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9a1cd96729
Setting date for HAKMEM 141 to be 1972-03-16. Setting date for HAKMEM 17 to be 1971-01-01, although this date is probably not correct, but we have no info to base its date on other than that it was "before" 1972.
1029 lines
39 KiB
Plaintext
1029 lines
39 KiB
Plaintext
*********************************************
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GEOMETRY, ALGEBRA, CALCULUS
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*********************************************
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ITEM (SCHROEPPEL):
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FUNCTIONS OF THE FORM AY'' + BY' + CY = D
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WHERE A, B, C, AND D ARE FUNCTIONS OF X
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CAN BE CONVERTED TO THE FORM W' = W^2 - E
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WHERE E = B^2/4A^2 - C/A + (AB' - BA')/2A^2.
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IN ADDITION, A PARTICULAR SOLUTION OF THE NEW FORM
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IS ENOUGH BECAUSE SOLUTIONS DIFFER BY A SOLVABLE AMOUNT.
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FOR EXAMPLE, BESSEL'S FUNCTION BECOMES
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W' = W^2 + 1 - (N^2 - 1/4)/X^2
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SIMILARLY, Y''' + AY'' + BY' + CY = D
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CAN BE REDUCED TO
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Y'' = Y^2 + A AND
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Y'' + 3YY' + Y^3 + AY + B = 0
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ITEM (SCHROEPPEL):
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GAMMA(1/3) AND GAMMA(2/3) ARE INTEREXPRESSIBLE.
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GAMMA(1/4) AND GAMMA(3/4) ARE INTEREXPRESSIBLE.
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THUS THESE TWO PAIRS ARE OF DIMENSIONALITY ONE.
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GAMMA(1/10) AND GAMMA(2/10) ARE SUFFICIENT TO EXPRESS GAMMA(N/10) FOR ALL N.
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GAMMA(1/12) AND GAMMA(2/12) ARE SUFFICIENT TO EXPRESS GAMMA(N/12) FOR ALL N.
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GAMMA(1/3) AND GAMMA(1/4) ARE SUFFICIENT TO EXPRESS GAMMA(N/12) FOR ALL N.
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THUS THE THREE CASES ABOVE ARE OF DIMENSIONALITY TWO.
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PROBLEM: FIND SOME ORDER TO THIS DIMENSIONALITY BUSINESS.
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ITEM (JAN KOK):
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PROBLEM: GIVEN A REGULAR N-GON WITH ALL DIAGONALS DRAWN,
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HOW MANY REGIONS ARE THERE? IN PARTICULAR, HOW MANY TRIPLE (OR N-TUPLE)
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CONCURRENCES OF DIAGONALS ARE THERE?
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ITEM (SCHROEPPEL):
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REGARDING CONVERGENCE OF NEWTON'S METHOD FOR QUADRATIC EQUATIONS:
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DRAW THE PERPENDICULAR BISECTOR OF THE LINE CONNECTING THE TWO ROOTS.
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POINTS ON EITHER SIDE CONVERGE TO THE CLOSEST ROOT. ON THE LINE:
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1 THEY DO NOT CONVERGE
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2 THERE IS A DENSE SET OF POINTS WHICH INVOLVE DIVISION BY ZERO
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3 THERE IS A DENSE SET OF POINTS WHICH LOOP, BUT ROUNDOFF ERROR
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PROPAGATES SO ALL LOOPS ARE UNSTABLE
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4 BEING ON THE LINE IS ALSO UNSTABLE (UNLESS, OF COURSE, THE ROOTS ARE
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IMAGINARY AND YOU ARE ON THE REAL AXIS)
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ITEM (SCHROEPPEL):
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THE MOST PROBABLE SUIT DISTRIBUTION IN BRIDGE HANDS IS 4-4-3-2, AS COMPARED
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TO 4-3-3-3, WHICH IS THE MOST EVENLY DISTRIBUTED. THIS IS BECAUSE THE
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WORLD TENDS TO HAVE UNEQUAL NUMBERS, WHICH IS SORT OF A THERMODYNAMIC EFFECT
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SAYING THINGS WILL NOT BE IN THE STATE OF LOWEST ENERGY, BUT IN THE STATE
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OF LOWEST DISORDERED ENERGY.
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ITEM (SCHROEPPEL):
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BY MATHLAB, THE DISCRIMINANT OF X^4 + FX^3 + GX^2 + HX + I IS
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(AS THE DISCRIMINANT OF AX^2 + BX + C IS B^2 - 4AC):
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- 27 H^4 + 18 FGH^3 - 4 F^3H^3 - 4 G^3H^2 + F^2G^2H^2
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+ I * [144 GH^2 - 6 F^2H^2 - 80 FG^2H + 18 F^3GH + 16 G^4 - 4 F^2G^3]
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+ I^2 * [- 192 FH - 128 G^2 + 144 F^2G - 27 F^4]
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- 256 I^3
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ITEM (SCHROEPPEL):
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IF A IS THE FIRST SYMMETRIC FUNCTION OF N VARIABLES = X + Y + Z + ...
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AND B IS THE SECOND SYMMETRIC FUNCTION OF N VARIABLES = XY + XZ + ... + YZ + ...
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(B = SUM OF PAIRS), THEN X^2 + Y^2 + Z^2 + ... = A^2 - 2B.
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FURTHERMORE, SIMILAR RULES EXIST FOR SUMS OF OTHER POWERS, SUCH AS
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X^3 + Y^3 + Z^3 + ... AND X^4 + Y^4 + Z^4 + ..., ETC.
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FOR EXAMPLE, X^4 + Y^4 + Z^4 + ... = A^4 - 4A^2B + 2B^2 + 4AC - 4D.
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ITEM (SCHROEPPEL):
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IF X^4 + BX^2 + CX + D = 0, THEN
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2X = +OR- (SQRT Z1) +OR- (SQRT Z2) +OR- (SQRT Z3) WHERE AN ODD NUMBER OF SIGNS ARE PLUS.
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Z1, Z2, Z3 ARE ROOTS OF Z^3 - 2BZ^2 + (B^2 - 4D)Z - C^2 = 0.
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PROBLEM: REPLACING C BY -C SHOULD TAKE X INTO -X, BUT THE CUBIC WILL HAVE THE SAME ROOTS.
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MAYBE THERE IS AN ERROR, OR MAYBE YOU TAKE AN EVEN NUMBER OF SIGNS
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TO BE PLUS?
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ITEM (SCHROEPPEL):
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SOLUTIONS TO F(X) = X^3 - 3BX^2 + CX + D = 0 ARE
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B - K * CUBE ROOT [F(B)/2 + SQRT [(F(B)/2)^2 + (F'(B)/3)^3]]
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- L * CUBE ROOT [F(B)/2 - SQRT [(F(B)/2)^2 + (F'(B)/3)^3]]
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WHERE OMEGA = CUBE ROOT [1],
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K = L = 1 GIVES THE FIRST ROOT
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K = OMEGA, L = OMEGA^2 GIVES THE SECOND ROOT
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K = OMEGA^2, L = OMEGA GIVES THE THIRD ROOT
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ITEM (GOSPER):
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THE MYTH THAT ANY GIVEN PROGRAMMING LANGUAGE IS MACHINE INDEPENDENT
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IS EASILY EXPLODED BY COMPUTING THE SUM OF POWERS OF TWO.
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IF THE RESULT LOOPS WITH PERIOD = 1 WITH SIGN +, YOU ARE ON A SIGN-MAGNITUDE MACHINE.
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IF THE RESULT LOOPS WITH PERIOD = 1 WITH SIGN -, YOU ARE ON A TWOS-COMPLEMENT MACHINE.
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IF THE RESULT LOOPS WITH PERIOD > 1, INCLUDING THE BEGINNING, YOU ARE ON A ONES-COMPLEMENT MACHINE.
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IF THE RESULT LOOPS WITH PERIOD > 1, NOT INCLUDING THE BEGINNING, YOU ARE ON A DECIMAL MACHINE.
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IF YOU RUN OUT OF MEMORY, YOU ARE ON A STRING SYSTEM/MACHINE.
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BY THIS STRATEGY, CONSIDER THE UNIVERSE, OR, MORE PRECISELY, ALGEBRA:
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LET X = THE SUM OF MANY POWERS OF TWO = ...111111
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NOW ADD X TO ITSELF; X + X = ...111110
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THUS, 2X = X - 1 SO X = -1
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THEREFORE ALGEBRA IS RUN ON A MACHINE (THE UNIVERSE) WHICH IS TWOS-COMPLEMENT.
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ITEM (SALAMIN):
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CONSTRUCT CONFORMAL MAPPINGS OF EUCLIDEAN N-SPACE INTO ITSELF BY
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STEREOGRAPHIC PROJECTION ONTO A N-SPHERE, ROTATING THE SPHERE, AND
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PROJECTING BACK.
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PROBLEM: GENERALIZE HOMOGRAPHIC TRANSFORMATIONS AND ANALYTIC FUNCTIONS.
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ITEM (GOSPER):
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PROBLEM: PICK A RANDOM ANGLE. TURN THAT ANGLE AND GO A UNIT STEP.
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TURN TWICE THE ANGLE AND GO A UNIT STEP. TURN FOUR TIMES ...
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FOR WHAT INITIAL ANGLES IS YOUR LOCUS BOUNDED?
