Advanced Programming Handout 5. Enter Okasaki. Persistent vs. Ephemeral. Functional Queues. Simple Example. Persistent vs.


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1 Avne Progrmming Hnout 5 Purel Funtionl Dt Strutures: A Cse Stu in Funtionl Progrmming Persistent vs. Ephemerl An ephemerl t struture is one for whih onl one version is ville t time: fter n upte opertion, the struture s it eiste efore the upte is lost. A persistent struture is one where multiple versions re simultneousl essile: fter n upte, oth ol n new versions n e use. Persistent vs. Ephemerl In impertive lnguges, most t strutures re ephemerl. It is generll epte tht persistent vrints, when the re possile t ll, will e more omple to oe n smptotill less effiient. In purel funtionl lnguges like Hskell, ll t strutures re persistent! Sine there is no ssignment, there is no w to estro ol informtion. When we re one with it, we just rop it on the floor n let the grge olletor tke re of it. So one might worr tht effiient t strutures might e hr or even impossile to oe in Hskell. Enter Okski Interestingl, it turns out tht mn ommon t strutures hve purel funtionl vrints tht re es to unerstn n hve el the sme smptoti effiien s their impertive, ephemerl vrints. These strutures hve een eplore in n influentil ook, Purel Funtionl Dt Strutures, Chris Okski, n in long series of reserh ppers Okski n others. Simple Emple To get strte, let s tke quite simple trik tht ws known to funtionl progrmmers long efore Okski... Funtionl Queues A queue (of vlues of tpe ) is struture supporting the following opertions: enqueue :: > Queue > Queue equeue :: Queue > (, Queue ) We epet tht eh opertion shoul run in O(1)  i.e., onstnt  time, no mtter the sie of the queue.
2 Nive Implementtion A queue of vlues of tpe is just list of s: tpe Queue = [] To equeue the first element of the queue, use the he n til opertors on lists: equeue q = (he q, til q) To enqueue n element, ppen it to the en of the list: enqueue e q = q ++ [e] Nive Implementtion This works, ut the effiien of enqueue is isppointing: eh enqueue requires O(n) ons opertions! Of ourse, ons opertions re not the onl things tht tke time! But ounting just onses tull iels prett goo estimte of the (smptoti) wlllok effiien of progrms. Here, it is ertinl ler tht the rel effiien n e no etter thn O(n). Better Implementtion Better Implementtion Ie: Represent queue using two lists: 1. the front prt of the queue 2. the k prt of the queue in reverse orer E.g.: ([1,2,3],[7,6,5,]) represents the queue with elements 1,2,3,,5,6,7 ([],[3,2,1]) n ([1,2,3],[]) oth represent the queue with elements 1,2,3 To enqueue n element, just ons it onto the k list. To equeue n element, just remove it from the front list......unless the front list is empt, in whih se we reverse the k list n use it s the front list from now on. Better Implementtion t Queue = Queue [] [] enqueue :: > Queue > Queue enqueue e (Queue front k) = Queue front (e:k) Effiien Intuition: equeue m require O(n) ons opertions (to reverse the k list), ut this nnot hppen too often. equeue :: Queue > (, Queue ) equeue (Queue (e:front) k) = (e, (Queue front k)) equeue (Queue [] k) = equeue (Queue (reverse k) [])
3 Effiien In more etil: Note tht eh element n prtiipte in t most one list reversl uring its lifetime in the queue. When n element is enqueue, we n hrge two tokens for two ons opertions. One of these is performe immeitel; the other we put in the nk. At ever moment, the numer of tokens in the nk is equl to the length of the k list. When we fin we nee to reverse the k list to perform equeue, we will lws hve just enough tokens in the nk to p for ll of the ons opertions involve. Effiien So we n s tht the mortie ost of eh enqueue opertion is two onses. The mortie ost of eh equeue is ero (i.e., no onses  just some pointer