離散對數問題(Discrete Logarithm Problem，DLP)
已知p，0p，GCD(p)=1，x(modp)，當p很大時，計算x值非常困難。

橢圓曲線離散對數問題(Elliptic Curve Discrete Logarithm Problem，ECDLP)
(1)令E為基於有限體Fp的橢圓曲線，PQ為橢圓曲線上的點坐標(PQE(Fp))且Q=kPkN，當p很大時，計算k值非常困難。
(2)令E為基於二元體F2n的橢圓曲線，PQ為橢圓曲線上的點坐標(PQE(F2n))且Q=kPkN，當n很大時，計算k值非常困難。
以下介紹解離散對數問題和橢圓曲線離散對數問題的各種方法。

1.解離散對數問題的方法-小步大步(Baby-step Giant-step)

將答案改寫成x=im+j，其中m=p ，0im，0jm
將原本離散對數問題改成
x(modp)
im+j(modp)
j(−m)i(modp)
計算j稱為Baby-step
計算(−m)i稱為Giant-step
當某個i和j值讓j和(−m)i相等時，就代表找到答案x=im+j

虛擬碼
1.m=Ceiling(p) 
2.for j=0~m-1
　計算[j(modp)j]數對放在List中
3.計算−m
4.設=
5.for i=0~m-1
　檢查目前值是否等於List中某個j(modp)
　若是，答案x=im+j
　若否，計算=−m

參考資料：https://en.wikipedia.org/wiki/Baby-step_giant-step



Baby_stepGiant_step副程式
(%i1)
Baby_stepGiant_step(alpha,beta,p):=block
([m,gamma,List,inv_am,i,j,Powerj,x:[]],
print("取Giant步長m=Ceiling(",sqrt(p),")=",m:ceiling(sqrt(p))),
print("Baby step，計算",alpha^"0",",",alpha^"1",",...,",alpha,""^(m-1),"(mod ",p,")"),
print("List=",List:create_list(power_mod(alpha,j,p),j,0,m-1)),
print("(",a^"-1",")"^m,"≡(",alpha^"-1",")"^m,"≡",inv_mod(alpha,p),""^m,"≡",inv_am:power_mod(alpha,-m,p),"(mod ",p,")"),
print("令γ=β=",gamma:beta),
print("Giant step，每次γ值再乘上(α"^"-1",")"^m),
for i:0 thru m-1 do
  (print("當i=",i,"時"),
   Powerj:sublist_indices(List,lambda([aj],aj=gamma))-1,/*檢查List中是否有gamma值*/
   if Powerj#[] then
      (print("List中有γ=",gamma,"值"),
       for j in Powerj do
         (print("j=",j,"次方，找到一解x=i*m+j=",i,"*",m,"+",j,"=",last(x:append(x,[i*m+j])))
         )
      )
  else
      (print("List中沒有γ=",gamma,"值")
      ),
  print("計算γ≡γ*(α"^"-1",")"^m,"≡",gamma,"*",inv_am,"≡",gamma*inv_am,"≡",gamma:mod(gamma*inv_am,p),"(mod ",p,")")
      
  ),
return(x)
)$

x(modp)，x=？
(%i4)
alpha:2;
beta:108;
p:181;
(alpha)　2
(beta)　108
(p)　181

利用Baby_stepGiant_step方法解離散對數問題
(%i5)　x:Baby_stepGiant_step(alpha,beta,p);
取Giant步長m=Ceiling(181)=14 
Babystep，計算2021213(mod181)
List=[1248163264128751501195711447]
(−1)14(2−1)14911452(mod181)
令==108
Giantstep，每次值再乘上(−1)14
當i=0時
List中沒有=108值
計算(−1)141085256165(mod181)
當i=1時
List中沒有=5值
計算(−1)1455226079(mod181)
當i=2時
List中沒有=79值
計算(−1)1479524108126(mod181)
當i=3時
List中沒有=126值
計算(−1)1412652655236(mod181)
當i=4時
List中沒有=36值
計算(−1)143652187262(mod181)
當i=5時
List中沒有=62值
計算(−1)1462523224147(mod181)
當i=6時
List中沒有=147值
計算(−1)1414752764442(mod181)
當i=7時
List中沒有=42值
計算(−1)144252218412(mod181)
當i=8時
List中沒有=12值
計算(−1)14125262481(mod181)
當i=9時
List中沒有=81值
計算\gamma\equiv\gamma*(\alpha^{-1})^{14}\equiv81*52\equiv4212\equiv49\pmod{181}
當i=10時
List中沒有\gamma=49值
計算\gamma\equiv\gamma*(\alpha^{-1})^{14}\equiv49*52\equiv2548\equiv14\pmod{181}
當i=11時
List中沒有\gamma=14值
計算\gamma\equiv\gamma*(\alpha^{-1})^{14}\equiv14*52\equiv728\equiv4\pmod{181}
當i=12時
List中有\gamma=4值
j=2次方，找到一解x=i*m+j=12*14+2=170
計算\gamma\equiv\gamma*(\alpha^{-1})^{14}\equiv4*52\equiv208\equiv27\pmod{181}
當i=13時
List中沒有\gamma=27值
計算\gamma\equiv\gamma*(\alpha^{-1})^{14}\equiv27*52\equiv1404\equiv137\pmod{181}
(x)　[170]

2.解離散對數問題的方法-小步大步(Baby-step giant-step)

相同原理提供另一種實作
虛擬碼
1.m=Ceiling(\sqrt{p})
2.計算\alpha^{-m}
3.for j=0~m-1
　計算[\alpha^j \pmod{p},j]數對放在BabyStep List中
4.for i=0~m-1
　計算[\beta(\alpha^{-m})^i \pmod{p},i]數對放在GiantStep List中
5.將BabyStep和GiantStep的List由小到大重新排序
6.兩個List互相比較找出相同值
　若有，答案x=im+j
　若沒有，答案無解

參考資料：Computational Number Theory and Modern Cryptography



Baby_stepGiant_step副程式
(%i1)
Baby_stepGiant_step(alpha,beta,p):=block
([m,inv_am,BabyStep,GiantStep,indexB:1,indexG:1,i,j,x:[]],
print("取Giant步長m=Ceiling(",sqrt(p),")=",m:ceiling(sqrt(p))),
print("(",a^"-1",")"^m,"≡(",alpha^"-1",")"^m,"≡",inv_mod(alpha,p),""^m,"≡",inv_am:power_mod(alpha,-m,p),"(mod ",p,")"),
print("BabyStep=",BabyStep:create_list([power_mod(alpha,j,p),j],j,0,m-1)),
print("GiantStep=",GiantStep:create_list([mod(beta*power_mod(inv_am,i,p),p),i],i,0,m-1)),
print("由小到大排序"),
print("BabyStep=",BabyStep:sort(BabyStep)),
print("GiantStep=",GiantStep:sort(GiantStep)),
print("BabyStep和GiantStep互相比較找出相同值"),
print("一開始indexB=1,indexG=1"),
while indexB<=m and indexG<=m do
   (if BabyStep[indexB][1]=GiantStep[indexG][1] then
      (do
          (i:GiantStep[indexG][2],j:BabyStep[indexB][2],/*第2個值是i和j*/
           print("BabyStep[",indexB,"]=",BabyStep[indexB],"第1個值等於",
                   "GiantStep[",indexG,"]=",GiantStep[indexG],"第1個值，得到一解x=i*m+j=",i,"*",m,"+",j,"=",last(x:append(x,[i*m+j]))),
           print("GiantStep移到下一組，indexG=indexG+1=",indexG:indexG+1),
           if indexG=m or GiantStep[indexG][1]#GiantStep[indexG-1][1] then
              (return())
          ),
        print("BabyStep移到下一組，indexB=indexB+1=",indexB:indexB+1)
      )
    else if BabyStep[indexB][1]<GiantStep[indexG][1] then
      (print("BabyStep[",indexB,"]=",BabyStep[indexB],"第1個值小於",
               "GiantStep[",indexG,"]=",GiantStep[indexG],"第1個值，BabyStep移到下一組，indexB=indexB+1=",indexB:indexB+1)
      )
    else
      (print("BabyStep[",indexB,"]=",BabyStep[indexB],"第1個值大於",
               "GiantStep[",indexG,"]=",GiantStep[indexG],"第1個值，GiantStep移到下一組，indexG=indexG+1=",indexG:indexG+1)
      )
   ),
return(x)
)$

\alpha^x \equiv \beta \pmod{p}，x=？
(%i4)
alpha:2;
beta:108;
p:181;
(alpha)　2
(beta)　108
(p)　181

利用Baby_stepGiant_step方法解離散對數問題
(%i5)　Baby_stepGiant_step(alpha,beta,p);
取Giant步長m=Ceiling(\sqrt{181})=14
(a^{-1})^{14}\equiv(2^{-1})^{14}\equiv 91^{14}\equiv52\pmod{181}
BabyStep=[[1,0],[2,1],[4,2],[8,3],[16,4],[32,5],[64,6],[128,7],[75,8],[150,9],[119,10],[57,11],[114,12],[47,13]]
GiantStep=[[108,0],[5,1],[79,2],[126,3],[36,4],[62,5],[147,6],[42,7],[12,8],[81,9],[49,10],[14,11],[4,12],[27,13]]
由小到大排序
BabyStep=[[1,0],[2,1],[4,2],[8,3],[16,4],[32,5],[47,13],[57,11],[64,6],[75,8],[114,12],[119,10],[128,7],[150,9]]
GiantStep=[[4,12],[5,1],[12,8],[14,11],[27,13],[36,4],[42,7],[49,10],[62,5],[79,2],[81,9],[108,0],[126,3],[147,6]]
BabyStep和GiantStep互相比較找出相同值
一開始indexB=1,indexG=1
BabyStep[1]=[1,0]第1個值小於GiantStep[1]=[4,12]第1個值，BabyStep移到下一組，indexB=indexB+1=2
BabyStep[2]=[2,1]第1個值小於GiantStep[1]=[4,12]第1個值，BabyStep移到下一組，indexB=indexB+1=3
BabyStep[3]=[4,2]第1個值等於GiantStep[1]=[4,12]第1個值，得到一解x=i*m+j=12*14+2=170
GiantStep移到下一組，indexG=indexG+1=2
BabyStep移到下一組，indexB=indexB+1=4
BabyStep[4]=[8,3]第1個值大於GiantStep[2]=[5,1]第1個值，GiantStep移到下一組，indexG=indexG+1=3
BabyStep[4]=[8,3]第1個值小於GiantStep[3]=[12,8]第1個值，BabyStep移到下一組，indexB=indexB+1=5
BabyStep[5]=[16,4]第1個值大於GiantStep[3]=[12,8]第1個值，GiantStep移到下一組，indexG=indexG+1=4
BabyStep[5]=[16,4]第1個值大於GiantStep[4]=[14,11]第1個值，GiantStep移到下一組，indexG=indexG+1=5
BabyStep[5]=[16,4]第1個值小於GiantStep[5]=[27,13]第1個值，BabyStep移到下一組，indexB=indexB+1=6
BabyStep[6]=[32,5]第1個值大於GiantStep[5]=[27,13]第1個值，GiantStep移到下一組，indexG=indexG+1=6
BabyStep[6]=[32,5]第1個值小於GiantStep[6]=[36,4]第1個值，BabyStep移到下一組，indexB=indexB+1=7
BabyStep[7]=[47,13]第1個值大於GiantStep[6]=[36,4]第1個值，GiantStep移到下一組，indexG=indexG+1=7
BabyStep[7]=[47,13]第1個值大於GiantStep[7]=[42,7]第1個值，GiantStep移到下一組，indexG=indexG+1=8
BabyStep[7]=[47,13]第1個值小於GiantStep[8]=[49,10]第1個值，BabyStep移到下一組，indexB=indexB+1=8
BabyStep[8]=[57,11]第1個值大於GiantStep[8]=[49,10]第1個值，GiantStep移到下一組，indexG=indexG+1=9
BabyStep[8]=[57,11]第1個值小於GiantStep[9]=[62,5]第1個值，BabyStep移到下一組，indexB=indexB+1=9
BabyStep[9]=[64,6]第1個值大於GiantStep[9]=[62,5]第1個值，GiantStep移到下一組，indexG=indexG+1=10
BabyStep[9]=[64,6]第1個值小於GiantStep[10]=[79,2]第1個值，BabyStep移到下一組，indexB=indexB+1=10
BabyStep[10]=[75,8]第1個值小於GiantStep[10]=[79,2]第1個值，BabyStep移到下一組，indexB=indexB+1=11
BabyStep[11]=[114,12]第1個值大於GiantStep[10]=[79,2]第1個值，GiantStep移到下一組，indexG=indexG+1=11
BabyStep[11]=[114,12]第1個值大於GiantStep[11]=[81,9]第1個值，GiantStep移到下一組，indexG=indexG+1=12
BabyStep[11]=[114,12]第1個值大於GiantStep[12]=[108,0]第1個值，GiantStep移到下一組，indexG=indexG+1=13
BabyStep[11]=[114,12]第1個值小於GiantStep[13]=[126,3]第1個值，BabyStep移到下一組，indexB=indexB+1=12
BabyStep[12]=[119,10]第1個值小於GiantStep[13]=[126,3]第1個值，BabyStep移到下一組，indexB=indexB+1=13
BabyStep[13]=[128,7]第1個值大於GiantStep[13]=[126,3]第1個值，GiantStep移到下一組，indexG=indexG+1=14
BabyStep[13]=[128,7]第1個值小於GiantStep[14]=[147,6]第1個值，BabyStep移到下一組，indexB=indexB+1=14
BabyStep[14]=[150,9]第1個值大於GiantStep[14]=[147,6]第1個值，GiantStep移到下一組，indexG=indexG+1=15
(%o5)　[170]

3-1.解離散對數問題的方法-小步和倍增步長的大步(Baby-step and double step width Giant-step)

原本的小步大步	小步和倍增步長的大步
將答案改寫成x=im+j 其中m=\lceil\;\sqrt{p}\rceil\;，0\le i<m，0\le j<m \alpha^x \equiv \beta \pmod{p} \alpha^{im+j}\equiv \beta \pmod{p} \alpha^j \equiv \beta(\alpha^{-m})^i \pmod{p} 計算\alpha^j稱為Baby step 計算\beta(\alpha^{-m})^i稱為Giant step 每次步長都是\alpha^{-m}	將答案改寫成x=2^kvq+r　(x除以2^kv，商數q，餘數r) 其中v為偶數，k\ge 0，1\le r \le 2^kv \alpha^x \equiv \beta \pmod{p} \displaystyle \alpha^{2^kvq+r}\equiv \beta \pmod{p} \displaystyle \beta \alpha^{-r}\equiv (\alpha^{2^kv})^q \pmod{p} 計算 \beta\alpha^{-r}稱為Baby step 計算(\alpha^{2^kv})^q稱為Giant step 每次步長\alpha^{2^kv}隨著k值倍增

Lemma 2.1　設v為正偶數，對每一個正整數x，存在唯一的k \ge 0，q和r，\lfloor\;4^{k-1}\rfloor\;v^2\le 2^kvq <4^kv^2和1\le r\le 2^kv，使得x=y+r,y=2^kvq。

以2^x \equiv 108 \pmod{181}，參數v=2為例。
－－－－－－－－
當k=0時
BabyStep：1\le r \le 2^0v，1\le r \le 2，r=1,2
GiantStep：\lfloor\;4^{-1}\rfloor\;v^2\le 2^kvq<4^0v^2，0\le 2^kvq<4，2^kvq=0,2，倍增步長為2
可能x=1,2,3,4
－－－－－－－－
當k=1時
BabyStep：1\le r \le 2^1v，1\le r \le 4，r=1,2,3,4
GiantStep：\lfloor\;4^0\rfloor\;v^2\le 2^kvq<4^1v^2，4\le 2^kvq<16，2^kvq=4,8,12，倍增步長為4
可能x=5,6,\ldots,16
－－－－－－－－
當k=2時
BabyStep：1\le r \le 2^2v，1\le r \le 8，r=1,2,3,4,5,6,7,8
GiantStep：\lfloor\;4^1\rfloor\;v^2\le 2^kvq<4^2v^2，16\le 2^kvq<64，2^kvq=16,24,32,40,48,56，倍增步長為8
可能x=17,18,\ldots,64
－－－－－－－－
當k=3時
BabyStep：1\le r \le 2^3v，1\le r \le 16，r=1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16
GiantStep：\lfloor\;4^2\rfloor\;v^2\le 2^kvq<4^3v^2，64\le 2^kvq<256，2^kvq=64,80,96,112,128,144,160,176,192,208,224,240，倍增步長為16
可能x=65,66,\ldots,256，已經超過全部可能答案x=1,2,\ldots,181

參考資料：https://www.ams.org/journals/mco ... 5718-97-00880-6.pdf



Baby_stepDoubleGiant_step副程式
(%i1)
Baby_stepDoubleGiant_step(alpha,beta,p,v):=block
([k:0,BabyStep,GiantStep,GiantPower,indexB:1,indexG:1,LengthB,LengthG,x:[]],
if primep(p)=false then
  (print("參數p=",p,"要是質數"),
   return()
  ),
if mod(v,2)=1 then
  (print("參數v=",v,"要是偶數"),
   return()
  ),
do
  (print("當k=",k,"時"),
   print("1≦r≦",2^"k","v，1≦r≦",2^k*v),
   print(alpha^"-1","=",inv_mod(alpha,p),"(mod",p,")"),
   print("BabyStep=[",beta,"*",alpha^"-r","(mod",p,"),",r,"]=",
            BabyStep:create_list([mod(beta*power_mod(alpha,-r,p),p),r],r,1,2^k*v)),
   print("floor(",4^"k-1",")v"^"2","≦",2^"k","vq<",4^"k","v"^2,"，",floor(4^(k-1))*v^2,"≦",2^"k","vq<",4^k*v^2),
   print(2^"k","vq=",GiantPower:create_list(2^k*v*q,q,floor(4^(k-1))/2^k*v,2^k*v-1),
          ",倍增步長為",2^k*v),
   print("GiantStep=[",alpha^(2^"k"*"v"*q),"(mod",p,")",",",2^"k"*"v"*q,"]=",
            GiantStep:map(lambda([power],[power_mod(alpha,power,p),power]),GiantPower)),
   print("由小到大排序"),
   print("BabyStep=",BabyStep:sort(BabyStep)),
   print("GiantStep=",GiantStep:sort(GiantStep)),
   print("BabyStep和GiantStep互相比較找出相同值"),
   LengthB:length(BabyStep),
   LengthG:length(GiantStep),
   while indexB<=LengthB and indexG<=LengthG do
     (if BabyStep[indexB][1]=GiantStep[indexG][1] then
        (do
            (print("BabyStep[",indexB,"]=",BabyStep[indexB],"和","GiantStep[",indexG,"]=",GiantStep[indexG],"的第一個值相同"),
             print("得到一解x=",BabyStep[indexB][2],"+",GiantStep[indexG][2],
                   "=",last(x:append(x,[BabyStep[indexB][2]+GiantStep[indexG][2]]))),
             indexG:indexG+1,
             if indexG>LengthG or GiantStep[indexG][1]#GiantStep[indexG-1][1] then
               (return())
            ),
         indexB:indexB+1
        )
      else if BabyStep[indexB][1]<GiantStep[indexG][1] then
        (indexB:indexB+1)
      else
        (indexG:indexG+1)
     ),
   if x=[] then/*沒找到解答*/
     (print("BabyStep和GiantStep沒有相同值"))
   else/*找到解答跳出去*/
     (print("將x=",x,"(mod",p-1,")=",x:mod(x,p-1)),
      print("再刪除相同值x=",x:unique(x)),
      return(x)
     ),
   if 4^k*v^2>p then
     (print("無解"),
      return()
     ),
   k:k+1,/*換下一個k值*/
   print("－－－－－－－－")
  )
)$

\alpha^x\equiv \beta \pmod{p}，初始步長v要是偶數
(%i5)
alpha:2;
beta:108;
p:181;
v:2;
(alpha)　2
(beta)　108
(p)　181
(v)　2

