Pseudocode & Problem Solving

— Computer Science · A Level (9618) Paper 2 · Part C: Syntax · Selection · Loops · Arrays · Files · Modules · Bubble Sort —

★ Exam Tip: Paper 2 now tests 7–9 mark pseudocode questions far more than theory. This is where your marks are. Forget Python syntax completely — pseudocode has its own keywords. Use the keywords exactly, indent, and declare every variable.

Note on arrows: the board writes assignment as "=" or "<-----". Thepseudocode symbol is ← (type <- ). Use ← in the exam. All code in these notes uses ←.

1. Basic Syntax

Operator	Meaning
*	multiplication
/	division
+	addition
−	subtraction
&	string concatenation (joins strings) — not +

Variables and input

Declaration is necessary: DECLARE name : TYPE. Question: input 2 numbers and print their sum.

DECLARE N1, N2, Sum :
         INTEGER
         INPUT N1
         INPUT N2
         Sum ← N1 + N2
         OUTPUT Sum

Prompt rule: In Python the prompt went inside input(). In pseudocode you cannot — you must print a message first, then input. "Prompt the user" = display a message, then INPUT.

PRINT ("Type two
         numbers")
         INPUT N1, N2

Careful: brackets after PRINT are compulsory (sir). Quotation marks around the message — without them it is treated as a variable and the message won't show. PRINT Sum shows the value of Sum. "Suitable prompt" = any clear message.

String concatenation: FullName ← FirstName & LastName (FirstName "Taha", LastName "Ali"). Python used "+".

Examiner always tells you exactly what to input / whether to prompt — read every word of the question.

2. Selection Statements

Two selection statements: IF … THEN … ELSE … ENDIF and CASE … OF … OTHERWISE … ENDCASE.

IF syntaxIF (condition) THEN
         instruction
         ELSEIF (condition) THEN
         instruction
         ELSE
         instruction
         ENDIF

AND / OR: AND — both conditions must be true. OR — any one condition must be true.

practicelargest of three numbers

DECLARE A, B, C, Large
            : INTEGER
            INPUT A, B, C
            IF A > B AND A > C
            THEN Large ← A
            ELSEIF B > C
            THEN Large ← B
            ELSE
            Large ← C
            ENDIF
            OUTPUT Large

questionpositive / negative / zero

INPUT Number
            IF Number > 0 THEN OUTPUT "Positive"
            ELSEIF Number < 0 THEN OUTPUT "Negative"
            ELSE OUTPUT "Zero"
            ENDIF

CASE … OF … OTHERWISE … ENDCASE

Use when there are a lot of choices / conditions.

CASE Marks OF
         > 80 : Grade ← "A"
         > 70 : Grade ← "B"
         < 50 : Grade ← "F"
         OTHERWISE Grade ← "U"
         ENDCASE

3. Built-in Functions & String Comparison

Function	What it does	Example
LENGTH(s)	number of characters	LENGTH("Eric") → 4
LEFT(s, n)	first n characters	LEFT("Papers", 3) → "Pap"
RIGHT(s, n)	last n characters	RIGHT("Papers", 3) → "ers"
MID(s, start, n)	n characters from position start (first character = 1)	MID("Papers", 2, 3) → "ape"
INT(x)	integer part (drops decimals)	INT(24.8) → 24
ASC(c) / CHR(n)	character ↔ ASCII code	ASC('C') → 67
NUM_TO_STR / STR_TO_NUM	convert between number and string	STR_TO_NUM("27") → 27
ISNUM(s)	TRUE if the string value is all numbers	ISNUM("123") → TRUE

String comparison: pseudocode lets you compare strings — "Z" > "A". To test a capital letter: Character >= "A" AND Character <= "Z".

built-in functions Qevaluate · 5 marks

Title = "101 tricks with spaghetti" · Version = 'C' · Author = "Eric Peapod" · PackSize = 4 · WeightEach = 6.2 · Paperback = TRUE. If invalid write ERROR.

Expression	Evaluates to
MID(Title, 5, 3) & RIGHT(Author, 3)	"tri" & "pod" = "tripod"
INT(WeightEach * PackSize)	INT(24.8) = 24
PackSize >= 4 AND WeightEach < 6.2	TRUE AND FALSE = FALSE
LEFT(Author, ASC(Version) − 65)	ASC('C') = 67 → LEFT(Author, 2) = "Er"
RIGHT(Title, (LENGTH(Author) − 6))	LENGTH = 11 → RIGHT(Title, 5) = "hetti"

practicepassword check

Password is correct if: ① 1st character is a capital ② 2nd–4th characters are numbers ③ length is 9. Take the password and print "Approve" if all conditions are met.

