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1.1 Switch on loss and switch on loss/tie

When pspace was first implemented on the hills, one of the first switching algorithms was the simple 'switch on loss' or 'switch on tie or loss'.
res     ldp.ab  _RESULT,   #0
str     ldp.a   _STRATEGY, str1     ;load strategy in use
        sne.ab  #0,        res      ;check result
lost    add.a   #1,        str1     ;lost change 
        mod.a   #X,        str1     ;X = number of strategies to switch
win     stp.ab  str1,      _STRATEGY
str1    jmp     @0,        strat0
        dat     0,         strat1
        dat     0,         strat2
        . . .
        dat     0,         stratX-1
Length: 6 + 1/strategy
Cycles to completion: after Win/Tie: 6
                            Loss   : 7
The code is short and decisions are made quickly. However it is definitely lacking in sophistication ( which you don't always need to be effective ). The mod instruction ensures that a valid strategy is chosen even if brainwashed. However, since there is no active brainwash detection, the strategy chosen is essentially random. The algorithm does average ( for a pspace warrior ) against non-pspace opponents, but tends to do better against more compex p-opponents that take multiple rounds to make decisions. Jack in the box used this type of p-logic quite successfully.

The following switcher is very similar to the above one, but with two changes: firstly, and less important, the add.a #1,pTable line has been changed to nop.f }pTable,}-1000: this has two advantages 1) If the b-field of this instruction happens to be hit very early in the round, the warrior will still behave properly and 2) It gets a free bomb in. Both minor advantages, but there are no disadvantages, so why not? The second, and more important, change is that the pspace value is not re-stored unless it has changed (note that it still always undergoes a mod before being used); this saves a cycle and leads to exactly the same behaviour. Disadvantages of this switcher: same as before, and also it can't be modified to a switch-on-loss or tie, without losing it's speed advantage. If the jmn.b pGood,pGo line is changed to jmn.b pTable,pGo, the switcher gets another cycle quicker on a win or a tie, but it then has the possibility of suicide if it is brainwashed and doesn't lose. Whether this is a good thing or not depends on the number of brainwashers on the hill, what they're brainwashing with, how quick your boot routine is and countless other things.
pGo     ldp.ab  #pResult,  #0
        ldp.a   #pSpace,   pTable
        jmn.b   pGood,     pGo
        nop.f   }pTable,   }-1000
        stp.ab  pTable,    #pSpace
pGood   mod.a   #pNum,     pTable
pTable  jmp.a   @0,        strat1
        dat.f   0,         strat2

Length: 6 + 1/strategy
Cycles: Win/Tie: 5
           Loss: 7
Another possible solution is shown below. It is one instruction smaller but it needs one cycle more (two cycles more for strategy 2) than the above mentioned code:
pGold   ldp.ab  #0,        #0          ; get results of last battle
        ldp.a   #pflag,    pGold       ; retrieve attempted strategy
        add.a   #1,        @pGold      ; if a loss, increment strategy
        mod.a   #2,        pGold       ; safeguard against brainwashing
        stp.ab  pGold,     #pflag      ; store current strategy
        jmz.a   strat1,    pGold       ; select strategy 1
        jmp     strat2                 ; select strategy 2

Length: 5 + 1/strategy
Cycles: Win/Tie: 6 for strat1 and 7 for strat2 (6 if strategy 2 immediately follows)
           Loss: 6 for strat1 and 7 for strat2 (6 if strategy 2 immediately follows)
A powerful adaptation of the routine can be made with no extra instructions. By increasing the MOD number we have an assymetric switcher, by which the second strategy is selected more often than the first. This can be very helpful in pairing up a strong all-purpose warrior with a special-purpose warrior.

The following two switcher have the same behaviour: on a loss, they will change to the next strategy in their table, whilst they will select a specific strategy after a tie. The first switcher is slower but smaller than the second.
pSelect equ     0               ; which to select on tie

pGo     ldp.ab  #pResult,   #0
        ldp.a   #pSpace,    pTable
        sne.ab  #0,         pGo
        add.a   #1,         pTable
        sne.ab  #2,         pGo
        mov.a   #pSelect,   pTable
        mod.a   #pNum,      pTable
        stp.ab  pTable,     #pSpace
pTable  jmp.a   @0,         strat1
        dat.f   0,          strat2

Length: 8 + 1/strategy
Cycles:      Win: 7
        Tie/Loss: 8
Switch on loss, select on tie number 2
pGo     ldp.a   #pResult,   pJump
        ldp.a   #pSpace,    pTable
pJump   jmp.a   @0,         pLoss
        dat.f   0,          pWin
        dat.f   0,          pTie

        [any number of instructions you like in here]

pTie    mov.a   #(pSelect-1),pTable
pLoss   nop.f   }pTable,    }-1000
        stp.ab  pTable,     #pSpace
pWin    mod.a   #pNum,      pTable
pTable  jmp.a   @0,         strat1
        dat.f   0,          strat2

Length: 9 + 1/strategy
Cycles: First round: 4
                Win: 5
                Tie: 8
               Loss: 7
A more complex switcher was first released in Kline's Gem of the Ocean. It is quite effective while maintaining simplicity. It uses an array to determine its next strategy thereby controlling the sequence of warriors. The key here is to select the proper sequence of warriors as well as good components.
parray    dat    0,           strat1         ;table for pswitching
          dat    0,           strat3         ;- stay after win
          dat    0,           strat1         ;- go to ptie after tie
          dat    0,           strat4         ;- move up one after loss
          dat    0,           strat1
          dat    0,           strat3
ptie      dat    0,           strat2
          dat    0,           strat4

presult   ldp.ab #0,          #0             ;pswitch code, nice and short
pselect   ldp.ab #PFLAG,      #0
          sne.ab #0,          presult        ;advance table selection after a loss   
          add.ab #1,          pselect             
          sne.ab #2,          presult        ;go to Sweep after a tie
          mov    #ptie-parray,pselect
 pselect,     pselect
          mod.ab #presult-parray,pselect
 pselect,     pchoice
pchoice   jmp    @parray,     {pselect       ;jump to selection