利用Baby_stepDoubleGiant_step方法解離散對數問題
(%i6)　x:Baby_stepDoubleGiant_step(alpha,beta,p,v);
當k=0時
1\le r\le 2^k v，1\le r\le 2
2^{-1} \equiv 91 \pmod{181}
BabyStep=[ 108 * 2^{-r} \pmod{181}, r ]= [[54,1],[27,2]]
floor( 4^{k-1} )v^2 \le 2^k vq< 4^k v^2，0 \le 2^k vq< 4
2^k vq= [0,2],倍增步長為 2
GiantStep=[ 2^{2^kvq} \pmod{181},2^kvq]=[[1,0],[4,2]]
由小到大排序
BabyStep= [[27,2],[54,1]]
GiantStep= [[1,0],[4,2]]
BabyStep和GiantStep互相比較找出相同值
BabyStep和GiantStep沒有相同值
－－－－－－－－
當k=1時
1\le r\le 2^k v，1\le r\le 4
2^{-1} \equiv 91 \pmod{181}
BabyStep=[ 108 * 2^{-r}\pmod{181},r]=[[54,1],[27,2],[104,3],[52,4]]
floor( 4^{k-1})v^2 \le 2^k vq< 4^k v^2，4\le 2^k vq< 16
2^k vq= [4,8,12],倍增步長為 4
GiantStep=[ 2^{2^kvq} \pmod{181},2^kvq]=[[16,4],[75,8],[114,12]]
由小到大排序
BabyStep=[[27,2],[52,4],[54,1],[104,3]]
GiantStep=[[16,4],[75,8],[114,12]]
BabyStep和GiantStep互相比較找出相同值
BabyStep和GiantStep沒有相同值
－－－－－－－－
當k=2時
1\le r\le 2^k v，1\le r\le 8
2^{-1}\equiv 91 \pmod{181}
BabyStep=[ 108 * 2^{-r} \pmod{181},r]=[[54,1],[27,2],[104,3],[52,4],[26,5],[13,6],[97,7],[139,8]]
floor( 4^{k-1})v^2 \le 2^k vq< 4^k v^2，16\le 2^k vq< 64
2^k vq= [16,24,32,40,48,56],倍增步長為 8
GiantStep=[ 2^{2^kvq} \pmod{181},2^kvq]=[[14,16],[145,24],[15,32],[39,40],[29,48],[3,56]]
由小到大排序
BabyStep=[[13,6],[26,5],[27,2],[52,4],[54,1],[97,7],[104,3],[139,8]]
GiantStep=[[3,56],[14,16],[15,32],[29,48],[39,40],[145,24]]
BabyStep和GiantStep互相比較找出相同值
BabyStep和GiantStep沒有相同值
－－－－－－－－
當k=3時
1\le r\le 2^k v，1\le r\le 16
2^{-1}\equiv 91 \pmod{181}
BabyStep=[ 108 * 2^{-r} \pmod{181},r]=[[54,1],[27,2],[104,3],[52,4],[26,5],[13,6],
[97,7],[139,8],[160,9],[80,10],[40,11],[20,12],[10,13],[5,14],[93,15],[137,16]]
floor( 4^{k-1})v^2 \le 2^k vq< 4^k v^2，64 \le 2^k vq< 256
2^kvq=[64,80,96,112,128,144,160,176,192,208,224,240],倍增步長為 16
GiantStep=[ 2^{2^kvq} \pmod{181},2^kvq]=[[44,64],[73,80],[117,96],[9,112],
[126,128],[135,144],[80,160],[34,176],[114,192],[148,208],[81,224],[48,240]]
由小到大排序
BabyStep=[[5,14],[10,13],[13,6],[20,12],[26,5],[27,2],[40,11],[52,4],[54,1],[80,10],
[93,15],[97,7],[104,3],[137,16],[139,8],[160,9]]
GiantStep=[[9,112],[34,176],[44,64],[48,240],[73,80],[80,160],[81,224],[114,192],
[117,96],[126,128],[135,144],[148,208]]
BabyStep和GiantStep互相比較找出相同值
BabyStep[10]=[80,10]和GiantStep[6]=[80,160]的第一個值相同
得到一解x=10+160=170
將x=[170] \pmod{180}=[170]
再刪除相同值x=[170]
(x)　[170]

3-2.解離散對數問題的方法-小步和倍增步長的大步(Baby-step and double step width Giant-step)

若能理解上一篇文章，再回頭看論中的演算法2.2計算g的order。
g^x \equiv 1\pmod{p}
g^{y+r}\equiv 1 \pmod{p}
g^y \equiv g^{-r}\pmod{p}
計算g^{-r}稱為Baby step
計算g^y稱為Giant step

演算法2.2
輸入：g\in G，|\;g|\;的下限C+1(若已知解答x不會在1\sim C之間)，初始步長v(v要是偶數)
輸出：x=|\;g|\;
(1) x = 0
(2) s = 1; y = v; u = v
(3) h = g^{−1}
(4) a = h^C; b = g^v; c = b　　　　/*a = g^{−C−r};b = g^y; c = g^u*/
(5) R=\phi;
(6) while (x == 0) do
(7) 　for (r = s,s + 1\ldots, u) do　/*new baby steps*/
(8) 　　a=a*h
(9) 　　if (s==1) then　　　　　/*check if 1\le x\le v*/
(10) 　　　if (a==1) then
(11) 　　　　x=r+C
(12) 　　　　return (x)
(13) 　　　　break while
(14) 　　　else
(15) 　　　　R = R\cup \{\;(a,r)\}\;
(16) 　　　fi
(17) 　　else
(18) 　　　R=R\cup \{\;(a,r)\}\;
(19)  　　fi
(20) 　od
(21) 　while(x==0 and y<u^2)do　　/*giant steps*/
(22) 　　if(there is a number r such that (b,r)\in R) then
(23) 　　　x=y+r+C
(24) 　　　return(x)
(25) 　　else
(26) 　　　y=y+u
(27) 　　　b=b*c
(28) 　　fi
(29) 　od
(30) 　s=u+1;u=2u　　　　　　　/*double step-width*/
(31) 　c=c^2
(32) od



計算一個元素g的Order副程式
(%i1)
Order(g,p,C,v):=block
([x,r,s,y,u,h,a,b,c,R],
print("1.變數初始化－－－－－－－"),
print("x=0"),x:0,
print("s=",s:1),
print("y=v=",y:v),
print("u=v=",u:v),
print("h=g"^"-1","=",g^"-1","≡",h:inv_mod(g,p),"(mod",p,")"),
print("從",g^concat("-",C),"(mod",p,")開始計算，a=h"^"C","=",h,""^C,"≡",a:power_mod(h,C,p),"(mod",p,")"),
print("b=g"^"v","=",g,""^v,"≡",b:power_mod(g,v,p),"(mod",p,")"),
print("c=b=",c:b),
print("R=",R:[]),
while x=0 do
  (print("2.計算BabyStep－－－－－－－"),
   for r:s thru u do
     (print("計算",g^concat("-",C+r),"(mod",p,")","，a=a*h=",a,"*",h,"≡",a:mod(a*h,p),"(mod",p,")"),
      if s=1 then
        (if a=1 then
           (print("解答x=r+C=",r,"+",C,"=",x:r+C),
            return(x)
           )
         else
           (R:append(R,[[a,r]]),
            print("R=R∪",[a,r])
           )
        )
      else
        (R:append(R,[[a,r]]),
         print("R=R∪",[a,r])
        )
     ),
   print("BabyStep=",R),
   print("3.計算GiantStep－－－－－－－"),
   while x=0 and y<u^2 do
     (if (r:assoc(b,R))#false then
         (print("計算",g,""^y,"≡",b,"(mod",p,")在R[",r,"]=",R[r],"裡面"),
          print("解答x=y+r+C=",y,"+",r,"+",C,"=",x:y+r+C),
          return(x)
         )
      else
         (print("計算",g,""^y,"≡",b,"(mod",p,")不在R裡面"),
          print("更新次方y=y+u=",y,"+",u,"=",y:y+u),
          print("b=b*c=",b,"*",c,"≡",b:mod(b*c,p),"(mod",p,")")
         )
     ),
    print("s=u+1=",s:u+1),
    print("u=2*u=",u:2*u),
    print("倍增步長c=c"^2,"=",c:c^2)
  ),
return(x)
)$

g^x \equiv 1 \pmod{p}，C+1下限，初始步長v要是偶數
(%i5)
g:2;
p:181;
C:10;
v:2;
(g)　2
(p)　181
(C)　10
(v)　2

利用Baby_stepDoubleGiant_step方法計算g的order
(%i6)　x:Order(g,p,C,v);
1.變數初始化-------
x=0
s=1
y=v=2
u=v=2
h=g^{-1}= 2^{-1}\equiv 91\pmod{181}
從2^{-10}\pmod{181}開始計算，a=h^C = 91 ^{10} \equiv 108\pmod{181}
b=g^v = 2 ^2 \equiv 4\pmod{181}
c=b=4
R=[]
2.計算BabyStep-------
計算2^{-11}\pmod{181}，a=a*h= 108 * 91 \equiv 54\pmod{181}
R=R\cup[54,1]
計算2^{-12}\pmod{181}，a=a*h= 54 * 91 \equiv 27\pmod{181}
R=R\cup[27,2]
BabyStep=[[54,1],[27,2]]
3.計算GiantStep-------
計算2^2 \equiv 4\pmod{181}不在R裡面
更新次方y=y+u=2+2=4
b=b*c= 4 * 4 \equiv 16\pmod{181}
s=u+1=3
u=2*u=4
倍增步長c=c^2=16
2.計算BabyStep-------
計算2^{-13}\pmod{181}，a=a*h= 27 * 91 \equiv 104\pmod{181}
R=R\cup[104,3]
計算2^{-14}\pmod{181}，a=a*h= 104 * 91 \equiv 52\pmod{181}
R=R\cup[52,4]
BabyStep=[[54,1],[27,2],[104,3],[52,4]]
3.計算GiantStep-------
計算2^4 \equiv 16\pmod{181}不在R裡面
更新次方y=y+u=4+4=8
b=b*c= 16 * 16 \equiv 75\pmod{181}
計算2^8 \equiv 75\pmod{181}不在R裡面
更新次方y=y+u= 8 + 4 = 12
b=b*c= 75 * 16 \equiv 114\pmod{181}
計算2^{12} \equiv 114\pmod{181}不在R裡面
更新次方y=y+u=12+4=16
b=b*c= 114 * 16 \equiv 14\pmod{181}
s=u+1= 5
u=2*u= 8
倍增步長c=c^2 = 256
2.計算BabyStep-------
計算2^{-15}\pmod{181}，a=a*h= 52 * 91 \equiv 26\pmod{181}
R=R\cup[26,5]
計算2^{-16}\pmod{181}，a=a*h= 26 * 91 \equiv 13\pmod{181}
R=R\cup[13,6]
計算2^{-17}\pmod{181}，a=a*h= 13 * 91 \equiv 97\pmod{181}
R=R\cup[97,7]
計算2^{-18}\pmod{181}，a=a*h= 97 * 91 \equiv 139\pmod{181}
R=R\cup[139,8]
BabyStep=[[54,1],[27,2],[104,3],[52,4],[26,5],[13,6],[97,7],[139,8]]
3.計算GiantStep-------
計算2^{16} \equiv 14\pmod{181}不在R裡面
更新次方y=y+u=16+8=24
b=b*c= 14 * 256 \equiv 145\pmod{181}
計算2^{24} \equiv 145\pmod{181}不在R裡面
更新次方y=y+u= 24 + 8 = 32
b=b*c= 145 * 256 \equiv 15\pmod{181}
計算2^{32} \equiv 15\pmod{181}不在R裡面
更新次方y=y+u= 32 + 8 = 40
b=b*c= 15 * 256 \equiv 39\pmod{181}
計算2^{40} \equiv 39\pmod{181}不在R裡面
更新次方y=y+u= 40 + 8 = 48
b=b*c= 39 * 256 \equiv 29\pmod{181}
計算2^{48} \equiv 29\pmod{181}不在R裡面
更新次方y=y+u= 48 + 8 = 56
b=b*c= 29 * 256 \equiv 3\pmod{181}
計算2^{56} \equiv 3\pmod{181}不在R裡面
更新次方y=y+u= 56 + 8 = 64
b=b*c= 3 * 256 \equiv 44\pmod{181}
s=u+1= 9
u=2*u= 16
倍增步長c=c^2 = 65536
2.計算BabyStep-------
計算2^{-19}\pmod{181}，a=a*h= 139 * 91 \equiv 160\pmod{181}
R=R\cup[160,9]
計算2^{-20}\pmod{181}，a=a*h= 160 * 91 \equiv 80\pmod{181}
R=R\cup[80,10]
計算2^{-21}\pmod{181}，a=a*h= 80 * 91 \equiv 40\pmod{181}
R=R\cup[40,11]
計算2^{-22}\pmod{181}，a=a*h= 40 * 91 \equiv 20\pmod{181}
R=R\cup[20,12]
計算2^{-23}\pmod{181}，a=a*h= 20 * 91 \equiv 10\pmod{181}
R=R\cup[10,13]
計算2^{-24}\pmod{181}，a=a*h= 10 * 91 \equiv 5\pmod{181}
R=R\cup[5,14]
計算2^{-25}\pmod{181}，a=a*h= 5 * 91 \equiv 93\pmod{181}
R=R\cup[93,15]
計算2^{-26}\pmod{181}，a=a*h= 93 * 91 \equiv 137\pmod{181}
R=R\cup[137,16]
BabyStep=[[54,1],[27,2],[104,3],[52,4],[26,5],[13,6],[97,7],[139,8],
[160,9],[80,10],[40,11],[20,12],[10,13],[5,14],[93,15],[137,16]]
3.計算GiantStep-------
計算2^{64} \equiv 44\pmod{181}不在R裡面
更新次方y=y+u= 64 + 16 = 80
b=b*c= 44 * 65536 \equiv 73\pmod{181}
計算2^{80} \equiv 73\pmod{181}不在R裡面
更新次方y=y+u= 80 + 16 = 96
b=b*c= 73 * 65536 \equiv 117\pmod{181}
計算2^{96} \equiv 117\pmod{181}不在R裡面
更新次方y=y+u= 96 + 16 = 112
b=b*c= 117 * 65536 \equiv 9\pmod{181}
計算2^{112} \equiv 9\pmod{181}不在R裡面
更新次方y=y+u= 112 + 16 = 128
b=b*c= 9 * 65536 \equiv 126\pmod{181}
計算2^{128} \equiv 126\pmod{181}不在R裡面
更新次方y=y+u= 128 + 16 = 144
b=b*c= 126 * 65536 \equiv 135\pmod{181}
計算2^{144} \equiv 135\pmod{181}不在R裡面
更新次方y=y+u= 144 + 16 = 160
b=b*c= 135 * 65536 \equiv 80\pmod{181}
計算2^{160} \equiv 80\pmod{181}在R[10]=[80,10]裡面
解答x=y+r+C= 160 + 10 + 10 = 180
s=u+1= 17
u=2*u= 32
倍增步長c=c^2 = 4294967296
(x)　180



－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－
論文中演算法3.1將計算order和解離散對數問題結合在一起
g^x \equiv d \pmod{p}
g^{y+r}\equiv d \pmod{p}
d^{-1}g^y\equiv g^{-r}\pmod{p}
計算g^{-r}稱為Baby step
計算d^{-1}g^y稱為Giant step

輸入：d,g \in G(d\ne 1)，初始步長v(v要是偶數)
輸出：執行結果t=1，若d \in \langle\;g \rangle\;，則離散對數問題的解x=log_g d
　　　執行結果t=0，若d \notin \langle\;g \rangle\;，則g的oder為x=|\;\langle\;g \rangle\; |\;
(1) t=2;x=0
(2) s=1;y=v;u=v
(3) h=g^{−1}
(4) a=1;b=g^v;c=b　　　/*a=g^{−r},b=g^y,c=g^u*/
(5) R=\phi
(6) f=d^{−1}
(7) while(t==2)do
(8) 　for(r=s,s+1\ldots,u) do　　/*new baby steps*/
(9) 　　a=a*h
(10) 　　if(s==1)then　　　　　/*check if 1\le x\le v*/
(11) 　　　if(a==f)then
(12) 　　　　x=r;t=1
(13) 　　　　break while
(14) 　　　else
(15) 　　　　if(a==1)then
(16) 　　　　　x=r;t=0
(17) 　　　　　break while
(18) 　　　　else
(19) 　　　　　R=R\cup \{\;(a,r) \}\;
(20) 　　　　fi
(21) 　　　fi
(22) 　　else
(23) 　　　　R=R\cup \{\;(a,r) \}\;
(24)　　　fi
(25) 　od
(26) 　while(t==2 and y<u^2) do　　/*giant steps*/
(27) 　　if(there is a number r such　/*checking for discrete log.*/
(28) 　　　that (f*b,r)\in R) then
(29) 　　　x=y+r;t=1
(30) 　　else
(31) 　　　if (there is a number r such　　/*checking for the order of g*/
(32) 　　　　that (b,r)\in R) then
(33) 　　　　x=y+r;t=0
(34) 　　　else
(35) 　　　　y=y+u
(36) 　　　　b=b*c
(37) 　　　fi
(38) 　　fi
(39) 　od
(40) 　s=u+1;u=2u　　　　　　　　　　　/*double step-width*/
(41) 　c=c^2　　　　　　　　　　　　　　　
(42) od
(43) return(t,x)



利用Baby_stepDoubleGiant_step方法解離散對數問題副程式
(%i1)
DiscreteLogarithm(d,g,p,v):=block
([t,x,s,y,u,h,a,b,c,R,f,r],
print("1.變數初始化－－－－－－－－－－－－"),
print("t=",t:2),
print("x=",x:0),
print("s=",s:1),
print("y=v=",y:v),
print("u=v=",u:v),
print("h=","g"^"-1","=",g^"-1","≡",h:inv_mod(g,p),"(mod",p,")"),
print("a=",a:1),
print("b=g"^"v","=",b:g^v),
print("c=b=",c:b),
print("R=",R:[]),
print("f=","d"^"-1","=",d^"-1","≡",f:inv_mod(d,p),"(mod",p,")"),
while t=2 do
  (print("2.計算BabyStep－－－－－－－－－－－－"),
   for r:s thru u do
     (print("計算",g^concat("-",r),"(mod",p,")，a=a*h=",a,"*",h,"≡",a:mod(a*h,p),"(mod",p,")"),
      if s=1 then
        (if a=f then
           (print("x=r=",x:r),
            print("t=1",t:1),
            return(x)
           )
         else
           (if a=1 then
              (print("x=r=",x:r),
               print("t=0",t:0),
               return(x)
              )
            else
              (R:append(R,[[a,r]]),
               print("R=R∪",[a,r])
              )
           )
        )
      else
        (R:append(R,[[a,r]]),
         print("R=R∪",[a,r])
        )
     ),
   print("BabyStep=",R),
   print("3.計算GiantStep－－－－－－－－－－－－"),
   while t=2 and y<u^2 do
     (if (r:assoc(mod(f*b,p),R))#false then
        (print("計算",d^"-1","*",g,""^y,"(mod",p,")，f*b=",f,"*",b,"≡",mod(f*b,p),"(mod",p,")在R[",r,"]=",R[r],"裡面"),
         print("解答x=y+r=",y,"+",r,"=",x:y+r),
         print("狀態t=",t:1),
         return(x)
        )
      else
        (print("計算",d^"-1","*",g,""^y,"(mod",p,")，f*b=",f,"*",b,"≡",mod(f*b,p),"(mod",p,")不在R裡面"),
         print("更新次方，y=y+u=",y,"+",u,"=",y:y+u),
         print("計算",g,""^y,"(mod",p,")，b=b*c=",b,"*",c,"≡",b:mod(b*c,p),"(mod",p,")")
        )
     ),
    print("s=u+1=",s:u+1),
    print("u=2*u=",u:2*u),
    print("倍增步長，c=c"^2,"=",c:c^2),
    k:k+1
  ),
return([t,x])
)$

g^x \equiv d\pmod{p}，初始步長v要是偶數
(%i5)
d:108;
g:2;
p:181;
v:2;
(d)　108
(g)　2
(p)　181
(v)　2

