Language agnostic XKCD中NP完全问题的求解
问题/漫画： 我不确定这是否是最好的方法，但以下是我到目前为止的想法。我使用的是CFML，但任何人都应该可以阅读Language agnostic XKCD中NP完全问题的求解,language-agnostic,np-complete,Language Agnostic,Np Complete,问题/漫画： 我不确定这是否是最好的方法，但以下是我到目前为止的想法。我使用的是CFML，但任何人都应该可以阅读 <cffunction name="testCombo" returntype="boolean"> <cfargument name="currentCombo" type="string" required="true" /> <cfargument name="currentTotal" type="numeric" requir
<cffunction name="testCombo" returntype="boolean">
<cfargument name="currentCombo" type="string" required="true" />
<cfargument name="currentTotal" type="numeric" required="true" />
<cfargument name="apps" type="array" required="true" />
<cfset var a = 0 />
<cfset var found = false />
<cfloop from="1" to="#arrayLen(arguments.apps)#" index="a">
<cfset arguments.currentCombo = listAppend(arguments.currentCombo, arguments.apps[a].name) />
<cfset arguments.currentTotal = arguments.currentTotal + arguments.apps[a].cost />
<cfif arguments.currentTotal eq 15.05>
<!--- print current combo --->
<cfoutput><strong>#arguments.currentCombo# = 15.05</strong></cfoutput><br />
<cfreturn true />
<cfelseif arguments.currentTotal gt 15.05>
<cfoutput>#arguments.currentCombo# > 15.05 (aborting)</cfoutput><br />
<cfreturn false />
<cfelse>
<!--- less than 15.05 --->
<cfoutput>#arguments.currentCombo# < 15.05 (traversing)</cfoutput><br />
<cfset found = testCombo(arguments.currentCombo, arguments.currentTotal, arguments.apps) />
</cfif>
</cfloop>
</cffunction>
<cfset mf = {name="Mixed Fruit", cost=2.15} />
<cfset ff = {name="French Fries", cost=2.75} />
<cfset ss = {name="side salad", cost=3.35} />
<cfset hw = {name="hot wings", cost=3.55} />
<cfset ms = {name="moz sticks", cost=4.20} />
<cfset sp = {name="sampler plate", cost=5.80} />
<cfset apps = [ mf, ff, ss, hw, ms, sp ] />
<cfloop from="1" to="6" index="b">
<cfoutput>#testCombo(apps[b].name, apps[b].cost, apps)#</cfoutput>
</cfloop>
#参数。currentCombo#=15.05
#arguments.currentCombo#>15.05（正在中止）
#参数.currentCombo#<15.05（遍历）
#testCombo（应用程序[b]。名称，应用程序[b]。成本，应用程序）#
是否有更好的算法得出正确的解决方案？我找到正确的解决方案了吗？请仔细阅读。事实上，我对算法进行了更多的重构。我遗漏了几个正确的组合，这是因为当成本超过15.05美元时我就回来了——我没有费心去检查我可以添加的其他（更便宜的）项目。这是我的新算法：
<cffunction name="testCombo" returntype="numeric">
<cfargument name="currentCombo" type="string" required="true" />
<cfargument name="currentTotal" type="numeric" required="true" />
<cfargument name="apps" type="array" required="true" />
<cfset var a = 0 />
<cfset var found = false />
<cfset var CC = "" />
<cfset var CT = 0 />
<cfset tries = tries + 1 />
<cfloop from="1" to="#arrayLen(arguments.apps)#" index="a">
<cfset combos = combos + 1 />
<cfset CC = listAppend(arguments.currentCombo, arguments.apps[a].name) />
<cfset CT = arguments.currentTotal + arguments.apps[a].cost />
<cfif CT eq 15.05>
<!--- print current combo --->
<cfoutput><strong>#CC# = 15.05</strong></cfoutput><br />
<cfreturn true />
<cfelseif CT gt 15.05>
<!--<cfoutput>#arguments.currentCombo# > 15.05 (aborting)</cfoutput><br />-->
<cfelse>
<!--- less than 15.50 --->
<!--<cfoutput>#arguments.currentCombo# < 15.05 (traversing)</cfoutput><br />-->
<cfset found = testCombo(CC, CT, arguments.apps) />
</cfif>
</cfloop>
<cfreturn found />
</cffunction>
<cfset mf = {name="Mixed Fruit", cost=2.15} />
<cfset ff = {name="French Fries", cost=2.75} />
<cfset ss = {name="side salad", cost=3.35} />
<cfset hw = {name="hot wings", cost=3.55} />
<cfset ms = {name="moz sticks", cost=4.20} />
