+Vorticon Posted April 17 Author Share Posted April 17 This is a fascinating topic. So I looked at my algorithm again, and I was actually mistaken as it's not really a gnome sort either because the latter goes through the entire list once the lowest swap is made in the current loop, whereas in my algorithm I jump directly to the location of the first swap for that loop iteration after the lowest swap for that iteration. Here's an example of a run for a worse case scenario: 5 4 3 2 1 4 5 3 2 1 4 3 5 2 1 3 4 5 2 1 3 4 2 5 1 3 2 4 5 1 2 3 4 5 1 2 3 4 1 5 2 3 1 4 5 2 1 3 4 5 1 2 3 4 5 That's 10 swaps for that run. Here are the swap numbers for each n from 2 to 7. To clarify, each time a swap situation exists, then 3 additional operations occur in the algorithm, otherwise no additional operations occur. So the number of swaps can be substituted for number of ops = swaps * 3. n swaps swaps_{n}-swaps_{n-1} _{2 1 0} _{3 3 2} _{4 6 3} _{5 10 4} _{6 15 5} _{7 21 6} Clearly it's a linear progression, so for the worst case scenario the algorithm has a growth of O(n) which is much better than even the combsort or the gnome sort which have a worst case scenario of O(n^{2})! Am I missing something here? The number of swaps for any n here (n>2) is: ( [ i = 3 to n ]SUM ( i - 1 ) ) + 1 1 Quote Link to comment Share on other sites More sharing options...

+FarmerPotato Posted April 17 Share Posted April 17 99'er magazine had an article on sorting methods in TI BASIC. My older brother studied that, implemented them, but it was beyond me then. Last summer I coded Heapsort in Forth. Two interesting properties of Heapsort: First, it doesn't need recursion. It iterates through an array, but skipping forward by powers of 2. Imagine adding up the series 1,2,4,8. How many steps will get you up to N? The answer is log2(N). (More precisely, the sum is N-1.) It does this loop on each of N elements. To move X to its destination, it has to compare-swap X with log2(N) other elements. Maybe that helps reinforce why we say N log N complexity. Outer loop N times, inner loop log2(N) times. And you can see where N log N is faster than Bubble Sort (which might compare every element against each other, N*N.) Second interesting property, Heapsort always takes N log N steps. No more, no less. I should mention that it takes two passes. Disadvantages. In contrast, quicksort has many opportunities to exit early, avoiding a lot of compare-swap. It too is O(N log N)--but imagine that , on average, a lot of elements are left alone after one or two swaps. Consider that there are special situations where one type of sort is advantageous to your kind of data, or for what you want to get out. And the fastest sort might not be the right choice. One situation is that you need a sorted list AND the ability to quickly insert new elements or remove some. While still keeping the list sorted. And not by shoving a giant block of memory one position to the right! Heapsort , after the first pass, gives you a "heap" which is ideal for further insertion/removal. (Guaranteed.) Sometimes that is the special property you need. Binary trees, in particular red-black trees, are another great choice for insertion/removal. 5 Quote Link to comment Share on other sites More sharing options...