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*********************************************
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BOOLEAN ALGEBRA
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*********************************************
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ITEM (SCHROEPPEL):
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IN AN ATTEMPT TO GET A HOLD ON THE PROBLEM OF MINIMAL LOGIC GATE SYNTHESIS,
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WHICH MAY (HOPEFULLY) LEAD TO USEFUL ALGORITHMS, SOME PARTIAL RESULTS ARE:
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DESCRIPTION OF RESTRICTIONS:
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2 BINARY INPUTS
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OUTPUTS COMPOSED OF 0, 1, OR DON'T CARE
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FOR FUNCTIONS OF 2 INPUT BITS, A MINIMAL SYNTHESIS EXISTS.
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RULES TO EVALUATE "MINIMAL":
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1 -- NOT FUNCTIONS ARE FOR FREE
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2 -- CHARGE 1 FOR EACH AND GATE OF 2 INPUTS
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3 -- THERE IS NO FEEDBACK
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4 -- AN XOR GATE COSTS (I.E., CAN BE MADE OF NO FEWER THAN) 3 AND GATES
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5 -- A MAJORITY OF 3 FUNCTION COSTS 4 AND GATES
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EXAMPLE: PRIME PREDICATE ON 4 BITS COSTS AT MOST 7
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ITEM (SPECINER):
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N NUMBER OF BOOLEAN MONOTONIC FUNCTIONS OF N VARIABLES
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0 2 (T, F)
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1 3 (T, F, P)
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2 6 (T, F, P, Q, P AND Q, P OR Q)
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3 20
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4 168
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5 7581
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6 7,828,354
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ITEM (SCHROEPPEL):
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APPLICABLE TO 2-NOTS PROBLEM (SYNTHESIZE A BLACK BOX WHICH COMPUTES
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NOT-A, NOT-B, AND NOT-C FROM A, B, AND C, USING ONLY 2 NOTS):
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FUNCTIONS SYNTHESIZABLE WITH ONE NOT ARE THOSE WHERE ANY UPWARD PATH
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HAS AT MOST ONE DECREASE (THAT IS, FROM T TO F).
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ITEM (ROGER BANKS):
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A VENN DIAGRAM FOR N VARIABLES WHERE THE SHAPE REPRESENTING EACH VARIABLE IS CONVEX
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CAN BE MADE BY SUPERIMPOSING SUCCESSIVE M-GONS (M = 2, 4, 8, ...),
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EVERY OTHER SIDE OF WHICH HAS BEEN PUSHED OUT TO THE CIRCUMSCRIBING CIRCLE.
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THE NESTED M-GONS MUST SHRINK SLIGHTLY; PROBABLY RADIUS = 2, 3/2, 5/4, ETC.
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IS SUFFICIENT.
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ITEM (SCHROEPPEL & WALTZ):
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PROBLEM: COVER THE EXECUPORT CHARACTER RASTER COMPLETELY WITH THE MINIMUM NUMBER
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OF CHARACTERS. THE THREE CHARACTERS I, H AND # WORKS. USING CAPITAL LETTERS ONLY,
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THE FIVE CHARACTERS B, I, M, V AND X IS A MINIMAL SOLUTION.
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FIND A GENERAL METHOD OF SOLVING SUCH PROBLEMS.
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ITEM (GOSPER):
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PROBLEM: GIVEN SEVERAL BINARY NUMBERS, HOW CAN ONE FIND A MASK
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WITH A MINIMAL NUMBER OF BITS ON WHICH, AND-ED WITH
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EACH OF THE ORIGINAL NUMBERS, WILL PRESERVE THEIR DISTINCTNESS FROM EACH OTHER?
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*********************************************
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RANDOM NUMBERS
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*********************************************
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ITEM (SCHROEPPEL):
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RANDOM NUMBER GENERATORS, SUCH AS ROLLO SILVER'S, WHICH USE SHIFTS AND XORS,
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AND GIVE AS VALUES ONLY SOME PART OF THEIR INTERNAL STATE, CAN BE DESCRIBED
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AND THEIR OPERATIONS INVERTED AND NUMBERS FROM THEM USED TO OBTAIN THEIR
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TOTAL INTERNAL STATE. FOR EXAMPLE, 2 CONSECUTIVE VALUES FROM ROLLO'S
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SUFFICE TO ALLOW PREDICTION OF ITS ENTIRE FUTURE. ROLLO'S IS:
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REGISTER A GETS LOADED WITH "HIGH" WORD
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REGISTER B GETS LOADED WITH "LOW" WORD
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REGISTER A GETS STORED IN "LOW" WORD
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PDP-6/10 INSTRUCTION LSHC A,35. IS EXECUTED
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REGISTER A XORED WITH "HIGH" WORD GETS STORED IN "HIGH" AND IS VALUE
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THIS SUGGESTS A SUSCEPTIBILITY TO ANALYSIS OF MECHANICAL CODE MACHINES.
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ITEM (SCHROEPPEL):
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PROBLEM: WRITE A NUMBER WHICH HAS THE SAME NUMBER OF LETTERS AS THE NUMBER.
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FOUR IS THE ONLY SOLUTION.
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*********************************************
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NUMBER THEORY, PRIMES, PROBABILITY
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*********************************************
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ITEM (SCHROEPPEL):
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AFTER ABOUT 40 MINUTES OF RUN TIME TO VERIFY THE ABSENCE OF ANY
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FURTHER FACTORS LESS THAN A PDP-10 WORD SIZE,
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THE 125TH MERSENNE NUMBER, 2^125 - 1, WAS FACTORED ON TUESDAY,
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JANUARY 5, 1971, IN 371 SECONDS RUN TIME AS FOLLOWS:
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2^125 - 1 = 31 * 601 * 1,801 * 269,089,806,001 * 4,710,883,168,879,506,001
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ITEM (SCHROEPPEL):
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FOR A RANDOM NUMBER X, THE PROBABILITY OF ITS LARGEST PRIME FACTOR BEING
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(1) GREATER THAN THE SQUARE ROOT OF X IS LN 2.
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(2) LESS THAN THE CUBE ROOT OF X IS ABOUT 4.86%.
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THIS SUGGESTS THAT SIMILAR PROBABILITIES ARE INDEPENDENT OF X;
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FOR INSTANCE, THE PROBABILITY THAT THE LARGEST PRIME FACTOR OF X
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IS LESS THAN THE 20TH ROOT OF X MAY BE A FRACTION INDEPENDENT OF X.
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RELEVANT DATA:
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RANGE COUNT CUM. SUM OF COUNT
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10^12 TO 10^6 7198 [6944] 10018
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10^6 TO 10^4 2466 2820
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10^4 TO 10^3 354 402 [487]
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10^3 TO 252 40 48 (252 = 10^(12/5))
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252 TO 100 7 8
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100 TO 52 1 1
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51 TO 1 0 0
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WHERE:
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"COUNT" IS THE NUMBER OF NUMBERS WHOSE LARGEST PRIME FACTOR IS IN "RANGE"
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FOR THE NUMBERS 10^12 + 1 TO 10^12 + 10018.
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[ ] DENOTE THE EXPECTED VALUE OF ADJACENT ENTRIES.
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THE NUMBER OF PRIMES IN 10^12 + 1 TO 10^12 + 10018 IS 335;
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THE PRIME NUMBER THEOREM PREDICTS 363 IN THIS RANGE.
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THIS IS RELEVANT TO KNUTH'S DISCUSSION OF FACTORING, VOL. 2, P. 351-354.
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ITEM (SCHROEPPEL):
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TWIN PRIMES:
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166,666,666,667 = (10^12 + 2)/6
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166,666,666,669
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THE PRIMES WHICH BRACKET 10^12 ARE 10^12 + 39 AND 10^12 - 11
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THE PRIMES WHICH BRACKET 10^15 ARE 10^15 + 37 AND 10^15 - 11
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THE NUMBER 23,333,333,333 IS PRIME.
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THE NUMBER 166,666,666,666,667 IS PRIME, BUT
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166,666,666,666,669 IS NOT.
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VARIOUS PRIMES, USING T = 10^12, ARE:
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40T + 1
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62.5T + 1
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500T - 1
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500T - 7
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200T - 3
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ITEM (SCHROEPPEL):
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RAMANUJAN'S PROBLEM OF SOLUTIONS TO 2^N - 7 = X^2 WAS SEARCHED TO
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ABOUT N = 10^40; ONLY HIS SOLUTIONS (N = 3, 4, 5, 7, 15) WERE FOUND.
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ITEM (SCHROEPPEL):
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TAKE A RANDOM REAL NUMBER AND RAISE IT TO LARGE POWERS; WE EXPECT THE
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FRACTION PART TO BE UNIFORMLY DISTRIBUTED.
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SOME EXCEPTIONS:
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1 -- PHI = (1 + SQRT 5)/2
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2 -- ALL -1 < X < 1
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3 -- SQRT 2 (HALF ARE INTEGERS, OTHER HALF ARE UNCLEAR)
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4 -- 1 + SQRT 2 -- PROOF:
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(1 + SQRT 2)^N + (1 - SQRT 2)^N = INTEGER (BY INDUCTION);
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THE (1 - SQRT 2)^N GOES TO ZERO.
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5 -- 2 + SQRT 2 -- SIMILAR TO 1 + SQRT 2
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NOW, 3 + SQRT 2 IS SUSPICIOUS; IT LOOKS NON-UNIFORM, AND SEEMS TO
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HAVE A CLUSTER POINT AT ZERO. PROBLEM: IS IT NON-UNIFORM?
