Codebusters C

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Re: Codebusters C

Postby hippo9 » October 18th, 2018, 4:18 pm

On the same topic, I'd assume that would be true also for any affine ciphers (although that might be harder to ensure) because they are also just monoalphabetic substitutions.
Last edited by hippo9 on October 18th, 2018, 6:04 pm, edited 1 time in total.
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Re: Codebusters C

Postby Name » October 18th, 2018, 5:06 pm

On the same topic, I'd assume that would be true also for any affine cipher's (although that might be harder to ensure) because they are also just monoalphabetic substitutions.
Rule g only states aristocrats, patristocrats, and xenocrypts
Affine is a math based cipher, and the point of it is to know the math, not to treat it as a monoaplhabetic. There's no reason why a letter shouldn't encrypt onto itself.
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Re: Codebusters C

Postby megrimlockawesom » October 19th, 2018, 3:48 am

can someone explain to me multiplicative inverse with euclidean algorithm? I have been able to get most of the affine cipher except for this bit.
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Re: Codebusters C

Postby UTF-8 U+6211 U+662F » October 19th, 2018, 4:34 pm

can someone explain to me multiplicative inverse with euclidean algorithm? I have been able to get most of the affine cipher except for this bit.
Sure!

Suppose you want to find the multiplicative inverse of 8 modulus 27.

(Little math side note: this is a solution for x in the Diophantine equation )

We proceed by using the regular Euclidean algorithm:
Divide 27 by 8.

Divide 8 by 3.

Divide 5 by 3.

Divide 3 by 2.

Now that we have a 1 at the end, we can proceed to the extended Euclidean algorithm. First, rearrange the equations.
Equation 1:
Equation 2:
Equation 3:
Equation 4:
Now.. we can substitute equation 3 into equation 4:

And then equation 2:

And then equation 1:

Rearrange a little...

Now, we have solved the Diophantine equation.
Here's the trick: we set the equation modulus 27.




So, the multiplicative inverse of 8 mod 27 is 17.
This won't take nearly as long once you get used to it.

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Re: Codebusters C

Postby Name » October 19th, 2018, 8:13 pm

can someone explain to me multiplicative inverse with euclidean algorithm? I have been able to get most of the affine cipher except for this bit.
Sure!

Suppose you want to find the multiplicative inverse of 8 modulus 27.

(Little math side note: this is a solution for x in the Diophantine equation )

We proceed by using the regular Euclidean algorithm:
Divide 27 by 8.

Divide 8 by 3.

Divide 5 by 3.

Divide 3 by 2.

Now that we have a 1 at the end, we can proceed to the extended Euclidean algorithm. First, rearrange the equations.
Equation 1:
Equation 2:
Equation 3:
Equation 4:
Now.. we can substitute equation 3 into equation 4:

And then equation 2:

And then equation 1:

Rearrange a little...

Now, we have solved the Diophantine equation.
Here's the trick: we set the equation modulus 27.




So, the multiplicative inverse of 8 mod 27 is 17.
This won't take nearly as long once you get used to it.

Well honestly I have no idea what you did so here's what I do
(Also generally it's mod 26 not 27)
Let's use 5 as a example
5x = 1mod26
Add 26 to one until it's divisable by 5
1, 27, 53, 79, 105
105 is divisable by 5
105/5 is 21
21 is the mod inverse of 5

Even using UTFs example- 8x=1mod27
1, 28, 55, 82, 109, 136
136/8 is 17
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Re: Codebusters C

Postby UTF-8 U+6211 U+662F » October 20th, 2018, 10:01 am

can someone explain to me multiplicative inverse with euclidean algorithm? I have been able to get most of the affine cipher except for this bit.
Sure!

Suppose you want to find the multiplicative inverse of 8 modulus 27.

(Little math side note: this is a solution for x in the Diophantine equation )

We proceed by using the regular Euclidean algorithm:
Divide 27 by 8.

Divide 8 by 3.

Divide 5 by 3.

Divide 3 by 2.

Now that we have a 1 at the end, we can proceed to the extended Euclidean algorithm. First, rearrange the equations.
Equation 1:
Equation 2:
Equation 3:
Equation 4:
Now.. we can substitute equation 3 into equation 4:

And then equation 2:

And then equation 1:

Rearrange a little...