+TheBF Posted April 17 Share Posted April 17 1 hour ago, Vorticon said: This is a fascinating topic. So I looked at my algorithm again, and I was actually mistaken as it's not really a gnome sort either because the latter goes through the entire list once the lowest swap is made in the current loop, whereas in my algorithm I jump directly to the location of the first swap for that loop iteration after the lowest swap for that iteration. Here's an example of a run for a worse case scenario: 5 4 3 2 1 4 5 3 2 1 4 3 5 2 1 3 4 5 2 1 3 4 2 5 1 3 2 4 5 1 2 3 4 5 1 2 3 4 1 5 2 3 1 4 5 2 1 3 4 5 1 2 3 4 5 That's 10 swaps for that run. Here are the swap numbers for each n from 2 to 7. To clarify, each time a swap situation exists, then 3 additional operations occur in the algorithm, otherwise no additional operations occur. So the number of swaps can be substituted for number of ops = swaps * 3. n swaps swaps_{n}-swaps_{n-1} _{2 1 0} _{3 3 2} _{4 6 3} _{5 10 4} _{6 15 5} _{7 21 6} Clearly it's a linear progression, so for the worst case scenario the algorithm has a growth of O(n) which is much better than even the combsort or the gnome sort which have a worst case scenario of O(n^{2})! Am I missing something here? How many times do you touch each element in the array? That will determine the 'n' factor. If your method is O(n) then you only have to scan the array once. In other words: a 5 element array would need a 1 to 5 FOR loop run only once to complete the sort. That's why my test example had a "passes" variable to count how many times I had to run through the array. 2 Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 17 Author Share Posted April 17 1 minute ago, TheBF said: How many times do you touch each element in the array? That will determine the 'n' factor. If your method is O(n) then you only have to scan the array once. In other words: a 5 element array would need a 1 to 5 FOR loop run only once to complete the sort. That's why my test example had a "passes" variable to count how many times I had to run through the array. And that's exactly what happens. I have only one loop that runs from 1 to n. Inside each iteration, it backtracks until the previous number is smaller than the current one then picks back up where it was initially in the array. Here's the code: 5000 REM ARRAY SORTING ROUTINE 5010 FOR I=1 TO R 5020 IF I=R THEN RETURN 5030 POINT=I 5080 V1=DLIST(POINT,1)::V2=DLIST(POINT+1,1) 5090 IF V1>V2 THEN DLIST(POINT,1)=V2::DLIST(POINT+1,1)=V1 ELSE 5140 5100 POINT=POINT-1::IF POINT>=1 THEN 5080 ELSE 5140 5140 NEXT I 2 Quote Link to comment Share on other sites More sharing options...

+TheBF Posted April 17 Share Posted April 17 18 minutes ago, Vorticon said: And that's exactly what happens. I have only one loop that runs from 1 to n. Inside each iteration, it backtracks until the previous number is smaller than the current one then picks back up where it was initially in the array. Here's the code: 5000 REM ARRAY SORTING ROUTINE 5010 FOR I=1 TO R 5020 IF I=R THEN RETURN 5030 POINT=I 5080 V1=DLIST(POINT,1)::V2=DLIST(POINT+1,1) 5090 IF V1>V2 THEN DLIST(POINT,1)=V2::DLIST(POINT+1,1)=V1 ELSE 5140 5100 POINT=POINT-1::IF POINT>=1 THEN 5080 ELSE 5140 5140 NEXT I Ok. So I guess the only thing to do is time it for an array of size 50 and then an array of size 100 and see if it the time increase is double or what the factor actually turns out to be. You may have a new sort algorithm. Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 17 Author Share Posted April 17 OK so I tested 2 arrays each initialized with the worst case scenario, i.e in descending order. One array is 50 elements and the second is 100 elements. Sort times: 50 elements: 57 seconds 100 elements: 229 seconds Definitely not quadratic! I wonder if @TheBF can compare this with his other sorts. Here's the test code. Timing is manual. 10 CALL CLEAR::OPTION BASE 1 20 DIM A(50),B(100) 30 REM FILL ARRAYS 35 PRINT "INITIALIZING ARRAYS...."::PRINT 40 FOR I=1 TO 100 60 B(I)=101-I::IF I<51 THEN A(I)=101-I 80 NEXT I 85 PRINT "READY! PRESS A KEY"::PRINT 86 CALL KEY(0,K,S)::IF S=0 THEN 86 90 CALL SORT(A(),50) 95 PRINT::PRINT "PRESS A KEY FOR NEXT ARRAY"::PRINT 100 CALL KEY(0,K,S)::IF S=0 THEN 100 110 CALL SORT(B(),100) 120 CALL KEY(0,K,S)::IF S<>0 THEN STOP 1000 SUB SORT(C(),N) 1005 PRINT "SORTING FOR ";N;" ELEMENTS" 1010 FOR I=1 TO N 1020 IF I=N THEN 1075 1030 P=I 1040 V1=C(P)::V2=C(P+1) 1050 IF V1>V2 THEN C(P)=V2::C(P+1)=V1 ELSE 1070 1060 P=P-1::IF P>=1 THEN 1040 1070 NEXT I 1075 PRINT "DONE SORTING!" 1080 SUBEND 1 Quote Link to comment Share on other sites More sharing options...