利用Baby_stepDoubleGiant_step方法計算g的order或解離散對數問題
(%i6)　[t,x]: DiscreteLogarithm(d,g,p,v);
1.變數初始化------------
t=2
x=0
s=1
y=v=2
u=v=2
h=g^{-1}=2^{-1} \equiv 91 \pmod{181}
a=1
b=g^v=4
c=b=4
R=[]
f=d^{-1}=108^{-1} \equiv 119 \pmod{181}
2.計算BabyStep------------
計算2^{-1} \pmod{181}，a=a*h= 1*91 \equiv 91 \pmod{181}
R=R\cup [91,1]
計算2^{-2} \pmod{181}，a=a*h= 91*91 \equiv 136 \pmod{181}
R=R\cup [136,2]
BabyStep=[[91,1],[136,2]]
3.計算GiantStep------------
計算108^{-1}*2^2 \pmod{181}，f*b= 119*4 \equiv 114 \pmod{181}不在R裡面
更新次方，y=y+u=2+2=4
計算2^4 \pmod{181}，b=b*c= 4*4 \equiv 16 \pmod{181}
s=u+1=3
u=2*u=4
倍增步長，c=c^2=16
2.計算BabyStep------------
計算2^{-3} \pmod{181}，a=a*h= 136*91 \equiv 68 \pmod{181}
R=R\cup [68,3]
計算2^{-4} \pmod{181}，a=a*h= 68*91 \equiv 34 \pmod{181}
R=R\cup[34,4]
BabyStep=[[91,1],[136,2],[68,3],[34,4]]
3.計算GiantStep------------
計算108^{-1}*2^4 \pmod{181}，f*b= 119*16 \equiv 94 \pmod{181}不在R裡面
更新次方，y=y+u=4+4=8
計算2^8 \pmod{181}，b=b*c=16*16 \equiv 75 \pmod{181}
計算108^{-1}*2^8 \pmod{181}，f*b= 119*75 \equiv 56 \pmod{181}不在R裡面
更新次方，y=y+u=8+4=12
計算2^{12} \pmod{181}，b=b*c=75*16 \equiv 114 \pmod{181}
計算108^{-1}*2^{12} \pmod{181}，f*b= 119*114 \equiv 172 \pmod{181}不在R裡面
更新次方，y=y+u=12+4=16
計算2^{16} \pmod{181}，b=b*c=114*16 \equiv 14 \pmod{181}
s=u+1=5
u=2*u=8
倍增步長，c=c^2=256
2.計算BabyStep------------
計算2^{-5} \pmod{181}，a=a*h= 34*91 \equiv 17 \pmod{181}
R=R\cup [17,5]
計算2^{-6} \pmod{181}，a=a*h= 17*91 \equiv 99 \pmod{181}
R=R\cup [99,6]
計算2^{-7} \pmod{181}，a=a*h= 99*91 \equiv 140 \pmod{181}
R=R\cup [140,7]
計算2^{-8} \pmod{181}，a=a*h= 140*91 \equiv 70 \pmod{181}
R=R\cup [70,8]
BabyStep=[[91,1],[136,2],[68,3],[34,4],[17,5],[99,6],[140,7],[70,8]]
3.計算GiantStep------------
計算108^{-1}*2^{16} \pmod{181}，f*b= 119*14 \equiv 37 \pmod{181}不在R裡面
更新次方，y=y+u=16+8=24
計算2^{24} \pmod{181}，b=b*c= 14*256 \equiv 145 \pmod{181}
計算108^{-1}*2^{24} \pmod{181}，f*b= 119*145 \equiv 60 \pmod{181}不在R裡面
更新次方，y=y+u=24+8=32
計算2^{32} \pmod{181}，b=b*c= 145*256 \equiv 15 \pmod{181}
計算108^{-1}*2^{32} \pmod{181}，f*b= 119*15 \equiv 156 \pmod{181}不在R裡面
更新次方，y=y+u= 32 + 8=40
計算2^{40} \pmod{181}，b=b*c= 15*256 \equiv 39 \pmod{181}
計算108^{-1}*2^{40} \pmod{181}，f*b= 119*39 \equiv 116 \pmod{181}不在R裡面
更新次方，y=y+u= 40 + 8=48
計算2^{48} \pmod{181}，b=b*c= 39*256 \equiv 29 \pmod{181}
計算108^{-1}*2^{48} \pmod{181}，f*b= 119*29 \equiv 12 \pmod{181}不在R裡面
更新次方，y=y+u= 48 + 8=56
計算2^{56} \pmod{181}，b=b*c= 29*256 \equiv 3 \pmod{181}
計算108^{-1}*2^{56} \pmod{181}，f*b= 119*3 \equiv 176 \pmod{181}不在R裡面
更新次方，y=y+u= 56 + 8=64
計算2^{64} \pmod{181}，b=b*c= 3*256 \equiv 44 \pmod{181}
s=u+1=9
u=2*u=16
倍增步長，c=c^2=65536
2.計算BabyStep------------
計算2^{-9} \pmod{181}，a=a*h= 70*91 \equiv 35 \pmod{181}
R=R\cup [35,9]
計算2^{-10} \pmod{181}，a=a*h= 35*91 \equiv 108 \pmod{181}
R=R\cup [108,10]
計算2^{-11} \pmod{181}，a=a*h= 108*91 \equiv 54 \pmod{181}
R=R\cup [54,11]
計算2^{-12} \pmod{181}，a=a*h= 54*91 \equiv 27 \pmod{181}
R=R\cup [27,12]
計算2^{-13} \pmod{181}，a=a*h= 27*91 \equiv 104 \pmod{181}
R=R\cup [104,13]
計算2^{-14} \pmod{181}，a=a*h= 104*91 \equiv 52 \pmod{181}
R=R\cup [52,14]
計算2^{-15} \pmod{181}，a=a*h= 52*91 \equiv 26 \pmod{181}
R=R\cup [26,15]
計算2^{-16} \pmod{181}，a=a*h= 26*91 \equiv 13 \pmod{181}
R=R\cup [13,16]
BabyStep=[[91,1],[136,2],[68,3],[34,4],[17,5],[99,6],[140,7],[70,8],
[35,9],[108,10],[54,11],[27,12],[104,13],[52,14],[26,15],[13,16]]
3.計算GiantStep------------
計算108^{-1}*2^{64} \pmod{181}，f*b= 119*44 \equiv 168 \pmod{181}不在R裡面
更新次方，y=y+u= 64 + 16=80
計算2^{80} \pmod{181}，b=b*c= 44*65536 \equiv 73 \pmod{181}
計算108^{-1}*2^{80} \pmod{181}，f*b= 119*73 \equiv 180 \pmod{181}不在R裡面
更新次方，y=y+u= 80 + 16=96
計算2^{96} \pmod{181}，b=b*c= 73*65536 \equiv 117 \pmod{181}
計算108^{-1}*2^{96} \pmod{181}，f*b= 119*117 \equiv 167 \pmod{181}不在R裡面
更新次方，y=y+u= 96 + 16=112
計算2^{112} \pmod{181}，b=b*c= 117*65536 \equiv 9 \pmod{181}
計算108^{-1}*2^{112} \pmod{181}，f*b= 119*9 \equiv 166 \pmod{181}不在R裡面
更新次方，y=y+u= 112 + 16=128
計算2^{128} \pmod{181}，b=b*c= 9*65536 \equiv 126 \pmod{181}
計算108^{-1}*2^{128} \pmod{181}，f*b= 119*126 \equiv 152 \pmod{181}不在R裡面
更新次方，y=y+u= 128 + 16=144
計算2^{144} \pmod{181}，b=b*c= 126*65536 \equiv 135 \pmod{181}
計算108^{-1}*2^{144} \pmod{181}，f*b= 119*135 \equiv 137 \pmod{181}不在R裡面
更新次方，y=y+u= 144 + 16=160
計算2^{160} \pmod{181}，b=b*c= 135*65536 \equiv 80 \pmod{181}
計算108^{-1}*2^{160} \pmod{181}，f*b= 119*80 \equiv 108 \pmod{181}在R[10]=[108,10] 裡面
解答x=y+r=160+10=170
狀態t=1
s=u+1=17
u=2*u=32
倍增步長，c=c^2=4294967296
(%o6)　[1,170]

4.解離散對數問題的方法-小步和線性增加步長的大步

原本的小步大步	小步和倍增步長的大步	小步和線性增加步長的大步
將答案改寫成x=im+j 其中m=\lceil\;\sqrt{p}\rceil\;，0\le i<m，0\le j<m \alpha^x \equiv \beta \pmod{p} \alpha^{im+j}\equiv \beta \pmod{p} \alpha^j \equiv \beta(\alpha^{-m})^i \pmod{p} 計算\alpha^j稱為Baby step 計算\beta(\alpha^{-m})^i稱為Giant step 每次步長都是\alpha^{-m}	將答案改寫成x=2^kvq+r　(x除以2^kv，商數q，餘數r) 其中v為偶數，k\ge 0，1\le r \le 2^kv \alpha^x \equiv \beta \pmod{p} \displaystyle \alpha^{2^kvq+r}\equiv \beta \pmod{p} \displaystyle \beta \alpha^{-r}\equiv (\alpha^{2^kv})^q \pmod{p} 計算 \beta\alpha^{-r}稱為Baby step 計算(\alpha^{2^kv})^q稱為Giant step 每次步長\alpha^{2^kv}隨著k值倍增	將答案改寫成x=t_j-i 其中t_0=2v，t_{j+1}=t_j+j+v+1 \alpha^x \equiv \beta \pmod{p} \alpha^{t_j-i}\equiv \beta \pmod{p} \alpha^{t_j} \equiv \beta \alpha^i \pmod{p} 計算\beta \alpha^i稱為Baby Step 計算\alpha^{t_j}稱為Giant Step 每次步長\alpha^{t_j}隨著j值線性增加(t_{j+1}=t_j+j+v+1)

以2^n \equiv 1\pmod{181}，參數v=2為例。
\matrix{　&BabyStep&GiantStep\cr 　&2^0\equiv1\pmod{181}&　\cr 　&2^1\equiv2\pmod{181}&　\cr j=0&2^2\equiv4\pmod{181}&t=2v=4\cr j=j+1=1&2^{j+v}=2^3\equiv8\pmod{181}&t=t+j+v=4+1+2=7，2^7\equiv128\pmod{181}\cr j=j+1=2&2^{j+v}=2^4\equiv16\pmod{181}&t=t+j+v=7+2+2=11，2^{11}\equiv57\pmod{181}\cr j=j+1=3&2^{j+v}=2^5\equiv32\pmod{181}&t=t+j+v=11+3+2=16，2^{16}\equiv14\pmod{181}\cr j=j+1=4&2^{j+v}=2^6\equiv64\pmod{181}&t=t+j+v=16+4+2=22，2^{22}\equiv172\pmod{181}\cr j=j+1=5&2^{j+v}=2^7\equiv128\pmod{181}&t=t+j+v=22+5+2=29，2^{29}\equiv115\pmod{181}\cr j=j+1=6&2^{j+v}=2^8\equiv75\pmod{181}&t=t+j+v=29+6+2=37，2^{37}\equiv118\pmod{181}\cr j=j+1=7&2^{j+v}=2^9\equiv150\pmod{181}&t=t+j+v=37+7+2=46，2^{46}\equiv143\pmod{181}\cr j=j+1=8&2^{j+v}=2^{10}\equiv119\pmod{181}&t=t+j+v=46+8+2=56，2^{56}\equiv3\pmod{181}\cr j=j+1=9&2^{j+v}=\color{blue}{2^{11}\equiv57}\pmod{181}&t=t+j+v=56+9+2=67，2^{67}\equiv171\pmod{181}\cr j=j+1=10&2^{j+v}=2^{12}\equiv114\pmod{181}&t=t+j+v=67+10+2=79，2^{79}\equiv127\pmod{181}\cr j=j+1=11&2^{j+v}=2^{13}\equiv47\pmod{181}&t=t+j+v=79+11+2=92，2^{92}\equiv177\pmod{181}\cr j=j+1=12&2^{j+v}=2^{14}\equiv94\pmod{181}&t=t+j+v=92+12+2=106，2^{106}\equiv167\pmod{181}\cr j=j+1=13&2^{j+v}=2^{15}\equiv7\pmod{181}&t=t+j+v=106+13+2=121，2^{121}\equiv83\pmod{181}\cr j=j+1=14&2^{j+v}=2^{16}\equiv14\pmod{181}&t=t+j+v=121+14+2=137，2^{137}\equiv76\pmod{181}\cr j=j+1=15&2^{j+v}=2^{17}\equiv28\pmod{181}&t=t+j+v=137+15+2=154，2^{154}\equiv137\pmod{181}\cr j=j+1=16&2^{j+v}=2^{18}\equiv56\pmod{181}&t=t+j+v=154+16+2=172，2^{172}\equiv70\pmod{181}\cr j=j+1=17&2^{j+v}=2^{19}\equiv112\pmod{181}&t=t+j+v=172+17+2=191，\color{blue}{2^{191}\equiv57}\pmod{181}}
找到2^{11}=2^{191}\equiv 57\pmod{181}，故n=191-11=180，2^{180}\equiv 1\pmod{181}



演算法2.1，計算g的order。
輸入：g \in G，v \in Z，v\ge 2
輸出：n=|\;g|\;
(1)if (g == 1) then　　　　　　　　　　　　/* trivial case */
(2)　return (1)
(3)else
(4)　n = 0; i = 1; R ={(1, 0),(g,1)}; a = g　　　/* initialization */
(5)　while (n == 0 and i < v) do
(6)　　a = g*a; i = i + 1　　　　　　　　　　/* baby steps */
(7)　　if (a == 1) then
(8)　　　n = i　　　　　　　　　　　　　　/* found order among baby steps */
(9)　　　return (n)
(10)　　else
(11)　　　R = R ∪ {(a,i)}　　　　　　　　　/* update hash table */
(12)　　fi
(13)　od
(14)　j = 0; b = a*a; t = 2*v　　　　　　　　/* initialize giant steps (t = t_j) */
(15)　while (n == 0) do
(16)　　if (there exists a number i such that (b; i) 2 R) then  /* table lookup */
(17)　　　n = t - i　　　　　　　　　　　　　/* order of g */
(18)　　　return (n)
(19)　　　break while
(20)　　else
(21)　　　a = g*a; j = j + 1; R = R∪{(a,j + v)}　/*baby step */
(22)　　　b = a*b; t = t + j + v　　　　　　　　/* giant step */
(23)　　fi
(24)　od
(25)fi

參考資料：https://www.ams.org/journals/mco ... 5718-99-01141-2.pdf


計算一個元素g的Order副程式
(%i1)
Order(g,p,v):=block
([n,i,R,a],
if g=1 then
  (return(1)
  )
else
  (print("1.變數初始化－－－－－－－"),
   print("n=",n:0),
   print("i=",i:1),
   print("R=",R:[[1,0],[g,1]]),
   print("a=",a:g),
   print("2.計算i=2~v的BabyStep－－－－－－－"),
   while n=0 and i<v do
      (print("i=i+1=",i:i+1),
       print("計算",g,""^i,"(mod",p,")，","a=g*a=",g,"*",a,"≡",a:mod(g*a,p),"(mod",p,")"),
       if a=1 then
         (print("解答n=",n:i),
          return(n)
         )
       else
         (print("R=R∪[a,i]=",R:append(R,[[a,i]]))
         )
      ),
   print("3.計算BabyStep和GiantStep－－－－－－－"),
   print("j=",j:0),
   print("b=a*a=",a,"*",a,"=",b:a^2),
   print("t=2*v=2*",v,"=",t:2*v),
   while n=0 do
      (if (i:assoc(b,R))#false then
         (print("R=",R),
          print("b=",b,"在R[",i+1,"]=",R[i+1],"裡面"),
          print("解答n=t-i=",t,"-",i,"=",n:t-i),
          return(n)
         )
       else
         (print("b=",b,"不在R裡面"),
          print("更新次方j=j+1=",j:j+1),
          print("計算BabyStep，",g^"j+v","=",g,""^(j+v),"(mod",p,")，","a=g*a=",g,"*",a,"≡",a:mod(g*a,p),"(mod",p,")"),
          print("R=R∪",[a,j+v]),R:append(R,[[a,j+v]]),
          print("更新次方t=t+j+v=",t,"+",j,"+",v,"=",t:t+j+v),
          print("計算GiantStep，g"^"t","=",g,""^t,"(mod",p,")，b=a*b=",a,"*",b,"≡",b:mod(a*b,p),"(mod",p,")")
         ),
       print("－－－－－－－－")
      )
  )
)$

g^x \equiv 1\pmod{p}，初始值v
(%i4)
g:2;
p:181;
v:2;
(g)　2
(p)　181
(v)　2

利用Baby_stepLinearGiant_step方法計算g的order
(%i5)　n:Order(g,p,v);
1.變數初始化-------
n=0
i=1
R=[[1,0],[2,1]]
a=2
2.計算i=2~v的BabyStep-------
i=i+1= 2
計算2^2\pmod{181}，a=g*a=2 * 2 \equiv 4\pmod{181}
R=R\cup[a,i]= [[1,0],[2,1],[4,2]]
3.計算BabyStep和GiantStep-------
j=0
b=a*a=4*4=16
t=2*v=2*2=4
b=16不在R裡面
更新次方j=j+1=1
計算BabyStep，2^{j+v}=2^3\pmod{181}，a=g*a=2*4 \equiv 8\pmod{181}
R=R\cup [8,3]
更新次方t=t+j+v=4+1+2=7
計算GiantStep，g^t=2 ^7\pmod{181}，b=a*b=8*16 \equiv 128\pmod{181}
--------
b=128不在R裡面
更新次方j=j+1=2
計算BabyStep，2^{j+v}=2^4\pmod{181}，a=g*a=2*8 \equiv 16\pmod{181}
R=R\cup [16,4]
更新次方t=t+j+v=7+2+2=11
計算GiantStep，g^t=2^{11}\pmod{181}，b=a*b=16*128 \equiv 57\pmod{181}
--------
b=57不在R裡面
更新次方j=j+1=3
計算BabyStep，2^{j+v}=2^5\pmod{181}，a=g*a=2*16 \equiv 32\pmod{181}
R=R\cup [32,5]
更新次方t=t+j+v=11+3+2=16
計算GiantStep，g^t=2^{16}\pmod{181}，b=a*b=32*57 \equiv 14\pmod{181}
--------
b=14不在R裡面
更新次方j=j+1=4
計算BabyStep，2^{j+v}=2^6\pmod{181}，a=g*a=2*32 \equiv 64\pmod{181}
R=R\cup [64,6]
更新次方t=t+j+v=16+4+2=22
計算GiantStep，g^t=2 ^{22}\pmod{181}，b=a*b=64*14 \equiv 172\pmod{181}
--------
b=172不在R裡面
更新次方j=j+1=5
計算BabyStep，2^{j+v}=2^7\pmod{181}，a=g*a=2*64 \equiv 128\pmod{181}
R=R\cup [128,7]
更新次方t=t+j+v=22+5+2=29
計算GiantStep，g^t=2 ^{29}\pmod{181}，b=a*b=128*172 \equiv 115\pmod{181}
--------
b=115不在R裡面
更新次方j=j+1=6
計算BabyStep，2^{j+v}=2^8\pmod{181}，a=g*a=2*128 \equiv 75\pmod{181}
R=R\cup [75,8]
更新次方t=t+j+v=29+6+2=37
計算GiantStep，g^t=2^{37}\pmod{181}，b=a*b=75*115 \equiv 118\pmod{181}
--------
b=118不在R裡面
更新次方j=j+1=7
計算BabyStep，2^{j+v}=2^9\pmod{181}，a=g*a=2*75 \equiv 150\pmod{181}
R=R\cup [150,9]
更新次方t=t+j+v=37+7+2=46
計算GiantStep，g^t=2^{46}\pmod{181}，b=a*b=150*118 \equiv 143\pmod{181}
--------
b=143不在R裡面
更新次方j=j+1=8
計算BabyStep，2^{j+v}=2^{10}\pmod{181}，a=g*a=2*150 \equiv 119\pmod{181}
R=R\cup [119,10]
更新次方t=t+j+v=46+8+2=56
計算GiantStep，g^t=2^{56}\pmod{181}，b=a*b=119*143 \equiv 3\pmod{181}
--------
b=3不在R裡面
更新次方j=j+1=9
計算BabyStep，2^{j+v}=2^{11}\pmod{181}，a=g*a=2*119 \equiv 57\pmod{181}
R=R\cup [57,11]
更新次方t=t+j+v=56+9+2=67
計算GiantStep，g^t=2^{67}\pmod{181}，b=a*b=57*3 \equiv 171\pmod{181}
--------
b=171不在R裡面
更新次方j=j+1=10
計算BabyStep，2^{j+v}=2^{12}\pmod{181}，a=g*a=2*57 \equiv 114\pmod{181}
R=R\cup [114,12]
更新次方t=t+j+v=67+10+2=79
計算GiantStep，g^t=2^{79}\pmod{181}，b=a*b=114*171 \equiv 127\pmod{181}
--------
b=127不在R裡面
更新次方j=j+1=11
計算BabyStep，2^{j+v}=2^{13}\pmod{181}，a=g*a=2*114 \equiv 47\pmod{181}
R=R\cup [47,13]
更新次方t=t+j+v=79+11+2=92
計算GiantStep，g^t=2^{92}\pmod{181}，b=a*b=47*127 \equiv 177\pmod{181}
--------
b=177不在R裡面
更新次方j=j+1=12
計算BabyStep，2^{j+v}=2^{14}\pmod{181}，a=g*a=2*47 \equiv 94\pmod{181}
R=R\cup [94,14]
更新次方t=t+j+v=92+12+2=106
計算GiantStep，g^t=2^{106}\pmod{181}，b=a*b=94*177 \equiv 167\pmod{181}
--------
b=167不在R裡面
更新次方j=j+1=13
計算BabyStep，2^{j+v}=2^{15}\pmod{181}，a=g*a=2*94 \equiv 7\pmod{181}
R=R\cup [7,15]
更新次方t=t+j+v=106+13+2=121
計算GiantStep，g^t=2^{121}\pmod{181}，b=a*b=7*167 \equiv 83\pmod{181}
--------
b=83不在R裡面
更新次方j=j+1=14
計算BabyStep，2^{j+v}=2^{16}\pmod{181}，a=g*a=2*7 \equiv 14\pmod{181}
R=R\cup [14,16]
更新次方t=t+j+v=121+14+2=137
計算GiantStep，g^t=2^{137}\pmod{181}，b=a*b=14*83 \equiv 76\pmod{181}
--------
b=76不在R裡面
更新次方j=j+1=15
計算BabyStep，2^{j+v}=2^{17}\pmod{181}，a=g*a=2*14 \equiv 28\pmod{181}
R=R\cup [28,17]
更新次方t=t+j+v=137+15+2=154
計算GiantStep，g^t=2^{154}\pmod{181}，b=a*b=28*76 \equiv 137\pmod{181}
--------
b=137不在R裡面
更新次方j=j+1=16
計算BabyStep，2^{j+v}=2^{18}\pmod{181}，a=g*a=2*28 \equiv 56\pmod{181}
R=R\cup [56,18]
更新次方t=t+j+v=154+16+2=172
計算GiantStep，g^t=2^{172}\pmod{181}，b=a*b=56*137 \equiv 70\pmod{181}
--------
b=70不在R裡面
更新次方j=j+1=17
計算BabyStep，2^{j+v}=2^{19}\pmod{181}，a=g*a=2*56 \equiv 112\pmod{181}
R=R\cup [112,19]
更新次方t=t+j+v=172+17+2=191
計算GiantStep，g^t=2^{191}\pmod{181}，b=a*b=112*70 \equiv 57\pmod{181}
--------
R= [[1,0],[2,1],[4,2],[8,3],[16,4],[32,5],[64,6],[128,7],[75,8],[150,9],
[119,10],[57,11],[114,12],[47,13],[94,14],[7,15],[14,16],[28,17],[56,18],[112,19]]
b=57在R[12]=[57,11]裡面
解答n=t-i=191-11=180
(n)　180

