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# D1 Problem Question watch

1. Hi guys. I just completed the D1 June 2014 IAL paper and have come across a problem question which I require some help on! I always get this type of question wrong and I'm not sure how to approach it. It is part D I cannot understand/get right. I don't understand the method of the markscheme so could you please give me a different explanation! Thank you, I won't be active for around another hour after this post as I will be completing an S1 paper but I'll be back after that! Many thanks

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2. Since there are only two odd nodes, and one of your nodes is where you start at and you don't need to finish there, there is no point repeating the vertex between the two odd nodes.

Therefore, you should finish at the OTHER odd node in order to minimise the route; such that you visit every arc at least once.

-

For questions with 4 odd nodes...
You only need to repeat one route of arcs, rather than the two shortest routes.

So now, you'd want to repeat only the shortest route of arcs - calculate these between every odd node and choose the smallest. You'll have two odd nodes left over; one will now be your starting point and the other your end point (it doesn't matter which so long as the question doesn't say otherwise; i.e starting at a specific node and ending anywhere).
3. (Original post by iMacJack)
Hi guys. I just completed the D1 June 2014 IAL paper and have come across a problem question which I require some help on! I always get this type of question wrong and I'm not sure how to approach it. It is part D I cannot understand/get right. I don't understand the method of the markscheme so could you please give me a different explanation! Thank you, I won't be active for around another hour after this post as I will be completing an S1 paper but I'll be back after that! Many thanks

Posted from TSR Mobile
Seems like trial and error to me. (You will most likely have to finish at a point with odd valency to minimise things, and then look at the repeated arcs and compare their lengths)
4. (Original post by JLegion)
Since there are only two odd nodes, and one of your nodes is where you start at and you don't need to finish there, there is no point repeating the vertex between the two odd nodes.

Therefore, you should finish at the OTHER odd node in order to minimise the route; such that you visit every arc at least once.

-

For questions with 4 odd nodes...
You only need to repeat one route of arcs, rather than the two shortest routes.

So now, you'd want to repeat only the shortest route of arcs - calculate these between every odd node and choose the smallest. You'll have two odd nodes left over; one will now be your starting point and the other your end point (it doesn't matter which so long as the question doesn't say otherwise; i.e starting at a specific node and ending anywhere).
Hi - I've highlighted the bits which I don't really follow!

So we start at G, there's still three other odd nodes, which of these odd nodes do I choose? You jumped straight into 'finishing at the other odd node' but I'm quite confused at this?

I'm still quite confused! Maybe a little bit more and I'll get it!
5. (Original post by iMacJack)
Hi - I've highlighted the bits which I don't really follow!

So we start at G, there's still three other odd nodes, which of these odd nodes do I choose? You jumped straight into 'finishing at the other odd node' but I'm quite confused at this?

I'm still quite confused! Maybe a little bit more and I'll get it!
Ah! I made a mistake - I thought your drawn on arc was actually involved in the system.
So you may as well ignore the first paragraph.
Wait a moment and I'll create a better response here.
6. Alright...
Since there are 4 odd nodes, and one node is where you start at, AND you can finish anywhere, you will only need to repeat one arc, rather than two.

You know where you start at, so between the 3 odd nodes left over (all but G) work out the shortest distance between any two nodes - that is the arc you want to repeat. The left over odd node will be your end point.

This minimises the distance such that you can finish anywhere and traverse each arc once.

Does that make sense?
7. (Original post by JLegion)
Alright...
Since there are 4 odd nodes, and one node is where you start at, AND you can finish anywhere, you will only need to repeat one arc, rather than two.

You know where you start at, so between the 3 odd nodes left over (all but G) work out the shortest distance between any two nodes - that is the arc you want to repeat. The left over odd node will be your end point.

This minimises the distance such that you can finish anywhere and traverse each arc once.

Does that make sense?
Is that because the left over node is then essentially the furthest away 'odd node' from our starting point (G) ?

Thank you
8. (Original post by iMacJack)
Is that because the left over node is then essentially the furthest away 'odd node' from our starting point (G) ?

Thank you
No. The left over node won't always be the farthest from G.

It's because then you only repeat the shortest arc, thus creating the shortest possible distance.
9. (Original post by JLegion)
No. The left over node won't always be the farthest from G.

It's because then you only repeat the shortest arc, thus creating the shortest possible distance.
Ah okay, gotcha! Thank you.

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Updated: June 7, 2016
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