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    (Original post by matthewdownloads)
    For question two, you just isolated the "red-blue" nodes.

    Then you switched the "red blue" around to "blue red" for each one.

    Then you replaced the other two colours (was it orange and yellow or something) with blue and red where they fitted (on the top it ended up with red BLUE red, where blue was a repalcment of orange).

    That's all they wanted! It threw me as well for 10 minutes, but once you worked it out they were easy marks!
    so what were we supposed to show on the empty graph to get marks?
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    Ok here's my take.

    1. Pretty piss. A,D critical activities.

    2. WTF. Raped.

    3. Hmmm. Graph question. Not sure about the last bit, was fairly Ok though. Added 6 arcs, did the definitions.

    4. LP pretty easy. Got 12,000, 9000 and 7500 in that order. Then got child seat to be around 66.67.

    5. Can't remember, but again pretty easy.

    6. Simulation, quite easy, got Day 6 and Day 3 although not 100% sure.


    Overal: TBH, with the exception of question 2, quite a straight forward paper. Although, because the grade boundaries are so feckin high, I have no clue what grade I will get.
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    Ah ha It confused me for ages
    I replaced the orange and green (or whatever they were) with a red and blue then just drew arrows for the ones i'd swap the colours. I got the final result just had no idea how to get there!
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    Question 2 was lame.

    The linear programming one threw me off - I got all the correct answers with two inequalities rather than three, and I was sat there trying to do it with three but then I thought "screw this, I got the answers anyway".

    Everything else was goooood. :3

    Here's what I got;
    1. AD critical activities
    2.i Noted blue/red nodes only on new graph, drew the arcs between the ones that were linked on the original graph.
    2.ii Wrote red/blue aternating on the graph so none of the same colour matched. (confused)
    2.iii Algorithm covered swapping the colours, not putting the colours in in the first place.
    3.i No loops, no alternate arcs
    3.ii Not all nodes were connected i.e. two overlapping trees
    3.iii Six arcs
    3.iv 9, adding one that links the original two trees. (confused)
    4.i Two constraints
    y = adult seat
    2x = kid+adult
    y+2x>=100
    y+2x<=120
    4.ii max = 12k // min full = 9k // min = 7.5k
    4.iii child seat = 66.6 recurring, rounded to £67
    5.i shortest = 26 // MST = 45
    5.ii shorten AD to 9 for shortest route// shorten AD to 10 for MST
    6.i 0-2 for falling, 3-8 for not falling, ignore 9
    6.ii 5 for first run, 3 for second.
    6.iii RUN MORE SIMULATIONS.
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    For question 5, it asked how much AD needed to be shorted by for it to be included in a shortest route and a minimum connector, so I think it only needed to be shortened by just enough for it to be able to be selected (I got 3 miles and 2 miles).
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    (Original post by Muffinz)
    Question 2 was lame.

    The linear programming one threw me off - I got all the correct answers with two inequalities rather than three, and I was sat there trying to do it with three but then I thought "screw this, I got the answers anyway".

    Everything else was goooood. :3

    Here's what I got;
    1. AD critical activities
    2.i Noted blue/red nodes only on new graph, drew the arcs between the ones that were linked on the original graph.
    2.ii Wrote red/blue aternating on the graph so none of the same colour matched. (confused)
    2.iii Algorithm covered swapping the colours, not putting the colours in in the first place.
    3.i No loops, no alternate arcs
    3.ii Not all nodes were connected i.e. two overlapping trees
    3.iii Six arcs
    3.iv 9, adding one that links the original two trees. (confused)
    4.i Two constraints
    y = adult seat
    2x = kid+adult
    y+2x>=100
    y+2x<=120
    4.ii max = 12k // min full = 9k // min = 7.5k
    4.iii child seat = 66.6 recurring, rounded to £67
    5.i shortest = 26 // MST = 45
    5.ii shorten AD to 9 for shortest route// shorten AD to 10 for MST
    6.i 0-2 for falling, 3-8 for not falling, ignore 9
    6.ii 5 for first run, 3 for second.
    6.iii RUN MORE SIMULATIONS.
    I did pretty much what you did bar Q3 and Q4.
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    (Original post by Phil_Waite)
    For question 5, it asked how much AD needed to be shorted by for it to be included in a shortest route and a minimum connector, so I think it only needed to be shortened by just enough for it to be able to be selected (I got 3 miles and 2 miles).
    Yup I got that, can't remember which order though.
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    (Original post by tomdebs)
    so what were we supposed to show on the empty graph to get marks?
    On the top graph (Q3) you drew in the isolated noes and edges only (the "subgraph") with the red/blue switched round.

    On the graph below that, you drew everything including the new switched colours AND the orange/yellow replaced with the appropriate red/blue colours.