DECLARE Pass : STRING
            PRINT ("Enter password")
            INPUT Pass
            IF LENGTH(Pass) = 9 AND LEFT(Pass, 1) >= "A" AND LEFT(Pass, 1) <= "Z"
            AND ISNUM(MID(Pass, 2, 3)) = TRUE
            THEN PRINT ("Approve")
            ELSE PRINT ("Reject")
            ENDIF

4. Loops (Repetition)

Repetition of instructions is called a loop. Three types:

FOR … TO … NEXT	WHILE … DO … ENDWHILE	REPEAT … UNTIL
count-controlled repetition known fixed iteration — always executes no condition · no Count ← Count + 1	pre-condition (conditional) repetition unknown runs while the condition is TRUE — may run 0 times use Count ← Count + 1	post-condition (conditional) repetition unknown runs until the condition becomes TRUE — always runs at least once use Count ← Count + 1

questioninput 500 names, print "Hello Name", with prompt — all three loops

FOR

DECLARE Count :
                  INTEGER
                  DECLARE Names, Final : STRING
                  FOR Count ← 1 TO 500
                  PRINT ("Enter name")
                  INPUT Names
                  Final ← "Hello " & Names
                  PRINT Final
                  ENDFOR

WHILE

Count ← 0
                  WHILE Count < 500 DO
                  Count ← Count + 1
                  PRINT ("Enter name")
                  INPUT Names
                  Final ← "Hello " & Names
                  PRINT Final
                  ENDWHILE

REPEAT

Count ← 0
            REPEAT
            Count ← Count + 1
            PRINT ("Enter name")
            INPUT Names
            Final ← "Hello " & Names
            PRINT Final
            UNTIL Count = 500

5. Programming Techniques

Technique	Outside the loop	Inside the loop
Sum	Sum ← 0	Sum ← Sum + Number
Counting	Count ← 0	Count ← Count + 1
Largest	Large ← 0	IF Number > Large THEN Large ← Number
Smallest	Small ← 9999	IF Number < Small THEN Small ← Number

Careful: initialise outside the loop — inside it would reset every time. Large starts at 0 (or lower than any possible value), Small starts very high. Print results outside the loop too.

questionsum of 10 numbers

DECLARE Sum, Count,
            Number : INTEGER
            Sum ← 0
            FOR Count ← 1 TO 10
            INPUT Number
            Sum ← Sum + Number
            ENDFOR
            OUTPUT Sum

questioninput 500 numbers — how many are positive?

Positive ← 0
            FOR Count ← 1 TO 500
            INPUT N1
            IF N1 > 0 THEN
            Positive ← Positive + 1
            ENDIF
            ENDFOR
            OUTPUT Positive

practice (board: 499 numbers)count positive, negative and zero

Positive ← 0 Negative
            ← 0 Zero ← 0
            FOR Count ← 1 TO 499
            INPUT Number
            IF Number > 0 THEN Positive ← Positive + 1
            ELSEIF Number < 0 THEN Negative ← Negative + 1
            ELSE Zero ← Zero + 1
            ENDIF
            ENDFOR
            PRINT ("Positive numbers count: ", Positive)
            PRINT ("Negative numbers count: ", Negative)
            PRINT ("Zero numbers count: ", Zero)

questioninput 5 numbers, print the largest

Large ← 0
            FOR Count ← 1 TO 5
            INPUT Num
            IF Num > Large THEN Large ← Num
            ENDIF
            ENDFOR
            OUTPUT Large

board stops after ENDFOR — remember to output Large outside the loop.

Flag looping

Whenever we need to stop the loop or make the code efficient, use a flag and a conditional loop.