+FarmerPotato Posted April 17 Share Posted April 17 2 hours ago, TheBF said: How many times do you touch each element in the array? This is the relevant count. The number of compares, regardless of whether each leads to a swap. Consider the problem as "how many steps does it take to sort and prove that it is sorted?" The proof requires comparing a lot more numbers along the way. Also, O() notation is an upper bound. Some algorithms require exactly that number of steps--better algorithms in the same O() class might avoid some of the compares or swaps. Getting an element to its final spot with fewer swaps. Quicksort and Heapsort are both O(N log N) but Quicksort avoids many swaps on average. While Heapsort always does the upper limit number of swaps! We also worry about computer implementations: which takes longer, compares or swaps? Does the computer stall when numbers are not near neighbors? (Computer waits on slow RAM.) 1 Quote Link to comment Share on other sites More sharing options...

+FarmerPotato Posted April 18 Share Posted April 18 57 minutes ago, Vorticon said: 50 elements: 57 seconds 100 elements: 229 seconds Definitely not quadratic! I wonder if @TheBF can compare this with his other sorts. But that is quadratic! 50^2/57 = 43.859 100^2/229 = 43.668 Which fits 44 * x^2 1 Quote Link to comment Share on other sites More sharing options...

+TheBF Posted April 18 Share Posted April 18 52 minutes ago, FarmerPotato said: But that is quadratic! 50^2/57 = 43.859 100^2/229 = 43.668 Which fits 44 * x^2 1 hour ago, Vorticon said: OK so I tested 2 arrays each initialized with the worst case scenario, i.e in descending order. One array is 50 elements and the second is 100 elements. Sort times: 50 elements: 57 seconds 100 elements: 229 seconds Definitely not quadratic! I wonder if @TheBF can compare this with his other sorts. Here's the test code. Timing is manual. 10 CALL CLEAR::OPTION BASE 1 20 DIM A(50),B(100) 30 REM FILL ARRAYS 35 PRINT "INITIALIZING ARRAYS...."::PRINT 40 FOR I=1 TO 100 60 B(I)=101-I::IF I<51 THEN A(I)=101-I 80 NEXT I 85 PRINT "READY! PRESS A KEY"::PRINT 86 CALL KEY(0,K,S)::IF S=0 THEN 86 90 CALL SORT(A(),50) 95 PRINT::PRINT "PRESS A KEY FOR NEXT ARRAY"::PRINT 100 CALL KEY(0,K,S)::IF S=0 THEN 100 110 CALL SORT(B(),100) 120 CALL KEY(0,K,S)::IF S<>0 THEN STOP 1000 SUB SORT(C(),N) 1005 PRINT "SORTING FOR ";N;" ELEMENTS" 1010 FOR I=1 TO N 1020 IF I=N THEN 1075 1030 P=I 1040 V1=C(P)::V2=C(P+1) 1050 IF V1>V2 THEN C(P)=V2::C(P+1)=V1 ELSE 1070 1060 P=P-1::IF P>=1 THEN 1040 1070 NEXT I 1075 PRINT "DONE SORTING!" 1080 SUBEND I had to go out to a rehearsal, but I can try comparing this to combsort which I have in BASIC. -or- I have a sort workbench in Forth so maybe I can translate your algorithm to Forth and then do comparisons on much bigger data. 2 Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 18 Author Share Posted April 18 1 hour ago, FarmerPotato said: But that is quadratic! 50^2/57 = 43.859 100^2/229 = 43.668 Which fits 44 * x^2 But processing time is not. In the context of sort performance, if a sort algorithm is O(n^{2}) shouldn't processing time increase quadratically? This is not the case here. Quote Link to comment Share on other sites More sharing options...

