Munchausen numbers: Difference between revisions

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Alternative that finds all 4 Munchausen numbers. As noted by the Pascal sample, we only need to consider one arrangement of the digits of each number (e.g. we only need to consider 3345, not 3435, 3453, etc.). This also relies on the non-standard 0^0 = 0.

<lang algol68># Find all Munchausen numbers - note 11*(9^9) has ~~only~~ 10 digits so there are no #

<lang algol68># Find all Munchausen numbers - note 11*(9^9) has oly 10 digits so there are no #

# Munchausen numbers with 11+ digits #

# table of Nth powers - note 0^0 is 0 for Munchausen numbers, not 1 #

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[ 0 : 9 ]INT d count := z count;

# as the digit ~~power~~ sum is independent of the order of the digits, we need only #

# as the digit powr sum is independent of the order of the digits, we need only #

# consider one arrangement of each possible combination of digits #

FOR d1 FROM 0 TO 9 DO

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digit power sum +:= nth power[ d7 ] + nth power[ d8 ];

digit power sum +:= nth power[ d9 ] + nth power[ da ];

⚫

# count ~~the~~ ~~occurrences~~ ~~of each digit (including~~ leading zeros #

d count := z count;

d count[ d1 ] +:= 1; d count[ d2 ] +:= 1; d count[ d3 ] +:= 1;

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d count[ d7 ] +:= 1; d count[ d8 ] +:= 1; d count[ d9 ] +:= 1;

d count[ da ] +:= 1;

⚫

# ~~subtract~~ ~~the~~ ~~occurrences~~ of ~~each~~ ~~digit~~ ~~in the power sum #~~

⚫

# ~~(also~~ ~~including leading zeros) - if all counts drop to~~ 0 ~~we #~~

# have a Munchausen number #

LONG INT number := digit power sum;

~~FOR~~ d TO 10 DO

INT leading zeros := 10;

⚫

WHILE number > 0 DO

d count[ SHORTEN ( number MOD 10 ) ] -:= 1;

⚫

leading zeros -:= 1;

number OVERAB 10

OD;

⚫

d count[ 0 ] -:= leading zeros;

IF d count[ 0 ] = 0 AND d count[ 1 ] = 0 AND d count[ 2 ] = 0

AND d count[ 3 ] = 0 AND d count[ 4 ] = 0 AND d count[ 5 ] = 0