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ITEM (SCHROEPPEL):
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NUMBERS WHOSE RIGHT DIGIT CAN BE REPEATEDLY REMOVED AND THEY ARE STILL PRIME:
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CONJECTURE: THERE ARE A FINITE NUMBER OF THEM IN ANY RADIX.
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IN DECIMAL THERE ARE 51, THE LONGEST BEING 1,979,339,333 AND 1,979,339,339.
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ITEM (SCHROEPPEL):
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PROBLEM: CAN EVERY POSITIVE INTEGER BE EXPRESSED IN TERMS OF 3 AND THE
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OPERATIONS FACTORIAL AND INTEGER SQUARE ROOT?
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ITEM (SCHUTZENBERGER):
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PROBLEM: USING N DIGITS, CONSTRUCT A STRING OF DIGITS WHICH AT NO TIME
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HAS ANY SEGMENT APPEARING CONSECUTIVELY TWICE.
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N = 2 => FINITE MAXIMUM STRING
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N = 10 => KNOWN INFINITE
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DETERMINE MAXIMUM STRING LENGTH FOR N = 3.
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SUB-PROBLEM: HOW MANY SEQUENCES EXIST OF ANY PARTICULAR LENGTH?
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ITEM (BEELER):
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THERE ARE NO ZEROS IN THE DECIMAL EXPRESSION OF POWERS OF TWO GREATER THAN 2^86,
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WHICH = 77,371,252,455,336,267,181,195,264; SEARCHED THROUGH 2^30,739,014.
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ITEM (SCHROEPPEL):
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TAKE AS MANY NUMBERS AS POSSIBLE FROM 1 TO N SUCH THAT NO 3 ARE
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IN ARITHMETIC PROGRESSION. CONJECTURE: THE DENSITY OF SUCH SETS APPROACHES ZERO
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AS N APPROACHES INFINITY; IN FACT, IT GOES AS N^((LN 2)/(LN 3)).
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XX.XX IS A KNOWN SOLUTION FOR N = 5
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XX.XX....XX.XX IS A KNOWN SOLUTION FOR N = 14
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CONJECTURE THAT XX.XX JUST KEEPS GETTING COPIED.
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IF THE N^((LN 2)/(LN 3)) CAN BE PROVED, IT FOLLOWS THAT THERE ARE INFINITELY MANY
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PRIMES P1, P2, P3 IN ARITHMETIC PROGRESSION, SINCE PRIMES ARE MUCH MORE
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COMMON THAN N^((LN 2)/(LN 3)).
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ITEM (SCHROEPPEL):
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PROBLEM: HOW MANY SQUARES HAVE NO ZEROS IN THEIR DECIMAL EXPRESSION?
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ITEM (GOSPER):
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THE FOLLOWING WAS CALCULATED WITH PETER SAMSON'S
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STRING HANDLING INTERPRETIVE TRANSLATOR PROGRAM, "STRING."
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THE PROBABILITY OF AT LEAST ONE OF THE DECIMAL DIGITS NOT APPEARING
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IN A RANDOM STRING OF DECIMAL DIGITS OF LENGTH N IS:
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N PROBABILITY
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0 THROUGH 9 1/1
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10 1,561,933/1,562,500
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11 3,118,763/3,125,000
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12 31,056,653/31,250,000
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13 61,608,109/62,500,000
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14 949,886,851/976,562,500
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15 2,981,405,549/3,125,000,000
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16 145,264,092,067/156,250,000,000
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17 281,220,490,739/312,500,000,000
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18 135,207,403,123,699/156,250,000,000,000
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19 258,374,516,368,781/312,500,000,000,000
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20 2,453,945,865,008,123/3,125,000,000,000,000
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21 4,635,475,839,633,883/6,250,000,000,000,000
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22 544,486,732,876,350,041/781,250,000,000,000,000
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23 509,277,073,221,163,409/781,250,000,000,000,000
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24 189,739,871,255,887,679/312,500,000,000,000,000
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25 70,424,508,530,924,423/125,000,000,000,000,000
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26 1,628,170,420,719,167,436,233/3,125,000,000,000,000,000,000
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27 3,002,290,440,846,272,623,291/6,250,000,000,000,000,000,000
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28 27,606,049,370,460,038,309,269/62,500,000,000,000,000,000,000
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50 APPROXIMATELY 0.0508976470723677636754403338329777
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ITEM (GOSPER):
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THE VARIANCE OF A PSEUDO-GAUSSIAN DISTRIBUTED RANDOM VARIABLE MADE BY ADDING
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T INDEPENDENT, UNIFORMLY DISTRIBUTED RANDOM INTEGER VARIABLES
|
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WHICH RANGE FROM 0 TO N-1, INCLUSIVE, IS T((N^2 - 1)/12).
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ITEM (SCHROEPPEL):
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THE JOYS OF 239 ARE AS FOLLOWS:
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PI = 16 ARCTAN (1/5) - 4 ARCTAN (1/239),
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WHICH IS RELATED TO THE FACT THAT 2 X 13^4 - 1 = 239^2.
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239 NEEDS 4 SQUARES (THE MAXIMUM) TO EXPRESS IT.
|
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239 NEEDS 9 CUBES (THE MAXIMUM, SHARED ONLY WITH 23) TO EXPRESS IT.
|
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239 NEEDS 19 FOURTH POWERS (THE MAXIMUM) TO EXPRESS IT.
|
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(ALTHOUGH 239 DOESN'T NEED THE MAXIMUM NUMBER OF FIFTH POWERS.)
|
||
1/239 = .00418410041841..., WHICH IS RELATED TO THE FACT THAT
|
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1,111,111 = 239 X 4,649.
|
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THE 239TH MERSENNE NUMBER, 2^239 - 1, IS KNOWN COMPOSITE, BUT NO FACTORS ARE KNOWN.
|
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239 = 11101111 BASE 2.
|
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239 = 22212 BASE 3.
|
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239 = 3233 BASE 4.
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AND 239 IS PRIME, OF COURSE.
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ITEM (GOSPER, SALAMIN):
|
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PI = 16 ARCTAN (1/5) - 4 ARCTAN (1/239)
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PI = 28 ARCTAN (3/79) + 20 ARCTAN (29/278)
|
||
PI = 48 ARCTAN (3/79) + 20 ARCTAN (1457/22049)
|
||
SEE ALSO RAMANUJAN'S COLLECTED PAPERS FOR A SERIES FOR
|
||
1/PI (6 DECIMAL PLACES PER TERM) AND A SERIES FOR
|
||
1/(PI SQRT 8) (8 DECIMAL PLACES PER TERM).
|
||
|
||
ITEM (BEELER):
|
||
THE "LENGTH" OF AN N-DIGIT DECIMAL NUMBER IS DEFINED AS THE NUMBER
|
||
OF TIMES ONE MUST ITERATIVELY FORM THE PRODUCT OF ITS DIGITS
|
||
UNTIL ONE OBTAINS A ONE-DIGIT PRODUCT (SEE TECHNOLOGY REVIEW PUZZLE CORNER,
|
||
DECEMBER 1969 AND APRIL 1970). FOR VARIOUS N, THE FOLLOWING SHOWS
|
||
THE MAXIMUM "LENGTH", AS WELL AS HOW MANY DISTINCT NUMBERS
|
||
(PERMUTATION GROUPS OF N DIGITS) THERE ARE:
|
||
N MAX L DISTINCT
|
||
2 4 54
|
||
3 5 219
|
||
4 6 714
|
||
5 7 2,001
|
||
6 7 5,004
|
||
7 8 11,439
|
||
8 9 24,309
|
||
9 9 48,619
|
||
10 10 92,377
|
||
11 10 167,959
|
||
12 10 293,929
|
||
ALSO, FOR N = 10, 11 AND 12, A TENDENCY FOR THERE TO BE MANY
|
||
FEWER NUMBERS OF "LENGTH" = 7 IS NOTED. OTHER THAN THIS,
|
||
THE FREQUENCY OF NUMBERS OF ANY GIVEN N, THROUGH N = 12,
|
||
DECREASES WITH INCREASING "LENGTH".
|
||
|
||
ITEM (BEELER):
|
||
THE FIBONACCI SERIES MODULO M HAS BEEN STUDIED. THIS SERIES HAS A CYCLE LENGTH L
|
||
AND WITHIN THIS CYCLE HAS SUB-CYCLES WHICH ARE BOUNDED BY ZERO MEMBERS.
|
||
THE LENGTH OF POWERS OF PRIMES SEEMS TO BE
|
||
L = (LENGTH OF PRIME) X (PRIME^(POWER - 1))
|
||
THE LENGTH OF PRODUCTS OF POWERS OF PRIMES SEEMS TO BE
|
||
L = LEAST COMMON MULTIPLE OF LENGTHS OF POWERS OF PRIMES WHICH ARE FACTORS
|
||
THERE CAN BE ONLY 1, 2 OR 4 SUB-CYCLES IN THE CYCLE OF A PRIME.
