MATH1081 Discrete Maths (2 Viewers)

Flop21

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Re: UNSW MATH1081 Discrete Maths

When creating the table to show a graph is isomorphic... how do I work out what vertex pairs with the second graph's vertex? Does it even matter, as long as their degrees are the same?
 

Flop21

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Re: UNSW MATH1081 Discrete Maths

When creating the table to show a graph is isomorphic... how do I work out what vertex pairs with the second graph's vertex? Does it even matter, as long as their degrees are the same?
Nevermind, I realise it does matter and you can find it by looking at the degree of what that vertex is connected to and then comparing.
 

KingOfActing

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Re: UNSW MATH1081 Discrete Maths

When creating the table to show a graph is isomorphic... how do I work out what vertex pairs with the second graph's vertex? Does it even matter, as long as their degrees are the same?
It does matter - consider these two non-isomorphic graphs with the same degree sequences.



In general I don't think there's a nice way to prove two graphs are isomorphic - the degrees sequences being equal is necessary but not sufficient, and to prove they are isomorphic you usually have to explicitly provide the bijection
 

Flop21

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Re: UNSW MATH1081 Discrete Maths

Show if G is self-complementary then G has 4k or 4k + 1 vertices for some integer k.


So I know for G, edges = [n(n-1)]/4. With n = vertices. Do I use this to show?
 

InteGrand

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Re: UNSW MATH1081 Discrete Maths

Show if G is self-complementary then G has 4k or 4k + 1 vertices for some integer k.


So I know for G, edges = [n(n-1)]/4. With n = vertices. Do I use this to show?
If we know the no. of edges is n(n-1)/4, since the no. of edges is an integer, what does this tell us about n?
 

Flop21

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edit nvm lol
 
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Flop21

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Any tips for easily redrawing graphs as planar maps?
 

KingOfActing

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I'm not sure if I've done this question correctly. "Find the number of 20 letter words using the capital English alphabet containing the subword 'MATHS'"

I did:


Since you have to place "MATHS" somewhere, and then choose the remaining 15 letters. But this overcounts when "MATHS" appears twice, so you have to subtract that, which undercounts, etc.
 

Flop21

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Help plssss

How do I simplify this:

SUM[k=2..N+1] k^3 - SUM[k=0..N-1] k^3

I need to show it = (N+1)^3 + N^3 - 1.


I am very confused.
 

InteGrand

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Help plssss

How do I simplify this:

SUM[k=2..N+1] k^3 - SUM[k=0..N-1] k^3

I need to show it = (N+1)^3 + N^3 - 1.


I am very confused.
Almost all the terms cancel. If you can't see it immediately, it might help you to write out the terms of each sum in long notation.
 

KingOfActing

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Help plssss

How do I simplify this:

SUM[k=2..N+1] k^3 - SUM[k=0..N-1] k^3

I need to show it = (N+1)^3 + N^3 - 1.


I am very confused.
write them with common upper lower bounds, and take everything else out of the sum

(sum(k=2..n-1)k^3 + n^3 + (n+1)^3) - (0^3 + 1^3 + sum(k=2..n-1)k^3 + 0^3 +1^3) = (n+1)^3 + n^3 - 1
 

leehuan

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I'm not sure if I've done this question correctly. "Find the number of 20 letter words using the capital English alphabet containing the subword 'MATHS'"

I did:


Since you have to place "MATHS" somewhere, and then choose the remaining 15 letters. But this overcounts when "MATHS" appears twice, so you have to subtract that, which undercounts, etc.
Yes, you need to use the inclusion exclusion principle.
 

Flop21

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write them with common upper lower bounds, and take everything else out of the sum

(sum(k=2..n-1)k^3 + n^3 + (n+1)^3) - (0^3 + 1^3 + sum(k=2..n-1)k^3 + 0^3 +1^3) = (n+1)^3 + n^3 - 1
Oh thanks lol. For some reason I just forgot how to do all sum questions, but after 2 hours I know get it (fml).
 

Flop21

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wtf am I doing wrong??

111x ≡ 75 (mod 321)


I find gcd(111,321) = 3. So divide original equation by 3 -> 37x ≡ 28 (mod 107).

Now applying Euclid's algo, I get, 1 = 9(107) - 26(37). So -26 * 28 = -728.

This does not give me the answer of 99,206,313 mod 321. Why???
 

InteGrand

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wtf am I doing wrong??

111x ≡ 75 (mod 321)


I find gcd(111,321) = 3. So divide original equation by 3 -> 37x ≡ 28 (mod 107).

Now applying Euclid's algo, I get, 1 = 9(107) - 26(37). So -26 * 28 = -728.

This does not give me the answer of 99,206,313 mod 321. Why???
75 divided by 3 is not 28.
 

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