5-1.低漢明權重(Low Hamming Weight)的離散對數問題-Heiman-Odlyzko方法

原本的離散對數問題\alpha^x \equiv \beta \pmod{p}的解x以二進位表示為
\displaystyle x=\sum_{i=0}^{m-1}x_i2^i，x_i \in \{\;0,1 \}\;，0\le i \le m-1，m=\lceil\;log_2 ord(\alpha)\rceil\;，ord(\alpha)代表\alpha的order
以wt(x)表示x的二進位中有多少個1，稱為x的漢明權重(Hamming Weight)。實務上以平方-相乘方法計算\alpha^x時，先將x表示成二進位，當遇到數字0時計算平方，遇到數字1時計算相乘，因為計算平方快於計算相乘，解x的二進位以較少的1來減少相乘運算，此時解x的漢明權重偏少，但x的漢明權重wt(x)=t太少時，有方法可以解出x。

以2^x\equiv 108 \pmod{181}，已知x的漢明權重為t=4為例。
m=\lceil\;log_2 ord(2)\rceil\;=\lceil\;log_2 180\rceil\;=8
Z_m=\{\; 0,1,2,3,4,5,6,7\}\;
取Z_m所有元素個數為2(\displaystyle |\;Y_1|\;=|\;Y_2|\;=\frac{t}{2}=2)的子集合Y_1和Y_2。
\matrix{Y_1或Y_2 & val(Y_1)或val(Y_2) & \alpha^{val(Y_1)}\pmod{181}& \beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;0,1\}\;&2^0+2^1=3&2^3\equiv8&108*2^{-3}\equiv104 \cr \{\;0,2\}\;&2^0+2^2=5&2^5\equiv32&108*2^{-5}\equiv26 \cr \{\;0,3\}\;&2^0+2^3=9&2^9\equiv150&108*2^{-9}\equiv160 \cr \{\;0,4\}\;&2^0+2^4=17&2^{17}\equiv28&108*2^{-17}\equiv159 \cr \{\;0,5\}\;&2^0+2^5=33&2^{33}\equiv30&108*2^{-33}\equiv76 \cr \{\;0,6\}\;&2^0+2^6=65&2^{65}\equiv88&108*2^{-65}\equiv174 \cr \{\;0,7\}\;&2^0+2^7=129&2^{129}\equiv71&108*2^{-129}\equiv78 \cr \{\;1,2\}\;&2^1+2^2=6&2^6\equiv64&108*2^{-6}\equiv13 \cr \{\;1,3\}\;&2^1+2^3=10&2^{10}\equiv119&108*2^{-10}\equiv80 \cr \{\;1,4\}\;&2^1+2^4=18&2^{18}\equiv56&108*2^{-18}\equiv170 \cr \{\;1,5\}\;&2^1+2^5=34&2^{34}\equiv60&108*2^{-34}\equiv38 \cr \{\;1,6\}\;&2^1+2^6=66&2^{66}\equiv176&108*2^{-66}\equiv87 \cr \{\;1,7\}\;&2^1+2^7=130&2^{130}\equiv142&108*2^{-130}\equiv39 \cr \{\;2,3\}\;&2^2+2^3=12&2^{12}\equiv114&108*2^{-12}\equiv20 \cr \{\;2,4\}\;&2^2+2^4=20&2^{20}\equiv43&108*2^{-20}\equiv133 \cr \{\;2,5\}\;&2^2+2^5=36&2^{36}\equiv59&108*2^{-36}\equiv100 \cr \{\;2,6\}\;&2^2+2^6=68&2^{68}\equiv161&108*2^{-68}\equiv67 \cr \{\;2,7\}\;&2^2+2^7=132&2^{132}\equiv25&108*2^{-132}\equiv55 \cr \{\;3,4\}\;&2^3+2^4=24&2^{24}\equiv145&108*2^{-24}\equiv178 \cr \{\;3,5\}\;&2^3+2^5=40&2^{40}\equiv39&108*2^{-40}\equiv142 \cr \{\;3,6\}\;&2^3+2^6=72&2^{72}\equiv42&108*2^{-72}\equiv106 \cr \{\;3,7\}\;&2^3+2^7=136&2^{136}\equiv38&108*2^{-136}\equiv60 \cr \{\;4,5\}\;&2^4+2^5=48&2^{48}\equiv29&108*2^{-48}\equiv166 \cr \{\;4,6\}\;&2^4+2^6=80&2^{80}\equiv73&108*2^{-80}\equiv180 \cr \{\;4,7\}\;&2^4+2^7=144&2^{144}\equiv135&108*2^{-144}\equiv37 \cr \{\;5,6\}\;&2^5+2^6=96&2^{96}\equiv117&108*2^{-96}\equiv168 \cr \{\;5,7\}\;&2^5+2^7=160&2^{160}\equiv80&108*2^{-160}\equiv119 \cr \{\;6,7\}\;&2^6+2^7=192&2^{192}\equiv114&108*2^{-192}\equiv20}
可找到
2^{136}\equiv 108*2^{-34}\equiv 38\pmod{181}，Y_1=\{\;3,7 \}\;,Y_2=\{\;1,5 \}\;，x=136+34=170
2^{40}\equiv 108*2^{-130}\equiv 39\pmod{181}，Y_1=\{\;3,5 \}\;,Y_2=\{\;1,7 \}\;，x=40+130=170
2^{34}\equiv 108*2^{-136}\equiv 60\pmod{181}，Y_1=\{\;1,5 \}\;,Y_2=\{\;3,7 \}\;，x=34+136=170
2^{160}\equiv 108*2^{-10}\equiv 80\pmod{181}，Y_1=\{\;5,7 \}\;,Y_2=\{\;1,3 \}\;，x=160+10=170
2^{10}\equiv 108*2^{-160}\equiv 119\pmod{181}，Y_1=\{\;1,3 \}\;,Y_2=\{\;5,7 \}\;，x=10+160=170
2^{130}\equiv 108*2^{-40}\equiv 142\pmod{181}，Y_1=\{\;1,7 \}\;,Y_2=\{\;3,5 \}\;，x=130+40=170
170=(10101010)_2漢明權重為4(4個1)。


演算法1
1. INPUT: \alpha,\beta \in G, an integer m and an even integer t
2. For all Y_1 \subseteq Z_m such that |\;Y_1|\;=t/2, compute \alpha^{val(Y_1)}
3. Sort the list of ordered pairs (val(Y_1),\alpha^{val(Y_1)}) by their second coordinates
4. For all Y_2 \subseteq Z_m such that |\;Y_1|\;=t/2, compute (\beta(\alpha^{val(Y_2)})^{-1})
5. Sort the list of ordered pairs (val(Y_2),\beta(\alpha^{val(Y_2)})^{-1}) by their second coordinates
6. If possible, find Y_1,Y_2 such that \alpha^{val(Y_1)}=\beta(\alpha^{val(Y_2)})^{-1}
7. If the previous step is successful, output log_{\alpha}\beta=(val(Y_1)+val(Y_2))\pmod{ord(\alpha)}. Otherwise, output fail.

參考資料：https://citeseerx.ist.psu.edu/vi ... p=rep1&type=pdf


從L中取n個元素的子集合副程式
(%i1)
Combination(L,n):=block
([c:[],s],
if length(L)>n then
  (for r in L do
     (s:delete(r,L),
      c:unique(append(c,Combination(s,n)))
     )
  )
else
  (c:[L]),
return(c)
)$

例子：從[0,1,2,3]中取2個元素的子集合，共C(4,2)=6個
(%i2)　Combination([0,1,2,3],2);
(%o2)　[[0,1],[0,2],[0,3],[1,2],[1,3],[2,3]]

解低漢明權重的離散對數問題Heiman-Odlyzko方法副程式
(%i3)
LowHammingWeightDLP1(alpha,beta,p,t):=block
([m,order,Zm:[],Y:[],valY:[],lenY,inv_alpha,ListY1:[],ListY2:[],indexY1:1,indexY2:1,x1:0,x2:0,x:0],
if mod(t,2)#0 then
  (print("t要是偶數"),return()),
print("ord(",alpha,")=",order:zn_order(alpha,p),"，",alpha^string(order),"≡1(mod",p,")"),
print("m=Ceiling(log2",order,")=",m:ceiling(log(order)/log(2))),
print("Zm=",Zm:makelist(i,i,0,m-1)),
print("從Zm取元素個數為","t"/2,"=",t/2,"個的子集合Y，共C(",m,",",t/2,")=",lenY:m!/((m-t/2)!*(t/2)!),"個"),
print("Y=",Y:Combination(Zm,t/2)),
print("val(Y)=Σ2"^"y","=",valY:create_list(apply("+",y),y,2^Y)),
print("計算[",alpha^"val(Y1)",",","val(Y1)]數對"),
print("ListY1=",ListY1:create_list([power_mod(alpha,valY[ i ],p),valY[ i ]],i,1,lenY)),
inv_alpha:inv_mod(alpha,p),
print("計算[",beta,"*",alpha^"-val(Y2)",",","val(Y2)]數對"),
print("ListY2=",ListY2:create_list([mod(beta*power_mod(inv_alpha,valY[ i ],p),p),valY[ i ]],i,1,lenY)),
print("由小到大排序"),
print("ListY1=",ListY1:sort(ListY1)),
print("ListY2=",ListY2:sort(ListY2)),
print("ListY1和ListY2互相比較找出相同值"),
while indexY1<=lenY and indexY2<=lenY do
  (if ListY1[indexY1][1]=ListY2[indexY2][1] then
     (x1: ListY1[indexY1][2],x2: ListY2[indexY2][2],
      print(alpha^string(x1),"≡",beta,"*",alpha^concat("-",x2),"≡",ListY1[indexY1][1],"(mod",p,")"),
      indexY1:indexY1+1,
      indexY2:indexY2+1
     )
   else if ListY1[indexY1][1]<ListY2[indexY2][1] then
     (indexY1:indexY1+1)
    else
     (indexY2:indexY2+1)
  ),
if x1=0 and x2=0 then
  (print("無解"),
   return()
  )
else
  (return(x:mod(x1+x2,order)))
)$

\alpha^x \equiv \beta\pmod{p}，已知x的漢明權重t=4
(%i7)
alpha:2;
beta:108;
p:181;
t:4;
(alpha)　2
(beta)　108
(p)　181
(t)　4

解低漢明權重的離散對數問題-Heiman-Odlyzko方法
(%i8)　x: LowHammingWeightDLP1(alpha,beta,p,t);
ord(2)=180，2^{180}\equiv 1 \pmod{181}
m=Ceiling(log_2 180)=8
Zm=[0,1,2,3,4,5,6,7]
從Zm取元素個數為\displaystyle \frac{t}{2}=2個的子集合Y，共C(8,2)=28個
Y=[[0,1],[0,2],[0,3],[0,4],[0,5],[0,6],[0,7],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7],
　　[2,3],[2,4],[2,5],[2,6],[2,7],[3,4],[3,5],[3,6],[3,7],[4,5],[4,6],[4,7],[5,6],[5,7],[6,7]]
val(Y)=Σ2^y = [3,5,9,17,33,65,129,6,10,18,34,66,130,12,20,36,68,132,24,40,72,136,48,80,144,96,160,192]
計算[ 2^{val(Y1)},val(Y1)]數對
ListY1= [[8,3],[32,5],[150,9],[28,17],[30,33],[88,65],[71,129],[64,6],[119,10],[56,18],[60,34],[176,66],[142,130],[114,12],
　　　　　[43,20],[59,36],[161,68],[25,132],[145,24],[39,40],[42,72],[38,136],[29,48],[73,80],[135,144],[117,96],[80,160],[114,192]]
計算[ 108 * 2^{-val(Y2)},val(Y2)]數對
ListY2= [[104,3],[26,5],[160,9],[159,17],[76,33],[174,65],[78,129],[13,6],[80,10],[170,18],[38,34],[87,66],[39,130],[20,12],
　　　　　[133,20],[100,36],[67,68],[55,132],[178,24],[142,40],[106,72],[60,136],[166,48],[180,80],[37,144],[168,96],[119,160],[20,192]]
由小到大排序
ListY1= [[8,3],[25,132],[28,17],[29,48],[30,33],[32,5],[38,136],[39,40],[42,72],[43,20],[56,18],[59,36],[60,34],[64,6],
　　　　　[71,129],[73,80],[80,160],[88,65],[114,12],[114,192],[117,96],[119,10],[135,144],[142,130],[145,24],[150,9],[161,68],[176,66]]
ListY2= [[13,6],[20,12],[20,192],[26,5],[37,144],[38,34],[39,130],[55,132],[60,136],[67,68],[76,33],[78,129],[80,10],[87,66],
　　　　　[100,36],[104,3],[106,72],[119,160],[133,20],[142,40],[159,17],[160,9],[166,48],[168,96],[170,18],[174,65],[178,24],[180,80]]
ListY1和ListY2互相比較找出相同值
2^{136} \equiv 108 * 2^{-34} \equiv 38\pmod{181}
2^{40} \equiv 108 * 2^{-130} \equiv 39\pmod{181}
2^{34} \equiv 108 * 2^{-136} \equiv 60\pmod{181}
2^{160} \equiv 108 * 2^{-10} \equiv 80\pmod{181}
2^{10} \equiv 108 * 2^{-160} \equiv 119\pmod{181}
2^{130} \equiv 108 * 2^{-40} \equiv 142\pmod{181}
(x)　170

5-2.低漢明權重(Low Hamming Weight)的離散對數問題-Coppersmith方法

從上一篇文章可以發現Heiman-Odlyzko方法的關鍵在於列舉解x=(10101010)_2中1的可能位置Y_1,Y_2，Y_1,Y_2各有C(m,\frac{t}{2})=C(8,2)=28種，當\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}時，解x=val(Y_1)+val(Y_2)\pmod{ord(\alpha)}。

Coppersmith方法將解x二進位切成兩等份，每個等份再分別列舉1的可能位置，
\{\;0,1,2,3 \}\;任取2個的子集合Y_1，\{\;4,5,6,7 \}\;任取2個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;0,1\}\;&2^0+2^1=3&2^3\equiv 8&\{\;4,5\}\;&2^4+2^5=48&108*2^{-48}\equiv 166\cr \{\;0,2\}\;&2^0+2^2=5&2^5\equiv 32&\{\;4,6\}\;&2^4+2^6=80&108*2^{-80}\equiv 180\cr \{\;0,3\}\;&2^0+2^3=9&2^9\equiv 150&\{\;4,7\}\;&2^4+2^7=144&108*2^{-144}\equiv 37\cr \{\;1,2\}\;&2^1+2^2=6&2^6\equiv 64&\{\;5,6\}\;&2^5+2^6=96&108*2^{-96}\equiv 168\cr \{\;1,3\}\;&2^1+2^3=10&2^{10}\equiv 119&\{\;5,7\}\;&2^5+2^7=160&108*2^{-160}\equiv 119\cr \{\;2,3\}\;&2^2+2^3=12&2^{12}\equiv 114&\{\;6,7\}\;&2^6+2^7=192&108*2^{-192}\equiv 20}
可找到
2^{10}\equiv 108*2^{-160}\equiv 119 \pmod{181}，Y_1=\{\;1,3 \}\;，Y_2=\{\;5,7 \}\;，x=10+160=170


再以2^{142}\equiv 79\pmod{181}為例，解x=(142)_{10}=(10001110)_2，1的分佈不平均。
\{\;0,1,2,3 \}\;任取2個的子集合Y_1，\{\;4,5,6,7 \}\;任取2個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;0,1\}\;&2^0+2^1=3&2^3\equiv 8&\{\;4,5\}\;&2^4+2^5=48&79*2^{-48}\equiv 165\cr \{\;0,2\}\;&2^0+2^2=5&2^5\equiv 32&\{\;4,6\}\;&2^4+2^6=80&79*2^{-80}\equiv 11\cr \{\;0,3\}\;&2^0+2^3=9&2^9\equiv 150&\{\;4,7\}\;&2^4+2^7=144&79*2^{-144}\equiv 136\cr \{\;1,2\}\;&2^1+2^2=6&2^6\equiv 64&\{\;5,6\}\;&2^5+2^6=96&79*2^{-96}\equiv 143\cr \{\;1,3\}\;&2^1+2^3=10&2^{10}\equiv 119&\{\;5,7\}\;&2^5+2^7=160&79*2^{-160}\equiv 139\cr \{\;2,3\}\;&2^2+2^3=12&2^{12}\equiv 114&\{\;6,7\}\;&2^6+2^7=192&79*2^{-192}\equiv 142}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

Coppersmitth方法將Y_1,Y_2各平移1個元素再重新列舉
\{\;1,2,3,4 \}\;任取2個的子集合Y_1，\{\;5,6,7,0 \}\;任取2個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;1,2\}\;&2^1+2^2=6&2^6\equiv 64&\{\;0,5\}\;&2^0+2^5=33&79*2^{-33}\equiv 69\cr \{\;1,3\}\;&2^1+2^3=10&2^{10}\equiv 119&\{\;0,6\}\;&2^0+2^6=65&79*2^{-65}\equiv 77\cr \{\;1,4\}\;&2^1+2^4=18&2^{18}\equiv 56&\{\;0,7\}\;&2^0+2^7=129&79*2^{-129}\equiv 47\cr \{\;2,3\}\;&2^2+2^3=12&2^{12}\equiv 114&\{\;5,6\}\;&2^5+2^6=96&79*2^{-96}\equiv 143\cr \{\;2,4\}\;&2^2+2^4=20&2^{20}\equiv 43&\{\;5,7\}\;&2^5+2^7=160&79*2^{-160}\equiv 139\cr \{\;3,4\}\;&2^3+2^4=24&2^{24}\equiv 145&\{\;6,7\}\;&2^6+2^7=192&79*2^{-192}\equiv 142}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

再將Y_1,Y_2各平移1個元素再重新列舉
\{\;2,3,4,5 \}\;任取2個的子集合Y_1，\{\;6,7,0,1 \}\;任取2個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;2,3\}\;&2^2+2^3=12&2^{12}\equiv 114&\{\;0,1\}\;&2^0+2^1=3&79*2^{-3}\equiv 123\cr \{\;2,4\}\;&2^2+2^4=20&2^{20}\equiv 43&\{\;0,6\}\;&2^0+2^6=65&79*2^{-65}\equiv 77\cr \{\;2,5\}\;&2^2+2^5=36&2^{36}\equiv 59&\{\;0,7\}\;&2^0+2^7=129&79*2^{-129}\equiv 47\cr \{\;3,4\}\;&2^3+2^4=24&2^{24}\equiv 145&\{\;1,6\}\;&2^1+2^6=66&79*2^{-66}\equiv 129\cr \{\;3,5\}\;&2^3+2^5=40&2^{40}\equiv 39&\{\;1,7\}\;&2^1+2^7=130&79*2^{-130}\equiv 114\cr \{\;4,5\}\;&2^4+2^5=48&2^{48}\equiv 29&\{\;6,7\}\;&2^6+2^7=192&79*2^{-192}\equiv 142}
可找到
2^{12}\equiv 79 *2^{-130}\equiv 114\pmod{181}，Y_1=\{\;2,3 \}\;，Y_2=\{\;1,7 \}\;，x=12+130=142


Splitting families
假設m和t都是偶數且0<t<m，定義(m,t)的splitting system是符合以下2點的數對(X,\cal{B})
1.集合X有m個元素，集合\beta有\displaystyle \frac{m}{2}個X的子集合，\cal{B}稱為區塊(blocks)。
2.對所有X的子集合Y，Y有t個元素，存在一個區塊B使得B和Y的交集個數為\displaystyle \frac{t}{2}。

Coppersmith也提供一種產生\displaystyle (\frac{m}{2};m,t)的splitting system方法。
令X=Z_m，定義B_i=\{\;i+j\pmod{m}:0\le j\le m/2-1\}\;，\cal B=\{\;B_i:0\le i \le m/2-1\}\;

以m=8,t=4為例
X=Z_m=\{\;0,1,2,3,4,5,6,7\}\;，B_0=\{\;0,1,2,3\}\;，B_1=\{\;1,2,3,4\}\;，B_2=\{\;2,3,4,5\}\;，B_3=\{\;3,4,5,6\}\;，\cal B=\{\;B_0,B_1,B_2,B_3\}\;
子集合Y=\{\;0,2,4,6\}\;有4個元素，存在B_0,B_1,B_2,B_3和Y交集2個元素。
子集合Y=\{\;1,3,5,7\}\;有4個元素，存在B_0,B_1,B_2,B_3和Y交集2個元素。
子集合Y=\{\;0,1,6,7\}\;有4個元素，存在B_0和Y交集2個元素。
子集合Y=\{\;0,2,5,7\}\;有4個元素，存在B_0和Y交集2個元素。
子集合Y=\{\;4,5,6,7\}\;有4個元素，存在B_2,B_3和Y交集2個元素。

演算法2
1. INPUT:\alpha,\beta \in G, even integers m and t, and an (N; m, t)-SS, (Z_m,\cal{B}), where \cal B= \{\;B_i : 0 \le i \le N − 1\}\;.
2. For i = 0,\ldots,N−1, perform the following steps:
(a) For all Y_1\subseteq B_i such that |\;Y_1|\; = t/2, compute \alpha^{val(Y_1)}
(b) Sort the list of ordered pairs (val(Y_1), \alpha^{val(Y_1)}) by their second coordinates
(c) For all Y_2 \subseteq Z_m\backslash\; B_i such that |\;Y_2|\; = t/2, compute \beta(\alpha^{val(Y_2)})^{−1}
(d) Sort the list of ordered pairs (val(Y_2), \beta(\alpha^{val(Y_2)})^{−1}) by their second coordinates
(e) If possible, find Y_1, Y_2 such that \alpha^{val(Y_1)} = \beta(\alpha^{val(Y_2)})^{−1}.
(f) If the previous step is successful, output log_{\alpha}\beta=val(Y_1)+val(Y_2) \pmod{ord(\alpha)}and QUIT. Otherwise, proceed to the next iteration of the FOR loop.

