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    What did people draw for the 2 semi-eularian graph theory questions? I guessed
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    (Original post by MrFantastisch)
    Did anyone get F and H for part 5b?
    Pretty sure i got that because they had a distance of 12 so you could finish at either. Hence addint in the 2 vertices with distance 10?
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    I was one of those unfortunate people who got P=£500 for the last part of the last question. I got the right objective line I think (y=-3/4X+P), so I'm not sure what I've done wrong, unless I just plotted the inequalities wrong (I'm pretty sure my inequalities were right). I had a few mins so I wish I'd checked it again now. :/ quite possibly drew the wrong line in exam pressure.
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    (Original post by OturuDansay)
    Ahh yeah I got the exact same thing.
    Pretty sure it was 7 but guess not lol
    I think I've realised now, it's because I've worked out the number of comparisons after pass six, not during.
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    (Original post by sufiyan1999)
    Pretty sure i got that because they had a distance of 12 so you could finish at either. Hence addint in the 2 vertices with distance 10?
    For the first part I did 10+12=24 but noticed lucky That would've haunted me forever!!
    Wasn't sure what to do for the last two parts. It said start and finish at any point, so I added 10 to the total time they gave us, and for the last part I added 12 and stated the one that he didn't finish at from the '10'
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    (Original post by yelash)
    For the first part I did 10+12=24 but noticed lucky That would've haunted me forever!!
    Wasn't sure what to do for the last two parts. It said start and finish at any point, so I added 10 to the total time they gave us, and for the last part I added 12 and stated the one that he didn't finish at from the '10'
    That is correct. It was start from B.
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    (Original post by BanterBus)
    Did you sketch your objective line on the graph wrong due to writing x and y the wrong way around? If so then it'll be a 1/2 marks for that and then 2 marks for finding the wrong x and y values. You should still pick up the final mark for subbing your x and y values into the P equation though - otherwise they'd be pretty harsh. Overall maybe 3 or 4?
    I think I did everything right apart from the last part where I didn't even look at my equation just did 20*24 + 15*8 to get 600
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    (Original post by -jordan-)
    That is correct. It was start from B.
    How many marks would I lose for adding 10 instead of 12 to get 177 again like in part b but saying that you would start at B (which was right) for the last part?
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    (Original post by SM-)
    How many marks would I lose for adding 10 instead of 12 to get 177 again like in part b but saying that you would start at B (which was right) for the last part?
    Just 1
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    (Original post by SM-)
    How many marks would I lose for adding 10 instead of 12 to get 177 again like in part b but saying that you would start at B (which was right) for the last part?
    I did this too!!
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    The maximum profit for the last question was definitely £520, just checked on Wolfram Alpha. I somehow ended up with £515 in the exam so I'm not too sure where I went wrong there!
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    For djikstran did everyone get minimum distance of 15
    And value of x to be 3
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    (Original post by sufiyan1999)
    For djikstran did everyone get minimum distance of 15
    And value of x to be 3
    I got 13
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    (Original post by sufiyan1999)
    For djikstran did everyone get minimum distance of 15
    And value of x to be 3
    x was 3. Minimum distance for what?
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    closest thing to unofficial mark scheme
    not sure how many marks for each question so if anyone remembers just say

    1a. complete matching question

    2a. 9
    b. 6
    c. 45

    3a. 26
    b. 10<x<14

    4a. 189
    b. 177
    c. 179
    d. B

    5a. x<4 x>>3
    b. x=3

    6. 14 edges

    7a. 36
    b. 62
    c.62 and 67
    d. optimal is 62 as upper = lower

    8a. x+y<<32
    x>>2y
    3y+2x<<72 (*or other way round cant remember)
    b.x=24 y=8
    c. profit =520
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    (Original post by yelash)
    x was 3. Minimum distance for what?
    dijkstras algorithm from A to J
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    I think a lot of people might have failed to spot in the Dijkstras's question that they weren't looking for the minimum distance to the last node you visit which is the usual. They just asked you to perform the algorithm and the next question was the path from A to G, not A to J as expected
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    (Original post by -jordan-)
    I think a lot of people might have failed to spot in the Dijkstras's question that they weren't looking for the minimum distance to the last node you visit which is the usual. They just asked you to perform the algorithm and the next question was the path from A to G, not A to J as expected
    The answer to the first part was 13 right?
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    (Original post by Hjyu1)
    The answer to the first part was 13 right?
    There was no answer, you just had to do the algorithm. But I did get AJ as 13. It said find the shortest distances to all the fair ground rides or something like that. The next question was 1 mark and was the shortest route from A to G.
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    Isn't AJ 15 and AG 13
 
 
 
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