我將一個現實世界的問題抽象為一個圖論問題。目前,我嘗試通過使用 networkx 庫中的 clique-algorithm 來解決這個問題。我需要想法如何調整我的演算法以提高運行時性能。
問題:
令G為無向彩色圖,令R為映射,其中每種顏色都映射到給定數量,例如
R = {"red": 3, "blue": 1, "green": 1, "magenta": 1, "orange": 1}
我需要找到一組節點V,具有以下屬性(或確定不存在這樣的集合):
對于R中的每個 (color,amount) ,我們至少需要V中的彩色節點數量。
(以上面的R為例,我們需要在V中至少三個紅色節點,一個藍色節點等)
V中的節點不允許成為G中的鄰居。
簡單示例:
有了這個彩色圖和上面給定的要求R,一個可能的解決方案是:V = {0, 1, 2, 8, 13, 15, 16},因為R所需的所有顏色量都包含在V中,并且沒有節點與另一個節點相鄰。

當前方法:
- 從G創建補圖CG
- 迭代地查找 cliques 并檢查 clique 中彩色節點的數量是否符合R的要求。
import networkx as nx
def find_result(G, R):
CG = nx.complement(G)
for V in nx.find_cliques(CG):
color_count = get_color_count(G, V)
if all([(color_count.get(color, 0) >= R[color]) for color in R.keys()]):
return V
# Helper method to count colors
def get_color_count(G, V):
colors = [G.nodes[v]["color"] for v in V]
count = dict()
for color in colors:
count[color] = count.get(color, 0) 1
return count
性能問題:
使用這種簡單的方法,我遇到了性能問題(即演算法對于R的特定配置花費的時間太長)。我的圖表可能會變得更糟,如下所示,有 282 個節點、4640 條邊和 7 種不同的顏色。
所以這是我的問題:如何自定義
編輯 1
這是上面大圖中的圖形內容(sparse6 格式)。可以將內容粘貼到檔案中,然后使用 networkx 讀取。
>>sparse6<<:~?CY_A??@_??OA_??OA?M??@?G?oC_??OA?K@?D_??OA?K@?D?Y??@?G?oC?S@_F_??OA?K@?D?W@oG_??OA?K@?D?W@oG?e??@?G?oC?S@_F?_AOI`CCGO@ED?R`WBOT`_B_V`gEW[@mEo[`qF?]`}DWV`eGwYAKHG[@wFw\AYI_??C?_B?O@OE?[A?H?gAwk???OA?K@?D?W@oG?cA_J_??OA?K@?D?W@oG?cA_Ja{FOf_??OA?K@?D?W@oG?cA_J_??OA?K@?D?W@oG?cA_JbO??@?G?oC?S@_F?_AOI?mM?__??OA?K@?D?W@oG?cA_JbkBw|AQB_Wc?Nw~CANp?CE??@?G?oC?UPOK?sLwK?sLpD_oBOvB{OPDCYB?LB[O@ACSP`F_oBPDCWPpG_oBPDCWPpGCeB?LB{OPDCWPpGCcQgK?sO@ACSP`FC_QPICmB?LB{O@DCWPpGCcQ`JCqB?LB{O@DCWPpGCcQ`JCoRWK?sNp?CCPPEC[Q@HCgQpKCsRgK?sNp?CGPPEC[Q@HCgQpKCsR`N_oBP@CGPPEC[Q@HCgQpKCsR`NDAB?LCCO`DCWPpGCcQ`JCoRPMC{S@P_oBO~CCO`DCWPpGCcQ`JCoRPMC{S@PDIB?LC?OPACSP`FC_QPICkR@LCwRpODCS`RdWTW~C?RPMC{S@TDYNp@C[QpNDKTPUD]TPUD[UHTDWTpWDeNp?CsR`ND?TPUD[U@XDiNp@C[QpNDKTPUD[U@XDgUwq_??OA?K@?D?W@oG?cA_JBCK`\bGVP]dsV`^_??OA?K@?D?W@oG?cA_JBCVP]D{WH\DwVp_EE??@?G?oC?SOw??C?_B?O@PBEM??@?G?oC?SOpbEQOpbEOXXbEOXPeeKX@dEWXxBEKX@dEWXpgeKX@dEWXpgEeWpcESX`fE_YPi_??OA?K@?D?W@oG?cA_JBKLG??C?_B?O@OE?[A?H?gAosEq??@?G?oC?S@_F?