## Codebusters C

hippo9
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### Re: Codebusters C

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

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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megrimlockawesom
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### Re: Codebusters C

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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UTF-8 U+6211 U+662F
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### Re: Codebusters C

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 $8x+27y=1$)

We proceed by using the regular Euclidean algorithm:
Divide 27 by 8.
$27 = 3(8) + 3$
Divide 8 by 3.
$8 = 1(3) + 5$
Divide 5 by 3.
$5 = 1(3) + 2$
Divide 3 by 2.
$3 = 1(2) + 1$
Now that we have a 1 at the end, we can proceed to the extended Euclidean algorithm. First, rearrange the equations.
Equation 1: $3 = 27 - 3(8)$
Equation 2: $5 = 8 - 1(3)$
Equation 3: $2 = 5 - 1(3)$
Equation 4: $1 = 3 - 1(2)$
Now.. we can substitute equation 3 into equation 4:
$1 = 3 - 1(5 - 1(3)) = 2(3) - 5$
And then equation 2:
$1 = 2(3) - (8 - 1(3)) = 3(3) - 8$
And then equation 1:
$1 = 3(27-3(8)) - 8 = 3(27) - 10(8)$
Rearrange a little...
$8(-10)+27(3)=1$
Now, we have solved the Diophantine equation.
Here's the trick: we set the equation modulus 27.
$8(-10)+27(3)\equiv1\ (\textrm{mod}\ 27)$
$8(-10)\equiv1\ (\textrm{mod}\ 27)$
$8^{-1}\equiv-10\ (\textrm{mod}\ 27)$
$8^{-1}\equiv17\ (\textrm{mod}\ 27)$
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

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 $8x+27y=1$)

We proceed by using the regular Euclidean algorithm:
Divide 27 by 8.
$27 = 3(8) + 3$
Divide 8 by 3.
$8 = 1(3) + 5$
Divide 5 by 3.
$5 = 1(3) + 2$
Divide 3 by 2.
$3 = 1(2) + 1$
Now that we have a 1 at the end, we can proceed to the extended Euclidean algorithm. First, rearrange the equations.
Equation 1: $3 = 27 - 3(8)$
Equation 2: $5 = 8 - 1(3)$
Equation 3: $2 = 5 - 1(3)$
Equation 4: $1 = 3 - 1(2)$
Now.. we can substitute equation 3 into equation 4:
$1 = 3 - 1(5 - 1(3)) = 2(3) - 5$
And then equation 2:
$1 = 2(3) - (8 - 1(3)) = 3(3) - 8$
And then equation 1:
$1 = 3(27-3(8)) - 8 = 3(27) - 10(8)$
Rearrange a little...
$8(-10)+27(3)=1$
Now, we have solved the Diophantine equation.
Here's the trick: we set the equation modulus 27.
$8(-10)+27(3)\equiv1\ (\textrm{mod}\ 27)$
$8(-10)\equiv1\ (\textrm{mod}\ 27)$
$8^{-1}\equiv-10\ (\textrm{mod}\ 27)$
$8^{-1}\equiv17\ (\textrm{mod}\ 27)$
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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UTF-8 U+6211 U+662F
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### Re: Codebusters C

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 $8x+27y=1$)

We proceed by using the regular Euclidean algorithm:
Divide 27 by 8.
$27 = 3(8) + 3$
Divide 8 by 3.
$8 = 1(3) + 5$
Divide 5 by 3.
$5 = 1(3) + 2$
Divide 3 by 2.
$3 = 1(2) + 1$
Now that we have a 1 at the end, we can proceed to the extended Euclidean algorithm. First, rearrange the equations.
Equation 1: $3 = 27 - 3(8)$
Equation 2: $5 = 8 - 1(3)$
Equation 3: $2 = 5 - 1(3)$
Equation 4: $1 = 3 - 1(2)$
Now.. we can substitute equation 3 into equation 4:
$1 = 3 - 1(5 - 1(3)) = 2(3) - 5$
And then equation 2:
$1 = 2(3) - (8 - 1(3)) = 3(3) - 8$
And then equation 1:
$1 = 3(27-3(8)) - 8 = 3(27) - 10(8)$
Rearrange a little...
$8(-10)+27(3)=1$
Now, we have solved the Diophantine equation.
Here's the trick: we set the equation modulus 27.
$8(-10)+27(3)\equiv1\ (\textrm{mod}\ 27)$
$8(-10)\equiv1\ (\textrm{mod}\ 27)$
$8^{-1}\equiv-10\ (\textrm{mod}\ 27)$
$8^{-1}\equiv17\ (\textrm{mod}\ 27)$
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.

l0lit
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### Re: Codebusters C

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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* Others coming soon...

mjcox2000
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### Re: Codebusters C

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.

There are pros and cons to heavier weighting for the timed question. On one hand, emphasizing solving the problem quickly means you have to have a firm grasp on what you’re doing in order to do well, so it selects strongly for the best-prepared teams. On the other hand, small mistakes that take time to fix incur huge penalties, and scores on the various test portions can be wildly out of proportion — if I’m not mistaken, my team’s time bonus was greater than the number of available points on the test by a factor of 2 to 3.
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drsparc
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### Re: Codebusters C

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?

Jacobi
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### Re: Codebusters C

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.