+TheBF Posted April 18 Share Posted April 18 For the comparison here is COMBSORT in BASIC using the VORTSORT chassis. (I named it) For the small amount of extra code COMBSORT is a lot of bang for the buck. 1 ! VORTSORT chassis with COMBSORT 2 ! 50 elements 9.5 seconds 3 ! 100 elements 23 seconds 10 CALL CLEAR::OPTION BASE 1 20 DIM A(50),B(100) 30 REM FILL ARRAYS 35 PRINT "INITIALIZING ARRAYS...."::PRINT 40 FOR I=1 TO 100 60 B(I)=101-I::IF I<51 THEN A(I)=101-I 80 NEXT I 85 PRINT "READY! PRESS A KEY"::PRINT 86 CALL KEY(0,K,S)::IF S=0 THEN 86 90 CALL COMBSORT(A(),50) 95 PRINT::PRINT "PRESS A KEY FOR NEXT ARRAY"::PRINT 100 CALL KEY(0,K,S)::IF S=0 THEN 100 110 CALL COMBSORT(B(),100) 120 CALL KEY(0,K,S)::IF S<>0 THEN STOP 2000 SUB COMBSORT(C(),N) 2100 PRINT "SORTING FOR ";N;" ELEMENTS" 2420 PASSES=0 2424 GAP=N 2440 SORTED=1 2442 GAP=INT(GAP/1.3) 2450 FOR I=1 TO N-GAP 2460 IF C(I)<C(I+GAP) THEN 2530 2470 rem swap the array items 2480 TEMP=C(I) 2490 C(I)=C(I+GAP) 2500 C(I+GAP)=TEMP 2520 SORTED=0 2530 NEXT I 2540 PASSES=PASSES+1 2550 REM test if sort is finished 2570 IF GAP=1 THEN 2600 2580 GOTO 2440 2600 SUBEND 1 Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 18 Author Share Posted April 18 Wow. The difference is impressive... So much for Vortsort I did track the number of comparisons instead for just swaps for arrays 50,100,150 and 200 elements and I got the following: Clearly not linear. And as @FarmerPotato noted, n^{2}/#comparisons is a constant = 2. So yeah, this sort is an O(n^{2}) type under the worst case scenario. Oh well, I guess fame will have to wait... 😝 3 2 Quote Link to comment Share on other sites More sharing options...

apersson850 Posted April 18 Share Posted April 18 19 hours ago, SteveB said: ...still some Quicksort implementation totally freak out when handed an already sorted list ... The language for the implementation should support recursions, I wouldn't do it in TI BASIC. When implementing it in assembly you need to implement the recursion manually. A whole new science is sorting datasets larger than your RAM, which has been a big issue in the early mainframe world... tape-to-tape sorting. The Quicksort I wrote once, when I learned about the algorithm, doesn't "freak out", but it does run for a longer time when the list is already sorted. You can "unwrap" the recursion even when writing it in e.g. Pascal. They claimed where the article was described that it would be more efficient. It may of course depend on how recursion is implemented in the language. Since I managed to sort a thousand integers in about half a second on the TI 99/4A, I was happy. Due to the limited memory capacity of such a small computer there's normally not the need to sort vast amounts of numbers either. 3 Quote Link to comment Share on other sites More sharing options...

SteveB Posted April 18 Share Posted April 18 7 hours ago, Vorticon said: Oh well, I guess fame will have to wait... 😝 There is a reason that there is not "THE ONE^{[tm]}" sorting algorithm. It depends very much on data, hardware, programming language ... you just need to find the niche where VORTSORT outperforms the others. The O() logic also looks at infinity ... most data we are dealing with is finite. A simple O(N^3) may outperform a O(n*log(n)) on a small dataset on a real machine. And coding time is also important 🙃 4 Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 18 Author Share Posted April 18 2 hours ago, apersson850 said: Since I managed to sort a thousand integers in about half a second on the TI 99/4A, I was happy. Due to the limited memory capacity of such a small computer there's normally not the need to sort vast amounts of numbers either. Would you happen to have the code for it? I assume this was done using UCSD Pascal which supports recursion. Quote Link to comment Share on other sites More sharing options...