    Maybe this will clear it (quick summary of it):

    Code:
    Basically if I said, get a string of 5 letters which can't have 2
    consecutive letters the same, using B and R:
    
    Original letters (original graph):
    RBYBR
    
    So isolate the relevant letters ("Draw the subgraph")
    RB  BR
    
    Switch them round ("Switch colours with 2 or more nodes")
    BR  RB
    
    Add the appropriate letter to replace Y ("Show the graph can be done in two colours")
    BRBRB
    Of course, the graph went vertically as well in the same way^ rather than just horizontally like I wrote above.
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    Could somebody post the wording and mark allocation of question 3 please? I think I might have managed to sneak a few marks in there despite making a mess of it, but I need to see the questions again .
    Thanks.
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    (Original post by DaGianni)
    Could somebody post the wording and mark allocation of question 3 please? I think I might have managed to sneak a few marks in there despite making a mess of it, but I need to see the questions again .
    Thanks.
    i) How do you know the graph is simple? [2]
    ii) How do you know the graph is not connected? [2]
    iii) Draw as many arcs as possible whilst keeping the graph simple and not connected. Give the number of arcs you drew. [3]
    iv) This one was too wordy for me to remember it all, but it was just [1] mark.
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    Okay, so f*cked this up.... !
    Quesions:
    1) - easy
    2) - i tried, like changed the colours and that, but no idea how I did it all, and just generally confused.
    3) Messed this up soo bad !
    Got simple graph definition wrong, added wrong number of arcs and did last bit bad.
    4, 5 & 6 - Pretty Standard, and think I did okay on them...

    I'm just going to try to forget about it.
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    (Original post by Phil_Waite)
    i) How do you know the graph is simple? [2]
    ii) How do you know the graph is not connected? [2]
    iii) Draw as many arcs as possible whilst keeping the graph simple and not connected. Give the number of arcs you drew. [3]
    iv) This one was too wordy for me to remember it all, but it was just [1] mark.
    Thanks for that, it seems as though my confusion of "connected" with "complete" may only have cost me two marks then [part ii)]. Do you happen to remember roughly what part iv) was about? My mind has gone completely blank .
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    Part iv) asked something like 'imagine a graph in which a vertex is connected to another vertex if there exists no connection, direct or indirect, on the original graph. How many arcs would it contain?'.
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    So all it was asking was for you to count how many connections there are between each vertex, and every other vertex on the other section of the graph. You can think of it as a bipartite graph with 4 vertices on one graph and 4 on the other, where each vertex is connected to every vertex on the other graph. This should give you 4*4=16 arcs.
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    (Original post by Phil_Waite)
    So all it was asking was for you to count how many connections there are between each vertex, and every other vertex on the other section of the graph. You can think of it as a bipartite graph with 4 vertices on one graph and 4 on the other, where each vertex is connected to every vertex on the other graph. This should give you 4*4=16 arcs.
    Thanks for that, that question went straight over my head but thankfully it's only worth one mark.
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    (Original post by Phil_Waite)
    So all it was asking was for you to count how many connections there are between each vertex, and every other vertex on the other section of the graph. You can think of it as a bipartite graph with 4 vertices on one graph and 4 on the other, where each vertex is connected to every vertex on the other graph. This should give you 4*4=16 arcs.
    Have I got it wrong then? On the back page (where it says "Do not write here" I drew the graph and counted 28 by connecting every node to each other and then counting the arcs drawn). You sure it wasn't 28?
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    I got 22, although I see many got 16.

    I drew a bipartite graph just to help me imagine the new arcs, which made 16. Then I had to original 6 arcs already there. 6 + 16 = 22

    Was the original question asking for the number of arcs or the number of new arcs?..
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    (Original post by matthewdownloads)
    Have I got it wrong then? On the back page (where it says "Do not write here" I drew the graph and counted 28 by connecting every node to each other and then counting the arcs drawn). You sure it wasn't 28?
    You'd get 28 if you made the graph into a complete graph, but all it was asking for was for you to connect vertices which have no connection between them, direct or indirect, in the original graph.
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    (Original post by EmperorMustard)
    I got 22, although I see many got 16.

    I drew a bipartite graph just to help me imagine the new arcs, which made 16. Then I had to original 6 arcs already there. 6 + 16 = 22

    Was the original question asking for the number of arcs or the number of new arcs?..
    It didn't matter about the fact that there were 6 new arcs, because even if they weren't there, there would still be an indirect connection between the vertices which make up the same part of the graph, so you still wouldn't be able to connect them.

    It wasn't asking for you to connect all the vertices which weren't connected by an arc in the original graph, just ones where the was no path at all connecting them.
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    (Original post by Phil_Waite)
    It didn't matter about the fact that there were 6 new arcs, because even if they weren't there, there would still be an indirect connection between the vertices which make up the same part of the graph, so you still wouldn't be able to connect them.

    It wasn't asking for you to connect all the vertices which weren't connected by an arc in the original graph, just ones where the was no path at all connecting them.
    I thought that just making one line between the two trees would fix the problem and give everything an indirect connection... so I put 9 o_O
 
 
 
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