|
||
PRIMES WITH 1 SUB-CYCLE SEEM TO HAVE LENGTHS
|
||
L = (PRIME - 1)/N, N COVERING ALL INTEGERS
|
||
PRIMES WITH 2 SUB-CYCLES SEEM TO HAVE LENGTHS
|
||
L = (PRIME - (-1)^M)/M, M COVERING
|
||
ALL INTEGERS EXCEPT 10 K + 5 FORM
|
||
PRIMES WITH 4 SUB-CYCLES SEEM TO ALWAYS BE OF FORM 4 K + 1,
|
||
AND SEEM TO HAVE LENGTHS
|
||
L = 2 (PRIME + 1)/R OR (PRIME - 1)/S,
|
||
R COVERING ALL INTEGERS OF FORM 10 K + 1, 3, 7 OR 9
|
||
S COVERING ALL INTEGERS
|
||
AT SCHROEPPEL'S SUGGESTION, THE PRIMES HAVE BEEN SEPARATED MOD 40,
|
||
WHICH USUALLY DETERMINES THEIR NUMBER OF SUB-CYCLES:
|
||
PRIME MOD 40 SUB-CYCLES
|
||
1, 9 USUALLY 2, OCCASIONALLY 1 OR 4 (ABOUT EQUALLY)
|
||
3, 7, 23, 27 2
|
||
11, 19, 31, 39 1
|
||
13, 17, 33, 37 4
|
||
21, 29 1 OR 4 (ABOUT EQUALLY)
|
||
2 (ONLY 2) 1
|
||
5 (ONLY 5) 4
|
||
ATTENTION WAS DIRECTED TO PRIMES WHICH ARE 1 OR 9 MOD 40
|
||
BUT HAVE 1 OR 4 SUBCYCLES. 25 X^2 + 16 Y^2 SEEMS TO EXPRESS THOSE
|
||
WHICH ARE 9 MOD 40; (10 X +/- 1)^2 + 400 Y^2 SEEMS TO EXPRESS
|
||
THOSE WHICH ARE 1 MOD 40.
|
||
PROBLEM: CAN SOME OF THE "SEEMS" ABOVE BE PROVED?
|
||
ALSO, CAN A GENERAL TEST BE MADE WHICH WILL PREDICT EXACT LENGTH FOR ANY NUMBER?
|
||
|
||
ITEM (BEELER):
|
||
IF S = THE SUM OF ALL INTEGERS WHICH EXACTLY DIVIDE N, INCLUDING 1 AND N,
|
||
THEN "PERFECT NUMBERS" ARE S = 2 N;
|
||
THE FIRST THREE NUMBERS WHICH ARE S = 3 N ARE:
|
||
120 = 2^3 X 3 X 5 = 1111000 BASE 2
|
||
672 = 2^5 X 3 X 7 = 1010100000 BASE 2
|
||
523,776 = 2^9 X 3 X 11 X 31 = 1111111111000000000 BASE 2
|
||
|
||
ITEM:
|
||
THERE IS A UNIQUE "MAGIC HEXAGON" OF SIDE 3:
|
||
3 17 18
|
||
19 7 1 11
|
||
16 2 5 6 9
|
||
12 4 8 14
|
||
10 13 15
|
||
*********************************************
|
||
AUTOMATA THHEORY
|
||
*********************************************
|
||
|
||
ITEM (SCHROEPPEL):
|
||
A 2-COUNTER MACHINE, GIVEN N IN ONE OF THE COUNTERS, CANNOT GENERATE 2^N.
|
||
(SATURDAY, SEPTEMBER 26, 1970)
|
||
|
||
ITEM (SCHROEPPEL):
|
||
WHAT EFFORT IS REQUIRED TO COMPUTE PI(X), THE NUMBER OF PRIMES < X?
|
||
*********************************************
|
||
GAMES
|
||
*********************************************
|
||
|
||
ITEM (SCHROEPPEL):
|
||
REGARDING "POKER COINS" GAME, WHOSE RULES ARE:
|
||
1 A PLAYER THROWS N; HE MUST THEN PUT ASIDE AT LEAST ONE AND RETHROW THE REST
|
||
2 THIS THROWING IS REPEATED UNTIL HE NO LONGER HAS ANY TO THROW
|
||
3 HIGHEST SCORE (DICE) OR MAXIMUM NUMBER OF HEADS (COINS) WINS
|
||
FOR POKER COINS, THE OPTIMAL STRATEGY, WITH N COINS THROWN, IS:
|
||
Z = NUMBER OF ZEROS (TAILS)
|
||
IF Z = 0, QUIT
|
||
IF Z = 1, THROW THE ZERO
|
||
IF 1 < Z < N, SAVE ONE ONE, THROW THE OTHER N-1 COINS
|
||
IF Z = N, SAVE A ZERO, THROW THE OTHER N-1 COINS
|
||
THE OPTIMAL STRATEGY FOR POKER DICE IS HARIER.
|
||
|
||
ITEM (SCHROEPPEL):
|
||
PROBLEM: SOLVE BLACKOUT, A GAME AS FOLLOWS:
|
||
TWO PLAYERS ALTERNATE PLACING X'S ON A RECTANGULAR GRID.
|
||
NO TWO X'S MAY APPEAR ADJACENT ALONG A SIDE OR ACROSS THE DIAGONAL AT A CORNER.
|
||
THE LAST X WINS.
|
||
KNOWN BEHAVIOR:
|
||
FOR LINES, A PERIOD OF 34 IS ENTERED AFTER LINE IS ABOUT 80 LONG.
|
||
FOR L'S, THERE IS A TENDANCY TOWARD LONGER INITIAL SETTLING FOR LARGER L'S,
|
||
THEN IT SEEMS TO HAVE PERIOD 34.
|
||
THE "NUMBER" FOR A POSITION IS:
|
||
MAKE ALL POSSIBLE MOVES, EVALUATING POSITION AFTER EACH
|
||
"NUMBER" IS MINIMUM NUMBER NOT THUS ENCOUNTERED
|
||
THE NULL GAME = "NUMBER" OF ZERO = 2ND PLAYER WINS
|
||
THIS "NUMBER" ANALYSIS IS SIMILAR FOR MANY OTHER TAKE-AWAY GAMES, SUCH AS NIM.
|
||
FOR 2 BOARDS (NOT TOUCHING) PLAYED TOGETHER, "NUMBER" IS XOR OF THE 2
|
||
NUMBERS FOR THE SEPARATE BOARDS.
|
||
ON AN ODD X ODD BOARD, THE 1ST PLAYER WINS.
|
||
ON A 4 X N BOARD, THE 2ND PLAYER WINS.
|
||
ON A 6 X 6 BOARD, THE 1ST PLAYER WINS BY PLAYING AT THE CENTER OF ONE QUARTER.
|
||
|
||
ITEM:
|
||
BERLEKAMP OF BELL LABS HAS DONE THE 9 SQUARES (16 DOTS) DOTS GAME;
|
||
THE 2ND PLAYER WINS.
|
||
|
||
ITEM (BEELER):
|
||
THERE IS ONLY ONE DISTINCT SOLUTION TO THE COMMERCIAL "INSTANT INSANITY"
|
||
COLORED-FACES CUBES PUZZLE, WHICH IS HOW IT COMES PACKED.
|
||
|
||
ITEM (BEELER):
|
||
A WINDOW-DICE GAME IS AS FOLLOWS:
|
||
1 THE PLAYER STARTS WITH EACH OF NINE WINDOWS OPEN, SHOWING THE DIGITS 1 - 9.
|
||
2 ROLL TWO DICE.
|
||
3 COVER UP ANY DIGITS WHOSE SUM IS THE SUM ON THE DICE.
|
||
4 ITERATE THROWING AND CLOSING WINDOWS UNTIL THE EQUALITY OF SUMS IS IMPOSSIBLE.
|
||
5 YOUR SCORE (MINIMUM WINS) IS THE TOTAL OF OPEN WINDOWS (UNCOVERED DIGITS).
|
||
AN OPTIMUM STRATEGY HAS BEEN TABULATED.
|
||
USUALLY IT IS BEST TO TAKE THE LARGEST DIGITS POSSIBLE, BUT NOT ALWAYS;
|
||
IT ALSO DEPENDS CRITICALLY ON THE REMAINING NUMBERS.
|
||
|
||
ITEM (BEELER):
|
||
SIM IS A GAME WHERE TWO PLAYERS ALTERNATELY DRAW LINES CONNECTING SIX DOTS.
|
||
THE FIRST PERSON TO FORM A TRIANGLE IN HIS COLOR LOSES.
|
||
THE SECOND PLAYER CAN ALWAYS WIN, AND WHETHER HIS FIRST MOVE CONNECTS WITH
|
||
THE FIRST PLAYER'S FIRST MOVE DOESN'T MATTER; FROM THERE ON, HOWEVER,
|
||
THE STRATEGY BRANCHES TO A RELATIVELY GRUESOME DEGREE.
|
||
PROBLEM: 6 DOTS IS MINIMUM TO ENSURE NO STALEMATE WHTH 2 PLAYERS;
|
||
HOW MANY DOTS ARE REQUIRED WITH 3 PLAYERS?
|
||
|
||
ITEM (BEELER):
|
||
THE 4 X 4 GAME OF NIM IS A WIN FOR THE SECOND PLAYER, WHO ON
|
||
HIS FIRST MOVE CAN REPLY CENTER-SYMMETRICALLY UNLESS THE FIRST
|
||
PLAYER'S FIRST MOVE WAS B1 AND B2 (ANALYZED ON RLE PDP-1).
|
||
|
||
ITEM (BEELER):
|
||
TRIANGULAR HI-Q (OR PEG SOLITAIRE) IS 15 PEGS IN A TRIANGLE.
|
||
ONE PEG IS REMOVED, AND THEREAFTER PEGS JUMP OTHERS, WHICH ARE REMOVED.