+Lee Stewart Posted April 18 Share Posted April 18 16 hours ago, FarmerPotato said: Two books that I learned from are Algorithms, by Sedgwick, and Numerical Recipes in C. Sedgwick comes in Pascal, C and other flavors. Numerical Recipes covers just enough sorting, and is a reference on many other mathematical computing topics. I have used many references for this and other topics of scientific computing (more for data analysis than sorting) over the years—including Numerical Recipes: The Art of Scientific Computing by Press, Teukolsky, Vetterling, and Flannery and various volumes of The Art of Computer Programming by D. E. Knuth, but probably none more than Numerical Recipes. ...lee 4 Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 18 Author Share Posted April 18 The final word... 5 Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 18 Author Share Posted April 18 On 4/18/2023 at 4:22 AM, SteveB said: And coding time is also important Indeed. Vortsort can be compressed into just 4 lines of code in XB and performance is likely comparable to Insertion or Gnome sort or perhaps even a little better. 1000 SUB VORTSORT(C(),N) 1010 FOR I=1 TO N-1::P=I 1040 V1=C(P)::V2=C(P+1)::IF V1<V2 THEN 1070 ELSE C(P)=V2::C(P+1)=V1::P=P-1::IF P>=1 THEN 1040 1070 NEXT I::SUBEND I'm keeping it for future use. 4 Quote Link to comment Share on other sites More sharing options...

Willsy Posted April 18 Share Posted April 18 <pedant> I think the ELSE 1070 at the end of line 1040 can be omitted - it's redundant. </pedant> 4 Quote Link to comment Share on other sites More sharing options...

+TheBF Posted April 18 Share Posted April 18 4 hours ago, apersson850 said: The Quicksort I wrote once, when I learned about the algorithm, doesn't "freak out", but it does run for a longer time when the list is already sorted. You can "unwrap" the recursion even when writing it in e.g. Pascal. They claimed where the article was described that it would be more efficient. It may of course depend on how recursion is implemented in the language. Since I managed to sort a thousand integers in about half a second on the TI 99/4A, I was happy. Due to the limited memory capacity of such a small computer there's normally not the need to sort vast amounts of numbers either. It would be very interesting to see how a non-recursive Quicksort performs in TI BASIC. The amazing speed of the sort would be offset by the extra BASIC instructions which are very slow. Below some size of the data there would be no benefit versus a simpler algorithm I suspect. I think I have textbook with the non-recursive version in Pascal. I might take a run at it. 2 Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 18 Author Share Posted April 18 18 minutes ago, Willsy said: <pedant> I think the ELSE 1070 at the end of line 1040 can be omitted - it's redundant. </pedant> Yup. Forgot to remove it when I reformatted the code. 3 Quote Link to comment Share on other sites More sharing options...

apersson850 Posted April 18 Share Posted April 18 (edited) 3 hours ago, Vorticon said: Would you happen to have the code for it? I assume this was done using UCSD Pascal which supports recursion. Here it is. The assembly version, that is. I did start with a Pascal version, but don't have that handy now. Quicksort It's still kind of recursive in prinicple, but not with a full scale recursive routine call. By mistake I wrote somewhere that it uses bubblesort to finish up, but it doesn't. It's insertionsort I use to finalize short lists. The program is based on a program written in Algol from the beginning. Edited April 18 by apersson850 1 1 Quote Link to comment Share on other sites More sharing options...