Length: 10+1/strategy
Cycles:     First round: 4
                    Win: 6
                    Tie: 5
        Loss:    switch: 12
              no switch: 10

1.2 Switch on loss or loss/tie with brainwash detection

To equip a switcher with a decent brainwash detection system is simple. One method is to simply load a number into pspace on the first round and check it every remaining round to see if it's been tampered with. You could reset it and run a special anti-wash warrior or stick with the anti-wash warrior for the rest of the battles. This could take a lot of extra p-code. A better method is may be this:
res     ldp.ab _RESULT, #0
str     ldp.a _STRATEGY, str1   ;load strategy in use

        mod.a #97, str1         ;valid strategies are all mod 97
        jmn.a washed, str1      ;do 'have been washed' code [then reset?]
        ldp.a _STRATEGY, str1   ;got to be a way to get rid of repeating this

        sne.ab  #0, res         ;check result
lost    add.a   #97, str1       ;lost change 
        mod.a   #X, str1        ;X = number of strategies to switch
win     stp.ab  str1, _STRATEGY
str1    jmp     @0, strat0
        dat     0, strat1
        dat     0, strat2
        . . .
        dat     0, stratX-1
Length: 9 + 1/strategy
Cycles: brainwashed: 4
                Win: 9
                Tie: 9
               Loss: 10
This is almost exactly the same p-code mentioned at first in chapter 1.1 with the addition of three extra lines of instructions to ensure that all valid strategies are mod 97. A brainwash would have to wash with a multiple of 97 (unfortunately zero is one of them) to escape detection. Of course the special number can vary as long as the mod.a #X instruction yields the proper series to switch your components.

1.3 Switch on two consecutive loss/tie

Quickness is especially important after the release of a new generation of superfast quickscans (Q^2, Q^3, etc). The downside is no matter what effective strategy you are using, a spurious loss will switch you to a less effective one. It may take even another loss to return to the optimal strategy if you are switching more than one component. To combat this phenomena, a switch on two consecutive tie/loss combinations was developed. Strategies will only change on a loss loss, tie loss, loss tie, or tie tie.

A loss or tie will advance the table pointer contained in _STRAT. If the _STRAT pointer is 0 or 2, that strategy will get its 'second chance'. If the index is 1 or 3, strategies will switch. A win should reset the pointer to either 0 or 3 depending on what strategy is currently in use. This was done by taking the index modulo 2 and subtracting the result from the current index.

result  ldp.ab _RES, wlt
strat   ldp.a _STRAT, choice
wlt     sne.ab #1, #0
        jmp win                 ; win has occured
        add.a #1, choice        ; loss or tie has occured
        mod.a #4, choice
        stp.a choice, _STRAT
choice  jmp @0, strat1          ; _STRAT = 0, choose strat1
        nop 0, strat1           ; _STRAT = 1, choose strat1
        nop 0, strat2           ; _STRAT = 2, choose strat2
        nop 0, strat2           ; _STRAT = 3, choose strat2
win     mov.ab choice, #0       ; reset index becuase of win
        mod.ab #2, win          win, choice      
        jmp choice-1            

Length: 11 + 2/strategy
Cycles:         Win: 8
                Tie: 7
               Loss: 7
All in all, the brain is much bigger (8 vs. 15) and a bit slower. But it should has the smarts necessary to defeat non-pspace programs quite handily. However this isn't the case. When switching the components the logic performe similarly or worse! The optimal strategy would win for awhile, but the occasional loss tie combination would cause it to switch. The secondary strategy (being a somewhat general purpose strategy) would not switch out as fast due to the new switching rules. This seemed to balance the extra wins gained. It seems that the p-logic would be much more effective if the secondary strategy is very specialized against one type of program and losses quickly against everything else.

1.4 Switch on n losses

If the warrior loses for pLosses consecutive rounds, then it will switch to the next strategy in its table. This means that a spurious loss for the optimal strategy won't kick in another strategy. Also, after a win or a tie, it is as quick as the above switchers, but is quite a bit slower after a loss, and also bigger.

One potential problem with this switcher is that it is very zero brainwash sensitive: if it is brainwashed with a zero, the first strategy in the table will be selected, and the loss counter reset, which could produce problems if this strategy lost reguarly to the brainwasher. If pLosses is 2, the switcher can be modified by changing pLoss to ldp.ab #pSpace2,#1 and pThink to jmz.b pWin,pLoss, to make it sensitive to a 1-brainwash rather than 0-brainwash, whilst if pLosses>2, then changing pLoss to ldp.ab #pSpace2,#1 and pThink to cmp.b pLoss,#1 will have the same effect, but increasing the running time by 1 cycle for a loss without switch.