|
||
WITH PEGS NUMBERED 1 AT THE TOP, 2 AND 3 IN THE NEXT ROW, ETC.,
|
||
REMOVE CAN END WITH ONLY THE PEG
|
||
1 1, 7 = 10, 13
|
||
2 2, 6, 11, 14
|
||
4 3 = 12, 4, 9, 15
|
||
5 13
|
||
REMOVING ONLY ONE, NO WAY EXISTS TO GET TO EITHER 1 + 11 + 15 (TIPS)
|
||
OR 4 + 6 + 13 (CENTERS OF SIDES). STARTING WITH PEG 1 REMOVED,
|
||
3,016 POSITIONS ARE ATTAINABLE (NOT TURNING BOARD);
|
||
THE SUM OF WAYS TO GET TO EACH OF THESE IS 10,306.
|
||
AN EXAMPLE IS: REMOVE PEG 1, THEN JUMP AS FOLLOWS:
|
||
6, 13, 10, 1, 2, 11, 14/13, 6, 12/13, 15, 7/4, 13, 4; LEAVING PEG 1.
|
||
*********************************************
|
||
PROPOSED (SCHROEPPEL) COMPUTER PROGRAMS,
|
||
IN ORDER OF INCREASING RUNNING TIME
|
||
*********************************************
|
||
|
||
PROBLEM: COUNT THE POLYOMINOS UP TO, SAY, ORDER 20.
|
||
FROM APPLIED COMBINATORIAL MATHEMATICS, PAGES 201 AND 213:
|
||
ORDER E. H. NOT ENCLOSING HOLES
|
||
1 1 1
|
||
2 1 1
|
||
3 2 2
|
||
4 5 5
|
||
5 12 12
|
||
6 35 35
|
||
7 108 107
|
||
8 369 363
|
||
9 1285 1248
|
||
10 4655 4271
|
||
11 17073
|
||
12 63600
|
||
|
||
PROBLEM: SOLVE "MINICHESS", CHESS PLAYED ON A 5 X 5 BOARD WHERE EACH SIDE
|
||
HAS LOST A ROOK, KNIGHT, BISHOP, AND 3 PAWNS FROM ONE SIDE AND THE OPPONENTS ARE SHOVED
|
||
CLOSER TOGETHER (1 EMPTY ROW INTERVENING).
|
||
|
||
PROBLEM: SOLVE THE TIGER PUZZLE, A SLIDING BLOCK PUZZLE MENTIONED IN
|
||
SCIENTIFIC AMERICAN FEBRUARY 1964, PAGES 122 - 130.
|
||
|
||
PROBLEM: FIND SMALLEST SQUARED SQUARE (A SQUARE COMPOSED ENTIRELY OF SMALLER,
|
||
UNEQUAL SQUARES). SMALLEST KNOWN HAS 24 SMALL SQUARES
|
||
(MARTIN GARDNER'S SCIENTIFIC AMERICAN BOOK, VOL. 2, PAGE 206).
|
||
|
||
PROBLEM: LIST (THAT IS, COUNT) THE SEMIGROUPS OF 7 ELEMENTS;
|
||
ALSO, THE GROUPS OF 256 ELEMENTS.
|
||
|
||
PROBLEM: COMPUTE AS A FUNCTION OF N THE LARGEST SIZE FOR N SMALL CIRCLES
|
||
TO BE PLACED INSIDE ONE OF UNIT SIZE. FOR EXAMPLE, THE SIZE IS THE
|
||
SAME FOR N = 6 AND N = 7 (GOSPER); ARE THERE ANY OTHER SUCH
|
||
PHENOMENA?
|
||
|
||
PROBLEM: SOLVE PENTOMINOS ON AN 8 X 8 CHECKERBOARD GAME(S). RULES:
|
||
1 THE CHECKERBOARD IS FOR AID IN ORIENTING ONLY; BLACK AND WHITE ARE THE SAME.
|
||
2 THE TWO PLAYERS MAY EACH HAVE A FULL COMPLEMENT OF 12 PENTOMINOS, OR
|
||
THEY MAY "CHOOSE UP" THEIR HALF OF ONE SET.
|
||
3 THE LAST PLAYER TO PLACE A PENTOMINO WINS.
|
||
|
||
PROBLEM: WITH REGARD TO DISSECTION THEOREMS, THE FOLLOWING ARE KNOWN:
|
||
A TRIANGLE INTO A SQUARE, 4 PIECES (PROVEN MINIMAL)
|
||
A PENTAGON INTO A SQUARE, 6 PIECES (BEST KNOWN)
|
||
ETC. ("GEOMETRIC DISSECTIONS" BY HARRY LINDGREEN, SCIENTIFIC AMERICAN NOVEMBER 1961).
|
||
A PROGRAM CAN PROBABLY CHECK THE KNOWN DISSECTIONS FOR MINIMALITY!
|
||
|
||
PROBLEM: FIND THE NUMBER OF DOMINO COVERINGS FOR VARIOUS OBJECTS.
|
||
FOR EXAMPLE, AN ASYMPTOTIC FORMULA IS KNOWN FOR RECTANGLES;
|
||
ALSO, ON A SQUARE BOARD, IF SIDE MOD 4 = 0, COVERINGS APPEARS TO BE A SQUARE;
|
||
ON A SQUARE BOARD, IF SIDE MOD 4 = 2, COVERINGS APPEARS TO BE TWICE A SQUARE.
|
||
PROBLEM: ANALYZE GIVEAWAY CHESS, WHICH IS AS FOLLOWS:
|
||
1 CAPTURES MUST BE MADE, ALTHOUGH YOU CAN CHOOSE WHICH CAPTURE TO MAKE
|
||
2 PAWNS MUST BE PROMOTED TO QUEENS
|
||
3 KING IS JUST ANOTHER PIECE
|
||
4 PLAYER TO GIVE AWAY ALL PIECES FIRST WINS
|
||
|
||
PROBLEM: ANALYZE "ESCALATION CHESS", WHERE WHITE GETS 1 MOVE, BLACK 2, WHITE 3, ETC.
|
||
|
||
PROBLEM: IN THE GAME "4 PAWNS", ONE SIDE HAS 4 PAWNS, A KING, AND
|
||
TWO MOVES TO THE OTHER'S ONE. PROVE THE PAWNS WIN.
|
||
|
||
PROBLEM: SOLVE TIC-TAC-TOE ON A 4 X 4 X 4 BOARD.
|
||
|
||
PROBLEM: SOLVE SCARNE'S GAME, "TECO," WHICH IS PLAYED ON A 5 X 5 BOARD BY TWO PLAYERS WHO
|
||
ALTERNATE PLACING, ONE AT A TIME, THEIR 4 COUNTERS EACH, AFTER WHICH
|
||
THE COUNTERS ARE MOVED AROUND (INCLUDING DIAGONALLY). 4 IN A ROW OR SQUARE WINS.
|
||
|
||
PROBLEM: SOLVE CHECKERS (COMPUTING TIME CURRENTLY ESTIMATED (SCHROEPPEL) AT 1 YEAR).
|
||
|
||
PROBLEM: SOLVE HEX ON LARGE BOARDS (11 TO 23 ON A SIDE); THROUGH ORDER 7 HAVE
|
||
BEEN ANALYZED BY HAND. THERE IS A PROOF THAT IN GAMES WHERE
|
||
HAVING AN EXTRA MOVE CAN NEVER (REPEAT: NEVER) HURT YOU, THE WORST
|
||
THE FIRST PLAYER CAN BE FORCED TO DO IS DRAW.
|
||
THUS, WITH HEX, IN WHICH THERE IS NO DRAW, THE FIRST PLAYER CAN ALWAYS WIN.
|
||
*********************************************
|
||
CONTINUED FRACTIONS
|
||
*********************************************
|
||
|
||
ITEM (SCHROEPPEL):
|
||
SIMPLE PROOFS THAT CERTAIN CONTINUED FRACTIONS ARE SQRT 2, SQRT 3, ETC.
|
||
PROOF FOR SQRT 2:
|
||
X = [1, 2, 2, 2, ...]
|
||
(X-1)(X+1) = [0, 2, 2, 2, ...] * [2, 2, 2, 2, ...] = 1
|
||
X^2 - 1 = 1
|
||
X = SQRT 2
|
||
PROOF FOR SQRT 3:
|
||
Y = [1, 1BAR, 2BAR]
|
||
(Y + 1)(Y - 1) = [2, 1BAR, 2BAR] * [0, 1BAR, 2BAR]
|
||
= 2 * [1, 2BAR, 1BAR] * [0, 1BAR, 2BAR] = 2
|
||
Y^2 - 1 = 2
|
||
Y = SQRT 3
|
||
SIMILAR PROOFS EXIST FOR SQRT 5 AND SQRT 6; BUT SQRT 7 IS HAIRY.
|
||
|
||
ITEM (SCHROEPPEL):
|
||
THE CONTINUED FRACTION EXPANSION OF THE POSITIVE MINIMUM OF THE GAMMA FUNCTION
|
||
(ABOUT 1.46) IS [1, 2, 6, 63, 135, 1, 1, 1, 1, 4, 1, 43, ...].
|
||
|
||
ITEM (SCHROEPPEL):
|
||
THE VALUE OF A CONTINUED FRACTION WITH PARTIAL QUOTIENTS INCREASING IN
|
||
ARITHMETIC PROGRESSION IS
|
||
[A+D, A+2D, A+3D, ...] = (I SUB A/D OF (2/D)/(I SUB 1+(A/D) OF (2/D)
|
||
WHERE THE I'S ARE BESSEL FUNCTIONS.
|
||
A SPECIAL CASE IS [1, 2, 3, 4, ...] = (I SUB 0 (2))/(I SUB 1 (2)).