<cfset sp = {name="sampler plate", cost=5.80} />
<cfset apps = [ mf, ff, ss, hw, ms, sp ] />
<cfset tries = 0 />
<cfset combos = 0 />
<cfoutput>
<cfloop from="1" to="6" index="b">
#testCombo(apps[b].name, apps[b].cost, apps)#
</cfloop>
<br />
tries: #tries#<br />
combos: #combos#
</cfoutput>
我认为这可能有所有正确的组合，但我的问题仍然存在：有更好的算法吗？您现在已经有了所有正确的组合，但您仍然需要检查更多的组合（结果显示的许多排列证明了这一点）。此外，您还忽略了最后一个达到15.05分的项目
function testCombo(minIndex, currentCombo, currentTotal){
var a = 0;
var CC = "";
var CT = 0;
var found = false;
tries += 1;
for (a=arguments.minIndex; a <= arrayLen(apps); a++){
combos += 1;
CC = listAppend(arguments.currentCombo, apps[a].name);
CT = arguments.currentTotal + apps[a].cost;
if (CT eq 15.05){
//print current combo
WriteOutput("<strong>#CC# = 15.05</strong><br />");
return(true);
}else if (CT gt 15.05){
//since we know the array is sorted by cost (asc),
//and we've already gone over the price limit,
//we can ignore anything else in the array
break;
}else{
//less than 15.50, try adding something else
found = testCombo(a, CC, CT);
}
}
return(found);
}
mf = {name="mixed fruit", cost=2.15};
ff = {name="french fries", cost=2.75};
ss = {name="side salad", cost=3.35};
hw = {name="hot wings", cost=3.55};
ms = {name="mozarella sticks", cost=4.20};
sp = {name="sampler plate", cost=5.80};
apps = [ mf, ff, ss, hw, ms, sp ];
tries = 0;
combos = 0;
testCombo(1, "", 0);
WriteOutput("<br />tries: #tries#<br />combos: #combos#");
<cfelse>
<!--- less than 15.50 --->
<!--<cfoutput>#arguments.currentCombo# < 15.05 (traversing)</cfoutput><br />-->
<cfset found = testCombo(CC, CT, arguments.apps) />
------- remove the item from the apps array that was just checked here ------
</cfif>
</cfloop>
MF (2.15)
MF (4.30)
MF (6.45) *
FF (7.05) X
SS (7.65) X
...
[MF removed for depth 2]
FF (4.90)
[checking MF now would be redundant since we checked MF/MF/FF previously]
FF (7.65) X
...
[FF removed for depth 2]
SS (5.50)
...
[MF removed for depth 1]
#灯
键入开胃菜={name:string；cost:int}
让菜单=[
{name=“fruit”；成本=215}
{name=“fries”；成本=275}
{name=“沙拉”；成本=335}
{name=“wings”；成本=355}
{name=“moz sticks”；成本=420}
{name=“sampler”；成本=580}
]
//选择：列表->列表->整型->列表
让rec选择剩余资金后选择的允许菜单=
如果剩余货币=0，则
//解决它，返回此解决方案
[pickedSoFar]
其他的
//还有更多的钱要花
[将允许的菜单与匹配]
|[]->yield！[]//没有更多项目可供选择，此分支没有解决方案
|项目：：休息->
if item.cost List.iter（有趣的解决方案->
解决方案|>List.iter（趣味项目->打印“%s”，项目名称）
打印fn“”
)
(*
2种解决方案：
水果，水果，水果，水果，水果，水果，水果，水果，
采样器，翅膀，翅膀，水果，
*)
！/usr/bin/perl
严格使用；
使用警告；
我的体重=（2.15,2.75,3.35,3.55,4.20,5.80）；
我的$total=0；
我的@order=（）；
迭代（$total，@order）；
次迭代
{
我的（$total，@order）=；
每个我的$w（@weights）
{
如果（$total+$w==15.05）
{
打印联接（“，”，（@order，$w）），“\n”；
}
如果（$total+$w<15.05）
{
迭代（$total+$w，（@order，$w））；
}
}
}
mf mf mf mf mf mf mf
mf hw hw sp
209
marco@unimatrix-01:~$./xkcd背包.pl
2.15,2.15,2.15,2.15,2.15,2.15,2.15,2.15
2.15,3.55,3.55,5.8
2.15,3.55,5.8,3.55
2.15,5.8,3.55,3.55
3.55,2.15,3.55,5.8
3.55,2.15,5.8,3.55
3.55,3.55,2.15,5.8
3.55,5.8,2.15,3.55
5.8,2.15,3.55,3.55
5.8, 3.55, 2.15, 3.55
function testCombo(minIndex, currentCombo, currentTotal){
var a = 0;
var CC = "";
var CT = 0;
var found = false;
tries += 1;
for (a=arguments.minIndex; a <= arrayLen(apps); a++){
combos += 1;
CC = listAppend(arguments.currentCombo, apps[a].name);
CT = arguments.currentTotal + apps[a].cost;
if (CT eq 15.05){
//print current combo
WriteOutput("<strong>#CC# = 15.05</strong><br />");
return(true);
}else if (CT gt 15.05){
