Write a GCD program for the HP-41C with chatGPT

[Thomas Klemm]

I thought to give it another try using copilot:

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Quote:Use the Euclidian algorithm to write a program for the old calculator HP-41C from Hewlett Packard that calculates the greatest common divisor of two numbers.

To write a program for the HP-41C calculator that calculates the greatest common divisor (GCD) of two numbers using the Euclidean algorithm, follow these steps in pseudocode:

1. Start: Begin the program.
2. Input A and B: Prompt the user to enter two numbers, A and B.
3. Check if B is 0: If B is 0, then the GCD is A. Go to step 7.
4. Compute Remainder: Calculate the remainder of A divided by B.
5. Set A to B and B to Remainder: Replace the value of A with B and the value of B with the remainder calculated in step 4.
6. Repeat: Go back to step 3.
7. Display GCD: The value of A is now the GCD. Display this value.
8. End: End the program.
   

Now, translating this pseudocode into a program for the HP-41C, which uses RPN (Reverse Polish Notation) and has a specific set of instructions:

Code:

01 LBL "GCD"       ; Label the program GCD
02 PROMPT          ; Prompt for first number, stores in X
03 STO 00          ; Store first number in register 00
04 PROMPT          ; Prompt for second number, stores in X
05 STO 01          ; Store second number in register 01
06 LBL 01          ; Loop label
07 RCL 01          ; Recall second number
08 X=0?            ; Check if second number is 0
09 GTO 02          ; If 0, go to label 02
10 RCL 00          ; Recall first number
11 RCL 01          ; Recall second number
12 MOD             ; Calculate remainder of first number divided by second
13 STO 00          ; Store result in register 00 (new A)
14 RCL 01          ; Recall second number
15 STO 01          ; Store previous A in register 01 (new B)
16 RCL 00          ; Recall new A to check in next iteration
17 GTO 01          ; Go back to start of loop
18 LBL 02          ; Label for GCD result
19 RCL 00          ; Recall the GCD result
20 PSE             ; Pause to display the GCD
21 RTN             ; Return from the program

This program uses the HP-41C's capabilities to calculate the GCD of two numbers entered by the user. It utilizes registers for storing intermediate values and loops until the second number becomes 0, at which point the first number is the GCD.

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Quote:Do not prompt for the numbers but assume that the two numbers are already on the stack.

To modify the program so it assumes the two numbers are already on the stack instead of prompting the user to enter them, we'll remove the PROMPT instructions and adjust the storage instructions accordingly. Here's the updated program:

Code:

01 LBL "GCD"       ; Label the program GCD
02 STO 01          ; Store second number (already on stack) in register 01
03 STO 00          ; Store first number (now on stack) in register 00
04 LBL 01          ; Loop label
05 RCL 01          ; Recall second number
06 X=0?            ; Check if second number is 0
07 GTO 02          ; If 0, go to label 02
08 RCL 00          ; Recall first number
09 RCL 01          ; Recall second number
10 MOD             ; Calculate remainder of first number divided by second
11 STO 00          ; Store result in register 00 (new A)
12 RCL 01          ; Recall second number
13 STO 01          ; Store previous A in register 01 (new B)
14 RCL 00          ; Recall new A to check in next iteration
15 GTO 01          ; Go back to start of loop
16 LBL 02          ; Label for GCD result
17 RCL 00          ; Recall the GCD result
18 PSE             ; Pause to display the GCD
19 RTN             ; Return from the program

This version assumes the user has already placed the two numbers on the stack with the first number entered being on the bottom (first to be stored in register 00) and the second number on top (to be stored in register 01).

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Quote:There's a problem with step 5: Set A to B and B to Remainder. The remainder has to be stored in register 01 (new B). The second number has to be stored in register 00 (new A). To do so, use the RDN (roll down) command.

