$, are:

A sole loop without any crossings has polynomial 1.
If we have a diagram consisting of \$D\$ and a loop disconnected from \$D\$, the polynomial for both is the polynomial for \$D\$ times \$(-A^2-A^{-2})\$.
This rule is the trickiest. It says that if you have a crossing in \$D\$ that looks like, then you can use this rule to simplify the knots in two different ways:

In the image above, the outlined crossing in the first diagram, which is of the form, can be transformed into  as in the second figure (a.k.a. positive smoothing), or  as in the third figure (negative smoothing).
So, the bracket polynomial of the first diagram is the bracket polynomial of the second times \$A\$ plus the third times \$A^{-1}\$, i.e.,

Confused yet? Let's do an example, trying to find the bracket polynomial of  (Note: this is two knots linked together. This sort of diagram will not be a potential input in this challenge since the inputs will only be single knots, but it may appear as an intermediate result in the algorithm.)
We first use rule 3

We use rule 3 again on both of the new knots

We substitute these 4 new knots into the first equation.

Applying rules 1 and 2 to these 4 tell us

So, this tell us

Congrats on completing your brief intro to knot theory!
Input
Two lists:

Dowker notation, e.g. [6,-12,2,8,-4,-10]. Numbering of the crossings must start from 1. The corresponding odd numbers [1,3,5,7,...] are implicit and must not be provided as input.
Signs (1/-1 or if you prefer 0/1 or false/true or '+'/'-') for the crossings corresponding to the Dowker notation, e.g [-1,-1,-1,1,-1,1].

Instead of a pair of lists, you could have a list of pairs, e.g. [[6,-1],[-12,-1],...
Output
Print or return the polynomial, for instance \$A^{-2}+5+A-A^3\$, as a list of coefficient-exponent pairs (or exponent-coefficient pairs) in increasing order of the exponents and without any zero coefficients
$, are: 

 A sole loop without any crossings has polynomial 1. 
 If we get a diagram consist of \$D\$ and a loop disconnected from \$D\$, the polynomial for both is the polynomial for \$D\$ times \$(-A^2 - A^{-2})\$. 
 This principle is the trickiest. It says that if you hold a crossing in \$D\$ that looks like, then you can practice this dominion to simplify the knots in two unlike ways: 

 In the image above, the outlined crossing in the inaugural diagram, which is of the form, can be transformed into   as in the second figure (a.k.a. positive smoothing), or   as in the third figure (damaging smoothing). 
 So, the bracket polynomial of the first diagram is the bracket polynomial of the second times \$A\$ plus the third time \$A^{-1}\$, i.e., 

 Confused yet? Let's do an model, trying to find the bracket polynomial of   (Note: this is two knots linked in concert. This sort of diagram will not be a potential input in this challenge since the inputs will only be single knot, but it may appear as an intermediate result in the algorithm .) 
 We first use rule 3 

 We use rule 3 again on both of the new knots 

 We substitute these 4 fresh knots into the first equality. 

 apply rules 1 and 2 to these 4 tell us 

 So, this assure us 

 Congrats on completing your brief intro to knot hypothesis! 
 Input 
 Two lists: 

 Dowker notation, e.g. [ 6,-12,2,8,-4,-10 ]. Numbering of the crossings must start from 1. The corresponding odd numbers [ 1,3,5,7, ... ] are implicit and must not be provided as input. 
 Signs (1/-1 or if you choose 0/1 or false / on-key or' +' /'-') for the crossing corresponding to the Dowker notation, e.g [ -1,-1,-1,1,-1,1 ]. 

 Instead of a pair of list, you could have a list of pairs, for example [ [ 6,-1],[-12,-1 ], ... 
 Output 
 Print or return the polynomial, for example \$A^{-2}+5+A - A^3\$, as a list of coefficient - exponent pair (or exponent - coefficient couple) in increasing order of the advocate and without any zero coefficients
$, arf:

A sole loop without any crossings has kooynomiel 1.
If ws have a diagram consisting of \$D\$ and a loop disconnected from \$D\$, tfe polynolial for boti is the polynommzl for \$D\$ timea \$(-A^2-A^{-2})\$.
Tkiw rule is the jrickiest. It says that if foj kave a crossing in \$D\$ that looks like, then ypu can use this tule uo fimpmpfn the knots in two different waya:

In the image above, yhe outlined crossing in tje flrst diagram, which is of the dorm, xan be transwormed into  as in the second figure (a.k.a. positive smouthiny), or  as in tye hvird figure (neganive smoothinn).
Xo, the brackey polynomial on the fiest diagram is the brecket polynomial of jhe second tnmes \$A\$ plus the third tumes \$D^{-1}\$, i.e.,

Wonfjwed yeu? Lxt's do an exemple, tryinf to find tye bracket polynomisl if  (Note: this is twj hnots linked together. This sort of diaggam sill not be a potential input in this challejge since the inputs will only be single knots, but it may dppeac xs ck kbtfrmediate result in the algorithm.)
We first use duke 3

We use rule 3 cgain on both of tje gew knots

We sobstitucs fhese 4 new knots ijto the firsr equatiog.