//since we know the array is sorted by cost (asc),
//and we've already gone over the price limit,
//we can ignore anything else in the array
break;
}else{
//less than 15.50, try adding something else
found = testCombo(a, CC, CT);
}
}
return(found);
}
mf = {name="mixed fruit", cost=2.15};
ff = {name="french fries", cost=2.75};
ss = {name="side salad", cost=3.35};
hw = {name="hot wings", cost=3.55};
ms = {name="mozarella sticks", cost=4.20};
sp = {name="sampler plate", cost=5.80};
apps = [ mf, ff, ss, hw, ms, sp ];
tries = 0;
combos = 0;
testCombo(1, "", 0);
WriteOutput("<br />tries: #tries#<br />combos: #combos#");
函数testCombo（minIndex、currentCombo、currentTotal）{
var a=0；
var CC=“”；
var-CT=0；
var=false；
尝试次数+=1；
对于（a=arguments.minIndex；a来说，即使背包是NP完全的，它也是一个非常特殊的问题：通常的动态程序实际上是非常优秀的（）
如果你做了正确的分析，结果是O（nW），n是
#light
type Appetizer = { name : string; cost : int }
let menu = [
{name="fruit"; cost=215}
{name="fries"; cost=275}
{name="salad"; cost=335}
{name="wings"; cost=355}
{name="moz sticks"; cost=420}
{name="sampler"; cost=580}
]
// Choose: list<Appetizer> -> list<Appetizer> -> int -> list<list<Appetizer>>
let rec Choose allowedMenu pickedSoFar remainingMoney =
if remainingMoney = 0 then
// solved it, return this solution
[ pickedSoFar ]
else
// there's more to spend
[match allowedMenu with
| [] -> yield! [] // no more items to choose, no solutions this branch
| item :: rest ->
if item.cost <= remainingMoney then
// if first allowed is within budget, pick it and recurse
yield! Choose allowedMenu (item :: pickedSoFar) (remainingMoney - item.cost)
// regardless, also skip ever picking more of that first item and recurse
yield! Choose rest pickedSoFar remainingMoney]
let solutions = Choose menu [] 1505
printfn "%d solutions:" solutions.Length
solutions |> List.iter (fun solution ->
solution |> List.iter (fun item -> printf "%s, " item.name)
printfn ""
)
(*
2 solutions:
fruit, fruit, fruit, fruit, fruit, fruit, fruit,
sampler, wings, wings, fruit,
*)
#!/usr/bin/perl
use strict;
use warnings;
my @weights = (2.15, 2.75, 3.35, 3.55, 4.20, 5.80);
my $total = 0;
my @order = ();
iterate($total, @order);
sub iterate
{
my ($total, @order) = @_;
foreach my $w (@weights)
{
if ($total+$w == 15.05)
{
print join (', ', (@order, $w)), "\n";
}
if ($total+$w < 15.05)
{
iterate($total+$w, (@order, $w));
}
}
}
function testCombo(minIndex, currentCombo, currentTotal){
var a = 0;
var CC = "";
var CT = 0;
var found = false;
tries += 1;
for (a=arguments.minIndex; a <= arrayLen(apps); a++){
combos += 1;
CC = listAppend(arguments.currentCombo, apps[a].name);
CT = arguments.currentTotal + apps[a].cost;
if (CT eq 15.05){
//print current combo
WriteOutput("<strong>#CC# = 15.05</strong><br />");
return(true);
}else if (CT gt 15.05){
//since we know the array is sorted by cost (asc),
//and we've already gone over the price limit,
//we can ignore anything else in the array
break;
}else{
//less than 15.50, try adding something else
found = testCombo(a, CC, CT);
}
}
return(found);
}
mf = {name="mixed fruit", cost=2.15};
ff = {name="french fries", cost=2.75};
ss = {name="side salad", cost=3.35};
hw = {name="hot wings", cost=3.55};
ms = {name="mozarella sticks", cost=4.20};
sp = {name="sampler plate", cost=5.80};
apps = [ mf, ff, ss, hw, ms, sp ];
tries = 0;
combos = 0;
testCombo(1, "", 0);
WriteOutput("<br />tries: #tries#<br />combos: #combos#");
item(X) :- member(X,[215, 275, 335, 355, 420, 580]).