One final modification that can be made is changing the first three lines to
pGo     ldp.ab  #pResult,pJump
        ldp.a   #pSpace1,pTable
pJump   jmp.a   @0,pLoss
        dat.f   0,pTie
which will lead to a switch after pLosses rounds without a win. Note that this modification is not compatible with the brainwash changes above, and changes the stats to: Length: 12+1/strategy. Cycles: Win: 6, Tie/Loss no switch:10, Tie/Loss switch: 12, First round: 4
pSpace1 equ     1               ; used to store the current strategy
pSpace2 equ     2               ; used to store the loss counter
pLosses equ     3               ; how many losses before we switch

pGo     ldp.ab  #pResult,#0
        ldp.a   #pSpace1,pTable
pJump   jmn.b   pWin,pGo
pLoss   ldp.ab  #pSpace2,#0
        nop.f   >pLoss,}-1000
        mod.ab  #pLosses,pLoss
pThink  jmn.b   pWin,pLoss
        nop.f   }pTable,}-2000
        stp.ab  pTable,#pSpace1
pWin    stp.b   pLoss,#pSpace2
        mod.a   #pNum,pTable
pTable  jmp.a   @0,strat1
        dat.f   0,strat2

        end     pGo

Length: 11 + 1/strategy
Cycles: Win/Tie:  6
        Loss: when no strategy switch occurs: 10
                          when switch occurs: 12
The following switcher is very similar to the switch on n consecutive losses switcher above, but has different enough behaviour to warrant a separate entry. The only difference here is that a tie will not reset the loss count, so a sequence of loss-tie-loss-loss will cause a switch, if pLosses is equal to 3.
pGo     ldp.ab  #pResult,pJump
        ldp.a   #pSpace1,pTable
pJump   jmp.a   @0,pLoss
        dat.f   0,pWin
        dat.f   0,pTie

        [any number of instructions here]

pLoss   ldp.ab  #pSpace2,#0
        nop.f   >pLoss,}-1000
        mod.ab  #pLosses,pLoss
pThink  jmn.b   pWin,pLoss
        nop.f   }pTable,}-2000
        stp.ab  pTable,#pSpace1
pWin    stp.b   pLoss,#pSpace2
pTie    mod.a   #pNum,pTable
pTable  jmp.a   @0,strat1
        dat.f   0,strat2

Length: 13+1/strategy
Cycles:     First round: 4
                    Win: 6
                    Tie: 5
        Loss:    switch: 12
              no switch: 10

A p-switcher with a more sophisticated scoring system is shown below:
start   ldp.ab _STRATEGY, check
        jmz.b switch, check     ; washed or first round, just do a switch
check   mod.ab #97, #0000       ; if not multiple of 97, have been washed!
        jmn.b scanner, check    ; washed --> do scanner (can do any plogic)
        ldp.a _RESULT, keeper
        ldp.ab _SCORE, adjust
keeper  jmp @0, adjust+1        ; result == 0 == loss
        jmp 0, adjust           ; result == 1 == win
        jmp 0, adjust+2         ; result == 2 == tie
adjust  add.ab #6, #0           ; win
        add.ab #-2, adjust      ; lose
        add.ab #-1, adjust      ; tie
        ldp.a _STRATEGY, select ; again.  can't seem to save it.
        slt.b adjust, #7000     ; did it go negative?
        jmp switch
return  stp.b adjust, _SCORE
        mod.a #3, select
select  jmp @0, scanner
        jmp 0, stone
        jmp 0, paper
switch  add.a #97, select
        stp.ab select, _STRATEGY
        mov.ab #10, adjust
        jmp return
Length: 21 + 1/strategy
Cycles: first round: 4
        brainwashed: 9 (4 if brainwashed with 0)
                Win: 20
                Tie: 18
               Loss: 19
This code creates a 'score' for each strategy starting at 10. For every win the score increases by 3, tie--decrease by 1, loss--decrease by 3. The strategy switches when the score goes negative. When fighting a non-pspace opponent, the logic should quickly find the proper strategy and not deviate from the optimum.

1.5 The P^2 switchers

Most of the above presented various switching routines have reasonably good response times. The disadvantage of all this routines is that they are not very flexible. A switching strategy that works well for one component may work poorly with another. Any attempt to rectify this usually means using a large, slow switching routine.

To solve this Anton Marsden wrote a general purpose switching routine (P^2) that is fast no matter what strategy the 'programmer' wanted to use.

The P^2 is a fast table-based switcher which can manage any conceivable strategy or combination of strategies using 6 cycles. However, the tables are potentially large and for this reason it may not be the best switcher to use in some cases, especially if you want to use a quickscan in your P-Warrior.

Below is a demonstration of a switch-on-3-consecutive-losses routine containing 2 components.

1. Draw a state transition diagram showing all possible states. Each state should have three outward arrows for win, loss, and tie inputs. Each arrow should identify a component to be run and point to the next state. Note that 'components' and 'states' are not equated in this logic as they are in many previous switchers. State numbers should start from 0.
  {n}=state n
  c1=component 1
  c2=component 2
The following diagram is a bit messy but you should get the idea.
                |                           T,W (c1)        |
                |   +-------------------+                   |
                |   |      T,W (c1)     |                   |
                |   v                   |                   |
                   | ^       L (c1)               L (c1)        L (c2)
                   | |
                   T,W (c1)

                |                           T,W (c2)        |
                |   +-------------------+                   |
                |   |      T,W (c2)     |                   |
                |   v                   |                   |
                   | ^       L (c2)               L (c2)        L (c1)
                   | |
                   T,W (c2)
2. Pick an initial state transition, eg. {2} -> {0} using c1. 3. Put all loss transitions into a data table where entry n-1 contains the state transition information for state n-1:
  dat 1,c1 ; State 0 goes to state 1 on a loss using component 1.
  dat 2,c1 ; State 1 goes to state 2 on a loss using component 1.
  dat 3,c2 ; State 2 goes to state 3 on a loss using component 2.
  dat 4,c2 ; State 3 goes to state 4 on a loss using component 2.
  dat 5,c2 ; State 4 goes to state 5 on a loss using component 2.
  dat 0,c1 ; State 5 goes to state 0 on a loss using component 1.
4. Do the same for wins and ties.
  dat 0,c1 ; State 0 goes to state 0 on a win using component 1.
  dat 0,c1 ; State 1 goes to state 0 on a win using component 1.
  dat 0,c1 ; State 2 goes to state 0 on a win using component 1.
  dat 3,c2 ; State 3 goes to state 3 on a win using component 2.
  dat 3,c2 ; State 4 goes to state 3 on a win using component 2.
  dat 3,c2 ; State 5 goes to state 3 on a win using component 2.