Flag ← TRUE
         WHILE Flag = TRUE DO
         INPUT Response
         IF Response = "No" THEN Flag ← FALSE
         ENDIF
         ENDWHILE

questioninput until the password is correct (1st character a capital)

DECLARE Correct :
            BOOLEAN
            DECLARE Pass, FirstCharacter : STRING
            Correct ← FALSE
            WHILE Correct = FALSE DO
            PRINT ("Enter your password")
            INPUT Pass
            FirstCharacter ← LEFT(Pass, 1)
            IF FirstCharacter >= "A" AND FirstCharacter <= "Z"
            THEN PRINT ("Approve") Correct ← TRUE
            ELSE PRINT ("Re-write your password")
            ENDIF
            ENDWHILE

practiceinput numbers until 0 is typed — find the average

Total ← 0 Count ← 0
            INPUT Number
            WHILE Number <> 0 DO
            Total ← Total + Number
            Count ← Count + 1
            INPUT Number
            ENDWHILE
            IF Count > 0 THEN
            OUTPUT Total / Count
            ENDIF

sentinel 0 is not added; check Count > 0 to avoid dividing by zero (a run-time error).

6. Arrays

Array: stores multiple values of the same datatype; each value sits at a position called an index.

Names[3] = "Bano" · Names[0] = "Taha" · Names[5] = "Pappan" · Names[6] would be out of bounds

Declare a 1D arrayDECLARE NameOfArray : ARRAY[LowerBound : UpperBound] OF DataType

e.g. array "FoodItems", bounds 1–6, strings: DECLARE FoodItems : ARRAY[1:6] OF STRING

question565 names input by the user

DECLARE Names :
            ARRAY[1:565] OF STRING
            FOR Index ← 1 TO 565
            INPUT Names[Index]
            ENDFOR

practice100 elements all set to "No Data" (ResultArray)

DECLARE ResultArray :
            ARRAY[1:100] OF STRING
            FOR Index ← 1 TO 100
            ResultArray[Index] ← "No Data"
            ENDFOR

First access the individual element (ResultArray[5] ← "No Data"), then wrap it in a loop.

Linear search

Each element is compared with the value to find, in order from lower bound to upper bound, until the item is found or the upper bound is reached. Use a selection statement inside the loop.

questionSTNames (500 names): at which index is "Bano"?

DECLARE Index :
            INTEGER
            FOR Index ← 1 TO 500
            IF STNames[Index] = "Bano" THEN
            PRINT Index
            ENDIF
            ENDFOR

practiceSearchBox (500 strings): how many times is "Empty" repeated?

DECLARE Index,
            EmptyCount : INTEGER
            EmptyCount ← 0
            FOR Index ← 1 TO 500
            IF SearchBox[Index] = "Empty" THEN
            EmptyCount ← EmptyCount + 1
            ENDIF
            ENDFOR
            PRINT EmptyCount

same idea as the Structured-English example in Part B.

2-dimensional arrays

A grid (matrix) of rows and columns; each element identified by its row and column values. Commonly used for images, game boards, tables.

[row, column] — first index = row, second = column

Declare a 2D arrayDECLARE NameOfArray : ARRAY[LB : UB, LB : UB] OF DataType
rows columns
DECLARE Numbers : ARRAY[1:4, 1:3] OF INTEGER

Slide typo: the practice answer for "545 rows × 250 columns of STRING" must be DECLARE Names : ARRAY[1:545, 1:250] OF STRING (the slide wrote INTEGERS). The data type is the singular keyword INTEGER / STRING.

Nested loop = loop within a loop. FOR Count ← 1 TO 10 → FOR Index ← 1 TO 5 → PRINT "PAPERSDOCK" runs 10 × 5 = 50 times. For 2D arrays: outer loop = rows, inner loop = columns.

e.g. Number[2,1] = 21 · Number[4,1] = 78 · Number[3,3] = 54 (row first, then column).

question500 × 45 string array EmptyBox — initialise all elements

DECLARE EmptyBox :
            ARRAY[1:500, 1:45] OF STRING
            FOR RowIndex ← 1 TO 500
            FOR ColumnIndex ← 1 TO 45
            EmptyBox[RowIndex, ColumnIndex] ← ""
            ENDFOR
            ENDFOR

questionSearchBox 500 × 30: how many "Empty"?

Count ← 0
            FOR Row ← 1 TO 500
            FOR Column ← 1 TO 30
            IF SearchBox[Row, Column] = "Empty" THEN
            Count ← Count + 1
            ENDIF
            ENDFOR
            ENDFOR
            PRINT (Count)

practiceSearchBox 500 × 356: replace "Empty" with "Bano"

FOR Row ← 1 TO 500
            FOR Column ← 1 TO 356
            IF SearchBox[Row, Column] = "Empty" THEN
            SearchBox[Row, Column] ← "Bano"
            ENDIF
            ENDFOR
            ENDFOR

slide used 30 columns and printed Count — the question says 356 columns and no output.