|
||
*********************************************
|
||
GROUP THEORY
|
||
*********************************************
|
||
|
||
ITEM (KOMOLGOROFF, MAYBE?):
|
||
GIVEN A SET OF REAL NUMBERS, HOW MANY SETS CAN YOU GET USING ONLY
|
||
CLOSURE AND COMPLEMENT? ANSWER: 14. IF MORE DIMENSIONS ARE ALLOWED,
|
||
MORE CAN BE GOTTEN.
|
||
|
||
ITEM (SCHROEPPEL):
|
||
AS OPPOSED TO THE USUAL FORMULATION OF A GROUP, WHERE YOU ARE GIVEN
|
||
1 THERE EXISTS AN I SUCH THAT A * I = I * A = A, AND
|
||
2 FOR ALL A, B AND C, (A * B) * C = A * (B * C), AND
|
||
3 FOR EACH A THERE EXISTS AN ABAR SUCH THAT A * ABAR = ABAR * A = 1, AND
|
||
4 SOMETIMES YOU ARE GIVEN THAT I AND ABAR ARE UNIQUE
|
||
IF INSTEAD YOU ARE GIVEN A * 1 = A AND A * ABAR = 1, THEN
|
||
THE ABOVE RULES CAN BE DERIVED. BUT IF YOU ARE GIVEN
|
||
A * 1 = A AND ABAR * A = 1, THEN
|
||
SOMETHING VERY MUCH LIKE A GROUP, BUT NOT(!) A GROUP, RESULTS.
|
||
FOR EXAMPLE, EVERY ELEMENT IS DUPLICATED.
|
||
|
||
ITEM (GOSPER):
|
||
THE HAMILTONIAN PATHS MADE ONLY OF SWAP (SWAP ANY SPECIFIC PAIR) AND ROTATE
|
||
FOR N ELEMENTS ARE AS FOLLOWS:
|
||
N PATHS + REVERSES
|
||
2 2 + 0
|
||
3 2 + 2
|
||
4 3 + 3, NAMELY:
|
||
SRR RSR SRR RSR RRS RSR RSR RR
|
||
RSR SRR RSR RRS RSR RRS RSR RR
|
||
SRR RSR RRS RRS RSR RRS RRR SR
|
||
PROBLEM: A QUESTIONABLE PROGRAM SAID THERE ARE NONE FOR N = 5; IS THIS SO?
|
||
|
||
ITEM (SCHROEPPEL):
|
||
ANY PERMUTATION ON 72 BITS CAN BE CODED WITH A ROUTINE
|
||
CONTAINING ONLY THE PDP-6/10 INSTRUCTIONS "ROT" AND "ROTC".
|
||
ORDER IAMONDS OMINOS HEXA'S SOMA-LIKE
|
||
1 1 1 1 1
|
||
2 1 1 1 1
|
||
3 1 2 3 2
|
||
4 3 5 7 8
|
||
5 4 12 22 29
|
||
6 12 35
|
||
7 24
|
||
8 66
|
||
9 160
|
||
10 448
|
||
POLYOMINOS OF ORDER 1, 2 AND 3 CANNOT FORM A RECTANGLE .
|
||
ORDERS 4 AND 6 CAN BE SHOWN TO FORM NO RECTANGLES BY A CHECKERBOARD COLORING.
|
||
ORDER 5 HAS SEVERAL BOARDS AND ITS SOLUTIONS ARE DOCUMENTED
|
||
(COMMUNICATIONS OF THE ACM, OCTOBER 1965):
|
||
BOARD DISTINCT SOLUTIONS
|
||
3 X 20 2
|
||
4 X 15 368
|
||
5 X 12 1010
|
||
6 X 10 2339 (VERIFIED)
|
||
TWO 5 X 6 -- 2
|
||
8 X 8 WITH 2 X 2 HOLE IN CENTER -- 65
|
||
|
||
AN ORDER-4 HEXAFROB SOLUTION:
|
||
|
||
A A A A B C C
|
||
D E B B C F C
|
||
D E E B F G G
|
||
D D E F F G G
|
||
ORDER-6 IAMOND BOARDS AND SOLUTION COUNTS:
|
||
SIDE 9 TRIANGLE WITH INVERTED SIDE 3 TRIANGLE IN CENTER REMOVED -- NO SOLUTIONS
|
||
TRAPEZOID, SIDE 6, BASES 3 & 3+6 -- NO SOLUTIONS
|
||
TWO TRIANGLES OF SIDE 6 -- NO SOLUTIONS
|
||
TRAPEZOID, SIDE 4, BASES 7 & 7+4 -- 76. DISTINCT SOLUTIONS
|
||
PARALLELOGRAM, BASE 6, SIDE 6 -- 156. DISTINCT SOLUTIONS
|
||
PARALLELOGRAM, BASE 4, SIDE 9 -- 37. DISTINCT SOLUTIONS
|
||
PARALLELOGRAM, BASE 3, SIDE 12. -- NO SOLUTIONS
|
||
TRIANGLE OF SIDE 9 WITH TRIANGLES OF SIDE 1, 2 AND 2 REMOVED FROM ITS CORNERS
|
||
(A COMMERCIAL PUZZLE) -- 5885. DISTINCT SOLUTIONS
|
||
WITH SOMA-LIKE PIECES, ORDERS 1, 2 AND 3 DO NOT HAVE INTERESTING BOXES.
|
||
ORDER 4 HAS 1390 DISTINCT SOLUTIONS FOR A 2 X 4 X 4 BOX.
|
||
1124 OF THESE HAVE THE FOUR-IN-A-ROW ON AN EDGE; THE REMAINING
|
||
266 HAVE THAT PIECE INTERNAL. 320 SOLUTIONS ARE DUE TO VARIATIONS
|
||
OF TEN DISTINCT SOLUTIONS DECOMPOSABLE INTO TWO 2 X 2 X 4 BOXES.
|
||
A SOMA-LIKE 2 X 4 X 4 SOLUTION:
|
||
|
||
AAAA BBHH
|
||
BCCC BHHC
|
||
DDDE FGGE
|
||
FDGE FFGE
|
||
EXAMPLES OF ORDER-6 IAMOND SOLUTIONS:
|
||
|
||
DD DDD DDF FFF FLL LLI III II
|
||
D DDD DDD DFF FFL LLL LII III I
|
||
AA ABD DJJ JFF FFL LLL LKK KKI II
|
||
A AAB BDJ JJF FFF FLL LLK KKK KII I
|
||
AA ABB BCJ JJJ JJE EEK KKK KGG GHH HH
|
||
A AAB BBC CJJ JJJ JEE EKK KKG GGH HHH H
|
||
AA ABB BBC CCC CCC CEE EEE EGG GGG GHH HH
|
||
A AAB BBB BCC CCC CCE EEE EEG GGG GGH HHH H
|
||
|
||
GG GGG GRR RRR RRG GGG G
|
||
G GGG GGR RRR RRR RGG GG
|
||
GG GYY YGG GGR RYG GGG G
|
||
G GGY YYG GGG GRY YGG GG
|
||
RR RRY YYG GGG GYY YYY Y
|
||
R RRR RYY YGG GGY YYY YY
|
||
GG GRR RYY YBB BBB BRY Y
|
||
G GGR RRY YYB BBB BBR RY
|
||
GG GYR RGG GGG GBB BRR R
|
||
G GGY YRG GGG GGB BBR RR
|
||
GG GYY YYY YYG GGR RRR R
|
||
G GGY YYY YYY YGG GRR RR
|
||
|
||
R RRR RCC CCC CCF FFF FTT TTG GGG GG
|
||
RR RRC CCC CCC CFF FFT TTT TGG GGG G
|
||
R RRR RUU UUC CFF FFT TDT TTG GGA AA
|
||
RR RRU UUU UCF FFF FTD DTT TGG GAA A
|
||
U UUU UEE EXX XXD DDD DSS SSS SBA AA
|
||
UU UUE EEX XXX XDD DDS SSS SSB BAA A
|
||
E EEE EEX XXX XDD DSS SBB BBB BBA AA
|
||
EE EEE EXX XXD DDS SSB BBB BBB BAA A
|
||
|
||
JF FFF FCG GGG GGA AA
|
||
J JFF FFC CGG GGG GAA A
|
||
JJ JFF FFC CCG GGH HHH HA
|
||
J JJF FFF FCC CGG GHH HHA A
|
||
J JJJ JEE EDC CCC CIH HAA A
|
||
JJ JJE EED DCC CCI IHA AA
|
||
E EEE EED DDD DII IAA A
|
||
EE EEE EDD DDI IIA AA
|
||
B BBB BKD DDI IIA A
|
||
BB BBK KDD DII IA
|
||
B BBK KKK KKI I
|
||
BB BKK KKK KI
|
||
B BLL LLK K
|
||
BL LLL LK
|
||
L LLL L
|
||
LL LL
|
||
*********************************************
|
||
TOPOLOGY
|
||
*********************************************
|
||
|
||
ITEM (SCHROEPPEL):
|
||
TESSELATING THE PLANE WITH POLYOMINOS:
|
||
THROUGH ALL HEXOMINOES, THE PLANE CAN BE TESSELATED WITH EACH PIECE
|
||
(WITHOUT EVEN FLIPPING ANY OVER). ALL BUT THE FOUR HEPTOMINOS BELOW CAN
|
||
TESSELATE THE PLANE, AGAIN WITHOUT BEING FLIPPED OVER. THUS, FLIPPING
|
||
DOES NOT BUY YOU ANYTHING THROUGH ORDER 7. (THERE ARE 108 HEPTOMINOS).