  dat 0,c1 ; State 0 goes to state 0 on a tie using component 1.
  dat 0,c1 ; State 1 goes to state 0 on a tie using component 1.
  dat 0,c1 ; State 2 goes to state 0 on a tie using component 1.
  dat 3,c2 ; State 3 goes to state 3 on a tie using component 2.
  dat 3,c2 ; State 4 goes to state 3 on a tie using component 2.
  dat 3,c2 ; State 5 goes to state 3 on a tie using component 2.
5. Set up the initial state.
init_state EQU (tie_table+2) ; {2} -> {0} using c1.
6. Put the tables and components into P^2. Remember to set STATES to the correct value (6 in this case). 7. Optimise the code - this is left as an exercise for the reader.

How it works: the result of the last fight is loaded as well as the state used for the last fight. From this information the next state is determined using the tables that were constructed from the state transition diagram. Then the component associated with the transition is executed.
;name P^2
;author Anton Marsden
;assert CORESIZE==8000

PSTATE    EQU 666 ; pspace location containing current state
STATES EQU 6      ; maximum number of states (for brainwash protection)

;NOTE: state values go from 0 to STATES-1

      dat    0,init_state-state
in    dat    0,loss_table-state
      dat    0,win_table-state
      dat    0,tie_table-state

think ldp.a  #0,in              ; get input value
load  ldp.a  #PSTATE,state      ; load old state
      mod.a  #STATES,state      ; brainwash protection *in,state          ; select correct state table
store stp.a  *state,load        ; store new state
state jmp    @0                 ; jump to warrior code

      ; You could also delete the #PSTATE value
      ; if you wish (or the stp.a instruction).

c1 EQU stone_boot
c2 EQU paper_boot

  dat 1,c1
  dat 2,c1
  dat 3,c2
  dat 4,c2
  dat 5,c2
  dat 0,c1

  dat 0,c1
  dat 0,c1
  dat 0,c1
  dat 3,c2
  dat 3,c2
  dat 3,c2

  dat 0,c1
  dat 0,c1
init_state dat 0,c1
  dat 3,c2
  dat 3,c2
  dat 3,c2

END think
The P^2b switcher is based upon the ideas presented already in the P^2 switcher. P^2 has a major weakness: it needs a very large table to implement even simple state transitions. Generally, the state transition rules we use for one component can be used with another - but P^2 doesn't take this into account. P^2b attempts to use this transition similarity property. P^2b takes 7 instructions to execute (as opposed to 6 for P^2). In order to keep the code size to a minimum, it is required that state 0 and state 1 execute the same warrior. Apart from this restriction, it should be possible to implement any state transition table that was possible in P^2. However, keep in mind that the initial state is a little bit tricky to set up properly.
;name P^2b
;author Anton Marsden
;strategy P-switcher
;assert CORESIZE==8000

PSTATE  EQU 666                 ; pspace location containing current state
STATES  EQU 9                   ; number of states

T1 EQU (t1-in)
T2 EQU (t2-in)
T3 EQU (t2-in)

think   ldp.a  #0,in            ; get input value
load    ldp.a  #PSTATE,state    ; load old state
        mod.a  #STATES,state    ; brainwash protection *state,in        ; select correct state table
in 0,state
store   stp.a  state,load       ; store new state

state   jmp    }0,T1            ; jump to warrior code
        dat    w1,T2            ; if state.A == 0, the A field of this 
                                ; instruction will be jumped to.
        dat    w1,T3
        dat    w2,T1
        dat    w2,T2
        dat    w2,T3
        dat    w3,T1
        dat    w3,T2
        dat    w3,T3

t2      dat    w1,1     ; loss - add 1 to the state
        dat    0,-1     ; win - subtract 1 from the state
        dat    0,0      ; tie - leave the state as it is
t1      dat    0,1      ; loss - add 1 to the state
        dat    0,0      ; win - leave the state as it is
        dat    0,0      ; tie - leave the state as it is

;Warrior code

w1:     jmp    #1,1
w2:     jmp    #2,2
w3:     jmp    #3,3

END think
T1, T2 and T3 correspond to different states a component can be in. In this case, T2 and T3 use the same win/loss/tie table to switch states.

Another P^2 variation was first introduced in Ian Oversby's Nomolos. It uses just one table instead of 3 as in the case of the original P^2 switcher. The advantage is that the switching pattern is easier to modify, because it is clearly seen in the code.
init	dat.f	#0,	#0
loss	dat.f	#0,	#1
win	dat.f	#0,	#-1
tie	dat.f	#0,	#0

pcode	ldp.a	#0,	loss
	ldp.a	#plc,	table	*loss,	table
	mod.a	#19,	table	; (-1 mod 19) = 0 :-)
	stp.ab	*table,	#plc
table	jmp.b	@0,	w1	; State 0

	dat.f	#1,	w1	; Stone
	dat.f	#2,	w1
	dat.f	#3,	w1
	dat.f	#4,	w1

	dat.f	#6,	w2	; Must always waste one line :-(
	dat.f	#6,	w2	; One-shot
	dat.f	#7,	w2
	dat.f	#8,	w2
	dat.f	#9,	w2
	dat.f	#10,	w2