_AOI?kL@dE_YpkEu??@?G?oC?S@_F?_AOI?kL@kEsZg??C?_B?O@OE?[A?H?gAosC?O`GCoS@SEoZPmE}KpkEsZ`nFAZ@lEwZpoFEXPgEkZ@lEwZpoFC[hkEsZ`nF?[PqFMO@AC_R@ODOZ@lEwZpoFC[`rFQKphEgYpkEsZ`nF?[PqFK\@tecY`jEoZPmE{[@pFG[psFS\hdE_YPiEkZ@lEwZpoFC[`rFO\PuF]YPiEkZ@lEwZpoFC[`rFO\PuF[]H?CGQ@KD?T@hEgYpkEsZ`nF?[PqFK\@tFW\pwFeKpkEsZ`nF?[PqFK\@tFW\pwFc]hkEsZ`nF?[PqFK\@tFW\pwFc]`zeSY@jEoZPmE{[@pFG[psFS\`vF_]PyFk^HkEsZ`nF?[PqFK\@tFW\pwFc]`zFo^X?CGQ@KD?T@kEsZ`nF?[PqFK\@tFW\pwFc]`zFo^P}bKOPADCS`RDOZ@lEwZpoFC[`rFO\PuF[]@xFg]p{Fs^`~cCO`PDGSpSEoZPmE{[@pFG[psFS\`vF_]PyFk^@|Fw^q?cCO`PDGSpSESY@jEoZPmE{[@pFG[psFS\`vF_]PyFk^@|Fw^q?GEOPADCS`RDOZ@lEwZpoFC[`rFO\PuF[]@xFg]p{Fs^`~G?_QAc?OPAC_R@ODCS`RDOZ@lEwZpoFC[`rFO\PuF[]@xFg]p{Fs^`~G?_QAGM??@?G?oE?[A?H_??OA?W@oGGU??@?G@_F?_Np?CsR`ND?TpZGS`g??K@_HGS`aFgS`aFGaNp?CsR`ND?TpZGS`aFG_aW??K@_HGS`aFG_aQIgS`aFG_aQIGmNp?CsR`ND?TpZGS`aFG_aQIGkbIN?G@OG?kbiMG}?_D?_AqMG{cGM?{E?zByB_WBwcgM@_NaQHMB_WBwcaRHQBozHGcqSHUcaRHOdQUhGcqSHSdaVhGcqSHSdaVHaBozCOcaRHOdQUH[eAXcOcaRHOdQUH[eAXHiPAQHKdATHWdqWHceaZcOcaRHOdQUH[eAXHgeq[_{MqQHKdATHWdqWHceaZHofYQHKdATHWdqWHceaZHofQ]hGcqSHSdaVH_eQYHkfA\HwfyQHKdATHWdqWHceaZHofQ]H{gIbEs[PqFK\@tF[^A@e{[PqFK\@tFc^aBIMO@AC_R@ODO[@pFG[psFS]`~GOgqces[`vFk^@|Fw^q@IKhAde{\@xFk^@|Fw^qBIKhAdIYO@AC_R@ODO[@tFg]p{Fs^`~GOgqcIShafcCO`PDGSpSEs[`vFo_A@GG_qCIKhAdIWhqgcCO`PDGSpSE{\@xFw_A@GG_qCIKhAdIWhqgIeO@@CGQ@KD?SPQDKT@oFS]`~G?_QAGK`AbIOhQeI[iAhIi??@?G?oE?[A?HGS`aFG_aw??C?_E?[A?~C?RPMC{S@VDk`QEG[aaLIq??B?WAQDG_aqKGsjAlb{O@LCwRpOD[UqFGgaqKGsjAlIy??@?G?oC?S@_F?_AOI?kIOkBY??@?G?oC?S@_F?_AOI?kIOkJA??@?G?oC?S@_F?_AOI?kIOkHwfq_ICkApacLaoJCkgE?[A?H?gAohJ?kQqJMIQ]H{gA`J?kQqJKlGhBWaqKGsjanJ?kQqJKlAt_W@oG?cA_JAcaqKGsjanJ?kQqJKlAtJYIQJGobQ]H{gA`IwjqoJCkarJOlQuJ]??@?G?oC?S@_F?_AOI?kJ?uH?cQoJCkarJOlQuJ[mG??C?_B?O@OE?[A?H?gAokH?cQoJCkarJOlQuJ[mAx_??OA?K@?D?W@oG?cA_JAocAPHwfq_ICkApJGkqsJSlavJ_mQybWcAPJ?kQqJKlAtJWlqwJcmaz_W@oG?cA_JH?cQoJCkarJOlQuJ[mAxJgmq{h?cQ]H{gA`J?kQqJKlAtJWlqwJcmazJonWuGkbALH?cQmI{kApJGkqsJSlavJ_mQyJknA|Jy@_F?_AOI?kaqKGscAPIwjqoJCkarJOlQuJ[mAxJgmq{Jsna~gkbALH?cQ]H{gA`IwjqoJCkarJOlQuJ[mAxJgmq{Jsna~KA??