Now, we have solved the Diophantine equation.
Here's the trick: we set the equation modulus 27.




So, the multiplicative inverse of 8 mod 27 is 17.
This won't take nearly as long once you get used to it.

Well honestly I have no idea what you did so here's what I do
(Also generally it's mod 26 not 27)
Let's use 5 as a example
5x = 1mod26
Add 26 to one until it's divisable by 5
1, 27, 53, 79, 105
105 is divisable by 5
105/5 is 21
21 is the mod inverse of 5

Even using UTFs example- 8x=1mod27
1, 28, 55, 82, 109, 136
136/8 is 17
That works well with relatively small numbers, but it's very tedious with larger numbers, especially with 4-function calculators. The extended Euclidean algorithm works for much larger numbers, and you can verify that you didn't make any mistakes along the way just by testing your latest equation.

For more information about the extended Euclidean algorithm, see http://www-math.ucdenver.edu/~wcherowi/ ... ucalg.html

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Re: Codebusters C

Postby l0lit » October 31st, 2018, 6:55 pm

Not a math question,

What do you guys think would be a good score on the timed question? Last year, I believe the nationals winner was sub-2 minutes, but it was a short cipher. Of course, thank goodness it is no longer weighted so heavily.
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Re: Codebusters C

Postby mjcox2000 » October 31st, 2018, 8:19 pm

Not a math question,

What do you guys think would be a good score on the timed question? Last year, I believe the nationals winner was sub-2 minutes, but it was a short cipher. Of course, thank goodness it is no longer weighted so heavily.
Nationals winner here. We were at 1 minute 51 seconds.

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Re: Codebusters C

Postby drsparc » November 9th, 2018, 7:15 pm

I'm having trouble finding a true 4-function calculator. Most of the simple ones also include percent and/or square root keys. Does anyone know if those are OK for this event?

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Re: Codebusters C

Postby Jacobi » November 10th, 2018, 4:19 pm

I'm having trouble finding a true 4-function calculator. Most of the simple ones also include percent and/or square root keys. Does anyone know if those are OK for this event?
I know that including a sqrt key harms nothing.
I have a feeling that a percent key would be OK.

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Re: Codebusters C

Postby Longivitis » November 11th, 2018, 7:06 am

I'm having trouble finding a true 4-function calculator. Most of the simple ones also include percent and/or square root keys. Does anyone know if those are OK for this event?
As long as it doesn't have trig functions you should be set.
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Re: Codebusters C

Postby nicholasmaurer » November 11th, 2018, 12:26 pm

I'm having trouble finding a true 4-function calculator. Most of the simple ones also include percent and/or square root keys. Does anyone know if those are OK for this event?
As long as it doesn't have trig functions you should be set.
Per the Calculator Policy, square root functionality is permitted. If the calculator includes trigonometric functions, etc. it would not be allowed. Any calculator with modulus capabilities would not be permitted.
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Re: Codebusters C

Postby UTF-8 U+6211 U+662F » November 11th, 2018, 1:41 pm

I'm having trouble finding a true 4-function calculator. Most of the simple ones also include percent and/or square root keys. Does anyone know if those are OK for this event?
As long as it doesn't have trig functions you should be set.
Per the Calculator Policy, square root functionality is permitted. If the calculator includes trigonometric functions, etc. it would not be allowed. Any calculator with modulus capabilities would not be permitted.
(Of course it's fairly easy to calculate the modulus anyway by just dividing, subtracting the integer part, and then multiplying again, e.g. 7/3 = 2.333333; 0.333333 * 3 = 1)

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Codebusters C

Postby someone1580 » November 12th, 2018, 2:13 pm

Do you guys have any tips or tricks on any of the codes in this event? For some reason in my area, they changed Game On B to Codebusters B.

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Re: Codebusters C

Postby Jacobi » November 12th, 2018, 4:20 pm

Do you guys have any tips or tricks on any of the codes in this event? For some reason in my area, they changed Game On B to Codebusters B.
Too many. See the Codebusters Wiki for information.


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