	dat.f	#12,	w3	; Must always waste one line :-(
	dat.f	#12,	w3	; Mini-HSA
	dat.f	#13,	w3
	dat.f	#14,	w3
	dat.f	#15,	w3

	dat.f	#17,	w4	; Paper
	dat.f	#17,	w4

	dat.f	#17,	w4	; Safety net
1.6 The P^3 switcher

Redcoders often design their pspace-based switching routines in the form of a state-transition diagram as already shown in Chapter 1.5:
       .->------.                   L = loss
       |        |                   W = win
   W,T |      __V______             T = tie
  (w0) |     |         |<---.
       `--<--| State 0 |     \      w0 = strategy A (e.g. scissors)
             `---------'      `.    w1 = strategy B (e.g. paper)
                  | L (w1)     |
                  |            |
              ____V____        |
       .---->|         |       |
       |     | State 1 |---->--'
   W,T |     `---------'
  (w1) |       |         L (w0)
Each round gives the program a new chance to change its mode of operation, depending on the previous state and the result of the previous battle. In addition, there is some action (or strategy) associated with each transition between states:
   State  Input  Next  Action

   0      L      1     w1
   0      W      0     w0
   0      T      0     w0
   1      L      0     w0
   1      W      1     w1
   1      T      1     w1
Note that this pspacer would run w0 upon all transitions to state 0, and that it would run w1 upon all transitions to state 1. Most interesting P^2 switchers execute the same program upon all transitions to the same state. Can you see why? If we assume that all transitions to a particular state run the same warrior, then we no longer need to specify an action for each transition; we simply specify an action for each state. That leads to a more compact representation of the table:
   State  Action  L  W  T

   0      w0      1  0  0
   1      w1      0  1  1
The corresponding diagram becomes:
       .->------.                   L = loss
       |        |                   W = win
   W,T |      __V______             T = tie
       |     |         |
       |     | State 0 |<---.
       `--<--|   (w0)  |     \      w0 = strategy A (e.g. scissors)
             `---------'      `.    w1 = strategy B (e.g. paper)
                  | L          |
                  |            |
              ____V____        |
       .---->|         |       |
       |     | State 1 |       |
       |     |   (w1)  |---->--'
   W,T |     `---------'
       |       |         L
The switcher for Electric Head has that property. Specifically, Electric Head implemented the following state-transition table:
   State  Action  L  W  T

   0      w0      1  0  1    <---- initial state
   1      w0      2  0  2
   2      w0      3  0  3
   3      w1      4  3  3
   4      w1      5  3  4
   5      w1      6  4  5
   6      w2      7  6  0
   7      w3      8  7  7
   8      w3      9  7  8
   9      w3      3  8  9
A P^3 switcher can implement the same table in less time and in less space than Electric Head's original code:
think ldp.a      #0, in
      ldp.a #PSTATE, table   *in, table
      stp.b  *table, #PSTATE

table jmp }0, 441 ; = (44*10) + 1 = (49*9) + 0 = (40*11) + 1 ;initial state
      dat w0, 882 ; = (88*10) + 2 = (98*9) + 0 = (80*11) + 2
      dat w0, 333 ; = (33*10) + 3 = (37*9) + 0 = (30*11) + 3
      dat w1, 894 ; = (89*10) + 4 = (99*9) + 3 = (81*11) + 3
      dat w1, 345 ; = (34*10) + 5 = (38*9) + 3 = (31*11) + 4
      dat w1, 346 ; = (34*10) + 6 = (38*9) + 4 = (31*11) + 5
      dat w2, 627 ; = (62*10) + 7 = (69*9) + 6 = (57*11) + 0
      dat w3, 898 ; = (89*10) + 8 = (99*9) + 7 = (81*11) + 7
      dat w3, 349 ; = (34*10) + 9 = (38*9) + 7 = (31*11) + 8
      dat w3,  53 ; = ( 5*10) + 3 = ( 5*9) + 8 = ( 4*11) + 9

      dat w3, 349 ; = (34*10) + 9 = (38*9) + 7 = (31*11) + 8 ;unreachable

in    dat 0, 10  ; must have non-zero b-field in the previous cell
      spl 1,  9
      spl 1, 11
For example, after a transition to state 4, the pspacer will remember the code 345. In the case of a loss, the next state will be decoded by 345 mod 10; after a win, the next state would be 345 mod 9; after a tie, the next state would be 345 mod 11. Each of these decodings gives the proper transition from one state (4) to the next (5, 3, or 4). Note that we cannot specify different warriors for two different transitions to the same state. However, as we have seen before, that is a very soft restriction. Another restriction is that the 0th state must use the same warrior as the 1st state. Now we show how to construct the codes that are used in the table. This requires some abstract algebra, but the results will be summarized below. First, we must introduce some notation. Operators listed from highest to lowest priority:
   gcd(a,b)   the greatest common divisor of a and b
   a =< b     a is less than or equal to b
   a >= b     a is greater than or equal to b
   a and b    logical and
   a -> b     a implies b (if a is true, then b is true)
"mod" will no longer refer to the MOD operation from redcode. Instead:
   a mod n            a modulo n
   a *' b (mod n)     the product of a and b modulo n
   a +' b (mod n)     the sum of a and b modulo n
   a == b (mod n)     a is congruent to b modulo n (also: n divides a-b)
I will borrow the % operation from the C language, which is similar to the MOD from redcode:
   a % b      the unique c such that 0 =< c < b  and  c == a (mod b)
One of the theoretical results requires some more notation:
   Z          the set of all integers
   Z mod n    the set of all congruence classes modulo n
   G x H      the group formed by the Cartesian product of groups G and H
Modular arithmetic should look familiar to Redcoders, since all arithmetic on the standard hills is done in Z mod 8000. Examples:
   1 == 4 (mod 3)  ->  3*k = 1-4 for some integer k