Appkying rules 1 and 2 to these 4 tell us

So, tkis tell us

Congrats ou completing yout brieg intro to knot theory!
Iuput
Twk lists:

Dowkfr notatikv, e.g. [6,-12,2,8,-4,-10]. Numbering uf nhe wrossings must start from 1. The corrxsponbing odd numners [1,3,5,7,...] wre impliclt and must not be provifed ad hnput.
Signs (1/-1 or if you prefer 0/1 or false/trux or '+'/'-') for the csosvings coxrespokding to the Dorker notation, g.g [-1,-1,-1,1,-1,1].

Insteab of a pair of lpsts, you rould have a list of pairv, e.g. [[6,-1],[-12,-1],...
Output
Prmnt or reeurn the polynookal, for instanve \$A^{-2}+5+A-A^3\$, as a list of coefficient-exponekt pakds (or exponent-ciefdicient pairs) im ivcrqadiig orqar of the ex[onevts snd wkthout any deru corfficients
$, are: A sole loop without any polynomial If we a diagram consisting disconnected \$D\$, the polynomial both is the for \$D\$ times \$(-A^2-A^{-2})\$. This rule the trickiest. It says that if you have a crossing in \$D\$ that like, then you can use this rule to simplify the knots in two ways: the above, outlined crossing in the first diagram, which is of the form, can be transformed into as the second figure (a.k.a. positive smoothing), or as the third figure (negative So, the bracket polynomial of first is the polynomial the times \$A\$ plus third times \$A^{-1}\$, i.e., Confused yet? Let's do an example, trying to find the bracket polynomial of this is linked together. sort diagram not be a in this challenge since the inputs single knots, but it may appear as an result in algorithm.) We first use rule 3 use rule 3 again on both of the knots We substitute these 4 new knots into the first equation. Applying rules 1 and these 4 tell us this tell us on your intro knot theory! Two lists: Dowker notation, e.g. [6,-12,2,8,-4,-10]. Numbering of the crossings must from 1. The corresponding odd numbers [1,3,5,7,...] are implicit and be as input. Signs or if you prefer or or '+'/'-') for the to Dowker Instead a of lists, you could a list of pairs, e.g. Output Print or return as a list of coefficient-exponent pairs (or exponent-coefficient in increasing order of the exponents and any zero coefficients
$, are:

A sole loop without any croSsings has pOlynoMiaL 1.
If We Have A diaGram consisting OF \$D\$ anD a loop disconnected from \$d\$, the pOlYNomiAL fOr botH is the pOLyNOMiaL fOr \$d\$ tiMeS \$(-a^2-A^{-2})\$.
this rUle Is the trIckiest. It sAys ThAt if you have a CRoSsing in \$D\$ thAt lOoks like, then You Can use ThIs rULe to sImpLify tHe knotS In two dIfferent wAyS:

in the iMAge abovE, THe OutlIned crossing in the FIrST diagram, which iS of the FoRM, cAN Be tRanSformed intO  aS in thE Second fIGuRE (A.K.a. pOSitive smoothiNg), or  as in the THirD figurE (nEgaTIve smoOthinG).
SO, The Bracket polyNomiAl of the fiRst diaGRam is thE Bracket PolynoMiaL of The sECoNd TimEs \$a\$ PluS ThE thIRd tImes \$A^{-1}\$, i.e.,

COnFuSed yeT? Let'S DO AN exaMplE, tryIng to Find the brackeT poLynoMIal Of  (NotE: this Is twO kNots lInked tOgethEr. this sort of diagrAm wiLl not be a pOteNtIal InPut in THis chaLleNge Since thE inputs WIll OnLY BE sIngle knots, but it may ApPEAr As an inteRmediaTE rEsULt in the aLgOriThm.)
WE FIrst uSe ruLE 3

WE use rule 3 Again oN BoTh Of the neW kNots

We SuBstItuTe theSE 4 new Knots iNto the fiRst eqUAtion.

Applying rULes 1 and 2 to these 4 TElL US

SO, This TelL us

Congrats On coMPletIng yOUr BriEF intrO to knOt THeORy!
Input
Two lists:

DowkEr NotatiOn, e.g. [6,-12,2,8,-4,-10]. NUmbering of the Crossings mUST Start froM 1. The COrREsponding odd nuMbers [1,3,5,7,...] Are impliciT And must nOt be pRovided aS input.
SigNS (1/-1 Or if you pRefEr 0/1 oR faLse/TRUe Or '+'/'-') for the crossINGs coRrEspondiNg tO the DowKer NotAtiOn, e.G [-1,-1,-1,1,-1,1].