|
||
H H HHH H H
|
||
HHHHH H H HHHH HHHH
|
||
HH H H
|
||
H H
|
||
|
||
ITEM:
|
||
ALTHOUGH NOT NEW, THE FOLLOWING COLORING NUMBER (CHROMATIC NUMBER)
|
||
MAY BE USEFUL TO HAVE AROUND:
|
||
N = [[(7 + SQRT (1 + 48 * HOLES))/2]]
|
||
WHERE N IS THE NUMBER OF COLORS REQUIRED TO COLOR ANY MAP ON AN OBJECT WHICH HAS
|
||
"HOLES" NUMBER OF HOLES (NOTE: PROOF NOT VALID FOR HOLES = 0).
|
||
FOR EXAMPLE:
|
||
A DONUT (HOLES = 1) REQUIRES 7 COLORS TO COLOR MAPS ON IT.
|
||
A 17-HOLE FROB REQUIRES 17 COLORS.
|
||
AN 18-HOLE FROB REQUIRES 18 COLORS.
|
||
|
||
ITEM (GOSPER):
|
||
THE FOLLOWING TRANSFORMATION MAPS A BIT STREAM INTO A POINT (X,Y)
|
||
ON THE INFINITE LIMIT OF A SPACE FILLING CURVE:
|
||
*********************************************
|
||
SERIES
|
||
*********************************************
|
||
|
||
ITEM (SCHROEPPEL & GOSPER):
|
||
THE SUM FROM N = 0 OF [N!N!/(2N)!] = 4/3 + 2*PI/9*SQRT 3.
|
||
PROBLEM: EVALUATE IN CLOSED FORM THE SUM FROM N = 0
|
||
OF [N!N!N!/(3N)!]. THIS SUM HAS BEEN SHOWN EQUAL TO THE FOLLOWING:
|
||
THE INTEGRAL FROM 0 TO 1 OF (P + Q ARCCOS (R))DT
|
||
WHERE
|
||
2 (8 + 7 T^2 - 7 T^3)
|
||
P = ---------------------
|
||
(4 - T^2 + T^3)^2
|
||
AND
|
||
4 (T - T^2) (5 + T^2 - T^3)
|
||
Q = ------------------------------------------------
|
||
(4 - T^2 + T^3)^2 SQRT ((4 - T^2 + T^3) (1 - T))
|
||
AND
|
||
T^2 - T^3
|
||
R = 1 - ---------
|
||
2
|
||
|
||
|
||
ITEM (HENRY COHEN):
|
||
GAMMA = - LN X + X - X^2/2*2! + X^3/3*3! - X^4/4*4! ... + ERROR
|
||
WHERE ERROR IS OF THE ORDER OF (E^-X)/X.
|
||
|
||
ITEM (SCHROEPPEL):
|
||
EULER'S SERIES ACCELERATION METHOD APPLIED TO PI/4 GIVES
|
||
PI/4 = SUM [2^(N-1)(N!)^2/(2N+1)!]
|
||
|
||
ITEM (SCHROEPPEL):
|
||
PROBLEM: SQUARE SOME SERIES FOR PI TO GIVE THE SERIES
|
||
PI/6 = 1/1^2 + 1/2^2 + ... = SUM [1/N^2].
|
||
|
||
ITEM (SCHROEPPEL):
|
||
CONSIDER SUM [1/N^2] =>
|
||
SUM [1/(N-1/2)-1/(N+1/2)] + SUM [1/N^2-1/(N^2-1/4)]
|
||
= 2 - SUM [1/((4N^2-1)*N^2)]
|
||
TAKE THE LAST SUM AND RE-APPLY THIS TRANSFORMATION.
|
||
THIS MAY BE A WINNER FOR COMPUTING THE ORIGINAL SUM.
|
||
FOR EXAMPLE, THE NEXT ITERATION GIVES
|
||
31/18 - SUM [9/(N^2)(4N^2-1)(25N^4+5N^2+9)]
|
||
WHERE THE DENOMINATOR ALSO = (N^2)(2N+1)(2N-1)(5N^2+5N+3)(5N^2-5N+3)
|
||
|
||
ITEM (SCHROEPPEL):
|
||
THE "PARITY NUMBER" IS THE SUM (0 TO INFINITY) OF (2^-N)*(PARITY OF N)
|
||
WHERE THE PARITY OF N IS 0 OR 1. THE PARITY NUMBER'S VALUE IS BETWEEN
|
||
0 AND 1. IT CAN BE WRITTEN IN STAGES BY TAKING THE PREVIOUS STAGE, COMPLEMENTING,
|
||
AND APPENDING TO THE PREVIOUS STAGE:
|
||
0.
|
||
0.1
|
||
0.110
|
||
0.1101001
|
||
0.110100110010110
|
||
0.1101001100101101001... RADIX 2
|
||
OR, FASTER (GOSPER), BY SUBSTITUTING IN THE STRING AT ANY STAGE:
|
||
THE STRING ITSELF FOR ZEROS, AND
|
||
THE COMPLEMENT OF THE STRING FOR ONES.
|
||
FOR HEXADECIMAL FREAKS, HALF THE PARITY NUMBER IS .6996966996696996...
|
||
IT IS CLAIMED (PERHAPS PROVEN BY THUE?) THAT THE PARITY NUMBER IS TRANSCENDENTAL.
|
||
IN ADDITION,
|
||
2*(PARITY NUMBER) = 2-(1/2)*(3/4)*(15/16)*(255/256)*(65535/65536)*...
|
||
*********************************************
|
||
FLOWS AND ITERATED FUNCTIONS
|
||
*********************************************
|
||
|
||
ITEM (SCHROEPPEL):
|
||
AN ANALYTIC FLOW FOR NEWTON'S METHOD:
|
||
DEFINE F(X) BY (X^2+K)/2X; THEN
|
||
(X+SQRT K)^2^N + (X-SQRT K)^2^N
|
||
F(F(F(...(X)))) = SQRT K -------------------------------
|
||
[N TIMES] (X+SQRT K)^2^N - (X-SQRT K)^2^N
|
||
|
||
ITEM (SCHROEPPEL):
|
||
SUPPOSE Y SATISFIES A DIFFERENTIAL EQUATION OF THE FORM
|
||
P(X)Y(NTH DERIVATIVE) + ..... + Q(X) = R(X)
|
||
WHERE P, ..... Q, AND R ARE POLYNOMIALS IN X
|
||
(FOR EXAMPLE, BESSEL'S EQUATION, X^2Y''+XY'+(X^2-N^2)Y = 0)
|
||
AND A IS AN ALGEBRAIC NUMBER
|
||
THEN Y(A) CAN BE EVALUATED TO N PLACES IN TIME PROPORTIONAL TO N(LN N)^C
|
||
WHERE IT SEEMS C ALWAYS = 1 PROBABLY.
|
||
FURTHER, E^X AND LN X CAN BE EVALUATED TO N PLACES IN SIMILAR TIME
|
||
FOR X A REAL NUMBER.
|
||
IF F(X) CAN BE EVALUATED IN SUCH TIME, SO CAN THE INVERSE OF F(X)
|
||
(BY NEWTON'S METHOD), AND THE FIRST DERIVATIVE OF F(X).
|
||
|
||
ITEM (SCHROEPPEL):
|
||
P AND Q ARE POLYNOMIALS IN X; WHEN DOES P(Q(X)) = Q(P(X)) ?
|
||
(THAT IS, P COMPOSED WITH Q = Q COMPOSED WITH P.)
|
||
KNOWN SOLUTIONS ARE:
|
||
1 VARIOUS LINEAR THINGS
|
||
2 X TO DIFFERENT POWERS, SOMETIMES MULTIPLIED BY ROOTS OF 1
|
||
3 P AND Q ARE EACH ANOTHER POLYNOMIAL R COMPOSED WITH ITSELF DIFFERENT NUMBERS OF TIMES
|
||
4 SOLUTIONS ARISING OUT OF THE FLOW OF X^2-2, AS FOLLOWS:
|
||
SUPPOSE X = Y+1/Y
|
||
THEN Y^N+1/Y^N CAN BE WRITTEN AS A POLYNOMIAL IN X
|
||
FOR EXAMPLE, P = THE EXPRESSION FOR SQUARES = X^2-2 (N = 2)
|
||
AND Q = THE EXPRESSION FOR CUBES = X^3-3X (N = 3)
|
||
5 REPLACE X BY Y-A, THEN ADD A TO THE ORIGINAL CONSTANTS
|
||
IN BOTH P AND Q. FOR EXAMPLE, P = X^2 AND Q = X^3
|
||
THEN P = 1+(Y-1)^2 = Y^2-2Y+2 AND Q = 1+(Y-1)^3
|
||
THEN P(Q) = 1+(Y-1)^6 = Q(P)
|
||
SIMILARLY, REPLACING X WITH AY+B WORKS.
|
||
6 THERE ARE NO MORE THROUGH DEGREES 3 AND 4 (CHECKED WITH MATHLAB);
|
||
BUT ARE THERE ANY MORE AT ALL?
|
||
|
||
ITEM (SCHROEPPEL):
|
||
PROBLEM: GIVEN F(X) AS A POWER SERIES IN X WITH CONSTANT TERM = 0,
|
||
WRITE THE FLOW POWER SERIES.
|
||
FLOW SUB ZERO = X
|
||
FLOW SUB ONE = F(X)
|
||
FLOW SUB TWO = F(F(X))
|
||
ETC.
