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The tail wags the dog. M1A1 A1 A1 We mark your best effort

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1 The tail wags the dog. M1A1 A1 A1 We mark your best effort
D1 stuff The tail wags the dog. M1A1 A1 A1 We mark your best effort

2 Minimum Spanning Trees
If you do Prims and it asks for Kruskals even though you get the same answer it is 0 marks. Remember Prims (P) for points, Kruskals has an A in it for arcs. You must write out the order you select the arcs (even in Prims from a matrix). The examiners report suggests you do this as you go through the algorithm, rather than at the end. In Prims, they always give you the point to start at. In Kruskals they need to see the word ‘reject’ at least once. Prims doesn’t have any rejections so if you accidently write the word ‘reject’ anywhere in a Prims you lose all marks.

3 Critical Path Analysis
If you have to draw an activity network from a precedence table all arcs must have arrows, in the right directions. Additional dummies lose the final accuracy mark. If it tells you the network has two dummies but you draw a network with no dummies you lose all say (five) marks. You need to draw at least one dummy. Qu is ‘explain why each of the dummies is necessary’ Answer is ‘ A and B must be able to be described uniquely in terms of the events at each end’ and ‘I depends on D only but J depends on D and G’ (use ‘event D’ here instead of ‘activity D’ is 0 marks) All boxes must be completed. Miss a number out and lose say 4 marks. We check your top numbers are generally increasing left to right (and vice versa for bottom)

4 more CPA Check carefully if they want a Gantt (cascade) chart or a scheduling diagram. If it is scheduling and there are four workers, if you draw more rows it’s 0 marks. If it is a Gantt chart then you can have as many rows as you like. For both graphs there are four marks and the first two marks come from drawing in the critical path. With both graphs we count to check you have the correct number of activities drawn. Missing out an activity loses two marks. If they ask you whether it is possible to complete the project with 3 workers on time (say on a Gantt chart) check carefully if the question says to explain or to calculate. If it wants an explanation then you need 3 things ‘at days activities D, C, F and G are needed so 4 workers needed, so it is not possible’ If it says calculate then do a lower bound eg 108/35 = 3.08 so 4 workers needed. If in doubt, do both. Be careful if you are using the ‘at days’ argument’ You must have some mention of a numerical time so ‘D,C,F and G are happening at the same time’ is no marks. Be super careful the time is a strict inequality so 12<time<13. A good idea is to always state a decimal. Accept Don’t accept during/on day 13 At/day/time 13 on the 13th day At day/time 12 At/during/on (time) 12.5, 12.1, 12.3 etc (any non integer between 12<time<13 At (anything that’s an integer) 12<time<13 During 12 On day 12 Between/during days 12≤time≤13

5 Route Inspection Problem
The first method mark is for three distinct pairings of the correct four odd nodes. Get this wrong and you lose all say (five) marks. State clearly which individual arcs you repeat. ‘Repeat DA, AE, FJ, FK, is accepted whereas ‘repeat DE, FK’ is not. We also don’t accept ‘DAE’. Don’t just point to the table with an arrow. If they ask for you to state the route count the number of edges (say 14), add on the repeated edges (say 3) then add 1 more to give 18. We count 18 letters in your route, then we check you have the repeats (underlining AB and BA helps), then we check all letters are included. Often at the end of a RIP (semi Eulerian qu). Qu is ‘the boy starts at K and finishes at a different vertex, traversing each path at least once. Determine where he finishes giving a reason for your answer’ (you’ll have checked the lengths of all arcs not including K and determined the shortest) so your answer is ‘Repeat DF (24) as it is the shortest that does not include K. Start at K, finish at E ‘ (you need 4 things)

6 Matchings If they ask you to show the matching on a diagram (say there are four matchings) don’t just highlight them with a thicker line on the bipartite graph. The examiner won’t search for them on your diagram. Redraw a sketch with only four matchings on it. Any extra lines is 0 marks. Qu is ‘explain why a complete matching is not possible’ Ans is ‘N and P can only be allocated to V’ (‘V can only be allocated to N and P’) or (‘only N and P can be allocated to V’) is 0 marks. The positioning of the word ‘only’ is important. You must have an indication you have changed the status. The safest way is to change the symbols. We accept ‘cs’ (but it is easy to miss it) or write ‘change of status’ . Best to rewrite symbols. P – V = F – T = A – C = J – D P = V – F = T – A = C – J = D You must write the final matching out at the end (or clearly draw it on a diagram at the end). We won’t search for lines shaded thicker on a working bipartite graph. It helps us if you start on the left hand side. A = C, F = T, J = D, P = V, R = G