solution([X|Y], Z) :- item(X), plus(S, X, Z), Z >= 0, solution(Y, S).
solution([], 0).
?- solution(X, 1505).
X = [215, 215, 215, 215, 215, 215, 215] ;
X = [215, 355, 355, 580] ;
X = [215, 355, 580, 355] ;
X = [215, 580, 355, 355] ;
X = [355, 215, 355, 580] ;
X = [355, 215, 580, 355] ;
X = [355, 355, 215, 580] ;
X = [355, 355, 580, 215] ;
X = [355, 580, 215, 355] ;
X = [355, 580, 355, 215] ;
X = [580, 215, 355, 355] ;
X = [580, 355, 215, 355] ;
X = [580, 355, 355, 215] ;
No
>>> from constraint import *
>>> problem = Problem()
>>> menu = {'mixed-fruit': 2.15,
... 'french-fries': 2.75,
... 'side-salad': 3.35,
... 'hot-wings': 3.55,
... 'mozarella-sticks': 4.20,
... 'sampler-plate': 5.80}
>>> for appetizer in menu:
... problem.addVariable( appetizer, [ menu[appetizer] * i for i in range( 8 )] )
>>> problem.addConstraint(ExactSumConstraint(15.05))
>>> problem.getSolutions()
[{'side-salad': 0.0, 'french-fries': 0.0, 'sampler-plate': 5.7999999999999998, 'mixed-fruit': 2.1499999999999999, 'mozarella-sticks': 0.0, 'hot-wings': 7.0999999999999996},
{'side-salad': 0.0, 'french-fries': 0.0, 'sampler-plate': 0.0, 'mixed-fruit': 15.049999999999999, 'mozarella-sticks': 0.0, 'hot-wings': 0.0}]
....
private void findAndReportSolutions(
int target, // goal to be achieved
int balance, // amount of goal remaining
int index // menu item to try next
) {
++calls;
if (balance == 0) {
reportSolution(target);
return; // no addition to perfect order is possible
}
if (index == items.length) {
++falls;
return; // ran out of menu items without finding solution
}
final int price = items[index].price;
if (balance < price) {
return; // all remaining items cost too much
}
int maxCount = balance / price; // max uses for this item
for (int n = maxCount; 0 <= n; --n) { // loop for this item, recur for others
++loops;
counts[index] = n;
findAndReportSolutions(
target, balance - n * price, index + 1
);
}
}
public void reportSolutionsFor(int target) {
counts = new int[items.length];
calls = loops = falls = 0;
findAndReportSolutions(target, target, 0);
ps.printf("%d calls, %d loops, %d falls%n", calls, loops, falls);
}
public static void main(String[] args) {
MenuItem[] items = {
new MenuItem("mixed fruit", 215),
new MenuItem("french fries", 275),
new MenuItem("side salad", 335),
new MenuItem("hot wings", 355),
new MenuItem("mozzarella sticks", 420),
new MenuItem("sampler plate", 580),
};
Solver solver = new Solver(items);
solver.reportSolutionsFor(1505);
}
...
7 mixed fruit (1505) = 1505
1 mixed fruit (215) + 2 hot wings (710) + 1 sampler plate (580) = 1505
348 calls, 347 loops, 79 falls
;; np-complete.clj
;; A Clojure solution to XKCD #287 "NP-Complete"
;; By Sam Fredrickson
;;
;; The function "items-with-price" returns a sequence of items whose sum price
;; is equal to the given price, or nil.