+FarmerPotato Posted April 18 Share Posted April 18 7 hours ago, Vorticon said: 1000 SUB VORTSORT(C(),N) 1010 FOR I=1 TO N::IF N=I THEN SUBEXIT ELSE P=I 1040 V1=C(P)::V2=C(P+1)::IF V1<V2 THEN 1070 ELSE C(P)=V2::C(P+1)=V1::P=P-1::IF P>=1 THEN 1040 ELSE 1070 1070 NEXT I::SUBEND I'm keeping it for future use. Agree this is valuable because it's so short. you can eliminate IF I=N by saying FOR I=1 TO N-1 1 Quote Link to comment Share on other sites More sharing options...

+TheBF Posted April 18 Share Posted April 18 6 hours ago, apersson850 said: Here it is. The assembly version, that is. I did start with a Pascal version, but don't have that handy now. Quicksort It's still kind of recursive in prinicple, but not with a full scale recursive routine call. By mistake I wrote somewhere that it uses bubblesort to finish up, but it doesn't. It's insertionsort I use to finalize short lists. The program is based on a program written in Algol from the beginning. And here is a version that assembles with the default TI Assembler. You will have to un-comment the BLWP line and the two below. Readme from my time working on it: This code is from a demo in UCSD Pascal by @apperson850 on artiage.com. The original has been modified to assemble with the stock TI-99 ASSEMBLER. The resulting object file can be linked with LINKEXE. Once linked it can be used on DATA in the Forth's memory, taking the address and size of an array from Forth's data stack and jumping into QUICKSORT. 1000 items can be sorted in under 2 seconds on a stock console. Spoiler * TITL "QUICKSORT FOR INTEGERS" * Procedure sorting integers in ascending order * Called from Pascal host by QSORT(A,N)* * Declared as PROCEDURE QSORT(VAR A:VECTOR* N:INTEGER)* EXTERNAL* * Vector should be an array [..] of integer* * Author: @APERSON850 on Atariage.com * Modified for TI Assembler, to be called from Camel99 Forth * 2022 Brian Fox *-------------- * Workspace registers for subroutine at START * * R0 Array base pointer * R1 Array end pointer * R2 L * R3 R * R4 I * R5 J * R6 KEY * R7 Temporary * R8 Temporary * R9 Subfile stack pointer * R10 Main stackpointer * R11 Pascal return address (P-system WS) * R12 Not used * R13 Calling program's Workspace Pointer * R14 Calling program's Program Counter * R15 Calling program's Status Register *----------------- * The actual vector in the Pascal-program could be indexed [1..n] * This routine assumes only that n indicates the number of elements, not the * last array index. DEF QSORT LIMIT EQU 16 *Quick- or Insertionsort limit * * Don't need this for Forth ** * BLWP @QSORT * AI R10,4 *Simulate pop of two words * B *R11 *Back to Pascal host QSORT DATA SORTWS,START *Transfer vector for Quicksort START MOV @12(R13),R10 *Get stackpointer (R6) from Forth's WP->R10 LI R9,ENDSTK *SUBFILE STACKPOINTER MOV *R10+,R1 * POP PARAMETER N MOV *R10+,R0 * POP ARRAY POINTER DEC R1 SLA R1,1 A R0,R1 *CALCULATE ARRAY ENDPOINT