7. Text Files

Purpose of a file: to store data permanently (not lost when the program ends).

READ mode	WRITE mode	APPEND mode
when you only want to read data from an existing file	to write data to a new file — creates the file, all previous data deleted, written from scratch	to add data to the end of an existing file

Write / AppendOPENFILE "File.txt" FOR WRITE
(or FOR APPEND)
WRITEFILE "File.txt", "String"/Variable
CLOSEFILE "File.txt"

ReadOPENFILE "File.txt" FOR READ
READFILE "File.txt", Variable
CLOSEFILE "File.txt"
READFILE reads one line into the variable

questionwrite 700 names from array School into a new file Names.txt

DECLARE Index :
            INTEGER
            DECLARE Name : STRING
            OPENFILE "Names.txt" FOR WRITE
            FOR Index ← 1 TO 700
            Name ← School[Index]
            WRITEFILE "Names.txt", Name
            ENDFOR
            CLOSEFILE "Names.txt"

practice54 employees input ID (4 chars) + name e.g. "2675Bano" — store only the ID in New.txt

DECLARE Value, ID :
            STRING
            OPENFILE "New.txt" FOR WRITE
            FOR Index ← 1 TO 54
            INPUT Value
            ID ← LEFT(Value, 4)
            WRITEFILE "New.txt", ID
            ENDFOR
            CLOSEFILE "New.txt"

questionadd one more ID "2356" to the existing New.txt

OPENFILE "New.txt" FOR
            APPEND
            WRITEFILE "New.txt", "2356"
            CLOSEFILE "New.txt"

questionprint all 500 lines of Sample.txt

OPENFILE "Sample.txt"
            FOR READ
            FOR Index ← 1 TO 500
            READFILE "Sample.txt", Filedata
            PRINT Filedata
            ENDFOR
            CLOSEFILE "Sample.txt"

EOF concept: to read from beginning to end when you don't know how many lines there are, use a conditional loop. A text file has a special End-Of-File marker; EOF("file") becomes TRUE when it is reached.

OPENFILE "Sample.txt" FOR
         READ
         WHILE NOT EOF("Sample.txt") DO
         READFILE "Sample.txt", Filedata
         PRINT Filedata
         ENDWHILE
         CLOSEFILE "Sample.txt"

Slide wrote "CLOSEWHILE" — the correct keyword is ENDWHILE, then CLOSEFILE.

practicemusic file: each line = Title (40 chars) · ArtistName (40 chars) · Location (8 chars)

User inputs a location; print the title and artist of every CD at that location and how many CDs there are.

Total ← 0
            OPENFILE "MyMusic" FOR READ
            PRINT ("Input location")
            INPUT CDSearch
            WHILE NOT EOF("MyMusic") DO
            READFILE "MyMusic", FileData
            IF RIGHT(FileData, 8) = CDSearch THEN
            CDTitle ← LEFT(FileData, 40)
            CDArtist ← MID(FileData, 41, 40)
            PRINT (CDTitle & " " & CDArtist)
            Total ← Total + 1
            ENDIF
            ENDWHILE
            PRINT (Total & " CDs found")
            CLOSEFILE "MyMusic"

fixed-width fields: LEFT 40 = title · MID start 41, length 40 = artist · RIGHT 8 = location.

8. Functions & Procedures

Function	Procedure
returns a value	does not return a value

Print vs Return: PRINT / OUTPUT only displays a value. RETURN sends the value back to the calling code so it can be stored or used in another calculation.

Function headerFUNCTION Name(Param : TYPE, Param : TYPE) RETURNS TYPE
…
RETURN value
ENDFUNCTION

Procedure headerPROCEDURE Name(Param : TYPE, Param : TYPE)
…
ENDPROCEDURE
called with CALL Name(...)

Note: the procedure-header slide shows "RETURN Datatype" — a procedure has no return type.

questionmodule Square — takes an integer and returns its square

FUNCTION Square(BYVAL
            x : INTEGER) RETURNS INTEGER
            Sq ← x * x
            RETURN Sq
            ENDFUNCTION

boat hire (past paper)ListAvailable() — 7–8 marks

Global arrays: HireTime (STRING, "10:15" or "Available") · Duration (INTEGER minutes) · Cost (REAL) — 10 boats. ListAvailable() is called with a STRING start time; it outputs each boat number that will be available at that time (currently not on hire, or due back before the start time), the number of boats available, or a suitable message if none. AddTime(time, minutes) returns the new time as a STRING.