演算法3則隨機產生集合B，其中包含\displaystyle \frac{m}{2}個子集合，其餘步驟和演算法1,2相同。
演算法 3
1. INPUT: \alpha,\beta \in G, and even integers m and t.
2. REPEAT the following steps:
(a) Let B be a random \displaystyle \frac{m}{2}-subset of X
(b) For all Y_1 \subseteq B such that |\;Y_1|\;=t/2, compute \alpha^{val(Y_1)}
(c) Sort the list of ordered pairs (val(Y_1), \alpha^{val(Y_1)}) by their second coordinates
(d) For all Y_2 \subseteq Z_m\backslash\;B such that |\;Y_2|\;=t/2, compute \beta(\alpha^{val(Y_2)})^{−1}
(e) Sort the list of ordered pairs (val(Y_2),\beta(\alpha^{val(Y_2)})^{−1}) by their second coordinates
(f) If possible, find Y_1, Y_2 such that \alpha^{val(Y_1)} =\beta(\alpha^{val(Y_2)})^{−1}.
(g) If the previous step is successful, output log_{\alpha}\beta=val(Y_1)+val(Y_2) \pmod{ord(\alpha)} and QUIT. Otherwise, proceed to the next iteration of the REPEAT loop.

參考資料：https://citeseerx.ist.psu.edu/vi ... p=rep1&type=pdf


從L中取n個元素的子集合副程式
(%i1)
Combination(L,n):=block
([c:[],s],
if length(L)>n then
  (for r in L do
     (s:delete(r,L),
      c:unique(append(c,Combination(s,n)))
     )
  )
else
  (c:[L]),
return(c)
)$

解低漢明權重的離散對數問題Coppersmith方法副程式
(%i2)
LowHammingWeightDLP2(alpha,beta,p,t):=block
([m,order,Beta,inv_alpha,i,Y1,Y2,valY1,valY2,lenY1,lenY2,ListY1,ListY2,indexY1:1,indexY2:1,x1:0,x2:0,x:0],
if mod(t,2)#0 then
  (print("t要是偶數"),return()),
print("ord(",alpha,")=",order:zn_order(alpha,p),"，",alpha^string(order),"≡1(mod",p,")"),
print("m=Ceiling(log2",order,")=",m:ceiling(log(order)/log(2))),
if mod(m,2)#0 then
  (print("m要是偶數"),return()),
print("Zm=",create_list(i,i,0,m-1)),
print("Bi={i+j (mod m):0≤j≤m/2-1}，β={Bi:0≤i≤m/2-1}"),
print("β=",Beta:create_list(create_list(mod(i+j,m),i,0,m/2-1),j,0,m/2-1)),
print("－－－－－－－－"),
inv_alpha:inv_mod(alpha,p),
for i:1 thru m/2 do
  (print("B",i-1,"=",Beta[ i ]),
   print("從B",i-1,"取元素個數為","t"/2,"=",t/2,"個的子集合Y1，共C(","m"/2,",","t"/2,")=C(",m/2,",",t/2,")=",lenY1: (m/2)!/((t/2)!*(m/2-t/2)!),"個"),
   print("Y1=",Y1:Combination(Beta[ i ],t/2)),
   print("val(Y1)=Σ2"^"y","=",valY1:create_list(apply("+",y),y,2^Y1)),
   print("計算[",alpha^"val(Y1)",",val(Y1)]數對"),
   print("ListY1=",ListY1:create_list([power_mod(alpha,valY1[ i ],p),valY1[ i ]],i,1,lenY1)),
   print("Zm\\B",i-1,"=",mod(Beta[ i ]+m/2,m)),
   print("從Zm\\B",i-1,"取元素個數為","t"/2,"=",t/2,"個的子集合Y1，共C(","m"/2,",","t"/2,")=C(",m/2,",",t/2,")=",lenY2: (m/2)!/((t/2)!*(m/2-t/2)!),"個"),
   print("Y2=",Y2:Combination(sort(mod(Beta[ i ]+m/2,m)),t/2)),
   print("val(Y2)=Σ2"^"y","=",valY2:create_list(apply("+",y),y,2^Y2)),
   print("計算[",beta,"*",alpha^"-val(Y2)",",","val(Y2)]數對"),
   print("ListY2=",ListY2:create_list([mod(beta*power_mod(inv_alpha,valY2[ i ],p),p),valY2[ i ]],i,1,lenY2)),
   print("由小到大排序"),
   print("ListY1=",ListY1:sort(ListY1)),
   print("ListY2=",ListY2:sort(ListY2)),
   indexY1:1,indexY2:1,x1:0,x2:0,
   while indexY1<=lenY1 and indexY2<=lenY2 do
     (if ListY1[indexY1][1]=ListY2[indexY2][1] then
        (print("ListY1和ListY2互相比較找到相同值"),
         x1: ListY1[indexY1][2],x2: ListY2[indexY2][2],
         print(alpha^string(x1),"≡",beta,"*",alpha^concat("-",x2),"≡",ListY1[indexY1][1],"(mod",p,")"),
         print("x=",x1,"+",x2,"=",x:mod(x1+x2,order)),
         indexY1:indexY1+1,
         indexY2:indexY2+1
        )
      else if ListY1[indexY1][1]<ListY2[indexY2][1] then
        (indexY1:indexY1+1)
       else
        (indexY2:indexY2+1)
     ),
   if x1=0 and x2=0 then
      (print("ListY1和ListY2互相比較沒找到相同值")),
   print("－－－－－－－－")
  ),
if x=0 then
  (print("無解"),
   return()
  )
else
  (return(x))
)$

\alpha^x \equiv \beta \pmod{p}，已知x的漢明權重t=4
(%i6)
alpha:2;
beta:79;
p:181;
t:4;
(alpha)　2
(beta)　108
(p)　181
(t)　4

解低漢明權重的離散對數問題-Coppersmith方法
(%i7)　x: LowHammingWeightDLP2(alpha,beta,p,t);
ord(2)=180，2^{180}\equiv 1\pmod{181}
m=Ceiling(log_2 180)=8
Zm= [\;0,1,2,3,4,5,6,7]\;
Bi=\{\; i+j \pmod{m}:0 \le j \le m/2-1 \}\;，\beta=\{\;Bi:0 \le i \le m/2-1 \};
\beta=[\;[\;0,1,2,3]\;,[\;1,2,3,4]\;,[\;2,3,4,5]\;,[\;3,4,5,6]\;]\;
－－－－－－－－
B0=[\;0,1,2,3]\;
從B0取元素個數為\displaystyle \frac{t}{2}=2個的子集合Y1，共\displaystyle C(\frac{m}{2},\frac{t}{2})=C(4,2)=6個
Y1=[\;[\;0,1]\;,[\;0,2]\;,[\;0,3]\;,[\;1,2]\;,[\;1,3]\;,[\;2,3]\;]\;
val(Y1)=Σ2^y = [\;3,5,9,6,10,12]\;
計算[\; 2^{val(Y1)},val(Y1)]\;數對
ListY1=[\;[\;8,3]\;,[\;32,5]\;,[\;150,9]\;,[\;64,6]\;,[\;119,10]\;,[\;114,12]\;]\;
Zm\backslash\; B0=[\;4,5,6,7]\;
從Zm \backslash\;B0取元素個數為\displaystyle \frac{t}{2}=2個的子集合Y1，共\displaystyle C(\frac{m}{2},\frac{t}{2})=C(4,2)=6個
Y2=[\;[\;4,5]\;,[\;4,6]\;,[\;4,7]\;,[\;5,6]\;,[\;5,7]\;,[\;6,7]\;]\;
val(Y2)=Σ2^y=[\;48,80,144,96,160,192]\;
計算[\;79 * 2^{-val(Y2)}, val(Y2)]\;數對
ListY2=[\;[\;165,48]\;,[\;11,80]\;,[\;136,144]\;,[\;143,96]\;,[\;139,160]\;,[\;142,192]\;]\;
由小到大排序
ListY1=[\;[\;8,3]\;,[\;32,5]\;,[\;64,6]\;,[\;114,12]\;,[\;119,10]\;,[\;150,9]\;]\;
ListY2=[\;[\;11,80]\;,[\;136,144]\;,[\;139,160]\;,[\;142,192]\;,[\;143,96]\;,[\;165,48]\;]\;
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B1 = [\;1,2,3,4]\;
從B1取元素個數為\displaystyle  \frac{t}{2}=2個的子集合Y1，共\displaystyle C(\frac{m}{2},\frac{t}{2})=C(4,2)=6個
Y1= [\;[\;1,2]\;,[\;1,3]\;,[\;1,4]\;,[\;2,3]\;,[\;2,4]\;,[\;3,4]\;]\;
val(Y1)=Σ2^y = [\;6,10,18,12,20,24]\;
計算[\; 2^{val(Y1)},val(Y1)]\;數對
ListY1=[\;[\;64,6]\;,[\;119,10]\;,[\;56,18]\;,[\;114,12]\;,[\;43,20]\;,[\;145,24]\;]\;
Zm\backslash\;B1 = [\;5,6,7,0]\;
從Zm\backslash\;B1取元素個數為\displaystyle  \frac{t}{2}=2個的子集合Y1，共\displaystyle C(\frac{m}{2},\frac{t}{2})=C(4,2)=6個
Y2= [\;[\;0,5]\;,[\;0,6]\;,[\;0,7]\;,[\;5,6]\;,[\;5,7]\;,[\;6,7]\;]\;
val(Y2)=Σ2^y = [\;33,65,129,96,160,192]\;
計算[\; 79 * 2^{-val(Y2)}, val(Y2)]\;數對
ListY2=[\;[\;69,33]\;,[\;77,65]\;,[\;47,129]\;,[\;143,96]\;,[\;139,160]\;,[\;142,192]\;]\;
由小到大排序
ListY1=[\;[\;43,20]\;,[\;56,18]\;,[\;64,6]\;,[\;114,12]\;,[\;119,10]\;,[\;145,24]\;]\;
ListY2=[\;[\;47,129]\;,[\;69,33]\;,[\;77,65]\;,[\;139,160]\;,[\;142,192]\;,[\;143,96]\;]\;
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B2=[\;2,3,4,5]\;
從B2取元素個數為\displaystyle  \frac{t}{2}=2個的子集合Y1，共\displaystyle C(\frac{m}{2},\frac{t}{2})=C(4,2)=6個
Y1=[\;[\;2,3]\;,[\;2,4]\;,[\;2,5]\;,[\;3,4]\;,[\;3,5]\;,[\;4,5]\;]\;
val(Y1)=Σ2^y=[\;12,20,36,24,40,48]\;
計算[\; 2^{val(Y1)},val(Y1)]\;數對
ListY1=[\;[\;114,12]\;,[\;43,20]\;,[\;59,36]\;,[\;145,24]\;,[\;39,40]\;,[\;29,48]\;]\;
Zm\backslash\;B2 = [\;6,7,0,1]\;
從Zm\backslash\;B2取元素個數為\displaystyle \frac{t}{2}=2個的子集合Y1，共\displaystyle C(\frac{m}{2},\frac{t}{2})=C(4,2)=6個
Y2= [\;[\;0,1]\;,[\;0,6]\;,[\;0,7]\;,[\;1,6]\;,[\;1,7]\;,[\;6,7]\;]\;
val(Y2)=Σ2^y = [\;3,65,129,66,130,192]\;
計算[\; 79 * 2^{-val(Y2)}, val(Y2)]\;數對
ListY2=[\;[\;123,3]\;,[\;77,65]\;,[\;47,129]\;,[\;129,66]\;,[\;114,130]\;,[\;142,192]\;]\;
由小到大排序
ListY1=[\;[\;29,48]\;,[\;39,40]\;,[\;43,20]\;,[\;59,36]\;,[\;114,12]\;,[\;145,24]\;]\;
ListY2=[\;[\;47,129]\;,[\;77,65]\;,[\;114,130]\;,[\;123,3]\;,[\;129,66]\;,[\;142,192]\;]\;
ListY1和ListY2互相比較找到相同值
2^{12}\equiv 79 * 2^{-130}\equiv 114 \pmod{181}
x=12+130=142
－－－－－－－－
B3 = [\;3,4,5,6]\;
從B3取元素個數為\displaystyle  \frac{t}{2}=2個的子集合Y1，共\displaystyle C(\frac{m}{2},\frac{t}{2})=C(4,2)=6個
Y1= [\;[\;3,4]\;,[\;3,5]\;,[\;3,6]\;,[\;4,5]\;,[\;4,6]\;,[\;5,6]\;]\;
val(Y1)=Σ2^y = [\;24,40,72,48,80,96]\;
計算[\; 2^{val(Y1)},val(Y1)]\;數對
ListY1=[\;[\;145,24]\;,[\;39,40]\;,[\;42,72]\;,[\;29,48]\;,[\;73,80]\;,[\;117,96]\;]\;
Zm\backslash\;B3 = [\;7,0,1,2]\;
從Zm\backslash\;B3取元素個數為\displaystyle  \frac{t}{2}=2個的子集合Y1，共\displaystyle C(\frac{m}{2},\frac{t}{2})=C(4,2)=6個
Y2= [\;[\;0,1]\;,[\;0,2]\;,[\;0,7]\;,[\;1,2]\;,[\;1,7]\;,[\;2,7]\;]\;
val(Y2)=Σ2^y = [\;3,5,129,6,130,132]\;
計算[\; 79 * 2^{-val(Y2)}, val(Y2)]\;數對
ListY2=[\;[\;123,3]\;,[\;76,5]\;,[\;47,129]\;,[\;38,6]\;,[\;114,130]\;,[\;119,132]\;]\;
由小到大排序
ListY1=[\;[\;29,48]\;,[\;39,40]\;,[\;42,72]\;,[\;73,80]\;,[\;117,96]\;,[\;145,24]\;]\;
ListY2=[\;[\;38,6]\;,[\;47,129]\;,[\;76,5]\;,[\;114,130]\;,[\;119,132]\;,[\;123,3]\;]\;
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
(x)　142

5-3.低漢明權重(Low Hamming Weight)的離散對數問題-可變參數的Splitting System

上一篇Coppersmitch方法將解x二進位切成兩等份，每個等份再分別列舉1的可能位置。可變參數的Splitting System方法則根據t_s值，調整集合\cal{B}的大小來列舉1的可能位置，所以說Coppersmitch方法可算是可變參數的Splitting System的特例。以下表格為兩種方法的比較。

Coppersmith Splitting System	可變參數的Splitting System
定義 假設m和t都是偶數且0<t<m，定義(m,t)的splitting system是符合以下2點的數對(X,\cal{B}) 1.集合X有m個元素，集合\cal{B}有\displaystyle \frac{m}{2}個X的子集合，\cal{B}稱為區塊(blocks)。 2.對所有X的子集合Y，Y有t個元素，存在一個區塊B使得B和Y的交集個數為\displaystyle \frac{t}{2}。	定義 假設m、t和t_s和都是整數且0<t<m、0\le t_s \le t，定義(m,t,t_s)的可變參數splitting system是符合以下2點的數對(X,\cal{B}) 1.集合X有m個元素，集合\cal{B}有\displaystyle \Bigg\lfloor\; \frac{t_s m}{t}\Bigg\rfloor\;個X的子集合，\cal{B}稱為區塊(blocks)。 2.對所有X的子集合Y，Y有t個元素，存在一個區塊B使得B和Y的交集個數為\displaystyle t_s。
產生\displaystyle (\frac{m}{2};m,t)的Coppersmith splitting system方法 令X=Z_m，定義B_i=\{\;i+j\pmod{m}:0\le j\le m/2-1\}\; \cal B=\{\;B_i:0\le i \le m/2-1\}\;	產生\displaystyle (m;m,t,t_s)的可變參數splitting system方法 令X=Z_m，定義B_i=\{\;i+j\pmod{m}:0\le j\le \Bigg\lfloor\;\frac{t_sm}{t}\Bigg\rfloor\;-1\}\; \cal B=\{\;B_i:0\le i \le m-1\}\;
以m=8,t=4為例(解x有8位元，其中有4個1，分成2個1和2個1分別列舉) X=Z_m=\{\;0,1,2,3,4,5,6,7\}\;，B_0=\{\;0,1,2,3\}\;，B_1=\{\;1,2,3,4\}\;，B_2=\{\;2,3,4,5\}\;，B_3=\{\;3,4,5,6\}\;，\cal B=\{\;B_0,B_1,B_2,B_3\}\; 子集合Y=\{\;0,2,4,6\}\;有4個元素，存在B_0,B_1,B_2,B_3和Y交集2個元素。 子集合Y=\{\;1,3,5,7\}\;有4個元素，存在B_0,B_1,B_2,B_3和Y交集2個元素。 子集合Y=\{\;0,1,6,7\}\;有4個元素，存在B_0和Y交集2個元素。 子集合Y=\{\;0,2,5,7\}\;有4個元素，存在B_0和Y交集2個元素。 子集合Y=\{\;4,5,6,7\}\;有4個元素，存在B_2,B_3和Y交集2個元素。	以m=8,t=4,t_s=1為例(解x有8位元，其中有4個1，分成1個1和3個1分別列舉) X=Z_m=\{\;0,1,2,3,4,5,6,7\}\;，B_0=\{\;0,1\}\;，B_1=\{\;1,2\}\;，B_2=\{\;2,3\}\;，B_3=\{\;3,4\}\;，B_4=\{\;4,5\}\;，B_5=\{\;5,6\}\;，B_6=\{\;6,7\}\;，B_7=\{\;7,0\}\;，\cal B=\{\;B_0,B_1,B_2,B_3,B_4,B_5,B_6,B_7\}\; 子集合Y=\{\;0,2,4,6\}\;有4個元素，存在B_0,B_1,B_2,B_3,B_4,B_5,B_6,B_7和Y交集1個元素。 子集合Y=\{\;1,3,5,7\}\;有4個元素，存在B_0,B_1,B_2,B_3,B_4,B_5,B_6,B_7和Y交集1個元素。 子集合Y=\{\;0,1,6,7\}\;有4個元素，存在B_0,B_1,B_5,B_6,B_7和Y交集1個元素。 子集合Y=\{\;0,2,5,7\}\;有4個元素，存在B_0,B_1,B_2,B_4,B_5,B_6,B_7和Y交集1個元素。 子集合Y=\{\;4,5,6,7\}\;有4個元素，存在B_3,B_4,B_5,B_6,B_7和Y交集1個元素。
	以m=8,t=4,t_s=2為例 X=Z_m=\{\;0,1,2,3,4,5,6,7\}\;，B_0=\{\;0,1,2,3\}\;，B_1=\{\;1,2,3,4\}\;，B_2=\{\;2,3,4,5\}\;，B_3=\{\;3,4,5,6\}\;，B_4=\{\;4,5,6,7\}\;，B_5=\{\;5,6,7,0\}\;，B_6=\{\;6,7,0,1\}\;，B_7=\{\;7,0,1,2\}\;，\cal B=\{\;B_0,B_1,B_2,B_3,B_4,B_5,B_6,B_7\}\; B_0,B_1,B_2,B_3和Coppersmith splitting system相同 B_4,B_5,B_6,B_7則是B_0,B_1,B_2,B_3的補集。

演算法和上一篇大致相同，僅集合\cal{B}大小不同。

以2^{142}\equiv 79\pmod{181}為例，解x=(142)_{10}=(10001110)_2，1的分佈不平均。

\{\;0,1 \}\;任取1個的子集合Y_1，\{\;2,3,4,5,6,7 \}\;任取3個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;0\}\;&2^0=1&2^1\equiv 2&\{\;2,3,4\}\;& 2^2+2^3+2^4=28&79*2^{-28}\equiv 36\cr \{\;1\}\;&2^1=2&2^2\equiv 4&\{\;2,3,5\}\;&2^2+2^3+2^5=44&79*2^{-44}\equiv 106\cr &&&\{\;2,3,6\}\;&2^2+2^3+2^6=76&79*2^{-76}\equiv 176\cr &&&\{\;2,3,7\}\;&2^2+2^3+2^7=140&79*2^{-140}\equiv 4\cr &&&\{\;2,4,5\}\;&2^2+2^4+2^5=52&79*2^{-52}\equiv 180\cr &&&\{\;2,4,6\}\;&2^2+2^4+2^6=84&79*2^{-84}\equiv 12\cr &&&\{\;2,4,7\}\;&2^2+2^4+2^7=148&79*2^{-148}\equiv 99\cr &&&\{\;2,5,6\}\;&2^2+2^5+2^6=100&79*2^{-100}\equiv 156\cr &&&\{\;2,5,7\}\;&2^2+2^5+2^7=164&79*2^{-164}\equiv 20\cr &&&\{\;2,6,7\}\;&2^2+2^6+2^7=196&79*2^{-196}\equiv 122\cr &&&\{\;3,4,5\}\;&2^3+2^4+2^5=56&79*2^{-56}\equiv 147\cr &&&\{\;3,4,6\}\;&2^3+2^4+2^6=88&79*2^{-88}\equiv 46\cr &&&\{\;3,4,7\}\;&2^3+2^4+2^7=152&79*2^{-152}\equiv 108\cr &&&\{\;3,5,6\}\;&2^3+2^5+2^6=104&79*2^{-104}\equiv 55\cr &&&\{\;3,5,7\}\;&2^3+2^5+2^7=168&79*2^{-168}\equiv 137\cr &&&\{\;3,6,7\}\;&2^3+2^6+2^7=200&79*2^{-200}\equiv 166\cr &&&\{\;4,5,6\}\;&2^4+2^5+2^6=112&79*2^{-112}\equiv 49\cr &&&\{\;4,5,7\}\;&2^4+2^5+2^7=176&79*2^{-176}\equiv 178\cr &&&\{\;4,6,7\}\;&2^4+2^6+2^7=208&79*2^{-208}\equiv 36\cr &&&\{\;5,6,7\}\;&2^5+2^6+2^7=224&79*2^{-224}\equiv 106}
可找到
2^2\equiv79*2^{-140}\equiv 4\pmod{181}，Y_1=\{\;1\}\;，Y_2=\{\;2,3,7\}\;，x=2+140=142

