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Math puzzle strategy finder
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import kotlin.math.* | |
/* | |
Player A gets N random coin flips. | |
Player B gets M random coin flips. | |
They don't see each other coins, but has agreed on strategy in advance: | |
- A names the # of B's coin | |
- B names the # of A's coin | |
If both their named rolls agree they win, otherwise they loose. | |
Question: What is their best strategy and chances for winning? | |
*/ | |
// Configuration parameters | |
const val n = 4 | |
const val m = 4 | |
val np = 1 shl n | |
val mp = 1 shl m | |
val a = IntArray(np) | |
val b = IntArray(mp) | |
fun computeB(j: Int, bj: Int): Int { | |
var cnt = np | |
for (i in 0 until np) { | |
val ai = a[i] | |
cnt -= ((j shr ai) and 1) xor ((i shr bj) and 1) | |
} | |
return cnt | |
} | |
fun display() = "[${a.joinToString("")}] x [${b.joinToString("")}]" | |
var best = 0 | |
fun findB() { | |
var sum = 0 | |
for (j in 0 until mp) { | |
var bc = 0 | |
var bk = 0 | |
for (k in 0 until min(j + 1, n)) { | |
val c = computeB(j, k) | |
if (c > bc) { | |
bc = c | |
bk = k | |
} | |
} | |
b[j] = bk | |
sum += bc | |
} | |
if (sum <= best) return | |
best = sum | |
var cur = sum | |
var div = np * mp | |
while (cur and 1 == 0) { | |
cur = cur shr 1 | |
div = div shr 1 | |
} | |
println("\rUpdated to $cur/$div: ${display()}") | |
} | |
var dots = 0 | |
var prevPct = 0 | |
fun findA(i: Int, prod: Int) { | |
if (i == np) { | |
findB() | |
return | |
} | |
// Periodically print progress report | |
if (i == 5) { | |
print(".") | |
dots++ | |
val pct = 10 * dots / prod | |
if (pct > prevPct) { | |
prevPct = pct | |
print("${100 * dots / prod}%") | |
} | |
} | |
// The actual search of cases | |
val to = min(i + 1, m) | |
val prod2 = prod * to | |
for (k in 0 until to) { | |
a[i] = k | |
findA(i + 1, prod2) | |
} | |
} | |
fun main() { | |
findA(1, 1) | |
} |
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