MOV R0,R2 *L:=1 MOV R1,R3 *R:=N MOV R1,R7 S R0,R7 CI R7,LIMIT JLE INSERT *FIGURE OUT IF QUICKSORT IS NEEDED MNLOOP MOV R2,R7 SRL R7,1 MOV R3,R8 SRL R8,1 A R8,R7 ANDI R7,>FFFE *R7:=INT((L+R)/2) MOV *R7,R8 MOV @2(R2),*R7 MOV R8,@2(R2) *A[(L+R)/2]:=:A[L+1] C @2(R2),*R3 JLT NOSWP1 MOV @2(R2),R8 MOV *R3,@2(R2) MOV R8,*R3 *A[L+1]:=:A[R] NOSWP1 C *R2,*R3 JLT NOSWP2 MOV *R2,R8 MOV *R3,*R2 MOV R8,*R3 *A[L]:=:A[R] NOSWP2 C @2(R2),*R2 JLT NOSWP3 MOV @2(R2),R8 MOV *R2,@2(R2) MOV R8,*R2 *A[L+1]:=:A[L] NOSWP3 MOV R2,R4 INCT R4 *I:=L+1 MOV R3,R5 *J:=R MOV *R2,R6 *KEY:=A[L] JMP INCLOP INRLOP MOV *R4,R8 *LOOP UNWRAPPING MOV *R5,*R4 MOV R8,*R5 *A[I]:=:A[J] INCLOP INCT R4 *I:=I+1 C *R4,R6 JLT INCLOP *A[I]<KEY DECLOP DECT R5 *J:=J-1 C *R5,R6 JGT DECLOP *A[J]>KEY C R4,R5 JLE INRLOP *IF I<=J THEN CONTINUE OUT MOV *R2,R8 MOV *R5,*R2 MOV R8,*R5 *A[L]:=:A[J] DEL1 MOV R5,R7 *Quicksort subfiles? S R2,R7 *R7:=J-L MOV R3,R8 S R4,R8 INCT R8 *R8:=R-I+1 CI R7,LIMIT JH DEL2 CI R8,LIMIT JH DEL2 CI R9,ENDSTK *LVL=0? JEQ INSERT *No more Quicksorting at all? MOV *R9+,R2 *POP L MOV *R9+,R3 *POP R JMP MNLOOP DEL2 C R7,R8 *Determine what is small and large subfile JL ELSE2 MOV R2,@LSFL MOV R5,@LSFR DECT @LSFR MOV R4,@SSFL MOV R3,@SSFR JMP DEL3 ELSE2 MOV R4,@LSFL MOV R3,@LSFR MOV R2,@SSFL MOV R5,@SSFR DECT @SSFR DEL3 CI R7,LIMIT *Is small subfile big enough to be sorted by JLE THEN3 *Quicksort? CI R8,LIMIT JH ELSE3 THEN3 MOV @LSFL,R2 *Don't Quicksort small subfile, only large MOV @LSFR,R3 JMP MNLOOP ELSE3 DECT R9 *Stack large subfile MOV @LSFR,*R9 *PUSH R DECT R9 MOV @LSFL,*R9 *PUSH L MOV @SSFL,R2 *Sort small subfile MOV @SSFR,R3 JMP MNLOOP * * Insertion Sort finishing up * INSERT MOV R1,R4 DECT R4 *I:=N-1 C R4,R0 JL LEAVE *Check if any looping at al FORI C @2(R4),*R4 JGT NEXTI *If next is greater than this, it's OK MOV *R4,R6 *KEY:=A[I] MOV R4,R5 INCT R5 *J:=I+1 WHILE MOV *R5,@-2(R5) *A[J-1]:=A[J] INCT R5 *J:=J+1 C R5,R1 JH ENDWHL *J>N? C *R5,R6 *A[J]<KEY? JLT WHILE ENDWHL MOV R6,@-2(R5) *A[J-1]:=KEY NEXTI DECT R4 C R4,R0 *Check if passed array base point JHE FORI LEAVE RTWP * Return Forth workspace * AI R6,4 * remove parameters from Forth stack B @R10 * Branch to Forth interpreter *-------------- * DATA AREA SORTWS BSS >20 *Workspace for sorting routine SUBSTK BSS >40 *Internal subfile stack ENDSTK EQU SUBSTK+>40 *End of that stack * variable used by quicksort LSFL DATA 0 *Large SubFile Left pointer LSFR DATA 0 *Large SubFile Right pointer SSFL DATA 0 *Small SubFile Left pointer SSFR DATA 0 *Small SubFile Right pointer END 1 Quote Link to comment Share on other sites More sharing options...

+Vorticon Posted April 18 Author Share Posted April 18 1 hour ago, FarmerPotato said: Agree this is valuable because it's so short. you can eliminate IF I=N by saying FOR I=1 TO N-1 Yes! Every little trim helps. 1 Quote Link to comment Share on other sites More sharing options...

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