\{\;1,2 \}\;任取1個的子集合Y_1，\{\;0,3,4,5,6,7 \}\;任取3個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;1\}\;&2^1=2&2^2\equiv 4&\{\;0,3,4\}\;&2^0+2^3+2^4=25&79*2^{-25}\equiv107\cr \{\;2\}\;&2^2=4&2^4\equiv 16&\{\;0,3,5\}\;&2^0+2^3+2^5=41&79*2^{-41}\equiv124\cr &&&\{\;0,3,6\}\;&2^0+2^3+2^6=73&79*2^{-73}\equiv141\cr &&&\{\;0,3,7\}\;&2^0+2^3+2^7=137&79*2^{-137}\equiv32\cr &&&\{\;0,4,5\}\;&2^0+2^4+2^5=49&79*2^{-49}\equiv173\cr &&&\{\;0,4,6\}\;&2^0+2^4+2^6=81&79*2^{-81}\equiv96\cr &&&\{\;0,4,7\}\;&2^0+2^4+2^7=145&79*2^{-145}\equiv68\cr &&&\{\;0,5,6\}\;&2^0+2^5+2^6=97&79*2^{-97}\equiv162\cr &&&\{\;0,5,7\}\;&2^0+2^5+2^7=161&79*2^{-161}\equiv160\cr &&&\{\;0,6,7\}\;&2^0+2^6+2^7=193&79*2^{-193}\equiv71\cr &&&\{\;3,4,5\}\;&2^3+2^4+2^5=56&79*2^{-56}\equiv147\cr &&&\{\;3,4,6\}\;&2^3+2^4+2^6=88&79*2^{-88}\equiv46\cr &&&\{\;3,4,7\}\;&2^3+2^4+2^7=152&79*2^{-152}\equiv108\cr &&&\{\;3,5,6\}\;&2^3+2^5+2^6=104&79*2^{-104}\equiv55\cr &&&\{\;3,5,7\}\;&2^3+2^5+2^7=168&79*2^{-168}\equiv137\cr &&&\{\;3,6,7\}\;&2^3+2^6+2^7=200&79*2^{-200}\equiv166\cr &&&\{\;4,5,6\}\;&2^4+2^5+2^6=112&79*2^{-112}\equiv49\cr &&&\{\;4,5,7\}\;&2^4+2^5+2^7=176&79*2^{-176}\equiv178\cr &&&\{\;4,6,7\}\;&2^4+2^6+2^7=208&79*2^{-208}\equiv36\cr &&&\{\;5,6,7\}\;&2^5+2^6+2^7=224&79*2^{-224}\equiv106}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;2,3 \}\;任取1個的子集合Y_1，\{\;0,1,4,5,6,7 \}\;任取3個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;2\}\;&2^2=4&2^4\equiv 16&\{\;0,1,4\}\;&2^0+2^1+2^4=19&79*2^{-19}\equiv151\cr \{\;3\}\;&2^3=8&2^8\equiv 75&\{\;0,1,5\}\;&2^0+2^1+2^5=35&79*2^{-35}\equiv153\cr &&&\{\;0,1,6\}\;&2^0+2^1+2^6=67&79*2^{-67}\equiv155\cr &&&\{\;0,1,7\}\;&2^0+2^1+2^7=131&79*2^{-131}\equiv57\cr &&&\{\;0,4,5\}\;&2^0+2^4+2^5=49&79*2^{-49}\equiv173\cr &&&\{\;0,4,6\}\;&2^0+2^4+2^6=81&79*2^{-81}\equiv96\cr &&&\{\;0,4,7\}\;&2^0+2^4+2^7=145&79*2^{-145}\equiv68\cr &&&\{\;0,5,6\}\;&2^0+2^5+2^6=97&79*2^{-97}\equiv162\cr &&&\{\;0,5,7\}\;&2^0+2^5+2^7=161&79*2^{-161}\equiv160\cr &&&\{\;0,6,7\}\;&2^0+2^6+2^7=193&79*2^{-193}\equiv71\cr &&&\{\;1,4,5\}\;&2^1+2^4+2^5=50&79*2^{-50}\equiv177\cr &&&\{\;1,4,6\}\;&2^1+2^4+2^6=82&79*2^{-82}\equiv48\cr &&&\{\;1,4,7\}\;&2^1+2^4+2^7=146&79*2^{-146}\equiv34\cr &&&\{\;1,5,6\}\;&2^1+2^5+2^6=98&79*2^{-98}\equiv81\cr &&&\{\;1,5,7\}\;&2^1+2^5+2^7=162&79*2^{-162}\equiv80\cr &&&\{\;1,6,7\}\;&2^1+2^6+2^7=194&79*2^{-194}\equiv126\cr &&&\{\;4,5,6\}\;&2^4+2^5+2^6=112&79*2^{-112}\equiv49\cr &&&\{\;4,5,7\}\;&2^4+2^5+2^7=176&79*2^{-176}\equiv178\cr &&&\{\;4,6,7\}\;&2^4+2^6+2^7=208&79*2^{-208}\equiv36\cr &&&\{\;5,6,7\}\;&2^5+2^6+2^7=224&79*2^{-224}\equiv106}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;3,4 \}\;任取1個的子集合Y_1，\{\;0,1,2,5,6,7 \}\;任取3個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;3\}\;&2^3=8&2^8\equiv 75&\{\;0,1,2\}\;&2^0+2^1+2^2=7&79*2^{-7}\equiv19\cr \{\;4\}\;&2^4=16&2^{16}\equiv 14&\{\;0,1,5\}\;&2^0+2^1+2^5=35&79*2^{-35}\equiv153\cr &&&\{\;0,1,6\}\;&2^0+2^1+2^6=67&79*2^{-67}\equiv155\cr &&&\{\;0,1,7\}\;&2^0+2^1+2^7=131&79*2^{-131}\equiv57\cr &&&\{\;0,2,5\}\;&2^0+2^2+2^5=37&79*2^{-37}\equiv174\cr &&&\{\;0,2,6\}\;&2^0+2^2+2^6=69&79*2^{-69}\equiv84\cr &&&\{\;0,2,7\}\;&2^0+2^2+2^7=133&79*2^{-133}\equiv150\cr &&&\{\;0,5,6\}\;&2^0+2^5+2^6=97&79*2^{-97}\equiv162\cr &&&\{\;0,5,7\}\;&2^0+2^5+2^7=161&79*2^{-161}\equiv160\cr &&&\{\;0,6,7\}\;&2^0+2^6+2^7=193&79*2^{-193}\equiv71\cr &&&\{\;1,2,5\}\;&2^1+2^2+2^5=38&79*2^{-38}\equiv87\cr &&&\{\;1,2,6\}\;&2^1+2^2+2^6=70&79*2^{-70}\equiv42\cr &&&\{\;1,2,7\}\;&2^1+2^2+2^7=134&79*2^{-134}\equiv75\cr &&&\{\;1,5,6\}\;&2^1+2^5+2^6=98&79*2^{-98}\equiv81\cr &&&\{\;1,5,7\}\;&2^1+2^5+2^7=162&79*2^{-162}\equiv80\cr &&&\{\;1,6,7\}\;&2^1+2^6+2^7=194&79*2^{-194}\equiv126\cr &&&\{\;2,5,6\}\;&2^2+2^5+2^6=100&79*2^{-100}\equiv156\cr &&&\{\;2,5,7\}\;&2^2+2^5+2^7=164&79*2^{-164}\equiv20\cr &&&\{\;2,6,7\}\;&2^2+2^6+2^7=196&79*2^{-196}\equiv122\cr &&&\{\;5,6,7\}\;&2^5+2^6+2^7=224&79*2^{-224}\equiv106}
可找到
2^8\equiv79*2^{-134}\equiv 75\pmod{181}，Y_1=\{\;3\}\;，Y_2=\{\;1,2,7\}\;，x=8+134=142

\{\;4,5 \}\;任取1個的子集合Y_1，\{\;0,1,2,3,6,7 \}\;任取3個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;4\}\;&2^4=16&2^{16}\equiv 14&\{\;0,1,2\}\;&2^0+2^1+2^2=7&79*2^{-7}\equiv19\cr \{\;5\}\;&2^5=32&2^{32}\equiv 15&\{\;0,1,3\}\;&2^0+2^1+2^3=11&79*2^{-11}\equiv103\cr &&&\{\;0,1,6\}\;&2^0+2^1+2^6=67&79*2^{-67}\equiv155\cr &&&\{\;0,1,7\}\;&2^0+2^1+2^7=131&79*2^{-131}\equiv57\cr &&&\{\;0,2,3\}\;&2^0+2^2+2^3=13&79*2^{-13}\equiv71\cr &&&\{\;0,2,6\}\;&2^0+2^2+2^6=69&79*2^{-69}\equiv84\cr &&&\{\;0,2,7\}\;&2^0+2^2+2^7=133&79*2^{-133}\equiv150\cr &&&\{\;0,3,6\}\;&2^0+2^3+2^6=73&79*2^{-73}\equiv141\cr &&&\{\;0,3,7\}\;&2^0+2^3+2^7=137&79*2^{-137}\equiv32\cr &&&\{\;0,6,7\}\;&2^0+2^6+2^7=193&79*2^{-193}\equiv71\cr &&&\{\;1,2,3\}\;&2^1+2^2+2^3=14&79*2^{-14}\equiv126\cr &&&\{\;1,2,6\}\;&2^1+2^2+2^6=70&79*2^{-70}\equiv42\cr &&&\{\;1,2,7\}\;&2^1+2^2+2^7=134&79*2^{-134}\equiv75\cr &&&\{\;1,3,6\}\;&2^1+2^3+2^6=74&79*2^{-74}\equiv161\cr &&&\{\;1,3,7\}\;&2^1+2^3+2^7=138&79*2^{-138}\equiv16\cr &&&\{\;1,6,7\}\;&2^1+2^6+2^7=194&79*2^{-194}\equiv126\cr &&&\{\;2,3,6\}\;&2^2+2^3+2^6=76&79*2^{-76}\equiv176\cr &&&\{\;2,3,7\}\;&2^2+2^3+2^7=140&79*2^{-140}\equiv4\cr &&&\{\;2,6,7\}\;&2^2+2^6+2^7=196&79*2^{-196}\equiv122\cr &&&\{\;3,6,7\}\;&2^3+2^6+2^7=200&79*2^{-200}\equiv166}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;5,6 \}\;任取1個的子集合Y_1，\{\;0,1,2,3,4,7 \}\;任取3個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;5\}\;&2^5=32&2^{32}\equiv 15&\{\;0,1,2\}\;&2^0+2^1+2^2=7&79*2^{-7}\equiv19\cr \{\;6\}\;&2^6=64&2^{64}\equiv 44&\{\;0,1,3\}\;&2^0+2^1+2^3=11&79*2^{-11}\equiv103\cr &&&\{\;0,1,4\}\;&2^0+2^1+2^4=19&79*2^{-19}\equiv151\cr &&&\{\;0,1,7\}\;&2^0+2^1+2^7=131&79*2^{-131}\equiv57\cr &&&\{\;0,2,3\}\;&2^0+2^2+2^3=13&79*2^{-13}\equiv71\cr &&&\{\;0,2,4\}\;&2^0+2^2+2^4=21&79*2^{-21}\equiv83\cr &&&\{\;0,2,7\}\;&2^0+2^2+2^7=133&79*2^{-133}\equiv150\cr &&&\{\;0,3,4\}\;&2^0+2^3+2^4=25&79*2^{-25}\equiv107\cr &&&\{\;0,3,7\}\;&2^0+2^3+2^7=137&79*2^{-137}\equiv32\cr &&&\{\;0,4,7\}\;&2^0+2^4+2^7=145&79*2^{-145}\equiv68\cr &&&\{\;1,2,3\}\;&2^1+2^2+2^3=14&79*2^{-14}\equiv126\cr &&&\{\;1,2,4\}\;&2^1+2^2+2^4=22&79*2^{-22}\equiv132\cr &&&\{\;1,2,7\}\;&2^1+2^2+2^7=134&79*2^{-134}\equiv75\cr &&&\{\;1,3,4\}\;&2^1+2^3+2^4=26&79*2^{-26}\equiv144\cr &&&\{\;1,3,7\}\;&2^1+2^3+2^7=138&79*2^{-138}\equiv16\cr &&&\{\;1,4,7\}\;&2^1+2^4+2^7=146&79*2^{-146}\equiv34\cr &&&\{\;2,3,4\}\;&2^2+2^3+2^4=28&79*2^{-28}\equiv36\cr &&&\{\;2,3,7\}\;&2^2+2^3+2^7=140&79*2^{-140}\equiv4\cr &&&\{\;2,4,7\}\;&2^2+2^4+2^7=148&79*2^{-148}\equiv99\cr &&&\{\;3,4,7\}\;&2^3+2^4+2^7=152&79*2^{-152}\equiv108}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;6,7 \}\;任取1個的子集合Y_1，\{\;0,1,2,3,4,5 \}\;任取3個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;6\}\;&2^6=64&2^{64}\equiv 44&\{\;0,1,2\}\;&2^0+2^1+2^2=7&79*2^{-7}\equiv19\cr \{\;7\}\;&2^7=128&2^{128}\equiv 126&\{\;0,1,3\}\;&2^0+2^1+2^3=11&79*2^{-11}\equiv103\cr &&&\{\;0,1,4\}\;&2^0+2^1+2^4=19&79*2^{-19}\equiv151\cr &&&\{\;0,1,5\}\;&2^0+2^1+2^5=35&79*2^{-35}\equiv153\cr &&&\{\;0,2,3\}\;&2^0+2^2+2^3=13&79*2^{-13}\equiv71\cr &&&\{\;0,2,4\}\;&2^0+2^2+2^4=21&79*2^{-21}\equiv83\cr &&&\{\;0,2,5\}\;&2^0+2^2+2^5=37&79*2^{-37}\equiv174\cr &&&\{\;0,3,4\}\;&2^0+2^3+2^4=25&79*2^{-25}\equiv107\cr &&&\{\;0,3,5\}\;&2^0+2^3+2^5=41&79*2^{-41}\equiv124\cr &&&\{\;0,4,5\}\;&2^0+2^4+2^5=49&79*2^{-49}\equiv173\cr &&&\{\;1,2,3\}\;&2^1+2^2+2^3=14&79*2^{-14}\equiv126\cr &&&\{\;1,2,4\}\;&2^1+2^2+2^4=22&79*2^{-22}\equiv132\cr &&&\{\;1,2,5\}\;&2^1+2^2+2^5=38&79*2^{-38}\equiv87\cr &&&\{\;1,3,4\}\;&2^1+2^3+2^4=26&79*2^{-26}\equiv144\cr &&&\{\;1,3,5\}\;&2^1+2^3+2^5=42&79*2^{-42}\equiv62\cr &&&\{\;1,4,5\}\;&2^1+2^4+2^5=50&79*2^{-50}\equiv177\cr &&&\{\;2,3,4\}\;&2^2+2^3+2^4=28&79*2^{-28}\equiv36\cr &&&\{\;2,3,5\}\;&2^2+2^3+2^5=44&79*2^{-44}\equiv106\cr &&&\{\;2,4,5\}\;&2^2+2^4+2^5=52&79*2^{-52}\equiv180\cr &&&\{\;3,4,5\}\;&2^3+2^4+2^5=56&79*2^{-56}\equiv147}
可找到
2^{128}\equiv79*2^{-14}\equiv 126\pmod{181}，Y_1=\{\;7\}\;，Y_2=\{\;1,2,3\}\;，x=128+14=142

\{\;7,0 \}\;任取1個的子集合Y_1，\{\;1,2,3,4,5,6 \}\;任取3個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;7\}\;&2^7=128&2^{128}\equiv 126&\{\;1,2,3\}\;&2^1+2^2+2^3=14&79*2^{-14}\equiv126\cr \{\;0\}\;&2^0=1&2^1\equiv 2&\{\;1,2,4\}\;&2^1+2^2+2^4=22&79*2^{-22}\equiv132\cr &&&\{\;1,2,5\}\;&2^1+2^2+2^5=38&79*2^{-38}\equiv87\cr &&&\{\;1,2,6\}\;&2^1+2^2+2^6=70&79*2^{-70}\equiv42\cr &&&\{\;1,3,4\}\;&2^1+2^3+2^4=26&79*2^{-26}\equiv144\cr &&&\{\;1,3,5\}\;&2^1+2^3+2^5=42&79*2^{-42}\equiv62\cr &&&\{\;1,3,6\}\;&2^1+2^3+2^6=74&79*2^{-74}\equiv161\cr &&&\{\;1,4,5\}\;&2^1+2^4+2^5=50&79*2^{-50}\equiv177\cr &&&\{\;1,4,6\}\;&2^1+2^4+2^6=82&79*2^{-82}\equiv48\cr &&&\{\;1,5,6\}\;&2^1+2^5+2^6=98&79*2^{-98}\equiv81\cr &&&\{\;2,3,4\}\;&2^2+2^3+2^4=28&79*2^{-28}\equiv36\cr &&&\{\;2,3,5\}\;&2^2+2^3+2^5=44&79*2^{-44}\equiv106\cr &&&\{\;2,3,6\}\;&2^2+2^3+2^6=76&79*2^{-76}\equiv176\cr &&&\{\;2,4,5\}\;&2^2+2^4+2^5=52&79*2^{-52}\equiv180\cr &&&\{\;2,4,6\}\;&2^2+2^4+2^6=84&79*2^{-84}\equiv12\cr &&&\{\;2,5,6\}\;&2^2+2^5+2^6=100&79*2^{-100}\equiv156\cr &&&\{\;3,4,5\}\;&2^3+2^4+2^5=56&79*2^{-56}\equiv147\cr &&&\{\;3,4,6\}\;&2^3+2^4+2^6=88&79*2^{-88}\equiv46\cr &&&\{\;3,5,6\}\;&2^3+2^5+2^6=104&79*2^{-104}\equiv55\cr &&&\{\;4,5,6\}\;&2^4+2^5+2^6=112&79*2^{-112}\equiv49}
可找到
2^{128}\equiv79*2^{-14}\equiv 126\pmod{181}，Y_1=\{\;7\}\;，Y_2=\{\;1,2,3\}\;，x=128+14=142

參考資料：
Sungwook Kim, Jung Hee Cheon:A Parameterized Splitting System and Its Application to the Discrete Logarithm Problem with Low Hamming Weight Product Exponents. Public Key Cryptography 2008: 328-343
https://www.iacr.org/archive/pkc2008/49390329/49390329.pdf


從L中取n個元素的子集合副程式
(%i1)
Combination(L,n):=block
([c:[],s],
if length(L)>n then
  (for r in L do
     (s:delete(r,L),
      c:unique(append(c,Combination(s,n)))
     )
  )
else
  (c:[L]),
return(c)
)$

解低漢明權重的離散對數問題可變參數Splitting System方法副程式
(%i2)
LowHammingWeightDLP3(alpha,beta,p,t,ts):=block
([m,order,SetBnumber,Beta,inv_alpha,i,Y1,Y2,valY1,valY2,lenY1,lenY2,SetBcomplement,ListY1,ListY2,indexY1:1,indexY2:1,x1:0,x2:0,x:0],
print("ord(",alpha,")=",order:zn_order(alpha,p),"，",alpha^string(order),"≡1(mod",p,")"),
print("m=Ceiling(log2",order,")=",m:ceiling(log(order)/log(2))),
print("Zm=",create_list(i,i,0,m-1)),
print("|Bi|=Floor(ts*m/t)=",SetBnumber:floor(ts*m/t)),
print("Bi={i+j (mod m):0≤j≤Floor(ts*m/t)-1}，β={Bi:0≤i≤m-1}"),
print("β=",Beta:create_list(create_list(mod(i+j,m),j,0,SetBnumber-1),i,0,m-1)),
print("－－－－－－－－"),
inv_alpha:inv_mod(alpha,p),
for i:1 thru m do
  (print("B",i-1,"=",Beta[ i ]),
   print("從B",i-1,"取元素個數為ts=",ts,"個的子集合Y1，共C(Floor(ts*m/t),ts)=C(",SetBnumber,",",ts,")=",lenY1: (SetBnumber)!/((ts)!*(SetBnumber-ts)!),"個"),
   print("Y1=",Y1:Combination(Beta[ i ],ts)),
   print("val(Y1)=Σ2"^"y","=",valY1:create_list(apply("+",y),y,2^Y1)),
   print("計算[",alpha^"val(Y1)",",val(Y1)]數對"),
   print("ListY1=",ListY1:create_list([power_mod(alpha,valY1[ i ],p),valY1[ i ]],i,1,lenY1)),
   print("Zm\\B",i-1,"=",SetBcomplement:sort(create_list(mod(i-1-j,m),j,1,m-SetBnumber))),
   print("從Zm\\B",i-1,"取元素個數為t-ts=",t-ts,"個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(",m-SetBnumber,",",t-ts,")=",lenY2: (m-SetBnumber)!/((m-SetBnumber-t+ts)!*(t-ts)!),"個"),
   print("Y2=",Y2:Combination(SetBcomplement,t-ts)),
   print("val(Y2)=Σ2"^"y","=",valY2:create_list(apply("+",y),y,2^Y2)),
   print("計算[",beta,"*",alpha^"-val(Y2)",",","val(Y2)]數對"),
   print("ListY2=",ListY2:create_list([mod(beta*power_mod(inv_alpha,valY2[ i ],p),p),valY2[ i ]],i,1,lenY2)),
   print("由小到大排序"),
   print("ListY1=",ListY1:sort(ListY1)),
   print("ListY2=",ListY2:sort(ListY2)),
   indexY1:1,indexY2:1,x1:0,x2:0,
   while indexY1<=lenY1 and indexY2<=lenY2 do
     (if ListY1[indexY1][1]=ListY2[indexY2][1] then
        (print("ListY1和ListY2互相比較找到相同值"),
         x1: ListY1[indexY1][2],x2: ListY2[indexY2][2],
         print(alpha^string(x1),"≡",beta,"*",alpha^concat("-",x2),"≡",ListY1[indexY1][1],"(mod",p,")"),
         print("x=",x1,"+",x2,"=",x:mod(x1+x2,order)),
         indexY1:indexY1+1,
         indexY2:indexY2+1
        )
      else if ListY1[indexY1][1]<ListY2[indexY2][1] then
        (indexY1:indexY1+1)
       else
        (indexY2:indexY2+1)
     ),
   if x1=0 and x2=0 then
      (print("ListY1和ListY2互相比較沒找到相同值")),
   print("－－－－－－－－")
  ),
if x=0 then
  (print("無解"),
   return()
  )
else
  (return(x))
)$