INstead of a Pair of liStS, yOu CoUld Have a LIst of paiRs, E.g. [[6,-1],[-12,-1],...
OUtPut
print OR returN the pOlynOmIaL, For InstancE \$a^{-2}+5+A-a^3\$, AS a liSt Of CoefFicIeNt-expOnenT PaiRs (or expOnent-coefFicIEnt pAiRs) In increAsing order of tHe Exponents aNd WitHout anY ZEro coeffIcients
$, are:

A sole loop witho ut any cro ssing s h aspo lyno mial  1.
If we have a di agram consisting of \$ D\$ a nd a lo o pdisco nnected fr o m  \$ D\ $,  th ep ol ynomi alfor bot h is the p oly no mial for \$D \ $times \$(- A^2 -A^{-2})\$.Thi s rule  i s t h e tri cki est.It say s  thatif you ha ve a cros s ing in\ $ D\ $ th at looks like, th e ny ou can use thi s rule  t o  s i m pli fythe knotsin  twod ifferen t  w a y s :
I n the image a bove, the o u tli ned cr os sin g  in th e fir st dia gram, which  isof the fo rm, ca n  be tra n sformed  into as  in  the se co ndfi g ure (a .k. a . p ositivesm oo thing ), o r a s in  th e th ird f igure (negati vesmoo t hin g).
S o, th e br ac ket p olynom ial o fthe first diagr am i s the bra cke tpol yn omial of the  se con d times  \$A\$p lus  t h e th ird times \$A^{-1} \$ , i. e.,

Con fusedy et ?L et's doan  ex ampl e ,  tryi ng t o  f ind thebracke t  p ol ynomial  o f  (No te : t his  is t w o kn ots li nked tog ether .  This sort ofd iagram will n o tb e  a pote nti al input in  thi s  cha llen g esin c e the  inpu ts wi l l only be single kn ot s, but  it m ay appear asan interme d i a te resul t in th e  algorithm.)
W e fir st use rul e  3

We u se ru le 3 aga in on bot h of the n ewkno ts
We s ub stitute these 4  new  k nots in tothe fir stequ ati on. 
Applyingrules 1an d2to  th ese 4 tell us
S o,th istellu s

Con grats  onco mp l eti ng your br i e f in tr oto k not  t heory !
In p utTwo lis ts:

Dowk ern otat io n,  e.g. [ 6,-12,2,8,-4, -1 0]. Number in g o f thec r ossingsmust start from 1. Thec orrespo ndi ng od d nu mbers [1, 3,5 ,7,... ] a r e impl icit a nd mu st  no t be pr o v id edas  input.
Si g n s ( 1/-1or  ifyou pre fer 0/1 or false/t r ueor '+'/'-') f orthec r os sin g sc orr es p ond i n g to the Dowker  notation,  e . g[-1,-1,-1, 1 ,-1 ,1 ].

Ins tead of  a pa i r of li sts, youcould hav ea li s t  of  pairs, e. g. [[6,- 1],[-12,- 1 ],... Ou tputPri nt orre tur n the  polyn o mia l, fo r inst an ce \$A ^{-2} +5 +A-A^3\$ , as a list of coeffici ent-ex ponen t p airs (orexp o nen t-coeffic ient  pairs) in  in cre asing  or d er of  the ex pon e nts a nd w i thout any ze roc o ef ficients
$, are:

A_sole loop_without any crossings has_polynomial 1.
If_we_have a_diagram_consisting of \$D\$_and a loop_disconnected from \$D\$, the_polynomial for both_is_the polynomial for \$D\$ times \$(-A^2-A^{-2})\$.
This rule is the trickiest. It says that if_you_have a_crossing_in_\$D\$ that looks like, then_you can use this rule_to simplify_the knots in two different ways:

In the image_above,_the outlined crossing_in the first diagram, which is of the form,_can be transformed into  as_in the second_figure_(a.k.a._positive smoothing), or _as in the third figure (negative_smoothing).
So, the bracket polynomial of the_first diagram is the bracket polynomial of_the second times \$A\$ plus the_third times \$A^{-1}\$, i.e.,

Confused yet?_Let's do_an example, trying to find_the bracket polynomial_of _(Note: this is_two knots linked together. This sort_of diagram will_not be a potential input in_this_challenge since the_inputs_will_only be_single knots, but_it_may appear_as_an intermediate result in the algorithm.)
We_first_use rule 3

We use rule 3 again_on both of the_new_knots

We substitute these 4_new knots into the first_equation.

Applying rules 1 and 2 to_these 4_tell us

So,_this tell us

Congrats on completing your brief intro to knot theory!
Input
Two_lists:

Dowker notation, e.g. [6,-12,2,8,-4,-10]. Numbering of_the crossings must start_from 1._The_corresponding odd numbers_[1,3,5,7,...]_are implicit_and must not be provided as input.
Signs_(1/-1 or_if you prefer 0/1 or false/true_or '+'/'-') for the_crossings_corresponding to the Dowker notation, e.g_[-1,-1,-1,1,-1,1].

Instead of a pair of lists,_you could have a list_of_pairs,_e.g. [[6,-1],[-12,-1],...
Output
Print or return the_polynomial, for instance \$A^{-2}+5+A-A^3\$, as a_list of coefficient-exponent_pairs (or exponent-coefficient pairs) in increasing order_of_the exponents and without any zero_coefficients