To correct the issue with step 5, where the remainder needs to be stored in register 01 (new B) and the second number in register 00 (new A), we can use the RDN (roll down) command to adjust the values accordingly. Here's the revised section of the program:

Code:

01 LBL "GCD"       ; Label the program GCD
02 STO 01          ; Store second number (already on stack) in register 01
03 STO 00          ; Store first number (now on stack) in register 00
04 LBL 01          ; Loop label
05 RCL 01          ; Recall second number
06 X=0?            ; Check if second number is 0
07 GTO 02          ; If 0, go to label 02
08 RCL 00          ; Recall first number
09 RCL 01          ; Recall second number
10 MOD             ; Calculate remainder of first number divided by second
11 X<>Y            ; Swap X and Y registers
12 STO 01          ; Store the remainder (now in Y after swap) in register 01 (new B)
13 RDN             ; Roll down the stack
14 STO 00          ; Store the original second number (now on X after RDN) in register 00 (new A)
15 GTO 01          ; Go back to start of loop
16 LBL 02          ; Label for GCD result
17 RCL 00          ; Recall the GCD result
18 PSE             ; Pause to display the GCD
19 RTN             ; Return from the program

This adjustment ensures that after calculating the remainder (step 10), we swap the X and Y registers so that the remainder can be stored in register 01 as the new B (step 12). Then, we use the RDN command to roll down the stack, bringing the original second number into the X register, which is then stored as the new A in register 00 (step 14). This corrects the logic to match the intended algorithm steps.

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Quote:We're close. The swap command (X<>Y) in line 11 has the be removed. And thus the comment in line 12 has to be adjusted.

Based on the feedback, here's the corrected excerpt without the swap command (X<>Y) in line 11, and with the adjusted comment in line 12:

Code:

01 LBL "GCD"       ; Label the program GCD
02 STO 01          ; Store second number (already on stack) in register 01
03 STO 00          ; Store first number (now on stack) in register 00
04 LBL 01          ; Loop label
05 RCL 01          ; Recall second number
06 X=0?            ; Check if second number is 0
07 GTO 02          ; If 0, go to label 02
08 RCL 00          ; Recall first number
09 RCL 01          ; Recall second number
10 MOD             ; Calculate remainder of first number divided by second
11 STO 01          ; Store the remainder in register 01 (new B)
12 RDN             ; Roll down the stack
13 STO 00          ; Store the original second number (now on X after RDN) in register 00 (new A)
14 GTO 01          ; Go back to start of loop
15 LBL 02          ; Label for GCD result
16 RCL 00          ; Recall the GCD result
17 PSE             ; Pause to display the GCD
18 RTN             ; Return from the program

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Quote:Almost there: we need another RDN command after line 02. Otherwise the second number is also stored in register 00 in line 03.

To ensure the correct order of storing the numbers in the registers and to avoid the second number being stored in both registers, an additional RDN command is needed after line 02. Here's the corrected code:

Code:

01 LBL "GCD"       ; Label the program GCD
02 STO 01          ; Store second number (already on stack) in register 01
03 RDN             ; Roll down the stack
04 STO 00          ; Store first number (now on stack) in register 00
05 LBL 01          ; Loop label
06 RCL 01          ; Recall second number
07 X=0?            ; Check if second number is 0
08 GTO 02          ; If 0, go to label 02
09 RCL 00          ; Recall first number
10 RCL 01          ; Recall second number
11 MOD             ; Calculate remainder of first number divided by second
12 STO 01          ; Store the remainder in register 01 (new B)
13 RDN             ; Roll down the stack
14 STO 00          ; Store the original second number (now on X after RDN) in register 00 (new A)
15 GTO 01          ; Go back to start of loop
16 LBL 02          ; Label for GCD result
17 RCL 00          ; Recall the GCD result
18 PSE             ; Pause to display the GCD
19 RTN             ; Return from the program

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Quote:Drop the PSE command in line 18. Replace the RTN command in line 19 by the END command.

Code:

01 LBL "GCD"       ; Label the program GCD
02 STO 01          ; Store second number (already on stack) in register 01
03 RDN             ; Roll down the stack
04 STO 00          ; Store first number (now on stack) in register 00
05 LBL 01          ; Loop label
06 RCL 01          ; Recall second number
07 X=0?            ; Check if second number is 0
08 GTO 02          ; If 0, go to label 02
09 RCL 00          ; Recall first number
10 RCL 01          ; Recall second number
11 MOD             ; Calculate remainder of first number divided by second
12 STO 01          ; Store the remainder in register 01 (new B)
13 RDN             ; Roll down the stack
14 STO 00          ; Store the original second number (now on X after RDN) in register 00 (new A)
15 GTO 01          ; Go back to start of loop
16 LBL 02          ; Label for GCD result
17 RCL 00          ; Recall the GCD result
18 END             ; End of the program

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Quote:Perfect. Now I want to test the program. Can you give me two numbers that I could use? And what is the expected result?

You can use the numbers 48 and 18.
The expected result for the Greatest Common Divisor (GCD) of 48 and 18 is 6.