@?G?oC?S@_F?_AOI?kJ?uHSeQ\ICkApJGkqsJSlavJ_mQyJknA|Jwnr?KE??@?G?oC?S@_F?_AOI?kJATHcfQ`J?kQqJKlAtJWlqwJcmazJonQ}J{oB@KI??@?G?oC?S@_F?_AOI?kJATHcfQ]H{gA`J?kQqJKlAtJWlqwJcmazJonQ}J{oB@KGowuHSeQ\ICkApJGkqsJSlavJ_mQyJknA|Jwnr?KCobBKQ@_F?_AOI?kdQXHsgQoJCkarJOlQuJ[mAxJgmq{Jsna~K?oRAKKpBDhSeQ\Hwfq_ICkApJGkqsJSlavJ_mQyJknA|Jwnr?KCobBKOpREbWaqKGsdQXHsgQmI{kApJGkqsJSlavJ_mQyJknA|Jwnr?KCobBKOpREK]@_F?_AOI?kaqKGsdQXHsgQmI{kApJGkqsJSlavJ_mQyJknA|Jwnr?KCobBKOpREK[qIJGobQTHcfQ]H{gA`IwjqoJCkarJOlQuJ[mAxJgmq{Jsna~K?oRAKKpBDKWprGKe??@?G?oC?S@_F?_AOI?kG?oB_OpbEWYW??C?_B?O@OE?[A?H?gAo_B?MADG_aqkIwqw??C?_B?O@OE?[A?H?gAo_B?M@UDgqrK_??OA?K@?D?W@oG?cA_JA?K?wEs[`vFo_QbIWiRJKorW??C?_B?O@OE?[A?H?gAo_B?MBJKorRM_??OA?K@?D?W@oG?cA_JA?M?xCKWpeEcqrKKsrbN_??OA?K@?D?W@oG?cA_JA?M?xGSaAJIojbJKorRMK{sG??C?_B?O@OE?[A?H?gAo_B_MPUDgqrKKsrbNL?sW??C?_B?O@OE?[A?H?gAo_B_MPlFG\p{GCgqeIcqrKKsrbNL?sRQ_??OA?K@?D?W@oG?cA_JA?M?xKkrBLKwrrOLCsbRa?M@BDST`VD_WpeEcqrKKsrbNL?sRQLKtG??K@_HA?M@TDWTpWGSaAJIojbJKorRMK{sBPLGsrSLUG?wDST`VD_UbJKorRMK{sBPLGsrSLStg_B_TPUD[U@lFG\p{GCgqeIcqrKKsrbNL?sRQLKtBTLWtw_B_TPUD[UBJKorRMK{sBPLGsrSLStbVLa??@?G?oC?S@_F?_AOI?kK@BDSUPbEWYRJKorRMK{sBPLGsrSLStbVL_uW??C?_B?O@OE?[A?H?gAooDSUQDG_aqkIwqrKKsrbNL?sRQLKtBTLWtrWLcug??C?_B?O@OE?[A?H?gAooDST`XDgqrKKsrbNL?sRQLKtBTLWtrWLcubZ_??OA?K@?D?W@oG?cA_JB?TPXEs[`vFo_QbIWiRJKorRMK{sBPLGsrSLStbVL_uRYLkvG??C?_B?O@OE?[A?H?gAooDSURJKorRMK{sBPLGsrSLStbVL_uRYLkvB\_??OA?K@?D?W@oG?cA_JBcOpTDcWpeEcqrKKsrbNL?sRQLKtBTLWtrWLcubZLovR]_??OA?K@?D?W@oG?cA_JBcTPXGSaAJIojbJKorRMK{sBPLGsrSLStbVL_uRYLkvB\Lwvw??C?_B?O@OE?[A?H?gAoxDST`XDgqrKKsrbNL?sRQLKtBTLWtrWLcubZLovR]L{wG??C?_B?O@OE?[A?H?gAoxDSUPlFG\p{GCgqeIcqrKKsrbNL?sRQLKtBTLWtrWLcubZLovR]L{wB`_??OA?K@?D?W@oG?cA_JBcTPXKkrBLKwrrOLCsbRLOtRUL[uBXLgur[Lsvb^M?wRacKTPUD[U@XEKX`hKkrBLKwrrOLCsbRLOtRUL[uBXLgur[Lsvb^M?wRaMM??B?WAPTDWTpWDc`QGGkjAmKkrBLKwrrOLCsbRLOtRUL[uBXLgur[Lsvb^M?wRaMKxHTDWTpWDcUbJKorRMK{sBPLGsrSLStbVL_uRYLkvB\Lwvr_MCwbbMOxXTDWTpWDcZPqF[^A@IKhahKkrBLKwrrOLCsbRLOtRUL[uBXLgur[Lsvb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要為圖形著色,請使用以下實作:
G = nx.read_sparse6("graph_file")
for i in range(0,12):
G.nodes[i]["color"] = "yellow"
for i in range(12,40):
G.nodes[i]["color"] = "red"
for i in range(40,63):
G.nodes[i]["color"] = "turquoise"
for i in range(63,85):
G.nodes[i]["color"] = "green"
for i in range(85,163):
G.nodes[i]["color"] = "blue"
for i in range(163, 176):
G.nodes[i]["color"] = "magenta"
for i in range(176, 282):
G.nodes[i]["color"] = "orange"
uj5u.com熱心網友回復:
形式上,我們有一個獨立集
問題,旁邊有一些覆寫約束。整數編程似乎很適合,但我不知道R具體是哪一個給你帶來了麻煩。
獨立集的直接整數規劃公式對每個節點都有一個 0-1 變數,對每個邊都有一個約束。然而,線性松弛很差,因為(例如)三角形圖有一個分數解,其中每個節點都是 0.5,對于“1.5 節點”分數獨立集。然而,由于G只有幾百個最大團,我們可以使用更好的公式,對每個最大團有一個約束(對于每個團,獨立集最多包含一個團節點)。
使用下面的 NetworkX 和 OR-Tools實作 :
import networkx as nx
G = nx.read_sparse6("graph_file")
for i in range(0, 12):
G.nodes[i]["color"] = "yellow"
for i in range(12, 40):
G.nodes[i]["color"] = "red"
for i in range(40, 63):
G.nodes[i]["color"] = "turquoise"
for i in range(63, 85):
G.nodes[i]["color"] = "green"
for i in range(85, 163):
G.nodes[i]["color"] = "blue"
for i in range(163, 176):
G.nodes[i]["color"] = "magenta"
for i in range(176, 282):
G.nodes[i]["color"] = "orange"
R = {
"red": 11,
"turquoise": 11,
"blue": 8,
"orange": 6,
"green": 4,
"magenta": 2,
"yellow": 1,
}
import collections
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver.CreateSolver("SCIP")
x = {v: solver.BoolVar(str(v)) for v in G.nodes}
for K in nx.find_cliques(G):
solver.Add(sum(x[v] for v in K) <= 1)
nodes_with_color = collections.defaultdict(list)
for v in G.nodes:
nodes_with_color[G.nodes[v]["color"]].append(v)
for color, bound in R.items():
solver.Add(sum(x[v] for v in nodes_with_color[color]) >= bound)
solver.Maximize(sum(x_v for x_v in x.values()))
status = solver.Solve()
if status == pywraplp.Solver.OPTIMAL:
solution = [v for (v, x_v) in x.items() if x_v.SolutionValue()]
print(*solution)
print(collections.Counter(G.nodes[v]["color"] for v in solution))
elif status == pywraplp.Solver.INFEASIBLE:
print("no solution")
else:
print("unknown status", status)
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