   2 *' 3 == 1 (mod 5)

   (2 * 3) % 5 = 1

   gcd(3,3) = 3
   gcd(2,3) = 1
   gcd(2,4) = 2
We begin with a semi-formal description of the switcher:
   |  P^3 switcher                                        |
   |  r,s,t are positive integers                         |
   |                                                      |
   |  For each state in the table, there are non-negative |
   |  integers p,a,b,c such that p < CoreSize and         |
   |                                                      |
   |  a < r  and  p == a (mod r)  and                     |
   |  b < s  and  p == b (mod s)  and                     |
   |  c < t  and  p == c (mod t)                          |
There will be different p,a,b,c for each state. However, since each state may be considered independently, we won't bother to put subscripts on those variables. In the example above, r = 10, s = 9, and t = 11. Depending on the state, p could be 441, 882, 333, ..., or 53. In the 0th state, a = 1, b = 0, and c = 1. To guarantee that we can compute any transition table, we need to choose r,s,t such that for all possible a,b,c, there is a value for p such that the above properties hold. In that case, there are r*s*t possible triplets (a,b,c). It would be nice if every p between 0 and r*s*t-1 resulted in a different triplet (p mod r, p mod s, p mod t). Then every possible triplet (a,b,c) could be computed by choosing one of the first r*s*t non-negative integers for p. That would guarantee that we can compute any transition table if r*s*t does not exceed CoreSize. Using some results from group theory, we can make those guarantees true if gcd(r,s)=1 and gcd(r,t)=1 and gcd(s,t)=1. For ambitious mathematicians who want to prove that statement, two of the relevant theorems are:
   Z mod m x Z mod n is cyclic if and only if gcd(m,n)=1.

   Every finite cyclic group of order n is isomorphic to Z mod n.
To express any possible transition table with n states, we must be able to compute all triples (a,b,c) such that a<n, b<n, and c<n. Therefore, we should choose r,s,t such that r >= n, s >= n, and t >= n. If we do that, then we can assume that either r>n, s>n, or t>n (otherwise, either n=1, gcd(r,s)>1, gcd(r,t)>1, or gcd(s,t)>1). The result is that we will have some extra states which are not reachable from the transitions specified by our n-state table. We should specify the transitions for those unreachable states just in case a lucky opponent brainwashes the switcher. That is why we added an extra table entry in the example from Electric Head. We chose the same transitions as state 8 since we hypothesized that Carbonite was the most brainwash-resistant component. Since the size of the table will be the maximum value of r,s,t, we should choose r,s,t to minimize max{r,s,t}. Since r,s,t are at least as large as n, we can't achieve any result that is better than s=n, r=n+1, and t=n+2. In fact, if n is odd, then that particular assignment does satisfy all of the constraints. If n is even, then gcd(r,t) = gcd(n,n+2) = 2, so we are forced to look for larger r,s, or t. In that case, we can simply add one state to the table (to make n odd) to achieve the next best possible result. That gives us the following rule for selecting r,s,t for an n-state table:
   If n is odd, then m = n+1.
   If n is even, then m = n+2.

   r = m, s = m-1, t = m+1.
Now that we have fixed r,s,t, we can compute the value p for each state given a,b,c. We use the following result:
 |  Theorem  |
Given a,b,c,m,n,i,j,p such that
   m = n+1 if n is odd
   m = n+2 if n is even

   0 =< a < m    and
   0 =< b < m-1  and
   0 =< c < m+1

   i = b+c-2*a

   j = (i+1+m)/2 if i is odd
   j = i/2       if i is even

   p = (j*(m-1)+b-a)*m+a
   p == a (mod m)
   p == b (mod m-1)
   p == c (mod m+1)

 |   Proof   |
Suppose that i is odd.
   j = (i+1+m)/2
   ->    j = (i+1)/2 + m/2        (since m is even and i is odd)
   ->  2*j = (i+1) + m
   ->  2*j == i+1+m (mod m+1)
   ->  2*j == i +' m+1 (mod m+1)
   ->  2*j == i +' 0 (mod m+1)
   ->  2*j == i (mod m+1)
Suppose that i is even.
   j = i/2  ->  2*j = i  ->  2*j == i (mod m+1)
In either case, 2*j == i (mod m+1).
   2*j == i (mod m+1)
   ->          2*j == b+c-2*a (mod m+1)
   ->       j *' 2 == b+c-2*a (mod m+1)
   ->    j *' (-2) == 2*a-c-b (mod m+1)
   -> j *' (m+1-2) == 2*a-c-b (mod m+1)
   ->   j *' (m-1) == 2*a-c-b (mod m+1)
   ->      j*(m-1) == 2*a-c-b (mod m+1)
   ->  j*(m-1)+b-a == a-c (mod m+1)
Substitute k for j*(m-1)+b-a.
   k == a-c (mod m+1)
   ->            k*(-1) == c-a (mod m+1)
   ->         k *' (-1) == c-a (mod m+1)
   ->      k *' (m+1-1) == c-a (mod m+1)
   ->            k *' m == c-a (mod m+1)
   ->               k*m == c-a (mod m+1)
   ->             k*m+a == c (mod m+1)
   -> (j*(m-1)+b-a)*m+a == c (mod m+1)
   ->                 p == c (mod m+1)
This proves 1/3 of the theorem. Furthermore,
   k = j*(m-1)+b-a  ->  j*(m-1) = k-(b-a)
   ->                 k == b-a (mod m-1)
   ->               k*1 == b-a (mod m-1)
   ->            k *' 1 == b-a (mod m-1)
   ->      k *' (m-1+1) == b-a (mod m-1)
   ->            k *' m == b-a (mod m-1)
   ->               k*m == b-a (mod m-1)
   ->             k*m+a == b (mod m-1)
   -> (j*(m-1)+b-a)*m+a == b (mod m-1)
   ->                 p == b (mod m-1)
For the last part, we simply observe:
   p = k*m+a  ->  k*m = p-a  ->  p == a (mod m)
Thus concluding the proof of this theorem.
 |  Summary  |
In summary, to construct a table of n states:
   m = n+1 if n is odd
   m = n+2 if n is even

