RE: memoizing function again
- To: mathgroup at smc.vnet.net
- Subject: [mg32510] RE: [mg32495] memoizing function again
- From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.de>
- Date: Thu, 24 Jan 2002 05:21:04 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
> -----Original Message----- > From: Erich Neuwirth [mailto:erich.neuwirth at univie.ac.at] To: mathgroup at smc.vnet.net > Sent: Wednesday, January 23, 2002 7:00 AM > To: mathgroup at smc.vnet.net > Subject: [mg32510] [mg32495] memoizing function again > > > i have the following function > > f[x_] /; x <= 2 := f[x] = 1 > f[x_] /; x > 2 := f[x] = f[x - 1] + f[x - 2] > > > it remembers what it already calculated > i want to be able to throw away rules with values > between calculations > > In[3]=DownValues[f] > produces > > Out[3]={HoldPattern[f[x_]/;x\[LessEqual]2]\[RuleDelayed](f[x]=1), > HoldPattern[f[x_]/;x>2]\[RuleDelayed](f[x]=f[x-1]+f[x-2])} > > > after f[4] > > we have > > In[5]:= > DownValues[f] > > Out[5]= > {HoldPattern[f[1]]\[RuleDelayed]1,HoldPattern[f[2]]\[RuleDelayed]1, > HoldPattern[f[3]]\[RuleDelayed]2,HoldPattern[f[4]]\[RuleDelayed]3, > HoldPattern[f[x_]/;x\[LessEqual]2]\[RuleDelayed](f[x]=1), > HoldPattern[f[x_]/;x>2]\[RuleDelayed](f[x]=f[x-1]+f[x-2])} > > > \[RuleDelayed] is ascii for :>, i think > > so > > DownValues[f] = Take[DownValues[f], -2] > > removes all the rules giving calculated values > but i would like to throw away the rules following the pattern > > HoldPattern[f[x_Integer]:>y_Integer > > but i have not been able wo write an expression using Cases and a > pattern > which gets rid of the rules i want to get rid of > > > > > > > > -- > Erich Neuwirth, Computer Supported Didactics Working Group > Visit our SunSITE at http://sunsite.univie.ac.at > Phone: +43-1-4277-38624 Fax: +43-1-4277-9386 > Erich, why not just Remove[f] and execute its definition again? However you are free to do funny things: c1 = 0; c2 = 0; Remove[f]; f[0] = 1; f[1] = 1; f[n_] := (++c1; Unevaluated[f[n] = Unevaluated[++c2; rhs]] /. rhs :> RuleCondition[f[n - 1] + f[n - 2]]) ?f (c1 = 0; c2 = 0; {f[#], c1, c2}) & /@ Range[10, 1, -1] ?f Scan[Unset, Take[First /@ DownValues[f], {3, -2}]] ?f (c1 = 0; c2 = 0; {f[#], c1, c2}) & /@ Range[10] ?f To go into your question for the pattern, look Remove[f]; f[0] = 1; f[1] = 1; f[n_] := f[n] = f[n - 1] + f[n - 2] f[10] ?f Now you may reset the definitions for f: DownValues[f] = DeleteCases[DownValues[f], rule_ /; IntegerQ[rule[[1, 1, 1]]] && (rule[[1, 1, 1]] > 1) && IntegerQ[rule[[2]]] ] You must be careful as not to execute the recursive definition for f when doing the test. This is achieved here by the non-strict evaluation of And. You also must be careful if you try to use names for parts of the rule: DownValues[f] = Select[DownValues[f], Apply[Function[{arg, rhs}, ! (IntegerQ[arg] && (arg > 1) && IntegerQ[rhs]), {HoldAll}], Extract[#, {{1, 1, 1}, {2}}, Unevaluated]] &] HoldAll and Unevaluated prevent the evaluation of rhs before the test IntegerQ[arg] has been made. If you have a recursive definition for an integer argument (not a pattern), which is possible in principle -- e.g. in my first example, if I had not used RuleCondition -- then more work is needed. See for example Map[ (Function[{rhs}, Replace[Unevaluated[rhs], {_Integer -> False, _ -> True}], {HoldAll}] @@ Extract[#, {2}, Hold] &) , DownValues[f]] or else Function[{rhs}, Replace[Unevaluated[rhs], {_Integer -> False, _ -> True}], {HoldAll}] /@ (Extract[#, {2}, Unevaluated] &) /@ DownValues[f] {False, False, False, False, False, False, False, False, False, False, False, True} -- Hartmut Wolf