mnipultion). Cvet: This effiien rgument is somewht rough n re  it is intene just to give n intuition for wht is going on. Mking everthing preise requires more work, espeill in l lnguge. Morl We n implement persistent queue t struture whose opertions hve the sme (smptoti, mortie) effiien s the stnr (ephemerl, oulepointer) impertive implementtion. Binr Serh Trees Suppose we wnt to implement tpe Set supporting the following opertions: empt :: Set memer :: Or => > Set > Bool insert :: Or => > Set > Set One ver simple implementtion for sets is in terms of inr serh trees... Binr Serh Trees t Set = E  empt T (Set ) (Set )  nonempt empt = E memer E = Flse memer (T ) < = memer > = memer True = True insert E = T E E insert (T ) < = T (insert ) > = T (insert ) True = T Quik Digression on Ptterns The insert funtion is little hr to re euse it is not immeitel ovious tht the phrse T in the o just mens return the input. insert E = T E E insert (T ) < = T (insert ) > = T (insert ) True = T
4 Quik Digression on Ptterns Hskell for suh situtions. insert E = T E E insert ) < = T (insert ) > = T (insert ) True = t The pttern ) mens hek tht the input vlue ws onstrute with T, in its prts to,, n, n itionll let t stn for the whole input vlue in wht follows... Blne Trees If our sets grow lrge, we m fin tht the simple inr tree implementtion is not fst enough: in the worse se, eh insert or memer opertion m tke O(n) time! We n o muh etter keeping the trees lne. There re mn ws of oing this. Let s look t one firl simple (ut still ver fst) one tht ou hve prol seen efore in n impertive setting: relk trees. ReBlk Trees Invrints A relk tree is inr serh tree where ever noe is itionll mrke with olor (re or lk) n in whih the following invrints re mintine... The empt noes t the leves re onsiere lk. Invrints The root is lws lk. Invrints From eh noe, ever pth to lef hs the sme numer of lk noes. Re noes hve lk hilren
5 Invrints Together, these invrints impl tht ever relk tree is pproimtel lne, in the sense tht the longest pth to n empt noe is no more thn twie the length of the shortest. From this, it follows tht ll opertions will run in O(log2 n) time. Now let s look t the etils... Tpe Delrtion The t elrtion is strightforwr moifition of the one for unlne trees: t Color = R B t ReBlkSet = E T Color (ReBlkSet ) (ReBlkSet ) Memership The empt tree is the sme s efore. Memership testing requires just trivil hnge. pronoune where empt = E memer E = Flse memer (T _ ) < = memer > = memer True = True Insertion Insertion is implemente in terms of reursive uilir funtion ins, whih wlks own the tree until it either gets to n empt lef noe, in whih se it onstruts new (re) noe ontining the vlue eing inserte... ins E = T R E E Insertion is more interesting... Insertion... or isovers tht the vlue eing inserte is lre in the tree, in whih se it returns the input unhnge: ins olor ) < =... > =... The reursive ses re where the rel work hppens... Insertion In the reursive se, ins etermines whether the new vlue elongs in the left or right sutree, mkes reursive ll to insert it there, n reuils the urrent noe with the new sutree. ins olor ) < =... (T olor (ins ) ) > =... (T olor (ins ))
6 Insertion Before returning it, however, we m nee to relne to mintin the relk invrints. The oe to o this is enpsulte in helper funtion lne. ins olor ) < = lne (T olor (ins ) ) > = lne (T olor (ins )) Blning The ke insight in writing the lning funtion is tht we o not tr to relne s soon s we see re noe with re hil. Inste, we return this tree sis n wit until we re lle with the lk prent of this noe. I.e., the jo of the lne funtion is to relne trees with lkrere pth strting t the root. Sine the root hs two hilren n four grnhilren, there re four ws in whih suh pth n hppen. Blning Blning Blning Blning All tht remins is to turn these pitures into oe...