|
||
|
||
ITEM (SCHROEPPEL, GOSPER, HENNEMAN & BANKS):
|
||
THE "3N+1 PROBLEM" IS ITERATIVELY REPLACING N BY N/2 IF N IS EVEN
|
||
OR BY 3N+1 IF N IS ODD. KNOWN LOOPS FOR N TO FALL INTO ARE:
|
||
1 THE ZERO LOOP, 0 => 0
|
||
2 A POSITIVE LOOP, 4 => 2 => 1 => 4
|
||
3 THREE NEGATIVE LOOPS (EQUIVALENT TO THE 3N-1 PROBLEM WITH POSITIVE N)
|
||
-2 => -1 => -2
|
||
-5 => -7 => -10 => -5
|
||
-17 => -25 => -37 => -55 => -82 => -41 => -61 => -91 => -136 => -68 => -34 => -17
|
||
IN THE RANGE -100 * 10^6 < N < 60 * 10^6, ALL N FALL INTO THE ABOVE LOOPS.
|
||
ARE THERE ANY OTHER LOOPS? DOES N EVER DIVERGE TO INFINITY?
|
||
|
||
ITEM (SCHROEPPEL):
|
||
TAKING ANY TWO NUMBERS A AND B, FINDING THEIR ARITHMETIC MEAN AND THEIR
|
||
GEOMETRIC MEAN, AND USING THESE MEANS AS A NEW A AND B, THIS PROCESS, WHEN REPEATED,
|
||
WILL APPROACH A LIMIT WHICH CAN BE EXPRESSED IN TERMS OF ELLIPTIC FUNCTIONS.
|
||
|
||
ITEM (SCHROEPPEL):
|
||
PROBLEM: ALTHOUGH THE REASON FOR THE CIRCLE ALGORITHM'S STABILITY IS UNCLEAR,
|
||
WHAT IS THE NUMBER OF DISTINCT SETS OF RADII? (NOTE: ALGORITHM IS
|
||
INVERTIBLE, SO ALL POINTS HAVE PREDECESSORS.)
|
||
CIRCLE ALGORITHM:
|
||
NEW X = OLD X - EPSILON * OLD Y
|
||
NEW Y = OLD Y + EPSILON * NEW(!) X
|
||
|
||
*********************************************
|
||
SET THEORY
|
||
*********************************************
|
||
|
||
ITEM (GOSPER):
|
||
ALEPH-TWO FUNCTIONS SATISFY F(F(X)) = SIN X.
|
||
*********************************************PROGRAMMING HACKS
|
||
*********************************************
|
||
|
||
ITEM (GOSPER):
|
||
TO SWAP THE CONTENTS OF TWO LOCATIONS IN MEMORY:
|
||
EXCH A,LOC1
|
||
EXCH A,LOC2
|
||
EXCH A,LOC1
|
||
NOTE: LOC1 MUST NOT EQUAL LOC2!
|
||
|
||
ITEM (GOSPER):
|
||
TO SWAP TWO BITS IN AN ACCUMULATOR:
|
||
TRCE A,BIT1
|
||
TRCE A,BIT2
|
||
TRCE A,BIT1
|
||
NOTE: LAST TRCE NEVER SKIPS, AND COULD BE A TRC, BUT TRCE IS LESS FORGETABLE.
|
||
ALSO, USE TLCE OR TDCE IF THE BITS ARE NOT IN THE RIGHT HALF.
|
||
|
||
ITEM (GOSPER):
|
||
PROVING THAT SIMPLE PROGRAMS ARE NEITHER TRIVIAL NOR
|
||
EXHAUSTED YET, THERE IS THE FOLLOWING:
|
||
TLCA 1,1(1)
|
||
ROT 1,11
|
||
JRST .-1
|
||
THIS MAKES PRETTY PATERNS (THAT IS, IT IS A DISPLAY HACK)
|
||
WITH THE LOW 9 BITS = Y AND THE NEXT HIGHER = X; ALSO, IT MAKES INTERESTING,
|
||
RELATED NOISES WITH A STEREO AMP HOOKED TO THE X AND Y SIGNALS.
|
||
RECCOMMENDED VARIATIONS INCLUDE:
|
||
CHANGE GOOD <1>
|
||
NONE 377767,,377767; 757777,,757757; MANY OTHERS
|
||
TLC 1,2(1) 373777,,0; 300000,,0
|
||
TLC 1,3(1) -2,,-2; -5,,-1; -6,,-1
|
||
ROT 1,1 7,,7; A0000B,,A0000B
|
||
ROTC 1,11
|
||
AOJA 1,.-2
|
||
|
||
ITEM (SUSSMAN):
|
||
TO EXCHANGE TWO VARIABLES IN LISP WITHOUT USING A THIRD VARIABLE:
|
||
(SETQ X (PROG2 0 Y (SETQ Y X)))
|
||
|
||
ITEM (GOSPER):
|
||
A MINIATURE SINE AND COSINE ROUTINE FOLLOWS.
|
||
COS: FADR A,[1.57079632689]
|
||
SIN: MOVM B,A
|
||
CAMG B,[.0002]
|
||
POPJ P,
|
||
FDVR A,[-3.0]
|
||
PUSHJ P,SIN
|
||
FMPR B,B
|
||
FSC B,2
|
||
FADR B,[-3.0]
|
||
FMPRB A,B
|
||
POPJ P,
|
||
*********************************************PROGRAMMING ALGORITHMS, HEURISTICS
|
||
*********************************************
|
||
|
||
ITEM (GOSPER):
|
||
THE "BANANA PHENOMENON" WAS ENCOUNTERED WHEN PROCESSING A TEXT OF NEWS
|
||
BY TAKING THE LAST 3 LETTERS OUTPUT, SEARCHING FOR A RANDOM
|
||
OCCURRENCE IN THE TEXT OF THAT SEQUENCE, TAKING THE LETTER FOLLOWING
|
||
THAT OCCURRENCE, TYPING IT OUT, AND ITERATING.
|
||
THIS ENSURES THAT EVERY 4-LETTER STRING OUTPUT OCCURS IN THE ORIGINAL.
|
||
THE PROGRAM TYPED BANANANANANANANA....
|
||
WE NOTE AN AMBIGUITY IN THE PHRASE, "THE NTH OCCURRENCE OF."
|
||
IN ONE SENSE, THERE ARE FIVE 00'S IN 0000000000; IN ANOTHER, THERE ARE NINE.
|
||
THE EDITING PROGRAM TECO MAKES AN ARBITRARY DECISION OF WHICH YOU MEAN
|
||
(IT FINDS FIVE). SUCH PROGRAMS AND ROUTINES, HOWEVER, MIGHT
|
||
WELL BENEFIT FROM AN OPTION ALLOWING ONE TO SPECIFY WHICH IS MEANT.
|
||
<03>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><><EFBFBD>0`<60><>):
|
||
P AND Q ARE POLYNOMIALS IN X; WHEN DOES P(Q(X)) = Q(P(X)) ?
|
||
(THAT IS, P COMPOSED WITH Q = Q COMPOSED WITH P.)
|
||
KNOWN SOLUTIONS ARE:
|
||
1 VARIOUS LINEAR THINGS
|
||
2 X TO DIFFERENT POWERS, SOMETIMES MULTIPLIED BY ROOTS OF 1
|
||
3 P AND Q ARE EACH ANOTHER POLYNOMIAL R COMPOSED WITH ITSELF DIFFERENT NUMBERS OF TIMES
|
||
4 SOLUTIONS ARISING OUT OF THE FLOW OF X^2-2, AS FOLLOWS:
|
||
SUPPOSE X = Y+1/Y
|
||
THEN Y^N+1/Y^N CAN BE WRITTEN AS A POLYNOMIAL IN X
|
||
FOR EXAMPLE, P = THE EXPRESSION FOR SQUARES = X^2-2 (N = 2)
|
||
AND Q = THE EXPRESSION FOR CUBES = X^3-3X (N = 3)
|
||
5 REPLACE X BY Y-A, THEN ADD A TO THE ORIGINAL CONSTANTS
|
||
IN BOTH P AND Q. FOR EXAMPLE, P = X^2 AND Q = X^3
|
||
THEN P = 1+(Y-1)^2 = Y^2-2Y+2 AND Q = 1+(Y-1)^3
|
||
THEN P(Q) = 1+(Y-1)^6 = Q(P)
|
||
SIMILARLY, REPLACING X WITH AY+B WORKS.
|
||
6 THERE ARE NO MORE THROUGH DEGREES 3 AND 4 (CHECKED WITH MATHLAB);
|
||
BUT ARE THERE ANY MORE AT ALL?
|
||
|
||
ITEM (SCHROEPPEL):
|
||
PROBLEM: GIVEN F(X) AS A POWER SERIES IN X WITH CONSTANT TERM = 0,
|
||
WRITE THE FLOW POWER SERIES.
|
||
FLOW SUB ZERO = X
|
||
FLOW SUB ONE = F(X)
|
||
FLOW SUB TWO = F(F(X))
|
||
ETC.
|
||
|
||
ITEM (SCHROEPPEL, GOSPER, HENNEMAN & BANKS):
|
||
THE "3N+1 PROBLEM" IS ITERATIVELY REPLACING N BY N/2 IF N IS EVEN
|
||
OR BY 3N+1 IF N IS ODD. KNOWN LOOPS FOR N TO FALL INTO ARE:
|
||
1 THE ZERO LOOP |