\alpha^x \equiv \beta \pmod{p}，已知x的漢明權重t=4，參數t_s=1
(%i7)
alpha:2;
beta:79;
p:181;
t:4;
ts:1;
(alpha)　2
(beta)　79
(p)　181
(t)　4
(ts)　1

解低漢明權重的離散對數問題-可變參數Splitting System方法
(%i8)　x: LowHammingWeightDLP3(alpha,beta,p,t,ts);
ord(2)=180，2^{180}\equiv 1\pmod{181}
m=Ceiling(log_2 180)=8
Zm=[0,1,2,3,4,5,6,7]
|\;Bi|\;=Floor(ts*m/t)=2
Bi=\{\;i+j\pmod{m}:0\le j\le Floor(ts*m/t)-1\}\;，\beta=\{\;Bi:0\le i\le m-1\}\;
\beta=[[0,1],[1,2],[2,3],[3,4],[4,5],[5,6],[6,7],[7,0]]
－－－－－－－－
B0=[0,1]
從B0取元素個數為ts=1個的子集合Y1，共C(Floor(ts*m/t),ts)=C(2,1)=2個
Y1=[[0],[1]]
val(Y1)=Σ2^y=[1,2]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[2,1],[4,2]]
Zm\backslash\;B0=[2,3,4,5,6,7]
從Zm\backslash\;B0取元素個數為t-ts=3個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(6,3)=20個
Y2=[[2,3,4],[2,3,5],[2,3,6],[2,3,7],[2,4,5],[2,4,6],[2,4,7],[2,5,6],[2,5,7],[2,6,7],
[3,4,5],[3,4,6],[3,4,7],[3,5,6],[3,5,7],[3,6,7],[4,5,6],[4,5,7],[4,6,7],[5,6,7]]
val(Y2)=Σ2^y=[28,44,76,140,52,84,148,100,164,196,56,88,152,104,168,200,112,176,208,224]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[36,28],[106,44],[176,76],[4,140],[180,52],[12,84],[99,148],[156,100],[20,164],[122,196],
[147,56],[46,88],[108,152],[55,104],[137,168],[166,200],[49,112],[178,176],[36,208],[106,224]]
由小到大排序
ListY1=[[2,1],[4,2]]
ListY2=[[4,140],[12,84],[20,164],[36,28],[36,208],[46,88],[49,112],[55,104],[99,148],[106,44],
[106,224],[108,152],[122,196],[137,168],[147,56],[156,100],[166,200],[176,76],[178,176],[180,52]]
ListY1和ListY2互相比較找到相同值
2^2\equiv 79*2^{-140}\equiv 4\pmod{181}
x=2+140=142
－－－－－－－－
B1=[1,2]
從B1取元素個數為ts=1個的子集合Y1，共C(Floor(ts*m/t),ts)=C(2,1)=2個
Y1=[[1],[2]]
val(Y1)=Σ2^y=[2,4]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[4,2],[16,4]]
Zm\backslash\;B1=[0,3,4,5,6,7]
從Zm\backslash\;B1取元素個數為t-ts=3個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(6,3)=20個
Y2=[[0,3,4],[0,3,5],[0,3,6],[0,3,7],[0,4,5],[0,4,6],[0,4,7],[0,5,6],[0,5,7],[0,6,7],
[3,4,5],[3,4,6],[3,4,7],[3,5,6],[3,5,7],[3,6,7],[4,5,6],[4,5,7],[4,6,7],[5,6,7]]
val(Y2)=Σ2^y=[25,41,73,137,49,81,145,97,161,193,56,88,152,104,168,200,112,176,208,224]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[107,25],[124,41],[141,73],[32,137],[173,49],[96,81],[68,145],[162,97],[160,161],[71,193],
[147,56],[46,88],[108,152],[55,104],[137,168],[166,200],[49,112],[178,176],[36,208],[106,224]]
由小到大排序
ListY1=[[4,2],[16,4]]
ListY2=[[32,137],[36,208],[46,88],[49,112],[55,104],[68,145],[71,193],[96,81],[106,224],[107,25],
[108,152],[124,41],[137,168],[141,73],[147,56],[160,161],[162,97],[166,200],[173,49],[178,176]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B2=[2,3]
從B2取元素個數為ts=1個的子集合Y1，共C(Floor(ts*m/t),ts)=C(2,1)=2個
Y1=[[2],[3]]
val(Y1)=Σ2^y=[4,8]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[16,4],[75,8]]
Zm\backslash\;B2=[0,1,4,5,6,7]
從Zm\backslash\;B2取元素個數為t-ts=3個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(6,3)=20個
Y2=[[0,1,4],[0,1,5],[0,1,6],[0,1,7],[0,4,5],[0,4,6],[0,4,7],[0,5,6],[0,5,7],[0,6,7],
[1,4,5],[1,4,6],[1,4,7],[1,5,6],[1,5,7],[1,6,7],[4,5,6],[4,5,7],[4,6,7],[5,6,7]]
val(Y2)=Σ2^y=[19,35,67,131,49,81,145,97,161,193,50,82,146,98,162,194,112,176,208,224]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[151,19],[153,35],[155,67],[57,131],[173,49],[96,81],[68,145],[162,97],[160,161],[71,193],
[177,50],[48,82],[34,146],[81,98],[80,162],[126,194],[49,112],[178,176],[36,208],[106,224]]
由小到大排序
ListY1=[[16,4],[75,8]]
ListY2=[[34,146],[36,208],[48,82],[49,112],[57,131],[68,145],[71,193],[80,162],[81,98],[96,81],
[106,224],[126,194],[151,19],[153,35],[155,67],[160,161],[162,97],[173,49],[177,50],[178,176]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B3=[3,4]
從B3取元素個數為ts=1個的子集合Y1，共C(Floor(ts*m/t),ts)=C(2,1)=2個
Y1=[[3],[4]]
val(Y1)=Σ2^y=[8,16]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[75,8],[14,16]]
Zm\backslash\;B3=[0,1,2,5,6,7]
從Zm\backslash\;B3取元素個數為t-ts=3個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(6,3)=20個
Y2=[[0,1,2],[0,1,5],[0,1,6],[0,1,7],[0,2,5],[0,2,6],[0,2,7],[0,5,6],[0,5,7],[0,6,7],
[1,2,5],[1,2,6],[1,2,7],[1,5,6],[1,5,7],[1,6,7],[2,5,6],[2,5,7],[2,6,7],[5,6,7]]
val(Y2)=Σ2^y=[7,35,67,131,37,69,133,97,161,193,38,70,134,98,162,194,100,164,196,224]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[19,7],[153,35],[155,67],[57,131],[174,37],[84,69],[150,133],[162,97],[160,161],[71,193],
[87,38],[42,70],[75,134],[81,98],[80,162],[126,194],[156,100],[20,164],[122,196],[106,224]]
由小到大排序
ListY1=[[14,16],[75,8]]
ListY2=[[19,7],[20,164],[42,70],[57,131],[71,193],[75,134],[80,162],[81,98],[84,69],[87,38],
[106,224],[122,196],[126,194],[150,133],[153,35],[155,67],[156,100],[160,161],[162,97],[174,37]]
ListY1和ListY2互相比較找到相同值
2^8\equiv 79*2^{-134}\equiv 75\pmod{181}
x=8+134=142
－－－－－－－－
B4=[4,5]
從B4取元素個數為ts=1個的子集合Y1，共C(Floor(ts*m/t),ts)=C(2,1)=2個
Y1=[[4],[5]]
val(Y1)=Σ2^y=[16,32]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[14,16],[15,32]]
Zm\backslash\;B4=[0,1,2,3,6,7]
從Zm\backslash\;B4取元素個數為t-ts=3個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(6,3)=20個
Y2=[[0,1,2],[0,1,3],[0,1,6],[0,1,7],[0,2,3],[0,2,6],[0,2,7],[0,3,6],[0,3,7],[0,6,7],
[1,2,3],[1,2,6],[1,2,7],[1,3,6],[1,3,7],[1,6,7],[2,3,6],[2,3,7],[2,6,7],[3,6,7]]
val(Y2)=Σ2^y=[7,11,67,131,13,69,133,73,137,193,14,70,134,74,138,194,76,140,196,200]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[19,7],[103,11],[155,67],[57,131],[71,13],[84,69],[150,133],[141,73],[32,137],[71,193],
[126,14],[42,70],[75,134],[161,74],[16,138],[126,194],[176,76],[4,140],[122,196],[166,200]]
由小到大排序
ListY1=[[14,16],[15,32]]
ListY2=[[4,140],[16,138],[19,7],[32,137],[42,70],[57,131],[71,13],[71,193],[75,134],[84,69],
[103,11],[122,196],[126,14],[126,194],[141,73],[150,133],[155,67],[161,74],[166,200],[176,76]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B5=[5,6]
從B5取元素個數為ts=1個的子集合Y1，共C(Floor(ts*m/t),ts)=C(2,1)=2個
Y1=[[5],[6]]
val(Y1)=Σ2^y=[32,64]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[15,32],[44,64]]
Zm\backslash\;B5=[0,1,2,3,4,7]
從Zm\backslash\;B5取元素個數為t-ts=3個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(6,3)=20個
Y2=[[0,1,2],[0,1,3],[0,1,4],[0,1,7],[0,2,3],[0,2,4],[0,2,7],[0,3,4],[0,3,7],[0,4,7],
[1,2,3],[1,2,4],[1,2,7],[1,3,4],[1,3,7],[1,4,7],[2,3,4],[2,3,7],[2,4,7],[3,4,7]]
val(Y2)=Σ2^y=[7,11,19,131,13,21,133,25,137,145,14,22,134,26,138,146,28,140,148,152]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[19,7],[103,11],[151,19],[57,131],[71,13],[83,21],[150,133],[107,25],[32,137],[68,145],
[126,14],[132,22],[75,134],[144,26],[16,138],[34,146],[36,28],[4,140],[99,148],[108,152]]
由小到大排序
ListY1=[[15,32],[44,64]]
ListY2=[[4,140],[16,138],[19,7],[32,137],[34,146],[36,28],[57,131],[68,145],[71,13],[75,134],
[83,21],[99,148],[103,11],[107,25],[108,152],[126,14],[132,22],[144,26],[150,133],[151,19]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B6=[6,7]
從B6取元素個數為ts=1個的子集合Y1，共C(Floor(ts*m/t),ts)=C(2,1)=2個
Y1=[[6],[7]]
val(Y1)=Σ2^y=[64,128]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[44,64],[126,128]]
Zm\backslash\;B6=[0,1,2,3,4,5]
從Zm\backslash\;B6取元素個數為t-ts=3個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(6,3)=20個
Y2=[[0,1,2],[0,1,3],[0,1,4],[0,1,5],[0,2,3],[0,2,4],[0,2,5],[0,3,4],[0,3,5],[0,4,5],
[1,2,3],[1,2,4],[1,2,5],[1,3,4],[1,3,5],[1,4,5],[2,3,4],[2,3,5],[2,4,5],[3,4,5]]
val(Y2)=Σ2^y=[7,11,19,35,13,21,37,25,41,49,14,22,38,26,42,50,28,44,52,56]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[19,7],[103,11],[151,19],[153,35],[71,13],[83,21],[174,37],[107,25],[124,41],[173,49],
[126,14],[132,22],[87,38],[144,26],[62,42],[177,50],[36,28],[106,44],[180,52],[147,56]]
由小到大排序
ListY1=[[44,64],[126,128]]
ListY2=[[19,7],[36,28],[62,42],[71,13],[83,21],[87,38],[103,11],[106,44],[107,25],[124,41],
[126,14],[132,22],[144,26],[147,56],[151,19],[153,35],[173,49],[174,37],[177,50],[180,52]]
ListY1和ListY2互相比較找到相同值
2^{128}\equiv79*2^{-14}\equiv126\pmod{181}
x=128+14=142
－－－－－－－－
B7=[7,0]
從B7取元素個數為ts=1個的子集合Y1，共C(Floor(ts*m/t),ts)=C(2,1)=2個
Y1=[[0],[7]]
val(Y1)=Σ2^y=[1,128]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[2,1],[126,128]]
Zm\backslash\;B7=[1,2,3,4,5,6]
從Zm\backslash\;B7取元素個數為t-ts=3個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(6,3)=20個
Y2=[[1,2,3],[1,2,4],[1,2,5],[1,2,6],[1,3,4],[1,3,5],[1,3,6],[1,4,5],[1,4,6],[1,5,6],
[2,3,4],[2,3,5],[2,3,6],[2,4,5],[2,4,6],[2,5,6],[3,4,5],[3,4,6],[3,5,6],[4,5,6]]
val(Y2)=Σ2^y=[14,22,38,70,26,42,74,50,82,98,28,44,76,52,84,100,56,88,104,112]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[126,14],[132,22],[87,38],[42,70],[144,26],[62,42],[161,74],[177,50],[48,82],[81,98],
[36,28],[106,44],[176,76],[180,52],[12,84],[156,100],[147,56],[46,88],[55,104],[49,112]]
由小到大排序
ListY1=[[2,1],[126,128]]
ListY2=[[12,84],[36,28],[42,70],[46,88],[48,82],[49,112],[55,104],[62,42],[81,98],[87,38],
[106,44],[126,14],[132,22],[144,26],[147,56],[156,100],[161,74],[176,76],[177,50],[180,52]]
ListY1和ListY2互相比較找到相同值
2^{128}\equiv79*2^{-14}\equiv126\pmod{181}
x=128+14=142
－－－－－－－－
(x)　142

5-4.低漢明權重(Low Hamming Weight)的離散對數問題-可變參數的Splitting System(2010版)

Kim和Cheon在2008年發表可變參數的Splitting System之後，又在2010年發表新版論文，新版方法僅小地方更新對解決整個問題改善不多。

可變參數的Splitting System(2008版)	可變參數的Splitting System(2010版)
定義 假設m、t和t_s和都是整數且0<t<m、0\le t_s \le t，定義(m,t,t_s)的可變參數splitting system是符合以下2點的數對(X,\cal{B}) 1.集合X有m個元素，集合\cal{B}有\displaystyle \Bigg\lfloor\; \frac{t_s m}{t}\Bigg\rfloor\;個X的子集合，\cal{B}稱為區塊(blocks)。 2.對所有X的子集合Y，Y有t個元素，存在一個區塊B使得B和Y的交集個數為\displaystyle t_s。	定義相同
產生\displaystyle (m;m,t,t_s)的可變參數splitting system方法 令X=Z_m，定義B_i=\{\;i+j\pmod{m}:0\le j\le \Bigg\lfloor\;\frac{t_sm}{t}\Bigg\rfloor\;-1\}\; \cal B=\{\;B_i:0\le i \le m-1\}\;	產生方法相同
以m=8,t=4,t_s=3為例(解x有8位元，其中有4個1，分成3個1和1個1分別列舉) X=Z_m=\{\;0,1,2,3,4,5,6,7\}\;，B_0=\{\;0,1,2,3,4,5\}\;，B_1=\{\;1,2,3,4,5,6\}\;，B_2=\{\;2,3,4,5,6,7\}\;，B_3=\{\;3,4,5,6,7,0\}\;，B_4=\{\;4,5,6,7,0,1\}\;，B_5=\{\;5,6,7,0,1,2\}\;，B_6=\{\;6,7,0,1,2,3\}\;，B_7=\{\;7,0,1,2,3,4\}\;，\cal B=\{\;B_0,B_1,B_2,B_3,B_4,B_5,B_6,B_7\}\; 子集合Y=\{\;0,2,4,6\}\;有4個元素，存在B_0,B_1,B_2,B_3,B_4,B_5,B_6,B_7和Y交集3個元素。 子集合Y=\{\;1,3,5,7\}\;有4個元素，存在B_0,B_1,B_2,B_3,B_4,B_5,B_6,B_7和Y交集3個元素。 子集合Y=\{\;0,1,6,7\}\;有4個元素，存在B_3,B_7和Y交集3個元素。 子集合Y=\{\;0,2,5,7\}\;有4個元素，存在B_0,B_2,B_3,B_4,B_6,B_7和Y交集3個元素。 子集合Y=\{\;4,5,6,7\}\;有4個元素，存在B_1,B_5和Y交集3個元素。	集合X相同，集合B_0,B_1,\ldots,B_7相同，集合\cal{B}相同
差異處 集合B_i有6個元素，列舉3個元素是1的位置，共C_3^6=20種	差異處 集合B_i有6個元素，假設第i個位元是1，再列舉2個元素是1的位置，共C_2^5=10種

以2^{142}\equiv 79\pmod{181}為例，解x=(142)_{10}=(10001110)_2，1的分佈不平均。

\{\;0,1,2,3,4,5 \}\;，假設第0個位元是1，再任取2個的子集合Y_1，\{\;6,7 \}\;任取1個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;0,1,2\}\;&2^0+2^1+2^2=7&2^{7}\equiv128&\{\;6\}\;&2^6=64&79*2^{-64}\equiv154\cr \{\;0,1,3\}\;&2^0+2^1+2^3=11&2^{11}\equiv57&\{\;7\}\;&2^7=128&79*2^{-128}\equiv94\cr \{\;0,1,4\}\;&2^0+2^1+2^4=19&2^{19}\equiv112&&&\cr \{\;0,1,5\}\;&2^0+2^1+2^5=35&2^{35}\equiv120&&&\cr \{\;0,2,3\}\;&2^0+2^2+2^3=13&2^{13}\equiv47&&&\cr \{\;0,2,4\}\;&2^0+2^2+2^4=21&2^{21}\equiv86&&&\cr \{\;0,2,5\}\;&2^0+2^2+2^5=37&2^{37}\equiv118&&&\cr \{\;0,3,4\}\;&2^0+2^3+2^4=25&2^{25}\equiv109&&&\cr \{\;0,3,5\}\;&2^0+2^3+2^5=41&2^{41}\equiv78&&&\cr \{\;0,4,5\}\;&2^0+2^4+2^5=49&2^{49}\equiv58}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;1,2,3,4,5,6 \}\;，假設第1個位元是1，再任取2個的子集合Y_1，\{\;7,0 \}\;任取1個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;1,2,3\}\;&2^1+2^2+2^3=14&2^{14}\equiv94&\{\;7\}\;&2^7=128&79*2^{-128}\equiv94\cr \{\;1,2,4\}\;&2^1+2^2+2^4=22&2^{22}\equiv172&\{\;0\}\;&2^0=1&79*2^{-1}\equiv130\cr \{\;1,2,5\}\;&2^1+2^2+2^5=38&2^{38}\equiv55&&&\cr \{\;1,2,6\}\;&2^1+2^2+2^6=70&2^{70}\equiv101&&&\cr \{\;1,3,4\}\;&2^1+2^3+2^4=26&2^{26}\equiv37&&&\cr \{\;1,3,5\}\;&2^1+2^3+2^5=42&2^{42}\equiv156&&&\cr \{\;1,3,6\}\;&2^1+2^3+2^6=74&2^{74}\equiv168&&&\cr \{\;1,4,5\}\;&2^1+2^4+2^5=50&2^{50}\equiv116&&&\cr \{\;1,4,6\}\;&2^1+2^4+2^6=82&2^{82}\equiv111&&&\cr \{\;1,5,6\}\;&2^1+2^5+2^6=98&2^{98}\equiv106}
可找到
2^{14}\equiv 79*2^{-128}\equiv 94\pmod{181}，Y_1=\{\;1,2,3\}\;，Y_2=\{\;7\}\;，x=14+128=142

\{\;2,3,4,5,6,7 \}\;，假設第2個位元是1，再任取2個的子集合Y_1，\{\;0,1 \}\;任取1個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;2,3,4\}\;&2^2+2^3+2^4=28&2^{28}\equiv148&\{\;0\}\;&2^0=1&79*2^{-1}\equiv130\cr \{\;2,3,5\}\;&2^2+2^3+2^5=44&2^{44}\equiv81&\{\;1\}\;&2^1=2&79*2^{-2}\equiv65\cr \{\;2,3,6\}\;&2^2+2^3+2^6=76&2^{76}\equiv129&&&\cr \{\;2,3,7\}\;&2^2+2^3+2^7=140&2^{140}\equiv65&&&\cr \{\;2,4,5\}\;&2^2+2^4+2^5=52&2^{52}\equiv102&&&\cr \{\;2,4,6\}\;&2^2+2^4+2^6=84&2^{84}\equiv82&&&\cr \{\;2,4,7\}\;&2^2+2^4+2^7=148&2^{148}\equiv169&&&\cr \{\;2,5,6\}\;&2^2+2^5+2^6=100&2^{100}\equiv62&&&\cr \{\;2,5,7\}\;&2^2+2^5+2^7=164&2^{164}\equiv13&&&\cr \{\;2,6,7\}\;&2^2+2^6+2^7=196&2^{196}\equiv14}
可找到
2^{140}\equiv 79*2^{-2}\equiv 65\pmod{181}，Y_1=\{\;2,3,7\}\;，Y_2=\{\;1\}\;，x=140+2=142