7 Blning lne (T B (T R (T R ) ) ) = T R (T B ) (T B ) lne (T B (T R (T R )) ) = T R (T B ) (T B ) lne (T B (T R (T R ) )) = T R (T B ) (T B ) lne (T B (T R (T R ))) = T R (T B ) (T B ) lne t = t One Finl Detil Sine we onl relne lk noes with re hilren n grnhilren, it is possile tht the ins funtion oul return re noe with re hil s its finl result. We n fi this foring the root noe of the returne tree to e lk, regrless of the olor returne ins. Finl Version insert t = mkerootblk (ins t) where ins E = T R E E ins olor ) < = lne (T olor (ins ) ) > = lne (T olor (ins )) mkerootblk (T _ ) = T B The Whole Bnn t Color = R B t ReBlkSet = E T Color (ReBlkSet ) (ReBlkSet ) empt = E memer E = Flse memer (T _ ) < = memer > = memer otherwise = True lne (T B (T R (T R ) ) ) = T R (T B ) (T B ) lne (T B (T R (T R )) ) = T R (T B ) (T B ) lne (T B (T R (T R ) )) = T R (T B ) (T B ) lne (T B (T R (T R ))) = T R (T B ) (T B ) lne t = t insert t = olorrootblk (ins t) where ins E = T R E E ins olor ) < = lne (T olor (ins ) ) > = lne (T olor (ins )) otherwise = s olorrootblk (T _ ) = T B For Comprison... /* This funtion n e lle onl if K2 hs left hil */ /* Perform rotte etween noe (K2) n its left hil */ /* Upte heights, then return new root */ stti Position SingleRotteWithLeft( Position K2 ) Position K1; K1 = K2>Left; K2>Left = K1>Right; K1>Right = K2; return K1; /* New root */ /* This funtion n e lle onl if K1 hs right hil */ /* Perform rotte etween noe (K1) n its right hil */ /* Upte heights, then return new root */ stti Position SingleRotteWithRight( Position K1 ) Position K2; K2 = K1>Right; K1>Right = K2>Left; K2>Left = K1; return K2; /* New root */ /* Perform rottion t noe X */ /* (whose prent is psse s prmeter) */ /* The hil is eue emining Item */ stti Position Rotte( ElementTpe Item, Position Prent ) if( Item < Prent>Element ) return Prent>Left = Item < Prent>Left>Element? SingleRotteWithLeft( Prent>Left ) : SingleRotteWithRight( Prent>Left ); else return Prent>Right = Item < Prent>Right>Element? SingleRotteWithLeft( Prent>Right ) : SingleRotteWithRight( Prent>Right ); stti Position X, P, GP, GGP; stti voi HnleReorient( ElementTpe Item, ReBlkTree T ) X>Color = Re; /* Do the olor flip */ X>Left>Color = Blk; X>Right>Color = Blk; if( P>Color == Re ) /* Hve to rotte */ GP>Color = Re; if( (Item < GP>Element)!= (Item < P>Element) ) P = Rotte( Item, GP ); /* Strt oule rotte */ X = Rotte( Item, GGP ); X>Color = Blk; T>Right>Color = Blk; /* Mke root lk */ ReBlkTree Insert( ElementTpe Item, ReBlkTree T ) X = P = GP = T; NullNoe>Element = Item; while( X>Element!= Item ) /* Desen own the tree */ GGP = GP; GP = P; P = X; if( Item < X>Element ) X = X>Left; else X = X>Right; if( X>Left>Color == Re && X>Right>Color == Re ) HnleReorient( Item, T ); if( X!= NullNoe ) return NullNoe; /* Duplite */ X = mllo( sieof( strut ReBlkNoe ) ); if( X == NULL ) FtlError( "Out of spe!!!" ); X>Element = Item; X>Left = X>Right = NullNoe; if( Item < P>Element ) /* Atth to its prent */ P>Left = X; else P>Right = X; HnleReorient( Item, T ); /* Color it re; me rotte */ return T;
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