(12C) Modulus - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Software Libraries (/forum-10.html) +--- Forum: General Software Library (/forum-13.html) +--- Thread: (12C) Modulus (/thread-6558.html) (12C) Modulus - Eddie W. Shore - 07-15-2016 02:37 AM HP 12C Modulus This program takes the modulus of two numbers: Y MOD X = X * FRAC(Y/X) In this program, X > 0 and Y > 0. Code: ``` STEP    CODE            KEY 01    10             ÷ 02    43, 36    LST x 03    34             X<>Y 04    43, 24    FRAC 05    20             * 06    43, 33, 00    GTO 00``` Input: Y [ENTER] X [R/S], Result: Y MOD X Test 1: 124 MOD 77 = 47 Test 2: 3862 MOD 108 = 82 RE: (12C) Modulus - Dieter - 07-15-2016 06:32 PM (07-15-2016 02:37 AM)Eddie W. Shore Wrote:  This program takes the modulus of two numbers: Y MOD X = X * FRAC(Y/X) That's not a good idea: roundoff errors will spoil the results. Consider your two examples: 124 mod 77 does not return 47 but 46,99999997 3862 mod 108 does not return 82 but 82,00000008 That's why you better use the relation y mod x = y – x * INT(y/x) instead. Code: ```01-       34  x<>y 02-     44 0  STO 0 03-       34  x<>y 04-       10  ÷ 05-    43 36  LSTX 06-       34  x<>y 07-    43 25  INTG 08-       20  x 09-  44 30 0  STO- 0 10-       33  R↓ 11-     45 0  RCL 0 12- 44,33 00  GTO 00``` That's a bit longer and requires one data register, but it works. The R↓ in line 11 was included to preserve the initiial values in T and Z. Dieter RE: (12C) Modulus - bshoring - 07-18-2016 10:12 PM (07-15-2016 06:32 PM)Dieter Wrote:   (07-15-2016 02:37 AM)Eddie W. Shore Wrote:  This program takes the modulus of two numbers: Y MOD X = X * FRAC(Y/X) That's not a good idea: roundoff errors will spoil the results. Consider your two examples: 124 mod 77 does not return 47 but 46,99999997 3862 mod 108 does not return 82 but 82,00000008 That's why you better use the relation y mod x = y – x * INT(y/x) instead. Code: ```01-       34  x<>y 02-     44 0  STO 0 03-       34  x<>y 04-       10  ÷ 05-    43 36  LSTX 06-       34  x<>y 07-    43 25  INTG 08-       20  x 09-  44 30 0  STO- 0 10-       33  R↓ 11-     45 0  RCL 0 12- 44,33 00  GTO 00``` That's a bit longer and requires one data register, but it works. The R↓ in line 11 was included to preserve the initiial values in T and Z. Dieter Here's a routine I wrote about a month ago. One step longer, but it uses only the stack and Last-X register. I wrote it for the HP-38C, but the steps are exactly the same for the HP-12C (but the keycodes would be different). For the two examples given by Dieter, it gets the right results. Code: ```01-     x<>y 02-     Enter 03-     Enter 04-     R↓ 05-     R↓ 06-     R↓ 07-     ÷ 08-     LSTX 09-     x<>y 10-     INTG 11-     x 12-     - 13-     GTO 00``` RE: (12C) Modulus - RobertM - 07-21-2016 11:36 PM From "ENTER: Reverse Polish Notation Made Easy", Jean-Daniel Dodin / Keith Jarett, (pg 115): Code: ```01-       36 ENTER 02-       36 ENTER 03-       30 - 04-       33 R↓ 05-       34 x<>y 06-    43 36 LSTx 07-       10 ÷ 08-    43 25 INTG 09-       20 x 10-       30 - 11- 43 33 00 GTO 00``` RE: (12C) Modulus - Dieter - 07-25-2016 09:17 AM (07-18-2016 10:12 PM)bshoring Wrote:  Here's a routine I wrote about a month ago. One step longer, but it uses only the stack and Last-X register. Great. In the meantime I got exactly the same solution. Many/most other HPs offer a R↑ which can replace the three consecutive R↓, making the program even shorter. (07-21-2016 11:36 PM)RobertM Wrote:  From "ENTER: Reverse Polish Notation Made Easy", Jean-Daniel Dodin / Keith Jarett, (pg 115): That's a nice one as well. Here's another 10/11 step solution: Code: ```01 ENTER 02 ENTER 03 - 04 + 05 / 06 LastX 07 x<>y 08 INT 09 x 10 - 11 GTO 00 or RTN``` The first four lines copy the content of Y to Z and T while X and Y are left unchanged. However, all these solutions destroy the stack, while the version I initially posted keeps the values of Z and T. That's why it requires one data register. I wonder if it is possible to do it only with the stack, i.e. without a data register, while Z and T are still preserved. Here's another challenge: what about returning y mod x as well as y div x (i.e. the integer quotient of y and x). There are many applications where both values are required at the same time. That's why I once suggested a DIVMOD command for the 34s that returns these two values. Now, what do you think ?-) Dieter RE: (12C) Modulus - Dieter - 07-25-2016 09:46 AM (07-25-2016 09:17 AM)I Wrote:  Here's another challenge: what about returning y mod x as well as y div x (i.e. the integer quotient of y and x). There are many applications where both values are required at the same time. That's why I once suggested a DIVMOD command for the 34s that returns these two values. I just realized that the code in the previous post can be extended easily to return both values at the same time – it's just two more steps: Code: ```01 ENTER 02 ENTER 03 - 04 + 05 / 06 LastX 07 x<>y 08 INT 09 ENTER 10 R↓ 11 x 12 - 13 GTO 00 or RTN``` The same can be done with the Dodin/Jarett program by inserting ENTER R↓ between step 08 and 09. The above version returns the remainder in X and y div x in Y. 3782 [ENTER] 72 [R/S] =>  38   [X<>Y]   52 There even is a third useful result: [LastX] => 3744 The largest number ≤ Y that is divisible by X. Dieter RE: (12C) Modulus - bshoring - 07-28-2016 11:01 PM (07-25-2016 09:46 AM)Dieter Wrote:   (07-25-2016 09:17 AM)I Wrote:  Here's another challenge: what about returning y mod x as well as y div x (i.e. the integer quotient of y and x). There are many applications where both values are required at the same time. That's why I once suggested a DIVMOD command for the 34s that returns these two values. I just realized that the code in the previous post can be extended easily to return both values at the same time – it's just two more steps: Code: ```01 ENTER 02 ENTER 03 - 04 + 05 / 06 LastX 07 x<>y 08 INT 09 ENTER 10 R↓ 11 x 12 - 13 GTO 00 or RTN``` The same can be done with the Dodin/Jarett program by inserting ENTER R↓ between step 08 and 09. The above version returns the remainder in X (and Z) and y div x in Y (and T). 3782 [ENTER] 72 [R/S] =>  38   [X<>Y]   52 There even is a third useful result: [LastX] => 3744 The largest number ≤ Y that is divisible by X. Dieter Nice program. When I run 3782 [ENTER] 72 [R/S], it leaves 38 in X only and 52 in Y, Z, & T. Works for me. RE: (12C) Modulus - Dieter - 07-29-2016 05:39 PM (07-28-2016 11:01 PM)bshoring Wrote:  When I run 3782 [ENTER] 72 [R/S], it leaves 38 in X only and 52 in Y, Z, & T. Hmmm... you're right. I am sure I had a version were the two results were in X and Z resp. Y and T. Anyway – I now corrected my previous post. Thank you for your feedback. Dieter