\{\;3,4,5,6,7,0 \}\;，假設第3個位元是1，再任取2個的子集合Y_1，\{\;1,2 \}\;任取1個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;3,4,0\}\;&2^3+2^4+2^0=25&2^{25}\equiv109&\{\;1\}\;&2^1=2&79*2^{-2}\equiv65\cr \{\;3,4,5\}\;&2^3+2^4+2^5=56&2^{56}\equiv3&\{\;2\}\;&2^2=4&79*2^{-4}\equiv152\cr \{\;3,4,6\}\;&2^3+2^4+2^6=88&2^{88}\equiv45&&&\cr \{\;3,4,7\}\;&2^3+2^4+2^7=152&2^{152}\equiv170&&&\cr \{\;3,5,0\}\;&2^3+2^5+2^0=41&2^{41}\equiv78&&&\cr \{\;3,5,6\}\;&2^3+2^5+2^6=104&2^{104}\equiv87&&&\cr \{\;3,5,7\}\;&2^3+2^5+2^7=168&2^{168}\equiv27&&&\cr \{\;3,6,0\}\;&2^3+2^6+2^0=73&2^{73}\equiv84&&&\cr \{\;3,6,7\}\;&2^3+2^6+2^7=200&2^{200}\equiv43&&&\cr \{\;3,7,0\}\;&2^3+2^7+2^0=137&2^{137}\equiv76}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;4,5,6,7,0,1 \}\;，假設第4個位元是1，再任取2個的子集合Y_1，\{\;2,3 \}\;任取1個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;4,0,1\}\;&2^4+2^0+2^1=19&2^{19}\equiv112&\{\;2\}\;&2^2=4&79*2^{-4}\equiv152\cr \{\;4,5,0\}\;&2^4+2^5+2^0=49&2^{49}\equiv58&\{\;3\}\;&2^3=8&79*2^{-8}\equiv100\cr \{\;4,5,1\}\;&2^4+2^5+2^1=50&2^{50}\equiv116&&&\cr \{\;4,5,6\}\;&2^4+2^5+2^6=112&2^{112}\equiv9&&&\cr \{\;4,5,7\}\;&2^4+2^5+2^7=176&2^{176}\equiv34&&&\cr \{\;4,6,0\}\;&2^4+2^6+2^0=81&2^{81}\equiv146&&&\cr \{\;4,6,1\}\;&2^4+2^6+2^1=82&2^{82}\equiv111&&&\cr \{\;4,6,7\}\;&2^4+2^6+2^7=208&2^{208}\equiv148&&&\cr \{\;4,7,0\}\;&2^4+2^7+2^0=145&2^{145}\equiv89&&&\cr \{\;4,7,1\}\;&2^4+2^7+2^1=146&2^{146}\equiv178}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;5,6,7,0,1,2 \}\;，假設第5個位元是1，再任取2個的子集合Y_1，\{\;3,4 \}\;任取1個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;5,0,1\}\;&2^5+2^0+2^1=35&2^{35}\equiv120&\{\;3\}\;&2^3=8&79*2^{-8}\equiv100\cr \{\;5,0,2\}\;&2^5+2^0+2^2=37&2^{37}\equiv118&\{\;4\}\;&2^4=16&79*2^{-16}\equiv122\cr \{\;5,1,2\}\;&2^5+2^1+2^2=38&2^{38}\equiv55&&&\cr \{\;5,6,0\}\;&2^5+2^6+2^0=97&2^{97}\equiv53&&&\cr \{\;5,6,1\}\;&2^5+2^6+2^1=98&2^{98}\equiv106&&&\cr \{\;5,6,2\}\;&2^5+2^6+2^2=100&2^{100}\equiv62&&&\cr \{\;5,6,7\}\;&2^5+2^6+2^7=224&2^{224}\equiv81&&&\cr \{\;5,7,0\}\;&2^5+2^7+2^0=161&2^{161}\equiv160&&&\cr \{\;5,7,1\}\;&2^5+2^7+2^1=162&2^{162}\equiv139&&&\cr \{\;5,7,2\}\;&2^5+2^7+2^2=164&2^{164}\equiv13}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;6,7,0,1,2,3 \}\;，假設第6個位元是1，再任取2個的子集合Y_1，\{\;4,5 \}\;任取1個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;6,0,1\}\;&2^6+2^0+2^1=67&2^{67}\equiv171&\{\;4\}\;&2^4=16&79*2^{-16}\equiv122\cr \{\;6,0,2\}\;&2^6+2^0+2^2=69&2^{69}\equiv141&\{\;5\}\;&2^5=32&79*2^{-32}\equiv138\cr \{\;6,0,3\}\;&2^6+2^0+2^3=73&2^{73}\equiv84&&&\cr \{\;6,1,2\}\;&2^6+2^1+2^2=70&2^{70}\equiv101&&&\cr \{\;6,1,3\}\;&2^6+2^1+2^3=74&2^{74}\equiv168&&&\cr \{\;6,2,3\}\;&2^6+2^2+2^3=76&2^{76}\equiv129&&&\cr \{\;6,7,0\}\;&2^6+2^7+2^0=193&2^{193}\equiv47&&&\cr \{\;6,7,1\}\;&2^6+2^7+2^1=194&2^{194}\equiv94&&&\cr \{\;6,7,2\}\;&2^6+2^7+2^2=196&2^{196}\equiv14&&&\cr \{\;6,7,3\}\;&2^6+2^7+2^3=200&2^{200}\equiv43}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

\{\;7,0,1,2,3,4 \}\;，假設第7個位元是1，再任取2個的子集合Y_1，\{\;5,6 \}\;任取1個的子集合Y_2
\matrix{Y_1&val(Y_1)&\alpha^{val(Y_1)}\pmod{181}&Y_2&val(Y_2)&\beta \alpha^{-val(Y_2)}\pmod{181}\cr \{\;7,0,1\}\;&2^7+2^0+2^1=131&2^{131}\equiv103&\{\;5\}\;&2^5=32&79*2^{-32}\equiv138\cr \{\;7,0,2\}\;&2^7+2^0+2^2=133&2^{133}\equiv50&\{\;6\}\;&2^6=64&79*2^{-64}\equiv154\cr \{\;7,0,3\}\;&2^7+2^0+2^3=137&2^{137}\equiv76&&&\cr \{\;7,0,4\}\;&2^7+2^0+2^4=145&2^{145}\equiv89&&&\cr \{\;7,1,2\}\;&2^7+2^1+2^2=134&2^{134}\equiv100&&&\cr \{\;7,1,3\}\;&2^7+2^1+2^3=138&2^{138}\equiv152&&&\cr \{\;7,1,4\}\;&2^7+2^1+2^4=146&2^{146}\equiv178&&&\cr \{\;7,2,3\}\;&2^7+2^2+2^3=140&2^{140}\equiv65&&&\cr \{\;7,2,4\}\;&2^7+2^2+2^4=148&2^{148}\equiv169&&&\cr \{\;7,3,4\}\;&2^7+2^3+2^4=152&2^{152}\equiv170}
找不到Y_1,Y_2使得\alpha^{val(Y_1)}\equiv \beta \alpha^{-val(Y_2)}\pmod{p}

參考資料：
Sungwook Kim, Jung Hee Cheon:Parameterized splitting systems for the discrete logarithm. IEEE Trans. Inf. Theory 56(5): 2528-2535 (2010)
https://ieeexplore.ieee.org/document/5452195


從L中取n個元素的子集合副程式
(%i1)
Combination(L,n):=block
([c:[],s],
if length(L)>n then
  (for r in L do
     (s:delete(r,L),
      c:unique(append(c,Combination(s,n)))
     )
  )
else
  (c:[L]),
return(c)
)$

解低漢明權重的離散對數問題可變參數Splitting System方法副程式
(%i2)
LowHammingWeightDLP4(alpha,beta,p,t,ts):=block
([m,order,SetBnumber,Beta,inv_alpha,i,Y1,Y2,valY1,valY2,lenY1,lenY2,SetBcomplement,ListY1,ListY2,indexY1:1,indexY2:1,x1:0,x2:0,x:0],
print("ord(",alpha,")=",order:zn_order(alpha,p),"，",alpha^string(order),"≡1(mod",p,")"),
print("m=Ceiling(log2",order,")=",m:ceiling(log(order)/log(2))),
print("Zm=",create_list(i,i,0,m-1)),
print("|Bi|=Floor(ts*m/t)=",SetBnumber:floor(ts*m/t)),
print("Bi={i+j (mod m):0≤j≤Floor(ts*m/t)-1}，β={Bi:0≤i≤m-1}"),
print("β=",Beta:create_list(create_list(mod(i+j,m),j,0,SetBnumber-1),i,0,m-1)),
print("－－－－－－－－"),
inv_alpha:inv_mod(alpha,p),
for i:1 thru m do
  (print("B",i-1,"=",Beta[ i ]),
   print("假設第",i-1,"個位元是1，再取元素個數為ts-1=",ts-1,"個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(",SetBnumber-1,",",ts-1,")=",lenY1: (SetBnumber-1)!/((ts-1)!*(SetBnumber-ts)!),"個"),
   print("Y1=",Y1:map(lambda([beta],append([i-1],beta)),Combination(rest(Beta[ i ],1),ts-1))),
   print("val(Y1)=Σ2"^"y","=",valY1:create_list(apply("+",y),y,2^Y1)),
   print("計算[",alpha^"val(Y1)",",val(Y1)]數對"),
   print("ListY1=",ListY1:create_list([power_mod(alpha,valY1[ i ],p),valY1[ i ]],i,1,lenY1)),
   print("Zm\\B",i-1,"=",SetBcomplement:sort(create_list(mod(i-1-j,m),j,1,m-SetBnumber))),
   print("從Zm\\B",i-1,"取元素個數為t-ts=",t-ts,"個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(",m-SetBnumber,",",t-ts,")=",lenY2: (m-SetBnumber)!/((m-SetBnumber-t+ts)!*(t-ts)!),"個"),
   print("Y2=",Y2:Combination(SetBcomplement,t-ts)),
   print("val(Y2)=Σ2"^"y","=",valY2:create_list(apply("+",y),y,2^Y2)),
   print("計算[",beta,"*",alpha^"-val(Y2)",",","val(Y2)]數對"),
   print("ListY2=",ListY2:create_list([mod(beta*power_mod(inv_alpha,valY2[ i ],p),p),valY2[ i ]],i,1,lenY2)),
   print("由小到大排序"),
   print("ListY1=",ListY1:sort(ListY1)),
   print("ListY2=",ListY2:sort(ListY2)),
   indexY1:1,indexY2:1,x1:0,x2:0,
   while indexY1<=lenY1 and indexY2<=lenY2 do
     (if ListY1[indexY1][1]=ListY2[indexY2][1] then
        (print("ListY1和ListY2互相比較找到相同值"),
         x1: ListY1[indexY1][2],x2: ListY2[indexY2][2],
         print(alpha^string(x1),"≡",beta,"*",alpha^concat("-",x2),"≡",ListY1[indexY1][1],"(mod",p,")"),
         print("x=",x1,"+",x2,"=",x:mod(x1+x2,order)),
         indexY1:indexY1+1,
         indexY2:indexY2+1
        )
      else if ListY1[indexY1][1]<ListY2[indexY2][1] then
        (indexY1:indexY1+1)
       else
        (indexY2:indexY2+1)
     ),
   if x1=0 and x2=0 then
      (print("ListY1和ListY2互相比較沒找到相同值")),
   print("－－－－－－－－")
  ),
if x=0 then
  (print("無解"),
   return()
  )
else
  (return(x))
)$

\alpha^x\equiv \beta \pmod{p}，已知x的漢明權重t=4，參數ts=3
(%i7)
alpha:2;
beta:79;
p:181;
t:4;
ts:3;
(alpha)　2
(beta)　79
(p)　181
(t)　4
(ts)　3


(%i8)　x: LowHammingWeightDLP4(alpha,beta,p,t,ts);
ord(2)=180，2^{180}\equiv1\pmod{181}
m=Ceiling(log_2 180)=8
Zm=[0,1,2,3,4,5,6,7]
|\;Bi|\;=Floor(ts*m/t)=6
Bi=\{\;i+j\pmod{m}:0\le j\le Floor(ts*m/t)-1\}\;，\beta=\{\;Bi:0\le i\le m-1\}\;
\beta=[[0,1,2,3,4,5],[1,2,3,4,5,6],[2,3,4,5,6,7],[3,4,5,6,7,0],[4,5,6,7,0,1],[5,6,7,0,1,2],[6,7,0,1,2,3],[7,0,1,2,3,4]]
－－－－－－－－
B0=[0,1,2,3,4,5]
假設第0個位元是1，再取元素個數為ts-1=2個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(5,2)=10個
Y1=[[0,1,2],[0,1,3],[0,1,4],[0,1,5],[0,2,3],[0,2,4],[0,2,5],[0,3,4],[0,3,5],[0,4,5]]
val(Y1)=Σ2^y=[7,11,19,35,13,21,37,25,41,49]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[128,7],[57,11],[112,19],[120,35],[47,13],[86,21],[118,37],[109,25],[78,41],[58,49]]
Zm\backslash\;B0=[6,7]
從Zm\backslash\;B0取元素個數為t-ts=1個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(2,1)=2個
Y2=[[6],[7]]
val(Y2)=Σ2^y=[64,128]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[154,64],[94,128]]
由小到大排序
ListY1=[[47,13],[57,11],[58,49],[78,41],[86,21],[109,25],[112,19],[118,37],[120,35],[128,7]]
ListY2=[[94,128],[154,64]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B1=[1,2,3,4,5,6]
假設第1個位元是1，再取元素個數為ts-1=2個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(5,2)=10個
Y1=[[1,2,3],[1,2,4],[1,2,5],[1,2,6],[1,3,4],[1,3,5],[1,3,6],[1,4,5],[1,4,6],[1,5,6]]
val(Y1)=Σ2^y=[14,22,38,70,26,42,74,50,82,98]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[94,14],[172,22],[55,38],[101,70],[37,26],[156,42],[168,74],[116,50],[111,82],[106,98]]
Zm\backslash\;B1=[0,7]
從Zm\backslash\;B1取元素個數為t-ts=1個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(2,1)=2個
Y2=[[0],[7]]
val(Y2)=Σ2^y=[1,128]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[130,1],[94,128]]
由小到大排序
ListY1=[[37,26],[55,38],[94,14],[101,70],[106,98],[111,82],[116,50],[156,42],[168,74],[172,22]]
ListY2=[[94,128],[130,1]]
ListY1和ListY2互相比較找到相同值
2^{14}\equiv79*2^{-128}\equiv94\pmod{181}
x=14+128=142
－－－－－－－－
B2=[2,3,4,5,6,7]
假設第2個位元是1，再取元素個數為ts-1=2個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(5,2)=10個
Y1=[[2,3,4],[2,3,5],[2,3,6],[2,3,7],[2,4,5],[2,4,6],[2,4,7],[2,5,6],[2,5,7],[2,6,7]]
val(Y1)=Σ2^y=[28,44,76,140,52,84,148,100,164,196]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[148,28],[81,44],[129,76],[65,140],[102,52],[82,84],[169,148],[62,100],[13,164],[14,196]]
Zm\backslash\;B2=[0,1]
從Zm\backslash\;B2取元素個數為t-ts=1個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(2,1)=2個
Y2=[[0],[1]]
val(Y2)=Σ2^y=[1,2]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[130,1],[65,2]]
由小到大排序
ListY1=[[13,164],[14,196],[62,100],[65,140],[81,44],[82,84],[102,52],[129,76],[148,28],[169,148]]
ListY2=[[65,2],[130,1]]
ListY1和ListY2互相比較找到相同值
2^{140}\equiv79*2^{-2}\equiv65\pmod{181}
x=140+2=142
－－－－－－－－
B3=[3,4,5,6,7,0]
假設第3個位元是1，再取元素個數為ts-1=2個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(5,2)=10個
Y1=[[3,4,0],[3,4,5],[3,4,6],[3,4,7],[3,5,0],[3,5,6],[3,5,7],[3,6,0],[3,6,7],[3,7,0]]
val(Y1)=Σ2^y=[25,56,88,152,41,104,168,73,200,137]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[109,25],[3,56],[45,88],[170,152],[78,41],[87,104],[27,168],[84,73],[43,200],[76,137]]
Zm\backslash\;B3=[1,2]
從Zm\backslash\;B3取元素個數為t-ts=1個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(2,1)=2個
Y2=[[1],[2]]
val(Y2)=Σ2^y=[2,4]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[65,2],[152,4]]
由小到大排序
ListY1=[[3,56],[27,168],[43,200],[45,88],[76,137],[78,41],[84,73],[87,104],[109,25],[170,152]]
ListY2=[[65,2],[152,4]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B4=[4,5,6,7,0,1]
假設第4個位元是1，再取元素個數為ts-1=2個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(5,2)=10個
Y1=[[4,0,1],[4,5,0],[4,5,1],[4,5,6],[4,5,7],[4,6,0],[4,6,1],[4,6,7],[4,7,0],[4,7,1]]
val(Y1)=Σ2^y=[19,49,50,112,176,81,82,208,145,146]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[112,19],[58,49],[116,50],[9,112],[34,176],[146,81],[111,82],[148,208],[89,145],[178,146]]
Zm\backslash\;B4=[2,3]
從Zm\backslash\;B4取元素個數為t-ts=1個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(2,1)=2個
Y2=[[2],[3]]
val(Y2)=Σ2^y=[4,8]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[152,4],[100,8]]
由小到大排序
ListY1=[[9,112],[34,176],[58,49],[89,145],[111,82],[112,19],[116,50],[146,81],[148,208],[178,146]]
ListY2=[[100,8],[152,4]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B5=[5,6,7,0,1,2]
假設第5個位元是1，再取元素個數為ts-1=2個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(5,2)=10個
Y1=[[5,0,1],[5,0,2],[5,1,2],[5,6,0],[5,6,1],[5,6,2],[5,6,7],[5,7,0],[5,7,1],[5,7,2]]
val(Y1)=Σ2^y=[35,37,38,97,98,100,224,161,162,164]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[120,35],[118,37],[55,38],[53,97],[106,98],[62,100],[81,224],[160,161],[139,162],[13,164]]
Zm\backslash\;B5=[3,4]
從Zm\backslash\;B5取元素個數為t-ts=1個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(2,1)=2個
Y2=[[3],[4]]
val(Y2)=Σ2^y=[8,16]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[100,8],[122,16]]
由小到大排序
ListY1=[[13,164],[53,97],[55,38],[62,100],[81,224],[106,98],[118,37],[120,35],[139,162],[160,161]]
ListY2=[[100,8],[122,16]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B6=[6,7,0,1,2,3]
假設第6個位元是1，再取元素個數為ts-1=2個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(5,2)=10個
Y1=[[6,0,1],[6,0,2],[6,0,3],[6,1,2],[6,1,3],[6,2,3],[6,7,0],[6,7,1],[6,7,2],[6,7,3]]
val(Y1)=Σ2^y=[67,69,73,70,74,76,193,194,196,200]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[171,67],[141,69],[84,73],[101,70],[168,74],[129,76],[47,193],[94,194],[14,196],[43,200]]
Zm\backslash\;B6=[4,5]
從Zm\backslash\;B6取元素個數為t-ts=1個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(2,1)=2個
Y2=[[4],[5]]
val(Y2)=Σ2^y=[16,32]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[122,16],[138,32]]
由小到大排序
ListY1=[[14,196],[43,200],[47,193],[84,73],[94,194],[101,70],[129,76],[141,69],[168,74],[171,67]]
ListY2=[[122,16],[138,32]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
B7=[7,0,1,2,3,4]
假設第7個位元是1，再取元素個數為ts-1=2個的子集合Y1，共C(Floor(ts*m/t)-1,ts-1)=C(5,2)=10個
Y1=[[7,0,1],[7,0,2],[7,0,3],[7,0,4],[7,1,2],[7,1,3],[7,1,4],[7,2,3],[7,2,4],[7,3,4]]
val(Y1)=Σ2^y=[131,133,137,145,134,138,146,140,148,152]
計算[2^{val(Y1)},val(Y1)]數對
ListY1=[[103,131],[50,133],[76,137],[89,145],[100,134],[152,138],[178,146],[65,140],[169,148],[170,152]]
Zm\backslash\;B7=[5,6]
從Zm\backslash\;B7取元素個數為t-ts=1個的子集合Y1，共C(m-floor(ts*m/t),t-ts)=C(2,1)=2個
Y2=[[5],[6]]
val(Y2)=Σ2^y=[32,64]
計算[79*2^{-val(Y2)},val(Y2)]數對
ListY2=[[138,32],[154,64]]
由小到大排序
ListY1=[[50,133],[65,140],[76,137],[89,145],[100,134],[103,131],[152,138],[169,148],[170,152],[178,146]]
ListY2=[[138,32],[154,64]]
ListY1和ListY2互相比較沒找到相同值
－－－－－－－－
(x)　142