PROCEDURE
            ListAvailable(StartTime : STRING)
            BoatNumAvailable ← 0
            FOR Index ← 1 TO 10
            Available ← FALSE
            IF HireTime[Index] = "Available"
            THEN Available ← TRUE
            ELSE
            TimeBack ← AddTime(HireTime[Index], Duration[Index])
            IF TimeBack < StartTime THEN Available ← TRUE
            ENDIF
            ENDIF
            IF Available = TRUE THEN
            OUTPUT "Boat ", Index, " is available"
            BoatNumAvailable ← BoatNumAvailable + 1
            ENDIF
            ENDFOR
            IF BoatNumAvailable > 0
            THEN OUTPUT "There are ", BoatNumAvailable, " boats available"
            ELSE OUTPUT "Sorry, no boats"
            ENDIF
            ENDPROCEDURE

board header wrote StartTime : INTEGER — the question says a STRING. Initialise Available ← FALSE inside the FOR loop (for each boat).

9. Bubble Sort

An algorithm to arrange an array in ascending or descending order.

Swapping — the classic mistake: you must use a temporary variable.

Incorrectarray ← [9, 0]array[0] ← array[1]array[1] ← array[0]both elements become 0 — the 9 is lost!

Correctarray ← [9, 0]temp ← array[0]array[0] ← array[1]array[1] ← temp

Two loops: the inner loop compares neighbours and swaps — after each pass one value is in its final position (the largest bubbles to the end). The outer loop repeats the passes (depends on number of elements).

Pass	Array after the pass	Note
start	10 5 6 3 7 8
1	5 6 3 7 8 10	10 bubbled to the end (5 swaps)
2	5 3 6 7 8 10	1 swap (6 ↔ 3)
3	3 5 6 7 8 10	1 swap (5 ↔ 3)
4	3 5 6 7 8 10	no swaps → NoSwaps = TRUE → stop

green = already in its final position · the efficient version stops after pass 4; the inefficient one would keep looping.

Inefficient

Two FOR loops — performs extra unwanted loops even if already sorted.

FOR X ← 1 TO 10
               FOR Y ← 1 TO 9
               IF arrayData[Y] > arrayData[Y+1] THEN
               temp ← arrayData[Y]
               arrayData[Y] ← arrayData[Y+1]
               arrayData[Y+1] ← temp
               ENDIF
               ENDFOR
               ENDFOR

Efficient

Outer loop = conditional loop with a flag; inner loop = FOR loop. Boundary shrinks each pass.

questionarray studentID of 10 elements — bubble sort, ascending

Boundary ← 9
            REPEAT
            NoSwaps ← TRUE
            FOR J ← 1 TO Boundary
            IF studentID[J] > studentID[J + 1] THEN
            Temp ← studentID[J]
            studentID[J] ← studentID[J + 1]
            studentID[J + 1] ← Temp
            NoSwaps ← FALSE
            ENDIF
            ENDFOR
            Boundary ← Boundary − 1
            UNTIL NoSwaps = TRUE

Boundary = number of elements − 1 (for 10 elements start at 9). The board example used Boundary = 4, i.e. a 5-element array — adjust to the question. For a descending sort change > to <.

✓ Quick Recap — Part C

Use ←, DECLARE every variable, indent, capitalise keywords · "&" joins strings · PRINT needs brackets + quotes
IF…THEN…ELSEIF…ELSE…ENDIF · CASE…OF…OTHERWISE…ENDCASE · AND = both, OR = either
LENGTH · LEFT · RIGHT · MID(s, start, n) · INT · ASC/CHR · ISNUM · STR_TO_NUM / NUM_TO_STR
FOR (known count) · WHILE (check first, may run 0 times) · REPEAT…UNTIL (runs at least once)
Sum, count, largest, smallest: initialise outside loop · flag looping for "until correct"
1D: ARRAY[LB:UB] OF TYPE · 2D: ARRAY[rows, cols] with nested loops (rows outer, columns inner)
Files: OPENFILE … FOR READ/WRITE/APPEND · READFILE / WRITEFILE · CLOSEFILE · WHILE NOT EOF()
FUNCTION … RETURNS … RETURN · PROCEDURE … CALL · BYVAL (copy) / BYREF (original changes)
Bubble sort: swap with temp; flag NoSwaps + shrinking Boundary = efficient