AQA A-level Computer Science Paper 1 - 11th June 2025 [Exam Chat]

Reply 60

5

When I type in 4*5 this is how my score changes using my answer to Part D

score: 0
score: -1
score: -2
score: -3
score: -9
score: -48
score: -73

Reply 61

acting-earthquak

9

Original post

by cdinsley7

It was the training game which you end up with getting 23 in the first spot so then 23 + 10 = 33
33 + 5 = 38
38+1 = 39
and you already had -10
so its -49

huh, that’s true, idk how i got -22 then

Reply 62

kofiW

How did people approach the last question on finding solutions with exactly three numbers?
I personally just used a brute force approach iterating through all numbers allowed and creating different expressions
But i thought about just randomly picking 3 numbers adding two operators and doing it that way as there was nothing saying it had to be efficient.

Reply 63

rgrant7

Anybody that got the last question. For the test did it output that there is a possible target using 3 numbers, then input the expression, then output that there are no possible targets using 3 numbers?

Reply 64

Mcfc23

4

Original post

by rgrant7

Anybody that got the last question. For the test did it output that there is a possible target using 3 numbers, then input the expression, then output that there are no possible targets using 3 numbers?

I got that

Reply 65

thevingetsthewin

5

Original post

by kofiW

How did people approach the last question on finding solutions with exactly three numbers?
I personally just used a brute force approach iterating through all numbers allowed and creating different expressions
But i thought about just randomly picking 3 numbers adding two operators and doing it that way as there was nothing saying it had to be efficient.

I did something like

for i in range(5):
for j in range(5):
for k in range(5):
for O1 in "+-*/":
for O2 in "+-*/":
result = EvaluateRPN(ConvertToRPN(#Concatenating the numbers together. too lazy to type it out #))
if i != j and i != k and j != k and result != -1 and result in Targets:
return True
return False

Reply 66

acting-earthquak

9

Original post

by rgrant7

Anybody that got the last question. For the test did it output that there is a possible target using 3 numbers, then input the expression, then output that there are no possible targets using 3 numbers?

did we have to enter an expression in the test? i only showed true and the number of expressions available

Reply 67

rgrant7

Original post

by acting-earthquak

did we have to enter an expression in the test? i only showed true and the number of expressions available

Yeah it said enter 8+3-2

Reply 68

acting-earthquak

9

Original post

by rgrant7

Yeah it said enter 8+3-2

lmao didn’t see that, oh well

Reply 69

username7152832

1

Did anyone know how to answer the vector question in section C. I put Targets but it was probably wrong. I had no idea what they wanted.

Reply 70

Conor Appleyard

1

i put numbers allowed could be represented as a 5 dimentional vector
but i think targets also works but it just has a higher dimention like 15 or something

Reply 71

Scoobby

Original post

by Isabella Great

Did anyone know how to answer the vector question in section C. I put Targets but it was probably wrong. I had no idea what they wanted.

Yh that’s what I said

Reply 72

username7152832

1

Original post

by Conor Appleyard

i put numbers allowed could be represented as a 5 dimentional vector
but i think targets also works but it just has a higher dimention like 15 or something

I don't think numbers allowed is right because after numbers are removed it has fewer than 5 for a moment.

Reply 73

sfhsrthjsrjh

4

Original post

by cdinsley7

It was the training game which you end up with getting 23 in the first spot so then 23 + 10 = 33
33 + 5 = 38
38+1 = 39
and you already had -10
so its -49

thats what i got

Reply 74

Razzywazzy

2

Original post

by dwdwadawdwadaw

you will still get 5/6 / 7, the only change was to add a and precedence[currentoperator] != 5 (or whatever u gave to ^ operator), in the precedence[operators[^1]] == precedence[currentOperator] line. Found this solution by accident when i deleted my old one .

The precedent one i put “^”:8 at the end of the dictionary is that okay?

Reply 75

PhoebeCradd

1

Original post

by thevingetsthewin

I said 99 cause if the list was fully reversed, once you have the second smallest sorted, the smallest item is sorted automatically cause it is swapped to the start

I said 100 cause i think it would still pass through them all again after the 99th pass to make sure everything’s in the right order. i might be wrong tho

Reply 76

PhoebeCradd

1

Original post

by thevingetsthewin

I did something like
for i in range(5):
for j in range(5):
for k in range(5):
for O1 in "+-*/":
for O2 in "+-*/":
result = EvaluateRPN(ConvertToRPN(#Concatenating the numbers together. too lazy to type it out #))
if i != j and i != k and j != k and result != -1 and result in Targets:
return True
return False

would that not use multiple of the same number twice? i added ifs after my for loop

Reply 77

PhoebeCradd

1

Original post

by Scoobby

100 x 100 =10,000

it was passes not comparisons

Reply 78

acting-earthquak

9

Original post

by Razzywazzy

The precedent one i put “^”:8 at the end of the dictionary is that okay?

yeah as long as it was bigger than all the others you’re good, i put 6 just cause the others went up in 2s

Reply 79

graphic-beer

1

Original post

by brokenhooper

Did the program finish on -35 for question 9.

It’s 48 from taking away from -9 the first -1 then -5 and - (10+23 )

AQA A-level Computer Science Paper 1 - 11th June 2025 [Exam Chat]

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