

Without being an expert, I remember to read something (Hey, Mr. Sedgewick, nice book!), where the O(n) notation was used to estimate average or typical cases too, which is very reasonable.

If you're not good in theoritical analysis, you could try empirical analysis.

Make

(perhaps we can agree to avoid #define at least on a java-page?)

Then use a counter, to count the inner loop.

Increase MAX to 1.000, 10,000, 100,000, 1,000,000 and protocoll the values.

Put them into a diagram, to see whether it looks logarithmic, linear or exponential.

I'm not sure whether I understood the purpose of your codesnipplet,

You generate a rand-value, and if it's not in the generated values so far, you generate a new one?

So you have to ensure N is smaller than or equal to (MAX-MIN)!

But is N perhaps allways equal (MAX-MIN)?

For an arraysize of 100 and (MAX-MIN)=100, the first value in the array is found for sure in one guess (100%) and the last by 1%.

So your inner loop is run by an average of 100 times, to find a new value.

Element i is found by an average of i runs.

1 + 2 + 3 + ... 100 =

(1 + 100) + (2 + 99) + (3 + 98) + ... =

101 + 101 + 101 + ... = about 50 * 101 = 5050.

For thousend elements it would be

1 + ...... + 1000

1001 + 1001 + ... = 500,000 (not exactly).

It's increasing faster than linear.

It's about N*N/2.

But how about Arraysize being smaller than the random-range?

Assume 100,000 random-candidates and an Array of 100.

The first is gratis again.

now we have a hit expection of

1 100

2 100

3 100

...

100 99

It would only need an average of something between 100 or 101 runs to fill the array.

For 200 values perhaps 203 runs.

If the arraysize is small, compared to (MAX-MIN), the algorithm is quite good.

If the arraysize is nearly or exactly (MAX-MIN), it's poor.

If you're not good in theoritical analysis, you could try empirical analysis.

Make

(perhaps we can agree to avoid #define at least on a java-page?)

Then use a counter, to count the inner loop.

Increase MAX to 1.000, 10,000, 100,000, 1,000,000 and protocoll the values.

Put them into a diagram, to see whether it looks logarithmic, linear or exponential.

I'm not sure whether I understood the purpose of your codesnipplet,

You generate a rand-value, and if it's not in the generated values so far, you generate a new one?

So you have to ensure N is smaller than or equal to (MAX-MIN)!

But is N perhaps allways equal (MAX-MIN)?

For an arraysize of 100 and (MAX-MIN)=100, the first value in the array is found for sure in one guess (100%) and the last by 1%.

So your inner loop is run by an average of 100 times, to find a new value.

Element i is found by an average of i runs.

1 + 2 + 3 + ... 100 =

(1 + 100) + (2 + 99) + (3 + 98) + ... =

101 + 101 + 101 + ... = about 50 * 101 = 5050.

For thousend elements it would be

1 + ...... + 1000

1001 + 1001 + ... = 500,000 (not exactly).

It's increasing faster than linear.

It's about N*N/2.

But how about Arraysize being smaller than the random-range?

Assume 100,000 random-candidates and an Array of 100.

The first is gratis again.

now we have a hit expection of

1 100

2 100

3 100

...

100 99

It would only need an average of something between 100 or 101 runs to fill the array.

For 200 values perhaps 203 runs.

If the arraysize is small, compared to (MAX-MIN), the algorithm is quite good.

If the arraysize is nearly or exactly (MAX-MIN), it's poor.

Francis Siu

Ranch Hand

Posts: 867

posted 11 years ago

- 0

Many thanks

Stefan Wagner

Yes

yes

You have answered the question

I require to compute expected values from the program and then estimate the worse case.

[ September 25, 2004: Message edited by: siu chung man ]

Stefan Wagner

**You generate a rand-value, and if it's not in the generated values so far, you generate a new one?**Yes

**But is N perhaps allways equal (MAX-MIN)?**yes

You have answered the question

I require to compute expected values from the program and then estimate the worse case.

[ September 25, 2004: Message edited by: siu chung man ]

Francis Siu

SCJP, MCDBA

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