7 Sorts and Bin Packing and Binary Search
If you do a bubble sort instead of a quick sort it’s 0 marks. Use lines. Common to miss a number out or misread a number. Check by counting at the end. A bubble sort needs a final statement saying you have stopped. Best to repeat the line or say ‘stop’ or ‘sorted’ A quick sort needs a final statement too. Once you have no sublists greater then 1 remaining it’s best to rewrite the line (or say ‘complete etc). We won’t look for your final line in your workings. Make it clear at the end. In Quick sort, the pivot is the middle right. If you go wrong you can still get full marks if you choose middle left. Choosing the first or last number to pivot on is 0 marks. The first M1 is for a pivot with smaller numbers to the left and larger to the right (if ascending). Have a rogue number in there and lose all marks. In Quick sort, if there are 2 sublists, do not do one pivot per line. You’ll lose all your marks except 1. If you sort in ascending order and the qu says descending, keep going BUT you must reverse the list at the very end (full marks)

8 more sorts If your bin pack follows your sort check you have the same amount of numbers. You need the first four items placed correctly and at least six items to score the first M1. Get this wrong and lose 4 easy marks Qu is ‘determine whether your answer shows the min amount of bins. Justify your answer’ They don’t want a ‘yes, I drew 4 bins’ they want a calculation. Eg Ans is ‘6.8/2 = 3.4 so yes, 4 bins is enough’ In binary search state what the pivot is and the reduced list each time. You always need a final statement, eg ‘Jenny is found’.

9 Dijkstras The start box doesn’t need zero in working values
The first method mark is for a smaller number replacing a larger number at least once in the main boxes. Miss that and lose all marks. All boxes must be labelled. Miss a number out loses the whole mark to that box. Safest to never write in larger values. If you do, they have to be in the right order. If any numbers are in the wrong order you lose the mark for that box. If you need more room list outside the box in a line. If they ask you to ‘justify your answer’ they expect a full backread (not an essay) so: T-H = 11 H-F = 35 F-E = 11 etc

10 Linear programming If they ask for the constraints, don’t forget the non negatitvity constraints (x,y≥0) If they ask you to indicate clearly the feasible region with R, that’s all we accept. Shading the region, outlining or pointing to it is not enough. Put R on. Be super careful plotting the lines for your region. You are allowed one square accuracy. If you don’t get all the lines correct you could lose up to four marks as technically R is incorrect and then you won’t get marks for the correct optimal solution later on (as your region is inaacurate). If the inequality is say less then they’ll want a dotted line. If they give you a choice about finding the optimal point it is safer to use the objective line method. Make sure your line is not too small. You don’t need to draw it hitting an optimal point. The correct line drawn is often worth two points. If you choose the point testing you need all four points correct and all four values calculated correctly to get the same 2 points. If in doubt, do both. If you draw your objective line wrong you will lose a mark only if you muddle up the plotting of the x and y axes. Any old line loses you all the marks. So plot carefully. You need to identify the optimal point by solving simultaneous equations. If you have chosen the points method you need to have calculated three points correctly (objective line is quicker and safer). If they want an integer solution you must show all four integer solutions around your coordinate properly tested into both inequalities and then into the OF. If you just guess the integer solution (because it’s obvious) - no marks. Check out the last two international (IAL) papers. You can find these on physicsandmathstutor.

11 Flow charts and definitions
If they ask you for a definition there are usually two marks. One for an attempt and the second mark is if it is perfect. In 2016 and 2017 they asked for ‘bipartite’ If you do a flow chart (June 2016) make sure you state the output. Be careful where your ‘no’ and ‘yes’ go. Writing more ‘no/yes’ will lose the last mark. The first M mark (out of 4 marks) is for a correct first row.


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