   For each state, choose integers a,b,c such that
      0 =< a < m    and
      0 =< b < m-1  and
      0 =< c < m+1

   i = b+c-2*a

   j = (i+1+m)/2  if i is odd
   j = i/2        if i is even

   p = (j*(m-1)+b-a)*m+a % m*(m-1)*(m+1)

   p % m   = a
   p % m-1 = b
   p % m+1 = c

 |  Example  |
For Electric Head, the number of states is n = 10. Normally, m = 12 to guarantee that all possible tables can be implemented. However, in this case, m = 10 will work if we choose to decode with m on loss, m-1 on win, and m+1 on tie. This implementation works because we know the particular a,b,c values for each state and they are within the bounds:
   0 =< a < 10  (a is the transition on loss)
   0 =< b <  9  (b is the transition on win)
   0 =< c < 11  (c is the transition on tie)
Now we must determine the p value for each state. For state 0,
   a = 1
   b = 0
   c = 1

   i = 0+1-2*1 = -1

   j = (-1+1+10)/2 = 5  (since i was odd)

   p = (5*(10-1)+0-1)*10+1 % 10*(10-1)*(10+1)
     =        (5*9-1)*10+1 % 10*9*11
     =                 441 % 990
So p = 441 is the value that will actually appear in the redcode. We verify that it is indeed correct:
   441 % 10 = 1
   441 %  9 = 0
   441 % 11 = 1
Note that m*(m-1)*(m+1) = 990 does not exceed CoreSize = 8000. Since 20*19*21 = 7980, any table with 19 states can be implemented in a CoreSize of 8000. Of course, larger tables are possible with tricky coding.

1.7 Addaptive switcher

The first apperance of an addaptive switcher was in Neogryzor's P-Key. His entry for the Redcoders Frenzy Round 7. In this round one had to construct a p-warrior from three given components (Cloner II, CLP and Frontwards), which cannot be modified. The Redcoding required was *only* to create the p-brain. Each entry was fight each other entry for 2000 rounds in a round robin tournament without self-fights. The scoring was one point for a win and no points for a tie or loss. The scoring encourages offensive switchers, so tieing was not an acceptable result.

The main idea was to change the switching method if it was not successful. Notice that a single one switches always in the same way, so it can be predictable. And all opponents in that round were p-warriors as well.

-loss -> switch to the next -> tie : The switch has not been successful so it must be modified.

It is necessary to define three variables wich contain the change of the strategy after every result: 'w','l' and 't'.
    -on loss:       strat=strat+l

    -on tie:        strat=strat+t

    -on win:        strat=strat+w
The P-brain starts working like a single switcher:
    l=1     -> next strat if loss
    t=2     -> previous strat if tie
    w=0     -> same strat if win
But these constants will be modified if they are not effective:
result two rounds before        last result     action
------------------------        -----------     ------

        tie                     tie             t=t+2   (same as -1)
        loss                    tie             l=l+2
        win                     tie             w=w+2

        tie                     loss            t=t+1
        loss                    loss            l=l+1
        win                     loss            w=w+1

        xxx                     win             no changes, all right
It is recommended to use MOD #3,X after this changes to make sure the constants are between 0 and 2, for brainwash detection can be done and so the trace is easier, but I think this is not a real necessity.

The working order is:
    1- adjust constants depending on the last result and two rounds before

2- Switch depending on the last result, using the modified constants

As you see, it is necesary to save into the p-space:

    - the last strategy selected

    - result two round before

    - l, w and t
The variables are stored in the p-space in this order just to make the code a little shorter:
        D_STRAT EQU 249
        D_PREV EQU 250
        D_LOS EQU D_PREV+1
        D_WIN EQU D_PREV+2
        D_TIE EQU D_PREV+3
When working with the previous results we find a problem: in the first round all variables are 0. Just checking the location 0 of the p-space seems to be the solution. If it is = -1, the values are saved in the p-space, but the problem doesn't ends here. The last result value is still -1. It can be incremented using '>', but remember that 0 means a loss in the previous fight and the switcher will change the 'l' constant, (the result two round before is 0). The initial value of 'l' must be 0 because it will be changed to 1 in the first round. An easier solution is to add 2 to the previous result, so it will be = 1 and the initial values will not be changed, but it needs one instruction more. P-key uses the first method, but it took a lot of tweaking because some instructions needed to be modified:
RESULT: LDP.B 1,#0              ;if RESULT==-1, the JMZ line detects a
DAT 0,0
        JMZ.F INIT,>RESULT      ;if RESULT==-1 then = 0 and jumps, else it is
                                ;              decremented later

(INIT loads the default values assuming it is the first round.)
Well, the most important thing is already done. Now one have to explain some important details:

The strategies are identified by comparison:
*       0 < strat < 2667        :Cloner II      (1000 by default)
*       2667 <= strat < 5334    :CLP
*       5334 <= strat < 0       :Frontwards
The switch is made easily just adding 2667 or 2*2667, (CORESIZE/3), to the previous strategy. Obviously CORESIZE%3 is not = 0, so after 3 switches, the strat has been incremented or decremented. The default starting value is 1000 for Cloner, so after 3*(1000) backward switches the strategy number gets wrong, (in the worst case). That's 3000 consecutive switches, but only 2000 rounds were fought, so there was no problem.