(defstruct item :name :price)
(def *items* #{(struct item "Mixed Fruit" 2.15)
(struct item "French Fries" 2.75)
(struct item "Side Salad" 3.35)
(struct item "Hot Wings" 3.55)
(struct item "Mozzarella Sticks" 4.20)
(struct item "Sampler Plate" 5.80)})
(defn items-with-price [price]
(let [check-count (atom 0)
recur-count (atom 0)
result (atom nil)
checker (agent nil)
; gets the total price of a seq of items.
items-price (fn [items] (apply + (map #(:price %) items)))
; checks if the price of the seq of items matches the sought price.
; if so, it changes the result atom. if the result atom is already
; non-nil, nothing is done.
check-items (fn [unused items]
(swap! check-count inc)
(if (and (nil? @result)
(= (items-price items) price))
(reset! result items)))
; lazily generates a list of combinations of the given seq s.
; haven't tested well...
combinations (fn combinations [cur s]
(swap! recur-count inc)
(if (or (empty? s)
(> (items-price cur) price))
'()
(cons cur
(lazy-cat (combinations (cons (first s) cur) s)
(combinations (cons (first s) cur) (rest s))
(combinations cur (rest s))))))]
; loops through the combinations of items, checking each one in a thread
; pool until there are no more combinations or the result atom is non-nil.
(loop [items-comb (combinations '() (seq *items*))]
(if (and (nil? @result)
(not-empty items-comb))
(do (send checker check-items (first items-comb))
(recur (rest items-comb)))))
(await checker)
(println "No. of recursions:" @recur-count)
(println "No. of checks:" @check-count)
@result))
class Solver(object):
def __init__(self):
self.solved = False
self.total = 0
def solve(s, p, pl, curList = []):
poss = [i for i in sorted(pl, reverse = True) if i <= p]
if len(poss) == 0 or s.solved:
s.total += 1
return curList
if abs(poss[0]-p) < 0.00001:
s.solved = True # Solved it!!!
s.total += 1
return curList + [poss[0]]
ml,md = [], 10**8
for j in [s.solve(p-i, pl, [i]) for i in poss]:
if abs(sum(j)-p)<md: ml,md = j, abs(sum(j)-p)
s.total += 1
return ml + curList
priceList = [5.8, 4.2, 3.55, 3.35, 2.75, 2.15]
appetizers = ['Sampler Plate', 'Mozzarella Sticks', \
'Hot wings', 'Side salad', 'French Fries', 'Mixed Fruit']
menu = zip(priceList, appetizers)
sol = Solver()
q = sol.solve(15.05, priceList)
print 'Total time it runned: ', sol.total
print '-'*30
order = [(m, q.count(m[0])) for m in menu if m[0] in q]
for o in order:
print '%d x %s \t\t\t (%.2f)' % (o[1],o[0][1],o[0][0])
print '-'*30
ts = 'Total: %.2f' % sum(q)
print ' '*(30-len(ts)-1),ts
Total time it runned: 29
------------------------------
1 x Sampler Plate (5.80)
2 x Hot wings (3.55)
1 x Mixed Fruit (2.15)
------------------------------
Total: 15.05
public class NPComplete {
private static final int[] FOOD = { 580, 420, 355, 335, 275, 215 };
private static int tries;
public static void main(String[] ignore) {
tries = 0;
addFood(1505, "", 0);
System.out.println("Combinations tried: " + tries);
}
private static void addFood(int goal, String result, int index) {
// If no more food to add, see if this is a solution
if (index >= FOOD.length) {
tries++;
if (goal == 0)
System.out.println(tries + " tries: " + result.substring(3));
return;
}
// Try all possible quantities of this food
// If this is the last food item, only try the max quantity
int qty = goal / FOOD[index];
do {
addFood(goal - qty * FOOD[index],
result + " + " + qty + " * " + FOOD[index], index + 1);
} while (index < FOOD.length - 1 && --qty >= 0);
}
}
9 tries: 1 * 580 + 0 * 420 + 2 * 355 + 0 * 335 + 0 * 275 + 1 * 215
88 tries: 0 * 580 + 0 * 420 + 0 * 355 + 0 * 335 + 0 * 275 + 7 * 215
Combinations tried: 88