Though it can be done by defining the strategies as 0, 1 and 2. The switch can be done by adding 1 or 2 and then using MOD #3 to ensure the strategy value is not greater than 2, so one more instruction is needed. Otherwise the selection and the strategy launch is easier and shorter.

Both methods were tested and the first one has shown to score better. I believe this probably happens because the first one is more robust when brainwashed with 0, but it depends on the opponent's brainwash.

P-key has partial brainwash detection capability by finding unexpected values in a variable. If the previous result is > 3, that means that the warrior has been brainwashed, and the default values are loaded into the p-space like in the first round. Though when it has been brainwashed with a number between 0 and 3, (most opponents were using 0), it is not detected. Fortunately the p-brain seems to be not so vulnerable after a brainwash.
        JMZ.F INIT,>RESULT      ;first round check
        SLT.AB #3,PREV          ;brainwash detection.There is a little
bug. It should be #2 not 3
        JMP CHLS,<RESULT        ;jumps to the switcher
INIT:   STP.AB #2,#D_TIE        ;loads default values into p-space
        STP.AB #0,#D_WIN
        STP.AB #0,#D_LOS        ;notice that this value will change to 1 in the first round
        MOV #1000,STRAT         ;STRAT=ClonerII, but it will change to CLP in the first round
In the first round RESULT=-1. It is incremented by the check. Then RESULT=0. It means that the last result is loss, so the strategy will change from Cloner II to CLP. The last detail is the selected starting strategy. Using Frontwards as first strat is riskly because it scores 50% against itself and it is beaten, (and brainwashed) by CLP. Cloner II beats CLP but it can be brainwashed too, (it's a paper). Being brainwashed at the beginning is a very bad start. The warrior could get stuck executing the default starting strategy, and being brainwashed repeatedly. If a warrior is caught in this situation, it is condemned. CLP looks to be the best default strat when starting and P-key uses it. Anyway, Cloner II is not so bad, and Frontwards is not recommended.

Finally I can say there are 4 characteristics wich make P-key a successful switcher:
    * It's adaptative switching, which allows it to beat any single

    * The default selected data.

    * Its switching engine, which showed to be less vulnerable than
      others when brainwashed.

    * The brainwash detection.
As you can see, the switch only depends on previous results, not strategies, so P-key is not so effective against some table based or semi-random switchers.
;name   P-Key
;author G.Labarga
;assert 1

;strategy       learning switcher
;strategy       It should defeat any single switcher
;strategy       Redcoders Frenzy, round 7 winner, Feb. 22th 2003

        STEP EQU 2667           ;switching step
        D_STRAT EQU 249         ;variables are placed in consecutive
p-space locations
        D_PREV EQU 250
        D_LOS EQU D_PREV+1      ;the 'l', 'w' and 't' variables
        D_WIN EQU D_PREV+2
        D_TIE EQU D_PREV+3

RESULT: LDP.B 1,#0              ;<- Last round result
STRAT:  LDP.AB #D_STRAT,#0      ;<- Last strategy
PREV:   LDP.AB #D_PREV,#0       ;<- result two rounds before. This
comment was wrong in the submited version
        JMZ.F INIT,>RESULT      ;First round check, RESULT is
        SLT.AB #3,PREV          ;brainwash detection. Very slight bug:
should be #2 instead of #3
        JMP CHLS,<RESULT        ;jumps to CHLS and tweaks RESULT if it
is not the first round
INIT:   STP.AB #2,#D_TIE        ;first round or brainwashed: Load the
default values
        STP.AB #0,#D_WIN
        STP.AB #0,#D_LOS
        MOV #1000,STRAT         ;CLP will be the first executed

CHLS:   JMN CHWI,RESULT         ;Check the last round result
        MOV.A #1,AWLT           ;loss-> last decision,(w/l/t) ++
        JMP ACT                 ;win: w,l,t are not modified. jumps to
the switcher
        MOV.A #2,AWLT           ;tie-> last decision,(w/l/t) --

ADJ:    ADD.BA PREV,1           ;Adjust w/l/t
WLT:    LDP.AB #D_PREV+1,#0     ;WLT: the variable that must be
AWLT:   ADD.AB #0,WLT           ;increment or decrement,(-1=+2)
        MOD.AB #3,WLT           ;reduce switch to mod3
        STP.BA WLT,WLT          ;and save the modified variable
ACT:    ADD.BA RESULT,1         ; Now switches the strategy depending
last result (W/L/T)
ACTS:   LDP.AB #D_PREV+1,#0     ;Loads the variable, (w, l or t)
        MUL.AB #STEP,ACTS       ;strat = strat + 2667 * w/l/t

SAVE:   STP.B RESULT,#D_PREV    ;last result saved in the p-space
        STP.B STRAT,#D_STRAT    ;the selected strat is saved in the

SLCT:   SLT #STEP,STRAT         ;Launches the selected strat
        JMP SRC                 ;Cloner II
        SLT #2*STEP,STRAT
        JMP slDodger            ;CLP
        JMP boot                ;Frontwards

; *** components size: 61 ins; P-brain: 32
But nevertheless keep in mind that adaptive switchers work best with specialist components, which beat the opponents components by a good percentage.

2002-2005 Logo C. Schmidt
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